geometry
parallelogram
geometric transformation
rotation
geometry proposed

Let $ ABC$ be a triangle where $ \angle ACB\equal{}90^{\circ}$. Let $ D$ be the midpoint of $ BC$ and let $ E$, and $ F$ be points on $ AC$ such that $ CF\equal{}FE\equal{}EA$. The altitude from $ C$ to the hypotenuse $ AB$ is $ CG$, and the circumcentre of triangle $ AEG$ is $ H$. Prove that the triangles $ ABC$ and $ HDF$ are similar.

Take K so that DCKE be a paralellogram, see that KE || CD, KE = GD and their angle is 2*<A, HE = HG and the angle between them is 2*<A too, therefore the triangles KEH and DGH are equal and a rotation of 2*<A around H maps K to D and E to G, hence KH = DH, the triangle KHD is isosceles, so KD and HF are perpendicular and the angle <FHD = 1/2*<KHD = <A, hence the triangles HDF and ABC are similar.

Note: As it can be seen, E - a point on (AC) and F the midpoint of CE suffice, there is no need of AE = EF = FC.

Best regards,
sunken rock


What is the probability that if you roll two fair, standard six-sided
dice, the difference between the two numbers rolled will
be 0? Express your answer as a common fraction.

We want the cases where the two numbers are equal. Since there are $ 6$ numbers, and the total number of ways to choose the two numbers is $ 6\times{6}\equal{}36$, the probability is $ \boxed{\frac{1}{6}}$.

The only way to do this is by rolling the same number. Since there are 6 numbers and there are 36 possibilities.
$6/36$=$1/6$.


calculus
integration
logarithms
function
trigonometry
algebra
partial fractions

Is it possible to calculate
$\int_{0}^{\infty}\frac{\sqrt{x}}{x^{2}+1}dx$

Of course. Two approaches:

The elementary calculus approach: Substitute $x=t^{2}$, getting $\int_{0}^\infty \frac{2t^{2}}{t^{4}+1}\,dt$. Factor the denominator as $(t^{2}+\sqrt{2}t+1)(t^{2}-\sqrt{2}t+1)$ and apply partial fractions to find an antiderivative.

The complex analysis approach:
Let $C$ be the curve in the upper half-plane consisting of the segments from $-R$ to $-r$ and from $r$ to $R$, as well as semicircles of radius $r$ and $R$ centered at the origin. We have $\int_{C}\frac{\sqrt{z}}{z^{2}+1}\,dz= \int_{-R}^{-r}\frac{i\sqrt{-x}}{x^{2}+1}\,dx+\int_{r}^{R}\frac{\sqrt{x}}{x^{2}+1}\,dx+$(integrals around semicircles).
The semicircle terms go to zero as $r\to0^{+}$ and $R\to\infty$, so the limit of $\int_{C}\frac{\sqrt{z}}{z^{2}+1}\,dz$ is $\int_{-\infty}^{0}\frac{i\sqrt{-x}}{x^{2}+1}\,dx+\int_{0}^\infty\frac{\sqrt{x}}{x^{2}+1}\,dx= \int_{0}^\infty\frac{(1+i)\sqrt{x}}{x^{2}+1}\,dx$.
The integral $\int_{C}\frac{\sqrt{z}}{z^{2}+1}\,dz$ can be calculated by the residue theorem; the only pole is at $z=i$.

The [hide="answer"]$\frac{\pi}{\sqrt{2}}$[/hide]

Hi,
Another method for calculating this improper integral would be via the substitution $x=t^{1/2}$ and by using the Gamma and Beta function properties as folows:
Thus we get that
$\int_{0}^{\infty}\frac{\sqrt{x}}{x^{2}+1}dx=\frac{1}{2}\int_{0}^{\infty}\frac{t^{3/4-1}}{t+1}dt=\frac{1}{2}B(\frac{3}{4},\frac{1}{4})=\frac{1}{2}\Gamma(1/4)\Gamma(3/4)=\frac{1}{2}\frac{\pi}{\sin \frac{\pi}{4}}=\frac{\pi}{\sqrt{2}}$.

[quote="Summerburn"][quote="kunny"]Let $x=\tan \theta ,$ we can rewite as

$\int_{0}^{\infty}\frac{\sqrt{x}}{x^{2}+1}\ dx=\int_{0}^{\frac{\pi}{2}}\sqrt{\tan \theta}\ d\theta .$[/quote]

Can you show how to continue?

because it seems I can't get the correct answer.[/quote]
kunny was hinting at the following:

First, show $\int_{0}^{a}f(x)\,dx = \int_{0}^{a}f(a-x)\,dx$; just use the substitution $y=a-x$ to evaluate the RHS)

Let $I = \int_{0}^{\frac{\pi}{2}}\sqrt{\tan t}\,dt$.
Then, $I = \int_{0}^{\frac{\pi}{2}}\sqrt{\cot t}\,dt$ ... (using the above identity)

$\Rightarrow 2I = \int_{0}^{\frac{\pi}{2}}(\sqrt{\tan t}+\sqrt{\cot t})\,dt$

$\Rightarrow 2I = \int_{0}^{\frac{\pi}{2}}\frac{\sin t+\cos t}{\sqrt{\sin t \cos t}}\,dt = \sqrt2 \int_{0}^{\frac{\pi}{2}}\frac{\sin t+\cos t}{\sqrt{1-(1-2\sin t \cos t)}}\,dt$

$\Rightarrow 2I = \sqrt2 \int_{0}^{\frac{\pi}{2}}\frac{\sin t+\cos t}{\sqrt{1-(\sin t-\cos t)^{2}}}\,dt$ ... (this is really the key step)

Now, use one last substitution $y = \sin t-\cos t.$ We obtain

$2I = \sqrt2 \int_{-1}^{1}\frac{dy}{\sqrt{1-y^{2}}}$

$\Rightarrow 2I = \sqrt2 \sin^{-1}y \Big|_{-1}^{1}$

$\Rightarrow \boxed{I = \frac{\pi}{\sqrt2}}.$


I would apply the distributive law to (-1)(-1 + 1).

I think it is just a mathematical definition.  It works, as Ravi showed.  In mathematics, you can make a definition, as long as it holds.

I am trying to think of a physics example where sign changes make results realistic.

Hmm... I was initially going to suggest a physics example of negative acceleration on a negative velocity, but that's additive, not multiplicative.

But I'll refer to the classic formula of d = rt (or for the physics type,  :tri:x = v:tri:t).  It can be reworked so that d/r = t.  So, if you travel at a negative velocity and cover a specific distance in the negative direction (bearing in mind that if you have a negative velocity, you're traveling in the negative direction), you will do this in a positive amount of time.

[quote="Yasha"]Multiplying something by -1 is the same as taking its additive inverse.[/quote]

Can you show this?  In other words, (-1) is defined to be the number such that (-1) + 1 = 1 + (-1) = 0.  In general, the additive inverse of a number x, (-x), is defined to be the number such that x + (-x) = (-x) + x = 0.  So, can you prove (-1)*x = (-x)?

Ravi's solution works nicely, although you already have to know that 0 times any number is 0.  (A new challenge: can anyone show that [i]that[/i] is true?  Remember, you have to avoid using the fact that (-1)*x = (-x), since that is what we're trying to prove to begin with.)

I find it terribly annoying that we use a compound symbol to represent the negative integers when they have every claim to being numbers in their own right.  This probably stems from 5th or 6th grade when I got a question wrong on the Middle School Math League contest which asked you to compute [tex]-3^2[/tex].

I don't remeber at the moment how to prove that x*(-1)=-x but I have seen it proven before and I bet I could figure it out soon enough.

How to prove -1*x=-1? I'm not sure, but...

[hide]-1*x is equal to (-1)+(-1)+(-1)+...+(-1), that is x -1's added together. Adding together the 1's gives x, and then since the 1's are all negative, you put a negative sign in front of the x, giving -x?[/hide]

I doubt I'm even on the right track...

How to prove 0*x=0?

[hide]Well, a*x is equal to a x's added together, so 0*x would be 0 x's added together, or nothing added together, or 0?[/hide]

Again, I highly doubt that's what you wanted, but...

If you give away 3 apples 3 times, you lose 9 apples. If you go back in time, you will see yourself gain 9 apples. You are effectively watching yourself take -3 apples -3 times.

Things such as 0a = 0, and ab = ba are defined to be true.

edit - testing...
[tex]\frac{dy}{dx}=2x^{2}[/tex]

Prove 0*a=0 for all real a. (0 is defined to be the additive identity)

Since x+0=x for all x, we know that 0+0=0. Now, mulitplying this by a and distributing gives a*0+a*0=a*0. Now, adding -(a*0) to both sides gives a*0=0.

-------------

Prove -a=(-1)*a for all real a. (-a is the additive inverse of a, -1 is the additive inverse of 1)

(-1)*a+a = (-1)*a+1*a = (-1+1)*a = 0*a = 0 = -a+a.

Thus we have (-1)*a+a = -a+a. We can add -a to both sides to get (-1)*a=-a as desired.

[quote="mikewd"]Things such as 0a = 0, and ab = ba are defined to be true.[/quote]

ab=ba is indeed assumed to be true. However 0a=0 is not assumed to be true and can be proven (see above). The only thing assumed about 0 is that it is the additive identity, that is, a+0=a.

A reason that a negative number squared is positive:

We will solve x^2 - a^2 = 0 where a is a positive real number.

This can be factored as

(x-a)(x+a)

The factorization implies that a^2 = (-a)^2.  Since a is positive, a^2 is positive.  The equation above implies that (-a)^2 is positive since a^2 is.

Doesn't your proof assume that -(-a) = a, which is close to what we need to prove in the first place?

Oops.  I didn't see that when I wrote it. :oops:

Generally proving basic stuff with more complicated stuff is a bad idea, because most of the complicated stuff uses all the basic stuff. Then you get circular logic.

[quote="Rep123max"]Does anybody have any good example of why [multiplying a negative number by a negative number is a positive number] is true?[/quote]

I'm surprised that no one has yet mentioned the treatment of  this issue in the book Algebra by I.M. Gelfand (my avatar) and Alexander Shen. I think Scrambled has been reading that book recently--it is excellent, and everyone on this forum should work through it. You can find the passage by Gelfand and Shen about multiplication of negative numbers quoted in a review of another book by mathematician Richard Askey.

http://www.aft.org/american_educator/fall99/amed1.pdf

(Read through the review, which is a good review of a good book, and you will see the sidebar about Gelfand and Shen's discussion of multiplying negative numbers.)

Hope this helps!

Multiplication is adding over and over again.

4*3 = 0+4+4+4 = 12
(-4) * 3 = 0+(-4)+(-4)+(-4) = -12

But when you try to do (-4)*(-3) you run into the problem that you can't add a negative amount of times.  But let's say you add 4+4 and get eight.  When you go back in time, adding a "negative" amount of times, you get 4.  So adding a negative amount of times is like subtracting a positive amount of times.

(-4)*(-3) = 0-(-4)-(-4)-(-4)

And subtracting a negative is like adding a positive

(-4)*(-3) = 0+4+4+4 = 12

And that's how multiplying two negative numbers get a positive number.

-(-x)=x is clear since -(-x) is the additive inverse of x which is defined by -(-x)+(-x)=0. but -x+x=0, so by commutivity we have -x+(-(-x))=0=-x+x. adding x from the left we get -(-x)=x as we wanted to. all that basic stuff is easy to proof and i think questions like "find a good reason" are nonsense, at least if they are answered with physical examples, but maybe i'm too much of a mathematician.

Peter

[quote="Peter Scholze"]but maybe i'm too much of a mathematician.

Peter[/quote]

is that saying the rest of us aren't?  :)

Well, take a and b two positive real numbers, then

[hide]

[hide]

and hence we conclude that:

[/hide][hide][/hide][/hide]


inequalities
function
probability and stats

X is nondegenerate positive random value with finite mathematical expectation. Proof:
   (1) 1/E[X]<E[1/X]
   (2) E[log(X)]<log(E[X])
I need arguments for strict inequality.

What you're looking for is this way to say Jensen's inequality:

Let $ I$ be an interval (which could be infinite). Suppose $ \varphi: I\to\mathbb{R}$ is [i]strictly convex[/i]. Suppose $ X$ is a random variable taking values in $ I$ that has an expected value. (That is, $ X\in L^1.$) Suppose that $ X$ is nondegenerate in the sense that there is no $ m$ such that $ P(X = m) = 1.$ Then
\[ \varphi(E(X)) < E(\varphi(X))
\]
I'll let you think about how you would define "strictly convex" and how you would test such a property. Whatever definition you settle on, you'll need this version of the [i]supporting line lemma[/i]:

Suppose $ \varphi$ is strictly convex on $ I$ and $ m\in I.$ Then there exists at least one linear function $ L(x) = ax + b$ (that's an affine function, if you're being picky) such that $ L(m) = \varphi(m)$ but for all $ x\in I\setminus\{m\},$ $ L(x) < \varphi(x).$

Now follow the standard proof of Jensen, using that version of the supporting line lemma and just pay close attention to the criteria for equality.

Comment on #1: $ \varphi(x) = \frac1x$ is strictly convex on $ (0,\infty)$ but not appropriately behaved outside that interval. Hence #1 would have to have the stipulation that $ X$ takes only positive values.


limit
calculus
integration
real analysis
real analysis unsolved

Problem 1:
Let $ p$ is natural number such that $ p\ne 0$. Compute the limit: $ \mathop {\lim }\limits_{n \to \infty } \frac {{\sum\limits_{i = 1}^n {{i^p}} }}{{{n^{p + 1}}}}$ (*)

Then apply the result (*) to compute the limit: $ \mathop {\lim }\limits_{n \to \infty } \frac {{{1^{2008}} + {2^{2008}} + ..... + {{(n - 1)}^{2008}} + {n^{2008}}}}{{{n^{2009}}}}$

Problem 2:
Prove that: $ {\lim }\limits_{n \to \infty } \frac {{\sum\limits_{i = 1}^n {\sqrt i } }}{{\sqrt {{n^3}} }} = \frac {2}{3}$.

Have fun,  :)

$ \lim_{n\rightarrow \infty }\frac {1}{n^{p \plus{} 1}}\sum\limits_{i \equal{} 1}^{n}i^{p} \equal{} \frac {1}{p \plus{} 1}$ is a well known result.

It can be deduced using the sum of a GP.

More generally, $ \frac {n^{p \plus{} 1}}{p \plus{} 1} < \sum\limits_{i \equal{} 1}^{n}i^{p} < \left( 1 \plus{} \frac {1}{n}\right) ^{p \plus{} 1}\frac {n^{p \plus{} 1}}{p \plus{} 1}$

Hence, $ \lim_{n\rightarrow \infty }\frac {1^{2008} \plus{} 2^{2008} \plus{} ..... \plus{} (n \minus{} 1)^{2008} \plus{} n^{2008}}{n^{2009}} \equal{} \frac {1}{2009}$

And so problem (2) is obvious.

You can bound the series using suitable integrals:

$ \int_{1}^{n\plus{}1}x^{p}dx\le\sum_{i\equal{}1}^{n}i^{p}\le1\plus{}\int_{1}^{n}x^{p}dx$

and problem 2 can be solved similarly.

All those problems can be solved just using the Limit of Riemman Sum. :wink:

\[ \boxed{\lim_{n\to\infty} \frac{1}{n}\sum_{k\equal{}1}^n f\left(\frac{k}{n}\right)\equal{}\int_0^1 f(x)\ dx}\]


function
algebra
rational function
real analysis
real analysis solved

According to my prof and the text book any rational function of multiple variables is continous on its Domain.
However, if the function is defined piecewise it may be discontinous.
For example,
$f(x,y)=\{\begin{array}[c]{c}\,\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\text{ if }(x,y)\neq(0,0)\ 0\text{ if }(x,y)=(0,0)\end{array}$
is discontinous because $lim_{(x,y)\rightarrow(0,0)}$ Does not exist.

What about:
$f(x,y)=\{\begin{array}[c]{c}\,\frac{x^{2}y^{3}}{2x^{2}+y^{2}}\text{ if }(x,y)\neq(0,0)\ 1\text{ if }(x,y)=(0,0)\end{array}$
Would this only be continous if $lim_{(x,y)\rightarrow(0,0)}=1$ or will it be continous is the limit exists, no matter what the value of that limit?

That would be discontinuous at the origin, even though the limit at the origin does exist (and equals 0.) One can say that the continuity of a function at a point requires three things: (1) the function is defined at that point, (2) the limit as one approaches that point exists, and (3) the value agrees with the limit. Your example has both (1) and (2) but not (3).

At any point at which the limit exists, a function has at worst a "removable discontinuity." Removable in that if we were to go back and redefine (or define in the first place) the function at that point so that is has the "correct" value, then we would make it continuous.


inequalities
logarithms
function
inequalities unsolved

Cho $ a,b,c$ be positive real numbers such that  $ abc\equal{}1$.Prove that:
 $ 3\leq \sum\frac{a(3a\plus{}1)}{(a\plus{}1)^2}\leq a\plus{}b\plus{}c$

[quote="math10"]Cho $ a,b,c$ be positive real numbers such that  $ abc \equal{} 1$.Prove that:
 $ 3\leq \sum\frac {a(3a \plus{} 1)}{(a \plus{} 1)^2}\leq a \plus{} b \plus{} c$[/quote]
The right inequality can be done by rewritting
$ \sum_{cyc}\frac{a(3a\plus{}1)}{(a\plus{}1)^2}\le a\plus{}b\plus{}c\quad\Leftrightarrow\quad\sum_{cyc}\left (\frac{a(3a\plus{}1)}{(a\plus{}1)^2}\minus{}a\plus{}\frac{1}{4}\ln (a)\right )\le 0.$

[quote="math10"]Why??I don't understand your solution.[/quote]
[1] Note that $ \sum_{cyc}\frac {1}{4}\ln (a)$ with $ abc \equal{} 1$ gives $ \frac {1}{4}\ln (abc) \equal{} 0$.

[2] Consider the function
$ f(x): \equal{} \frac {x(3x \plus{} 1)}{(x \plus{} 1)^2} \minus{} x \plus{} \frac {1}{4}\ln (x),\qquad x > 0.$
Then
$ f^{\prime}(x) \equal{} \frac {1}{4x} \minus{} \frac {x(x^2 \plus{} 3x \minus{} 2)}{(x \plus{} 1)^3} \equal{} \minus{} \frac {(x \minus{} 1)(4x^3 \plus{} 15x^2 \plus{} 4x \plus{} 1)}{(x \plus{} 1)^3}.$
Hence, $ \max_{x > 0}(f(x)) \equal{} f(1) \equal{} 0$. Finally, this implies that for all $ x > 0$ the inequality $ f(x)\le f(1)\equal{}0$ holds. The proof is finished by assummptions on the numbers $ a,b,c$, namely that $ a,b,c\in\mathbb{R}_ \plus{}$ s.t. $ abc \equal{} 1$.

I think here is a correct solution, but not beautiful because it must performance $ f'(x)$

[quote="math10"]Cho $ a,b,c$ be positive real numbers such that  $ abc \equal{} 1$.Prove that:
 $ 3\leq \sum\frac {a(3a \plus{} 1)}{(a \plus{} 1)^2}\leq a \plus{} b \plus{} c$[/quote]

Hint :

$ (1)\ \ \frac {a(3a \plus{} 1)}{(a \plus{} 1)^2}\leq \frac{3}{4}a\plus{}\frac{1}{4}$

$ (2)\ \ (\sum \frac {a(3a \plus{} 1)}{(a \plus{} 1)^2})(\sum \frac{(3a\plus{}1)(a\plus{}1)^2}{a}) \ge 9(a\plus{}b\plus{}c\plus{}1)^2$

I must agree with Mathias DK, it is extraordinarily fabulous when you established that remarkable inequality

[quote="math10"]Cho $ a,b,c$ be positive real numbers such that  $ abc \equal{} 1$.Prove that:
 $ 3\leq \sum\frac {a(3a \plus{} 1)}{(a \plus{} 1)^2}\leq a \plus{} b \plus{} c$[/quote]
Ineq in LHS is obviously trues by Am-Gm. With ineq in RHS, using Am=Gm, we obtain an old result:
$ \sum\ \frac{a^2}{(a\plus{}b)^2}\plus{} \frac{2abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq\ 1$ (obviously trues by Am-Gm again) :)

[quote="Mathias_DK"][quote="M@ri@n"][quote="math10"]Why??I don't understand your solution.[/quote]
[1] Note that $ \sum_{cyc}\frac {1}{4}\ln (a)$ with $ abc \equal{} 1$ gives $ \frac {1}{4}\ln (abc) \equal{} 0$.

[2] Consider the function
$ f(x): \equal{} \frac {x(3x \plus{} 1)}{(x \plus{} 1)^2} \minus{} x \plus{} \frac {1}{4}\ln (x),\qquad x > 0.$
Then
$ f^{\prime}(x) \equal{} \frac {1}{4x} \minus{} \frac {x(x^2 \plus{} 3x \minus{} 2)}{(x \plus{} 1)^3} \equal{} \minus{} \frac {(x \minus{} 1)(4x^3 \plus{} 15x^2 \plus{} 4x \plus{} 1)}{(x \plus{} 1)^3}.$
Hence, $ \max_{x > 0}(f(x)) \equal{} f(1) \equal{} 0$. Finally, this implies that for all $ x > 0$ the inequality $ f(x)\le f(1) \equal{} 0$ holds. The proof is finished by assummptions on the numbers $ a,b,c$, namely that $ a,b,c\in\mathbb{R}_ \plus{}$ s.t. $ abc \equal{} 1$.[/quote]
Very, very, very cool! :)[/quote]

The solution is often used by arqady, silouan and by me. :lol:


Calculate the total charge withing each of the indicated volumes a) $ 0.1\le|x|,|y|,|z|\le0.2$: $ \rho_v \equal{} \frac{1}{xyz}$, b) $ 0\le\rho\le, 0\le\phi\le\pi, 2\le{z}\le4$: $ \rho_v \equal{} \rho^{2}z^{2}sin0.6\phi$, c) universe: $ \rho_v \equal{} \frac{e^{\minus{}2r}}{r^2}$

Answers: 0;1.018;6.28 C

In b) it's missing the superior limit of $ \rho$.

(Seems to be ~ 0.562 ???)


Show that if the smallest prime factor $ p$ of the positive integer $ n$ exceeds $ \sqrt[3]{n}$ then $ \frac{n}{p}$ must be prime or $ 1$.

if $ \frac{n}{p}>1$ is not a prime
then $ \frac{n}{p}$ have at last $ 2$ prmie factor $ \ge p$
so $ \frac{n}{p}\ge p^{2}$ gives $ p\le \sqrt[3]{n}$ (contradiction !!)
then $ \frac{n}{p}=1$ or it's a prime


modular arithmetic
number theory proposed
number theory

This is completely straightwforward: if $ 7$ divides any factor, it divides all of them.
The generalisation is obvious:
Let $ p$ be prime and let $ S\equal{} \prod_{k\equal{}1} ^{\frac{p\minus{}1}2} (kx \plus{} (p\minus{}k)y)$ for some $ x,y$. Then if $ p|S$, then $ p^{\frac{p\minus{}1}2}|S$.

[b]Lemma:[/b]  If $ p$ is prime, then $ p | ab \implies p | a \text{ or } p | b$.

[b]Corollary:[/b]  $ p | a^n \implies p | a \implies p^n | a^n$.

Note that $ a$ only has to be defined $ \bmod p$.  Here $ a \equiv x \minus{} y \bmod 7$.  The given expresson is $ 6(x \minus{} y)^3 \bmod 7$.

Notice that $ x\plus{}6y\equiv 2x\plus{}5y \equiv 3x\plus{}4y \pmod7$ (since $ x\plus{}6y\equiv x\minus{}y \pmod7$). But $ 7$ must divide at least one of $ (x\plus{}6y)$, $ (2x\plus{}5y)$, $ (3x\plus{}4y)$, so it divides all of them. Therefore $ 7\times7\times7\equal{}343$ divides $ (x\plus{}6y)(2x\plus{}5y)(3\plus{}4y)$. Q.E.D.

[quote=vedran6]Let $ x,y,z$ be positive integers. If $ 7$ divides $ (x\plus{}6y)(2x\plus{}5y)(3x\plus{}4y)$ than prove that $ 343$ also divides it.
(I am searcing for some kind of generalisation  :D )[/quote]

$Lemma$:$(a+b)|(ax + by)$ $\implies$ $(x-y) \equiv 0$ $(mod$ $a+b)$ where $a,b>0$.
$Proof$:$ax + by \equiv 0$ $(mod$ $a + b)$ $\implies$ $ax + (a + b - a)y \equiv 0$ $(mod$ $a + b)$ or $ax - ay \equiv 0 $ $(mod$ $a+b)$ and since $a$ doesn't divide $(a+b)$ we get $(x-y) \equiv 0$ $(mod$ $a+b)$.
We apply this $Lemma$ and we get that $x \equiv y$$(mod$ $7)$ $\implies$ $7$ divides $(x+6y)$,$(2x+5y)$,$(3x+4y)$ $\implies$ $7^3 = 343$ divides $(x+6y)(2x+5y)(3x+4y)$ and we are done.

$x+6y\equiv 2x+5y\equiv 3x+4y\equiv (x–y) ,2(x–y), 3(x–y) (mod7)$ all bears $x–y$
But$ 7| P(x,y,z)$
 So it must divide at least one of them if it divides one of the given no it will divide all since same$ mod7$ ,claim
$7|abc\implies 7|a,7|b,7|c$ if same mods

x+6y mod 7 is 6(y-x) mod 7, similarly the others are 5(y-x) and 4(y-x) mod 7. Then at least one of these must be 0 mod 7, or y-x must be 0 mod 7, so all of them are multiples of 7, product is 0 mod 343.


inequalities
function
algebra proposed
algebra

Find all $a\in\mathbb{R}$ for which there exists a non-constant function $f: (0,1]\rightarrow\mathbb{R}$ such that \[a+f(x+y-xy)+f(x)f(y)\leq f(x)+f(y)\] for all $x,y\in(0,1].$

$ x \equal{} y \equal{} 1$ gives $ a\leq f(1) \minus{} f(1)^{2} \equal{} {1\over 4} \minus{} (f(1) \minus{} {1\over 2})^{2}\leq{1\over 4}$.

For $ a < {1\over 4}$ we can take $ f: (0,1]\to[{1\over 2},\min\{{3\over 4} \minus{} a,1\}]$ arbitrary.
Then
$ a \plus{} f(x \plus{} y \minus{} xy) \plus{} f(x)f(y) \minus{} ( f(x) \plus{} f(y)) \\
\hspace*{0.5in} \equal{} a \plus{} f(x \plus{} y \minus{} xy) \plus{} (1 \minus{} f(x))(1 \minus{} f(y)) \minus{} 1\leq a \plus{} ({3\over 4} \minus{} a) \plus{} {1\over 2}\cdot{1\over 2} \minus{} 1 \equal{} 0$
 for all $ x,y\in (0,1]$.

For $ a \equal{} {1\over 4}$ we get $ (f(1) \minus{} {1\over 2})^{2}\leq 0$ and $ f(1) \equal{} {1\over 2}$.
Then $ y \equal{} 1$ gives $ f(x)\geq{1\over 2}$ for all $ x\in(0,1]$.
$ x \equal{} y$ gives $ f(2x \minus{} x^{2})\leq 2f(x) \minus{} f(x)^{2} \minus{} {1\over 4} \equal{} {3\over 4} \minus{} (1 \minus{} f(x))^{2}\leq{3\over 4}$ for all $ x\in(0,1]$.
But $ 2x \minus{} x^{2} \equal{} 1 \minus{} (1 \minus{} x)^{2}$ runs through all of $ (0,1]$ for $ x\in(0,1]$.
So $ t \equal{} \sup\{f(x): x\in(0,1]\}\in [{1\over 2},{3\over 4}]$ exists.
But then $ t \equal{} \sup_{x\in(0,1]}f(2x \minus{} x^{2}) \leq \sup_{x\in(0,1]}({3\over 4} \minus{} (1 \minus{} f(x))^{2}) \equal{} {3\over 4} \minus{} (1 \minus{} t)^{2}$
and $ t \equal{} {1\over 2}$.
So $ f\equiv{1\over 2}$ constant. Contradiction!

[quote="prowler"]Nice solution, I am just wondering if this problem were proposed by O. Mushkarov ?[/quote]
No, it was proposed by Nikolay Nikolov.


trigonometry
Euler
function
LaTeX
complex analysis
complex analysis solved

Prove that any real number x satisfies :
1/sin(x) = ... + 1/(x - 2\pi) - 1/(x - \pi) + 1/x - 1/(x + \pi) + 1/(x + 2\pi) - ...

Hi Arne,

It seems that the Euler product fro sinus function will help, here an intuitive proof

sinx = x(1-x/ \pi )(1-x/(2 \pi ))(1-x/(3 \pi ))......

(*) 1/sinx = ...+ A(-2)/(x-2 \pi )+A(-1)/(x- \pi )+ A(0)/x + A(1)/(x+ \pi )+A(2)/(x+2 \pi ) +....

to find A(k) multiply (*) both side by x+k \pi  and evaluate x = -k \pi

chers

oubi

my friends try to write with $lateX$

thank you:D


This will be my first and last year in mathcounts because I only heard about mathcounts this summer.  Unfortunately my school doesn't have a mathcounts team.  I'm not sure whether I should try to start a mathcounts team at my school or enter as an individual.

1>  I'm not sure my school team will make it past chapter.  If I make it past chapter will I still get to advance to state if the rest of my team doesn't?

2>  If you enter as an individual and advance to state,  can you be on the state team?

One other school in the district has a mathcounts team.  I need advise please!

1.Um...try to convince your school to make a team. Then in case you don't move on as an individual, you can still move on as a team.
2.Almost no national teams (if any) have ever consisted of only students from one school, so yes.

The top 4 in each state (depending on your state, countdown may determine this) will make up the state team at nationals.

Most chapters do send top individuals to state competitions. You will then compete at state as an individual along with other participants who are there as an individual. If you are at state as an individual, you will take the sprint and target rounds individually - just as everyone else does. You will then likely be grouped up with other individual participants to work on the team round. However, these teams full of individuals are not eligible for team awards - only individual awards. Those participating in the countdown round at your state will be based on the individual test scores, so you could make that as an individual.

Each chapter is different. Contact your chapter coordinator for details. You can find out who your chapter coordinator is by going to the MATHCOUNTS website at http://www.mathcounts.org and clicking the Find a Coordinator link.

in my state, the top 4 ppl in every chapter who's teams did not qualify form a chapter individual team.  When I was in 6th grade, one of the ppl who went to nats with me made it that way, so it's definitely possible.  I'm not sure if this is tru for all states, my state is very small, and has only 4 chapters, a state with many chapters might not be able to do this

I would say in your suma_milli try to create a team but if can't do that just enter as an indiv. Also, try out for PMWC cause the problems (i think) are about the same level and aren't very hard but the time limit just kills.  :mad:

Thanks everyone!

How do you try to get a team started?

just put up posters, talk to teachers, talk to your friends, talk to principal for funding and so on.

Yeah well if your school doesn't have MC then that's very difficult to do... :wink:

But good luck! :D

suma you're in 8th grade right?

umm if you really want more info on how to go to Mathcounts if your school doesn't offer it: check this link: it has been asked before:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1916034886&t=203712[/url]

first things first- Make sure first talk to your teachers/principal about the program and what you want to do. They can probably help you, then go from there. Good Luck

I would say try to smart a team if you want to. Even though you might not be a good team in the first year, who knows? Maybe you will volunteer with MC or otherwise just hear about the team at your school doing awesome! You would know that you started a tradition that carried on and developed. :D I dunno, its totally your decision (but if it were me, someone without a MC program already at their school)...


In der  Zeitschrift '' Wurzel '' habe ich diesen Geometrie - begriff   gesehen :

[color=red][b]  Tebaultpunkt[/b][/color]

 Was versteht ihr in Deutschland unter diesem ''term'' ?

 Leider , haben wir in Griechenland diesen Punkt nicht.

  Vielen Dank !

 Babis Stergiou , Griechenland

Babis, ich habe mal der Wurzel-Redaktion eine Nachricht gesandt und nach der Definition des Punktes gefragt.

Hallo,

ich bin gerade ein paar hundert Kilometer weg von meinen $ \sqrt {\text{Wurzeln}}$, aber ich [i]glaube[/i], der Thebaultpunkt bezeichnet (in Kulanins Artikeln) folgenden Punkt:

Seien X, Y und Z die Fußpunkte der Höhen des Dreiecks ABC. Dann schneiden sich die Eulergeraden der Dreiecke AYZ, BZX und CXY in einem Punkt auf dem Feuerbachkreis von Dreieck ABC. Vermutlich ist dies der Thebaultpunkt.

Ich habe vergessen, welcher Punkt in Kimberling das war, aber ich glaube, einer um X(113) herum ("centers 113-139 lie on the nine-point circle.")

  darij


vector
induction
linear algebra
combinatorics solved
combinatorics

I took this problem from an unofficial Italian local contest (February 2004), but I'm pretty sure it was not original and was taken from some book. None of the contestants could solve it.

Let $n>1$, and let $A_1, A_2, \ldots, A_{n+1}$ be non-empty subsets of $X=\{1, 2, \ldots, n\}$. Prove that there exist non-empty disjoint subsets $I$ and $J$ of $\{1, 2, \ldots, n+1\}$, such that $\displaystyle \bigcup_{i \in I} A_i = \displaystyle \bigcup_{i \in J} A_i$.

Actuall, I know that this problem was proposed at a Chinese math competition in 2002 or 2003.
It is a hard problem, for a solution ask "iura"

Nice problem :)
(Nice Avatar too !)

Any answer ?

Thanks!   :cool:
Unfortunately, I had to resize the picture because it was too large to fit in my profile. However, I've just improved the quality of the image (brighter colours, lower compression, and so on): the previous version was just experimental!

About the problem, I must say that nobody I know has found an elementary solution, not even the organizers of the contest, who "stole" it from that Chinese competition. They just thought it was [i]much[/i] easier, and gave a wrong solution... After discovering their mistake (some days later), they removed the problem.   :D    :D    :D

This question can be solved by use of linear algebra method. We can identify in an obvious way every subset of {1,...,n} to a (0,1)-vector in the vector space $ \mathbb{Q}^n$. The vectors representing the subsets $ A_1,..., A_{n\plus{}1}$ are Q-linearly dependant. By writing down a non trivial Q-linear relation among those vectors, we get
$ \sum_{i\in I}a_iA_i\equal{}\sum_{j\in J}a_jA_j$ for two disjoint and non empty subsets I and J and with all the $ a_k$'s positive integers.

I found the above  at the url
[url]http://les-mathematiques.u-strasbg.fr/phorum/list.php?f=2[/url]

Trivector

Hi!
Could you please be a little more explicit, because you are both french,(I suppose), and the linki above is in french :) :D

call degree of an element the number of subsets the elements belongs to.
(1)Suppose every element in the set has degree >1. The we prove a stronger result:
There exist two disjoint groups of subsets whose union give A.
Call an element weak if it has degree 2.
We claim there is a subset containing at most one weak element.
Proof:suppose every subset contains at least 2 weak elements. Summing over n+1 subsets and diving by 2 (every weak element belongs to 2 subsets) we get at least n+1 weak elemnts, which is absurde.
So let A_1 be the set containing at most one weak element. Call the weak element x. Let A_2 be the other subset containing x.
By removing A_1,x, and putting A_i'=A//{x} we use induction.
So there are two groups of subsets A and B s.t. union {sets from A}=union{sets from B}=A/{x}.
Now, if none of groups contain x, we add A_1 to one of them, A_2 to another.
If one of them contains x, we add A_1 to another.
If both of them contain x, we are done.
(1) is proved (note that is a subset contains one element, we can remove it and its element directly, without finding A_1)
Now pass to the general case. Let Z be the set of elements having degree at least 2.
I claim there are at least |Z|+1 subsets contained in Z.
Suppose not. Then there are at least n+1-|Z| subsets containing an element with degree 1.
Summing we get at least n+1-|Z| elements of degree 1, so a total of n+1 elements, impossible.
Hence there are at least |Z|+1 subsets contained in Z. If |Z|=n, then we apply (1).
If not, we apply induction for Z which has less elements

To Dust :

Here is a link to a solution more explicit and written by GeorgesZ from the above-mentionned french forum.

[url]http://minilien.com/?9iN1rADVOM[/url]

Don't hesitate to ask if you feel this solution don't clear things up.

Trivector

[quote="iura"]We claim there is a subset containing at most one weak element.
[/quote]
And what about case $A_i=\{1,2,...,n\}$ for all i?

Any element belongs to n+1 subsets, so  it's not weak.
So we could tahe A_1.
I think I should correct the proof:
There is AT LEAST ONE subset containing AT MOST one weak element.
If this subset contains no weak element, then we choose x to be any element of A_1

Thanks, it wasn't that hard, after all!  :maybe:
I proved that we could wlog suppose that every element had degree > 1, but then I got stuck. So, the key ideas are that in this case we have at least a subset containing at most one element with degree = 2, and that we can take our subsets so that their union is the whole set. Then, it becomes trivial.

Note, however, that your "general case" is unnecessarily messy. If an element has degree = 0, we forget about this element and about a random subset. If an element has degree = 1, we forget about this element and about the subset containing it. In the end, we have all elements with degree > 1.

[quote="Gillian_Seed"]The proof with vectors is much funnier ^^[/quote]
Maybe it's shorter, but it uses the concept of linear dependency, so it's not an olympic proof. However, iura's proof can be rewritten in a much shorter and clearer way, and I think it gives more insight into the problem than the other one. I mean, it really explains how things work, instead of hiding everything behind linear algebra.

Here's a cool related problem which, according to what I've read, has no known elementary solution (the non-elementary one uses linear algebra very similar to what was done here):

Given $n+2$ subsets of an $n$-element set, prove that there are two disjoint groups of such sets s.t. the both the unions and the intersections of the sets in the groups are equal.

It's called Lindstrom's Theorem, I think.

[quote="MindFlyer"]
About the problem, I must say that nobody I know has found an elementary solution, not even the organizers of the contest, who "stole" it from that Chinese competition. They just thought it was [i]much[/i] easier, and gave a wrong solution... After discovering their mistake (some days later), they removed the problem.   :D    :D    :D[/quote]
Yes, this problem originally comes from cwmo 2002. It was proposed by a chinese MO trainer (趙敦, sorry, i dunno his eng name... :blush: )
Another elegant solution using linear algebra was done by a chinese professor (潘承彪), who was also a committee member of cwmo 2002.


If V is an n-dimensional vector space and 0 <= m <= n, show that the number of m dimensional subspaces of V is the same as the number of n-m dimensional vector spaces.

Here is how I think this is solved. Extend a basis for the m-dimensional subspace U to a basis Vfor V (we assume m < n). The elements of B which do not belong to U form an n-m dimensional subspace and this subspace is obviously uniquely determined by U. Hence there is an injection f between m dimensional subspaces of V and n-m dimensional subspaces of V. Repeating the procedure with m replaced with n-m shows that f is a bijection.

Does this sufficiently answer the question? The wording of the question implies that there's a finite number of m dimensional subspaces of V but I don't see any way of counting them.

[quote="§outh§tar"]this subspace is obviously uniquely determined by U.[/quote]"Obviously" is a dangerous word.  In particular, a basis of a subspace can be extended to a basis of the entire space in many ways.

The number of subspaces of a given space will be finite if the underlying field is finite.  If the underlying field is infinite, for any $ k$, $ 1\leq k\leq n\minus{}1$, there are infinitely many $ k$-dimensional subspaces of an $ n$-dimensional field.  For example, if $ n\geq 2$ then we get a distinct 1-D subspace for each element of the field by considering the span of the vector $ (a, 1,\ldots)$ where the other slots are filled arbitrarily.

In the case of a finite field, counting the number of subspaces is not that hard -- just build up one degree at a time and be careful of the problem you've already stumbled across, namely nonuniqueness of bases.

If you don't want to count them directly, here's another approach: choose an "inner product" and look at orthogonal complements. Let $ f$ be a bilinear map from $ V\times V$ to the field $ k$, $ f(x,y)\equal{}f(y,x)$, and suppose that $ f(a,y)\equal{}0$ for all $ y$ only if $ a\equal{}0$. (This can always be done; on $ k^{n}$, use $ f(x,y)\equal{}\sum_{i}x_{i}y_{i}$)

For each set $ S$, the orthogonal complement $ S^{\perp}$ of $ S$ is $ \{x: f(x,y)\equal{}0\text{ for all }y\in S\}$. If $ S$ is also a subspace of dimension $ n$, $ S^{\perp}$ is a subspace of dimension $ n\minus{}m$, and $ (S^{\perp})^{\perp}\equal{}S$. That's a bijection.

Note: $ S^{\perp}$ and $ S$ may have nontrivial intersection.


中日关系已十分恶劣，如果长此以往，必然导致政治危机。大家怎么看待这个问题？

[quote="wuhao_jshj"]中日关系已十分恶劣，如果长此以往，必然导致政治危机。大家怎么看待这个问题？[/quote]
多增强贸易上的往来.

Well, I think a war(could be political, or a real one) is a must-be between China and Japan, unless China changes itself so rapidly that it can overcome all Japanese influences within very few years , at that time, Japan will stop acting against China.

A real war is not very likely to break out between China and Japan. What if China uses nuclear weapons? In the end, only both parts will have an immense economic strike and enter a recession in almost every areas. Of course, an economic war may and very likely to break out in the foreseeable future.

:!:  I don't think that would be a war, maybe we can get a better method to deal with the friction.


[quote="The Enigma"]hey friends this is a nice one-
for all $ x,y,z$ positive reals such that $ xyz \equal{} 1$ .
P.T $ x \plus{} y \plus{} z \ge \frac {2xy}{1 \plus{} y} \plus{} \frac {2yz}{1 \plus{} z} \plus{} \frac {2zx}{1 \plus{} x}$[/quote]
Let $ x \equal{} \frac{a}{b},y \plus{} \frac{b}{c},z \equal{} \frac{c}{a}$
then we obtain
$ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \ge \frac{{2a}}{{b \plus{} c}} \plus{} \frac{{2b}}{{a \plus{} c}} \plus{} \frac{{2c}}{{a \plus{} b}}$
which is very well known :)

[quote="TRAN THAI HUNG"]
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {{2a}}{{b \plus{} c}} \plus{} \frac {{2b}}{{a \plus{} c}} \plus{} \frac {{2c}}{{a \plus{} b}}$
which is very well known :)[/quote]
Write it down, Hung, don't just say it  :wink:

dear humg i got the same solution! but i am looking other kinds of solutions.
something without transformations. :lol:

[quote="TRAN THAI HUNG"]
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {{2a}}{{b \plus{} c}} \plus{} \frac {{2b}}{{a \plus{} c}} \plus{} \frac {{2c}}{{a \plus{} b}}$
which is very well known :)[/quote]
Indeed, it's well-known. But it's only trues with $ a, b, c$ are three sides of a triangle  :wink:
With $ a, b, c > 0$, we only have: $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \sum\ \frac{a}{b\plus{}c}\plus{}\frac{3}{2}$

[quote="nguoivn"][quote="TRAN THAI HUNG"]
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {{2a}}{{b \plus{} c}} \plus{} \frac {{2b}}{{a \plus{} c}} \plus{} \frac {{2c}}{{a \plus{} b}}$
which is very well known :)[/quote]
Indeed, it's well-known. But it's only trues with $ a, b, c$ are three sides of a triangle  :wink:
With $ a, b, c > 0$, we only have: $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \sum\ \frac {a}{b \plus{} c} \plus{} \frac {3}{2}$[/quote]
I have solved this inequality long time ago so I don't know exactly the condition. However, according to your condition, the original inequality have to have some more condition :) .Anyway I think it is true for any x,y,z.I will try to sovle it in another way.
To nguoivn [hide]Have you solved my inquality? [/hide]

[quote="The Enigma"]hey friends this is a nice one-
for all $ x,y,z$ positive reals such that $ xyz \equal{} 1$ .
P.T $ x \plus{} y \plus{} z \ge \frac {2xy}{1 \plus{} y} \plus{} \frac {2yz}{1 \plus{} z} \plus{} \frac {2zx}{1 \plus{} x}$[/quote]
This is not true, for counter-example,  $ \ \ \ \ \ \  x\equal{}5,\, y\equal{}2,\ z\equal{}\frac{1}{10}.$


induction
number theory unsolved
number theory

Find all sets $A$ and $B$ that satisfy the following conditions:

a) $A \cup B= \mathbb{Z}$;

b) if $x \in A$ then $x-1 \in B$;

c) if $x,y \in B$ then $x+y \in A$.

[i]Laurentiu Panaitopol[/i]

I believe it has been discussed before on the forum. It was used in the first Romanian TST in 2002, if I am not mistaken.

[quote="socrates"]A is the set of the even and B of the odd numbers![/quote]
$A=B=\mathbb{Z}$ is also the answer.

If 0 was in B, by c) 0 is also in A, so 0 is in A. By b) -1 is in B, by c) -2 is in A, by c) -3 is in B. By induction if
-2k+1 is in B, by c) -2k is in A and -2k-1 is in B, so all negative odd numbers are in B. By b) , all negative even numbers are in A.
If 1 iwas not in B, 1 would be in A, by b) 0 would be in B, so by c) if x is in B, x is also in A, so all numbers would be in A. Because of that all numbers would be in B to, contradiction. So 1 is in  B. By c) 2 is in b. By induction it's simple to show that all positive odds are in B and all positiv evens in A.
If there were no other numbers in the sets, the solution wolud be set of evens and set of odds.
If there was an odd number in A, there would be by b) an even number in A. So in any other solution then set of evens and set of odds there must be an even number in B. But then 1 is in A, 0 in B so all the numbers from B are in A too and by b) all are in A as well.
So the only one solutions are (evens,odds), ($\mathbb Z$,$\mathbb Z$).


geometry
circumcircle
incenter
3D geometry
prism
trapezoid
quadratics

Let $ABCD$ be a bicentric quadrilateral with the circumcircle $(O,R)$ and the incenter $I$. If $P$ is the cut point of $AC$ and $BD$ besides $IO=d$. Prove that:$OP= \frac{2R^{2}d}{R^{2}+d^{2}}$

Let $(I)$ be the quadrilateral incircle and $r$ the inradius. Since $O, I, P$ are collinear, $PO = PI+d$ and we can calculate $PI$ instead. Because of Poncelet's porism, it is sufficient to calculate $PI$ for a bicentric (isosceles) trapezoid. Let $AB \parallel CD$ be the trapezoid bases, $AB > CD,\ BC = DA.$ Let $E, F, G, H$ be the tangency points of $AB, BC, CD, DA$ with the incircle $(I).$ The tangency points $E, G$ are the midpoints of the bases $AB, CD$. Let $M, N$ be the midpoints of the shoulder sides $BC, DA$. Since $MN = \frac{AB+CD}{2}$ for a trapezoid and $AB+CD = BC+DA$ for a tangential quadrilateral, the trapezoid midline is equal to its shoulder sides, $MN = BC = DA.$ The incenter $I$ is the midpoint of the midline $MN$, the median $IM$ of the triangle $\triangle BCI$ is equal to half of its side $BC$, hence the angle $\angle BIC = 90^\circ$ is right. Then the angles $\angle BIE+\angle CIG = 90^\circ$ add up to a right angle, the right angle triangles $\triangle BIE \sim \triangle ICG$ are similar and $\frac{BE}{IE}= \frac{IG}{CG}.$ By Pythagorean theorem,

$BF^{2}= BE^{2}= OB^{2}-OE^{2}= OB^{2}-(IB-IO)^{2}= R^{2}-(r-d)^{2}$

$CF^{2}= CG^{2}= OC^{2}-OG^{2}= OC^{2}-(IC+IO)^{2}= R^{2}-(r+d)^{2}$

$\left(\frac{BE}{IE}\right)^{2}= \left(\frac{IG}{CG}\right)^{2}\ \ \Longrightarrow\ \ \frac{R^{2}-(r-d)^{2}}{r^{2}}= \frac{r^{2}}{R^{2}-(r+d)^{2}}$

$(R^{2}-r^{2}-d^{2})^{2}-4r^{2}d^{2}= r^{4}\ \Longrightarrow\ r^{2}= \frac{(R^{2}-d^{2})^{2}}{2(R^{2}+d^{2})}$

In addition, $\frac{PI}{GI}= \frac{FM}{CM},$ where $GI = r,\ FM = \frac{BF-CF}{2},\ CM = \frac{BF+CF}{2},$ so that

$\frac{PI}{r}= \frac{BF-CF}{BF+CF}= \frac{(BF-CF)^{2}}{BF^{2}-CF^{2}}$

$BF^{2}-CF^{2}= 4rd$

$(BF-CF)^{2}= 2[R^{2}-(r^{2}+d^{2})]^{2}-2\sqrt{[R^{2}-(r-d)^{2}][R^{2}-(r+d)^{2}]}=$

$= 2[R^{2}-(r^{2}+d^{2})]^{2}-2\sqrt{[R^{2}-(r^{2}+d^{2})]^{2}-4r^{2}d^{2}}$

Substituting for $r^{2}:$

$R^{2}-(r^{2}+d^{2}) = \frac{2(R^{2}+d^{2})(R^{2}-d^{2})-(R^{2}-d^{2})^{2}}{2(R^{2}+d^{2})}=$

$= \frac{(R^{2}+d^{2})^{2}-4d^{4}}{2(R^{2}+d^{2})}= \frac{(R^{2}+3d^{2})(R^{2}-d^{2})}{2(R^{2}+d^{2})}$

$\sqrt{[R^{2}-(r^{2}+d^{2})]^{2}-4r^{2}d^{2}}= \sqrt{\frac{(R^{2}+3d^{2})^{2}(R^{2}-d^{2})^{2}}{4(R^{2}+d^{2})^{2}}-\frac{2d^{2}(R^{2}-d^{2})^{2}}{R^{2}+d^{2}}}=$

$= \frac{R^{2}-d^{2}}{2(R^{2}+d^{2})}\sqrt{(R^{2}+3d^{2})^{2}-8d^{2}(R^{2}+d^{2})}= \frac{(R^{2}-d^{2})^{2}}{2(R^{2}+d^{2})}$

Substituting these expressions into $(BF-CF)^{2}$ and $\frac{PI}{r}:$

$(BF-CF)^{2}= \frac{(R^{2}+3d^{2})(R^{2}-d^{2})-(R^{2}-d^{2})^{2}}{R^{2}+d^{2}}= \frac{4d^{2}(R^{2}-d^{2})}{R^{2}+d^{2}}$

$\frac{PI}{r}= \frac{(BF-CF)^{2}}{BF^{2}-CF^{2}}= \frac{4d^{2}(R^{2}-d^{2})}{4rd(R^{2}+d^{2})}\ \Longrightarrow\ PI = \frac{d(R^{2}-d^{2})}{R^{2}+d^{2}}$

$PO = PI+d = \frac{d(R^{2}-d^{2})+d(R^{2}+d^{2})}{R^{2}+d^{2}}= \frac{2dR^{2}}{R^{2}+d^{2}}$

Nice Vladimir's proof based on Poncelet's porism, I'll present an alternate approach.

Let  $ M \equiv{} AD \cap BC$ and $N \equiv{} AB \cap DC.$ Then $ MN$ is the polar of $P$ with respect to $ (O)$ and the polar of $P$ with respect to $(I)$ $\Longrightarrow$ $ P,I,O$ are collinear and $PIO \perp MN.$ If $ X \equiv IO \cap MN,$ then $ X$ is the inverse of  $ P$ under inversion WRT $(O)$ and $(I).$ Therefore

$ OP \cdot OX \equal{} R^2$ and $ IP \cdot IX \equal{} (OP \minus{} d)IX \equal{} r^2$

Since $ OX \minus{} IX \equal{} d,$ we get

$ \frac {R^2}{OP} \minus{} \frac {r^2}{OP \minus{} d} \equal{} d  \Longrightarrow  \frac {1}{r^2} \equal{} \frac {OP}{(R^2 \minus{} OP \cdot d)(OP \minus{} d)}$

From Fuss theorem we have the relation

$ \frac {1}{r^2} \equal{} \frac {1}{(R \plus{} d)^2} \plus{} \frac {1}{(R \minus{} d)^2}$

$ \Longrightarrow  \frac {OP}{(R^2 \minus{} OP \cdot d)(OP \minus{} d)} \equal{} \frac {1}{(R \plus{} d)^2} \plus{} \frac {1}{(R \minus{} d)^2}$

Expanding the latter expression gives the quadratic equation

$ 2d(R^2 \plus{} d^2)OP^2 \minus{} OP(R^4 \plus{} d^4 \plus{} 6R^2d^2) \plus{} 2dR^2(R^2 \plus{} d^2) \equal{} 0$

This quadratic equation has two positive roots, namely

$ OP \equal{} \frac {2R^2d}{R^2 \plus{} d^2}$ and $ OP \equal{} \frac {R^2 \plus{} d^2}{2d}  ,$  but since $ R>OP > 0$

$\Longrightarrow  OP \equal{} \frac {2R^2d}{R^2 \plus{} d^2},$ as desired.

Thank you for the kind words. My proof is too long with too many formulas :blush: and I would definitely try something else now. Your proof is [b]really[/b] nice.


abstract algebra
Ring Theory
superior algebra
superior algebra solved

Let be $R$ a ring with unity $1_{R}$ and let be $M$ a left $R$-module. The projective dimension of $M$ on $R$ is the small positive integer $n\in \math{N}\cup\left\{\infty\right\}$ such that there exist $n+1$ projective $R$-modules $P_{0}, P_{1}, \ldots, P_{n}$ and the sequence
$0\to P_{n}\to \cdots\to P_{1}\to P_{0}\to M\to 0$
is exact.

For $R=\mathbb{Z}_{4}$, prove that the projective dimension of $\mathbb{Z}_{4}/2\mathbb{Z}_{4}$, is  $+\infty$.

We have a projective resolution

\[\cdots\longrightarrow \mathbb{Z}_{4}\longrightarrow \mathbb{Z}_{4}\longrightarrow \mathbb{Z}_{4}\longrightarrow \mathbb{Z}_{4}/2\mathbb{Z}_{4}\rightarrow 0\]

Each of $\mathbb{Z}_{4}\longrightarrow \mathbb{Z}_{4}$ is a multiplication by $2$ and $\mathbb{Z}_{4}\longrightarrow\mathbb{Z}_{4}/2\mathbb{Z}_{4}\rightarrow 0$ is the canonical map.

[quote="ficus2002"][quote="barra"]We have a projective resolution
\[\cdots\longrightarrow \mathbb{Z}_{4}\longrightarrow \mathbb{Z}_{4}\longrightarrow \mathbb{Z}_{4}\longrightarrow \mathbb{Z}_{4}/2\mathbb{Z}_{4}\rightarrow 0 \]
Each of $\mathbb{Z}_{4}\longrightarrow \mathbb{Z}_{4}$ is a multiplication by $2$ and $\mathbb{Z}_{4}\longrightarrow\mathbb{Z}_{4}/2\mathbb{Z}_{4}\rightarrow 0$ is the canonical map.[/quote]

Ok, but you must prove that for all $n\in \mathbb{N}$ there isn't projective modules $P_{0}, P_{1}, \ldots, P_{n}$ such that the sequences
\[0\longrightarrow P_{n}\longrightarrow \cdots\longrightarrow P_{1}\longrightarrow P_{0}\longrightarrow \mathbb{Z}_{4}/2\mathbb{Z}_{4}\rightarrow 0 \]
is exact.[/quote]

I think it is included in the definition of projective dimension that the dimension does not depends on how you choose your projective resolution.

I mean, we need to have this property in order to make projective dimension well defined.

[quote="barra"]I think it is included in the definition of projective dimension that the dimension does not depends on how you choose your projective resolution.[/quote]
Yes, but the definition on my text book says: "if no finite resolution exist, we set the projective dimension of M equal to $\infty$". So i must prove that don't extist a finite projective resolution of $M$.

[quote="barra"]
I think it is included in the definition of projective dimension that the dimension does not depends on how you choose your projective resolution.

I mean, we need to have this property in order to make projective dimension well defined.[/quote]

The finite projective dimension is defined as the length of the shortest finite projective resolution, if one exists (starting the count at 0 as in ficus2002's post). Only if no finite projective resolution exists, we set $pd = \infty$.

LiMa

Let $R=\mathbb Z/4\mathbb Z$, $m=2R$ and $F=R/m$.
$R$ is a local ring with maximal ideal $m$ and residual field $F$.

Generally, if a finite resolution exists, arbitrarily long resolutions exist. Just make a/many direct sum/s with redundant resolution like $\ldots\to0\to R\to R\to 0\to 0\to\ldots\to 0$.

There is a theorem that states however that a free resolution $\ldots\to F_{n}\to\ldots\to F_{0}\to M$ for a finitely generated module $M$ over a local ring is minimal if and only if the entries in the syzygy matrices (i.e. the matrices that give the aplications $F_{i+1}\to F_{i}$ between free modules) are contained in $m$, the maximal ideal of the ring.
There is also a theorem that states that this can be done using only finitely generated free modules.
Another theorem proves that in this case, free dimension is the same as projective dimension. Probably the fact that over noetherian local rings, projective=free


group theory
abstract algebra
function
superior algebra
superior algebra solved

If a group $G$ contains a subgroup($\neq G$) of finite index, then it contains a normal subgroup($\neq G$) of finite index.

This is like the fourth time this problem appears :). You look at the intersection of all the conjugates of $H$. This is called the normal interior of $H$ in $G$. Try to show that there is an injective function from the index of this group to $k!$, where $k$ is the index of $H$.

When I find a link, I'll post it.

Edit:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=normal&t=22796[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=normal&t=19724[/url]


I have a question about probabilities.
John tosses consecutively a coin. If it appears a head he wins a dollar, if it appears a tail he loses a dollar. Game is over when John loses all his money or earns [b]y[/b] dollars (y has previously been defined). At the beginning John has [b]x[/b] dollars. What's the probability that John will lose.

I'm not sure that's the right place to post that, but please help me! :roll:
And I excuse if my English is not very good :oops:

well the probability for the coin to land on heads or tails is $\frac{1}{2}$

.. and from there im not sure where to proceed using the x and y values which are unknown.. trick q?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141405
[hide]
By the logic/generalization in that thread, the probability is
$\frac{x}{x+y}$
[/hide]


Let $ A,B\in{M_n(\mathbb{C})}$. If $ rank(AB\minus{}BA)\equal{}1$ prove that $ (AB\minus{}BA)^2\equal{}0$


group theory
abstract algebra
real analysis
real analysis solved

hi evrone,

Here is a nice problem: Does there exist an uncountable subgroup of R (<>R of course) ??

bye.

Let B be a basis of $\mathbb{R}$ over $\mathbb{Q}$.
It is known that B is uncountable (otherwise $\mathbb{R}$ would be countable).

Let V be the $\mathbb{Q}$ vectorspace generated by $B\backslash \{x_0\},\ x_0\in B$. Then, because V is a vectorspace, V is a subgroup of $\mathbb{R}$, obviously different from $\mathbb{R}$ and also obviously uncountable.


[quote="DaReaper"]How many natural numbers from 1-800 contain the digit 7 at least once?[/quote]
[hide]There are 10 numbers 1-100 that at least contain one 7.
1-100    10
100-200 10
200-300 10
300-400 10
400-500 10
500-600 10
600-700 10
700-800 100
So total is 170.[/hide]
I'm not good at these problems, so can someone check this?

i.....i dont think that is right because the answer choices are:

$80,152,251,233,$ and $NOTA$

and with those choices i doubt the answer would be $NOTA$...

Your preceudre was correct...

but...i think this is right...not sure

1-100=9 (you counted 107 as 1-100)
100-200=9
200-300=9
300-400=9
400-500=9
500-600=9
600-700=9
700-800=9

=72 but i dont think this is right either

Here is my solution:
[hide] There are 10 numbers with a seven in the units digit (for all possible hundreds) and 10 numbers with seven in the ten's digit.  There are a total of 20 numbers.  However, one of them has a 77.  This implies we have to take one off of 20 to get 19 numbers per hundred (except for 700).  There are 7 hundreds (including 0 as a hundreds digit) other than 700 (and 800 doesn't count because there is no 7 in 800 and that is the only 800 shown), therefore 7*19=133 for everything other than 700.  Now for 700, there are 100 digits which have 7 (since all of them have 7).  This adds to 133+100=$233$, which is our answer.  [/hide]

oops.. i forgot to count 77,277,377, and etc... :oops:

[hide="Am I right?"][b]Numbers with exactly one 7[/b]

If the hundreds' digit is $7$, then there are $9\cdot 9=81$ ways.
If the tens' digit is $7$, then there are $7\cdot 9=63$ ways.
If the ones' digit is $7$, then there are $7\cdot 9 = 63$ ways.

Total: $207$ ways.

[b]Numbers with exactly two 7's[/b]

If the hundreds' and tens' digits are both $7$, there are $10$ ways.
If the tens' and units' digit are both $7$, there are $6$ ways.
If the hundreds' and units' digit are both $7$, there are $6$ ways.

Total: $25$ ways.

[b]Numbers with three 7's[/b]

Obviously, there is only $1$ such number.

I get a total of $233$.[/hide][hide="Simpler method (more efficient)"]If there are $0$ $7$'s, there are $7\times 9\times 9=567$. Subtract this from $800$ to get $233$.[/hide]

[hide="solution"]We use complementary counting.

7 choices for the hundreds digit because we don't count 7,8,or 9.
9 choices for the tens digit
9 choices for the units digit

7*9*9=567

subtract 1 since we included 0, and add 1 since we didn't count 800.

800-567=233[/hide]


geometry
incenter
parallelogram
perimeter
angle bisector
geometry proposed

Assume that the lengths of $ PA,PB,PC$ are fixed, $ P$ is a fixed point, and $ P$ is in the interior of triangle $ ABC$.
Choose $ A,B,C$ such that $ AB\plus{}BC\plus{}CA$ is maximum, then $ P$ is the incenter of $ ABC$.

This is the greatest solution I have come up with in my life.

[b]Given:[/b] A point $ P$ is fixed in the plane. We look at all triangles $ ABC$ with point $ P$ in their interior and lengths $ PA, PB, PC$ fixed. Prove that among all these triangles, the triangle with $ AB+BC+CA$ maximum has $ P$ as its incenter.

[b]Proof:[/b]
[b]1.[/b] Let us look at $ \triangle{ABC}$ such that $ AB+BC+CA$ is maximum.
[b]2. [/b]Construct point $ M$ on ray $ BA$ such that $ AM = AC$. Construct point $ N$ on ray $ AB$ such that $ BN = BC$.
[b]3.[/b] Construct point $ Q$ such that $ \triangle{MQA}\cong\triangle{APC}$ so that $ Q$ lies on the same side of line $ AB$ as $ P$. Construct point $ R$ such that $ \triangle{NRB}\cong\triangle{BPC}$ so that point $ R$ lies on the same side of $ AB$ as $ P$.
[b]4.[/b] Contruct point $ S$ such that $ \angle{TNS}=\angle{AMQ}$ and $ NS = MQ$.
[b]5.[/b] Because $ NS = MQ$ and $ NS\parallel MQ$ (as $ \angle{TNS}=\angle{AMQ}$), $ NSQM$ is a parallelogram, so $ SQ = NM$.
[b]6.[/b] $ NM = NB+BA+AM = BC+BA+BC$ (as $ NB = BC$ as $ \triangle{NRB}\cong\triangle{BPC}$ while $ AM = AC$ as $ \triangle{APC}\cong\triangle{MQA}$). $ SQ = NM\Rightarrow SQ = BC+BA+AC$.
[b]7. [/b]Assume points $ S, P, R, Q$ are not collinear. Then, $ SR+RP+PQ > SQ = BC+BA+AC$.
[b]8. [/b]$ \angle{SNR}+\angle{RBP}+\angle{PAQ}= (180^{o}-\angle{TNS}-\angle{RNB})+$ $ +(180^{o}-\angle{NBR}-\angle{PBA})+(180^{o}-\angle{PAB}-\angle{QAM}) =$
$ = 540^{o}-(\angle{AMQ}+\angle{RNB}+\angle{NBR}+\angle{PBA}+\angle{PAB}+\angle{QAM}) =$
$ = 540^{o}-(\angle{PAC}+\angle{PBC}+\angle{PCB}+\angle{PBA}+\angle{PAB}+\angle{PCA})$ (because $ \angle{AMQ}=\angle{PAC},\angle{MAQ}=\angle{PCA}$ as $ \triangle{MQA}\cong\triangle{APC}$; $ \angle{NBR}=\angle{PCB},\angle{BNR}=\angle{PBC}$ as $ \triangle{NRB}\cong\triangle{BPC}$).
So, $ \angle{SNR}+\angle{RBP}+\angle{PAQ}= 540^{o}-$ $ -(\angle{PAC}+\angle{PBC}+\angle{PCB}+\angle{PBA}+\angle{PAB}+\angle{PCA}) =$
$ = 540^{o}-(\angle{PAC}+\angle{PAB}+\angle{PBA}+\angle{PBC}+\angle{PCB}+\angle{PCA}) =$
$ = 540^{o}-(\angle{BAC}+\angle{CBA}+\angle{ACB})$
$ = 540^{o}-180^{o}= 360^{o}$.
Therefore, $ \angle{SNR}+\angle{RBP}+\angle{PAQ}= 360^{o}$.
Also, as $ \triangle{MQA}\cong\triangle{APC}$ and $ \triangle{NRB}\cong\triangle{BPC}$ so $ MQ = AP, NR = PB, BR = AQ = PC$. Also, as $ NS = MQ$ from 4, have $ NS = MQ = AP$.
[b]9.[/b] Therefore, we can construct a triangle $ \A'B'C'$ with $ \triangle{A'PB'}\cong\triangle{SNR}$; $ \triangle{A'PC'}\cong\triangle{PAQ}$; $ \triangle{B'PC'\cong\triangle{PBR}}$, because then $ \angle{A'PB'}+\angle{B'PC'}+\angle{A'PC'}=\angle{SBR}+\angle{PAQ}+\angle{PBR}= 360^{o}$ (see 8) and $ NS = PA = PA = PA'$; $ NR = BP = PB = PB'$; $ AQ = BR = PC'$. Therefore, we obtain a triangle $ A'B'C'$ with point $ P$ in its interior, $ A'P = AP, B'P = BP, C'P = CP$ and $ A'B'+B'C'+A'C' = SR+PR+PQ > AB+BC+CA$ by the assumption in 7. Therefore, we get a new triangle $ A'B'C'$ with a larger perimeter than the perimeter of $ \triangle{ABC}$ contradicting the definition of $ \triangle{ABC}$ (see 1). So, the assumption in 7 is impossible, therefore points $ S, P, Q, R$ are collinear.
[b]10.[/b] Points $ S, P, Q, R$ are collinear, and because $ SQ\parallel AB$, get $ PQ\parallel AM$ and $ RP\parallel NB$.
[b]11.[/b] $ AP = MQ, AM\parallel PQ$ (from 8 and 10) so $ MAPQ$ is a parallelogram, so $ AP\parallel MQ\Rightarrow\angle{BAP}=\angle{AMQ}$ (corresponding angles), but $ \angle{AMQ}=\angle{PAC}$ (as $ \triangle{AMQ}\cong\triangle{CAP}$) so $ \angle{BAP}=\angle{PAC}$.
[b]12.[/b] $ NR = BP, NB\parallel RP$ (from 8 and 10) so $ BNRP$ is a parallelogram, so $ NR\parallel BP\Rightarrow\angle{ABP}=\angle{BNR}$ (corresponding angles), but $ \angle{BNR}=\angle{PBC}$ (as $ \triangle{BPC}\cong\triangle{NQB}$) so $ \angle{ABP}=\angle{PBC}$.
[b]13.[/b] From 11, 12 get $ \angle{BAP}=\angle{PAC}$ and $ \angle{ABP}=\angle{PBC}$, so $ AP$ and $ BP$ are angle bisectors of angles $ \angle{A}$ and $ \angle{B}$, respectively, so $ P$ lies on the angle bisectors of angles $ \angle{A}$ and $ \angle{B}$ so $ P$ also lies on the angle bisector of $ \angle{C}$ in $ \triangle{ABC}$ as the angle bisectors of a triangle are concurrent. But because the angle bisectors of a triangle intersect at the incenter, and $ P$ is the intersection of the angle bisectors of $ \triangle{ABC}$, $ P$ is the incenter of $ \triangle{ABC}$ when it has $ AB+BC+CA$ maximized, QED.


[quote="behdad.math.math"]consider k is a natural number and k>1.
prove that k has a multiple that is less than k^4 and this multiple has at most for different digit.[/quote]
Just to be sure. Do you mean four different digits ? :)


Σε συνέχεια του προβλήματος (γενίκευση) που πρότεινα στο [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276607[/url], θέλω να προτείνω ένα άλλο (παρεμφερές θα μπορούσαμε να πούμε) πρόβλημα, που κατασκεύασα.

Εστω σταθερή γωνία $ xAy$ και σημεία M,N που κινούνται στις $ Ax, Ay$ ώστε $ \frac {1}{AM} \minus{} \frac {1}{AN} \equal{} c$ (σταθερό). Θεωρούμε $ P$ το ίχνος της καθέτου απ' το Α στην $ MN$. Να βρεθεί ο γεωμετρικός τόπος του $ P$ (καθώς τα $ M, N$ μεταβάλλονται)!

Υ.Γ. Eλπιζω να σας αρέσει η άσκηση!

Επειδή καταλαβαίνω ότι η άσκηση μάλλον είναι αρκετά δύσκολη, θα δώσω ένα σημαντικό hint - άσκηση.

Αποδείξτε το ισοδύναμο:
[hide]Εστω ότι έχουμε σταθερή γωνία $ xAy$ και $ M,N$ που κινούνται στις $ Ax, Ay$ αντιστοίχως, ώστε $ AM \minus{} AN \equal{} c$. Αν $ P$ είναι το αντιδιαμετρικό του $ A$ ως προς τον περιγεγραμμένο κύκλο του $ AMN$, να δείξετε ότι το $ P$ κινείται σε ευθεία[/hide]

Eπειτα απλά δείτε γιατί τα παραπάνω προβλήματα είναι ισοδύναμα!!

Αρχικά συγχαρητήρια για το [b]χρυσό[/b] και είναι τιμη μας που σε έχουμε στην παρέα μας
Mηπως κανω κάπου λάθος (είναι και αργά);
Aν η γωνία $ xAy$ είναι ορθη και $ M(a,0),N(0,b)$ τότε η $ AB: x/a \plus{} y/b \equal{} 1$  και απ τη υποθεση εύκολα βγαίνει οτι η ΑΒ διέρχεται απ το σταθερό σημείο $ K(1/c, \minus{} 1/c)$. Αφού η γωνία $ APK$ είναι ορθή τότε ο γ.τ. του $ P$ είναι κύκλος διαμέτρου $ AK$.
H απορία μου είναι πως βγαίνει ευθεία; εκτος αν εννοείς του αλλου προβλήματος αλλά πάλι λες ισοδυναμο κ.λ.π.

k patila den kanete lathos dioti o geometrikos topos einai kuklos (mallon pio sosta tha eprepe na po i grammi pano stin opoia kineitai to P- gia na min blekoume me austirotites klp).
i alitheia einai oti den exo diavasei to deutero post tou dimitri, alla pisteuo re mim oti i askisi einai arketa apli an gnorizeis kati!
auto einai to theorima ton simeion emmerich. dustuxos epsaksa to web gia na doso link kai den vrika tpt. eimai omos sigouros oti uparxei ston skiada kathos kai se alla klasika vivlia geometrias os vasiki askisi (xoris omos na anaferetai to onoma). auto to theorima leei oti an duo shmeia $ B,C$ kinountai pano se mia gonia me ton tropo pou egrapse o dimitris h i idia sxesh alla me prosthesi tote i $ BC$ dierxetai apo stathero simeio $ Z$(to opoio einai i tomi tis $ BC$ me tin exoteriki i tin esoteriki dixotomo antistoixa). tora h apodeiksi autou tou theorimatos einai straightforward an thimithei kaneis oti h exoteriki dixotomos tis gonias $ A$ einai isi me $ {\delta_A}' \equal{} \frac {2bc}{c \minus{} b}\cdot\sin{\frac {A}{2}}$ eno gia tin esoteriki dixotomo allazoume to - me + to imitono me sunimitono.
e opote to provlhma teleiose. profanos to $ P$ vlepei tin $ AZ$ upo orthi gonia.

ps. dimitri ti lush exeis? elpizo na mhn kaneis kamia antistrofi etsi???

Κατ' αρχας σας ευχαριστώ πολύ κ. Πατήλα :)

[quote="Nick Rapanos"]
ps. dimitri ti lush exeis? elpizo na mhn kaneis kamia antistrofi etsi???[/quote]

Χαχα! Με κατάλαβες! Βέβαια και η λύση μου είναι με αντιστροφή! Και μετά την αντιστροφή μπορούμε να οδηγηθούμε στο ισοδύναμο πρόβλημα που έγραψα στο 2ο post.
Γιατί μετά την αντιστροφή $ AM'\minus{}AN'\equal{}\frac{r^2}{AM}\minus{}\frac{r^2}{AN}\equal{}c$ ενώ το P πάει στο αντιδιαμετρικό, όπως έγραψα. Αρκεί λοιπόν να αποδείξουμε ότι αυτό το $ P$ (του δεύτερου post) κινείται σε ευθεία. Για να είμαι απολύτως ειλικρινής  δεν πολυκαταλαβαίνω τι σας φαίνεται λάθος, αλλά εύχομαι τώρα να έγινα πιο ακριβής (ίσως υπήρχε ένα μπέρδεμα που χρισημοποίησα πάλι το $ P$ στο 2ο ποστ).

Αυτό το 2ο ποστ ήρθε έπειτα από τη βοήθεια του Σιλ, που πρότεινε αυτή τη λύση! Εγώ μετά την αντιστροφή απλά χρησιμοπίησα το γεγονός ότι αν $ AM'\minus{}AN'\equal{}c$ τότε ο περιγεγραμμένος του $ AM'N'$ περνάει από σταθερό σημείο της εξωτερικής διχοτόμου, οπότε (αφού έχουμε κάνει αντιστροφή), η $ MN$ του προβλήματός μας διέρχεται από σταθερό σημείο κ.τ.λ!

Υ.Γ Εννοείται με τον ίδιο τρόπο η άσκηση ισχύει αν αντί για - είχα + !

[quote="bilot4"]Κατ' αρχας σας ευχαριστώ πολύ κ. Πατήλα :)

[quote="Nick Rapanos"]
ps. dimitri ti lush exeis? elpizo na mhn kaneis kamia antistrofi etsi???[/quote]

Χαχα! Με κατάλαβες! Βέβαια και η λύση μου είναι με αντιστροφή! Και μετά την αντιστροφή μπορούμε να οδηγηθούμε στο ισοδύναμο πρόβλημα που έγραψα στο 2ο post.
Γιατί μετά την αντιστροφή $ AM' \minus{} AN' \equal{} \frac {r^2}{AM} \minus{} \frac {r^2}{AN} \equal{} c$ ενώ το P πάει στο αντιδιαμετρικό, όπως έγραψα. Αρκεί λοιπόν να αποδείξουμε ότι αυτό το $ P$ (του δεύτερου post) κινείται σε ευθεία. Για να είμαι απολύτως ειλικρινής  δεν πολυκαταλαβαίνω τι σας φαίνεται λάθος, αλλά εύχομαι τώρα να έγινα πιο ακριβής (ίσως υπήρχε ένα μπέρδεμα που χρισημοποίησα πάλι το $ P$ στο 2ο ποστ).

Αυτό το 2ο ποστ ήρθε έπειτα από τη βοήθεια του Σιλ, που πρότεινε αυτή τη λύση! Εγώ μετά την αντιστροφή απλά χρησιμοπίησα το γεγονός ότι αν $ AM' \minus{} AN' \equal{} c$ τότε ο περιγεγραμμένος του $ AM'N'$ περνάει από σταθερό σημείο της εξωτερικής διχοτόμου, οπότε (αφού έχουμε κάνει αντιστροφή), η $ MN$ του προβλήματός μας διέρχεται από σταθερό σημείο κ.τ.λ!

Υ.Γ Εννοείται με τον ίδιο τρόπο η άσκηση ισχύει αν αντί για - είχα + ![/quote]

ma kai oi duo sas xrimopoihsate purhnika gia na skotosete to murmhgi??

[quote="bilot4"]Ελα ρε Νίκο! Κάτσε... Εξάλλου αυτήν ήταν η ιδέα μου για να κατασκευάσω την άσκηση!!
Γιατί και αυτή η φόρμουλα που λες για την εξωτερική διχοτόμο (την οποία αγνοούσα παντελώς) μη μου πεις ότι είναι και τόσο γνωστή πια...[/quote]

koita gnosti einai-ego omos den tin boro pote na ti thimitho kai tin koitaksa sto web.

alla opos kai na exei einai gnosta ta simeia emmerich opote den xreiazesai ti formula. tora gia tin apodeiksi tou simeiou emmerich nomizo uparxei kai mia pio omorfi apodeiksi xoris metrikes sxeseis-apla ego edosa tin apodeiksi me ti metriki gt itan pio amesi.

Oντως το μπέρδεμα ηταν το κοινό γραμμα $ P$ .....


For every positive integer $n$, find the largest constant $C(n)$, so that
\[(n+1)\sum_{j=1}^{n}a_{j}^{2}-\left(\sum_{j=1}^{n}a_{j}\right)^{2}\geq C(n) \]
for all pairwise distinct integers $a_{1},\ldots,a_{n}$.


Let $X_{1}, X_{2},\ldots$ be i.i.d  with mean 0 and variance 1. Prove that $\frac{\sum_{i=1}^{n}X_{i}}{\sqrt{\sum_{j=1}^{n}X_{j}^{2}}}$converges in distribution to N(0,1)

[quote="Dannyboy990"]Let $X_{1}, X_{2},\ldots$ be i.i.d  with mean 0 and variance 1. Prove that $Z_{n}= \frac{\sum_{i=1}^{n}X_{i}}{\sqrt{\sum_{j=1}^{n}X_{j}^{2}}}$converges in distribution to N(0,1)[/quote]
$T_{n}= n^{-1/2}\sum_{i=1}^{n}X_{i}$ converges in distribution to $N(0,1)$, and $Y_{n}= n^{-1}\sum_{i=1}^{n}X_{i}^{2}$ converges a.s. to 1 by Kolmogorov's Law of Large Numbers.

Hence also $\frac{T_{n}}{\sqrt{Y_{n}}}= Z_{n}$ converges in distribution to $N(0,1)$.  :)


Twelve distinct points are located on a circle. If the twelve points are connected in order, how many diagonals can be drawn in the resulting polygon?

The number of diagonals in any polygon is $ \frac{n(n\minus{}3)}{2}$, where $ n$ is the number of sides. In a 12-sided polygon (which, by the way, is a called a dodecagon), this would be $ \frac{12(12\minus{}3)}{2}\equal{}\frac{12(9)}{2}\equal{}6(9)\equal{}\boxed{54}$.

You could also do it by reasoning that there are $ \dbinom{12}{2}\equal{}66$ ways to pick $ 2$ points, and $ 12$ of them are the sides of the polygon, so the answer is $ 66\minus{}12\equal{}\boxed{54}$


group theory
abstract algebra
probability and stats

[quote="a3creek"]I am analyzing results from a survey, using ANOVA to determine differences between the means of different subgroups (sorted by variables such as Ethnicity, Job level, etc).  My total sample size is 400, and the subgroup sample sizes range from 6 to 79.  Does ANOVA take into consideration the size of the samples?

A colleague recommended that I use weighted means before running ANOVA.  Would that produce accurate results?

The concern is that the group with 6 people only represents 1.5% of my total sample - yet their mean is different to a statistically significant amount.  Is the subgroup size large enough to be significant?  Are there guidelines as to how big a subgroup has to be before it's significant?

Thanks!![/quote]

Degrees of freedom is based on sample size.


Are there exist [b]infinite[/b] primes:p  that 2 is the  primitive root mod p.
  Waiting for your answer :)

Does 2 holds in Artin conjecture?Or there is not a single integer known hold.Can you give detail.Thank you :)

It's known that one out of any three nonsquares works (e.g. at least one of $ 2,3,5$ does). Not much more is known.

If $ p,q\equal{}2p\plus{}1$ primes and $ p\equal{}1\mod 4$, then 2 is primitive root for q.
Therefore if there are infinite Jermen's primes $ q\equal{}3\mod 8 (p\equal{}1\mod 4)$, then it is true.


geometry
circumcircle
incenter
inradius
trigonometry
geometry unsolved

Let $O$ be the circumcenter and $I$ be the incenter of the $\triangle ABC$. And $\angle OIC=30$. Prove that $\angle B=60$ or $\angle C=60$.

The problem should be $\angle B=\pi/3$ or $\angle A=\pi/3$, or else setting $\angle A=\pi/3$, $\angle B=\pi/2$ would be a counter example.

Let $r$ be the inradius, $R$ be the circumradius. $\angle OCI=\angle {C \over 2} + \angle A - \pi/2$. $CI = {r \over \sin{\angle C \over 2}}$. By sine law, ${{\sin \angle IOC}\over{\sin \pi \over 6}} = {r \over R\sin {\angle C \over 2 }}$. Using the product and sum formula several times and the fact that ${r \over R} = 4\sin{A\over 2}\sin{B\over 2}\sin{C\over 2}$, you simplify to get $\sin(\pi/3 - B)\sin(\pi/3 - A) = 0$, from which the result follows.


geometry
3D geometry
modular arithmetic
number theory unsolved
number theory

If a is a perfect cube, a= n^3, for some integer n, and p is a prime with p is congreunt to 1 mod 3, then show that a cannot be a primitive root mod p, tat is ep(a) is not equal to p - 1 :?:

Because $a^{\frac{p-1}{3}}\equiv n^{p-1}\equiv 1\pmod p$. We are done! :D


geometry
geometric transformation
reflection
videos

I just thought I share a [url=http://antwrp.gsfc.nasa.gov/apod/astropix.html]link[/url], if one's interested in seeing some astronomy pictures from NASA (note, they give a new picture per day). Also, after some digging, I found this interesting event that took place in 2003.

A [url=http://antwrp.gsfc.nasa.gov/apod/image/0310/fireball_burnett_big.jpg]fireball[/url] was found over South Wales. This was [url=http://antwrp.gsfc.nasa.gov/apod/image/0310/fireball2_burnett_big.jpg]another[/url] picture taken roughly 4 minutes after the first picture was taken.

[quote]"[b]Explanation[/b]: Jon Burnett, a teenager from South Wales, UK, was photographing some friends skateboarding last week when the sky did something very strange. By diverting his camera, he was able to document this rare sky event and capture one of the more spectacular sky images yet recorded. Roughly four minutes later, he took another picture of the dispersing trail. What is it? Experts disagree. The first guess was a sofa-sized rock that exploded as a daytime fireball, but perhaps a better hypothesis is an unusual airplane contrail reflecting the setting Sun. Bright fireballs occur over someplace on Earth nearly every day. A separate bolide, likely even more dramatic, struck India only a few days ago."[/quote]
Does anyone want to try to speculate, describe, or discuss this rare event?

[url=http://antwrp.gsfc.nasa.gov/apod/ap031001.html]Here's the link.[/url]

Nice picture-find.  :)

Both explainations that the caption described seem to be the most probably ones.
Fireballs are just rare cases of the Earth's everyday encounter with debris from space that make it through the atmosphere, though fireballs are just very large.  There are even rarer cases where it hits the Earth's surface.  The most frequent of these events are meteor showers created by very small debris hitting the atmosphere, though even on nights where no shower is expected, observors can see meteors.

I doubt that it's an unusual airplane contrail; it seems [i]significantly[/i] bright for it to be a reflection from the setting sun. Plus I don't see evidence that could suggest an aftertrail, or some other contrail formation that would proceed after it to indicate if it's from a jet.

I'm also curious, how hot, in terms of scale temperature, can the heat produced from the friction while entering the atmosphere be?
$ > 1,000^0 F$ hot or hotter?
-----------------------------------------------------------------
Also, here are some other interesting images (relating to other topics in astronomy):
[url=http://antwrp.gsfc.nasa.gov/apod/ap991108.html]A short video on the lunation of the moon.[/url]
A big [url=http://antwrp.gsfc.nasa.gov/apod/ap000303.html]dust storm[/url] that once occurred on Earth. Some of these dusts may even travel as far as to Mexico. That's nothing compared to the [url=http://antwrp.gsfc.nasa.gov/apod/ap011017.html]aftermath[/url] of Mars after a huge dust storm.

[url=http://antwrp.gsfc.nasa.gov/apod/image/0504/WaterOnMars2_gcc_big.jpg]Water[/url] on Mars!
[hide] :P [/hide]

There is probably not a specific value for the temperature, though I have heard ranges before, it depends on what the meteor is made of, how fast it is going, and various atmospheric specifications.  :wink:

If it were indeed an explosion of this size, a big boom would have been heard and reported.
Also, the radars would have picked up this explosive event.

regardless, it's an interesting scene.

According to the caption, the fireball never made it to the ground.

hey dude that was a nice pic could  u refer more of them u see i got on astronomy of the archive  but it is so vast that i just couldnt get the right pic...............

This topic is older than the two others and, hence, this is kind of too old to revive.  :wink:


function
limit
advanced fields
advanced fields unsolved

Find, with proof, the cardinalities of:
a) the set of infinite subsets of N
b) the set of subsets A in R, where |A|=c.
c) the set of smooth functions (C-infinity) from R to R.
----------------------------

Here's what I did so far:
a)The cardinality of the set of infinite subsets of N = (the power set of N)- (the set of finite subsets of N)
= P(N) - N_o (since we can list the finite subsets of N)
= c -N_o
= c

is that enough???? I'm super bad at the bijection type of proof. Can someone tell me the correct way to do it?

b) A is an interval, in the form (a,b)
since |R| = c, we have c choices of the beginning number a.
and for every fixed a, we have another c choices of the ending number b.
so the cardinality of subsets A in R is c x c = c.

That's my logic, it seems that I'm missing something, as http://en.wikipedia.org/wiki/Cardinality_of_the_continuum#Sets_with_cardinality_greater_than_c
stated that it should be greater than c.

c) I still have no idea about it, can someone help as well??

Thanks

a)  Yes, that's correct.  It is easier to do problems like this by constructing two injections than by constructing a single bijection.  One injection is obvious.  The injection from subsets to infinite subsets is less obvious, but can be done like this:  take a subset $ \{ a_1, a_2, ... \}$ and construct the subset $ \{ 1, 3, 5, ... \} \cup \{ 2a_1, 2a_2, ... \}$.

b)  Why are you assuming that $ A$ is an interval?  This is very similar to the above problem; the answer is the cardinality of the powerset of $ \mathbb{R}$.

(b) Show that there are at most c many countable subsets of R, assume the continuum hypothesis, and you're done!  :)

(c) Gut feeling is c, can't think of a proof right now, there was some neat proof somewhere though, I'll tell you if I can get hold of any.

[quote="bubka"](c) Gut feeling is c, can't think of a proof right now, there was some neat proof somewhere though, I'll tell you if I can get hold of any.[/quote]

well, any such function is determined completely by its values on $ \mathbb{Q}$, so ...

I got a proof for c) finally:

It's less than or equal to c, since {C-infinity from R to R} is a subset of {Continuous functions from R to R}.
It's greater than or equal to c, since constant function belongs to {C-infinity}, and there are c constant in the real line.

But can someone tell me why {continuous functions from R to R} has cardinality c?????
As blahblahblah said: "...any such function is determined completely by its values on Q, so ..."
I totally don't understand what does "is determined by the values on Q" mean.....????

given a function $ f: \mathbb{Q}\to\mathbb{R}$, there is at most one [i]continuous[/i] function $ \widehat{f}: \mathbb{R}\to\mathbb{R}$ such that $ \widehat{f}|_{\mathbb{Q}}\equal{}f$ (that is, the restriction of $ \widehat{f}$ to $ \mathbb{Q}$ is simply $ f$).

This is easy to see, since by continuity, for all $ t\in\mathbb{R}\setminus\mathbb{Q}$, we must have $ \mathop{\lim_{x\to t}}_{x\in\mathbb{Q}} f(x) \equal{} \widehat{f}(t)$.  This means that $ \widehat{f}$ is completely determined on $ \mathbb{R}\setminus\mathbb{Q}$ by $ f$.


I am an undergraduate math major who is currently pursuing math research as a future career option and will soon be applying to grad school. My question is how do research mathematicians stay motivated on their work? I enjoy learning math in every field but when I look at some of the things that my professors publish, they seem to be on very specific problems. I am not saying their work is not applicable or not beautiful but just too esoteric and specific for me to justify putting in the time into. I feel that I may enjoy math now when I am still learning the general theory behind the branches, but may lose interest in the future when I have to choose a field. I have participated in an REU program before working on a problem that I found somewhat interesting but not something I can see myself really caring about. Also, some of my friends have expressed similar feelings.

Serious researchers in any field generally stay motivated about their work because they are passionate about their field, are deeply interested in finding out the answers to their research questions, and enjoy the very process of doing research.

If you're seriously considering going into research mathematics, but not sure how passionate you are about mathematics research, you should try exploring a few things.  One thing that can make individual research results seem esoteric is not knowing how they relate to the "big questions" of that field, and they might seem more interesting to you if you had a broader understanding of the framework of the field.  Another thing is that you may not have yet hit on a branch of mathematics that really grabs you; you could try reading papers or going to talks on as many different subjects as you can find.

Also-- how much time do you spend playing around with mathematical ideas outside of what's required for your classes and such?  Which areas of mathematics do you spend the most time on for fun?  That would probably give you the best idea of where your real interests lie.

I write this as someone who spent a lot of time trying to figure out what kind of research I could really spend the rest of my life working on.  I went through some pretty unrelated areas-- the final and decisive switch was from geology to linguistics.  The deciding factor for me was that I was spending almost all of my free time playing around with interesting linguistic questions, but had little or no motivation to go to talks or do work for my geology lab outside of what was strictly required.  On the other hand, I'd taken a summer off and put in 8 and 10 hour days working on a linguistics problem I'd thought up.  I love geology, don't get me wrong-- but I wasn't passionate about doing geology research in the same way that I was passionate about linguistics research, and that was what decided the matter for me.

I do not really know if I can qualify as a "research mathematician" (because I take too light view on my "research"; for me it is just "problem solving") but, I can tell that there are problems that do raise my curiosity and problems that don't. The ones in the first category may keep me occupied for years and the ones in the second category just get dismissed. I noticed no correlation between this partition and the "importance" of the questions measured in any objective way (like being a part of a bigger question, being long-standing, having real life applications, or whatever else). Sometimes I had borderline cases when a noticeable part of enthusiasm came from my collaborators.

Anyway, once you get really interested in a question, you do not need to maintain your interest; it maintains itself and strengthens with every attempt. It is akin to how you keep interested in a jigsaw puzzle. You just see more and more of the picture revealed and get more and more eager to finish. The published paper is just the puzzle completed and hung on the wall. Anybody looking at the result may ask what was so special about this particular picture on a piece of cardboard and why putting it together was interesting enough to occupy anybody for hours and days. The answer is that it is not the picture but the very process of creating it with false steps, dead end moves, brilliant guesses, etc. that kept one eager to continue.

What's the moral? Don't try to find something that would motivate you in what other people publish. Unless you spent noticeable amount of time thinking of something similar to the problem yourself, reading a ready solution will, most likely, just make you shrug your shoulders and no amount of explanations of why it is interesting will change your reaction.

The key test is whether you have ever experienced the situation when you lost sleep over some question or project and when your idea of a good time was to think of it, to discuss it with your friends, to search the books or the web for the answer or related things, etc. I do not care what that question/project was about but if you had such an experience, you have the "research virus" in your blood; if not, it is either not there, or hasn't manifested itself yet. In the first case, you can answer your question yourself (just make a small leap of faith and imagine that some people may really lose sleep over some ridiculous problem about positive integers or whatever else it was that your professors published about); in the second, nobody can give you a satisfactory answer.

There are at least a few possible causes for the "symptoms" you are feeling:

1)  A career in mathematics research is not right for you.  This is not meant as a knock against you in any way.  The fact is that a relatively small number of people enjoy mathematical research enough to dedicate their careers to it.

2)  Your REU problem was simply not that interesting.  Because of the limited backgrounds of undergraduates, the problems assigned in REUs are often not very interesting.

3)  Your REU problem was interesting, but not to you, for the reason that different people have different tastes, and maybe the project didn't fit your tastes.

I hesitate to mention a 4th reason:  You might not have sufficient mathematical background and context to understand why the research was worth doing.  (I hesitate to mention this one, because I'm not sure it matters much.  People who love math often enjoy working on research problems before knowing much about what the point is.)

I think you should take possibility 1 seriously, though.  Look for corroborating evidence for possibility 1.


Given ${1}<{n}\in{N}$. Find all positive integers$({a}_{1},{a}_{2}...{a}_{n})$ wich is coprime to each other such that for every $i\in N$ ( ${1}\le{i}\le{n}$ )holds ${a}_{1}+{a}_{2}+...+{a}_{n}\mid{a}^{i}_{1}+...+{a}^{i}_{n}$.


[quote="MAT"]Hello :)

Let $ xi$ some real positive numbers

Let  A£R+*, Find the maximum of: $ x_{1}x_{2}+x_{2}x_{3}+........x_{n-1}x_{n}$
Where :
$ A=sum xi$, means: $ A=x_{1}+x_{2}+...+x_{n}$[/quote]
See here the Mildorf's post:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30787


Functional Analysis
advanced fields
advanced fields unsolved

It's easy to prove that for a finite-dimensional Hilbert space, the set of invertible operators is dense in the space of operators with the norm topology. How about infinite-dimensional (complex) Hilbert spaces?

It is no longer true for infinite-dimensional Hilbert spaces. Indeed, consider the space $\ell^2$ of sequences $\{x_n\}_{n\ge 1}$ summable with square. Define the operator $S$ (the so-called forward shift) by $S(x_1,x_2,\dots)=(0,x_1,x_2,\dots)$.
Then the distance from $S$ to the set of invertible operators is $1$. To see it, define $S^*$ (the backward shift) by $S^*(x_1,x_2,\dots)=(x_2,x_3,\dots)$. Note that $S^*S=I$ (the identity operator). Suppose that $S+T$ is invertible and $\|T\|<1$. Then $S^*(S+T)=I+S^*T$ is invertible as a sum of $I$ and an operator of norm less than $1$. But then, since $S+T$ is invertible, $S^*$ must be invertible, which is false.

I found the exact same example :). How about the set of operators which have either a left or a right inverse? Is it dense?

Yes, it is dense. Indeed, let $T$ be any operator in a Hilbert space $H$. Let $A$ be the self-adjoint positive square root of $T^*T$. Then $\|Tx\|=\|Ax\|$ for every $x\in H$ and, therefore, we can define an isometry $U_0: \text{Clos}\,AH\to\text{Clos}\,TH$ such that $T=U_0A$. Then we can extend $U_0$ to an operator $U$ that either maps the whole $H$ isometrically onto some subspace of $H$ (and, thereby, is left invertible) or maps some subspace of $H$ isometrically onto $H$ (and, thereby, is right invertible). We still have $T=UA$.  It remains to note that the self-adjoint operator $A$ is the limit of invertible operators $A+i\delta I$ as $\delta\to 0+$.


Anyone know how to get an account? (I'd be very very extremely grateful :-) )

I think confuted has one b/c it's in his profile.

its invite only. at least that's what i read in the article about it.

google's mail account. one gb. which is what caused yahoo!mail to bump up their account to 100mb (which is way shy of 1 gb, obviously, but I still don't know what to do with all of this space I now have). and for those who want an account, here's some suggestions on how to get one:

http://www.suntimes.com/output/tech/cst-nws-gmail07.html

*grin* To be taken lightly, don't go trying to sell your sister or anything for one.

[quote="MathFiend"]what exactly [i]is[/i] g-mail???[/quote]

http://www.gmail.com

Yay! Bragging time! :lol:

The way to get one is to contact someone you know who has the least degree of separation between the people at Google. The less degrees, the more likely they'll have accounts to give away.

http://www.gmailswap.com

Mathfanatic was gracious enough to hook me up, but don't go bugging him for invites.

You could always try the crazy contest over here:

http://www.runescapecommunity.com/index.php?showtopic=148096

(basically you have to go on runescape.com and try to guess the password to the account "peenzor.")  Been trying all day and I've gotten up to it (maybe) having something to do with "Habt ihr eine frage?", "Barbie," Sibelius's Finlandia and some other stuff...

[quote="Ragingg"]Ha!  Gmail's invitations have been RUINED.  Have any of you looked on Ebay?  They were going for 20 dollars and now... A measly 5 cents.  If you want one so bad just buy it![/quote]
Or...... we could run a contest of some sort here on AoPs, with a gmail invite as the prize, provided by... me.

edit: As promised:
http://artofproblemsolving.com/Forum/viewtopic.php?t=6946&sid=84c2f3d74e6b7d2e76d9adf085b9e4be

I want to clarify that I have only one gmail account and that is for myself. So I don't have any extras to give away. (Not high enough in the hierarchy) :-).

I can't even find the safes inside the Rune thing! LOL... NM I found it :)

Here's one way: if you've had a Blogger account for a while, then when you log in, you should see a link asking you to try gmail out towards the left side on the first screen afterwards (since Blogger is apparently owned by Google). Once you've used it though, it won't show the link again and also new screen names (ones my friends made last week still...) don't show the ad because they're smart like that.

You can also purchase popular e-mail account names on places like e-bay as mentioned above if you're really desperate. One account went for almost $150!! o_O

Easiest way: I bought an invitation in eBay for $3.80.

just use something like http://www.gmailswap.com

You'll probably be able get a gmail invite in a few hours.

[b]Edit:[/b] Whoo hoo, I just got one!

[b]Edit 2:[/b] All I needed to offer was eternal gratitude. :)

David, may I have one? My email is stevenliao91@yahoo.com- please give one to me if you get a gmail invite (in your new account, of course)

I have extra accounts to give away..please PM me with your information..and the invites will be sent on a first come first serve basis :)

Are u sure..i dont have any new PM's... :?

[quote="bob123"]Giving out free g-mail to people. I have 18 invites. Just PM me with why I should give you an account.[/quote]

Well I have 36 invites just waiting to go out. (well, 35, since one is out)

Send to chinesefastfood@gmail.com

or, cooldemocrat@gmail.com

Gmail might be infected, though.

I just got six invites, PM me if you want one...

Guys please don't spam this thread.

If you have no idea what Gmail is, visit http://www.gmail.com or just read previous posts. It's been explained multiple times. "I'm too lazy" is not a good excuse.

If you need an invitation, email Lun: mumu.kau@gmail.com and she'll be gracious enough to send you an invitation. :-P (Otherwise, lemme know: mithsapprentice@gmail.com ...I have some too) There's also others offering invitations and they named themselves...just look back over old posts.

huh? hey pakagawa, when i click on your site, it says: The link you followed to create a Gmail account has already been used to create an account for nntrkien@gmail.com. Now, its account creating powers are all gone. To create another Gmail account, you'll need a shiny new account creation link. We apologize for the inconvenience.
um...so what am i supposed to do?

That means somebody else has used the invite already. If you still want a Gmail account, PM me with your email.

ah rats! everyone's been takin. oh well, i gues i don't really need it...i'll find someone else's and try...


Let [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex] be non-negative real numbers.  Prove that
[tex]a^3 + b^3 + c^3 + 3abc \ge ab \sqrt{2a^2 + 2b^2} + ac \sqrt{2a^2 + 2c^2} + bc \sqrt{2b^2 + 2c^2}.[/tex]

See the topic "An interesting inequality" posted by Anh Cuong in Proposed and Own problems Section.

Sorry, how can I write directly the address of the topic?

I think you mean this topic:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=4700[/url]

I checked it, but I don't see how it helps.

Anh Cuong's "An interesting inequality." is at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6317

[quote="Anh Cuong"]Give $a,b,c\geq0$.Prove that:
[tex]a^3+b^3+c^3 + 3abc \geq ab\sqrt{2(a^2+b^2)}+bc\sqrt{2(b^2+c^2)}+ca\sqrt{2(c^2+a^2)}[/tex] :)[/quote]

harazi has a even stronger one at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6558

[quote="harazi"]Prove that for all nonnegative reals a,b,c we have:
$ (a^3+b^3+c^3+3abc)^2\geq 2(ab+bc+ca)( ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)) $.[/quote]

There are too many interesting inequalities. :D

Thanks.

[quote]Sorry,who can tell me what is MOP?[/quote]

MOP is the USA's Olympiad training camp.

[quote="harazi"]Prove that for all nonnegative reals $ a,b,c$ we have:

$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)^2\geq 2(bc \plus{} ca \plus{} ab)\left[bc\left(b^2 \plus{} c^2\right) \plus{} ca\left(c^2 \plus{} a^2\right)\plus{}ab\left(a^2 \plus{} b^2\right)\right]$.[/quote]$ (a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)^2\minus{} 2(bc \plus{} ca \plus{} ab)\left[bc\left(b^2 \plus{} c^2\right) \plus{} ca\left(c^2 \plus{} a^2\right)\plus{}ab\left(a^2 \plus{} b^2\right)\right]$

$ \equal{}\sum\frac{a}{504(a\plus{}b\plus{}c)}\Big[\left(3a^2\minus{}2b^2\minus{}c^2\plus{}3bc\minus{}2ca\minus{}ab\right)^2\left(28a^2\plus{}20b^2\plus{}c^2\plus{}64bc\plus{}79ca\plus{}24ab\right)$

$ \plus{}\left(3a^2\minus{}2c^2\minus{}b^2\plus{}3cb\minus{}2ba\minus{}ac\right)^2\left(28a^2\plus{}20c^2\plus{}b^2\plus{}64cb\plus{}79ba\plus{}24ac\right)\Big]$

$ \geq0.$

See also here : http://www.mathlinks.ro/viewtopic.php?t=216948


geometry
trapezoid
LaTeX
algebra
system of equations

A trapezoid has legs $ 4$ and $ 5$, bases $ 10$ and $ x$. Where $ x<19$. What can $ x$ NOT be?

A $ 6$ digit number is in the form $ ABCDE6$. Let This number be $ x$. If 4 times $ x$ is equal to $ 6abcde$. Find all $ x$.

Given: $ (x^{3}\minus{}3x^{2}\plus{}7x\minus{}5)^{\minus{}1}$= $ \frac{5}{6}$
Compute $ (x^{3}\minus{}3x^{2}\plus{}7x\minus{}4)^{\minus{}1}$

1: EDIT - oh, haha, you wanted an interval

[hide="2"]Let $ u \equal{} \overline{abcde}$. We have $ 10u \plus{} 6 \equal{} x$ and $ 600000 \plus{} u \equal{} 4x$. Solving the system of equations yields $ x \equal{} \boxed{153846}$[/hide]

[hide="3"]Let $ u \equal{} \text{that thing involving }x$. Then $ u \equal{} 6/5$, so $ u \plus{} 1 \equal{} 11/5$, and $ (u \plus{} 1)^{ \minus{} 1} \equal{} \boxed{\frac 5{11}}$[/hide]

Edit: Yes, it's 5/11, typo.

[hide="1"]Obviously, $ x<19$. So then swing both legs around from there while keeping the bases parallel, and you stop when the shorter leg is perpendicular to the bases. At this point, the $ 5$ leg forms a 3-4-5 triangle so $ x\equal{}13$, so one range of x is $ 13\le x <19$. Now, if you move the $ 5$ leg around, you get that the base can also be $ x\equal{}7$ and of course, swinging it around, $ x$ approaches $ 1$. So the second range is $ 1 <x \le7$.
So the possibilities of x are $ 1 <x \le7$ and $ 13\le x <19$.[/hide]

Uhhh... wasn't this competition just yesterday?

[quote="pascal12"]
So the possibilities of x are $ 1 < x \le7$ and $ 13\le x < 19$.[/quote]

Here is the question asked by the original poster:
"What can  $ x$ NOT be?"

You did not answer the question, especially as the word "not" was/is
emphasized in the question.  Being a competition problem, for instance,
you should not get full credit then.

So, the answer would be that (positive side length) $ x$ cannot be in
$ 0 < x \le 1$ or $ 7 < x < 13$ or $ x \ge 19$.

(Note that I used the word "or," as in $ x$ is in this, or $ x$ is in that.
Otherwise, using "and" between intervals would mean that $ x$ is
simultaneously in the intervals.)

Alternatively written, $ x$ cannot be in $ (0, 1]$ U $ (7, 13)$ U $ [19, oo)$.

[quote="Arrange your tan"]Alternatively written, $ x$ cannot be in $ (0, 1]$ U $ (7, 13)$ U $ [19, oo)$.[/quote]

$ x\not\in(0,1]\cup(7,13)\cup[19,\infty)$ is the proper $ \text{\LaTeX}$ for that.

can someone post a full solution to number $ 1$? I have NO idea how to do it. AND dont understand the previous solution  :(

[quote="pascal12"][hide="1"]Obviously, $ x < 19$. So then swing both legs around from there while keeping the bases parallel, and you stop when the shorter leg is perpendicular to the bases. At this point, the $ 5$ leg forms a 3-4-5 triangle so $ x \equal{} 13$, so one range of x is $ 13\le x < 19$. Now, if you move the $ 5$ leg around, you get that the base can also be $ x \equal{} 7$ and of course, swinging it around, $ x$ approaches $ 1$. So the second range is $ 1 < x \le7$.
So the possibilities of x are $ 1 < x \le7$ and $ 13\le x < 19$.[/hide][/quote]If I get what you're saying, and I'm not entirely sure I do, you're saying that both legs have to be going either "out" or "in". I'm pretty sure that isn't true of a trapezoid, so x can be anything from 1-19 excluding 10 obviously.

Sorry for the incomplete solution, and sorry for not making it too clear, I'm not sure how to draw diagrams.

Ihatepie: It does have to be "in" or "out" because if the longer leg dropped straight down, then the shorter leg would be too short to reach the other base. So the longer leg has to be at an angle where the shorter leg can meet the base.

It seems we're misunderstanding each other. I'm basically trying to say that this can be a trapezoid:

[url=http://img6.imageshack.us/i/picture8lu.png/][img]http://img6.imageshack.us/img6/9660/picture8lu.th.png[/img][/url]


Hi peoples, i m an Alaska mathelete and i have state competition next week

I m pretty good at everything but i still got tons of problem on algebra and probablity, is there anybody can give me some fundimental lessons at probability and algebra? Or just some brief tips on those things?

thx..... :blush:  :blush:

SOS??? Dosnt that stand for Save Our Ship? Anyway, do you have specific problem?

save our ship...nice one..

I m always ambiguous on probablity problems in the competition, mostly i just guess them or skip them, i know that ani't good.... Samples? i have some

Heres one:

20 plps ub a room and they each shaked each others hand, how many handshakes there is....

I know i m being newbie but i really have no idea how to solve this kinda of problems...

SOS again :(  :blush:

Ps: if u have any sample problems( like most often occured in competition), let me know, please! Only u guys can help me get into national......  :P

[quote="kidwithshirt"]
Heres one:

20 plps ub a room and they each shaked each others hand, how many handshakes there is....

[/quote]

I dont know if this is probability but sure i'll solve it for you. There are a couple of ways you can do this, but I like to use the handshake formula which is n(n-1)/2= (20*19)/2=190, So there will be 190 handshakes.

now thats a greaet formula for my collection... thx...

wat about some of those poker problems, do u have any?

i really think it works... lol

i tried it

[quote="lightspeed"]Does that formula work for all handshake problems that ask how many handshakes there are?  Like if it specifically said they could only shake each other's hand once...  Does it still work?  Thanks for the formula, you should post it on my thread "What we really need to know."

Thanks :D[/quote]

This formula ONLY works if each person shakes hands ONCE with every other person.

Don't think of this as 'the handshake formula'!  Understand why it works.  I can ask this problem in 50 different ways.  If you just memorize formulas, you'll have to memorize 50 formulas.  Learn why this works, and you'll be able to do these 50 problems (and thousands others) without ever learning any formulas.

Let's try this series of questions.

There are 20 people in a room.  Each person pokes kicks each other person once.  Our goal is to count the number of kicks.

A) How many ways can we pick a person to do the kicking for a given kick?

B) How many ways can that person pick someone to kick?

C) How can we combine our answers to A + B to get the total number of kicks?

Let's now change the problem.  Each person gives each other person a high-five once.  High-five-giving is mutual, so if I give you a high-five, you are giving me one at the same time.

D) How many ways can I pick someone to give the first high-five?

E) How many ways can I pick the second person to pair with that first person to have the first high-five?

F) When I combine parts D+E the same way I combined A+B in part C above, how many times will I count each high-five?

G) How do I get the answer to this problem, and what does it have to do with your handshake problem?

ouch i got kicked :(

The most common problems i ve seen is hand shake once lol....

Okay GoBraves.  Thanks. It only works for it if each person shakes each other person's hand ONCE. (the formula is n(n-1)/2, the more I type it, the better I learn) I see.  And as for you rrusc, thanks for your explanation.  It is a lot better if you actually get the concept, but I'm trying to memorize a few formulas to make me faster.  Do you have any you can contribute to me?  Go to my thread, "What we really need." and post it there please.

Thanks again! :D

P.S.  I would love to have everyone P.M. with formulas and their uses!

[quote="kidwithshirt"]lightspeed i contributed my heros 4mula to u lol[/quote]
Yeah, I just saw it.  Thanks!  I wish more ppl would share their formulas!  lol It would be cool if i had a genie, and then when I made the previous wish,  he'd go, your wish is my comand, and 1000 posts came up on my thread!  That would be awesome!!!!!!!
Hey, I can dream, can't I?