How many permutations $ (w_1, \cdots, w_n)$ of $ (1,...,n)$ such that $ w_i \not\equal{} i$ or $ i\minus{}1$?

The sequence is [url=http://www.research.att.com/~njas/sequences/A000271]A000271[/url] in the OEIS, which has a simple recursion that one could probably prove similarly to the proof for derangements.  Similar to [url=http://www.research.att.com/~njas/sequences/A000179]A00179[/url].


Let a, b, c and
[img]http://www.mathlinks.ro/Forum/latexrender/pictures/edf22308435cfdd8054e75beb6ea7eeb.gif[/img]
Prove that
[img]http://www.mathlinks.ro/Forum/latexrender/pictures/45564c1892677ae2016aba9cdd9a50ff.gif[/img]
 :(

Please do not post the same topic in different forums.


geometry
circumcircle
similar triangles
geometry proposed

Let a triangle $ ABC$ inscribed in the circle $ (O)$. The bisectors of angles $ ABC$, $ ACB$ meet $ (O)$ at again $ M$, $ N$ respectively. Let $ D$ on the opposite ray of the ray $ BC$ and $ E$ on the opposite ray of the ray $ CB$ such that $ BD \equal{} BA$, $ CE \equal{} CA$. Denote $ O_1,O_2$ be the circumcenter of triangles $ NBD$, $ MCE$ respectively. Prove that $ \angle O_1AB \equal{} \angle O_2AC$.

P.s. Keepsake for 2 years on MathLinks.  :lol:
26th January, 2010.  :)

We will prove the proposed problem, in its more general formulation a follows :

[b][size=100][color=DarkBlue] A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup ABD,\ \bigtriangleup AMC$ be, two similar triangles with $ \angle BAD \equal{} \angle MAC$ and $ \angle ADB \equal{} \angle ACM,$ erected on its side-segments $ AB,\ AC$ respectively, outwardly to it. Also, let $ \bigtriangleup ABN,\ \bigtriangleup AEC$ be, two other similar triangles with $ \angle BAN \equal{} \angle EAC$ and $ \angle ABN \equal{} \angle AEC,$ erected by the same way. Prove that $ \angle O_{1}AB \equal{} \angle O_{2}AC,$ where $ O_{1},\ O_{2}$ are the circumcenters of the triangles $ \bigtriangleup BDN,\ \bigtriangleup CEM$ respectively.[/color][/size][/b]

( In the schema of the solution by [b]livetolove212[/b], we remind that $ \angle BAD \equal{} \angle ADB \equal{} \frac {\angle B}{2} \equal{} \angle CAM \equal{} \angle ACM$ and $ \angle ABN \equal{} \angle BAN \equal{} \frac {\angle C}{2} \equal{} \angle CAE \equal{} \angle AEC$ ).

[b]PROOF OF THE GENERALIZED PROBLEM.[/b] - From the similar triangles $ \bigtriangleup ABD,\ \bigtriangleup AMC,$ we have that $ \frac {AD}{AC} \equal{} \frac {AB}{AM}$ $ ,(1)$

From the similar triangles $ \bigtriangleup ABN,\ \bigtriangleup AEC,$ $ \Longrightarrow$ $ \frac {AN}{AC} \equal{} \frac {AB}{AE}$ $ ,(2)$

From $ (1),\ (2)$ $ \Longrightarrow$ $ \frac {AD}{AE} \equal{} \frac {AN}{AM}$ $ ,(3)$

From $ (3)$ and because of $ \angle DAN \equal{} \angle EAM,$ we conclude that the triangles $ \bigtriangleup AND,\ \bigtriangleup AME$ are similar and so, we have that $ \angle ADN \equal{} \angle AEM$ $ ,(4)$

$ \bullet$ We denote the points $ G\equiv (O_{1})\cap AD,\ D'\equiv (O_{1})\cap AN$ and then, we have also the similarity of the triangles $ \bigtriangleup AGD',\ \bigtriangleup AME$.

We denote the point $ H\equiv (O_{2})\cap AE$ and it is easy to show that $ \angle AGN \equal{} \angle DBN \equal{} \angle ABD \minus{} \angle ABN \equal{} \angle AMC \minus{} \angle AEC$

$ \Longrightarrow$ $ \angle AGN \equal{} \angle AMC \minus{} \angle HMC \equal{} \angle AMH$ $ ,(5)$

From $ (5),$ in the configuration of the similar triangles $ \bigtriangleup AGD',\ \bigtriangleup AME,$ we have the circles $ (O_{1}),\ (O_{2}),$ as the circumcircles of the triangles $ \bigtriangleup GD'N,\ \bigtriangleup MEH$ respectively, the vertices of which are homologous points.

$ \bullet$ So, we conclude that $ \angle O_{1}AG \equal{} \angle O_{2}AM$ $ ,(6)$ because these angles are formed by the lines connecting homologous points ( = are formed by homologous lines ), of the  similar triangles $ \bigtriangleup AGD',\ \bigtriangleup AME.$

Hence, from $ (6)$ and because of $ \angle BAD \equal{} \angle CAM,$ we conclude that $ \angle O_{1}AB \equal{} \angle O_{2}AC$ and the proof is completed.

Kostas Vittas.

Two interesting solutions, livetolove212 and Kostas Vittas.  :wink:
Congrats!  :lol:


There are three piles of matches on the table: one with $ 100$ matches, one with $ 200$, and one with $ 300$. Two players play the following game. They play alternatively, and a player on turn takes one of the piles and divides each of the remaining piles into two nonempty piles. The player who cannot make a legal move loses. Who has a winning strategy?

May not belong in Pre-Olympiad.

[hide="solution"]Note that each game will last the same number of moves, as each move increases the number of piles by one, and we need $ 100\plus{}200\plus{}300\equal{}600$ piles. The game will end after $ 600\minus{}3 \equal{} 597$ moves. Thus the first player to move wins.[/hide]

[quote="leoxnlin"] as each move increases the number of piles by one,  [/quote]

Not true - read the problem carefully  :wink:

the second player has a winning strategy.
The player who start is $ \alpha$ the other is $ \beta$.
The strategy for $ \beta$ is to move always in the same pile of $ \alpha$.


[img]http://img204.imageshack.us/img204/1460/janyooj3.png[/img]

Squares $ABDE$ and $BCFG$ are constructed on the sides $\overline{AB}$ and $\overline{BC}$ of a $\triangle ABC$, $O$ is the intersection of $\overline{AF}$ and $\overline{CE}$. Prove that $\overline{BO}\perp \overline{AC}$.

Also true, for similar rectangles instead of squares. The ''isogonic'' theorem is coming back...

Kostas Vittas.


LaTeX
induction
inequalities
inequalities theorems

For any positive reals $x_1,x_2,..,x_n$ we have

$(n-1)(x_1^n+\cdots+x_n^n)+nx_1\cdots x_n \geq (x_1+\cdots+x_n)(x_1^{n-1}+\cdots+x_n^{n-1}) $

For $n=3$ it is Shur.

To define beautiful this one is not so much !!

I guess I know a little better LaTeX than you. :)  Use \ldots or \cdots instead of "..." I learned a lot of LaTeX tricks by simply looking at the quotes of other people's posting (I am still learning  :) ).

For any positive reals $x_1,x_2,\ldots,x_n$ we have
$(n-1)(x_1^n+\cdots+x_n^n)+nx_1\cdots x_n \geq (x_1+\cdots+x_n)(x_1^{n-1}+\cdots+x_n^{n-1}). $

Cheers!  :D  :D

Thank you very much Fuzzylogic for your advice.

Well, the only solution I know for this one is using induction, but the induction step is highly nontrivial. I'm in Athens now and I hope I will post the solution next week, when I'm home.

Oh, yes, I apologise, Im home for a lot of time, but I simply forgot about the proof of this inequality. I will put a Word file with the solution, since its too long to write here.

[quote="harazi"]Oh, yes, I apologise, I�m home for a lot of time, but I simply forgot about the proof of this inequality. I will put a Word file with the solution, since it�s too long to write here.[/quote]
The proof is false, because we must prove that
\[(\sum x_{i})(\sum x_{i}^{n-1})-%Error. "sumx" is a bad command.
_{i}^{n}+n(n-1)(n-2)\prod x_{i}\geq \prod x_{i}(n+(n-2)(\sum x_{i})(\sum x_{i}^{n-1})) \]
Suranyi's inequality not works here... :mad: Please check it.

Hey guys, could anyone explain me this? I dont understand this step  :blush:

[hide][img]http://i.gyazo.com/e868d4360b36b66e3f4eada661fef77e.png[/img][/hide]


Find all integer solutions $(x^{2}+y^{2})^{2}=(x+y)^{3}$

2nd solution :)
Cleraly $x+y\geq 0$ and we can write that $x+y=a^{2},x^{2}+y^{2}=a^{3}(a\in Z^{+}\cup \{ 0\})$.So, $y=a^{2}-x$.

$x^{2}+(a^{2}-x)^{2}=a^{3}\Longrightarrow 2x^{2}-2a^{2}x+a^{4}-a^{3}=0$ Let us examine the discriminant:
$\bigtriangleup =4a^{4}-8(a^{4}-a^{3})=4a^{3}(2-a)\geq 0$.Hence, $a$ may equal to $0,1,2$.
If $a=0\Longrightarrow \left\{\begin{array}{cc}& x^{2}+y^{2}=0^{3}\\ & x+y=0^{2}\end{array}\right.\Longrightarrow (x,y)=(0,0)$
If $a=1\Longrightarrow \left\{\begin{array}{cc}& x^{2}+y^{2}=1^{3}\\ & x+y=1^{2}\end{array}\right.\Longrightarrow (x,y)=(1,0),(0,1)$
If $a=2\Longrightarrow \left\{\begin{array}{cc}& x^{2}+y^{2}=2^{3}\\ & x+y=2^{2}\end{array}\right.\Longrightarrow (x,y)=(2,2)$

Lokman GÖKÇE


Given three external circles $ C_a,C_b,C_c$ with three external common tangents of them $ d_a,d_b,d_c$,respectively, prove that exist four circles tangent to $ d_a,d_b,d_c$ and show the contruction of them!

Sorry but how is this an extension of incircle/excircle ?

Aren't we just constructing circles tangent to 3 given lines, which is just the incircle and the 3 excircles of the formed triangle?

ohh yes of course, there are some my ideas in my thingking, only but you are right, it is only a circle tagent to three lines, but my idea, when three circle have the radius zero then it is the incircle or excircles!


function
integration
algebra
domain
search
real analysis
real analysis unsolved

We know :

$ \left(L^1[0,1],\|\cdot\|_1\right)$ is a normed space , where :

$ L^1 \cong$  Collection of equivalence classes according to the relation R , we say $ f \ R \ g$ iff $ f = g , a.e$ .

and $ \parallel{} f\parallel{}_1 = \int\limits_{0}^1 |f(t)| dt$ . (Leb Int )

Also , $ \left(L^\infty [0,1],\|\cdot\|_\infty\right)$ is a normed space , where :

$ \parallel{}f\parallel{}_\infty = \mathop {essup }\limits_{t\in [0,1]} \ |f(t)|$ , (Essential Supremum )

Note that : $ f = [f]$ , the class of functions which they equal a.e ( for both previous spaces .)

The Super question is :

for any measurable function $ g$ on $ [0,1]$ , then $ g\in L^\infty[0,1] \Longleftrightarrow g.L^{1}[0,1] \subseteq L^{1}[0,1]$

IN FACT I NEED THE DIRECTION :"$ \Longleftarrow$"

 :)

Consider the contrapositive: If $ g\not\in L^{\infty},$ then there exists $ f\in L^1$ such that $ fg\not\in L^1.$

Let's look at a plan of attack. The first thing we can do is to assume WLOG that $ g\ge 0$ and that the $ f$ we will pick is also nonnegative. We can do that by choosing the size of $ f$ and then multiplying that by $ \operatorname{sgn}(g).$ (Or $ \overline{\operatorname{sgn}(g)}$ if we're using complex-valued functions.)

Divide the domain into disjoint sets on which $ k\le g<2^{k\plus{}1}.$ Or maybe $ 2^k\le g<2^{k\plus{}1}$ - maybe that would be easier to manipulate. Choose a (constant) value for $ f$ on each of these sets. Can we arrange it so that $ \int f<\infty$ but $ \int fg\equal{}\infty?$ Fiddle around with it until you can make it work.

what do you mean by the size of $ f$? :|

I couldn't do it yet  :huh:

Any help will be appreciated .


Would someone mind giving me a list of every1 who made countdown and also possibly their AOPS account?

I've posted the first. The latter is up to those participants; revealing it without their consent is against privacy rules.


trigonometry
parameterization
function
Pythagorean Theorem
geometry

Which of the following rectangular equations is equivalent to the parametric equations $ x\equal{}sintcost$, $ y\equal{}cos^{2}(2t)$?
A) $ y\equal{}\sqrt{1\minus{}x^{2}}$
B) $ y\equal{}1\minus{}4x^{2}$
C) $ y^{2}\equal{}4\minus{}4x^{2}$
D) $ y\equal{}2x^{2}$
E) NOTA=None of the Above

Note: I have absolutely no idea what parametric equations are so can someone clarify what this means?

[quote="JRav"]$ x \equal{} \frac {1}{2}\sin{2t}$ and $ y \equal{} \cos^{2}{2t}$.  Then $ (2x)^{2} \plus{} y^{2} \equal{} 1 \equal{} 4x^{2} \plus{} y^{2}$[/quote]I would think then that
$ 2x \equal{} \sin{2t}$
$ 4x^2 \equal{} \sin^{2}{2t}$
$ 4x^2 \plus{} y \equal{} \sin^{2}{2t} \plus{} \cos^{2}{2t}$
$ 4x^2 \plus{} y \equal{} 1$
$ y \equal{} 1 \minus{} 4x^2$
but with so many $ 2$'s dancing in front of our eyes, we may see more or less of them.

My idea of parametric equations is that they are equations that give us the variables we care about (x, y, ..) in terms of one variable that we call a parameter (t) because we are not particularly interested in it at the time. For example, we may want to know an equation describing the trajectory of a projectile shot at a certain angle. We go through some parametric equations to get to the answer we want. We can somewhat easily determine the (parametric) functions relating the height (y) and the horizontal distance covered (x) to the time since launching (t). From those parametric equations, it takes some algebra to get y as a function of x to determine the shape of the trajectory. Occasionally, we may not be able to get a reasonable y=f(x) function and we may have to keep the parametric equations.

Basically you define x and y in  terms of some other variable t.

[quote="JRav"]$ x \equal{} \frac {1}{2}\sin{2t}$ and $ y \equal{} \cos^{2}{2t}$.  Then $ (2x)^{2} \plus{} y^{2} \equal{} 1 \equal{} 4x^{2} \plus{} y^{2}$[/quote]

Also, if the above confused you, it was a mistake. Since $ y$ was defined as $ \cos^2 2t$, we should have $ (2x)^2 \plus{} y \equal{} 1 \implies y \equal{} 1 \minus{} 4x^2$.

hello, we have $ x \equal{} 1/2\cdot \sin(2t)$ and from here $ 4x^2 \equal{} \sin^2(2t)$ and by the Pythagorean theorem we get $ 4x^2 \plus{} y\equal{} 1$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we have $ x \equal{} 1/2\cdot \sin(2t)$ and from here $ 4x^2 \equal{} \sin^2(2t)$ and by the Pythagorean theorem we get $ 4x^2 \plus{} y^2 \equal{} 1$.
Sonnhard.[/quote]

You also made the same mistake as JRav. $ y$ is defined as $ \cos^2 2t$, so it should be $ 4x^2 \plus{} y \equal{} 1$.

hello, yes, you have right, i have corrected it.
Thank you very much,Sonnhard.

[quote="Yongyi"]Also, if the above confused you, it was a mistake.[/quote]
Oops.  I squared the $ y$-term by accident.


inequalities
geometry
circumcircle
trigonometry
geometry proposed

Given triangle $ ABC$ and $ O$ is circumcenter $ P$ is an arbitrary point in plane prove that
\[ \frac{\sin A}{PA}+\frac{\sin B}{PB}+\frac{\sin C}{PA}\ge 4\sqrt{\frac{OP}{PAPBPC}}\sin A\sin B\sin C \]
Equality iff $ ABC$ is equilateral and $ P\to\infty$.
Gook luck all mathlinkers in IMO.

Just a question regarding this inequality. Is it correct the way it is written?

Is it $ \frac{\sin C}{PA}$ or $ \frac{\sin C}{PC}$?

Sorry you are right it must be
Given triangle $ ABC$ and $ O$ is circumcenter $ P$ is an arbitrary point in plane prove that
\[ \frac{\sin A}{PA}+\frac{\sin B}{PB}+\frac{\sin C}{PC}\ge 4\sqrt{\frac{OP}{PAPBPC}}\sin A\sin B\sin C \]
Equality iff $ ABC$ is equilateral and $ P\to\infty$.


is there a coloring of $\mathbb{R}^2$ with countable color for the angel $\theta$ such that there are not any $A,B,C$ st $\widehat{ABC}=\theta $


Prove that for every polynomial $P(x)$ with real coefficients there exist a positive integer $m$ and polynomials $P_{1}(x),\ldots , P_{m}(x)$ with real coefficients such that
\[P(x) = (P_{1}(x))^{3}+\ldots +(P_{m}(x))^{3}\]


Let

        M = X/(Y + aZ) + Y/(Z+aX) + Z/(X+aY),

Find the minimun value of M.
 If X,Y,Z are positive real numbers and "a" a real number.

I have found a partial solution for this one:
first we write:[tex]
M= \frac{x^2}{xy+azx} + \frac{y^2}{yz+axy} + \frac{z^2}{zx+ayz}[/tex]
and by cauchy's inequality:
[tex] (xy+yz+zx)(a+1)M \ge (x+y+z)^2 \ge 3(xy+yz+zx)[/tex]
then   [tex] M \ge \frac{3}{1+a} [/tex]
if [tex] a \succ -1[/tex]


geometry
3D geometry
superior algebra
superior algebra unsolved

Let E1/F and E2/F be Galois extensions. If E1 and E2 are finite, I know how to show that the composite field E1E2 is finite and Galois. What about when E1/F and E2/F are infinite?

Also, a classmate conjectured that if E1/F and E2/F are infinite and Galois, then E1E2 = E1 or E2. Is this true?

The composition field is clearly seperable (it's generated by such elements) and normal, the latter since being true by the following definition of normal (equivalent to the others): $E|F$ is normal iff every homomorphism to an algebraic closure is already an automorphism/a homomorphism into $E$ itself.
And it's also infinite since a infinite subfield is contained in it

No, the conjecture is clearly false, e.g. take those extensions of $\mathbb{Q}$ by addjoining square roots (giving $E_{1}$) and adjoining cube roots (also the complex ones; giving $E_{2}$).


induction
algebra
polynomial
algebra proposed

here is my solution:
let $x_0=1,x_(n+1)=x_n^2+1$, then by induction we obtain $p(x_n)=x_n$,by polynomial identity theorem, we obtain $p(x)=x$.

Your solution to what?  :?:  ;)

Pierre.

[quote="pbornsztein"]Your solution to what?  :?:  ;)

Pierre.[/quote]
He must have clicked "New topics" instead of "Post reply" when he was typing his solution... :D


calculus
integration
parameterization
function
vector
absolute value
calculus computations

I suppose it could be if you use a reversed parametrization.

If by "arc length" you mean "the length of an arc" then the answer is obviously "no," as lengths are nonnegative.  If by "arc length" you mean "value of an integral meant to give arc length with respect to some parametrization," then the answer is "yes, if you have the bounds of integration in an unusual order."

There are two attitudes about integrals here.

One is the attitude, strongest in first-semester calculus, that emphasizes the algebra of antiderivatives, and has - and needs - the theorem $ \int_a^bf(x)\,dx=-\int_b^af(x),\,dx.$

An example of this thinking: do the integral $ \int_0^2x\sqrt{4-x^2}\,dx$ by making the substitution $ u=4-x^2.$ That gives us $ du=-2x\,dx$ and leaves us with

$ -\frac12\int_4^0\sqrt{u}\,du$ which we may then turn around to $ \frac12\int_0^4\sqrt{u}\,du=\frac83.$

JRav's post is rooted in this thinking.

There's a second attitude, which starts to really take root around the first multivariable calculus course. De-emphasize the algebraic and antiderivative aspects, and emphasize that we integrate functions over [i]sets.[/i] The notion of a multivariable change of variables, with its attendant Jacobian, forces us to think that way. With this attitude, return to the same problem.

$ \int_0^2x\sqrt{4-x^2}\,dx$ is the integral of a positive function, hence positive. (Well, the function is positive except for two points, but two points don't count.)

Make the substitution $ u=4-x^2.$ We note that the mapping $ x\mapsto 4-x^2$ sends the set $ [0,2]$ to the set $ [0,4].$ Sure, it flips the set around (because that's a decreasing function), but we don't care. As for the Jacobian: we always take the absolute value of that anyway. Hence $ du=2x\,dx.$ The integral become

$ \frac12\int_0^4\sqrt{u}\,du=\frac83.$

Now, I'm going to take that second attitude - the attitude that integrals are over sets, the vector calculus-analysis attitude - and I'll answer ThetaPi's question.

No, an arclength can't be negative. It's given by the integral over a set (and it's not an oriented integral) of a nonnegative function, and the result will be nonnegative. (In fact, positive except for trivial "curves" that consist of a single point.)


When each side of a square is decreased by $25\%$, the area of the square decreases by 7 square meters. What is the number of square meters in the area of the original square?

Yes this is a different solution

[hide]chances are they give you integers. Doesn't hurt to try the two that differ by 7, 3&4. Almost instant answer of $4^{2}= 16$[/hide]

I know that ur new, but please hide solutions. This is done by surrounding the solution w/ hide tags.

[hide]
$x^{2}-(\frac{3}{4}x)^{2}=7$

don't know how to solve that, but since it's obvious that 8 doesn't work, 4 does (maybe 6, but chances are that 4 works)

16-9=7

answer is 16[/hide]

jorian


We define a primitive solution to $ x^2 \plus{} y^2 \equal{} z^2$ such that $ (x,y,z) \equal{}1$. Further more two solutions is essentialy alike if we can get from one to the other by changing signs of $ x,y,z,$ or by changing $ x$ and $ y$.

Given $ z \in \mathbb{Z}$ Prove that there are either $ 0$ or $ 2^n$ essentialy different solutions for some $ n \in \mathbb{N}_0$ to the equation $ x^2 \plus{} y^2 \equal{} z^2$ where $ x,y \in \mathbb{Z}$

Generally, this is not true. The smallest counterexample is $ z\equal{}25$ for which there are 3 essentially different solutions:
$ 0^2\plus{}25^2\equal{}25^2$
$ 7^2\plus{}24^2\equal{}25^2$
$ 15^2\plus{}20^2\equal{}25^2$

In general, the number of solutions is given by the [url=http://mathworld.wolfram.com/SumofSquaresFunction.html]formula (17)[/url].

The number of solutions with coprime $ x,y,z$ is a power of $ 2$ (or $ 0$). Seeing this is rather easy using arithmetic of $ \mathbb Z[i]$. In fact, one already gets the some result for the solutions of $ x^2 \plus{} y^2 \equal{} n$ and the idea is that the number of ways to write a number as product of coprime numbers is a power of $ 2$.
Formula 17 is in my oppinion not the nicest way to state the result ;)
(and well, due to excessive errors, often very stupid ones, I've some bad attitude towards MathWorld in general; that formula is correct, but I wouldn't trust them on anything I don't have a second source or proof for)

[quote="maxal"]Generally, this is not true. The smallest counterexample is $ z \equal{} 25$ for which there are 3 essentially different solutions:
$ 0^2 \plus{} 25^2 \equal{} 25^2$
$ 7^2 \plus{} 24^2 \equal{} 25^2$
$ 15^2 \plus{} 20^2 \equal{} 25^2$

In general, the number of solutions is given by the [url=http://mathworld.wolfram.com/SumofSquaresFunction.html]formula (17)[/url].[/quote]
Remember that $ (x,y,z) \equal{} 1$. Indeed $ (0,25,25) \equal{} 25 \neq 1$ and $ (15,20,25) \equal{} 5 \neq 1$.

I have too proved it in $ \mathbb{Z}[i]$ so I was just looking for a proof using classic number theory :)


Four positive integers have a product of 504 and a sum of 20. What is their arithmetic mean?

Since the integers have a sum of 20, and there are 4 integers, the arithmetic mean is $ 20 \div 4 \equal{} 5$.


function
inequalities
algebra unsolved
algebra

determine all functions f defined from the set of positive real numbers to itself such that
$ f(g(x))\equal{} \frac{x}{xf(x)\minus{}2}$ and $ g(f(x))\equal{} \frac{x}{xg(x)\minus{}2}$ for all positive real numbers x :roll:

Let's denote :   $ u(x)\equal{}xf(x) ,\ v(x)\equal{}xg(x)$ . Manifestly $ u(x)>2 ,\ v(x)>2$ and $ u(g(x))\equal{}\frac{v(x)}{u(x)\minus{}2}>2,\ v(f(x))\equal{}\frac{u(x)}{v(x)\minus{}2}>2$  therefore
$ v(x)>2u(x)\minus{}4,\ u(x)>2v(x)\minus{}4$, so $ u(x)\plus{}v(x) < 8$ and  it's easy to see that $ u(x),v(x)$ are bounded.
Let's denote : $ U_i \equal{} \inf (u(x)),U_s\equal{}\sup (u(x)) ,\ V_i \equal{} \inf (v(x)),V_s\equal{}\sup (v(x))$ the inferior,superior limit bounds of the functions.
$ u(g(x))\equal{}\frac{v(x)}{u(x)\minus{}2}\ \Longrightarrow U_i\leq \frac{v(x)}{u(x)\minus{}2}\leq U_s$ $ \ \Longrightarrow \frac {V_s}{U_i}\geq u(x)\minus{}2 \geq \frac {V_i}{U_s} \ \Longrightarrow  \frac {V_s}{U_i}\plus{}2 \geq U_s\geq U_i\geq \frac {V_i}{U_s}\plus{}2$ $ \Longrightarrow V_s\plus{}2U_i\geq U_iU_s\geq V_i\plus{}2U_s$
In the same way, from $ v(g(x))\equal{}\frac{u(x)}{v(x)\minus{}2}$ we obtain  $ U_s\plus{}2V_i\geq  U_i\plus{}2V_s$
Now , by adding the two inequalities, we obtain : $ V_s\plus{}2U_i\plus{} U_s\plus{}2V_i\geq V_i\plus{}2U_s \plus{} U_i\plus{}2V_s\ \Longrightarrow\ 0\geq (U_s\minus{}U_i)\plus{}(V_s\minus{}V_i)$
Because of $ U_s\geq U_i,\ V_s\geq V_i$ we can write $ U_s\equal{}U_i\equal{}U ,\ V_s\equal{}V_i\equal{}V$ and $ U\equal{}\frac VU\plus{}2,\ V\equal{}\frac UV\plus{}2$
Finally we obtain $ U\equal{}V\equal{}3$ so $ f(x)\equal{}g(x)\equal{}\frac 3x$
 :cool:


solve the diophantine equation
$(x-y-z)(x-y+z)(x+y-z)=8xyz$
where x,y,z are relatively prime


Wow :)
Cool solution :)
I knew it was kind of weak,because in my solution I used a very weak Am-Gm.
I have a soln with Cauchy for this ineq..

it is very similar with Levinson's inequality :)

Also ways by rearrangement:
$\sum_{cyc}\frac{b+c}{a}\geq4\sum_{cyc}\frac{a}{b+c}\geq4\sum_{cyc}\frac{a}{a+b}$
 $\sum_{cyc}\frac{b+c}{a}\geq\sum_{cyc}\frac{a+b}{b}\geq4\sum_{cyc}\frac{a}{a+b}$

[quote="andyciup"]Let $a,b,c$ be positive real numbers.Prove that the following inequality occurs: \[\sum\frac{b+c}{a}\geq4\sum_{cyc}\frac{a}{a+b}\] [/quote]

apply:

$\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}$ (cauchy)

so we have:

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

$\Rightarrow 2\sum{\frac{c+b}{a}}\geq 2\sum{\frac{4c}{a+b}}$, and now rearrangement.

The square root inequality is familiar with somebody. But it is still new with some others. So I post it (just for fun). \[\sqrt{b+c \over a}+\sqrt{c+a \over b}+\sqrt{a+b \over c}\ge 2\left(\sqrt{a \over a+b}+\sqrt{b \over b+c}+\sqrt{c \over c+a}\right).\]

Here is my solution for the initial problem:
By AM-GM we have $a+b\geq2\sqrt{ab}$ and $c^{2}+ab\geq2c\sqrt{ab}$,which means that \[(a+b)(c^{2}+ab)\geq4abc\] ,which is equivalent to \[\frac{b}{a+b}\leq\frac{c^{2}+ab}{4ac}\] ,which means that \[\frac{c}{a}+\frac{b}{c}\geq4\frac{b}{a+b}\] Summing up the other two cyclic ones gives us the requested inequality.

Also,here is a proof with cauchy,but it`s not a big deal:
By cauchy we have $(a+b)(ab+c^{2})\geq b(a+c)^{2}$,which means that \[\frac{b}{a+b}\leq\frac{c^{2}+ab}{(a+c)^{2}}\] But $(a+c)^{2}\geq4ac$,which means that $\frac{b}{a+b}\leq\frac{c^{2}+ab}{4ac}$,which is what we were looking for..


and the penguin comes back :D:D:D with enhanced features this time :D

i got exactly 750.... :D
and the best record is 1156.x after 1 min....
now the lastest record is 1215.8?

I'm not so sure about enhanced features...it's still "click, click again when you want to hit the penguin", right?

I think the distances were scaled so 320 in the previous version should be around 1200 or so. :) [to reorientate ourselves to the record to beat] :D

previous record: 321.7
 Now: 1203.3. I aggree with Valiowk.

yeah but I love to hit it to go further than 320 :)
more of the penguin coming soon :D

This is more fun  :D .Best rec:1215 after 2 mins

Which is the direct link to the swf?

Edit:Nevermind.It is:
[url]http://www.mathlinks.ro/files/pingu3.swf[/url]

A friend played it a few hours and got 4 times 1224.1 (I think thats the max possible achievable distance)

wt's the shortest distance you can get?
I get -56.3  :D , it's -56.3

Yup PDP,  1224.1 on several occasions - Not a tenth higher. Why have you done this to us Valentin?!?!?!?!?!?!

[quote="tripleJ"]Why have you done this to us Valentin?!?!?!?!?!?![/quote]
what, what, what have I done?  :)

[quote]what, what, what have I done?[/quote]
This is what god said when he created Adam [img]http://www.smiliedb.de/s/sdb49878.gif[/img]

Here's my record after a few minutes of playing:)

Indeed, what have you done to us ? :D

Maybe it's time to go back to good ol' penguin v. 1 (and try cracking Valentin's 327.1, because I don't think any of us can do the 383...) :D

[quote="PiDeltaPhi"]Ahahahaha!I was thinking,what the heck,the poor pinguin gets kicked with the baseball bat,then poked...but NO,the pinguing strikes back.GO GO Pinguin! [img]http://community.the-underdogs.org/smiley/happy/thumb.gif[/img][/quote]
speaking of strike-ing have fun with the new version of pingu.

it's [url=http://www.mathlinks.ro/fun/pingslap.swf]HERE!!!![/url] :D

[url]http://www.mathlinks.ro/fun/pingslap.swf[/url]

The link doesnt work,sorry

[quote="Valentin Vornicu"]damn. I've played to much Pingu' and my comps' starting getting weird errors :?. The last one I've got looks like this:[/quote]
some bugs in your picture.
the fonts in the buttons are different...
heh, you need to use "MS Sans Serif" for the two rightmost buttons.

the link is fixed. liyi there is no problem with the fonts, it was just a joke :D.

1199 is my new score.This is really nonsense...


Given $ a_i\ge0$ and $ \sum_{i \equal{} 1}^n a_i \equal{} 1$.Find the maximum value of

$ P \equal{} \sum_{i \equal{} 1}^n \frac1{\sqrt {1 \plus{} a_i^3}}$

[quote="dduclam"]Given $ a_i\ge0$ and $ \sum_{i \equal{} 1}^n \equal{} 1$.Find the maximum value of

$ P \equal{} \sum_{i \equal{} 1}^n \frac1{\sqrt {1 \plus{} a^3}}$[/quote]

Hi. [b]dduclam[/b] $ \sum_{i \equal{} 1}^n \equal{} 1$   ?  $ \sum_{i \equal{} 1}^n{a_{i}} \equal{} 1$ or ...

[quote="zaya_yc"][quote="dduclam"]Given $ a_i\ge0$ and $ \sum_{i \equal{} 1}^n \equal{} 1$.Find the maximum value of

$ P \equal{} \sum_{i \equal{} 1}^n \frac1{\sqrt {1 \plus{} a_i^3}}$[/quote]

Hi. [b]dduclam[/b] $ \sum_{i \equal{} 1}^n \equal{} 1$   ?  $ \sum_{i \equal{} 1}^n{a_{i}} \equal{} 1$ or ...[/quote]

Yes, $ \sum_{i \equal{} 1}^n{a_{i}} \equal{} 1$, i edited my post,sorry :(


Disculpen pero alguien podria postear toda la prueba de la OMR 2006, ó pasarme un link para ver la prueba. por favor.
chau

tengo los dos primeros problemas de nivel 2
el primero esta en el topic de geo :

http://www.mathlinks.ro/Forum/viewtopic.php?t=123246

el segundo problema es :
carlitos escribio todos los subconjuntos de ${1,2,...,2006}$ en los que la diferencia entre la cantidad de numeros pares y de numeros impares es multiplo de 3
¿cuantos subconjuntos escribio carlitos?

P_3) se han elegido varios segmentos en una recta (con posibles superposiciones) si a cada segmento se le colorea de azul sus dos terceras partes de la izquierda, el conjunto de los puntos azules es un segmento de longitud 31. si a cada segmento se le colorea de rojo sus dos terceras partes de la derecha, el cunjunto de los puntos rojos es un segmento de longitud 23. sean $M$ y $m$ la maxima y minima longitud de un segmento de la coleccion, respectivamente.
¿cual es el valor minimo que puede tener $M-m$ ?

Ya que soy muy malo contando tengo una solucion horrible como esta.

Primero notemos que si un subconjunto tiene $a$ elementos pares y $b$ elementos impares, es necesario y suficiente para que Carlitos lo escriba que:
$a \equiv b (mod 3)$.
Entonces si Carlitos escribio $n$ subconjuntos tenemos que:
$\\ n= \big[ \binom{1003}{0}+\binom{1003}{3}+...+\binom{1003}{1002}\big]^{2}+\big[ \binom{1003}{1}+\binom{1003}{4}+...+\binom{1003}{1003}\big]^{2}+\big[\binom{1003}{2}+\binom{1003}{5}+...+\binom{1003}{1001}\big]^{2}= a^{2}+b^{2}+c^{2}$

Porque primero escogemos la cantidad de pares $a$ y luego la de impares $b$ que sea congruente con $a$ modulo 3.

Consideremos a $\omega \not= 1$ tal que $\omega^{3}-1=0$, tenemos que:
$\\ \omega^{2}+\omega+1=0 \implies \\ \omega^{2i}+\omega^{i}+1=0 \forall i: 3\not|i$ (*)
Además $(1+\omega)(1+\omega^{2})=1$ y $\omega^{a}= \omega^{b}\iff a\equiv b (3)$

Ahora consideremos $f(x)=(1+x)^{1003}$ expandimos y nos queda:
$f(x)= \sum_{i=0}^{1003}\binom{1003}{i}x^{i}$
Ahora usemos las propiedades de $\omega$ para hallar que:
$f(1)+f(\omega)+f(\omega^{2}) = \sum_{i=0}^{1003}\binom{1003}{i}(1+\omega^{i}+\omega^{2i}) = 3a$ pues se cancelan todos los terminos que no tengan a $i$ mulitplo de 3.
Además:
$f(1)+\omega f(\omega)+\omega^{2}f(\omega^{2}) = \sum_{i=0}^{1003}\binom{1003}{i}(1+\omega^{i+1}+\omega^{2(i+1)})= 3b$

$\\ f(1)+\omega^{2}f(\omega)+\omega f(\omega^{2})= \\ f(1)+\omega^{2}f(\omega)+\omega^{4}f(\omega^{2})= \sum_{i=0}^{1003}\binom{1003}{i}(1+\omega^{i+2}+\omega^{2(i+2)})= 3c$

Esto quiere decir que:
$\\ n= a^{2}+b^{2}+c^{2}= \\ \frac{[f(1)+f(\omega)+f(\omega^{2})]^{2}+[f(1)+\omega f(\omega)+\omega^{2}f(\omega^{2})]^{2}+[f(1)+\omega^{2}f(\omega)+\omega f(\omega^{2})]^{2}}{9}$
Expandimos la expresion anterior y obtenemos despues de agrupar convenientemente algunos terminos:

$n= \frac{3f^{2}(1)+(f^{2}(\omega)+f^{2}(\omega^{2})+2f(1)f(\omega)+2f(1)f(\omega^{2})) (1+\omega+\omega^{2})+3\omega^{3}f(\omega)f(\omega^{2})}{9}=$
$\frac{3f^{2}(1)+3f(\omega)f(\omega^{2})}{9}=$
$3\frac{[(1+1)^{1003}]^{2}+2(1+\omega)^{1003}(1+\omega)^{1003}}{9}=$
$\frac{2^{2006}+2}{3}$

Con lo que queda terminado el problema.

Como veran es una solucion muy tecnica, pero el truco de $\omega$ es muy bueno.


Simplify and express your answer as a common fraction:
$ \frac{2\cdot(3)^{\minus{}2}\plus{}3\cdot(2)^{\minus{}3}}{2\cdot(3)^{\minus{}2}\minus{}3\cdot(2)^{\minus{}3}}$

Check out this alternate method of getting rid of "fractions within factions":

$ \frac{2\cdot(3)^{\minus{}2}\plus{}3\cdot(2)^{\minus{}3}}{2\cdot(3)^{\minus{}2}\minus{}3\cdot(2)^{\minus{}3}}$

Multiply the top and bottom by $ 3^2$ and $ 2^3$, distribute and cancel:

$ \frac{3^2\cdot 2^3\cdot(2\cdot(3)^{\minus{}2}\plus{}3\cdot(2)^{\minus{}3})}{3^2\cdot 2^3\cdot (2\cdot(3) ^{\minus{}2}\minus{}3\cdot(2)^{\minus{}3})}\equal{}\frac{2^4 \plus{}3^3}{2^4\minus{}3^3}\equal{}\frac{16\plus{}27}{16\minus{}27}\equal{}\frac{43}{\minus{}11}\equal{}\minus{}\frac{43}{11}$

Would that be finding the LCM of the numerator and denominator?

Nice trick, btw. :wink:

Yeah ernie, that's right... the LCM of the denominators you want to wipe out.

So in
$ \frac{\frac{1}{3}\plus{}\frac{5}{6}}{\frac{3}{4}\minus{}\frac{2}{3}}$

you could multiply top and bottom by 12, reducing the problem to

$ \frac{4\plus{}10}{9\minus{}8}\equal{}14$

Standard algebra trick :)

Yeah.

[quote="RightAnnihilator"]"fractions within factions"[/quote]

Complex fractions. :wink:


if A and B are two nxn matrices  such that  $ AB \equal{} I_{n}$     then prove that $ BA \equal{} I_{n}$  and prove that $ B \equal{} A^{ \minus{} 1}$

We have that $ BA\equal{}(BA)(BB^{\minus{}1})\equal{}B(AB)B^{\minus{}1}\equal{}BI_{n}B^{\minus{}1}\equal{}I_{n}$.

So... how do you know that $ B$ has an inverse? You can't prove this by just manipulating operations, since it's not true in a general ring.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=277173]this thread[/url].

He said to prove that $ B\equal{}A^{\minus{}1}$ so I assumed the existence of inverses.  But if this isn't a division ring, then the statement shouldn't be true in general.

Surprise: the statement becomes truer if you actually read it.

  darij


Hi, when I run LaTeX on an emacs text file, I get this error:

blazes > latex chap3.tex
This is pdfeTeX, Version 3.141592-1.21a-2.2 (Web2C 7.5.4)
entering extended mode
(./chap3.tex
LaTeX2e <2003/12/01>
Babel <v3.8d> and hyphenation patterns for american, french, german, ngerman, b
ahasa, basque, bulgarian, catalan, croatian, czech, danish, dutch, esperanto, e
stonian, finnish, greek, icelandic, irish, italian, latin, magyar, norsk, polis
h, portuges, romanian, russian, serbian, slovak, slovene, spanish, swedish, tur
kish, ukrainian, nohyphenation, loaded.
! Undefined control sequence.
l.1 \section
            {Introduction}
?

I'm new to LaTeX. I have checked the syntax, which seems to be correct, and as far as I know the correct packages have already been installed for me, so what could be causing this error? Any advice appreciated, Thanks.

If \section is indeed line 1 of your document (as your error message implies), then that's your problem.  The first line should be the \documentclass line, and before you can start doing stuff like \section you have to have \begin{document}

chap3.tex is part of a number of chapters in which are all put into one document, main.tex, when I run latex main.tex.

In main.tex I have \begin{document} and \end{document}. This should not be required again in the individual chapters, as I used the same format last year and I didn't have this problem. (My supervisor gave me the code) So do you think this problem might be with the package? Thank you.

See below:

\begin{document}

\title{}

\author{}
\date{September 2004}
\maketitle

\chapter*{Declaration}

I hereby declare that this thesis is entirely my own work and that it
has not been submitted as an exercise for a degree at any other
university.

\begin{center}
\vspace*{2in}

\underline{\hspace*{3in}} \today

Myriam

\end{center}

\chapter*{Acknowledgements}
\input{ack}

\pagestyle{headings}
\tableofcontents
\listoftables

\begin{abstract}
\noindent

\end{abstract}

\chapter{Introduction}
\include{chap1}

\chapter{}
\include{chap2}

\chapter{Programs Utilised for Empirical Research}
\include{chap4}

\chapter{}
\include{chap5}

\chapter{}
\include{chap6}

\chapter{}
\include{chap7}

\chapter{Conclusion}
\include{chap8}

\bibliography{bibliography}
\bibliographystyle{egapa}

\appendix

\chapter{}
\include{}

\chapter{}
\include{}

\chapter{}
\include{}

\chapter{}
\include{}

\chapter{}
\include{}

\chapter{}
\include{}

\end{document}

But the output you posted above (in your first message) was for the command "latex chap3.tex", not "latex main.tex"...maybe that's your problem?

And \documentclass is still required somewhere, before \begin{document}.

[65]
Underfull \hbox (badness 10000) in paragraph at lines 221--222

[66]) [67] [68]
Chapter 7.
[69] (./chap8.tex [70]) [71] (./main.bbl [72]) [73]
Appendix A.
[74]latex: Not writing to .aux (openout_any = p).

! I can't write on file `.aux'.
\@include ...\immediate \openout \@partaux #1.aux
                                                  \immediate \write \@partau...
l.155 \include{}

Please type another output file name:


Korisni linkovi:

[url]http://www.mathlinks.ro/viewtopic.php?t=70684[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=27757[/url]

Napomena moderatora: Dio diskusije je preseljen ovdje sa teme Off-off-Broadway, radi lakseg snalazenja.

P. S. Otvaranje besmislenih tema se ne tolerise, timmysa.

Da bi ova tema dobila neki smisao, evo nešto što me zanima (to sam se danas na tulumu zapitao): ima li $ \frac {a^p \minus{} 1}{a \minus{} 1} \equal{} p^k$ rješenja gdje su $ a, p, k$ prirodni brojevi, $ p \ge 3$ prost broj i $ a \equal{} 1$  $ (mod$  $ p)$.

[url]http://www.math.olympiaadid.ut.ee/eng/archive/prob0607.pdf[/url]

Eto vidi tu na stranici 12 imas taj 3 zadatak koji mi se donekle cini slicnim ovom tvom pa ti to mozda da neku ideju.

Jedino mislim da se tu dokazuje obrnuti smjer:

Ako je izraz potencija prostog broja ---> p je prost.

Ma to je trivijalno ako znas sledecu teoremu:

Ako je $ p>2$ prost tada $ p^k\parallel a\minus{}1$ ako i samo ako $ p^{k\plus{}1}\parallel a^p\minus{}1$ ($ k\geq 1$). ($ p^k\parallel x$ ako i samo ako $ p^k\mid x$ i $ p^{k\plus{}1}\nmid x$)

To znači da jednadžba nema rješenja, dakle zadatak s tuluma je riješen!

Gobline hvala ti. Je li teško taj teorem dokazati?

Nije tesko... Ali to i nije teorema bas , to je vise lema ...

[quote="hana"]U vezi teoreme.. Recim da sam je ja koristila na izbornom bez dokaza i da mi nisu oduzeli poene, ali su nam rekli da bi to trebalo dokazati na medjunarodnom taknicenju, jer ocigledno nije bas toliko poznato, a i nema ime ...[/quote]

Ma, ta teorema ili lema, kako god, je toliko poznata (svaka malo ozbiljnija knjiga iz teorije brojeva je mora imati, jer se preko nje dokazuje par mnogo bitnih teorema), da bih bio veoma iznenadjen ako bi vam na bilo kom ozbiljnijem takmicenju oduzeli poene. Kako nema ime mozda je treba navesti, tacnije napisati nesto u stilu koristicemo sledecu teoremu, i tada skoro sigurno nece biti nikakvih problema.

Sto se naziva tice ovde imate nesto http://www.mathlinks.ro/viewtopic.php?t=70684 .

Ovo je znano kao Hensels lemma dokazuje se indukcijom, a baš ne znam ili ovo lahko na natjecanju samo napišeš da je teorem.

Otkrio sam i ja ovaj podforum :D
Ali kad sam znao da mornik i hsiljak moraju biti ovde, kao i hana kad se cima oko nekih gluposti :P

Takodje korisna lema:

-postoji prost broj q>p, gde je p takodje prosto, tako da q deli a^p-1, a ne deli a-1.


今天我的数学?师问我们:"?知?为什么?"诱导公?"?"我们摇头,等待答案,于是,?师?紧?慢地说:"我也?知?......"

哪?能帮帮忙,告诉我到底为什么?诱导公?? :?

maybe because you are adding the period, so its adding or inducing.

in what context?  what area of math are you doing in class?

[quote="liyi"]I've only heard "诱导公?" in trignometry...
which is something like $\sin(x+n\frac{\pi}{2})=\pm\sin x(\cos x)$...[/quote]

就是三角里的那"一大砣"公?呀......


Hello,

What is your favorite pet? Mine is either dogs or cats, but I voted for dogs. I tried to think in all the options I could, but it seems like it only accept up to 10 options.

Best,

hmmm...between a cat and a dog, i'd have to go with...a dog. I never really had a pet before but my neighbors were pet crazy. They probably had every kind of pet listed in the survey. The cats were real nice and they can think for themselves. Dogs will follow you around EVERYWHERE. You can play catch with them. We taught his dog how to play soccer, but then it popped my soccer ball :(

[quote="Treething"]Little kids [img]http://forums.finalhit.org/html/emoticons/whistling.gif[/img] .[/quote]

Little kids can be pretty nice. However, one time, i was taking care of my neighbors and their son (who's real little) started "cutting" me up with his toy lightsaber. Thats freaky!!! :P

[quote="Treething"]Little kids [img]http://forums.finalhit.org/html/emoticons/whistling.gif[/img] .[/quote]

gosh darnit...u beat me to it. i was going to say my little brother *evil grin*

[quote="Valentin Vornicu"]Clearly dogs are the best pets ever :P ... I am amazed that Gyan hasn't yet replied in this topic :)[/quote]

I do think that dogs are great pets; very loyal, usually they like to play. But also they require a lot of time, at least if you really want to have a happy dog. On the other hand, if you don't have too much time, usually a cat is great. They don't need that much attention, are very independent, but they are less prone to play with you.

Best,

[quote="DarkKnight"][quote="yif man12"][quote="miraculouspostmaster"]Pet rocks, anyone?[/quote]

Right here![/quote]

Lol... Rocks. I chose a dog because my friend has a dog that plays with us all the time.[/quote]

Cool! My friend has a pet rock and he plays with us all the time too!  :D  ;)

I have a pet snake, houdina, and she is great. She is a corn snake that is somewhat olive colored. She is pretty big, about 1.5 meters, and as the name suggests, she is an escape artist. :lol:

I used to have another snake, snowpowder, but she contracted cancer and it spread. A very, sad and painful death :(
but u have to move on

pet rocks are cool.... i know someone that got bit by one once

cats are better than dogs!!!!!!!!  but... my parents hate the "mess" pets make so i don't have one.

elvenchamp777

[quote="Valentin Vornicu"]What's a pet rock anyway? (I only know that rock is a kind of stone (or vice-versa) or a musical genre - and an adverbe meaning something in the lines of "cool" - e.g. "they rock")  :?  :?[/quote]

there really is nothing special about a pet rock. it's basically a stone that sometimes gets features glued or drawn onto it and usually has a name.

pet rocks were a huge fad in the seventies. it is what it says it is- a pet ROCK. that is, a rock that someone has painted and is now a pet. you dont have to feed them, walk them, or clean up after them.

but anywho. you need to add an "all of the above" option to the poll.

gosh, school was great when we had class pets. "sigh", now all the science classes has fake human skeletons. Turtles are supercool!!!

[quote="Valentin Vornicu"]ok so it is a [b]stone[/b] in the end :) ... I'm not sure, but last time I checked, pets ment [b]live[/b] animals, right? :? :)[/quote]

Oh, you should meet my pet rock.  :D

Chickens  :D . Not only are they fun to watch run around their little coop, but they lay eggs which you can sell for big *inster your currency symbol here*. Though when it comes to a house pet, nothing beats a good dog.

[quote="filletwho"]My math book...[/quote]

did you name it?  :D

[quote="mcalderbank"]does it (for lack of better word) have a name?[/quote]

His name is Bill.

[quote="filletwho"]My math book...[/quote]

I beg your pardon, but I think it is worst than having a Pet Rock. C'mon, I love math, but everybody needs a life too.

Best,

Cats are awesome!

I have 3 cats, one of them is a ghost cat, hehehe.  Its white with blue eyes that glow red.  She actually doesn't show up in any pictures we took of her, only as an extremely faint outline if you look for it.  Its really kinda scary.

Ever heard the quote by Churchill?  Not about pets, but close enough.  "I like Pigs. Dogs look up to us.  Cats look down on us.  Pigs treat us as equals."

I voted for dogs. :) . The one pictured on the left is actually a graduate of  the  Oregon campus of the famous .[url=http://www.guidedogs.com/]Guide Dogs.[/url]  and  some of you may know him from   [url=http://www.thefacebook.com/]" The Face Book "[/url] as a graduate student in MIT.  (That explains why he spends most of his spare time socialising with his friends online)  Besides, he is my friend and a family member  so he is more than a pet.:) (His eating habits does inspire mathcounts problems: eg: [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=25885]this ..:)[/url]

I have a pet rock named Mike the Mineral

also, i feed my pet cat-flavored dog food

You totally forgot about stuff like [hide]crawfish[/hide]

[quote="djimenez"]It seems like no body likes birds. Nobody have voted for parrots or singing birds  :([/quote]

Birds are great! I used to own one and it's actually a very good pet. The only annoying part is shutting it up when you're trying to go to sleep, but it didn't disturb my sleeping after about a week.

On those lines, I think passenger pigeons would be very cool. But the coolest thing to do with a bird is to teach it to poop on other people. That would be pretty funny, but a little mean too.

I luv little bunnies. Dwarf rabbits are cool. Mine will respond to name and seems to like drinking tomato juice. I dunno why

mine may be a horse or an eagle, i love both!

[quote="filletwho"]My math book can show you tricks that a dog would never master...[/quote]

my pet rock (mike the mineral) can eat your magical math book.

[quote="jackroland"]I voted for dogs but I really like all of them.
After all I do have:
4 dogs
4 cats
2 hedgehogs :D
1 cockateel
and soon to be...
2 horses
3 more dogs
2 more cats
and possibly a snake.
For a grand total of 19 pets. :D  :D[/quote]

thats a lot...my neighbors have...

1 dog
3-6 puppies
3 cats
5 fish
4 ducks
4 chickens
maybe more...

[quote="jackroland"][quote="plokoon51"]
3-6 puppies
[/quote]
"3-6"? does that mean it changes?
We used to have chickens but they were all eaten by various other animals.
(and I'm sure not complaining! ;) )[/quote]

well, i don't know how many exactly. they (neighbors) might've given some away, or kept them, i'm not quite sure.

on the other hand, the puppies like to deceive people by changing their numbers :D


I've seen these type of problems and I'm posting in Pre-Olympiad forum because I want to know the general way to solve this. (I consider Intermediate to find the answer rather than regular way)

Like how woud you estimate the value of $\sqrt[3] 7$? How about $\sqrt[3] 4$? Or $\sqrt[8] 91$? Also, how would you find the ones like this $\sqrt[3] 2$ and $\sqrt[3] 3$ where they must be very close..

How would you do those estimations?

I mean without using calculator.

Thank you.

How close do you want your estimation to be?  Anyways the best way is usually just to multiply the number by 1000 or something and guess and check.

There is a way to do it...but it will be somewhat hard with cube roots
Write out the number.
Place lines every three places (for instance, place one between the hundreds and thousands, the hundred thousands and millions, etc.  You may want to write some zeros to the right of the decimal point and place lines every three for those as well.)
Now draw a radical sign over the number.  Look at the group of digits on the very left (could be 1, 2 or 3 digits.).  Write down the largest number whose cube is less that the cube of that number.  Subtract the cube of the number from the number formed by the group.  Now bring down the next three digits.  If the first digit you wrote on top was n, then you must now find the greatest x such that
300xn^2+30nx^2+x^3
is less than this number.  Subtract.  Repeat.  This basically uses the principle that (a+b)^3=a^3+3ba^2+3ab^2+b^3.  As you get larger numbers, the process will become tedious, but you can get an exact number to as many decimal places as you wish.


logarithms
real analysis
real analysis unsolved

Let the Taylor series $ e^{e^{x}}\equal{}\sum_{n \equal{} 1}^{\infty}a_{n}x^{n}$
Prove:$ a_{n}\ge e (\gamma\ln n)^{\minus{}n}$
wher $ \gamma$ is a constant number greater than $ e$.
I find $ a_{n}\equal{}\frac{1}{n!}\sum_{k\equal{}0}^{\infty}\frac{k^{n}}{k!}$
which is related the sum of the Stirling Number of second kind.
But I don't know how to do next.

Hi
your question is equivalent to the next question :
we must limit $ \sum\frac{k^{n}}{k!}$ or find asymptotical expression for this.
I am not sure but I think that $ a_{n}> a_{n\minus{}1}$ if it is true then your problem solved.
in this forum was posted some particular case of this sum
$ \sum\frac{k^{n}}{k!}$
to be exact this sum $ \sum\frac{k^{4}}{k!}$
The idea was in next :
$ k^{n}\equal{} Ak(k\minus{}1)..1\plus{}B(k\minus{}1)(k\minus{}2)..\plus{}..\plus{}Z$
we find this coeficient and after can calculate this sum
or
your sum is equal to
$ x(...(xe^{x}\minus{}x)'...)'$ in point zero.
I think somewhere must be calculated this sum
$ \sum\frac{k^{n}}{k!}$ or asymptotical expression

The coefficients are (up to a multiple of $ e$) the Bell numbers.  The [url=http://www.research.att.com/~njas/sequences/A000110]OEIS[/url] has a lot of information on them, as does [url=http://mathworld.wolfram.com/BellNumber.html]mathworld[/url].  I don't know if any of it is useful for solving your question, however.

Thank you,two.But I don't think it always be true that $ a_{n}>a_{n\minus{}1}$
In fact you can try some small number you will find it is wrong.
Yes,it is really the Bell numbers,but it appeared in an eximine,so I think there may exist an acute solution.

Yes.
I find some information about Bell number and I also calculate that when $ n\to+oo$ then
$ a_{n}$~${ Cn!e^{1+n(\frac{1}{\ln n}-\ln\ln n)}}$
where $ C$ some constant


Patrick drove to work on Monday at an average speed of 40 mph and arrived one minute late. He left at the same time on Tuesday, drove an average speed of 45 mph, and arrived one minute early. How many miles does he drive to work each day?

Watch the units :wink:

\begin{align*}
40\left(t+\frac{1}{60}\right)=45\left(t-\frac{1}{60}\right) &\Rightarrow 40t+\frac{40}{60}=45t-\frac{45}{60} \\
&\Rightarrow 5t=\frac{85}{60} \\
&\Rightarrow t=\frac{17}{60} \\
&\Rightarrow d=40\left(\frac{17}{60}+\frac{1}{60}\right)=\boxed{12}
\end{align*}


Show that 2 <= (1 - x2)2 + (1 - y2)2 + (1 - z2)2 <= (1 + x)(1 + y)(1 + z)
for non-negative reals x, y, z with sum 1.

For the problem with x+y+z=1 etc.:

For the first part:
(1-x^2)^2+(1-y^2)^2+(1-z^2)^2>=2 iff x^4+y^4+z^4+1 >= 2(x^2+y^2+z^2) iff x^4+y^4+z^4+(x+y+z)^4 >= 2(x+y+z)^2*(x^2+y^2+z^2) iff x^4+y^4+z^4+2(xy+yz+zx)(x+y+z)^2 >= (x+y+z)^2*(x^2+y^2+z^2) iff x^4+y^4+z^4+4(xy+yz+zx)^2 >= (x^2+y^2+z^2)^2 iff 2(xy+yz+zx)^2 >= x^2*y^2+y^2*z^2+z^2*x^2, which is obviously true, with equality iff x=y=0 (or y=z=0 or z=x=0).

The second part:
(1-x^2)^2+(1-y^2)^2+(1-z^2)^2 <= (x+1)(y+1)(z+1) iff x^4+y^4+z^4+1 <= 2(x^2+y^2+z^2)+xy+yz+ zx+xyz iff x^4+y^4+z^4+(x+y+z)^4 <= 2(x+y+z)^2*(x^2+y^2+z^2)+(xy+yz+zx)(x+y+z)^2+ xyz(x+y+z). We keep simplifying this (by subtracting common terms from both sides until we are left with 2xyz(x+y+z) <= x^3*(y+z)+y^3*(x+z)+z^3*(x+y). For proving this we just add the inequalities of the type x^3*y+y^3*x >= 2x^2*y^2 and then we use the fact that x^2*y^2+y^2*z^2+z^2*x^2 >= xyz(x+y+z).


search
number theory unsolved
number theory

Problem 2:
Let $ n$ be a natural number such that $ n \equal{} a^{2}\plus{}b^{2}\plus{}c^{2}$, for some natural numbers $ a, b, c$. Prove that
$ 9n \equal{} (p_{1}a \plus{} q_{1}b \plus{} r_{1}c)^{2} \plus{} (p_{2}a \plus{} q_{2}b \plus{} r_{2}c)^{2} \plus{} (p_{3}a \plus{} q_{3}b \plus{} r_{3}c)^{2}$,
where $ p_{j} 's, q_{j} 's, r_{j} 's$ are all nonzero integers. Further, of $ 3$ does not divide at least one of $ a, b, c$, prove that $ 9n$ can be expressed in the form $ x^{2} \plus{}y^{2} \plus{}z^{2}$, where $ x, y, z$ are natural numbers none of which is divisible by $ 3$.

my solution isn't complete but...here is what i did:
part 1-the answer is $ (p_1,p_2,p_3)\equal{}(2,2,\minus{}1),(q_1,q_2,q_3)\equal{}(\minus{}1,2,2),(r_1,r_2,r_3)\equal{}(2,\minus{}1,2)$
part 2-if two of them(i mean $ a,b,c$) are divisable by $ 3$ then the answer will be the same,
it remains two cases:
none of them is divisable by $ 3$
one of them is divisable by $ 3$
if none of them is divisable by three and we want to present
$ 9n \equal{} (p_1a\plus{}q_1b\plus{}r_1c)^2\plus{}(p_2a\plus{}q_2b\plus{}r_2c)^2\plus{}(p_3a\plus{}q_3b\plus{}r_3c)^2$ but this time we r allowed to choose the coefficients between integers, then u can prove that none of them is $ 0$ because if it happens then one of the coefficients must be $ 3$ and...(the rest is trivial).


Does it cost AoPS money to create new communities?  If it does, I see why there wouldn't be a community for every state.  However, if it does not, what are the disadvantages of having 50 state communities, besides a cluttered interface?

It costs them server space (I doubt it's substantial, though)... not sure who'd they pay the money too.... PhpBB is a open-source, free, forum package. (Run-on sentences for the win!)

Clutter, and time.  Also, we'd have to police forums that don't have moderators, which we don't want to do.  So, we only create forums when there's someone who wants to take care of them.


[quote="puuhikki"]Assume $x,y>0, t\in (0,1)$. Prove that
$a) t/x+(1-t)/y \geq 1/(x^{t}+y^{1-t})$ and
$b) 1/(tx+(1-t)y) \leq 1/(x^{t}\cdot y^{1-t}) \leq t/x+(1-t)/y$[/quote]

a) isn't true, put $x=y=100$ and $t=1/2$.

b) Jensen, of course.


For each positive integer $ {m}>{1}$, let P(m) denote the greatest prime factor of m.  For how many positive integers n is it true that both P(n)=$ \sqrt{n}$ and P(n+48)=$ \sqrt{n\plus{}48}$?

Thanks for the help!

[hide="Solution"]$ P(n) \equal{} \sqrt {n}$ implies $ n$ is a perfect square. So, $ n$ and $ n \plus{} 48$ are perfect squares. We want to solve the diophantine equation $ x^2 \minus{} y^2 \equal{} 48$, namely $ (x \plus{} y)(x \minus{} y) \equal{} 48$. There are several cases.

Case 1: $ x \plus{} y \equal{} 48$ and $ x \minus{} y \equal{} 1$. No solution.

Case 2: $ x \plus{} y \equal{} 24$ and $ x \minus{} y \equal{} 2$. $ x \equal{} 13$ and $ y \equal{} 11$.

Case 3: $ x \plus{} y \equal{} 16$ and $ x \minus{} y \equal{} 3$. No solution.

Case 4: $ x \plus{} y \equal{} 12$ and $ x \minus{} y \equal{} 4$. $ x \equal{} 8$ and $ y \equal{} 4$.

Case 5: $ x \plus{} y \equal{} 8$ and $ x \minus{} y \equal{} 6$. $ x \equal{} 7$ and $ y \equal{} 1$. However, $ y \neq 1$, so this doesn't count.

Therefore, there are $ \boxed{2}$ such integers.[/hide]

[hide="Small change to above"]
$ P(n) \equal{} \sqrt{n}$ does imply that $ n$ is a perfect square, but more: that $ n$ is the square of a prime. (Note: it says the greatest prime factor.) So unfortunately Case 4 above doesn't work: $ P(16) \neq \sqrt{16}$, which means only one solution left.
[/hide]


Just to let everyone know (I hope this hasn't already been posted), but the relly awards (where we find out if kevin chen won) will be on regis and kelly this friday

Klebian, I don't think this topic should be posted. There already has been many other topics about this Kevin Chen thing.

Mods please lock this.

He's just telling us when it will be. There's nothing wrong with that.

It is useful to know; however, there is already a topic about this still on the front page. Please post there :) .

http://www.mathlinks.ro/Forum/viewtopic.php?t=166113

Btw, if you don't know if a topic is already posted, searching the forum with the website's search is very easy.


analytic geometry
vector
Euler
geometry
trigonometry
trig identities
Law of Cosines

I know that all AMC questions can be solved without a calculator, but there are just some that either you have to (1) think of an insanely simple and intelligent way or (2) brute force it taking like 20 minutes on one problem or (3) know how to add, divide, square root everything perfectly in a very short amount of time.

for example this problem: http://www.mathlinks.ro/viewtopic.php?p=383159#383159
I can easily do it a simple way--assigning coordinates to the triangle and then using Hebron's formula/law/whatever by finding the length of each side and then calculating sqrt.S(S-A)(S-B)(S-C). but that's pretty much impossible in the short, short amount of time that you get, plus a little carelss mistake will kill the answer. now on my calculator i got it in like 10 seconds! so what now, is the new AMC going to have problems with messy work like this ? (yes, i know that there's probably some easy peasy way to do it, so...is the reason why we don't have calculators so we all have to find the intelligent way? :( )

and yes, i'm probably stupid, but how was i supposed to know, in one of the problems, that 29 was a factor of 667 without a calculator? =P

[quote="Narcissa"]is the reason why we don't have calculators so we all have to find the intelligent way? :( [/quote]

Yes.  Think of this as incentive to improve!  I'm sure you feel better when you find a quick way to solve a problem than when you brute-force it for 20 minutes.  (Coordinate bashing is [i]never[/i] a good idea, and personally I don't find it any fun at all.)

[quote="Narcissa"]and yes, i'm probably stupid, but how was i supposed to know, in one of the problems, that 29 was a factor of 667 without a calculator? =P[/quote]

I don't usually give this advice, but memorizing the prime factorizations of the year numbers is a good idea for competitions.

[quote]I don't usually give this advice, but memorizing the prime factorizations of the year numbers is a good idea for competitions.
[/quote]

that will probably be a very smart thing to do =O

[quote="t0rajir0u"]I don't usually give this advice, but memorizing the prime factorizations of the year numbers is a good idea for competitions.[/quote]
Heh, I do this too.

[quote="t0rajir0u"](Coordinate bashing is [i]never[/i] a good idea, and personally I don't find it any fun at all.)[/quote]
Says you :P

[quote="t0rajir0u"]I distinguish "coordinate bashing" from "using vectors," if that's what you mean.   :?:[/quote]
No, that isn't. I mean giving every point a coordinate, then using the distance formula/shoelace excessively :D

[quote="mwpl11"]So will the problems be slightly different when compared to previous years (more clever solutions; not as much calculation), or will they write the problems to be just as they were before?[/quote]

If its just as before then it shouldnt be any difference because the best solutions as the years before required no calculator (very limited calculation).  Only those "out of the box" solutions required a calculator.

[quote="firecricket91"]Only those "out of the box" solutions required a calculator.[/quote]

I guess you're right.  :blush:

But by doing that, they're basically restricting people to solving the problems in a very limited number of ways ("the best solutions"). That's no fun.

[quote="t0rajir0u"]If you like computational methods of solving problems, try the USACO or [url=http://projecteuler.net/]Project Euler[/url].  But the AMC is about mathematics - [i]good mathematics[/i] - and the best scores should go to the best solutions.[/quote]

I just have to say, USACO usually requires more thought than "co-ordinate bashing" on this problem.

Well, yes; I just meant that the USACO has a different focus compared to the (competitions leading up to) USAMO.

[quote="Sly Si"]I think you can expect to see some problems in a slightly different style now, problems that weren't used before because calculators rendered them trivial.  Take a look at pre-1994 (or is it 1995?) AHSME's to see what I mean.  You'll see problems on there that are definitely interesting if you don't have calculators allowed, but trivial with them.  As I understand it, including problems like that was one of the big motivations for removing calculators in the first place.[/quote]

I agree. That's probably going to be the only major change; a question that occurred on any previous AMC could still occur now even without calculators (except, obviously, they'd be different questions -- you know what I mean). The question # might be slightly different, but that would be a minor change that no one would notice anyways.

[quote="t0rajir0u"](Coordinate bashing is [i]never[/i] a good idea, and personally I don't find it any fun at all.)[/quote]

So, is there always a better solution than coordinate geometry, or can that sometimes be the most elegant method?

Like I said, I distinguish coordinate bashing from the use of vectors.  While the use of vectors can often be quite illuminating (because we set certain quantities as vectors without computing them), the use of coordinate bashing in particular requires calculations that are never (in my experience) the most elegant way of doing a problem.  At least try Law-of-Cosines bashing first, since that [i]can[/i] occasionally be a decent way of doing a problem.

Edit:  I'm talking about geometry problems that aren't stated in terms of coordinates.  Even when problems are stated in terms of coordinates, coordinate-bashing (except as far as Pick's Theorem or the Shoelace Theorem are concerned) is rarely the best method.

someone suggested memorizing "prime factors of the year numbers" -- can someone tell me what is meant by that? By Prime factors of the year?  Thanks

Memorizing the prime factorization of the year - for example, in year 2002, competitors may have memorized $ 2002 \equal{} 2\cdot7\cdot11\cdot13$, so that if the problem "find the sum of the factors of $ 2002$" came up at a competition, they could find the answer more quickly.

I don't know if someone already said this, but I like to split up things like $ 796^2$ into 2 binomials if you will.
$ 796^2\equal{}(800\minus{}4)(800\minus{}4)\equal{}640000\minus{}6400\plus{}16\equal{}633616$

I actually did that one in my head today. I suggest not using a calculator for a bit, and that should help you adjust to manual arithmetic.

My math teacher says that because of the no calculator rule, there will be more clever-based questions that can be solved very quickly by paper, but require some knowledge of, say, correct formulas. But hey don't worry: if this new rule is gonna hurt you, then it's gonna hurt us all! *gulp*

[quote]but require some knowledge of, say, correct formulas.[/quote]GULP! don't know any formulas =O

[quote="hunter34"][quote="AoPS book"]
It is not important to remember formulas, but how they are derived. If you happen to forget that formula you can prove it.
[/quote][/quote]
Or you could take the other route and just not forget formulas ;)

[quote="MellowMelon"][quote="zephyredx"]but require some knowledge of, say, correct formulas.[/quote]
[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]This site's webmaster disagrees.[/url]

Better to say is "but require some mathematical ingenuity." Granted, formulas do work on AMCs, but that approach won't get you much farther.[/quote]

I didn't mean "memorize formulas right-on" i meant "know them well", which would imply knowing to derive them, and also know what kind of problem they appear in. I actually don't trust many of the formulas I can't derive unless it's an absolute emergency.

^whoa, yeah i've seen that before but not used to it. i guess i'll have to do more problems that way!! thanks for sharing


So, if anyone has heard of those famous open problems (Millennium Problems) that includes: Riemann Hypothesis; P vs NP; Yang-Mills Theory etc.

I have a query pertaining to these problems, and it is: Why are these problems so difficult to solve, secondly what does one require to solve these problems, is it advanced knowledge in their respective fields?

Try solving one of them and you'll see.  You're not solving them without some really deep knowledge of the fields and very deep insight.

What does insight mean with reference to problem-solving? Is it intuition or something else?  :maybe:

:o !!!! Insight is the most basic concept of problem solving!!!! It is what separates the US (and probably most other places) public school mathematics education from mathematics competitions.

Insight is that feeling when you (often suddenly) understand or see through something, when you suddenly can view a problem or something else in a new way.

Thus to solve these problems, one not only needs knowledge in the advanced, technical fields in which they are posed (the tools to solve the problem), but also experience and insight (how to use those tools). Both of these are very difficult, since the subject matter is often quite abstract, but the insight is much more valuable and desired. Any old computer can be programmed the tools of mathematics, but what they lack is insight.

As for the relation with intuition, when solving a difficult problem, one first develops an intuition, which may be vague, and then insight to verify and sharpen the idea proposed by the intuition.

I have no such resolutions of solving any of those open problems, but one can get tempted by the reward one may get for solving any one of those open problems, but I sometimes feel it is not worth the effort; On other times, I feel maybe one should pursue those problems (any one of them) in earnest  :D ,but even when someone pursues them in earnest the problems are not solved (consider the instance of 'The Riemann Hypothesis' which has remained an "open problem" for the last 150 years.

I don't know -----what these problems demand from a human being?

They demand insight.

Usually good insight comes with experience. As one becomes familiar with a subject one can more easily get insight. Thus the problem demands experience (i.e. years of work) from a human being.

On a side note: sometimes it is said that a problem is not likely to be solved by an already well-known, experienced mathematician, because those ideas ingrained in his/her head have already been tested, and this mathematician may not have the fresh, new mind and imagination to come up with the novel ideas that may be needed to solve a problem. In that case, talent is demanded. (The talent/work debate is one that often seems to come up.)

But, I feel the person who is going to solve the Riemann Hypothesis , that is , provided it is  solvable ( mostly, it is going to be proved, rather than being disproved) will cover himself with glory, I wonder who that person might be.

If  the biography "A Beautiful Mind" is taken into consideration, it is stated in it that John Nash has indeed tried his hand at conquering the 'Riemann Hypothesis' but due to some reason, he couldn't make further progress.

Hmm, coming to imagination, how much does role does this play in solving these problems ( or solving problems for that matter). :lol:

in general, it's kind of hard to explain why a problem is hard without already having some reasonable knowledge of the field in question, so I'm probably not going to be able to do that. Terry Tao can, though! Here's a link to his thoughts on why the Navier-Stokes problem is hard -- see what you can get out of it: http://terrytao.wordpress.com/2007/03/18/why-global-regularity-for-navier-stokes-is-hard/

Arguments I've seen for why P vs NP is hard (as presented in most introductory computational complexity books) usually cite that diagonalization arguments cannot possibly work by talking about oracles and whatnot, so this here is a very specific tactic that can be shown to fail, but I'm sure there's a much firmer understanding of the problem that I can't provide.

I don't really know much about any of the other millennium problems (besides poincare), but Tao's article might give you some ideas.

It's a good question, certainly -- understanding why a problem is hard and what makes it hard is always a good thing to be thinking about at any level, both in terms of possible tactics to approaching the problem if you end up working on it and to make sure your eventual proof bypasses all the initial difficulties instead of having a hole in it somewhere.

[quote="JRav"]The one I'm particularly interested in (and who isn't?) is Riemann.  I would love to understand why this problem is so difficult but my knowledge of number theory and complex analysis is rather limited.[/quote]

eh why what is so exciting about RH

I agree. The Hodge Conjecture is certainly the most interesting.  :P

A few days ago there was published a proof of Goldbach Conjecture. It's being tested now. What do you think aobut it?

http://arxiv.org/pdf/0812.0930

I don't understand anything in that paper but it would be cool if it turned out to be correct.

Is it humanely possible to solve not one , but two of these problems? ( A two-million dollar prize in the offing.  :lol: )

What if Dr Perelman manages to solve the Navier-Stokes problem? It would be great, ain't it.

I think they demand knowledge of vast amount of literature across different fields of Mathematics in order to solve two of these Millennium problems.

And by the way, It's almost 150 years since the 'Riemann Hypothesis' was proposed by Riemann himself.

I also personally feel P!=NP ( '!=' denotes not equal to), but I  might be mistaken.

I think if one could come up with an instance that shows-- A problem that is easy to check (NP) is also easy to solve (P), and Eureka!, P=NP problem is all but solved, but I don't think it is possible to come up with such an instance.

Is it possible?  Yes.  Is it likely?  Not even remotely.

Firstly, I would like to apologize to JBL for reviving this thread after such a long time (I just completely forgot about this thread, so I couldn't ask this particular question at that time)

I mean-- Is anyone on AoPS trying to solve any of those millennium problems? (I think one of them is resolved by Dr Perelman)

Yes, Grisha Perelman already solved Poincare, but by his own admission, he's completely withdrawn himself from mathematics.  I have no idea if he has really stopped math altogether, but that's what he's claimed.

If someone on this site is working on one of them, they haven't done a lot of advertising.


Support
linear algebra
linear algebra unsolved

Let $ A, B\in M_n(\emph{C})$ be positive definite matrices, and $ p\ge1$. Prove or disprove that \[ Tr[(A\plus{}B^{\frac{1}{2}}AB^{\frac{1}{2}})^p]\le Tr[(A\plus{}A^{\frac{1}{2}}BA^{\frac{1}{2}})^p].\]

[quote="transseries"]Can you do p=1?[/quote]
When $ p\equal{}1$, the equality holds. It is not difficult to see the case when $ p\equal{}2$.

I think the equality always holds. Since $ \text{Tr}(AB)\equal{}\text{Tr}(BA)$, we expand $ (A\plus{}A^{1/2}BA^{1/2})^p$, group the terms that contains $ i$ numbers of $ A$ and $ p\minus{}i$ numbers of $ A^{1/2}BA^{1/2}$, take trace and change the order of multiplication, so to obtain $ \text{Tr}[(A\plus{}A^{1/2}BA^{1/2})^p]\equal{}\sum_i\binom{p}{i}\text{Tr}[A^i (A^{1/2}BA^{1/2})^{p\minus{}i}]\equal{}\sum_i\binom{p}{i}\text{Tr}(A^p B^{p\minus{}i})$. Similarly the right-hand side has the same expression.

[quote="sofeshue"]I think the equality always holds. Since $ \text{Tr}(AB) \equal{} \text{Tr}(BA)$, we expand $ (A \plus{} A^{1/2}BA^{1/2})^p$, group the terms that contains $ i$ numbers of $ A$ and $ p \minus{} i$ numbers of $ A^{1/2}BA^{1/2}$, take trace and change the order of multiplication, so to obtain $ \text{Tr}[(A \plus{} A^{1/2}BA^{1/2})^p] \equal{} \sum_i\binom{p}{i}\text{Tr}[A^i (A^{1/2}BA^{1/2})^{p \minus{} i}] \equal{} \sum_i\binom{p}{i}\text{Tr}(A^p B^{p \minus{} i})$. Similarly the right-hand side has the same expression.[/quote]

$ \text{Tr}[(A \plus{} A^{1/2}BA^{1/2})^p] \equal{} \sum_i\binom{p}{i}\text{Tr}[A^i (A^{1/2}BA^{1/2})^{p \minus{} i}]$  is not correct.

Numerical experiment supports this inequality. It is really difficult to give a proof.


Are there any events associated with pi day?

Does anyone actually "celebrate" it?

can  u gimme any link where i can get information about thhis  pi  day.       i am hearing about it for the very first time. :!:

hello, here is a link for you
http://www.exploratorium.edu/pi/
Sonnhard.

[quote="isabella2296"]Today, 3/14, is Pi Day because of the date. $ \pi \approx 3.14$. Of course, it won't really be Pi Day until 3/14/15 at 9:26 AM and 54 seconds.  :P[/quote]

couldn't it also be at 9:26 PM? :)
aw dang
it's only 7:27 *sits and waits*

9:26PM=21:26 in military time

we had pie today

[quote="zamboni34"]I had MATHCOUNTS on Pi day, and all the coaches wore t-shirts that said stuff like "Happy Pi Day, 3/14/09" on them.[/quote]

Hello zamboni, and welcome to AoPS. I see you are new.

In the future, I'd recommend trying to refrain from "reviving" old topics. It's commonly regarded on AoPS as "spam".

well, you just do alot of math, go on FTW, and eat pie!!! :lol:  :icecream:
 (sorry it's icecream couldn't find pi)

Well, IN State Mathcounts was hosted on Pi Day(didn't give me any extra luck for my 6th grade year at doing well.  I got 63rd.  :()  But according to my sister, at our local high school, some people give teachers pies.  That's about it though.


geometry
rectangle
ratio
invariant
geometry solved

We have a rectangle ABCD, and M and N are the midpoints of AD and BC (respectively!!!). Extend CD through D to any old point P. If Q is the intersection of PM and AC, prove that NM bisects angle QNP.

[b]EDIT:[/b] I have moved your two other problems into separate topics:
http://www.mathlinks.ro/Forum/viewtopic.php?t=141467
http://www.mathlinks.ro/Forum/viewtopic.php?t=141464

Nice problem...

[i]Solution:[/i]

Let the lines NQ and AB meet at T. By Menelaos' theorem, applied to triangle ACD with the collinear points Q, M and P, we have

$\frac{AQ}{QC}\cdot \frac{CP}{PD}\cdot \frac{DM}{MA}=-1$,

or simply

$\frac{AQ}{QC}\cdot \frac{CP}{PD}=-1$,

since DM / MA = 1. On the other hand, by Menelaos' theorem, applied to triangle ABC with the collinear points Q, N and T, we have

$\frac{AQ}{QC}\cdot \frac{CN}{NB}\cdot \frac{BT}{TA}=-1$,

so that

$\frac{AQ}{QC}\cdot \frac{BT}{TA}=-1$,

because CN / NB = 1. Comparing this with

$\frac{AQ}{QC}\cdot \frac{CP}{PD}=-1$,

we imply that

$\frac{BT}{TA}=\frac{CP}{PD}$.

In other words, the points T and P divide the segments BA and CD in the same ratio. Now, since these two segments BA and CD are equal in length, we can conclude BT = CP. Together with BN = CN and < TBN = < PCN = 90�, this leads to the conclusion that the triangles TBN and PCN are congruent, so that < BNT = < CNP. Now, the line NM is perpendicular to BC (remember that ABCD is a rectangle!); therefore, < TNM = 90� - < BNT and < PNM = 90� - < CNP, so that from < BNT = < CNP we obtain < TNM = < PNM, and the line NM bisects the angle QNP.

  Darij

[quote="grobber"]Here's a cute approach to the problem:

The lines $CD,CM,CA,CB$ form a harmonic cross-ratio, so the same can be said about the lines $NP,NM,NQ,NB$. This means that the map $NP\leftrightarrow NQ$ is an involution on the pencil of lines through $N$. Such an involution is defined by two pairs of corresponding lines, so take $P\to \infty,P=D$, and the property we want to prove holds in these two cases, so it always holds.[/quote]

hmm. I am very much afraid that I only understood "Here's a cute approach to the problem:".

- what is a harmonic cross ratio? Kedlaya's packet says that for four concurrent lines, the cross ratio can be defined as the cross ratio for the four points another line intersects the four concurrent lines at, and it is harmonic if it has ratio -1, but I don't see how you find that the ratio is -1...we don't know where line PMQ intersects CB

- I think I understand how you deduce the second part of the first sentence, i.e. that NP, NM, NQ, NB are harmonic cross ratio, because we can draw the same line PMQ and use the definition of cross ratio on that (it being the same as the cross ratio of the previous four lines CD, CM, CA and CB)

- I don't understand the rest very much. Mathworld defines line involution as "pairs of points on a line, the product of whose distances from a fixed point is a given constant", which I think is the definition you are using, but I don't quite understand how that follows from NP, NM, NQ, NB being a harmonic cross-ratio (because that only relates the distances PQ, PB, etc., not the distances involving N ?).

Could you explain your solution or recommend me to a resource that explains all this please ?

If you cut four lines with another line, then the cross-ratio of the four points on that line will be invariant, so you can define the cross-ratio of the four lines. The lines $CD,CM,CA,CB$ have a harmonic cross-ratio because if you cut these four lines with $AD$, you get $A,D,M$ and the point at infinity, and these four form a harmonic cross-ratio (this really is well-known). Now cut the four lines with the line $PQ$. The points of intersection are $P,M,Q$ and $CB\wedge PQ\in NB$, so the lines $NP,NM,NQ,NB$ also form a harmonic cross-ratio because they pass through the four points.

An involution is, more generally, a transformation between points on the same line (or, as in this example, between lines through the same point), which conserves the cross-ratio and which is, of course, an involution :) ($f(f(d))=d$). The map I defined is an involution. Try to figure out why an involution is uniquely determined by two pairs of lines which correspond to each other.


geometry
incenter
circumcircle
geometric transformation
reflection
geometry solved

give $ABC$triangle. and $AA'$, $BB'$ , $CC'$ intersectin point $o$.prove:
$A''$, $B''$ and $C''$ intersect in some common point .

[quote="M4RI0"]Do you mean: $AA''$, $BB''$ and $CC''$ intersect in some common point :?:[/quote]

Probably. But I think that this concurrency is true, only when the segment lines $A'A'',$ $B'B'',$ $C'C'',$ are concurrent, where $B'',$ $C'',$ are the orthogonal projections of $O,$ on $A'C',$ $A'B',$ respectively $($ opposite than in [b][size=109]Extremal's[/size][/b] figure $).$

The segment lines $A'A'',$ $B'B'',$ $C'C''$ now, are concurrent only in some special cases of $O,$ with respect to the triangle $\bigtriangleup A'B'C'.$

As I know this concurrency is true, when the point $O,$ is coincided with the orthocenter $H',$, or the incenter $I'$, or the circumcenter $O'$ of $\bigtriangleup A'B'C'$ and also with the points $H'',$ $I'',$ as the isotomic conjugates of $H',$ $I'$ respectively, with respect to the triangle $\bigtriangleup A'B'C'$ $($ easy to prove $).$

I am interesting if someone knows other points in the plane of a reference triangle, whose their pedal triangle is also a cevian triangle and what is the locus of such that points.

Kostas Vittas.

[quote="Extremal"][quote="M4RI0"]Do you mean: $AA''$, $BB''$ and $CC''$ intersect in some common point :?:[/quote]
Yes,  $AA''$, $BB''$ and $CC''$ intersect in some common point.
to vittasko
Trust me it's true for all point's in traiangle. I have solved it, My sulution is very big. therefore maybe someone can solve it more best way than my.
When $o=H$ this is particular case of my fact  :wink:[/quote]

Sorry, but I doupt that this result is true generally. Please see the figure t=144160(a).

However in the figure t=144160(b), we have the configuration of the particular case when the concurrency point of $AA',$ $BB',$ $CC',$ is the incenter $I'$ of the triangle $\bigtriangleup A'B'C'$ and, as well known, the segment lines $A'A'',$ $B'B'',$ $C'C'',$ are concurrent at the [b][size=109]Gergone's[/size][/b] point of $\bigtriangleup A'B'C'.$

Hence, based on the Lemma I mentioned at http://www.mathlinks.ro/Forum/viewtopic.php?t=131476 , we conclude that the segment lines $AA'',$ $BB'',$ $CC'',$ are concurrent at one point, so be it $P.$

The proof of the above Lemma I have in mind, is not so difficult and as I have been promised, I will post it here soon.

Kostas Vittas.

I agree with vittasko; in general, the three lines are not concurrent.

[quote="Extremal"][quote="M4RI0"]Do you mean: $AA''$, $BB''$ and $CC''$ intersect in some common point :?:[/quote]
Yes,  $AA''$, $BB''$ and $CC''$ intersect in some common point.
to vittasko
Trust me it's true for all point's in traiangle. I have solved it, My sulution is very big. [/quote]

[quote="Extremal"]I'll try post my sulution very soon  :wink:[/quote]

OK. I am looking forward to your approach.

Kostas vittas.

There is no post after 1 year  :)
I think the first problem is not true, but if we denote $ A''$ be the reflection of $ O$ through $ B'C'$ ( instead of projection, as initial problem), the new problem is true, and I think it's nice :)

Indeed, PDatK40SP's version is true. This is called the Begonia theorem, and it is due to Floor van Lamoen.
For a proof using inversion, see Darij Grinberg, Begonia points and coaxal circles, available at [url]http://de.geocities.com/darij_grinberg/[/url].


im planning on not buying the intro to counting and probability book and just doing alcumus to learn it. is this a good idea? i have volume 1, so i have some background on counting and probability, so, yea...

It really depends on how you learn. I think Alcumus does cover lots of the Intro to C&P book, but the book has a lot more details.

I recommend you buy the book, because it contains lots of details and strategies, and then focus on the problems in Alcumus. Alcumus has over 1000 challenging problems, and it is a very good practice.

Don't really on Alcumus to learn, but it is good practice. You should buy the book though.

I agree with remy1140 and AwesomeToad.  The book actually [i]teaches[/i] you the topics.  First get the book, then do Alcumus for extended practice (1000+ [more] problems).  And whenever you can't do a problem, review the topic covered and look at the solution, re-solving it in great detail.

Personally, i have to admit, i am not bragging. I am a lot smarter than i was before i did alcumus. I also have the book.  They are both excellent. Combined, i could be be the smartest peson in the world. :rotfl:

Lol, but remember that there's intermediate series and [i]beyond[/i] (such as olympiad). But I agree that AoPS has the best resources for math competitions.

Probability used to be my worst subject of math. I did Alcumus. Now it's my best.

No, I don't mean to sound like an advertisement. I'm not kidding.

Actually, me too. Once I took a counting/probability quiz in school and I got about 10/30 correctly, and my teacher was pretty frustrated. Over the summer I worked on Intro to CP and I learned things I would never have learned anywhere else. Combinatorics is so fun now (and that quiz I failed seems too easy now that I am almost finishing Intermediate CP.)

There's also a lot of stuff you can find in Alcumus you never learn in school. Personally, I have found Geometric Probability most fascinating and helpful, while at the same time elegant.

I use alcumus to help me with my advanced 9th grade math! It rocks. I really like how it tracks your progress and the program won't give you a problem out of your range.

Many of us are probably attempting to unearth the secrets in the special recipe for Alcumus score, maybe? I am (failing at it).

[quote="AdithyaGanesh"]The Alcumus score is a number calculated by looking at relative performance of everyone, isn't it?[/quote]

Yes.  Passing and mastery is not relative to others, but your overall Alcumus score is.  So, you can boost your Alcumus score by getting members of your state legislature to use it!

Alcumus is a great learning tool, but you should at least try the book.  Maybe a friend of yours has it?

Alcumus was intended as a supplement to the book, and I believe it makes a very good supplement. After you get the book and you do some Alcumus, it's really only a matter of practice left until mastery of probability.

I think Alcumus is very helpful, but only as a supplement for the book. It covers other problems and acts like a review test for those who have finished the Counting and Probability book.

It isn't really a test as in a sort of assessment thing, in my opinion; the main focus is for additional practice, extending ideas, with just the "progress report" as an additional bonus.


search
number theory
relatively prime
number theory proposed

Find all positive integers $ n$ such that $ \phi(n)\plus{}\delta(n)\equal{}2n$ , where $ \delta(n)$ denotes sum of positive divisors of $ n$ and $ \phi(n)$ number of positive integers less then $ n$ and relatively prime to $ n$.

This problem is already posted [hide="here"][url]http://www.mathlinks.ro/viewtopic.php?search_id=14841832&t=276352&start=40[/url][/hide]


The number $916238457$ is an example of a nine-digit number which conatins each of the digits $1$ to $9$ exactly once. It also has the property that the digits $1$ to $5$ occur in their natural order, while the digits $1$ to $6$ do not. How many such numbers are there?

[hide="duno"]nvm, i'll ponder this  [/hide]

No, because you haven't taken into account that the digits 1-5 must be in the right order (or that 1-6 aren't)

Well, we know that there are 9! total possibilities.
The six simply must go before the five for the number to be valid.
Of course, the number is clearly less than 9!, but I don't know by how much.

[hide="a lot less than 9!"]
We begin with 9 ways to place the 9, 8 ways to place the 8, and 7 ways to place the 7. Then there are only 5 ways to place the 6 (anywhere except the rightmost remaining spot) and the order of 1-5 is fixed. So the answer is just $9\cdot 8\cdot 7\cdot 5=\boxed{2520}$.
[/hide]

[hide]There are ${9\choose 6}$ ways to choose where the digits 1-6 go. From these, given the placement of 6, the placements of 1-5 are determined. 6 can go in any of the six places, except the last, so $\times 5$. There are $3!$ ways to place the remaining 7-9.

${9\choose 6}\times 5\times 3!=2520$.[/hide]

[hide]First place the 9,8,7, no restrictions on that, so $9\cdot8\cdot7=504$. Then, we have 6 spots left. There are 5 places to put the 6, since we can't put it in the last spot, since it would imply that 1-6 are in order. Thus, there are 5 more places, then the 1, 2, 3, 4, 5 are forced, so the final answer is $504\cdot5=2520$. [/hide]


geometry
perimeter
ratio
Pythagorean Theorem

[quote="ch1n353ch3s54a1l"][quote="ch*yeuk"]Each right triangle has a hypotenus 7 cm and the shortest side 3 cm. Determine the perimeter of the figure.

[hide="view image"][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=545[/img][/hide]

a) (28 + 2√10 - 3) cm
b) (28 + 8√10 - 3) cm
c) (28 + 2√10) cm
d) (28 + 8√10) cm
e) (16 + 8√10) cm[/quote]
[hide]
E. there are four 7s so it's 28. then there is the small portions of $\sqrt{49-9}=\sqrt{40}$ and the portion of $\sqrt{40}$ is 3 so the length that is in the perimeter is $\sqrt{40}-3$, and that is there four times so the perimeter of the whoel thing is $28+4(\sqrt{40}-3)$ or $16+4\sqrt{40}=\boxed{16+8\sqrt{10}}$[/hide][/quote]

Ahhh so you have to use Pythagorean Theorem first! I tried the 30 60 90 theory first.. ahh. Thanks!

[quote="The Master"]the 30 60 90 ratios of a triangle only works if the other angles in that triangle are 30 and 60, but they are not.  so ya, you gotta use pythagorean theorem......[/quote]

Yeah I noticed after trying to figure it out from that theory.. ^_^;;


If $ \theta$ is a constant such that $ 0 < \theta < \pi$ and $ x \plus{} \frac{1}{x} \equal{} 2\cos{\theta}$. then for each positive integer $ n$, $ x^n \plus{} \frac{1}{x^n}$ equals

$ \textbf{(A)}\ 2\cos{\theta}\qquad
\textbf{(B)}\ 2^n\cos{\theta}\qquad
\textbf{(C)}\ 2\cos^n{\theta}\qquad
\textbf{(D)}\ 2\cos{n\theta}\qquad
\textbf{(E)}\ 2^n\cos^n{\theta}$

[hide="Fairly Large Hint"]Note that a solution is $ x \equal{} \text{cis } \theta$.[/hide]

[hide]$ \implies x^2\minus{}2x\cos(\theta)\plus{}1\equal{}0 \implies x \equal{} \cos\theta\plus{}i\sin\theta \equal{} e^{i\theta}$, or $ x^n \equal{} e^{ni\theta}$

If $ f(e^{i\theta}) \equal{} 2\cos\theta$, then $ f(e^{ni\theta}) \equal{} 2\cos(n\theta)$, D.[/hide]


Farz koni $D$ noghtei dar mosalase haddatozavieye $ABC$ bashad agar
$DA.DB.AB+DB.DC.BC+DC.DA.CA=AB.BC.CA$ basahd mahalle noghteye $D$ ra biabid.
[color=red][Moderator edit : LaTeX Formula edited][/color]


function
calculus
inequalities proposed
inequalities

With $ a,b,c > 0$. Prove that
  1. $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \sqrt {\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \geq 5/2$
  2. $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \sqrt {\frac {2(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2}} \geq 3$

(1) is obviously trues by CS and Am-Gm
(2) is only a case in my old problem: (see it here)  :)
http://www.mathlinks.ro/viewtopic.php?t=279762
Will your way work for this general problem, 2424?  :)

[quote="anas"]in (1) the best constant that can replace the exponent of $ \frac {ab \plus{} ac \plus{} bc}{a^2 \plus{} b^2 \plus{} c^2}$ is $ k_0 \equal{} 0.9568...$ :wink: ,  we can obtain it by noting that the expression in function of $ abc$ is increasing , with the same way we can obtain the best value in (2) :)[/quote]
Of course, if use uvw of pqr or ABC... we only need to check these ineqs with $ a \equal{} b$ and $ c \equal{} 0$.
But I think the case $ a \equal{} b$ is not nice because we'll have to use calculus.
Example, with the general problem:
Given $ a, b, c > 0, k \ge 1, n > 0$. Prove that:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} 2\sqrt [n]{2}k.\sqrt [n]{\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \ge 2(k \plus{} 1)$
We also can solve it easily by Am-Gm and not to use any calculus  :)
@To2424: The trues problem is:
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\plus{}2\sqrt{\frac{2(ab\plus{}bc\plus{}ca)}{a^{2}\plus{}b^{2}\plus{}c^{2}}}\geq 4$
And $ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\plus{}\sqrt{\frac{2(ab\plus{}bc\plus{}ca)}{a^{2}\plus{}b^{2}\plus{}c^{2}}}\geq \frac{3}{2}\plus{}\sqrt{2}$
It's also my mistake when posted the general problem. I edited it  :oops:


[quote="perfect_radio"]In a paralelipiped  filled with water up to $H=250cm$ we make two holes $50cm$ apart (they are one under the other). Compute the distance between the surface of the water and hole $1$ s.t. the water reaches the ground in the same place.[/quote]

Can you explain the problem?

you have something which holds water ( paralelipiped form, or whatever). On the side we make two holes through which water comes out. (they are on the same line - $\perp$ to the earth) We must find the placement of these holes in terms of height such that the water which comes out of them reaches the ground in the same place (not necesarilly in the same time)

sorry if you don't understand. i'm not good at explaining things  :(

Let us consider a vessel filled with a liquid to a height $H$ and a hole in it a a depth $h$ from the liquid surface.
Then, by Bernoulli's Principle the speed of the liquid coming out from the orifice $v = \sqrt{2gh}$. let the horizontal range of the liquid coming out from the orifice be $d$. Let the time taken by the liquid to reach the bottom be $t$.
The $\frac 1{2}gt^2 = H-h$
$i.e, t = \sqrt{\frac{2(H-h)}{g}}$
So, $d = v.t$
$i.e, d = 2\sqrt{h(H-h)}$
For d to be maximum or minimum
$\frac{d [h(H-h)]}{dh} = 0$
$i.e, h = \frac H{2}$
also  $\frac{d^2 [h(H-h)]}{dh^2} = -2$
So, $d$ is maximum at $h = \frac H{2}$
Let us now find the value of $d$ for an orifice located a distance $k$ to the top and bottom of the centre of the portion filled with liquid i.e, $h = \frac H{2}$.
$d_{upper} = 2\sqrt{[(\frac H{2} - k)(H - (\frac H{2} - k))]}$
$i.e, d_{upper} = 2\sqrt{[(\frac H{2} - k)(\frac H{2} + k)]}$
and,
$d_{lower} = 2\sqrt{[(\frac H{2} + k)(H - (\frac H{2} + k))]}$
$i.e, d_{lower} = 2\sqrt{[(\frac H{2} + k)(\frac H{2} - k)]}$
So, $d_{upper} = d_{lower}$

So, you can remember it as a result.
i.e, the horizontal range first increases to become maximum at$h = \frac H{2}$ and then decreases. Also, the points having the same horizontal range are symmetrically located about $h = \frac H{2}$.

According to given problem the holes are separated by a distance of $50 cm$. Maximum horizontal range of the liquid is at a distance $\frac{250}{2} cm = 125 cm$ from top as well as bottom. So, the position of the holes is $(1.25 - 0.25)m$ and $(1.25 + 0.25)m$ from top as well as bottom the liquid surface.
Thus, the first hole should should be at a distance of $1m$ from top of the liquid surface.

That's a very very useful result!!!

However, in a more classic way:

$v_1=\sqrt{2gh} ; v_2=\sqrt{2g(h+l)} ; t_1=\sqrt{\frac{2(H-h)}{g}} ; t_2=\sqrt{\frac{2(H-h-l)}{g}}$

Solving the equation $v_1 t_1 = v_2 t_2$:

$h=\frac{H - l}{2}$ therefore $h = 1 m$.


Let $a,b,c$ be three positive reals such that $abc=2$.
Prove that \[a^{3}+b^{3}+c^{3}\geq a \sqrt{b+c}+b \sqrt{c+a}+c \sqrt{a+b}.\]

By Cauchy Chebychev and again AM-GM we have
$a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\leq\sqrt{2(a^{2}+b^{2}+c^{2})(a+b+c)}\leq\sqrt{6(a^{3}+b^{3}+c^{3})}\leq a^{3}+b^{3}+c^{3}$

 :wink:

This problem is from JBMO Shortlist 2002 and problem 9 in "Old and new inequalities"

[quote="pohoatza"]
but from AM-GM:
$a \sqrt{b+c}+b \sqrt{c+a}+c \sqrt{a+b}\geq 6$
[/quote]

Is there a shorter way to arrive at this than:

$abc\sqrt{b+c}\sqrt{c+a}\sqrt{a+b}= \sqrt{a^{2}b^{2}c^{2}(b+c)(c+a)(a+b)}\geq \sqrt{8a^{3}b^{3}c^{3}}=8$. I feel like I must be missing something more obvious.


[hide="click text"]October 16 2009?[/hide][/hide]

[quote="pianoforte"]What day of the year will it be 1500 days from now, as of Sept.7 2005?[/quote]

[hide]$\frac{1500}{365}=4$ remainder $40$ But in those $4$ years, there will be $1$ leap year, $2008$.

So we subtract $1$ from the remainder, making it $39$

So it would be $39$ days after September $7$ in the year $2009$.

39 days after it would be October $16$.

So it would be October $16$, $2009$[/hide]

[hide]
$\frac{1500}{365}=4 r 40$
There will be one leap year also, 2008.
So, you have 39 days after September 7th:
October 16th
(2009)[/hide]

[hide]you take 1500 and divide it by 365. You find that its 4r40. You know that every 4 years theres a leap year and there has to be 1 in there so you subtract a day. So find whats 39 days after september 7th, and you get October 16th, then add in the 4 yeras you got to get October 16th, 2009.[/hide]

1500 days from now...hmm...let me check the calendar...

[hide]Just kidding. What day of the week was September 7? It was Wednesday, okay.

[hide]We divide 1500 days by 7 to find out how many weeks (and thus how many Wednesdays) passed in that 1500 days. Since 1500 doesn't divide 7 equally, we have a remainder. The remainder is the number of days we must count forward to get our answer.

1500/7= 214 r2

So there are 215 Wednesdays in the 1500 days. We then count forward to[/hide] [hide]Friday.[/hide][/hide]