I am having a slight problem in terms of finding the LCD of rational expressions.

Sample:

Find the LCD of (7/3x) + (x - 5)/(27(2x + 1)^4 + (x +2)/x^3(8x + 4)^2

The book's answer is LCD = 432x^3(2x + 1)^4

Why is that the answer?

$\frac{7}{3x}+{\frac{(x-5)}{27(2x+1)^{4}}+\frac{x+2}{x^{3}(8x+4)^{2}}}$

Is this it?

[hide="Hint?"]
Unfortunately the only way for me is to expand it all out.
[/hide]

[hide="why expand?"]
Astrophys: Expansion is unnecessary and tedious

$\frac{7}{3x}+{\frac{(x-5)}{27(2x+1)^{4}}+\frac{x+2}{x^{3}(8x+4)^{2}}}$

$\frac{7}{3x}+{\frac{(x-5)}{27(2x+1)^{4}}+\frac{x+2}{x^{3}(4(2x+1))^{2}}}$

$\frac{7}{3x}+{\frac{(x-5)}{3^{3}(2x+1)^{4}}+\frac{x+2}{2^{4}x^{3}(2x+1)^{2}}}$

now the denominators are all powers of the same non-factorable values of 3, x, (2x+1), and 2
take the largest of each and multiply together yielding the common denominator

$2^{4}\cdot 3^{3}\cdot x^{3}\cdot (2x+1)^{4}$
[/hide]


i'm reading some cryptographic stuff and i got to some corollary's of fermat's little theorem.
($\widehat{x^{p-1}}=1$ in $Z_p$ where $(x,p)=1$ and $x,p\in N$)

i don't remember exactly how they where
something like this
$1)r-1=k mod (p-1) \Rightarrow X^p=X(modp)$
and $2)e\cdot d-1=k(p-1)(q-1)\Rightarrow X^{ed}=X^{de}=X(mod p) $also $ X{}^X{}^{ed}=X^{de}=X(mod q)$

the latter i understand but not completely and the first is totally unclear to me

it's about the RSA algorithm
what is the source of what u found ?
i'got security profil in my engineering studies ;)

its from a book
its romanian so you cant know it
anyway give me a good link or pdf to read from if you can :roll:

can anyone tell me what mods are? I'm in 7th grade math.


sometimes, after i login, i check my messages, but when i click on the messages button, it logs me out.  but, if i go to the mathcounts forum (or any other forum) and then to my messages, i stay logged in. what's happening?

What skin and browser are you using? Do you have cookies enabled?


modular arithmetic
algebra
polynomial
number theory
prime numbers
number theory unsolved

Suppose $ a, b , c \in Z $ . Prove that : For all prime numbers p , the congruence expression $ ax^2+by^2+cz^2 \equiv 0 [ $ mod $ p] $ has a root $ (x,y,z) $ which is different from $ (0,0,0) $

The number of solutions to that equation $\pmod p$ is precisely the residue $\pmod p$ of the expression $\sum_{x,y,z\in\overline{0,p-1}}(1-(ax^2+by^2+cz^2)^{p-1})$. Try to show that that expression is $0\pmod p$, so the number of solutions of our equation is divisible by $p$. Since we have the solution $(0,0,0)$, there must be at least another one (there can't be only one, because $p\not |1$).

More generaly you can use the theorem of Chevalley-Warning (which in fact can be proofed exactly the way grobber says)

Can you please show me the content of Chevalley-Warning theorem ?

If you have a set of polynomials in $n$ variables over a finite field (with $q=p^k$ elements, where $p$ is prime), and the sum of the degrees of the polynomials is $<n$, then the number of common roots of the polynomials is $0\pmod p$.

You can see that this was a very particular case: $q=p$, and there was only one polynomial here, of degree $2$ and in $3>2$ variables.

[url=http://math.rice.edu/~hassett/teaching/465spring03/antlec5.pdf]Here[/url]'s a PDF which contains pretty much exactly this: the statement of the theorem and the proof, which, as has been noticed, is basicly the same as that given here.


In a single elimination tournament, the better player always won.  The winner of Round 3 was the champion, and the loser of Round 3 was the runner-up.  Eight players were randomly assigned slots in Round 1.  What is the probability that the runner-up was not the second-best player in the tournament?

[hide]
For the second best person to be runner up, they must have not faced the best person in either of the first two rounds. That means that they have to be on the other half of the bracket.
4 working spots
7 total spots
$\boxed{\boxed{\boxed{\frac 47}}}$[/hide]

Unfortunately, your answer is very close, but incorrect.

[hide="The correct answer is"]$\frac37$.  Biffanddoc, you answered if the second best was the runner-up, but the questions asks the probability if the second best is not runner-up. [/hide]

Be careful...in 7th grade I had 4th place in chapter, bumped to 5th.  I made some careless errors and otherwise would have been bumped 3rd to 4th and gone to state.  But this year, I'm going to be very careful at state.  ( got 1/8 on 2001 target due to careless errors)

do you realize how pointless it is to revive year and a half old topics to give advice? This person probably isn't even in MATHCOUNTS anymore!


geometry
geometric transformation
homothety
power of a point
radical axis
geometry unsolved

A line $d$ is given in the plane. Let $B\in d$ and $A$ another point, not on $d$, and such that $AB$ is not perpendicular on $d$. Let $\omega$ be a variable circle touching $d$ at $B$ and letting $A$ outside, and $X$ and $Y$ the points on $\omega$ such that $AX$ and $AY$ are tangent to the circle. Prove that the line $XY$ passes through a fixed point.

[i]Proposed by F. Bakharev [/i]

Let $T$ be the midpoint of $XY,\ S=AB\cap XY$, and $P=d\cap XY$. If we are to prove that $XY$ passes through a fixed point, then that point lies on the limit of $XY$ as the radius of $\omega$ goes to $0$, and that line is the perpendicular $\ell$ through $B$ to $AB$. Proving that $XY\cap\ell$ is constant is equivalent, because $AT\perp XY$, to proving that $T$ moves on a circle arc passing through $A,B$ (and then the desired fixed point will be the antipode of $A$ in this circle). This means that it suffices to prove that $\angle ATB$ is constant, or, equivalently, that $\angle PTB$ is constant.

$BS$ is the $B$-symmedian of $BXY$, so $\angle YBS=\angle TBX$. This means that $\angle PBS=\angle PBY+\angle YBS$, which equals $\angle BXT+\angle TBX=\angle PTB$. This shows that $\angle PTB=\angle PBS=\angle PBA$, which is, of course, constant.

during the contest i managed to solve it using coordinates. the computations are very easy...
i took $B$ as origin and $d$ as y-axis, and the equation of $XY$ was a first degree expresion, with one parameter. after i managed to remember the equation of tangent to circle, the problem took me 5 minutes...

Let $Q$ be the intersection of the perpendicular at $B$ to $AB$ with $XY$.
It's interesting to notice that the midpoint of $AQ$ lies on $d$.
This follows from the fact that $\triangle{OTB}\sim\triangle{OBA}$ and hence $\angle{AQB}=\angle{BTO}=\angle{OBA=\angle{PBQ}}$.

Let $ M $ , $ N $ be the midpoints of $ AX $ , $ AY $ respectively.Analogically $ M' $ , $ N' $ for $ X'Y' $ then $ MN $ , $ M'N' $ , $ d $ intersect at one point ( radical axis for A,w,w') now let the homothety at center A and coefficient 2.then $ MN $ goes to $ XY $ , $ M'N' $ goes to $ X'Y' $ and $ d $ goes to $ d' $
because of $ d' $ is constant line and we have that all such $ XY $ intersect at the $ d' $ then we have that all $ XY $ goes through fixed point


inequalities
trigonometry
inequalities unsolved

1)Given $ a,b,c>0$ show that :

\[ \frac{a^3\plus{}b^3\plus{}c^3}{3abc}\plus{}\frac{9abc}{(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}\ge 2\]

2)Let $ x,y,z\in \mathbb{R}$ such that $ x\plus{}y\plus{}z\equal{}xyz$ . Prove that :

\[ \frac{\sqrt{1\plus{}x^2}}{4x^4\plus{}5}\plus{}\frac{\sqrt{1\plus{}y^2}}{4y^4\plus{}5}\plus{}\frac{\sqrt{1\plus{}z^2}}{4z^4\plus{}5}< \frac{3}{8}\]

Solution for the first one:
By muirhead $ 3(a^3 \plus{} b^3 \plus{} c^3) \ge (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$. Use this inequality and then apply $ \frac {x}{y} \plus{} \frac {y}{x} \ge 2$.

use trigo substitution for the second  :wink:

$ x \equal{} \tan(\alpha)$
...

[quote="Georg-A."]use trigo substitution for the second  :wink:

$ x \equal{} tan(\alpha)$
...[/quote]

and after this substitution how do i proceed ?

[quote="Spribo"]Solution for the first one:
By muirhead $ 3(a^3 \plus{} b^3 \plus{} c^3) \ge (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$.[/quote]I think Muirhead is too ugly for such an easy ineq:
$ 3(a^3 \plus{} b^3 \plus{} c^3) \ge (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \ge (ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)$

[quote="Pain rinnegan"][quote="FelixD"]If $ x$, $ y$, $ z$ are reals, the second one is wrong.[/quote]

I'm sorry :blush:  , can you prove it for positives ?[/quote]
Actually I can prove it, but my prove is very ugly. Maybe someone finds a nicer proof. :)

I proved $ f(x) \plus{} f(y) \plus{} f(z) \le f(x) \plus{} f(\frac {1}{x}) \plus{} f(\sqrt {3}) < \frac {3}{8}$, where $ z \ge y \ge x$.


A number, abc in base y, where a, b, and c are digits (for example, bc means 10b + c), equals bc + 10 in base 10. Find y.

Note that $2^x$ = bc. (If $2^x < 10$, b = 0. Ignore the restriction for b.)

a, b, c, x, and y are all positive integers.

[hide="ansarrrr"]
edit: woops forgot to edit my answer, it's 5
[/hide]

are you sure about that? I don't think you can do it if by "abc"in base y, you mean that
a,b, and c are all integers between 0 and y-1, with "a" greater than 0

***NOTE: this is now not true, since the problem has since been reworded: in what follows, I had the original formulatinon in which
abc in base y  had to equal bc in base 10.

working out the details -- have I misunderstood you, or erred?:

[hide]abc in base y  has value is a*y^2 + b*y + c,
and if  ay^2 + by + c = 10b + c, then ay^2 = b*(10-y)
which means a = b*(10-y)/y^2
if yis an integer greater than 5, you couldn't have an integer value for a
since the right hand side immay be written as (b/y)*(10-y)/y and both of those fractions are less than 1
similarly if y=5, a would have to equal b/5, which is less than 1

and if y=3, then a = 7b/9 which won't be a positive integer if you plug in b=0,1, or 2

all that's left is y=2 or y=4 [sparing considerations of base 1, which wouldn't help, anyway]
y=2 gives a=b*8/4 = 2b  ... a couldn't be 1
y=4 gives a = b*6/16 = 3b/8. you can't plug in b=0,1,2,3 and get a positive integer out for a.

now if you DO allow a=0, you could have y=10 (and any b and c would work, of course)
or a=b=0 and you could use ANY value for y and ANY c less than both 10 and y would work.
[you could add that 2^x restriction to handle these cases, but there are still multiple values of y][/hide]

sorry, I made a little booboo.

there's a nice treat in the problem for you fans of a certain book

[hide]The problem contradicts itself by saying that b is a positive integer and b can be 0. I wll just assume b must be a positive integer. Therefore, bc can be 16, 32, or 64.

For 16, the equation is $ay^2+y=20$. Solving yields 1 for a and y=$\boxed{4}$, like Treething's ansarrrr.
$116_4=26$

For 32, the equation is $ay^2+3y=40$. Solving this equation yields another possible value for y when a=1, $\boxed{5}$. :P
$132_5=42$

For 64, the equation is $ay^2+6y=70$ which yields no solutions in the set of positive integers.[/hide]

To limit the values of bc.

btw I edited my answer and yes there's only one correct answer

stupid me. :fool:  There's no 6 in base 4! :rotfl:


trigonometry
algebra
function
domain
quadratics
polynomial
quadratic formula

[hide="Another solution"]$\sin^{2}x+\cos^{2}x=1$
$\sin^{8}x+\cos^{8}x+4\sin^{2}x \cos^{2}x (1-2\sin^{2}x \cos^{2}x)+6\sin^{4}x \cos^{4}x=1$
Let $y=\sin^{2}2x$:
$y(1-\frac{y}{2})+\frac{3y^{2}}{8}=\frac{31}{128}$.
$8y-y^{2}=\frac{31}{16}$
$y^{2}-8y+\frac{31}{16}=0$
$y=\frac{1}{4},\frac{31}{4}$
Using the double angle formula, I got a quadratic polynomial in terms of $\sin^{2}x$, so I used the quadratic formula to get real values for $\sin x$. This should give the same answers given previously.
[/hide]

how did you know that $\theta=\frac{5\pi}{12},\frac{\pi}{12}$?

i know u could possible use the sine half angle formula but is it just one of those things that you memorize?

i know how to find $\sin\left(\frac{\pi}{12}\right)$ if i need to but i would be clueless if i had to go the other way.  :blush:


[quote="ordinary boy"]Find the smallest possible value of $ |x-1|+|y-x|+|y-2007|$ for all real numbers $ x$ and $ y$[/quote]

[hide]If X<Y<2007 then equation simplifies to |X-1|+|X-2007| with lowest possible answer 2006

If Y=2007 then |X-1|+|2007-X| which again becomes 2006

I'm too tired to prove all cases, but I'm almost positive answer is 2006[/hide]

Use the 1st triangle inequality :
Notice that |a+b|<=|a|+|b| for any reals,
|x-1|+|y-x|+|y-2007|=|x-1|+|y-x|+|2007-y| by simetry
So :2006 =|x-1+y-x+2007-y|<=|x-1|+|y-x|+|y-2007|
The minimum value is 2006. (I find it happens when x=y=1)


Gauss
number theory solved
number theory

prove every non-negative integer can be written as a^2+b^2+c^2 or a^2+b^2+2*c^2, for non-negative integers (a,b,c)

: This result is a consequence of a Theorem of Gauss whose proof I don?t know. It goes like this: an integer n can be written as a^2+b^2+c^2 iff n is not on the form 4^k*(8m+7). (1)
So if n is not on the form (1) then n is a sum of 3 squares.
If n is on that form, then 2n is not, so 2^(2k+1)*(8m+7)=2n=a^2+b^2+c^2. It is easy to see that a,b,c are all even or two of them are odd. Anyhow 4n=(a+b)^2+(a-b)^2+2*c^2. Assuming a,b,c even or c even and a,b odd one gets n=((a+b)/2)^2+((a-b)/2)^2+2*(c/2)^2 . So n can be written as x^2+y^2+2*z^2.

Lets prove that any number of the form: 4^n(8k-1) where n>=0 and k>0 (pozitive integers) can't be written as a sum of three squares! :D

[b]Solution:[/b]

Lets consider that 4^n(8k-1)=X^2+Y^2+Z^2 where X,Y and Z are integers <>0.

If n=0 then the last equality is false!!! ( the square of an odd number is M(8)+1 and the square of an even number is M(4)), and we have that the right member is M(8)+3 if all numbers are odd, M(4)+1 is only one is odd, and in the left we have that it is M(8)-1.

If n>0 then the left member being even, and so in the right member we have two odd numbers and one even which gives M(4)+2 and the equality is impossible because the left member is M(4) or we can have all numbers even and taking X=2X1,Y=2Y1 and Z=2Z1 and the equality is equivalent to:
4^(n-1) *(8k-1)=X1^2+Y1^2+Z1^2 and we do the same until we get that:
8k-1=a^2+b^2+c^2 which we proved before....

cheers! :D
:D :D :D :D :D


geometry
3D geometry
sphere
function
complex analysis
complex analysis unsolved

Consider the meromorphic function $ f(z)\equal{}\frac{z^3}{1\minus{}z^2}$ on the Riemann sphere.
a) Find all the ramification and branch points of $ f$.
b) Check Riemann-Hurwitz formula for $ f$.

Did you check on the charts $ z\mapsto z$ and $ z\mapsto 1/z$?


Well, just by inspection, some trivial solutions include
$ b\equal{}0$ and $ a\in Z$
$ a\equal{}0$ and $ b\equal{}9k$ with $ k\in Z$
$ a\equal{}b\equal{}9$

[quote="Chalao"]Find all ordered pairs (a,b) >> positive integers <<, such that

$ \frac {a^2 b \plus{} b}{ab^2 \plus{} 9}$ is an integer number[/quote]

[quote="mismatchtea"]
$ b \equal{} 0$ and $ a\in Z$
$ a \equal{} 0$ and $ b \equal{} 9k$ with $ k\in Z$
[/quote]
These don't count because the problem states "...(a,b) positive integers..."

Sorry i forgot to write that condition. That`s why he just wrote that solution. :P

[quote="Chalao"]Find all ordered pairs (a,b) positive integers, such that

$ \frac {a^2 b \plus{} b}{ab^2 \plus{} 9}$ is an integer number[/quote]

[hide]
I did long division, and I got a quotient of $ \frac ab$ with a remainder of
$ b \minus{} \frac {9a}{b}$.  I set the remainder equal to zero.

$ b^2 \minus{} 9a \equal{} 0$

$ a \equal{} \frac {b^2}{9}$

If b = 9, then a = 9.
If b = 81, then a = 729.
If b = 729, then a = 59049.
And so on.

The ordered pairs $ (a, b)$ for positive $ a$ and positive $ b$ are:

$ (9^{2k \minus{} 1}, 9^k)$ for $ k$ belonging to the set of positive integers.

[/hide]

Those are valid solutions, but aren´t ALL the solutions. In example (32,1) and (73,1) for (a,b) satisfy the given conditions

Someone knows how to solve this problem?


The sides $a,b,c$ of a triangle $ABC$ are natural numbers and the triangle is inscribed in a circle of diameter $2R=6,25$. Find the $a,b,c$

[quote="kunny"]$abc=\frac{25}{2}\triangle ABC$[/quote]

kunny can you elaborate please??

I will help a little
[hide]Use the fact that$ E=\frac{abc}{4R}$ where $E$ the area of the triangle.[/hide]

[quote="ashwath.rabindranath"]Isn't it rather vague?[/quote]

No. I think that all are good. What you didnot understand?


function
limit
topology
real analysis
real analysis unsolved

let $ f_k$ be a sequence of measurable functions on [0,1] . If $ |f_k(x)|\leq M_x$ for all k, then show that given $ \epsilon >0$ there is a closed set $ F\in [0,1]$ and a finite M such that m([0,1]-F)<$ \epsilon$ and $ |f_k(x)\leq M$ for all k and all x n F.

Let $ g(x)\equal{}\sup_k|f_k(x)|.$ Defining it that way, we know that $ g$ is a measurable function, and the hypothesis is that $ g$ is finite everywhere on $ [0,1].$

Consider the sets $ E_n\equal{}\{x: g(x)\le n\}.$ This is an increasing sequence of sets whose union is all of $ [0,1].$ Hence $ \lim_n|E_n|\equal{}1.$ (Note: I'm using $ |E|$ for the measure of $ E;$ it is the notation in Wheeden and Zygmund, after all.)

Find an $ n$ such that $ |E_n|>1\minus{}\frac{\epsilon}2.$ We also know that we can find a closed $ F\subset E_n$ such that $ |E_n\minus{}F|<\frac{\epsilon}2.$ This gives us the $ F$ that was asked for.

Note that we did explicitly use that $ [0,1]$ has finite measure. The theorem would be false on a set of infinite measure.

thanks.i did the same thing ,except that i used lusins theorem to get the closed set F ( here it is also compact) and since a cont fn on a compact set is bounded we get the result. i hope its correct.

Where is there a continuous function? I don't see one.

from lusin's theorem. g is measurable and finite , so i can find a closed set F with the given condition such that g restricted to F is continuous. here we have a closed set of a compact set, so F is also compact. so on F, g is bounded , hence we have the result. is it right?


Hi Maverick,

is there really a greater-equal sign in 3(a+b+c) >= bc+ca+ab+2 or is there an equal sign instead?

It's  \geq . This problem is troubling me for quite some time.
By the way nice picture! :D


Real cryptic title... :ninja: :D

Is there a symbolic way to represent "the set of all real numbers except 42"?

Just wondering :)

[hide="?"]
${S=\{x|x\not=42,x\text{('element of' sign thingy)}\mathbb{R}}$?[/hide]

Wouldn't it just be X is greater than 42, or X is less than 42?

Many such ways.  The most direct is ${\mathbf R}-\{42\}$.  (Some people use a "\" symbol in place of a "-" for set differences.)

junggi, the symbol you want is "\in," as in $x \in \bf R$


real analysis
inequalities
calculus
calculus computations

Show that if $ a_n>0$ for every $ n$, then

$ \liminf{a_{n\plus{}1}/a_n}\leq \liminf{(a_n)^{1/n}} \leq \limsup{(a_n)^{1/n}} \leq \limsup{a_{n\plus{}1}/a_n}$


number theory
relatively prime
number theory unsolved

Let $ a_1,a_2,...,a_n$ be the sequence of all positive integers that are relatively prime to $ 75$,where $ a_1<a_2<a_3<....$.(The first 5 terms of the sequence are $ 1,2,4,7,8$).Find the value of $ a_{2008}$

[quote="SUPERMAN2"]Let $ a_1,a_2,...,a_n$ be the sequence of all positive integers that are relatively prime to $ 75$,where $ a_1 < a_2 < a_3 < ....$.(The first 5 terms of the sequence are $ 1,2,4,7,8$).Find the value of $ a_{2008}$[/quote]

These are all the integers not divisible by 3 or 5. So these are :
$ 15n\plus{}1,15n\plus{}2,15n\plus{}4,15n\plus{}7,15n\plus{}8,15n\plus{}11,15n\plus{}13,15n\plus{}14$ for any $ n\geq 0$

Since $ 2008\equal{}8*251$ : $ a_{2008}\equal{}15*250\plus{}14\equal{}3764$

IMHO


Estoy colgando el examen de las olimpiadas de un nivel aceptable para esta seccion
 que es de nive de inciacion en olimpiadas .
espero les agrade...

Carlos Bravo ;)
----
[b][size=150][color=blue]V Olimpiada del Cono Sur
 Uruguay - 1994[/color][/size] [/b]

$\blacktriangleright$  [color=blue]Problema 1.[/color]
El entero positivo $n$ tienen 1994 d\'igitos. De esos, 14 son
iguales a $0$ y las cantidades de veces que los digitos $1, 2, 3, 4, 5, 6, 7, 8, 9$
aparecen son respectivamente proporcionales a
$1, 2, 3, 4, 5, 6, 7, 8, 9$. Mostrar que $n$ no puede ser un
cuadrado perfecto.

$\blacktriangleright$  [color=blue]Problema 2.[/color]
Se considera una circunferencia $C$, de di\'ametro $AB = 1$. Se
toma un punto cualquiera $P_0$ en $C$, distinto de $A$, y a partir
de ella se costruy\'o una secuencia de puntos $P_0$, $P_1$,
$P_2$,... en $C$ del siguiente modo: $Q_n$ es el sim\'etrico de
$A$ en relaci\'on a $P_n$ y la recta que contiene a $B$ y $Q_n$
intersecta a $C$ en $B$ y $P_(n+1)$ (no necesariamente distintos).
Muestre que es posible escojer $P_0$ de manera que las dos
condiciones a seguir sean satisfechas:
(a). $P_0\hat{A}B < 1^\diamond$.
(b). Existen enteros $0\leqslant r < s$ tales que el tri\'angulo
$AP_rP_s$ sea equil\'atero.

$\blacktriangleright$ [color=blue]Problema 3.[/color]
Sea $a$ un real positivo dado. Determine el valor m\'inimo posible
de $x^3 + y^3$, donde $x$ y $y$ son reales positivos tales que
$xy(x+y) = a$.

$\blacktriangleright$ [color=blue]Problema 4.[/color]
Pedro y Cecilia participan de un juego con las siguientes reglas:
Pedro escoge un entero positivo $a$ y Cecilia gana si puede
encontrar un entero positivo $b$, primo relativo con $a$ y tal que
la descompocición en factores primos de $a^3 + b^3$ contenga al
menos tres primos distintos. Muestre que Cecilia siempre puede
ganar.

$\blacktriangleright$ [color=blue]Problema 5.[/color]
Halle todas las ternas $(x, y, z)$ de enteros, con $x$ e $y$
primos entre s\'i - PESI y tales que $x^2 + y^2 = 2z^2$.

$\blacktriangleright$  [color=blue]Problema 6.[/color]
Sea $ABC$ un triángulo restángulo en $C$. Sobre el lado $AB$
tomamos un punto $D$ de modo que  $CD = k$ y los radios de las
circunferencias inscritas en los triángulos $ACD$  y $BCD$ sean
iguales. Mostrar que $S_{ABC} = k^2$.

$\text{\LaTeX}{}ed$ by Carlos Bravo 2006.

Haber ..
aki cargo la version en pdf
y el source en latex ..

para q  les sea mas facil descargarlos...

Saludos

Carlos Bravo ;)
Lima - Peru

[quote="carlosbr"]Sea $ ABC$ un triángulo rectángulo en $ C.$ Sobre el lado $ AB$ tomamos un punto $ D$ de modo que  $ CD \equal{} k$ y los radios de las circunferencias inscritas en los triángulos $ ACD$  y $ BCD$ sean iguales. Mostrar que $ [ABC] \equal{} k^2.$[/quote]
Por conveniencia denotemos $ AC \equal{} b,$ $ CB \equal{} a,$ $ AB \equal{} c,$ $AD \equal{} m,$ $ BD \equal{} n,$ $r$ el inradio, $ h$ la altura sobre la hipotenusa $ BC,$ $ p$ el semiperímetro y $ \varrho$ el inradio igual de $\triangle ADC$ y $\triangle  BDC.$ Usando el [url=http://garciacapitan.auna.com/problemas/sakabe/]Teorema de Sakabe Kohan[/url] para $ n \equal{} 2$ se tiene:
\[1 \minus{} \frac {c}{p} \equal{} \left (1 \minus{} \frac {2\varrho}{h} \right)^2 \ . (1)\]
Como el área un triangulo es igual al producto de su inradio por su semiperímetro, el área $ [ABC]$ puede ser expresada

$ [ABC] \equal{} \frac {_1}{^2}\varrho(b \plus{} k \plus{} m) \plus{} \frac {_1}{^2}\varrho(a \plus{} k \plus{} n) \equal{} \varrho(p \plus{} k) \ \Longrightarrow \ \frac {2\varrho}{h} \equal{} \frac {c}{p \plus{} k} \ . \ (2)$

Teniendo en cuenta que $ \triangle ABC$ es rectángulo en $ C$ $ \Longleftrightarrow$ $ r \equal{} p \minus{} c.$  Combinando las expresiones $ (1)$ y $ (2)$ tenemos

$ \frac {r}{p} \equal{} \left (1 \minus{} \frac {c}{p \plus{} k} \right)^2  \Longrightarrow \ k \equal{} \frac {p\sqrt {r} \minus{} r\sqrt {p}}{\sqrt {r} \minus{} \sqrt {p}} \ \Longrightarrow$

$k^2 \equal{} \frac {rp^2 \plus{} pr^2 \minus{} 2pr\sqrt {rp}}{r \plus{} p \minus{} 2\sqrt {rp}} \equal{} \frac {rp(r \plus{} p \minus{} 2\sqrt {rp})}{r \plus{} p \minus{} 2\sqrt {rp}} \equal{} rp \equal{} [ABC]$


algebra
polynomial
modular arithmetic
number theory unsolved
number theory

Let $ P(x)\in \mathbb{Z}[x]$ be an irreducible polynomial. Prove that for any given positive integer $ r$, we can find $ k\in \mathbb{Z}$ such that there exist different primes $ p_1,p_2,...,p_r$ such that $ p_i\parallel{}P(k)$ for each $ i\equal{}1,2,...,r$

There are exist different primes $ p_i,i\equal{}1,2,...r$, suth that exist $ k_i: \ p_i|P(k_i)$.
We can chose (By china theorem) k, suth that $ k\plus{}i\equal{}k_i mod p_i$. It give $ p_i|P(k\plus{}i)$.

Rust have you read the problem statement carefully? These primes have to divide value of the polynomial exactly in the first power.

We want to prove that for a given irreducible (and obviously non constant) polynomial $ P$ there exist infinitely many primes $ p$ such that $ p \parallel{} P(x)$ for some integer $ x$ (and then the thesis follows directly from the CRT).
It's well known that set of primes $ p$ such that $ p|P(x)$ for some integer $ x$ is infinite. We will show that subset of this set, consisting of primes $ p$ such that $ p|P(x)$ implies $ p^2|P(x)$, is finite and then the result will folow.
So, suppose that $ p$ is such prime that $ p|P(x)$ implies $ p^2|P(x)$ and $ p|P(k)$ for certain $ k$. Then of course $ p|P(p \plus{} k)$ and so $ p^2|P(p \plus{} k)$. But if $ P(x) \equal{} a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \ldots \plus{} a_1x \plus{} a_0$ then
\[ 0 \equiv P(p \plus{} k) \equal{} a_n(p \plus{} k)^n \plus{} a_{n \minus{} 1}(p \plus{} k)^{n \minus{} 1} \plus{} \ldots \plus{} a_1(p \plus{} k) \plus{} a_0 \equiv
\]

\[ \equiv p(na_nk^{n \minus{} 1} \plus{} a_{n \minus{} 1}(n \minus{} 1)k^{n \minus{} 2} \plus{} \ldots \plus{} a_1) \plus{} (a_nk^{n} \plus{} a_{n \minus{} 1}k^{n \minus{} 1} \plus{} \ldots \plus{} a_1k \plus{} a_0) \equal{}
\]

\[ \equal{} pP'(k) \plus{} P(k) \equiv pP'(k) \pmod{p^2},
\]
where the second congruence follows from the binomial expansion.
It follows that $ p|P'(k)$. But since $ P$ is irreducible the polynomials $ P$ and $ P'$ are co-prime in the ring of the integer polynomials, so there exist polynomials $ a, b \in \mathbb{Z}[x]$ such that
\[ a(x)P(x) \plus{} b(x)P'(x) \equal{} c
\]
where $ c$ is some non-zero integer. After plugging $ k$ in this equality we get that $ p|c$, so every $ p$ satisfying mentioned conditions is a divisor of fixed, non zero integer, so it is clear that there are only finitely many of them.

Let $ p_1, p_2, ...$ be an enumeration of the primes such that
- $ \exists n : p_i | P(n)$
- $ P'(x) \not \equiv 0 \bmod p_i$.

It's well-known that the set of primes satisfying the first condition is infinite and the second condition throws away at most finitely many primes.  Then by Hensel's lemma we can solve $ P(x) \equiv p_i \bmod p_i^2$ for every $ i$ and the conclusion follows by CRT.

[quote="0714446459923"]Also, what does $ a\parallel{}b$ mean?

Does it mean $ a|b , a^2\not|b$[/quote]
Yes

[quote="linker"][quote="TomciO"]
It's well known that set of primes $ p$ such that $ p|P(x)$ for some integer $ x$ is infinite.[/quote]

Sorry for bringing this post back after such a long time, but somebody can explain why this is true? Thank you[/quote]

I'm pretty sure that you can find many different approaches on this forum. But for example such method comes to my mind.  Of course we assume that the polynomial is non constant (otherwise it doesn't have to be true). Let $ a_0$ be a constant term. If $ a_0\equal{}0$ then there is nothing to prove, so assume not. Consider $ Q(x) \equal{} \frac{P(a_0x)}{a_0}$. It is easily seen that such polynomial has integer coefficients and the constant term equal to $ 1$. It is also clear that it is enough to prove the thesis for $ Q$. So, suppose that there are only finitely many $ p_1, \ldots, p_r$ which divide certain value of $ Q$. Consider $ x \equal{} p_1p_2 \ldots p_r t$ where $ t$ is an integer. Then $ Q(x)$ is not divisible by any of $ p_1, p_2, \ldots p_r$. Since it is non constant, for suitaible $ t$ we have also $ Q(x) > 1$. It is clearly a contradiction.


The problem is like this:

We have a triangle ABC, a step is defined as moving from the present vertex to one of the other two. A person starts at A ; in how many ways can he reach back to point A in n steps.

[hide]
Is the recursive formula
P(k)=P(k-1) + 2*P(k-2)  correct ??

I did it like this.
Let P(k) denote no of ways in reaching A in k steps,starting from A.
Let Q(k) denote no of ways in reaching B in k steps,starting from A.
By symmetry,Q(k) denotes no of ways in reaching C in k steps,starting from A.

In the (k-1)th step he has to be in either B or C.
So,
P(k) = 2*Q(k-1)              --------I

Now,consider starting from A and ending in B in k steps.
In (k-1)th step he should be in A or C.
So,Q(k) =P(k-1) + Q(k-1)
So, Q(k+1) = P(k) + Q(k)      ----II

I + II
Q(k+1) =2*Q(k-1) + Q(k)
So,Q(k-1)=2*Q(k-3) + Q(k-2)

By I,
P(k)=P(k-1) +2*P(k-2)

[/hide]

I think its perfect. Excellent. :)

Are there any other ways?


Solve the inequation: $2x^2-3x \lfloor x-1 \rfloor + \lfloor x-1 \rfloor^2 \leq 0$

Can you tell me which set x belong to?

Math is my life :!:

im not sure if this is legal:
[hide]
$2x^2-3x \lfloor x-1 \rfloor + \lfloor x-1 \rfloor^2 \leq 0$
make substitutions $a=x$, $b=\lfloor x-1 \rfloor$
$2a^2-3ab+b^2\leq 0$
$(2a-b)(a-b)\leq 0$
$a\leq b \leq 2a$
$x\leq \lfloor x-1 \rfloor \leq 2x$
EDIT: oops yeah.
there are no solutions for $x\geq0$
but for $x<0$, $2x\leq \lfloor x-1 \rfloor \leq x$
the set of real x that satisfies this is $\left(-\infty,-\frac{3}{2}\right)$ and $-1$, since the greatest integer function is defined on the left side. The function $y=2x$ and $y=\lfloor x-1 \rfloor$ intersect at $-1$ and $-\frac{3}{2}$ and $2x\leq \lfloor x-1 \rfloor$ for all $x\leq -\frac{3}{2}$
[/hide]

[quote="minsoens"]im not sure if this is legal:
[hide]
$2x^2-3x \lfloor x-1 \rfloor + \lfloor x-1 \rfloor^2 \leq 0$
make substitutions $a=x$, $b=\lfloor x-1 \rfloor$
$2a^2-3ab+b^2\leq 0$
$(2a-b)(a-b)\leq 0$
$a\leq b \leq 2a$
$x\leq \lfloor x-1 \rfloor \leq 2x$
but $x> \lfloor x-1 \rfloor$, so no solutions(?)  :?
[/hide][/quote]

What if $a < 0$? Then we have $2a \le b \le a$, which might have some solutions... ;). For example,

$2(-1)^2-3(-1)(-2)+(-2)^2 = 0$.

In fact, we can show that

[hide]
all negative numbers satisfying $2x \le \lfloor x-1 \rfloor$ work, i.e., $x = -1, x \le -\frac{3}{2}$.[/hide]


When 14 apple trees are planted per acre in a certain orchard, the average yield per tree is 480 apples per year.  For each additional tree planted per acre in the same orchard, the annual yield per tree decreases by 100 apples.  How many additional trees should be planted in the orchard so that the maximum number of apples will be produced per year?

***How would you use graphical representations to solve this problem?***

[hide="your apple trees are on steroids"]
Let $ x$ be the number of apple trees planted in the orchard
Let $ y$ be the total yield

we want to find $ y$ in terms of $ x$ so we can find when $ y$ is the biggest

$ y = (x)(480-100x) =-100x^{2}+480x$

x is the number of trees
480-100x is the yield per tree

now, the graph of that function is a parabola that opens downward. The optimum number of trees to plant to get the greatest yield is at the peak of that parabola, at the vertex.

The vertex is at $ \frac{-b}{2a}= \frac{480}{200}= 2.4$
since you can only plant a whole number of trees, the best number of trees to have is
2 trees

so to optimize the number of apples, you want to chop down 12 of your trees
[/hide]


induction
limit
algebra proposed
algebra

Sequence $ (a_{n})$ is defined by $ a_{1}\equal{}a_{2}\equal{}1$ and $ a_{n\plus{}2}\equal{}\frac {1}{a_{n}} \plus{} \frac {1}{a_{n\plus{}1}}$. Find $ lim a_{n}$, where $ n$ tends to infinity.

I think it's $ \sqrt{2}$ but I can't prove it. Please post a solution :oops:

I don't know the solution. But if it has limit it's easy to show that your answer is right.
P.S. Sorry that I posted it in proposed problems and not in ''Unsolved'' ones. :)

By induction, we can show $ 1\le a_n\le 2$. So the series are bounded.
$ a_{n \plus{} 2} \equal{} \frac {a_na_{n \minus{} 1}}{a_n \plus{} a_{n \minus{} 1}} \plus{} \frac {a_{n \minus{} 1}a_{n \minus{} 2}}{a_{n \minus{} 1} \plus{} a_{n \minus{} 2}} \\
\le\frac {a_n \plus{} a_{n \minus{} 1}}4 \plus{} \frac {a_{n \minus{} 1} \plus{} a_{n \minus{} 2}}4 \\
\equal{} \frac {a_n \plus{} 2a_{n \minus{} 1} \plus{} a_{n \minus{} 2}}4$  (*)

Define $ b_n \equal{} a_{n \plus{} 2} \plus{} a_{n \plus{} 1} \plus{} \frac34a_n \plus{} \frac14a_{n \minus{} 1}$, we have
$ b_n\le b_{n \minus{} 1}$.

Since $ b_n$ is bounded and monotone, we have $ b \equal{} \lim b_n$.
Hence, $ \lim (b_n \minus{} b_{n \minus{} 1}) \equal{} 0$, or
$ \lim (\frac {a_n \plus{} 2a_{n \minus{} 1} \plus{} a_{n \minus{} 2}}4 \minus{} a_{n \plus{} 2}) \equal{} 0$.

or

$ \lim (\frac {a_n \plus{} 2a_{n \minus{} 1} \plus{} a_{n \minus{} 2}}4 \minus{} \frac {a_na_{n \minus{} 1}}{a_n \plus{} a_{n \minus{} 1}} \minus{} \frac {a_{n \minus{} 1}a_{n \minus{} 2}}{a_{n \minus{} 1} \plus{} a_{n \minus{} 2}}) \equal{} 0$,

or

$ \lim (\frac {(a_n \minus{} a_{n \minus{} 1})^2}{4(a_n \plus{} a_{n \minus{} 1})} \plus{} \frac {(a_{n \minus{} 1} \minus{} a_{n \minus{} 2})^2}{4(a_{n \minus{} 1} \plus{} a_{n \minus{} 2})}) \equal{} 0$

Since $ 1\le a_n\le 2$, it is not hard to get
$ \lim (a_n \minus{} a_{n \minus{} 1}) \equal{} 0$.

Combined with $ b \equal{} \lim b_n \equal{} \lim (a_{n \plus{} 2} \plus{} a_{n \plus{} 1} \plus{} \frac34a_n \plus{} \frac14a_{n \minus{} 1})$, we can easily show the existence of $ \lim a_n$. Then $ \lim a_n \equal{} \sqrt {2}$.


geometry
circumcircle
induction
LaTeX
geometry proposed

Prove that every convex polygon with the property that every 3 consecutive vertices determine a triangle with circumradius 1 is cyclic.

I know that this problem is trivial for most of you, but I'm still waiting to see a nice and short solution. Now let me tell you why I like it. I gave it in a regional contest held in my town and some contestants gave a "complete" solution, but which was wrong. However, they got the full mark because the coordinators did the same mistake  :P .

I'm not very sure why $A_2A_1$ and $A_3A_4$  have to meet on the side of $A_2A_3$ which does not contain $A_1,A_4$. Maybe I'm missing something obvious, but I think it is not true in any convex polygon.

I think this could be done using (strong) mathematical induction, i.e. supposing that the statement is true for $3,4,\ldots,n-1$ and proving it for $n$.

The case $n=3$ is clear, but the case $n=4$ is not THAT clear. (BTW, it took me a while to notice that this case isn't trivial) Once we prove it for $n=4$, the induction step should be pretty clear. But I can think only of tedious computations  :? Can someone find a nice solution?

I think my solution is not messy at at, but I'd like to see another one  :D . Case $n=4$ follows from grobber's idea, but I don't see how to apply induction here.

You're right, this problem is out of my league.  :oops: I'll think twice before posting next time.

I am sure you are wrong. It is nothing more than a cheap trick, as most of my problems are. I could post the solution, but that would spoil the fun. If you want, I can pm you my solution, just give me a sign.
P.S. Sorry for so many useless posts in this topic.

I am almost certainly being thick, but why does this not follow from the fact that there are at most two circles of radius one passing through a pair of points, one of which wil cause the convexity condition to be violated?

Make a drawing: Two points $A,B$ and two circles, $\mathcal C_{1,2}$, symmetrical wrt $AB$. I think you cand find two points $P_1 \in \mathcal C_1$ and $P_2 \in \mathcal C_2$ which form a convex quadrilateral.

To do what I said, ignore everything that's above $AB$.

:?: by  cyclic you mean that the polygon is inscribed ?  :roll:  if so i think i have a solution but i'm not sure  about it because is not that hard .i'll try to post it

here it goes:
 let $A_1A_2A_3...A_n$ be the convex polygon .let  $O_1$  be the circumcenter of triangle $A_1A_2A_3$ ,thus $OA_1=OA_2=OA_3=1$,(1)IF  $O_2$  is circumcenter of triangle $A_2A_3A_4$, and M the midpint of $A_2A_3$ then $O_1M\perp A_2A_3$ and
$O_2M\perp A_2A_3$   so , $O_1,M,O_2$ are collinear ,neither $O_1$ or $O_2$ can not  be outside of the interior of the polygon, because if there are distinct are symethricv wr to M so that 1 is true, one is placed in the exterior of the polygon and can not be the circumcenter of a triangle formed with the vertices of the polygon because otherwise the polygon would not be convex,wich is  not true. so the circumcenters of every two consecutive triangles are the same, thus the polygon is inscribed.
 hope i didn't make  mistakes (is  my first msg in latex :blush: )

It is possible that the circumcenter of some triangle is outside the polygon but the polygon is still convex.

here it goes:
 let $A_1A_2A_3...A_n$ be the convex polygon .let  $O_1$  be the circumcenter of triangle $A_1A_2A_3$ ,thus $OA_1=OA_2=OA_3=1$,(1)IF  $O_2$  is circumcenter of triangle $A_2A_3A_4$, and M the midpint of $A_2A_3$ then $A_2A_3\perp O_1M$ and
$O_2M\perp A_2A_3$   so , $O_1,M,O_2$ are collinear ,neither $O_1$ or $O_2$ can not  be outside of the interior of the polygon, because if there are distinct are symethricv wr to M so that 1 is true, one is placed in the exterior of the polygon and can not be the circumcenter of a triangle formed with the vertices of the polygon because otherwise the polygon would not be convex,wich is  not true. so the circumcenters of every two consecutive triangles are the same, thus the polygon is inscribed.
 hope i didn't make  mistakes (is  my first msg in latex :blush: )

i think the triangle $A_2A_3A_4$  can not have  $O_2$  exterior because otherwise thre angle $A_3A_4A_2$ shuld be grater than 90 degrees,and for a n quite big (n>4)is not cuz  $A_3 and A_4$ are consecutive vertics, isn't?

No it isn't. Angle $A_3A_4A_2$ can be obtuze for any $n$.

Since the problem already got more attention than it deserves, I think it is time to post my solution and end this topic.
The hypothesis implies the fact that each side is seen from the neighbouring vertices under either equal or supplementary angles. For each side, compute the sum of the measures of these 2 angles. By straightforward calculation, these quantities add up to $2\pi$. Then, the case when the two angles are supplementary can occur at most once. However, if the angles are equal, the circumcircles of the two triangles determined by consecutive vertices coincide, which means that all such circumcircles coincide, so the polygon is cyclic.


induction
function
LaTeX
limit
floor function
geometry
geometric transformation

Let F(n) = 2n+2/3 - G(n)e^n where G(n) = Sum[ ( (e^(-k)) (k-n)^k)/k!, {k,0 n-1}].

Goal: Prove Limit[F(n), n-->Infinity] = 0.

I've already tried a few things, including using Stirling's formula.

One approach will be to show that |F(n)|<(1/e)^n which has numerical evidence behind it.  I've found this to be hard but maybe I'm stupid and/or it's false and I just haven't tried it for enough n.

Induction seems to be useless but I could be wrong.

Possibly of note is that I can rewrite G as G(p)+S(n) where p>n/2 (or Floor(n/2) if you like) and
S(n) = G(n) - G(p).  I know how to make S(n) smaller than E(n) where
E(n)=( (n - an)^(an) e^(-an) ) / (an)! where 1/2<a<1, like a=4/5.  E(n) -->0 as n-->Infinity.

I was thinking that since G(n) < G(p) + E(n), we can use complete induction if we were to have a function d of n such that G(p)<d(p) and G(p)+E(n)<d(n) would then imply that for all n, G(n) < d(n).  (I think.)

I've had a lot of ideas and think I'm stupid for not being able to solve this.  If you know how to solve it, please just give me a hint.  If you don't have a clue or close to it, then we can work together.

1. "Source" means "Where did this problem come from"? If it's from a particular contest, or a particular textbook, this is where you can mention that, or if it's your own problem, you can say that.

2. For notation-heavy problems, you'll find that you get a quicker response on this site, and from potentially more people, if you avail yourself of the the $\LaTeX$ resources of this site to write your problem.

I'll do that for you in this case:

Let $F(n)=2n+\frac23-G(n)$ where $G(n)=\sum_{k=0}^{n-1}\frac{e^{-k}(k-n)^k}{k!}$

Goal: Prove $\lim_{n\to\infty}F(n)=0.$

Hmm... I [b]know[/b] this problem. I'll have to go looking for a topic I think I started here some months ago.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20848]This[/url] is the topic I was looking for, with references back to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20753]this previous topic.[/url]

You weren't going to get anywhere manipulating the sum and approximating things- this involves extremely delicate cancellation. The correct approach involves interpreting the exact sum so you can use completely different methods.

Hi all.

Thank you for replying.  Well, I wouldn't have thought of that approach.  I guess that's why I asked about it.  I'm in the process of digesting the solution as I am not super familiar with Laplace transforms nor (especially) the inverse Laplace transform.  I get the basic idea of the proof (I think) but I'm not sure how I'd know the Laplace transform of $g$ when for me it was given by $g(y)=\sum_{k=0}^{\lfloor y\rfloor}\frac{(-1)^k(y-k)^k}{k!}e^{y-k}$.

As for my source, it was a problem that no one seemed able to totally do over at [url=http://www.physicsforums.com]physicsforums.com[/url].  That has been my main-stay for math problems but now I think I'll switch over to here.

I wasn't sure how to input latex code until I quoted someone.  Now I know and will use latex in the future.

Back to the solution...  Should I post any questions about it here or there?  I would think there, unless you want to preserve the fact that the solution is the last item in the thread.

So if $g(y)=\sum_{k=0}^{\lfloor y\rfloor}\frac{(-1)^k(y-k)^k}{k!}e^{y-k}$, the basic idea I'm gathering is to write $g$ as $g\left( y\right) =L^{-1}\left( L\left( g\left( y\right) \right) \right) $ where $L$ is the Laplace transform operator.  Is that right?  If so, then how do I find the Laplace transform of $g$?  Are the steps sort of there in reverse in that other thread?

Thank you for you assistance.  This problem has really bothered me...

Yes, it's basically the same steps in reverse.

The $k$th term only contributes if $k\le y.$ So we introduce the Heaviside function $H$ which is 1 for positive values and zero for negative values.

$g(y)=\sum_{k=0}^{\infty}\frac{(-1)^k(y-k)^k}{k!}e^{y-k}H(y-k).$

One of the major indenties of the Laplace transform is this:

If $f$ is a function supported on $[0,\infty)$ (that is, we go ahead and consider $f(x)$ to be zero for $x<0$) and the Laplace tranform of $f$ is $F,$ then if $g(x)=H(x-a)f(x-a),$ the Laplace transform of $g$ is $e^{-as}F(s).$

Some of the basic rules of action for the whole Laplace-Fourier family of transforms:

Translation on one side $\longleftrightarrow$ multiplication by an exponential on the other side.

(That's the one we just used.)

Differentiation on one side $\longleftrightarrow$ multiplication by a monomial on the other side.

(In the Laplace transform case, there are some boundary details at zero.)

Pointwise multiplication on one side $\longleftrightarrow$ convolution on the other side.

[edit - fixed typo noted by BrianT, below.]

Excellent.  Nice explanation.

I will now try to sit down and work out the Laplace transform of $g$ as you defined it.  Seems like this approach would have a lot of use in asymptotic analysis....

This will tell you how much I don't know because if I did know, then it would be obvious.

[quote="Kent Merryfield"]

If $f$ is a function supported on $[0,\infty)$ (that is, we go ahead and consider $f(x)$ to be zero for $x<0$) and the Laplace tranform of $f$ is $F,$ then if $g(x)=H(x-a)f(x-a),$ the Laplace transform of $f$ is $e^{-as}F(s).$[/quote]

This this quote, did you mean, "the Laplace transform of $g$ is $e^{-as}F(s)$?

Thanks.

Now I'm suddenly interested in this stuff and I'm trying to find the asymptotic behavior of $g\left( y\right) =\sum_{k=1}^{\infty }\left( 1-\frac{y}{k}\right) ^{k}H\left( y-k\right) $.   So do I say that the Laplace transform of an infinite sum is the infinite sum of the Laplace transforms?  That's like interchanging limit and (improper) integral, right?  Hmm...

Hi All

My goal is to find the asymptotic behavior of $g\left( y\right) =\sum_{k=1}^{y-1}\left( 1-y/k\right) ^{k}$.

I found what I believe to be the Laplace transform of $\left( 1-y/k\right) ^{k}H(y-k)$: $k!\left( -1/k\right) ^{k}\frac{1}{s^{k+1}e^{ks}}$.

So then is the Laplace transform of $\sum_{k=1}^{\infty }\left( 1-y/k\right) ^{k}H(y-k)$ equal to $\sum_{k=1}^{\infty }k!\left( -1/k\right) ^{k}\frac{1}{s^{k+1}e^{ks}}$?  And what about finding the inverse Laplace transform?

Any help will be appreciated.


How do I include the diagram from the problem in my solution?

thank you for your help

In the future, please submit similar questions to the [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]LaTeX and asymptote forum[/url].

What I consider the easiest way to include a diagram in a LaTeX file is to create it using [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_%28Vector_Graphics_Language%29]Asymptote[/url].

If you have a file, say diagram.pdf, just say
[code]\includegraphics{diagram}[/code]

[quote="emant777"]How do I include the diagram from the problem in my solution?
[/quote]

It used to be easier when they included the latex source for the problems.  Now I think the easiest way would be to open up the problem file, take a screenshot, copy it into some graphics program (mspaint would be good enough), crop it, and then save it as a jpeg or png in the same folder as your solutions.tex file.  Then you can include the image into your solutions document using the command
\includegraphics{image.png}
(if you saved the diagram as image.png)

Thanks!

Now I got all of the pictures to work  :D


Let $ a,b,c \geq 0$  and  $ a\plus{}b\plus{}c\equal{}3$   Prove that

$ \frac{a}{ab\plus{}1} \plus{} \frac{b}{bc\plus{}1} \plus{} \frac{c}{ca\plus{}1} \geq \frac{3}{2}$

I think you can apply Cauchy's to find
$ LHS\equal{}\sum\frac{a^2}{a^2b\plus{}a}\ge\frac{9}{(\sum a^2b)\plus{}3}$,
so it suffices to show that
$ a^2b\plus{}b^2c\plus{}c^2a\le 3$.

$ [(ab\plus{}1)\plus{}(bc\plus{}1)\plus{}(ca\plus{}1)](\frac{a}{ab\plus{}1}\plus{}\frac{b}{bc\plus{}1}\plus{}\frac{c}{ca\plus{}1})\geq {9}$
=>
$ \frac{a}{ab\plus{}1}\plus{}\frac{b}{bc\plus{}1}\plus{}\frac{c}{ca\plus{}1}\geq \frac{9}{ab\plus{}bc\plus{}ca\plus{}3}$

but $ ab\plus{}bc\plus{}ca<\equal{}\frac{(a\plus{}b\plus{}c)^2}{3}$ so $ ab\plus{}bc\plus{}ca<\equal{}3$ because $ a\plus{}b\plus{}c\equal{}3$

so $ \frac{9}{ab\plus{}bc\plus{}ca\plus{}3}>\equal{}9/6\equal{}3/2$

=> $ \frac{a}{ab\plus{}1}\plus{}\frac{b}{bc\plus{}1}\plus{}\frac{c}{ca\plus{}1}\geq\frac{3}{2}$
 and the case of equality when a=b=c=1

[quote="hanschinaman"]so it suffices to show that
$ a^2b \plus{} b^2c \plus{} c^2a\le 3$.[/quote]
which wrong: $ a \equal{} b \equal{} 1.5$ and $ c \equal{} 0.$  :wink:

[quote="setheus_soul"]$ [(ab \plus{} 1) \plus{} (bc \plus{} 1) \plus{} (ca \plus{} 1)](\frac {a}{ab \plus{} 1} \plus{} \frac {b}{bc \plus{} 1} \plus{} \frac {c}{ca \plus{} 1})\geq {9}$
[/quote]

Unfortunately this is incorrect.

$ [(ab \plus{} 1) \plus{} (bc \plus{} 1) \plus{} (ca \plus{} 1)](\frac {a}{ab \plus{} 1} \plus{} \frac {b}{bc \plus{} 1} \plus{} \frac {c}{ca \plus{} 1})\geq (\sqrt{a} \plus{} \sqrt {b} \plus{} \sqrt {c})^{2}$

[quote="iverson_h3"]hi!!!!!!!!
we have by chebechev : S>= 1/3(a+b+c)(1/ab+1 +1/ac+1 +1/bc+1) = 1/ab+1 +1/ac+1 +1/bc+1
[/quote]

It 's also not true  :(  Any idea???

[quote="ntt289"]Let $ a,b,c \geq 0$  and  $ a \plus{} b \plus{} c \equal{} 3$   Prove that

$ \frac {a}{ab \plus{} 1} \plus{} \frac {b}{bc \plus{} 1} \plus{} \frac {c}{ca \plus{} 1} \geq \frac {3}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21689


abstract algebra
calculus
integration
algebra
function
domain
Ring Theory

Let $R$ be an integral domain and $I\subset R$ is a nonzero ideal.  Show that $M=\frac{R}{I}$ is not a flat module.

I know that $M\otimes_R \frac{R}{I}\cong\frac{M}{IM}$.  I just cannot think of an example.

Generally, if $R/I$ was flat, then we would have an exact sequence $0\to I\otimes_RR/I\toR\otimes_RR/I\to R/I\otimes_RR/I\to 0$.
Up to isomorphisms, this is $0\to I/I^2\to R/I\to R/I\to 0$. It is not that hard to check that the map $R/I\to R/I$ in this induced sequence is the identity map, so it is an isomorphism. But then $I/I^2=0$. This does not generally happen, but it is possible. For example if $p$ is a prime ideal in the integral closure of $\mathbb Z$ in $\mathbb C$.

So what I proved is that if $I^2\neq I$ then $R/I$ is not a flat $R$-module.

Assume that $R/I$ is flat.
Let $J$ be an arbitrary ideal of $R$. Then we have a short exact sequence $0\to J\to R\to R/J\to 0$.
Upon tensoring with $R/I$, according to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76587[/url], we have a short exact sequence
$0\to J/IJ\to R/I\to R/(I+J)\to 0$. The maps are $J/IJ\to R/I: j+IJ\to j+I$ and $R/I\to R/(I+J): r+I\to r+(I+J)$. Since the first map must be injective, its kernel $\frac{I\cap J}{IJ}$ must be $0$. This happens iff $I\cap J=IJ$.
So if $R/I$ is flat, then $I\cap J=IJ$ for every ideal $J$ in $R$.

Let $J=xR$ with $x\in I$, $x\neq 0$. Then $x\in I\Rightarrow J\subset I$. Hence we must have $xR=xI$ for every $x\in I$, $x\neq 0$. $xR=xI\Rightarrow x\in xI$. Since $R$ is a domain, this implies $1\in I$ which contradicts something which should have been in the hypothesis i.e. that $I$ is a proper (besides from being nonzero) ideal of $R$. After all what can be more flat than $0$ :D? except maybe when you exclude the $0$ case by definition.


I watched it two years ago. I think they just showed the countdown, along with some interviews, random stuff, etc. Maybe they've changed it? I remembering seeing Mr. Rusczyk on it :D.

Mr. Rusczyk did do the broadcast last year for ESPN, although he didn't have a telestrater(sp?) like the first year where he could draw on the screen.  They were more advanced having digital pictures and typed explanations of select problems.

Otherwise with the exception of the format of the countdown and having 12 competitors, it was basically same as the year before.  I would like to see them show more of the competition, but I guess time constraints restrict them to only one or two glimpses of the competition.

Yes, the time constraints are certainly a problem.  Probably the bigger problem, though, is that it's just not good television to watch people take a written test.  Stick them on stage and show what they can do quickly - that makes better TV, even if it doesn't make as good math.

[quote="felmo"]I think there should be more math on ESPN. It makes me very sad, like the time my boyfriend walked out on me and threw a pitcher of Tang at me (the new kind with more vitamin C).[/quote] :?  :? I don't get this.

You guys should consider yourself lucky. There's absolutely nothing like this (maths live on TV) here in Romania. Not even ESPN :(

You can get me!  :blush:  That will be better than Romania

[quote="Valentin Vornicu"]You guys should consider yourself lucky. There's absolutely nothing like this (maths live on TV) here in Romania. Not even ESPN :([/quote]To be fair, it's not going to air [i]live[/i] here either.  I think it will probably air sometime this summer.

My guess is that part of the reason that ESPN decided to start showing MATHCOUNTS is due to the popularity of ESPN's broadcasts of the National Spelling Bee (which I believe they do air live).  Also another cable channel (I can't remember which one) airs the national Geography Bee.  And many local cable systems air local math contests or other quiz bowl type events.

The idea is that there is no such contest here :(

I think they should expand the Countdown Round.  Having only the Top 12 middle school students in the NATION compete leaves out some VERY talented students.  I think that if the Nationals countdown round was changed to 32 people, it would be much more fun, and of course, much longer.  Believe it or not, a lot of people who watch ESPN for Sportscenter saw the MathCounts Competition last year.  I remember two years ago I was in fifth grade with no clu what MathCounts was. :o But I still watched the competition on ESPN, by pure luck!

I think the countdown round should remain as it is (for the most part--I didn't like the bracket thingy), but they should add another oral round, maybe unofficial, between the written and countdown rounds.

I think they should have the masters' round on ESPN too. Do competitors who aren't the top four get to watch it at all?

Yeah, others get to watch it....

except last year the top 12 had some trouble getting to watch it because the ESPN people were interviewing (I waited probably a whole hour, they claimed it would be 5 minutes...)

....I missed the whole thing :(

[quote]towersfreak2006
PostPosted: Tue Apr 19, 2005 9:58 pm    Post subject: Mathcounts on ESPN
What exactly does ESPN show? I know it broadcasts the countdown round, at least, but anything else? Also, will it ever air past years again? I wasn't able to watch it last year, and one of my friends did really well.

And I sure hope ESPN does it again this year. Gotta cheer on the Texas team. ^_^[/quote]

i only saw it two years ago and i only watched part of it because i didnt care much for mathcounts because at that time it was my sisters thing and btw is your "friend" Jeffrey Chen and you better cheer on the texas team  :D

Probably during the 1 hour lunch break of the spelling bee.

"In problem solving, visualization is a key concept.  Here is the net of the cube, and this is what he saw...."- 2003-2004 National Mathcounts countdown commentor

"What lesson is next? adding 1+2?"- fellow teammate commenting while watching at practice.

[quote="sophia"][quote="DPatrick"]To be fair, it's not going to air [i]live[/i] here either.  I think it will probably air sometime this summer.

My guess is that part of the reason that ESPN decided to start showing MATHCOUNTS is due to the popularity of ESPN's broadcasts of the National Spelling Bee (which I believe they do air live).  Also another cable channel (I can't remember which one) airs the national Geography Bee.  And many local cable systems air local math contests or other quiz bowl type events.[/quote]

Yes, spelling bee does get live coverage and the taped MATHCOUNTS gets run on ESPN during the lunch break of the live spelling bee coverage.  Some of the spellers were actually watching it in the hallway outside the spelling bee ballroom during their lunch break!

And the Geo Bee runs on PBS, not on a cable network actually.  It isn't done live either, but they do get Alex Trebek!  And they only show the top 10 on the one-hour Geo Bee broadcast.

There's just something about spelling that makes for riveting live TV (at least for some people!)  People in the viewing audience can actually spell along in real time (and have SOME shot at actually getting it right before the correct spelling is revealed.)

I suspect that there aren't many people in the viewing audience who could get a MATHCOUNTS countdown problem right  in real time.

A nontrivial number of people watching the spelling bee on ESPN are somewhat inebriated (in sports bars!), by the way.  They apparently enjoy betting on the spellers.[/quote]

I think the Geo Bee is live on the National Geographic Channel.

The National Geographic Channel?????
Never Heard of it.

Must be a premium channel.

I didn't know we had the lunch break of the spelling bee!!! Was it like that last year? (I remember some weird show called cold pizza or something like that being on before it.)

Mathcounts winners get plenty of money. The winner of the Putnam (undergraduate olympiad-level problem solving--probably the most well known undergraduate contest) only gets 1000. Besides, phelpedo, do you actually expect to get any of that money? :)

[quote="Phelpedo"]The spelling bee has a lot more glamorous prizes too. Isn't first place like 25,000 and second 15,000 (or even more)? I think someone like Bill Gates should give a portion of that multi-billion fund to Mathcounts to boost the popularity.[/quote]

I agree.

Hey, be nice to that kid who fainted.

[quote]The last Spelling Bee was awesome, because that kid fainted and i taped it and kept watching it and laughing everytime....who else saw that?[/quote]

my tutor told me :P

Here's a tape of the faint:
http://www.ebaumsworld.com/spellingbeefaint.html

That's funny. Did he still get it right?

yeah! he still got it right... that's remarkable!

[url=http://www.mathcounts.org/webarticles/articlefiles/536-Program_At_A_Glance.pdf]www.mathcounts.org/webarticles/articlefiles/536-Program_At_A_Glance.pdf[/url]
It will be broadcast June 3&7th at 6:00 ET for one hour

Will you be one of the commentators again this year, Mr. Rusczyk?

I just realized that those hobbies they show for the competitors are the ones we put on that registration form. I'm slow. :?

Well, if you did okay in Countdown, then you're faster than a lot of people....

I came in third in the Ohio countdown (bracket with top 16) and lost to the person who eventually won the countdown in the semifinals, winning the consolation match to take third. Countdown was unofficial. I was second written, the winner was third written, second place was first written. Bit of an interesting cycle there.

Don't think so, but I really have no idea. I'm know every square inch of my tennis racket but can't figure out my computer. :P

[quote="Monkey"]Know who that first written, second countdown kid was?  :D  :lol:

[hide]Me!!  :D  :lol:  :D  :lol: [/hide]

Dang it, I don't have LaTex; my emoticons are just stupid letters on the keyboard.

Btw, are the emoticons determined by whether or not you have LaTex? :?[/quote]

Monkey, judging from your Math Jam performance, I'd think u were a more countdown oriented person.

Monkey did great on the countdown, trust me. The final was very close.


conics
parabola
analytic geometry
search
geometry unsolved
geometry

Let points A , B , C  be on a parabola with its focus P
tangent line at A to the parabola=k , tangent line at B to the parabola=m , tangent line at C to the parabola=n
D=k∩m , E=m∩n , F=n∩k
Prove that D, E, F, P are on a circle.

Is there a synthetic proof not using coordinate system ?
May I ask you a favor ? thanks

It has been posted before as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=142495]Parabola[/url], by [b][size=100]Amir.S[/size][/b]

Kostas Vittas.

thanks , vittasko. next time I first will search posts related to my questions  :)
excuse me , please give me the synthetic proof or hints of poncelet theorem. :blush:

See also [url]http://www.mathlinks.ro/viewtopic.php?t=123080[/url] (the 2nd reply).

so beautiful proof , yetti
your nice answer inspires me though it is not easy for me . thanks


Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail.  However, the only available scales is not accurate for weights less than $ 100$ pounds or more than $ 150$ pounds.  So the boxes are weighed in pairs in every possible way.  The results are $ 122, 125, 127$ pounds.  What is the combines weight in pounds of the three boxes?

$ \textbf{(A)}\ 160\qquad
\textbf{(B)}\ 170\qquad
\textbf{(C)}\ 187\qquad
\textbf{(D)}\ 195\qquad
\textbf{(E)}\ 354$

Let $ x,y,z$ be the three boxes.  WLOG, let $ x\leq y\leq z$.  Then $ x\plus{}y\equal{}122$, $ x\plus{}z\equal{}125$ and $ y\plus{}z\equal{}127$.  Hence
$ (x\plus{}y)\plus{}(x\plus{}z)\plus{}(y\plus{}z)\equal{}2(x\plus{}y\plus{}z)\equal{}122\plus{}125\plus{}127\equal{}374\Rightarrow x\plus{}y\plus{}z\equal{}187$.


Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup.  Joe drank 2 ounces of his coffee and then added 2 ounces of cream.  JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces.  What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?

$ \textbf{(A) } \frac 67 \qquad \textbf{(B) } \frac {13}{14} \qquad \textbf{(C) } 1 \qquad \textbf{(D) } \frac {14}{13} \qquad \textbf{(E) } \frac 76$

[hide="Answer"]Joe has $10$ ounces of coffee and $2$ ounces of cream: $\frac{2}{10+2}=\frac{1}{6}$. JoAnn had $12$ ounces of coffee and $2$ ounces of cream: $\frac{2}{12+2}=\frac{1}{7}$. Since it was well-mixed, she still has the same ratio. Therefore, $\frac{\frac{1}{6}}{\frac{1}{7}}=\frac{7}{6}\Rightarrow \boxed{E}$.[/hide]


[quote="mathgeek2006"]Hide your answers!!![/quote]
I'll make you a deal, mathgeek.  I'll hide her answers if you show your work.   :P   Just kidding.

Seriously, though, I wish you would show your work.  You seem to know how to solve many problems.  But nobody learns from you, if you don't show your work.

Showing your work will also help you grow, mathematically.

I usually don't show my work because I'm still learning LaTeX, so it's still hard for me to show my work.  When I can, I try to show it, unless it takes too much time.  I'll try to show it in the future as much as I can.

[hide="Click to reveal hidden content...if you dare!"]Seems pretty easy to me, but that means I'm probably waaay off. 6.25-5.98=0.27 difference in pounds per gallon. Since there are 10,000 gallons, 10,000*0.27=2700 pounds.[/hide]

extension,

VERY EASY!!

[quote=extension]What part of information in this passage could be changed and still have the same answer and the whole basic concept?[/quote]

you could change the units... maybe... instead of gallons you could have quarts

[quote="MCrawford"]A tanker carries 10,000 gallons of gasoline.  Gasoline weighs 5.98 pounds per gallon at 60 degrees Fahrenheit and 6.25 pounds per gallon at zero degrees Fahrenheit.  How many pounds heavier is the tanker when it carries a full load at the colder temperature than at the warmer temperature?[/quote]

0.27 pounds heavier if there's only 1 pound

Now, 0.27 is more than 0.1.

0.27 times 10,000 would be less than 10,000 and begin with 27. [b]2,700[/b].

[hide="da answer?"]10,000 gallons of gasoline at 60 degrees fahrenheit weighing 5.98 pounds per gallon
= 59800 pounds (10,000 * 5.98)
The 10,000 gallons of gasoline at 0 degrees fahrenheit weighing 6.25 pounds per gallon
= 62500 pounds (10,000 * 6.25)
62500 pounds - 59800 pounds = 2700 pounds[/hide]


Could anyone explain the rating that's on my profile?  What's that, and how is that calculated?  Just curious.

The fact that you posted such a question without bothering to notice [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=227261]this currently active topic[/url] is at least a partial indication of why your rating is as low as it is.


number theory
least common multiple
prime numbers

[quote="aidan"]If $ r$, $ s$, $ t$ are prime numbers and $ p$, $ q$, are positive integers such that the LCM of $ p$ and $ q$ is $ r^{2}t^{4}s^{2}$, how many possible ordered pairs $ (p, q)$ are there? ($ p$ and $ q$ are expressed in terms of nultiples and powers of $ r$, $ s$ and $ t$)[/quote]
[hide]
Well consider each prime separately: the exponent of $ r$ in $ p$, say, is two, then the exponent of $ q$ can take three possible values: $ 0,1,2$. Then multiply by two because it's an ordered pair, then subtract one in the case both $ p$ and $ q$ have factors $ r^{2}$. Do the same for the others, then multiply all them together and you get $ (5)(9)(5)\equal{}225$?[/hide]

Nice  :) I listed everything out and over counted.  :P


As the title says, name a person and something (s)he invented without which your life would be very different now. I'll start it up with the Czech chemist Otto Wichterle which invented the Soft Contact lenses, without which I would not be able to play basketball  :)

Thomas Edison, who invented the light. Without a light,
I wouldn't be able to do math in the night.

David Bradley, the inventor of ctrl+alt+del

Nicholas Amatti, who invented the modern violin as we know it today.

Anybody who invented anything technologically related (specifically, TVs, computers, stoves, air conditioning, lightbulbs)

I could not live without technology.

nostradumas

i can't spell
without him, who would i make fun of when the world doesn't end

kevin pittsnogle

without whom, WV would've lost to TExas Tech in the NCAA Tourney two years ago...

the cave man who found a stick that got struck by lightning, threw it into a bush and found Fire

FWEE FWEE. MY FWEE FWEE. FWEE FWEE GOOD...

Yeah I believe that Thomas Edison and Alexander Graham bell have impacted the world profoundly with their inventions. The telephone I believe is one of the most useful ways of communicating. And just imagine life without a light bulb.

The founding fathers.

Without them, no america.  :D

marc anderseen FTW!
otherwise, i wouldnt be typing this right now
and i wouldnt be using the internet
which would really suck

[quote="asdff"][quote="vicseb"]Albert Einstein

If he didn't invent the nuclear bomb, how else can we destroy whole cit-... I mean, how else could we have won World War 2 (or was it the Civil War?).[/quote]

lol?  are you joking me civil wars?[/quote]
this i find amusing. after all where WOULD we be? i mean...in another world war or would there be more innocent people on the face of the earth?

we have to question who was responsible for that tho...hitler? the president? the people who researched further? the people who listened to hitler and burned down Einstein's house? Einstein himself?....

one can never know who really changed there life.

um aops admins?
if not for aops i would hate math right now?

[quote="alkjash"]George Bush.
He invented massacring huge numbers of people, which the video game makers took up.
 :rotfl:  :rotfl:  :rotfl:[/quote]
WOW MAN YOU'RE A FUNNY GUY

Richard Rusczyk without him we wouldn't have AOPS and a bunch of cool math competitons and USAMO expansions

Matt Groening, for without him, we wouldn't be watching the Simpsons right now.

[quote="usaha"]Anna Nicole Smith-for the good times[/quote]

you mean Anna Nicole Smith and Mr. Right Hand for the good times

uhhh on a more G-rated note, Jesus is pretty important

I nominate Richard Feynman for being a brilliant physicist


If you know the answer, please hide your post.

Find all the whole numbers that have an odd number of factors, and explain why those numbers have an odd number of factors.  :)

I guess I have found a simple metod to find it all.

[hide]
We can group the factors of the number $ n$ in pairs if them product is $ n$, if $ n$ have odd factors so we will have a factor $ p$ that dont have a pair, then the vaule of $ \frac {n}{p}$ can only bem $ p$, so $ n$ have to be a perfect square
[/hide]

indeed this is very simple.

[hide="dont click here"]
because all square numbers have an odd number of factors, for any real number (well, i think imaginary numbers would work too) [i]n[/n], n^2 will have an odd number of factors, with n being the lonely factor without a "partner", so to speak.
[/hide]

[hide="The Answer"]
Textangle asked for all whole numbers with an odd number of factors, so the answer is:

$ \{x| x \text{is a perfect square}.\}$
[/hide]

EDIT: @fishy: Gah, I can't believe I did that! Where on earth did my brain go? Thanks, fishy.  :)

$ \{x \ | \ x\text{ is a perfect square}\}$ (Latex cleaned up, izzy :) )

Or $ \{x^2 \ | x\in\mathbb{R}\}$

@fishythefish: did you mean $ \{ x^2 |x\in \mathbb{N}\}$?

5849206328x is right...I think it has to be a subset of all NATURAL numbers...not real.
Real is too generic.

Only the numbers that are perfect squares have an odd number of factors and that is because the square root of those numbers don't have a different number that is the quotient.

Oops. Actually, I meant $ x\in\mathbb{I}$

This is assuming that factors cannot be imaginary.

[quote="Wickedestjr"]Obviously when Textangle puts a question in Classroom Math titled 'Very Simple Problem :)' he is not intending for it to involve imaginaries.[/quote]
I think it's Natural...then...

Yeah, usually the broadest group of numbers that middle schoolers use are rational numbers. Except for pi.

so 5849206328x was right
okay so it makes sense now =]

Did you notice some posts can get so off topic? (like this one)


geometry
circumcircle
parallelogram
perpendicular bisector
geometry proposed

Let convex quadrilateral $ ABCD$ be inscribed in a circle centers at $ O.$ The opposite sides $ BA,CD$ meet at $ H$, the diagonals $ AC,BD$ meet at $ G.$ Let $ O_{1},O_{2}$ be the circumcenters of triangles $ AGD,BGC.$ $ O_{1}O_{2}$ intersects $ OG$ at $ N.$ The line $ HG$ cuts the circumcircles of triangles $ AGD,BGC$ at $ P,Q$, respectively. Denote by $ M$ the midpoint of $ PQ.$ Prove that $ NO \equal{} NM.$

It is easy to prove that O1OO2G is a parallelogram
So GN=NO
If we can prove OM⊥PQ then the problem is solved  =>  all we have to do is to prove M is the midpoint of XY => we need XP = YQ

∠YCB+∠XAD=∠YGB=∠XGD=∠PAD=∠XAD+∠XAP  =>  ∠YCB=∠XAP   =>  ∠XAD=∠QCY
PX/sin∠PAX=AP/sin∠AXY
PX=AP/sin∠AXY* sin∠PAX=AP/sin∠GBY*sin∠YCB=AP/GY*BY
YQ/sin∠QCY=YC/sin∠YQC
YQ= YC/sin∠YQC* sin∠QCY=YC/sin∠CAD*sin∠XAD=YC/CD*XD
AP/GY*BY=AP/GD*XD     YC/CD*XD
PX/YQ=(AP/GD)/ (YC/CD)
But we have AP/GD= sin∠PGA/ sin∠CAD      YC/CD= sin∠YDC/ sin∠CYD
∠CAD=∠CYD    ∠PGA=∠YDC=∠YBC
=>    PX=YQ
=>  M is the midpoint of XY  => OM⊥PQ
DONE!

I think the idea for this problem isn’t new. We can solve this problem by using a well know lemma (It’s similar a problem in IMO 1977): Let (O) and (O’) have common AB, a line throughs A, intersect two circles resepective at M, N. We have MN always pass a fixed point T, T is midpoint of CD ( C in (O), D in (O’) and CD is perpendicular to AB at B)

I think the idea for this problem isn’t new. We can solve this problem by using a well know lemma (It’s similar a problem in IMO 1977): Let $ (O)$ and $ (O^')$ have common $ AB$, a line throughs $ A$, intersect two circles resepective at $ M, N$. We have $ MN$ always pass a fixed point $ T$, $ T$ is midpoint of $ CD$ ( $ C$ in $ (O)$, $ D$ in $ (O^')$ and $ CD$ is perpendicular to $ AB$ at $ B$)

Can't we draw $ {XN }$ ⊥$ {QP}$ then try to prove $ {GX\equal{}MX}$ $ {\rightarrow}$ $ {MO\parallel XN}$ to get it ?Dear Mathlinkers ?
By this way I find that :when $ {H}$ is not on $ {AB  \cap CD}$  we also have :$ {MN\equal{}ON}$[See the figure ,I think the conclusion is right   :maybe: ]

By angle chasing we have ∠P’AX’=∠Y’CB
P’X’=AP’*sin∠P’AX’/ sin∠AX’Y’=AP’* sin∠Y’CB/ sin∠AX’Y’=AP’*BY’/AY’
Y’Q’=CY’* sin∠Y’CQ’/ sin∠Y’Q’C=CY’* sin∠X’AD/ sin∠DBC=CY’*X’D/CD
P’X’/Y’Q’= (AP’*BY’/AY’) / (CY’*X’D/CD)
Notice that ⊿X’CD∽⊿Q’CY’ =>  CD/CY’=CX’/CQ’  =>
P’X’/Y’Q’=( AP’ *BY’* CX’)/( CQ’*X’D*AY’)
Again notice ⊿BCQ’∽⊿DAP’ => AP’/CQ’=AD/BC  => P’X’/Y’Q’=(AD*BY’* CX’)/( BC *X’D*AY’)
It is well-known if BD,AC and X’Y’ are concurrent in a circle , then we have
AD*BY’* CX’= BC *X’D*AY’  (It can be proved by sin law)
Thus we acquire P’X’=Y’Q’=> M’is the midpoint of X’Y’ => OM’⊥ X’Y’
From the above , we have O1OO2G is a parallelogram  So GN=NO
Which means M’N=NO

we can apply this result (these points are different from those of the problem):
let $ XYZW$ a parallelogram and let $ l$ be a line through $ X$. consider the intersections $ P,Q$ of the circles of center $ Y$ and $ W$ of radius $ XY$ and $ XW$, resp, with $ l$. then $ Z$ lies on the perpendicular bisector of $ PQ$.
proof:
let $ l$ be the x-axis, $ X \equal{} (0,0)$, $ Y \equal{} (a,b)$, $ W \equal{} (c,d)$, then, $ P \equal{} (2a,0)$, $ Q \equal{} (2c,0)$ and $ Z \equal{} (a \plus{} c,b \plus{} d)$, and we're done...

applying the lemma to parallelogram $ OO_1GO_2$ and any line $ l$, it follows that $ OP \equal{} OQ$... this implies that $ OM$ is perpendicular to $ l$... this implies that triangle $ OMG$ is right-angled, and we conclude that $ ON \equal{} MN$...  :D

(there's no need for $ H$ to be on $ l$)

This well-known result came from [b]Russia 1995, Romanian TST 1996[/b] and others :
Let $ T$ be the second intersection of $ (O_1)$ and $ (O_2)$ then $ HO\perp GT$ and $ H,T,O$ are collinear.
Since $ OO_2\perp BC, O_1G\perp BC$ then $ OO_2//O_1G$, similarly we get $ GO_1OO_2$ is a parallelogram thus they intersect at the midpoint $ N$ of $ OG$.
We only need to show that the perpendicular bisector of $ PQ$ passes through $ O. (*)$
Let $ HO\cap (O_1)\equal{}\{J\}. \cap (O_2)\equal{}\{F\}. OO_2//GJ$ then $ OO_2$ is midline parallel to $ GJ$ of triangle $ GJF.$
It follows that $ O$ is the midpoint of $ JF$ and $ (*)$ is right.

As mentioned above $N$ is the midpoint of $OG$.

Let $(GAB),(GCD)$ intersect at $M'$, then since $AB,M'G,CD$ are concurrent, $M'$ lies on $HG$. It is well known that $\angle OM'G=90^{\circ}$, and from [url=http://www.artofproblemsolving.com/community/c6h105631p596309]here[/url], we see that $M'$ is the midpoint of $PQ\implies M\equiv M'$. Hence $N$ is circumcenter in right triangle $\triangle OMG\implies NO=NM$.


Prove that there exist uncountable subfamily of subsets of $ \mathbb N$ with property that the subsets are uncoparable by inclusion.

A stronger version: there exists an uncountable (in fact, of continuum cardinality) family of infinite subsets $ E_{\alpha}\subset\mathbb N$ such that $ E_{\alpha}\cap E_{\beta}$ is finite whenever $ \alpha\ne\beta$.
Idea of proof: replace $ \mathbb N$ by $ \mathbb Q$ and consider sequences of rationals converging to an irrational number.

I didn't make any progress. Can you write down the whole proof? Thank you.

Take a sequence for each irrational number. What can you say about the intersection of two sequences with different limits?

BTW: For a different construction, not involving any choices, take $ \mathbb Q\cap[\alpha,\alpha+1]$ for $ \alpha\in\mathbb R$. However, this does not satisfy mlok's stronger condition. owk

Intersections of sequences of rationals  with different irrational limits are finit. But I don't understand how that solves my problem.

My guess would be that you label the rationals $ \{r_{1}, r_{2}, r_{3}, \ldots \}$ in one of the standard ways. Now you can identify $ r_{i}$ with $ i$ which is in $ \mathbb{N}$.

The sci.math thread [url=http://groups.google.com/group/sci.math/browse_thread/thread/366b294aae1c414/9ae0edcfe0e387c7]uncountable family of subsets with finite intersection[/url] (four years ago) contains other solutions as well.

I assume that this propsition is true for any omega-sequence under ZFC.  However, can it still be proved using ZF or ZF+AD?
I'm not sure.


Nice solution, arqady! I solved it by using Cauchy-Schwarz.

[quote="Lovasz"]Let $x,y,z \geq 0$ such that $x+y+z=1$. Prove that:
$\left(x \over 1+x\right)^{2}+\left(y \over 1+y\right)^{2}+\left(z \over 1+z\right)^{2}\geq{3 \over 16}$.
[/quote]
$\sum\left(x \over 1+x\right)^{2}\geq \frac 1{12}\sum\frac 1{1+x}\geq{3 \over 16}$.

[quote="arqady"][quote="Lovasz"]Let $a,b,c \geq 0$ such that $a+b+c=1$. Prove that:
\[\left(a \over 1+a\right)^{2}+\left(b \over 1+b\right)^{2}+\left(c \over 1+c\right)^{2}\geq{3 \over 16}. \]
[/quote]
$\left(a \over 1+a\right)^{2}+\left(b \over 1+b\right)^{2}+\left(c \over 1+c\right)^{2}\geq{3 \over 16}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(\frac{a^{2}}{(1+a)^{2}}-\frac{1}{16}-\frac{9}{32}\cdot\left(a-\frac{1}{3}\right)\right)\geq0\Leftrightarrow\sum_{cyc}\frac{(3a-1)^{2}(1-a)}{(1+a)^{2}}\geq0.$ :)[/quote]

For godsake how did you come up with this? :read:

[quote="Lovasz"]Why did you come to the first part?[/quote]
[hide="The first part is also nice and easy."]
Let $f(x)=\left(\frac{x}{1+x}\right)^{2}-\frac{1}{12(1+x)}$, then $f''(x)=\frac{11-25x}{6(1+x)^{4}}$.

$f(x)$ is convex on  $\left[0,\frac{11}{25}\right]$ and  convave on $\left[\frac{11}{25},1\right]$.

Assuming $x \le y \le z$. By the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64933]Half concave and half convex theorem[/url], we only to prove

$f\left(\frac{1-z}{2}\right)+f\left(\frac{1-z}{2}\right)+f(z)=\frac{(4z^{2}-5z+3)(3z-1)^{2}}{12(z-3)^{2}(z+1)^{2}}\geq0$.
[/hide]


calculus
integration
greatest common divisor
algebra
difference of squares
special factorizations

What is the largest number by which the expression n^3 - n is divisible for all possible integral values of n.

it's num. 157 in aops v1....I don't understand the explanation though: why does one of the factors, after gcf and difference of squares, have to be divisible by 3?...can anyone explain the whole problem and that specific part thoroughly?

the answer is 6 btw

I don't think I understand the problem. Are you asking for the largest number such that $ n^3\minus{}n$ is divisible by $ n$? And the only thing I can see is that if you factor $ n^3\minus{}n$ to $ n(n^2\minus{}1)$ and then this factors even more nicely into $ n(n\minus{}1)(n\plus{}1)$. Then from here you can see that you have 3 consecutive numbers, so one of them has to be divisible by $ 3$. Also you know that at least one of these integers are even. So I see the $ 6$ part.

i think it's asking for the largest divisor of n^3 - n no matter what integer value of n you put into n^3 - n

yea it's obvious one of the factors is even, so 2 is a divisor...but if there are three consec integers why does one of them have to be divisible by 3?

Find the largest integer $ k$ such that $ \forall n, \; k | n^3 \minus{} n$.

In a basic sense, you have a multiple of 3 every 3 integers. There's a multiple of three in any 3 consecutive integers for the same reason why there is "obviously" a multiple of 2 in any two consecutive integers. Here are the first few sets of three consecutive positive integers:
1 2 [color=blue][b]3[/b][/color]
2 [color=blue][b]3[/b][/color] 4
[color=blue][b]3[/b][/color] 4 5
4 5 [color=blue][b]6[/b][/color]
5 [color=blue][b]6[/b][/color] 7
...

oh ok that makes sense...never thought of it that way - thanks

so is the answer [b]6[/b] because for any 3 consecutive numbers (n-1), n, and (n+1) the only thing you can determine for sure is that there is, at least, one multiple of 2 and one multiple of 3 among them, so when those numbers are mutliplied together, 6, at the least, is a divisor?


Hi,
Show that if S is a convex set which is bounded and open then S=(m,M) where m=Min(S) and M=MAX(S).
Cheers

Thanks for your reply .Yes S is a subset R.


So we have yet another exciting news for you guys: you can post pictures created with GeoGebra on the Forum and the Wiki now.

It's very simple to use: click on the GeoGebra button (it's the last to the right in both edit boxes, Forum and wiki). A popup window will appear. You can draw a picture there (it's pretty obvious how to do that, as it's a WYSIWYG geometry program).

When you are done, click on the "Insert Picture" button (the big one). Wait a couple of seconds until the picture is processed on the server. A new popup will apear with two options: OK to close the edit window, and come back to the message box (with the geogebra code attached that looks something like below
[code][geogebra]8ff8e58406cb897e651a8311e8a4ff74c33cd053[/geogebra][/code] or click Cancel if you are not finished drawing and you pressed the Insert Picture by mistake.

Then you submit the message and a button like the one below appears:
[geogebra]8ff8e58406cb897e651a8311e8a4ff74c33cd053[/geogebra]

Click on it, and you will see the drawing (you can pull the vertices away, and do all sorts of neat stuff with it).

Disclaimers: you will need Javascript and Java enabled on your browsers in order for this feature to work. It has been tested on Firefox 3.0, IE 7, Google Chrome and Opera 9.5. I know it's not yet working properly on Firefox 2.0 (and I strongly recommend you upgrade to 3.0 anyway) and I have not even tried to see if it works on IE 6 (I recommend all IE 6 users to upgrade quickly to IE 7!).

Please post any bugs you notice right now in this thread. [b]DO NOT TEST IN THIS THREAD[/b]. Please post your Browser and Operating system. Also note that if you don't run FF 3.0, Opera 9.5, IE 7, Google Chrome or Safari 3.1 I won't be able to help you! Upgrade your browser!

The only minor bug I've noticed is that the mouse-over text for the Geogebra button is empty, so when you mouse over it just says whatever it said before. I'm talking about the text that shows up right above the text edit box, which says things like "[youtube]YouTube[/youtube]" when you mouse over the YouTube button.

(Browser: Google Chrome, OS: WinXP sp2 but I don't think it matters much).

Feature request: would be nice to be able to use [geogebra="Text"] to alter the default "Click to view applet" text.

This is a really, really great feature. Way to go guys.

Sorry there won't be any text. I have something else planned, but more on that later.

Awesome. Now I don't have to buy the Geometer's Sketchpad...

One thing I found missing, though: altitudes. I can construct these with semicircles, which was tedious. It sounds pretty simple as a command: you need a point and a line, and if it could show the right angle that would be great.

Can't wait to get to use this feature...

Awesome man just marvelous. It would have been a pain to buy Geometer Sketchpad!

The thing about this tool is not that it would somehow replace Geometer's Sketchpad but the fact that you just need a browser to use it. Log in and post your solution using nice drawings. No need to learn Asymptote anymore, just point and click.

Of course we will be still supporting Asymptote on the site in parallel with Geogebra.

[quote="Valentin Vornicu"] No need to learn Asymptote anymore, just point and click.[/quote]
Hmmm... Asymptote allows you much, much more (if used with some skill, of course). Still, geogebra is a nice addition to the site too. Thanks, Valentin! :)

[quote="#H34N1"]Thanks a lot ===) Now I don't have to use paint, upload to imageshack and quote image here :)[/quote]
lol:
Have you ever heard about the other soft like GeoNext or Assymptote?
But uploading to imageshack and learning Asymptote do really annoy,so Thanks a lot,Valentin...
P.S:It reminds GeoNext a bit...Java is pretty good in creating a such helpful geometry soft.

I believe Asymptote is already installed.
I looked this up and the full version seemed to have a litter more capabilities?
Yes Asymptote is much more accurate and powerful than this, but I think this tool is great and has it's uses, like when you can't figure out an asymptote, just sketch one with this!

Great tool!!
~Dojo

SamE: Are you trying to construct altitudes?  The fourth button in does that.

The full version of Geogebra does mark little right angles.  We'll see about adding that to the Forum embedded version of Geogebra.

I should add a special thanks to Markus Hohenwarter for creating Geogebra, which you can download at [url]http://www.geogebra.org[/url].  I'd also like to thank Professor Steve Dunbar of the AMC for telling me about the flexibility of Geogebra a couple months ago.

The geogebra tool will improve soon (maybe show a thumbnail of the image instead of a "click me button", and other stuff in the future.)

Big problem =(

Tested on Mozilla Firefox 3.1, Vista x32.  Result: [color=red]Did not work.[/color]
Tested on Google Chrome, Vista x32.  Result: [color=red]Did not work.[/color]
Tested on IE8, Vista x32.  Result: [color=green]Did work,[/color] but I don't like IE.

Java Version 1.7 alpha.  Reinstalled to 1.6_7, still did not work.

I told you I do not support those browsers ... a for chrome it works, but you haven't installed the correct java version. It's a chrome problem (as well as many MANY MANY other chrome bugs - I would take IE 6 over chrome any day of the week right now).

This does work on IE6 (at least you can view  the applet)

[quote="SamE"]altitudes[/quote]

Draw a circle with center at one of the vertices tangent to the opposite side. Connect the vertex and the point of tangency. It might not look right, but it's right.

He already said he constructed them, but didn't want to have to.

Anyway I was fooling around with it and it sometimes doesn't let you label everything...It's sort of sporadic. In the same picture, it lets you label (and otherwise affect) certain points, etc. but not others. XP,
Firefox 3.0.

I can't do anything about the java applet itself, if there are bugs inside it. If you can see it and draw a couple of points, my part of the deal is done.

you can fix that by first drawing points then lines.

No. It's <geogebra> .... </geogebra>. There is a button there (it has the geogebra logo on it, it's the first one I think right now).

it works on ff3 but not on opera 9.5  :(

hey man that is really awesome ( i mean geogabra)

Is there any reason why the Forum version of Geogebra is so, erm..., "truncated"? The full version allows you to define your tools (I just defined incenter and incircle to test this option and it was rather cute) and shows a command line that is often needed if you want to work with more complicated objects than just lines, points, and circles. If there is a compelling reason for this severe truncation, we should think a bit about how to mitigate its effects. There are two obvious options: the first one is to create a wide assortment of standard tools (i.e., start a special thread where the requests will be made and let the people Valentin trusts do the "programming"; it seems like all we need is to create a "template" file which has all necessary tools but no picture and use it at the applet start instead of the default one), which is completely safe and, I think, will be appreciated by many; the second one is to allow uploading geogebra files, which may pose some security problems because, unlike Asymptote code, the geogebra files cannot be easily checked for threats, but I am not an expert here.

Interesting idea.  We'll look into it.  We definitely stripped down a bit to get *something* working to see if we could.  But this does look like an interesting extension if we can get it to go.

[quote="archimedes1"]This is great!  I especially like that viewers can manipulate the diagrams.

one thing: I use IE7 on XP.  After spending some time making a fancy diagram, I pressed Insert Picture and pressed the button that closes the window and that usually puts the picture in the post.  Then nothing happened at all.

Also, accidentally pressing Reset Picture is annoying.[/quote]

Also using IE7 on XP: pressing Insert Picture does nothing at all.

The IE 7 bug is fixed. I will upload the new version files in a couple of minutes.

A little upgrade: a few menu options were added. Also if you right click you will have the option of show all objects if you somehow zoom in/out too much.

Also double-clicking on the applet (in the EDIT window) will create a full-fledged version of the applet (so you can open your own custom tools for example, or have the input algebra window for those coders that can't work with point and click like the rest of us :) ).

Also use refresh (F5) in your browser window to get the new feature enabled: reset now has a confirmation window in the editing box, so accidental pressed should be avoided ;).

[quote="Valentin Vornicu"]
Also double-clicking on the applet (in the EDIT window) will create a full-fledged version of the applet [/quote]
Thanks a lot! :) Now, if one decides to work in the full version mode, then, after finishing the picture one has to go to the File menu, choose "Close file" and the picture will be automatically transferred to the edit window, right? (at least, this works, but I'm not sure it is the way it was intended to be :))

You can also click the big X on the top-right corner  :P  :icecream:

Hmm...
Geogebra worked for me before.
Now all it says is Applet not found.
Opera 9.5, XP.

The first time I tried geogebra I got a blank screen, but I tried again and it's pretty awesome for an applet! I remember when I took the AMC 12 class here it was kind of hard to do the geometry problems posted in the class forum, cause one had to explain everything in words. A bit difficult to keep track of everything. This will definitely make things a lot easier. Thanks for the new feature.

GEOGEBRA is AMAZING!!!  :lol: thanks Valentin and all the ones who were involved in this!!! I guess this was the thing that I was looking for a long time to have more fun with in MathLinks!!! Actually we were having fun even before this but now its more FUN!!! :lol:
Thank You Guys Very Much, GEOGEBRA is really Cute! :)

New tools added to Geogebra. Thanks goes to fedja for implementing them and actually taking the time to draw little cute images for them. Enjoy! :D

Geogebra is awesome, but if I use the RHS commands, it says Unknown command: "insert command here". Would you look into this please? TYiA

RHS meaning the additional tools? If you use them from the command line, type everything in lowercase letters (standard GeoGebra tools understand both formats but I didn't provide for this option with additional tools and, since I normally type in lowercase,...); if it is really irritating, we can easily fix that :). The click mode should work fine.

The geogebra work too slow and sometimes more than I espect

It's a java-based applet. If you're computer is really old, then yes it will work slow. But that's why you have the option of not loading it ;)

There is big button saying Insert Picture :) Then press OK at the promt and voila the picture is inserted:

 [geogebra]3b7cdb5780d458896229aae0553695c804df7ecb[/geogebra]

some one help me !!!
everytime i draw figure and click insert picture the whole thing just closes!! i dont kno wat the reason is
i hope someone can help


My physical science textbook has this article on the fifth force of nature.

[quote="Science Gazette, Prentice Hall Physical Science Textbook"] Until recently, practically all scientists would have said that there is no force that works against gravity. Now, however, some scientists are not so sure. In fact, these scientists have even suggested the existence of such a force. [/quote]

The article goes on to say things about a force called "hypercharge," a possible 5th force that is very weak.

This theory was proposed by:

Ephrain Fischbach of the University of Washington
Daniel Sudarsky, Aaron Szafer, and Carrick Talmadge of Purdue University
and Samuel Aronson of Brookhaven National Laboratory

Most physicists believe in the existence of only 4 basic forces of nature, and hypercharge isn't one of them. The four basic forces are gravity, weak nuclear force, strong nuclear force, and electromagnetism.

According to Dr Fischbach, hypercharge is a relatively weak force that works against gravity when objects are within about 200 meters of each other.

Do you think that there is a fifth force in nature?

I'm pretty sure I have no information to support an informed choice.

[quote="wikipedia"]Nonetheless, in the late 1980s a fifth force, operating on municipal scales (i.e. with a range of about 100 meters), was reported by researchers (Fishbach et al.) who were reanalyzing results of Loránd Eötvös from earlier in the century. The force was believed to be linked with hypercharge. Over a number of years, other experiments have failed to duplicate this result, and physicists now believe that there is no evidence for a fifth force.[/quote]

[url=http://en.wikipedia.org/wiki/Fifth_force]en.wikipedia.org/wiki/Fifth_force[/url]

Also, if there was such a force, surely there would be evidence of it within it in the Standard Model. As that has yet to be adapted, and most particle physicists revere its ability to explain the forces of nature ;) , I'm inclined to think it doesn't exist.

[quote="mysmartmouth"][u]Fifth Force[/u] would be an excellent movie title.[/quote]

Similar to "The Sixth Sense." I like it  :)