chesspro [color=green]confirmed[/color]
MathWhIz2004 [color=green]confirmed[/color]
-Terminator- [color=green]confirmed[/color]
WindSlicer [color=green]confirmed[/color]
GoBraves [color=green]confirmed[/color]
MithsApprentice [color=green]confirmed[/color]
jmadsen [color=green]confirmed[/color]
ricebelly
Sudonym
LynelleYe [color=green]confirmed[/color]
Lucky707 [color=green]confirmed[/color]
bubala [color=green]confirmed[/color]
Chigr [color=green]confirmed[/color]
h_s_potter2002 [color=green]confirmed[/color]
Sly Si [color=green]confirmed[/color]
paladin8 [color=green]confirmed[/color]

16 people, 4 mafia, 2 inspector, 1 doctor, 9 citizens. The night will begin. The PM's will be sent out in about 5-10 minutes, so hold tight. Choose your victim, Mafia!

I forgot to  say... the game won't start until i get the responses.

Ooh, hello everybody!

Finally the time has come. :)

lol ... we'll try not to lynch you (first)

[quote="Chigr"]im alive, unless sum1 didnt tell me i already got lynched[/quote]

well the game officially doesnt start until i get everyone's response back, telling me that they will be playing. When that is all done, i will simply post, "the game begins." Since the night comes first, though, i dunno what will happen.

I think they can choose a person to save during the night. And if the mafia happens to choose that exact same person - the person doesn't die but stays alive, therefore preventing the mafia from succeeding in a killing.

Sodonym and MithsApprentice are still yet to confirm (i think) and where is our host?

er yeah...I'm good. Okay let's go whenever we're ready.

uglyman621 hasnt replied either....this is taking way too long..can we just start?

hmm..... well let's make this short. chesspro is dead.

please accuse someone.

Darn... I died. At least it was fast.

I know 3 of the 4 mafia...

[EDIT]: I'm a citizen, btw.

[quote="chesspro"]I know 3 of the 4 mafia...[/quote]
Err...don't we all?
So we should get either mathwhiz or jmadsen. I don't know who's more likely but if one's innocent the other is probably mafia. We have a fair amount of room for error at the moment which is good.

I [color=red]accuse sly si[/color]. He's definitely mafia. Unless LynelleYe is. I'm inspector. I know Chigr, mathwhiz, and jmadsen are all innocent. Therefore either sly si or lynelleye is mafia. But I think it's more likely to be sly si.

i [color=red]second sly si[/color] and if its not him then we can deal with that tomorow

[quote="xxreddevilzxx"]Sly Si on trial!!!

Chigr [color=green]aye[/color]
chesspro
lynelleye
sly si
mathwhiz
jmadsen
paladin8 [color=green]aye[/color]
mithsapprentice

need three more votes to kill![/quote]

Lol, reddevil. I'm dead already. I can't vote.

Sly Si, I am (was?) a citizen.

LynelleYe, I know everyone knows 3 of the 4 mafia. I just wanted to say that for no reason.

As the other inspector, I confirm that what paladin8 said is true. I vote [color=red]aye[/color] on sly si.

[quote="Sly Si"]Well, I vote NAY, naturally.  I'm an ordinary citizen, which means either paladin isn't inspector, or LynnelleYe is mafia.

Is the other inspector still around to confirm or refute paladin's statement?  Or was chesspro the other inspector?

Anyway, I hate to leave in the middle of my trial, but I leave town in about half an hour.  So this will be my last post for this game.  If you don't kill me, I'll need to be replaced, as I will have no internet access for the next week.

Bye, and good luck![/quote]

Well we aren't SURE you are mafia; there's just a high chance. Anyway, later man.

Actually, since chesspro was dead, that will do......and the citizens win!!!!!!!

Good job you guys.

Glad I could help out by saving a few of you :)

Awesome! Now to wait for game 9 to start...


integration
calculus
logarithms
calculus computations

1. Factor the denominator: $x^{4}+1=(x^{2}+\sqrt{2}x+1)(x^{2}-\sqrt{2}x+1).$

2. Use partial fractions.

3. Be very patient, and have a substantial quantity of scratch paper available.

And I think we should consider a generalization of this problem:

Find all $\alpha \in \mathbb{R}$ such that the following integration can be solved:

$\int \frac{dx}{1+x^{\alpha}}$

[quote="marshell"]Find all $\alpha \in \mathbb{R}$ such that the following integration can be solved:
$\int \frac{dx}{1+x^{\alpha}}$[/quote]
Is $x\,\,{}_{2}F_{1}(1/\alpha,1,1+1/\alpha,-x^{\alpha})$ acceptable?


MATHCOUNTS
modular arithmetic
number theory
least common multiple

I was doing Mathcounts Chapter Sprint 2002, and could understand how to solve all questions but this one.

28. The length of a year on planet Mars is exactly 697 days. If Mars has a calendar with a 12-day week, and year 0 begins on the first day of the week, what is the next year that will begin on the first day of the week?

Thanks in advance.

[hide]Since $ 697\equiv 1 \pmod {12}$, the first day of every year will be the day after the day that the previous year started on.
So, the pattern is like this.
Year 0-Day 1
Year 1-Day 2
Year 2-Day 3
etc.
So, the year that you will get the 12th day of the week first is Year 11. The next year, Year 12, will be the year where the first day of the week is the first day of the year, so the answer is Year 12. [/hide]

I believe you need to find the lcm.  Let $ n$ be a non-negative integer.  Then $ 697n$ represents the number of days that have passed after $ n$ years.  In order for a year to start on the first day of the week, some multiple of 12 days must have passed.  So find the lcm, which is $ 8364$.  This gives a value of $ n\equal{}12$, which would mean 12 years.

697/12= blah remainder 1
Every year the day the year starts off with will be moved forward one. Since there are 12 days in a week, that takes 12 years until you can come back on the same day. The year 12 would be the answer


geometry rectangle integration vector calculus function LaTeX

If $ S_1$ is a rectangle with area 1 and $ S_2$ is a rectangle with area 2, then

$ 2 \int\limits_{S_1} \vec{F} \cdot d\vec{A} \equal{} \int\limits_{S_2} \vec{F} \cdot d\vec{A}$

why is this not true?

The orientation of the rectangles in the vector field matters.

The flux through a rectangle is defined as the integral over the region in question of the DOT PRODUCT of F and dA.

You're using a normal product, and so you miss the orientation aspect.

[quote="t0rajir0u"]What does the "normal" product of two vector-valued functions even mean?  I think it's reasonably clear that the OP meant dot products and just didn't know how to $ \text{\LaTeX}$ them.

Anyway, I don't understand the question.  If $ \vec{F}$ is nonconstant then there's no reason the two integrals should be related at all.[/quote]

I've modified the original question so it clearly says a dot product, hope it makes more sense


Just like Volumes 1 and 2 prepare students for like AMCs, AIMEs, is there a book that prepares students for USAMO? Is this book the Olympiad-equivalent of Volumes 1 and 2?

http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=5003

I'm not sure if I can take WOOT, and if not, I'd like to have a kinda textbook-like book.

Also, if anybody has this book, could you give some examples of how this book is like?

There is no such book. Mainly because it would be almost impossible to make one. The topics for USAMO/IMO are so widely diversificated that not even a collection of books would be enough to thoroughly cover them. The best you can do is check the Olympiad forums and the resources section (contests) for problems and try to solve them.

Problem solving strategies by Engel is close though


A $ 12^{\prime \prime}$-diameter pizza and a $ 16^{\prime \prime}$-diameter pizza are each cut into eight congruent slices. Jane ate three slices of the $ 12^{\prime \prime}$ pizza. Mark ate three slices of the $ 16^{\prime \prime}$ pizza. How many more square inches of pizza than Jane did Mark eat? Express your answer as a common fraction in terms of $ \pi$.

So in this problem, we want to find $ \frac{3}{8} \cdot 256\pi \minus{} \frac{3}{8} \cdot 144\pi$

We can easily use the distributive property:

$ \frac{3\pi}{8}(256 \minus{} 144) \equal{} \frac{3\pi}{8}(112) \equal{} \boxed{42\pi}$.

Actually, it says diameters of 12 and 16, so your answer is off by a factor of 4. The real answer should be 10.5pi.

No, it shouldn't be $ \frac{21}{2}\pi$; $ 42\pi$ is correct. Why do you need to divide by 4 anyways?

Because the question gave diameters, not radii. Therefore, the radii, of each pizza are $ 6$ and $ 8$, respectively. Hence,

$ \frac{3\pi}{8}(8^2\minus{}6^2)\equal{}\frac{3\pi}{8}(28)\equal{}\boxed{10.5\pi}$.

Alcumus solution:They both ate $\frac{3}{8}$ of a pizza. Therefore, the quantity that Mark ate in excess of Jane is simply $\frac{3}{8} \times$ the difference in total area of the pizzas. The 16'' pizza has $64\pi$ area, and the 12'' pizza has $36\pi$ area, making for a difference of $28\pi$.  $\frac{3}{8} \times 28\pi = \boxed{\frac{21}{2}\pi}$


[quote="randomdragoon"]There are two kinds of people: Those who like to group people into two different groups, and those that do not.[/quote]
It's not the solution  :D
The solution is veeeeery simple  :D

[quote="mz94"]obvios: He cant count[/quote]

obvious: you have already seen my enigma somewhere  :wink:

[quote="BanishedTraitor"]There are $ 10$ types of people, those who can count in binary and those who can't.

$ 4$ out of $ 3$ people can't do fractions.[/quote]
mz94's answer is right  :wink:

[quote="miyomiyo"]Um,
(1) They weren't posting solutions, they were posting things like the one you said.
(2) Your engima is obvious because it references the reason that he can't count in the enigma.

Stop trying to turn bumper stickers into AoPS threads.[/quote]

[quote="i_like_pie"]Quoted for truth.[/quote]

though thisi s G&FF, so I suppose a certain amount of leniency should be exercised.

Lol, I remember when my older brother told me the binary one when I was in like 1st grade. I'm all like, what, but that's only 2 people!

[quote="FOURRIER"]The Enigma Is Simple  :wink: :

Some one said:
[quote="X"]
There are [b][u][color=red]3[/color][/u][/b] kinds of people : people who [b]know how to count[/b]  and people who [b]does not[/b]  .
[/quote]
Why did he said that there are 3  :gathering:  kinds, and he just listed 2 :ddr:[/quote]

He used bad grammer: is should have been

"There are [b][u][color=red]3[/color][/u][/b] kinds of people : people who [b]know how to count[/b]  and the person who [b]does not[/b]  ."

It's obvious from there.

[quote="1=2"]He used bad grammer[/quote]

He should have said "There are 3 kinds of people: those who know how to count and those who don't."

[quote="Nerd_of_the_Ages"]Btw, who's rating all our posts one?[/quote]

It doesn't really matter here, at least.  :)

[quote="FOURRIER"]The Enigma Is Simple  :wink: :

Some one said:
[quote="X"]
There are [b][u][color=red]3[/color][/u][/b] kinds of people : people who [b]know how to count[/b]  and people who [b][u]does[/u] not[/b]  .
[/quote]
Why did he [u]said[/u] that there are 3  :gathering:  kinds, and [u]he just[/u] list[u]ed[/u] 2 :ddr:[/quote]

1. He cant count
2. I underlined the places with spelling/grammer mistakes
3. "There are 10 kinds of people in the world, the ones who understand binary and the ones who use base 9" is better


geometry
trigonometry
inequalities open
inequalities

Let $ \triangle ABC$ be a triangle with area $ \frac{1}{2}$ in which $ a$ is the length of the side opposite to vertex $ A$. Prove that:

$ a^{2}\plus{}\csc (A) \geq \sqrt {5}$

[quote="delegat"]Let $ \triangle ABC$ be a triangle with area $ \frac {1}{2}$ in which $ a$ is the length of the side opposite to vertex $ A$. Prove that:

$ a^{2} \plus{} \csc (A) \geq \sqrt {5}$[/quote]
do you mean $ cos(A)$  :?:  :|

No  $ \csc (A) \equal{} \frac{1}{\sin (A)}$

[quote="delegat"]Let $ \triangle ABC$ be a triangle with area $ \frac {1}{2}$ in which $ a$ is the length of the side opposite to vertex $ A$. Prove that:

$ a^{2} \plus{} \csc (A) \geq \sqrt {5}$[/quote]

sorry i am late i had some problemes with this old pC  :oops:

we know that $ S(ABC) \equal{} \frac {1}{2}bc*sin(A) \equal{} \frac {1}{2}$ so $ bc \equal{} csc(A)$

let prouve :$ a^2 \plus{} bc > \sqrt(5)$

using some triangle formulas $ a^2 \equal{} b^2 \plus{} c^2 \minus{} 2bc*cos(A)$

and $ cos(A) \equal{} \sqrt(1 \minus{} (sin(A)^2))$  so $ a^2 \equal{} b^2 \plus{} c^2 \minus{} 2bc*(\sqrt(1 \minus{} \frac {1}{(bc)^2}))$

$ a^2 \plus{} bc \equal{} b^2 \plus{} c^2 \minus{} 2\sqrt((bc)^2 \minus{} 1) \plus{} bc$

and we know tha $ b^2 \plus{} c^2 > 2bc$  so $ a^2 \plus{} bc > 3bc \minus{} 2\sqrt((bc)^2 \minus{} 1)$

the inq <=> $ 3bc \minus{} 2\sqrt((bc)^2 \minus{} 1) > \sqrt(5)$  (NB:$ bc > 1$)

supposing $ bc \equal{} x$ $ f(x) \equal{} 3x \minus{} 2\sqrt(x^2 \minus{} 1) \minus{} \sqrt(5)$

$ f(x) > 0$ <=>  $ 5(x^2) \minus{} 6\sqrt(5)x \plus{} 9 \equal{} (\sqrt(5)x \minus{} 3)^2 >0$

so  $ a^2 \plus{} bc > \sqrt(5)$ that include $ a^{2} \plus{} \csc (A) \geq \sqrt {5}$

so what do you think   :wink:

[quote="delegat"]Let $ \triangle ABC$ be a triangle with area $ \frac {1}{2}$ in which $ a$ is the length of the side opposite to vertex $ A$. Prove that:

$ a^{2} + \csc (A) \geq \sqrt {5}$[/quote]
$ h_a = 1/a$;
1. if  $ h_a \le a/2$  $ ( a \ge \sqrt {2} )$ then : $ \frac {1}{sin(A)} \ge \frac {1}{Sin({\pi/2})} = 1$ , so for $ a \ge \sqrt {2}$ : $ a^{2} + \csc (A) \geq 3$;
2. if  $ h_a \ge a/2$  $ ( a \le \sqrt {2} )$ then : $ \frac {1}{sin(A)} \ge \frac {a^2}{4} + \frac {1}{a^2}$ , so for $ a \le \sqrt {2}$ : $ a^{2} + \csc (A) \geq a^2 + \frac {a^2}{4} + \frac {1}{a^2} = \frac {5}{4}a^2 + \frac {1}{a^2} \ge \sqrt {5}$;
equality is for $ \triangle$ with sides : $ a = \sqrt [4]{\frac {4}{5}}, b = c = \sqrt {\frac {3\sqrt {5}}{5}}$; :)

[quote="castigioni"]as "a" is dimensional and csc(A) is dimensionless the proposition is incorrect[/quote]

the proposition is incorrecrt !!!!!!!!!  :D  so show me something wrong in my solution  :evilgrin:

Thank you very much for these nice proofs Problem is of course correct and I do not understand this comment posted by [b]castigioni[/b]


Fie a,b,c reale pozitive...sa se demonstreze ineq:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{9}{(a+b+c)}\geq  4 (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})$

Se aplica inegalitatea lui Popoviciu, \[f(a)+f(b)+f(c)+3 f \left( \frac{a+b+c}3 \right) \geq 2 f \left( \frac{b+c}2 \right)+2 f \left( \frac{c+a}2 \right)+2 \left( \frac{a+b}2 \right) ,\] pentru functia convexa $f(x) = \frac1{x}$.

aaa...mijto teorema :| eu o facusem altfel mai "muncitoreste" :) dar mijto asta...:)

Putem presupune ca avem ordonarea $\ a\geq\ b\geq\ c$ sau $\ c\geq\ a\geq\ b$.Se stiu inegalitatile $\frac{1}{a}+\frac{1}{b}\geq\frac{4}{a+b}$ si $\frac{1}{c}+\frac{9}{a+b+c}\geq\frac{4}{b+c}+\frac{4}{c+a}$, ultima dintre ele fiind echivalenta cu $\ 2c((c-a)(c-b)-a^{2}+2ab-b^{2})+(a+b)(c-a)(c-b)\geq\ 0$, care e adevarata in ipotezele neristrictive de mai sus.Dar e mai usor cu Popoviciu...

[quote="stancioiu sorin"]$\frac{1}{c}+\frac{9}{a+b+c}\geq\frac{4}{b+c}+\frac{4}{c+a}$, ultima dintre ele fiind echivalenta cu

$\mathbf{2c((c-a)(c-b)-a^{2}+2ab-b^{2})+(a+b)(c-a)(c-b) \geq 0}$,

care e adevarata in ipotezele neristrictive de mai sus[/quote]

Poti sa explici mai bine de ce e adevarata inegalitatea in [b]bold[/b]?

Cred ca a uitat si el! Desi solutia cu Popoviciu este la indemana iata si o alta solutie!

  Problema.

   Fie $a, b, c$ trei numere reale strict pozitive. Aratati ca \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{9}{a+b+c}\geq 4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right).\] Solutie.

  Inmultind cu $a+b+c$ inegalitatea devine: \[\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}+9\geq 4\left(\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\right)\Leftrightarrow\]  \[\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}+12\geq 12+4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\Leftrightarrow\]  \[\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\geq4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right).\] Aasta ultima e usurica rau avand in vedere ca iese imediat din \[\frac{1}{x}+\frac{1}{y}\geq\frac{4}{x+y}.\]

Foarte frumoasa solutia ta Cezar  :) . Felicitari!

[quote="Cezar Lupu"]

\[ \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {a \plus{} b}{c}\geq4\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right).
\]
Aasta ultima e usurica rau avand in vedere ca iese imediat din
\[ \frac {1}{x} \plus{} \frac {1}{y}\geq\frac {4}{x \plus{} y}.
\]
[/quote]

Buna !

eu am inlocuit $ x\equal{}\frac{a}{b}$ si $ y\equal{}\frac{a}{c}$ si imi da ca $ \frac{b\plus{}c}{a}\geq\frac{4bc}{ab\plus{}ac}$ :oops:

Let $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}3$ .Prove that
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{2}{9}(bc+ca+ab)\geq\frac{13}{6} .$$


Έστω $T$ το σύνολο των τριγώνων ενός επιπέδου. Θεωρούμε τη συνάρτηση $f: T\rightarrow(0,+\infty)$ με τύπο
$f(ABC)=min\left(\frac{b}{a},\frac{c}{b}\right),$ για κάθε τρίγωνο ABC με $a\leq b\leq c.$ Να  βρείτε το πεδίο τιμών της συνάρτησης αυτής.

Λύση:

 Προφανώς min(β/α , γ/β)>=1.

 Από θεώρημα συνημιτόνων έχουμε α^2 = β^2 + γ^2 -2βγcosA =>

  (α/β)^2 = 1 + (γ/β)^2 - 2(γ/β)cosA. με γ/β>=1.

  Θέλουμε να βρούμε το μέγιστο της min(y,x) όπου y=β/α, x=γ/β.
  =>
   (1/y)^2 = 1 + x^2 - 2xcosA.      Η συνάρτηση g(x)=1 + x^2 - 2xcosA (για οποιοδήποτε Α,το Α δεν εξαρτάται από το
   x) είναι γν.αύξουσα (x>=1 και cosA<=1)).Eπίσης η (1/y)^2 είναι γν.φθίνουσα. Άρα το μέγιστο της min(x,y)
  επιτυγχάνεται όταν x=y , δηλαδή 1= x^2 + x^4 - 2(x^3)cosA. => cosA = (x^4 + x^2 - 1) /(2(x^3)) = f(x) είναι
γν. αύξουσα για x>=1 (πρώτη παράγωγος θετική) και f(x)<1 , λύνουμε την x^4 + x^2 -1 - 2x^3 = 0 =>

    (x^2 - x - 1)*(x^2 - x + 1)=0  => x0=(1+sqrt(5))/2 . Άρα το πεδίο τιμών είναι το [1,x0) . (Eίναι εύκολο να δείξει κανείς ότι θα πάρει και τις ενδιάμεσες τιμές μιας και η παραπάνω συναρτήσεις είναι συνεχείς...)


group theory
abstract algebra
superior algebra
superior algebra solved

since $Z \triangleleft Q$, then quotient group $Q/Z$ contains cosets $q+Z$ for $q \in Q$,if i have $(p+Z)(q+Z)=(pq)+Z$  for $p,q \in Q$ how would i show that this is not well defined,  do i just show that $(p+Z)(q+Z)\neq(pq)+Z$ ?

well-defined means that for different ways of writing the same cosets... the resulting product should come out to the same coset either way...

however, $(1/2+\mathbb{Z})(1/3+\mathbb{Z}) = 1/6+\mathbb{Z}$, whereas $(3/2+\mathbb{Z})(4/3+\mathbb{Z}) = 2+\mathbb{Z} = \mathbb{Z}$

(the coset $1/2+\mathbb{Z}$ is the same as the coset $3/2+\mathbb{Z}$ and the coset $1/3+\mathbb{Z}$ is the same as the coset $4/3+\mathbb{Z}$)

Of course, $\mathbb{Q}/\mathbb{Z}$ makes perfect sense as an additive structure.

It doesn't work under multiplication because $\mathbb{Z}$ is merely an additive subgroup and not an ideal.


compute the area of the largest square inscribed in a triangle which has sides $5,6,7$
[with proof]

WOw this problem is so computational heavy so Im not going to bother solving all those radicals.

The basic idea is to find the height and then use similar triangles (use the triangle above the square to the whole triangle)
The ratio should be

$\frac{h}{s}=\frac{h-d}{d}$

where the height you can find (h)
the side length (s) is 5,6,or 7 (you have to decide which gives biggest area) :wink:
and d is the side length of square .

you only hvae 1 variable (which is d) but the height is a messy radical so im not really going to do it..too lazy

EDIT: huh what do you mean prove (referring to bottom post)...isnt prove basically doing computational work since this isnt really a proof?

[quote="da ban man"]The basic idea is to find the height and then use similar triangles (use the triangle above the square to the whole triangle)[/quote]
Well, the main problem (once you see which side will give the largest square of the three) is to [b]prove[/b] that it's the largest square possible (if it is). :)

as he has got the idea i dont think that its hard to prove.... :wink:
[this can be done by showing that no area can be larger than this if two vertices of the square isnt on the base]

wait im confused..how cna you have an inscribed square without two of its vertices touching the base?

[quote="Moonmathpi496"]oh i forgot to say...when u use the idea of similar triange u'll find that all the 4 vertices are not on the sides of the triangle so u need to enlarge that and use the idea of limit....and a bit more computation.[/quote]

im kinda confused here...i thought it was inscribed so all 4 vertices must be on the sides of the triangle...?

Here u were asked to find the largest area so not necessary that all the vertices are on the sides of the triangle...
and note that:here inscribed means any square inside the triangle....[probably i mistook to write]

and about ur last post...if u check u'll find that all the vertices are not on the sides but u have to make the square which has 4 vertices on the sides if u use the fomula u mentioned

[quote]The basic idea is to find the height and then use similar triangles (use the triangle above the square to the whole triangle)
The ratio should be

\frac{h}{s}=\frac{h-d}{d}

where the height you can find (h) [/quote]

[quote="Moonmathpi496"]

and about ur last post...if u check u'll find that all the vertices are not on the sides but u have to make the square which has 4 vertices on the sides if u use the fomula u mentioned

[/quote]

yes tahts correct but I only mentioned that post because I thought inscribed means have to be touching the sides (which my way would work)..but I guess I was mistaken (line inscribed circle..)


you can see that   :
  http://www.mathlinks.ro/Forum/viewtopic.php?t=78481

http://www.mathlinks.ro/Forum/viewtopic.php?t=21405
http://www.mathlinks.ro/Forum/viewtopic.php?t=78481

  dg

b is not appearing on the RHS, so we can write :
LHS>(a^2)^(1/2)+(c^2)^(1/2)=a+c>(a^2+ac+c^2)^(1/2)
we just put that b=0 and that's it :D


What comes next in the following sequence:

15,23,18,6,9,24...

See, I can do it too!!  :P

[hide="Hint"]No math required. At all.[/hide]

Can people who do not live in your area understand the sequence?

And how about...

31, 41, 59, 26, 53, 58, 97, .... ? :D

My sequence is very simple. I can't believe only one person got it.


function
algebra
domain
real analysis
real analysis solved

I found this [url=http://www.math.columbia.edu/~zare/prob_all.html]here[/url]. Check out problem $22$. I think it can also be found in Yaglom, generalized a little bit (if I'm not mistaken, the problem there concerns continuous curves, not necessarily functions, but I can't remember the exact statement).

If $g(x)=g(x+c)$ for some $x$, call $c$ a chord of $g$. Which real numbers are chords of all continuous functions $f$ with domain [$0,1$] such that $f(0)=f(1)$?

Levi's theorem states that only the reciprocals of positive integers are solutions. This seems to be a classical problem given at oral de ENS and X.

Wow! I had no idea there was a theorem stating this. Indeed, these are the solutions. Once you realize this, a proof shouldn't be too difficult to find.

Indeed, the proof is not that difficult. Anyway, I do not find it absolutely trivial!  ;)

harazi, could you please give some more explanation either link concerning Levi's theorem-i haven't heard about this one


a) let $1\le a<2$ be a real number.
Prove that if $f: R^+\to R^+$ is a function satisfying the conditions:
i) $2f(x)\le af(\frac{x}{a})+x \forall x>0$.
ii) $f(x)<2005 \forall x<\frac{1}{2005}$.
then $f(x)\le x \forall x>0$
b) let $a>1$ prove that exists a function $f: R^+\to R^+$ satisfying condition (i) and $f(x)>x \forall x>0$.

I think the problem is nice. No one interested in proving it ? :?:

I think the problem is nice. No one is interested in proving it ? :?:

yes,it is nice. :)
for part a) if there exist $x_0$ such that $f(x_0) >x_0$
first we can prove $a>1$
because if $a=1$ then we have $f(x_0) \geq 2f(x_0) -x_0$ with i) ,contradiction.
let  $f(x_0) -x_0=b,b>0$
we have $f(\frac{x_0}{a}- \frac{x_0}{a}\ge \frac{2b}{a}$

$f(\frac{x_0}{a^2} - \frac{x_0}{a^2}\ge \frac{4b}{a}$
so ....
$f(\frac{x_0}{a^n} - \frac{x_0}{a^2}\ge \frac{2^n b}{n a}$
so when $x \to 0$,$f(x) \to \infty$,contradiction with ii).
for part b)
let $f(x)=\frac{1}{x^n}$  n such that $a^n >2$

[quote="Hong Quy"][quote="zhaobin"]yes,it is nice. :)
for part b)
let $f(x)=\frac{1}{x^n}$  n such that $a^n >2$[/quote]

I think you are wrong because $\frac{1}{x^n}<x\forall x>1$[/quote]
why, I can't see...
but I think it is right ;)
with $f(x)=\frac{1}{x^n}$   then $af(\frac{x}{a})=\frac{a^{n+1}}{x^n}$
and $2f(x)=\frac{2}{x^n}$
and we can easily see $2f(x)\le af(\frac{x}{a})+x \forall x>0$


Let $ a, b \in R$ and $ \sin{\frac{a}{2}} \neq 0$.

Find $ S\equal{}\cos{b}\plus{}\cos{(a\plus{}b)}\plus{}\cos{(2a\plus{}b)}\plus{}...\plus{}\cos{(na\plus{}b)}$

We use
\[ \sum_{k \equal{} 0}^n \cos(\alpha \plus{} kh) \equal{} \left\{\begin{array}{ll}\frac {\cos\left(\alpha \plus{} \frac {n}{2}h\right)\sin\frac {n \plus{} 1}{2}h}{\sin\frac {h}{2}} & (h \neq 2m\pi) \\
(n \plus{} 1) \cos \alpha & (h \equal{} 2m\pi) \end{array}\right.
\]

Thank you very muck. Can you give me other solution?

other solution?  as in what?

anyway, Kouichi's formula can be proved simply by induction, with a few sum to product and product to sum identities.


What is $ \frac{x}{y}$ in the following equation?

$ x^3\minus{}4x^2y\minus{}11xy^2\plus{}30y^3 \equal{} 0$

[hide]
Factor $ x^3 \minus{} 4x^2y \minus{} 11xy^2 \plus{} 30y^3 \equal{} 0$ into
$ (x \minus{} 2y)(x \minus{} 5y)(x \plus{} 3y) \equal{} 0$
Then setting each part equal to zero gives
$ x \equal{} 2y$
$ x \equal{} 5y$
$ x \equal{} \minus{} 3y$

Thus, $ \frac {x}{y} \equal{} 2, \frac {x}{y} \equal{} 5,$ and $ \frac {x}{y} \equal{} \minus{}3$.

Blarg, sorry, corrected
[/hide]

If $ y \not\equal{} 0$, then denote by $ t\equal{}\frac{x}{y} \Rightarrow t^3\minus{}4t^2\minus{}11t\plus{}30\equal{}0 \Rightarrow (t\minus{}2)(t\minus{}5)(t\plus{}3)\equal{}0$. The rest of problem can be deduced easily.

[quote="27r"][hide]
Factor $ x^2 \minus{} 4x^2y \minus{} 11xy^2 \plus{} 30y^3 \equal{} 0$ [b]<----Typo[/b]: $ x^3$
into
$ (x \minus{} 2y)(x \minus{} 5y)(x \plus{} 3y) \equal{} 0$
Then setting each part equal to zero gives
$ x \equal{} 2y$
$ x \equal{} 5y$
$ x \equal{} \minus{} 3y$

Let $ y \equal{} 1$ in each and plug the values of $ x$ and $ y$ back into the original equation.
The only ratio of $ x/y$ that works is $ x \equal{} 5y$, or $ \frac {x}{y} \equal{} 5$.
[/hide][/quote]Are you sure that's the only ratio? I'm pretty sure the typo messed up your calculations.


[color=red]Trying to clear a broblem , I have formulated these questions:[/color]

Let $ a ,b$ positive integers with $ (a,b) = 1$.
[b]a)[/b]	Prove that for any positive integer $ c$ which is greater than $ ab-a-b$ we can find  natural numbers $ x ,y$ such that $ c= xa+yb$
[b]b)[/b]	Can we find a general formula (with proof)which gives the number of the numbers $ c$ between

$ 0,ab-a-b$, which can not be expressed in the form  $ xa+yb$ ?

 [color=red]Edited [/color].

Sorry  :( .I misunderstood the original english text where it was said : the product minus the sum of these numbers is the maximum....

 Thanks for the correction and the  solutions.

 [u]Babis[/u]

[quote="stergiu"][color=red]Trying to clear a broblem , I have formulated these questions:[/color]

Let $ a ,b$ positive integers with $ (a,b) = 1$.
[b]a)[/b]	Prove that for any positive integer $ c$ which is greater or equal to $ ab-a-b$ we can find  natural numbers $ x ,y$ such that $ c= xa+yb$
[b]b)[/b]	Can we find a general formula (with proof)which gives the number of the numbers $ c$ between

$ 0,ab-a-b$, which can not be expressed in the form  $ xa+yb$ ?[/quote]

That's wrong :

Let $ a=2$, $ b=3$, $ c=1\geq ab-a-b=1$, then it does not exist any natural numbers $ x$ and $ y$ such that $ 1=2x+3y$

[quote="pco"]That's wrong :

Let $ a=2$, $ b=3$, $ c=1\geq ab-a-b=1$, then it does not exist any natural numbers $ x$ and $ y$ such that $ 1=2x+3y$[/quote]
Exactly for all $ c>ab-a-b$ exist $ x\ge 0,y\ge 0$, suth that $ c=ax+by, \ (a,b)=1$.
From $ 1=ax_{0}-by_{0}$ we get a).
b) Let $ S_{0}=ab-a-b$, then $ N=S_{0}+1-\sum_{y=0}^{[S_{0}/b]}([1+\frac{S_{0}-yb}{a}])$/

Follow the link - it's just the [url=http://mathworld.wolfram.com/CoinProblem.html]coin problem[/url] in the case $ n = 2$. :wink:


How many numbers between 1 and 2005 are integer multiples of 3 or 4 but not
12?

This question is too hard.
How are we supposed to know exactly how many multiples of 3 and 4 are between 1 and 2005?!

From $ 1$ to $ n$, there are $ \left \lfloor \frac{n}{k} \right \rfloor$ multiples of $ k$.  Thus:

There are $ 668$ multiples of $ 3$, $ 501$ multiples of $ 4$, and $ 167$ multiples of $ 12$ less than $ 2005$.  Our desired answer is thus $ (668\minus{}167)\plus{}(501\minus{}167)\equal{}835$.

First:  How many numbers between 1 and 12 are integer multiples of 3 and 4?
       Answer: 5  ( They are 3, 6, 9, 4, and 8).
   Therefore, there are 5 integers that are multiples of 3 and 4, but not 12 for every cycle of 12.
Second:  How many cycles of 12 are in 2005?
       Well, 2004/12 =167
   So, we will get 5 integers 167 times.
Answer:  167 * 5 = 835

[hide="Explanation"]When we found the number of multiples of 3, we also included those that were multiples of 12. This is where the first subtracting of the multiples of 12 occurs. Similarly, when we found the multiples of 4, we included the multiples of 12 as well, so we must subtract those again.

Alternately, using the Principle of Inclusion-Exclusion, we find the number of numbers that are either multiples of 3 or 4. From PIE, we have to subtract the multiples of 12 to prevent overcounting. However, since this sum includes the multiples of 12 once, we must subtract yet again.

Hope this made sense.[/hide]

[hide]It took me a few minutes to figure out this one...

1. 2005 is not divisible by 3, but 2004 is.  2004/3=668

2.1/4 of the numbers that are multiples of 3 are also multiples of 4.  (3*1 is not divisible by 4, nor is 3*2, or 3*3, but 3*4 is)

3. so there are 668*3/4=501 numbers that are multiple of 3, but not 12

4. let's now look at the multiples of 4, there are 2004/4=501

5.1/3 of the numbers that are multiples of 4 are also multiples of 3

6. so there are 501*2/3=334 that are multiple of 4, but not 12

7.  so we add them together: 334+501=835

so 835 is your answer!
 :lol:  :lol:  :lol:  :lol:[/hide]

[hide=Solution] Since $\frac{2005}{3} = 668\frac13$, there are 668 multiples of 3 between 1 and 2005. Since $\frac{2005}{4} = 501\frac14$, there are 501 multiples of 4 between 1 and 2005. Since $\frac{2005}{12} = 167\frac{1}{12}$, there are 167 multiples of 12 between 1 and 2005.

Every multiple of 12 is also a multiple of 3 and of 4, so there are $668-167 = 501$ multiples of 3 that are not multiples of 12 and $501-167 = 334$ multiples of 4 that are not multiples of 12. That leaves $501 + 334 = \boxed{835}$ numbers that are multiples of 3 or 4 but not 12. (Note: no number can be a multiple of 3 and 4 without also being a multiple of 12. So, no number is included twice in our count $501+334$.)[/hide]

[hide=ALTERNATIVE SOLUTION]
By counting, there are $5$ multiples of $3$ and $4$ in every multiple of $12$ (excluding the multiple of $12$). And there are $2005\div12\approx167$ multiples of $12$ in the range of $1\text-2005$. So that's $5\times167=\boxed{835}$ multiples of $3$ or $4$ but not $12$ that are between $1$ and $2005$.
[/hide]


I don't know, it becomes almost trivial after applying induction.

What is induction? And I don't see why you would use it, there are already two good proofs out there

induction is just proving that if it works for some k, it also works for some k+1. you also need a base case.

so suppose your base case is 1. then you prove k works => k+1 works.

so then 1 works => 2 works => 3 works => 4 works => etc. so it works for all natural numbers.

it does really trivialize the problem if you know what you're doing.

Well, I don't see the good of it here if it can trivialize it because this question isn't that trivial just long algebra or thinking algebra

Yeah, the proofs provided are good. Induction is just an easy way to prove it without understanding very much :P. You don't learn much from using it on this proof.

Induction proofs are important, and this formula is a great application of the inductive principle. Sure, it doesn't tell you where this formula came from - but it does tell you why it works.

So if you use induction, would you just prrove that x=1 works by plugging 1 in, and then proove that $\frac{x(x+1)(2x+1)}{6} + {(x+1)}^2 = \frac{(x+1)(1(x+1)+1)(2(x+1)+1)}{6}$?

Because it that's right, then inducton is really easy.  You can probably tell that I'm not really familiar with induction  :blush:

[quote="h_s_potter2002"]Because it that's right, then inducton is really easy.[/quote]

The concept of induction is simple, but that inductive step can be very difficult in some problems. And there's all kinds of cool stuff you can do with induction, such as backwards induction.

Now, my challenge: Prove, using induction:

Given $a_1 \leq a_2 \leq ...  \leq a_n$ with $a_1,...,a_n$ positive integers and $1/a_1 + 1/a_2 + ... + 1/a_n=1$ that $a_n \leq 2^{n!}$


Look around the forum, its everywhere.

20

[b]17[/b]

i remmebr jongao started at 45 once and the girls still won!! XD

[b]16[/b]
hmm this is good for us so i'll do it

There are too many people posting at the same time so we don't see each others' posts.

umm

you added 5 like every single time!!

PowerOfPi is only about 1000 posts from 5.

What? I got an error and couldn't post. Darn  :wallbash_red:  :wallbash_red:

haha nice try

the score is like 8 1/2 to 2 1/2 girls winning


If x is a positive integer and x(x+1)(x+2)(x+3)+1=3792, compute x.

Any ideas?

Fierytycoon

There is no such x because x(x+1)(x+2)(x+3)+1=3792 is the same as x(x+1)(x+2)(x+3)=3791 and if x is an odd integer then x+1 is clearly even, the converse gives that x is even, then 3791 must be even, which it's not. Also, x(x+1)(x+2)(x+3) is one less than a perfect square since x(x+1)(x+2)(x+3) + 1 = (x^2 + 3x + 1)^2 an odd perfect square. But 3792 is not odd and neither a perfect square.

oh, whoops: i copied the problem incorrectly. It is actually:

If x is a positive integer and x(x+1)(x+2)(x+3)+1=379^2, compute x.

The exponent didn't get copied correctly. Sorry!

But now, I think the problem is not bad. now x(x+1)(x+2)(x+3) = (379-1)(379+1)=(378)(380), so x must be very near Sqrt(379). A bit of guess-and-check should yield the result.

btw, I found the problem at http://incolor.inebraska.com/sabalka/arml.html , which has some nice short-answer problems.

Fierytycoon

[quote="Fierytycoon"]But now, I think the problem is not bad. now x(x+1)(x+2)(x+3) = (379-1)(379+1)=(378)(380), so x must be very near Sqrt(379). A bit of guess-and-check should yield the result.[/quote]

Since sqrt(379) :approx: 19.5, we need (x+1) and (x+2) to be around 19 and 20.  Indeed, if x is 18, x(x+1)(x+2)(x+3) = 18*19*20*21 = 378*380, which is the desired result.  Hence x = 18.

Also, factoring (378)(380) gives (2*3^3*7)(2^2*5*19) = (2^3*3^3*5*7*19).  Since there's one 19, I'd guess that 19 must be one of the numbers (since 38, 57, etc. would be way too big).  Now we want to find numbers around 19.  2*3^2 makes 18.  2^2*5 makes 20.  3*7, the last two prime factors left, make 21.  So we have (18)(19)(20)(21), which works.  Hence x = 18.

[quote="ComplexZeta"]I assume this is from ARML, but I forget which year. x(x+1)(x+2)(x+3)+1 is something that comes up from time to time. The thing to remember is that x(x+1)(x+2)(x+3)+1=(x^2+3x+1)^2. You can prove this by reordering the terms to x(x+3)(x+1)(x+2)+1=(x^2+3x)(x^2+3x+2)+1=(x^2+3x+1)^2.[/quote]

Oh-ho!  So in that case, (x^2+3x+1)^2 = 379^2
so x^2+3x+1 = 379
x^2 + 3x - 378 = 0
(x-18)(x+21) = 0
x = 18

Nice...I didn't realize that factored so nicely.  Must remember that one.

[color=cyan]You got rather lucky there, because you might have missed out on roots.

(x^2+3x+1)^2 = 379^2
so

x^2+3x+1 = 379   [u][i][b]or[/b][/i][/u]   x^2+3x+1 = -379.  The second equation just happens not to have any positive whole number solutions.  Or real solutions.  But that's not important![/color]

Well, true, but I didn't bother writing out the -21, either.  You're right, I need to be more careful.


Let $p>5$ be prime number. Prove: \[ \binom{qp}{p} \equiv q\ (mod \ p^3)  \]

It is Luka theorem \[ ord_p( \binom{ap}{bp}-\binom{a}{b})=ord_p(\binom{a}{b})+ord_p(a)+ord_p(b)+ord_p(a-b)+v(p),p>2  \]  were $v(3)=2,v(p)\ge 3,p>3$. Primes with $v(p)>3$ called Wolstenholm's primes. There are only 2 Wolstenholm's primes p<1 000 000 000. There are p=16483,2124679 v(p)=4.

Is $ord_p(a)$ the gratest power of $p$ that divides $a$. What is $v(p)$.


A ball is dropped from a height of 10 feet. At each bounce it rises to two-thirds of its previous height. How far will it have traveled if it bounces "forver"?

Thanks

Ahah!
[hide="Solution"]
Geometric Series Time.
The first drop is 10 feet.
It then comes up 6.6666 and back down 6.666
Up 40/9 feet, down 40/9 feet.
We wish to evaluate the infinite sum:
$10+2\sum_{n=0}^{\infty}\frac{20}{3}\cdot(2/3)^{n}$

By the formula for solving an infinite series,
we get
$\frac{a}{1-r}\longrightarrow 10+2 \cdot\frac{20/3}{1-\frac{2}{3}}\longrightarrow 10+2\cdot \frac{20/3}{1/3}\longrightarrow \boxed{50}$ feet.
[/hide]


Find all function $ f: R\rightarrow R$ so that:
i) $ f(x)\leq 4\plus{}2009x$
ii) $ f(x\plus{}y)\leq f(x)\plus{}f(y)\minus{}4$

[quote="tdl"]Find all function $ f: R\rightarrow R$ so that:
i) $ f(x)\leq 4 \plus{} 2009x$
ii) $ f(x \plus{} y)\leq f(x) \plus{} f(y) \minus{} 4$[/quote]

i) $ \implies$ $ f(0)\leq 4$
ii) $ \implies$ $ f(x\plus{}0)\leq f(x) \plus{} f(0) \minus{}4$ and so $ f(0)\geq 4$

So $ f(0)\equal{}4$

ii) $ \implies$ $ f(x\plus{}(\minus{}x)) \leq f(x) \plus{} f(\minus{}x) \minus{} 4$ and so $ f(x) \plus{} f(\minus{}x) \geq 8$

i) $ \implies$  $ f(x)\leq 4 \plus{} 2009x$
i) $ \implies$  $ f(\minus{}x)\leq 4 \minus{} 2009x$
Adding these two lines, we get $ f(x) \plus{} f(\minus{}x) \leq 8$, so (since  $ f(x) \plus{} f(\minus{}x) \geq 8$) :  $ f(x) \plus{} f(\minus{}x) \equal{} 8$ and the two lines are equalities.

So $ f(x)\equal{}2009x \plus{} 4$ and it is easy to verify that this solution fit the initial inequations


superior algebra
superior algebra solved

Study the convergence of the sequence (x_n)n>=0 where x_0 is a real number and x_(n+1)=[1+2^p*(x_n)^p]^(1/p)-x_n, for every n natural, p is an even natural number.

Manifestly there is an increasing sequence : x(n+1) = (1+(2*|x(n)|)^p)^(1/p)  - x(n)  >  x(n)
Suppose a finite limit L for x(n) ===>  L = (1+(2*|L|)^p)^(1/p)  - L  >  L  (contradiction)
Increasing sequence and no finite limit   ===> lim(n-->+oo) = +oo


Farz konid $ a,b,c>0$ va $ ab+bc+ca=3$ sabet konid :
$ a^{3}+b^{3}+c^{3}+7abc \geq 10$ .

in ye rahe hal dare ke ziad ghashango khosh das nis,vali be har hal:

aval az hame midoonim $ (a+b+c)^{2}\geq 3(ab+bc+ca)=9$ pas $ a+b+c\geq 3$
----------------------------
tebghe namosavie shur midoonim:

$ a^{3}+b^{3}+c^{3}+3abc\geq a^{2}b+ab^{2}+b^{2}c+bc^{2}+c^{2}a+ca^{2}$

ya be ebarati:

$ a^{3}+b^{3}+c^{3}+3abc\geq (a+b+c)(ab+bc+ca)-3abc$

pas darim:

$ a^{3}+b^{3}+c^{3}+6abc\geq (a+b+c)(ab+bc+ca)$

ke mishe intori neveshtesh:

$ (a+b+c)^{3}+9abc\geq 4(a+b+c)(ab+bc+ca)$

hala ba tavajoh be sharte $ ab+bc+ca=3$ in rabete be shekle zir dar miad:

$ (a+b+c)^{3}+9abc\geq 12(a+b+c)$

pas darim:

$ abc\geq \frac{12(a+b+c)-(a+b+c)^{3}}{9}$

hala age gharar bedim $ S=a+b+c$ mishe:

$ abc\geq \frac{12S-S^{3}}{9}$
-----------------------------------
khob hala dobare be masale bargardim,soorate hokmemoon moadele ine:

$ (a+b+c)^{3}+10abc-3(a+b+c)(ab+bc+ca)\geq 10$

ke ba tavajoh be sharte $ ab+bc+ca=3$ moadele ine:

$ \boxed{S^{3}+10abc-9S-10\geq 0}$

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

khob hala age az namosavi ke dar avale esbatemoon be das ovordim estefade konim mishe:

$ S^{3}+10abc-9S-10\geq S^{3}+10\frac{12S-S^{3}}{9}-9S-10\geq 0$

ke age sadash konim mishe:

$ -S^{3}+39S-90\geq 0$

hala ino mishe injoori neveshtesh:

$ (S-3)(30-S^{2}-3S)\geq 0$

hala hamantor ke dar avale esbat gofte shod darim $ S\geq 3$ pas age paranteze dovomo taine alamat konim mibinim ke age

$ S\leq \frac{-3+\sqrt{129}}{2}$ bashe masale hal shodast...

pas mimoone baghie halata...

amma baghie halata azin kheili asoontare chera ke darim:

$ S^{3}+10abc-9S-10\geq S^{3}-9S-10=(S+2)(S^{2}-2S-5)$

pas kafie sabet konim:

$ (S+2)(S^{2}-2S-5)\geq 0$

ke age ino taine alamt konim mibinim ke vaseye $ S\geq \frac{2+\sqrt{24}}{2}$ in namosavi bargharare...

hala chon $ \frac{2+\sqrt{24}}{2}\leq \frac{-3+\sqrt{129}}{2}$ pas hokmemoon be tore kamel esbat shode...

Can you solve this problem in English. :lol:
I think you have just proved in this case:$ 3\leq S\leq \frac{\minus{}3\plus{}\sqrt{129}}{2}$ then the inequality is true,and in the case:$ S\geq\frac{2\plus{}\sqrt{24}}{2}$ it's true too but in the case $ S>\frac{\minus{}3\plus{}\sqrt{129}}{2}$,you haven't done it

read the post completely:
we want to prove that:

$ S^3 \plus{} 10abc \minus{} 9S \minus{} 10\geq 0$ (*)

now as we said before,we know that $ S\geq 3$ now note that if $ S\leq \frac { \minus{} 3 \plus{} \sqrt {129}}2$ then we have:

$ S^3 \plus{} 10abc \minus{} 9S \minus{} 10\geq \minus{} S^3 \plus{} 39S \minus{} 90 \equal{} (\underbrace{S \minus{} 3}_{\geq 0})(\underbrace{30 \minus{} S^2 \minus{} 3S}_{\geq0})\geq 0$
----------------------------
so it remains to prove (*) for $ S > \frac { \minus{} 3 \plus{} \sqrt {129}}2$

which is pretty easy because we have:

$ (*) \equal{} S^3 \plus{} 10abc \minus{} 9S \minus{} 10\geq S^3 \minus{} 9S \minus{} 10 \equal{} (S \plus{} 2)(S^2 \minus{} 2S \minus{} 5)$

so it's sufficient to show that:

$ (S \plus{} 2)(S^2 \minus{} 2S \minus{} 5)\geq 0$ for $ S > \frac { \minus{} 3 \plus{} \sqrt {129}}2$

but note that $ S\geq 3\Rightarrow S \plus{} 2\geq 0$ so we have to show that:

$ S^2 \minus{} 2S \minus{} 5\geq 0$ (**)

but the roots of (**) are $ \frac {2\pm\sqrt {24}}{2}$ so for $ S\geq \frac {2 \plus{} \sqrt {24}}2$ (**) holds...
but in the other hand we have:

$ S\geq\frac { \minus{} 3 \plus{} \sqrt {129}}2\geq\frac {2 \plus{} \sqrt {24}}2$

thus:

$ S^2 \minus{} 2S \minus{} 5\geq 0$

QED


Sequences  {U_n}   {V_n}   satisfy the following conditions

U_(n+2)  = a.U_(n+1)  + b.U_(n)

V_(n+2)  = c.V_(n+1)  + d.V_(n)   where a, b, c  and  d are constant.

U-(0) , U_(1) , V_(0) , V_(1)  are arbitrary numbers subject to  V_(0) = U_(0)   and V_(1)  = U_(2)   =a.U_(1)  + b.U_(0)

Find necessary and sufficient conditions on a, b, c and  d  to ensure that  U_(2n) = V_(n) for all n.

Solve the characteristic equation... and equate the two numbers?


Solve this cryptarithm, where each of the letters represents a different decimal digit from 0 to 9, and S is not zero.

SAME + SEAM = 2961

[b]SAME + SEAM = 2961[/b]

$ 2000S \plus{} 110A \plus{} 11M \plus{} 101E \equal{} 2961$.

Clearly $ S \equal{} 1$. Therefore:

$ 110A \plus{} 11M \plus{} 101E \equal{} 961$. Factor 11, we get:

$ 11(10A \plus{} M) \plus{} 101E \equal{} 961$

Therefore, $ 961 \minus{} 101E$ must be a multiple of 11. Note that $ 961\equiv 4 \pmod {11}$ and $ 101\equiv 2\pmod {11}$. Therefore $ E \equal{} 2$.

Thus, $ 11(10A \plus{} M) \equal{} 759 \implies 10A \plus{} M \equal{} 69 \implies A \equal{} 6,M \equal{} 9$.

[b]Answer:[/b]
[b]A[/b]=6
[b]E[/b]=2
[b]M[/b]=9
[b]S[/b]=1
[b]SAME[/b]=1692
[b]SEAM[/b]=1269
[b]SAME+SEAM[/b]=2961

Obviously S is 1. Assume there are carries for the other three place values in the addition (you can check it is not possible otherwise). Then,
e+m=11
m+a=15
a+e=8
Adding them and cutting in half, a+e+m=17, so we find a=6, e=2, m=9

Hm why are my methods so rigorous... and complicated... :blush:

@Yongyi781: Maybe that's just your way. Professionalism is, in most cases, a good thing. However in the practice, shortcuts are always nice, as long as they work.


The marked price of a book was $30%
$ less than the suggested retail price. Alice purchased the book for half the marked price at a Fiftieth Anniversary sale.

[quote="4everwise"]The marked price of a book was [30%] less than the suggested retail price. Alice purchased the book for half the marked price at a Fiftieth Anniversary sale.[/quote]

Err.. is there a question to this?
 :lol:


If $ x\plus{} |y| \minus{} y \equal{} 12$ and $ |x| \plus{} x \plus{}y \equal{} 10$, find $ x\plus{}y$.

[hide]I have tried isolating $ x\minus{}y$ of one equation and $ x\plus{}y$ for the other, and get $ 2x \equal{} 22 \minus{} |x| \minus{} |y|$. Nothing comes to mind then. [/hide]

[quote="jeez123"]If $ x \plus{} |y| \minus{} y \equal{} 12$ and $ |x| \plus{} x \plus{} y \equal{} 10$, find $ x \plus{} y$.[/quote]
[hide="Solution"]Adding the two equations yields $ 2x \plus{} |x| \plus{} |y| \equal{} 22$.

[color=darkred][b]Case 1:[/b][/color] $ y\geq0$

$ x \equal{} 12\Longrightarrow |x| \plus{} |y| \equal{} \minus{} 2$

Obviously, this is not possible.

[color=darkred][b]Case 2a:[/b][/color] $ y < 0,\,x\geq0$

$ x \plus{} 2y \equal{} 12,\,2x \plus{} y \equal{} 10\Longrightarrow\boxed{x \plus{} y \equal{} \tfrac{22}{3}}$

[color=darkred][b]Case 2b:[/b][/color] $ y < 0,\,x < 0$

$ x \plus{} 2y \equal{} 12,\,y \equal{} 10\Longrightarrow\boxed{x \plus{} y \equal{} 2}$[/hide]

[hide="Hint"]Split into four cases - both x,y non-negative; x negative, y non-negative; x non-negative, y negative; both x,y negative - and discuss the system (it's really simple enough for casework).[/hide]

[hide="Correct answer"]$ x \equal{} {32\over 5}, y \equal{} \minus{} {14\over 5}, x \plus{} y \equal{} {18\over 5}$[/hide]

[hide="Where i_like_pie is wrong"]Case 2a should be $ x\boxed{ \minus{} }2y \equal{} 12$
Case 2b can't have both $ y<0$ and $ y\equal{}10$[/hide]

[quote="Farenhajt"]Case 2a should be $ x\boxed{ \minus{} }2y \equal{} 12$
Case 2b can't have both $ y < 0$ and $ y \equal{} 10$[/quote]
I definitely agree with your second correction, but how can $ x \plus{} |y| \minus{} y \equal{} 12$ become $ x \minus{} 2y \equal{} 12$? The way I see it, you're suggesting the absolute value becomes negative. :huh:

Remember the definition of absolute value? $ |y| \equal{} y$ for non-negative $ y$ and $ |y| \equal{} \minus{} y$ for negative $ y$ (notice that $ \minus{}y$ is actually positive if $ y$ is negative). Since you assumed $ y < 0$ in case 2a, then $ |y| \equal{} \minus{} y$

Ah, I see. Thanks for correcting me. :)

I forgot that $ y$ would be negative and it would result in a positive outcome. :oops:

[quote="i_like_pie"][quote="jeez123"]If $ x \plus{} |y| \minus{} y \equal{} 12$ and $ |x| \plus{} x \plus{} y \equal{} 10$, find $ x \plus{} y$.[/quote]
[hide=Solution]Adding the two equations yields $ 2x \plus{} |x| \plus{} |y| \equal{} 22$.

[color=darkred][b]Case 1:[/b][/color] $ y\geq0$

$ x \equal{} 12\Longrightarrow |x| \plus{} |y| \equal{} \minus{} 2$

Obviously, this is not possible.

[color=darkred][b]Case 2a:[/b][/color] $ y < 0,\,x\geq0$

$ x \minus{} 2y \equal{} 12,\,2x \plus{} y \equal{} 10\Longrightarrow\boxed{x \plus{} y \equal{} ?}$

[color=darkred][b]Case 2b:[/b][/color][/quote]

I understood up to those corrected parts above. How do you calculate for $ x\plus{}y$ in Case 2a? I added both of the equations and got $ 3x\minus{}y\equal{}22$ and cannot find the $ x\plus{}y$ part. Could you finish up this solution? Thanks!

You're on the right track. You should add the second equation again:
$ (3x \minus{} y) \plus{} (2x \plus{} y) \equal{} 22 \plus{} 10 \implies 5x \equal{} 32 \implies x \equal{} \frac {32}{5}$

We can plug this into a known equation: $ x \plus{} y \equal{} (2x \plus{} y) \minus{} x \equal{} 10 \minus{} \frac {32}{5} \equal{} \frac {18}{5}$


I couldnt find this one in forum
factor:
$ {(a\plus{}b\plus{}c)}^3\minus{}(a^3\plus{}b^3\plus{}c^3)$

[hide]Simplifying yields $ 3\sum_{cyc}^{}a^2b \plus{} 6abc$.  Noticing the cyclic terms, we try cyclic factors.  $ (a\plus{}b)(a\plus{}c)(b\plus{}c)\equal{} \sum_{cyc}^{}a^2b \plus{}2abc$.  Multiply by three and we are done.[/hide]


In the context of elliptical polarization,every book writes out this formula:

(E_y/E_0y)^2 + (E_x/E_0x)^2 -2(E_y/E_0y)(E_x/E_0x) cos ε= (sinε)^2

where ε is the initial phase difference.

After that they write this is the equation of an ellipse making an angle α with the E_x-E_y co-ordinate system such that

tan 2α=[[2E_0x*E_0y]cos ε]/[(E_0x)^2 - (E_0y)^2]

This is NOt a homework problem;but I want to clarify how does the 2nd relation concering tan 2α comes.

I tried to understand by expanding tan 2α and by dividing RHS by (E_0x)^2 throught.
But the factor cos ε is annoying.

Can anyone please tell how the relation comes?

Come on kolahalb, use LaTeX please. :wink:  its almost impossible to figure out what you write. :huh:

In $\text{\LaTeX}$: (I think  :) )
$\hline$

In the context of elliptical polarization,every book writes out this formula:

$\left(\frac{E_{y}}{E_{0y}}\right)^{2}+\left(\frac{E_{x}}{E_{0x}}\right)^{2}-2\left(\frac{E_{y}}{E_{0y}}\right)\left(\frac{E_{x}}{E_{0x}}\right) \cos \epsilon= (\sin\epsilon)^{2}$

where $\epsilon$ is the initial phase difference.

After that they write this is the equation of an ellipse making an angle $\alpha$ with the $E_{x}$-$E_{y}$ co-ordinate system such that

$\tan 2\alpha=\frac{\left[2E_{0x}\cdot E_{0y}\right]\cos \epsilon}{[(E_{0x})^{2}-(E_{0y})^{2}]}$

This is NOt a homework problem;but I want to clarify how does the 2nd relation concering $\tan 2\alpha$ comes.

I tried to understand by expanding $\tan 2\alpha$ and by dividing $RHS$ by $(E_{0x})^{2}$ throught.
But the factor $\cos \epsilon$ is annoying.

Can anyone please tell how the relation comes?

Thanks color,I exactly meant that.

Now can you hep something about it?


Hi, I can't solve the following problems, please help.

1.  A rectangle with its dimensions 5m by 12m is surrounded by a path of uniform width.  How wide will this path be if the area of the rectangle is the same as the area of the path?

2. A number is the quotient of 5.52 and 11.5, decrease by 2.79.  What is the number?

For 1. draw a diagram. Call the width w and split up the path into rectangles, and write the areas of the rectangles in terms of 5, 12, and w. (if this doesn't make sense I can try explaining more)

2 seems straightforward...
[hide="2"]
$ \frac{5.52}{11.5}\minus{}2.79$
[/hide]

[quote="chuchen"]1.  A rectangle with its dimensions 5m by 12m is surrounded by a path of uniform width.  How wide will this path be if the area of the rectangle is the same as the area of the path?[/quote]
[hide="Hint"]Set up an equation and solve for the width.[/hide]
[hide="Solution"]Let the width of the path be $ w$.

The area of the rectangle is $ 5\cdot12$, and the area of the path is $ (5 \plus{} w)(12 \plus{} w) \minus{} 5\cdot12$.

We know that the two areas are equal, so we can solve for $ w$ by setting those two expressions equal to each other.

Expanding the RHS, we get $ 60 \equal{} w^2 \plus{} 17w \plus{} 60 \minus{} 60$.

Simplifying and putting everything over on one side gives $ w^2 \plus{} 17w \minus{} 60 \equal{} 0$.

Factoring, $ (w \minus{} 3)(w \plus{} 20) \equal{} 0$.

The width cannot be negative, so $ \boxed{w \equal{} 3\ \text{m}}$.[/hide]


I work with three problem,and found nice solution(I think).
Problem 1:(Vasile cirtoaje)
Give $ a,b,c \ge 0$ and no two which of them are zero.Prove
$ \frac{1}{a^2\plus{}ab\plus{}b^2}\plus{} \frac{1}{b^2\plus{}bc\plus{}c^2}\plus{} \frac{1}{c^2\plus{}ca\plus{}a^2} \ge \frac{9}{(a\plus{}b\plus{}c)^2}$
First solution:(by Bui Tuan Vu):
This ineqlity can be written :
$ \frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}ab\plus{}b^2}\plus{} \frac{(a\plus{}b\plus{}c)^2}{b^2\plus{}bc\plus{}c^2}\plus{} \frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}ac\plus{}c^2} \ge 9$
or :
$ \sum\frac{(c\minus{}a)(c\minus{}b)}{a^2\plus{}ab\plus{}b^2}\plus{}3\plus{} \sum\frac{3a(b\plus{}c)}{b^2\plus{}bc\plus{}c^2} \ge 9$
Which is true by schur's inequality and
$ \sum\frac{a(b\plus{}c)}{b^2\plus{}bc\plus{}c^2} \ge 2$(vasile cirtoaje)
Second solution(by Bui Ngoc Anh) we use AM-GM and Iran TST 96
$ \sum\frac{1}{a^2\plus{}ab\plus{}b^2}\equal{} \sum\frac{ab\plus{}bc\plus{}ca}{(ab\plus{}bc\plus{}ca)(a^2\plus{}ab\plus{}b^2)} \ge \sum\frac{4(ab\plus{}bc\plus{}ca)}{(a\plus{}b)^2(a\plus{}b\plus{}c)^2} \ge \frac{9}{(a\plus{}b\plus{}c)^2}$
Problem 2:
$ \frac{a^2\plus{}bc}{b^2\plus{}bc\plus{}c^2}\plus{} \frac{b^2\plus{}ac}{a^2\plus{}ac\plus{}c^2}\plus{} \frac{c^2\plus{}ab}{a^2\plus{}ab\plus{}b^2} \ge 2$
Solution(by Bui Ngoc Anh)
We use :
$ \sum\frac{a(b\plus{}c)}{b^2\plus{}bc\plus{}c^2} \ge 2$(vasile cirtoaje)
and we can prove the stronger:
$ \frac{a^2\plus{}bc}{b^2\plus{}bc\plus{}c^2}\plus{} \frac{b^2\plus{}ac}{a^2\plus{}ac\plus{}c^2}\plus{} \frac{c^2\plus{}ab}{a^2\plus{}ab\plus{}b^2} \ge \frac{a(b\plus{}c)}{b^2\plus{}bc\plus{}c^2}\plus{} \frac{b(a\plus{}c)}{a^2\plus{}ac\plus{}c^2}\plus{} \frac{c(a\plus{}b)}{a^2\plus{}ab\plus{}b^2}$
or
$ \sum\frac{(c\minus{}a)(c\minus{}b)}{a^2\plus{}ab\plus{}b^2} \ge 0$
Which is true by schur's inequality.
problem 3:
Give $ x\plus{}y\plus{}z\plus{}xyz\equal{}4$ and $ x,y,z \ge 0$.Prove that:
$ x\plus{}y\plus{}z \ge xy\plus{}yz\plus{}zx$
Solution(by Bui Ngoc Anh)
if $ xyz >1$ and by AM-GM we have:$ x\plus{}y\plus{}z >3$.
and $ x\plus{}y\plus{}z\plus{}xyz>4$.
So we have $ xyz \le 1$ and from $ x\plus{}y\plus{}z\plus{}xyz\equal{}4$ we get $ x\plus{}y\plus{}z \ge 3$
If $ xy\plus{}yz\plus{}zx \le 3$,this inequality is obivious true,because $ x\plus{}y\plus{}z \ge 3 \ge xy\plus{}yz\plus{}zx$
and we only need to consider $ xy\plus{}yz\plus{}zx \ge 3$
$ x\plus{}y\plus{}z \ge xy\plus{}yz\plus{}zx$ or $ (x\plus{}y\plus{}z)^2 \ge (x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)\equal{}(4\minus{}xyz)(xy\plus{}yz\plus{}zx)$
and we will to show that
$ (x\plus{}y\plus{}z)^2\plus{}xyz(xy\plus{}yz\plus{}zx) \ge 4(xy\plus{}yz\plus{}zx)$
or $ x^2\plus{}y^2\plus{}z^2\plus{}xyz(xy\plus{}yz\plus{}zx) \ge 2(xy\plus{}yz\plus{}zx)$
from $ xy\plus{}yz\plus{}zx \ge 3$ and $ x\plus{}y\plus{}z \ge 3$,we have
$ xyz(xy\plus{}yz\plus{}zx) \ge 3xyz \ge \frac{9xyz}{x\plus{}y\plus{}z}$
and finally,we will to show that:
$ x^2\plus{}y^2\plus{}z^2\plus{} \frac{9xyz}{x\plus{}y\plus{}z} \ge 2(xy\plus{}yz\plus{}zx)$
But it is schur's inequality.
Our proof is completed,equality occur if and only a=b=c.

problem #1 can be done easily by Titu's Lemma
[hide="#1"]

$ LHS \geq \frac{(1\plus{}1\plus{}1)^2}{2a^2 \plus{}2b^2 \plus{}2c^2 \plus{}ab \plus{}ac\plus{}bc}$

By AM-GM
 $ {2a^2 \plus{}2b^2 \plus{}2c^2 \plus{}ab \plus{}bc\plus{}ca } \geq a^2 \plus{}b^2 \plus{}c^2 \plus{}2ab\plus{}2bc \plus{}2ca\equal{}(a\plus{}b\plus{}c)^2 \implies LHS \geq \frac{9}{(a\plus{}b\plus{}c)^2}$

[/hide]

[quote="enndb0x"]problem #1 can be done easily by Titu's Lemma
[hide="#1"]

$ LHS \geq \frac {(1 \plus{} 1 \plus{} 1)^2}{2a^2 \plus{} 2b^2 \plus{} 2c^2 \plus{} ab \plus{} ac \plus{} bc}$

By AM-GM
 $ {2a^2 \plus{} 2b^2 \plus{} 2c^2 \plus{} ab \plus{} bc \plus{} ca } \geq a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2bc \plus{} 2ca \equal{} (a \plus{} b \plus{} c)^2 \implies LHS \geq \frac {9}{(a \plus{} b \plus{} c)^2}$

[/hide][/quote]

You 're wrong :)  :P

[quote="enndb0x"]Can you become kind and tell us where is my mistake,instead of just saying "you're wrong" ?[/quote]

your denominator decreased with AM-GM, so the value increased   :D

[quote="enndb0x"]Can you become kind and tell us where is my mistake,instead of just saying "you're wrong" ?[/quote]
$ \frac {9}{2(a^2 \plus{} b^2 \plus{} c^2)\plus{}ab\plus{}bc\plus{}ca} \le \frac {9}{(a \plus{} b \plus{} c)^2}$  :)

By AM-GM
$ 4(a^2\plus{}ab\plus{}b^2)(ab\plus{}bc\plus{}ca) \le (a^2\plus{}ab\plus{}b^2\plus{}ab\plus{}bc\plus{}ca)^2\equal{}(a\plus{}b)^2(a\plus{}b\plus{}c)^2$:)

Are you sure the second solution is yours, Bui Ngoc Anh? I think you had read here before ;) (see #11): http://www.maths.vn/forums/showthread.php?t=17775&page=2

@secrets: Anh nghi la chung ta nen ton trong nhung nguoi giai. Chu neu lay loi giai cua nguoi khac roi nhan lai la cua minh la khong hay! Ok? Anh thay cung co nhieu ban o VN minh nhu the, that tiec! Hay tu tim loi giai cua chinh minh nhe! (Anh noi nhu the vi anh thay sau bai post cua anh o mathsvn co bai post cua em, nen anh nghi em doc loi giai cua anh truoc!)

Indeed, it's an old proof of can_hang, secrets.
@can_hang: [hide]Chac la` do secrers vo tinh` + ko de? y' thoi anh a, nhung dung' la` em thay' gan` day co nhieu` truong` hop nhu anh noi' that :-<. Du` sao cung~ ko fai? viec cua minh` nen em ko quan tam lam'.
Hiz, ma` may' tinh' nha` em bi hong? may hom nay nen ko go~ dc them gi`, luc nao` anh ol nhay' may' cho em nhe', em lo qua', giua~ thang' 6 roi`ma` cong viec van~ chua dau vao` dau  :( [/hide]

@canhang_2007 and nguoivn.
Em xin lỗi,thực sự là em không cố ý,tầm hiểu biết của em thấp,em cũng không rõ là lời giải này xuất hiện ở đâu,em hứa là em sẽ không bao giờ post những topic như thế này nữa.Cám ơn anh đã nhắc nhở

Sorry,but I don't read your solution,can hang.In the future,I don't learn inequalities.Thank you.:)
@ canhang_2007
sr anh nha,em không cố ý đâu,vì trí nhớ em rất tệ.Em cũng biết mathlinks là 1 trang web rất nổi tiếng,và chắc chắn rằng người biết lời giải của anh rất nhìu,vì vậy em đâu dám công khai lấy lời giải của anh:D.Anh nặng lời mất rùi,đây là lần thứ hai anh nhắc nhở em.Cám ơn anh nhìu.Anh đừng cho rằng em lấy lời giải của anh nha.:) do em không biết mà thui.

Trong tương lai,em cũng sẽ không học được bất đẳng thức nữa,có lẽ em chỉ cố gắng nốt tháng này thui,em sẽ phải tập trung vào các môn khác,nếu không pa má em sẽ la em.:D

Anh nhớ reply nhanh nha,em không muốn bị ai giận cả.:(Hì,anh nhớ đọc kĩ hai bài post của em rùi hãy reply,rùi em sẽ nhờ mod khoá topic này lại.Coá lẽ từ nay em sẽ không làm bất đẳng thức nữa,bi quan chút :D:).

Để giải thích tại sao em lại tìm ra lời giải trùng với lời giải của anh.Đó là do em "học" được AM-GM từ anh Dương Đức Lâm
$ 4(a^2\plus{}ab\plus{}b^2)(ab\plus{}bc\plus{}ca) \le  (a\plus{}b)^2(a\plus{}b\plus{}c)^2$(anh có thể xem tại problem of day trên vnineqmath) rùi em cộng lại.Thật tuyệt,em thấy đó là bất đẳng thức Iran96 có trong đó.Dùng Iran 96,ra bài của Vasc.Một cậu nhóc chưa biết gì về bất đẳng thức lại có thể tìm ra 1 lời giải khá tuyệt như vậy,em cảm thấy rất vui.:)Vì dù sao đó cũng là 1 chút thành quả nhỏ.
Lúc anh post lời giải của anh trên maths.vn,thú thực,em không quan tâm,vì lúc đó đến một chút về AM-GM em còn chưa vững,bài Iran96 là 1 bài mà em rất sợ.Vì vậy,lời giải của anh em không đọc kĩ,và không lưu nhìu ấn tượng.
Dù sao,em xin lỗi anh,và mong anh đừng kết luận 1 rằng em "lấy" lời giải của anh.:)Cám ơn anh.

[quote]
(a+b+c)^2(a+b)^2-4(ab+bc+ca)(a^2+ab+b^2)=(a^2+b^2-ac-bc)^2
[/quote]

Kids, I know of such solution - not mine, just in case you want to know - long before you get your ears dry. You're all pathetic !!

Em nghĩ anh cẩn nặng lời rùi,không thể từ lời giải giống nhau mà suy ra chép lời giải được.Còn  serec cũng kô cần giải thích nhiều,chỉ cần mình biết lời giải là của mình,thế là ổn không cần phài viết những 3 bài để làm việc ko đâu(nếu có thừa thời gian thì hãy chăm chỉ học tip BĐT để người khác công nhận mình :D )


Cute!

[hide]For the sum of proper factors to be n, n must be a perfect number.
For the product of proper factors to be n, n must have 4 factors (1, itself, and 2 others).
The only number that satisfies both conditions is 6.[/hide]

sky9073 - proof for both claims? (and especially that no other numbers exist?)

Ok. Proofs..... I'll try.

[hide]For the sum of proper factors to be n, n must be a perfect number.
- definition of perfect number.. i think.

For the product of proper factors to be n, n must have 4 factors (1, itself, and 2 others).
- The product of proper factors of n = n^(number of factors/2 - 1).[/hide]

I'll have to think about how to prove the last bit.

Factors other than the number itself and 1.

Proof of the original (less precise) version of sky's second claim (i.e. for the product of n's proper factors to be n, n must have four factors):

[hide]Say n has at least three distinct proper factors.  Then if the prime factorization of n is n=(p_1^a)(p_2^b). . . then a+b+. . .>=3.  For if it is equal to 2, then n=pq or p^2, p and q prime, then n only has two proper factors, a contradiction.  So, other than p_1^a, p_2^b, etc, (p_1)*(p_2) must also be a proper factor of n because the sum of the exponents of primes in its factorization is >=3 so that p_1*p_2<n.  Therefore, if n has three or more distinct proper factors, their product is >n, because the product contains all the prime factors of n with their respective exponents, as well as extra repeats of at least two of those primes.[/hide]

I can prove that version of sky's claim, but I'm still thinking about how to prove this more precise version of her statement:
[quote="sky9073"]
The product of proper factors of n = n^(number of factors/2 - 1).
[/quote]

[quote="tetrahedr0n"]Here's a fun problem I thought of last night:

Find all positive integers $n$ such that the proper factors of $n$ have sum and product $n$.[/quote]
a1+a2+a3+a4+... none of these are 1, and its equal to a1a2a3a4... and then, we cant have more than 3 factors, because a1a2a3... increases exponentially while a1+a2+... increases arithmetically. Therefore (since 2*2*2>2+2+2) it can only have 2 factors. Since 2*3>2+3, both terms are 2. so 4 is the only number

I think I sort of got the last bit about there being no other numbers..

[hide]We could start by the fact that all the known perfect numbers are even. (I saw this in Intro NT... someone would have to prove there ARE odd perfect numbers before we can work on those)

So all known perfect numbers are divisible by 2.

Denoting a number that satisfies both conditions (k is perfect and has 4 factors) k, the proper factors of k are 1, 2, and k/2.
From "the sum of proper factors of k = k," 1 + 2 + k/2 = 3 + k/2 = k. k/2 = 3, k = 6
Therefore k must equal 6 and no other number....[/hide]

[quote="Scrambled"][quote="goodyfresh741"]Good proof scrambled.  But I still think we should work on proving sky's claim that (product of proper factors of n) = n^((number of factors)/2-1), which would be a much stronger result than anything we've proven in this discussion so far.[/quote]isnt that rather obvious because two proper factors of n multiply to get n. so we have n^[(n-2)/2]=n^[(n)/2-1][/quote]

But doesn't that only work if the number of proper factors is even?  Because if the number of factors is odd, then they may not necessarily be able to be paired up the way you stated.  Well, the way to do it, I just realized, is that n has an even number of proper factors if and only if n is not a perfect square.  Because, any factor a|n such that a<sqrt(n) has n/a as a distinct factor of n so that there are two factors total for every factor less than
sqrt(n), giving even number of factors for n.  If n=c^2 for some integer c, than c cannot be "paired up" with n/c but if you multiply c=sqrt(n) along with all the other proper factors, you again get n^(number of factors/2-1), so that after analyzing all possible cases this way, the proof is complete.

Yep, thats one way to think about it - the other way, which is very similar  but doesn't need you to worry about an odd/even number of factors, is to multiply out all the factors *twice*: eg for n=16 we have 1*16 * 2*8 * 4*4 * 8*2 * 16*1, or for n = 6 we have 1*6 * 2*3 * 3*2 * 6*1. That way, whether n is a square or not, we always get a product of $n^{\text{factors}}$, and we can then take the square root to calculate the product of the factors (and then divide by n to make it proper factors. Or you could have ignored the 1*n to begin with.)

I am sort of digging up an old post.  :blush:

But (I think) the question has not been completely answered yet, so hopefully no mods will come and jail me :D.

[hide]I think we have established that n has exactly four factors, so I will not bother to prove that.  We then have n = pq where p and q are primes.  We have
p + q + 1 = n = pq
(p+1)(q+1) = 2pq
Due to the fact that p and q are prime, the number of possible solutions is greatly reduced.  Assume p and q are both odd; it follows that LHS = 0 (mod 4) and RHS = 2 (mod 4).  Therefore, at least one of p and q is 2, the only even prime.  WLOG p = 2 --> 3(q+1) = 4q --> q = 3

n = pq = 6[/hide]

probability1.01, I hereby sentence you to....

Actually, I'm glad you posted, especially since its my question.

But if proper factors are defined as factors other than 1 and the number, then the proper factors of 6 are 2 and 3, which don't sum to 6...

According to the definition I am familiar with, the proper factors of n are all the (positive) factors except n (but not excluding 1).  I have sometimes seen the term nontrivial factors used to describe the (positive) factors other than 1 and n.

but couldn't n have only the factors p and p^2 (where p is prime) so that it also satisfies the condition that the product of the proper divisors equals n (which is p^3)?

Then n would be p^3, and the proper factors 1, p, and p^2.
Remember that the proper factors also have to ADD UP to n..
p^2 + p + 1 = p^3 doesn't give integer values for p.


Express $ \frac{1}{1\plus{}\frac{1}{1\plus{}\frac{3}{4}}}$ as a common fraction.


Functional Analysis
real analysis
real analysis unsolved

suppose $ H$ is Hilbert space.
Let $ L$ is belonging  $ H$ and is linear.suppose we have $ f: L \to R$ linear continuous map. then exist $ !$ extpanded  $ F$ on $ H$, such that  $ F|_{L} \equal{} f$ and $ \parallel{}F\parallel{} \equal{} \parallel{}f\parallel{}$

This is Hahn-Banach, only quite a bit easier because of the Hilbert space structure. If $ p$ is orthogonal projection onto $ H$, define $ F\equal{}f\circ p$.


calculus
integration
function
limit
derivative
algebra
functional equation

For arbiterary integers $n,$ find the continuous function $f(x)$ which satisfies the following equation. \[\lim_{h\rightarrow 0}\frac{1}{h}\int_{x-nh}^{x+nh}f(t) dt=2f(nx).\] Note that $x$ can range all real numbers and $f(1)=1.$

Note that the limit on the LHS of the given equation has $0/0$ form; naturally the first thing I thought was we should probably use L'Hospital's rule and see where that leads us to.

So, using L'Hospital's rule and differentiation under the integral sign, we get

$\lim_{h \to 0}\left( f(x+nh) \cdot n-f(x-nh) \cdot (-n)\right) = 2f(nx)$ ... [Note we are differentiating w.r.t. $h$]

$\Rightarrow 2nf(x) = 2f(nx)$

$\Rightarrow nf(x) = f(nx)$ ... $(*)$

The above is a nice functional equation, which resembles a lot like Cauchy's equation, viz. $f(x+y) = f(x)+f(y)$ (if I remember it correctly.) We suspect that $f(x) = x$ for all $x \in \mathbb{R}.$ Now, there is of course a well-known technique to handle such equations, and this is shown below.

[b]Claim 1[/b]: $f(x) = x$ for all $x \in \mathbb{Z}.$

Just put $x=1$ in $(*)$ to get $nf(1) = f(n) \Rightarrow f(n) = n\cdot 1 = n$ for all $n \in \mathbb{Z}.$
Also, for any integer $m > 1, f(0) = f(m \cdot 0) = m\cdot f(0) \Rightarrow f(0) = 0.$

Thus, our claim is verified.[b]

Claim 2[/b]: $f(x) = x$ for all $x \in \mathbb{Q}.$

Let $x = \frac{p}{q}$ where $p, q \in \mathbb{Z}$ and $q \not =0.$
Then, $p = p \cdot 1 = pf(1) = f(p\cdot 1) = f(qx) = qf(x) \Rightarrow f(x) = \frac{p}{q}= x.$

Thus, our claim is again verified.

[b]Claim (3)[/b]: $f(x) = x$ for all $x \in \mathbb{R}.$
Now for any $x \in \mathbb{R}$, let ${x_{n}}$ be a sequence of rational numbers such that $x_{n}\to x.$ Then, by continuity of $f,$ we have $f(x) = f(\lim_{x_{n}\to x}x_{n}) = \lim_{x_{n}\to x}f(x_{n}) = \lim_{x_{n}\to x}x_{n}\cdot f(1) = \lim_{x_{n}\to x}x_{n}\cdot 1 = x.$

The last claim gives us our answer, viz. $f(x) = x$ for all $x \in \mathbb{R}.$ And, we are done.

I rather imagined
$F(x) = \int f(x) dx$, so the LHS becomes
$\frac{F(x+nh)-F(x-nh)}{h}$
setting $x' = x-nh$ (and $dx' = dx$)
$2n \frac{F(x'+2nh)-F(x')}{2nh}= 2n F'(x') = 2n f(x')$
since we're ultimately thinking of $h \to 0$, $2n f(x') = \boxed{2n f(x)}$

[quote="FieryHydra"]So, using L'Hospital's rule and differentiation under the integral sign, we get

$\lim_{h \to 0}\left( f(x+nh) \cdot n-f(x-nh) \cdot (-n)\right) = 2f(nx)$ ... [Note we are differentiating w.r.t. $h$]
[/quote]

Well, I didn't get how you did it with L'Hospital's. Can you do it explicitly?


Find all solutions to this equation:

$ \sin 3x \cos x - \cos 3x \sin x = \displaystyle \frac{\sqrt2}{2} $

any quick or efficient solutions?  :D  thanks

[hide="Hint"]
The LHS is equal to $\sin(3x-x)$
[/hide]

I suppose I'll finish it off.

[hide]

sin(2x) = sqrt(2)/2

sqrt(2)/2 = 45 degrees and 135 degrees and 405 degrees and 495 degrees

Assuming 0 <= x <= 2pi...

x = 22.5 degrees, 67.5 degrees, 202.5 degrees, and 247.5 degrees.

[/hide]

ahhhhhh! i see! very slick, i didnt catch the sin(a-b) formula until you pointed it out. thanks a bunch.  :lol:


calculus
algorithm
videos
linear algebra

Is calculus important to the people who specialize in computer science and technology? If your answer is "yes", please explain why.

In my mind, I think other math subjects with "discreteness", such as discrete math, linear algebra, and algorithm, are more important than calculus, which mainly focus on continuous object. However, the students majoring in computer science and technology should have a strong background on calculus. I would like to know the reason. (I major in math)

I would say it is important to 'many' undergraduates who specialize in computer science.  Insofar as there is no way to tell who will or will not wind up needing it, it's reasonable to give all CS students at least some background in calculus.  For graduate students, however, I would change 'many' to 'most'.

A few examples of computer science fields where calculus, or at least principles similar to those underlying calculus, are used extensively: Computer graphics, robotic motion/vision, image/voice recognition, audio and video data compression, digital signal processing.

But I think the biggest one, and one that deserves its own category, is numerical computing.  Essentially every field of applied science or engineering eventually needs computer algorithms to numerically simulate all sorts of models of continuous (or approximately continuous) phenomena.  An understanding of calculus and differential equations in this context is absolutely necessary to develop stable and efficient algorithms (though of course a solid understanding of linear algebra is also crucial).  A few examples: [url=http://en.wikipedia.org/wiki/Computational_chemistry]computational chemistry[/url], [url=http://en.wikipedia.org/wiki/Computational_physics]computational physics[/url], [url=http://en.wikipedia.org/wiki/Computational_neuroscience]computational neuroscience[/url], [url=http://en.wikipedia.org/wiki/Bioinformatics]bioinformatics[/url], [url=http://en.wikipedia.org/wiki/Neuroimaging]neuroimaging[/url]; though I'm sure others could mention many more.


An ice cube tray is in the form of a 2 by 8 rectangle, as shown. In how many different ways can you remove half of the ice cubes in a full tray such that none of the ice cubes remaining in the tray are next to each other horizontally or vertically?

[asy]draw((0,0)--(8,0), linewidth(1));

draw((0,1)--(8,1), linewidth(1));

draw((0,2)--(8,2), linewidth(1));

draw((0,0)--(0,2), linewidth(1));
draw((1,0)--(1,2), linewidth(1));
draw((2,0)--(2,2), linewidth(1));
draw((3,0)--(3,2), linewidth(1));
draw((4,0)--(4,2), linewidth(1));
draw((5,0)--(5,2), linewidth(1));
draw((6,0)--(6,2), linewidth(1));
draw((7,0)--(7,2), linewidth(1));
draw((8,0)--(8,2), linewidth(1));[/asy]

the only way to do that is if the icecubes are removed diagonally

therefore there are only $ 2$ ways

Sort of "diagonally", if diagonally means picking cubes out in a checkerboard pattern.  Basically you must pull one cube out of each column, but that determines how you pick the adjacent column all the way across.  So, depending on how you pick the first cube (top or bottom) you'll get one of the two patterns.


ARML email geometry calculus AMC USA(J)MO USAMO

yeah so basically i enjoyed it, the competition and the 7-hour bus rides

someone threatened to do me in my sleep but i don't think he did

and fanatic/mihika kinda owned me/jeffrey in kemps alsdjfim yeah we're gonna beat you next year

and uhh i got a cold from playing ultimate frisbee in the rain for an hour so bye and see you all next year

to be fair, mihika and I owned SEVERAL people in kemps, not just you and jeffrey :p. Although I didn't keep track all the way, at one point our record was 25 wins, 7 losses. In the end it was probably like 54-15 or so... not counting the free wins we gave you guys when playing to 10 or 15.

/bragging rights. lol... taking pride in kemps.

ERS... cut underneath my fingernail!! nasty... stung all day for two straight days because I cut it again the second day.

Except lyndon ignoring me and running away when I walked by, the three days were great (even playing ultimate frisbee in the rain and stepping in puddles)!

Of the people that live in Indiana, I think I know like everyone's score... asked all around.

[quote="cxywe"]This is pretty rediculous.  I was gone for 3 days and I havent even come to this site but I cant log on to my other account.  I havent even logegd on so I couldnt have done anything, so I find this really weird.  Also, Richard, pm me on everything rebirth said.[/quote]

Send me an email, as I explained in a different thread.  There is a great deal of evidence that you have formed many, many accounts on the site.  This is very clearly against our rules.  Furthermore, some of the conduct from some of these accounts has been very inappropriate.  Perhaps some of this conduct is coming from a sibling or someone else in your household, but at this point, I will continue banning the accounts I see you form (including this newest one) until I have a explanation for the behavior we have seen, and an assurance that it will not continue.

[quote="Fanatic"]to be fair, mihika and I owned SEVERAL people in kemps, not just you and jeffrey :p. Although I didn't keep track all the way, at one point our record was 25 wins, 7 losses. In the end it was probably like 54-15 or so... not counting the free wins we gave you guys when playing to 10 or 15.

/bragging rights. lol... taking pride in kemps.[/quote]

haha, that was great. I still don't think anyone found out our signal (or lack thereof). remember our strategy for next year so we can beat everyone all over again.

I thought the trip was great. the competition, however...not so much. I did really badly. It was surprising cuz i was getting around 4's consistantly on the practice tests i took (95-03) and then i bombed the actual thing. i sucked. it was shameful

Does anyone know how they pick the teams? And does it have anything to do with going to the practices (which i conveniently didn't figure out about till they were all over)? Thanks.

They pick the teams based off scores on other contests and... drumroll... practices!

You had no shot for being on gold if you didn't do practices =/. the coaches decide touchy-feely, I don't think there's some special formula for deciding who gets on which team.

thanks. when do practices start? and they are in terre haute right?

did anyone get number 2 on the first relay and if anyone did, how did they do it. it was the one which was like "what is the sum of all the values of x if sin x = -1/4"

assuming $x$ is measured in radians and is defined as $0 \leq x < 2\pi$

the terminal sides of the two possible values of $x$ ends up in quadrants III and IV.

think of the two angles separately (drawing would help), and subtract $\pi$ from both angles. It becomes obvious the remaining angles add up to $\pi$, so $2\pi+\pi = \boxed{3\pi}$

[quote="indianamath"]omg the bus was like oxygen-deprived.

i had a 7-hour headache on the bus each way. the bus was nice, but the bus ride was torture for me. i felt like throwing up so bad.

so....anybody agree with me?[/quote]
uh
did you go through 5 hours of INTENSE HEAT IN THE DESERT WITH NO AIR CONDITIONING?
I THINK NOT
YOU
GOT
PWNED

[quote="indianamath"]omg the bus was like oxygen-deprived.

i had a 7-hour headache on the bus each way. the bus was nice, but the bus ride was torture for me. i felt like throwing up so bad.

so....anybody agree with me?[/quote]

it was really hot on the way back. u were lucky u slept through most of it. i swear, temperature must've broken 90 somewhere along the way. the bus was air-conditioned on the way there, so it wasn't that bad.

i had a headache the whole ride both ways.

[quote="indianamath"]did you play ultimate frisbee ina thunderstorm for 40 minutes?

[hide=" :lol: "]I
THINK
NOT
OWNT.[/hide][/quote]
well we did play ultimate frisbee in the sprinklers
does that count
NOT TO MENTION THE INTENSE HEAT OF LAS VEGAS

PWNED

dudeeee.. playing ultimate in the storm was like the highlight of my week

and yeah the bus ride was torture and i also had a stomach ache.

well the last time i've been to IUPUI (science fair asldifjlke) i think it took a little more than an hour, but i could be wrong

uhh by "hard math" do you mean beyond ARML level? cause i was hopeless at last thursday's practice.. :blush:

[quote="Fanatic"]Just wondering -- why're you visiting indiana's forum?

Because we rock?

lol

we're a bunch of 7th, 8th, and 9th graders =/[/quote]
No 6th graders? Come on :(  If I had gone to ARML, I would have gotten 0, so I didn't go. I don't think any 6th graders went from MI this year, so I'm fine. Maybe next year :ninja:

is it just me, or did the ARML individual have WAY more geometry than normal?

geometry is cool. ish.

sorry if that sounds hypocritical, but it's how i feel right now.

aaaand i just took the 2001 AMC 10 without a calculator and got a 138 alsidjflkefmasdf NOT GOOD ENOUGH

let's go exeter!

haha sorry. i was on the indiana team last year but then changed schools to exeter.  who was on the gold team this year?

Like, no one...

Nathan Benjamin (9th)
Phil Thomas (10th)
Sachin Shinde (9th)
Samuel Dittmer (10th? 11th?)
E.G. Wright (11th, I think)
Ashley Neagle (9th? 10th? 11th?)
Tyler Zhou (10th? 11th?)
Lauren Stephens (9th)
Jeffrey Shen (8th)
Richard Ni (8th)
Soo-Hyun Yoo (8th)
Yuqing Zhang (7th)
Lyndon Ji (7th)
Shawn Qian (9th)
??? can't think of last person

two 7th
three 8th
four 9th

so what are all the subjects you need to know to do well in ARML? I saw some of mathcrazed's posts that you need to know all of high school math (like up to calculus), but then i also heard (from what i consider a pretty reliable source) that most of it is under precalc level, with the occasional precalc problem.

so i'm planning to finish the geometry and hopefully algebra II honors (precalc) texbooks this summer. will that be enough to have a good knowledge base for ARML?

i just took some amc 12 without a calc and got a 103.5(ouch ouch ouch ouch)

i think i got like 15 correct, 1 wrong, and 9 ommited.

You guys sent a recruitment letter to me via my school. That was pretty shady...  :ninja:

[quote="Fanatic"]Just wondering -- why're you visiting indiana's forum?

Because we rock?

lol

we're a bunch of 7th, 8th, and 9th graders =/[/quote]
because its spamtastic

[quote="Fanatic"]Just wondering -- why're you visiting indiana's forum?

Because we rock?

lol

we're a bunch of 7th, 8th, and 9th graders =/[/quote]
No 6th graders?? :o

i think we had one.

the guy from honey creek named Alex.

Mark Botros is a 7th grader (going into 8th).

Alex is a 6th grader, was on the silver team.

[quote="indianamath"]the last person was yixin.

and sam is going into 11th.[/quote]
im pretty sure sam is going into 12th

ah i was just glad i was on the gold team.

[quote="mwpl11"][quote="Fanatic"]Well, technically, there wasn't any thunder...

West Lafayette people, how long do you think it'll take for you guys to drive to IUPUI? We have math practices there at times, and it's free (not for ARML, just really hard math in general). I'll talk to Dr. Ji later?[/quote]

it takes about an hour to get to indianapolis. when do these practices start? is there some sort of schedule that they follow?

and when do the arml practices for this year start (if anyone knows)?[/quote]

hmm ill just bump this back up.

after USAMO is where it starts.  I think first couple  of weeks is at Dr. Ji's class (IUPUI) and then its at Terre Haute at Rose Hulman.
I dont think ill be able to go to most of them (only the ones after like May 15ish) so I hope they dont use your practice scores for placement.

And btw, did any of you guys sign up for ICTM.  I dont think I did, but i was wondering can you just show up and take the test (i probably wont go-and i forgot to sign up anyways-but just in case im not busy, can i just show up to it or something?)

i think i will, if i am allowed to go by the rest of indiana.

somemod should make this 2008 so no one is confused.

I'm going. I jokingly mentioned to Dr. Ji that I wanted to bring a couple non-mathnerd friends and he was like "yeah! the more the merrier!"... so I'm pretty sure that there'll be a spot for you and your (I forget what it was, but it was pretty good) score on the AIME.


algebra
polynomial
Rational Root Theorem
complex numbers
number theory

hi !!
find the number of roots$ \in{\mathbb{Q}}$ for this equation :

$ x^{4}+x^{3}+x^{2}-6=0$

thanx.

Just use this ([url=http://en.wikipedia.org/wiki/Rational_root_theorem]Rational root theorem[/url]): if polynomial $ a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \ldots \plus{} a_1x \plus{} a_0$ with integer coefficients has a rational root $ \frac {k}{l}$ ($ gcd(k,l) \equal{} 1$), then $ k|a_0$ and $ l|a_n$. In your case, it will be particularily easy to check all the possible roots - there will not be many of them and they will all be integers.

[size=75](Just a note: I don't think this question belongs in this subforum.)[/size]

Checking the possibilities:$ \pm 1, \pm2, \pm3, \pm6$, we find that there are no integer roots and therefore, since the polynomial is monic, no rational roots.

Differentiating, we get $ 4x^{3} \plus{} 3x^{2} \plus{}2x$. Setting this equal to zero, we see that there is only one turning point:  a local minimum at $ (0, \minus{}6)$.

Also since, for $ f(x) \equal{} x^{4} \plus{} x^{3} \plus{} x^{2} \minus{}6$,  $ f(\minus{}2) \equal{} 6, f(\minus{}1) \equal{} \minus{}5; f(1) \equal{} \minus{}3, f(2)\equal{} 28$, we find that there are two real roots, one between $ \minus{}2$ and $ \minus{}1$ and one between $ 1$ and $ 2$ and therefore two complex (conjugate) roots.

[hide="Number Theory solution"]Assume that there is indeed a rational solution. Since it isn't 0, we know that it can be expressed as a fraction $ \frac{p}{q}$, where $ \gcd(p,q)\equal{}1$, and $ \frac{p^4}{q^4}\plus{}\frac{p^3q}{q^4}\plus{}\frac{p^2q^2}{q^4}\minus{}\frac{6q^4}{q^4}\equal{}0$, which means $ p^4\plus{}p^3q\plus{}p^2q^2\minus{}6q^4\equal{}0$. Since $ \frac{p}{q}\neq \pm 1$, we know that one of $ |p|$ or $ |q|$ is greater than 1.

[b]Case 1: $ |p|>1$[/b]
Then we know that there exists a prime $ p_1$ such that $ p_1|p$. Note that $ p_1^2|p^4\plus{}p^3q\plus{}p^2q^2$. Since $ p^4\plus{}p^3q\plus{}p^2q^2\minus{}6q^4\equal{}0$, $ p_1^2|6q^4$. This means that $ p_1|q$. But this violates the assumption that $ \gcd(p,q)\equal{}1$, and hence there is no rational root such that $ |p|>1$.

[b]Case 2: $ |q|>1$[/b]
Then we know that there exists a prime $ q_1$ such that $ q_1|q$. Note that $ q_1|p^3q\plus{}p^2q^26q^4$. Since $ p^4\plus{}p^3q\plus{}p^2q^2\minus{}6q^4\equal{}0$, $ q_1|p^4$. This means that $ q_1|p$. But this violates the assumption that $ \gcd(p,q)\equal{}1$, and hence there is no rational root such that $ |q|>1$.

This means that there are no rational roots period.[/hide]

Er, 1=2, I think that's essentially how you prove the rational root theorem...


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

$p$ is prime.
Is it true that $\forall A \in \mathcal{M}_{n}(\mathbb{Z}),\ \mu_{A^{p}}\equiv\mu_{A}~[p] \textnormal{,}$ where $\mu_{A}$ is the minimal polnomial of $A$?

Note: It is true for the characteristic polynomial $\chi_{A}$.

I don't recognize the notation- is that meant to be the minimal polynomial of $A$ considered as a matrix mod $p$, compared to $\mu_{A}$ considered mod $p$?

If so, it's trivially false; let $M$ be any nonzero nilpotent matrix, and let $A=pM$. The two minimal polynomials have different degrees.

jmerry,

You're right, it was obvious, I should have thought about it first:)
Try to prove it for the characteristic polynomial, it's less obvious (and true this time).

one can trigonalize $A$ in an algebraic closure $K$ of $\mathbb{F}_{p}$ and use the fact that $(a+b)^{p}= a^{p}+b^{p}$ for $a,b \in K$.
$\chi_{A}(x)=\prod (x-\alpha_{i}) = \sum a_{k}x^{k}$ and $\chi_{A^{p}}= \prod (x-\alpha_{i}^{p}) = \sum b_{k}x^{k}$.

There are symetric polynomials $P_{k}$ such that $P_{k}(\alpha_{1}, \ldots, \alpha_{n}) = a_{k}$ and $P_{k}(\alpha_{1}^{p}, \ldots, \alpha_{n}^{p}) = b_{k}$. But $b_{k}=P_{k}(\alpha_{1}^{p}, \ldots, \alpha_{n}^{p}) = P_{k}(\alpha_{1}, \ldots, \alpha_{n})^{p}= a_{k}^{p}$. And because $a_{k}\in \mathbb{F}_{p}$, $a_{k}^{p}= a_{k}$. This shows $\chi_{A}= \chi_{A^{p}}$ modulo $p$.

I prefer this one-liner:
The coefficients of the characteristic polynomial are polynomial (with integer coefficients) in the entries of $A$. These polynomials commute with ring homomorphisms.

I don't realy get it  :D  but maybe, if you elaborate too much it will achieve 4 lines  :)


I am the moderator for the group.  I started it to increase communication in the state, and to give the national team member somewhere to ask each other questions.

Gee, can't Oklahoma be in the spotlight for just a smidgeon?

oh dude I remember oklahoma! that's with you and neal and the crazy indian guy OH MAN IT'S OKLAHOMA hmm was the school jenkins? I believe it was jenkins? perhaps?

hey i know neal and his family really well

Dis is probably a dead topic, but there is like no one here. I MUST CREATE POPULARITY FOR THE OKLAHOMA FORUM!!!!  :D


function
analytic geometry
graphing lines
slope
calculus
derivative

If a fucntion is x^3 - x^2 -4x+4, and there is a point (a,b) on the graph on which the tangent line passes through the point (0,-8), what are the values of a and b.

Another one:
With the graph of x^2-xy+y^2=9, at the point (0,3) find the rate of change of the slope of the curve with respect to x

i didn't quite understand your first problem but the second problem seems very easy:
just figure out the second derivative of the curve wtith respect to x, i got:
second derivative: [((y-2x)/(2y-x))-2]/[((2y-4x)/(2y-x))-1]
plugging in the values of x and y, you get:
second derivative equals:  -1.5/0. which means that, the rate of change of slope of the curve is infinity.
in orhter vords, the slope of the curve is not differentiable at that point.

Anyway I dont really understand the question but shouldn't it be moved to the Calculus section.

For the first, b = f'(a)*a - 8.  $f'(x) = 3x^2 - 2x - 4$.  Thus $b = 3a^3 - 2a^2 - 4a - 8$.  Now we also have b = f(a) $\implies b = a^3 - a^2 - 4a + 4$.  Now we have $2a^3 - a^2 - 12 = 0$.  Go solve the cubic on your own :P.

Edit--

$2a^3 - a^2 - 12 = (a - 2)(2a^2 + 3a + 6)$

So a = 2.

$2a^3-a^2-12 = (a-2)(2a^2+3a+6) \Rightarrow a=2$ is the only real solution.

EDIT - probability's edit beat me to it.


Given a positive integer $a_{0}$, we construct a sequence as follows: If the unit digit of $a_{i}$ does not exceed $5$, then $a_{i+1}$ is obtained by deleting this digit (If nothing remains upon this deletion, the sequence ends). Otherwise, $a_{i+1}=9a_{i}$. Can the sequence be infinite?

[hide]
The sequence cannot be infinite. Suppose there exists such an infinite sequence. Let $L$ be the least positive integer such that $a_{0}=L$ produces infinite sequence. The units digit of $L$ must exceed 5, since the sequence ends automatically if $1\leq L\leq 5$, and if $L>5$ but has units digit at most 5, then
\[a_{1}=\left\lfloor \frac{L}{10}\right\rfloor \leq \frac{L}{10}<L\]
This means the sequence $a_{n+1}$ for nonnegative integers $n$ starts at a positive integer less than $L$ and is infinite, a contradiction.

Now that the units digit of $L$ is at least 6, we have $a_{1}=9L$. In particular, $a_{1}\equiv-L\pmod{10}$ so the units digit of $a_{1}$ is at most 4. Therefore, we get
\[a_{2}=\left\lfloor \frac{a_{1}}{10}\right\rfloor =\left\lfloor \frac{9L}{10}\right\rfloor\]
and the sequence starting with $a_{2}$ is infinite, again a contradiction. So no infinite sequence exists.
[/hide]


I don't understand your question
Do you mean $x^4 + 3x^3 + 6x^2 + ax + 4 =0$ has root $x_1, x_2,x_3$ and $x_4$, Find an $a$ such that  $x_1 = 1/x_2 + 1/x_3 + 1/x_4$?

[quote="Arne"]Seems like it.

In that case, notice that $1 = \frac{1}{x_1x_2} + \frac{1}{x_1x_3} + \frac{1}{x_1x_4} = \frac{x_2x_3 + x_2x_4 + x_3x_4}{x_1x_2x_3x_4}$.

Now, $\frac{x_2x_3 + x_2x_4 + x_3x_4}{x_1x_2x_3x_4} = \frac{\left(x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4\right) - x_1\left(x_2 + x_3 + x_4\right)}{x_1x_2x_3x_4}$.

Using Vieta's relations, we get $1 = \frac{6 - x_1\left(-3 - x_1\right)}{4} \Rightarrow x_1 \in \left\{-1,-2\right\}$.

Hence

$a = -x_1x_2x_3x_4\left(\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \frac{1}{x_4}\right) = -4 \left(\frac{1}{x_1} + x_1\right)$

so $a = 8$ or $a = 10$.

I probably made some mistake in my calculations, but oh well, you get the idea.

Can some mod move this to the right section?[/quote]

can you please  explain what is vieta's relation

Vieta's relations give us the connection between the roots of a polynomial and its coefficients.

[url=http://mathworld.wolfram.com/VietasFormulas.html]Vieta's relations (Mathworld)[/url]


MATHCOUNTS
search
number theory
relatively prime

5. A relatively prime date is a date for which the number of the month and the number of the day are relatively prime. How many relatively prime dates are in the month with the fewest relatively prime dates?

you could always test all 365 days...but whats a faster way

10. ABCD is a square. How many squares have two or more vertices in the set {A,B,C,D}?

I keep getting 9...I only count square ABCD once though.

number 10 is 9, although the answer key says like 13.

Er...no? Consider the center of ABCD and choose any two adjacient vertices of ABCD. That's four. Now, there are four where you just choose any two adjacient vertices as a side and then two more points outside of ABCD. There's also four by using diagonals, two per diagonal. Then there's the original square. Hence, 4+4+4+1=13.

[hide="solution for 5"] The number for the month is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, or 12. Clearly neither 1 nor any of the primes provide the least number of relatively prime days. We're left with 4, 6, 8, 9, 10, or 12. The numbers relatively prime to 4 or 8 are the same as those to 2, so neither of those two work. Same reasoning with 9. Now just check 6, 10, and 12. 6 gives 30-15-10+5=10. 10 gives 31-15-6+3=13. 12 also has 2 and 3 as factors, but since it has mroe days than 6, we know that it can have 10 at best. Thus, our answer is 10.[/hide]

Well your Mathcounts class was wrong.

EDIT: A quick search gives the topic that is Week 7 Problem 2 in the Mathcounts Problem Series, where the answer is 13.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1500295517&t=217457]Some may not be able to see[/url].


modular arithmetic
number theory unsolved
number theory

a)Prove between each 200 integer we can find hundred nums taht the sum of them is divisible by the hundred
b)generalize it.

do you mean to say "prove that among any 200 integers one can find 100 of them whose sum is divisible by 100"?

if so, this is a consequence of the well-known result of Erdos, Ginzburg, and Ziv:

[i]Among any $ 2n \minus{} 1$ integers, one can find $ n$ whose sum is divisible by $ n$.[/i]

This result is sharp in that $ 2n \minus{} 1$ cannot be replaced with $ 2n \minus{} 2$: we may pick $ n \minus{} 1$ numbers that are $ \equiv 0\pmod{n}$ and another $ n \minus{} 1$ that are $ \equiv 1\pmod{n}$, and then no $ n$ of them have a sum div. by $ n$.

A proof can be found here http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1730989#1730989


Find $ z \in C^*$ s.t exist $ r\in(2,\infty)$ and$ \displaystyle |z\plus{}\frac{1}{z}|\equal{}|z^3\plus{}\frac{1}{z^3}|\equal{}r$.

[quote="posabogdan"]Find $ z \in C^*$ s.t exist $ r\in(2,\infty)$ and$ \displaystyle |z \plus{} \frac {1}{z}| \equal{} |z^3 \plus{} \frac {1}{z^3}| \equal{} r$.[/quote]

$ z^3\plus{}\frac 1{z^3}\equal{}(z\plus{}\frac 1z)^3\minus{}3(z\plus{}\frac 1z)$. So, let $ z_1\equal{}z\plus{}\frac 1z$ and the equation becomes $ |z_1|\equal{}|z_1^3\minus{}3z_1|>2$. So $ |z_1|\neq 0$ and this is equivalent to :

$ |z_1^2\minus{}3|\equal{}1$ and $ |z_1|>2$. So $ z_1^2$ is on the circle centered on $ (3,0)$ and whose radius is $ 1$.

But for all these points, we have $ |z_1|^2\leq 4$ and so $ |z_1|\leq 2$

So no solution.


topology
real analysis
real analysis unsolved

Let $(X,d)$ be a metric space, and $f:X\to X$ a map s.t. $d(f(x),f(y))<d(x,y),\ \forall x\ne y\in X$. Assume there is an $a\in X$ s.t. $a,f(a),(f\circ f)(a),\ldots$ has a subsequence converging to $\ell\in X$. Prove that $\ell$ is the unique fixed point of $f$.

Let me revitalize this very nice topic. Let us consider the sequence defined by $a_{0}=a$ and $a_{n+1}=f(a_{n})$. The assumption is that $a_{n}$ has a subsequence $a_{n_{k}}$ converging to $l$.

The sequence $d_{n}=d(a_{n+1},a_{n})$ is non-increasing, hence it has a limit $\lambda\geq 0$. Now for every $k$ we have that
\[d(l,f(l))\leq d(l,a_{n_{k}})+d(a_{n_{k}},a_{n_{k}+1})+d(a_{n_{k}+1},f(l))\]
(note that the index is $n_{k}+1$ and not $n_{k+1}$)  and
\[d(l,f(l))\geq d(a_{n_{k}},a_{n_{k}+1})-d(l,a_{n_{k}})-d(a_{n_{k}+1},f(l)).\]

Letting $k\to +\infty$ we have that $d(l,a_{n_{k}})\to 0$, and also $d(a_{n_{k}+1},f(l))=d(f(l),f(a_{n_{k}}))\to 0$ because it is smaller, and by definition $d(a_{n_{k}},a_{n_{k}+1})\to\lambda$. So we have proved that $d(l,f(l))= \lambda$. If $\lambda=0$, we have proved that $l$ is a fixed point.

Assume now that $\lambda>0$. Arguing as before we have that
\[d(f(l),f(f((l)))\leq
d(f(l),a_{n_{k}+1})+d(a_{n_{k}+1},a_{n_{k}+2})+d(a_{n_{k}+2},f(f(l))),\]
and
\[d(f(l),f(f((l)))\geq d(a_{n_{k}+1},a_{n_{k}+2})-d(f(l),a_{n_{k}+1})-
d(a_{n_{k}+2},f(f(l))),\]
and letting $k\to +\infty$ we deduce that $d(f(l),f(f(l)))= \lambda=d(l,f(l))$, and this is absurd if $f$ is a weak contraction.

Let me add two more problems inspired by grobber's post.

1 - Assume that $f$ is a weak contraction with a fixed point $l$. Is it true that Grobber's sequence has a subsequence converging to $l$ for every $a$?

2 - Assume that $f$ is a weak contraction and that, for a given $a$, Grobber's sequence has a subsequence converging to $l$. Is it true that the whole sequence converges to $l$?

The answer to the second one is 'yes', since we know that $\ell$ is a fixed point, so we have $d(f^{(n+1)}(a),\ell)=d(f^{(n+1)}(a),f(\ell))<d(f^{(n)}(a),\ell)$, i.e. the distance from $f^{(n)}(a)$ to $\ell$ decreases as $n\to\infty$.

As for the first one, I don't think it's true. We can take $X$ to be the subspace of $\mathbb R$ consisting of $0$ and $[-2,-1)\cup(1,2]$. We can now put $f(0)=0$ and $f(x)=\frac{1+x}2,\ \forall x>1,\ f(x)=\frac{-1+x}2,\ \forall x<-1$.

[quote="grobber"] i.e. the distance from $f^{(n)}(a)$ to $\ell$ decreases as $n\to\infty$.[/quote]

But this doesn't mean that the distance tends to 0 ...

[quote="grobber"]As for the first one, I don't think it's true. We can take $X$ to be the subspace of $\mathbb R$ consisting of $0$ and $[-2,-1)\cup(1,2]$. We can now put $f(0)=0$ and $f(x)=\frac{1+x}2,\ \forall x>1,\ f(x)=\frac{-1+x}2,\ \forall x<-1$.[/quote]

Right.

[quote="Xamog"][quote="grobber"] i.e. the distance from $f^{(n)}(a)$ to $\ell$ decreases as $n\to\infty$.[/quote]

But this doesn't mean that the distance tends to 0 ...

[/quote]

It does here, since it tends to $0$ for a subsequence (remember, we're talking about your second problem here, and you said there was a subsequence converging to $\ell$).


quadratics
modular arithmetic
number theory unsolved
number theory

can we choose some numbers from n, n+1, n+2, n+3 ,..., n+27, product of chosen numbers equals to product of other numbers?
n is natural number.

[quote="Evariste-Galois"]Very easy!
If 29 divide not one of the terms
n(n+1).......(n+27)=-1mod29 or -1 isn't a quadratic residue mod 29.
If 29 divide one of the terms , 29 is a factor only for one of the terms! so 29 can't be a factor of the other hand of the equality.
So there is no solution.[/quote]

$ \minus{}1$ is a quadratic residue $ \pmod{29}$ since $ 12^2 \equiv \minus{}1 \pmod{29}$

that's right!! I did a mistake in calculation. :blush:
I will check this later!

[quote="pco"][quote="Evariste-Galois"]Very easy!
If 29 divide not one of the terms
n(n+1).......(n+27)=-1mod29 or -1 isn't a quadratic residue mod 29.
If 29 divide one of the terms , 29 is a factor only for one of the terms! so 29 can't be a factor of the other hand of the equality.
So there is no solution.[/quote]

Nice :)[/quote]

Not so nice, ...  :blush:


Do the following:
/ignore LanguageFilter

You actually don't receive the warning "Hey! No...language", but sadly, you still get kicked.


calculus
derivative
function
calculus computations

[quote="iamhe"]How do you find the derivative of an inverse?
For example, if given that [i]f'(-2)=-7[/i], how do you find the original function?

Thanks[/i][/quote] Do you know about the inverse rule theorem?

[hide]I think it is something like this. Let $ g$ be the inverse of $ f$ i.e if $ y\equal{}f(x)$, then $ x\equal{}g(y)$. Suppose that $ f$ is differentiable at $ g(a)$ and the derivative is non zero. Then $ g$ is differentiable at $ a$ and $ \frac{dg}{dy}(a)\equal{}\frac{1}{\frac{df}{dx}(g(a))}$. In other words: $ g'(y)\equal{}\frac{1}{f'(g(y))}$. Can you use this?[/hide]

Quite simply. We go by the rule $ f(f^{\minus{}1}(x)) \equal{} x$ and use implicit differentiation (don't forget the chain rule). The result is pretty clear from there.

Uhhh,
Could you please elaborate on the implicit differentiation and the chain rule? I am a beginner at calculus and don't understand what you are talking about.
I think I know a little about the chain rule.
Isn't that the derivative of a composition?

Thanks for the help. :)

Yes, the chain rule is how you differentiate a composition. It's all you need to calculate the derivative of an inverse- if you assume that inverse is differentiable.
There are theorems that imply such functions are differentiable when they "should" be- those theorems belong to multivariable calculus.


Έστω $ n>2$ ένας ακέραιος και έστω $ x_{1},x_{2},...,x_{n}$ θετικοί αριθμοί με άθροισμα $ 1$. Να αποδείξετε ότι $ \prod^{n}_{i\equal{}1}\left(1\plus{}\frac{1}{x_{i}}\right)\geq \prod^{n}_{i\equal{}1}\left(\frac{n\minus{}x_{i}}{1\minus{}x_{i}}\right)$.

Prepei na ginei xrisi tis anisotitas tou Huygens...otan exo xrono tha valo tin lysi..nomizo tin exo dei sto Crux tin askisi

Βαζω μια λυση στα γρηγορα οποτε μπορει να μου εχει ξεφυγει κατι

[hide]Απλα $ n \minus{} 1$ φορες Jensen στην $ f(x) \equal{} ln\left(1 \plus{} \frac {1}{x}\right)$ για τους $ n$ αριθμους ανα $ n \minus{} 1$  και μετα προσθεση κατα μελη.[/hide]