Ring Theory
function
integration
algebra
polynomial
superior algebra
superior algebra unsolved

Sigh... The saga of the Romanian olympiad continues... :D

(1)
Let A be a ring

a)Show that Z(A)={a in A | ax=xa for all x in A} is a subring of A.
b)Show that if every commutative subring  of A (=/=A) is a field then so is A.

(2)
Let f:[0,1]->R be a Riemann integrable function s.t. 0<| \int{0->1}f(x)dx |<=1. Show that there exist x_1=/=x_2 in [0,1] s.t  \int {x_1->x_2}f(x)dx = (x_1-x_2) 2002 .

(3)
Let f:R->R be a continuous and bounded function. If x \int {x->x+1}f(t)dt =  \int {0->x}f(t)dt for all real x, then f is constant.

(4)
Let K be a field with q=p n elements, where p is a prime and n>=2. For a in K define the polynomial f_a(x)=x q -x+a. Show that:

a)f_1 divides f=(x p - x) q - (x p - x).
b)f_a has at least p  n-1 irreducible divisors in K[x]; here polynomials f and a*f with a in K* are not considered to be distinct.

Sorry I don't have the authors' name.

(1) Ion Savu
(2) Radu Gologan
(3) Mihai Pitcari
(4) Marian  Andronache

ah let's see 1 is pretty easy, 1a) is trivial actually, as for 1b) just take the commutative ring to be A(x) for an element x in A, where A(x) is the ring having elements f(x) where f is a polynomial with coefficients in Z(A). Obviously this is a sub-ring of A, and furthermore it is commutative, hence a field, thus x (which is a non-zero element in the field) is invertible, and because x was arbitrarly chosen  among the elements of A we are done.

[quote="grobber"]

(4)
Let K be a field with q=p n elements, where p is a prime and n>=2. For a in K define the polynomial f_a(x)=x q -x+a. Show that:

a)f_1 divides f=(x p - x) q - (x p - x).
[/quote]

a)
K is a finite field , it is commutative (theorem Wedderburn) useful
for product of polynomial.
K* is a multiplicative group of order q=p^n - 1 Lagrange theorem =>
x^(q-1)=1 => x^q=x
...


algebra
polynomial
induction
greatest common divisor

1. A sequence is defined by a_0 = 1 and a_(n+1) = (a_n)^2 + 1 for each n>=0.  Find the greatest common divisor of a_999 and a_2004.

2.  There exists a polynomial P of degree 5 with the following property:  if z is a complex number such that z^5 + 2004z = 1, then P(z^2) = 0. Calculate P(1)/P(-1).

How can $\ P(x)$ be a fifth degree polynomial if it has $\ 10$ roots?

It has five roots.  The problem statement means that if $z$ is a root of some other fifth degree polynomial, then $z^{2}$ is a root of $P$.  That's five roots.


calculus
integration
function
limit
real analysis
real analysis unsolved

Let $ f: [0,1] \rightarrow R$ be a Riemann-integrable function s.t. for any $ x \in [0,1)$, $ \lim_{n \rightarrow \infty} n \int_{x}^{x \plus{} \frac {1}{n}} f(x) dx \equal{} 0$.
Prove that for any $ a,b \in (0,1), \int_{a}^{b} f(x)dx \equal{} 0$.

I'm stuck on this annoying problem too. If anybody solves it, please post the proof. It is proposed by Cristinel Mortici btw.

It was also discussed [url=http://www.mathlinks.ro/viewtopic.php?t=260443]here[/url] and , in romanian, [url=http://mateforum.ro/viewtopic.php?t=3499]here[/url].


geometry
ratio
trigonometry
trig identities
Law of Cosines
area of a triangle

Side BC of triangle ABC is extended through C to X such that BC=CX. Similarly, side CA is extended through A to Y such that CA = AY, and side AB is extended through B to Z such that AB = BZ. Find the ratio of the area of triangle XYZ to that of ABC.

I solved this problem by arbitrarily letting ABC be equilateral, having side s. Then to find angle ZBX, I used the law of cosines, knowing that BZ = s, BX = 2s, and ZBX = 180-60=120. As triange XYZ is equilateral, I could thus find the needed ratio easily. However, this solution seems sorta cheap, with the arbitrary setting of ABC's equilateral property. Is there a better way to solve this problem?

Here is one way: recall that the area of a triangle is 1/2ab*sin(C). Consider triangles ABC and BXZ. BX=2BC, and BZ=BA, and sin(ABC)=sin(XBZ) since they are supplementary. Thus [BXZ]=2[ABC]. Similarly, [AYZ]=2[ABC], and [CXY]=2[ABC]. Thus [XYZ]=7[ABC]. Here [t] denotes the area of triangle t.


trigonometry
induction
quadratics
algebra
quadratic formula

If $\theta$ is a constant such that  $0<\theta<\pi$ and $x+\frac{1}{x}=2\cos \theta$, then for each positive integer n, find $x^n+\frac{1}{x^n}$ in terms of $n$ and $\theta$.

This is at least intermediate, could a mod bump it up?

[hide="Hint"]
\[\cos x=\frac{e^{ix}+e^{-ix}}{2}.\][/hide]
[hide="Answer"]

$2\cos n\theta$

Now, you can use induction to prove this formula.  Or it would be more enlightening to derive this formula.[/hide]

Yeah, I shouldn't have posted this problem in this forum.

Can anybody prove it not using induction?

[hide="Proof"]

First, I will prove that

\[\cos x =\frac{e^{ix}+e^{-ix}}{2}.\]

By Euler's identity we have:

\begin{eqnarray}
e^{ix} &=& \cos x +i\sin x\\
e^{-ix}=\cos(-x)+i\sin(-x) &=& \cos x -i\sin x
\end{eqnarray}

Adding $(1)+(2)$ and dividing by 2 produces the desired result.

Hence we move on to the problem at hand.

Let $x=re^{i\theta}.$  Then we have

\[\frac{re^{i\theta}+re^{-i\theta}}{2}=\cos \theta.\]

But the left hand side is $r\cdot \cos \theta$!  Thus $r=1$.  We can find $\cos n\theta$ from this:

\[\cos n\theta=\frac{e^{i(n\theta)}+e^{-i(n\theta)}}{2}.\]

We sub $e^{i\theta}=x$ back in and get

\[x^n+\frac 1{x^n}=2\cos n\theta.\] $\text{QED}$[/hide]

[quote="warut_suk"]Thanks!

One small note: You refer to De Moivre's, but
$e^{ix} = \cos x +i\sin x$
is not De Moivre's, is it?[/quote]

It's Euler's identity, sorry.  I'll edit it.

Ahh.. yes. Euler's. I couldn't recall its name at first.

i dunno how similar my solution is to joml88's.. but yeah:

rewrite the expression as a quadratic:

$x+\frac{1}{x} = 2\cos \theta$

$x^2 - 2x\cos \theta + 1 = 0$

Solving for x using the quadratic formula:

$x = \cos\theta \pm \sqrt{\cos^2\theta-1} = \cos\theta \pm i\sin\theta$

Therefore, by DeMoivre's: $x^n = \cos n\theta \pm i\sin n\theta$

And $\frac{1}{x^n} = \frac{1}{\cos n\theta \pm i\sin n\theta} = \cos n\theta \mp i\sin n\theta$

Therefore, $x^n + \frac{1}{x^n} = 2\cos n \theta$.


What kind of career do you dream of pursuing? What do you do for fun?

[quote="The Sigma"]I am currently a sophomore in high school.

-I have a 3.96 GPA
-2nd alternate for state violin competition
-3rd place oratory
-(started)independent research

Do I have a good chance of getting into my dream college: MIT and is it really worth going to a good school for an Undergrad degree or should I stay home and get my Undergrad degree at a local university?[/quote]

First, you're a sophomore.  You still have two years to make or break yourself.  Second, 4 statistics about you is not nearly enough to decide whether or not you stand a good shot of MIT.  My advice is to not worry about your chances of getting in right now.  You need to focus on school and push yourself right now.  In my opinion, your junior year is your most important year.  If you want a decent shot of getting into an elite college you need to step it up here.

Of course it's worth it to go to a good school for your undergrad degree. In gener


trigonometry
function
symmetry
inequalities
absolute value
puzzles

I don't think that is solvable... She could have shaken hands with anywhere from 0 to 8 people. Unless there is some trick, it's impossible to solve. Not enough information.

[quote="EnderX"]I'm gonna need a couple of genius's or math wiz to do this one, not cuz its hard, but i need the answer to it in like an hour or so, and i figure i need as many minds as possible.

Suppose I invite 4 married couples to a party at my house. After a while, I ask everyone present, including my wife, how many people they shook hands with. It turns out that everyone shook hands with a different number of people. It is also the case that no one shook hands with his or her spouse. How many people did my wife shake hands with? (I did not ask myself any questions.)[/quote]
[hide="Hint"]There are 9 people other than you. What are the fewest number and largest number of handshakes they could have made? Since each of them have to have made a different number of handshakes, what does this tell you? What conditions does this situation force to happen?[/hide]

@laughinghead505: The problem has enough information.

there is enough information. :wink:
Let $P_{i}$ denote the person who shook hands with $i$ people.
Since $P_{9}$ shook hands with everyone except for his/her partner and $p_{0}$ shook hands with no one at all, they must be partners. You can continue this way and the one who is left is your wife.

[quote="laughinghead505"][quote="K81o7"]I've done this one before (or similar at least), and it IS possible (if I remember correctly).

Let me give you a hint.  Each of the 9 people shook a different number of hands.  But each person shook hands with anywhere from 0 to 8 people.  Then, assign each person a number based on the number of hands that person shook.  You then have 0 1 2 3 4 5 6 7 8.  Now whose hands did 8 shake?  Then based on that information, can you figure out whose hands 7 shook?  Work your way down, and you'll figure it out  :)[/quote]

I understand why 8 and 0 have to be spouses, but how do you go from there? You would have to try to figure out 7's spouse...  Oh wait! 7 is spouses with 1 because 8 shook with one, so no one else did and 7's spouse has to be someone he didn't shake hands with in the set {1,2,3,4,5,6}. So 7's spouse is 1? And then 6 will be with 2 and 5 with 3. So did his wife shake hands with 4 people?????[/quote]
yeah that's how you would do it.  :)

That's a very interesting problem. I like it a lot. :lol: It's interesting how you look at it and think "that's unsolvable!" but then actually find a way to do it... or maybe that's just me.  :D

Where did you get this problem? :huh:

Its pretty awesome :D

well he says it was for homework  :)
and it's in ACoPS(ugh the evil book) chapter 1 and solution was later in the book.

I wish I had homework like that :(

Right now, my trig homework is like, simplify $(x+2)(x-2)$ or is 13 a rational number? IT IS DRIVING ME INSANE!!!!!

Yeah. Really. I just had a chapter 1 test and it's like "Test the function y=x^3 for symmetry about the x-axis" and "Billy Bob has chemistry scores of 85, 86, 67, and 90 on his previous tests. What does he need to get on the next test to raise his average to a B?"

That sounds so much more interesting than mine :rotfl:

Today we learned to graph a number line for the absolute value of $x>3$

Since we've changed the topic:

I'm in Algebra 2, and we're solving inequalities in word problems this week.  This question didn't even fit in, but teach gave it to us anyway yesterday:

If John has \$800, wants to save \$300, and will give \$200 to his mom, what is the most he can spend while shopping?

I remember OUR WHOLE CLASS solving these problems in third grade.

Yay I'm in math seminar.  Okay, so the class isn't extremely interesting yet, but right now my task is to find a sketchpad construction using basic tools of the radial axis of two circles...

[quote="mustafa"]Today we learned to graph a number line for the absolute value of $x>3$[/quote]
Sounds like AP Calculus. :D
Well, after a week it's getting slightly challenging. :lol:

i'm taking abstract algebra. its a very interesting subject, but the professor waters it down


[quote="LoboSolitario"][quote="felipesa"][hide]b[/hide][/quote]

Could you please show your work felipesa? I know the answer  :P[/quote]

[hide]
$y \in \Re$   and  $0<x<2\pi$
$\\ - 2 + \csc \sec x \ge 0 \\ \csc \sec \ge 2 \\ \sin x \le \frac{1}{2} \\ \\ \sqrt{1 - \cos ^2 x} \le \frac{1}{2} \\ 1 - \cos ^2 x \le \frac{1}{4} \\ \cos ^2 x \ge \frac{3}{4} \\ \cos x \ge \frac{{\sqrt 3 }}{2} \\ \\ 1 + \tan ^2 x = \sec ^2 x \\ \tan x \le \frac{\sqrt 3}{3} \\$
But for -1 dont work
A  possible alternative that satisfy is b
[/hide]

[quote="ifai"][quote="LoboSolitario"]If $y$ is real where $y=\sqrt{-2 + \textrm{cossec} x}$, $0 < x < 2 \pi$, then we can have:

a) $\sin x = 0,939$
b) $\tan x = 0,247$
c) $\cos x = 0,374$
d) $\cos x = - 0,2$
e) $\tan x = -1$[/quote]
I don't know what you mean.[/quote]

Excuse me for my english.

If $y$ is real, then you must find which of following is true (or which of following can happens).

a) $\sin x = 0,939$
b) $\tan x = 0,247$
c) $\cos x = 0,374$
d) $\cos x = - 0,2$
e) $\tan x = -1$

[quote="felipesa"][quote="LoboSolitario"][quote="felipesa"][hide]b[/hide][/quote]

Could you please show your work felipesa? I know the answer  :P[/quote]

[hide]
$y \in \Re$   and  $0<x<2\pi$
$\\ - 2 + \csc \sec x \ge 0 \\ \csc \sec \ge 2$
[/hide][/quote]

[hide]I agree till there. But $\textrm{cossec}x \geq 2$ do not implies in $0<\sin x \leq \frac{1}{2}$?

Hence $0<x \leq \frac{\pi}{6}$ or $\frac{5\pi}{6} \leq x < \pi$ (since $0<x<2\pi$) [/hide]

[quote="scorpius119"][quote="LoboSolitario"]If $y$ is real where $y=\sqrt{-2 + \textrm{cossec} x}$, $0 < x < 2 \pi$, then we can have:

a) $\sin x = 0,939$
b) $\tan x = 0,247$
c) $\cos x = 0,374$
d) $\cos x = - 0,2$
e) $\tan x = -1$[/quote]

Does that mean $y=\sqrt{-2+\csc x}$ and are the commas decimal points? If so then...

[hide][b](B)[/b]: $\tan x=0.247\Rightarrow \cot x=0.247^{-1}\Rightarrow \csc^2 x=1+0.247^{-2}$. Since $0.247<\frac{1}{4}$, $\csc^2 x>17$. Choose the first-quadrant angle $x$ such that $\tan x=0.247$; then $\csc x>\sqrt{17}>2$. Then $y^2$ is positive, so $y$ is real.[/hide][/quote]

I don't think so scorpius119.


Rolling three die, what is the probability of their sum being 16?

Also, if someone could explain a general method doing there dice problems. Thanks  :|

We want the number of solutions to $ a \plus{} b \plus{} c \equal{} 16$ where $ 1\leq a,b,c\leq6$.

Negating both sides and adding $ 18$, we see that $ (6 \minus{} a) \plus{} (6 \minus{} b) \plus{} (6 \minus{} c) \equal{} 2$, where the only restriction on $ 6 \minus{} a,6 \minus{} b,6 \minus{} c$ is that they are nonnegative.

This is simply balls and urns, and it has $ \dbinom{2 \plus{} 3 \minus{} 1}{3 \minus{} 1} \equal{} 6$ solutions.

Finally, our probability is $ \frac6{6^3} \equal{} \boxed{\frac1{36}}$.

As for a general method, it just involves finding the number of solutions to

$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} k$ where $ 1\leq a_1,a_2,\ldots,a_n\leq6$, which represents "rolling $ n$ dice, what is the probability of their sum being $ k$?" (Balls and urns can be applied, but remember to be careful! Try $ a_1 \plus{} a_2 \plus{} a_3 \equal{} 8$.)

OR we could solve this by writing 16 as 1 + 1 + 1... + 1. Now we just need to put two "dividers" that will tell us how these are divided. For example, a possibility is 1-1-1-1-1-1 | 1-1-1-1-1-1 | 1-1-1-1. Now for the first divider, we have 6 possible choices (since it must be at least 4 it can be on either end). For the second, we can have (depending on the first divider) 6-6, 5-6, 6-5, 4-6, 6-4, giving us 6 choices, also. However, in this calculation we assumed that the order the dices are rolled matters, so we divide by 3! to get 6 possible ways. Now, our probability is this over $ 6^3$, giving us $ \frac {6}{6^3} \equal{} \frac {1}{6^2} \equal{} \boxed{\frac {1}{36}}$.

Sorry if that wasn't clear.

Btw, how exactly did you go from the second part to the third part? Can you explain where you got those numbers? Sorry again.

Thanks.

So, for the dice, they aren't distinct right?

Would rolling a 5, 4, 6 be different from 4, 6, 5 (in this question)?

Yes--it weights the cases.

For example, there's one case in which there is a $ 5,5,5$, but there are $ 6$ cases with $ 4,5,6$, so the latter is more likely.

If we were to ignore order, than the weights would be incorrect.


Suppose
$ \sin(x_1)\plus{}\sin(x_2)\plus{}\sin(x_3)\equal{}0$ and
$ \cos(x_1)\plus{}\cos(x_2)\plus{}\cos(x_3)\equal{}0$.

Prove that
$ \sin(2x_1)\plus{}\sin(2x_2)\plus{}\sin(2x_3)\equal{}0$ and
$ \cos(2x_1)\plus{}\cos(2x_2)\plus{}\cos(2x_3)\equal{}0$.

2sin(x1+x2/2)cos(x1-x2/2)=-sinx3

2cos(x1+x2/2)cos(x1-x2/2)=-cosx3

squarring and adding

4cos^2(x1-x2/2)=1

solve for x1,x2

hence etc

Let $ z_1 \equal{} \cos x_1 \plus{} i\sin x_1$ etc.

We have $ z_1\plus{}z_2\plus{}z_3 \equal{} 0$. Also $ |z_i| \equal{} 1 \Rightarrow \bar{z} \equal{} \frac{1}{z}$.

$ \because \ z_1\plus{}z_2\plus{}z_3 \equal{} 0$ we also have $ \bar{z_1}\plus{}\bar{z_2}\plus{}\bar{z_3} \equal{} 0 \Rightarrow \frac{1}{z_1}\plus{}\frac{1}{z_2}\plus{}\frac{1}{z_3} \equal{} 0$ or $ z_1z_2\plus{}z_2z_3\plus{}z_3z_1 \equal{} 0$

Now, $ z_1^2\plus{}z_2^2\plus{}z_3^2 \equal{} (z_1\plus{}z_2\plus{}z_3)^2 \minus{} 2 (z_1z_2\plus{}z_2z_3\plus{}z_3z_1) \equal{} 0$

$ z_i^2 \equal{} \cos 2x_i \plus{} i \sin 2x_i$ etc.

And hence we conclude that $ \cos 2x_1\plus{}\cos 2x_2\plus{} \cos 2x_3 \equal{} \sin 2x_1\plus{}\sin 2x_2\plus{}\sin 2x_3 \equal{} 0$


probability
abstract algebra
geometry
geometric transformation
rotation
superior algebra
superior algebra unsolved

Prove that in a finite and nonabelian group [b]G[/b], the probability that for any two elements , say $a,b$ , we have $a\ast b=b\ast a$ is not greater than $\frac 58$.

P.S Can you extend your idea to infinite groups?

In fact the bound above can be tighter. To show the above fact, we need the following claim:

If $G/Z(G)$ is cyclic, then $G$ is abelian.

Proof: Let $aZ(G)$ be the generator of $G/Z(G)$. Let $x$, $y\in G$. Then $x=a^nx_0$ and $y=a^my_0$ where $x_0, y_0 \in Z(G)$. It is then not difficult to check that $xy=yx$.

So, since $G$ is non-abelian, then $[G: Z(G)]\ge 4$. Because if it is 2 or 3, then $G/Z(G)$ is cyclic.

Therefore,

\[ P(ab=ba) = P(a\in Z(G))\ \rm{or}\ P(b\in Z(G)) = 1-P(a\ \rm{and}\ b \notin Z(G)) = 1-P(a\notin Z(G))P(b\notin Z(G))\le 1 -(3/4)^2 =7/16 < 5/8.  \]

I do not think I use the fact that $G$ is finite anywhere.

[quote="mao_zai"]So, since $G$ is non-abelian, then $[G: Z(G)]\le 4$. Because if it is 2 or 3, then $G/Z(G)$ is cyclic.
 $\[ P(ab=ba) = P(a\in Z(G))\ \rm{or}\ P(b\in Z(G))$.[/quote]

Did you mean $[G: Z(G)]\ge 4$ and $\[ P(ab=ba) = P(a\in Z(G)\ \rm{or}\ b\in Z(G))$ ?
 :arrow: How did you deduce that $a\in Z(G)\ \rm{or}\ b\in Z(G)$?
Maybe my statement was misleading."[b]Any[/b]" here means "[b]Random[/b]".Did I use it in a wrong way?

Yup yup. My mistakes. $[G: Z(G)]\ge 4$.

Sorry. My proof is wrong. It's possible that $a$ and $b$ not in $Z(G)$ and yet $ab=ba$.

Moreover, the bound is tight, and $Q_8$ attains the bound.

[quote="mao_zai"]$[G: Z(G)]\ge 4$.[/quote]

Then we're almost done, for the centralizer of any element not in $Z(G)$ must have index at least 2. So after choosing one element, the probability that the next element chosen will commute with it is bounded above by 1/4 + 1/2(3/4) = 5/8, as required.

Thank you guys.But it seems that we have talked about finite groups so far .'Cos
$[G: Z(G)]\ge 4$ yields$\frac{Card (G)}{Card(Z(G))}\ge 4$ .So what's your idea about infinite ones?

How do you define probability for infinite groups¿

You don't - unless the group has a natural measure of finite total mass (any compact topological group works using Haar measure).

Trying this for $SO_3(\mathbb{R})$, we see that two rotations commute iff one is the identity or they have the same axis. This has probability zero.


Let $ a;b;c$ be real numbers satisfying $ 0 \leq a;b;c \leq 1$
Prove that:
$ \frac{x}{7\plus{}y^3\plus{}z^3}\plus{}\frac{y}{7\plus{}z^3\plus{}x^3}\plus{}\frac{z}{7\plus{}x^3\plus{}y^3} \leq \frac{1}{3}$

See hear
http://www.mathlinks.ro/viewtopic.php?p=182171#182171   :lol:

[quote=chien than]Let $ a;b;c$ be real numbers satisfying $ 0 \leq a;b;c \leq 1$
Prove that:
$ \frac{x}{7+y^3+z^3}+\frac{y}{7+z^3+x^3}+\frac{z}{7+x^3+y^3} \leq \frac{1}{3}$[/quote]
$$ \frac{x}{7+y^3+z^3}+\frac{y}{7+z^3+x^3}+\frac{z}{7+x^3+y^3} \le  \frac{x+y+z}{6+x^3+y^3+z^3}\le \frac{1}{3}$$
[url]https://artofproblemsolving.com/community/c6h1378697p7632965[/url]


Let there be a circle with points A, B, C, D on it (in that order). Let lines AB and CD intersect at a point P outside the circle. Why is angle BPC = (1/2)(arc AD - arc BC)? What if instead the points were ordered A, C, B, D so that point P is inside the circle?

Thanks.

draw lines from A to C and B to D, then mess around with the inscribed angles formed. try from there.

You have to use the fact that the measure of a central angle is equal to the measure of the arc.  Then afterwards it is a matter of angle chasing.  It's easier if you start with a line through the center of the circle.


Two nice proofs.

1)  If p≥5 is a prime number show that (p^2)+2 is composite.
      [ hint: p takes one of the forms 6k+1 or 6k+5]

2)  If p ≠ 5 is an odd prime prove that either [(p^2)-1] or [( p^2) +1] is divisible by 10 .

[hide="1"]
Using the hint, $ p$ can either take the form $ 6k \plus{} 1$ or $ 6k \plus{} 5$.
When $ p \equal{} 6k \plus{} 1$, we need to prove that $ ((6k \plus{} 1)^2 \plus{} 2$ is composite, expanding, this is $ 36k^2 \plus{} 12k \plus{} 3$, which can be factored as $ 3(12k^2 \plus{} 4k \plus{} 1)$. So $ 3$ is a factor and $ p$ is composite.
For $ p \equal{} 6k \plus{} 5$, we can use the same method to find it is $ 3(12k^2 \plus{} 20k \plus{} 9)$. So it is also composite.
[/hide]

[hide="2"]
Any odd prime number other than $ 5$ has to end in $ 1$, $ 3$, $ 7$, or $ 9$. The square of any of these will give a number ending in either $ 1$ or $ 9$. Adding or subtracting $ 1$ will then give a multiple of $ 10$.
[/hide]

EDIT:Thanks Arrange your tan :blush:

Quoted from brightzhu . . . Part of the hidden to #2

"Adding or subtracting 1 will then give a power of 10."

No.  Adding or subtracting 1 will then give a MULTIPLE of 10.

[hide="For the first one..."]

...you can also just use modular arithmetic to note that for all primes $ p$ besides $ 3$, $ p \equiv 1 \mod{3}$ or $ p \equiv 2 \mod{3}$ In the first case, $ p^2 \plus{} 2 \equiv 1^2 \plus{} 2 \equiv 3 \mod{3}$, and in the second case $ p^2 \plus{} 2 \equiv 2^2 \plus{} 2 \equiv 6 \mod{3}$, so in both cases $ p^2 \plus{} 2$ is divisible by $ 3$[/hide]

[hide="1."]
$ p^2 \equiv 1 \; (mod \;3)$ (Fermat's little theorem) $ \implies p^2\plus{}2 \equiv 3 \equiv 0 \; (mod \;3) \qquad QED$
[/hide]

[hide="2."]
$ (p^2\minus{}1)(p^2\plus{}1)\equal{}p^4\minus{}1 \equiv 0 \; (mod \; 5)$ Since $ p^4\minus{}1$ is even the expression is clearly divisible by $ 10. \qquad QED$[/hide]


algorithm
number theory
greatest common divisor
Euclidean algorithm

Hi all. I hope this is the right section to put this in. I was reading through the part on linear diophantine equations in the second AOPS book, and there was one question I had. Basically, the book has you solve the equation like this:
$ax+by = c$
$ax\equiv c \bmod{b}$
Through trial and error, you can find a solution for x in terms of an arbitrary integer k. Then you can easily solve for y in terms of k.
My question is, is there any better way to solve the equation
$ax\equiv c \bmod{b}$
than just adding multiples of b and checking whether it is divisible? For large numbers, this would take a really long time. Thank you in advance for you help.

It's called the Euclidean algorithm. The Euclidean algorithm allows you to compute the GCD of two integers by repeating division algorithm over and over until you get remainder 0.
Say you want to find the GCD of the numbers 547 and 632, it's written like so:
632=[color=orange]547[/color] x 1 + [color=darkred]85[/color]
[color=orange]547[/color]=[color=darkred]85[/color] x 6 + [color=brown]37[/color]
[color=darkred]85[/color]=[color=brown]37[/color] x 2 + [color=green]11[/color]
[color=brown]37[/color]=[color=green]11[/color] x 3 + [color=olive]4[/color]
[color=green]11[/color]=[color=olive]4[/color] x 2 + [color=blue]3[/color]
[color=olive]4[/color]=[color=blue]3[/color] x 1 + [color=violet]1[/color]
[color=blue]3[/color]=[color=violet]1[/color] x 3 + [color=indigo]0[/color]
Then we take the last remainder, [color=violet]1[/color], and this is the GCD. I will not provide a proof here, but you could look one up in an elementary NT book or on the net.
So then say we want to find r and s such that $547r+632s=1$. This moves on to your question. Well, we can express the last remainder ([color=violet]1[/color]) in terms of 3 and 4, and then express 3 in terms of 4 and 11, etc. until we express it transitively in terms of 632 and 547. The work goes as follows:
[color=violet]1[/color]=[color=olive]4[/color]-[color=blue]3[/color]=[color=olive]4[/color]-([color=green]11[/color]-[color=olive]4[/color] x 2)=[color=olive]4[/color] x 3 - [color=green]11[/color]. Now I'm going to stop color coding because it gets very tiresome, you can follow along with the pattern though. So
$4 \times 3 - 11 = (37-11 \times 3) \times 3 - 11 = 37 \times 3 - 11 \times 10$
$37 \times 3 - (85-37 \times 2) \times 10 = 37 \times 23 - 85 \times 10$
$(547-85 \times 6) \times 23 - 85 \times 10 = 547 \times 23 - 85 \times 148$
$547 \times 23 - (632-547) \times 148 = 547 \times 171 - 632 \times 148$
So $r=171$ and $s=-148$.
To find $547r+632s=t$ just multiply 171 and -148 by t.
One last thing: after using the Euclidean algorithm and working backwards to express the GCD in terms of the two numbers in a linear combination, you can express only multiples of the GCD.


The sum of two numbers is 352. One number is three times the
other. What is the value of the smaller number?

Let x be the smaller number. Then we have

x + 3x = 352

4x = 352

x = 88


analytic geometry
geometry
geometric transformation
rotation
complex numbers

Sorry, this might be too easy, but i just wanted to know is there any easy way of knowing the coordinates for C, if A is $(a, b)$, B is $(c, d)$ and $AB=BC$ and $\angle ABC = 90$

We will use notation $A,B,C$ as for points as for complex numbers corresponding with these points.
Your definition of $C$ means that $C$ can be created by rotating $A$ about $B$ through angle $90^{o}$, so $C-B$ can be created by rotating $A-B$ about $0$ through angle $90^{o}$.
1. Rotation is clockwise
We have \[C-B=(A-B)i\iff C=Ai+(1-i)B = (a+bi)i+(c+di)-(c+di)i=\]  \[=ai-b+c+di-ci+d=c+d-b+i(a+d-c),\] so $C=(-b+c+d,a-c+d)$.
2. Rotation is unclockwise
We have \[A-B=(C-B)i\iff C=\frac{A+(1-i)B}{i}=-i(A+(1-i)B)=\]  \[=-Ai-(1+i)B=-i(a+bi)-(c+di)-i(c+di)=-ai+b-c-di-ci+d=\]  \[=b-c+d+i(-a-c-d),\] so $C=(b-c+d,-a-c-d)$.
The answer is: \[C_{1}=(-b+c+d,a-c+d)\]  \[C_{2}=(b-c+d,-a-c-d).\]


MATHCOUNTS
number theory
modular arithmetic

Consider x^y where x is a non-0 integer and y is a non-negative integer. What percentage of x has the same units (last) digit regardless of what y is?
(Restatement (with modular arithmetic): Consider x^y (mod10) where x is an integer and y is a non-negative number. What percentage of x gives us the same answer regardless of the value of y?)

Consider x^y where x is an integer and y is a positive integer. What percentage of x gives us an answer with a pattern? (For example 2 will give us the pattern 2, 4, 8, 6, 2, 4, 8, 6...)

u mean non-negative integers right? not number... cause y can be 1/4 in your question.

Oh whoops, you're absolutely right. I'll go fix it right now.

You're wrong...*changes problem from integer to non-0 integer*...now you're right! :lol:

Cool trick: Type in a number (integer please), then the multiplication sign (usually *), then that same number, and keep pressing Enter (press and hold)
If the unit's digit is undisturbed, then it is one solution :lol:

I got 0, 1, 5, 6... .what else is there???

oh.... ok...
everything's fine then... including the i at the end :lol:

good...I hope you weren't offended  :lol:  :lol:  :lol:

lol (no offense taken.... if that was actually a question... lol)

can 5 5 5 5 5 be a pattern???

what about 4 6 4 6 4 6... if so... 100%

if not... 40%

You're correct![size=9]There'll always be a pattern no matter what.[/size] ([b]not [/b]very important) Odd, si?

And yes, 5555555555555... is a pattern.

not really though... well... for the beginners... might be.

not really if you're in pre-calc... no offense... i don't really think you're a beginner... considering you have so many problems you can post... and the fact that you're mod.

Er, you're not helping :lol:
Most of my problems are either unsolvable, has non-real solution, or has misspelling so no one solves them, and I'm a mod in a math forum since I didn't pass the qualifications to become a non-math forum mod...oh well :)

At least luck is on my side :lol:


calculus
search
function
geometry
geometric transformation
reflection

Given a triangle ABC with AB = AC, BC = 2 cm and $\angle BAC = 15^{o}$. A curve is drawn to divide triangle ABC into two regions of equal areas. What is the minimum length of the curve?
(Answer should be expressed in surd form.)

also is there a unique triangle defined?

and isn't the "curve" just a line segment??

1.surd form = answer with root sign
2.triangle is uniquely determined but the curve is free to draw.
3.we didn't say that the curve must be a line segment.

I meant, doesn't it seem that if the curve is shortest, it will be a line? Anyway, if it wasn't a line how could you determine it?
Also, is "surd form" an abbreviation for anything or is that just wat it is called

I don't know how to prove it generally but if you make the following assumptions:
1. The curve is a line
2. The line is parallel to BC
then it is easier and also seems like it might be right.

Then call the length of the line x.  The line cuts the triangle into a smaller triangle and a quad.  The smaller triangle is similar to ABC so:
$A_{small}=(x/2)^2 A_{big}$
$(x/2)^2=\frac{A_{small}}{A_{big}}=1/2$
$x=\sqrt{2}$

I know my answer is mostly just assuming the problem into an easier one, so I hope someone can solve it.

this should be the correct answer, I was asking those kind of questions earlier (if its a line)

nope, you can also draw lines that are smaller. :)

EDIT: surd is an english word, and even if it is a curve, you can figure out the length of it using calculus at least.

me@home, rather than posting many small posts in a row, use the Edit button to put those questions into a single post.  Also, many of your questions (not just in this thread, but in general) can be answered by appealing to [url=http://mathworld.wolfram.com]Mathworld[/url], the [url=http://www.artofproblemsolving.com/Forum/search.php]forum search function[/url], or any search engine.

The curve [i]might[/i] be a line segment -- presumably, a solution to this problem will show whether or not it is.

The circular arc centered at point A gives you an arclength smaller than $\sqrt2$.  In fact, this is the shortest such arc.  I provide a proof, but I invite anyone to figure out why it is for themselves before they read, because the realization is fun and my presentation is less than ideal.
[hide]It's very clear that the curve won't have endpoints on BC -- to prove this is not perhaps trivial, but it's also really, really uninteresting.  So, the curve has an endpoint on AB and another on AC.  At this point, you're probably going to want a picture: Reflect the whole triangle across side AC.  Now, take twelve copies of the resulting figure and set them next to each other so that they all share point A and don't overlap.  This gives you a regular 24-gon.  Note that this process has the effect of joining together 24 copies of the arc we're looking for into a closed loop of some sort.  This arc encloses a fixed area equal to half of the area of the 24-gon.  We want to minimize its arclength.  The curve of minimal arclength containing a given area is a circle (this can be proven through nice elementary means, but I'll just take it as granted here), so it's a circle that we want.  At this point, we've done the interesting part.  To actually compute its arclength requires finding the area of the triangle we began with, but I'll spare you the calculations and just give the answer, $2\sqrt{\frac3{\pi}(2+\sqrt2+\sqrt3+\sqrt6)}$.[/hide]

Question: this solution has an obvious generalization where $15^o$ is replaced with any of a variety of other values -- however, it's not clear if it works for most values -- for instance, $25^o$.  (Irrational degree lengths add another level of complexity.)  So, can anyone handle those cases?

Well I think your method's right (and pretty crazy, how'd you think of that?) but I think your answers wrong because it is greater than 2.  So I basically followed your method and came to a different answer:
[hide]
So following everything as in JBL's post gets you a 24-gon, and a curve going around it, and you want a circle with center A and area of:
$\frac{1}{2} A_{24-gon} = 12*A_{triangle} = 12(2+\sqrt2+\sqrt3+\sqrt6)$
Then you want to find the circumference of this circle, because the desired curve's lengh is 1/24 the circumference of the circle.
$\pi r^2=12(2+\sqrt2+\sqrt3+\sqrt6)$
$r = \sqrt{\frac{12}{\pi}(2+\sqrt2+\sqrt3+\sqrt6)}$
$\frac{2 \pi r}{24}=\frac{\pi}{12}\sqrt{\frac{12}{\pi}(2+\sqrt2+\sqrt3+\sqrt6)}=\sqrt{\frac{\pi}{12}(2+\sqrt2+\sqrt3+\sqrt6)}$

=1.41016445

Which makes more sense because it is very close to the answer of $\sqrt2$ that followed from the assumptions earlier.
[/hide]

For rational angles, you can just have the ngon go around itself in a spiral, and the same calculations hold.

Yes, you're right -- I copied down the wrong number :blush: .  As for how I thought of it, I'm not really sure -- I couldn't find a simple convincing reason that a line segment would be the shortest curve, and that arc seemed like another good possibility.  It was only after I calculated it's length that I realized how to prove it.  I didn't have a real process for coming up with the method of solution -- I guess that thinking about the circles was what led me to it.

Okay, sure, spiraling works nicely.  In fact, I think you can add a little more argumentation to get it for irrational angles, too.  (The only real problem is that it doesn't close off, but it really [i]should[/i], since otherwise a later copy will give you a shorter curve than an earlier copy, or something.)

that is a really nice way to aproach the problem, thinking outside the box. what do you mean when you say that the ngon spirals?


How about the metathread about 'FIITJEE PETHALS'?

dei know how much u hate fiitjee' solution da ....so if others agree ok

I give my assent to this  :D  There are many places where FIITJEE bugs me wid his answers  :mad: . I think he should atleast give detailed solns for his answers

How about one thread for all discussions pertaining to FIITJEE All India Test Series. We cant create a thread everytime after a Full test and before the result. That is waste of space. Pardesi, why don't you merge all FIITJEE tests.

actually i store all our discussions a su know in the Stickies...but it  owuld be a mess if put all of them together,,,
so willl make a thread for FIITJEE SOLUTIONS only :)

i moved our discussions [url=http://www.mathlinks.ro/viewtopic.php?t=186505&start=140]here[/url]


analytic geometry
graphing lines
slope
calculus
calculus computations

How can I translate a f'(x) graph to a f(x) graph by hand?
Right now I have a f'(x) graph, but the question is asking for the line tangent to the graph f(x) at x=a, any ideas?

Look at the value of $ f'(a)$ in the $ f'(x)$ graph. This will be the slope of the tangent line.

Now you need to know the ordered pair $ (a, f(a))$.

Then, you have $ y\equal{}f'(a) \cdot (x\minus{}a)\plus{}f(a)$ is the tangent line by the point-slope formula


quadratics
algebra
number theory
relatively prime
algebra proposed

Find all pairs [i]a[/i],[i]b [/i]of positive numbers for which [i]a [/i]divides [i]b [/i]2+[i]b[/i]+1 and [i]b [/i]divides [i]a [/i]2+[i]a[/i]+1

I recognize this problem. :)

I'm not sure I have the right answer or not.

[hide="answer"]
(1,3)
(1,1)
(3,1)
[/hide]

The reason is that one of the numbers has to be 1 or the other part would have too small denominator or not divisible. For instance, if I have (2,3)

2^2+2+1 = 7/3 = 2 R 1

3^2+3+1 = 13/2 = 6 R 1

And after my answer's minimum, the one part would have larger denominator so it won't be divisible.

I hope it's right.. ;)

I'm afraid there are infinitely many solutions, Silverfalcon. Many problems such as this one have been discussed here. You have to imagine a procedure which gets you infinitely many solutions (and then worry about proving that those are all of them)  ;)

Just to give another example, $(3,13)$. And another one :): $(13,61)$. Should I write another one? :).

Do you mean something like $5a_n-a_{n-1}-1$?

Here's how I found those (and it's trivial to show that we can find infinitely many): put $a_n\le b_n$ (we name our solutions $(a_n,b_n)$). Then use the recurrence $a_{n+1}=b_n,\ b_{n+1}=\frac{b_n^2+b_n+1}{a_n}$.

The recurrence you observed is probably correct as well, I didn't check it.

[P.S. I don't know if all of them are of this form right now, but we are pretty close; I remember solving this one before, but maybe it wasn't on the forum  :? ]

Indeed, it is the same reccurence.
I found it from following quadratic equation:
\[a^2-a(5b-1)+b^2+b+1=0.\]
(if we start from $(1,1)$)

There's a standard way of solving these. Now we can look at $T_n=\frac{a_n^2+a_n+b_n^2+b_n+1}{a_nb_n}$. It shouldn't be too hard (by using a process similar to what I did, only backwards, going from large solutions to small ones) to show that $T_n$ has only a few values ($1$ or $2$; like I said, I didn't check the details of the problem).

Each such value then corresponds to a sequence of solutions derived as I derived the $(a_n,b_n)$ in my previous post.

And my approach immediately gives us descent algorithm.
Just write $a^2+a+b^2+b+1=kab$ and let $a>b$, then $\frac{b^2+b+1}{a}<a$ is also solution for given $k$!
So we will descend to (1,1)!

Wow. :(

And I know you guys know all this but what is exactly recurrence meant here?

 ;)

Expressing the members of a sequence in terms of previous members. What other meanings did you have in mind? :)

I had no idea. I saw that term before in the sequence problem like, sequences are given recursively(maybe?) :P

Any pair of consecutive terms represents a solution, since the pair (1,3) is a solution.
a and b are relative prime, since a divides the number b 2+b+1 which is relatively prime to b. Thus the only solution with a=b is 1.

How does b^2+b+1 relatively prime to b? How do you know that?

I think it is but I don't know how you got it.

Also, is this forum really "high-school" algebra? I doubt it... ;) Maybe this is like world-intelligent society forum or something. :P

The materials covered are really high.

[quote="Silverfalcon"]How does b^2+b+1 relatively prime to b? How do you know that? I think it is but I don't know how you got it.[/quote]
If $(b^2+b+1,b)=d\neq 1$, then $d\mid b^2+b+1=b(b+1)+1$, but $d\mid b(b+1)$, so we conclude $d\mid (b(b+1)+1-b(b+1))=1$, i.e. $d\mid 1$ -- contradiction.

Hmm...

*trying to translate it into the english*

What does d| means?

[quote="grobber"]It means "$d$ divides".[/quote]
Are you sure you said it in english?
 :lol:  :lol:  :lol:


Four married couple went to the movies together.  They found one row of eight seats to sit in.  If each couple must sit together, how many ways can the eight people be seated?

i think that the answer would be $24$ because the $8$ people have to be split into couples.  This means that there are really $4$ different sets of couples.  I just did $4!$ to get $24$

Am I Right?

24 people... right?
the first two seats can be filled in four ways, the second two can be filled in three, the third two in two, and the last two in one way.  Therfore, the result is four factorial or 24 ways.  :ddr:  :?:

[hide]sorry, not 24.  Keep thinking[/hide]

It would be [hide]4!*(2^4) because each couple can be arranged in two different ways, which gives you 384[/hide]

[hide]Count each pair as one person (Let's say... A,B,C,D).  There are 4! ways of arranging these "people". BUT.... each couple can be arranged two ways. $4! \cdot 2^{4}= 24 \cdot 16 = 384$ [/hide]

[quote="alanchou"]Extension: If this same group of 4 couples sat down at a round table with 8 chairs, how many arrangements are there if each couple must sit with each other?  (Actually, I tried this one and got confused  :oops: )[/quote]

[hide]If we assume the 4 couples to be A,B,C, and D, there are 3 different arrangements for the couples to sit in a round table. "A' sits next to 2 other couples, BC, BD, or CD. Since the couples can switch, multiply 3 by 2^4=3x16=48.[/hide]

Double check how many ways there are to arrange the people around the table....

[quote="erdogankerem123"][quote="Moldytape"][quote="erdogankerem123"][hide] maybe there are 6 ways to arrange the couples, since if the couples can switch so BD would be different from DB, as D would then switch from left of A to right of A. Then there would be 6 *2^4= 96 ways????  [/hide] Btw, is there some sort of formula for this besides simply listing?[/quote]

I don't think so, because every couple still sits next to the same couple if you switch them up, like if A sits with BD, and switch to DB, A and C still sits to DB and D and B still sits with AC[/quote]
The question asks for position, though, right? If so, I think left and right would matter, because though they're still with the same couples, the relation between the couples are still different.[/quote]

[hide="How about this?"]

First, label the couples as $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}, D_{1}, D_{2}$

To fix the circle, let $A_{1}$ sit at the top of the circle. His couple has $2$ places to sit.

Then, the 3 other couples sit in $3! = 6$ different arrangements.

Then, each of those couples can be switched ($B_{1}, B_{2}$ can become $B_{2},B_{1}$) so you have $2^{3}=8$ combinations.

So, there are $8*6*2=96$ different combinations, right?

[/hide]


Which ones do u have?

I dont have any currently...

Don't look at me, I only wear a wristwatch, dude...

Livestrong bands? Those are so awesome. And it's not just a rubber band bracelet. In case somebody doesn't know what livestrong bands are...

Livestrong bands are yellow bracelets or bands with the word "livestrong" imprinted on it. They are an item of the American Cancer Society being sold for a dollar each. They were introduced early this year, spring to summer, but were really popularized after Lance Armstrong won his unprecedented sixth consecutive Tour de France. All profits from the Livestrong bands go to the American Cancer Society to find a cure for cancer.
Right now, they are in extremely high demand, since the official website, http://www.wearyellow.com has run out of the bands, and is back-ordered about 5-6 weeks. So far, they have raised about 12-13 million dollars to find a cure for cancer. Many famous people wear them as well, one of the most well-known, John Kerry.

I used to have two, but lost one and gave the other to my cousin. (before I lost the other, of course.)

I wear the LiveStrong bracelet as well. It really spreads the awareness of cancer etc. In the first week I had mine, I had at least 10 ppl ask me what it was about. When I told them, they all wanted to kno where they could get their own. It was pretty cool.

Yeah, my friend bought in bulk (about 30-50) and I chipped in for the payment. I'm seeing if he has any more, because I want one again.

i have a livestrong and a breast cancer one =]

o are the breast cancer ones  the pink ones?

if so, how do u get them?

i'd appreciate it if you told me :D

Yeah, I'd like to know where to get those too. Lots of the people at my school wear them, and I'm jealous that they can make a difference without any effort and I can't. Same with the breast cancer stamps (though I never use snail mail, so it wouldn't matter).

I think you can buy the pink ones at Krogers.

[quote="uglyman621"]o are the breast cancer ones the pink ones?

if so, how do u get them?

i'd appreciate it if you told me [/quote]

Yeah, they're the pink ones. I actually am not sure how to get them. One of my friends ordered a bulk purchase, and I bought one from him.

I got my LiveStrong (the yellow cancer bracelets) one the same way, but through another friend.. lol

my sister bought 10 livestrong bracelets.  she gave one to me and to herself.  then she gave the rest all away :(

okay...now that's just weird.  i posted something like 5 mins ago and its not here

yea almost everyone i no wears them. where do u buy them? besides...that place that was already mentioned.

don't post things that are useless and that people already talked about them

I have one, got it from a friend 4 Christmas ( the cancer one) she searched everywhere and only found 3, which is sad b/c she needed like 10

When we went 2 this tournament in TN, these guys from my school all bought one, but didn't know about them then, so didn't buy one.

Every1 in the school has or wants one, ppl r selling them for like 5 dollars each, which is a rip-off

Yeah, five dollars each is a rip-off, but I've got to admit, if somebody offered me one for five dollars, I'd take it in a heartbeat.

You can get the LiveStrong bracelets at your local bike shop.  That's were I got mine.  My other friends also got theirs at sporting tournaments...
I bet you could get one at the Tour de France.... I wish I could go.
I don't think most places are running out, yet. :?

almost everyone...EVERYONE...at my school has a livestrong band(i do too :P ) We had this drive supporting the organization so we could buy bands at our school

I want one of them -_- ...
I used to have one of those pink breast cancer ones, but I lost it.

They're good for whacking people.

Hmmm...anyone know if they sell the livestrong ones in the mall or something?

One of the people who goes to my school, his dad bought a thousand of the Livestrong Bands.

He brought like ahundred to school.

He was selling them for a buck-fifty until someone pushed him and he dropped all the bands.

Everybody stole them and he started crying.

the next day, he bought 100 more.

lol, where did his dad get 1000 armbands... unless you order them from the internet, they are pretty rare now, at less in Alabama... but ordering from the internet it kinda not worth it unless you are buying lots at a time.


inequalities
LaTeX
algebra
system of equations
algebra unsolved

Hello everyone,

Math is not my strong point so it seems. If someone can solve these problems by 6:00 am EASTERN TIME, I will send 10 dollars via paypal to the user who successfully completes the problems below. Only terms are that the problems have to be accurate and YOU MUST SHOW WORK!

Thank you.

1) Solve the system of equations by using substitution:
y=x+1
4x+y=6

2) Solve the system of equations by using elimination:
3x+2y=9
-x+3y=8

3) Solve the following system of equations, shade your solution.
y (LESS-THAN OR EQUAL TO) x+1
y>x-1

4)Graph the following inequality, shade your solution.
y (LESS-THAN OR EQUAL TO) 2x-3

Wrong idea, wrong attitute, wrong forum, and wrong price.  :roll:

Can you elaborate on that? How is everything all "wrong" in your mind?

Sure:

1&2: You need to do your homework by yourself if you want to learn anything. We can help you a bit in some cases but don't expect us to give you ready solutions to copy.

3: Such questions belong to middle school subfora, not to the Olympiad section

4: Well, if you had donated $ \$1,000,000$ to AoPS in advance and promised the same sum upon completion, it would be tempting to consider your offer. Anything less won't work (but you can get help for free here if you change your attitude, learn basic $ \text{\LaTeX}$ and show your own attempts to do the problems).


Suppose you have two power series
$ a(x) \equal{} a_0 \plus{} a_1 x \plus{} a_2 x^2 \plus{} \ldots$
$ b(x) \equal{} b_0 \plus{} b_1 x \plus{} b_2 x^2 \plus{} \ldots$
which converge for every value of $ x$ in an interval $ I \subset \mathbb{R}$. If there exists an open interval $ J \subset I$ such that $ a(x) \equal{} b(x)$ for every $ x \in J$, does it follow $ a(x) \equal{} b(x)$ for every $ x \in I$?

(sorry if this question is not well-formed)

The question is fine as it is. The answer is "Yes, it does". I wonder if we can see it without appealing to the notion of analyticity (which, I'm afraid, is unfamiliar to OP) :?.


I'm sick, and im the only one at my school taking the AIME, so can i take it on the 22nd rather than the 8th because im afraid that im not feeling very good?  My teacher already got all the AIME materials for the 8th, but that doesnt mean i have to take on the 8th right?

You MUST take AIME I on the 8th. Or your coach and you must check the rules and see whether being sick on the 8th is a ground for ordering the AIME II, which you MUST take on its appointed day. I feel for you, because I am sick today (and hoping like crazy that my son isn't BADLY sick on the 8th, because he is sick today), but rules are rules.


geometry
geometric transformation
reflection
\/closed

For those of us in the USA, Daylight Savings finally ended this last weekend. The clock on the forum doesn't reflect this.


With Rate the Avatar and Rate the Signature, it was inevitable....
RATE! THE! USERNAME!!!

on a scale from 1 (Bad)- 10 (Good!), rate the username of the person who posted before you.

awesome qwerty backwards | 8

pingala was a indian mathematician like in the 300 BCE.. i kinda forgot

Google is ur friend use it to find out stuff that u dont noe what it means!!!

Fly Agaric is a Mushroom, and I like mushrooms. Goes with the avatar too.

9.5!!!

STOP THAT  :mad:  :mad:  :mad:

Anyhow, 0, for mistaking my name for tcjy.\

And whats a carneal?

TCJY stands for:

The Crazy Joking Yaks (yes I made that up)

TCJV is

The Crazy Jiggling Vectors.

8 | at least you didn't make it like mathboy or mathfreak or ilikemath.

8, half of my clothes are nike.

tcjy, hjoon is nothing. its just cool like that.

8.8 | hjoon sounds like a Lune (its a type of bird thats really rare to find)

Pretty cool!

 :play_ball:

6, I absolutely despise numbers with all my heart. (And now you're probably wondering, what the heck is this person doing on AoPS?)

And BTW, Mines has numbers 'cause "Dreambold" was my original, but then I screwed up that account so I had to make a new one.  :D

uh, 7.

the message is too small. please make the message longer before submitting.

9.829387, cus it just looks cool.

ALEEEEEEEEEPPPPPPHHHHHHHHH...0

yay this message is long


Does anyone here read comics in the newspaper? Post any ones that you really like.
My favorites are: Garfield, Foxtrot, Peanuts and Calvin & Hobbes.

ahh...GARFIELD!!!...lol...

calvin and hobbes is goodlys too... :-D

cant wait for the garfield movie :-) :-)

you cud make that a poll..... my favorites are garfield, foxtrot, and zits. ^___^

Garfield... (since i was a wee titchy kid)

Peanuts... ok now, used to LOVE them...

Foxtrot... recently...

Calvin & Hobbes... just like them... for about 5 yrs...

Zits.... just read them....

(only listing a few)

hmm, almost the same with everyone else: Garfield, Foxtrot, Peanuts, Calvin and Hobbes, Crayon shinchan, Doraemon, and that's about it.

Don't really like them, but my sister does so I know many of her favorite ones from memory.  They are okay I guess.

I like donut man lol! My imaginary friend who I write comics about. I also like garfield and peanut. I dont really like any others just those. I like political cartoons the best. Ones about Al Gore are the best (not to hurt anyones feelings about Gore but I disliked him extremely.) Have any of you ever read dirty comics :lol:? Those prove to be quite interesting :lol:.

[hide]Here's a rated R comics (very funny): Purple "Cat" [/hide]

mmm, where do you people get the comics? I only know ucomics.com...

[quote="MCrawford"]It's a shame Gary Larson retired the Far Side.[/quote]

Yes, I enjoyed those too.

On another note, Charlie Brown!! :-D

FOXTROT RULES!!!

I stopped reading Garfield after a while because it was no longer funny (rarely, I mean).

And Charlie Brown, its good, but not as good as foxtrot.

good idea, a poll would be great.
And the far side was great. simple, but great.

I like Foxtrot the best out of the choices...my favorite is Pickles though...

The Far Side is really fabulous.  Peanuts is pretty good, too.  I'm also a big fan of Doonesbury, Boondocks, and the work of Tom Tomorrow (which probably doesn't appear in most regular newspapers, although I don't really know).  I also read the New Yorker cartoons, which generally are pretty good.

I like Calvin and Hobbes and Peanuts, but my favorites are the Far Side and Frank and Ernest.

[img]http://www.comics.com/comics/franknernest/archive/images/franknernest2004457520401.jpg[/img]

On another note, the man who draws Pickles came to our school and gave a presentation.

And what about Dilbert?  In my opinion, this is one of the funniest Dilberts ever:
[img]http://www.comics.com/comics/dilbert/archive/images/dilbert2004032348581.jpg[/img]

Re: the Dilbert cartoon posted by Mathfanatic, I thought that was so funny I was afraid I'd laugh out loud with enough volume to wake the baby.

All my friends from high school who are engineers in big corporations now say that Dilbert is BY FAR the most accurate description of corporate life in print.

My favorite is peanuts. I find it cute and funny but mostly cute. I dont know I like The O's comics (yall wouldnt know it's a kid at school who does really hilarious comics about dancing babies and odd ball stuff such as that) They are quite hilarious he does a good job of them. Ill have to see if he will let me borrow them so I can scan some of them in and share but hed have to copyright them first I wouldnt want his ideas to go to waste and stuff.

For the sake of preserving an interesting thread, and to display some more Dilbert:

[img]http://www.dakine.net/dilbert/images/dilbert-goals-injuries.gif[/img]

Garfield is my favorite...

I also like calvin and hobbes...

Anyway, how many comics can you think of????????????

I love Foxtrot, simply because Amend has brilliant comedic timing and generally hilarious illustration work.

I grew up with Calvin and Hobbes, and even though I've started reading a lot of Foxtrot, Calvin and Hobbes is still my favorite.  After a while Foxtrot becomes too predictable.

Foxtrot and Dilbert are my favorite comics. Peanuts is also pretty good.


Let $ a,b,c$ be positive real numbers. Prove that: $ a^a b^b c^c \ge (abc)^{\frac{a\plus{}b\plus{}c}{3}}$.

[quote="moldovan"]Let $ a,b,c$ be positive real numbers. Prove that: $ a^a b^b c^c \ge (abc)^{\frac {a \plus{} b \plus{} c}{3}}$.[/quote]
It's one old problem . i don't miss : Where is it ?
using chebyshev :
$ a.ln(a) \plus{} b.ln(b)\plus{}c.ln(c) \ge 1/3 .(a\plus{}b\plus{}c)(ln(a)\plus{}ln(b)\plus{}ln(c) )$  :lol:


algebra
function
domain
geometry
3D geometry
Ring Theory
Diophantine equation

Determine all pairs of integers $ (x,y)$ for which the equation
\[ x^{3}\plus{}3^{x}\equal{} y^{2}\]
is satisfied.

(My guess is that only $ (x,y) \equal{} (0,\pm{1})$ and $ (1,\pm{2})$ are the answers.)

I don't know if my idea is right.  It is not complete but, hopefully
 there is a
way to continue from what I have done.

It is clear that $ x\equal{}0$ and $ 1$ are good.  Now, suppose that $ x>1$ (obviously,
 $ x$ can't
be negative).
Note that if $ x$ is even and $ x\geq 12$, $ 2\cdot3^{x/2}> x^{3}$.  Therefore,
\[ [3^{x/2}\plus{}1]^{2}> x^{3}\plus{}3^{x}> [3^{x/2}]^{2},\]
which is clear that this does not give any solutions.  As a result, if
 $ x$ is even, $ x \equal{} 2,4,6,8,$ or $ 10$.  We can check that all these values are bad.

Hence, $ x$ is odd (and hence, $ y$ is even).  Therefore, $ x\equal{}2k\plus{}1$ for some
 positive integer $ k$ (recall that $ x>1$).  Thus,
\[ x^{3}\equal{} [y\plus{}3^{k}\sqrt{3}][y\minus{}3^{k}\sqrt{3}].\]
Since $ \mathbb{Z}[\sqrt{3}]$ is a Euclidean domain, it is also a unique
 factorization domain.   Consider
\[ \gcd(y\plus{}3^{k}\sqrt{3},y\minus{}3^{k}\sqrt{3}) \equal{}\gcd(y\plus{}3^{k}\sqrt{3},2y) \equal{}\gcd(y\plus{}3^{k}\sqrt{3},y),\]
since $ y$ is even.
Hence,
\[ \gcd(y\plus{}3^{k}\sqrt{3},y\minus{}3^{k}\sqrt{3}) \equal{}\gcd(y,3^{k}\sqrt{3}).\]

1)
If $ y$ is not divisible by $ 3$, $ y\plus{}3^{k}\sqrt{3}$ and $ y\minus{}3^{k}\sqrt{3}$ must both
 equal units times cubes, namely, there is a unit $ u$ and $ z\in\mathbb{Z}[\sqrt{3}]$ such that
\[ y\plus{}3^{k}\sqrt{3}\equal{} uz^{3}.\]
It is known that $ u \equal{}\pm(2\plus{}\sqrt{3})^{t}\equal{} p\plus{}q\sqrt{3}$ for some integer $ t$
 (this is immediate, using some knowledge about the Pell equation $ x^{2}\minus{}3y^{2}\equal{}\pm1$).  We can assume safely that t can either be $ 0, 1,$ or $ 2$ (otherwise, adjoint ($ 2\plus{}\sqrt{3})^{3}$ to $ z^{3}$).  If $ z \equal{} a\plus{}b\sqrt{3}$, then
\[ z^{3}\equal{} a(a^{2}\plus{}9b^{2})\plus{}3\sqrt{3}b(a^{2}\plus{}b^{2}).\]

Hence, $ y \equal{} pa(a^{2}\plus{}9b^{2})\plus{}9qb(a^{2}\plus{}b^{2})$ and $ 3^{k}\equal{} qa(a^{2}\plus{}9b^{2})\plus{}3pb(a^{2}\plus{}b^{2})$.  The
second equation implies that $ 3$ divides $ qa$.  Since $ 3$ does not divide $ y$,
 $ a$ is not a multiple of $ 3$.  Hence, from $ 3^{k}\equal{} qa(a^{2}\plus{}9b^{2})\plus{}3pb(a^{2}\plus{}b^{2})$, it means
 $ 3$ divides $ q$.  Therefore, the only legal $ q$ is $ 0$.  (Do not forget that  $ t\equal{}0,1,$ or $ 2$.)

Consequently, $ p\equal{}\pm1$, and $ y\equal{}\pm a(a^{2}\plus{}9b^{2})$ and $ 3^{k}\equal{}\pm 3b(a^{2}\plus{}b^{2})$.
  The latter tells that $ a$ must be zero.  Therefore, $ y \equal{} 0$, a contradiction.

2) $ y$ is divisible by $ 3$.  Write $ y \equal{} 3^{r}s$ such that s is not divisible
 by $ 3$.  If $ r\leq k$, $ \gcd(y,3^{k}\sqrt{3}) \equal{} 3^{r}$.  Therefore, if $ d \equal{} k\minus{}r$, then
$ x^{3}\equal{} 3^{2r}(s\plus{}3^{d}\sqrt{3})(s\minus{}3^{d}\sqrt{3})$.

Hence, if $ 2r \equal{} 0 (\textrm{mod }3)$, we can write
\[ v^{3}\equal{} (s\plus{}3^{d}\sqrt{3})(s\minus{}3^{d}\sqrt{3}).\]
If $ d > 0$, this will bring us back to the last case.
If $ d \equal{} 0$, $ s^{2}\minus{}4 \equal{} v^{3}\minus{}1$.  I am not sure, but I believe the solution is
 only $ v\equal{}1$.  If it is true, this will imply $ x\equal{}1$, about which we don't care.

If $ 2r \equal{}\minus{}1 (\textrm{mod }3)$, we can write
\[ 3v^{3}\equal{} (s\plus{}3^{d}\sqrt{3})(s\minus{}3^{d}\sqrt{3}).\]
This gives $ s\plus{}3^{d}\sqrt{3}\equal{}\sqrt{3}uz^{3}$ for some unit $ u$ and $ z\in\mathbb{Z}[\sqrt{3}]$.
This is, however, impossible, since s would be divisible by $ 3$!

If $ 2r \equal{}\minus{}2 (\textrm{mod }3)$, we can write
\[ 3^{2}v^{3}\equal{} (s\plus{}3^{d}\sqrt{3})(s\minus{}3^{d}\sqrt{3}).\]
Therefore, $ s\plus{}3^{d}\sqrt{3}\equal{} 3uz^{3}$ for some unit $ u$ and $ z\in\mathbb{Z}[\sqrt{3}]$.
  This is, again, ruled out, since s would be divisible by $ 3$!

3) $ y$ is divisible by $ 3$. Write $ y \equal{} 3^{r}s$ such that $ s$ is not divisible by
 $ 3$.  If $ r > k$, $ \gcd(y,3^{k}\sqrt{3}) \equal{} 3^{k}\sqrt{3}$.  Consequently, if $ d \equal{} r\minus{}k\minus{}1$,
\[ x^{3}\equal{}\minus{}3^{2k\plus{}1}(1\plus{}3^{d}s\sqrt{3})(1\minus{}3^{d}s\sqrt{3}).\]

Similar to 2), $ 2k\plus{}1$ must be divisible by $ 3$.  That is,
\[ v^{3}\equal{}\minus{}(1\plus{}3^{d}s\sqrt{3})(1\minus{}3^{d}s\sqrt{3}) \equal{} 3L^{2}\minus{}1,\]
for some $ L$.

Thus, $ (v\plus{}1)(v^{2}\minus{}v\plus{}1) \equal{} 3L^{2}$.  If $ \gcd(3,v\plus{}1)\equal{}1$, then $ 3$ does not divide
 both $ v\plus{}1$ and $ v^{2}\minus{}v\plus{}1 \equal{} (v\plus{}1)^{2}\minus{}3v$, a contradiction.  Hence, $ v\plus{}1 \equal{} 3w$ for some $ w$.  Thus,
\[ 9w(3w^{2}\minus{}3w\plus{}1) \equal{} 3L^{2}.\]
Therefore, $ w \equal{} 3m^{2}$ and $ 3w^{2}\minus{}3w\plus{}1\equal{}n^{2}$ for some $ m,n$.  This leads to
\[ (2n)^{2}\minus{}3(6m^{2}\minus{}1)^{2}\equal{} 1.\]
The above is claimed to have a solution when $ m\equal{}0$ only.

First a simple lemma and its proof:

Lemma $ x^{3}\plus{}3^{1\plus{}2z}\equal{}y^{2}$ has no integer solutions with $ gcd(x,y)\equal{}1,z>0$

Proof: We write $ x^{3}\equal{}(y\plus{}\sqrt{3}\,3^{z})(y\minus{}\sqrt{3}\,3^{z})$ and note, that $ xy\not\equiv 0\mod{3}$ holds.

Claim: $ y\pm\sqrt{3}\,3^{z}\equal{}(a\plus{}\sqrt{3}\,b)^{3}$ cannot be solved over $ \mathbb{Z}[\sqrt{3}]$ and all divisors (different from units) of $ y\pm\sqrt{3}\,3^{z}$ lie in $ \mathbb{Z}[\sqrt{3}]\setminus\mathbb{Z}$

Proof of the claim: Because of the homomorphism $ A\plus{}\sqrt{3}\,B\rightarrow A\minus{}\sqrt{3}\,B$ of $ \mathbb{Z}[\sqrt{3}]$ we can restrict to the "+" case. Direct calculation gives $ (a\plus{}\sqrt{3}\,b)^{3}\equal{}a[a^{2}\plus{}(3b)^{2}]\plus{}\sqrt{3}\,3b(a^{2}\plus{}b^{2})$. Now a comparison gives $ 0\not\equiv y\equiv a^{3}\mod{3}$ and $ 3^{z}\equal{}3b(a^{2}\plus{}b^{2})$. This yields (i) $ b\equiv a^{2}\plus{}b^{2}\equiv 0\mod{3}$ or (ii) $ b\equal{}1\,,\,a^{2}\plus{}b^{2}\equal{}3^{z\minus{}1}$ or (iii) $ a^{2}\plus{}b^{2}\equal{}1\,,\,b\equal{} 3^{z\minus{}1}$. (i) triggers the contradiction $ a\equiv 0\mod{3}$. If $ z\equal{}1$ holds, then (ii),(iii) yield the contradiction $ a\equal{}0$ and if $ z>1$ is valid, (ii) is equivalent to $ a^{2}\equiv\minus{}1\mod{3}$ and (iii) to $ a^{2}<0$. Both relations are not possible. As for the second part of the claim assume $ y\plus{}\sqrt{3}\,3^{z}\equal{}pa\plus{}\sqrt{3}\,pb$ for some integer divisor $ p\not \equal{}\pm 1$. But then $ 3$ must divide $ p$ and therefore $ y$, too, a contradiction.

If $ y\plus{}\sqrt{3}\,3^{z}$ has the factorization $ \prod_{i\equal{}1}^{k}(a_{i}\plus{}\sqrt{3}\,b_{i})^{t_{i}}$ modulo units, then $ y\minus{}\sqrt{3}\,3^{z}$ has the factorization $ \prod_{i\equal{}1}^{k}(a_{i}\minus{}\sqrt{3}\,b_{i})^{t_{i}}$. Note, that no factor $ a_{i}\mp\sqrt{3}\,b_{i}$ can divide $ y\pm\sqrt{3}\,3^{z}$, since otherwise $ a_{i}^{2}\minus{}3b_{i}^{2}$ would divide $ y\pm\sqrt{3}\,3^{z}$ contradicting the second statement of the claim. Let the factorization of $ x$ be given by $ \prod_{j\equal{}1}^{l}(A_{j}\plus{}\sqrt{3}\,B_{j})^{s_{j}}$. Then by the uniqueness of the factorization and the previous preparations, there is a non-empty index set $ J\subseteq\{1,\ldots,l\}$ such that $ y\plus{}\sqrt{3}\,3^{z}\equal{}\prod_{j\in J}(A_{j}\plus{}\sqrt{3}\,B_{j})^{3s_{j}}$, a contradiction to the first statement of the claim. This completes the proof of the lemma.

Now back to the proof of the original problem. For that we recognize the trivial solutions driven by $ x\equal{}0,1$ and that we can exclude negative and even choices of $ x$. From now on assume $ x>2$ and odd, i.e. $ y$ is even. This yields $ x^{3}\equal{}y^{2}\minus{}3^{x}\equiv 1\mod{4}\Longrightarrow x\equal{}1\plus{}4m\geq 5$.

Case 1: $ gcd(x,y)\equal{}1$ is solved by application of the lemma with $ z\equal{}2m>0$

Case 2: $ gcd(x,y)>1$. Here we can obviously write $ x\equal{}3^{2a}x'\,,\,y\equal{}3^{3a}y'\,,\,a>0\,,\,x'y'\not\equiv 0\mod{3}$. Substitution gives $ x'^{3}\equal{}y'^{2}\minus{}3^{t}$ with $ t\equal{}3^{2a}x'\minus{}6a$. Since $ t\equiv 1\mod{2}$ and $ t\geq 3^{2a}\minus{}6a\geq 3$ we can apply the lemma again setting $ x\equal{}x',y\equal{}y',z\equal{}\frac{t\minus{}1}{2}$.

[quote]
Claim: $ y\pm\sqrt{3}\,3^{z}\equal{}(a\plus{}\sqrt{3}\,b)^{3}$ cannot be solved over $ \mathbb{Z}[\sqrt{3}]$ and all divisors (different from units) of $ y\pm\sqrt{3}\,3^{z}$ lie in $ \mathbb{Z}[\sqrt{3}]\setminus\mathbb{Z}$
[/quote]

Why can we disregard the units of the ring $ \mathbb{Z}[\sqrt3]$?  It may occur that $ y\plus{}\sqrt{3}\equal{} (2\plus{}\sqrt{3})(a\plus{}b\sqrt{3})^{3}$ and $ y\minus{}\sqrt{3}\equal{} (2\minus{}\sqrt{3})(a\minus{}b\sqrt{3})^{3}$

Sorry, for not coming back earlier, I was on a holiday trip.

A less sloppy and more direct proof of the lemma would have concluded that each $ t_{i}$ is divisible by $ 3$, hence, $ y+\sqrt{3}3^{z}=\left[\prod_{i=1}^{k}(a_{i}+\sqrt{3}b_{i})^{\frac{t_{i}}{3}\right]^{3}}$ yielding the same contradiction as in the original proof.


(a) Choose $N_0$ to be a three digit number, such that not all three digits are equal.
(b) Set $i=0$
(c) Arrange the digits of $N_i$ in descending order.  Call this number $A_i$.
(d) Arrange the digits of $N_i$ in ascending order.  Call this number $B_i$.
(e) Set $N_{i+1}=A_i-B_i$
(f) Set $i=i+1$, and return to step (c)

Find $N_6$.

Bonus (I don't know the answer to this part):
Why does this work?

i've made some progress on demonstrating why this works, but it's nothing elegant at all.  quite simply, if you take a,b,c as distinct digits with a>b>c and subtract cba from abc (i.e., c *100 + b*10 + a from a*100 + b*10 + c) then you get either 198, 297, 396, 495, .... 891.  All of these eventually iterate to 495.

the proof i'm working on isn't elegant at all, as i've said.

would someone else please post any progress they've made on this?

I noticed that 9 | N1 and 11 | N1 (using the usual tests of divisibility) so N1 must be in 99*n for n = 2 .. 10, i.e. 198 297 396 495 594 693 792 891 990

Now there are few cases to test by hand since 198 and 891 gives same sequences, ....


1)Solve the  equation:

sum(sqrt(a^x+b^x))>=sum(sqrt(2a^x)), with a,b,c >0.

2)Solve the ecuation:
2^(x^4+1/x^4)+2^(sqrt(x)+1/sqrt(x))=8 (G.M. 5/1995)

3) Solve the equation:

lg(x^2-8)lg(2-x)=log_5 (x^2-8)/ log_5 (2-x)

4) Solve the equation:
(log_4(1+sin^2(x)))(log_4 (1+cos^2(x))=(log_4)^2 (3/2)

2) we know that x^4+1/x^4>=2 and sqrt(x)+1/sqrt(x)>=2 so we get that:
2^(x^4+1/x^4)+2^(sqrt(x)+1/sqrt(x))>=2^2+2^2=8 so we have equality iff x=1.

cheers! :D :D

3)Put the condition: x in (-oo,-2sqrt(2)) and you rewrite the eqs like this:
lg(x^2-8)lg(2-x)=lg(x^2-8)/lg (2-x) and from here you have three cases:
lg(x^2-8)=0 and lg(2-x)=+-1 and from here you get the solutions! :D

cheers! :D

1)use that sqrt(a^x+b^x)>=(sqrt(2a^x)+sqrt(2b^x))/2 and the other ineqs and you get that equality holds iff a^x=b^x=c^x.
1)a<>b or b<>c or a<>c then x=0
2)a=b=c we have that x is real.

cheers! :D :D

4) here all we have to do is to apply the mean inequalities:
(log_4)^2 (3/2)<=[(log_4(1+sin^2(x)))+(log_4 (1+cos^2(x))/2]^2=1/4 (log_4)^2 (2+sin^2xcos^2x)<=1/4 (log_4)^2(2+1/4)=(log_4)^2 (3/2). so we get equality in the ineq iff sinx=+-cosx which gives x=(2k+1)m/4, with m,k in Z.

cheers! :D :D :D

there you go! they weren't very hard! :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D


AMC
AIME
USA(J)MO
USAMO
AMC 12
MATHCOUNTS

I don't know... in 2004, the cutoff was 210, and aren't there going to be more people accepted into USAMO this year?

...I hope?

[quote="pieterminate"]

The highest index possible given that one scored 6 on aime is 60+150=210, and 210 is definitely not qualifying, plus no perfect scorer or near perfect scorer on the amc 12 is going to score a 6.  In other words, if one scores a 6, it's game over :([/quote]

Actually you will be quite surprised that some people scoring near perfect on AMC12 don't even qualify for USAMO, there were a couple last year, so there should be a couple this year too. But in any case the floor cannot be 6 because too many people got 6 or above and they can only take 120 people anyway.

[quote="pieterminate"][quote="trickwave"]For everyone who took the AMC 12 and also got a six on the AIME, this is an attempt to find the highest index among us, and see if there's a possibility of qualifing for USAMO.

Mine's 130/6, so my index is 190.[/quote]

The highest index possible given that one scored 6 on aime is 60+150=210, and 210 is definitely not qualifying, plus no perfect scorer or near perfect scorer on the amc 12 is going to score a 6.  In other words, if one scores a 6, it's game over :([/quote]

I don't think a 210 is definetley not qualifying.  It probably won't happen, but I wouldn't be too suprised if it did.

And it's possible that a prefect scorer might get a 6.  There are always a couple each year that don't even make it to USAMO.

Actually I think the floor might possibly be a 6.  And I'm not saying that just to make people feel better, I seriously think so.

[quote="DPopov"]6 is possible by my guess would be 7 or 8 though I do know quite a few people that scored in the 6-7 range with high AMC scores.[/quote]

[b][i]So is 8 definitely qualifying?  [/i][/b]

I heard people say 9 being the floor and one of my non-friends got an 8 (so lets hope it's 9+).  Im so annoyed, stupid mistakes when I could have gotten at least 10... :([/b]

Don't think about contest scores. Think about national ranking  ;) That will get you on par.

[quote="pieterminate"][quote="amirhtlusa"]not necessarliy
and lets not start another thread here about what the floor will be, as there are other, longer, threads that are discussiong the same issue ;)[/quote]

Indicate the other existing threads? :)[/quote]

there is a sticky thread about index/floor with 11 pages ;)

Also I think nationals is something you will enjoy more than USAMO (though I never qualified) :whistling:

[quote="daine"][quote="trickwave"]For everyone who took the AMC 12 and also got a six on the AIME, this is an attempt to find the highest index among us, and see if there's a possibility of qualifing for USAMO for the sophomores and below.

Mine's 130/6, so my index is 190.[/quote]
I got a 129 on AMC 12 (highest 7th grade score I think) and a 6 on the AIME (only one better than last year  :( )  so my index is 189.

When my mom and I were discussing my chances of qualifying for the USAMO v. my chances of qualifying for Mathcounts nationals, we thought that USAMO would be much more likely, but instead I qualified for Mathcounts Nationals! (I think I actually prefer it this way because instead of USAMO but not nationals because I have much more time left to qualify for USAMO than for nationals.)[/quote]Mathcounts Nationals should be easier to qualify for.  There are 200 something middle schoolers that qualify for mathcounts nationals, but only a handful of middle schoolers qualifying for USAMO.

[quote="daine"][quote="trickwave"]For everyone who took the AMC 12 and also got a six on the AIME, this is an attempt to find the highest index among us, and see if there's a possibility of qualifing for USAMO for the sophomores and below.

Mine's 130/6, so my index is 190.[/quote]
I got a 129 on AMC 12 (highest 7th grade score I think) and a 6 on the AIME (only one better than last year  :( )  so my index is 189.

When my mom and I were discussing my chances of qualifying for the USAMO v. my chances of qualifying for Mathcounts nationals, we thought that USAMO would be much more likely, but instead I qualified for Mathcounts Nationals! (I think I actually prefer it this way because instead of USAMO but not nationals because I have much more time left to qualify for USAMO than for nationals.)[/quote]
How'd you do that on the 12???!!!

[quote="hockeyadi23"][quote="pieterminate"][quote="09husbri"]I think that as a seventh grader you should definitely cherish your time at Nationals. With that AMC-12 score you'll definitely qualify for the USAMO four or even five times, which will give you plenty of opportunities to excel in that. But Nationals is something special. Take it from me - I went from being 42nd in the so-so state of Ohio in 6th to qualifying for Nationals in 8th, and it could not have come at a better time - it was an incredible weekend. Enjoy![/quote]

Yep! Nats is definitely better, even if one does horrible :( ......

Stupid mistakes matter way too much in Mathcounts though......[/quote]

tell me about it...[/quote]

I probably made a record with over 10 stupid mistakes in the sprint :(  :(  :( : oops:, or one can just call it incompetence.........

[quote="Xantos C. Guin"]yes, Nationals is way more fun and the problems are way more doable.
I tied the 12th place guy but 6 target meant that I lost the tiebraker and got 16th.[/quote]

Hey, was that me? Are you a sophomore?

Oh well, I just remember someone else asking me how many Targets I got since they got the same score as me at nationals two years ago. Otherwise, how do you find out that kind of stuff?


The government has increased the sales tax from $ 7\%$ to $ 8\%$. If you buy a car that costs $ \$20,000$ before taxes, what is the percent, to the nearest whole percent, of increase of the taxes that you can expect to pay?

[hide="Solution"]Before the increase in taxes, sales tax on the respective car would've amounted $ \$1400$. After the increase in taxes, it would've amounted to $ \$1600$. Thus, the increase in percentage is $ \frac{1600\minus{}1400}{1400}\cdot100\equal{}\frac{200}{1400}\cdot100\equal{}\frac{1}{7}\cdot100\approx\boxed{14}$.[/hide]


Let $n$ be a natural number greater than $1$, Denote $M=\{1,2,...,n\}$
For each $k\in\{1,2,...,n\}$, define
\[ x_k=\frac{1}{n+1}\sum_{A\subset M,|A|=k}(minA+maxA) \]
Calculate \[ \sum_{k=1}^nx_k \]

[quote="Agrippina"]$\frac{1}{n+1}\sum_{A\subset M,|A|=k}\max A= \frac{1}{n+1}\sum_{l=1}^nl{n-l\choose k-1}$ and $\frac{1}{n+1}\sum_{A\subset M,|A|=k}\min A= \frac{1}{n+1}\sum_{l=1}^nl{l-1\choose k-1}$, so
$x_k=\frac{1}{n+1}\sum_{l=1}^nl[{n-l\choose k-1}+{l-1\choose k-1}]$.

Then
\begin{eqnarray*} \sum_{k=1}^nx_k&=&\frac{1}{n+1}\sum_{k=1}^n\sum_{l=1}^nl[{n-l\choose k-1}+{l-1\choose k-1}]\\ &=&\frac{1}{n+1}\sum_{l=1}^nl\sum_{k=1}^n[{n-l\choose k-1}+{l-1\choose k-1}]. \end{eqnarray*}

$\sum_{k=1}^n{n-l\choose k-1}=2^{n-l}$ and $\sum_{k=1}^n{l-1\choose k-1}=2^{l-1}$, so
\begin{eqnarray*} &&\frac{1}{n+1}\sum_{l=1}^nl\sum_{k=1}^n[{n-l\choose k-1}+{l-1\choose k-1}]\\ &=&\frac{1}{n+1}\sum_{l=1}^nl[2^{n-l}+2^{l-1}]\\ &=&\frac{1}{n+1}\sum_{r=1}^{n-1}[(r+1)+(n-r)]2^r\\ &=&\sum_{r=1}^{n-1}2^r\\ &=&2^n-1. \end{eqnarray*}[/quote]
Your solution is very algebraic and long. Is there any one do with another argument

uh, nice!
consider, for any set $A = \{a_1, \dots, a_k\} \subset M$ it's "complement" ${A' = \{n+1-a_1, \dots, n+1-a_k} \subset M$.
it's clear that you can re-write your sum as:\[ \sum_{|A|=1}^n \frac{\min A + \max A}{n+1} = \sum_{|A|=1}^{n} \frac{\min A + \min A' + \max A + \max A'}{2(n+1)} = \sum_{|A| = 1}^n 1 = 2^n-1 \]


trigonometry
geometry
AMC
AIME
analytic geometry
trig identities

For some reason I really like making study guides.  I have made a whole bunch of them for my math team (matrices, geometry formulas, analytic geometry, sequences) and now I want to type up a lot of them, and make one on polar graphs.   I also want to make one on tough trig equations.

So if anyone wants to post some hard trig equations on this thread that would be very helpful!  Also, if they are from a certain place like (this book, taken from this contest like AMC, AIME, etc, got this one off a test I had in school, etc) then please note this so I can put the source.

(note: you don't need to solve them, but I might ask you to later)

By study guide, do you mean a crib sheet (aka a cheat sheet) with a bunch of formulas on it, or a challange sheet (with problems to solve)?

Well I make all of them I guess.  I would say that there are 3 types of study guides:

(1) set of challenging problems on a certain topic (w/solutions)
(2) a cheat sheet for formulas
(3) a tutorial

Any good study guide would be a combination of all 3, w/ a focus on one of them.  My trig equation guide will have a list of trig identities and basic formulas for refrence, maybe I will teach trig (who knows, I havent made it yet) in tutorial form, and then I will have a lot of solved problems, which are tough trig equations.  I am looking for donations of tough trig equations( (1) type) from AoPS members, so that I can make this study guide.


limit
real analysis
real analysis solved

$f$ from $R^n$ to $R^m$ such that

the image of every compact set is compact and
the image of every connected set is connected

Prove that $f$ is continuous

For every point $x_{0}$ take the family of closed balls $B(x_{0},\frac{1}{n})$.Denote by $K_{n}=f(B(x_{0},\frac{1}{n}))$.We have that $K_{n}$ is compact for all natural $n$-s.Besides $K_{1}{\supset}K_{2}{\supset}\cdots{\supset}K_{n}{\supset}\cdots$.Denote by $K=\bigcap_{n=1}^{\infty}{K_{n}}$.$K$ is compact and moreover since $K_{n}$ are compact $\lim_{n{\rightarrow}\infty}{diam(K_{n})}=diam(K)$,but $f^{-1}(K)=f^{-1}(\bigcap_{n=1}^{\infty}{K_{n}})=\bigcap_{n=1}^{\infty}{f^{-1}(K_{n})}=\bigcap_{n=1}^{\infty}{B(x_{0},\frac{1}{n})}=x_{0}$,so
$K=\{ f(x_{0}) \}$ so $diam(K)=0$ so $\lim_{n{\rightarrow}\infty}{diam(K_{n})}=0$ so $f$ is continious.
Probably I have a mistake becasuse I didn't use that the image of connected set is connected  ,but I can't find it.    :?

[quote="LevonNurbekian"]$\bigcap_{n=1}^{\infty}{f^{-1}(K_{n})}=\bigcap_{n=1}^{\infty}{B(x_{0},\frac{1}{n})}=x_{0}$[/quote]

I believe this is where it falls apart. You don't know that the preimage of $K_n$ is the ball $B(x_0,\frac 1n)$. Maybe there are some points outside this ball which are also mapped into $K_n$.

Moubinool, the last few problems you've posted (this one included) come from some Miklos Schweitzer competitions. Why did you put * as the source? :)

How about this:

Take, as Levon did, that descending sequence of balls $B_n=B(x_0,\frac 1n)$ around $x_0$. $K_n=f(B_n)$ is a connected and compact set, so let $K=\bigcap K_n$. Assume there's a point $y_1\in K$, different from $y_0=f(x_0)$, and let $d=d(y_0,y_1)$. Since each $K_n$ is connected, we can find points in $K_n$ at any distance from $y_0$ in the range $(0,d)$. Let $a_n\in B_n$ with the property that $d(y_0,f(a_n))=d-\frac 1n$.

The image $F$ of the compact set $\{x_0\}\cup (a_n)_n$ is a bounded sequence of points and it's also compact, meaning that it contains all its limit points. However, from the sequence $f(a_n)$ we can extract a convergent subsequence, which must converge to a point whose distance to $x_0$ is $d$, but no such point exists in $F$. We thus have a contradiction.

Is it correct?

[quote="grobber"][quote="LevonNurbekian"]$\bigcap_{n=1}^{\infty}{f^{-1}(K_{n})}=\bigcap_{n=1}^{\infty}{B(x_{0},\frac{1}{n})}=x_{0}$[/quote]

I believe this is where it falls apart. You don't know that the preimage of $K_n$ is the ball $B(x_0,\frac 1n)$. Maybe there are some points outside this ball which are also mapped into $K_n$.

[/quote]
Yes of course grobber you are absolutely right.I'll think on your new solution.Thank you.


function
logarithms
integration
real analysis
real analysis unsolved

A)Prove that $ \Gamma ''(x)>0$
B)Is there a simple way to compute $ \Gamma '(x)$ ?

everyona know that $ \Gamma (x) > 0$
and now we need to prove that $ \Gamma ' (x) > 0$
it's enough to show that $ \ln \Gamma (x)$ is convex function.
 Ok Let's try to prove it:
$ \ln \Gamma$ is convex . means that
if for all $ \alpha, \beta > 0$ such as  $ \alpha \plus{} \beta \equal{} 1$ and for all $ x_{1},x_{2} \in < a,b >$ (we need to prove convexity on interval $ < a,b >$) then
$ \ln (\Gamma (\alpha x_{1} \plus{} \beta x_{2})) \leq \alpha \ln \Gamma (x_{1}) \plus{} \beta \ln \Gamma (x_{2})$
OK
or
$ \Gamma (\alpha x_{1} \plus{} \beta x_{2}) \leq \Gamma^{\alpha}(x_{1})\Gamma^{\beta}(x_{2})$
hence :
$ \Gamma (\alpha x_{1} \plus{} \beta x_{2}) \equal{} \int_{0}^{\infty}u^{\alpha(x_{1} \minus{} 1) \plus{} \beta (x_{2} \minus{} 1)}e^{ \minus{} \alpha u}e^{ \minus{} \beta u}du$=
$ \int_{0}^{\infty}(u^{x_{1} \minus{} 1}e^{ \minus{} u})^{\alpha}(u^{x_{2} \minus{} 1}e^{ \minus{} u})^{\beta}du\leq ...Holder...\leq (\int_{0}^{\infty} u^{x_{1} \minus{} 1}e^{ \minus{} u}du)^{\alpha}(\int_{0}^{\infty} u^{x_{2} \minus{} 1}e^{ \minus{} u}du )^{\beta}$ Done.
about 2.
I know something but I'm not sure will be it simplest or not..   :maybe:


[b]Can anyone give me the proof of the following:
If a and b are coprime...Then the largest number which cant be created using a and b is (a.b-a-b).[/b]
(We have infinite supply of a and b)

What do you mean by "created"? What do you mean by $(a.b-a-b)$?

Ok.....I describe it more effectively..The largest sum that can NOT be created..As an eg. If the numbers are 8 and 3..then the we cant have 1 frm it we cant have 2 frm it we can have 3( as 0*8+1*3)..we cant have 4..we cant have 5 and 7..we can have 6 as 0*8+2*3,so amongst all the numbers the lagrest sum which we cant get is 8*3-8-3=24-11=13.....after 13 we can get any number using 8's and 3's

[b]So if a and b are coprime.then the largest sum that cant be had using a's and b's is a*b-a-b[/b]

I think you mean the largest integer not representable in the form $ax+by$ for $x,y\geq0 \in \mathbb{N}$ is $ab-a-b$

Edit: Thank you Varun!

[quote="seamusoboyle"]I think you mean the smallest integer not representable in the form $ax+by$ for $x,y\geq0 \in \mathbb{N}$ is $ab-a-b$[/quote]
On the contrary the [b]Largest integer [/b]not representable in the form $ax+by$ for $x,y\geq0 \in \mathbb{N}$ is $ab-a-b$
[b]whenevr GCD(a,b)=1[/b]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=105458&highlight=mcnugget+chicken

amcavoy provides a solution near the bottom.

But he hasn still proved that it is the [b]largest [/b]number....I guess after that induction can be used we just need to prove that all the numbers after ab-a-b can be formed using a nd b.


topology
superior algebra
superior algebra unsolved

Let $ A$ be a commutative ring with identity.Let $ Min \{A\}$ be the subspace of $ Spec \{A\}$ with Zariski Topology consist of all the minimal prime ideal.
Prove that $ Min \{A \}$ is a Hausdorff space.

Thanks in advance.

Given two different minimal prime ideals $ P_1,P_2$ in $ A$, construct an element $ a\in A$ becoming a unit in $ A_{P_1}$ and 0 in $ A_{P_2}$. Then the standard affine subsets $ D_a$ and $ D_{1 \minus{} a}$ will do the separation job.
[hide="One way to construct $ a$:"]For this consider $ S^{ \minus{} 1}A$ with $ S \equal{} (A\setminus P_1)\cap (A\setminus P_2)$. Then $ Spec(S^{ \minus{} 1}A)$ is a two point Hausdorff space, apply the Chinese remainder theorem and lift things from $ S^{ \minus{} 1}A$ to $ A$.[/hide]
[b]Remark:[/b] The technique uses the fact that the topology of the open subsets of $ P_1$ and $ P_2$ becoming sufficiently small is coded in the topology of  $ Spec(S^{ \minus{} 1}A)$, because  $ Spec(S^{ \minus{} 1}A)$ is the intersection of all open subsets containing $ P_1$ and $ P_2$. This generalizes to every multiplicative subset $ S$.

I don't see why $ D_a$ and $ D_{1\minus{}a}$ have to be disjoint. this is equivalent to: $ a(1\minus{}a)$ is nilpotent.

This is simply the remark: Localize to assume wlog that $ a$ is unity in $ A_{P_1}$. This cuts off any uncomfortable intersections. The whole story is about local properties. Being nilpotent is a global property, since, in general, this property needs to be checked on every single irreducible component, i.e. in every point of the prime spectrum.

ps: Wlog assume that $ A$ is reduced, since the closed immersion $ Spec(A/nil(A))\to Spec(A)$ is a homeomorphism, i.e. the topology cannot distinguish between nilpotent elements. This shows that nilpotency cannot be a relevant point in any argument for the main result.

Thank you,But I still don't understand why $ a(1\minus{}a)$ is nilpotent?Can you explain it more explicitly?

Sorry for delay.

Oh,sorry,I think you may misunderstand what I mean,or there is something unclear in my expression,I am sorry.

I want to use $ a(1\minus{}a)$ is nilpotent to prove that $ D(a) \cap D(1 \minus{} a)\equal{}\emptyset$,because it is obviously they are equivalent.

Or can you tell me have can I know $ D(a) \cap D(1 \minus{} a)\equal{}\emptyset$ from the construction of "Third Edition"'s proof.
Thank you.
[quote="yannick"]If $ D(a) \cap D(1 \minus{} a) \equal{} \{\mathfrak{p} \in \textrm{Spec}(A): a \notin \mathfrak{p} \} \cap \{\mathfrak{p} \in \textrm{Spec}(A): 1 \minus{} a \notin \mathfrak{p} \} \equal{} \emptyset$ then we must have that either $ a \in \mathfrak{p}$ or  $ 1 \minus{} a \in \mathfrak{p}$ for every prime ideal. But then we can just say that $ a(1 \minus{} a) \in \mathfrak{p}$ for every $ \mathfrak{p} \in \textrm{Spec}(A)$ or, even better, that $ a(1 \minus{} a) \in \mathcal{N}(A)$ (the nilradical, i.e. the intersection of all prime ideals), so $ a(1 \minus{} a)$ is nilpotent.[/quote]


Let $ V$ be the set of real numbers. Regard $ V$ as a vector space over $ \mathbb{Q}$, the set of rationals. Prove that this vector space is not finite dimensional.

I tried it by assuming that $ \alpha_1,...,\alpha_r$ is a basis for $ V$ and then tried to find $ \alpha\in\mathbb{R}$ such that $ \alpha\neq\sum \alpha_i q_i$ for any set of rationals $ q_i$

This hard result is a special result of H. Cartan and Choquet, the proof use Axiom of Choice. Its cardinality is $ 2^{{\mathcal N}_0}$ but I don't know the details of the proof  :(

You certainly do not need the axiom of choice just for showing that the dimension is not finite. But considering cardinalities is probably one of the easiest arguments. owk

But this is an excercise in the second chapter of Hoffman & Kunze, which is considered an intro. to linear algebra book :|

EDIT: Posted before reading owk's message.

the idea is to show that any finite dimensional space over a countable field is itself countable (which R is not).

Induction on the dimension on the space will do the trick.

[b]@§outh§tar[/b] Thanks.
To show that finite dimensional space over a countable feld is countable, we could use the fact that the set of ordered n-tuples of members of a countable set is countable.


topology
trigonometry
advanced fields
advanced fields unsolved

In the book I'm reading, it says that when a surface $ S$ and a simple curve $ C$ (1-manifold in $ R^3$ intersects traversely, then the set $ S \cap C$ is discrete set in $ R^3$. I don't think it is true. I think there would be some problem when one considers something like topologist sine curve. I think $ S \cap C$ is only discrete in C. Am I right?

This may depend on what is a surface and what is a simple curve. For example, if a simple curve is a homeomorphic image of $ [0,1]$, then the sine curve is not allowed. Also, the set $ \{1/n\colon n\ge 1\}$ is discrete (although not uniformly separated).


geometry
incenter
geometric transformation

Prove that the incenter of a triangle is Nagel's point for the triangle determinated by the midpoints of the sides.

Nice property, thanks for reminding me of the beauty of the Nagel line!

The proof is very easy if you know the Nagel line, or if you consult Mathworld:
http://mathworld.wolfram.com/NagelPoint.html for Nagel Point
http://mathworld.wolfram.com/NagelLine.html for Nagel line

The triangle and its medial triangle are similar, with $G$ as their point of similarity. So the spiral Similarity with center $G$, angle $180^{\circ}$ and factor $\frac{1}{2}$ maps $\Delta ABC$ to its medial triangle $DEF$.

We also know that $I,G, Na$ lie on one line, the Nagel line. We also know that $2 \left|GI \right| = \left|GNa\right|$, so $Na$ is mapped onto $I$ by this spiral similarity. Therefore, $I$ is the Nagel point of the medial triangle.


[quote="Demon"]I just recieved an e-mail saying that I got a new private message, but in my AoPS Inbox the last message is dated January 5. What's going on?[/quote]Simple: someone wanted to sent you pm, changed his mind, and deleted the message from his outbox (where the message lies until you actually read it). There is nothing wrong. Also it might have been a late email :)


vector
trigonometry
inequalities
triangle inequality
linear algebra
linear algebra theorems

Hey all,

Is anyone aware of a theorem to handle the following question, which seems related to linear dependence of a set of vectors:

Suppose we are given a set of n positive real numbers, S. Is it possible to come up with a set of n vectors (in the two-dimensional real Euclidean space), where vector i has magnitude S[i], and summing all n vectors gives 0? (where the notation V[i] means the ith element of the set V). I don't care about finding the actual vectors in the solution, I just need to know whether a solution exists, given a specific S.

This can be equivalently phrased as: Is it possible to construct a polygon of n sides, where side i has length S[i]?

I imagine there must be some necessary law as to the relationship of the sides of a polygon, i.e. for triangles we have the cosine law.

Any ideas?
Thanks,
Sev

Assume that $ n\ge3;$ else the problem makes no sense. WLOG, assume that $ S_j$ is the largest number - that is, $ S_j\ge S_i$ for all $ 1\le i\le n.$

Then there is the following necessary and sufficient condition: $ S_j\le \sum_{i\ne j}S_i.$

Given that that's the condition, can you see how you might try proving it? One hint: there's no real difference between $ n\equal{}3$ and $ n>3,$ since you can always string $ n\minus{}2$ edges together in the same direction.

(By the way, "Linear Dependence" is a bad title for this topic.)

Thanks for your response.

Hmm, I gave it some quick thought, sounds like a glorified triangle inequality. I managed to prove "most" of it, a few issues:

1) The triangle inequality also has a required lower bound (as per Wikipedia). Specifically, suppose $ n\equal{}3$ and $ a$ is the largest of 3 sides, $ a$, $ b$, and $ c$. Then we need $ |b\minus{}c|\leq a \leq b\plus{}c$. Does this need to factor into your nec/suff. condition? Given that $ a \leq b\plus{}c$, I'm having difficulty seeing that the lower bound should implicitly hold?

2) For $ n\geq 3$, from your hint it seems clear that if $ S_j \leq \sum_{i\neq j}S_i$ (where $ S_j$ is the largest side), then we can just create a larger triangle with $ n\minus{}2$ of the sides combined as one side. Is this truly a necessary condition though for $ n>3$? (ie if we allow non-triangular shapes, can we relax this constraint?). I guess this question arises since I'm still not sure what the full proof of your statement would be.

Thanks,
Sev


$ \lim_{x\to0^{ \plus{} }} \frac {arccos(\frac {1}{1 \plus{} x})}{\sqrt {x}}$

not using L'Hospital's

$ \lim_{x\to0^{ \plus{} }} \frac {arccos(\frac {1}{1 \plus{} x})}{\sqrt {x}}\equal{}\lim_{x\to0^{ \plus{} }} \frac{arcsin\sqrt{\frac{x(x\plus{}2)}{(x\plus{}1)^2}}}{\sqrt{\frac{x(x\plus{}2)}{(x\plus{}1)^2}}} \cdot \sqrt{\frac{(x\plus{}2)}{(x\plus{}1)^2}}\equal{}\sqrt{2}$

or

$ \lim_{x\to0^{ \plus{} }} \frac {arccos(\frac {1}{1 \plus{} x})}{\sqrt {x}}\equal{}H\equal{}\lim_{x\to0^{ \plus{} }} \frac{\frac{1}{\sqrt{1\minus{}x^2}(1\plus{}x)^2}}{\frac{1}{2\sqrt{x}}}\equal{}0$

??

You applied the chain rule incorrectly while computing $ \frac {d}{dx} \, \arccos x$.  Fix that and your two methods agree.

\arccos looks nicer than arccos, likewise \arcsin


geometry
rectangle
number theory proposed
number theory

Prove that we can not title  the rectangles of the sizes (4m+1).(4n+1) by the finite numbers of small rectangles with 1.(2k) and (2k).1 rectangles such that only the square in the center of the table does not belong to any rectangle ( k,m,n are the positive integers more than 1) :D


[quote="C'est lavie"]let $f [0,1] \to [0,1]$ satisfy
$|f(x1)-f(x2)|<|x1-x2|$,  $\forall x1,x2 \in [0,1] ,x1\neq x2$
prove that   $\exists$ only one $x_{0}\in [0,1] : f(x_{0})=x_{0}$[/quote]
Isn't that the Lifshitz condition.
Therefore its continous.
Now use IVT to prove existence.

[quote="C'est lavie"][quote="sylow_theory"][quote="C'est lavie"]let $f [0,1] \to [0,1]$ satisfy
$|f(x1)-f(x2)|<|x1-x2|$,  $\forall x1,x2 \in [0,1] ,x1\neq x2$
prove that   $\exists$ only one $x_{0}\in [0,1] : f(x_{0})=x_{0}$[/quote]
Isn't that the - condition.
Therefore its continous.
Now use IVT to prove existence.[/quote]
what IVT???can you express more and more???[/quote]
Oh, I see, the forum automatically removed a four letter word which appear in a mathemations name, and you did not see it.  :rotfl:

[b]Lifsh1tz Condition:[/b] If a function $f$ is defined on some non-empty set $S$ such that:
$|f(x_{2})-f(x_{1})|<M\cdot |x_{2}-x_{1}| ,\, \forall x_{1},x_{2}\in S ,\, x_{1}\not = x_{2}$
For some $M>0$, then $f$ is continous on $S$.

This shows us that your function, $f: [0,1] \mapsto [0,1]$ is continous.

There are to ways to show $f(x_{0})=x_{0}$, one is more elementary which is what I assume.

[b]Intermediate Value Theorem:[/b] If $f$ is a continous function on a closed interval $[a,b]$ such that there is a real numer $r$ strictly in between $f(a), f(b)$ (meaning $f(a)<r<f(b)$ or $f(b)<r<f(a)$) then there exists $x_{0}\in (a,b)$ such that $f(x_{0})=r$. This is also called IVT for short.

[b]Answer to Question:[/b] Define a function $g(x)=f(x)-x$, this function is continous on $[0,1]$ because both $f(x)$ and $x$ are continous on $[0,1]$.
Note,
$g(0)=f(0)-0=f(0)\geq 0$
$g(1)=f(1)-1\leq 1-1=0$
Then by IVT (a slight different version),
$g(x_{0})=0\to f(x_{0})=x_{0}$ for some $x_{0}\in [0,1]$.

Here is a quicker approach from topology.

[b]Brouwer Fixed Point Theorem:[/b] Every continous map from itself to itself must contain a fixed point.

[b]Answer to Question:[/b] You are mapping $[0,1]$ to itself, hence it must have a fixed point.

If you spell Lipschitz correctly, it doesn't get deleted.

That does seem like an overactive censor.

Read the correct spelling that jmerry gave: there's a c.  If you leave out the c, you need to put a 1 in place of the i to avoid the censor.  mlok was remarking on the absence of the c, not the presence of a 1 in place of an i.

[quote="mlok"]sylow_theory is not the only one to make that mistake. One respectable mathematical journal recently published a paper titled [url=http://www.springerlink.com/content/kg5665u2l1208341/]"Lipsh1tz maps from surfaces"[/url]  :rotfl: The name is misspelled throughout the 38-page paper.[/quote]

Sweet  :rotfl:

Apparently, I thought it was,
http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Lifshitz.html

But really it was,
http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Lipschitz.html


[quote="socrates"][quote="tamtrantrong"]Let a,b,c>=0.Prove that:
   Sum a^2:(2a^2+b^2) >=(a+b+c)/3[/quote]

Do you mean:

Let $a,b,c\geq0$.
Prove that: $\sum \frac{a^2}{2a^2+b^2}\geq \frac{a+b+c}{3}$ ??

 :D[/quote]

Probably no, because it is obviously false  ;)

Let $a,b,c\geq0$.
Prove that: $\sum \frac{a^2}{2a^2+b^2}\geq \frac{a+b+c}{3}$ ??

 :D[/quote]

                          But I think :  the left must be     (  <= ) the right

[quote="tamtrantrong"]No,i think you are wrong,babylearnmath294
Prove that Sum a^3/(2a^2+b^2)>=(a+b+c)/3[/quote]
  let$a,b,c \geq 0$
  $\sum \frac{a^3}{2a^2+b^2} \geq \frac{a+b+c}{3}$

It have been post [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%5C%5Cfrac%7Ba%5E3%7D%7B2a%5E2%2Bb%5E2%7D&t=23111]here[/url],but what a pity there is no solution.

All these inequalities are solved in the same way as Vasc and Peter  Scholze taught us. Again I will post the beautiful solution of Vasc in that type of ineqs

It's a problem of my friend, so long ago. I solved it, of courseso long ago. Use the way "sum of square". I had sent my solution to Manlio, where is he now?

The other problem, I think it's not belong of Vasc. He only has a nice solution for it.


LaTeX
inequalities unsolved
inequalities

Consider a  triangle  ABC. Find  all  $ x,y,z \epsilon\Re $  such that
$ xa^2 + yb^2 + zc^2 \leq 0  $

hi nickolas :) ... a few latex tips for you today: use \in for $\in$ not \epsilon ;) also use \mathbb{R} for $\mathbb{R}$, also \mathbb{Z} for $\mathbb{Z}$ etc :)

as for the problem itself, it seems to me that there is something missing from the statement.  Otherwise, we easily see that at least one of $x,y,z$ is negative. If all are negative, we are done. If $x\leq 0$ and $y\geq 0$ then if $z\leq 0$, then $y \leq -\frac { xa^2+zc^2}{b^2}$, and so on ...

Hmm, Nickolas, it's not at all clear to me what you want...

  Darij


How many two digit integers $ x$ exist such that $ x+1$ evenly divides into $ 1+2+\cdots+x$?

[hide]
The sum of the first $ x$ integers is $ \frac{x(x+1)}{2}$, so all two digits number evenly divide into that sum
$ \boxed{ 99-10+1 = 90 }$.
[/hide]

[quote="AstroPhys"][hide="If x=83..."]
$ \frac{x(x+1)}{2}=84 \frac{83}{2}\Rightarrow 84 \frac{83}{2}\equiv 0 \mod (83+1) \equiv 0 \mod 84$.
[/hide][/quote]

Ehh... by that logic, $ 3 \equiv 6\mod0$, cuz $ 3=6\cdot\frac12$.

[hide="Then do you mean... EDIT Typed the wrong number getting the wrong answer"]
that the final fraction must reduce to an integer?
Assumed you meant the that the sum of integers up to $ x$ has a factor of $ x+1$
So then it would be all even two digit integers or $ \boxed{ \frac{99-10+1}{2}=45 }$.
[/hide]

we have $ 1+2...x=\frac{x(x+1)}{2}$ we want to divide by x+1. then we have $ \frac{x}{2}$ must be an integer. so x must be even. there are 90 2 digit integers so there are 45 even 2 digit integers.

$ 45$


induction
factorial
inequalities
function
calculus
derivative
inequalities open

What about this one? Again, there might be a trivial counterexample...

Prove or disprove that ${{a+1}\choose{b+1}} \leq (\frac a b)^{2b}$, where $1 \leq b \leq \frac a 2$.

Still no answers? The formula is true for all the values I tried and for the trivial cases, but any kind of induction has failed so far. Even the strategy described in Binomial bounding #1 seems useless here.
Any ideas?

Maybe some Mathematica freaks out there could look for counterexamples...
...Please...  :(

Matlab says you're good through a=169
(170 is highest factorial it will take)

Also, b=1, a=2 produces the closest to equality for a<170.

Many thanks!   :)
We should definitely look for a proof, then.

I think this inequality is true. Let us suppose that $ b>2 $ since otherwise is trivial. Let us take $ a=b+x $, where $ x\geq b $. Then the inequality becomes $ (1+\frac{x}{1})\cdot\cdot (1+\frac{x}{b+1})\leq (1+\frac{x}{b})^{2b} $. Now, take the function $ f(x)=\sum_{i=1}^{b+1} {ln(1+\frac{x}{i})}-2bln(1+\frac{x}{b}) $ defined for $ x\geq b$. I claim that it is decreasing. Its derivative is $\sum_{i=1}^{b+1} {\frac{1}{x+i}}-\frac{2b}{x+b} $. So, we have to prove that $\sum{i=1}^{b+1} {\frac{x+b}{x+i}}\leq 2b$. Its enough to prove that $\sum_{i=1}^b {\frac{x+b}{x+i}}\leq 2b-1 $. This is the same as $\sum{i=1}^b {\frac{b-i}{x+i}}\leq b-1 $. Of course, it is enough to prove this when $ x=b $, case in which it becomes $\sum{i=1}^b {\frac{2i}{b+i}}\geq 1 $ and its trivial (look at the last term of the summation). Thus, $f$ is decreasing and it suffices to prove the inequality for $ a=2b $, which isnt hard. Hope I didnt do again idiot mistakes.

Yes, sure!
With a similar argument, one can also prove that ${{a+2}\choose{b+2}} \leq (\frac a b)^{2b}$, with $2 \leq b \leq \frac a 2$.

How to prove in a simple way that for any $n\geq0$ integer we have

$(1+n)(1+\frac{n}{2})(1+\frac{n}{3}).....(1+\frac{n}{n+1}) \leq 2^{2n} $

Thank you very much.

Manlio: just by induction.
Your statement is equivalent to ${{2n+1}\choose{n+1}} \leq 2^{2n}$.
For $n=0$ we get $1 \leq 1$, true.
Now, let's suppose ${{2n-1}\choose{n}} \leq 2^{2n-2}$, and claim that ${{2n+1}\choose{n+1}} \leq 2^{2n}$. We just have to prove that $\frac{(2n+1)(2n)}{(n+1)(n)} \leq 4$, which is trivial and holds for every $n \geq 1$.

Here's another proof that ${2n + 1 \choose n + 1} \le 2^{2n}$.  Note that ${2n + 1 \choose n + 1} = {2n + 1 \choose n}$.  Also, the sum of all the $2n + 1 \choose k$ is $2^{2n+1}$.  Combine those two facts and we are done.