Perhaps it sounds absurd, but I came upon this thought about the presence of extra electrons in a sample of pure water; a negative static charge is produced by an accumulation of electrons in an object; so, I imagined that something similar to this might occur, if we gave a sample of water a substantial negative charge:

$ \ce{2e^ \minus{} } \plus{} \ce{2H \plus{} } \rightarrow \ce{H_2}$

($ \ce{H \plus{} }$ is present due to the autoionization of water:
$ \ce{H2O} \rightleftharpoons \ce{H \plus{} } \plus{} \ce{OH^ \minus{} }$)

So, with $ \ce{H \plus{} }$ removed from the sample, the hydroxide anion $ \ce{OH^ \minus{} }$ will begin to accumulate in significant quantities. This would lead to a substantial increase in the solution's pH, effectively creating a "superbasic" solution which would allow the study of reactions in such a environment, as long as the reactants are provided with a similar charge prior to placing them in solution.

(Plus, if the charge is neutralized we would have a ton of $ \ce{OH}\cdot$ free radicals floating around and probably forming hydrogen peroxide, which decays to water and diatomic oxygen. "Electrolysis" of pure water with static charge ftw?)

Of course, I recognize that this will probably not work, but I do not understand why it would be the case. If it is not, and you have an adequate explanation, please do post! I would really appreciate it.

Part I'm not sure about is how you're going to introduce the actual electrons in the water. :maybe:

First, how would it be possible to force a reduction reaction without occuring at the same time an oxidation reaction?
Second, even if you devise an electrochemical method to decrease H3O+, since there is always the equilibrium

$ 2\,H_2O \rightleftharpoons H_3O^ \plus{} \plus{} HO^ \minus{}$

Le Chatelier Principle tells us that this equilibrium would develop to the right in order to produce more H3O+.

If you look at the Kelvin Water Dropper (William Thomson, Lord Kelvin of Largs, 1869) it is a device that through electrostatic induction stores both positive and negative charge in separate vessels of water which discharge when the potential difference becomes large enough.  The charge carrying species are probably hydrated hydrogen ions and hydrated hydroxide ions rather than hydrated electrons.  As water is a conductor, albeit a poor one when near pure, physics dictates that the excess charge accumulates at the surface so only the surface would have "interesting" properties and the internal bulk would be just water, making the net average concentration of excess charge quite low.  The excess charge per unit mass would be much greater in bubbles, that is, hollow drops.   In a strong electric field, water (and bubbles) form Taylor cones that would allow the carrying away of charge as charged water clusters.  This mechanism is probably the reason why the Kelvin Water Dropper does not achieve more than about 15 kV.

If you were to create a surplus of OH- without oxidizing the H atoms wouldn't you end up evolving a huge amount of H2. Just a sidenote

Also Carcul I don't understand your logic regarding Le Chatelier's principle. Wouldn't you simply end up having the H+ concentration diminish to maintain Kw?

Carcul your logic would simply imply $OH$ increases in concentration as well.


factorial
number theory
number theory proposed

Solve in non- negative integers the following equation

$ y^2 \plus{} 2009! \equal{} x!$

Daniel

If I'm not wrong, $ 2009!\equal{}2^{2003} \cdot z$, where $ z$ is an odd integer.

If $ x\equal{}2009$ then $ y\equal{}0$

If $ x>2009$ then $ y^2\equal{}2009!(2010 \cdot 2011 \cdot ... \minus{}1)$.

Since $ 2009!\equal{}2^{2003} \cdot z$, where $ z$ is an odd integer, and $ 2010 \cdot 2011 \cdot ... \minus{}1$ is an odd integer $ \Rightarrow  y \in \emptyset$

[quote="moldovan"]If I'm not wrong, $ 2009! \equal{} 2^{2003} \cdot z$, where $ z$ is an odd integer.

If $ x \equal{} 2009$ then $ y \equal{} 0$

If $ x > 2009$ then $ y^2 \equal{} 2009!(2010 \cdot 2011 \cdot ... \minus{} 1)$.

Since $ 2009! \equal{} 2^{2003} \cdot z$, where $ z$ is an odd integer, and $ 2010 \cdot 2011 \cdot ... \minus{} 1$ is an odd integer $ \Rightarrow y \in \emptyset$[/quote]

You are almost right since $ \sum_{n\equal{}1}^{\infty} \ [\frac{2009}{2^{n}}]\equal{}1004\plus{}502\plus{}251\plus{}125\plus{}62\plus{}31\plus{}15\plus{}7\plus{}3\plus{}1\equal{}2001$ :wink:

You're right... I should be more careful.

Well when I created this problem I didn't think it was so easy, but having seen Moldovan's solution (similar to a friend's solution which also made me realize it was easy) there's not too much to say about it. Nevertheless I want to share the idea I had in mind. Hope you like it

It's clear that for $ n\equal{}2009$ there is a solution, and for $ n\equal{}2010$ there are several (easy) ways to prove there aren't solutions.
If $ n>2010$   $ y!$ is divisible by $ 2011$ (which is a prime) so the left side of the equation is divisible by $ 2011$ also. By wilson's theorem $ 2009!\equiv1mod{2011}$ So this implies that $ y^2\equiv\minus{}1\mod2011$ But this is imposible since $ 2011$ is a prime of the form $ 4k\plus{}3$ implying that $ \minus{}1$ is a non residue $ mod 2011$

So for $ n>2010$ there aren't any more solutions

Daniel


A cylindrical tank with an open top has a diameter of 3.00m and is completely filled with water. When the setting SUn reaches an angle of 28.0 degress above the horizon, sunlight ceases to illuminate the bottom of the tank. How deep is the tank?

yes, that is what it says in the back of the book. How did  you get that? I'm not sure how to approach this problem. I figured it might be a critical angle problem but you are given the initial angle so D:

wait i think i got it. I'm not sure if this is right or if it's a miscalculation that just happens to be really close to the right answer. So here is my solution.

Since the Angle of incidence must be measured from the normal, it is 90 - 28 = 62 degrees.

Now using snell's law n1 sin angle 1 = n2 sin angle 2

so (1.0)sin 62 = (1.333) sin angle 2

sin angle 2 = sin 62/1.333 = 0.66

angle 2 (aka angle of refraction) = sin ^-1 (0.66) = 41.3 degrees

so now since the diameter of the tank is 3.00 m, you set up a right triangle diagram in which the bottom right-hand angle is 90 degrees, the top angle is 41.3 degrees, the length opposite of 41.3 is 3.00 m, and the adjacent length is the height (h)

So using triangle trigonometry, tan 41.3 = 3.00m/ h

so h = 3.00m/ tan 41.3 = about 3.4 m = height of tank

So is my solution correct with slight rounding differences with the book or am i wrong but my answer just happens to be coincidentally close with the right answer?

Thanks for your help


group theory
abstract algebra
superior algebra
superior algebra unsolved

show that a group with no proper nontrivial subgroups is cyclic

is'nt this trivial as for any member of a group there exists a cyclic group $ \{ a^n|n \in Z\}$. so if a group has only itself as a non-trivial subgroup it must be this cyclic group? this answer is a bit sus to me as this is the last question of the chapter and they tend to be longer answers/reasoning.. so please some1 confirm/correct it?

This is incomplete. Let $ G$ be a group with no proper nontrivial subgroups.The essential thing is not that there exists a group generated by every element but that these groups, having only elements in $ G$, are subgroups of $ G$. So if you take a nontrivial element, say $ a \in G$, the subgroup generated by $ a$ will be nontrivial thus will be all of $ G$ making $ G$ cyclic.


Who is going take part in APMO
Let's talk  about APMO

I will be writing it on Monday March 9, but I don't think that we can discuss it for a while after that... the website says not to discuss until the problems are up on the website... :D

Good luck to all writing it!

I will also do it Monday afternoon.

Anyone know when the result will be out?

I expect to do poorly on Monday. I am very excited to do so. Hopefully I can beat my performance from last year though :(

Me on Tuesday morning (due to time differences)

Good luck to you all Mathlinkers :)

Me too ! On Tuesday Morning .  :) Goodluck Everyone !

hm apparently I did not actually reply to the APMO email so... no APMO for me :(

very

very

unfortunate

Hey, I've just participated in the APMO a few minutes ago... obviously i can't talk about the problems, it was cool.
GeRaRd

i sat it a few days ago.
it was quite hard (at least for me).

http://www.kms.or.kr/competitions/apmo/

The questions and solutions are posted to the official website now~~

Let's discuss:
Q1: quite hard to think of the "squares of the reciprocals of the numbers"... didn't write much during the test
Q2: This is quite a classical problem in fact as we get 10 roots for a degree 10 polynomial, but I calculated the leading coefficient wrongly URGH!
Q3: quite easy if you know quite well about pole and polar stuff, but really hard if you dun... (co-geom-able though)
Q4: too easy.... many ways to construct the a_k, b_k.
Q5: i proved 4 | l-r at that time like most HK constestants did, but couldn't think of a way to model it, like using complex in the official solution X@

How do ppl from other countries do? guess the cutting pls~`

Hello everyone, I think it was a nice and difficult test

I solved 1 and 4 and did something in problem 5 , yet I;m not expecting any points in that problem

Daniel

Can anyone write last year's medal limits?

Does anyone know anything about the APMO results? Shouldn't they be on the APMO website by now?  :maybe:

[quote="varunrocks"]what is APMO?[/quote]

Asian Pacific Mathematical Olympiad. http://www.mathlinks.ro/resources.php?c=1&cid=20

[quote="blackbelt14253"]How do you register for APMO?
I want to take it next year, though I missed it this year. :(
Looking at the results, the competition looks pretty scary...(when you have black MOPpers making bronze...uh...)[/quote]

In the US, MOPpers from the previous year are emailed ~a month before. You sign up through the AMC, and your teacher fills out some forms.

The medaling is weird for APMO - each country has at most 1 gold, 2 silver, 3 bronze, 3 honorable mention or something like that. So "bronze" just means like 4th best in the country :P

[url]http://www.cms.math.ca/Competitions/APMO/mreg/index[/url]

[quote]b) For a particular country the number of

i)
gold awards $ \le$ 1,
ii)
gold + silver awards $ \le$ 3, and
iii)
gold + silver + bronze awards $ \le$ 7.[/quote]


linear algebra
matrix
linear algebra unsolved

find all $n$ such that there exists a matrix $M \in M_n(R)$ such that:
*$|M_{i,j}|=1$
*$MM^{t}$ and $I_n$ are not independant.

Clearly $n=2^k$ works. Then ??

What exactly does "not independent" mean in this context?

sorry for my english.
It does exist $(a,b) \neq (0,0)$ such that : $aM+bI_n=0_$

I think you assume we are to find non-trivial example?

The elements of the matrix should all be 1 or -1.

And what is $I_n$ here? Diagonal matrix?

No, in fact I think alekk made a mistake here. The statement should be, in my opinion:
  Find n such that there exists a matrix with 1 and -1, nxn such that there exist a,b not both 0 with $aI_n+bM^{t}M=0$.

The word I'll blame here is "orthogonal". Matrices like this should be called orthogonal, and the ones we call orthogonal should be orthonormal.

Of course, that "not independent" condition could be written as $M^TM=nI.$

I would prefer the other question, with $aI_n+bM=0$. :P  ;)  This one is really hard.

i've just seen it is called hadamard matrices, and the question I asked is open  :D
nevertheless there is a result:
Let $p=4*k-1$ be a prime number. Prove that there is
such a matrix for $n=4k$.

[quote="alekk"]nevertheless there is a result:
Let $p=4k-1$ be a prime number. Prove that there is such a matrix for $n=4k$.[/quote] :( I spent quite many time trying your initial problem.
For this one I have the following example:
1. $M_{i,j}=\left(\frac{i+j}{p}\right)$ (Legendre's symbol), where $0\leq i,j\leq p-1$;
2. $M_{i,p}=-1$, $0\leq i\leq p-1$;
3. $M_{p,j}=1$, $0\leq j\leq p$.

Read about Legendre's symbols [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=legendre&t=34664#]here[/url].


Express  the value of $ \frac{0.\overline{13}\plus{}0.\overline{5}}{0.\overline{16}\minus{}0.\overline{2}}$ as a  mixed number.

Uh, I'm not exactly a math genius, but am I right in saying that (5/9) is (50/90), not (55/90) as the dude above me says?

If true, than wouldn't the answer be:

(13+50)/(16-22)

(63)/(-6)

(-10.5)
???
am I right?

No.... You see, the guy above you said that $\frac{5}{9}=\frac{55}{99}$ (base of $99$ which he used all the time).... Therefore his answer is correct still.


[quote="Drunken_Math"][quote="236factorial"]When I see who's online, I sometimes see "math forum". Does that mean blogs and classes?[/quote]

I believe that means the Math Forum Index.[/quote]

So there are that many people on the index?!  :o Why does it never say "blogs"?

[quote="236factorial"]So there are that many people on the index?!  :o Why does it never say "blogs"?[/quote]Well at this point the blogs are included in the index count.


Solve for $ n$, where $ n$ is any natural number

$ n^2\minus{}2(n!)\plus{}n\equal{}0$

(not hit and trial)

[hide]$ n$ is natural, hence different form zero, so we can divide both sides with $ n$. That gives us

$ n\minus{}2[(n\minus{}1)!]\plus{}1\equal{}0$

Notice the following:

1.$ n \equal{} 1$

$ n \plus{} 1 \equal{} 2(n \minus{} 1)!$ because $ 2 \equal{} 2$

2. $ n \equal{} 2$

$ n \plus{} 1 > 2(n \minus{} 1)!$ because $ 3 > 2$

3. $ n \equal{} 3$

$ n \plus{} 1 \equal{} 2(n \minus{} 1)!$ because $ 4 \equal{} 4$

4. $ n \equal{} 4$
$ n \plus{} 1 < 2(n \minus{} 1)!$ because $ 5 < 12$

The last case can be inductively generalized for all $ n \geq 1$, hence the only solutions are $ 1$ and $ 3$.[/hide]

[hide="solution"]
$ n^2\plus{}n\equal{}2(n!)$
For $ n\equal{}1$ and $ n\equal{}3$ it's O.K.
For $ n\geqslant 4$ we can easy show that:
$ 2(n!)>n^2\plus{}n$
[/hide]

[quote="hsiljak"][hide]$ n$ is natural, hence different form zero, so we can divide both sides with $ n$. That gives us

$ n \minus{} 2[(n \minus{} 1)!] \plus{} 1 \equal{} 0$

Notice the following:

1.$ n \equal{} 1$

$ n \plus{} 1 \equal{} 2(n \minus{} 1)!$ because $ 2 \equal{} 2$

2. $ n \equal{} 2$

$ n \plus{} 1 > 2(n \minus{} 1)!$ because $ 3 > 2$

3. $ n \equal{} 3$

$ n \plus{} 1 \equal{} 2(n \minus{} 1)!$ because $ 4 \equal{} 4$

4. $ n \equal{} 4$
$ n \plus{} 1 < 2(n \minus{} 1)!$ because $ 5 < 12$

The last case can be inductively generalized for all $ n \geq 1$, hence the only solutions are $ 1$ and $ 3$.[/hide][/quote]

I think you have not seen the last line.

Can't I use trials for just three cases? Please? :)

OK, take it this way.

Dividing with $ n$ shows $ n$ is odd.

What next?

$ \frac {n \plus{} 1}{n \minus{} 1} \equal{} 2(n \minus{} 2)!$, if n is greater than one (if it isn't, well, that's one case we ought to test, and it works.)

It's obvious that $ \frac{n \plus{} 1}{n \minus{} 1}\leq 2$ (yes, I know, for $ n\equal{}2$, it is three, but we already know $ n$ is odd, so I'm taking $ n \geq 3$) so $ (n \minus{} 2)! \equal{} 1$, i.e. $ n \plus{} 1 \equal{} 2n \minus{} 2$, $ n \equal{} 3$. No more solutions.

I found another way:

$ n^2\minus{}2(n!)\plus{}n\equal{}0$
$ \Rightarrow n\minus{}2(n\minus{}1)!\plus{}1\equal{}0$
$ \Rightarrow 2(n\minus{}1)!\equal{}n\minus{}1\plus{}2$
$ \Rightarrow (n\minus{}1)(2(n\minus{}2)!\minus{}1)\equal{}2$

Now we can easily show that 1 and 3 are the two solutions.


Fifty cards, numbered 1- 50, are placed in a box. One card is
randomly selected. What is the probability that the number on
the card is prime and is a multiple of 7? Express your answer
as a common fraction.

The only prime multiple of $ 7$ is $ 7$.  The probability of choosing the $ 7$ is $ \boxed{\frac{1}{50}}$.


In trapezoid ABCD, AB is parallel to DC, and AB=BC= 5 cm, BD=12cm, and angle DBC is a right triangle. What is the number of square centimeters in the area of triangle ABD? Express your answer as a decimal to the nearest tenth.

so, its a trapezoid with the 2 parallel sides congruent with each other?

[hide="Answer"]Since $\angle DBC=90^\circ$, we know that $\triangle DBC$ is a right triangle, meaning that $CD=13$. Next, we find the length of the perpendicular from $B$ to $CD$. We form two right triangles that are both similar to $\triangle DBC$ by $AA$, so we can see that the height is $\frac{60}{13}$. Therefore, the total area of the trapezoid is $\frac{5+13}{2}\cdot \frac{60}{13}=\frac{540}{13}$. $[DBC]=\frac{5\cdot 12}{2}=30$, so $[ABD]=\frac{540}{13}-30=\frac{150}{13}\approx \boxed{11.5}$.[/hide]

Forgot similar triangles again  :blush:  Yes, that's the way I would like to see. thanks for the solution.


Does $ f: {R}\to{R}$ and $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)   \forall x,y\in{R}$ imply that f is continuous?

Hello

f(0)=0
let $ x\equal{}x_0$ , and  $ y\equal{} x\minus{}x_0$

So, $ f(x)\equal{}f(x_0)\plus{}f(x\minus{}x_0)$

and $ \lim_{x\rightarrow x_0}f(x)\equal{}f(x_0)$

[quote="*210*"]$ \lim_{x\rightarrow x_0}f(x) \equal{} f(x_0)$[/quote]
This is true if and only if $ \lim_{x \to 0} f(x) \equal{} f(0)$, and nothing given in the problem suggests this.  The non-continuous solutions to this functional equation are extremely pathological - continuous nowhere, dense in the plane...


In the addition problem below, each letter stands for a digit from 1 to 9 and different letters stand for different digits. What digits do the letters H, E, and A each represent?

                          H E
                          H E
                          H E
                      +  H E
                 ____________
                          A H

H   =      E  =      A  =

I'm confused, what do you mean by the last statement H=E=A?

Just based on the addition problem, since the sum is only 2 digits, we know that $ H<3$ or else it would be 3 digits. Guess and check, try that H is 2, E is 3, and A is 9, which works. If H was 1, than there couldn't be a 4E=1 or E=11, and same if H was 0.


linear algebra
matrix
superior algebra
superior algebra unsolved

Prove that all the ideals in $M_n(K)$, where $K$ is a ring, are the sets of matrices with elements from the ideals of $K$.

A bit of notation: in every algebraic exposition I know, $K$ is used for fields and never for generic rings.

Let $R$ be an arbitrary ring (with identity). Given a matrix $A$ with elements $a_{ij}$. If $E_{ij}$ is the matrix with $1$ in the $ij$ position and 0 elsewhere, $E_{ac}\cdot A\cdot E_{db}$ has $a_{cd}$ in the $ab$ position and zeros elsewhere. By taking combinations of these, if $I$ is an ideal of $M_n(R)$, $I$ contains every matrix with entries in the ideal of $R$ generated by the entries of matrices in $I$, and is therefore of the form described.
Conversely, it is easy to see that all sets of the form described are ideals in $M_n(R)$.


If you're given 3 points on a plane, how can you determine the equation of a circle that passes through them?

Every circle in the plane can be written as a form $(x-x_0)^2+(y-y_0)^2=r^2$, where $(x_0,y_0)$ is the center of the circle and $r$ is the radius of the circle. Now substitute the coordinates to this equation and you get three equations with three unknows. Then just solve $x_0,y_0$ and $r$.

Method:

[hide]
The general equation for a circle is $(x-h)^2+(y-k)^2=r^2$, where (h,k) is the center and r is the radius.

Note that if we substitute an x and a y coordinate, we have 3 unknowns (h, k, and r); thus, we know that three equations will be needed to solve a given system, and thus we know that three points define a circle.

(minor but relevent tangent: by the same logic, we observe the general equation for a conic...

$ax^2+bxy+cy^2+dx+ey+f=0$

... and note that if we substitute for x and y there are 6 unknowns, a through f; thus, a general conic is defined by 6 points, because we need 6 equations to solve for 6 unknowns)

So for the circle, we simply solve the linear system.

For example, consider the some circle with points (1,3) (4,6) and (7,3) (a little obvious, I know, but go with it). Substituting the three points into the general circle equation we get a linear system (rearranged with the constants by themselves):

$h^2+k^2-r^2-2h-6k=-10$

$h^2+k^2-r^2-8h-12k=-52$

$h^2+k^2-r^2-14h-6k=-58$

In a circle, the h squared, k squared, and r squared coefficients will always be 1, and will be present in each equation as above; this makes generating linear systems easy. Just subtract the equations from each other to get three linear systems.

There are any number of ways to proceed. In this case, my first two attempts gave me
$12h=48$ so $h=4$
$h-k=1$ so $k=3$
Then $r=3$

and the general equation is thus

$(x-4)^2+(y-3)^2=9$

This works for any circle.

[/hide]


With Maple.  We assume that the $x_{i}$ are real unknowns.
$x_{1}$ is a root of a polynomial of degree 276 of the form: $P_{1}(y)P_{2}(y)P_{3}(y)P_{4}(y)$ where $y=x_{1}^{4}$, the $P_{i}$ are irreducible of degree 8,8,8,45 and have 4,4,4,5 positive roots. Thus $x_{i}$ may take 34 real values.
There exist many solutions: for example $x_{1}\approx-1.64,x_{2}\approx-0.714,x_{3}\approx0.117,x_{4}\approx-0.790,x_{5}\approx1.02$.
We obtain particular solutions with the supplementary condition: $x_{2}=x_{3}$. Then $x_{1}$ is a root of the polynomial:
$121x_{1}^{32}+1-2062x_{1}^{28}+7451x_{1}^{24}-10365x_{1}^{20}+5113x_{1}^{16}-818x_{1}^{12}+1135x_{1}^{8}-215x_{1}^{4}$ and moreover $x_{1}=x_{4}$.

Yes, i really try to solve by maple, but that is an unexpected solution


vector
calculus
integration
linear algebra

what are projections and how are they used? any example problems? especially how do they relate to the dot product of vectors?

Here's a good piece: http://en.wikipedia.org/wiki/Vector_projection

If you pick up a basic linear algebra textbook, it should have plently of useful exercises.  They come in handy when you're working with bases and linear transformations.

[quote="Poincare"]i thought linear alg was done after Calc 3...[/quote]

Dad was doing it while he was reviewing calc 1...

im in BC calculus so Ill probably take it next year

Doesn't really matter the order.  Once you finish calc I and II, you can take multivariable calculus, linear algebra, or both.  I took multivariable calculus first, because I thought it would be more interesting, and for me, it was.

The fundamentals of linear algebra require no calculus whatsoever. If you know the basics of 1-variable calculus, and have a strong foundation in linear algebra, you can cruise through courses that involve both vectors/matrices and calculus, such as multivariable calculus and any class in differential equations.

[b]Note:[/b] [i]Please stop saying "calc 1" "calc 2" "calc 3" etc. These phrases mean very little out of context. For example, in one college I've seen, calc 2 was basic integration, and in another college, calc 2 was multivariable calculus.

On the other hand, "Calc BC" or "basic integration", etc. seem to mean the same thing for everyone.[/i]

A first class in linear algebra is pretty self-contained, as is a first course in multivariable calculus, meaning they don't depend on one another.  Differential Equations takes a little of each, but more from linear algebra.


calculus
derivative
induction
real analysis
real analysis unsolved

Assume the contrary and choose an interval $[a,b]$ such that $P(x)>Q(x)$ and both strictlyincreasing. This fact is posible since thaier derivatives have $\infty$ limit at infinity, and $P$ and $Q$ have at  most m points in common.

Now we have $PoP(x)>PoQ(x)> QoQ$ and by induction $PoPoPo..P>QoQo..Q$ for all x in the specified interval .


calculus
derivative
trigonometry
function
real analysis
algebra
rational function

For example $ f(x) \equal{} \frac1{1 \plus{} x^2}$ is uniformly continuous on $ \mathbb R$ because its derivative is bounded, but, how do I realize that its derivative is bounded? How should I see this in a fast way?

Graphically.  $ f$ is smooth, attains a maximum of $ 1$ at $ x \equal{} 0$, and decays slowly at infinity.

$ f'(x)\equal{}\minus{}\frac{2x}{1\plus{}x^{2}}$ and $ f''(x)\equal{}\minus{}\frac{2\plus{}2x^{2}\minus{}4x^{2}}{(1\plus{}x^{2})^{2}}\equal{}2\frac{x^{2}\minus{}1}{(1\plus{}x^{2})^{2}}$.  Set this equal to 0 to get $ x^{2}\equal{}1\Rightarrow x\equal{}\pm 1$, and $ f'(1)\equal{}\minus{}1$, $ f'(\minus{}1)\equal{}1$, so $ \left|f'(x)\right|\leq 1$.

As t0rajir0u said, graph it and see.  And for good measure, graph the derivative.

Graphing takes time, I prefer the analytic method you gave. Thanks for your answer.

[quote="Null"]Graphing takes time, I prefer the analytic method you gave.[/quote]
Plugging an equation into a calculator and pressing "Graph" takes more time than computing a derivative?  I usually don't interpret "see" as "prove."

If you want a proof, a slick way to see that the derivative is bounded is to make the substitution $ x \equal{} \tan \theta$.

The derivative of any proper rational function is a proper rational function. A proper rational function goes to zero at $ \infty$, so it is bounded on $ \mathbb{R}$ unless it has poles (zeros of the denominator) there. The denominator in this case is a power of $ \frac1{1\plus{}x^2}$, which has no real zeros.


Hello everyone!
I have 300 problems of Chinese Mathematical Contest for little children, but they all are in Chinese. I wish to find them in English or to find the way to translate them. I would like to find other problems also. If anyone can help me with my problem, please let me know and write me a message.  :)

Please, send me a message with your e-mail  :)

I'm an American Born Chinese. My parents teach Chinese in college. I can help translate if you sent me an email. :)


[quote="Makaveli"]Does somebody know the exact dates for the Balkan Mathematics olympiad 2006?[/quote]Not quite, but I think it is in the beginning of May.

tomorrow we have the final TST for BMO!!!!!!

Okay, the BMO is over and we should be expecting results in a couple of days. [i]Apparently[/i] iura and sasha have solved all the problems, while 5 of the Romanian team have solved at least 3 problems, none though expecting 10 points on the 4th problem. From Bulgaria 2 kids with the same expectations, while a kid from Albania seems to claim to have solved all ... :) Again all these are rumors based on what some of the students claimed they did after exiting the competition hall.

Here are the problems: http://www.mathlinks.ro/Forum/resources.php?c=1&cid=19&year=2006 (click on the problem number to get to the discussion topic for each).

[quote="Valentin Vornicu"], while a kid from Albania seems to claim to have solved all ... :) [/quote]

I think he is a mathlinker ....... :) Maybe Albanian Eangle

Could someone post the geometry problem ?? :!:

[quote="Valentin Vornicu"], while a kid from Albania seems to claim to have solved all ... :) Again all these are rumors based on what some of the students claimed they did after exiting the competition hall.
[/quote]

And why not? Why " :) "? Why this expression "seems to claim?"  :o  :|

You surprise me.

Well, good luck to everyone waiting for the official scores :) (are they available already? I'm curious :)). And remember-- the official score and the color of the medals are the things that matter least in this event. To anyone who's reading this, my wishes for a wonderful and memorable time in Cyprus :).

--Vesselin

[quote="vess"]And why not? Why " :) "? Why this expression "seems to claim?"  :o  :| [/quote]Well because they were claims (what the students say they did, not always matches the final score ;) ). The smile is because I think that it is a MathLinker, but I cannot be sure until he posts ;)

The results are now official: Iura and Sasha with 35 points on 1-2 position. Mihai Dumitrescu from Romania and a student from Albania (most probably the  students who said he solved all of them) with 34 on 3-4 positions. On 5th position Bazavan Eduard from Romania with 32 points. Dzeta follows with 30 points, while ikap scored 27. Turea Lucian scored 12 points.

I got these results by phone, so I really do not know any others  :( I will have an official list by next week though ...

PS About the good time, I hear they are having a lot of it down there in Cyprus :D

:)

Gongrats to Iura and Sasha  :)  (but how come they lost 5 points?)  BTW Valentin, are these all the results up to 27 points? (if so, how come the Bulgarian team did so poorly  :( ...). Also, are they absolutely final? Any contestations to follow? Thanks :)

Congratulations to the winners, especially to Iura and Sasha :)

[quote="dule_00"]Is Romania first?[/quote]

Well, Valentin said he doesn't know any other results, so... But, looking at the first few results, it's quite likely that Romania is first -- or Moldova, this year :). Still, let's wait for the official results :) Maybe Bulgaria didn't do as poorly, after all :). We shall see... I don't have any information.

I'm not sure (I might be talking nonsense) but I think that there was a couple of bulgarians in the range 30-32. Again I am not sure, as I got the results by phone. The medals are not officially set yet, but the first 4 will get gold medals surely. Perhaps the students with 32 points (if there is more than one) will also get one. The scores I wrote I belive are final.

I have no idea if Romania is first, but it has a good chance. Moldovia also has a good chance of making top 3, but I am not sure about the other 4 scores from them, so it depends ...

I heard that Marko Jevremovic from Serbia has gold and Marijana Smailagic has silver, Also three bronze are won by Serbia. Congratulations to all expecial ti Sasha, Iura and MARKO...

These should be scores:

1. Romania 165
2. Moldova 122
3. Albania 116
4. Bulgaria 107
5. Greece 106
6. Serbia and Montenegro 103
7. Cyprus 43
8. FYRO Macedonia 39
and Turkey didn't participate.

Serbian team scores:

Marko Jevremovic 32 Gold
Marijana Smailagic 23 Silver
Aleksandar Trokicic 16 Bronze
Mladen Radojevic 15 Bronze
Igor Ninkovic 14 Bronze
Marija Jelic 3

Congratulations to all, expecially gold medalist and Romanian team....

Wow... (no comments ) :o  :|  :?  Well, I join in congratulating the Romanian team. :) And special congratulations to the Moldovans and the Albanians for their all-time high rank in the BMO (am I right? I think that this is a very good progress, and by all means a success that is not to be belittled by the abnormally low general level -- and of course, the IMO is to come, and I wish both countries an even better performance there!) And, well, of course it is sad to see the results, but it doesn't matter so much -- I wish our young and inexperienced Bulgarian team better results in Slovenia ;). And there is nothing to worry about -- for half of our team, this is the first international contest.

Again, special congrats to Sasha and Iura -- winners of the Olympiad :).

--Vesselin

The British results (unofficial):

The UK team in the Balkan Mathematical Olympiad in Cyprus have achieved the following scores (out of 40).

Jos Gibbons - 14 - Bronze
Tim Hennock - 1
Jonathan Lee - 12 - Bronze
Daniel Lightwing - 16 - Bronze
Julia Robson - 2
Dominic Rowland	- 6

Team score - 51.

The Greek team :

Kofogiannis Fragiskos  28 pt  (silver)

Nikos Rapanos          26  pt   (silver)

Nikos  Bountas           25 pt   (silver)

Konstantinos  Phsyhas 19 pt  (bronze)

Despotakis Stylianos

Papagianakis Nikos

Congrats to all my friends  :D

3 bronzes...same as last year.

Shame none of the team got the inequality which appeared on the UK correspondence course last year.

congrats to all the participants!..

what about dragosh michnea (i see it's not in the 5 members that Valentin mentioned)?
also.. do you know the results of the other members of the Moldavian team? (except Iura and Sasha  :D )

Sorry, the phone is quite expensive from there :) I tried to get the questions also, but I couldn't stay too long. I forgot to ask about Michnea but presuming that the country scores are correct, subtracting the sum of the other 5 contestants we find that Dragos scored 30 points :)

In the Geometry section I posted the second problem of the contest . :D
See your solutions....... :lol:

[quote="ics"].. do you know the results of the other members of the Moldavian team? (except Iura and Sasha  :D )[/quote]

???

Hi, this would be my first message from this nick :lol:

And thanx to everybody who congrated for my gold!

Mr Vornicu I posted the geomery problem (problem 2) on the Geometry Proposed and Own problems . Please add it also to the Resourses page

I see that you have been talking here huh?

 :D

congrats to all students who took medals specially the ones with gold (including me :D )

also to BUL6 for the special prize on problem 4. and to romania for the first place

I did solve all problems but I took 10, 6, 10, 8

I knew that I had done a writing mistake in the second problem (mistakenly wrote $A$ instead of $A_1$ a couple of times eventhough my reasoning was true but anyway)
but I still don't know why I got 8 on the fourth :?:  :!:

Could you post here the solution of the student BUL6  of the problem 4 that gave him a special prize ??

I haven't seen the solution myself but I hope he will post it himself as he is a mathlinker here too. ( imortal )
he took silver medal with 30 points.
It was strange, the bulgarians couldn't get a gold medal even this year!

Hi, I am imortal. I promise to post my solution soon. I also got the original papers so I will try to scan them and attach to some post.
It is not a good tendance that there is no bulgarian with gold, but I hope that next year there will be (unfortunately, I won't be able to do this since this is my last grade).
And I made stupid mistake on problem No.3 which costed me 3 points and the gold medal. Still there is an IMO and I hope to have magna cum laude.

Congratulations to those who have gold medals, and to the Romanian team who was 1st.

And finally, in detail the Bulgarian results are:
1.Boris Strandjev-26(Silver) (10/6/0/10)
2.Anton Bikov-3 (0/0/1/2)
3.Emil Baronov-18(Bronze) (0/4/10/4)
4.Tzvetelina Tzeneva-19(Bronze) (0/10/2/7)
5.Deian Simeonov-11(Bronze) ( 0/0 /10/1 )
6.Dimitar Simeonov(me)-30(Silver) (3/10/7/10)

[quote="Sailor"][quote="AndersoN"]Errr... do we get a link or a full list of results any time soon? I mean it's May the 3rd, evening, when will they be available?!

Thanks to anyone, in advance.[/quote]

Our results are:

1. Iurie Boreico (aka Iura) -35 (Gold)
2. Alexandru Zamorzaev (aka Sasha) -35 (Gold)
3. Liubomir Chiriac (aka Sailor)- 24 (Silver)
4. Patrinica Vitalie(aka Vetali)- 13 (Bronze)
5. Ivan Borsenco (aka Prowler)-11(Bronze)
6. Vladimir Vanovschi - 4[/quote]

Good job Liubomir!!

I think it should have been :
Iurie, Alexandru, Liubomir, and Ivan = Gold!!!!

[quote="RaduSorici"]Good job Liubomir!!

I think it should have been :
Iurie, Alexandru, Liubomir, and Ivan = Gold!!!![/quote]

Good post, Radu!  I hope Vanea scores much better at IMO.

ALBANIAN EAGLE,
...it suits!
CONGRATULATIONS

What's up?
CAn s.o post the Bulgarian results?

The official results of the BMO2006 are posted at http://www.cms.org.cy. Don't be surprised with the plethora of medals! The BMO rules are different than the IMO rules. If IMO rules were to apply, the results would have been: COUNTRY:(Gold,Silver,Bronze):
ALB: (1,0,3), BUL:(0,2,2), CYP:(0,0,1), HEL:(0,2,2), MCD:(0,0,1), MLD:(2,0,1), ROM:(1,4,0), SCG:(0,1,1)
GUESTS: KAZ:(1,2,2), UNK:(0,0,0)
Compare this with the official BMO results:
ALB: (1,2,2), BUL:(0,2,3), CYP:(0,1,0), HEL:(0,3,1), MCD:(0,0,1), MLD:(2,1,2), ROM:(2,3,1), SCG:(1,1,3)
GUESTS: KAZ:(2,3,1), UNK:(0,0,3)

SUCCESS RATE (OFFICIAL PARTICIPANTS ONLY): PROBLEM 1: 68,33%, PROBLEM 2: 37,08%, PROBLEM 3: 43,96%, PROBLEM 4: 23,75%

When will you post your solution, imortal?

I just posted my solution at:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=517592#517592

There still may be some bugs because Im not good in Latex but if I see any I will try to fix them :)


Show that, for $ a>1$,

$ \frac{1}{\sqrt{a}} < \sqrt{a\plus{}1}\minus{}\sqrt{a\minus{}1}$.

hello, rewriting your inequality in the form
$ \frac{1}{\sqrt{a}}<\frac{2}{\sqrt{a\plus{}1}\plus{}\sqrt{a\minus{}1}}$ we get
$ \sqrt{a\plus{}1}\plus{}\sqrt{a\minus{}1}<2\sqrt{a}$
$ 2\sqrt{a^2\minus{}1}<2a$
$ a^2\minus{}1<a^2$ which is true.
Sonnhard.

The inequality is also true for $ a\equal{}1$.


Find all continuous functions $f: \mathbb R \to \mathbb R$ satisfying \[ f(x \plus{} y) \equal{} \frac{{f(x) \plus{} f(y) \plus{} 2f(x)f(y)}}{{1 \minus{} f(x)f(y)}},\]
for all $x, y \in \mathbb R$.

[quote="Xaenir"]Find all continuous functions$ f: R \to R$
 satisfying:$ f(x \plus{} y) \equal{} \frac {{f(x) \plus{} f(y) \plus{} 2f(x)f(y)}} {{1 \minus{} f(x)f(y)}}$,$ \forall x,y$[/quote]

The keypoint here is that, in order this equation be defined $ \forall x,y$, we need to have $ f(x)f(y)\neq 1$ $ \forall x,y$.

As a consequence, we know that $ |f(x)|\neq 1$ $ \forall x$

Let $ x\equal{}y\equal{}0$. We have $ f(0)(1\minus{}f^2(0))\equal{}2f(0)\plus{}2f^2(0)$ and so $ f(0)(f(0)\plus{}1)^2\equal{}0$

Since $ f(0)\equal{}\minus{}1$ is impossible ($ |f(x)|\neq 1$ $ \forall x$), we get $ f(0)\equal{}0$.

So, since continuous, $ \minus{}1<f(x)<1$ $ \forall x$

Then $ y\equal{}x$ implies $ f(2x)\equal{}\frac{2f(x)}{1\minus{}f(x)}$

Now consider $ a < f(x) < b$ $ \forall x$ with $ \minus{}1\leq a < 0 < b\leq 1$.

then $ a < f(2x) < b$ implies $ a < \frac{2f(x)}{1\minus{}f(x)} < b$ and, since $ 1 \minus{} f(x)>0$ :

$ a \minus{} af(x) < 2f(x) < b \minus{} bf(x)$ and so $ \frac{a}{a\plus{}2} < f(x) < \frac{b}{b\plus{}2}$

So $ u_n < f(x) < v_n$ $ \forall x\in \mathbb R$ $ \forall n\in \mathbb N$ where the sequences $ u_n$ and $ v_n$ are :
$ u_1\equal{}\minus{}1$ and $ u_{n\plus{}1}\equal{}\frac{u_n}{u_n\plus{}2}$
$ v_1\equal{}\plus{}1$ and $ v_{n\plus{}1}\equal{}\frac{v_n}{v_n\plus{}2}$

And, since these two sequences converge towards 0, $ f(x)\equal{}0$ $ \forall x$ and we immediatly check that this function fits the initial equation.


geometry
search
trigonometry
number theory
geometry proposed

[size=150]We are given a circle K and a diameter CD.Take the points A and B on the circle at the same side of CD.Let F be the intersection point of AB and CD and let J be the intersection point of AD and BC.G isthe projection of J on CD, and M is the intersection point of CD with the tangent of the circle at A.Find a necesary and sufficient condition regarding the points A and B, in order M to be the midpoint of FG.[/size]

                                                                                [i]Proposed by Nick Rapanos[/i]

This ia a search them ! $MF=MG\Longleftrightarrow OB\perp CD$, where the point $O$ is the centre of the circle $K$.

I am a short syntetical solution with the harmonical division (pol-polar w.r.t. a circle) and a short and nice analitical solution $(\ K: \ x^2+y^2=1$; $A(\cos u,\sin u)$, $B(\cos v,\sin v)$, $0<v<u<180^{\circ}$ $\Longrightarrow$ $x_F=\frac{\cos \frac{u-v}{2}}{\cos \frac{u+v}{2}}$, $x_M=\frac{1}{\cos u}$, $x_G=\frac{\cos \frac{u+v}{2}}{\cos \frac{u-v}{2}}$; $MF=MG$ $\Longleftrightarrow$ $2x_M=x_F+x_G$ $\Longleftrightarrow$ $\cos v=0$ $\Longleftrightarrow$ $v=90^{\circ}$ $\Longleftrightarrow$ $BO\perp CD\ .$

I have a quite same analitic geometry solution but I searched for a classic pure geometric solution without any succes. Could someone post one?

Hi!Congratulations to Virgin and to my friend Silouan!!Very good solution but there is a <<pure geometric solution>>....and i am waiting for it!Otherwise i will post my solution!
Virgin how did you find the problem??

[hide="Here is the first (syntetical) method."][color=darkred]The division $(C,D,F,G)$ is harmonically $^{(*)}$, i.e. $\frac{GC}{GD}=\frac{FC}{FD}$. Thus, $AC\perp AD$ $\Longleftrightarrow$ $\widehat {CAF}\equiv \widehat {CAG}$, i.e. the ray $[AC$ is the bisector of the angle $\widehat {FAC}$.  From the power of the point $M$ w.r.t. the circle $K$ results $MA^2=MC\cdot MD$. Therefore, $MF=MG$ $\Longleftrightarrow$ $MF^2=MC\cdot MD$ $\Longleftrightarrow$ $MF=MG=MA$ $\Longleftrightarrow$ $m(\widehat {FAG})=90^{\circ}$ $\Longleftrightarrow$ $m(\widehat {LAB})=m(\widehat {FAC})=45^{\circ}$ $\Longleftrightarrow$ $m(\widehat {CDB})=m(\widehat {LAB})=45^{\circ}$ $\Longleftrightarrow$ $BO\perp CD$ $\Longleftrightarrow$ $BC=BD$.[/color][/hide]
[hide="Here is the answer for the MStoenescu-question (from a private message)."]
[color=darkgreen][b]Remark[/b] $(*)$. Define the intersections $H\in AD\cap BC$ and $L\in AC\cap BD$. Then the point $H$ is the orthocentre of the triangle $CLD$. From the Menelaus' theorem applied to the transversal $\overline {BAF}$ in the triangle $CLD$ and from the Ceva's theorem applied to the point $H$ in the same triangle results that the division $(C,D,F,G)$ is harmonically.
[u]Generally[/u], if the points $M,N,P$ belong to the sides $BC,CA,AB$ respectively and denote the intersection $S\in BC\cap NP$, then the division $(B,C,S,M)$ is harmonically (the proof is similarly with the above solution).[/color][/hide]
[hide="Here is the answer for the Silouan-question (from a private message)."][color=brown]
[b]Lemma.[/b] Let $A,C,B,D$ be four points wich belong to the line $d$ in this order and let $P$ be a point which don't belong to the same line. Prove that if two from the following statements are truly then and the third statement is truly:
$1^{\circ}.\blacktriangleright$ The division $(A,B,C,D)$ is harmonically, i.e. $\frac{CA}{CB}=\frac{DA}{DB}\ ;$
$2^{\circ}.\blacktriangleright\ PC\perp PD\ ;$
$3^{\circ}.\blacktriangleright\ \widehat {CPA}\equiv \widehat {CPB}\ .$[/color][/hide]
[hide="Here is the second (analytical) method."][color=darkblue]
Suppose w.l.o.g. $K=C(O,1)$. Let $x^2+y^2=1$ be the equation of the circle $K$. Denote $0<v<u<180^{\circ}$ so that $A(\cos u,\sin u)$, $B(\cos v,\sin v)$. Thus,
$\blacktriangleright$  $\left| \begin{array}{ccc} x_F & 0 & 1\\\ \cos u & \sin u & 1\\ \cos v & \sin v & 1\ \end{array}\right|=0$ $\Longleftrightarrow$ $x_F=\frac{\sin u\cos v-\sin v\cos u}{\sin u-\sin v}=$ $\frac{\sin (u-v)}{\sin u-\sin v}=$ $\frac{2\sin \frac{u-v}{2}\cos \frac{u-v}{2}}{2\sin \frac{u-v}{2}\cos \frac{u+v}{2}}$ $\Longleftrightarrow$ $\boxed {x_F=\frac{\cos \frac{u-v}{2}}{\cos \frac{u+v}{2}}}\ .$
$\blacktriangleright$ Denote the intersection $H\in AD\cap BC$ and observe that $x_G=x_H$. Thus, the equation of the line $BC$ is $y=\frac{\sin v}{1+\cos v}\cdot (x+1)$ and the equation of the line $AD$ is $y=-\frac{\sin u}{1-\cos u}\cdot (x-1)$. Therefore, $\frac{\sin v}{1+\cos v}\cdot (x+1)+\frac{\sin u}{1-\cos u}\cdot (x-1)=0$ $\Longleftrightarrow$ $\sin u(1+\cos v)\cdot (x-1)+\sin v(1-\cos u)\cdot (x+1)=0$ $\Longleftrightarrow$ $2\sin \frac v2\cos \frac v2\cdot 2\sin^2\frac u2\cdot (x+1)+2\sin \frac u2\cos \frac u2\cdot 2\cos^2\frac v2\cdot (x-1)=0$ $\Longleftrightarrow$ $\left(\cos \frac u2\cos \frac v2+\sin \frac v2\sin \frac u2\right)x=\cos \frac u2\cos \frac v2-\sin \frac u2\sin \frac v2$ $\Longleftrightarrow$ $\boxed {x_G=\frac{\cos \frac{u+v}{2}}{\cos \frac{u-v}{2}}}\ .$
$\blacktriangleright$ The equation of the tangent $AA$ to the circle $K$ is $x\cos u+y\sin u=1$ and $y=0$ $\Longleftrightarrow$ $\boxed {x_M=\frac{1}{\cos u}}\ .$
$||\blacktriangleright$ Therefore, $MF=MG$ $\Longleftrightarrow$ $2x_M=x_F+x_G$ $\Longleftrightarrow$ $\frac{2}{\cos u}=\frac{\cos \frac{u-v}{2}}{\cos \frac{u+v}{2}}+\frac{\cos \frac{u+v}{2}}{\cos \frac{u-v}{2}}$ $\Longleftrightarrow$ $\frac{2}{\cos u}=\frac{\cos^2\frac{u-v}{2}+\cos^2\frac{u+v}{2}}{\cos \frac{u+v}{2}\cos \frac{u-v}{2}}$ $\Longleftrightarrow$ $\frac{2}{\cos u}=\frac{1+\cos (u-v)+1+\cos (u+v)}{\cos u+\cos v}$ $\Longleftrightarrow$ $\frac{1}{\cos u}=\frac{1+\cos u\cos v}{\cos u+\cos v}$ $\Longleftrightarrow$ $\cos v=\cos^2 u\cos v$ $\Longleftrightarrow$ $\cos v=0$ $\Longleftrightarrow$ $BO\perp CD$ $\Longleftrightarrow$ $BC=BD\ .$[/color][/hide]

Fantastic mr Virgil!I had the same solution!Thank you for trying my problem although i understood that it was very easy for you!!!
Ok i will propose my number theory problem....See it if you want!


modular arithmetic
superior algebra
superior algebra theorems

Let $ \sigma\equal{}(1\,2\,3\cdots n)$ be a cycle of length $ n\geq 2$.  If $ n\equal{}p$ is prime, prove that $ \sigma^{m}$ is a cycle of length $ p$ for each $ m\equal{}1,2,\ldots,p\minus{}1$.

Here is what I have: We accept that if $ 1\leq m\leq n$, then $ \sigma^{m}(k)\equiv k\plus{}m\pmod{n}$.  If $ m\mid n$, then $ n\equiv 0\pmod{n}$ so $ n\plus{}1\equiv 1\pmod{n}$.  Hence $ 1\mapsto 1\plus{}m\mapsto\cdots\mapsto 1$.  This will be a complete $ n$-cycle if $ 1\plus{}km\equiv i\pmod{m}$ for all $ 1\leq i\leq n$, for some $ k$.  In other words, if $ m$ does not divide $ n$ for all $ 1<m<n$.  Since $ p$ is prime, $ 2,3,\ldots,p\minus{}1$ don't divide $ p$ so $ \sigma^{m}$ will be a $ p$-cycle.

Is this even correct, and/or is there a way to make it more concise or clear?

[quote="JRav"]Let $ \sigma \equal{} (1\,2\,3\cdots n)$ be a cycle of length $ n\geq 2$.  If $ n \equal{} p$ is prime, prove that $ \sigma^{m}$ is a cycle of length $ p$ for each $ m \equal{} 1,2,\ldots,p \minus{} 1$.

Here is what I have: We accept that if $ 1\leq m\leq n$, then $ \sigma^{m}(k)\equiv k \plus{} m\pmod{n}$.  If $ m\mid n$, then $ n\equiv 0\pmod{n}$ so $ n \plus{} 1\equiv 1\pmod{n}$.  Hence $ 1\mapsto 1 \plus{} m\mapsto\cdots\mapsto 1$.  This will be a complete $ n$-cycle if $ 1 \plus{} km\equiv i\pmod{m}$ for all $ 1\leq i\leq n$, for some $ k$.  In other words, if $ m$ does not divide $ n$ for all $ 1 < m < n$.  Since $ p$ is prime, $ 2,3,\ldots,p \minus{} 1$ don't divide $ p$ so $ \sigma^{m}$ will be a $ p$-cycle.

Is this even correct, and/or is there a way to make it more concise or clear?[/quote]

It's correct...

If $ \sigma$ is a $ p$-cycle then $ \sigma^m$ is a $ p$-cycle iff $ m\ne sp$ for some $ s\in \mathbb{Z}$. I don't understand why you consider only the $ p$-cycle $ (1\,2\,3\cdots n)$.

There exist an isomorphism between $ <\sigma>$ and $ \mathbb{Z}/p\mathbb{Z}$ such that $ \sigma \mapsto 1$. If $ f$ is such isomorfism then $ f^{\minus{}1}$ is an isomorfism too. Then $ f(\sigma^m) \equal{} f(\sigma)^m \equal{} [m1] \equal{} [m]$ (you can use the mod notation if you want). Now... $ o([m]) \equal{} p$ iff $ [m]\ne[0]$ and so if $ [m]\ne[0]$ and $ 0<t<p$ then $ \sigma^{tm} \equal{} f^{\minus{}1}(t[m]) \ne f^{\minus{}1}(0) \equal{} e$ ($ e$ is the identity in $ S_n$) and so $ o(\sigma^m)\equal{}p$
$ p$ is prime and so $ \sigma^m$ contains a $ p$-cycle. But $ \sigma^m$ fix every elements that $ \sigma$ fix and so $ \sigma^m$ move only $ p$ elements and $ \sigma^m$ is a $ p$-cycle.

The problem asked to consider only $ n$-cycles when $ n$ is a prime number.  Technically, I'm not supposed to use isomorphisms with this problem (I know what they are and how to use them but they are introduced later on).  But what I have is correct?  I'm not happy with it because I think it may sound unclear.


If $a,b,c>0$ and $abc=1$ then $(a+b)(b+c)(c+a) \leq  ( \frac{a+b+c}{2})^{6}$.

cheers! :D :D :cool:

[quote="Lagrangia"]If $a,b,c>0$ and $abc=1$ then{a+b+ $(a+b)(b+c)(c+a) \leq  ( \frac c}{2})^{6}$.

cheers! :D :D :cool:[/quote]
 Dear lagrangia .I think this Inequalities is  (a+b)(b+c)(c+b)<[(a+b+c)/2]^6
  thank very much :?


Let $ (x_{n})$ a sequence such that:
$ x_{n\plus{}1}\equal{}\sqrt{x_{n}\plus{}6\sqrt{x_{n\minus{}1}}}$
$ x_{0}\equal{}1$
$ x_{1}\equal{}2$

define $ x_{n}$

[quote="reason"]Let $ (x_{n})$ a sequence such that:
$ x_{n \plus{} 1} \equal{} \sqrt {x_{n} \plus{} 6\sqrt {x_{n \minus{} 1}}}$
$ x_{0} \equal{} 1$
$ x_{1} \equal{} 2$

define $ x_{n}$[/quote]
[hide="Hint"]
Prove by induction that $ x_{n \plus{} 1} \equal{} 2\sqrt {x_n}$ and the conclusion follows immediatly  :wink:
[/hide]

by induction:
for any $ 0 \le n$
$ x_{n\plus{}1} \equal{} 2^{1\plus{}\frac{1}{2}\plus{}\frac{1}{4}\plus{}..\plus{}\frac{1}{2^n}}$


function
limit
real analysis
real analysis unsolved

Let $f: \mathbb{R}\rightarrow\mathbb{R}$ be a function such that $\forall x_0 \in \mathbb{Q} ,\exists \lim_{x \to x_0,x<x_0}f(x)$ and $\lim_{ x \to x_0,x<x_0}f(x)=f(x_0)$. Prove that f is continuous.
This problem was proposed by harazi I think.

What about \[ f(x)=I_{[\sqrt{2},+\infty)}  \] (i.m. $f(x)=0$ if $x<\sqrt{2}$ and $f(x)=1$ if $x\ge \sqrt{2}$)
or \[ f(x) = h(x) = I_{\{x>0\}}  \] or simply $f$ continuous from left, but not continuous at all.

I think that it was to prove that f is continuous on a dense subset of $\mathbb{R}$.

Here is my proof:
I used the following Lemma:
For any closed interval [a,b] and any $\epsilon>0$ there exists an interval $[c,d]\subset [a,b]$ such that $|f(x)-f(y)|<\epsilon, \forall x,y \in [c,d]$.
Proof: Choose a rational point $t$ from $[a,b]$. Since $\lim_{x \to t,x<t}f(x)=f(t)$ there exist $\delta$ such that for any $x<t$ with $|x-t|\le \delta$ we have that $|f(x)-f(t)|<\frac \epsilon 2$. Now choose $x,y$ such that $0<t-x\le \delta \text{ and } 0<t-y \le \delta$ and we have that  $|f(x)-f(t)|<\frac \epsilon 2$ and  $|f(y)-f(t)|<\frac \epsilon 2$. Now we have that $|f(x)-f(y)|\le|f(x)-f(t)|+|f(y)-f(t)|<\epsilon$. So, for any $x,y \in [t-\delta,t]$ we have that $|f(x)-f(y)|<\epsilon$. q.e.d.

Now choose an arbirary interval $[x_0,y_0] \subset \mathbb{R}$. Using the lemma for all $n\in \mathbb{N}$ we define the sequence of closed intervals $[x_n,y_n]\subset [x_{n-1},y_{n-1}]$ such that $|f(x)-f(y)|<\frac 1 n,\forall x,y \in [x_n,y_n]$. We know that a family of closed intervals , every interval included in the precedent have nonempty intersection. Let $a \in \cap_{n=0}^\infty[x_n,y_n]$ be a real number. We have that $f$ is continuous in $a$.
I just proved that from every closed interval someone can find a point in which $f$ is continuous. So, $f$ is continuous on a dense subset of $\mathbb{R}$.
Am I right? Any other solutinons? Is there a function which is not continuous in any rational point and respects the condition of problem?
In fact $\mathbb{Q}$ can be changed with any dense subset of $\mathbb{R}$.


Is the answer rational?What do you think of that question?

The sum is irrational, since its binary expansion is not periodic. Can't tell if it's algebraic or not - the rate of convergence falls short of [url=http://en.wikipedia.org/wiki/Liouville_number]Liouville-type series[/url].


I've had Mathlinks skin since forever, but today it mysteriously turned into the AoPS skin. And I couldn't find a place to switch back in my profile, so...  :maybe: did I just forget how to switch skins, or is there something wrong?

the option to change the skin is gone...


Forumda Samanyolu kolejden (Ankara) yazan var mi :?:

var bizim 'metehan' da pek ugramıyor bu aralar.

Delegat, ben Ankaradan yazıyorum.. Gerçi Ankara Samanyolundan değilim de.. Ama ordan sayılırım :)  :roll:
Başka bir yerde o kurumdan mezun oldum...
Zaten rus forumunda ermenilerin arasında bıkmıştım :P

Ben varım ama ben çok fazla yazmıyorum.

Hubeyb Gürdoğan

[quote]Ben de Ankaradan yazıyorum.
İrfan sensin herhalde.
Napiyorsun nasil gidiyor isler

Fatih Haltaş[/quote]

Hayir abi Irfan degilim Adim Salem

Irfan ile ayni okuldayim Yani Irfan bu sene Universiteye girdi Ben lise son (12.) siniftayim.

Bu sene ikimiz Dunya Matematik olimpiyatina katildik.

Islerimiz masaallah iyi gidiyor Hersey guzel
Bu hafta, bizim (kolejde) ek derslerimiz basliyor. Irfanin Universitede sinavlari bu ayin sonunda baslayacak.

Saray Bosnadan Selam

Salem Malikic


To prepare Alkane by Wurtz' reaction it says that AlkileHalide has to be mixed with dry ether.Why so?Does Alkilehalide's properties change in the prtesence of Dry ether?Also what would Alkilehalide's state be written(s,l,g,aq??)in the reaction?Also what is dry ether?In preparing ether H2SO4 takes away water so is all ether dry ether?
Thanks?

1. Dry ether is ethyl ether from which water has been removed. This can be done by refluxing ether with a sodium wire inside it for several hours (together with benzophenone).

2. The Wurtz reaction consists in mixing an alkyl halide (or two different ones) with sodium metal. Not only the reaction involves an alkaline metal, but also, if you look at the mechanism, there is the formation of carbanions. I think you know what can happen when sodium meets water, and, with water present, the carbanions would be converted to the alkanes of the alkyl halides, reducing product formation. On the other hand, note that this reaction is nowadays seldom used, with the CPWH reaction being a much more important alternative.

3. The choice of ether of solvent lyes on two properties: it is not an acidic solved, and alkyl halides are generaly soluble in it.

But why Ether?Couldn't we use other solvent in which Alkylhalides are soluble? Also why dry ether?
P.S.What is CPWH reaction??

I have already answered your first three questions, read carefully my post above.

CPWH reaction is Corey-Posner, Whitesides-House reaction.

Oh,thanks. :)
Could you please give me the definition of CPWH reactions or an example?
Thanks.


geometry rectangle MATHCOUNTS counting distinguishability

Someone tell me where this problem should go please

Problem: Construct the four perpendicular bisectors of the sides of the rectangle below, using 4 arcs and 2 lines.

Thanks

Not MATHCOUNTS material, but it's fine.

So basically we utilize that there are only two distinguishable perpendicular bisectors in the rectangle (since opposite sides are parallel). The rest is pretty self-explanatory (the circles are "arcs").


Given any positive even integer, x, the positive difference between the smallest odd number greater than 7x-2 and the largest odd number less than 3x+5 can be written in the form ax+b.  What is a+b?

Here subbing numbers is easiest.  Let x equal 2.  The smallest odd number greater than 7x-2 is 7x-1.  The largest odd number less than 3x+5 is 3x+3.  (7x-1)-(3x+3)=4x-4. a=4 b=-4 a+b=0

You need to be careful that your substitution does not skew up the rest of the problem...if $ x\equal{}2$, then $ 4x\minus{}4 \equal{} 4 \equal{} 2x\plus{}0 \equal{} 6x \minus{} 8 \equal{} \dots$.

Because $ x$ is even, $ 7x$ is even, and $ 7x\minus{}2$ is even.  Hence, the smallest odd number greater than $ 7x\minus{}2$ is $ 7x\minus{}1$.  Similarly, $ 3x\plus{}5$ is odd, so the largest odd number less than $ 3x\plus{}5$ is $ 3x\plus{}3$.  Then, $ (7x\minus{}1)\minus{}(3x\plus{}3)\equal{}4x\minus{}4$, so $ a\plus{}b\equal{}0$ as you said.


Prove that the equation $ y^2\equal{}x^5\minus{}4$ has no integer solutions

Got it! Because $ y^2\plus{}4\equal{}x^5$ we have $ [2,5]\equal{}10$ So we work with modulo 11.

If $ y\equal{}u (m0d11)$ then $ y^2\equal{}u^2 (mod11)$ for $ u\equal{}0,1,2,3,4,5,6,7,8,9,10$
So $ y^2\plus{}4\equal{}r (m0d11)$ for $ r\equal{}2,4,5,7,8,9$     (1)

If $ x\equal{}u (mod11)$ then $ x^5\equal{}u^5 (mod11)$ for $ u\equal{}0,1,2,3,4,5,6,7,8,9,10$
So $ x^5\equal{}t (mod11)$ for $ t\equal{}\minus{}1,0,1$      (2)

From (1) and (2) we conclude that the equations has no integer solutions. :)

Am i right?? :maybe:


MATHCOUNTS geometry videos Support Euler

Good Luck to everyone at this year's State Competitions. The first competitions are tomorrow, and they continue all month long (later if there are weather delays). Remember, no discussing difficulty, problems, or score. All the Best, everyone! :lol:

Haha mine is the first competition, then! Or maybe tied for first.

I'm soo nervous for tomorrow.

Delaware sure has an early State date. NC's is March 20.

Good luck!

Once again Michigan is last... March 28th...

Illinois is perhaps the second earliest...March 7.

Good luck Delawarians tomorrow! :D

Haha, delaware is actually first in something! that's so unusual.

Good luck to Delaware people. Also, if you are reading this and are from Delaware, you should probably be getting ready to sleep soon  :P .

If you're reading this and your state is on Saturday, go be more productive. If you're reading this and are from Ohio, keep reading this again and again.

Yeah. Mine is on this saturday too. I'm so nervous... But I live in OK so I won't be overkilled. :D

Haha, DE's got canceled today! Apparently, out of the 10 schools in it (I know, so many schools XD), all of them got a snow day, so it'll be rescheduled.

Oh, so now, what's the first state to hold its competition?

Kansas, on March 7 :D

meh... time to get beat up by a kid who got first chapter, first spelling bee, first geography bee, etc.

(thats true)

a good luck would be nice!! (i'll use it wisely lol)

Somehow I doubt 3 people already got first in their state....

How many people were in your guys' states?  KS had a nice, small number of 94.

michigan had like 117, with around 1500 participating in chapters.

Our state had approximately 180 people in Norcal, and I suspect a similar number for Socal, for a total of around 350-360.

Our state, PA, had about 170 participants

Thanks to the school I got to go to the state. I ended up beating a few members on the team.

How many people do they generally take from each chapter? It seems like top team and top two individuals but I've noticed some chapters have really good people while other chapters have really weak people. Because of this my chapter sends top 4 teams and top 4 individuals.

It varies for each state's chapters.
But the minimum is top team plus top two individuals not from the winning team, according to the MATHCOUNTS handbook.
In mine, they send the top 4 teams and top 5 individuals.

I've got a real strong chapter compared to the rest of the state, so they took 30+ indivs and i don't know but a lot of teams (10?). They probably should have taken more, because my chapter had 13 of the 16 in the state unofficial cd.

[quote]I'm looking at the NoCal results at the moment, and 12 of the first 14 positions are in the SCV (Santa Clara Valley) chapter.[/quote]

well SCV is arguably one of the toughest chapters in the U.S.
it had 5 perfect scores, and eight 45's (i think? somebody correct me)
some guy from harker with a 45 didnt even make states.

anyways... CA has a lot of people.

CT had 5 (:)) chapters, and 175 people at states.

Pennsylvania had 389 schools, 2651 students, and 23 local competitions  :D

@Euler: I already knew that SCV was the toughest chapter in CA :wink:

And there were 6 45's.


Hi everybody!
Sadly we got only 1 right :( .

#8. The mark-up on an item is the difference between what a store charges for an item and what the store originally paid for the item. A certain store buys table soccer games for $ 95 each and sells each of them for$399.99. How many times the original cost to the store is the store's mark-up on each game? Express your answer to the nearest hundredth.

8. Subtract $ 95.00 from$399.99
    Then divide the answer by $ 95.00
    Your answer is$3.21 after you have rounded to the nearest hundredth.

YAY!!!!!!!!!! We are done!!!!!!!! :P  :rotfl:

One thing I noticed when I graded your papers was that most of you put 4.21...

You probably put that because you did how many times of the orgininal price is the new price instead of figuring how many times of the original price is the MARK-UP.

So read the question carefully. You should underline the question stem when you're reading the question the first time.

hahaha you people are so smart...one right...lol just kidding we only got 2 or 3


geometry
geometric transformation
homothety

(C)and(C')and(C") are three circles  .each two of them are tangent to each other.

 the outher centers of homothecy betwin each two of them will make three points K,L,M.

prove that:K,L,Mare colinier.

This is the Mongé theorem.

Hint

[hide]You can use the Menelaos theorem...
[Or you can use some 3-dimensional interpretation][/hide]


The sum of two prime numbers is 20. What is the largest possible product of these two primes?

The highest possible product would be 91. The two prime numbers with this product is 13 and 7.

Note that numbers with a certain sum have a larger product when their difference is smaller.  This theory is known as AM-GM, which states that the arithmetic mean of two numbers is larger than the geometric mean, aka (a+b)/2 is greater than or equal to sqrt ab.  Equality holds when a=b.  13 and 7 is closer than 17 and 3 (the other possible choice).


In how many ways can we list the digits {1, 1, 1, 2, 2, 3, 4, 5} so that two identical digits are not in consecutive positions?

Let`s construct all the good permutations.
With the numbers 2;3;4;5 we can do 4! permutations.
Then we add a 2 at all of this permutations.
In the new permutations there are 4! that have two consecutive 2($ A$) and 3*4! that are good($ B$).
1- Let`s add the three 1s in ($ A$) permutations.
This permutations could be good, then a 1 is in the middle of the two consecutive 2s.
Then the number of good permutations is $ 4!\binom{5}{2}\equal{}240$.
2- Let`s add the three 1s in ($ B$) permutations.
The number of good permutations is $ 3*4!\binom{6}{3}\equal{}1440$
So the total of good permutations is 1680.

Now I get 720. What is wrong with what I did?
$ \frac {8!}{2!3!} \minus{} 7 \cdot \frac {6!}{3!} \minus{} (\frac {8!}{3!2!} \minus{} \frac {5!}{2!} \cdot {6 \choose 3}) \plus{} (5! \plus{} 4! \cdot {5 \choose 2}) \equal{} 720$.
Here's how I did it:
Answer=
Total-(# with consecutive 2s)-(# with consec. 1s)+(# with both consec 2s and consec 1s)=
Total-(# with consecutive 2s)-(Total-# without consec 1s)+(# with two consec 2s and exactly 2 consec 1s + #with two consec 2s and 3 consec 1s)

[hide="Solution"]

First line up all the 1s, and label the number of elements in the gap inbetween as follows:

$ a_1  1  a_2  1  a_3  1  a_4$

We have the follow constraints: $ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \equal{} 5$ with $ a_1, a_4 \ge 0$ and $ a_2, a_3 \ge 1$. Which is a well known balls in urns problem with the number of solutions $ \binom{6}{3}$. There are also $ \frac{5!}{2!}$ permutations of the non-1 elements.

However, we overcounted since we included cases where there are two adjacent 2s. We can count these then remove them:

We can treat "22" as a grouped element. Then we have the following constraints: $ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \equal{} 4$ with $ a_1, a_4 \ge 0$ and $ a_2, a_3 \ge 1$. With the number of solutions: $ \binom{5}{3}$. In this case there are $ 4!$ permutations of the non-1 elements.

This gives a total: $ \frac{5!}{2!} \binom{6}{3} \minus{} 4! \binom{5}{3} \equal{} 960$

[/hide]

[b]dgreenb801[/b]: $ a_1$ to $ a_4$ are not the individual elements themselves, they are the number of elements in that spacing or gap. For example, your first case would be considered by the fact that I can let my variables $ a_1 \equal{} a_4 \equal{} 0$.

There are two types of sequences, (A) those which contain 2,1,2 consecutively and (B) those that do not.

For (A), there are 3! = 6 ways to order {3,4,5}, and for each of those, there are 4 places to insert the (2,1,2), and then there are C(5,2) = 10 places for the remaining two 1's. So that makes 6 x 4 x 10 = 240 of type (A).

For (B), the 2's are always separated by a 3, a 4, or a 5, so there are 3! = 6 ways to order the {3,4,5}, and then C(4,2) = 6 ways to insert the two 2's into 4 places, followed by C(6,3)=20 ways to insert the three 1's into the six places. So 6 x 6 x 20 = 720 of type (B).

So, I get 240 + 720 = 960 also.


Let $A$ and $B$ be two strictly positive (i.e. minimal eigenvalue is strictly positive) hermitian (symmetric) matrices of the same dimension. Prove that
\[\mathrm{Tr}\,(AB) > 0.\]

// Here $\mathrm{Tr}\,A$ stands for the trace of matrix $A$.

you know that there exists an invertible matrix $P$, and a diagonal matrix $D$ with positive entries such that $A=P^{t}P$ and $B=P^{t}DP$. From here $Tr(AB)=..=Tr(PP^{t}PP^{t}D)=Tr(M^{2}D)$ where $S=PP^{t}$ is a symetric matrix. Now look at the elements on the diagonal of $M^{2}D$.

Good proof.

Let me modernize the problem: Prove that
\[\mathrm{Tr}\,(AB) \ge \mathrm{Tr}\,A \cdot \min\sigma(B) \]
if both $A$ and $B$ are hermitian (symmetric) and $A\ge0$.

// $\sigma(B)$ := spectrum of $B$.


Let $ a_n$ be an infinite sequence of reals. Show that if $ a_n$ is convergent then we must have $ a_n\minus{} a_{n\minus{}1}\rightarrow 0$. If $ a_n\minus{} a_{n\minus{}1}\rightarrow 0$, must $ a_n$ be convergent?

[quote="Pythagorean12"]Let $ a_n$ be an infinite sequence of reals. Show that if $ a_n$ is convergent then we must have $ a_n \minus{} a_{n \minus{} 1}\rightarrow 0$. If $ a_n \minus{} a_{n \minus{} 1}\rightarrow 0$, must $ a_n$ be convergent?[/quote]

No. If $ a_n \equal{} \sum_{i\equal{}1}^{n} \frac{1}{i}$  then $ \lim_{x \minus{}> \infty} a_n \minus{} a_{n \minus{} 1} \equal{} 0$ but the series is divergent because it is the harmonic series.

[quote="Pythagorean12"][quote="Geniuscide"][quote="Pythagorean12"]Let $ a_n$ be an infinite sequence of reals. Show that if $ a_n$ is convergent then we must have $ a_n \minus{} a_{n \minus{} 1}\rightarrow 0$. If $ a_n \minus{} a_{n \minus{} 1}\rightarrow 0$, must $ a_n$ be convergent?[/quote]

No. If $ a_n \equal{} \sum_{i \equal{} 1}^{n} \frac {1}{i}$  then $ \lim_{x \minus{} > \infty} a_n \minus{} a_{n \minus{} 1} \equal{} 0$ but the series is divergent because it is the harmonic series.[/quote]

It is not a series, but a sequence.[/quote]

$ a_n$ is a sequence: the sequence of partial sums of the harmonic series.

you should probably sit down and think through these things first before jumping to conclusions. the facts are pretty basic.


Given triangle $ ABC$ and $ P$ is an arbitrary point, a line $ d$ is through $ P$( not through vertex) intersect side $ BC,CA,AB$ at $ A',B',C'$ and $ A'',B'',C''$ are symetric points of $ A',B',C'$ through $ P$, prove that $ AA'',BB'',CC''$ are concurrent.

http://www.mathlinks.ro/viewtopic.php?t=38178

  darij


I really need to know this before taking the 10B
Thanks to whoever answers

Assuming 5 of the answers to the 25 questions are As, 5 are Bs, 5Cs, 5Ds, and 5Es,
if I pick five questions at random what should I guess on them in order to maximize my chance of getting at least two right?

Assuming you haven't found out the answers to the other questions and haven't eliminated any answer choices, it doesn't matter.  The chances are the same no matter what you pick.

If you leave the 5 qeustions blank, you get 5*2.5=12.5 points.

If you guess, you get 1 of 5 right, you get 6 points.

If you are lucky, you may get 2 of 5 right, you get 12 points.

For me, i won't answer them!

You need to be thinking about the problems on the test when taking an AMC test, not making possibly (likely) wrong assumptions about the frequency of answer choices.

Good luck. Think hard about the problems you work and answer, and leave the rest blank. That's the best strategy for a first-time test-taker.


I think similar problems (and harder, for that matter :)) have been discussed before, but this sounds really nice:

Prove that for arbitrary infinite natural sequences $\{a_n\},\ \{b_n\},\ \{c_n\}$ we can find $p\ne q$ s.t. $a_p\ge a_q,\ b_p\ge b_q,\ c_p\ge c_q$.

Additionally, here's a treasure trove: [url]http://members.tripod.com/~PertselV/RusMath.html[/url]

[Edit: I've replaced the word "integer" by "natural"; see Pierre's post below :); thanks, Pierre]

Hmmm... but what if $\{a_n \}$ is increasing and $\{b_n \}$ is decreasing?
Are you sure that you don't miss another hypothesis?

Pierre.

It's been edited, thanks for noticing :).

For natural numbers, the result is easy since for any infinite sequence of natural numbers either the sequence is bounded and it takes the same value an infinite number of times or it is not bounded and we may find an increasing subsequence.
Thus, in any case, we may extract a non-decreasing infinite subsequence from any given infinite natural sequence.
Use that for the sequence $\{a_n \}$ which leads to the non-decreasing infinite subsequence $\{a_{\phi_1(n)} \}$.
Now, use the same result for the sequence $\{b_{\phi_1(n)} \}$ to extract an infinite non-decreasing subsequence $\{b_{\phi_2(n)} \}$. Note that the sequence $\{a_{\phi_2(n)} \}$ is non-decreasing too as a subsequence of $\{a_{\phi_1(n)} \}$.
Then, use the result again for the sequence $\{c_{\phi_1(n)} \}$ to get an infinite subsequence $\{c_{\phi_3(n)} \}$.
Then :
The sequences $\{a_{\phi_3(n)} \}, \{b_{\phi_3(n)} \}, \{c_{\phi_3(n)} \}$ lead to the desired conclusion.

Pierre.


limit
integration
calculus
real analysis
real analysis unsolved

I think repeated partial integration of $ \int_{0}^{1}e^{x^{2}}dx$ will help in both cases :)
(possibly wrong, but does not have time to check..)

[quote="Cezar Lupu"]Hmm, I expect a full solution. My solution uses only Lagrange's theorem. The original solutions are not quite "elementary".[/quote]

I mean that by partial integration it is easy to get formula: $ \int_{0}^{1}e^{x^{2}}dx = e \sum_{k}^{n}(-1)^{k}\frac{2^{k}}{1 \cdot 3 \cdot ... \cdot (2k+1)}+(-1)^{n+1}\frac{2^{n+1}}{1\cdot3\cdot .. \cdot(2(n+1)+1)}\int_{0}^{1}e^{x^{2}}dx^{2(n+1)+1}$ and remainder goes to zero when $ n \to \infty$..


okay
so it would be pretty awesome if we could all get a plane ride together to get to MOP
so like
post here if you wanna take the same plane ride
say if you're sure/unsure/stuff like that
if someone's got more to add then
add

by the way
I am NOT organizing this
see my binder and you'll get why
some dude gotz to be the organizer here

I want to go on the same plane with other BA MOPpers.
-Vishal

dude you're going to be together for like A WHOLE FRICKIN MONTH

two-hour plane ride is not that important

nah that's not the picture

the idea is to like make me/my parents/everyone else's parents have to organize less for plane tickets and stuff whatever

yayayayayay
i nominate lynnelle to be the organizer
(you have previous experience)

[quote="bookaholic"]I want to be on this plane ride!

San Francisco Airport, I presume? Or are people closer to San Jose?

And I still don't have an email. WHAT THE HECK.[/quote]

I think your parents should have received the email. Junggi says that "you had previous experience", so did you organize BA MOP Plane ride last year? (or organize something else?)

It's really important that if we go together on a plane ride, then we have to start planning early (like [b]RIGHT NOW[/b]).

EDIT: While it may not be as important, I think that Quatto should change the name of the topic because it doesn't sound like it has to do with the plane rides.

Dude we are seriously going to need an entire plane. At least the entire economy section.

I'm pretty much going to take up the entire center aisles, and then you guys will have to squeeze in on the sides.

My parents didn't get the email either. Colleen also hasn't heard anything, so I think they're just ignoring the girls. Previous experience=my parents managed to put Kevin Yang and me on the same flight last year. That's about it.

Practical points: we WILL have to change planes. Lincoln's airport flies to Denver, Chicago, Minneapolis, and possibly some mysterious fourth city. It is not a busy airport. So we fly from San Francisco (San Jose?) to Denver, then we get on the same Denver-->Lincoln flight as every other MOPper from the western half of the U.S.

I suggest that we have as long a gap between connecting flights as reasonably possible. Last year I had a two-hour gap, my first flight was delayed by two hours, and I had to run from one end of the Denver airport to the other hoping that they wouldn't take off without me. (They didn't. Yay.) Denver's airport is really, really long. It was annoying. I would like to lower the risk of something like that happening again, even at the cost of getting up earlier in the morning.

That's just as bad as some airports I've been to in rural China. As well as Ulaanbaatar, now that I think of it.

[quote="jb05"][quote="Ubemaya"]Dude we are seriously going to need an entire plane. At least the entire economy section.[/quote]

Correction: the entire first-class section.

I doubt Denver is anywhere near as bad as Chicago. I don't know that I have ever gone through Chicago without a delay, and I nearly missed the plane to Lincoln two years ago. But then, the only time I've been through Denver, that was my destination. Hmm.

Lincoln airport has two gates!!![/quote]
dude yeah
Chicago Airport pissed me off
so like
we had a plane from New Jersey to Chicago to San Jose
then
like
the plane from NJ to Chicago was delayed like 15 minutes or something
then we got to Chicago
the plane was like broken or messed or something
so
they like made us wait for two hours
and then
another hour
then they canceled the flight
cause there were like no other ways to get to San Jose for 4 people (Dad, sister, Mom, me)
and I think that was for 1-2 days (no flight available)
then
like
they found a way to make us about 12 hours late
they made my dad take a flight to Oakland
and we had to go to LA and get a hotel
we got in the hotel at around 1 and had to get up at 5
and then we got a flight from LA to SJ FINALLY
dangit I hate that

anyways, I am also going on the SJ to Denver to Lincoln

Uh... I think my mom already made flights for me already. I'm not sure who/when/where/why, but uh... yeah.

[quote="CircleSquared"]shoot i barely missed red mop

ok have fun at mop anyway[/quote].
What. You're an 8th grader, you can't make red mop.

oh noes bowei isn't a moplet
join the submoplet club!

I think we might be better off doing this as an email list. Consider the following sets of people.

A={people who want to fly to MOP with the Bay Area group}
B={people who have not posted in this thread}
C={people who have posted in this thread and prefer an email address that is not the address their AoPS account uses}

If you belong to the set A ∩ (B U C), please PM me with your preferred email address. Otherwise, I already have it.

group A

my e-mail should be on this site but PM me if it's not there

[quote="bowei"]omg i fail
i didnt make mop
gah[/quote]
Bowei =(( I love you...

heres something funny
so quatto's mom emailed my mom and ubey's mom bout a $325 flight and tells us to book it so we do and it ends up quatto's parents forget to book, and two days later the price is$475

lol

anyways were all (me quatto Ubey) on United 1220 from SJC to DEN
then United 5880 from DEN to LNK

hopefully theyll be other moppers on this flight

[quote="davidlizeng"]heres something funny
so quatto's mom emailed my mom and ubey's mom bout a [hide]$ 325 flight and tells us to book it so we do and it ends up quatto's parents forget to book, and two days later the price is$[/hide]475

lol

anyways were all (me quatto Ubey) on United 1220 from SJC to DEN
then United 5880 from DEN to LNK

hopefully theyll be other moppers on this flight[/quote]

When using dollar signs not for LaTeX, remember to check the "Disable BBCode in this post" option. What you really meant was:

[quote="davidlizeng"]heres something funny
so quatto's mom emailed my mom and ubey's mom bout a $ \$325$ flight and tells us to book it so we do and it ends up quatto's parents forget to book, and two days later the price is $ \$475$

lol

anyways were all (me quatto Ubey) on United 1220 from SJC to DEN
then United 5880 from DEN to LNK

hopefully theyll be other moppers on this flight[/quote]

I'm on the same flight to MOP, but I'll be separated from you guys when I leave.  :(

Dudez, we need to have some kind of party on the airplane. Preferably with Meggy-Meg.

[quote="vishalarul"][quote="davidlizeng"]heres something funny
so quatto's mom emailed my mom and ubey's mom bout a [hide]$ 325 flight and tells us to book it so we do and it ends up quatto's parents forget to book, and two days later the price is$[/hide]475

lol

anyways were all (me quatto Ubey) on United 1220 from SJC to DEN
then United 5880 from DEN to LNK

hopefully theyll be other moppers on this flight[/quote]

When using dollar signs not for LaTeX, remember to check the "Disable BBCode in this post" option.

[/quote]

lol, completely forgot about latex and dollar signs

nah
she didn't forget
she was just thinking that we should get this organized with my teachers first
and I was like nah why don't you get the tickets first
and then she was like okay fine
and the pricez shot up 150 dollars
what the heck dangit BA MOPpers

[quote="bookaholic"]Wow, I've spent like no time on MOP homework. Yay for not all of the red problems being hard. This gives me a slight amount of hope for myself.[/quote]
dude i got owned by blue hw
wtf
im gonna really die at mop
wow


real analysis
limit
real analysis unsolved

Let $f: (0,1)\to\mathbb{R}$ a derivable function. Prove that $f'$ is Lebesgue measurable. ;)

uhm..
we've done that in class, so it wouldn't be fair if i posted it :D
but i have something to add:

show that $f'$ is not necessarily riemann-measurable.

There's a proof in post #5 from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=pointwise&t=63280]this topic[/url].

$f'(x) = \lim_{n\to\infty} n(f(x+\frac 1n)-f(x))$ is the limit of a series of Lebesgue-measurable functions. It is Lebesgue-measurable, too.

Incidentally, that's exactly what I proved in that other article :).


A telephone company charges $ 21.0$ cents for the first minute of phone use
to a certain area and 13.6 cents for each minute or portion of a minute
thereafter. How many cents, rounded to the nearest cent, would a
phone call that lasts 11 minutes 18 seconds cost?

Substituting into the formula, we simply have $ 21\plus{}11(13.6)\equal{}170.6 \approx \boxed{171}$ cents.

Actually, wouldn't the answer be 161 cents? We have 21 cents for the first minute, leaving 10 minutes and 18 seconds. The cost for this portion is just (10 + (18/60)*13.6)=140.08. Adding 21 cents yields in 161.08 or 161 cents.

[quote="GameBot"]A telephone company charges $ 21.0$ cents for the first minute of phone use to a certain area and 13.6 cents for each minute [b][size=200]or portion of a minute thereafter[/size][/b]. How many cents, rounded to the nearest cent, would a phone call that lasts 11 minutes 18 seconds cost?[/quote]

Using this logic, we get: $21 + 11(13.6) = 170.6$

When rounded, this gives $\boxed{171\text{ cents}}$.

Sorry for reviving, just trying to clarify...

Ahh true, good catch. I really have to start reading these mathcounts questions...