We have two halfcircles with common foots as on the picture
[img]http://www.gjar-po.sk/~prusak2c/pictures/sezam3.jpg[/img]

Points $A, B$ are on bigger halfcircle such that $AB$ is touching smaller halfcircle and $|AB|=24\ cm$. Determine the area of red part (area of bigger except area of smaller).

Sorry - Don't want to spoil the fun, so will not give a hint,  (yet)  but there is  a "physicst" (means - "lazy") way to solve this old problem.

[hide]
It seems to me that the shaded area is dependent on where line AB intersects the smaller semicircle, but assuming that AB is parallel to the foot of the semicircles, as is suggested by the picture, we set R equal to the radius of the larger semicircle, and r equal to the radius of the samller semicircle.   Thus:
[tex]
\indent A_{big} = \frac{1}{2} \pi R^2
[/tex]
The line through the center C of the bigger semicircle perpindicular to AB bisects AB at k.    The line through the center c of the smaller semicircle and the point of tangency t between AB and the smaller semicircle is perpindicular to AB.  Clearly segment ct and segment Ck are of equal length.
[tex]
\overline{Ck}^2+12^2=R^2\\
\overline{Ck}=\sqrt{R^2-144}
[/tex]
Ck = ct = r, so:
[tex]
A_{small} = \frac{1}{2} \pi (R^2-144)\\
A_{shaded} =\frac{1}{2} \pi R^2 -  \frac{1}{2} \pi (R^2-144)\\
A_{shaded} = 72 \pi
[/tex]
[/hide]

I think I know what Gyan is referring to.[hide]
If we center the smaller semicircle,
the problem is similar to "The Hole in the Sphere" problem.
The required area is a function of the length of the chord (only).[/hide]

Heh just use the Pythagorean theorem and R^2 - r^2 = 12^2.  So then :pi:(R^2 - r^2)/2 = 72 :pi:

[quote="Treething"]Off topic:

Isn't the plural of "foot" "feet" ?[/quote]

When I say foot I refer to the line that contains the two indivual feet, in which case plural isn't neccesary.


function
limit
topology
real analysis
real analysis unsolved

1. Suppose that the sequence (f_n),f:X->R (and X is a metric ) is equicontinious and that (f_n) converges
pointwise to an f . This => that f is continuous? Is this true? Any proof?

I think that is true . But i need a proof for this. I think is true   when   X is  a compact  metric space, but what happens  when X  is just a metric space?

2. Suppose that the sequence (f_n), f:X->R (and X is a metric ) is Lipschitz functions and that (f_n) converges
pointwise to an f . This => that f is continuous? Is this true? Any proof?

3. Is true that if (f_n) is Lipschitz, f:X->R (and X is a metric )then
the sequence (f_n) is equicontinuous?

For the other two, I believe the answers are negative:

For number (3), we can take $f_n:\mathbb R\to\mathbb R,\ f_n(x)=nx,\ \forall x\in\mathbb R,\ \forall n\in\mathbb N$.

For number (2), we could take the sequence $f_n:\mathbb R\to\mathbb R$ s.t. $f_n(x)$ equals $0$ for all $n$ and all $x\le 0$, it equals $\frac 1x$ for all $x\ge \frac 1n$, and it equals the linear function which connects $(0,0)$ to $\left(\frac 1n,n\right)$ for $x\in\left(0,\frac 1n\right)$. It's pretty clear that all $f_n$ ate Lipschitz, and the sequence converges pointwisely to $f$, which equals $0$ for $x\le 0$ and $\frac 1x$ for $x>0$.

Please  correct me if i am wrong.
number 3
I think that the sequence  f_n=nx aren't lipschitz

Why not? :? You have $|f_n(x)-f_n(y)|=n|x-y|$. Look here: [url]http://mathworld.wolfram.com/LipschitzFunction.html[/url]

I understood that you want each $f_n$ to be a Lipschitz function, and, if this is true, then the example is Ok. Maybe you meant something else?

Because for x=0 and y= n  we have n^2<=k . But the N   is not a bounded set.
Am i right?

Each $f_n$ is Lipschitz, and $f_n$ is Lipschitz with the constant $n$. You seem to want more than this: you want the functions to be Lipschitz with the same constant, in which case the answer to the question is obviously yes. I assumed this is not what you meant.

1. Suppose that the sequence (f_n),f:X->R (and X is a metric ) is equicontinious and that (f_n) converges
pointwise to an f . This => that f is continuous? Is this true? Any proof?
I think that  this is not true
finnaly.
But i  have another one  question:

 Suppose that the sequence of uniformly continuous  (f_n),f:X->R (and X is a metric ) converges
pointwise to an f :X->R. This implies  that f is continuous?

Assume additionally that $X$ is compact. Then you may omit the condition of uniform continuity. And of course, the point-wise limit of continuous function may be not continuos (say, $f_n=0$ on $[1/n,1]$, $f_n(0)=0$ and $f_n$ is linear on $[0,1/n]$). I tried to say the same in my previous post.

What  about, then,nnumber 1  ?
is true? or not?
i don't  know :s

[quote="mistake"]is  true that if  the sequence  of (f_n) is equicontinuous  then  each f_n is uniformly continuous?[/quote]

Of course not. You can take all the $f_n$ equal to some continuous but not uniformly continuous function, for example.

[quote="Fedor Petrov"]As for 1, if you mean that "for any $\epsilon>0$ there exists $\delta>0$ such that $d(x,y)<\delta$ implies $d(f_n(x),f_n(y))<\epsilon$ (where $d$ is a distance)", than of course a pointwise limit is (uniformly) continuous: $d(f(x),f(y))=\lim d(f_n(x),f_n(y))<\epsilon$ provided $d(x,y)<\delta$. If you mean that each $f_n$ is uniformly continuous, the pointwise limit may have discontinuities. Grobber's example for no.2 works.[/quote]

I don't think this is what equicontinuity means.

As far as I know, it's this: the family $(f_n)$ is caled equicontinuous in $x_0$ if for all $\varepsilon>0$ we can find $\delta>0$ s.t. $d(x,x_0)<\delta$ implies $d(f_n(x),f_n(x_0))<\varepsilon,\ \forall n$. We call a family equicontinuous if its equicontinuous in all the points of the domain. The condition you mentioned is stronger. It's some kind of "uniform uniform continuity" :). The solution to the problem still works along the same lines though.

Yes, grobber, I think, you are right. I just never used equicontinuity on not-compact spaces (where it is a basic concept for Arcela-Ascoli theorem). If $X$ is compact, equicontinuity is equivalent to my definition (easy excersise).


calculus derivative trigonometry LaTeX function limit algebra

is it zero? i think it is not or else it seemed to easy

actually it is, !!!!! Can you post your solution!!!

ok, I don't know latex but here's my solution..

we can use the chain rule on the function for x>0 and x!=integer and get

d/dx f(x)= -sin((pi/2)*floor(x)-x^3)*-3x^2

evaluating this at x=(pi/2)^(1/3) we get 0

How did you find derivative of $[x]$ Isn't that need in chain rule???

yeah, the derivative of the floor is zero at every point it is differentiable at- that is everywhere besides integers.

Can you explain a bit more!!! I don't quite understand!

[quote="vidyamanohar"][quote="galeos"]Set $g(t)=[t]$. Let $x_0$ be an integer. Then $\lim_{x\rightarrow x_{0^+}}\frac{g(x)-g(x_0)}{x-x_0}=0$ whereas $\lim_{x\rightarrow x_{0^-}}\frac{g(x)-g(x_0)}{x-x_0}=\lim_{x\rightarrow x_{0^-}}\frac{x_0-1-x_0}{x-x_0}\rightarrow\infty$,so $g$ is not differentiable on the set of integers. Similarly you can show that the derivative at any other point equals $0$.
But i do not think we need to get that far to find the original derivative. Since $1<x<2$, $[x]=1$ and then differentiate using the standard methods.[/quote]

why do we need all this here.....given 1<x<2 means [x]=1

and the given function simply becomes $sinx^3$
 after differentiation and substituting the given value for x, one has to get zero.

And the explanation given by arp is abosolutely fine....

[x], wherever differentialble, is a constant. hence its derivative is zero.[/quote]

that was exactly what i said! i just wrote a strict proof why the floor function is not differentiable at integer points.

[quote="galeos"]

that was exactly what i said! i just wrote a strict proof why the floor function is not differentiable at integer points.[/quote]

f(x)=[x] is not a continuous function at integers, hence is not differentiable would have been enough.

there are many many ways to prove this! :)


Prove that if $a,b\geq0$, then
\[\frac{a+b}2\leq\sqrt{ab}+\frac12|a-b|.\]

[quote="pvthuan"]Prove that if $a,b\geq0$, then
\[\frac{a+b}2\leq\sqrt{ab}+\frac12|a-b|. \]
[/quote]
Assume $a \geq b$ your inequatility becomes:
$\frac{a+b}2\leq\sqrt{ab}+\frac12(a-b)$
or $\frac{a+b}{2}-\frac{1}{2}(a-b) \leq \sqrt{ab}$
or $b \leq \sqrt{ab}$
It is too obvious, Thuan. So what is your general result so far?

Yes, it is quite obvious. But it is starting point. More results can be obtained as inspirations from this nice inequality.

For example,
\[\frac{a+b}2+\Bigl(\frac{\sqrt2-1}2\Bigl)|a-b|\geq\sqrt{\frac{a^{2}+b^{2}}2}.\]

and more ...... :P

This one is tougher and of a quite different nature:
 $\frac{a+b}2\leq\sqrt{ab}\cdot e^{\frac{(a-b)^{2}}{4ab}}$  :D

For the second...

assume as before $a\geq b$ (I supposed they are both positive, otherwise an easy counterexample can be found for $a=b=-1$)

Ok, after some algebraic passages on the LHS we get $\frac{b}{\sqrt{2}}(\sqrt{2}-1)+\frac{1}{\sqrt{2}}a\geq \sqrt{\frac{a^{2}+b^{2}}{2}}$

Multiply $\sqrt{2}$ to both sides to obtain

$b(\sqrt{2}-1)+a\geq \sqrt{a^{2}+b^{2}}$. Since both sides are positive we can square and after some easy algebraic passages we get $ab\geq b^{2}$ which is true for the initial conditions. The equality holds for $a=b$

 :)

The following is again more difficult
\[\frac{a+b+c}3\leq\sqrt[3]{abc}+\frac16(|a-b|+|b-c|+|c-a|).\]
But of course the even more general inequality is still a challenge.

But are you sure that the last ineq is correct?

[hide="click"]

I've plugged in some value $a=6,b=5,c=\frac{1}{30}$ and I've got

$\frac{331}{90}\leq \frac{269}{90}$

Am I mistaking something?  :wink:

[/hide]

[quote="spanferkel"]I think it should be
\[\frac{a+b+c}3\leq\sqrt[3]{abc}+\frac13(|a-b|+|b-c|+|c-a|). (*) \]
assuming wlog $a\ge b\ge c$, this is equivalent to $\frac{3c+b-a}3\leq\sqrt[3]{abc}$ or $27abc\ge(3c+b-a)^{3}$ which is obvious. Or am [i]I[/i] missing something this time?
Note that the denominator of the RHS in (*) cannot be bigger than $3$ as shows the example $(a,b,c)=(1,1,0)$.  :maybe:[/quote]
Unfortunately, this is inequality already appeared in Hojoo Lee's collection. :D

[quote="The soul of rock"]I conjectured that the following non-symmetrical inequality hold
$\frac{\sum_{1}^{n}a_{i}}{n}\le \sqrt[n]{a_{1}a_{2}...a_{n}}+\frac{\sum_{1}^{n}|a_{i}-a_{i+1}|}{n}$ where $a_{n+1}=a_{1}$. :?:[/quote]
No.  :( Put $(a_{i})=(1,...,1,0)$ and you'll see that the best we can get for the RHS is
$\frac{\sum_{1}^{n}a_{i}}{n}\le \sqrt[n]{a_{1}a_{2}...a_{n}}+\frac{n-1}2 \frac{\sum_{1}^{n}|a_{i}-a_{i+1}|}{n}$ which is quite weak.

It is easy to see that it's sufficient to prove that for $a_{1}\le...\le a_{n}$. (Otherwise, the RHS can only become bigger). We then have to show

\[\frac{\sum_{1}^{n}a_{i}}{n}\le \sqrt[n]{a_{1}a_{2}...a_{n}}+\frac{n-1}n(a_{n}-a_{1})\]  :) And the proof of this is straightforward, exactly the same as the $n=3$ case (see my first postin this thread) :wink:


I was just doing problem 1-1 from ARML 2005 Individual round and i solved it using guess and check.  But after solving it i noticed that a prime square, let n be prime, $n^2$, multiplied by a non-prime square is equal to another square.  I was wondering why this is the case and would greatly appreciate an algebrain soltuion.  By the way i would enjoy an algebraic solution to the problem.  It is: Compute the least positive prime p greater than 2 such that $p^3+7p^2$ is a perfect square.

Answer:
[hide]i got 29 for p[/hide]

i mean i got $(p^3)+7(p^2)=(p^2)(p+7)$ and i know the p+7 must equal a non-prime square number due to analysis after solving it.  But i just am missing something critical... i can feel it

[quote="demji"]also i seem to be having trouble with the tex on here... i went under the latex page and did the n^2 liek it said btu it didn't come out right[/quote]

$\$n^2$\$ comes out to $n^2$

[hide]
$p^2(p+7)$ so $p+7$ must be a perfect square, but $p>2$. So $p+7 = n^2, n > 3.$ Solutions: $(n,p)=(4,9),(5,18),(6,29)$. 29 is prime, so that's the answer.
[/hide]

yeah thx... i just went to another page did quote and saw that they did.  I haven't posted on here in a while so my skills are a bit shabby

I don't think you're missing anything. You want the first square of the form $p+7$, and you can quickly see that $p=29$ is the first.

Why is the product of two squares a square? $(x^2)(y^2)=(xy)^2$.

Why do you need to have $p+7$ a square if $p^2(p+7)$ is a square? Well, any number can be factored into primes. If a square is divisible by $p$, it must be divisible by $p^2$. So we would get a square as the quotient ($x^2/p^2) = (x/p)^2$.

oh duh... i always seem to overcomplicate things.  Just wasn't exactly thinking of $p+7=n^2$ where n is whatever satisfys that equation.  Its simple now...  Thanks

Wow... the whole time I thought you were stupid and the answer was three, but then I realized it said x^3 + 7x^2, instead of x^4 + 7x^2...

But yeah, your method is simple, and what I did to solve it :) .


i'm sorry.
p is a prime number ,how to prove 2^(p-1)-1 can be divided by p(that's Fermat's Little theory,everyone can prove) but can't be divided by p^2

...you're so nice.that's what i should say.

this problem is not true.
link: http://www.mathlinks.ro/viewtopic.php?t=217133

thanks very much.but i've never hear about Weieferich prime.i'm trying to understang.


How many different two-person sub-committees can be selected
from a committee of six people (the order of choosing the people does not matter)?

This just becomes $ \binom{6}{2}\equal{}15$ so the answer is $ \boxed{15}$.

We are calculating $\binom{6}{2}=\boxed{15}.$

[hide=Solution]
We have $6$ people for the first committee member that $5$ for the second. But since the order doesn't matter, (AB is the same as BA), we have to divide by $2$. Thus, the number of ways to select the committee is $\frac{6\times5}{2}=\frac{30}{2}=\boxed{15}$ ways.
[/hide]


proof that the number $121^n-25^n+1900^n-(-4)^n$ is divible by 2000 if n is natural.

Please Pavel, do not post you problems directly into the solved section....

Pierre.

You have to let people time to solve it....
So, let's solve it :

We verify that the conclusion holds by direct checking for $n=0$ and $n=1$. Thus, we assume that $n \geq 2$.
We have :
$S_n = 121^n - 25^n +1900^n - (-4)^n = (-4)^n  - 0 + 0 - (-4)^n \mod [125]$, so that $S_n = 0 \mod [125]$. (1)
Moreover :
$S_n = (-7)^2 - (-7)^n + 0 - 0 = 0 \mod [16]$. (2)

From (1) and (2), we deduce that $S_n = 0 \mod [16 \times 125] = 0 \mod [2000]$.

Pierre.


How do you determine which carbons are chiral centers? Identify all of them in Lovastatin, found here http://pubs.acs.org/cen/coverstory/83/8325/8325lovastatin.html

A Chiral Carbon (or Asymetric carbon atom) is one that has 4 covalent bonds joined to 4 different groups.

So would the carbon on the top ring of Lovastatin (see diagram at URL above) right above the dashed line be considered chiral?

The carbon connected to the Hydroxyl group ($-OH$)  would be a chiral carbon


Solve for $ \overline{ABCDE}$:
\[ A^5 \plus{} B^5 \plus{} C^5 \plus{} D^5 \plus{} E^5 \equal{} \overline{ABCDE}
\]

What are A,B,C,D,E?
Integers, positive integers, digits, rational numbers?

Sorry for not being specific. (I wasn't told either...) I edited my post.


Between the circles: 2 points
Between the lines: 3 points
1 circle 1 line: 12 points
17 points in total?

In other words, there are $ \dbinom22\equal{}1$ pairs of two circles, each with $ 2$ intersections, $ \dbinom32\equal{}3$ pairs of two lines, each with $ 1$ intersection, and $ \dbinom21\dbinom31\equal{}6$ pairs of a circle and a line, each with $ 2$ intersections, for a total of $ 1\cdot2\plus{}3\cdot1\plus{}6\cdot2\equal{}\boxed{17}$ intersections.

[quote="mathemagician1729"]Without actually drawing the whole thing out, how would you solve this problem?

Two circles and three straight lines lie in the same plane. If neither the circle nor the lines are coincident, what is the maximum number of points of intersection?[/quote]

I know you don't want to draw it out, but to me, that's the fastest way, unless you are awesomely amazing at these problems and can do them in 1 second.

another random way is that for 2 circles, there can only be 2 intersections
a lines gives 4 intersections
the second line has 4 intersections (circles) and another intersection.(lines)
the last line has 4 circle intersections and 2 line intersections

2+4+5+6 = 17 intersections


Let $ a,b,c$ be positive real numbers. Prove that
\[ \left(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\right)^2 \ge (a\plus{}b\plus{}c)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right).
\]

i have a solution only AM-GM
[hide]assuming abc=1 .
we know $ \frac {a}{b} \plus{} \frac {a}{b} \plus{} \frac {b}{c}\geq3\sqrt [3]{\frac {a^2}{bc}} \equal{} 3\sqrt [3]{\frac {a^3}{abc}} \equal{} 3a$
do this cyclicly 3 times and adding we get
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq (a \plus{} b \plus{} c)$(1)
but we also get in the same way
$ \frac {c}{a} \plus{} \frac {c}{a} \plus{} \frac {b}{c}\geq3\frac{1}{a}$
and adding these new relations
we get
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$(2)
(1)*(2)=>QED :) [/hide]

[hide]By AM-GM,

$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\ge3$

Also, by Chebychev, we have

$ 3(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})\ge(a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})$

Combining the two gives

$ (\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})^2\ge(a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})$
[/hide]


What is your religon and why do you believe in it?
I am a christian because God has done so much for me.

I think it would help if you included other, and because atheism isn't a religion, it would be no religion.

[b] @monkeygirl13: [/b]  :D

There's already an existing thread called: "Religion" that was created in the past. Check the link below:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=70830837&t=33938

Also, it becomes quite difficult to take your thread seriously if you don't take enough care while creating a thread which has the correct title or name.


$ A$, $ B$, $ C$ are three points on a circle and triangle $ ABC$ is acute-angled. $ AD$ is drawn parallel to $ BC$ and $ CD$ is drawn parallel to $ AB$. Prove that $ D$ cannot lie on the circle. ($ D$ is the point of intersection of the two parallels)

See, i think you would have to examine the case if it was. But then i dont know how to show that the lines are not parallel. also, if it was on the circle, i think the only way to have it is if the triangle is not acute.

Think of it as like: if D is on the circle, then you've got a cyclic quadrilateral. Also, you have parallel sides, so it must be some kind of parallelogram, and if it's cyclic quad, it's a rectangle, so ABC is right-angled, which gives you contradiction.

Thanks for that. See, my problem was that i wasn't considering special quadrilaterals. I drew a diagram of the question and then put d on the circle, so the 'inscribed' quadrilateral looked nothing like a parallelogram. And infact, in the exercise i was working on there was a question to show that a parallelogram was a rectangle if it was inscribed ><"

Cheers


$S_{x}O_{y}$
What is x and y if S is 40% in 1 mole of compound?

I think $x=2$ and $y=3$  :maybe: that means we have to write $S_{2}O_{3}$...

Hello,
    It is quite simple.The compound is SO3 - Sulphur trioxide.Taking the atomic weights of sulphur and oxygen as 32 and 16 respectively,sulphur clearly constitutes 40% by weight in one mole.
Sathej

Oh sorry :oops:  this proves that my chemistry is still very weak...


geometry
circumcircle
trigonometry
geometry unsolved

Let $ A$ and $ I_{1}$ be the circumcenter and $ A$-excenter of triangle $ ABC$, and $ A_{1}$ the intersection of $ OI_{1}$ and $ BC$. Similarly define $ B_{1}$ and $ C_{1}$. The lines $ AA_{1}$, $ BB_{1}$ and $ CC_{1}$ concur at the Schiffler point $ S$.

I have already solved this problem, but it isn't happy if I post solution of its. I hope somebody want to semina it with me!

See the FG papper by Lev Emelyanov and Tatiana Emelyanova, [i]A note on the Schiffler Point[/i]: [url]http://forumgeom.fau.edu/FG2003volume3/FG200312index.html[/url]

However, I will not lock this thread since I'm also interested in a synthetic solution.

Oh! My solution in this case is a synthentic! I only use Xeva sines theorem

[img]http://i225.photobucket.com/albums/dd18/nktp/Schiffler.png[/img]
Denote $ S$ be the Schiffler point.
We will prove that $ A,S,A_{1}$ are colinear!
We have:
$ \frac{sin\widehat{AJG'}}{sin\widehat{OJG'}}.\frac{sin\widehat{JOG}}{sin\widehat{AOG}}.\frac{sin\widehat{OAA_{1}}}{sin\widehat{JAA_{1}}}\equal{}\frac{sin\widehat{IJG'}}{sin\widehat{MJG'}}.\frac{sin\widehat{MOG}}{sin\widehat{AOG}}.\frac{sin\widehat{OAA_{1}}}{sin\widehat{I_{1}AA_{1}}}$
$ \equal{}\frac{\frac{2S_{IJG'}}{IJ.JG'}}{\frac{2S_{MJG'}}{MJ.JG'}}.\frac{\frac{2S_{MOG}}{MO.OG}}{\frac{2S_{AOG}}{AO.OG}}.\frac{\frac{2S_{OAA_{1}}}{OA.AA_{1}}}{\frac{2S_{I_{1}AA_{1}}}{I_{1}A.AA_{1}}}\equal{}\frac{2MJ}{IJ}.\frac{AO}{2MO}.\frac{OA_{1}}{I_{1}A_{1}}.\frac{I_{1}A}{OA}$
$ \equal{}\frac{2MJ}{BJ}.\frac{BO}{2MO}.\frac{OM}{I_{1}P}.\frac{I_{1}P}{OAsin\frac{A}{2}}\equal{}2sin\frac{A}{2}.\frac{1}{2sin\frac{A}{2}}\equal{}1$
Apply Xeva sines of the triangle $ AOJ$ we have $ AA_{1},OG,JG'$ concurent.
It mean $ A,S,A_{1}$ are colinear!
Now $ AA_{1},BB_{1},CC_{1}$ concurent at Schiffler point $ S$

Dear Mathlinkers,
an article concerning the “Schiffler point and a result by L. and T. Emelyanov’’ has been put on my website
with a new approach.
http://perso.orange.fr/jl.ayme    vol. 9
You can use Google translator
Sincerely
Jean-Louis

I'm pretty sure there's a complex number solution and have located all the necessary points, though I can't get the expression for the schiffler point to be symmetric. Could some one please tell me how to get k symmetric where:

$\frac{k(a+b+c)+ab}{-bc-ac}$ = $\overline{\frac{k(a+b+c)+ab}{\overline{-bc-ac}}}$

(and a,b,c,-ab,-bc,-ac lie on the unit circle.)

*doing so proves the statement.


quadratics
number theory proposed
number theory

For a prime p and a k such that d=(p-1;k)>1, let $n_k(p)$ denote the smallest k-th power non-residue mod p. Prove that:
a. $n_k ( p ) < \sqrt p + \frac{1}{2} $ for all p.
b. $n_k ( p ) = o ( p ^ { \frac {1} { e ^ {1 - \frac{1} {d} } } + \epsilon } )$.

here is the best bound for nonquadratic residues as a form of the extended riemann hypothesis(not proven): $n_2(p) < 2(log(p))^2$. maybe someone here is so clever  :D

Peter

[Edit: nonsense removed]

for a) (i can only prove $n_k(p)\leq |\sqrt{p}|+1$):
let x be a k-th power nonresidue. let M be set of $\{a+bx\}$ where $a,b=0,1,...,|\sqrt{p}|+1$(or something in that direction). then there are more than p elements in M, thus two are equal mod p: $a+bx=a'+b'x$, this means $x=\frac{a'-a}{b-b'}$. both numbers $a'-a$ and $b-b'$ are from $-|\sqrt{p}|-1,-|\sqrt{p}|,...,-1,1,...|\sqrt{p}|+1$ and one of them is a nonresidue. now if they are both positive or both negative, so wlog positive(just multiply both by -1 in the latter case), thus result. thus one of them is negative the other one positive. if -1 is a k-th power residue, this is problem, just multiply both sides by -1(since in that case -x is a k-th power residue as well). thus -1 must be a k-th power nonresidue. but if the result is wrong, all of the numbers $1,...,|\sqrt{p}|+1$ are k-th power residues, thus all of the number $-|\sqrt{p}|-1,...,-1$ are k-th power nonresidues(and since 2 is a k-th power residue, we may extend that result to the doubled range). now consider the integer $-|\sqrt{p}|$: it is a nonresidue, but its square(a k-th power residue) is in one of the numbers $-2|\sqrt{p}|-1,...,-1$, contradiction.

i think one can with a litle better analysis get the desired bound.

Peter

here it is: if $1,...,|\sqrt{p}|+1$ are all nonresidues, we must have that -1 is a nonresidue because we can write every number as a quotient of numbers from $-|\sqrt{p}|-1,...,-1,1,...|\sqrt{p}|+1$. we can now conclude as before.
this even shows the bound $n_k(x)\leq |\sqrt{p}|$, a slightly better one than desired :D

Peter

there's a little mistake in it, as the counterexample p=7, k=2 shows...the problem is that not all of the numbers $-2|\sqrt{p}|-1,...,-1$(only the even ones) need be k-th power nonresidues if -1 is a k-th power nonresidue. do you see how we can handle that case?

Peter

get the correct result for a)
Suppose -1 is a k power mod p
Let m be the largest number s.t. 1..m are not  powers.
Let g be a primitive root so not a power.
Consider the numbers g,2g,3g..mg mod p be a_1<a_2,..<a_m
The difference between any of them is ug mod p or p-ug mod p so not k power residue and hence a_i+1-a_i>=m+1 so a_m>=a_1+m^2-1>=m^2+m.
So m^2+m<p so m<sqrt(p)-1/2.
Suppose -1 is not a k power mod p. Then k is even and is sufficient to consider k=2.
Then 1..m are quadratic p-m..p-1 are not and so i(p-j) is not quadratic .
Now if m>sqrt(p)-1/2 then choose i=j=[sqrt(p)] and so p-[sqrt(p)]^2 is non-quadratic.
Now if p-1/2<[sqrt(p)] this number is smaller than m and we are done.
If not choose i=j-1=[sqrt(p)] and again get our result
QED


If $ f(x) \equal{} \frac{{e^x  \minus{} e^{ \minus{} x} }}{2}$
,then find $ f^{ \minus{} 1} (x)$ in terms of $ x$

$ \left(f(x)\right)^2\equal{}\frac{e^{2x}\plus{}e^{\minus{}2x}\minus{}2}{4}$
So, $ \sqrt{\left(f(x)\right)^2\plus{}1}\equal{}\sqrt{\frac{e^{2x}\plus{}e^{\minus{}2x}\plus{}2}{4}}\equal{}\frac{e^x\plus{}e^{\minus{}x}}{2}$.
Therefore,
$ f(x)\plus{}\sqrt{\left(f(x)\right)^2\plus{}1}\equal{}e^x$.
Thus, $ x\equal{}f^{\minus{}1}(x)\equal{}\ln\left(f(x)\plus{}\sqrt{\left(f(x)\right)^2\plus{}1}\right)$.


A hydrofoil made a round trip of 144 km in 4 h.  Because of head winds, the average speed on the returning was 15 km/h less than the average speed going out.  Find the two speeds.

[hide]ok, so the harmonic mean of x+15 and x-15 is 36.
$\frac{x^{2}-225}{2x}=36$
$x^{2}-72x-225=0$
x=75
So one way they went 45, the other way they went 30.[/hide]


Prove that the ecuation $ 8x^4\plus{}4y^4\plus{}2z^4\equal{}t^4$ doesn't have any roots in positive integers.

[hide]Trivial by infinite descent[/hide]

I'll just write it out.

[hide]Let $ 2^a$ be the largest power of $ 2$ dividing $ x,y,z$.  Then it is obvious that $ 2^a$ divides $ t$.  Then we have the equation

$ 8\left(\frac{x}{2^a}\right)^4\plus{}4\left(\frac{y}{2^a}\right)^4\plus{}2\left(\frac{z}{2^a}\right)^4\equal{}\left(\frac{t}{2^a}\right)^4$

Since $ a$ is maximal, we must have at least one of $ \frac{x}{2^a},\frac{y}{2^a},\frac{z}{2^a}$ odd.  Consider the equation $ \mod 16$.  The only fourth powers $ \mod 16$ are $ 0$ and $ 1$.  Notice that

$ 8\left(\frac{x}{2^a}\right)^4\plus{}4\left(\frac{y}{2^a}\right)^4\plus{}2\left(\frac{z}{2^a}\right)^4$

is even, so $ \frac{t}{2^a}$ is even.  Thus, we must have

$ 8\left(\frac{x}{2^a}\right)^4\plus{}4\left(\frac{y}{2^a}\right)^4\plus{}2\left(\frac{z}{2^a}\right)^4\equiv 0\pmod{16}$

A quick analysis yields no solutions if at least one of $ \frac{x}{2^a},\frac{y}{2^a},\frac{z}{2^a}$ is odd.  The only number that when divided by it's highest power of two is even is $ 0$, thus, there are no solutions in positive integers.[/hide]


How many factors of 3003 are prime numbers?

We first find the prime factorization of $ 3003.$ (This would be a very short post if I only showed the prime factorization, so I'll "work" through it.)

We see $ 3003$ is divisible by $ 3.$ Factor it out to get $ 3\times1001.$ Now we see $ 1001$ isn't divisible by $ 3$ or $ 5,$ so we divide by $ 7$ to get $ 3\times7\times143.$ $ 143$ isn't divisible by $ 3, 5$ or $ 7,$ so we divide by $ 11$ to get $ 3\times7\times11\times13.$ Now it becomes clear there are $ \boxed{4}$ prime factors of $ 3003.$

$3003$ is divisible by $3$ because the sum of it digits is divisible by $3$.

$3003=3\cdot1001$

$1001$ is divisible by $11$ because the difference of the sum of alternating digits is divisible by $11$.

$3003=3\cdot1001=3\cdot11\cdot91$

$91$ is $7\cdot13$ so it's divisible by $7$ and $13$.

$3003=3\cdot1001=3\cdot11\cdot91=3\cdot11\cdot7\cdot13$

The prime factors of $3003$ are $3$, $7$, $11$, $13$ so the answer is $\boxed{4}$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a finite group and $U$ a subgroup such that $U\neq G$. Prove that
\[ G\neq \displaystyle\ \cup_{g\in G} gUg^{-1} \]
By absurd, I have tried to construct a sequence $(g_n)$:
Let be $g\in G$, $g_0=g$ and $g_i$ is defined by $g_i=g_{i+1} u_i g_{i+1}^{-1}$. $G$ is finite hence there is $n$ such that $g_n=g_0$, I have found that for every $g_i$, $g_0$ is equal to a product of element of $U$ and $g_i$ but I have not found explicit formula which could solve the problem (taking $i=n$).
I think I'm not too far and there is certainly a faster solution with action group I guess ...

Another methode can be found in http://www.mathlinks.ro/Forum/viewtopic.php?highlight=conjugate&t=22729 .


A caterer offers five different appetizers, three different drinks and four different sandwiches. How many combinations of two appetizers, two drinks and two sandwiches can Scott choose for his party?

There are $ 5 \times 4/2 \equal{} 10$ ways to choose the two appetizers, $ 3 \times 2/2 \equal{} 3$ ways to choose the two drinks, and $ 4 \times 3/2 \equal{} 6$ ways to choose the two sandwiches. Thus, there are $ 10 \times 3 \times 6 \equal{} 180$ ways to choose two appetizers, two drinks and two sandwiches.

a little touchup :)

What PerpetualMotion did was use [url=http://www.artofproblemsolving.com/Wiki/index.php/Combinations]combinations.[/url]
Since he wants to choose $ 2$ out of the $ 5$ possible, there are $ \binom52\equal{}\frac{5!}{3!2!}\equal{}10$
Since he also wants $ 2$ out of the $ 3$ possible, there are $ \binom32\equal{}\frac{3!}{1!2!}\equal{}3$
Similarly, he wants $ 2$ out of the $ 4$ possible, there are $ \binom42\equal{}\frac{4!}{2!2!}\equal{}6$.
Since these are independent of each other, we multiply to get $ 10\times3\times6\equal{}\boxed{180}$


Witam, mam pytanie ile zadań udało wam sie zrobić na tegorocznym OM? Ja mam 4 lub 5 w zależności czy mi uznają kombinatorykę.

Zrobiłem cztery (1,2,3,5). Mógłbym zrobić jeszcze czwarte, gdyby nie to, że przez większość drugiego dnia robiłem nierówność.


algebra
polynomial
trigonometry
MIT
college
calculus
calculus computations

So I've been trying to get sleep-deprived freshmen with short attention span to realize the difference between convergent sequences and convergent series. Many of them learn by associating mathematical procedures with keywords, which works OK except when a keyword has two meanings. I imagine that things would be slightly better if I (and [url=http://www.stewartcalculus.com/media/2_home.php]The Book[/url]) talked about "convergent sequences" and "summable series". Is there a reason not to do so?

Another example: the remainder of a series is defined as the difference between its total sum and its partial sum. Except when the series is a Taylor series of $f$, in which case the remainder is the difference between $f(x)$ and a partial sum. Question: do I emphasize this point in tomorrow's lecture, or hope that nobody notices?

Now I notice that "the $n$th degree Taylor polynomial" $T_{n}(x)=\sum_{i=0}^{n}\frac{f^{(i)}(a)}{i!}(x-a)^{i}$ does not necessarily have degree $n$. I have to agree with my students: math does not make sense.

"Summable"... hmm. It's got possibilities.

Just this morning, in dealing with a problem asking for an estimate of the accuracy of the 3rd degree Taylor polynomial about zero for $f(x)=\tan x,$ [Edition 5e, Early Trancendentals, section 11.2, page 783, problem #17], I invoked the fact that that 3rd degree polynomial was equal to its 4th degree polynomial (doing that to get a better remainder expression.)

As for the mixed meaning of "remainder" - I don't think I'd ever worried about that until you just pointed it out.

I also tried to use and advocate "summable" when teaching from that awful Book.
Success was limited, as it is hard to introduce terminology contrary to the textbook.
Advocacy would have been better spent on convincing department to eliminate the textbook.

[quote="mlok"]I am not going to say "summable" if my textbook says "convergent". What text would you rather use for Calc II?[/quote]

I don't know much about the current US textbook market, but what was wrong with the old(est) editions of the MIT calculus book by Thomas?


induction
integration
calculus
calculus computations

If $ I_{n}$ denotes $ \int_{0}^{\infty} t^{n}e^{\minus{}at}\ dt$, where $ a$ is a positive constant, prove that $ aI_{n} \equal{} nI_{n\minus{}1}$ provided that $ n>0$. Use the method of mathematical induction to deduce that, if $ n$ is a positive integer, $ I_{n} \equal{} \frac{n!}{a^{n\plus{}1}}$.

Just integrate by parts, in pretty much the only reasonable way, and see what happens.

Thanks Kent. I get the reduction formula by letting $ u\equal{}t^{n}, dv\equal{} e^{\minus{}at}\ dt$ and integrating by parts. I just thought that the mathematical induction bit might interest someone.


Circles $ c_1,c_2$ with respective centers $ O_1,O_2$ intersect at $ P$. Circle $ c_2$ intersects $ O_1O_2$ at $ A$. Prove that there exists a circle which is tangent to $ c_1$ at $ P$ and tangent to $ O_1O_2$ at $ A$ if and only if $ \angle O_1PO_2\equal{}90^{\circ}$

[hide][quote="Johan Gunardi"]Circles $ c_1,c_2$ with respective centers $ O_1,O_2$ intersect at $ P$. Circle $ c_2$ intersects $ O_1O_2$ at $ A$. Prove that there exists a circle which is tangent to $ c_1$ at $ P$ and tangent to $ O_1O_2$ at $ A$ if and only if $ \angle O_1PO_2 \equal{} 90^{\circ}$[/quote]
It think this problem is rather easy. We make a similar enunciation:
[quote="mr.danh"]Circles $ c_1,c_2$ with respective centers $ O_1,O_2$ intersect at $ P$. Circle $ c_2$ intersects $ O_1O_2$ at $ A$. The perp line to $ O_1O_2$ at A cuts $ O_1P$ at B. Prove that AB=BP if and only if $ \angle O_1PO_2 \equal{} 90^{\circ}$.[/quote]
Since $ O_2A\equal{}O_2P, \angle BAO_2 \equal{} 90^o$, hence result.
[/hide]


AMC
AIME
function
parameterization
linear algebra
matrix
2009 AIME I

The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from $ 1$ to $ 15$ in clockwise order. Committee rules state that a Martian must occupy chair $ 1$ and an Earthling must occupy chair $ 15$. Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling. The number of possible seating arrangements for the committee is $ N\cdot (5!)^3$. Find $ N$.

[hide="solution"]
We see the number of ways to seat people with only regard to race is equal to N. Starting with chair 1 going clockwise, we see that the seating must go as follows: group of Martians, group of Venusians, group of earthlings, group of Martians, etc. We now divide this into cases based upon how many groups of each there are.
Case 1: 1 group of each race.
Each group must be the 5 people from each race sitting together. Thus, there is only  1 arrangement in this case.
Case 2: 2 groups of each race.
Since each group has at least one person, give each group one person from the given race. There are 3 people left from each race to put into the groups. With balls and urns(3 balls and 1 urn), we find that for each race, there are 4C1=4 ways to place the people into groups. Thus, there are 4^3=64 arrangements in this case.
Case 3: 3 groups of each race.
Similar to above. There are 6^3=216 arrangements.
Case 4: 4 groups of each race.
Similar to above. There are 4^3=64 arrangements in this case
Case 5: 5 groups of each race.
Give each group one person. There are no people left, so there is only 1 in this case.
Thus, there are $ 1\plus{}64\plus{}216\plus{}64\plus{}1\equal{}\boxed{346}$ seating arrangements
[/hide]

ARrggg.... :wallbash_red: I got this one wrong because, with my inherent anthropocentric tendencies, I labelled the Earthling in seat 15 as first on my work. I was going to get the right answer, but realising that a Martian came immediately "second", I somehow decided that I would be unable to put any more Earthlings in the first group and essentially gave myself an extra restriction.

This didn't work, but I assumed that any seating of the Venusians produced a unique seating for Earthlings and Martians.  Thus, $ \binom{12}{5}\equal{}792$.  How could we use this somehow to get 346?  I did notice that 792/2=346...

[quote="t0rajir0u"]Can anyone generalize?  The two main constraints here, if considered separately, lead to a linear recurrence (with only the seating restraint) and binomial coefficients (with only the five-of-each-race constraint); the two constraints together are troublesome.  In particular, the generating function (if $ 5$ is replaced by $ n$) is likely to be complicated (although probably still algebraic).[/quote]
Hmmm...well wouldn't it be, if n is number of people per race, m is number of races, $ (n!)^m(\sum_{i \equal{} 0}^n (C(n,i))^m)$

Note to self: #10 on the AIME probably will never involve tricky bijections. :ewpu:

[quote="AIME15"]This didn't work, but I assumed that any seating of the Venusians produced a unique seating for Earthlings and Martians.  Thus, $ \binom{12}{5} \equal{} 792$.  How could we use this somehow to get 346?  I did notice that 792/2=346...[/quote]

In fact, $ 792/2\equal{}396$. Oh well.

[quote="Yongyi781"][quote="AIME15"]This didn't work, but I assumed that any seating of the Venusians produced a unique seating for Earthlings and Martians.  Thus, $ \binom{12}{5} \equal{} 792$.  How could we use this somehow to get 346?  I did notice that 792/2=346...[/quote]

In fact, $ 792/2 \equal{} 396$. Oh well.[/quote]

Oh. Wow. Fail.

This is exactly the same type of arithmetic I used yesterday...

[quote="GANITEE"]Seating of V's doesn't mean unique seating of E's and M's. And why 12? V's have 13 places to sit. It's just a coincidence that 792/2 = 346.

[quote="AIME15"]This didn't work, but I assumed that any seating of the Venusians produced a unique seating for Earthlings and Martians.  Thus, $ \binom{12}{5} \equal{} 792$.  How could we use this somehow to get 346?  I did notice that 792/2=346...[/quote][/quote]

not a coincidence as that statement is false as well. (792/2=396, as someone earlier has posted).

[quote="AIME15"]This didn't work, but I assumed that any seating of the Venusians produced a unique seating for Earthlings and Martians.  Thus, $ \binom{12}{5} \equal{} 792$.  How could we use this somehow to get 346?  I did notice that 792/2=346...[/quote]

You can't have 2 Venusians (or any other race'lings, for that matter) sitting adjacent to each other with one empty seat between them, because no one else other than a Venusian can sit between them.

Thus not all arrangements of the 5 Venusians will work.

[quote="pascal12"]I paused on this one because I thought that seat 15 was to the left of 1 and therefore N would be 0... it was only when I came back to it I realized my mistake...  What a stupid mistake[/quote]

That's exactly what I did! Unfortunately, I still don't understand why its wrong  :lol:

they are assuming that martians/venusians don't sit backwards :/

[quote="E^(pi*i)=-1"]What sort of generalization are you thinking of?  It's clear from the method that the answer for $ r$ races and $ n$ of each race will be the sum of the $ r$th powers of the $ n$th row of Pascal's triangle.[/quote]
Alright, so it's actually [i]quite[/i] hard to figure out this sum in closed form, which is unfortunate.  I asked because this relates to a conjecture of mine about "local" vs. "global" patterns in counting words.  It's the case that words that avoid "local" patterns (that is, that do not have particular substrings) are described by walks on a particular graph, hence by linear recurrences.  On the other hand, some "global" patterns such as the number of two different kinds of letters used being the same can't be described in this manner and seem to be much more difficult to understand in the general case, although certain types of global patterns such as "use a number of $ A$s divisible by $ 3$" are still manageable using the graph technique.

Edit:  Alright, so not [i]much[/i] more.  Global patterns that require keeping track of an unbounded parameter can be described by walks on [i]infinite[/i], rather than finite, graphs (that is, lattices).  Here, the relevant lattice is the [url=http://en.wikipedia.org/wiki/Eisenstein_integer]Eisenstein integers[/url]:  we are counting walks on a triangular lattice with no right turns that begin and end at the origin.

It really depends on if you meant 'generalization' to mean

a) "A person from Planet $ i$ can only sit to the left of a person from Planet $ i$ or $ i\plus{}1$ (numbered cyclically)" (in which case the answer is the sum of the $ r$th powers of the $ n$th row of Pascal's triangle)
b) "A person from Planet $ i\minus{}1$ cannot sit to the left of a person from Planet $ i$ (numbered cyclically)" (which is much harder)

[quote="Sly Si"]Rats, I was hoping this problem would actually be something self-referential about the AIME committee.  But the real committee only has 10 people, and as far as I know they're all Earthlings.[/quote]

are you sure about that? :wink:
@AIME15: we've all done something like that once LOL

[quote="herefishyfishy1"]It really depends on if you meant 'generalization' to mean

a) "A person from Planet $ i$ can only sit to the left of a person from Planet $ i$ or $ i \plus{} 1$ (numbered cyclically)" (in which case the answer is the sum of the $ r$th powers of the $ n$th row of Pascal's triangle)
b) "A person from Planet $ i \minus{} 1$ cannot sit to the left of a person from Planet $ i$ (numbered cyclically)" (which is much harder)[/quote]
I'd like to consider both.  The first is already difficult enough to evaluate in closed form, but I wonder if anyone has any thoughts on the second, which is just as easy to translate into the language of walks on lattices, but that doesn't give any indication of its difficulty.

Is there a generating function solution to this problem?

I was considering the number of seating positions that each race contributes; for example, a Martian contributes 2 ways to seat other Martians, one way to seat an Earthling, and one way to seat a Venusian. However, I don't know how to incorporate the 5-of-each-race factor into this.

The best I can say is that the answer is given by the $ a^5 b^5 c^5$ coefficient of

$ \left[ \begin{array}{ccc} a & b & 0 \\ 0 & b & c \\ a & 0 & c \end{array} \right]^{15}_{1,3}$

but in general extracting the generating function $ f_{n,n,n}$ from a multivariate generating function $ f_{n,m,k}$ is difficult.  That's why I asked about a generalization.

Sorry, but what exactly do the $ 15$ and the $ 1,3$ there mean?
Also, where exactly does that come from?

You'll have to forgive me, I'm just starting to figure out advanced combinatorics...

$ A^{15}_{1,3}$ is the entry in the first row and the third column of the fifteenth power of the matrix $ A$.  That computation comes from thinking about a permissible seating as a walk on the graph with vertices $ M, V, E$, where the permitted directed edges are $ M \to M, M \to V$, etc. starting from $ M$ and ending at $ E$ (hence the $ 1, 3$).  The matrix I wrote down is a weighted [url=http://en.wikipedia.org/wiki/Adjacency_matrix]adjacency matrix[/url] that records which vertices such a walk travels to, and the fifteenth power counts walks of length $ 15$.  (Also, I think I actually want the coefficient of $ a^4 b^5 c^5$ in the fourteenth power.)

Ah, that's really interesting. I didn't associate this with graph theory at first, but your explanation makes a lot of sense. Thanks!

[quote="w squared parabola"]Sorry to revive the post,
How can you have only 1 group of each race? Since a Martian need to occupy seat 1 and an Earthling need to occupy seat 15. Then the formation will look like this (starting with seat 1): MMMMMVVVVVEEEEE. Which the rule states it is not possible. I am missing something? Thanks.[/quote]

That arrangement is possible. When you visualize that as a circular pattern, the last Martian is to the right of the first Venetian.

If you can, can you give me a diagram to show it? I don't know how to put up a diagram, so I cannot put one up.

sorry for yet another revive
does anyone know how to find a bijection between this problem and some graph which is easily enumerable by deck/hand enumerator functions? The computation of the series expansion doesn't need to be nice, since we could use a recursive equation to find the hand counts.

The problem says you cant have $MVE$ then the solution is contradictory?

[hide=Nice Problem]
The question places restrictions using "left", but we really know (based on the fact that E is 15, M is 1) that the question means that:
M cannot be right of V
V cannot be right of E
E cannot be right of M
Thus, the question is asking for the number of M's, V's, and E's to a "sequence" of:
M, V, E, M, V, E,......M,V,E.
Note that each cycle must contain {M,V,E} and the sequence should start with M and end with E (since 1 is M and 15 is E. Thus, we have converted the problem into 1D, where we basically converted the circumference into a "string" with 1 on the left, and 15 on the far right.

We further divide this into cases.
Case 1: M,V,E: 1 way (5,5,5)
Case 2: M,V,E,M,V,E: $(\binom{4}{1})^3$
Case 3: cycles 3 times: $(\binom{4}{2})^3$
Case 4: cycles 4 times: $(\binom{4}{3})^3$
Case 5: cycles 5 times: $(\binom{4}{4})^3$
Careful: cannot exceed 5 cycles since each of M, V, E must be a positive solution or else it will not follow the restrictions given. For example, if V is 0 for some step in a cycle, then an M would be left of E which is not allowed.

Add all these, and we get $346.$
[/hide]

Observe that the problem is equivalent to finding the number of possible seating arrangements of a line of people satsfiying the conditions, since the end points are already given (Those being the Earthling and the Martian). Now, observe that any time there is a change in races from one seat to another (say Earthling to Martian, et. al), We must either have $MV, VE,$ or $EM$. Since the first person is a Martian, we start with $M$. Now we go by casework depending on how many $MVE$s there are (that is, the number of times we return to a Martian).

[b]Case 1: [/b] $MVE$
Clearly, there is only $1$ way to do this.

[b]Case 2: [/b] $MVEMVE$
Observe for each race, we must distribute $5$ of that race's entity to $2$ strings of people, that is $x+y=5$. There are $4$ ways to do this (namely $(1,4),(2,3),(3,2),(4,1)$ so we have $4^3$.

[b]Case 3: [/b] $MVEMVEMVE$
Similarly, we get the equation $x+y+z=5$ which has $6$ solutions. Thus, we have $6^3$.

[b]Case 4: [/b] $MVEMVEMVEMVE$
Once more, we get $4$ solutions. Thus, $4^3$

[b]Case 5: [/b] $MVEMVEMVEMVEMVE$
This is dual to case 1, $1$ solution.

Summing, we get $346$. Thus, there are $346*(5!)^3$ possible seating arrangements (as we can permute any of the races) and thus, $\boxed{N=346}$

Wait what the question says no earthling can sit immediately to the left of a martian but seat 15 is to the immediate left of seat 1??

No if you look from the committee's perspective, seat 15 is to the immediate right of seat 1, because they are all facing inwards towards the center of the circle.

Great - thanks! I now understand the problem, and the solution too (due to post #2) :)

Can someone please tell me what is wrong with the following solution?

Assign Mars the number $0$, Venus $1$, and Earth $2$. Label the chairs in clockwise order $C_{1}, C_{2}, \ldots, C_{15}$. Let $x_{i}=C_{i+1}-C_{i}$ for $1\leq i\leq 14$. Then $\sum_{i=1}^{14}x_{i}\equiv 2\pmod{3}$ based on the conditions $C_{1}=0$ and $C_{15}=2$; also, each $x_{i}$ is either $0$ or $1$ (difference of $2$ would violate the conditions that no Earth can sit immediately to the left of Mars, etc.) The number of different sequences now possible is $N$, as there are $(5!)^{3}$ ways for the writers to sit down under each sequence.

Now consider the generating function $(1+x)^{14}$, where we choose the $1$ if the difference is $0$, and $x$ if it's $1$. Our answer would then simply be the sum of all coefficients of powers $x^{n}$ where $n\equiv 2\pmod{3}$. By a roots of unity filter, this sum equals \[N=\dfrac{2^{14}-1}{3}=5461.\]

[quote name="r00tsOfUnity" url="/community/p26850522"]
Can someone please tell me what is wrong with the following solution?
[/quote] I think you're ignoring the condition there are exactly five writers from each planet.


Let $ S$ denote the set of subsets of $ \mathbb{N}$; we say that $ a \le b$ for $ a, b \in S$ if $ a$ contains $ b$.  A [i]chain[/i] in $ S$ is a [url=http://en.wikipedia.org/wiki/Total_order]totally-ordered[/url] subset under this order.

Show that $ S$ contains an uncountable chain.  (Hints are available in the post in the description.)

Just randomly thought about this .. while none other than preparing for prom

[hide="more hint"]
Correspond each real number $ r$ with a certain subset $ s(r)$ of $ \mathbb{Q}$ [which of course bijects to $ \mathbb{N}$] with $ r_1 >  r_2 \in \mathbb{R} \Longleftrightarrow s(r_1) \subset s(r_2)$
[/hide]


Let $y_1,y_2,\ldots, y_n$ and $a_1,a_2,\ldots, a_n$ be positive reals, with the additional condition that $\sum_i a_i=1$.  Prove that

$\prod_i y_i^{a_i}\leq \sum_i a_iy_i$

sorry but i have a problem, if one of the $a_i$s be a not rational number then what is the meaning of $y_i^{a_{i}}$?


Can't it be proved by contradiction?? I mean,
$ \frac{x^2}{(x\minus{}1)^2}\plus{}\frac{y^2}{(y\minus{}1)^2}\plus{}\frac{z^2}{(z\minus{}1)^2}<{1}$
$ \Leftrightarrow\frac{xy}{(x\minus{}1)(y\minus{}1)}\plus{}\frac{yz}{(y\minus{}1)(z\minus{}1)}\plus{}\frac{zx}{(z\minus{}1)(x\minus{}1)}<1$
$ \Leftrightarrow\frac{3}{(x\minus{}1)(y\minus{}1)(z\minus{}1)}<{1}$
$ \Leftrightarrow{(x\minus{}1)(y\minus{}1)(z\minus{}1)}>{3}$
$ \Leftrightarrow{xyz\plus{}x\plus{}y\plus{}z\minus{}xy\minus{}yz\minus{}zx\minus{}1}>{3}$
$ \Leftrightarrow{xyz\plus{}x\plus{}y\plus{}z\minus{}xy\minus{}yz\minus{}zx}>4$
Now, $ {x\plus{}y\plus{}z}\geq\frac{3(\sqrt{3}\minus{}1)}{2}$, which implies $ {xy\plus{}yz\plus{}zx}\leq\frac{(x\plus{}y\plus{}z)^2}{3}$.
Thus putting the marginal values to obtain $ {xyz\plus{}x\plus{}y\plus{}z\minus{}xy\minus{}yz\minus{}zx}<{4}$......is my process $ OK$ :?:


Find the 100.000th number from this increasing sequence of numbers:

$ 123456789$
$ 123456798$
$ 123456879$
$ 123456897$
$ 123456978$
$ 123456987$
......
$ 987654321$

PS: I'm not sure this is  a number theory problem...
Where was this problem proposed?

I didn't understand anything  :D

can you explain this step

100000-2(8!)=19360
Sequence is 3--------.


Suppose $ (X_n)_{n \ge 1}$ are independent, normally distributed with $ E(X_n)\equal{}a_n$ and $ Var(X_n)\equal{}b_n^2$.
Show that $ \sum_{n\equal{}1}^{\infty}X_n$ converges almost surely iff $ \sum_{n\equal{}1}^{\infty}a_n$ converges and $ \sum_{n\equal{}1}^{\infty}b_n^2$ converges.

If the series converge a.s., then the sequence of partial sums converges in distribution, so it is enough to look at the characteristic function of the weak limit, which will give you the convergence for the means and variances.
The converse is a standart exercise: look at $ \sum (X_i\minus{}\mathbb{E}X_i)$:
$ \mathbb{E}\left(\sum (X_i\minus{}\mathbb{E}X_i)\right)^2\equal{}\sum \mathbb{E}\left(X_i\minus{}\mathbb{E}X_i\right)^2<\infty$, so $ \sum (X_i\minus{}\mathbb{E}X_i)$ converges absolutely, hence $ \sum X_i<\infty$ a.s. iff $ \sum a_i<\infty$.


Determine all pairs of positive integers (a, b) such that:
$ \frac{a^2}{2ab^2 \minus{} b^3 \plus{} 1}$ is a positive integer.
I found that for $ b\equal{}1$ and $ a$ even is a solution. And for $ a\equal{}7$ and $ b\equal{}2$ is also a solution.
But I think there are more solutions.

Aditional solution for $ b\equal{}2$ and $ a\equal{}1$

IMO 2003 problem 2  :wink:
[url]http://www.mathlinks.ro/viewtopic.php?t=94[/url]


calculus
derivative
function
symmetry
limit
algebra
functional equation

Find some $f(x)$ so that
$f(x) = \frac{1}{2}f(2x-1) + \frac{1}{2}f(2x+1) + C$.

We can consider $C=0$ and $C\ne 0$ separately.

This is a little tangent to some research that a group at my summer program is working on. If you post a solution and we use it in a paper (which may reach publication), we will of course give you full credit for it.

edit: A couple of further comments -- the important result is proving that such $f$ exist when $C \ne 0$. We don't care what is $C$ (i.e., a solution for a particular $C$ is fine). Also, we only really need proof that such an $f$ exists, we don't need to see it. Taking the derivative and examining limiting behavior suggests that $f(x)$ will behave like $\ln(x)$ for large $x$ (and we see in that case we want $C = -\ln(2)$).

The $C=0$ case is still interesting but not critical. There are probably several, if not a large class, of solutions ($f(x)$ constant works, and linear functions too probably).

Just 2 quick questions:

1) Do you want your equation to hold on the entire real line?

2) Do you want $f$ to belong to some nice class (continuous, smooth, analytic, etc.)?

1) yes (or $\mathbb{R}^+$ is ok and then make the argument in abs value).

2) It doesn't matter. If you want to assume a particular condition so that you can do a certain thing, it is probably fine. For example I wanted to make some characterization of its behavior for large $x$ and I took a derivative. I also tried Fourier transform but it was not so successful an approach (at first glance at least).

Also note this is a particular case of a more general problem. We want to show there exist such functions that satisfy

$f(x) = C + \sum_j c_jf(2x-j)$

This is a property my roommate defined as "almost refinable" when $C \ne 0$. When $C=0$ it reduces to the definition of 2-refinable function. We are trying to answer the following question: What are the functions that are almost 2-refinable, but are not of the form $g(x) + k$ where $g(x)$ is refinable?

It is not too difficult to see that such a function exists only for $\sum_j c_j = 1$. I have found
$f(x) = \ln(x)$
is such a function, and then $C = -\ln(2)$. We have $c_0 = 1$ and all others 0.

I find that if we want to find such $f(x) = f(2x) + C$ in terms of $C$, it becomes much more difficult. Also we have interest in set of $\{c_j\}$ other than just $c_0 = 1$. We thought the symmetry of the above example makes it more tractable than other possibilities, but it still seems very difficult.

I will post full solution for $f(x) = f(2x) + C$ if I can obtain it (right now I think my approach will work).

[quote="Xevarion"]
I find that if we want to find such $f(x) = f(2x) + C$ in terms of $C$, it becomes much more difficult. [/quote]
What do you mean? :? Just plug $x=0$ in and get $C=0$. And for $C=0$ it is just a periodicity condition on the logarithmic scale... Anyway, I think I've got an idea of what you are after. :)

Oh, you are correct. So I see we can't have $f(0)$ defined if we are to have $C \ne 0$.

The general case I thought I solved seems difficult too :-/

I must go but I'll post more on this later

Another trivial remark: if you do not care about how bad your function may be and don't mind using axiom of choice, the existence result in your problem from post #1 is obvious: all you need to do is to notice that there is no linear dependence between your equations for different $x_1<\dots<x_n$, which is clear because, if $x_n\ge 0$, then $f(2x_n+1)$ appears only once with non-zero coefficient and if $x_1\le 0$, then $f(2x_1-1)$ has this property. Thus, you'd better restrict yourself to some reasonable class of functions if you want to make the problem interesting.

It is actually quite interesting.

We are investigating this question of whether there are functions that are almost refinable but are not the sum of a refinable function plus a constant.

$\ln(x)$ is one particular solution, as I mentioned. Here you show there are many (uncountable, eh?) even if they are not nice functions (can't even be sure they're measurable I guess?). Nevertheless it is an useful result. Surely we can ask more specific questions: can we find such functions in $L^2$? Can we explicitly write any more general solution? Can we write an explicit solution that is not very similar to $\ln(x)$?

It is interesting to note that very little is known about the structure of the class of refinable functions. This arose out of a hope in obtaining some results of this kind by allowing more functions. But even though almost refinable functions are a nontrivially larger class of functions, I am not certain the problem gets any easier!

I realize we can solve $f(x) = f(2x) + C$ rather generally in the following way:

Pick any function $g(x)$ on $[a, 2a)$ and any other function $h(x)$ on $[-2b, -b)$, with $a, b > 0$. For $n \in \mathbb{Z}$, $x \in [2^na, 2^{n+1}a)$ define $f(x) = g(2^{-n}x) + nC$. Similarly for $b, h(x)$.

$f$ will not be defined at $x=0$.

We can make $f$ continuous by stipulating $g, h$ continuous and $\lim_{x \to 2a} g(x) = g(a)$ (and similarly for $b, h(x)$).

Taking $\lim_{a, b \to 0}$ in the continuous case gives $\ln(x)$, I think (though if I wanted to put it in a paper I'd probably have to prove it instead of saying it's obvious).

Question: Can we satisfy any other such functional equation as I mentioned above in this way, or does having more than 1 term mess it up?
Also, can we make $f$ differentiable in the above construction as easily as I made it continuous? (I think yes but all of a sudden I ran out of motivation to think about it. Maybe it's because I'm hungry.)


If you're in Mathcounts, are you also taking the AMCs?

Coaching both this week (AMC 12 and MATHCOUNTS chapter). Busy week.

Im in MATHCOUNTS, and I've been invited to take the AMC 10 (i'm in 7th grd, so I've already taken the AMC8)

I'd like to take AMC's 10 and 12, but no, just one or the other. I scored 95.5 on an old(2000) 12.

I am in MathCounts

I took the AMC-8 (I am in 8th grade) and did good enough to take the AMC-12 which I just took earlier today but I wont discuss it since we cant yet right?

eah, you can't dicuss the test right now.

And the AMC 10 B, I think, is on Febuary 16th.

I'm in 8th grade and I'm talking mathcounts. I'm at private school, 8-12, so I'm also talking AMC 10

[quote=Treething wrote]MathCounts and AMC12.

[/quote]

aren't you in 7th grade?

how can you take AMC12?

can anyone give me some tips to do good in AMC8?

Tips? Here's one:
PRACTICE!!! You can order practice AMC tests from AMC, you can do MATHCOUNTS problems, etc. Just practice! That how I learned it.

[quote="SoccerBrainy40"]Our middle school doesn't offer the AMC 8, 10, or 12. :([/quote]

If it doesn't, then you should try to get a teacher, or principal to offer it next year.

[quote]

aren't you in 7th grade?

how can you take AMC12?
[/quote]

You can take the AMC 10 in any grade 10 and under and the same on the AMC12 and grade 12th and under.  Many students take them at the junior high level

[quote="LynnelleYe"][quote="shinwoo"]aren't you in 7th grade?

how can you take AMC12? [/quote]
It's not invitational--a kindergardener could take it if they/their parents wanted and if a reasonably near host let them, though they probably wouldn't do too well.[/quote]

They could get a guaranteed 62.5  :lol:

[quote="tokenadult"][quote="Hexahedron"]They could get a guaranteed 62.5  :lol:[/quote]

Only if they follow the proper test-taking strategy. I'm really convinced that the downfall of most people who take AMC tests is answering questions beyond their level of ability, which is another way of saying overestimating how well they know math.[/quote]

They could also get a 150. That has a *counts on fingers* 1/(5^25) chance of happening, though :P .


geometry
geometric transformation
function
counting
distinguishability

Prove that the equation $x^2+y^2-z^2=1997$ have infinitly integrer solutions.

Work $\mod 4$ and get that the possibilities for $(x^2, y^2, z^2)$ are $(0, 1, 0)$, $(1, 0, 0)$ and $(1, 1, 1)$.
So for example we have $4k + 1 + 4h - 4m = 1997$ or $k + h - m = 499$. So infinetely triples of integers.

I was thinking on this problem for a while, but ruled out a solution with mods because it would limit the answer (i.e. the modded equation would not be completely congruent to the first equation - not all its solutions would be the original equations solutions). There must be another solution...

Is there a nice cancellation of this product:
$(x^2 + y^2 - z^2)(x^2 + y^2 + z^2)$
also, what are the (factoring or otherwise) tricks for expressions containing 3 squares?
also, what would happen if we replaced each square with a variable?

Aha! I have done it finally!
[hide="here is the proof:"]
First notice that when $z=0$, we have $(x,y) = (34,29)$ in either order. Now we see that if we can find a relation $f(x,y,z) \rightarrow (x_1,y_1,z_1)$
in which both triplets satisfy the equation $x^2 + y^2 - z^2 = 1997$, then we are done. This is because we would be able to keep apply $f(x,y,z)$ over and over up to infinity.
In trying to uncover this translation, we want $(x + n)^2 + (y + m)^2 - (a)^2$ to equal $x^2 + y^2 - z^2 = 1997$. Since we have found a case in which $z = 0$, we want to set the translation function $(x + n)^2 + (y + m)^2 - (a)^2 = x^2 + y^2$ (we have 0 set as z in the lowest triple of the function). Expanding the left hand side yields:
$x^2 + 2nx + n^2 + y^2 + 2my + m^2 - (a)^2 = x^2 + y^2$
$2nx + n^2 + 2my + m^2 - (a)^2 = 0$
$2nx + n^2 + 2my + m^2 = (a)^2$
So this just means we want the left hand side to be a perfect square. Therefore,
$2(nx + my) = 2(nm)$ so for the simplest solution (actually the integer restriction makes it the only solution) gives us:
$x = m/2, y = n/2$
so we solve: $2x = m, 2y=n$
also solve the original 'a' term by plugging in the n,m expressions:
$\sqrt{2n(m/2) + n^2 + 2m(n/2) + m^2} = \sqrt{(n + m)^2}$
so $a = n + m = 2x + 2y$
thus the original function $f(x,y,z) \rightarrow (x_1,y_1,z_1)$ is $f(x,y,z) \rightarrow (x + 2y, y + 2x, 2x + 2y)$
Therefore for any satisfying triplet (starting with $(29,34,0)$) we can create the next satisfying triplet $(x + 2y, y + 2x, 2x + 2y)$
so there are infinite solutions over the integers.
I have to thank my sister for her *very* long haftorah this morning...
[/hide]

[quote="me@home"]Aha! I have done it finally!
[hide="here is the proof:"]
First notice that when $z=0$, we have $(x,y) = (34,29)$ in either order. Now we see that if we can find a relation $f(x,y,z) \rightarrow (x_1,y_1,z_1)$
in which both triplets satisfy the equation $x^2 + y^2 - z^2 = 1997$, then we are done. This is because we would be able to keep apply $f(x,y,z)$ over and over up to infinity.
In trying to uncover this translation, we want $(x + n)^2 + (y + m)^2 - (a)^2$ to equal $x^2 + y^2 - z^2 = 1997$. Since we have found a case in which $z = 0$, we want to set the translation function $(x + n)^2 + (y + m)^2 - (a)^2 = x^2 + y^2$ (we have 0 set as z in the lowest triple of the function). Expanding the left hand side yields:
$x^2 + 2nx + n^2 + y^2 + 2my + m^2 - (a)^2 = x^2 + y^2$
$2nx + n^2 + 2my + m^2 - (a)^2 = 0$
$2nx + n^2 + 2my + m^2 = (a)^2$
So this just means we want the left hand side to be a perfect square. Therefore,
$2(nx + my) = 2(nm)$ so for the simplest solution (actually the integer restriction makes it the only solution) gives us:
$x = m/2, y = n/2$
so we solve: $2x = m, 2y=n$
also solve the original 'a' term by plugging in the n,m expressions:
$\sqrt{2n(m/2) + n^2 + 2m(n/2) + m^2} = \sqrt{(n + m)^2}$
so $a = n + m = 2x + 2y$
thus the original function $f(x,y,z) \rightarrow (x_1,y_1,z_1)$ is $f(x,y,z) \rightarrow (x + 2y, y + 2x, 2x + 2y)$
Therefore for any satisfying triplet (starting with $(29,34,0)$) we can create the next satisfying triplet $(x + 2y, y + 2x, 2x + 2y)$
so there are infinite solutions over the integers.
[/hide][/quote]

Actually, the proof I had posted above is partially incorrect.
The correct equation includes a $z$.
For instance, given a triplet $(x,y,z)$ which satisfies $x^2+y^2-z^2=1997$ we can formulate the next triplet $(x_1,y_1,z_1)$ by taking $x_1=x+2y+2z, y_1=y+2x+2z,z_1=2x+2y+3z$. Notice that the previously posted equation for this transformation holds when $z=0$, which was in effect the only case proved. The cancellation goes as follows:
From the previous proof we see that $(x+2y)^2+(y+2x)^2-(2x+2y)^2=x^2+y^2$.
We can use this fact to simply our equations, replacing each of the terms with $a,b,c$ respectively.
Then we have:
$(x_1)^2+(y_1)^2-(z_1)^2=(x+2y+2z)^2+(y+2x+2z)^2-(2x+2y+3z)^2$
$=[(x+2y)^2 + (2z)^2 + 2(2z)(x+2y)] + [(y+2x)^2 + (2z)^2 + 2(2z)(y+2x)] - [(2x+2y)^2+(3z)^2+2(3z)(2)(x+y)]$
$=a+b+4z(3)(x+y)+8z^2-[c+9z^2+12z(x+y)]$
$=a+b-c+(12z-12z)(x+y)+(8-9)z^2=x^2+y^2-z^2$.
Everything cancels out nicely, this is a nice problem.
It is a consequence of the genererality that any number that can be written as the sum of two squares can replace 1997 (which is 34^2+29^2).
I wonder if it can be made even more general to all integers??

The way this transformation was derived goes as follows:
We let $l$ represent some function of $z$ and $m$ another.
Trying to find the general relationship from the initial case found (the x+2y/y+2x/2x+2y case which cancels when there is no z), we apply $l$ and $m$ in the equation. By hypothesis the value set to $y_1$ and the $x_1$ value are equal to (for some function $f$ in $x,y,z$) $f(x,y,z) =x_1, f(y,x,z)=y_1$ because $x$ and $y$ ( and x1 y1) are indestinguishable in the equations. Therefore the solution to the problem will be of the form:
$(x+2y+l)^2+(y+2x+l)^2-(2x+2y+m)^2$ with $l$ and $m$ functions of $z$ and $z$ only. We then multiply out and simplify knowing that $(x+2y)^2+(y+2x)^2-(2x+2y)^2=x^2+y^2$ and find that if $l$ is $a*m$ and $m$ is $b*m$ then:
[list] $2(l^2)-m^2=-1$[/list]
[list] $6l-4m=0$[/list]
so we solve for $l$:
$m=(\frac32)*l$
$2(l^2)-(\frac32)^2(l^2)=-1\rightarrow -\frac{l^2}{4}=-1\rightarrow l=2$
since $l=2,m=3$
which is how the nice cancelling equations where contrived.[/list]

[quote="drunner2007"]Andreas, wouldn't your choices of $k$, $h$, and $m$ be limited though? We know that $y=2\sqrt{h}$ and $z=2\sqrt{m}$ and $x=\sqrt{4k+1}$. Then how would we be guaranteed that $k+h-m=499$ has infinitely many solutions if h and m are restricted to perfect squares and k must correspond to an integer x?[/quote]

hmm i think this solution works? since $k,h,$ and $m$ can be any positive integer, the possibilities for this are infinite....

[quote="José"][quote="molina"][quote="José"]An Argentinian Competition[/quote]

Cono Sur-1997, I think.[/quote]

Where are you from, molina???? Well, the real thing is that this problem I took it from a "pretorneo de ciudades"...[/quote]

I`m brazilian. There are a math brazilian magazine that yet post this problem.

My friends solve it by Pell.

[quote="José"]Maybe you say this problem, from Cono Sur, 1997.......... Prove that there are infinite natural numbers $a,b,c$ such that $2a^2+3b^2-5c^2=1997$...... Well, BTW, try to solve it.......[/quote]

Doing $\large{x=31}$:

$\large{3y^2-5z^2=75\,\,\Longrightarrow{y=5v\,\,\Longrightarrow{15v^2-z^2=15\,\,\Longrightarrow{z=5u\,\,}}}}$
$\Longrightarrow{3v^2-5u^2=3\,\,\,\Longrightarrow{u=3a\,\,v=b\,\,\,\Longrightarrow{b^2-15a^2=1}}}$

It`s a Pell`s equation with infinitely solutions.

See:

$\large{b^2-15a^2=(b-a\sqrt{15})(b+a\sqrt{15})}$ and doing

$\large{b-a\sqrt{15}=1}$

Then, we have others solutions :

$\large{(b_0-a_0\sqrt{15})^n=b_1+a_1\sqrt{15}}$

Where $\large{b_1,a_1}$ are the new solutions.

I see the brilliant work in your proof, you say there is a method using a Pell equation. In other words, is there a generalized way to make the amazing smiplifications you did? And how did you see to get down to 75 from 1997? I thought it was the fact that $1997=29^2+34^2$ that was the key to this problem, that is my (equally valid?) proof uses. Is mine correct also?


function
number theory proposed
number theory

Find all positive integers $k$ for which there exists a function $f: \mathbb{N}\to\mathbb{Z}$ satisfying $f(1997)=1998$ and $f(ab)=f(a)+f(b)+kf(\gcd{(a,b)})\forall a,b$.

epress f(n) on k then  QED. It is posted before


[quote="mychrom"]Show that:
a*a*b(a-b)+b*b*c(b-c)+c*c*a(c-a)>0, a,b,c different real nr.[/quote]
I think you mean the 6th problem from IMO 1983,don't you?
See here
[url]http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1983[/url] :wink:


$ABCD$ is a quadrilateral with right angles at $A$ and $C$. Points $E$ and $F$ are on $AC$, and $DE$ and $BF$ are perpendicular to $AC$. If $AE=3$, $DE=5$, and $CE=7$, then $BF=$

$\text{(A)}\ 3.6 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 4.2 \qquad \text{(D)}\ 4.5\qquad \text{(E)}\ 5$

[hide="solution"]
From a little angle chasing, we notice that $AED$ is similiar to $AFB$ and $DEC$ is similiar to $BFC$. From this, we have:

$AE/ED=BF/AF$ and $DE/EC=FC/BF$

So $3/5=BF/(3+x)$ and $5/7=(7-x)/BF$

Solving for $BF$ gives $21/5$ or $4.2$
[/hide]

[quote="Hellstar"][quote="4everwise"][hide="Cool Method"]Because the opposite angles of the quadrilateral are supplementary, this quadrilateral is cyclic.

Extend DE such to a point $P$ on the circle, such that $BF=PE$. We apply power of a point.

$(PE)(DE)=(AE)(EC)$ and plugging in the values, our answer is $4.2$.[/hide][/quote]

How can you be sure that $EP =BF$ and is collinear with $ED$ at the same time?[/quote]Oh, sorry. I'm kind of in a rush when archiving problems. I'll just reword the second part of the solution. :)

[hide]Extend $DE$ to $P$. We know that $\angle BAD$ is right, so that $BD$ is the diameter of the circle. Notice that after extending $DE$, we form a right triangle $BPD$ with $\angle BPD=90^\circ$. (It's hypotenuse is the diameter)

Now, $BF=ME$ because three right angles determine a rrectangle and BLAH!  :D [/hide]


algorithm
vector
real analysis
real analysis unsolved

Let $\sum_{n=1}^{\infty}a_n$-absolutely converges series. Prove that the set of sums, obtained from the given series with deleting finitely or infinitely many terms(if all the terms are deleted then the sum supposed to be zero) is closed.

Edit:

Deleted because everything I had written was plain wrong :)

For every $ P \subset \mathbb{N} $ denote
\[ S_{P} := \sum_{n \in P} a_{n} \]. Then we have to prove that the
set $D$, consisting of all $ S_{P} $ is closed. Suppose we have a
sequence of $ P_{k} \subset \mathbb{N} , k=1,2,\ldots $ such that
there exists a limit $ x = lim_{k \rightarrow \infty }S_{P_{k}} $.
Let us show that $ x \in D $. Really, by compactness argument we
can find a subsequence $ \{ Q_{m} \} $ of $ \{ P_{k} \} $ , which
stabilizes, i.e for every $ n \in \mathbb{N} $, there exists
$m_{0}$, such that $ n $ belongs or not belongs simultaneously to
every $ Q_{m} $ for $ m > m_{0} $ (to do it, observe that $ 1 $
belongs or not belongs simultaneously to infinite number of $
P_{k} $'s, so consider this infinite subsequence of $ P_{k} $'s,
now , similarly, take a subsequence of our subsequence, with the
property that $ 2 $ belongs or not belongs simultaneously to this
subsequence, and so on, and at the end take the diagonal
sequence). Now take a subset $ Q \subset \mathbb{N} $ which is the
"limit" of $ \{ Q_{m} \} $, i.e. $ n \in \mathbb{N} $ belongs to $
Q $ if for $ m $ large enough, $ n \in Q_{m} $. I claim that $
lim_{k \rightarrow \infty} S_{P_{k}} = lim_{m \rightarrow \infty}
S_{Q_{m}} = S_{Q} $. Indeed, take some $ n \in \mathbb{N} $, and
there exists some $ m_{0} \in \mathbb{N} $, such that for every $
1 \leqslant j \leqslant n $ we have for every $ m \geqslant m_{0}
, j \in Q_{m} $ or for every $ m \geqslant m_{0} , j \notin Q_{m}
$. Therefore, because of this and of absolute convergence of \[
\sum_{j=1}^{\infty} a_{j} \], we have for every $ m \geqslant
m_{0} $ that \[ | S_{Q_{m}} - S_{Q} | \leqslant
\sum_{j=n+1}^{\infty} |a_{j}| \], which tends to $ 0 $, when $ n $
goes to infinity.

in this context i would like to ask Can the sum of a conditionally convergent series can be made equal to any real no however large or small it is.
by the way Leva you & I share the same interest & the fate is also the same.
The sol. is great...

[quote="the game"]Can the sum of a conditionally convergent series can be made equal to any real no however large or small it is.[/quote]
Yes. This is well known. Many analysis books should have the proof, the essence of which is the algorithm for the rearrangement:

First, take positive terms until the partial sum is greater than the targeted limit. Then take negative terms until the partial sum is less than the targeted limit. Repeat, going back and forth.

In this context, there is one interesting problem, which is not
trivial. Suppose we have a sequence of vectors $ x_{k} \in
\mathbb{R}^{n} , k=1,2,\ldots $. Consider now a series
\[ \sum_{k=1}^{\infty} x_{k} \] ( it may not exist ), and consider
all sums which are rearrangement of this series, and consider in $
\mathbb{R}^{n} $ the set of all values of such series, which are
convergent. In other words for each bijection $ \sigma :
\mathbb{N} \rightarrow \mathbb{N} $, consider the sum \[
\sum_{k=1}^{\infty} x_{\sigma (k)} \], and if it is convergent,
include it into the set $ X $. Prove that $ X $ is a shifted
linear subspace of $ \mathbb{R}^{n} $.

This is also a famous result.  I cannot remember the solution but it is in (e.g.) Glazman and Ljubic's problem book on finite dimensional linear analysis.  It is not very difficult, I think.


Anne writes down the nine consecutive integers from $-4$ to $4$. She then performs a series of operations. In each operation she identifies two numbers that differ by two, decreases the largest by one, and increases the smaller by one so that the two numbers are now equal. After a while she has nine zeros left and stops. How many operations did she perform?

Well, this is going to be messy, but perhaps someone can help me develop a full solution describing the exact pattern. Please tell me if there are any unnecessary operations. My final answer is then
[hide]$\framebox{30}$,[/hide]
but it seems like a strange number to be the answer to such a problem.

EDIT: I accidently counted my first step as an operation, when it was really just copying the problem.

consider the sum of squares at each step.

before an operation,
(x-1)^2+(x+1)^2 = x^2 + 2

after the operation those two become
x^2 + x^2 = 2x^2

so the sum of the squares of the numbers decreases by 2 after each operation.

the initial sum of squares was 60, so it takes 30 moves to get back to 0,0,0,0,0,0,0,0,0 (sum of squares is 0)

[quote="gotztahbeazn"]eh this problem seems like it would involve monovariants or something but i couldnt figure out how to apply it...

but just doing the algorithm myself i get $28$[/quote]
You probably did a couple of double-operations in your steps somewhere. I had to edit those doubles out of my original steps.


find all three digit numbers such that ,when it is divided by 11,the number is equal tosquare of the digits of the first no.?

I think he means the squares of the digits of the original number is equal to the original number divided by 11, but that would make the problem way too easy. It would just be simple casework with a little bit of brute force.

[hide="Solution, assuming the question is interpreted that way"]We know trivially that if a 3-digit number is divided by 11 evenly, then the result must be a 2-digit number.
We also know that when you multiply a 2-digit number by 11, you get a 3-digit number that results from adding a middle digit to the original 2-digit number that is the sum of the 2 digits. (Also trivially shown through long multiplication.) Also note that if the sum of the 2 digits is greater than 10, it carries over.

Knowing this, we can split this problem into 2 cases.

Suppose the 2-digit number (original 3-digit number divided by 11) is $ ab$, where $ a$ and $ b$ are the digits.

[b]Case 1:[/b] $ a \plus{} b < 10$

The 3-digit number would then be $ acb$, where $ c \equal{} a \plus{} b$.
The equation would then be

$ a^2 \plus{} b^2 \plus{} (a \plus{} b)^2 \equal{} 10a \plus{} b$

$ 2a^2 \plus{} 2b^2 \plus{} 2ab \equal{} 10a \plus{} b$

$ a^2 \plus{} b^2 \plus{} ab \equal{} 5a \plus{} b/2$

It is now evident that $ b$ must be even. We can now, with a bit of brute force, find the possible values of $ a$ and $ b$.

It is soon clear that the only values of $ a$ and $ b$ that work are
$ a \equal{} 5, b \equal{} 0$

So the only number for this case would be $ 50*11 \equal{} \boxed{550}$

[b]Case 2:[/b] $ a \plus{} b\ge10$

We will apply the same method this time, except our 3-digit number is now $ deb$, where $ d \equal{} a \plus{} 1$ (This is the only possible value of $ d$ because $ a \plus{} b$ cannot be 20 or higher) and $ e \equal{} a \plus{} b \minus{} 10$.

$ (a \plus{} 1)^2 \plus{} (a \plus{} b \minus{} 10)^2 \plus{} b^2 \equal{} 10a \plus{} b$

$ 2a^2 \plus{} 2ab \minus{} 28a \plus{} 2b^2 \minus{} 21b \plus{} 101 \equal{} 0$

We now see that $ b$ must be odd, or else there will be a contradiction. We can soon determine that no values work, and so the only 3-digit number that satisfies the given statements is

$ \boxed{550}$[/hide]


Let x , y , z and w represent four distinct positive integers such that $x^2 - y^2 = z^2 -w^2 = 81$
Find the value of $xz+yw+xw+yz$
Again , a detailed explanation please....
 :(

[hide]
rewriting the equation, we have:
$81+y^2=x^2$
$81+w^2=z^2$

this means that 81 is in two different sets of Pythagorean triple. It happens to be:
$(9,12,15)$
$(9,40,41)$

so then WLOG, let $y=12$. Then $x=15$. Then by the other Pythagorean triple,
$w=40$, and $z=41$

going back to the actual question,
$xz+yw+xw+yz = (xz+xw)+(yz+yw) = x(z+w)+y(z+w)=(x+y)(z+w)=27 \cdot 81 = 2187$
[/hide]

Hmm, from the form of the answer, it looks like there might be a more clever way to do this, but
[hide]you can rearrange them into the following-

$9^2+y^2=x^2$
$9^2+w^2=z^2$

And since they're all integers, you just need to find Pythagorean Triples, and since you're given a low number (9) tbe triples are fairly obvious (9,12,15) and (9,40,41).  So then you no w,x,y, and z and can substitute, but again, since it asks for xz+yw+xw+yz, there's probably some cleverly intuitive way to do this, in case a larger number had been presented.
[/hide]

edit: dang, you beat me to it.  and by 3 minutes?  I type slow  :P

We are looking for $(x+y)(w+z)$.

Since $(x+y)(x-y)=x^2-y^2=81=3^4$ and $x+y>x-y$ we get that $x+y \in \{ 27 , 81 \}$ (and corresponding to that $x-y \in \{ 1,3 \}$. ) .
Now the same holds for $w+z$, but you can't have $x+y=w+z$, because this would give $x=z , y=w$, so you get $(x+y)(w+z)=3^3 \cdot 3^4 = 3^7$


geometry
inequalities unsolved
inequalities

For $a,b,c$ sides of a triangle of area $S$ we know that $a^x,b^x,c^x$ are the sides of a triangle of area $S_x$ for $0 \leq x \leq 1 $. Prove that

$ \frac{4}{\sqrt{3}}S_x \geq (\frac{4}{\sqrt{3}}S)^x $

Thank you very much for the solver of this problem.

This is really a very hard problem.

No idea ??  ;)


geometry
inradius
trigonometry
area of a triangle
IMO Shortlist

A point $ M$ is chosen on the side $ AC$ of the triangle $ ABC$ in such a way that the radii of the circles inscribed in the triangles $ ABM$ and $ BMC$ are equal. Prove that
\[ BM^{2} \equal{} X \cot \left( \frac {B}{2}\right)
\]
where X is the area of triangle $ ABC.$

In http://www.mathlinks.ro/Forum/viewtopic.php?t=32586 , it was proven that $BM=\sqrt{s\left(s-b\right)}$, where $s= \frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. Thus, $BM^2=s\left(s-b\right)$. On the other hand, by the Heron formula, the area X of triangle ABC equals $X=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$, and by the half-angle formulas, $\tan\frac{B}{2}=\sqrt{\frac{\left(s-c\right)\left(s-a\right)}{s\left(s-b\right)}}$. Thus,

$BM^2=s\left(s-b\right)=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}: \sqrt{\frac{\left(s-c\right)\left(s-a\right)}{s\left(s-b\right)}}$
$=X: \tan\frac{B}{2}=X\cot\frac{B}{2}$,

completing the proof.

By the way, Orl, this is not an inequality...

  Darij


Let $ a_i\geq0,b_i>0, i\equal{}1,2,...,p$, and $ k,p\in N, \ k,p\geq 1$, such that: $ a^k_1\plus{}a^k_2\plus{}\cdots\plus{}a^k_p\geq b^k_1\plus{}b^k_2\plus{}\cdots\plus{}b^k_p$.
For $ n\geq k$, prove that the following inequality holds

$ \left.\frac{a^{n\plus{}1}_1}{b^{n\minus{}k\plus{}1}_1}\plus{}\frac{a^{n\plus{}1}_2}{b^{n\minus{}k\plus{}1}_2}\plus{}\cdots\plus{}\frac{a^{n\plus{}1}_p}{b^{n\minus{}k\plus{}1}_p}\geq \frac{a^n_1}{b^{n\minus{}k}_1}\plus{}\frac{a^n_2}{b^{n\minus{}k}_2}\plus{}\cdots\plus{}\frac{a^n_p}{b^{n\minus{}k}_p}\right.$

Very nice your proof, [b]bakerbakura[/b]! You are really good at using Holder's inequality. My solution is different than yours. I'm waiting for new ideas.

It's time to rock Marin Bancos's Problems. :)

Here is my solution

We may take an individual term and investigate its property before applying for other in the same manner.

It is clear due to AM-GM that
\[ n\times \frac{a_1^{n\plus{}1}}{b_2^{n\minus{}k\plus{}1}} \plus{} b_1^{k} \ge (n\plus{}1)\times\frac{a_1^{n}}{b_1^{n\minus{}k}}\]

and

\[ k\times\frac{a_1^{n}}{b_1^{n\minus{}k}} \plus{} (n\minus{}k)b_1^k \ge na_1^k\]

Using the given condition, we easily come up with a conclusion.

Hence, the proof is completed.


Let $S = \{x_0, x_1,\dots , x_n\}$ be a finite set of numbers in the interval $[0, 1]$ with $x_0 = 0$ and $x_1 = 1$. We consider pairwise distances between numbers in $S$. If every distance that appears, except the distance $1$, occurs at least twice, prove that all the $x_i$ are rational.

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=283904&hilit=distance+twice[/url]. [b]fedja[/b]'s comment that "the condition that the numbers are distinct is missing" is somehow made moot by the fact that both posts speak of a [b]set [/b]$S$ of numbers, and as such implicitely assuming they are distinct; but nevertheless it is worth making this observation - indeed, otherwise we can just take all $x_2=x_3=\cdots = x_n$ (for $n\geq 4$) to be some irrational number.

We consider the basis of $x_i$. Let that be $\mathbb{Q}[r_1,r_2,\ldots,r_m]$. Then choose any $r$ let us say $r_1$. Since $r_1$ is in the basis thus there must exist two $x_i$ such that there difference contains $r_1$ (This ensures that the component of $r_1$ in the difference we'll define isn't $0$) . Let

$X_1$ = Those $x_i$ who have the maximum component of $r_1$ and
$X_2$ = Those $x_i$ having minimum component of $r_1$.

Then consider the difference of "rightmost" (i.e. greatest) member of $X_1$ and "leftmost" ( i.e. smallest ) member of $X_2$. If this has to have a friend then one vertex must be in $X_1$ and other in $X_2$. But if you choose any such pair (other than this one) then its length will be strictly less than the earlier chosen pair, the most preposterous statement! So all lengths must be rational (as we must dustbin any $r_i's$ if they pop up just like that quantum fluctuations anyone?) $\square$.
IMPORTANT:This stunt has been performed under the supervision of biomathematics. Do not try at home.
Patreon supporters: 1. (to be disclosed) 2. (to be disclosed)

Very instructive approach using Hamel basis. It so happened that I recently commented on two other problems using the same method. One may see all three of them (including this one) [url=https://dgrozev.wordpress.com/2021/01/11/hamel-bases-part-3-application-on-an-olympiad-problem/]in my blog[/url].


Analise's resting heart rate is 75 beats per minute. When she sleeps, it slows to 60 beats per minute. What percent decrease is this?

Her heartbeat decreased by 15 from 75.  So do $ \frac{15}{75}\equal{}\frac15\equal{}20\%$

The percent change is:

$ |\frac{\text{Original Value} \minus{} \text{ Final Value}}{\text{Original Value}}|$

The formula for these kinds of problems with original number $ a$ and new number $ b$ is:

$ \frac {b \minus{} a}{a} \equal{} \frac {x}{100}$

EDIT: Beaten...
EDIT 2: Actually, Ernie, no absolute value. You can have negative percent decreases.


conics
ellipse
analytic geometry
graphing lines
slope
algebra
polynomial

If $ax^2+by^2=1$, with $a,b,x,y \in\ Q$, has a rational solution $(x,y)$. then it has many rational solutions.

The geometric method works for such problems:
The equation defines an ellipse (well, every curve of second degree with rational coefficients will do), and has a rational point $P$ (both coordinates rational) on it.
Thus take a line with rational slope through $P$. As it follows either by simple calculating or Vieta, the second point of intersection is rational, too (and one can get all but one rational points by this).

ZetaX are you saying that on a ellipse there is only one point with rational coordinates ?
if so is there a proof (elementary if possible)?

I say: when there is one point, thene there are $\infty$ many.

Let $q(x,y)=0$ be the equation of a nondegenerated conic (ellipse, hyperbola, parabola).
Assume there is one rational point $P=(a,b)$ on it.
Then consider a line given by $cx+dy+e=0$ through $P$ (thus with $e=-(ac+bd)$) with rational $c,d,e$ and $c^2+d^2 >0$. Then the second intersection ($\neq P$) of this line with the conic is a rational point, too:

When you solve the system of equations (one of them with degree $1$, the other one with degree $2$), you get an equation of second degree with rational coefficients (this fact is important!) in e.g. $x$ (when $d=0$, you get an equation in $y$ only, but by changing $y \leftrightarrow x$ this makes no difference). Let $x_1,x_2$ be the (at first only complex) solutions.
But you know already that one solution is given by $x_1=a$ since $P$ is one point of intersection. Thus let $f$ be the linear coefficient of the equation, then you have by Vieta that $x_1+x_2 = -f$, thus $x_2 = -a-f$, which is rational.
Now it's also not hard to see that the $y_2$ belonging to this $x_2$ is rational: $cx_2 + dy_2 + e =0$ gives for $d \neq 0$ (this was what we excluded above) that $y_2$ is rational.

Is there Any simpler solution I could not understand If someone hes please post it

You could tell instead what you didn't understand...

ZetaX i read what you wrote,you actually say that a line can intersect a ellipse in 2 points of rational coordinates or no points of rat coord. at all.
How does that proove that there are allot($\infty$) of these points on the ellipse?

When you choose the line such that it goes through the one given point, then it has (at least) one intersection. By the above, it must have a second one (when the line is  tangent, both points coincide).
But for every different rational slope, the second point is different since two lines (with one common point) and different slope have only one common point; there is also only one tangent.
Thus, because there are $\infty$ many diffferent rationals to choose from, there are $\infty$ many different solutions.

thank you ZetaX,i should have tghought more.


calculus
integration
trigonometry
calculus computations

Let $\alpha ,\ \beta$ be the distinct positive roots of the equation of $2x=\tan x$.

Evaluate the following definite integral.
\[\int_{0}^{1}\sin \alpha x\sin \beta x\ dx \]

Let $I$ be the integral in question. Integrate by parts twice, and solve for $I$ to get

$\left(1-\frac{\beta^{2}}{\alpha^{2}}\right)I =-\frac1{\alpha}\cos\alpha\sin\beta+\frac{\beta}{\alpha^{2}}\sin\alpha\cos\beta$

But $\cos\alpha=\frac{\sin\alpha}{2\alpha}$ and $\cos\beta=\frac{\sin\beta}{2\beta}$ From that,

$\left(1-\frac{\beta^{2}}{\alpha^{2}}\right)I =-\frac1{2\alpha^{2}}\sin\alpha\sin\beta+\frac1{2\alpha^{2}}\sin\alpha\sin\beta$

So $I=0.$

[quote="sos440"][hide="My poor solution"]One can easily check that

$\int_{0}^{1}\sin \alpha x \sin \beta x \, dx = \frac{1}{2}\left[ \frac{\sin(\alpha-\beta)x}{\alpha-\beta}-\frac{\sin(\alpha+\beta)x}{\alpha+\beta}\right]$

holds. Since

$\sin(\alpha-\beta)x = \cos \alpha \cos \beta (\tan \alpha-\tan\beta) = \cos \alpha \cos \beta (2\alpha-2\beta)$ and
$\sin(\alpha+\beta)x = \cos \alpha \cos \beta (\tan \alpha+\tan\beta) = \cos \alpha \cos \beta (2\alpha+2\beta)$,

we have

$\int_{0}^{1}\sin \alpha x \sin \beta x \, dx = 0$[/hide][/quote]

This is not a poor solution, which is a standard solution.


Here's my proposal for a new competition format, some apply to everything, some apply to just Nationals (like countdown, masters)

Sprint: 20 questions, 40 minutes (harder than what they have now), 1 point each
Target: 10 questions in pairs, 6 minutes per pair (same as now), 2 points each
Team: 10 questions, 20 minutes (same as now), with the following point distribution: 1122334455
Relays: Three relays, all problems are approximately the same difficulty (between the four team members). 7 minutes per relay, first, second, third answers count as one point each, fourth answer counts as two points.

Team score is equivalent to the average of the individual scores plus the team round score and relay round score. Maximum Individual score is 40, Maximum team score is 85.

(Nationals only, SEPERATE COMPETITION NOT AFFECTING TEAM SCORE) Guts (as proposed by SplashD): One hour, 20 sets of 3 problems, given to teams one at a time. Score distribution as follows: 111111111111222222222222333333333333444444444444555555555555. When team finishes on set, it turns it in and recieves the next set, 60 minutes to complete everything. Real time scoring ftw.

Countdown: Top 16 bracket, just for fun (and prizes of course)
Masters: Top 4 Written, winner is national champion.

Yeah if you want, you can devise your own tiebreaker system.

Please comment, I know this is pretty big, and probably at nationals it would have to be split over two days (maybe Sprint/Target/Team/Countdown on day 1, Relay/Guts/Masters on day 2)

Also if you want you can come up with your own proposals

btw, I do not actually think Mathcounts will read this and say oh! let's do this!, but I just want to share my opinion...I think it would be pretty cool this way.

[quote="kyyuanmathcount"]Man, we'll need some major funding to keep us at the competition site for 2 days.  That would be cool tho...we should get week-longs like the spelling bee, where you go there, do fun stuff for 2 days, competition, do more fun stuff, then leave.

The relays will require each of the team members to be strong in order for the team to succeed - I think that's a good idea.[/quote]

The big difference with the LIAN relays and the ARML relays is that you get credit for anything else. So basically if you have a team of Neal, Gauthier, Kevin Chen, and Lebron James, you stick Lebron in chair 4 and still get your 3 points every time.

[quote="mnmath"]I agree with your proposal, but the Masters Round should take the top 10-16 students, instead of 4.[/quote]
Consider the time, most people got bored after watching an hour of masters round, almost everyone would've been bored to death after 4 hours. Keep in mind that people couldn't get up and leave without being rude.

I definitely support this, even though this was my last MC year. :yup:


Is it possible to find out the minimum permitted energy of an electron in a Hydrogen atom and its corresponding mean distance from the nucleus using Uncertainty principle?

I know that if the mean distance is known we can find the energy... But,according to the question,we are to find both...

The potential energy of the electron:
$ \phi \equal{} \minus{}k\frac{e^2}{r}$
Where $ r$ is the distance from the proton.
We can approximate the of the uncertainty of the position of the electron with the $ r$.
So the kinetic energy of the electro+n:
$ E_{kin} \equal{} \frac{p^2}{2m}\geq \frac{\hbar ^2}{2m(dx^2)}\approx \frac{\hbar ^2}{2mr^2}$
So the total energy of the electron:
$ E\equal{}\phi \plus{} E_{kin}\equal{} \minus{}k\frac{e^2}{r}\plus{}\frac{\hbar ^2}{2mr^2}\equal{}\minus{}\frac{\alpha}{r}\plus{}\frac{\beta}{r^2}$
Where $ \alpha \equal{}ke^2$ and $ \beta \equal{} \frac{\hbar ^2}{2m}$

The minimal energy of the electron is thus when $ \frac{dE}{dr}\equal{}0$
That is $ r_{min}\equal{}\frac{2\alpha}{\beta}$
now the energy:
$ E_{min}\equal{}\minus{}\frac{\alpha ^2}{4\beta}$


There are puzzle games such as Snowy Puzzle Islands, which are too difficult math and not enough fun. Not recommended.

Dizzy 11 for DOS is similar to Snowy and is a lot easier to get into.

I personally think [url=http://www.playfirst.com/game/cubology]Cubology[/url] is a gem among the match-3 cliches. It's an example of puzzles that are mildly challenging, and very fun to play. There is time limitation for each puzzle.

That's just an example, many more are out there.

Blokus seems underappreciated too, compared to the *viral* Brain Age, Sudoku phenomenon.

When it comes to board games, would you please stop saying chess? Since I'm Chinese, I got tired of Chinese chess as well. There are other board games out there, that are smart too, and in my opinion much more fun to play.

For example, Dominoes is twice as popular as Chess on Pogo.com, there are rule variations. (maybe it's not that smart. I can't think of examples right now.)

I sort of adore people who know how to play card and casino games, because I haven't learnt any yet, except the common Blackjack. Not for the money, just for the maths.

go is an awesome and computationally intensive game. check out my "Go game" thread to find out more.


Could someone give me a hint on approaching this problem?

Let (p_n) be a sequence and f: N onto N a bijection. The sequence (q_k) with q_k= p_f(k) where k is in N, is a rearrangement of (p_n)
1) Are limits of a sequence unaffected by rearrangement?
2) What if f is an injection?
3) What if f is a surjection?

Does (p_n) already have a limit since the first question seems to assume that? How is p_f(k) interpreted in this problem? I have some troubles with this.

Any rearrangement of a convergent sequence is convergent to the same limit, and any rearrangement of a Cauchy sequence is Cauchy. The "all but finitely many" way of stating things is well-suited here.

On the first, note that a rearrangement has an inverse; you're going to get an "if and only if", since the argument can also be run backwards.

A pair of useful examples (in $ \mathbb{R}$):
$ f(k)\equal{}2k$, $ p_n\equal{}(\minus{}1)^n$.
$ p_{f(k)}\equal{}1$ for all $ k$.

$ f(k)\equal{}\begin{cases}\frac k2& k\text{ is even}\\1& k\text{ is odd}\end{cases}$, $ p(n)\equal{}\frac1n$.
$ p_{f(k)}\equal{}\begin{cases}\frac 2k& k\text{ is even}\\1& k\text{ is odd}\end{cases}$.


Jenks had five people in the top 10 and 11 in the top 22.
University would have still won if we put the top four from Jenks together--We've got to work harder!  (For those who were not there, we took 2nd, 3rd, and 5th out of 14 teams.)
With our current scores, it is very possible that we have 8 people go to state!

All but one missed the problem about perfect numbers.  "A perfect number is a counting number that is equal to the sum of its proper factors.  What is the sum of the two smallest perfect numbers?"
--Proper factors are all the number's factors except itself.
Now can you solve it?  Want more info?  See: http://primes.utm.edu/glossary/page.php?sort=PerfectNumber

If you want your papers before break, stop by my room.

Interesting.
I wrote up a Mock mathcounts that used perfect numbers as Master's round.

I only got 6 as my perfect number... I thought 1 would be the other one..

Anyways, I think the answer should be [hide="this"]34. (6+28)[/hide]

Are there a formula for finding perfect number?

That's the second part of my mock mathcounts masters, prove all perfect numbers are of the form $ 2^k(2^{k\plus{}1}\minus{}1)$.

[quote="xpmath"]That's the second part of my mock mathcounts masters, prove all perfect numbers are of the form $ 2^k(2^{k \plus{} 1} \minus{} 1)$.[/quote]

You mean all [u]known[/u] perfect numbers?

Cause otherwise nobody would wonder if there were any odd perfect numbers...

[quote="perfect628"]You all should have just read my username...[/quote]

I really haven't noticed it.. Nice username. :lol:


Please ,
 can someone give me a link- or probably a solution - on Flett's theorem ? I had seen it somewhere but I can not find it again.
 Thank you  !

 babis


How many possible distinct arrangements are there of the letters in the word DOG?

The [i]factorial[/i] of a natural number $ n$ is the product of all the natural numbers from $ 1$ to $ n$.

Factorial is also defined on non-natural numbers, but that is much more complex.

Um, $ n!\equal{}n(n\minus{}1)(n\minus{}2)\cdots(2)(1)$. So in this case $ 3!\equal{}\boxed6$.

The obvious answer is $ 3!\equal{}6$ but you could try listing:

DOG
DGO
ODG
OGD
GOD
GDO

[hide=" solution nowadays"]mem the fact that the way to arrange 3 letters is 6. [/hide]

[quote=k8ds]solution never defined what factorial means[/quote]

factorial means multiplying by every number below it.

$3!$ means $3*2*1$


Let f: D(0,1)-->C (where D(0,1) is the unit disk and C the complex field) be an analytic function such that |f(z)|<1 for all z in D(0,1).

Show that

              |f'(z)|/(1-|f(z)|^2)[u]<[/u]1/(1-|z|^2).

(a) Next, suppose there exist two distinct points a,b in D(0,1) such that f(a)=a and f(b)=b. Show that f(z)=z for all z in D(0,1).

(b) Finally, suppose there exist a in D(0,1), a not equal to 0, such that f(a)=0=f(-a). Show that
|f(0)|[u]<[/u]|a|^2

What can we conclude if |f(0)|=|a|^2

I was going to reply, but then noticed that you double-posted. (The other copy is deleted).
So I will only give a hint: apply the Schwarz lemma (and its equality statement) to $ f$ composed with a Mobius transformation of $ D(0,1)$.