USAMTS
AMC
AIME
analytic geometry
function
geometry
inradius

How did you do on round 2 of the USAMTS?

I got a 20 with 2 commended. The graders were [b]much [/b]more strict than in round 1, I thought, and I noticed there were a whole lot less 25s.

I have absolutely no idea why I have a 23. I belive my solutions are precise enough for a 25. Gotta talk to somebody about this.

23 last round, 23 this round.

23 was exactly the score i thought i would get, except it is from a different breakdown (i predicted -2 on #2...in reality it was -1 on #2 and -1 on #3...hmmm...)

I'm a bit disappointed with my score for question 3.  My final set was {{2,3,6},{2,3,9,18}}, and the required set was {2,3,6,9,18}.   So, in that respect, not as well as I thought.

I was right.  I turned in the wrong document that had only two sketchy solutions!  Did all that work making solutions to my satisfaction and nothing to show.  It was still fun.

I managed to get a 25, although I got no comments on problem 1, probably because my solution was too long and convoluted and not fun at all to read  :P My goal for the next round is to make all of my problems concise and elegant, because they certainly weren't this round (only 1 commended, for my only short solution).

I dont know if the didn't get a page or what but problem one only gave my a 3 because I supposedly didn't show solutions for 17,19,20, and 22, but I reviewed what I sent in, and they're right there. I'd say that with those, my problem was worth a 5, but what can you do? Oh well...

I got a 21.
I was given a 2 on #4 because I (mistakenly) assumed that a side that passed through 3 points was illegal.
[quote="The grader"] Just because a quadrilateral passes through a given point doesn't mean that the point is a vertex--It could just be on a side of the quadrilateral!
Good case work aside from that[/quote]

3 off for one lousy assumption seems a bit much.  We'll see.

1 commended (#3)
[/quote]

I got a 20, no commendations, but I got a 3 on #2 when I felt I should've gotten a 5.  The grader comment was "SHOW THESE APPROX ARE EXACTLY THE TINY #'S" when I started my proof with showing how the respective $\frac{a}{b}$ for tiny numbers were the best rational approximations for $\sqrt{2}$ : $a - b\sqrt{2} \approx 0 \Rightarrow \frac{a}{b} \approx \sqrt{2}$ and stating that the best rational approximations for $\sqrt{2}$ were the convergents of the continued fraction for it, the latter being a well-known result in number theory (Lagrange).  I'm a little disappointed that my grader seems to have done a cursory scan rather than a thorough reading of this problem and depending on what the official rubric says, I might consider challenging this, loath though I am to do it.

I completely agree with the rest of the grading of my paper.

Thoobik, I just read your solution to Question 2, and I think you should be pleased with the score.  Your solution uses the term "best possible approximation" without defining what that means.  Then, most seriously, the solution uses the relation $\approx$ without definition.  I think you mean "approximate equality", but then how do you know the conclusion isn't "close to the best approximation" instead of "the best approximation"?  Finally, you state the result about convergents being the best approximations, but without a name or citation or even saying "well known".  In your post, you say it's well known, but I used to be a mathematician, and I don't think I could have quoted that result off the top of my head.

Don't take my post too harshly.  I just think there was much you left off.

Will someone please take a look at my solution for problem 2 and tell me how I got a 3? I can see a 4, but 3 doesn't make sense to me. The grader said I needed to prove that my characterizstion of tiny numbers actually works . . . I thought that my proof came from the definition of tiny numbers along with obvious intuition regarding the first equation in my solution. How is this inadequate justification?

[quote="thoobik"]I got a 20, no commendations, but I got a 3 on #2 when I felt I should've gotten a 5.  The grader comment was "SHOW THESE APPROX ARE EXACTLY THE TINY #'S" when I started my proof with showing how the respective $\frac{a}{b}$ for tiny numbers were the best rational approximations for $\sqrt{2}$ : $a - b\sqrt{2} \approx 0 \Rightarrow \frac{a}{b} \approx \sqrt{2}$ and stating that the best rational approximations for $\sqrt{2}$ were the convergents of the continued fraction for it, the latter being a well-known result in number theory (Lagrange).  I'm a little disappointed that my grader seems to have done a cursory scan rather than a thorough reading of this problem and depending on what the official rubric says, I might consider challenging this, loath though I am to do it.[/quote]

Problem 2/2/16 was graded at the NSA on Monday. I already sent the grading rubric we used to the AoPS, so it will probably be posted in the Grading Disputes section soon.

The caution with using a well-known result is that without an actual statement of the theorem, it might be misremembered or misinterpretted. The convergents of the continued fraction for $\sqrt{2}$ are best approximations of $\sqrt{2}$, but not all best approximations are convergents (unless "best approximation" is used in the strongest sense, which has to do with the reciprocal of the square of the denominator). The best approximations are:
\begin{eqnarray*}
1 = \sqrt{2} - 0.414213\ldots\\
3/2 = \sqrt{2} + 0.085786\ldots\\
4/3 = \sqrt{2} - 0.080880\ldots\\
7/5 = \sqrt{2} - 0.014213\ldots\\
17/12 = \sqrt{2} + 0.002453\ldots\\
24/17 = \sqrt{2} - 0.002448\ldots\\
41/29 = \sqrt{2} - 0.000420\ldots
\end{eqnarray*}
Two of those numbers, 4/3 and 24/17, are not convergents.

We did not intend for the problem to have this hidden trap inside of it, but it did. You and many others lost two points for relying on a false theorem.

[quote="ZennyK"]The grader said I needed to prove that my characterization of tiny numbers actually works . . . I thought that my proof came from the definition of tiny numbers along with obvious intuition regarding the first equation in my solution. How is this inadequate justification?[/quote]

I looked over your solution. Definitely one point off for assuming without proof that $a^2-2b^2$ had to be $\pm 1$. Your grader chose to take another point off, possibly due to your sloppy use of $\approx$ in which you essentially say $1\approx 0$. We prefer more rigor in the proofs.

Erin Schram
USAMTS grader

Can someone tell me why I got a 3 for #2 since my solution for #2 matches the 5-point solution on the rubric exactly?

Meh... I got a 19... That's a 4 point DE-provement. But, this is my first time taking these, so I can't really say I expected much better. Mostly, judges questioned the rigor of my solutions, and really I have no arguments because I agree with them for the most part.

Same here.  I really can't argue with any of the rulings and it seems as though the graders have moved to more stringent standards for rigor (not a bad thing).  My de-provement was also 4.

I got 19 without any commendation.

I got 25 for the first one.  Do you think it is possible for me to qualify for AIME through USAMTS this year?

This is the worst ever.

I got 2 on the 4th problem because I made an arithmatic mistake while adding all possible triangles.  I got a wrong answer.

Getting 3 points off because of my arithmatic mistake is a liittle bit too tough for me.  And I also am wondering why the reader said I overcounted.....it is very obvious that I made an arithmatic mistake instead of a logic failure.  Did the reader really "read" my work?

[quote="white_horse_king88"]You have a shot maybe. Me, I have no hopes whatsoever. I think the cutoff is around 95, so you can possibly make it. I already have a total of 42, so that just drops me out permanently. I'm not sure I even want to continue, since it is now just hopeless.  :([/quote]

the cutoff can't be 95 because they only count the first 3 rounds(Round 4 is due after AIME I think!). I did horrible also, 21 with 2 commended. Learned my lesson: don't procrastinate although I think I learned that lesson many times and I still make the same mistake.

Is 71 good enough to qualify for AIME?

71 will definitely qualify for the AIME, unless everyone gets 25s on round 3 or something. The cutoff score is generally around 68 or 69.

Oh, ok. Didn't know that. But still, I have no chance, so it doesn't matter.

[quote="Erin J. Schram]I looked over your solution. Definitely one point off for assuming without proof that $a^2-2b^2$ had to be $\pm 1$. Your grader chose to take another point off, possibly due to your sloppy use of $\approx$ in which you essentially say $1\approx 0$. We prefer more rigor in the proofs.[/quote]

I guess I don't quite understand what is sloppy about my use of the approximation symbol. When 1 is as close to 0 as we can get, why is it incorrect to write $1 \approx 0$?

Thanks.

Um... maybe because that means 1 is approzimately 0.

In the set of integers, 1 is approximately zero. As far as I know, there is no formal definition describing when it is appropriate to use the approximation symbol.

23 second round and 25 first round. I got points off on the two that I thought were the easiest:1 and 3.  :blush:

[quote="ZennyK"]As far as I know, there is no formal definition describing when it is appropriate to use the approximation symbol.[/quote]
Right, that's why the approximation symbol should be avoided in a proof.  It doesn't have a definition.

[quote="mathfanatic"]I got a 25 with four commendations.  I feel bad about #2, because I did essentially the same thing as thoobik, except didn't cite the 'false theorem' that Dr. Schram noted (and I used approximation signs too!) and somehow got 5 + commendation on that one.

I guess I was the only one who made pictures for #4.[/quote]

Yeah, we were all too lazy and just listed the vertex coordinates instead.  :lol:

12...
i knew how to do htem but i procastinated till end

[quote="zanttrang"]I'm just curious, if I got a 5 with no comments, does that mean that my solution was fine and needed no commenting on, or does it mean "I can't fault your solution, but I didn't like reading through it and you should make it more concise," or does it just mean the grader was buy and didn't want to bother with making a comment?  Am I just reading into this too much?[/quote]

Some graders feel that if a solution is correct then they have nothing to say about it. Gene Berg, who runs the NSA grading sessions, tells us graders that leaving no comment is rude, but he has not convinced everybody. And he has influence over only the NSA graders (Problems 1 and 2 in Round 2).

Erin Schram
USAMTS grader

Although my score decreased from last round, I am glad the graders were stricter this round. The grader had obviously read the entire proof, and deducted a point for one bad line in the middle of it. I got some comments which were quite useful this time, unlike round 1. I am very impressed and thankful that people are willing to set up a service like this without any payment from the users. Thanks a lot.

[quote="sam3.14"]For that problem, I was going to define my own symbol, so that $f(a,b) \triangle n$ means "$f(a,b)$ is tiny if $f(a,b)$ is closer to $n$ than any number $f(c,d)$, such that $c<a$ and $d<b$. From there I could manipulate it until I had $\frac{a^2}{2b^2} \triangle 1$, so $a^2\pm 1=2b^2$. However, I decided not to use this method, because I couldn't prove that properties we use all the time with an = sign would still work with $\triangle$[/quote]

Interesting idea, but I am glad you did not submit it. Whoever graded your solution would have had to learn your notation, which is hard to read. And your $\triangle$ relation does not follow all the laws equality does. In particular, we cannot preserve the tinyness relation when we multiply both sides by the denominator.

The best approximations to $\sqrt{2}$ are
\begin{eqnarray*}
1 = \sqrt{2} - 0.414213\ldots\\
3/2 = \sqrt{2} + 0.085786\ldots\\
4/3 = \sqrt{2} - 0.080880\ldots\\
7/5 = \sqrt{2} - 0.014213\ldots
\end{eqnarray*}
so $f(1,1)\triangle \sqrt{2}$, $f(3,2)\triangle \sqrt{2}$, $f(4,3)\triangle \sqrt{2}$, and $f(7,5)\triangle \sqrt{2}$, where the function $f$ for this instance of $\triangle$ is division, $f(a,b)=a/b$.

But when we convert the function to $g(a,b) = a-b\sqrt{2}$ to get $g(a,b)\triangle 0$, not all of the above (a,b) pairs stay tiny,
\begin{eqnarray*}
1-1\sqrt{2} = -0.414213\ldots\\
3-2\sqrt{2} = +0.171572\ldots\\
4-3\sqrt{2} = -0.242640\ldots\\
7-5\sqrt{2} = -0.071067\ldots
\end{eqnarray*}
The (4,3) pair drops out of the list of tiny numbers because 3(0.080880) is bigger than 2(0.085786).

Erin Schram
USAMTS grader

I got a 23/25, no commendations. - 2 from last time. It's an understandable score, since I wasn't too rigorous with #1 or #2.

[quote="rcv"]I posted this solution to 5/2/16, case (iii) before the board crash.  I think this solution is distinct from any of those posted elsewhere.  So, here is a repost.

[Of course, if this were a real submitted solution, you might provide more than 6 words, and you would have to show the solution for cases (i) and (ii).]  ...

Inscribe.  Scale.  Cut.  Slide.  Reinscribe.  Solve.

$13s+4s=13$

$s=\frac{13}{17}$

$r=2s=\frac{26}{17}$[/quote]
Regardless, that's still my favorite solution to 5/2/16.  :)

The terseness and the sheer simplicity made me want to weep.  It's not often a proof can do that.

Actually, rcv's solution is general- it doesn't need the right angle, so it applies to all cases. The answer depends on the inradius and the side.

Here's my closely related solution: Let $ABC$ be a triangle with sides $a,b,c$ ($BC=a$) and inradius $r$. We inscribe two congruent circles $\Gamma_1,\Gamma_2$ in the triangle so that $\Gamma_1$ and $\Gamma_2$ are both tangent to $BC$, $\Gamma_1$ and $\Gamma_2$ are tangent to each other, $\Gamma_1$ is tangent to $AB$, and $\Gamma_2$ is tangent to $AC$. Let the common radius of $\Gamma_1$ and $\Gamma_2$ be $x$.

Let $DE$ be parallel to $BC$ at distance $2x$, with $D\in AB$ and $E\in AC$; $DE$ is tangent to both circles. Dissecting the sides of trapezoid $BCED$ and applying the equal tangent rule, $BC-2x+ED-2x=BD+CE$.
Let $\frac{AD}{AB}=t$. The previous equation becomes $(1+t)a-4x=(1-t)(b+c)$, or $2a-4x=(1-t)(a+b+c)$.
Now we work with area. The area of the whole triangle is $\frac12r(a+b+c)$, so the area of $ADE$ is $\frac12r(a+b+c)\cdot t^2$ and the area of $BCED$ is $\frac12r(a+b+c)\cdot(1-t^2)$. We can also find the area of $BCED$ directly; it has bases $BC=a$ and $DE=ta$ and altitude $2x$, so its area is $ax(1+t)$. Therefore, equating the two expressions for this area, $x=\frac{(1-t)(a+b+c)r}{2a}$ and $(1-t)(a+b+c)=\frac{2ax}{r}$. Substituting into the original equation, we have $2a-4x=\frac{2ax}{r}$, so $(2a+4r)x=2ar$ and $x=\frac{ar}{a+2r}$.

For the specific problem, $r=2$, so the answers are $\frac{5\cdot 2}{5+4}=\frac{10}{9}$, $\frac{12\cdot 2}{12+4}=\frac32$, and $\frac{13\cdot 2}{13+4}=\frac{26}{17}$

rcv's solution gets to the answer much faster. With a few more words, including to say that it works for all three cases, it would be the best solution I know; commended and probably published as well.

I got a 9/1 commended.  My solution to #5 was pretty much the same, but sadly, I turned in a document with solutions to only 1 and 3.  I liked the easy way to #5, too, and that's why I said it was my favorite problem on the poll.  I want to know what I likely would have gotten had I sent in the right document and what I did wrong on #3 to get a score of 4.  Check it out.

Either the judge didn't like it, or you weren't rigorous in proving {2,3,6} is indeed the smallest recipricolly whole set. THat's what I got docked for.

[quote="gaussianprime"]I made all of my solutions as long as was necessary to be rigorous.

So, I got a 25 but no commendations.

I'll try to make them shorter and more elegant for the next round, but I can only do so much.  I don't want to make them any less rigorous.[/quote]

Don't gamble by reducing rigor to increase elegant conciseness. Elegance can also be created by presenting your ideas well.

Furthermore, it is possible to earn commendations by writing up extra insights about the problems. For example, in problem 2/2/16 in Round 2, some people proved that all the tiny numbers are powers of $1-\sqrt{2}$. They earned commendations for that extra insight. (Several other people assumed without proof that all the tiny numbers are powers of $1-\sqrt{2}$ and then used that unproven fact to answer the questions. That cost them 3 points.)

Though if your solutions are over three pages each, then please shorten them in order to reduce the workload on us eye-strained graders.

Erin Schram
USAMTS grader

Yeah, I shortened my proofs and my grader said i needed to be more clear. Though I don't understand how he couldn't understand it... owell. ooo santa clause :D


calculus
integration
function
inequalities
Cauchy Inequality
complex analysis
real analysis

It is well known that if f,g:[a,b] ->R are continuous functions, the following inequality holds:

( \int _a^b f(x)g(x) dx)^2  \leq  ( \int _a^b f^2 (x) dx ) ( \int _a^b g^2(x) dx )

What is the equality case ? (I'm suspecting f/g=constant) and, more importantly, how do you prove that it's the only equality case?

Louis Augustin Cauchy-Schwarz inequality is
( \int _[a,b]f(x)g(x)dx)^2  \leq \int _[a,b]f(x)^2 dx.\int _[a,b]g(x)^2 dx

Suppose that we have equality
 (\int _[a,b]f(x)g(x)dx)^2  = \int _[a,b]f(x)^2 dx.\int _[a,b]g(x)^2 dx  (1)

with f not identically 0, since f is
continuous  \int _[a,b] f(x)^2dx >0

With (1) for any k \in R,
 \int _[a,b](kf(x)+g(x))^2dx =
\int _[a,b] f(x)^2dx.( k + \int _[a,b]f(x)g(x)dx/\int_[a,b]f^2(x)dx )^2
then you take
k= - \int _[a,b]f(x)g(x)dx/\int_[a,b]f^2(x)dx  we get
\int _[a,b](kf(x)+g(x))^2dx = 0
(kf(x)+g(x))^2 is positive continuous with integral 0 then
kf(x)+g(x) = 0
f, g are colinear.

the full name of the inequality is Cauchy-Buniakowski-Schwarz?
Is the last person's name Schwarz or Schwartz?

Yes, the full name of the inequality is Cauchy-Buniakowsky-Schwartz, and as you can see the last name is Schwartz! :D

cheers!

[quote="Lagrangia"]Yes, the full name of the inequality is Cauchy-Buniakowsky-Schwartz, and as you can see the last name is Schwartz! :D
[/quote]

I think it's Schwarz and not Schwartz.

Laurent Schwartz (french mathematician who wrote nice Analysis books and invented "distributions" (not sure it called like that in English) !)  and  Karl Hermann Amandus Schwarz (german) are two different persons.

yes, it is Schwarz

see
[url]http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Schwarz.html[/url]

I've never really tried this, but I think Cauchy's ineq in integral form can be proved by using the regular Cauchy inequality for each Riemann sum.

let h(x)=(xf(t)-g(t)) 2 .
Integrate from a to b we have  x 2  \int (a to b) f 2 (t)dt - 2x \int (a to b)f(t)g(t)dt+ \int (a to b)g 2 (t)dt  \geq 0
So the discriminent d \leq 0, hence the result.

note that equality can be obtained only when d=0, that is, there exists an x such that xf(t)-g(t)=0.
So we have the equality condition.

Many famous inequalities, such as Jensen's, Hlder's, Minkowski's are generalized in normed space.

How could Jensen's Inequality be generalized into normed space?

maybe i should not say normed space...

let (\Omega, \mu) be a measure space with \mu(\Omega) < \infty
and let f:\Omega->R be measurable. Denote

<f> := \int_\Omega f d\mu

Let g:R->R be a convex function then
<g o f> \geq g(<f>)

You also need mu(Omega)=1=<1>.
(counterexample : f=1,g(x)=x², omega=[0,2] : <f>=<gof>=2, g(<f>)=4>2)
First the result is true for affine functions : g(x)=ax+b.
(g(<f>)=a<f>+b=a<f>+b<1>=<af+b>=<gof>)
If g is convex, then g(x)=sup{h(x), h affine, h <=g}
Let h <=g be affine, we have h(<f>)=<h(f)><= <g(f)>.
So g(<f>)<= <gof>, as said.

[quote="Gillian_Seed"]If g is convex, then g(x)=sup{h(x), h affine, h <=g}.[/quote]

What was exactly meant here?

If that's not particularly clear, you can look in Rudin's Real and Complex analysis for a proof (roughly at the beginning of chapter 3) the idea of which is exactly the same, e.g, you take the sup of quotients and integrate over the entire space.


14 E
15 B
16 D
17 D
18 A
19 C
20 B
21 D
22 C
23 B
24
25

For #14, the mean is (25+x)/7, the mode is 2, and the median is:

4 if x > 4
x if 2 < x < 4
2 if x < 2

Since the arithmetic progression is non-constant, the median cannot be 2, so x >= 2.  If 2 < x < 4, then the sequence is either mode, median, mean or mode, mean, median, so
$\frac12 \left(2 + \frac{25+x}{7}\right) = x$
$2 + \frac{25+x}{7} = 2x$
$14 + 25 + x = 14x$
$x = 3$

OR

$\frac12 \left(2 + x\right) = \frac{25+x}{7}$
$14 + 7x = 50 + 2x$
$5x = 36$
$x = 7.2$

but 2 < x < 4, so only x = 3 works.

If x > 4, then the mean is > 29/7 > 4, so the sequence must go mode, median, mean, i.e.

$\frac12 \left( 2 + \frac{25+x}{7}\right) = 4$
$2 + \frac{25+x}{7} = 8$
$\frac{25+x}{7} = 6$
$25+x = 42$
$x = 17$

so the answers are x = 3 or 17.  3+17=20 (E).

I'm still working on #24 and #25.


function
inequalities
algebra
difference of squares
special factorizations

Find the smallest positive number from the numbers below
$\text{(A)} \ 10-3\sqrt{11} \qquad \text{(B)} \ 3\sqrt{11}-10 \qquad \text{(C)} \ 18-5\sqrt{13} \qquad \text{(D)} \ 51-10\sqrt{26} \qquad \text{(E)} \ 10\sqrt{26}-51$

Obviously no calculators allowed...

[hide="Answer"]We use the method of approximating square roots where we approximate it by an integer, divide the number inside the square root by that number, and then average the result of that with our original number, repeating the process as many times as desired. We see that $\sqrt{11}\approx 3.3166$, $\sqrt{13}\approx 3.6057$, and $\sqrt{26}\approx 5.099$. Multiplying each answer choice out, we have $A\approx 0.0502$, $B\approx -0.0502$, $C\approx -0.0285$, $D\approx 0.01$, and $E\approx -0.01$, so our answer is $\boxed{D}$.[/hide]

There has to be a better way to do that. :huh:

Hint:

Consider 10 and $3 \sqrt{11}$. Which one is larger? By squaring each expression, you should see 100 > 99 and from here, the some of the choices get eliminated and you can create a function to determine the answer.

Ah, I see. For one, you could eliminate all choices but A and D, then just find the values for those.

[hide]Take the reciprocal of each number. A becomes $\frac{1}{10+3\sqrt{11}}$ and D becomes $\frac{1}{51+10\sqrt{26}}$, of which D is the smaller. The other three are negative (because the difference of squares is). $\boxed{D}$.[/hide]

you can eliminate most of the choices, and then just set them in an inequality, and show that the inequality faces a certain direction

$ f(x)\equal{} \sqrt{x\plus{}1}\minus{} \sqrt{x}$ is strictly decreasing over the positive reals. A is $ \sqrt{100}\minus{}\sqrt{99}$, D is $ \sqrt{2601}\minus{}\sqrt{2600}$, and C is $ \sqrt{324}\minus{}\sqrt{325}$ which is just negative, so it has to be D.


logarithms
limit
trigonometry
inequalities
real analysis
real analysis unsolved

For what $x$ does the series $\sum{\frac{(-1)^n}{\ln n+2\cos(xn)}}$ converge?

This has already been posted by alekk: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19407[/url]. The best I can do is to show that the series converges for almost all $x$ (actually, for all $x$ that cannot be approximated by rational multiples of $\pi$ too well). Do you or your teacher have a full solution?

No, fedja, I don't have a full solution (actually, I don't have a solution at all) and my teacher said something about Fourier series, but I don't think he has a full solution.  ;)

I have a very loose argument (very far from what you would call a proof) that suggests that the answer is the following. Let $\gamma_s$ ($s=0,1,2,\dots$) be the distance from $\pi+sx$ to the nearest integer multiple of $2\pi$. Let $\Gamma_s=\frac 1{\gamma_s}\left(\log\frac9{\gamma_s}\right)^{-(s+1)}$ (if $\gamma_s=0$, then $\Gamma_s=+\infty$). Then the series converges if and only if $\lim_{s\to \infty}\Gamma_s=0$. What does your teacher think about such an answer, Harazi?   :)

Indeed, he told me he wrote in an article very long time ago and the authors were speaking about this magical sequence. So, this must be true. But unfortunately he has no longer the article...  :(

[quote="harazi"] So, this must be true. But unfortunately he has no longer the article...  :([/quote] Well, we can try to figure it out then. It helps a lot to know that we are on the right track  :) . I'll try to make my loose argument into something more rigorous. If I succeed, I'll post it in a few days.

Here are my ideas. They don't amount to a rigorous proof yet, but now we are a bit closer. Also I'll just indicate the main steps, not every technical detail. As usual, questions and comments are welcome. I apologize for the sloppiness of the text below. That sloppiness comes from two sourses: firstly, I do not have a full proof yet, so sometimes I just have to wave hands instead of saying something rigorous and, secondly, I tried to make this post relatively short, so some things got omitted even though I knew how to do them.

Let's try to solve a bit simpler problem about convergence of the series $\sum_n \text{Re\,} \frac{(-1)^n}{\ln n+e^{inx}}$. The original problem can be approached in the same way but the technicalities become even more complicated. First of all, write
\[ \sum_{n=N_1}^{N_2}\text{Re\,}\frac 1{\ln n+e^{inx}}=\sum_{s\ge 0}(-1)^s\sum_{n=N_1}^{N_2}\frac{\cos n\gamma_s}{(\ln n)^{s+1}}\,. \]
Now, suppose that one of $\gamma_s$ is $0$. Let $s$ be the smallest index with this property. Then it is fairly easy to see that, for large $N_2$, the sum corresponding to this $s$ is large and dominates all other sums (the previous ones because the corresponding series converge and the next ones because it has much bigger terms). Also, in this case, $\Gamma_{ks}=+\infty$ for every odd $k$. So, $\Gamma_s$ does not tend to $0$. That reduces our task to considering the case when all $\gamma_s>0$. We'll need two classical lemmas about Fourier series.

Lemma 1:
Let $\gamma>0$. Let $\sigma(N)=\sum_{n=0}^N a_n \cos {n\gamma}$ where $a_n$ is a decreasing sequence of positive numbers. Then $|\sigma(N)|\le C\sum_{n=0}^M a_n$ for all $N$ where $M$ is the least integer that is greater than $\frac \pi\gamma$.

Lemma 2:
If, in conditions of Lemma 1, we also assume that $a_n$ is convex and $a_n\to 0$, then the series $\sum_{n=0}^\infty a_n \cos{n\gamma}$ converges and its sum $\sigma(\infty)$ satisfies $|\sigma(\infty)|\le C\sum_{n=0}^M(a_n - a_M)$.

Let $M_s$ be the least integer greater than $\frac \pi{\gamma_s}$. Put $\widetilde\Gamma_s=\sum_{n=10}^{M_s}\frac 1{(\ln n)^{s+1}}$. Since $\ln$ changes very slowly, $\widetilde\Gamma_s$ is approximately the same as $\Gamma_s$. In particular, $\widetilde\Gamma_s$ and $\Gamma_s$ do or do not tend to $0$ simultaneously. Our first task will be to try to show that, if $\Gamma_s$ does not tend to $0$, then the series diverges. So, let us assume that for some $\delta>0$ there is an arbitrarily large $s$ with $\Gamma_s\ge \delta$. Then the corresponding $\gamma_s$ must tend to $0$ very fast, faster than $\frac 1{(s\ln s)^s}$. Let us try to take $N_1=M_s$, $N_2=\frac32 M_s$. The logarithm does not change much within these limits, so I'll treat it like a constant (this is a sloppy place where I cheat a bit, but if you take $N_1=s^3M_s$ and $N_2=\left(s^3+\frac12\right)M_s$, the change in logarithm over the interval becomes really negligible and $(\ln M_s)^{s+1}$ is still almost the same as $(\ln (s^3M_s))^{s+1}$). Then the term $(-1)^s\sum_{n=N_1}^{N_2}\frac{\cos n\gamma_s}{(\ln n)^{s+1}}$ contributes about $\Gamma_s$ to the sum of the original series between $N_1$ and $N_2$. This contribution, clearly, cannot be cancelled by the terms corresponding to larger $s$ but still there is a danger that it will be cancelled by the terms with smaller $s$. So here we come to the

Problem:
Let $b_1>b_2>\dots>b_s>0$ be a fast decaying sequence of numbers (more precisely, we can assume that $b_{j+1}<\frac 1{s}b_j$ for all $j$. Let $\gamma_1,\dots,\gamma_s>0$. Assume that $I$ is an interval of length about $\frac\pi\gamma_s$ on which $\cos (n\gamma_s)$ preserves sign. Then $\max_{N_1,N_2\in I}\sum_{n=N_1}^{N_2} \left(\sum_{j=1}^s b_j\cos n\gamma_j\right)\ge c \frac{b_s}{\gamma_s}$ with some absolute $c>0$.

This problem is the only thing that separates us from the full proof of the necessity of the condition $\lim_{s\to\infty} \Gamma_s=0$ for the convergence of the series. If somebody solves it, I can supply all other details (most of them are just boring technicalities).

Now let us try to prove that this condition is sufficient for the convergence. Again, we shall use the Cauchy criterion and try to prove that the sum from $N_1$ to $N_2$ is small if $N_1$ is large enough. To this end, let us note that Lemma 1 allows us to neglect the terms with $s$ such that $\widetilde\Gamma_s<\frac 1{s^2}$, say. Now, if $\Gamma_s>\frac 1{s^2}$, we must have $\gamma_s<2^{-s}$ at the very least. This means that $|sx-\ell\pi|<2^{-s}$ for some odd integer $\ell$. My next claim is that the fraction $\frac \ell s$ must be irreducible then (for large enough $s$). Indeed, if it were reducible and we had $s=s'm$,$\ell=\ell'm$ for some $m$, then $\gamma_{s'}$ would be less than $\gamma_s$ and $\widetilde\Gamma_{s'}$ would be at least $2^{(m-1)s'}\widetilde\Gamma_s>2^{s/2}\widetilde\Gamma_s$. Since $\widetilde\Gamma_s$ is a bounded sequence, we would have then $\widetilde\Gamma_s<C2^{-s/2}<\frac 1{s^2}$ for large enough $s$. Now let us see how often we may have the inequality $|sx-\ell\pi|<2^{-s}$. If you have it for the pair $(s,\ell)$ and also for the pair $(S,L)$ with $S>s$, we have $\left|\frac\ell s-\frac LS\right|<\frac 1{\pi s 2^s}+\frac 1{\pi S 2^S}<\frac 1{s 2^s}$. Since the fractions we subtract are irreducible, the left hand side of the last inequality is at least $\frac 1{sS}$. Hence, $S>2^s$. So, we see that our sequence of $s$ for which $\widetilde\Gamma_s>\frac 1{s^2}$ is extremely sparse: as sparse as a tower of exponents. Moreover, the corresponding $\gamma_s$ must satisfy $1>\Gamma_s>\frac 1{s^2}$, which determines them up to a factor about $s^2$. It is not hard to show that the corresponding sequence $M_s$ is also as sparse as a tower of exponents. Now let us look at the sums $\sum_{n=N_1}^{N_2}\frac{\cos n\gamma_s}{(\ln n)^{s+1}}$ for such $s$. I claim that almost all of these sums are very small. First, we can easily get rid of the sums with $M_s>N_2$. Indeed, they contribute only about $\frac{N_2}{M_s}\Gamma_s$ which is easy to sum because $M_s$ is sparse and $\Gamma_s<1$ for large $s$. Secondly, let us observe (and this is a key step) that Lemma 2 allows to conclude that the full sum $\sum_{n=10}^{\infty}\frac{\cos n\gamma_s}{(\ln n)^{s+1}}$ is much less than $\widetilde\Gamma_s$ (about $\frac 1{\ln s}\widetilde\Gamma_s$, to be exact). So, the sums for which $N_1<M_s$, $N_2>M_s$ can also be safely disposed with (they are almost full sums). At last, the sums with $N_1>M_s$ are tails of convergent series and can also be estimated by something much smaller than $\Gamma_s$ (something like $\frac{\ln M_s}{\ln N_1}\widetilde\Gamma_s$). So, the moral of this story is that only a few $s$ can contribute as much as $\widetilde\Gamma_s$ to any fixed sum and all other $s$ have to contribute much less. That's why I expect the condition $\lim_{s\to 0}\Gamma_s=0$ to be sufficient already (using Lemma 1 alone, it is not hard to show that the condition $\sum_s\Gamma_s<+\infty$ would suffice). Actually, I believe that I have a full proof in this part though to write it down accurately will be a headache.

Again, please accept my apologies for the sloppiness of this text. If anybody gets interested, I'll gladly try to answer all questions that he/she may have.     :blush: :P


Using Geometry for Probability Part 1

We use length to solve a probability problem that our usual counting methods can't solve.

He's pretty funny...."No one has ever one!" I'm pretty sure I would've won cause I'm am a lucky person......

This was the first video I watched. I must say AoPS did a great job putting together all of these videos. Kinda reminds me of some children's show as he is talking directly to us though the camera lol.

wat topic do u have to pass, so then u can get this video???

He's funny :D . Also, he picked $ 4 $ and I picked $ 5 $. I one!

I still don't understand why its 9. Can someone explain it to me?


For how many positive integers $ x$ is $ x^2 \plus{} 6x \plus{} 9$ between 20 and
40?

Simply and get (x+3)^2
Since x is an integer, (x+3)^2 must be an integer.
(x+3)^2 can equal 25 or 36, so x+3 equals 5 or 6.
x=2 and 3.
There are 2 values.


Despite the obvious statistics of the board that tell us that you would be the 4256th registred user on the site, your user ID is in fact 4399. :D

That is, assuming that your "real" username is indeed wassup471 :)

Good luck on the competition.

it's the last number you see when moving over your own profile button: ...viewprofile.php?u=4399 in your case.

[quote="wassup471"]I've got to have my USAMTS User ID and my AoPS Community Username for 3 forms for school so I can do the competitions. Problem is, I can't seem to find the User ID (assuming my AoPS username is wassup471)

Can anyone help?[/quote]

Also, the User ID is listed on the My USAMTS Profile page when you are logged in.


function
calculus
derivative
analytic geometry
graphing lines
slope
algebra

Hi guys,

My friend and I found this interesting problem about (implicit) differentiation:

-------------------
Find the equation of the tangent in $ ( \minus{} 1; \minus{} 1)$ of

$ \frac {x}{y} \plus{} \left(\frac {y}{x}\right)^3 \equal{} 2$
--------------------

We found that the derivative is $ f'(x) \equal{} \frac {f(x)[3f^6(x) \minus{} x^2]}{x[3f^6(x) \minus{} x^2]} \equal{} \frac {y}{x}$ (really strange, isn't it?!), so, in $ ( \minus{} 1; \minus{} 1)$, the slope of the tangent line is $ \frac { \minus{} 1}{ \minus{} 1} \equal{} 1$,
and the equation is $ y \equal{} x$.When we plotted the function, we got very surprised, because we imagined everything except two lines.

So, we tried to obtain an explicit function from $ \frac {x}{y} \plus{} \left(\frac {y}{x}\right)^3 \equal{} 2$, and we only found a factorization with the Ruffini's rule.
We got:

$ (x \minus{} y)(x^3 \minus{} x^2y \minus{} xy^2 \minus{} y^3) \equal{} 0$

So, one line is y=x... but the other? How can I factorize $ (x^3 \minus{} x^2y \minus{} xy^2 \minus{} y^3)$?

What would the slope of the second line be? Differentiating $ (x^3 \minus{} x^2y \minus{} xy^2 \minus{} y^3)$ is a good idea?

Thanks in advance

Gabriele  :)

Let $ z \equal{} \frac{x}{y}$.  The polynomial $ z^3 \minus{} z^2 \minus{} z \minus{} 1$ has one positive root and either two or zero negative roots by Descartes' law of signs.  But if $ z < 0$ then $ z^3 < 0 < \frac{3}{4} < z^2 \plus{} z \plus{} 1$, so exactly one of the roots of this polynomial is real.  The second line you want has slope (the inverse of) this root.

Thank you very much t0rajir0u !!
(I'm the Gabuntu94's friend who worked with him about this problem).

L. :)


Suppose two distinct integers are chosen from between 5 and 17, inclusive.  What is the probability that their product is odd?

They need to both be odd.  There are $ 7$ odd numbers and $ 6$ even numbers in that set.  Thus, the answer is $ \frac{7}{13} \times \frac{6}{12}\equal{}\boxed{\frac{7}{26}}$.


Let $ x;y;z;t$ be positive real numbers .Prove that:
$ 3(x^2\plus{}y^2\plus{}z^2\plus{}t^2)\plus{}4\sqrt{xyzt} \geq (x\plus{}y\plus{}z\plus{}t)^2$

Easier problem:
Let $ x;y;z$ be positive real numbers.Prove that:
$ 2(x^2\plus{}y^2\plus{}z^2)\plus{}3(xyz)^{2/3} \geq (x\plus{}y\plus{}z)^2$

[quote="chien than"]Easier problem:
Let $ x;y;z$ be positive real numbers.Prove that:
$ 2(x^2 \plus{} y^2 \plus{} z^2) \plus{} 3(xyz)^{2/3} \geq (x \plus{} y \plus{} z)^2$[/quote]
Only Schur and Muirherd :wink:

general form
$ 3\minus{}t\plus{}t(abc)^{t} \plus{} \sum_{cyc} a^{2} \geq 2 \sum_{cyc}ab$
$ t \in (0,3]$ $ a,b,c \geq0$


vector
linear algebra
matrix
superior algebra
superior algebra solved

Let M[size=59]2[/size](Q) denote the ring of 2 x 2 matrices with rational entries and let S be a sub-ring of M[size=59]2[/size](Q). If (all) the units of M[size=59]2[/size] are contained in S, then prove that S=M[size=59]2[/size](Q).

For a more general case it is enough to prove that the invertible elements of $M_n(k)$, where $k$ is an infinite field, span $M_n(k)$ as a vector space.

Let $E_{ij}$ be the matrix that has $1$ on the $(i,j)$ entry and $0$ in the rest.
It is then easy to prove that the matrices $I_n+E_{ij}$ for all $i\neq j$ and $diag(a_1,\ldots,a_n)$ with $a_i\in k^*$ for all $i=\overline{1,n}$ span $M_n(k)$ as a vector space. To see this, take $A=(a_{ij})_{i,j=\overline{1,n}}\in M_n(k)$. If $A=0$, then we have nothing to prove.
Otherwise, $A=\sum_{i\neq j}a_{ij}(I_n+E_{ij})+(b+\sum_{i\neq j}a_{ij})I_n-bI_n+diag(c+a_{11},\ldots,c+a_{nn})-cI_n$, where $b$ and $c$ are chosen such that $bc(b+\sum_{i\neq j}a_{ij})(c+a_{11})\ldots(c+a_{nn})\neq 0$. This choices are possible because $k$ is infinite by hypothesis.

Does the statement of the problem hold in finite fields?

Yes it does. It is enough to write $E_{ij}$ as a linear combination of invertible matrices, because the $E_{ij}$s form a vector space basis for $M_n(k)$ over $k$.
For $i\neq j$, we have $E_{ij}=(I_n+E_{ij})-I_n$.
For $i=j$ I will only treat here the case $i=j=1$. The others can be treated similarly.
We have $E_{11}=(I+E_{21})-I+(I+E_{12})-(I-E_{11}+E_{12}+E_{21})$.


A building is 1225 feet tall and is situated about 1 miles inland from a shore. Ritter is in his pleasure boat in a lake directly in front of the building and takes an angle of elevation and found it to be 7 degrees. How far offshore is the boat?

This does not belong in intermediate topics because it is simple triangle trig

[hide]
$ \tan{7}= \frac{1225}{1+d}$

$ d = \frac{1225}{\tan{7}}-1$
[/hide]


Since Yasha posted a prisoner problem, I decided I would post another.

There are 100 prisoners in a large room. The jailkeeper walks into the room and tells the prisoners that they will play a game. In one hour, all the prisoners will be brought to separate rooms and will be unable to communicate with each other. Each day, the jailkeeper will select exactly one of the prisoners and will bring that prisoner back to the common room, where there is exactly one light switch. The prisoner can then choose whether to change the position of the light switch (the only positions are ``on'' and ``off'') or to leave it alone. At any point, one of the prisoners can decide to tell the jailkeeper that every prisoner has returned to the room. If this prisoner is correct, all will be set free. Otherwise, they will all die. How can they be set free?

But can't the jailkeeper keep bringing in the same guy?

Ok so assume that the jailkeeper will bring in everyone infinitely many times or as many times as needed. For example, suppose he picks randomly.

Will the prisioners know what the original position of the light switch will be?

Ugh.... u pick this problem. Did you get it from William's site?

Also, will the prisioners know how many days the game has been going on, or will they be in a dark cell and not aware of the time?

Oops. At the beginning the light is on. No, I did not get it from William's site (who is William?). The prisoners will know how many days the game has been going on.

[hide]Designate one of the prisoners as the 'counter.' When someone who isnt the counter goes into the room and sees the light on, he 'announces his presence' by turning it off. Then, no one else touches the off switch until the counter goes in again. The counter flips the switch on and adds 1 to his count. Everyone announces their presence exactly once. When the counter counts to 99, he knows that all of the others have been in there.[/hide]

Harder problem...... what if they didnt know if it would start on or off?

[hide]then the first guy just flips it on[/hide]

The strategy that zabelman offered is the better of the two strategies I know. I think there's even a better one though.

probability.... the first guy doesnt know hes the first guy

Zabelman's solution only works if the jailkeeper chose the prisioners randomly. Spoiler: [hide]If however, he eavesdropped on the prisioners' conversation and only led the counter in once, the jailkeeper could then lead every prisioner into the room and yet the prisioners would never know they've all been there.[/hide]

Yasha, that's not true if all prisoners visit the room infinitely often.

[quote="ComplexZeta"]The first guy definitely knows he's the first guy, but it doesn't help because no one else knows.[/quote]

It doesn't matter.  On the first day, whoever gets picked turns the light on.  Then, continue with the original strategy starting from day 2.

I've heard this problem before, and I'm pretty sure there's a solution in which the prisioners know they've all been in the room when the last prisioner enters it for the first time, so I'll try to figure it out.

Ok, for my 'harder' problem, assume that the jailkeeper doesnt announce the day that the game starts, so that the first person DOES NOT know that he is the first person. All other rules are the same. Oh, and they DONT know if the lightbulb starts on or off. How do they win?


algebra
polynomial
group theory
superior algebra
superior algebra unsolved

Describe the field $ F\equal{}\frac{\mathbb{F}_3[x]}{(p(x))}$ [$ p(x)$ is an irreducible polynomial in $ \mathbb{F}_3[x]$].  Find an element of $ F$ that generates the cyclic group $ F^*$ and show that your element works.

[$ p(x)\equal{}x^2\plus{}1$ is irreducible in $ \mathbb{F}_3[x]$ if that helps]

[quote="numbertheory12"]Describe the field $ F \equal{} \frac {\mathbb{F}_3[x]}{(p(x))}$ [$ p(x)$ is an irreducible polynomial in $ \mathbb{F}_3[x]$].  Find an element of $ F$ that generates the cyclic group $ F^*$ and show that your element works.

[$ p(x) \equal{} x^2 \plus{} 1$ is irreducible in $ \mathbb{F}_3[x]$ if that helps][/quote]

Using $ x^2 \plus{} 1$, elements are of the form $ ai \plus{} b$, with $ a,b \in \mathbb{F}_3$, and $ i^2 \equal{} 2$.

The element $ i \plus{} 1$ has order $ 8 \equal{} |\mathbb{F}_3^{\times}|$. We have $ (i \plus{} 1)^2 \equal{} 1 \plus{} 2 \plus{} 2i \equal{} 2i$, and $ (i \plus{} 1)^4 \equal{} (2i)^2 \equal{} i^2 \equal{} 2$, so $ i \plus{} 1$ has order $ 8$ and is a generator. The other generators are $ (i \plus{} 1)^3 \equal{} 1 \plus{} 2i$, $ (i \plus{} 1)^5 \equal{} 2 \plus{} 2i$, and $ (i \plus{} 1)^7 \equal{} 2 \plus{} i$.

If $ deg(p(x))\equal{}n$, then $ \frac{\mathbb{F}_3[x]}{(p(x))}\cong \mathbb{F}_{p^n}$.

[b]calc rulz[/b], what is $ i$ in your proof? I am not following your proof.

I know that when we form $ \mathbb{F}_3[x]/(x^2\plus{}1)$ we get $ \{ a\plus{}bx\plus{}(x^2\plus{}1): a,b\in \mathbb{F}_3\}$.
Now, $ \exists$ eight elements in the multiplicative group.  So, we look at each one of the eight and finds its order.  I think this is what you were doing.  I just don't know what $ i$ is in your proof.

$ i$ is $ x$; that is, it's a primitive second root of unity embedded into your field.  Note that $ i^2 \equal{} 2$ is equivalent to $ i^2 \equal{} \minus{}1$ in characteristic $ 3$.


calculus
integration
email
calculus computations

hello, send me your email adress and i will send you a Maple file with the solution of your integral.
Sonnhard.


Given the equation
\[ \left(x^2\minus{}3x\minus{}2\right)^2\minus{}3(x^2\minus{}3x\minus{}2)\minus{}2\minus{}x\equal{}0\],
prove that the real roots of the equation $ x^2\minus{}4x\minus{}2\equal{}0$ are roots of the initial equation and find all real roots of the given equation.

[hide="Non-Ohio-solution...:)"]

one line solution: expand the 1st equation, divide the LHS with LHS of the 2nd equation. As it is divisible, first part is proven. Dividing will give you a quadratic polynomial as the result, and it's easy to find its roots, so the second part is solved.
[/hide]

hsiljak, your solution is almost identical to mine. :)

Believe it or not, this was an olympiad problem from like 15 years back

Well, I believe... I've seen more than few trivial olympiad problems in my problem-solving career... (Even IMO level... I just wondered how come I never got them on the actual competitions, but only in my textbooks) :D


probability
expected value
superior algebra
superior algebra unsolved

if$(a_1,...,a_n)\in R^n$ prove that these are equivalent:
i) for any Variable $X>0,\prod_{i=1}^{n}E(X^i)^{a_i}\geq 1$
ii)$\sum _{i=1}^{n}i.a_i=0$ for $k=1,2,...,n-1$ ,$\sum_{i=1}^{k}i.a_{n-k+i}\geq 0$

oh ! any idea, I'm looking for easy access sol because the official one is realy terrible.

I don't quite understand the problem text.
What is $E(X^i)^{a_i}$?

amfulger ,E means the expected value of x ,xis variable, for example we have $E(x^2)\geq E(x)^2$ because $E((x-E(x))^2)=E(x^2)-E(x)^2$so we have $E(x)^{-2}.E(x^2)\geq 1$

Beats me  :( . Find yourself another solver for this problem!

oh no! you're just active and briliant problem solver in college play ground ,plz help.

it is more than three month that no one get back to this problem who wants to try?


Let a(n), n>=1 be an increasing sequence of positive real numbers.  Show that

(1/a(1))+(2/(a(1)+a(2))+...+(n/(a(1)+a(2)+...+a(n)) <= 4((1/a(1))+(1/a(2))+...+(1/(a(n))).

I've tried several things and I'm not getting anywhere...

http://www.mathlinks.ro/Forum/viewtopic.php?t=18543
http://www.mathlinks.ro/Forum/viewtopic.php?t=18542

  darij


search
function
calculus
integration
college contests

Could anyone be so kind and copy the problems from 16th competition here ?
After searching in the web I've found nothing.  :huh:

[quote="José"][quote="Moubinool"]Could we have the problems of Vojteck Jarnik 2006 ?

thank you
m.o.[/quote]

maybe not yet[/quote]

Better set up an easy internet search instead making lots of uninformed guesses, e.g. look [url=http://prf.osu.cz/kma/dokumenty/vjaimc/j16second.html]here.[/url] Be careful, the entropy of your postings has been pretty low recently.

as i see there is nobody willing to post the problems, so i will take the responsibility.
Category 2.

Problem1. a).Recall we have a commutative ring and u and v are two nilpotent elements. show that u+v is nilpotent
 b)  show an example of two elements u,v which are nilpotent in a non-commutative ring, such that u+v is not nilpotent
Problem2) Show that a finite grup has the property that every element is a square if and only if  its rank is odd
Problem 3) for a  continuous function defined on an interval [a,b] we will say that it crosses its graph for (c,d) , c,d are from the interval [a,b] if there exists an e in (c,d) such that  the point (e,f(e)) lies on the line through (c,f(c)) and (d,f(d))
a) find the set of all continuous functions f , such that f crosses its graph for every two points in the interval [a,b]
b) does there exist a continuous function f , which is not of the type described in a) such that for every choice of two rational numbers c,d in the interval [a,b] the function crosses its graph for (c,d)
Problem 4. the function f is defined on [0,\infinity] is strictly convex, continuous(sic!!) and f(x)/x tends to infinity if x tends to infinity  (as i remember these were the conditions).
Show that  the improper integral from 0 to infinity of the function sin(f(x)) is convergent but it is not absolutely convergent.

These were the problems. my personal impression is that this competition is becoming more and more a grotesque of a math competition.
 as you see there are in fact eight problems to solve(every one contains two points). the first two problems were solved by all but the very least prepared. The fourth problem is in fact pure technical counting. many big names became underdogs, whereas the category was dominated by Czechs from Prague.

Well, since there is no reply to my last post, i think that you were not interested in the problems and i spent my time writing them for nothing. Ok if this is the case i won't post the first category. Bye!

Thank you very much qw for posting the Vojtech Jarnik problems 2006
I think many mathlinks members and I, appreciate a lot you post them on
the forum.

Very sorry for this late answer.

thank 's again for the problems qw

Moubinool


search
number theory unsolved
number theory

[i]For a positive real number $ \alpha$ , define

  $ \mathbb{S} ( \alpha ) \ \equal{} \ \{ [n \cdot \alpha] | n \equal{} 1 ;2 ; ... \}$

  Prove that $ \mathbb{N}$ cannot be expressed as the disjoint union of three sets $ \mathbb{S} ( \alpha ) ; \mathbb{S} ( \beta ) ; \mathbb{S} ( \gamma )$[/i]

it's probably been posted before. search Beatty's Theorem.

if you don't find it, here's a hint anyway. assume that such a partition exists, and show that this implies $ \alpha, \beta, \gamma$ are irrational and $ \frac{1}{\alpha}\plus{}\frac{1}{\beta}\plus{}\frac{1}{\gamma}\equal{}1$.


[quote="L_Euler"]Giải pt: $ 2^x \equal{} x^2$

 :o[/quote]
Dễ thấy x=2 là no.
Ta c\m $ f(x)\equal{}2^x\minus{}x^2$ đơn điệu là xong.không biết đúng ko nữa bạn tự kiểm tra xem sao. :)

Bài này nếu vẽ đồ thị của hai hàm số sẽ thấy cọn có một nghiệm âm nữa. Khó ghê :!:

bài nay là tìm nghiệm nguyên thì ta quy nạp còn no hữu tỉ và vô tỉ thì em chịu  :roll:

Bài toán này chưa tổng quát đâu. Bài toán tổng quát hơn là:
$ x^y\equal{}y^x$
Xem chi tiết cách giải bài này trên web sau đây nhé:
http://mathforum.org/library/drmath/view/66166.html


inequalities
function
inequalities unsolved

Let $ A,B,C$ be three angles of a triangle, show that
\[ \frac {sinA}{sinB \plus{} sinC}(\frac {sinB}{sinC})^2(\frac {1 \minus{} cosC}{1 \plus{} cosB})^2 \plus{} \frac {sinB}{sinC \plus{} sinA}(\frac {sinC}{sinA})^2(\frac {1 \minus{} cosA}{1 \plus{} cosC})^2
\]

\[ \plus{} \frac {sinC}{sinA \plus{} sinB}(\frac {sinA}{sinB})^2(\frac {1 \minus{} cosB}{1 \plus{} cosA})^2 \leq\frac {1}{6}.
\]

$ \frac{sinA}{sinB\plus{}sinC}\equal{}\frac{a}{b\plus{}c}$ and $ cosC\equal{}\frac {a^2\plus{}b^2\minus{}c^2}{\ {2ab}}$ ...
then the inequality becomes
$ \sum_{cyc}\frac {a}{b\plus{}c}(\frac {b\plus{}c\minus{}a}{\ {a \plus{} b \plus{} c}})^2\leq \frac {1}{6}$  (1)
Since this inequality is homogeneous , we may assume that $ a\plus{}b\plus{}c\equal{}1$
With the substitution $ x\equal{}b\plus{}c\minus{}a , y\equal{} c\plus{}a\minus{}b ,  z\equal{}a\plus{}b\minus{}c$
the inequality (1) becomes$ \sum_{cyc}\frac {1\minus{}x}{\ {1\plus{}x}}x^2\leq \frac {1}{6}$   wich is true because the function $ \frac {1\minus{}x}{\ {1\plus{}x}}x^2$  is concave and $ x\plus{}y\plus{}z\equal{}1$

[quote="Srhiri"]wich is true because the function $ \frac {1 \minus{} x}{\ {1 \plus{} x}}x^2$  is concave and $ x \plus{} y \plus{} z \equal{} 1$[/quote]

it's not concave  :|   $ f''(x)\equal{}\frac{\minus{}2}{(x\plus{}1)^3}(x^3\plus{}3x^2\plus{}3x\minus{}1)$


Ring Theory
function
induction
number theory
prime numbers
superior algebra
superior algebra unsolved

Let $ R$ be an infinite ring such that every subring of $ R$ different from $ \{0 \}$ has a finite index in $ R$. (By the index of a subring, we mean the index of its additive group in the additive group of $ R$.) Prove that the additive group of $ R$ is cyclic.

[i]L. Lovasz, J. Pelikan[/i]

Solution:

Part 1:

In this part we will prove that there exists $\xi \in R: R=\{r_1\xi+r_2\xi^2+...r_k\xi^k+...\}_{r_1,r_2,...,r_k,...}$.

Denote $\wp$ as prime set. $\lambda R=\{\lambda*a| a\in R\}$.

Lemma 1:

$\forall p\in \wp, pR\not= \{ 0\}$

From Lemma 1 we can prove that $\forall a\in R\setminus \{ 0 \}, n\in Z\setminus 0$, $an\not=0$ (because if $pa=0$ for prime p, then $R_p=\{ a\in R| pa=0 \}$ is subring of $R$, so $|R/R_p|<\infty$ and $|R_p|=\infty$, $pR=\{ 0 \}$ and use Lemma 1).

Lemma 2:

$\forall p\in \wp, pR\not= R$

Next Lemmas will work when $R:$Lemma 2 holds, for other rings $R$ we will use modifications of this Lemmas wich are in Part 2.

From Lemma 2 we get that $\cap_{i\in Z\backslash 0}p^iR=\{0\}$ (if $\cap_{i\in Z\backslash 0}p^iR=R'\not=\{0\}$, then $pR'=R'$).

Denote function $\nu_p(a)=max\{i\in Z|a\in p^iR\}$, $\cap_{i\in Z\backslash 0}p^iR=\{0\}$, so $\forall a\in R, 0 \leq \nu_p(a)<\infty$.

Lemma 3:

$\forall p\in \wp , |R/pR|=p^n$, for some positive integer $n$.

$\bigtriangleup$

Construct rings $pR=R_0\subseteq R_1\subseteq R_2 \subseteq ...\subseteq R_{k-1}\subseteq R_k=R$, such that $\forall i, \forall R^*: R_i \in R^*\in R_{i+1}$, $R^*=R_i$, or $R^*=R_{i+1}$, $\forall i,\exists a_i\in R_{i+1}\backslash R_i$. $\alpha_i=a_i+R_i$, from defenition of $R_i$, $R_{i+1}/R_i=\{x_1\alpha_i+x_2\alpha_i^2+...+x_h\alpha_i^h+...\}_{x_1,x_2,...,x_h,...\in Z}$, $\forall r\in R_{i+1}/R_i, p*r\in pR\subset R_{i}$, so $R_{i+1}/R_i$ is ring over field $Z/pZ$, so if $f^*(x)=u_1x+u_2x^2+....+x^m$ is minimal polynom were $f^*(a)=0$ (he exist because $|R_{i+1}/R_i|<\infty$), then $R_{i+1}/R_i\cong Z[x]/(px,x^m)$, so $|R_{i+1}/R_i|=|Z[x]/(px,x^m)|=p^{m-1}$, $n=m-1$. $\Box$

Lemma 3':

$\forall \lambda \in Z, q||R/\lambda R|\Leftrightarrow q|\lambda, q\in \wp$.

$\bigtriangleup$

Let $\lambda = p_1^{k_1}p_2^{k_2}...p_i^{k_i}$, we will use Induction by $k_1+k_2+...+k_i$. $\lambda R=p_1R'$, were $R_1=\frac{\lambda}{p_1}R$, from Lemma 3 $|R/\lambda R|=|R/R'|*|R'/p_1R'|$ and from Lemma 3 and Induction we get Lemma 3'. $\Box$

Lemma 4:

$\forall p \not= q, p,q \in \wp$, $pR\cap qR\not= pR$.

$\bigtriangleup$

Let $pR\subseteq qR$, then $p^{n_1}=|R/pR|=|R/qR|*|qR/pR|=q^{n_2}|qR/pR|$. Contradiction. $\Box$

Lemma 5:

$\forall a\in R, \exists N>0: \forall p\in \wp , p>N, a\in R\backslash pR$.

$\bigtriangleup$

Let exists $\{p_i\}_{i=1}^{\infty}:p_i\in \wp$, $p_1<p_2<p_3<...$, $\forall i\in Z, a\in p_iR$, then $\cap_i p_iR=R^{\wp}\not= \{0\}$. $R^{\wp}$ is ring, $|R/R^{\wp}|<\infty$, $\forall i, R^{\wp}\subseteq p_iR$, so $|R/R^{\wp}|=|R/p_iR|*|p_iR/R^{\wp}|\geq p_i\to \infty (i\to \infty)$. Contradiction. $\Box$

Lemma 6:

$\forall p \not= q, p,q \in \wp$, $pR\cap qR= pqR$.

$\bigtriangleup$

Easy to see that $pqR\subseteq pR\cap qR$, from using ideas from Lemma 3 and from Lemma 4, we get that $|pR/(pR\cap qR)|=q^{l}$, where $l\not=0$. From Lemma 3, $|pR/pqR|=p^{f}$, so $(pR\cap qR)/pqR=p^{f-l}$, like the same $(pR\cap qR)/pqR=q^y$, so $y=f-l=0$, $pR\cap qR=pqR$. $\Box$

Lemma 6':

$m,n\in Z: (m,n)=1$, then $mR\cap nR= mnR$.

$\bigtriangleup$

Like in Lemma 6 $|(mR\cap nR)/mnR|=H$, from using Lemma 3', $p\in \wp$, $p|H\Leftrightarrow p|(m,n)$, so $H=1$. $\Box$

We will say that $k|a\in R \Leftrightarrow \exists b\in R:a=k*b$, $k\in Z$. We say that $a\in P_R\Leftrightarrow a\in R, \forall p\in \wp, p\nmid a$. From property $p|a, b=\lambda*a\Rightarrow p|b$, $a,b\in R$ and Lemma 5 we get that $P\setminus P_R\not=R$ and $\forall a\in R, \exists e\in P_R, p_1<p_2<...<p_i\in \wp , k_1,k_2,...,k_i\in Z_{+}: a=p_1^{k_1}p_2^{k_2}...p_i^{k_i}e$.

Lemma 7:

If $e_1,e_2\in P_R$, $a\in R$ and $p_1<p_2<...<p_i\in \wp , k_1,k_2,...,k_i\in Z_{+}: a=p_1^{k_1}p_2^{k_2}...p_i^{k_i}e_1$, $q_1<q_2<...<q_j\in \wp , l_1,l_2,...,l_j\in Z_{+}: a=q_1^{l_1}q_2^{l_2}...q_i^{l_i}e_2$, then $e_1=e_2$, $i=j$, $q_v=p_v$, $l_v=k_v$.

$\bigtriangleup$

From Lemma 6' we get that if $ p \in \wp \lambda \in Z, p\nmid \lambda a\in R$, $p|a, \lambda |a\Rightarrow p\lambda |a$, so if $q_u\not= p_v, v\in [1,i]$, then $q_u|a$, $p_1^{k_1}p_2^{k_2}...p_i^{k_i}|a\Rightarrow q_vp_1^{k_1}p_2^{k_2}...p_i^{k_i}|a \Rightarrow \exists b\in R:a=q_vp_1^{k_1}p_2^{k_2}...p_i^{k_i}b=p_1^{k_1}p_2^{k_2}...p_i^{k_i}e_1$, from Lemma 1 $p_1^{k_1}p_2^{k_2}...p_i^{k_i}(e_1-q_vb)=0$, so $q_v|e_1$ and $e_1\in P\setminus P_R$. Contradiction.
So $i=j$ $p_v=q_v, v\in [1,i]$, and so easy to see that $l_v=k_v, v\in [1,i]$, so $a=p_1^{k_1}p_2^{k_2}...p_i^{k_i}e_1=p_1^{k_1}p_2^{k_2}...p_i^{k_i}e_2$, so $e_1=e_2$. $\Box$

Lemma 8:

If $p\in \wp$, ring $R^*:pR\subseteq R^*\subseteq R$, then $R^*=pR$, or $R^*=R$.

$\bigtriangleup$

Let $R^*\not= R$, from using all previous construction to $R^*$, we get that $\forall a\in R^*, \exists !\lambda \in Z, \xi^* \in P_{R^*}: a=\lambda*\xi^*$, $\exists !\mu \in Z, \xi \in P_{R}: \xi^*=\mu*\xi$, so $a=\lambda \mu \xi$, $\mu*\xi=\xi^*\in R^*$, so from ideas of Lemma 3 we get that $R/R^*\cong Z[x]/(px,f^*(x))$ and so $p|\mu$, $a=\lambda \mu \xi\in pR$, so $pR\subseteq R^*\subseteq pR$, so $R^*=pR$. $\Box$

Lemma 9:

$\forall p\in \wp, |R/pR|=p$.

$\bigtriangleup$

Take ring $R/pR$ if there exists $f\in R/pR, f\not= 0, f^2=0$, then from Lemma 8, $R/pR=\{0,f,2f,...,(p-1)f\}$ and $|R/pR|=p$, if $\forall g\in R/pR, g^2\not= 0$, then take multiplicative group (maybe without 1) $G=R/pR\setminus 0$, well known  that from $|G|<\infty$ then $\exists e\in G: e^2=e$, so $R/pR=\{0,e,2e,...,(p-1)e\}$ and also $|R/pR|=p$. $\Box$

Easy to see that from Lemma 9 by Indicton we get that $\forall n\in Z, |R/nR|=n$ ($|R/abR|=|R/aR|*|aR/abR|=ab$).

Lemma 10:

Let $R'$ is ring in $R$, then there exists $n\in Z: R'=nR$.

$\bigtriangleup$

Let $n=|R/R'|$, then $nR\subseteq R'$ and from Lemma 9 $|R/nR|=n=|R/R'|$, so $R'=nR$. $\Box$

Now let's take $\xi \in P_R$, and ring $R_1=\{r_1\xi+r_2\xi^2+...r_k\xi^k+...\}_{r_1,r_2,...,r_k,...}$, from Lemma 10 $\exists n\in Z: R_1=nR$, so exists $b\in R: \xi=nb$, so $n=1$ and $R=R_1$. It finishes Part 1.

Part 2:

In this part we edited Lemmas of previous part (all solutions of them are same!) and reduce Problem to smal cases (we also prove result of Part 1 for more $(without some small cases)$ general types of ring $R$).

Denote $\widetilde{\wp}$ as subset of prime numbers such that $pR\not=R$.

Lemma $1^* =$ Lemma 1

From defenition of $\widetilde{\wp}$ we get that $\forall p\in \widetilde{\wp}$, $\cap_{i\in Z\backslash 0}p^iR=\{0\}$ (if $\cap_{i\in Z\backslash 0}p^iR=R'\not=\{0\}$, then $pR'=R'$, from Lemma 1 and defenition of $\widetilde{\wp}$, $p^2R\not=pR$ and $|R/R'|<\infty$, so $\exists i\in Z: p^iR\cap R'\not= R'$ and $pR'\not=R'$).

Denote function for $p\in \widetilde{\wp}$, $\nu_p^*(a)=max\{i\in Z|a\in p^iR\}$, $\cap_{i\in Z\backslash 0}p^iR=\{0\}$, so $\forall a\in R, 0 \leq \nu_p(a)<\infty$.

Lemma $3^*$:

$\forall p\in \widetilde{\wp} , |R/pR|=p^n$, for some positive integer $n$.

Lemma $3^{'*}$:

$\forall \lambda \in Z, q||R/\lambda R|\Leftrightarrow q|\lambda, q\in \widetilde{\wp}$.

Lemma $4^*$:

$\forall p \not= q, p,q \in \widetilde{\wp}$, $pR\cap qR\not= pR$.

Lemma $5^*$:

$\forall a\in R, \exists N>0: \forall p\in \widetilde{\wp} , p>N, a\in R\backslash pR$.

Lemma $6^*$ $=$ Lemma 6, Lemma $6^{'*}$ $=$ Lemma 6'

We say that $a\in P^*_R\Leftrightarrow a\in R, \forall p\in \widetilde{\wp}, p\nmid a$.

Lemma $7^*$:

If $e_1,e_2\in P^*_R$, $a\in R$ and $p_1<p_2<...<p_i\in \widetilde{\wp} , k_1,k_2,...,k_i\in Z_{+}: a=p_1^{k_1}p_2^{k_2}...p_i^{k_i}e_1$, $q_1<q_2<...<q_j\in \widetilde{\wp} , l_1,l_2,...,l_j\in Z_{+}: a=q_1^{l_1}q_2^{l_2}...q_i^{l_i}e_2$, then $e_1=e_2$, $i=j$, $q_v=p_v$, $l_v=k_v$.

Lemma $8^* = $ Lemma 8

Lemma $9^*$:

$\forall p\in \widetilde{\wp}, |R/pR|=p$.

Lemma $10^* = $ Lemma 10

Denote rings $\widetilde{Z_p[x]}=\{k_1x+k_2x^2+...+k_nx^n+...\}_{k_i\in Z/pZ}$, $\widetilde{Q[x]}=\{k_1x+k_2x^2+...+k_nx^n+...\}_{k_i\in Q}$, $\widetilde{Z[x]}=\{k_1x+k_2x^2+...+k_nx^n+...\}_{k_i\in Z}$ . Let's look on Lemma 1 if for some $p \in \wp$, $pR=\{0\}$, then $R$ is ring over field $Z/pZ$, so $R\cong \widetilde{Z_p[x]}$, or $R\cong \widetilde{Z_p[x]}/(f^*)$.
We need to look also on case when $\widetilde{\wp}$ is empty, so $pR=R$ for all $p\in \wp$, so $nR=R \forall n \in Z\setminus 0$, so for every ring $R=R^*\subseteq R$, $|R/R^*|R\subseteq R^*\subseteq R$ and $R^*=R$ and we get top result of Part 1.

If there exists $\xi \in R: R=\{r_1\xi+r_2\xi^2+...r_k\xi^k+...\}_{r_1,r_2,...,r_k,...}$, then construct homomorfism $\phi : \widetilde{Z[x]}\to R$, $\phi (r_1x+r_2x^2+...r_kx^k+...)=r_1\xi+r_2\xi^2+...r_k\xi^k+...$, so $R\cong \widetilde{Z[x]}/I$, where $I=Ker\phi$.

So we need to prove Problem in cases:

Case 1:

$R\cong \widetilde{Z_p[x]}$

Case 2:

$R\cong \widetilde{Z_p[x]}/(f^*)$

Case 3:

$R\cong \widetilde{Z[x]}/I$

So we finish Part 2.

Part 3:

In Case 1 note that $\{ x,x^3,x^5,... \}\subseteq \widetilde{Z_p[x]}/(x^2)$, so $|\widetilde{Z_p[x]}/(x^2)|=\infty$.

In Case 2 note that if $f^*=k_1x+...+k_rx^r$, then $|R|=p^{r-1}<\infty$.

So we only need to look on Case 3. $I=Ker\phi$ is ideal in $\widetilde{Z[x]}$, so if for some positive integer $n$, $nx\in I$, then $\phi(nx)=n\phi(x)=0$ and from Lemma 1 $\phi(x)=0$, so $x\in I$, $R\cong \{ 0 \}$. So for every $n\in Z\setminus 0, nx\in \widetilde{Z[x]}\setminus I$. And if $I\not=(x^2,x^3,...)$, then $\exists K\subset R: R/K\cong \widetilde{Z[x]}/(x^2,x^3,...)$, $x,2x,3x,...\in \widetilde{Z[x]}/(x^2,x^3,...)$, so $|R/K|=\infty$. So we get that $R\cong \widetilde{Z[x]}/(x^2,x^3,...)=0,x,-x,2x,-2x,3x-3x,...$ is cyclic. Done


Let $n$ be an integer, $n \geq 3$. Let $f(n)$ be the largest number of isosceles triangles whose vertices belong to some set of $n$ points in the plane without three colinear points. Prove that there exists positive real constants $a$ and $b$ such that $an^{2}< f(n) < bn^{2}$ for every integer $n$, $n \geq 3$.

First consider a convex regular $2k$-gon. It is clear that no three vertices are collinear, and for each vertex $A$ there are $k-1$ isosceles triangles at $A$.
Thus, since an isosceles triangle is counted at most three times in this way, the total number of isosceles triangles is at least $\frac{2k(k-1)}3.$
We have the same kind of estimation for a regular convex $2k+1$-gon.
It follows that $f(n) > an^{2}$ for some appropriate positive constant $a$.

In another hand, let's consider any set of $n$ points in the plane, no three collinear. There are $\frac{n(n-1)}2$ choices for two distinct points among the $n$. For each, say $A$ and $B$, there are at most two points $M$ among the given ones such that $AMB$ is isosceles at $M$, because such points $M$ belong to the perpendicular bissector of $[AB]$ and no three of the points can belong to this line. Thus the total number of isosceles triangles is at most $n(n-1) <n^{2}$.
Therefore $f(n) < n^{2}.$

Pierre.

Well, what is the "best" constants $a$ and $b$?

It seems that "the best" constants $ a$ and $ b$ are 1.
$ f(n)<n^2$ is already proved.

Lets consider an example. Regular $ 2n\minus{}1$-gon with a point in its center. It is easy to see that we get $ (2n\minus{}1)(2n\minus{}2)$ isosceles triangles for such configuration (every 2 points of $ 2n\minus{}1$-gon forms two triangles - one with another point and one with the center point. These triangles have two points as a base, that is all triangles are different). So we have $ 2n$ points and $ (2n\minus{}1)(2n\minus{}2)$ triangles, i.e.
$ (2n\minus{}1)(2n\minus{}2) < f(2n) < (2n\minus{}1)(2n)$.

For the lower bound we claim we can take $a=\frac{1}{20}$. We find an explicit construction that yields $>an^2$ isosceles triangles for some set of $n$ points with no three collinear, which proves $f(n) > an^2$.
-For $n=3, 4$ we can take an isosceles triangle, which yields $1 > \frac{4^2}{20}> \frac{3^2}{20}$ isosceles triangles.
-For $5 \le 10 \le 9$ we can take $5$ of the $n$ points to be a regular pentagon and appropriately place the other points, which yields at least$5 > \frac{9^2}{20}$ isosceles triangles.
Now suppose $n \ge 10$. Let $k = 2\lfloor \frac{n}{2} \rfloor$ and consider the following construction on $k$ of the $n$ points. Take one of the $k$ points to be the center $O$ of a regular $k-1$-gon and space the other $k-1$ points to be vertices of the regular $k-1$ gon. Assuming the other $n-k$ points are properly placed it follows that there are no three collinear points, since $k-1$ is odd. Every choice of 2 of the $k-1$ points and $O$ yields a different isosceles triangle, so letting $x=n/2$, this construction gives
\[ \dbinom{k-1}{2} > \frac{1}{2}(2x-3)(2x-4) = \frac{(n-3)(n-4)}{2} \]
isosceles triangles. But for $n \ge 5$ we can check that
\[ 4 > \frac{35}{9} \implies (n-\frac{35}{9})^2 \ge \frac{35^2}{9^2}-\frac{120}{9} \implies n^2 - \frac{70}{9}n+\frac{120}{9} \ge 0 \implies 9n^2 - 70n+120 \ge 0 \implies \frac{(n-3)(n-4)}{2} = \frac{n^2-7n+12}{2} \ge \frac{n^2}{20} \]
so this construction indeed works.

For the upper bound we claim we can take $b=1$. Note for every pair of two points $p,q$, there are at most 2 points that lie on the perpendicular bisector of $p,q$; else there would be three collinear points. So $f(n) \le 2\dbinom{n}{2}=n(n-1)<n^2$.

For any set of $n$ points in general position on the plane, observe that for each pair of points there are at most $2$ distinct points lying on its perpendicular bisector. Hence, for each pair of points, it is the basis of at most $2$ isosceles triangles. Therefore, we have at most $2 \binom{n}{2}= n^2-n<n^2$, so $b=1$ works.

Now, we claim that $a= \frac{1}{12}$ works. Observe that if $n$ is odd, then we can take a regular $n-$agon. It will have at least $\binom{n}{2}-\frac{2n}{3}$ isosceles triangles (when $3|n$ we count $\frac{n}{3}$ equilateral triangles twice). Observe that $\binom{n}{2}-\frac{2n}{3}= \frac{3n^2-7n}{6}>\frac{n^2}{12} \iff 3n^2-7n > \frac{n^2}{2}$, which is clearly true for $n \geq 3$. If $n$ is even, we can take a regular $(n-1)-$agon and its center. Clearly, the number of isosceles triangles in this set is greater than the number of isosceles triangles in a regular $(n-1)-$agon, so $f(n)>\frac{n^2}{12}$. Hence, $a=\frac{1}{12}, b=1$ works.

$\blacksquare$


vector
geometry
3D geometry
sphere
topology
real analysis
real analysis unsolved

As it was used in this topic http://www.mathlinks.ro/Forum/viewtopic.php?t=46799 I'll post here the following problem:
Prove that if the space $R$ is finite-dimensional then every it's norms $||\cdot ||_1$, $||\cdot||_2$ are equivalent

We have to show that $\|x_n\|_1\to 0\iff \|x_n\|_2\to 0$. Assume one of the norms, $\|\cdot\|_1$, is actually the usual $\|\cdot\|_{\infty}$.

Assume first that we have $\|\sum a_i^{(n)}e_i\|_{\infty}\to 0$, which means that for all $i$ we have $|a_i^{(n)}|\to 0$. We then have $\|\sum a_i^{(n)}e_i\|_2\le\sum\|a_i^{(n)}e_i\|_2=\sum |a_i|\cdot\|e_i\|_2\to 0$, so the implication $\|x_n\|_1\to 0\Rightarrow \|x_n\|_2\to 0$ is proven.

Suppose now that we have $\|\sum a_i^{(n)}e_i\|_2\to 0\ (*)$. If there are infinitely many vectors in this sequence far away from $0$ in $\|\cdot\|_1$, then we may assume WLOG that $|a_1^{(n)}|=1\ge |a_i^{(n)}|,\ \forall n,\ \forall i$. This means that from the sequence $(a_1^{(n)},\ldots,a_k^{(n)})_n$ ($k$ is the dimension of the space we're working in) we can extract a subsequence converging to some $(a_1,\ldots,a_k),\ |a_1|=1$. For simplicity, let's assume the whole sequence is convergent to that limit. This means that $\|\sum (a_i^{(n)}-a_i)e_i\|_1\to 0$, which, from the other implication, gives $\|\sum (a_i^{(n)}-a_i)e_i\|_2\to 0\ (**)$. $(*),(**)$ imply that, in fact, $a_i=0,\ \forall i$, which contradicts $|a_1|=1$.

Another proof: consider a base $ e_1,...,e_n$ in $R$ and define the norm 2 as follows: if $ x=x_1 e_1+...+x_n e_n$ then $||x||_2=\sqrt{x_1^2+...+x_n^2}$. We will prove that any norm $||$ is equivalent to this one. It is clear from Cauchy-Schwarz that there is a constant $C=\sqrt{||e_1||^2+...+||e_n||^2}$ such that $||x||\leq C ||x||_2$ for all $x$. It follows that $ ||$ is continuous in the metric space $ (R,||)$ and since $||kx||=|k|||x||$ it remains to prove that $ ||x||\geq p$ for a certain $p>0$ and all $ ||x||_2=1$. But the set of those $x$ with $||x||_2=1$ is compact (Borel-Lebesque theorem) in $ (R,||_2)$ and since the norm is continuous, it has a minimal value of the sphere of $R$ and the existence of $p$ follows.

I's sorry for reviving old threads, but i'm interested in one thing:
It is an example in $\mathbb{Q}[\sqrt{2}]$,(which has dimension $2$ over $\mathbb{Q}$) of two norms, which are not equivalent: $N_1(a+b\sqrt{2})=|a+b\sqrt{2}|$ and $N_2(a+b\sqrt{2})=|a|+|b|$.
This fact must be true when the basic field is complete, which is not the case when it is $\mathbb{Q}$.
Where is the flaw in,say, harazi's approach when the field isn't complete ?

The relevant property of the field here is local compactness. I used it: [quote="grobber"]This means that from the sequence $(a_1^{(n)},\ldots,a_k^{(n)})_n$ we can extract a subsequence converging to some $(a_1,\ldots,a_k),\ |a_1|=1$.[/quote] and harazi used it: [quote="harazi"]the set of those $x$ with $||x||_2=1$ is compact[/quote]


find infinitely many solutions:
$ x^2 \plus{} y^2 \plus{} z^2 \equal{}xyz$

Put $ x\equal{}1$

$ y^2\plus{}z^2\plus{}1\equal{}yz$

Solve the quadratic for $ y$, and you get a family of solutions

By AM-GM $ y^2\plus{}z^2 \ge 2yz$ so that doesn't work. It is hard to find a family of solutions, give a specific example.

What's wrong with that?

Write the equation as:
$ x^2 \minus{} (yz)x \plus{} (y^2 \plus{} z^2) \equal{} 0$

Thus, given [i]any[/i] $ y$ and $ z$, there is an $ x \in \mathbb{C}$:

$ x \equal{} \frac {yz \pm \sqrt {y^2 z^2 \minus{} 4(y^2 \plus{} z^2)}}{2}$

A more interesting problem happens when we restrict $ x,y,z \in \mathbb{R}$, or $ x,y,z \in \mathbb{Q}$, or $ x,y,z \in \mathbb{Z}$ (note that the last two equations are [b]not[/b] essentially the same, since the equation is not homogeneous).

Since the title was "diophantine" it is probably intended that solutions are in integers, positive integers, rationals, etc. but the OP needs to say which.
Meanwhile assume it is positive integers.

The first solution is $ (3, 3, 3)$.  New solutions can be obtained by letting the triple be $ (x, y, z)\rightarrow(xy \minus{} z, x, y)$.

This generates the sets $ (6, 3, 3), (15, 6, 3), (87, 15, 6)...$

I'll let someone else show that these will always work and that the sequence never repeats.

Nice!

Of course it works, as $ x^2\plus{}y^2\plus{}(xy\minus{}z)^2\equal{}x^2\plus{}y^2\plus{}z^2\minus{}2xyz\plus{}x^2y^2\equal{}x^2y^2\minus{}xyz\equal{}xy(xy\minus{}z)$ and we can just let z be the minimal element of (x,y,z) so the greatest number is strictly increasing.


Prove that $r=\sin\theta$ (on a standard polar graph) is a circle.

[hide="Maybe..."]Convert to cartesian form and use the definition of the equation of a cartesian circle?[/hide]

[quote="Altheman"][quote="cincodemayo5590"][hide="Feels like math class :P"]
$r^2=r \cdot sin \theta$
$x^2+y^2=y$
$(x-0)^2 + (y-\frac{1}{2})^2 = (\frac{1}{2})^2$, which is the equation of a circle with radius $\frac{1}{2}$.

It cant be that easy alex... [/hide][/quote]

i know it is easy, but there are some interesting things associated with it, also there are some other proofs that i liked better, and this easy lemma proves that the image of a circle becomes a line under inversion, and vice versa

p.s. sam how is the history hw going, i cannot get myself to do it... :P[/quote]

I havent started it either....
Darn Ms. Gerhardt for giving us hw... :diablo: .

Well, as long as i'm on AoPS, my not doing the hw is rationalized :D  yay

[quote="cincodemayo5590"]

Well, as long as i'm on AoPS, my not doing the hw is rationalized :D  yay[/quote]

that is how i feel too, but it still has to be done, i started it, but then i got an e-mail from my arml coach, and have been doing math since then


trigonometry quadratics geometry circumcircle algebra

Source: Komal
Given that the positive numbers $x, y, z$ satisfy
$x^2 + xy + y^2 = 9$,  $y^2 + yz + z^2 = 16$,  $z^2 + zx + x^2 = 25$,
determine $xy + yz + zx$.

In a (right angle) triangle $\triangle ABC$ with the sides a = BC = 4, b = CA = 5, c = AB = 3, the 1st Fermat point is constructed by drawing equilateral triangles of the sidec BC, CA, AB and then joining their outside vertices with the opposite vertices of the triangle $\triangle ABC$. These lines meet at the triangle 1st Fermat point F, wich has the minimum sum of distances x + y + z = AF + BF + CF from the triangle vertices. Since the angles $\angle AFB = \angle BFC = \angle CFA = 120^\circ$ and $\cos 120^\circ = -\frac 1 2$, the given equations are the cosine theorem formulas for the triangles $\triangle BFC, \triangle CFA, \triangle ABC$:

(1)[color=white] . . . [/color]$x^2 + y^2 + xy = c^2 = 9$
(2)[color=white] . . . [/color]$y^2 + z^2 + yz = a^2 = 16$
(3)[color=white] . . . [/color]$z^2 + x^2 + zx = b^2 = 25$

Subtracting the 1st equation from the 2nd one, the 2nd equation from the 3rd one, and the 1st one from the 3rd one, we get

$z^2 - x^2 + yz - xy = (z - x)(z + x) + y(z - x) = (z - x)(x + y + z) = 16 - 9 = 7$
$x^2 - y^2 + zx - yz = (x - y)(x + y) + z(x - y) = (x - y)(x + y + z) = 25 - 16 = 9$
$z^2 - y^2 + zx - xy = (z - y)(z + y) + x(z - y) = (z - y)(x + y + z) = 25 - 9 = 16$

Dividing the 1st equation by the 2nd one yields

$\frac{z - x}{y - x} = \frac 7 9,\ \ 9z - 9x = 7x- 7y,\ \ 9z = 16x - 7y$

Dividing any other pair of these equations yields the same thing.
_____________________________________________________________________

We can eliminate z by substituting $9z = 16x - 7y$ into the 2nd equation (and ignoring the 3rd one):

(1a)[color=white] . . . [/color]$x^2 + y^2 + xy = 9$

$81y^2 + (16x - 7y)^2 + 9y(16x - 7y) = 1296$

(2b)[color=white] . . . [/color]$256x^2 + 67y^2 - 80xy =1296$

Multiplying equation (1a) by 256 and subtracting form (2b) yields

$189y^2 + 336xy = 1296,\ \ \ 9y^2 + 16xy = 48$

Similarly, multiplying equation (1a) by 67 and subtracting form (2b) yields

$189x^2 - 147xy = 693,\ \ \ 9x^2 - 7xy = 33$

We can also eliminate x by substituting  $16x = 7y + 9z$ into the 3rd equation (and ignoring the 1st one):

(2a)[color=white] . . . [/color]$y^2 + z^2 + yz = 16$

$256z^2 + (7y + 9z)^2 + 16z(7y + 9z) = 6400$

(3b)[color=white] . . . [/color]$49y^2 + 481z^2 + 238yz = 6400$

Multiplying equation (2a) by 49 and subtracting from (3b) yields

$432z^2 + 189yz = 5616,\ \ \ 16z^2 + 7yz = 208$

Similarly, multiplying equation (2a) by 481 and subtracting from (2b) yields

$432y^2 + 243yz = 1296,\ \ \ 16y^2 + 9yz = 48$

Finally, we can also eliminate y by substituting  $7y = 16x - 9z$ into the 1st equation (and ignoring the 2nd one):

(3a)[color=white] . . . [/color]$z^2 + x^2 + zx = 25$

$49x^2 + (16x - 9z)^2 + 7x(16x - 9z) = 441$

(1b)[color=white] . . . [/color]$81z^2 + 417x^2 - 351zx = 441$

Multiplying equation (3a) by 81 and subtracting from (1b) yields

$336x^2 - 432zx = -1584,\ \ \ 7x^2 - 9zx = -33$

Similarly, multiplying equation (3a) by 417 and subtracting from (1b) yields

$336z^2 - 768zx = 9984,\ \ \ 7z^2 + 16zx = 208$
_____________________________________________________________________

Now we have to solve for $x^2, y^2, z^2$. For example, from $9x^2 - 7xy = 33$, we can calculate y in terms of x and substitute into the original equation (1), which leads to a quadratic equation for $x^2$:

$7y = \frac{9x^2 - 33}{x} = 9x - \frac{33}{x}$

$49x^2 + \left(9x - \frac{33}{x}\right)^2 + 7x\left(9x - \frac{33}{x}\right) = 441$

$49x^2 + 81x^2 - 594 + \frac{1089}{x^2} + 63x^2 - 231 = 441$

$193x^4 - 1266x^2 + 1089 = 0$

$x^2 = \frac{633 \pm 252 \sqrt 3}{193}$

Similarly, from $9y^2 + 16xy = 48$, we can calculate x in terms of y and substitute into the original equation (1):

$16x = \frac{48 - 9y^2}{y} = \frac{48}{y} - 9y$

$\left(\frac{48}{y} - 9y\right)^2 + 256y^2 + 16y \left(\frac{48}{y} - 9y\right) = 2304$

$\frac{2304}{y^2} - 864 + 81y^2 + 256y^2 + 768 - 144y^2 = 2304$

$193y^4 - 2400y^2 + 2304 = 0$

$y^2 = \frac{1200 \pm 576 \sqrt 3}{193}$

Finally, from $16z^2 + 7yz = 208$, we can calculate y in terms of z and substitute into the original equation (2):

$7y = \frac{208 - 16z^2}{z} = \frac{208}{z} - 16z$

$\left(\frac{208}{z} - 16z\right)^2 + 49z^2 + 7z \left(\frac{208}{z} - 16z\right) = 784$

$\frac{43264}{z^2} - 6656 + 256z^2 + 49z^2 + 1456 - 112z^2 = 784$

$193z^4 - 5984z^2 + 43264 = 0$

$z^2 = \frac{2992 \pm 448 \sqrt 3}{193}$

All roots are positive and we have to decide the signs in front of the square roots. Assume that the $y^2$ root has a positive sign in front of $\sqrt 3$. Then

$y^2 > \frac{1200 + 574 \cdot \frac 3 2}{200} = \frac{2064}{200} > 10 > 9$

hence, equation (1) would not hold for positive x, y, z. Likewise, assume that the $z^2$ root has a positive sign in front of $\sqrt 3$. Then

$z^2 > \frac{2992 + 448 \cdot \frac 3 2}{200} = \frac{3664}{200} > 18 > 16$

and equation (2) would not hold for positive x, y, z. So the roots $y^2, z^2$ have both negative signs in front of $\sqrt 3$. Assume that the $x^2$ root also has a negative sign in front of $\sqrt 3$ Then

$x^2 + y^2 = \frac{1833 - 828 \sqrt 3}{193} < \frac{1833 - 828 \cdot \frac 5 3}{193} = \frac{453}{193} < \frac{475}{190} = \frac 5 2$

$xy = \sqrt{x^2y^2} \le \frac{x^2 + y^2}{2} < \frac 5 4$

and equation (1) would not hold at all. So the $x^2$ root has a positive sign in fron of the $\sqrt 3$. Finally, adding equations (1), (2), (3) leads to

$xy + yz + zx = 50 - 2(x^2 + y^2 + z^2) = 50 - 2 \cdot \frac{633 + 1200 + 2992 + (252 - 576 - 448) \sqrt 3}{193} =$

$= 50 - \frac{9650 - 1544 \sqrt 3}{193} = 8 \sqrt 3$

The roots $x^2, y^2, z^2$ with the opposite signs in front of $\sqrt 3$ correspond to the distances x' = AF', y' = BF', z' = CF' from the 2nd Fermat point F' to of the triangle $\triangle ABC$ to its vertices. The 2nd Fermat point is constructed similarly as the 1st one, except drawing the equilateral triangles on the sides of the triangle $\triangle ABC$ inward instead of outward, in which case the 2 angles $\angle AF'B = BF'C = 60^\circ$ and $\cos 60^\circ = \frac 1 2$.

Since you know it's a 3-4-5 right triangle and x,y,z are distances to the vertices from a point in the center, why not just solve like this:

$\frac{(xy + yz + xz)\sin{120^{\circ}}}{2} = [ABC] = 6$
$(xy+yz+xz)\sqrt{3} = 24 \implies xy+yz+xz = 8\sqrt{3}$

Thank you agolsme, it is very nice, but I did not notice.

[quote="chess64"]:lol: You really don't need to know any Fermat point stuff or anything. That's probably the longest solution I've ever seen to a problem on this board. :)[/quote]
I did it without the Fermat point. How would you do it without the Fermat point :?:

Yetti, how did you get $\angle AFB = \angle BFC = \angle CFA = 120^\circ$?  Is that because you made a right triangle, or is that a special property of the fermat point?  I checked mathworld, I didn't see anything there.

It has nothing to do with the right angle, except making sure that the point F is inside of the triangle $\triangle ABC$. Start with the circumcircles of the equilateral triangles $\triangle BCA'$ on BC, $\triangle CAB'$ on CA, which intersect at points C, F. As long as $\text{max}(\angle A, \angle B \angle C) < 120^\circ$, the angles $\angle BFC = \angle CFA = 120^\circ$ (because the angles spanning the oppposite arcs BC, CA are $60^\circ$). Then the angle $\angle AFB = 360^\circ - 2 \cdot 120^\circ = 120^\circ$, so F lies on the circumcircle of the 3rd equilateral triangle $\triangle ABC'$ on AB. Then draw the segments AF, FA'. The angle $\angle AFA' = \angle AFC' + \angle C'FB + \angle BFA' = 3 \cdot 60^\circ = 180^\circ$, because the chords AC', C'B, BA' are all sides of equilateral triangles. So the points A, F, A' are collinear and similarly, the points B, F, B' and C, F, C' are collinear - AA', BB', CC' concur at F.  See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=44241]Another Geometry Problem[/url].


Let $ f: [0;1] \minus{} > [0;1]$x$ [0;1]$ be a two times continuous differentiable function.
Show that if $ A$ is subset of $ [0;1]$ and the Lebesque measure of $ A$ is equal to zero then the  Lebesque measure of $ f(A)$ is also equal to zero.

It's $ C^1$, so it's Lipschitz. Since it's Lipschitz, the image of an interval of size $ x$ is a curve with arclength at most $ Mx$. The image of a set with outer measure $ y$ is thus rectifiable, with arclength at most $ My$. This is zero when the outer measure is zero.
Also, any set with finite arclength has 2-dimensional measure zero.

A lot of thanks Jmerry!
My knowledge is poor and this problem is hard for me, but I understand your solution!


Let $ a,b,c$ be positive real numbers .

Prove that :

$ \sum \sqrt[3]{\frac{ab\plus{}c^2}{a^2\plus{}b^2}} \geq \frac{3}{2}\plus{}\frac{3(ab\plus{}bc\plus{}ca)}{2(a^2\plus{}b^2\plus{}c^2)}$

Cauchy Schwartz kills it gently.  :wink:


We have an unitary operator $ V$ and a positive definite operator $ H$. Show, that $ HV = VH$.

I have one solution, however I am not sure it is correct.

$ HV = VH \Leftrightarrow HV - VH = 0 \Leftrightarrow H(HV - VH) = 0.$ Since $ H$ is positive definite, all eigenvalues of $ %Error. "trace" is a bad command.
{(HHV-HVH)}$ are nonnegative. But since

 ${{ \mathrm{trace}{(H(HV - VH))} = \mathrm{trace}((HH)V) - \mathrm{trace}(H(VH)))} = \mathrm{trace}(VHH ) - \mathrm{trace}(VHH) = 0 = \sum^n}_i \lambda_i$

all eigenvalues must be $ 0$. Therefore $ HV - VH = 0$.

Neither the problem nor the solution makes any sense to me...

  darij

Ok. This is a part of some other task, but I thought it is independent from other parts.

Original problem:

We have two unitary matrices $ U$ and $ V$ and a positive definite operator $ H$. We know that $ UH\equal{}HV$. Show that:

a.) $ HU\equal{}VH$,
b.) $ H^2 V\equal{}V H^2$,
c.) $ HV \equal{} VH$,
d.) $ U\equal{}V$.

Let $ A\equal{}UH\equal{}HV$. Then $ A^*A\equal{}H^2\equal{}AA^*$. Thus $ A$ is a normal matrix and $ H$ is the unique positive square root of $ AA^*$; $ A$ is diagonalizable in an orthonormal basis, thus we may assume $ A\equal{}diag(\lambda_i)$ and $ H\equal{}diag(|\lambda_i|)$. Then $ U\equal{}AH^{\minus{}1}\equal{}H^{\minus{}1}A\equal{}V$ is a diagonal unitary matrix.

Very nice solution, loup blanc. Thank you.


$ L{(t^2\plus{}2t)U(t\minus{}3)}$

how can i transform $ t^2 \plus{} 2t$ into t-3 so it becomes $ L{(t\minus{}3)U(t\minus{}3)}$ ??

$ f(t\minus{}3)\equal{}t^{2}\plus{}2t$

$ f(t)\equal{}\allowbreak t^{2}\plus{}8t\plus{}15$

$ L\left[ f(t)\right] \equal{}\frac{2}{s^{3}}\plus{}\frac{8}{s^{2}}\plus{}\frac{15}{s}$

$ L\left[ (t^{2}\plus{}2t)U(t\minus{}3)\right] \equal{}e^{\minus{}3s}L\left[ f(t)\right] \equal{}e^{\minus{}3s}\left( \frac{2}{s^{3}}\plus{}\frac{8}{s^{2}}\plus{}\frac{15}{s}\right)$

Hi

thankk you for the reply

I was wondering WHY AND HOW you went from $ f(t \minus{} 3)$ to $ f(t)$ ?

$ f(t \minus{} 3) \equal{} (t \minus{} 3)^2 \plus{} 2(t \minus{} 3)$

OK, because Laplace 2nd shifting theorem tells us that for non negative values of a,
$ L(f(t\minus{}a)U(t\minus{}a))\equal{}e^{\minus{}as}F(s)$
Hence I let $ f(t\minus{}3)\equal{}t^{2}\plus{}2t$
This leads to the answer.


search
combinatorics unsolved
combinatorics

The points in space are colored with five colors being used. Prove that some plane contains four points of different colors.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1420017221&t=42710

Pierre.

that solution isn't quite complete, since the said line might be parallel to the said plane. here's my solution. assume the contrary, as does grobber in that post. then observe (like a religious holiday) that no line contains 3 colors, or else such a line and a point of a 4th color would form a plane with 4 colors. take $ A,B,C,D,E$ to be 5 points of different colors. if the line $ AB$ meets the plane $ CDE$, we're done because the point of intersection must be either type $ A$ or $ B$. ($ CDE$ is a legitimate plane because $ C,D,E$ aren't collinear, by the previous observation.) otherwise, $ AB$ is parallel to $ CDE$. but $ AB$ can't be parallel to all three lines $ CD,DE,EC$; say it's not parallel to $ CD$. then the plane $ ABE$ meets plane $ CDE$ in a line that's parallel to $ AB$ (since $ AB\parallel CDE$) and thus *not* parallel to $ CD$. $ CD$ therefore meets $ ABE$, so we have a contradiction, as before.


function
number theory unsolved
number theory

find function f:N->N satisfying (for all n is integer,n>-1)
f(n)+f(n+2)=f(n+4)*f(n+6)-2004 :?

[quote="Thanhliem"]find function f:N->N satisfying (for all n is integer,n>-1)
f(n)*f(n+2)=f(n+4)*f(n+6) +2004 :?[/quote]


calculus
integration
real analysis
real analysis unsolved

Can you solve this?

[url=http://www.imagehosting.com][img]http://www.imagehosting.com/out.php/i1809822_TTintegral.gif[/img][/url]

Have A Great Day!


function
linear algebra
matrix
invariant
vector
algebra
polynomial

Let $ B$ be a fixed invertible $ n\times n$ matrix.
Consider the linear operator $ T: \mathrm{M}_n\left(\mathbb{C}\right)\to\mathrm{M}_n\left(\mathbb{C}\right)$ defined by $ T\left(A\right)\equal{}BA\minus{}AB$ for every $ A\in\mathrm{M}_n\left(\mathbb{C}\right)$.
Find the eigenvalues of $ T$ in terms of those of $ B$.

We use the Kronecker product (cf Planetmath, Kronecker product or tensor product): if  $U,V$ are ($n,n$ ) and ($m,m$) matrices then $U\bigotimes{V}: M\in\mathcal{M}_{n,m}(\mathbb{C})\rightarrow{U^{T}{M}V}$ . Moreover  $spectrum(U\bigotimes{V})=\{\lambda_{i}\mu_{j}|i,j\}$ where  $spectrum(U)=\{\lambda_{i}|i\}$ and $spectrum(V)=\{\mu_{j}|j\}$.
Here  $m=n,spectrum(B)=\{\lambda_{i}|i\}$ and the studied linear function is  $f=I\bigotimes{B}-B^{T}\bigotimes{I}$. We may assume that $B$ is upper triangular therefore $f$ is a lower triangular block matrix with diagonal: $B-\lambda_{1}{I},\cdots,B-\lambda_{n}{I}$.
Conclusion: $spectrum(f)=\{\lambda_{i}-\lambda_{j}|i,j\}$.

Here is an elementary proof that doesn't use tensor products.
Consider $\phi_{A,B}(X)=AX+XB$, where $X, A, B \in \mathcal{M}_{n}(\mathbb{C})$. ($T$ in the initial question is a particular case: $\phi_{B,-B}$).
Denote by $\sigma(A)$ the spectrum of $A$. We know that the spectrum is invariant under transposition.
Let $(a,b)\in (\sigma(A),\sigma(B))$, and $V,W$ be associated non zero vectors such that $AV=aV$ and $B^{T}W=bW$. $\phi_{A,B}(VW^{T})=(a+b)VW^{T}$.
So \[\sigma(A)+\sigma(B) \subset \sigma(\phi_{A,B})\]

Let $\alpha\in \sigma(\phi_{A,B})$, [i]i.e.[/i] $\phi_{A,B}(X)=\alpha X$, then for all $P \in \mathbb{C}[X]$, $P(A)X=XP(\alpha I_{n}-B)$.
Denote by $\mu_{A}$ the minimal polynomial of $A$, $\mu_{A}(x)=\prod_{k=1}^{m}(x-a_{k})$ (some of the $a_{k}$ could be equal). We have $X\mu_{A}(\alpha I_{n}-B)=0$,and since $X\not=0$, one of the $(\alpha-a_{k})I_{n}-B$ is noninvertible, [i]i.e.[/i] $\alpha-a_{k}\in \sigma(B)$.
So \[\sigma(\phi_{A,B}) \subset \sigma(A)+\sigma(B)\]
Hence \[\sigma(\phi_{A,B}) = \sigma(A)+\sigma(B)\]


inequalities
algebra
system of equations
inequalities unsolved

For all pair of $k,n$ positive integers find all solutions $(x_1,x_2,\ldots,x_n)$ where $x_i\ge0$ for every $1\le i\le n$ satisfying the following system of equations:

$x_1^k+x_2^k+\ldots+x_n^k=1$
$(1+x_1)(1+x_2)\cdots(1+x_n)=2$

(I know there is a method using inequalities)

conceive

It seems more like an algebra problem to me.

Who wants to tackle this problem with Newton's sums?  :D

[quote="conceive"]For all pair of $k,n$ positive integers find all solutions $(x_1,x_2,\ldots,x_n)$ where $x_i\ge0$ for every $1\le i\le n$ satisfying the following system of equation:

$x_1^k+x_2^k+\ldots+x_n^k=1$
$(1+x_1)(1+x_2)\cdots(1+x_n)=2$

(I know there is a method using inequalities)

conceive[/quote]
Yes.there is,and luckily I think I find it.
first it is easy to get:$0 \le x_i \le 1$
thus:$1=x_1^k+x_2^k+\ldots+x_n^k \le x_1+x_2+\ldots+x_n$
so:$(1+x_1)(1+x_2)\cdots(1+x_n) \ge 1+ x_1+x_2+\ldots+x_n \ge 2$
and it is easy to when does the equality hold. :)


Recently the difficulty of problems posted in the Intermediate Forum has skyrocketed due both to Intermediate Forum getting more skilled and the former MathLinks participants no longer having a Beginner's Corner to post in.

We have therefore converted the old AoPS Advanced Section into an 'Other Problems' forum for problems which either don't fit the subjects in the Advanced Section or are a little too tough for Intermediate but not quite Advanced.

As a rule of thumb, the solution to a problem placed in Intermediate should be apparent to an Olympiad-level student in 5 minutes or so.  If it isn't, then the problem should live in the [url=http://www.artofproblemsolving.com/Forum/viewforum.php?f=151]Other Problems[/url].  When in doubt, assume that your problem is harder than you think it is!

Moderators will be moving the recent too-challenging problems to Other Problems forum.

Thank you so much Mr.  Rusczyk!  The intermediate forum was getting way too difficult.

[quote="rrusczyk"]As a rule of thumb, the solution to a problem placed in Intermediate should be apparent to an Olympiad-level student in 5 minutes or so.  If it isn't, then the problem should live in the [url=http://www.artofproblemsolving.com/Forum/viewforum.php?f=151]Other Problems[/url].[/quote]

Hmm that methods seems kind of useless:
There are high-level problems in the advanced section I solve in much less 5 minutes...
And there are easy problems (from the multiple choice round even) that cost me much more than 5 minutes...

Anyway the idea to make something inbetween isn't bad, but can't you make it a separate category, just like the intermediate ones? That would attract the difficult problems more that way I guess :cool:

Hey admin, there's something you'd like doing before I post in the Other problems section...

[quote]Sorry, but only [b]moderators[/b] can post topics in this forum.[/quote]

... :maybe: ...

[quote="rrusczyk"]Recently the difficulty of problems posted in the Intermediate Forum has skyrocketed due both to Intermediate Forum getting more skilled[/quote]

I observe that as moderators of forums become better problem-solvers, they lose their sense of what is easy and what is difficult.  My "secret" in staying acquainted with the true beginning level is that I have too little time to work many problems, and thus I am not getting better very rapidly. I try to use published books of problems for YOUNG students to remind me what a true beginner level, an intermediate level, and an advanced level are in international terms.

[quote="rrusczyk"]and the former MathLinks participants no longer having a Beginner's Corner to post in.[/quote]

I recall that the premerger MathLinks site, of which I was a lurking member, had a considerably higher overall level of difficulty than premerger AoPS. The MathLinks idea of "Beginner's Corner," in practice, seemed almost like first-time-IMO-participant level, which is much higher than what I would call Getting Started level as I know that from AoPS and my own coaching of a middle school math team.

[quote="rrusczyk"]When in doubt, assume that your problem is harder than you think it is![/quote]

That's a very important rule. Most problem-posers, at all levels, underestimate how difficult a problem will be for someone who hasn't seen the solution. And student problem-posers often make mistakes in posing problems with ambiguous statements or involving issues of higher math that they don't fully understand. That's why I like to find problems in PUBLISHED sources with carefully labeled levels. I know that anything found in a higher-numbered AMC 10 problem, for instance, and most lower-numbered AIME problems, fit well as Intermediate Topics level problems. There are many more problems at that level in published sources.

[quote="rrusczyk"]Moderators will be moving the recent too-challenging problems to Other Problems forum.[/quote]

I would rather be spending less time promoting problems from Getting Started to Intermediate (or even Advanced) and more time finding suitable problems, from dead-tree books, for posting to Getting Started myself. And I think it's a total waste of time to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15347]debate this issue[/url] with participants who can't exert themselves to look in the next higher forum for problems at the level they like. Something has to be done to make this site more inviting for the newest and youngest devotees of problem-solving.


probability
expected value
real analysis
real analysis unsolved

For $n$ distinct points chosen randomly and independently in the unit interval, what is the expected length of the smallest sub-interval?

Two points of interpretation:

1. I assume by "randomly ... in the unit interval" you mean the uniform distribution.

2. I assume that if the points are arranged as $0<x_1<x_2<\dots<x_n<1$, then the "smallest subinterval" is the minimum of
{$x_1.x_2-x_1.x_3-x_2,\dots,x_n-x_{n-1},1-x_n$}. In other words, the distance from the first or last point to the end of the interval counts as a subinterval.

Preliminary calculations:

For $n=1$, the answer is the expected value of the minimum of $x$ and $1-x$, namely $\frac14$.
For $n=2$, the answer is $\frac19$.

I don't have the details nailed down, but I conjecture that the general answer is $\frac1{(n+1)^2}$.


abstract algebra
superior algebra
superior algebra unsolved

Let R be a ring. prove that if M is a free R-module, then Ann_R M =0.

Suppose that $ M$ is annihilated by some element in $ R$, show that this gives rise to a dependence relation.


probability
counting
distinguishability
LaTeX
probability and stats

We have $ 7$ balls ($ 1,2,\cdots,7$) and $ 4$ boxes ( all are similar ).

Now we need to put these balls in these boxes , each box may has $ 7$ , $ < 7$ or $ 0 \equal{} empty$ of balls .

if we choose one box randomly , what is the probability to have :

1) empty box ??

2) box has all balls ?

 :D

Let $ \Omega \equal{} \{(x_1, x_2, x_3, x_4) | 0 \leq x_1, x_2, x_3, x_4 \in \mathbb{Z}, x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \equal{} 7\}$ ($ x_i$ is the number of balls in i-th box), $ B_i$ denotes the event that we have chosen the i-th box, $ C$ - randomly chosen box is empty (1), $ D$ - randomly chosen box contains 7 balls (2). Then we have

\[ P(C) \equal{} P(C|B_1)P(B_1) \plus{} ... \equal{} \frac{1}{4}(P(C|B_1) \plus{} ...) \\\equal{} \frac{1}{4}\cdot 4\cdot \frac{\#\{(x_1, x_2, x_3, x_4) \in \Omega | x_1 \equal{} 0\}}{\#\Omega} \equal{} \frac{{10 \choose 2}}{{10 \choose 3}},\]

\[ P(D) \equal{} P(D|B_1)P(B_1) \plus{} ... \equal{} \frac{1}{4}(P(D|B_1) \plus{} ...) \\\equal{} \frac{1}{4}\cdot 4\cdot \frac{\#\{(x_1, x_2, x_3, x_4) \in \Omega | x_1 \equal{} 7\}}{\#\Omega} \equal{} \frac{1}{{10 \choose 3}}.\]

[quote="ssheep"]\[ P(C) \equal{} P(C|B_1)P(B_1) \plus{} ... \equal{} \frac {1}{4}(P(C|B_1) \plus{} ...) \\
\equal{} \frac {1}{4}\cdot 4\cdot \frac {\#\{(x_1, x_2, x_3, x_4) \in \Omega | x_1 \equal{} 0\}}{\#\Omega} \equal{} \frac {{10 \choose 2}}{{10 \choose 3}},
\]

[/quote]

Both answers are true , since I have the answer of them in my papers.

Just need from you clarify more please about how did you get $ \# \Omega\equal{} {10\choose 3 }$.

 :)

Why are the answers not $ \left(\frac34\right)^7$ for the empty box and $ \left(\frac14\right)^7$ for the full box?

As I interpret this, for each ball, you make an equal-probability decision about which box to put this in, and each ball's placement is independent of each other ball. That makes the whole scheme a multinomial distribution, or if you concentrate on just one box, a binomial distribution.

Or perhaps there's something about the rules for distributing the balls that you haven't told us.

[quote="Kent Merryfield"]
As I interpret this, for each ball, you make an equal-probability decision about which box to put this in, and each ball's placement is independent of each other ball. That makes the whole scheme a multinomial distribution, or if you concentrate on just one box, a binomial distribution.
[/quote]

it's interested thinking .But,  ssheep answer's is identical with the answer in my paper.

[quote]

Or perhaps there's something about the rules for distributing the balls that you haven't told us.
[/quote]

I told you everything in the problem .Please feel free to write an extra information if we need it.

 :)

$ \#\Omega$ is the number of nonnegative integer solutions of equation $ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_3 \plus{} x_4 \equal{} 7$ which is equivalent to the number of ways in which we can put 7 similar balls in 4 drawers. Moreover, this is equivalent to the number of 10 terms binary sequences consisting of 7 zeros (=balls) and 3 ones (=walls between drawers). For example

the solution $ (2, 1, 3, 1)$ corresponds to the sequence $ 0010100010$;
the solution $ (0, 2, 5, 0)$ corresponds to the sequence $ 1001000001$.

There are $ { 10 \choose 3}$ such sequences.

Kent Merryfield, to my mind, you treat the balls as if they were distinguishable and this is the reason for the different result.

And why shouldn't they be distinguishable? If you say you are putting two balls into two boxes, I would think that the probability that you put exactly one ball in each is $ \frac12,$ and you're telling me it is $ \frac13.$

You are right. I am confused and I don't really know which model is appropriate:(

b)total posibilities :4^7  the posibilities we are looking for are: all balls in box one, all in 2 all in three or all in four
So the probability of putting all the balls in the same box is 4/4^7=  1/4^6
a)this is maybe a bit more difficult.... suppose that the box 1 is empty..so we have to put seven balls in the other three boxes. there are (7+3-1)CHOOSE 7 ways of do that. Because we can leave box 1,2,3 or 4 empty we have to multiply by four. then we have
11!4/7!4!(4^7)  hope is it correct and sorry for the lack of latex :P


What's the most secure algorithm for encrypting strings letter by letter?
I need a CGI/PHP function to encrypt/decrypt a string.

Well, the most secure method would be to generate a 100% random password, equal in length to the original string. Then for each character in the original string, add the ASCII value of the corresponding character in the password, and if the new value is greater than 255 (or 127, if you're using 7-bit ASCII) then subtract 256. To decrypt, simply subtract the ASCII value of each corresponding password character, and add 256 if the result is less than zero. The resulting encrypted string will be 100% random, and the only information one could derive from it without the password would be the string length (you could add extra characters if that's really a concern).

But obviously, that's not always a viable or appropriate solution. There's one-way encryption, two-way encryption, and public-private encryption. For two-way and public-private, there are also methods that use passwords, and methods that use keys. And sometimes the password or key is static, sometimes it is dynamic (e.g., a server generates a new password each day), and sometimes it is defined by the user of a program.

So basically, it depends on a lot of variables. If you have a specific issue that demands encryption, please describe how encryption should work in more detail. If you're just interested in exploring different techniques, Wikipedia has some pretty good articles and links on the subject.

Hmmm... could this be done using Javascript?

It sure can. :) It's not the best option if you need to encrypt very large amounts of data since Javascript is an interpreted language and thus executes rather slowly, but it's fine for many applications.

What are the functions for translating ASCII values to text/viceversa in JS?