Did anyone think it was really easy? There were no tough counting/geomretry problems, and I got a score of 5, answered 8 in total and also had two dumb mistakes, way above my average score ( usually I have a 3, 3 dumb mistakes). When reading the solutions guide, I even understood the solutions to more than 10 problems, which is a first for me.  Hopefully, if the 2005 AIME is this easy, I can hopefully get a score of 7 by checking my answers carefully, which will hopefully be the floor for sophomores.
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Or maybe this AIME was really easy. :(

I thought it was easy too. I got 3 above my average :)

That's odd... I actually thought 2000I was one of the harder tests. Numbers 1 to 11 were easy-normal, but 12, 13, 14 and especially 15 were hard. I scored about 2 points below my average on this one.

Although, if the 2005 AIME is really easy, the floor might be 8 or 9. Which is what I'm worried about, given the relatively easy AMCs.

tekno...if an (my) average score is low enough, then it is easy b/c 1-11 were easy :)

AIME Difficulty is subjective to who's taking it.

If you've just gotten to an AIME level, then difficulty depends on problems 1-3.

If you're still getting used to AIME type math, then the difficulty depends on problems 4-7.

If you're looking to qualify for USAMO, difficulty depends on 8-10.

If you're looking for 250+ index, difficulty depends on 11-13.

If you're USAMO top 30 material, difficulty depends on 14-15.

I honestly think that's a good spread on the problems. #15 is sometimes fairly hard and sometimes very hard considering the time constraints. A lot of years #14 is right up there. On the other hand, I haven't had trouble with 11 through 13 for almost any AIME, so there's a big dropoff there.

On the other side of the spectrum, #4 is usually where the first semi-tricky problem comes up for someone who's used to this kind of math. And #8 is where they start to get complicated. (Consider the multi-pointed star problem last year that MANY people missed because there were so many ways you could interpret something wrong or overlook something).

Was 1995 AIME generally considered easy, medium, or hard? I got a 10 on 1995, which is a little higher than my average(although I only solved 11 problems)

Hmm... I tried this and I got a 10, though I attempted 13. Minor errors on the other 3. That's 3 points above my average.

2000 aime for me was easy also. I got 11. WAY ABOVE my average. about 4-5 points

Does anyone know where I can find a transcript of this years AIME?

Tonight, there's this Math Jam of this year's AIME I, and they'll post the questions and discuss the solutions. After, you can get the transcript of the Math Jam - just click on "Math Jams" on the tabs above and click "Transcripts."


Let $ ABC$ be an equilateral triangle inscribed in circle $ O$. $ M$ is a point on arc $ BC$. Lines $ \overline{AM}$, $ \overline{BM}$, and $ \overline{CM}$ are drawn. Then $ AM$ is:
[asy]defaultpen(linewidth(.8pt));
unitsize(2cm);

pair O = origin;
pair B = (1,0);
pair C = dir(120);
pair A = dir(240);
pair M = dir(90 - 18);

draw(Circle(O,1));
draw(A--C--M--B--cycle);
draw(B--C);
draw(A--M);
dot(O);

label("$A$",A,SW);
label("$B$",B,E);
label("$M$",M,NE);
label("$C$",C,NW);
label("$O$",O,SE);[/asy]$ \textbf{(A)}\ \text{equal to }{BM + CM}\qquad \textbf{(B)}\ \text{less than }{BM + CM}\qquad \textbf{(C)}\ \text{greater than }{BM + CM}\qquad$
$ \textbf{(D)}\ \text{equal, less than, or greater than }{BM + CM}\text{, depending upon the position of }{ {M}\qquad}$
$ \textbf{(E)}\ \text{none of these}$

[hide="Solution"]
Obviously $ ABMC$ is cyclic. Let $ AB\equal{}BC\equal{}CA\equal{}x$.

By Ptolemy's theorem, $ x\cdot MB\plus{}x\cdot CM\equal{}x\cdot AM\implies MB\plus{}MC\equal{}MA$.

$ \boxed{\textbf{(C)}\ A}$.
[/hide]


inequalities
rearrangement inequality
inequalities unsolved

let $a,b,c >0$and $abc=a+b+c$
      prove that :
$\sqrt{27}/4 \leq \frac{bc}{a(1+bc)} +\frac{ab}{c(1+ab)}+ \frac{ac}{b(1+ac)}$

It 's easy.
Without loss of generalization, we can suppose $a \geq b \geq c$.
We have:
$bc \leq ca \leq ab$
$a(1+bc) \geq b(1+ca) \geq c(1+ab)$
then using the rearrangement inequality, we have:
$LHS \geq 3.\frac{ab+bc+ca}{a(1+bc)+b(1+ca)+c(1+ab)}$
We only need to prove that:
$ab+bc+ca \geq \sqrt{3}(a+b+c)$
Notice that
$(ab+bc+ca)^2 \geq 3abc(a+b+c)=3(a+b+c)^2$
so that $ab+bc+ca \geq \sqrt{3}(a+b+c)$

this is one of the problem in THTT magazine of vietnam.
@tran thanh nam:You shouldn't post in here


Calculate the following sum:$\sum_{s=2}^{\infty}\sum_{n=2}^{\infty}{\frac{1}{n^{s}}}$.

For whose, who considered this problem as an easy one, take a look to http://www.mathlinks.ro/Forum/viewtopic.php?t=5101 :)


inequalities
function
integration
absolute value
real analysis
real analysis unsolved

Let $f: [a,b]\to \mathbb{R}$ be an absolutely continuous function and $f_{avg}(x)=\frac{1}{b-a}\int_{a}^{b}f(x)\text{ d}x$ be the average of $f$ on $[a,b]$. Show that
\[\left |\int_{a}^{b}(f(x)-f_{avg}(x))\text{ d}x\right |\leq |b-a|\int_{a}^{b}|f'(x)|\text{ d}x.\]

Hi,
i think that the average function should be defined as
$f_{avg}(x)=\frac{1}{x-a}\int_{a}^{x}f(t)dt$.

Now, $f$ is absolutely continuous then $f$ is differentiable almost everywhere and $f$ is of bounded variation from which it follows that your function $f$ is a difference of two monotonic increasing functions, thus without loosing generality we assume that your function is monotonic increasing and of course differentiable.

Then the problem is simple since $f'(x)>0$ , f being an increasing function.

$f(a)\leq f_{avg}(x)\leq f(x)\leq f(b)$ thus

$f(a)-f(b)<f(x)-f(b)\leq f(x)-f_{avg}(x)\leq f(x)-f(a)<f(b)-f(a)$
Integrating we get that
$(f(a)-f(b))(b-a)<\int_{a}^{b}(f(x)-f_{avg}(x))dx<(f(b)-f(a))(b-a)$
thus the absolute value of the middle term is less than or equal to $f(b)-f(a))(b-a)=(b-a)\int_{a}^{b}|f'(x)|dx=(b-a)\int_{a}^{b}f'(x)dx$.

[quote="didilica"]thus without loosing generality we assume that your function is monotonic increasing[/quote]
I can't see why can we assume that. I was thinking a proof based on Poincaré's inequality.

I think $f_{\mathrm{avg}}$ does not depend on $x$.

Let $g=f-f_{\mathrm{avg}}$. There is $x_{0}\in [a,b]$ such that $g(x_{0})=0$. Since $g(x)=\int_{x_{0}}^{x}g'(x)dx$, it follows that $\sup_{[a,b]}|g|\le\int_{a}^{b}|f'(x)|dx$.

A function $f$ is of bounded variation if and only if it is the difference of two monotone increasing functions.
I think it is a classical result.


geometry inequalities AMC USA(J)MO USAMO function symmetry

area 10...
each patch is > 5...total area of patches is therefore > 25...
if two patches overlapped completely area of patches is >20...but this is still bigger than the coat itself ...so we have proved that at least two patches overlap...

to prove that there is a common are of two... if all patches overlapped each other (to subtract most area) and common area < 2...total area is still at least 17 (5*3)+2 common area...so there is the proof...is it correct or horribly wrong?

You can get a lot closer to 10 than 17.. for example if we have the 5 patches in a loop with each one intersecting in an area  just under 2 with its two neighbours, the total area would be about 1+2+1+2+1+2+1+2+1+2 = 15.. just as an example.

Lets see what I can cook up out of the blue here..

If the coat has area 10, then you have to lose at least 15...sooo...

Suppose the case that there is not a common area of 2 anywhere. That would mean the minimum you could get to is a loop, as TripleM stated, because everything is overlapping another thing....

Or is it the minimum way?

I'll have to play around with it a bit more, but I'm pretty sure proof is by contradiction.

My loop definitely won't be the best way, I was just saying that as an exampl for under 17... the best way will end up being just over 10 (as close to ten as you like..)
I'm not sure how contradiction would work. Going the other way, the first patch must take up area 5, the next one at least 3 more, and the third one at least 1 more. Now from here all we have to prove is the last two must take up at least 1 more in total.. so now we could go to contradiction and assume both of the last two patches use at least area 4 that has already been covered by the original three.. and here i'm a tiny bit stuck. I can sorta see whats going on but its impossible to visual exactly whats happening with 5 things intersecting..

[quote="rrusczyk"][img]http://www.artofproblemsolving.com/Community/Images/Problems/P2.gif[/img]

Post your solutions here...[/quote]

I have to ask you something...

Is the 2 at the end of the problem referring to units of area or number of patches?

Is "common area" referring to overlapping patches?

Do you have to use all the patches?

 :mrgreen:

The 2 at the end is area, not number of patches. [I made this clarification in the problem statement for you.]

Common area refers to the area of the coat covered by both patches.

All patches are used.

I think you can use graph theory on this one. Put down 5 dots. If two dots can be connected with a line without intersecting any other lines, then those two patches can intersect each other. I'm not sure how you generalize this to three patches intersecting each other, though. Am I on the (one of the) right track(s) though?

Does graph theory help you say anything about areas for this problem?

Graph theory is basically just to do with points and edges. Its very different to geometry but can be used to prove some famous results - eg among any 6 people either there are three who know each other or three who do not know each other.
But it doesn't deal with areas of any kind and can't really be used here.

No,  I mean that if two points can be connected, the two patches intersect, and you can chop of 2 sq. units of your total. Ah well, maybe it won't work anyway. It's just that you need some sort of geometric background to this problem because if you just algebraically assume that all possible intersections are possible, then that is 2*(5 choose 2) =20 sq. units that can be subtracted from your original 25 sq. units of patches and that is under 10, so it would work. Obviously, all the patches cannot all intersect one another, so you need some sort of geometry to show how many intersections are possible.

I'm pretty sure you can assume all intersections are possible. It doesn't tell you the shape of the coat or patches or anything anyway.. so I assume its nothing to do with geometry.. this doesn't mean we can subtract 20 units, because of the restriction on the size of each patch. If, say, one patch intesects with the other 4 we can't assume each of these sizes is 2 as the patch is only size 5. Of course it depends on intersections of three or more. And so on. I'm pretty sure the problem could be finished off from what I had in one of my other posts and I'm really close to doing it that way but I keep ending up using too many variables, it should have a nice simple finish to it..

[img]http://www.artofproblemsolving.com/Community/Images/Problems/P2.gif[/img]

:D

Assume each patch has an area of [color=#CC9933]5 square units (the lowest possible number)[/color].

Then the area of the [color=#99CC33]5 patches combined[/color] would be [color=#33CC99]25 square units[/color].

[color=#33CC99]25[/color] - [color=#3399CC]10 (size of coat)[/color] = [color=#9933CC]15, the total common area.[/color]

For no two patches to exist with a common area of [color=#CC3399]2 square units (minimum)[/color], that number would have to be [color=#CC3333]less than 10.[/color]

[color=#9933CC]15[/color] <= [color=#CC3333]10[/color] , thus [color=#33CC33]2 patches exist with a common area of at least 2 square units.[/color]

Does that work?

It should be possible to get 5 choose 2 = 10 overlaps. But that still only comes to 20, and I don't know what to do if there are three patches overlapping.

This is a stab at giving more definition to the approach sketched in Ians earlier postings, proceeding forwards instead of backwards.
We can consider the case where the patches making up the coat are all of area P = 5, rather than of area >= 5, since enlarging a patch beyond the minimum area allowed will not allow a reduction of its pairwise overlaps with other patches.
We want to show that some pair of patches has an overlap area of 2 or more. Let's assume that, as considerations of symmetry suggest, the greatest overlap is minimal if all patch pairs, triplets, etc. have the same overlap; if so, a proof for that case is enough.
The area of each patch can be partitioned into: Regions of total area A1, that have no overlap with other patches; regions of total area A2, that overlap in the coat with just one other patch (twofold overlap); regions of total area A3, that overlap in the coat with just two other patches (threefold overlap); regions of total area A4, that overlap in the coat with just three other patches (fourfold overlap); and regions of total area A5, that overlap in the coat with all four other patches (fivefold overlap).
By the symmetry assumption above, these quantities are the same for all patches. Thus,
P = A1 +A2 +A3 +A4 +A5 = 5, with A1 etc. non-negative

The total overlap between each pair of patches, O, is A2/4 from twofold overlap (since the total twofold overlap of A2 for any patch is equally shared among the 4 other patches), plus A3/2 from threefold overlap, because this always involves half of the 4 other patches, plus 3*A4/4 from fourfold overlap, which involves 3 of the 4 other patches, plus A5:

O = A2/4 +A3/2 +3*A4/4 + A5

The equation for P above gives one constraint, and another is set by the coat area. The no overlap areas for all five patches contribute 5*A1 to coat area. Coat regions with twofold overlap are covered by 2 patches, so their contribution to coat area is half the total twofold-overlapping patch area, or 5*A2/2. Likewise the coat area with triple overlap is 5*A3/3, etc., and the total coat area is

C = 5*(A1 +A2/2 +A3/3 +A4/4 +A5/5)

The coat area loss due to overlap, per patch, is

L = P  C/5 = A2/2 + 2*A3/3 +3*A4/4 + 4*A5/5 = 15/5 = 3

This equation shows that a coat with only twofold overlap is impossible, since it would require A2 = 6, in violation of the equation for P; but a coat with only threefold overlap is possible (A3 = 4.5), and so is is a coat with some threefold and some twofold overlap.
    Next we note that for the given loss due to overlap, the pairwise patch overlap, O, will be minimized by using as much twofold overlap as possible, and no four- of five-fold overlap. Comparing the equations for O and L shows that twofold overlap contributes twice as much to L as to O, and the ratio of contributions tapers down for threefold  overlap, etc.
In the equation for L, setting A4 and A5 to zero leads to
        A3 = 4.5  0.75*A2
The resulting experession for patch pair overlap in terms of A2 becomes
        o =  2.25 A2/8,
and the non-overlap area A1 is 5-A2-A3 = 0.5 - A2/4.
    Since A1 can not be negative, A2 <= 2, so o >=2.
We have blithely ignored all geometrical constraints. But incorporating those constraints could only prevent achievement of the derived minimum overlap of 2, so this doesnt invalidate the argument.

I think this solution works given the restrictions you added; however, to complete the problem, you'd have to prove that the optimal solution does involve symmetry as you suggest.  It is possible that another equal (or better in terms of reducing pairwise overlap) solution exists that does not have every patch overlapping every other in exactly the same way.  Perhaps each patch overlaps two others instead of all 4.

(As an aside, could you draw the coat that has every possible pair overlapping and every possible triplet overlapping in exactly the same area?)

I'm leaving this problem 'sticky' (at or near the top of the message list) until someone finishes off the solution...

Here is a proof of the patches problem using probabilistic methods.

  Let p be a random point on the coat.  Let n be the number of patches that contain p.  Because the sum of the areas of the patches is at least 25, the expected value of n is at least 25/10 = 2.5.  Because n is an integer, we have n(n-1)/2 >= 2n - 3.  Taking the expected value of both sides, we get
     E(n(n-1)/2) >= 2E(n) - 3 >= 2*2.5 - 3 = 2.
That means that the sum of the pairwise common areas is at least 2*10 = 20.  Thus the largest of the 10 pairwise common areas is at least 20/10 = 2.

Very intriguing method.

Can you elaborate how you went from here:

[quote]E(n(n-1)/2) >= 2E(n) - 3 >= 2*2.5 - 3 = 2[/quote]

to here:

[quote]That means that the sum of the pairwise common areas is at least 2*10 = 20[/quote]

The expression n(n-1)/2 is the number of pairwise intersections that contain the point p.  Thus the expected value E(n(n-1)/2) is the sum, over each of the 10 pairwise intersections I, of the probability that p is in I.  But the probability that p is in I is just area(I)/10.  In other words, the sum of the areas of the pairwise intersections is 10 E(n(n-1)/2), which is at least 20.

  Using your previous notation, we have shown that B >= 2A - 30.

Very nice approach - know any other non-obvious problems that can be attacked this way?

Well, here is a similar problem.  Suppose you have a coat of area 1 with a bunch of patches on it.  Let A be the sum of the areas of the patches.  Let B be the sum of the areas of the pairwise intersections of patches.  Prove that B >= A(A-1)/2.

  If I can think of problems in other domains that can be solved with this method, I will let you know.

Also, you may want to check out the book "The Probabilistic Method" (Second Edition) by Noga Alon and Joel Spencer (Wiley).  They present many examples of using probabilistic methods to attack non-probabilistic, mostly discrete problems.  You can find a table of contents and the first chapter at http://www.cs.nyu.edu/cs/faculty/spencer .

I found this wonderful problem through a link in Ravi's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23507]post[/url].

Here is an approach using characteristic function and integration of characteristic functions.

The characteristic function $\lambda_S$ for set $S$ is defined such that $\lambda_S(x)=1$ if $x\in S$, $\lambda_S(x)=0$ if $x \not\in S$. A simple fact is $\lambda_S\lambda_T=\lambda_{S\cap T}$. In particular $\lambda_S^2 = \lambda_S$.

Denote $C$ to be the coat and $P_i, i=1,2,3,4,5$ to be the patches.

Consider the function $F = \sum \lambda_{P_i}$, then $(F-2)(F-3)\ge0$ for any point in $C$ since $F$ takes on integer values only. That is $F^2 \ge 5F - 6$ over $C$.

Integrate over $C$, we get

$\int_C F^2 \ge 5 \int_C \sum \lambda_{P_i} - \int_C 6 = 5 \sum \int_C \lambda_{P_i}- 60$.

On the other hand,

$\int_C F^2 = \int_C (\sum \lambda_{P_i})^2 = \sum \int_C \lambda_{P_i}^2 + 2 \sum_{i<j} \int_C \lambda_{P_i}\lambda_{P_j} =\sum \int_C \lambda_{P_i} + 2 \sum_{i<j} \int_C \lambda_{P_i}\lambda_{P_j}$.

Hence

$2\sum_{i<j} \int_C \lambda_{P_i}\lambda_{P_j} \ge 4 \sum \int_C \lambda_{P_i}- 60\ge40$.

Therefore there exists a pair $P_i$ and $P_j$ such that $\int_C \lambda_{P_i\cap P_j} = \int_C \lambda_{P_i}\lambda_{P_j} \ge 2$ which means the overlap of the patches $P_i$ and $P_j$ has area at least $2$. Q.E.D.

[quote="Ravi B"] Suppose you have a coat of area 1 with a bunch of patches on it.  Let A be the sum of the areas of the patches.  Let B be the sum of the areas of the pairwise intersections of patches.  Prove that B >= A(A-1)/2.
[/quote]

This problem can be proved in similar fashion as above. In fact, if $F$ is the sum of the characteristic functions of all patches, then the integral of $F^2$ over $C$ (the coat) is $A+2B$. On the other hand, by Cauchy-Schwarz, $\int_C F^2=(\int_C F^2)(\int_C 1^2) \ge (\int_C F)^2 = A^2$.

[quote="Ravi B"]Another approach to the original problem is to use the following variant on the Principle of Inclusion-Exclusion:
\[ \Bigl| \bigcup_i P_i \Bigr| \geq \frac{2}{3} \sum_i \left| P_i \right| - \frac{1}{3} \sum_{i < j} |P_i \cap P_j| \, .  \]
Here $P_i$ is the $i$th patch, and $|S|$ means the area of region $S$.[/quote]

Wow! Very nice inequality! I bet not many people on this forum have seen this before.

This can also be proved using the inequality $F^2 \ge 5F - 6$.

If $C= \bigcup_i P_i$ is the union of all patches. Then:

$\int_C F^2 \ge 5 \int_C \sum \lambda_{P_i} - \int_C 6 = 5 \sum_i \left| P_i \right| - 6\Bigl| \bigcup_i P_i \Bigr|$.

On the other hand,

$\int_C F^2 = \int_C (\sum \lambda_{P_i})^2 = \sum \int_C \lambda_{P_i}^2 + 2 \sum_{i<j} \int_C \lambda_{P_i}\lambda_{P_j} =\sum \int_C \lambda_{P_i} + 2 \sum_{i<j} \int_C \lambda_{P_i \cap P_j}$

$=\sum_i \left| P_i \right| + 2\sum_{i < j} |P_i \cap P_j|$.

The desired inequality follows immediately.

By the way, I think [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=405456#p405456]Mediterranean Mathematics Olympiad 2005, Problem 3[/url] is also true for area of patches.

I know that this is quite a revive, but can you please repost the original image? There are some new topics that link to this topic in a chain of 2-3 links, and the first diagram seems not to exist. Does this have anything to do with the server crash almost a decade ago?

Thanks; I remember that problem. In fact, it appears twice: once in the PIE chapter, and another time somewhere else that I cannot remember right now...

Thank you very much!

EDIT: I canot spell correctly.


Mọi ng` ơi, có ai có đề thi China MO 2008 và đề thi China TST 2008 thì cho em xin với ạ, em toàn thấy bản tiếng tàu đọc mà rõ là ngộ chữ, ai có bản tiếng Anh hoặc tiếng Việt thì cho em xin với nha, em cảm ơn nhiều ạ

Anh Fire_dragon tieng Trung o dau vay.Anh post len duoc khong a(hay duong link cung duoc).

Why don't you google it? do not use http://www.google.com.vn because you wouldn't find any good information. Use http://www.google.com, it is better. Hope this help!


My question: how does one signify a line by placing a short line over two capital letters? (For example, the line between points X and Y would be noted XY with a line over it: how is that shown in LaTeX?)

Thanks.

\overline{XY} which gives $\overline{XY}$


algebra
polynomial
search
algebra proposed

It is known that polynomial $ (x\plus{}1)^{n}\minus{}1$ is divisible by $ x^{2m}\plus{}c_{2m\minus{}1}x^{2m\minus{}1}\plus{}\dots\plus{}c_0$,where $ c_{2m\minus{}1},c_{2m\minus{}2},\dots,a_0$ are integer,odd numbers.Prove that $ n$ is divisible by $ 2m\plus{}1$.
Actually this problem was posted before,but nobody solved it. :( I know the official proof,but i'm looking for another...To be honest,i don't like the official solution

hasn't been solved here
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1003100212&t=86564
but has been solved here
http://www.mathlinks.ro/viewtopic.php?t=103844


Let $ a,b,c>0$,with $ a\plus{}b\plus{}c\equal{}3$. Prove that:

$ a^2\plus{}b^2\plus{}c^2\plus{}3abc\ge 2(ab\plus{}bc\plus{}ca)$

Equivalent to Schur:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3abc\ge 2(ab \plus{} bc \plus{} ca)$ iff
$ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)/3 \plus{} 3abc \ge 2(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)/3$ iff
$ \sum a^3 \plus{} \sum a^2b \plus{} 9abc \ge 2\sum a^2b \plus{} 6abc$ iff
$ \sum a^3 \minus{} \sum a^2b \plus{} 3abc \ge 0$ iff
$ \sum a(a \minus{} b)(a \minus{} c) \ge 0$

Yes that's the solution.I found it interesting because the inequality $ a^2\plus{}b^2\plus{}c^2\plus{}2abc\plus{}1\ge 2(ab\plus{}bc\plus{}ca)$ holds for all $ a,b,c>0$.

for proof:[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=19666[/url]


Find all ordered triplets $(x, y, z)$ such that $x$, $y$, and $z$ are (positive) primes, and $x^y+1=z$.

[hide]The only one is (2,2,5).

x can't be odd, because any odd raised to any power, even or odd, is odd, and thus adding a 1 will make it even, so z wont be prime then.  So since x has to be even, and 2 is the only even prime, x can only be 2.  Moving on, we find that 2 raised to an odd number plus 1 gives us a number that is divisible by 3, although I'm not sure why.  So y can't be odd and y must be even.  The only even prime is 2, so x and y must both be 2.  Thus, 2^2 + 1 = 5, so our only ordered triple is (2,2,5).[/hide]

What kind of methods could we use to find out why $3|(2^{2n+1} + 1)$?

[hide]I don't quite know if my explanation is correct, but NightFlarer is right. (2,2,5). However, to prove 2^(y) + 1 isn't ever odd, where y > 2. We can start by noticing that for 2^(y) + 1, where y is odd, that it can be factored, with root -1, which means you obtain (n+1) as a factor. Since, n = 2, that number is always 0(mod 3), which means it can never be a prime. Thus, (2,2,5) is the answer. For example - to answer NightFlarer's "why", 2^3 + 1 = (2+1)(2^2-2+1). You use x^(2n+1)+1=(x+1)(x^(2n)-x^(2n-1)+x^(2n-2)+ - + -...+1).[/hide]

[hide]I havent' looked at the other solutions yet, but here's mine.

Let's say that x is an odd prime.  Then, x^y would also be odd since an odd number times and odd number euqals an odd number.  If we add 1 to that, then we get an even number, which isn't prime (except for 2, but that wouldn't work since 2-1 = 1 and no prime number to the power of a  prime number equals 1).

Therefore, x has to be 2, the only even prime number.  We also know that y has to be either 2 or odd.  I found out that if x is 2 and y is odd, then x^y+1(mod 3) = 0.  Since that is divisble by 3, and 2 is the only even prime number, then y has to be 2 if the equation will work for any number.

So if x =2 and y=2, then x^y+1 = 2^2+1 = 4+1 = 5, which is prime.

So, (2, 2, 5) is the only possible solution.[/hide]


geometry
rectangle
symmetry
rotation
combinatorics unsolved
combinatorics

Given real numbers $a > b > 0$. If a rectangle with sides $a$ and $b$ can be completely put into the interior (including the border) of a square, find the minimum side length of the square.

well... for symmetry reasons one should only consider 2 cases... horizontally, and rotated 45 degrees.

In the first case it's max(a,b), in the second case it is (a+b)/sqrt(2). And it's easy to check when eachone is bigger :)

Am I right?


ABCD... is a regular convex polygon. Determine the number of diagonals of polygons, knowing that the angle $ B\hat{D}E$ measures 144 degrees.

[img]http://img402.imageshack.us/img402/6289/calus.jpg[/img]

[hide="Solution"]Angle $ DBA$ is also $ 144$ degrees, so by looking at triangle $ BCD$, we know that, calling the internal angle of the polygon $ x$:

$ 2(x\minus{}144)\plus{}x\equal{}180$

$ 3x\equal{}468$

$ x\equal{}156$.

Using the formula for the size of an internal angle in a regular polygon:

$ \frac{180(n\minus{}2)}{n}\equal{}156$

$ 156n\equal{}180n\minus{}360$

$ 24n\equal{}360$

$ n\equal{}15$.

A convex n-gon has $ \frac{n(n\minus{}3)}{2}$ diagonals, so a 15-gon has:

$ \frac{15 \cdot 12}{2}\equal{} 90$.

So the polygon has 90 diagonals.[/hide]


Find $ n$:

$ (1\plus{}\tan 1^\circ)(1\plus{}\tan 2^\circ)(1\plus{}\tan 3^\circ)\cdots(1\plus{}\tan 45^\circ)\equal{}2^n$

[hide]Let's say that $ (1 \plus{} \tan(n))(1 \plus{} \tan(45^\circ \minus{} n)) \equal{} 2$
Proof: $ (1 \plus{} \tan(n))(1 \plus{} \tan(45^\circ \minus{} n)) \equal{} (1 \plus{} \tan(n))(1 \plus{} \frac {\tan(45^\circ) \minus{} \tan(n)}{1 \plus{} \tan(45^\circ)(\tan(n))})$$ \equal{} (1 \plus{} \tan(n))(1 \plus{} \frac {1 \minus{} \tan(n)}{1 \plus{} \tan(n)}) \equal{} (1 \plus{} \tan(n))(\frac {2}{1 \plus{} \tan(n)}) \equal{} 2$

We don't have $ (1 \plus{} \tan(0^\circ))$ but let's include it, anyways it's $ 1$.
$ \displaystyle\prod_{n \equal{} 0}^{22}(1 \plus{} \tan(n))(1 \plus{} \tan(45^\circ \minus{} n)) \equal{} 2^{23}$[/hide]

Thank you for the solution, I appreciate it


Sorry if this was posted already:

Determine the smallest perimeter for a convex 32-gon such that its vertices are all lattice points.

Unless I'm mistaken, the answer is $4(1 + \sqrt{2} + 2\sqrt{5} + 2\sqrt{10} + 2\sqrt{13})$.

Proof follows.

Necessity: since our polygon is convex, no three sides are parallel. Therefore we need 16 different gradients. A gradient $\displaystyle \frac{y}{x}$ in its lowest terms requires a minimum side length of $\sqrt{x^2 + y^2}$. It's easy to see by inspection that the 16 gradients that will minimise this are (in the order they will be placed to make the polygon convex):
$\displaystyle \frac{0}{1}, -\frac{1}{3}, -\frac{1}{2}, -\frac{1}{1}, -\frac{3}{2}, -\frac{2}{1}, -\frac{3}{1}, \frac{1}{0}, \frac{3}{1}, \frac{2}{1}, \frac{3}{2}, \frac{1}{1}, \frac{2}{3}, \frac{1}{2}, \frac{1}{3}$.
Call these gradients $g_i$, $1\leqq\leq16$.
They have total length half that shown above, but we need two of each. So the formula above is a lower bound for the required perimeter.

Sufficiency: Construct a set of 16 lines $l_i$, $1\leqq\leq16$, fitting the following conditions.

(1) Each line has one endpoint co-ordinate $(a_i, b_i)$ and the other at $(c_i, d_i)$, both w.r.t. a fixed origin.

(2) $c_i = a_{i+1}, d_i = b_{i+1}$

(3) $\displaystyle \frac{d_i-b_i}{c_i-a_i} = g_i$

(4) The length of each $l_i$ is minimal

Now superimpose the figure obtained with the figure given by rotating it a half-turn about $\displaystyle(\frac{a_1+c_{16}}{2},\frac{b_1+d_{16}}{2})$. The result is a 32-gon with minimum perimeter.

Don't be surprised if I've made a horribly stupid mistake. It happens rather often.


[quote="lord12"]how does 2 + 9y (congruent)3(mod7) = 2y(congruent)1(mod7)??[/quote]

$9y+2 \equiv 3 \mod 7$

$9y \equiv 1 \mod 7$

$2y+7y \equiv 1 \mod 7$

$2y \equiv 1 \mod 7$

Because $7y \equiv 0 \mod 7$

What does this have to do with calculus  :?:


Let $ n$ denote the minimum number of squares that must be removed from a 4 by 4 checker board so that there does not remain a 1 by 3 configuration of squares. What is $ n$?

Um, my intuition tells me the answer is $ 4$, but I'm not sure at all.

x o x x
x o x o
o x o x
x x o x

6 works.

Answer must be at least 4 because every row/column must have a space. All spaces must be in the middle two rows to block 1x3 formation. Furthermore answer must be even, otherwise it is rotationally asymmetric.

Therefore we need at least 6, 6 is attainable, so 6.


ratio
inequalities
number theory unsolved
number theory

For any nonempty set $S$ of real numbers, let $\sigma(S)$ denote the sum of the elements of $S$. Given a set $A$ of $n$ positive integers, consider the collection of all distinct sums $\sigma(S)$ as $S$ ranges over the nonempty subsets of $A$.  Prove that this collection of sums can be partitioned into $n$ classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2.

[hide="solution"]
WLOG assume:  $a_1<a_2<...<a_n$.  For $i=1,2,...,n$, let $s_i=\sum_{j=1}^{i}a_j, s_0=0$.  All of the sums are less than or equal to $s_n$ and if $\sigma$ is one of them we have:
$s_{i-1}<\sigma\le s_i$ for some $i$.
Divide thes ums into $n$ groups by letting $C_i$ denote the group of sums satisfying the above inequality.  We claim that these groups have the desired property.  to prove this, it is sufficient enough to prove that the following inequality implies that:
$\frac{1}{2}s_i<\sigma\le s_i$

Suppose $s_{i-1}<\sigma\le s_i\Rightarrow a_1+a_2+...+a_{i-1}=s_{i-1}<\sigma$  and since  $a_1<a_2<...<a_n$, we can conclude that $\sigma$ contains at least one term $a_k$ such that $k\geq i$.  Then since $a_k\geq a_i$ we have:
$s_i-\sigma<s_i-s_{i-1}=a_i\le a_k\le \sigma$
and since $\sigma \le s_i$ we are done.
Q.E.D. [/hide]

I tried this approach, but I'm not sure if it would work.  :D

[hide="Approach"]The approach is somewhat inductive. The base case n=1 obviously works. Now we assume that if there are $n$ positive integers ($a_{1}<a_{2}<a_{3}\cdots<a_{n}$) in $A,$ then the collection of sums can be partitioned into $n$ classes.

Adding an $a_{n+1}$ term (not necessarily greater than $a_{n}$), we now need to show that the new set can be partitioned into $n+1$ classes. To do this, we analyze two cases: when $a_{n+1}>a_{1}+a_{2}+\cdots+a_{n}$  and when $a_{n+1}\le a_{1}+a_{2}+\cdots+a_{n}.$

When $a_{n+1}>a_{1}+a_{2}+a_{3}+\cdots+a_{n},$ everything works. (all subsets can fit into a single class, which can be split into n+1 classes in numerous ways)

If this approach does work, I'm not sure how to finish the second case.  :roll:
[/hide]

Continuation of 4everwise's approach:

[hide]If all the elements are of different value, then you divide the inductive step into cases based on the value of $ a_{n \plus{} 1}$.  The case where $ a_{n \plus{} 1}$ is greater than the sum of $ a_1$ through $ a_n$ was already shown. If $ a_{n \plus{} 1}$ is less than $ a_{1}$, put it in the smallest class, moving all the other classes up by one.  If not all the classes have the maximum ratio of 2, then there is room to shift sums into lower classes so that eventually, the smallest sum of the largest class can be moved down, so that $ a_{n \plus{} 1}$, which is smaller than $ a_1$, can be added to the previous largest sum without exceeding the ratio restriction of 2.  We then must show that if all of the previously existing classes (from when $ A$ had n elements) already had the maximum ratio of 2, then there were only $ n \minus{} 1$ of them, so that we have room for a second new class to hold the sum of all the elements (if it's even possible to have all the classes with a ratio of 2).  If all the $ n$ elements of $ A$ are of different value, and all the classes have their maximum sum equal to twice their smallest sum, we can also show by contradiction that this isn't possible, if that's the case.

If $ a_{n \plus{} 1}$ is between $ a_1$ and the sum of all the elements of $ A$, inclusive, just create a new class if necessary- you won't need more than one since all the new sums will be no more than twice the other new sums which go into the new class.
[/hide]
"Given a set $ A$ of $ n$ positive integers, consider the collection of all distinct sums..."

By "all distinct sums", I think this just means that if you have 2,3,7,8 as elements of $ A$, you can't use the sum $ 10 \equal{} 2 \plus{} 8 \equal{} 3 \plus{} 7$ twice as the maximal sum of one class AND the minimal sum of the next class, even though you can find it in two different ways.

Does anyone have a solution that works if some, but not all, of the $ a$'s are equal? I have not found a solution that works in this case.(It's trivial when they're all equal)[/hide]

Does this work (it is kind of along the lines of 4everwise's approach)?

We proceed by induction. For a set of size 1, it is clearly true. Assume the statement holds for n=k. Now we prove it for n=k+1. Consider the set $\{a_1, a_2, \dots, a_k, a_{k+1}\}.$ Let us pick an element, WLOG, $a_{k+1}.$ The remaining elements, by the inductive hypothesis can be split into k classes with the desired condition. Now, we just place $a_{k+1}$ into the (k+1)st class.

Am I missing something?

@limac: We are not to place the elements, we are to place the [i]sums[/i] of the elements.

@LawOfSigns: Yes, that is what is meant by distinct sums. So if we have say $a_1=a_2$, then we may just eliminate $a_2$ because it doesn't give us a new subset sum.

Let the elements of $A_n$ be $a_1$, $a_2$...$a_n$. We claim that letting $a_1+a_2...+a_k$ for all $1\leq k\leq n$ be the maximum element of each class, gives a construction that satisfies the problems conditions. For any subset $S_i$ of $A$, $S_i$ either belongs in one of the classes or doesn't. Suppose there exists some $k$ such that for suitable choice of $i$, we have $\sum_{p=1}^{i} a_p < S(k) < \frac{1}{2}\sum_{p=1}^{i+1} a_p$, where $S(k)$ is the sum of the elements of $S_k$. In other words, $S(k)$ is "in-between" classes. It follows that $a_1+a_2+\dots +a_i <a_{i+1}$. However this produces a contradiction. If $a_{i+1}\in S_k$ then we have that $S(k)>\frac{1}{2}\sum_{p=1}^{i+1} a_p$. If $a_{i+1}\not \in S_k$, then we see that $S(k)<\sum_{p=1}^{i} a_p < S(k)$. This also produces a contradiction, hence, we have that every subset sum belongs to a class, and we are done. $\square$

Can you Please explain with say  set with 10 elements.I am having hard time understanding the class or partition of the sum into n class

Suppose $A = \{a_1 < \dots< a_n\}$ and consider partial sums $a_1$, $a_1+a_2$, $\dots$, $a_1+\cdots+a_n$. For $i = 1, \dots, n$, form the $i$th class as follows. If $a_i > a_1+\cdots+a_{i-1}$, it will be $[a_i, a_1+\cdots+a_i]$, and otherwise (if $a_i \le a_1+\cdots+a_{i-1}$) it will be $[a_1+\cdots+a_{i-1}, a_1+\cdots+a_i]$. It is easy to see that the ratio condition holds for this construction.

To show that this works, let $S = \{a_{i_1} < \cdots < a_{i_{\ell}}\}$ be some non-empty subset of $A$. There is some $j \le i_{\ell}$ such that $a_1+\cdots+a_{j-1} < \sigma(S) \le a_1+\cdots+a_j$. Then $\sigma(S)$ lies in both ranges $[a_1+\cdots+a_{j-1}, a_1+\cdots+a_j]$ and $[a_j, a_1+\cdots+a_j]$ (the latter being the case since $\sigma(S) \ge a_{i_{\ell}} \ge a_j$).

Solution from [i]Twitch Solves ISL[/i]: By induction on $n$ with $n=1$ being easy.

For the inductive step, assume \[ A = \left\{ a_1 > a_2 > \dots > a_n \right\}. \] Fix any index $k$ with the property that \[ a_k > \frac{\sigma(A)}{2^k} \] (which must exist since $\frac{1}{2}+\frac14+\dots+\frac{1}{2^k} < 1$). Then [list] 	[*]We make $k$ classes for the sums between $\frac{\sigma(A)}{2^k}$ 	and $\sigma(A)$; this handles every set 	which has any element in $\{a_1, \dots, a_k\}$. 	[*]We make $n-k$ classes via induction hypothesis 	on $\{a_{k+1}, \dots, a_n\}$. [/list] This solves the problem.

Assume all sets with $n$ elements satisfy the problem conditions. We want to prove that all sets with $n+1$ elements work as well.

Note that we are essentially adding an element greater than or equal to all elements in a set with $n$ elements to achieve a set with $n+1$ elements, let this number be $x$.

Now consider the collection $C$ of all  $\sigma(S)$ for all nonempty subsets $S$ in set $A$. Denote the least and greatest numbers in $C$ as $L$ and $G$ respectively.

Also note that the new collection of the set with $n+1$ elements is $C$ + every element in $C+x$. We also need to have another class.

We have the following two cases:

$\textbf{Case 1: } x \ge G$

Then $2(F+x) \ge G+x$, so all elements in $C+x$ can be held in a class.

$\textbf{Case 2: } x < G$

Then the maximum is $G+x$, which is also less than $2G$. Therefore, all elements within $[G,G+x]$ can be held in a class.

We also know that $n=1$ works for all sets

We have proved that it is possible to go from any set with $n$ elements to any set with $n+1$ elements, so we are done. $\blacksquare$

We will use induction. This is clearly true of $n=1$. Consider a set $A$ of size $n$, and suppose that this is true for all sets smaller than $n$.\\

Let $G$ be the sum of all elements. If the largest element of $A$ is at least $\frac{1}{2}G$, we can simply resolve the $n-1$ smallest elements with $n-1$ classes by induction, and put all subsets containing the largest element in their own class since they will all be between $\frac{1}{2}G$ and $G$.\\

Similarly, if the second largest element is at least $\frac{1}{4}G$, we can resolve the $n-2$ smallest elements with $n-2$ classes, and all subsets containing at least one of the two largest elements goes into either the $\frac{1}{4}G$ to $\frac{1}{2}G$ class or the $\frac{1}{2}G$ to $G$ class.\\

We can repeat the same argument if the third largest is at least $\frac{1}{8}G$, by putting everything containing one of the 3 largest into one of 3 classes that collectively span from $\frac{1}{8}G$ to $G$. Therefore, if there exists any $k$ such that the $k$th largest element is at least $\frac{1}{2^k}G$, we would be done by induction. However, such a $k$ clearly exists, since if not, the sum of all elements cannot be $G$ as even if the geometric series is summed infinitely we only get $G$, hence done.

Very magical problem, if you ask me. Or maybe I just don't have combinatorics intuition...

Let $a_1 < a_2 < \cdots < a_n$ be the elements of $S$. I claim that we can set the largest element of partition $A_k$ to be $p_k = a_1+a_2+\cdots+a_k$ for $1 \leq k \leq n$.

Suppose there exists a sum $\sigma(A)$ that is not in one of the partitions, and suppose that $p_k < \sigma(A) < p_{k+1}$. It follows that $\sigma(A) < \frac 12 p_{k+1}$ and thus $p_{k+1} > 2p_k$, or $a_{k+1} > p_k$. But then $A$ must contain $a_{k+1}$ or a larger $a_i$, so it follows that $\sigma(A) > a_{k+1} > \frac 12 p_{k+1}$, as needed.


If you were at Nats I met a lot of you, but I don't know who is going to Mathpath except one person who made it high up in the competion (don't want to say their ranking so as not to give away their identity) - who else is going and is this your first or second yr?  Maybe some of us going could chat prior to getting to the camp? Or maybe we already met? If you went last year, did you enjoy it? Would you recommend going? Advice?

I'm not going to the camp, but it's happening like on my doorstep.  I can say that UVM and vermont is a lovely place!!

thanks - Vermont weather has to be more consistent than Colorado's was this weekend - heck, we even had HAIL fall on us! sun, rain, hot, cold, Hail - all within 24 hours! ok, I guess this is slightly off topic - especially since I created the topic...

I wish I was going to MathPath...

I didn't meet you, karatemagic7...

[quote="alanchou"]me and xiaoyu call it snail :P

there is a thread on who's going to mathpath in that forum.   you could ask who there went to nats.[/quote]

Alan, don't forget me! I also called it snail. :D

[quote="karatemagic7"]Vermont weather has to be more consistent than Colorado's was this weekend[/quote]

You know, MathPath was in Colorado last year. =)

you know, it also hailed at least twice last year at mathpath

arg, this is getting too spammy.

I'm [i]Planning[/i] to go to MathPath, if I make it.

I can guarantee you that I would have made Nats if I was old

enough (5th grade  :( )

i am going to mathpath, but did not make it to nationals mathcounts...

yeah, yeah, I checked Mathpath Forum - Problem is in the "who is going"
 post, there are a zillion replies from Last Year (old post) but few, if any comments from people planning on going this yr. I know they take about 80 -100 and a LOT must use AoPS - so come out of hiding! Who are you??


Hi. I was wondering if you guys could help me solve this problem. I have been looking at it for two days now and I am not getting the right answer.

Complete combustion of 19.48 g of a particular hydrocarbon yielded 62.93 g CO 2 and 20.61 g of H 2 O. What is the empirical formula of this hydrocarbon?
Molar masses (in g mol -1 )
H, 1.00794 	C, 12.011 	O, 15.9994

The empirical formula is C_H_

I've tried almost everything but the only answers I come up with are C1H1, C2H2 and C1H4.. and they are all wrong.

I don't even know what the equation of such a problem would be.. so far all I've gotten is

C02 +h20+02--->cxhx?

Perhaps this will help.  Here is the reaction for complete combustion of a hydrocarbon:

$\text{C}_x\text{H}_y+\text{O}_2\longrightarrow \text{CO}_2 +\text{H}_2\text{O}$

Here is the balanced form:
[hide]$\text{C}_x\text{H}_y+(2x+\dfrac{y}{2})\text{O}_2\longrightarrow x\text{CO}_2 +\dfrac{y}{2}\text{H}_2\text{O}$[/hide]

Here is a step by step analysis to help you out:
[hide]
The first thing you'll want to do is write the reaction:
$\text{C}_x\text{H}_y+\text{O}_2\longrightarrow \text{CO}_2 +\text{H}_2\text{O}$

Next you need to find the moles of Carbon and Hydrogen in your products
[hide]
To convert to moles, you must divide by the Gram Formula Mass of the compound (the mass of 1 mol of the comound).  You find this by adding up the atomic weights of each of the elements in the compound.
So you have
$62.93 \mbox{ g CO}_2 * \dfrac{1 \mbox{ mol CO}_2}{44.01 \mbox{ g CO}_2} = 1.430 \mbox{ mol CO}_2$
And thus, $1.430 \mbox{ mol C}$

Do the same with the water to find moles of hydrogen
$20.61 \text{ g H}_2\text{O} * \dfrac{1 \text{ mol H}_2\text{O}}{18.015 \text{ g H}_2\text{O}} = 1.144 \text{ mol H}_2\text{O}$
And thus, $1.144 * 2 = 2.288 \text{ mol H}$

Now you should be able to write the formula
[hide]
$\text{C}_{1.430}\text{H}_{2.288}$
Now, you divide by the smallest value giving
$\text{C}_{1}\text{H}_{1.6}$
Now you need to multiply this until you obtain whole numbers in both subscripts
Multiplying by 5 gives
$\text{C}_{5}\text{H}_{8}$
[/hide]
[/hide]
[/hide]

thank you all of you. in particular, i am greatful to snowstorm.. it was easier for me to learn in the detailed step by step solution you provided me. thank you for all of your help. I really appreciate it.


a comment on this problem, I can't understand the sense of putting $ 1 \plus{} 2 \plus{} 3 \plus{} 4..........n \equal{} kn^2$ ... Can you tell me a little about this problem?


[quote="MCrawford"]If x+y = 117, y+z = 223, and z+x = 314, what is x+y+z?[/quote]

i found two ways to solve this problem, the short way and the long way. ill do both starting with the short way

add all 3 equations together to get 2x+2y+2z=654. divide both sides by 2 to get x+y+z=327.

now onto the second way

subtract x+y = 117 from the second equation (y+z = 223), hence
z-x=106, add that to the third equation
(z+x = 314), hence get 2z=420. dividing both sides by 2, hence z=210. plug that into the equation z+x=314 to get 210+x=314. hence x=104. plug x into x+y=117 to get 104+y=117. hence (dont you just love that word) y=13

now to delve ourselves into what the question is asking (no, i dont want fries with that) (oh nevermind him, he just gets out of control a lot) (no i dont) (yes you do)

person at counter says to him or herself, "uu"

now to actually delve ourselves into what the question is asking for (what is x+y+z?)
x=104
y=13
z=210

soooooooo........104+13+210=327 once again

P.S.
YYYYYYYAAAAAAAAAYYYYYYYYYYYYYY, WOOOOOOOOOOOOOOOTTTTTTTT

P.P.S.
i will not be held accountable to using hence too many times (even when it wasnt supposed to be). i blame the media :lol:

I did it the same way (The add them up way.) Well, gtg. My physics teacher is going to call me back soon. bye


Okay, you're taking it too far now, theani. Just stop.

And by the way, who was the second vote for "Yes"? I have my suspicions....

Stop making double accounts just to do this, ani!

dude what the...?

I am not!

If you want an admin can check records for you and prove my innocence  :P

YOU'D BETTER BE TALKING ABOUT THE SHAGGY FROM SCOOBY DOO!

ummmmmmm.......

okay, so [b]why[/b] did u make a poll to ask if im cool........?

Because this is fun and games and therefore all logic, rationality, and common sense go out the window.

Paladin is correct, but this kind of poll can be very offensive to the person in question no matter where it is posted.

@ Anirudh - No other accounts, you say? Then who are "#H34N1bot" and "hBot", both of which are contributors on your blog? Suspicious, eh?


Nothing special here about rank 1.

The key is Guassian elimination. Suppose that $B$ has rank $k.$ This means we can row reduce $B$ until it becomes a matrix $R$ in row echelon form with $k$ pivots. (And, necessarily, $n-k$ rows of zeros across the bottom.) That is, there exists an invertible $P$ such that $PB=R.$

Then  $(PA)A^{-1}B=R,$ the same row echelon form, so $A^{-1}B$ also has rank $k.$

I'm relying on one of the fundamental theorems of Gaussian elimination, namely that all row echelon forms row-equivalent to a given matrix have the same number of pivots.

Also: nothing special about $\mathbb{R}.$ This result would be the same over any field.

[quote="rgep"]Alternatively, let $\alpha$ and $\beta$ be the linear maps represented by those matrices.  Then $\alpha$ and $\alpha^{-1}$ are invertible, so for any subspace $U$, $\dim \alpha^{-1}U = \dim U$.  Applying this to ${\mathrm{im}} \beta$, whose dimension is ${\mathrm{rk}} \beta$ we have the result.[/quote]

this is exactly the same method I used for solving this question :)  :)

[quote="Michael"]Hi,

this is an easy question I would like to share with you:

let $A,B\in {\cal M}_n({\mathbb R})$ such that $A$ is invertible and $rank(B)=1$. Prove that $rank(A^{-1}B)=1$.

Michael[/quote]

You can use this :
If $A,B \in M_n({\mathbb R})$,$detA \neq 0$,$rank(B)=r$ then $rank(AB)=rank(B)=r$

What I posted in #2 is a proof of precisely the theorem jianovidiu just quoted.


[quote="chess64"]$xy-nx-ny+n^2=(x-n)(y-n)$

$(x-n)(y-n)=n^2$

$n=\sqrt{(x-n)(y-n)}$. You need the exact value of $n$ to do more than that, I think. :?[/quote]

This is reminiscient of Simon's Fav Factoring Trick  :lol:
and to find integer values of x and y, we can set $x-n=n^2$,$ y-n=1$ from $(x-n)(y-n)=n^2$ or any other factors of $n^2$

what are the restricitons on a,b, and n?

[quote="amirhtlusa"]the solutions are in a form of:
$x=k\cdot a\cdot (a+b)$
$y=k\cdot b\cdot (a+b)$
$n=k\cdot a\cdot b$

i had posted the solution to this long time ago
i can write it out again, if you want.[/quote]

Yes please, I would like to see it.

Amirhtlusa I knwo that your answer is true, but I don't know what is wrong with my proof!
Lemma $1$)if $(x,y)=1 \Rightarrow x+y \nmid xy$.
Now if $(x,y)=k \Rightarrow k(a+b)|k^2ab$ so $(a+b)|kab, (a,b)=1 \Rightarrow a+b|k$
but $k<a+b$ so I think we can conclude $x=y$.
What is wrong with me? :?

this is how to prove the Lemma:
assume $(x,y)=1$
suppose $\exists k>1 $ such that $k|x+y; k|xy$
$\Rightarrow k|x+y\Rightarrow k|x^2+xy\Rightarrow k|x^2$
similarly $k|y^2$ .
assume $k=p_1^{\alpha_1}...{p_i^{\alpha_i}}\Rightarrow p_j|x^2,y^2\Rightarrow p_j|x,y$ contradiction.

for the second part you finished the solution :)  this is exactly what i had in mind:
so we said $(x,y)=k$ then $x=k\cdot a$, $y=k\cdot b$ then by your proof we proved that $(a+b)|k\cdot a\cdot b$
now since $k,|a|,|b|$ must be relatively prime, then by the Lemma that we prove this is true iff, $(a+b)|k$ so put $k=c\cdot (a+b)$
that gives the solutions:
$\displaystyle x=c\cdot a\cdot (a+b)$
$y=c\cdot b\cdot (a+b)$
$n=c\cdot a\cdot b$

I've seen this problem so many times :D

[hide]
$n= \frac{xy}{x+y}$
Let $\gcd(x,y) = d$, so that $x = da$, $y=db.$  So we have $n = \frac{dab}{a+b}$ with $\gcd(a,b) = 1$.
Lemma 1: $\gcd(a+b,ab) = 1$ if $\gcd(a,b) = 1$.
Proof:Let k be any prime factor that divides ab.  Then it must be a factor of either a or b; wlog let it be a.  So ab = 0 mod k, but a+b =b mod k.  Since $\gcd(a,b) = 1$, b is not 0 mod k, and thus a+b is not 0 mod k.  So, given any prime factor of ab, that factor does not appear in a+b.  qed.

Back to proof: So we want to find when $\frac{dab}{a+b}$ is an integer.  We know that $\gcd(a+b,ab)=1$, we we must have $a + b | d$; or $d=c(a+b)$ for any integer c.  However, we know that $x=da$, and $y=db$, so we have $x=a \cdot (a+b) \cdot c$ and $y=b \cdot (a+b) \cdot c$  From this, we also find that $n = a\cdot b \cdot c$.[/hide]

As an exercise:
Find integers $x_1, x_2, \ldots, x_n,a$ such that
\[
\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} = \frac{1}{a}
\]

[quote="a_vakilian"]For each integer $n>1$, find [i]distinct[/i] positive integers $x$ and $y$ such that
			$\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$[/quote]
Well by what chess64 sad we see the obvious solutions to be $x=n^2+n$ and $y=n+1$ or vise-versa because $(x-n)(y-n)=n^2$.  Then there can be mor solutions based on the nuber of factors of $n$.

now try $\frac1x+\frac1y+\frac1z=\frac1n$ for distinct $x,y,z$

[quote="amirhtlusa"]assume $k=p_1^{\alpha_1}...{p_i^{\alpha_i}}\Rightarrow p_j|x^2,y^2\Rightarrow p_j|x,y$ contradiction.
[/quote]

I don't get this  :(

you see !! this problem seems familiar to me !! i think one of my nice friends told me this problem this morning!!
i haven't seen your solution but i will try to solve it !!!

[quote="amirhtlusa"]we just said let $p$, be a prime devisor of $k$.
now if $p|x^2$ then $p|x$.[/quote]

Oh, I see. Thanks ! :P


hello folks. :D

based on Carcul's recent question on $ \Gamma(x)$, here is a question i decided to pose for today's homework in class.

show

$ \Gamma^{(4)}(1)\;\equal{}\;\boxed{\gamma^4\plus{}6\gamma^2\zeta(2)\plus{}8\gamma\zeta(3)\plus{}\dfrac{27}{2}\zeta(4)}$

where $ \Gamma^{(4)}(1)\;\equal{}\;\dfrac{\textbf d^4}{\textbf dx^4}\;\Gamma(x)\;\bigg|_{x\equal{}1}$

[quote="misan"]hello folks. :D

based on Carcul's recent question on $ \Gamma(x)$, here is a question i decided to pose for today's homework in class.

show

$ \Gamma^{(4)}(1)\; \equal{} \;\boxed{\gamma^4 \plus{} 6\gamma^2\zeta(2) \plus{} 8\gamma\zeta(3) \plus{} \dfrac{27}{2}\zeta(4)}$

where $ \Gamma^{(4)}(1)\; \equal{} \;\dfrac{\textbf d^4}{\textbf dx^4}\;\Gamma(x)\;\bigg|_{x \equal{} 1}$[/quote]
not anly $ \Gamma ^{(4)}(1)$ we can find also $ \Gamma ^{(m)}(1)$ by some recurrence formula , and also find $ \Gamma ^{(m)}(p)$ by knowing that
$ \Gamma'(1) \equal{} \minus{} \gamma$
hint consider
$ \frac {\Gamma'(z)}{\Gamma(z)}$ so called Digamma function , and consider
$ \frac {d}{dz}\frac {\Gamma'(z)}{\Gamma(z)}$ so called Trigamma function  see here http://en.wikipedia.org/wiki/Trigamma_function
It's easy to note that Trigamma function is equal to $ \zeta(z,2)$ , I mean $ ... \equal{} \sum_{0}^{\infty}\frac {1}{(z \plus{} n)^{2}}$ it's so called particular case of  Hurwitz zeta function  see here http://en.wikipedia.org/wiki/Hurwitz_zeta_function
also note that $ (\zeta(z,n))^{(m)} \equal{} ( \minus{} 1)^{m}\frac {(m \plus{} n \minus{} 1)!}{(m \minus{} 1)!}\zeta(z,n \plus{} m)$ thus from
$ \frac {d}{dz}\frac {\Gamma'(z)}{\Gamma(z)} \equal{} \zeta(z,2)$ we can find $ \Gamma''(1)$and from here
$ \frac {d^{2}}{dz^{2}}\frac {\Gamma'(z)}{\Gamma(z)} \equal{} \minus{} 2\zeta(z,3)$ we can find $ \Gamma'''(1)$
and from here
$ \frac {d^{3}}{dz^{3}}\frac {\Gamma'(z)}{\Gamma(z)} \equal{} 6\zeta(z,4)$ we can  easy find $ \Gamma''''(1)$


Jose invested $ \$50,000$ for $ 2$ years at an annual interest rate of $ 4$ percent compounded yearly. Patricia invested $ \$50,000$ for the same period of time, at the same interest rate, but the interest was compounded quarterly. To the nearest dollar, how much more money did Patricia's investment earn than that of Jose?

Simply put, Jose has 2 compounding intervals at 4%, or (1.04)^2=1.0816
Patricia has 8 compounding intervals at 1%, or (1.01)^8=1.08285671

In other words Patricia should expect 0.00125671*50000 more money, or 63 dollars.


I was wondering, I was trying to join a different forum (on the stock market) and in the description, it says that your IP address can be found. This can be seen from:

http://einvesting.com/profile.php?mode=register

Do you people think this can be dangerous? I dont know if I should sign up.

I don't know, it all depends on a matter of trust.

If they have a legal contract and stuff, try reading all of the disclaimers and warrants...it might take a while, but it will help you feel a bit safer.

If you trust if from the reading that you did, you may.  But as all contracts and legal documents say, they will not use it for anything bad.

If you really suspect them I suppose you could run some virus scans for keyloggers and data miners.

Just be careful on what you choose. If they would steal ip's and hack people, why would they start with you and risk a huge legal involvement.

But if it's a small company, stay away!

Heed my warnings on the subject, but do what you want...

Every forum gets your IP; it's a matter of whether they use it for bad things. Even AoPS logs your IP address. If you have a good firewall, you're just as safe if you join as if you don't.

well i know that "whatismyip.com" is a good one.. havent been there in a while but i think its trustworthy. Some programs can do it too..

honestly, an IP address isnt enough to hack your system, especially with XP SP2. The old easiest way would be netbios hackicng, which cant be done cuz file and printer sharing is disabled by default. If this site MUCH more info than that (and using firefox or opera wouldnt let em!) then it could pose a security risk.


geometry
perpendicular bisector
geometry unsolved

Construct the center of the given square by only compass.

[hide="Construction"]Draw the circle with center $ A$ and through $ B$, Then another with center $ C$ through $ A$. Let them intersect at $ X$. Draw another with center $ X$ through $ A$. Repeat everything symmetric to the reflexion axis as the perpendicular bisector of $ AB$.The intersection of the last two circles in each procedure is the required center.[/hide]
[hide="Proof"]Suppose that the real center is $ O$ but not necessarily on the circle with center $ X$ or $ Y$. We will prove that $ XO$=$ XA$. Then $ YO \equal{} YB$ will be similar case and proved. Since $ XO$ is the median in circle $ AXC$, By appolonias'(I know the spelling is wrong :P ) theorem $ 2(XO^2 \plus{} AO^2) \equal{} XA^2 \plus{} XC^2\implies 2(XO^2 \plus{} (\frac {AD}{\sqrt {2}})^2) \equal{} AD^2 \plus{} (\sqrt {2}AD)^2$
$ \implies XO \equal{} AD\implies XO \equal{} AX$[/hide]


Determine the distance between the centroid and in-center of a $ 3$, $ 4$, $ 5$ triangle.

Mass coordinate bash!

[hide]Consider the triangle with vertices $ (0,0)$, $ (4,0)$, and $ (0,3)$.

The centroid is two-third the way from $ (0,0)$ to $ (2,1.5)$:

The centroid is $ \left(\frac {4}{3},1\right)$

The radius of incircle is 1, since $ 6 \equal{} 6r$.

The incenter is $ (1,1)$.

So the distance between the 2 points is $ \frac {1}{3}$[/hide]

That's not especially massive  :)

Quick Note: The centroid of a triangle is the arithmetic mean of its vertices.

I know it's not especially massive now...but my original solution was. I found more clever ways to find the points.


function
number theory
prime numbers
number theory unsolved

Let k be a positive integer. Find all functions f : N → N such that f(x)+f(y)
divides $ x^k \plus{}y^k$ for all x,y are integers

We will prove that $ f(x)\equal{}x^n$, where $ \frac kn$ is an odd number.
1) For any prime $ p$, we have $ f(p)\equal{}p^n$, where $ n\in N$ and $ \frac kn$ is an odd. Note $ n$ might depend on $ p$.
We have $ 2f(x) | 2x^k$, or $ f(x) | x^k$. So $ f(1)\equal{}1$. In addition, we have $ f(p)\equal{}p^n$. Now we claim $ \frac kn$ is an odd.
Suppose it is not true. Then we have $ k\equal{}an\plus{}r$ with $ 0< r<n$.
If $ a$ is even, $ p^k\plus{}1\equal{}p^r\plus{}1\mod p^n\plus{}1$. But $ p^r\plus{}1<p^n\plus{}1$ contradicts with $ p^n\plus{}1 | p^k\plus{}1$.
If $ a$ is odd, $ p^k\plus{}1\equal{}\minus{}p^r\plus{}1\mod p^n\plus{}1$. But $ |\minus{}p^r\plus{}1|<p^n\plus{}1$ contradicts with $ p^n\plus{}1 | p^k\plus{}1$.

2) $ f(x)\equal{}x^n$ for any $ x$.
We fix $ x$ and denote $ f(x)\equal{}t$. Find a prime $ p>x^{k^2}$. From 1), we have $ f(p)\equal{}p^n$, where $ \frac kn\equal{}a$ is some odd.
$ x^k\plus{}p^k\equal{}x^k\plus{}p^{an}\equal{}x^k\plus{}(\minus{}t)^a\equal{}x^k\minus{}t^a \mod t\plus{}p^n$.
We notice that $ x^k\leq x^{k^2}<p$, $ t^a\leq t^k\leq x^{k^2}<p$. So $ |x^k\minus{}t^a|<p< t\plus{}p^n$.
The only way to make sure $ f(x)\plus{}f(p)| x^k\plus{}p^k$ is to let $ x^k\minus{}t^a\equal{}0$.
Hence, $ f(x)\equal{}t\equal{}x^{\frac ka}\equal{}x^n$.

3) $ n$ in 2) does not depend on $ x$.
Suppose $ f(x)\equal{}x^a$, $ f(y)\equal{}y^b$, and $ a\neq b$. We can find a prime $ p>\max(x^{k^2},y^{k^2})$ and $ f(p)\equal{}p^n$. Using the argument in 2), we can show $ a\equal{}n\equal{}b$, a contradiction.

Excellent ! It's similar to my solution that I sent to Awesome ! . For step (3) we only need to use the fact that : there exist an integer  ${ m\in\{0,..,k}$ that $ f(p) = p^m$ holds for infinite prime numbers p .


geometry
geometric transformation
geometry proposed

Given are $\triangle ABC$, $I$ is any point in plane,$H_{a},H_{b},H_{c}$ is three orthocenter of $\triangle IBC,\triangle ICA,\triangle IAB$, resp prove that $S_{ABC}=S_{H_{a}H_{b}H_{c}}$ , more two triangle $\triangle ABC,\triangle H_{a}H_{b}H_{c}$ always have the same direction!

I was asked to post an elementary proof of this, so here it goes (it's the translation of a part of my [url=http://www.cip.ifi.lmu.de/~grinberg/Dreigeom/QEDMO4g.zip]4th QEDMO geometry proposed solutions[/url]):

We denote by $ \left[P_1P_2P_3\right]$ the directed area of any triangle $ P_1P_2P_3$.

We will use the following facts about directed areas:
- [i]Directed parallelism rule:[/i] Let $ P_1$, $ P_2$, $ P_3$, $ P_4$ be four points with $ P_1\neq P_2$ and $ P_3\neq P_4$. Then, $ P_3P_4\parallel P_1P_2$ if and only if $ \left[P_1P_2P_3\right] \equal{} \left[P_1P_2P_4\right]$.
- Any three points $ A$, $ B$, $ C$ satisfy
$ \left[ABC\right] \equal{} \left[BCA\right] \equal{} \left[CAB\right] \equal{} \minus{} \left[ACB\right] \equal{} \minus{} \left[BAC\right] \equal{} \minus{} \left[CBA\right]$.
- [i]Triangle summation rule:[/i] Any four points $ A$, $ B$, $ C$, $ D$ satisfy
$ \left[ABC\right] \equal{} \left[DBC\right] \plus{} \left[DCA\right] \plus{} \left[DAB\right]$.

Now we prove:

[color=blue][b]Theorem 1.[/b] If $ A$, $ B$, $ C$, $ D$, $ E$, $ F$ are six points with $ AB\parallel DE$, $ BC\parallel EF$ and $ CD\parallel FA$, then $ \left[ACE\right] \equal{} \left[BDF\right]$.[/color]

[i]Proof.[/i] By the directed parallelism rule, $ \left[ EFC\right] \equal{} \left[ EFB\right]$ (since $ CB\parallel EF$) and $ \left[ FAC\right] \equal{} \left[ FAD\right]$ (since $ CD\parallel FA$), and finally $ \left[ ABE\right] \equal{} \left[ ABD\right]$ (since $ ED\parallel AB$). Thus,

$ \left[ ACE\right] \equal{} \left[ FCE\right] \plus{} \left[ FEA\right] \plus{} \left[ FAC\right]$ (by the triangle summation rule)
$ \equal{} \left[ EFC\right] \plus{} \left[ EAF\right] \plus{} \left[ FAC\right] \equal{} \left[ EFB\right] \plus{} \left[ EAF\right] \plus{} \left[ FAD\right]$
$ \equal{} \left[ FAD\right] \plus{} \left( \left[ EAF\right] \plus{} \left[ EFB\right] \plus{} \left[ EBA\right] \right) \minus{} \left[ EBA\right]$
$ \equal{} \left[ FAD\right] \plus{} \left[ BAF\right] \minus{} \left[ EBA\right]$ (since $ \left[ BAF\right] \equal{} \left[ EAF\right] \plus{} \left[ EFB\right] \plus{} \left[ EBA\right]$ by the triangle summation rule)
$ \equal{} \left[ ADF\right] \plus{} \left[ AFB\right] \minus{} \left( \minus{} \left[ ABE\right] \right) \equal{} \left[ ADF\right] \plus{} \left[ AFB\right] \plus{} \left[ ABE\right]$
$ \equal{} \left[ ADF\right] \plus{} \left[ AFB\right] \plus{} \left[ ABD\right] \equal{} \left[ BDF\right]$ (by the triangle summation rule),

proving Theorem 1.

[i]Remark.[/i] Theorem 1 appeared as exercise 76 in §12.3.2 of Arthur Engel, [i]Problem-Solving Strategies[/i], New York 1998.

[color=blue][b]Corollary 2.[/b] Let $ ABC$ be a triangle and $ P$ be a point. Let $ A^{\prime}$, $ B^{\prime}$, $ C^{\prime}$ be the orthocenters of triangles $ BPC$, $ CPA$, $ APB$. Then, $ \left[ABC\right] \equal{} \left[A^{\prime}B^{\prime}C^{\prime}\right]$.[/color]

[i]Proof.[/i] Since $ A^{\prime}$ is the orthocenter of triangle $ BPC$, the line $ BA^{\prime}$ is an altitude of this triangle. Thus, $ BA^{\prime}\perp CP$. Similarly, $ AB^{\prime}\perp CP$. Thus, $ BA^{\prime}\parallel AB^{\prime}$. Similarly, $ CB^{\prime}\parallel BC^{\prime}$ and $ AC^{\prime}\parallel CA^{\prime}$. So the six points $ B$, $ A^{\prime}$, $ C$, $ B^{\prime}$, $ A$, $ C^{\prime}$ satisfy $ BA^{\prime}\parallel B^{\prime}A$ (this is just a rewording of $ BA^{\prime}\parallel AB^{\prime}$), furthermore $ A^{\prime}C\parallel AC^{\prime}$ (this is just a rewording of $ AC^{\prime}\parallel CA^{\prime}$), and finally $ CB^{\prime}\parallel C^{\prime}B$ (this is just a rewording of $ CB^{\prime}\parallel BC^{\prime}$). Applying Theorem 1 to these six points, we thus get $ \left[BCA\right] \equal{} \left[A^{\prime}B^{\prime}C^{\prime}\right]$. In other words, $ \left[ABC\right] \equal{} \left[A^{\prime}B^{\prime}C^{\prime}\right]$. Corollary 2 is proven.

  darij


AMC
AIME
SFFT
number theory
greatest common divisor

hello, under the condition that $ x$ and $ y$ are integers i have found the following solutions
$ (x\equal{}\minus{}99810090\land y\equal{}9990)\lor (x\equal{}\minus{}1019082\land y\equal{}9894)\lor (x\equal{}\minus{}959136\land y\equal{}9888)\lor (x\equal{}\minus{}618\land y\equal{}582)\lor (x\equal{}582\land y\equal{}\minus{}618)\lor (x\equal{}9888\land y\equal{}\minus{}959136)\lor
   (x\equal{}9894\land y\equal{}\minus{}1019082)\lor (x\equal{}9990\land y\equal{}\minus{}99810090)\lor (x\equal{}9992\land y\equal{}99830072)\lor (x\equal{}10088\land y\equal{}1039064)\lor (x\equal{}10094\land y\equal{}979118)\lor (x\equal{}19400\land
   y\equal{}20600)\lor (x\equal{}19982\land y\equal{}19982)\lor (x\equal{}20600\land y\equal{}19400)\lor (x\equal{}979118\land y\equal{}10094)\lor (x\equal{}1039064\land y\equal{}10088)\lor (x\equal{}99830072\land y\equal{}9992)$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, under the condition that $ x$ and $ y$ are integers i have found the following solutions
$ (x \equal{} \minus{} 99810090\land y \equal{} 9990)\lor (x \equal{} \minus{} 1019082\land y \equal{} 9894)\lor (x \equal{} \minus{} 959136\land y \equal{} 9888)\lor (x \equal{} \minus{} 618\land y \equal{} 582)\lor (x \equal{} 582\land y \equal{} \minus{} 618)\lor (x \equal{} 9888\land y \equal{} \minus{} 959136)\lor (x \equal{} 9894\land y \equal{} \minus{} 1019082)\lor (x \equal{} 9990\land y \equal{} \minus{} 99810090)\lor (x \equal{} 9992\land y \equal{} 99830072)\lor (x \equal{} 10088\land y \equal{} 1039064)\lor (x \equal{} 10094\land y \equal{} 979118)\lor (x \equal{} 19400\land y \equal{} 20600)\lor (x \equal{} 19982\land y \equal{} 19982)\lor (x \equal{} 20600\land y \equal{} 19400)\lor (x \equal{} 979118\land y \equal{} 10094)\lor (x \equal{} 1039064\land y \equal{} 10088)\lor (x \equal{} 99830072\land y \equal{} 9992)$
Sonnhard.[/quote]

You need to STOP doing this.  :mad:
Anyone can stick numbers into a computer algebra system and paste in a bunch of answers. Even if you found all those by hand, unless you're going to show your method you're not really helping at all.

[/rant]

As for the problem, it reminds me of the Harold Tanya Ulysses AIME problem (paintable integers). Try isolating one variable and then applying some ideas of continued fractions and divisibility. I'll try to post with more detail later.

you can just multiply out denominators, then use sfft
then bruteforce  :P


inequalities
algebra
polynomial
inequalities open

$x,y,z$ are positive real numbers and $n$ is a positive integer, then we have:
\[x^3+y^3+z^3+n(x^2y+y^2z+z^2x) \ge (n+1)(xy^2+yz^2+zx^2).\]
The case $n=1$ is obviously true by AM-GM.
The case $n=2$ was proved by Harazi in the post [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5901]www.mathlinks.ro/Forum/viewtopic.php?t=5901[/url]

I just want to know if it is true for all positive integer $n$.  :)

Using the method harazi used, we have
\[b^3+c^3 \ge (n-1)b^2c + (n-2)bc^2 \]
which is not true for large n.
even for n=3, it is not true for $c \le b \le 2c$

Thanks siuhochung. Then what is the largest real $n$ such that the inequality is true?

The best possible number is the unique positive root of the polynomail,

\[ n^4+2\,n^3-5\,n^2-6\,n-23\,, \]
approximately,       $ 2.484435332\ldots$

my answer is:the min of $\frac{a^2}{2}+\frac{3a}{2}+\frac{1}{2a}+1$
where a is a positive number.
if somebody want to see my solution,I'll post it. :D

The minimum of  $ \displaystyle{\frac{a^2}{2}+\frac{3\,a}{2}+\frac{1}{2\,a}+1}$  is  $\displaystyle{\frac{23}{8}}$,  but it does not satisfy the inequality, that is, the inequality
\[\displaystyle{ x^3+y^3+z^3+\frac{23}{8}\,(x^2y+y^2z+z^2x) \geq (\frac{23}{8}+1)(xy^2+yz^2+zx^2)}\]
does not hold. So your answer is not correct, is it?

now I post my solution(or not a solution,if there is something wrong)
let $f(x,y,z)=x^3+y^3+z^3+n(x^2y+y^2z+z^2x)-(n+1)(xy^2+yz^2+zx^2)$
we have the minimum of $f(x,y,z) \geq 0$
$WLOG,x \geq z \geq y$(if $x \geq y geq z$,the inequality always still true)
and $3f(x,y,z)=3x^3+3y^3+3z^3-3(xy^2+yz^2+zx^2)+3n(x^2y+y^2z+z^2x-xy^2+yz^2+zx^2)$
$=(x+2y)(x-y)^2+(y+2z)(y-z)^2+(z+2x)(z-x)^2+3n(x-y)(x-z)(y-z)$
$=(x+2y)(x-y)^2+(y+2z)(y-z)^2+(z+2x)(z-x)^2-3n(x-y)(x-z)(z-y)$
so we can easily find $f(x,y,z) \geq f(x-y,0,z-y)$
so we should only let  $f(x-y,0,z-y) \geq 0$
let $x-y=a,z-y=b$,we have $f(x-y,0,z-y)=a^3+b^3+nb^2a-(n+1)ba^2 \geq 0$
let b=1(I think it can),we have $f(x-y,0,z-y)=a^3+1+na-(n+1)a^2 \geq 0$ for all $a \geq 0$
let $g(a)=a^3+1+na-(n+1)a^2$
we have when  $g'(a)=3a^2+n-2(n+1)a$
we can get the $a$when $g'(a)=0$
then have the result.
but the answer may as $Guest$ said,
and there may be something wrong with my yesterday's calcultation. :D
BTW,my solution can be useful？？

I'am not sure that we can assume $x\ge y \ge z$. :blush:

[quote="Guest"]:lol: You can't, because the inequality is not full symmetric.[/quote]
We can! Because in the other case we have $x^2y+y^2z+z^2x \ge xy^2+yz^2+zx^2$ which is true only in the other case.($x \ge y \ge z$)

I saw this inequality 2 months ago :
[tex]x^3+y^3+z^3+\sqrt{2}(x^2y+y^2z+z^2x) \ge (\sqrt{2}+1)(xy^2+yz^2+zx^2)[/tex]
But I can't prove it. May be it's false, i think that.


Let ABCD be a quadrilateral. E,F are points on sides AD,BC,resp such that \frac{AE}{ED}=\frac{BF}{FC}. Ray FE meets rays BA,CD at S,T resp. Show that the circumcircles of triangles SAE,SBF,TCF,TDE pass through a common point.
( Have it been posted ? Sorry if it have posted. )


It is given that $ x,y$ are positive numbers such that $ 10x^2\plus{}12y^2\equal{}23xy$. Find the possible values of $ \frac{x^2\minus{}3xy\plus{}2y^2}{x^2\minus{}5xy\plus{}3y^2}$.

One hint:

[hide]From the first quadratic equation,$ x\equal{}4y/5 or 3y/2$

Then put the value of x in terms of y in the second system of equations.Simplify.
Had two terms in the numerator and denominator canceled each other out after factorisation,it would been better but that's not the case here.[/hide]

[quote="muks"]One hint:
From the first quadratic equation,$ x \equal{} 4y/5 or 3y/2$

Then put the value of x in terms of y in the second system of equations.Simplify.
Had two terms in the numerator and denominator canceled each other out after factorisation,it would been better but that's not the case here.[/quote]

as he got $ x \equal{} 4y/5$ put this value in the expression
and take $ \frac{x^{2}\minus{}3xy\plus{}2y^{2}}{x^{2}\minus{}5xy\plus{}3y^{2}} \equal{}z$ now cross multilply and take each and every term on LHS to get quadratic equation  and then apply $ D\ge0$ to get the range of the expression

[quote="21 7 15"]Sorry i did not publish my solution.(x,y)=(9/4,3/2) and (16/25,4/5) so one of the anwsers to that expression is 1/9[/quote]

but how can u say that  ?


geometry
incenter
perimeter
trigonometry
geometry unsolved

Given $\Delta ABC$ with $AB < AC$, let I be the incenter and M be the mid-point of BC. The line MI meets AB and AC at P and Q respectively.A tangent to the incircle meets sides AB and AC at D and E respectively.
  Prove that : \[ \frac{AP}{BD}+\frac{AQ}{CE}=\frac{PQ}{2MI} \]

Denote a = BC, b = CA, c = AB the triangle sides, $\alpha = \angle A,\ \beta = \angle B,\ \gamma = \angle C$ its angles, s the semiperimeter and r the inradius. Furthermore, denote the segment lengths d = |BD|, e = |CE|, f = |DE|, p = |AP|, q = |AQ|. The incenter I of the triangle $\triangle ABC$ is the Nagel point of its medial triangle $\triangle MNO$ (where M, N, O are the midpoints of BC, CA, AB). Hence, the cevian MI of the medial triangle is its perimeter splitter and consequently, the line through the vertex A parallel to MI is the perimeter splitter of the original triangle $\triangle ABC$. If this cevian meets the opposite side BC at a point X, then AB + BX = c + BX = s, BX = s - c and CA + CX = b + CX = s, CX = s - b. Since the triangles $\triangle BPM \sim \triangle BAX,\ \triangle CQM \sim \triangle CAX$ are similar,

$\frac{2(b + q)}{a} = \frac{CQ}{CM} = \frac{CA}{CX} = \frac{b}{s - b}$

$\frac{2(c - p)}{a} = \frac{BP}{BM} = \frac{BA}{BX} = \frac{c}{s - c}$

Calculating the segment lengths p, q from the above formulas,

$p = c - \frac{ca}{a + b - c} = \frac{c(b - c)}{2(s - c)},\ \ q = \frac{ab}{c + a - b} - b = \frac{b(b - c)}{2(s - b)}$

Since BI, CI bisect the angles $\angle B, \angle C$,

$\frac{PI}{MI} = \frac{BP}{BM} = \frac{2(c - p)}{a},\ \ \frac{QI}{MI} = \frac{CQ}{CM} = \frac{2(b + q)}{a}$

$\frac{PQ}{MI} = \frac{QI}{MI} - \frac{PI}{MI} = \frac{2(b + q) - 2(c - p)}{a} = \frac{b}{s - b} - \frac{c}{s - c} = \frac{s(b - c)}{(s - b)(s - c)}$

Now we should resort to oriented line segments, because the correct proposition (for oriented line segments) holding for an arbitrary tangent DE of the triangle incircle is

$\left| \frac{AQ}{CE} - \frac{AP}{BD} \right| = \frac{PQ}{2MI}$

Taking the directions of the rays $(AB,\ (CA$ to be positive,  we have q = AQ > 0, p = AP > 0, because c  = AB < CA = b by the problem condition. Considering the case, when the tangential quadrilateral BCED is convex, -d = BD < 0, e = CE > 0, and we have to show that

$(?)\ \ \frac q e + \frac p d = \frac{s(b - c)}{2(s - b)(s - c)}$

$(?)\ \ \frac{b(b - c)}{2e(s - b)} + \frac{c(b - c)}{2d(s - c)} = \frac{s(b - c)}{2(s - b)(s - c)}$

$(?)\ \ \frac{b(s - c)}{e} + \frac{c(s - b)}{d} = s$

Since the quadrilateral BCED is tangential, we have d + e = a + f or $f^2 = (d + e - a)^2$. Using the cosine theorem for the triangle $\triangle ADE$,

$f^2 = (b - e)^2 + (c - d)^2 - 2(b - e)(c - d) \cos \alpha$

$b^2 - 2be + e^2 + c^2 - 2cd + d^2 - 2(bc - bd - ce + de)\cos \alpha =$
$= d^2 + e^2 + a^2 + 2de - 2ad - 2ae$

$be + cd - (bd + ce - de)\cos \alpha = ad + ae - de$

$e(c \cos \alpha + a - b) + d(b \cos \alpha + a - c) = de (1 + \cos \alpha)$

$\frac{c \cos \alpha + a - b}{d} + \frac{b \cos \alpha + a - c}{e} = 1 + \cos \alpha$

Substituting

$c \cos \alpha + a - b = \frac{b^2 + c^2 - a^2 + 2b(a - b)}{2b} = \frac{c^2 - (a - b)^2}{2b} = \frac{2(s - a)(s - b)}{b}$

$b \cos \alpha + a - c = \frac{b^2 + c^2 - a^2 + 2c(a - c)}{2c} = \frac{b^2 - (c - a)^2}{2c} = \frac{2(s - c)(s - a)}{c}$

$1 + \cos \alpha = \frac{2bc + b^2 + c^2 - a^2}{2bc} = \frac{(b + c)^2 - a^2}{2bc} = \frac{2s(s - a)}{bc}$

we arrive to

$\frac{2(s - a)(s - b)}{bd} + \frac{2(s - c)(s - a)}{ce} = \frac{2s(s - a)}{bc}$

$\frac{c(s - b)}{d} + \frac{b(s - c)}{e} = s$

which is exactly the last equation in question, which we were supposed to prove. The remaining cases, when the the tangential quadrilateral BCED is concave or reflex, can be proved similarly, with the appropriate signs changed.


calculus
integration
function
calculus computations

Let $a,b$ be constant numbers such that $0<a<b.$ If a function $f(x)$ always satisfies $f'(x) >0$ at $a<x<b,$ for $a<t<b$ find the value of $t$ for which the following the integral is minimized.

\[ \int_a^b |f(x)-f(t)|x\ dx. \]

[quote="anik_andrew"]$t=\sqrt{a^2+b^2}$?[/quote]

Your answer is incorrect. :(

[quote="maokid7"]could someone please provide a proof.
i have no idea how they got that answer.  :oops:  :blush:[/quote]
$f' > 0 \Longrightarrow f$ is increasing , so :
 $F(t) = \int_a^b |f(x)-f(t)|x^c\ dx = \int_a^t (f(t)-f(x))x^c\ dx + \int_t^b (f(x)-f(t))x^c\ dx=$$f(t)\frac{2t^{c+1}-a^{c+1}-b^{c+1}}{c+1} - \int_a^t f(x)x^c\ + \int_t^b f(x)x^c$
Now it's easy : $F'(t)=f'(t)\frac{2t^{c+1}-a^{c+1}}{c+1}$,  so the minimum is reached for $t_m=(\frac{a^{c+1}+b^{c+1}}2)^{\frac 1{c+1}}$
 :cool:


x_n is the solution of the equation
1+x/(x-1)+...+x/(x-n+1)=ln(x^n/n!)  with x>n
   Find lim(x_n/n)

  I unsolve it,help me.

What is it $\lim \frac{x_n}{n}$ ? I don't understand, when we use $\lim$ we write; for example $\lim_{x\rightarrow a}(...)$ and $a\in R$ :(
Please explane more ????? ;)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a group such that each infinite subset of $G$ contains at least two elements that commute. Prove that $[G: Z(G)]<\infty$.

I'll use the following lemma (to which JC_math posted a proof [url=http://www.mathlinks.ro/viewtopic.php?p=667775#667775]here[/url]):

If a group $ G$ is covered by finitely many subgroups $ H_i\le G,\ i\equal{}\overline{1,n}$, then it's covered by those $ H_i$ which have finite index.

First, I want to show that every element $ x\in G$ has finitely many conjugates in $ G$, or, equivalently, that the centralizer $ C_G(x)$ has finite index in $ G$. In order to see this, consider a set cntaining $ x$, consisting of pairwise non-commuting elements of $ G$, and maximal with respect to this last property. The hypothesis on $ G$ says that this set is finite, say $ \{x_1\equal{}x,\ldots,\ x_n\}$. Since it's maximal, the centralizers $ C_G(x_i)$ cover $ G$. If $ [G: C_G(x)]\equal{}\infty$, the lemma above says that $ G$ is covered by $ C_G(x_i),\ i\ge 2$, so in particular $ x$ commutes with one of the elements $ x_i,\ i\ge 2$. This contradicts the fact that the $ x_i$ are pairwise non-commuting.

Now I'll show that $ G$ has an abelian subgroup of finite index. Assume the contrary. We can then find $ x_1\in G\setminus Z(G)$. This means that $ C_G(x_1)\ne G$. $ C_G(x_1)$ and the subgroup $ H_1\le G$ generated by all $ x$ outside $ C_G(x_1)$ cover $ G$ but $ C_G(x_1)$ does not, so, according to the lemma, $ H_1$ has finite index in $ G$. By our assumption, $ H_1$ can't be abelian, so we can find two non-commuting elements outside $ C_G(x_1)$. This means that there is an $ x_2\not\in C_G(x_1)$ such that $ C_G(x_1)$ and $ C_G(x_2)$ don't cover $ G$, and repeat the argument above using the group $ H_2$ generated by the elements outside $ C_G(x_1)\cup C_G(x_2)$ instead of $ H_1$. In this manner we get an infinite sequence $ x_1,x_2,\ldots$ of pairwise non-commuting elements, contradicting the hypothesis on $ G$. This means that there is indeed an abelian subgroup $ H\le G$ of finite index $ n$.

Now let $ a_1,\ldots,a_n$ be representatives for the cosets of $ H$ in $ G$. The group $ H\cap\bigcap_{i\equal{}1}^nC_G(a_i)$ is clearly central and of finite index in $ G$ because according to the first paragraph the centralizers $ C_G(a_i)$ all have finite index in $ G$.


Let ABC is an triangle.Prove that
(sin(A+B/2)*sin(B+C/2)*sin(C+A/2))/(sin(A/2)*sin(B/2)*sin(C/2))   \geq   8+  (3/sqrt(10))*(sum(  (sinA -sinB) 2)/(snA*sinB )  )
I had a mistake  :( .Sorry. I have  just changed it

hey minhthang, i found that
(sin(A+B/2)*sin(B+C/2)*sin(C+A/2))/((sin(A/2)*sin(B/2)*sin(C/2)) = 8+ sum( (sinA -sinB)  2 )/(snA*sinB )
so your problem is easy  :D


function
induction
real analysis
real analysis unsolved

For $ 0<x<1$, express $ \displaystyle\sum_{n\equal{}0}^{\infty}\frac{x^{2^n}}{1\minus{}x^{2^{n\plus{}1}}}$ as a rational functions of $ x$.

I have an answer but am not sure if it correct. Can someone please tell me how they would solve this problem.
[hide="my answer"] $ \frac{x}{1\minus{}x}$[/hide]

i tried that but was not able to get far. what i ended up doing was for each fixed $ n$ i wrote the expression as an infinite geometric series. and then showed that each positive integer could be uniquely written as $ \beta 2^t$ where $ \beta$ is an odd integer. then did some rearrange of terms.

Let $ S_N\equal{}\sum_{n\equal{}0}^N\frac{x^{2^n}}{1\minus{}x^{2^{n\plus{}1}}}$ define the sequence of partial sums.

It is easy to check via induction that $ S_N\equal{}\frac{1}{1\minus{}x}\minus{}\frac{1}{1\minus{}x^{2^{N\plus{}1}}}$ holds, hence $ \sum_{n\equal{}0}^{\infty}\frac{x^{2^n}}{1\minus{}x^{2^{n\plus{}1}}}\equal{}\frac{x}{1\minus{}x}$


calculus
integration
trigonometry
calculus computations

I have two questions :

1) What is the appropriate substitution for the this integral $ \int\frac {\sqrt {x^2 \minus{} x^4}}{x^5} \ dx$.

2) Is the information in this question true ?

if $ \int\limits_{1}^e xf(x) dx \equal{} 4$ , what is the value of $ \int\limits_0^2 f(\sqrt {e^x})dx$.

 :)

For 2) what do you mean ?
It's just that we have f such that this integral is 4.
And for solving this, just substitute $ t^2\equal{}e^x$...

For 1) substitute $ t^2\equal{}1\minus{}x^2$

When I substitute $ t^2\equal{}1\minus{}x^2$ in (1) , there still problem in denominator.

 :)

For (1), first clear out the extraneous powers of $ x$ to have

$ \int\frac{\sqrt{1\minus{}x^2}}{x^4}\,dx.$

Had that been an odd power in the denominator, I would have gone with $ t\equal{}1\minus{}x^2,\ x^2\equal{}1\minus{}t,\ dt\equal{}\minus{}2x\,dx.$ But with the even power, I don't like where that is going.

Hence, the trigonometric substitution: $ x\equal{}\sin\theta,\ \sqrt{1\minus{}x^2}\equal{}\cos \theta,\ dx\equal{}\cos\theta\,d\theta.$ Try that.

[quote]Hidden Scofield wrote
1) What is the appropriate substitution for the this integral $ \int\frac {\sqrt {x^2 - x^4}}{x^5} \ dx.$[/quote]

$ \int\frac {\sqrt {x^2 - x^4}}{x^5} \ dx.$ = $ \int\frac {\sqrt {1-x^2 }}{x^4} \ dx.$

= $ \int\frac {(\sqrt {\frac{1-x^2}{x^2 }}(2x)}{2x^4} \ dx.$

put $ {\sqrt {\frac{1-x^2}{x^2 }} = t^2}$ and solve


AMC AIME HMMT USAMTS search Asymptote LaTeX \/closed

The Contests section works like this:
1. Someone posts a problem on the forums. Each problem, to go into the resources section, [b]must have its own post[/b] because of technical restrictions.
2. An RManager takes the problem and puts it into the Contests section - each problem in there is linked to one post on the forums, as is each solution.
3. A PDF file is autogenerated for that contest on those dates.

If you want to get the problems you post put into the resources section:
1. Avoid any extraneous comments in the post that contains the information; reply with comments later.
2. Use the following format for answer choices for multiple-choice contests:
[code]$\textbf{(A)}\ AnswerA \qquad
\textbf{(B)}\ AnswerB \qquad
\textbf{(C)}\ AnswerC \qquad
\textbf{(D)}\ AnswerD \qquad
\textbf{(E)}\ AnswerE$[/code]3. Use LaTeX.
4. When making diagrams, use LaTeX's picture environment or (preferably) Asymptote. Don't link to an external site which will inevitably go down, and attachments won't show up in the resources section. If you absolutely can't reproduce the diagram using those two, upload it to a gallery and use that. However, these won't show up in the PDF.
5. Use the following format for problem sources:
[code]Contest Year Day/Grade Round Problem X[/code]So for example:
AMC 10 2008A Problem 1
AIME 2008I Problem 1
HMMT 2008 Algebra Problem 1
USAMTS 2008 Round 1 Problem 1
This makes it much, much easier to use Search to collect all the posts from a certain contest/year/day/grade in a single page, and thus much easier to put the contest into the Contests section. If you use something like 2008 AIME I, and I search for all problems with that in the title/subject description, the search will discard the I and only search for all posts talking about the "2008 AIME," not just the problems. It's a lot faster to enter problems from the search page than from a forum.

If you want to get your solutions posted:
1. Again, avoid extraneous comments; these can be posted in a separate reply.
2. (As always) hide your solution; the parser automatically strips out the hide tag, so it isn't an issue. The parser won't have trouble with unhidden solutions, so you should hide just for common courtesy.
3. Use LaTeX.
4. For diagrams, use Asymptote or LaTeX, for the same reasons as above.
5. Be clear and thorough.

[quote="worthawholebean"]The Contests section works like this:
1. Someone posts a problem on the forums. Each problem, to go into the resources section, [b]must have its own post[/b].
[/quote]

I'll repeat what I said. But I find this point extremely illogical.

There are pretty good reasons why every problem needs it's own post. This is not only a technical issue for the ressources section, but also a simple matter of clarity. You should not post more than one problem per thread as otherwise discussion gets messy.

And for non-managers, I see no reason not to PM some manager as this is surely less work than inserting it into the ressources section. The works of a "helper" and a manager have only one difference: the helper sends a PM, the manager inserts some numbers.
And the main reasons for not allowing all people to insert into the ressources section are simple: it's a more complicated process then you may think, it is not possible for a "moderator" to easily hold controll of stupid changes, and it is pretty hard to rechange idiotic changes.
I see no reason to allow everyone to insert.
But I don't get what your problems are here (speaking of Cyber-Shreyas and pardesi).

By the way, you can PM me for insertion, too.

[quote="pardesi"]yes me too and and each time do we have to pm an Rmanager i mean why isn't it open to all and with some moderators to keep a strict vigil just like other forums and it's so difficult to be collecting everything and putting each problem in a seperate thread and pm ing someone and wait for  him/her to upload[/quote]
If you want access, PM Valentin. It's not some exclusive club or anything. Especially if you have a record of posting lots of problems, you have a good chance of him giving you access.

These are the questions of the Indian National Maths Olympiad 2008. Just because each question does not have a separate thread, it doesn't mean that this will lose its fundamental property of being the questions of the olympiad. And if it is present in the resources section, then I'm sure it is easily accessible to all, which is much better than people searching all forums for it. This is only a question paper, why should each question be in a separate thread??

Is the reverse conversion: PDF -> Olympiad Section possible? For IIT JEE papers, its gonna be a total pain typing each question.

Our only problem is
Indian Section needs to be updated more.

:)

Technical restrictions mean we can't put a problem into the resources section unless it has its own post. It's not that we don't want to, it's that we simply cannot.

Yes, you have to retype the problems.

[quote="pardesi"]well an standard IIT-JEE paper has about 80 problems with some comprehensions and some multiple choice questions too...and creating 80 threads is a huge task and then pming it even bigger that was one thing we both wanted to address and we also have this jee thread in out section which  prently has some 160 problems and still growing

also we have had both prelims and finals so about 100 problems  each year that's a  lot to be posted in seperate threads :)[/quote]Well if you want to help nobody said it will be a piece of cake :)

You have to come up with those threads. Nobody is rushing you so take your time. Write one a day if you must. Keep track of them and then submit them to a RManager ( http://www.mathlinks.ro/groupcp.php?g=417 )

pardesi, this is the reason I'm so pedantic about the formatting of the "subject description" field. If every problem can be found easily by a single search term, all you need to do is PM the RManager the search term. So, for example, have IIT-JEE 2008 Multiple Choice Problem X. An RManager can then search for "IIT-JEE 2008 Multiple Choice" and get all the problems easily.

Oh! Okay I'm sorry. I didn't understand the meaning of 'technical restrictions'. I thought it was some guideline followed by you people. :D

1) Is it necessary that all the separate threads should be under the same sub-forum?
2) http://www.mathlinks.ro/viewtopic.php?t=195835 - Is this how you want us to format? (just an example)

Thanks

http://www.mathlinks.ro/viewtopic.php?p=1076671#1076671

Is this okay??

Yes, it is. (I am an RManager by the way, so you can ask me as well.)

  darij

[quote="worthawholebean"][quote="Valentin Vornicu"]Yes but don't put anything BUT the problem statement in the post body. Nothing else (see in that post there is a (5) appended to the beginning).

If you want put the problem number and other info in the subject description field[/quote]That's actually a point value. Should I take those out?[/quote]Yes.

[quote="Valentin Vornicu"]Yes.[/quote]Meh, the posts are too old now. I'd need to get a mod to do the menial task of taking out all of those.

I was looking at the CMO section ([url]http://www.artofproblemsolving.com/Forum/resources.php?c=33&cid=51[/url]) and many years were missing, so i feel like uploading all the questions from the book i have, as i go through them. This i because the CMO website ([url]http://www.cms.math.ca/Competitions/CMO/[/url]) has the questions in a very bad quality (excluding the recent ones, they are unreadable). So basically i have to type up the question following the rules, and then i PM an RManager? Whats an RManager? And do i have to do it for every question? Thanks

Just PM me. As soon as I have time I'll add them.

Is there anybody in particular who I should PM or just anybody in that group? A.k.a. are there some people who have more time these days and are more likely to do it if I PM them?

Are these guidelines and procedures still relevant? The features (e.g. "resources section") described in the OP seem outdated.

This is mostly accurate, a "resources section" is a collection, and the "Contests section" is the Contest Collections

There is another guide here [url]https://artofproblemsolving.com/community/c40244_contest_collections_discussion[/url]


ratio calculus integration trigonometry logarithms function Euler

i asked to show this fairly [u]easy [/u]integral yesterday and [b]not [/b]a [i]single [/i]student
within the integral calculus class could do it....  :noo:

i was quite disappointed, you know... :|

$\int_{0}^\frac{\pi}{2}\; \left( \left( \frac{\frac{1}{2}x}{\beta{(1)}}\right) \left( \;1\;-\;\left( \frac{\frac{1}{2}x}{\beta{(1)}}\right)^{2}\;\right) \cdot\tan x\;\right)\;\;\textbf dx\;=\;\boxed{\frac{\zeta{(3)}}{\zeta{(2)}}}$

I suppose that $\beta (1) = \frac{\pi}4$ :maybe:  :huh:  :what?:  :surrender:

We basically have to prove that $\int_{0}^\frac{\pi}2 \left[ x \cdot \left( \frac{\pi}2 \right)^{2}-x^{3}\right] \cdot \tan x \, dx = \frac{3 \pi}4 \cdot \zeta (3)$.

Integrating by parts, it suffices to prove that
\[\int_{0}^\frac{\pi}2 \left[ \left( \frac{\pi}2 \right)^{2}-3 x^{2}\right] \cdot \log \cos x \, dx = \frac{3 \pi}4 \cdot \zeta (3) . \]
Break this into a difference of integrals. The first is well-known, whilst the second was discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=151878]here[/url].

Unfortunatelly there is no much logic on that Radio..

http://mathworld.wolfram.com/BetaFunction.html

$ \mathrm{\beta}(x,y)=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}= \pi/4$

we can conclude that

$ p,q$ canot be integers!!!

they need to be half integer. remember duplication formula http://en.wikipedia.org/wiki/Gamma_function

$ \Gamma(x).\Gamma(y) = K \pi$ only if $ x,y$ be half integers

if on is integer the factor pi will cancelate. of corse.

si, $ Beta(1)$ make no sense, and put is $ pi/4$ make less sense.

since x or y = 1. we will have $ \beta(1, y)$ and this result is rational.

dump

[quote="yagaron"]Unfortunatelly there is no much logic on that Radio..[/quote]
Actually there is quite some logic! Look closely... is misan's integral even convergent? If $ \beta (1) \neq \frac{\pi}4$, then it's obviously not. But if $ \beta (1) = \frac{\pi}4$, then that integral equals $ \frac{\zeta(3)}{\zeta(2)}$ (you can check it with Mathematica, if you still don't believe me and/or didilica).
Hence, my statement has logic, qed!

[quote="yagaron"]Unfortunatelly there is no much logic on that Radio..

http://mathworld.wolfram.com/BetaFunction.html

...
dump[/quote]

Dump, for sure.  Why do you decide that there can only be one function which goes by the name of $ \beta$?

http://mathworld.wolfram.com/DirichletBetaFunction.html

See in particular on that page equation (6).

dump...

the zeta function above is clearly riemann zeta function, but nobody said that.

so why riemann? because is the most common zeta function.

so i used the most common beta function.

if was another, the Poster needed to tell..

and radio, i just said that that beta function cannot be never Pi/4. i dont said that you are wrong, i just said that was impossible for that beta be this value. looke gamma function, the euler one ;)