geometry
power of a point
angle bisector

Very funny, scrambled. I'm still working on it.

Can I use a calculator?

Sure you can.. it may not help though. Heres a hint:

[hide]

Think about how the number would have to end.

[/hide]

[quote="mpn"]Very funny, scrambled. I'm still working on it.
[/quote]
what i meant by"no" was that i dont think it is possible.

Heres some mini-problems, in spoiler, once you go through all of them you should have the answer.

1

[hide]

Can you have a square ending in one zero? Why/why not?

[/hide]

2

[hide]

What happens if a square ends in 2 or more zeroes? Can the smallest possible solution to our problem have this property? Why/why not?

[/hide]

3

[hide]

What must the smallest possible solution end in? Can this be a square?

[/hide]

i didn't use the mini-problems, even though it looks like it: [hide]

first of all, a perfect square cannot end in one zero.  It must end in a number of zeros that is a multiple of two.  If it does, you can take away the zeros by factoring out ten squared, and you must still be left with a perfect square.  Also, all squares must either be 1 (mod 4) or 0 (mod 4).  Any digits beyond the hundreds place do not matter, since 100 is a multiple of four.  The last two digits are either 66 or 06 (it can't end in zero).  66 and 06 are both 2 (mod 4), therefore there cannot be a perfect square consisting of only 0 and 6[/hide]

[quote="TripleM"]Can you find a perfect square whose only digits are 0 and 6? (not including just 0 by itself).[/quote]

Why wouldn't you consider this an intermediate problem?

[quote="TripleM"]I was waiting for someone to say that.

Because this is a perfect example of a problem for beginners at problem solving. It is not an intermediate problem.[/quote]

Did you begin with problems like this? In what kind of competition program? At what age?

Yep. Problems like these are perfect for beginning problem solving. Remember this doesn't mean you are starting math - thats a completely different story.
I didn't start with competitions.. I just found cool puzzles to do. If you don't think this is easy, can you give an example of a genuine problem-solving question that is easier?

[quote="TripleM"]Yep. Problems like these are perfect for beginning problem solving.[/quote]

You didn't answer my questions.

What would be an example of a typical intermediate level problem?

which grade level is getting started supposed to be for?

Its not really a grade level thing.. I'm not sure about in USA, but in New Zealand problem solving isn't really thought at schools at all (apart from occasional competitions). So I've always considered it a different subject to what you learn at school.

[quote="tokenadult"]
You didn't answer my questions.

What would be an example of a typical intermediate level problem?[/quote]

Yes I did, except for the age one, which didn't really apply, as I said I just liked solving problems.

You never answered my question though :P

Its hard to say what an intermediate type question is, because problem solving is more of a gradual learn from examples type thing, but have a look in the intermediate topics, most of the problems there aren't too bad, although a some are a bit easy for fear of people complaining if they're posted in here :)

In terms of my own problem-solving experience, here is how I see the Getting Started, Intermediate, and Advanced problems (or what I think they should be):

Getting Started problems are problems I could have solved in middle school without exerting an especially large amount of effort.

Intermediate problems are problems that I could have solved in high school without exerting an especially large amount of effort.

Advanced problems are problems that I'll have to work hard on to solve or problems that I still cannot solve.

That's basically meaningless to anyone else though since everyone can solve different problems. I think this square problem is on the tough side for Getting Started (I would have put it in Intermediate myself), but I don't think it's too hard.

OK, so it possibly is on the tough side. In any case, I think it is a good problem for beginners to play around with. It doesn't require any knowledge (you know what I mean :P) to understand, and seeing the solution may help you learn something. Actually, thats what problem solving is all about, anyway.. even if you can't solve a problem, trying to solve it can help a lot.

[quote="ComplexZeta"]In terms of my own problem-solving experience, here is how I see the Getting Started, Intermediate, and Advanced problems (or what I think they should be):

Getting Started problems are problems I could have solved in middle school without exerting an especially large amount of effort.[/quote]

I don't know how much effort is "an especially large amount of effort" for you, Complex Zeta, but I was rather stunned to learn in another thread that both you and JBL learned about the power of a point theorem in middle school. That is DEFINITELY unusual. (That is, that is definitely unusual in the United States. It may be routine in east Asia.)

So one thing we are up against in determining the level of problems is the very large variance in curriculum level at different schools.

By the way, this was the thread that prompted my interest in LOOKING UP just [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7254]what Mr. Rusczyk said about the intended level[/url] of problems for this Getting Started forum. And that inquiry led to me becoming a new moderator of this forum.

TripleM constructively  :)  convinced me that this problem is approachable to beginners by his set of mini-questions (helpfully posted in spoiler text), which certainly ought to be understandable to someone who has completed study of elementary school mathematics. I think it was another reply in the thread that first got me doubting the proper level of this question, as it used notation that I think would not be understandable to most people at the Getting Started level. Sometimes how a solution is presented can be "easy" for someone who is experienced with a particular form of notation, but hard for someone else who isn't shown the steps along the way in elementary terms.

Another thing that got me snarled up in this thread originally is that the title didn't have "Challenge Problem" in it, and the question was posed as "Can you . . . " rather than as "Is it possible to . . . " or "Is there any number that . . . " or something else that would make clear what the thrust of the question was. Since this thread was posted, TripleM and I have had a little bit of conversation (on board and by PM) about how the board is run, and I definitely welcome him posting more challenge problems, so titled, of this nature and level on this board in days to come.

[quote="tokenadult"]I don't know how much effort is "an especially large amount of effort" for you, Complex Zeta, but I was rather stunned to learn in another thread that both you and JBL learned about the power of a point theorem in middle school.[/quote]
I think I first learned Power of a Point about 6 months ago from AoPS volume 1, and I am 40 years old...Maybe I just forgot it all, but I don't recall learning several of the geometry theorems in AoPS in high school (let alone middle school).  For example, the Angle Bisector Theorem, Stewart's Theorem, Ceva's Theorem, and Menelaus's Theorem seemed new.  (Even now, I don't know if I can quote Ceva or Menelaus without peeking.  I had to peek at the book just now for the spelling of Menelaus.)  Maybe that explains why I used to solve most of the non-geometry questions in math contests, but fewer than half of the geometry questions.

When I was a mathematician, I hardly used geometry.  But I don't know if that is because Euclidean geometry does not arise much in higher math, or if I just chose research fields that were as far from geometry as possible.


symmetry
inequalities unsolved
inequalities

Let $ a,y,x$ are real numbers such that $ x\plus{}y\plus{}z\equal{}0,x^2\plus{}y^2\plus{}z^2\equal{}6$
Find max of $ x^2y\plus{}y^2z\plus{}z^2z$

[quote="Allnames"]Let $ a,y,x$ are real numbers such that $ x \plus{} y \plus{} z \equal{} 0,x^2 \plus{} y^2 \plus{} z^2 \equal{} 6$
Find max of $ x^2y \plus{} y^2z \plus{} z^2z$[/quote]
I don't remember The solution of this problem and General issues  :(

But I remember I used to solve them  :blush:

[b]Allnames[/b],Could you solve the General issues?

I hope you can solve it.

Good luck :brush:

[quote="Allnames"]Let $ a,y,x$ are real numbers such that $ x \plus{} y \plus{} z \equal{} 0,x^2 \plus{} y^2 \plus{} z^2 \equal{} 6$
Find max of $ x^2y \plus{} y^2z \plus{} z^2z$[/quote]

i think it's $ x^2y\plus{}y^2z\plus{}z^2x$

hello, here you can use the method of Lagrange multipliers, see here:
http://en.wikipedia.org/wiki/Lagrange_multiplier
Sonnhard.

[quote="y-a-s-s-i-n-e"][quote="Allnames"]Let $ a,y,x$ are real numbers such that $ x \plus{} y \plus{} z \equal{} 0,x^2 \plus{} y^2 \plus{} z^2 \equal{} 6$
Find max of $ x^2y \plus{} y^2z \plus{} z^2z$[/quote]

i think it's $ x^2y \plus{} y^2z \plus{} z^2x$[/quote]
Oh,yes,here is my mistake,thank you
Thank Dr Sonnhard Graubner,but that method is too hard for me,I remember it is problem in BMO,I solved  it two year ago,but now ,I faile,Who can help me with a simple technique :blush:

[quote="Allnames"][quote="y-a-s-s-i-n-e"][quote="Allnames"]Let $ a,y,x$ are real numbers such that $ x \plus{} y \plus{} z \equal{} 0,x^2 \plus{} y^2 \plus{} z^2 \equal{} 6$
Find max of $ x^2y \plus{} y^2z \plus{} z^2z$[/quote]

i think it's $ x^2y \plus{} y^2z \plus{} z^2x$[/quote]
Oh,yes,here is my mistake,thank you
Thank Dr Sonnhard Graubner,but that method is too hard for me,I remember it is problem in BMO,[b]I solved  it two year ago[/b],but now ,I faile,Who can help me with a simple technique :blush:[/quote]
I think you are a spirit  :blush:  :blush:

The basic observation is the symmetry of $ x, y, z$.

Denote $ A \equal{} \sum {x^2 y}$ and $ B \equal{} \sum {x y^2}$, then:
$ A \plus{} B \equal{} \minus{}3 xyz$
and
$ AB \equal{} 6 (xyz)^2 \minus{} \frac{27}{2}$

Notice: $ A, B$ are the roots of $ X^2 \plus{} (3xyz) X \plus{} (6 [xyz]^2 \minus{} \frac{27}{2}) \equal{} 0$. Because of symmetry, $ A$ can obtain any values that $ B$ can obtain and vice versa. Find the range of $ xyz$ and find the maximum of:

$ \frac{\minus{}3xyz\plus{}\sqrt{\Delta}}{2}$

over that range.

It is the maximum value of $ A$.


geometry
3D geometry
prism
AMC
AIME
AIME I
AMC 12 A

Points $ A$, $ B$, $ C$, $ D$, and $ E$ are located in 3-dimensional space with $ AB \equal{} BC \equal{} CD \equal{} DE \equal{} EA \equal{} 2$ and $ \angle ABC \equal{} \angle CDE \equal{} \angle DEA \equal{} 90^\circ.$ The plane of $ \triangle ABC$ is parallel to $ \overline{DE}$. What is the area of $ \triangle BDE$?

$ \textbf{(A)}\ \sqrt2 \qquad \textbf{(B)}\ \sqrt3 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \sqrt5 \qquad \textbf{(E)}\ \sqrt6$

lol, I can't believe I got this wrong..

Draw a tright triangle prism with ABE with one of the triangles, and EBCD is a square with side length 2.

So area of triangle EBD=$\frac{1}{2}*ED*EB=\boxed{2}$

damn lol.. why can't they just draw the picture out  :rotfl:

I got a 99 on the AMC 12 A and paid $\$$4 to go to the AMC 12 B.  Then I promptly forgot to go to it.  Seeing some of the problems on it, it looks harder than the A.  I'm kind of glad I forgot to go to it.....  (By the way, I got into the AIME - I was in the top percentile.)

[quote="Allan Z"]lol, I can't believe I got this wrong..

Draw a tright triangle prism with ABE with one of the triangles, and EBCD is a square with side length 2.

So area of triangle EBD=$ \frac {1}{2}*ED*EB \equal{} \boxed{2}$

- lol.. why can't they just draw the picture out  :rotfl:[/quote]

sorry to revive an old topic but I cannot picture it  :(

can someone draw it out please?

EDIT: it's ok now, I went to http://www.artofproblemsolving.com/Wiki/index.php/2007_AMC_12B_Problems/Problem_25 and I managed to draw it out. I really have no spacial awareness  :blush:

One of the main flaws with the solutions both given above and in the AoPS Wiki is that they make some assumptions about what the diagram looks like/are unrigorous.  Here is a method that (almost) completely avoids any assumptions and which deduces the structure of $ABCDE$ from scratch.  (I might draw a picture later, but drawing 3D diagrams in Asymptote is pretty hard.)

Note that from $AB=BC=2$ and $\angle ABC = 90^\circ$ we have $AC = 2\sqrt 2$.  Similarly, we get $BC=AD=2\sqrt 2$.  Thus $\triangle BAC\cong\triangle DCA$ by SSS.

Let $M$ be the midpoint of $\overline{AC}$.  By the median length formula in $\triangle ADC$, \[2(AD^2+DC^2) = AC^2 + (2DM)^2\quad\implies\quad DM = 2.\] Similarly, $EM = 2$.  Therefore $\triangle DEM$ is equilateral with side length $2$.

Let $Q_D$ and $Q_E$ denote the projections of $D$ and $E$ respectively onto the plane $ABC$, and WLOG let $Q_E$ lie on the same side of $AC$ as $B$ (it's not hard to convince yourself that exactly one of $Q_E$ or $Q_D$ must satisfy this).  The condition that $DE$ is parallel to the plane $ABC$ implies $DQ_D=EQ_E$ and $\angle DQ_DM = \angle EQ_EM = 90^\circ$.  Thus $\triangle DQ_DM\cong\triangle EQ_EM$, so $MQ_D = MQ_E$.  Furthermore, $AE=ME$ implies $AQ_E = MQ_E$, and similarly $MQ_D = MC$.  When combined with $AM=MC$ this implies $\triangle AQ_EM\cong\triangle CQ_DM$, so $\angle Q_EMA=\angle Q_DMC$ and $M\in Q_DQ_E$.  As a result, $DEQ_EQ_D$ is a rectangle, so $AQ_E=Q_EM= 1$ meaning that $\triangle AQ_EM$ is a 45-45-90 triangle.  In turn, $Q_E\in AB$.

This means that $\triangle AEB$ is isosceles, so $BE = 2$.  Furthermore, $BQ_EQ_DC$ is a rectangle, so \[BE^2 + ED^2 = (BQ_E^2 + Q_EE^2) + ED^2 = Q_EE^2 + BQ_D^2 = DQ_D^2 + BQ_D^2 = BD^2.\] Thus $\triangle BED$ is an isosceles right triangle, meaning that its area is $\tfrac12\cdot 2\cdot 2 = \boxed{2}$.


superior algebra
superior algebra solved

Show that $Z(S_{n})=\{ e \}$ for all n>2.
Thanx in advance :lol:

let G be a group with a permutation representation (faithful) on a set X, with at least three elements

suppose it works two transitive

then the center of the group is trivial

proof :

suppose g is in the center, and not trivial
suppose it sends a to b, with a and b different elements (remember the action is faithful)
now pick a third different element c
and let h be an element of G that sends a to c, and fixes b (remember two transitivity implies that there is such an h)

well then consider $g^{h}=h^{-1}g h = g$ as g is in the center
remember we are working from left to right here, very important!! just like in my other response to you

however $h^{-1}g h$ sends c to b (work that out, it is just like the hint in my other post) and as a is not equal to c, we have contradiction


[quote="nonie"]Three circles are externally tangent to each other. The radius of each of the circles is 2 units. A belt fits tightly around the three circles. Find the number of units in the lenght of the belt. Express your answer in terms of pi.[/quote]

Maybe I'm wrong, but the answer is: $4 \pi + 12$

[quote="M4RI0"][quote="ch1n353ch3s54a1l"]I got [hide]$2\pi+12$[/hide][/quote]

Did you consider that: $r=2$ (r:radium)  :?:[/quote]
Yes.
So the length of the arc is radians times radius. Since the radians is 120 degrees or $\frac{\pi}{3}$ and radius is 2, the length of arc is $\frac{2\pi}{3}$ and three of those is a total length of $2\pi$. Then the other lengths are 4, so $4\times3=12$ and the belt is length $2\pi+12$.

[quote="M4RI0"]$120$ degrees = $\frac{2 \pi}{3}$. There's your mistake[/quote]

Oops.  :blush: Accidentally used $\frac{\pi}{360}$ as the conversion factor...


inequalities
calculus
calculus computations

For any positive integer $ n$, show that
$ (1\plus{}\frac{1}{n})^n>\sum_{k\equal{}0}^n{\frac{1}{k!}}\minus{}\frac{3}{2n}$

[quote="hanschinaman"][quote="Soarer"]$ \equal{} 1 \plus{} 1 \plus{} \frac {1}{2!}\left(1 \minus{} \frac {1}{n}\right) \plus{} ... \plus{} \plus{} \frac {1}{n!}\left(1 \minus{} \frac {1}{n}\right)...\left(1 \minus{} \frac {n \minus{} 1}{n}\right)$
$ \equal{} 1 \plus{} 1 \plus{} \frac {1}{2!}\left(1 \minus{} \frac {1}{n}\right) \plus{} ... \plus{} \plus{} \frac {1}{n!}\left(1 \minus{} \frac {1 \plus{} ... \plus{} (n \minus{} 1)}{n}\right)$
[/quote]
It is wrong.[/quote]
It was a typing mistake.


Try to compute $ (1 \minus{} \frac{1}{3}) \times$ the product.


Is it easy?

[img]http://i15.tinypic.com/5x6p56t.jpg[/img]

I think you shouldn't post this problem here because it's a problem issuse in Mathematic and Youth.


modular arithmetic
superior algebra
superior algebra solved

Would appreciate any help on this problem.

Let p be prime. p does not divide a and p does not divide b.
Prove if a2 :equiv: b2 mod p, then a :equiv:  :pm: b mod p.

[quote="liyi"]$a^2-b^2=(a+b)(a-b)\equiv 0\pmod p$
$p|(a+b)(a-b)$
thus $p|(a+b)$ or $p|(a-b)$.[/quote]

This is what I thought at first, but it doesn't use the fact that p doesn't divide a and p doesn't divide b. Are those 2 facts irrelevant to this problem?

The fact that p does not divide a or b is not important.  Once you have the stated congruence the desired congruence is an immediate consequence of Euclid's lemma.

[quote="msbaa672"]
This is what I thought at first, but it doesn't use the fact that p doesn't divide a and p doesn't divide b. Are those 2 facts irrelevant to this problem?[/quote]
Well..I don't think we need the condition.


algebra
polynomial
geometry
3D geometry
inequalities
factorization
sum of cubes

International Zhautykov Olympiad 2008
                                         Almaty, Kazakhstan
                                           January 16,2008.
                                                   Day 1
   1.Points $ K,L,M,N$ are respectively the midpoints of sides $ AB,BC,CD,DA$ in a convex quadrilateral $ ABCD$.Line $ KM$ meets diagonals $ AC$ and $ BD$ at points $ P$ and $ Q$ respectively.Line $ LN$ meets diagonals $ AC$ and $ BD$ at points $ R$ and $ S$ respectively.
Prove that if $ AP \cdot PC = BQ \cdot QD$ then $ AR \cdot RC = BS \cdot SD$.
   2.A polynomial $ P(x)$ with integer coefficients is called $ good$ if it can be represented as a sum of cubes of several polynomials (in variable $ x$) with integer coefficients.For example, the polynomial $ x^{3} - 1$ and $ 9x^{3} - 3x^{2} + 3x + 7 = (x - 1)^{3} + (2x)^{3} + 2^{3}$ are $ good$.
a) Is the polynomial $ P(x) = 3x + 3x^{7}$ $ good$?
b) Is the polynomial $ P(x) = 3x + 3x^{7} + 3x^{2008}$ $ good$?
   3.Let $ A = {(a_{1},...,a_{8}) | a_{i}}$ belongs to $ N$, $ 1<=a_{i}<=i+1$ for each ${ i = 1,...,8}$$ .$.A subset $ X \subset A$ is called $ sparse$ if for each two distinct sequences $ (a_{1},...,a_{8}),(b_{1},...,b_{8}) %Error. "belong" is a bad command.
X$, there exist at least three indices $ i$ such that $ a_{i}$ does not equal to $ b_{i}$.
   Find the maximal possible number of elements in a $ sparse$ subset of set $ A$.
                                           January 17,2008
                                                  Day 2
   4.For each positive integer $ n$ denote by $ s(n)$ the sum of all digits in the decimal representation of $ n$.
   Find all positive integers $ n$ such that $ n = 2s(n)^{3} + 8$.
   5.Nonintersecting circles $ w_{1}$ and $ w_{2}$ with centers $ O_{1}$ and $ O_{2}$ touch line $ l$ at points $ A_{1}$ and $ A_{2}$,respectively (the circles lie on the same side of $ l$). Point $ K$ is the midpoint of segment $ A_{1}A_{2}$. Points $ B_{1}$ and $ B_{2}$ are chosen on circles $ w_{1}$ and $ w_{2}$ respectively such that lines $ KB_{1}$ and $ KB_{2}$ touch $ w_{1}$ and $ w_{2}$ respectively (point $ B_{1}$ is distinct from $ A_{1}$, and point $ B_{2}$ is distinct from $ A_{2}$).Lines $ A_{1}B_{1}$ and $ A_{2}B_{2}$ meet in point $ L$, and lines $ KL$ and $ O_{1}O_{2}$ meet in point $ P$.
Prove that points $ B_{1},B_{2},P$ and $ L$ are concyclic.
   6.Prove that for every positive real numbers $ a, b, c$ such that $ abc = 1$, the inequality
             $ \frac {1}{(a + b)b} + \frac {1}{c(b + c)} + \frac {1}{a(c + a)}>=\frac {3}{2}$
holds.
P.S. Sorry I have wrote the similar post several days before and I have made some serious typos. Now they are corrected. I am the participant of this olympiad and I hadn't got much time to write them.Good luck in solving these problems. :)

I've already posted the problems.See here
http://www.mathlinks.ro/viewtopic.php?t=183537
Best regards,Erken.


logarithms
geometry
3D geometry
calculus
integration

[quote="ckck"]What is the sum of all real X such that $(2^{X}-4)^{3}+(4^{X}-2)^{3}=(4^{X}+2^{X}-6)^{3}$[/quote]
[hide="first"]Let $2^{x}=y$[/hide]
[hide="second"]Factor the RHS[/hide]
[hide="third"]$a^{3}+b^{2}=(a+b)(a^{2}+b^{2}-ab)$[/hide]
[hide="last"]After finding y, solve for x[/hide]

That's a clever solution!

I would have never thought to have used fermat's last theorem

The case for cubes alone is, of course, much easier than the general theorem.  Can anyone give a proof?

[hide="In any case, here is a simpler solution"] The problem in fact has the form $a^{3}+b^{3}= (a+b)^{3}\Leftrightarrow ab(a+b) = 0$. [/hide]

[quote="lkryptonicl"]It is one of Fermat's theorems. The one that states that the sum of 2 cubes can not equal another cube. Therefore, one of the three terms have to be 0 and just split them up into cases. Well anyway doing this i get that the sum of the values of x is $\frac{7}{2}$.[/quote]
.
That states that the sum of two cubes of INTEGERS cannot equal another integer cubed.

Because we're dealing with real X, the values in the parentheses aren't necessarily integral, and thus you cannot use that theorem.

Well for the purpose of this problem I thought it would be o.k. to assume that the numbers were integral and besides that, even if it says real, you can use Fermat's theorem to find the integral solutions and then I guess you could go back and try to find any other solutions with logs or something.

[quote="lkryptonicl"]Well for the purpose of this problem I thought it would be o.k. to assume that the numbers were integral[/quote]

It's not.  The factorization I gave is the real reason why your method produces the correct results.


algebra polynomial calculus inequalities integration function logarithms

I get stuck with the following beautiful problem:
  If a monic polynomial of degree $ n$ (with complex coefficients) has the sum of the squares of the absolute values of all the coefficients $ 2$, then the sum of the squares of the absolute values of its roots is at most $ n$. Can anybody help me?

[quote="harazi"]
  ..  a monic polynomial of degree n has the sum of the squares of the absolute values of all the coefficients 2 ...[/quote]

Help, what does it mean  :!: :?:

:cool:

It is a polinomial of the form $ x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ such that $ f(a_0)+...+f(a_n)=1$ and $ f(x)$ is the square of the absolute value of the complex number x ( it is incredible, can anyone say how we write the modulus of a commplex number?)

Thanks, now it's ok : $a_n=1 \ ,  \ \displaystyle \sum_{k=0}^{n-1}|a_k|^2 = 1 $
:cool:

Well, anybody a useful idea? This problem is so difficult!

Are coefficients $a_i$ real, or complex?

[quote="harazi"]I get stuck with the following beautiful problem:
  If a monic polynomial of degree n has the sum of the squares of the absolute values of all the coefficients 2, then the sum of the squares of the absolute values of the roots is at most n. Can anybody help me?[/quote]
I am [b]not absolutely sure[/b] but I think is not too hard if we use the fact that $\max_{i=1,\dots,n}\{|\alpha_i|\}\le \max_{i,2,\dots,n}\{|a_i|\}$ then is easy because $|a_i|^2\le 1$ for every $i$, then $\sum_{i=1}^n |\alpha_i^2|\le n \max \{|a_i|^2\}\le n$.

RX

That's it, the virus is spreading. Soon, Mathlinks will be totally destroyed.  ;)

[quote="harazi"]I get stuck with the following beautiful problem:
  If a monic polynomial of degree n has the sum of the squares of the absolute values of all the coefficients 2, then the sum of the squares of the absolute values of the roots is at most n. Can anybody help me?[/quote] I tried different approaches. Unfortunately, all elementary ways failed in my hand. Therefore, mostly this reply should be placed in the Calculus Section.
I don't reckon my proof will be appreciated here.

First of all, I will prove some weak results on this problem. And then I will present my final solution.

Let $f(z)=z^n+a_1z^{n-1}+...+a_n$ be given polynomial and $z_1,...,z_n$ be its roots.

[size=125][b]Some weak results[/b][/size]

[b]Theorem 1. (I. Schur, C.Siegel, 1921)[/b] [i]The following estimate holds \[(1+|z_1|)(1+|z_2|)...(1+|z_n|)\leq 2^n\sqrt{1+|a_1|^2+...+|a_n|^2}.\][/i]
[i]Proof.[/i] Indeed, we can write $1+|z_k|\leq 2\max(1,|z_k|)\leq 2\exp\left(\frac{1}{2\pi}\int_0^{2\pi}\ln|e^{i\varphi}-z_k|d\varphi\right)$. Thus multiplying and applying Jensen's inequality for the convex function $\ln x$ we obtain the following inequality \[\prod_{k=1}^n(1+|z_k|)\leq 2^n\exp\left(\frac{1}{2\pi}\int_0^{2\pi}\ln|f(e^{i\varphi})|d\varphi\right)\leq 2^n\sqrt{\frac{1}{2\pi}\int_0^{2\pi}\ln|f(e^{i\varphi})|d\varphi}=2^n\sqrt{1+|a_1|^2+...+|a_n|^2}.\] $\blacksquare$

Unfortunately, this theorem gives too weak estimate for $|z_1|^2+|z_2|^2+...+|z_n|^2$. You can easily derive it yourself. I was hoping for some time, that I can use these ideas to establish the required result.

[b]Theorem 2. (Myth, 2005)[/b] [i]The following estimate holds \[|z_1|^2+|z_2|^2+...+|z_n|^2\leq n+|a_1|^2+...+|a_n|^2.\][/i]
[i]Proof.[/i] And again we will use complex analysis to establish this results. I will state the well known theorem here (without a proof).

[b]Jensen's Theorem.[/b] [i]Let $f(z)$ be an analytic function in the disk $|z|<R$. Suppose, that $f(0)\neq 0$, and $r_1,r_2,r_3,...$ is a nondecreasing sequences of roots' absolute values of $f(z)$. Then for $r_n\leq r\leq r_{n+1}$ we have \[\ln\frac{r^nf(0)}{r_1...r_n}=\frac{1}{2\pi}\int_0^{2\pi}\ln|f(re^{i\varphi}|d\varphi.\][/i]

Note, that we can WLOG assume that $a_n\neq 0$, so $|z_k|\neq 0$ for all $k$, $|f(0)|=|a_n|=|z_1...z_n|$. Suppose $z_1,...,z_m$ lies in the unit disk and all other roots lie outside of it. Let $r_k=|z_k|$. Applying Jensen's Theorem for $f(z)$ and $r=1$ we obtain \[\ln (r_{m+1}...r_n)=\frac{1}{2\pi}\int_0^{2\pi}\ln |f(e^{i\varphi})|d\varphi\leq \frac{1}{2}\ln \left(\frac{1}{2\pi}\int_0^{2\pi}|f(e^{i\varphi})|^2d\varphi\right)=\frac{\ln(1+|a_1|^2+...+|a_n|^2)}{2}.\] (we used Jensen's inequality for $\ln(x)$ here).
Therefore $r_{m+1}^2...r_{n}^2\leq 1+|a_1|^2+...+|a_n|^2$. Taking into consideration $r_{m+1},...,r_n\geq 1$, we conclude that $r_{m+1}^2+...+r_{n}^2\leq (n-m-1) + (1+|a_1|^2+...+|a_n|^2)$. Also we know that $r_1,...,r_m\leq m$, i.e. $r_1^2+...+r_m^2\leq m$. $\blacksquare$

[size=125][b]The proof to the initial problem[/b][/size]

[size=109][b]Auxilary results.[/b][/size]

[b]Schur's theorem (Schur Decomposition).[/b] [i]For an arbitrary square matrix $A$ there exists an unitary matrix $U$ s.t. $U^HAU$ is an upper triangular matrix.[/i] (See [url=http://mathworld.wolfram.com/SchurDecomposition.html]Eric W. Weisstein. "Schur Decomposition." From MathWorld--A Wolfram Web Resource.[/url]

[i]Proof.[/i] By induction. If $A=A_{1\times 1}$, then it is true. Suppose it is true for matrix sizes 1, 2, ..., $n-1$.
Let $A=A_{n\times n}$. Consider an arbitrary eigenvalue $\lambda_1$ and the corresponding normed eigenvector $x_1$. We can extend it to an orthonormal basis by the Gram-Schmidt orthogonalization, say $(x_1,u_2,...,u_n)$. Set a unitary matrix $U_1=[x_1,...,u_n]$ with these basis vectors as its columns. Note, that $U_1$ is an unitary matrix. We can easily show that \[U_1^HAU_1=\left( {\begin{array}{*{20}c}
   {\lambda _1 } & {...} & {...}  \\
   0 & B & {}  \\
   0 & {} & {}  \\
\end{array}} \right),\] where $B=B_{(n-1)\times(n-1)}$. By the inductive hypothesis there exists an $U_2$ s.t. $U_2^{H}BU_2$ is an upper triangular matrix with diagonal entries $\lambda_2,...,\lambda_n$. Define \[U=U_1 \left( {\begin{array}{*{20}c}
   1 & 0  \\
   0 & {U_2 }  \\
\end{array}} \right).\] Then it is easy to check that $U$ is also a unitary matrix and satisfies the condition. $\blacksquare$

[b]Remark.[/b] The upper triangular matrix $U^HAU$ contains the eigenvalues of $A$ on its diagonal.

Now some easy direct consequence of the Schur's Theorem.

[b]Schur's Inequality (1909).[/b] [i]For eigenvalues of an arbitrary square matrix $A_{n\times n}$ the following inequality holds \[\sum_{i=1}^n|\lambda_i|^2\leq \sum_{i,j\leq n}|a_{ij}|^2.\][/i]
[i]Proof.[/i] Indeed, an unitary transformation doesn't change the value $\sum_{i,j\leq n}|a_{ij}|^2$ and finally $\sum_{i=1}^n|\lambda_i|^2$ is just a part of the whole sum of entries's squares of $U^HAU$. $\blacksquare$

[size=109][b]The main result.[/b][/size]

[b]Theorem 3. (Myth, 2005)[/b]  [i]The following estimate holds \[|z_1|^2+|z_2|^2+...+|z_n|^2\leq n-1+|a_1|^2+...+|a_n|^2.\][/i]
[i]Proof.[/i] Consider the following matrix \[A=
\left( {\begin{array}{*{20}c}
   {a_1 } & {a_2 } & {...} & {a_n }  \\
   1 & 0 & 0 & 0  \\
   0 & 1 & 0 & 0  \\
   0 & 0 & 1 & 0  \\
\end{array}} \right).\] You can show that its characterstic polynomial is $z^n\pm a_1z^{n-1}\pm a_2z^{n-2}\pm...\pm a_n$ (we can choose appropriate signs for $a_i$ in the matrix in order to obtain $f(z)$). Using Schur's inequality we immediately obtain $|z_1|^2+|z_2|^2+...+|z_n|^2\leq n-1+|a_1|^2+...+|a_n|^2$.

[b]Remark.[/b] Compare it with Theorem 2 above.

[b]P.S.[/b] [i]The theorems, which stated as my results, are just the statements I invented without any exterior help. It is highly possible that they are known.[/i]

The proof using matrices is absolutely beautiful!! Thank you very much. And all the other results have so beautiful proofs. Really great.

That is a beautiful result. Incidentally, $\sum |a_{ij}|^2$ can be interpreted as the trace of $A^*A$. This inequality becomes a standard comparison between singular values and eigenvalues.

Yes, the tools used are standard, but what is great is to think to use them! ;)

[quote="Myth"]Proof. Indeed, an unitary transformation doesn't change the value $\sum_{i,j\leq n}|a_{ij}|^2$ and finally $\sum_{i=1}^n|\lambda_i|^2$ is just a part of the whole sum of entries's squares of $U^HAU$. $\blacksquare$ .[/quote]

Perhaps my question is rather trivial, but I dont understand why the unitary transformation does not change the value $\sum_{i,j\leq n}|a_{ij}|^2$. Could somebody please explain?

Thanks for your time,

In fact, $\sum_{i,j}{|a_{ij}|^2}=tr(A^{s}A)$ where $A^{s}_{ij}=c(a_{ji})$ and $c(x)$ is the conjugate of x. Now, using the fact that $Tr(AB)=Tr(BA)$ and that $A^{s}B^{s}=(BA)^{s}$ it is not difficult to conclude.

[quote]why the unitary transformation does not change the value $\sum_{i,j\leq n}|a_{ij}|^2$. [/quote]

Because the sum is (proportional to) the average of $|Ax|^2$ over the unit sphere $|x|=1$.


Let $a,b,c,d$ be positive integers such that $ab\equal{}cd$. Prove that $a\plus{}b\plus{}c\plus{}d$ is a composite number.

Nice one but I was looking for another one!

a+b+c+d=(1/a)*(a^2+ab+ac+ad)=(1/a)*(a^2+ac+ad+cd) = (1/a)*(a+c)*(a+d)
As c,d are positive a+c,a+d are greater than a. So, a+b+c+d is composite.

lemma.if $ m|xy , 0<|x|,|y|<m$ then $ m$ is composite.
$ ab\equal{}cd\Rightarrow{a(a\plus{}b\plus{}c\plus{}d)\equal{}(a\plus{}c)(a\plus{}d)}\Rightarrow a\plus{}b\plus{}c\plus{}d|(a\plus{}c)(a\plus{}d)$ and obviously $ 0<a\plus{}c , a\plus{}d<a\plus{}b\plus{}c\plus{}d$ therefore $ a\plus{}b\plus{}c\plus{}d$ is composite.
@praneeth:is this the same as your solution?
another solution:
$ ab\equal{}cd\Rightarrow (a\plus{}b\plus{}c\plus{}d)(a\plus{}b\minus{}c\minus{}d)\equal{}(a\minus{}b\plus{}c\minus{}d)(a\minus{}b\minus{}c\plus{}d)$
if $ a\minus{}b\plus{}c\minus{}d\equal{}0$ or $ a\minus{}b\minus{}c\plus{}d\equal{}0$ then $ a\plus{}b\plus{}c\plus{}d$ is even and greater than $ 2$,so is composite.
and if $ a\minus{}b\plus{}c\minus{}d\neq0$ and $ a\minus{}b\minus{}c\plus{}d\neq0$ then:
$ a\plus{}b\plus{}c\plus{}d|(a\minus{}b\plus{}c\minus{}d)(a\minus{}b\minus{}c\plus{}d) , 0<|a\minus{}b\plus{}c\minus{}d|,|a\minus{}b\minus{}c\plus{}d|<a\plus{}b\plus{}c\plus{}d$ so $ a\plus{}b\plus{}c\plus{}d$ is composite.

Assume $p=a+b+c+d$ where $p\ge 4$ is a prime number.
Then working mod $p$ we have $ab=cd\equiv c(-a-b-c)\implies 0\equiv ab+ac+bc+c^2=(c+a)(b+c)$
So $p|a+c\vee p|b+c$ but this is impossible, since $b+c,a+c<a+b+c+d=p$ Contradiction!

[quote=tjhance][hide]We can find positive integers $ w,x,y,z$ such that $ ab \equal{} cd \equal{} wxyz$ with

$ a \equal{} wx$
$ b \equal{} yz$
$ c \equal{} wy$
$ d \equal{} xz$

Then

$ a \plus{} b \plus{} c \plus{} d \equal{} wx \plus{} yz \plus{} wy \plus{} xz \equal{} w(x \plus{} y) \plus{} z(y \plus{} x) \equal{} (w \plus{} z)(x \plus{} y)$.

$ w,x,y,z\ge 1\implies w \plus{} z,x \plus{} y > 1$ so $ a \plus{} b \plus{} c \plus{} d$ is composite.[/hide][/quote]

Could you kindly show how to prove the existence of $w,x,y,z$?

It's Very simple just observe $\frac{a}{c}=\frac{d}{b}\implies\frac{a+c}{c}=\frac{d+b}{b}$

so we get $a+b=\frac{(d+b)c}{b}$ hence

$a+b+c+d=\frac{(d+b)(b+c)}{b}$

 which is always composite. We are done !
@below because $b|cd$

@above,
Would you enlighten me how $\frac{(d+b)(b+c)}{b}$ is always composite??
Thank you!


What is the greatest common multiple of 8 and 12 that is less than 90?

$ 8$ $ \equal{}$ $ 2^3$ and $ 12$ $ \equal{}$ $ 2^2$ * $ 3$, so the LCM of 8 and 12 is 8*3=24. Some common multiples of 8 and 12 are 24, 48, 72, and 96. So the greatest common multiple less than 90 is $ \boxed{72}$

Counting back from 90 in multiples of 8 looking for multiple of 12:

88 NO
80 NO
72 YAYNESS!


Let [tex]N>2[/tex] be a integer, [tex]N-1=FR[/tex], and we know that
[tex]\forall p|F, \exists a=a(p)[/tex] such that [tex]a^{N-1}\equiv 1 (mod n)[/tex] and [tex](a^{\frac{N-1}{p^k}}-1,N)=1.[/tex](here [tex]p^k\|F-1[/tex])

Prove that:
1) Every divisor of [tex]N[/tex] have the form [tex]kF+1[/tex];
2) If [tex]F>N^{1/2}[/tex], then [tex]N[/tex] is prime;
3) If [tex]N^{1/2}>F>N^{1/3}[/tex], and we have [tex]N=pF^2+qF+1[/tex], where [tex]0\leq p,q\leq F-1[/tex], then [tex]N[/tex] is a prime iff [tex]q^2-4p[/tex] is not a perfect square.

By $mod n$ do you by any chance mean $mod\ N$?

Could anyone write again the correct form of this problem? It's very interesting, but I don't understand what is with k, n.

Well, try this:

[url]http://www.utm.edu/research/primes/prove/prove3_1.html[/url]

I didn't see wpolly's problem, but I saw some similar ones.

I think that by $p^k\|N-1$ he means that $p^k$ is the largest power of $p$ which divides $N-1$. I'm saying it because I've seen this notation before. Also note that I've replaced $F-1$ with $N-1$ because that's what I think he meant. This is becoming really confusing :).

And p is any divisor of F or is it a prime factor of F?

Here's how I think it should be:

For any prime $p|F$ there is an $a=a(p)$ s.t. $N|a^{N-1}-1$ and $(a^{\frac{N-1}p}-1,N)=1$. Let $q$ be a prime divisor of $N$. Let $m$ be the order of $a$ mod $q$. Then $m|N-1$ but $m$ doesn't divide $\frac{N-1}p$, so $p^k|m$, where $p^k\|N-1$ (with the notation mentioned in my previous post). At the same time, $m|q-1$, so $q=M_{p^k}+1$, so $q=M_F+1$, which solves $(1)$. I'm not sure about it, of course.

That's exactly the answer :lol:

Now, Who'll solve problem (2) and (3)?

P. S. (1) is known as Pocklington's theorem and (3) is AKA Theorem of BLS(Brillhart, Lehmer and Selfridge)

I've deleted my last post because I had written something completely off-track :). Ok, here's what I meant: $(2)$ is easy because each prime factor of $N$ has form $M_F+1$, so if $N$ has at least two prime factors (not necessarily distinct), then $N=(aF+1)(bF+1),\ a,b\in N^*$, but in this case $N>F^2$, which is false.

$(3)$ doesn't seem that hard either. Assume $q^2-4p$ isn't a square. In this case, let's also assume $N=(aF+1)(bF+1)=abF^2+(a+b)F+1,\ a,b>0$. $ab<F$ (because $N<F^3$) $\Rightarrow a+b<F$, so $ab=p,\ a+b=q\Rightarrow q^2-4p=(a-b)^2$, which is a contradiction. I think the converse is equally easy to prove.


If anybody could help me solve this problem, that would be GREATLY appreciated!

The interior bisector of < BAC of triangle ABC meets BC at D.  Suppose that
1/(BD)² + 1/(CD)² = 2/(AD)²

Prove that <BAC= 90º

Thanks soo much for your help![/code]

[color=darkblue]By bisector theorem

$ \frac {BD}{CD} \equal{} \frac {AB}{AC} \Longleftrightarrow \frac {BD \plus{} CD}{CD} \equal{} \frac {AB \plus{} AC}{AC} \Longleftrightarrow \frac {1}{CD} \equal{} \frac {AB \plus{} AC}{AC\cdot BC}$ (1)

and

$ \frac {1}{BD} \equal{} \frac {AB \plus{} AC}{AB\cdot BC}$ (2)

By Stewart

$ AD^2 \equal{} \frac {4bc}{(b \plus{} c)^2}\cdot p(p \minus{} a) \Longleftrightarrow \frac {2}{AD^2} \equal{} \frac {(b \plus{} c)^2}{2bc\cdot p(p \minus{} a)}$

Therefore

$ \frac {(b \plus{} c)^2}{b^2\cdot a^2} \plus{} \frac {(b \plus{} c)^2}{c^2\cdot a^2} \equal{} \frac {(b \plus{} c)^2}{2bc\cdot p(p \minus{} a)}$

$ b^2 \plus{} c^2 \equal{} \frac {a^2b^2c^2}{2bc\cdot p(p \minus{} a)} \equal{} \frac {2a^2bc}{(b \plus{} c)^2 \minus{} a^2}$

$ (b^2 \plus{} c^2)((b \plus{} c)^2 \minus{} a^2) \minus{} 2a^2bc \equal{} 0$

$ (b \plus{} c)^2(a^2 \minus{} (b^2 \plus{} c^2)) \equal{} 0$

Therefore $ a^2 \equal{} b^2 \plus{} c^2$. So triangle $ ABC$ is $ A$ rightangled.[/color]

Solution:
  Let $ 2\alpha,2\beta,2\gamma$ is angles of triangle ABC respectively.
 We have:
   1/(BD)² + 1/(CD)² = 2/(AD)²
  or
  $ \frac{CD}{BD}\plus{}\frac{BD}{CD}\equal{} \frac{2\cdot BD\cdot CD}{AD^2}$
  or
  $ \frac{AC}{AB}\plus{}\frac{AB}{AC}\equal{} \frac{2\cdot BD\cdot}{AD}\cdot\frac{CD}{AD}$
  or
  $ \frac{sin2\beta}{sin2\gamma}\plus{}\frac{sin2\gamma}{sin2\beta}\equal{}\frac{2\cdot sin^2\alpha}{sin2\beta\cdot sin2\gamma}$
  then
  $ 2\cdot sin^2\alpha\equal{}sin^2 2\beta\plus{}sin^2 2\gamma$
 or
  $ (sin\alpha\plus{}sin2\beta)\cdot(sin2\beta\minus{}sin\alpha)\equal{}(sin\alpha\plus{}sin2\gamma)\cdot(sin\alpha\minus{}sin2\gamma)$
or
  $ sin(2\beta\plus{}\alpha)\cdot sin(2\beta\minus{}\alpha)\equal{}sin(2\gamma\plus{}\alpha)\cdot sin(\alpha\minus{}2\gamma)$
 and finally we have
  $ sin(2\beta\minus{}\alpha)\equal{}sin(\alpha\minus{}2\gamma)$
   $ 2\beta\minus{}\alpha$ and $ \alpha\minus{}2\gamma)$ both less than 180º and more than -180º , so that's why we have
  $ 2\beta\minus{}\alpha\equal{}\alpha\minus{}2\gamma$
And we can conclude that <BAC= 90º

Thank you for helping, but I have one question.  In the first solution, where did the p and (p-a) come from?

Thanks Again!

Hello all,

Susan here i understood all the problem except the (p-a) is it some kinda formula we need to add in the sum or something else lease help.

Cheers!!!


Two cyclists enter different ends of a tunnel at the same time and travel toward each other. The tunnel is 9 miles long. One cyclist is traveling 10 miles per hour and the other at 8 miles per hour. A fly touches one biker as she enters the tunnel and then proceeds to fly back and fourth between the bikers  at 15 miles per hour as they move toward each other. How far has the fly flown as the bikers meet? ( Assume the fly never stops)

[quote="akhil0422"]It will take $ \frac {9}{(10 \plus{} 8)} \equal{} 0.5$ hours for the bikers to meet. Thus, the fly travels $ 0.5\cdot15 \equal{} 7.5$ miles.[/quote]
Nice job dude


geometry
combinatorics solved
combinatorics

A calculator may be used to add two numbers, substract one number from another, divide  number by any positive integer, and raise any number to the tenth power. Prove that it may be used to find the product of any ten numbers.

Pierre.

Excuse me to interfere in this area of combinatorics, but I think that the problem follows from the identity $\sum {x_i^{10}}-\sum {(x_i+x_j)^{10}}+\sum {(x_i+x_j+x_k)^{10}}+...=10! x_1\cdot...\cdot x_{10}$.

Yeaaah! I remember I posted it for one of Putnam's 2004 problems.

Hmmm... I think that too :D (and thought about something like that while writing the statement).
Anyway, welcome into a new world Harazi :D Is that a wish for the new year to come?

Pierre.


A quarter is placed on the table. What is the number of quarters that can be placed around the original quarter so that each quarter is tangent to the original quarter and to two other quarters?

To make the quarters tangent to each other, we must circumscribe a hexagon about the circle.  Then, we place a quarter on each edge of that hexagon.  Thus, the answer is simply the number of sides on a hexagon, or $ \boxed{6}$.

I don't know much geometry, but why do you circumscribe a hexagon and not another arbitrary shape?


Let $p_{1}, p_{2},...,p_{n},...$ be a sequence of primes defined by $p_{1}=2$ and for $n \geq 1, p_{n+1}$ is the largest prime factor of $p_{1}p_{2}\cdot \cdot \cdot p_{n}+1.$ (Thus $p_{2}=3, p_{3}=7.$) Prove that $p_{n}\neq 5$ for any $n.$

[hide](strong) induction, working mod 5.

2 and 3 are multiplicative inverses mod 5.

if $p_{k-1}\equiv \pm 2$(mod 5), $p_{k}\equiv \mp 2$(mod 5) then $p_{k+1}\equiv \pm 2$(mod 5). For even k, all pairs $p_{1}.p_{2}$, ... $p_{k-1}.p_{k}$ multiply to 1 (mod 5), then adding 1 gives 2 (mod 5). If k is odd, then pairs $p_{2}.p_{3}$, ... $p_{k-1}.p_{k}$ multiply to 1 (mod 5), with total product $p_{1}p_{2}...p_{k}\equiv p_{1}\equiv 2$ (mod 5), then $p_{k+1}\equiv 2+1=3$(mod 5).

Inductive step- $p_{1}\equiv 2$(mod 5), $p_{2}= p_{1}+1 \equiv 3$(mod 5).  Then by PMI the cycle continues with $p_{i}\equiv 2$ (for odd i) or $3$ (for even i).

Then for no n would $p_{n}\equiv 0$(mod 5).[/hide]

im not sure but if $p_{1}\cdot p_{2}\cdot p_{3}\cdots p_{k}+1 \equiv \pm 2 \mod 5$

i dont think it implies that $p_{k+1}\equiv \pm 2 \mod 5$ it merely implies $p_{k+1}$ cannot be divisible by 5 as $p_{k+1}$ is the greatest prime divisor of $p_{1}\cdot p_{2}\cdot p_{3}\cdots p_{k}+1$

am i correct in saying this?

I don't really get what you're saying...

Wait.. yeah. I got confused. Sorry. Effectively, if $5 \nmid p_{1}p_{2}..p_{k}+1$ then $p_{k+1}$ cannot be 5.

See [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=442333#p442333[/url].
Lots of Indian Olympiad problems are in the Resources.  :)


Anyone have any good tutorials on making games with Flash 5?  I looked on the internet, followed everything they said, but it would end up with errors.  Please give me links to tutorials or tutorials that ACTUALLY work...

Thank you very much! (BTW, I'm just beggining with flash 5, I need some simple stuff to work with, thanks again)

I have to build up my skills.   :weightlift:

Do you know to animate in Flash ?? If you do , then I have a few good tutorials....
BTW , you need to seriously read something on actionscript before you start building games....try some book.
And as far as the internet is concerned , you have tutorials mainly for action script 2......whise code doesn't run properly on action script.


trigonometry
geometry
trig identities
Law of Cosines

Hello!

[img]http://img.photobucket.com/albums/v140/milwalt/distance.jpg[/img]

Please refer the above diagram.

There are two possible positions for Venus for a given position of Earth.  This is an SSA combination with the sun, earth, venus as the vertices of the triangle.
The radius of earth's orbit is 1.5 * 10^8 kilometers, and the radius of venus' orbit is 1.1 * 10^8 kilimeters.

Note the angle marked 30 degree.

With all these information, how do I find out the distance from earth to venus for both positions of venus? (The answers are 2 * 10^8km, and 0.5 * 10^8km)

Please answer it from a ninth grader's level.

Thanks in advance.[/url]

I'm not sure if you already know the law of cosines.
Here is a link that might help you:
http://en.wikipedia.org/wiki/Law_of_cosines

is the law of cosines like ummm..

c squared = a squared + b squared - 2ab cos C?

I think I've heard of it even though I haven't learned any trig before.

Thank you.

But is there any way to solve this without using trigonometry?  This is from the congruence chapter in the geometry book.    I am trying to help out a ninth grader who doesn't know the law of cosines.

Denote M the midpont between Venus1 and Venus2 ($V_{1}$ and $V_{2}$. And E the earth, S the sun. We can calculate $SM$ (half of $ES$) and then by Pythagorean theoreme $EM^{2}$. Let A and B be the intersections of ES with venus orbit and denote $MV_{1}=MV_{2}=x$. Then by power of the point theoreme we have $EA*EB=EV_{1}*EV_{2}=(EM-x)(EM+x)=EM^{2}-x^{2}$. From this we can find x.


algebra
polynomial
Rational Root Theorem
Galois Theory
superior algebra
superior algebra unsolved

Find the zeros of the following polynomial, over $ \mathbb{Q}$, $ \mathbb{R}$, and $ \mathbb{C}$: $ t^5\plus{}t\plus{}1$.

By the rational root theorem, this has no zeros over $ \mathbb{Q}$.

This factors as $ (t^3\minus{}t^2\plus{}1)(t^2\plus{}t\plus{}1)$. How do I find the real and complex roots?

Use the formulas for equations of degree $ 2,3$ on the factors. The first factor has maximal possible Galois group ($ S_3$), so finding it's roots will not really get simpler than that.


How many ordered pairs m, n of positive integer with m < n have harmonic mean 6^20?

All of these AIME problems you've been posting can be found in the AIME part of the contest section, [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]here[/url].

They each have their own thread where solutions are given, so you can just check those. Additionally, all problems related to AMCs/AIMEs should probably be posted in the AMC forums, I'm guessing.

[hide="solution"] We have $ \frac{2}{\frac{1}{m}\plus{}\frac{1}{n}}\equal{}\frac{2mn}{m\plus{}n}\equal{}6^{20}$.

Let $ k\equal{}6^{20}$. We now have $ 2mn\equal{}km\plus{}kn$ or $ (2m\minus{}k)(n\minus{}\frac{k}{2})\equal{}\frac{k^2}{2}$ or $ (2m\minus{}k)(2n\minus{}k)\equal{}k^2$. Since $ k$ is even, both of $ (2m\minus{}k)$ and $ (2n\minus{}k)$ are even. $ k^2\equal{}6^{40}\equal{}2^{40}3^{40}$. We can "assign" a factor of two to both $ 2m\minus{}k$ and $ 2n\minus{}k$ because they are already even.

That leaves us with $ 2^{38}3^{40}$, which has $ 39\cdot41\equal{}40^2\minus{}1\equal{}1599$ factors. One of these results in $ m$ being equal to $ n$, leaving us with $ 1588$ possibilities. By symmetry, half of these have $ m>n$, so our answer is $ \frac{1588}{2}\equal{}799$.[/hide]

[quote="jin thynj"][quote="xpmath"]All of these AIME problems you've been posting can be found in the AIME part of the contest section, [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]here[/url]. [/quote]

Wow, very thanks, i didn't know such things!

Now, do you think i should erase my questions?[/quote]Don't bother trying to reverse anything on the forums. Mods/admins will decide what to do with them.


graph theory
combinatorics proposed
combinatorics

The organizers of a mathematical congress found that if they accomodate any participant in a room the rest can be accomodated in double rooms so that 2 persons living in each room know each other. Prove that every participant can organize a round table on graph theory for himself and an even number of other people so that each participant of the round table knows both his neigbours.

[i]Proposed by S. Berlov, S. Ivanov[/i]

[quote="grobber"]Cute problem :). I hope the following is correct.

The hypothesis says that we have $2n+1$ vertices in a graph $G$, and for each vertex $x$ of $G,\ G-x$ has a perfect matching. The conclusion should be that every vertex lies on some odd cycle.

Let $x$ be a vertex of $G$, and let $x_1,x_2,\ldots,x_t$ be its neighbours. If we eliminate $x$, the remaining graph has a perfect matching, so let $v_1,\ldots,v_t$ be the neighbours of $x_1,\ldots,x_t$, respectively, in this perfect matching ($2t=n$, of course).

Now eliminate $x_1$. The neighbour of $x$ in the new perfect matching we get is one of the $x_i$'s, with $i\ge 2$. Suppose WLOG that it's $x_2$. The remaining vertices are now $x_3,\ldots,x_t$ and $v_1,v_2,\ldots,v_t$, which form a perfect matching. We also have edges between $x_i,v_i,\ \forall i\ge 3$, from the previous perfect matching, when we eliminated $x$.

Start from $v_1$, go to its neighbour in the new matching, then to that vertex' neighbour in the old matching, then to this new vertex' neighbour in the new matching, and so on, until you reach $v_2$. We have started with an edge in the new matching, alternating the old matching with the new one, the last edge also being from the new matching, which means tht we have used an odd number of edges. These edges, together with $xx_1,x_1v_1,xx_2,x_2v_2$, form an odd cycle on which $x$ lies.[/quote]

Grobber, when you say $2t=n$ you are asuming that there are no more vertex, but there could be, implying that in other matchings $x$ need not be conected to some other $x_j$??? I dont now who is wrong but i really don get that part but the other part seems clear....

I've deleted my previous reply, where I was saying that I was too busy to look into this (I actually was very busy :)).

That $2t=n$ thing makes absolutely no sense. I don't know what I was thinking. Of course, there can be other vertices. However, the neighbour of $x$ in every matching is always one of the $x_i$'s, since those are all its neighbours, so I don't really know what you meant by:

[quote="Pascual2005"]
implying that in other matchings $x$ need not be conected to some other $x_j$??? [/quote]

:?

Another thing that needs to be clarified: I was trying to show that $x$ lies in an odd cycle, so I assumed that none of the $x_i$'s are connected to each other. This means that the $v_i$'s are distinct from the $x_i$'s. I just wanted to point this out, so that there's no confusion.

Is it Ok now (the procedure is the same: we construct an alternating path with edges from both the new and old matchings)?

yes it is clear now, that $2t=n$ got me a little confused...but as i told you i understood the rest of the idea... :)

Here is an overkill.A graph doesn't contain an odd cycle iff it is bipartite(very well known).Now,we can divide our graph $G$ into two groups,$A$ and $B$ such that if two edges are in the same group they don't have an edge beetwen them.Now,since the $G$ has ann odd number of vertices,$A$ and $B$ can't have the same number of vertices,so pick an arbirtary edge from $B$(WLOG $B$ has less elements than $A$) and it is easy to see that we don't have a perfect matching for him,so done.
It is pretty strange this problem was in the contest,cause if you heard
 about bipartite graphs this is very natural and you are done pretty much instant.

No, your solution is completely wrong: "there exists an odd cycle" and "every vertex belongs to an odd cycle" are very different. (This is not ambiguous in the question and is mentioned in the earlier solution as well!)

Ok,I was in a rush so I didn't mention one more thing,the graph is connected.Assume it is not,then we will have $2$ groups $A$ and $B$ so that there isn't any edge from $A$ to $B$.Now,at least one will be odd and that is a contradiction.Now,consider a 'tree' from any vertex(I mean,if we can get to some vertex by minimum $i$ steps then it is at $i+1$-st level) and now color them alternate the starting vertex black,the vertices at the second level(its neighbours white) etc.Now,if we don't have an edge which vertex isn't the same colour we have a bipartite graph so contradiction,if not then we will have an odd cycle which contains our starting vertex.As JBL mentioned,my proof is wrong without this.

I have no idea about the relationship between what you have written and the problem -- as far as I can tell you do not use the conditions of the problem anywhere.  You appear to claim that in every connected non-bipartite graph, every vertex belongs to an odd cycle.  But this is simply false (e.g., if there is a vertex of degree 1).


At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, list these items:\\
$ 1\frac{1}{2}$ cups of flour, 2 eggs, 3 tablespoons butter,
$ \frac{3}{4}$ cups sugar, and 1 package of chocolate drops. They
will make only full recipes, no partial recipes.

They learn that a big concert is scheduled for the same night and
attendance will be down $ 25\%$. How many recipes of cookies should
they make for their smaller party?

Only 75% of the 108 students will be attending, and each person takes two cookies, so they need $ 108 \times 0.75 \times 2 \equal{} 162$ cookies. Each recipe makes 15 cookies, so $ 162 \div 5 \equal{} 10.8$ They only make whole recipes, so $ \boxed{11}$.

since 108 ppl gets 2 cookies

$ 108*2$

since attendence will down by 25%

$ 108*2*\frac{3}{4}$

soooo $ 108*2*\frac{3}{4}\div15\equal{}162\div15\simeq11$

answer : 11

EDIT : meh beaten to it

[hide="THIS IS MY EXPLANATION"]Ill answer this in a series of steps.

1.first of all we know that the attendence will be down by $ 25%
$ so only $ 75%
$ will be attending

2.$ 108 $ people eat 2 cookies so we get the equation $ 108*2 $

3.since only $ 75%
$ will be attending we multiply it out $ 108*2*\frac{3}{4} $

4.and each pan hold 15 cookies so we divide $ 108*2*\frac{3}{4}\div15=162\div15\simeq11 $

5.[hide="ANSWER"]11[/hide][/hide]

[hide="."][color=#800000][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH![/u][/i][/b][/color][/hide]

:starwars:

may the force be with you.


trigonometry
calculus
algebra
function
domain
inequalities
rearrangement inequality

[quote="t0rajir0u"]Find the minimum value of $\frac{\sin^3 x}{\cos x} + \frac{ \cos^3 x}{\sin x}$ for $0 < x < \frac{\pi}{2}$.

(Try not to use Calculus  :) )[/quote]

[hide="solution"]in the domain, they are both positive, using am-gm

$\frac{\sin^3 x}{\cos x} + \frac{ \cos^3 x}{\sin x}\ge 2 * \sqrt{\sin^2{x}\cos^2{x}}=2\sin x\cos x$
$\sin^4x-2\sin^2 x\cos^2 x + \cos^4 x \ge 0$
$(\sin^2 x-\cos^2 x)^2\ge 0$

minumum when $\sin^2 x-\cos^2 x =0$, occurs if $\sin{x}=\pm \cos x$, but we have only positive values, so $\sin x= \cos x$, $x=45$, then our answer is $\boxed{1}$[/hide]

[hide="Another solution"]By rearrangement inequality,
$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\geq\frac{\sin^3 x}{\sin x}+\frac{\cos^3 x}{\cos x}=1$. :D[/hide]

[hide]$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}=\frac{\sin^4 x+\cos^4 x}{\sin x\cos x}$.

Complete the square in the numerator: it becomes $\left( \sin^2 x+\cos^2 x \right)^2-2\sin^2 x\cos^2 x=1-2\sin^2 x\cos^2 x$.

So the fraction becomes $\frac{1}{y}-2y$ where $y=\sin x\cos x$. If $z=2y=\sin 2x$, then this is $\frac{2}{z}-z$.

Since $0<x<\frac{\pi}{2}$, $0<2x<\pi$ so $0<z\leq 1$. Then $\frac{2}{z}\geq 2$ and $-z\geq -1\Rightarrow \frac{2}{z}-z\geq \boxed{1}$. There is equality when $z=1$, which is at $x=\frac{\pi}{4}$.[/hide]


The five concentric circles shown have radii of 1, 2, 3, 4 and
5 centimeters. What is the number of square
centimeters in the total area of the three
shaded regions? Express your answer in
terms of $ \pi$.

[asy]fill(Circle((0,0),20),gray(0.7));
fill(Circle((0,0),16),white);
fill(Circle((0,0),12),gray(0.7));
fill(Circle((0,0),8),white);
fill(Circle((0,0),4),gray(0.7));[/asy]

The area of the outer ring is $ 5^2 \pi \minus{} 4^2 \pi$, or $ 9\pi$.

The area of the inner ring is $ 3^2 \pi \minus{} 2^2 \pi$, or $ 5\pi$.

The area of the circle is $ 1^2 \pi$ or $ \pi$.

Thus, the answer is $ 9\pi \plus{} 5\pi \plus{} \pi$ or $ \boxed{15\pi}$.


Show that if $a,b,c$ are positive real numbers and that for an arbitrary value of $n$, $a^{n}, b^{n}, c^{n}$ can always be the sides of a triangle then the triangles must be isosceles.

[hide]
WLOG C is at least tied for largest

a = rc
b = xc     0< r, x <= 1

so if a^n b^n and c^n are the sides of the triangle then:

(r^n)(c^n) + (x^n)(c^n) > c^n    by triangle inequality
r^n + x^n > 1 for all n

but since the limit of x^n and r^n as n goes to infinity is zero if they are less than one and greater than zero, so if they are both less than one, we can pick n large enough such that x^n ,  r^n < 1/2 and thus x^n + r^n < 1, so at least one of x or r must be 1 making it isosceles.[/hide]


[quote="nttu"]Find the greatest value of the expression :
 $T=\frac{a^2+1}{b^2+1}+\frac{b^2+1}{c^2+1}+\frac{c^2+1}{a^2+1}$ , where $a,b,c \geq 0$ and $a+b+c =1$[/quote]

It is not hard.

We have in $\frac{a^2+1}{b^2+1},\frac{b^2+1}{c^2+1},\frac{c^2+1}{a^2+1}$  two number are lager or smaller than 1: $\frac{a^2+1}{b^2+1},\frac{b^2+1}{c^2+1}$ are those

$(\frac{a^2+1}{b^2+1}-1)(\frac{b^2+1}{c^2+1}-1) \geq 0$

And we have if $x \in [1,2]$ then

$x+ \frac{1}{x} \geq \frac{5}{2}$

It looks better than my solution . This problem is in M&Y 2004


abstract algebra
linear algebra
linear algebra unsolved

Prove that if T in L(V) is normal, then $ nullT^k\equal{}nullT$ and $ rangeT^k\equal{}rangeT$ for every positive integer k.

1) It's enough to consider $ k\equal{}2$. (why?)
2) if $ TTx\equal{}0$, then $ \langle TTx,TTx\rangle\equal{}0$, which (using normality) can be written as $ \langle T^*Tx,T^*Tx\rangle\equal{}0$. Hence $ T^*Tx\equal{}0$. Since $ \langle T^*Tx,x\rangle\equal{}\langle Tx,Tx\rangle$, we have $ Tx\equal{}0$. This proves that $ T^2$ has the same kernel as $ T$.
3) Apply 2) to $ T^*$ in place of $ T$, and use the relation between range of $ T$ and kernel of $ T^*$.


I was wondering, is there a formula or way of finding the x and y of a point on a circle, given the radius and angle of the point (in reference to it's center of course)?

Sure, it follows directly from the trigonometric functions. If you let cX and cY be the x- and y-coordinates of the center, r be the radius, and A be the angle, then the point (x,y) would be (cX + rcos(A),cY + rsin(A)).

so in solving for x and y

x = rcos(A)
y = rsin(A)

?

Yes.  You can define each coordinate by $(r;\theta)$.  This is called polar coordinates.  And then you have the relations:

$x=r\cos \theta \\ y=r\sin \theta \\ x^2+y^2=r^2 \\ \tan{\theta}=\frac yx$

These are easily derived from trig and you can convert between rectangular (basic $(x,y)$ coordinates) and polar using them.

[hide]Assume center is $(x_0, y_0)$

$(x,y)=(r\cos \theta+x_0, r\sin \theta+y_0)$[/hide]


limit
logarithms
number theory
calculus
calculus computations

$ \pi \equal{} \displaystyle\lim_{n \rightarrow \infty}{\sqrt{\displaystyle\frac{6\cdot \ln{F_0F_1F_2\cdots F_n}}{\ln{[F_0,F_1,F_2,\dots ,F_n]}}}}$

By the way, what's the difference in using the displaystype code?  Just curious.

If you have access to jstor, this should help:

A New Formula for $ \pi$
Yuri V. Matiyasevich
The American Mathematical Monthly, Vol. 93, No. 8 (Oct., 1986), pp. 631-635

http://www.jstor.org/stable/2322322

If not, I'll try to post the outline of the proof later (but you should know that the proof isn't quite simple :lol: ).

EDIT: After reading the Matiyashevich's proof two more times, I gave up posting it here. Too advanced number theory for me...

It was introduced to [url=http://www.amazon.com/exec/obidos/ASIN/0471399698/]Fibonacci and Lucas Numbers with Applications[/url] P.95.


Find all natural numbers $ n$ for which the inequality
$ a\left(a \plus{} b\right)^{n} \plus{} b\left(b \plus{} c\right)^{n} \plus{} c\left(c \plus{} a\right)^{n}\ge 0$
holds for all real numbers $ a$, $ b$, $ c$.

if n is even, a = -1, b=0, c=0. i've been trying to look for the source instead of solving n odd  :P its so familiar its taunting me!

As far as I remember the answer is $n\in\left\{1,3,5\right\}$. I know that Vasc invented it for 3 and the proof is not nice, but for the case n=5 I never saw a solution. This is really a criminal problem!

This is the solution when n=3
Put $a+b=2z$
$b+c=2x$
$c+a=2y$
Then
$a(a+b)^3+b(b+c)^3+c(c+a)^3$
$=4[(x^2-y^2-xy)^2+(y^2-z^2-yz)^2+(z^2-x^2-zx)+x^2y^2+y^2z^2+z^2x^2]$
If this is also the solution you mention,Harazi,I don't think it's not nice

[quote="harazi"]I know that Vasc invented it for 3 and the proof is not nice, but for the case n=5 I never saw a solution. This is really a criminal problem![/quote]

I think this is a solution... substitute $a+b \rightarrow x, b+c \rightarrow y, c+a \rightarrow z$. We get: \[(x+z-y)x^{5}+(x+y-z)y^{5}+(y+z-x)z^{5}\ge 0\] Now we split this sum:
$\frac{x^{6}}2+x^{5}z-xz^{5}+\frac{z^{6}}2 = \frac12 (x^{2}+z^{2}) (x^{2}-xz-z^{2})^{2}\ge0$
$\frac{y^{6}}2+yz^{5}-y^{5}z+\frac{z^{6}}2 = \frac12 (y^{2}+z^{2}) (z^{2}-yz-y^{2})^{2}\ge0$
$\frac{x^{6}}2+y^{5}x-x^{5}y+\frac{y^{6}}2 = \frac12 (x^{2}+y^{2}) (y^{2}-xy-x^{2})^{2}\ge0$

Now I will finish the problem. We have left to prove that it's false for all odd numbers (even numbers are trivial), well here's a general counterexample:
$a=\frac14, b=-\frac94, c=1$.

Then we have that the LHS is $\frac{13}4\left(\frac54\right)^{n}-2^{n-2}$, which is negative from $k=6$ onwards. QED.

[quote=harazi]... but for the case \(n=5\) I never saw a solution?[/quote]

Its solution was posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=297032]here[/url].


Hoorah, I'm a thread starter!

5 5 5 5 5 = 25

Looked like I was about to not get a 5 on #4 and then the grader scribbles out "Nice Try" and then puts in "Very Nice" under the comments.

Happy Hannukah!

5 [b]5[/b] 5 5 5

Weird, what happened to my other thread? I messed up the poll, then it disappeared, then I added a new one, then it got messed up  :huh:

I like this poll much better. It has an option for my [size=75]15354[/size]

[b]5[/b] 5 4 5 5

I was happily surprised about my commendation for number 1. However, I am kind of confused as to why I didn't get a 5 for number 3.

-, 5, 5, -, 2= 12  :|

I'm surprised I got #3 though...

[quote="13375P34K43V312"]4 5 5 1 5

Still no commended? Give me a break.

Well for #1, my solution apparently had a flaw, and I thought I was going to get 4 in #3 because it had a minor hole. And I thought I was going to get 2 on #4, but oh well.

UGH I MISSED THE LEADERBOARD BY 1 WTF[/quote]
Okay, you really need to chill.
By the way, you don't expect a commended unless you somehow one line one of these. It's hard to get a commended, an ordinary common and correct solution won't do it.

can you still join in this competition? or is it too late

You can join the USAMTS at any point in the year :)

It's a really great competition, I suggest you try it.

[quote="now a ranger"]Okay. Do you think I have enough time to complete the problems by January 8 for Round 3?

It seems really fun, even though my overall score will be very low since I didn't do round 1 nor round 2.  :lol:[/quote]
Well it depends on how good you are at the topics. Since you are in 8th grade it would be very impressive if you got all the problems, the only one to have done that consistantly is Wenyu Cao and he's a genius.

I'm doing USAMTS but I forgot to type Round 1, and forgot to send ROund 2 :mad:  :mad:  :mad:

I am doing round 3 and typing it up alltogether. :)

[quote="chess64"]Whoa I actually got commended for my trig brute force on #5! It's strange how I never get commended for the solutions I think are nice...[/quote]

it's because of your natural talent  :lol:

[quote="bpms"]A crazy stat
[hide]The average score for 8th and below is more than the average for 9th, 10th, and 11th :o.[/hide][/quote]

That's not entirely surprising.  More marginal competitors are likely to compete at higher grades, whereas only the best 8th graders will attempt the USAMTS.  It's a competition that appeals to high school students mostly I suppose.

[quote="calc rulz"][quote="chess64"]Whoa I actually got commended for my trig brute force on #5! It's strange how I never get commended for the solutions I think are nice...[/quote]

Hey, I've noticed the same thing...I always seem to get the right number of commendations but on the wrong problems... :|[/quote]

Just do what your instinct tells you to do might help you get more commendations, with an odd but elegant way.  :rotfl:

NOOOOOO

[b]5[/b] 4 [b]5[/b] [b]5[/b] 5

How, HOW did I get so lazy on my #2 explanation that I LOST A POINT!!?!!  :(

[quote="bpms"]5 3 [b]5[/b] 2 5
Apparantly I should've proved that M lied on ENA :(.
Still I'm happy about the commended :)[/quote]

Yeah I forgot to do that too...   :oops_sign:

[b]5[/b] 3 5 [b]5[/b] 5

What makes a commended solution?

The only one I ever got was 5 pages full of my messy handwriting before I learned to latex.

5 5 4 5 5  :(   :|

Will there be a Math Jam for Round 3??

Oh, no!  Only a 24.  You know some of us might be very happy to get a 24.

I did not see USAMTS on the math jam schedule, they would not have a math jam after March when we already know our score...

5 3 5 5 5

Should have proved collinear-ty on #2

I am also very happy.
5 [color=red]5[/color] 5 5 5
I didn't think I would recieve a 5 for problem 4, but I did.  :D

@vardanch: Did you submit a permission form for just participation or for releasing your name?

EDIT: I also noticed that my graders comments last time were 45 words long, but this time they were 7 words long. Probably since I didn't find a great solution for one problem and mess up another this round.

[quote="vardanch"]I'm very happy  :lol:

5 5 5 5 $5$

 :huh:  But why my name is not written in "Usamts Leaders"  ?[/quote]

Sometimes they accidently don't put people's names in, you should probably PM one of the administrators.

[quote="renatzu"]5[b]5[/b]5[b]5[/b]5

I was very proud of my solution to number 4.[/quote]

What was your solution to number 4?  Everyone that posted their up last time had basically the same method.  Was it well written, or did you come up with some other trick we didn't see?

In any case, I ended up with

5 [b]5 5[/b] 5 5

5 5 5 5 5

so i'm happy about that.  but no commended yet.

 :ninja:

55505 (didn't do 4 haha..)

Man I should've at least attempted #4.. Then I would've gotten my name up there for the FIRST time. But oh well. I'm happy with what I got.  :D

34535

Comments:

On #1, I'm dead sure my grader just skimmed over a part of my argument which was pressent, so I protested that, and I'm pretty sure that will get bumped up to a 5.

On #4, I'm amazed I got 3 points, I could barely justify what was the correct answer

Then on #2 I had a typo: 114 --> 144.  Grrrrrrr...  first stupid mistake on USAMTS.

Overall, not a very good round for me.  I was slightly suprised that 5 didn't get a commend, and my grader for #2 actually wrote "This would have been a commended solution, but $\angle ANM = 180-66 = 114$... grrrrr.  I had written $180-66 = 144$.

[quote="calculuslover800"]
My grader for #2 actually wrote "This would have been a commended solution, but $\angle ANM = 180-66 = 114$... grrrrr.  I had written $180-66 = 144$.[/quote]

That kind of thing comes down to grader discresion (graders are allowed to remove one point for any non-rubric reason they see fit).  I had a similar typo on round 1 year 17, but the grader put down "Answer should be 8, not 10" and gave a 5.  Bad luck, man, but I wouldn't fret it too much (though you don't seem to be worried... just thought I'd toss in my experience on a similar thing).  Unless you wind up one point away from something, the score really doesn't matter.  (Cliche, but...) The journey is the reward.  Think of it as balancing out your score on #3.

[quote="eyefragment"][quote="calculuslover800"]
My grader for #2 actually wrote "This would have been a commended solution, but $\angle ANM = 180-66 = 114$... grrrrr.  I had written $180-66 = 144$.[/quote]

That kind of thing comes down to grader discresion (graders are allowed to remove one point for any non-rubric reason they see fit).  I had a similar typo on round 1 year 17, but the grader put down "Answer should be 8, not 10" and gave a 5.  Bad luck, man, but I wouldn't fret it too much.  Unless you wind up one point away from something, the score really doesn't matter.  (Cliche, but...) The journey is the reward.  Think of it as balancing out your score on #3.[/quote]
last year i had this one proof that was about 6 pages of summing various series...with a small algebra error somewhere, but the grader wrote "you have a small algebra error but i can't find it" and gave me a 5. :ninja:

[quote="eyefragment"]What was your solution to number 4?  Everyone that posted their up last time had basically the same method.  Was it well written, or did you come up with some other trick we didn't see?[/quote]It was well written- short but still rigorous.

5 5 [b]5[/b] 5 5

I'd rather have more commended solutions though...

Still, pretty happy!  :D

My scores so far

Round 1: 5 [b]5[/b] 5 [b]5 5[/b]
Round 2: 5 [b]5 5[/b] 5 5

I'm really on a roll this year  :D


pigeonhole principle
modular arithmetic
combinatorics unsolved
combinatorics

Hi Guys first of all i will thanks all people that try to help me in this problem, i work on it from 3 days
(also because i have other things to do :( ).

The problem says:
Prove that there exists a positive integer n so that 44^n — 1 is divisible
by 7.

So i know from the pigeonhole principle that i have 7 possible remainder 0...6 so for sure with n i have some number with the same remainder but the problem is how can tell that this remainder is 0?

so if the claim is true, is true also

44^n -1 mod 7 = 0
and so
44^n -1  = 7k.

Ok I suppose this is the right way to solve with pigeon but i not able to figure it out how continue in this proof.

Thanks a lot .

Best Regards[/code]

This has nothing to do with the pigeonhole principle. $ 44 \equiv 2 \pmod{7}$, so what is $ 44^2$ mod 7? Can you do this without actually calculating $ 44^2$? What about $ 44^3$?

Why not ? if you thinks about remainder of 7 you are sure that you have 7 hole 0...6 and n are pigeon.

So i can tell exist an n such that 44^n -1 mod 7 is 0. without computing and keep n.

I know from test that n=0 and n=3 works fine.

In the end this is an exercise on a book in pigeonhole principle section. so "This has nothing to do with the pigeonhole principle[..]" is partially false.

Thanks a lot for the answer !!!
 :)

OK, maybe I was a little harsh  :) How do you deduce that $ 44^n \equiv 1 \pmod{7}$ for some $ n$ from the observation that two of those residues are the same?

[quote="joksy"]Mhh yes in fact the my first way to think is this : I have a sequence of this number 44^n -1 44^(n-1) -1 44^(n-2) -1 and so on. and if i divide all of them by 7 for sure some remainder are the same. But in this way i not able to construct somethings that I can tell at the end : well exist an n st 44^n-1 is divisible by 7.

This problem make me crazy....  :wink:

Again thanks for help me !!!

;)[/quote]
By pigeon hole principle there exists $ n<m$ such that $ 44^n \equiv 44^m \bmod 7$ But $ 44^n$ is uniquely dertermined by $ 44^{n\minus{}1} \bmod 7$. Thus $ 44^{n\minus{}1} \equiv 44^{m\minus{}1} \bmod 7$, $ 44^{n\minus{}2} \equiv 44^{m\minus{}2} \bmod 7$, ...,  $ 44^{n\minus{}n} \equiv 44^{m\minus{}n} \bmod 7$. And thus $ 7 \mid 44^{m\minus{}n}\minus{}1$

[quote="joksy"]Thanks a lot for the replay.
 :lol:

I have a question for you:  i don t understand the equivalence, 44^n is equivalent to 44^m mod 7.
and  what you mean for 44^n  is uniquely determined by  44^(n-1) mod 7.

Anyway a big THANKS. I m try to study your proof again!![/quote]
That $ 44^n$ is equivalent to $ 44^m$  $ \bmod 7$, means that they have the same remainder when divided by $ 7$.

That $ 44^n$ is uniquely dertemined by $ 44^{n \minus{} 1}$ Means that if we know $ 44^n$, we also know $ 44^{n \minus{} 1}$. Let's just have a short look at the lemma:

Lemma: If $ 44 \cdot a \equiv 44 \cdot b \bmod 7$ then $ a \equiv b \bmod 7$.
Proof: Then $ 7 \mid 44(a \minus{} b)$ and thus $ 7 \mid a \minus{} b$ and thus $ a \equiv b \bmod 7$.

Thus if $ 44^n \equiv 44^m$ we know that $ 44 \cdot 44^{n \minus{} 1} \equiv 44 \cdot 44^{m \minus{} 1}$ and thus $ 44^{n \minus{} 1} \equiv 44^{m \minus{} 1} \bmod 7$.(According to the lemma)

How much do you know about congruences? If it was to fast, then just ask :)

Hi mathias , about congruences i don t know so much. But anyway if you tell me more this would be nice. really i don t know how to thanks you,

you are great!


On a chessboard, each of thirty-two dominoes is of such a size that it covers exactly two adjacent squares.  The thirty-two dominoes can thus cover all sixty-four of the chessboard squares.  But suppose we cut off two squares at diagonally opposite corners of the board and discard one of the dominoes.  Can all thirty-one dominoes be placed on the chessboard so that all sixty-two squares ar covered?  If so, show how this can be done.  If not, prove that it is impossible.

Posted by MrMcGeek

[hide]

They can not all be covered.

On a chessboard, each red square is adjacent to four black squares, and the other way around. Each domino, therefore, covers one black and one red square. There are 4*8=32 red squares, and 4*8=32 black squares.

When you remove the two opposite corner squares, you remove two squares of the same color. There will be 30 red and 32 black, or 30 black and 32 red. Since each domino must cover a red and a black square, this leaves 2 squares left uncovered (of the opposite color as the squares that were removed)[/hide]


linear algebra
matrix
analytic geometry
geometry
3D geometry
sphere
vector

[quote="liyi"]$A$ is called a positive matrix if all of its elements are positive reals. Show that the largest eigenvalue of $A$ is a positive real.[/quote]

Does largest eigenvalue mean that all eigenvalues of A are reals? If so then take Trace(A). This is a positive number and also the sum of all eigenvalues. Thus the eigenvalues of A can't be all negative.

But I gues it only means that A has some real eigenvalues.

Of course, there may be some complex eigenvalues.

Hi

The statement has a well known topological proof. You will find it in Munkres' book as an application of Brouwer's fixed point theorem. Let A be the matrix. We shall prove for the case n = 3, but the proof follows in a more general case n >= 3.

Consider the intersection of the octant in R^3 where all coordinates are positive, with a unit sphere centred at the origin. Call it B. Define the following transformation T on all vectors X in this intersection, as follows : - T(X) = AX/||AX|| where ||t|| is the magnitude of t. As AX also has all its entries positive, T(X) = AX/||AX|| lies on B.

Now T is a continuous map on B. As the intersection is homeomorphic to a circular disk, any continuous map has a fixed point and hence there is an X = T(X).

So AX = ||AX||X

And thus ||AX|| is a positive real eigenvalue.

The proof can be generalised as Brouwer's fixed point theorem is valid in higher dimensions.

And how do you show that this eigenvalue has the biggest module?

This the Perron-Frobenius theorem: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1734772401&t=34219.

I don't have an answer to your question, but harazi proves in my link that the biggest absolute value of an eigenvalue is itself an eigenvalue. Moreover, I have a whole chapter in a book dedicated to "non-negative" matrices.


let ABCD is a  trapezoid.(AB||CD)
consider F and E are 2point on AD and BC respectively such that <BFC and AED are Maximum.prove that <BFC=<AED ...


pigeonhole principle
number theory
prime factorization

Suppose that you have a collection of 101 whole numbers that lie in the range between 1 and 200, inclusive. Show that there is at least one number in the collection which divides another one in the collection.

[hide="Solution"]Let these numbers be $a_{1},\ldots,a_{101}$ and $a_{n}= 2^{k_{n}}\cdot l_{n}$ for $2\nmid l_{n}$. There are only $100$ odd numbers in the range $[1,200]$, so from the pigeonhole principle there exist $1\leqslant n,m\leqslant 101$ such that $l_{n}= l_{m}$ and WLOG $k_{n}<k_{m}$. Then obviously $a_{n}\mid a_{m}$ which ends the proof. [/hide]

[b]Solution.[/b] Let $\mathbb{A}\subset B=\{1,\hdots,2n\}\ni |A|=n+1$.  Select the $n$ odd numbers from $B$ and for each $x$ create the set $\{x2^{k}\}_{k=0}^{\infty}$.  Since every natural number has the prime factorization  of the form $2^{p}(2q-1),\ p,q\in\mathbb{N}$, at least two of the $51$ numbers are in the same set among the $n$ sets.  Since the two numbers are distinct, one must be smaller than than the other, that is, $2^{a}x|2^{b}x\ni a<b$. $\blacksquare$

EDIT: Did not see the post above since I was in course of posting.


linear algebra
matrix
vector
calculus
integration
linear algebra unsolved

Prove that given any numbers $m,n$ there is a real matrix $mxn$ A and a constant $c>0$ such that
   $max_{1\leq i\leq m} {|\sum_{1\leq j\leq n}{a_{ij}x_j}|}\geq c(max_{1\leq i\leq n} |x_i|)^{-\frac{m}{n}}$.

You meant this should hold for all vectors $(x_i)_{1\le i\le n}$? Because I don't think it's true in this case, since it fails if we fix a vector, and consider multiples of it by arbitrarily small scalars. Maybe you were referring to integral vectors or something like that (that's what "diophantine approximation suggests")? :?

Yes, of course, the vectors have integral components. :)

Ok, thanks for clarifying that, but I still have some doubts about it:

For instance, assume you have shown this assertion for any $n$ and $m=1$. Then, for larger $m$ and some arbitrary $n$, we can prove the assertion with the exponent $-\frac 1n$, which is better than $-\frac mn$, because $\max|x_i|\ge 1$. We can do this by choosing one of the rows of the matrix $(a_{ij})_{i,j}$ to be the row vector resulting from the proof for $m=1$ and that respective $n$. There's nothing explicitly wrong with this, but it makes the whole thing look suspicious.. :?

Or maybe I'm just overlooking something? :?

Actually, there was a typo in the statement. In the RHS, at the exponent we have $\frac{-n}{m}$


geometry
geometric transformation
reflection

When I fail to solve a geometry problem, I sometimes read the solution and find that it involves a construction or reflection that the student must perform. However, in many cases, I find the construction or reflection to seem quite contrived, and I often wonder how I would ever come up with such a thing myself. So my question is: what generally inspires you to construct additional lines or reflect a particular figure in a geometry problem?

After many many years of trying to solve geometry problems, I can tell you: I have absolutely no clue. Purely inspiration?

heh, glad I'm not the only one Stephen :).  Sometimes I think it's obvious -- billiards questions, for example, almost always call for reflections of the board.  Other than that, though, who knows?

Draw everything you can think of. Something has to work, right? In other words, I have no clue either.

...

Uh oh..This isn't good. The three smartest kids who are among the top 10 posters just replied saying that they have no clue.

Could you give an example?

Some of them I have no idea; but others actually make sense... its hard to give a concrete example, though.

Often, you might want to "move" an angle or side length somewhere so that you can use it...

exactly what do you mean by "construction?" Like drawing in a line or marking off a point? That way I might be able to give an example...

[quote]exactly what do you mean by "construction?" Like drawing in a line or marking off a point? That way I might be able to give an example...[/quote]

Yeah, like drawing an auxiliary line or something...

Hmm...
if you're given a condition such as something + something = something else, try to move the two somethings together to use the condition.  Also, making stuff cyclic is good.  And satisfy your sense of symmetry.

Well one thing I do know how to do is draw accurate diagrams. (I drew about 50 of them at the last IMO I had to keep asking for more paper :D). That often give hints of things you should be noticing.


integration
logarithms
calculus
trigonometry
calculus computations

Substitute $ x$ to $ \sinh{t}$, we get:

$ \int\frac{dx}{x^2\plus{}x\sqrt{1\plus{}x^2}}\equal{}\int\frac{\cosh{t}dt}{\sinh^2{t}\plus{}\sinh{t}\cosh{t}}$
$ \equal{}\int\frac{\cosh{t}}{\sinh{t}(\sinh{t}\plus{}\cosh{t})}dt\equal{}\int\frac{\cosh{t}(\cosh{t}\minus{}\sinh{t})}{\sinh{t}}dt$
$ \equal{}\int\frac{\cosh^2{t}}{\sinh{t}}dt\minus{}\int\cosh{t}dt$
$ \equal{}\int\frac{1}{\sinh{t}}dt\plus{}\int\sinh{t}dt\minus{}\int\cosh{t}dt$
$ \equal{}\int\frac{1}{\sinh{t}}dt\plus{}\cosh{t}\minus{}\sinh{t}$.

$ \int\frac{1}{\sinh{t}}dt\equal{}\int\frac{\sinh{t}}{\cosh^2{t}\minus{}1}dt\equal{}\int\frac{1}{2}\left(\frac{1}{\cosh{t}\minus{}1}\minus{}\frac{1}{\cosh{t}\plus{}1}\right)\sinh{t}dt\equal{}\frac{1}{2}\ln\frac{\cosh{t}\minus{}1}{\cosh{t}\plus{}1}\plus{}C$.
So, the answer is $ \boxed{\frac{1}{2}\ln\frac{\sqrt{1\plus{}x^2}\minus{}1}{\sqrt{1\plus{}x^2}\plus{}1}\plus{}\sqrt{1\plus{}x^2}\minus{}x\plus{}C}$.

hello, or alternatively you can write
$ \frac{1}{x^2\plus{}x\sqrt{1\plus{}x^2}}\equal{}\minus{}1\plus{}\frac{\sqrt{1\plus{}x^2}}{x}$,
the first integral is simple, the second one gives
$ \int\frac{\sqrt{1\plus{}x^2}}{x}\,dx\equal{}\sqrt{1\plus{}x^2}\minus{}artanh\left(\frac{1}{\sqrt{1\plus{}x^2}}\right)\plus{}C$
Sonnhard.

$ \int\frac{1}{x^{2}+x\sqrt{1+x^{2}}}\,dx = \int \frac{\sec^2u}{\tan^2u + \tan u \sec u}\,du = \int \frac{1}{\sin^2u + \sin u}\,du$

$ = \int \frac{1 + \sin u - \sin u}{\sin u (1 + \sin u)}\,dx = \int \csc u \,du - \int \frac{1 - \sin u}{(1+ \sin u)(1 - \sin u)}\,du$

$ \ln \left|\frac{1 - \cos u}{\sin u}\right| - \int \frac{1 - \sin u}{\cos^2u}\,du = \ln \left|\frac{\sqrt{1+ \tan^2u} - 1}{\tan u}\right| - \tan u + \sqrt{1 + \tan^2u} + C$

$ = \ln \left|\frac{\sqrt{1+ x^2} - 1}{x}\right| - x + \sqrt{1 + x^2} + C$.


Prove that $ \frac {1}{2}(a \plus{} b)^{2} \plus{} \frac {1}{4}(a \plus{} b) \geq b\sqrt {a} \plus{} a\sqrt {b}$

$ \sqrt{a}$ becomes larger than all of the other terms as $ a,b \rightarrow 0^\plus{}$, so something is wrong.

Are you mean $ \frac{1}{2}(a\plus{}b)^{2}\plus{}\frac{1}{4}(a\plus{}b)\geq b\sqrt{a}\plus{}a\sqrt{b}$ ?
[hide="Solution"]$ LHS \equal{} (a\plus{}b)(\frac{a\plus{}b}{2}\plus{}\frac{1}{4}) \geq \sqrt{ab}(a\plus{}b\plus{}\frac{1}{2})
\equal{} \sqrt{ab}[(a\plus{}\frac{1}{4})\plus{}(b\plus{}\frac{1}{4})] \geq a\sqrt{b}\plus{}b\sqrt{a}$ by Am-Gm
[/hide]

I am sorry ! I mean $ \frac {1}{2}(a \plus{} b)^{2} \plus{} \frac {1}{4}(a \plus{} b) \geq a\sqrt {b} \plus{} b\sqrt {a}$

Here is my proof:

[hide] $ \frac {1}{2}(a \plus{} b)(a \plus{} b \plus{} \frac {1}{2})\geq \sqrt {ab}(\sqrt {a} \plus{} \sqrt {b})$

$ \frac {1}{2}(a \plus{} b)\geq\sqrt{ab}$ is true by AM-GM

Than :

 $ a \plus{} b \plus{} \frac {1}{2} \minus{} \sqrt {a} \minus{} \sqrt {b}\geq 0$

$ (\sqrt {a} \minus{} \frac {1}{2})^{2} \plus{} (\sqrt {b} \minus{} \frac {1}{2})^{2}\geq 0$ [/hide]


What do you think about it?

D,E,F are any points on the sides of the triangle ABC.
If the intersection point of AD,BE,CF is O then prove that
i) CO/OF=CD/DB + CE/EA         ii) DO/AD + EO/EB +FO/FC =1

Sol: Using the group theory:
*Take AF/FB=a , AD/DC=b and the masses on the vertices A,B,C are 1,a,b
*Let's do D,E,F the centers of mass(gravity) of each side.
 Hence The masses on the vertices will be a+b,b+1,a+1
*Since the sum of the masses of A+D and C+E are equal to each other
then O is the center of mass(gravity) of all points.
So;
......i)  CO/OF =a/b + 1/b and CD/DB + CE/EA =a/b + 1/b (proved)
.....ii)  Since O is the center of mass(gravity)
..........DO=x, AD=(a+b+1)x, EO=az, EB=(a+b+1)z, FO=by, FC=(a+b+1)y
..........DO/AD + EO/EB +FO/FC will be proved by the same way

[quote="kim77"]D,E,F are any points on the sides of the triangle ABC.
If the intersection point of AD,BE,CF is O then prove that
i) CO/OF=CD/DB + CE/EA         ii) DO/AD + EO/EB +FO/FC =1[/quote]

That's a well-known and often useful problem, I don't understand why you call it "silly", and you can find a solution on http://cut-the-knot.com/triangle/Gergonne.shtml .

  darij


abstract algebra
Ring Theory
advanced fields
advanced fields unsolved

Prove that $S^{-1}R/S^{-1}p$ is isomorphic with $T^{-1}(R/p)$.

Commutative algebra is still algebra -> superior algebra forum ;)

And I think we have given:
$R$ a ring, $p$ prime ideal, $S$ a multiplicatively closed subset of $R$ and $T$ it's image in $R/p$.

If $I$ is any ideal in $R$ we have that $S^{-1}R/S^{-1}I \simeq S^{-1}(R/I) \simeq \bar{S}^{-1}(R/I)$ where $\bar{S}$ is the image of S in R/I.

To see this, consider the map $\phi : R \to R/I \to \bar{S}^{-1}(R/I)$ and notice that it takes S to units, hence factors through $S^{-1}R$ and the kernel of this latter is just $S^{-1}J$ where J is the kernel of $\phi$.

The kernel of $\phi$ is all $x \in R$ such that $xs \in I$ for some $s \in S$. But the elements of $S^{-1}J$ are of the form $x/t$ for $x \in J$ and $t \in S$. And $x/t = (sx)/(st) \in S^{-1}I$.


[quote="red_dog"]Suppose that $ a_{1}\geq a_{2}\geq\ldots\geq a_{n}$. Then
[/quote]

You cannot suppose that!  :D


ARML probability geometry trigonometry rotation

1.
When the sum of the first $k$ terms of the series

\[ 1^2 + 2^2 + 3^2 + \cdots + n^2 + \cdots \]
is subtracted from the sum of the first $k$ terms of the series

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) + \cdots, \]
the result is 210.  Find the numerical value of $k$.

2.
Three coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30.  A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles.  Find the length of the shortest line segment that could contain a point on each square.

3.
Find the numerical value of $k$ for which

\[ \frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}. \]

4.
Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies.  When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had.  After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had.  With how much money did Sam leave home that day?

5.
Two cards are randomly selected from a standard 52 card deck of playing cards, without replacement.  Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).

6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}. \]

7.
The length of the base of an isosceles triangle is 20.  In the plane of the triangle, a point 4 units from this base is 10 units from each leg.  Find the area of the triangle.

8.
If Arc denotes the principle value, find all values of $x$ which satisfy

\[ \sin{(4\text{Arc}\tan{x})} = \frac{24}{25}. \]

I never knew these were here :O

[hide=" #3"]Its well known that:

\[ \frac{7}{x+y}=\frac{k}{x+z}=\frac{11}{z-y} = \frac{7}{z+y}+\frac{11}{z-y} = \frac{18}{x+z}\]

Therefore, $\boxed{k=18}$[/hide]

[hide="#2"]
It is not hard to see that the orientation that will give us the shortest line segment is if the largest square is rotated $ 45^{\circ} $ with respect to the orientation of the smallest and medium square.

Thus, our length is simply:

$ \frac{30}{\sqrt{2}} - 10\sqrt{2} = 15\sqrt{2} - 10\sqrt{2} = \boxed{5\sqrt{2}} $

[/hide]


linear algebra
matrix
geometry
3D geometry
quadratics

If $ A,B,C \in \ M_n({ \mathbb{R}})$ which comute among them  $ AB\equal{}BA,AC\equal{}CA,BC\equal{}CB$ then $ det(A^2\plus{}B^2\plus{}C^2\minus{}AB\minus{}BC\minus{}CA) \ge 0$.

[hide="Hint"]
1. Note that because $ AB \equal{} BA,AC \equal{} CA,BC \equal{} CB$, then the matrix is simultaneously triangularizable.
2. Now, supposed $ A \equal{} UP{U}^{ \minus{} 1},B \equal{} UQ{U}^{ \minus{} 1},C \equal{} UR{U}^{ \minus{} 1}$, where all of $ R,Q$ and $ P$ is upper triangular, then the element in the main diagonal of $ A^2 \plus{} B^2 \plus{} C^2 \minus{} AB \minus{} BC \minus{} CA$ is greater or equal than $ 0$
(By using the fact that $ a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca \equal{} \frac {1}{2}((a \minus{} b)^2 \plus{} (a \minus{} c)^2 \plus{} (b \minus{} c)^2) \geq 0$
3. Because all the elements in the main diagonal of the matrix is greater than $ 0$ and because the matrix is an upper triangular matrix, then the determinant is greater than or equal to $ 0$
[/hide]

Hi Qruni,
unfortunately your reasoning is false.

Curiously the definitive idea is in the Qruni's post:
the question is equivalent to $ det((A\minus{}B)^2\plus{}(B\minus{}C)^2\plus{}(C\minus{}A)^2)\geq{0}$ where these $ 3$ matrices commute. The answer is in
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274989
The main idea was due to yuri.

The basic problem with Qruni's argument is that you can't assume a real matrix is triangularizable over $ \mathbb{R}$.

Here's a fix:
Work over $ \mathbb{C}$. We can write $ A^2\plus{}B^2\plus{}C^2\minus{}AB\minus{}BC\minus{}CA\equal{}(A\plus{}\omega B\plus{}\omega^2 C)(A\plus{}\omega^2 B\plus{}\omega C)$ where $ \omega$ is a cube root of unity. Write that as $ Z\overline{Z}$; we still have that they commute. Now, $ \overline{Z}$ has the same determinant as its transpose $ Z^*$, and $ ZZ^*$ is Hermitian positive semidefinite. Thus $ ZZ^*$ has a positive real determinant, and so does $ Z\overline{Z}$.

The jmerry's post shows that the quadratic form $ A^2\plus{}B^2\plus{}C^2\minus{}AB\minus{}BC\minus{}CA$ is degenerate. Its decomposition in squares is $ (A\minus{}(B\plus{}C)/2)^2\plus{}3/4(B\minus{}C)^2\equal{}U^2\plus{}V^2\equal{}(U\plus{}iV)(U\minus{}iV)$ whose determinant is $ |det(U\plus{}iV)|^2$.


Find all real numbers $x,y$ greater than $1$, satisfying the condition that the numbers $\sqrt{x-1}+\sqrt{y-1}$ and $\sqrt{x+1}+\sqrt{y+1}$ are nonconsecutive integers.

Hello Pohoatza,

Let f(x) = sqrt(x+1) - sqrt(x-1) for x >= 1

It's easy to see that f(x) is a decreasing function from sqrt(2) to 0.
The distance between the two numbers we are looking for is f(x) + f(y).
So f(x) + f(y) <= 2sqrt(2) < 3 and we are looking for x and y such that f(x)+f(y) = 2 (nonconsecutive)

Hence we have to solve :
(i) sqrt(x+1) + sqrt(y+1) = n+1
(ii) sqrt(x-1) + sqrt(y-1) = n-1
let A = sqrt(x+1) + sqrt(x-1) and B = sqrt(y+1) + sqrt(y-1)
(i) + (ii) ==> A+B = 2n
(i) - (ii) ==> 1/A + 1/B = 1 ==> AB = A + B = 2n

This gives  A = n+sqrt(n^2-2n) and B = n-sqrt(n^2-2n)

The last thing is to solve sqrt(x+1) + sqrt(x-1) = A and sqrt(y+1) + sqrt(y-1) = B
This is only possible if A and B >=sqrt(2) and, for B, this leads to n =2, unique value possible.

==> A = B = 2
==> x = y = A^2/4 + 1/A^2 = 5/4

So the unique solution is :

x = y = 5/4

sqrt(x+1) + sqrt(y+1) = 3
sqrt(x-1) + sqrt(y-1) = 1

--
Patrick

What about if they were also consecutive ?  :wink:  I have a nice solution for this


inequalities
LaTeX
inequalities proposed

Dear friends,

Last two years, due my business, I have no time to visit and contribute to mathlinks regularly.

Now, my mission is completed and I can join with you.

Firstly, I want to say hello with all of my friends (old and new), especially: Valentin, Gil, Harazi, Manlio, Orl. You are best friends.

Secondly, for memory, I send here my small collection, collected by me in 2004 from mathlinks. I hope old friends can see themself here. New friends also will be interested in provided material.

Last thing, I post here a promblem of Vo Quoc Ba Can, proposed for a Test of our mathematical club

           Let a, b, c are nonegative numbers not all are zero. Prove that

$ \frac {1}{3} \le \frac {a^2}{3a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b^2}{3b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c^2}{3c^2 \plus{} (a \plus{} b)^2} \le \frac {1}{2}$

[quote="arqady"][quote="Namdung"]
           Let a, b, c are nonegative numbers not all are zero. Prove that

$ \frac {1}{3} \le \frac {a^2}{3a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b^2}{3b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c^2}{3c^2 \plus{} (a \plus{} b)^2} \le \frac {1}{2}$[/quote]
The left inequality.
$ \sum_{cyc}\frac {a^2}{3a^2 \plus{} (b \plus{} c)^2} \equal{} \sum_{cyc}\frac {a^4}{3a^4 \plus{} a^2(b \plus{} c)^2}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum(3a^4 \plus{} 2a^2b^2 \plus{} 2a^2bc)}.$
Id est, it remains to prove that
$ 3(a^2 \plus{} b^2 \plus{} c^2)^2\geq\sum_{cyc}(3a^4 \plus{} 2a^2b^2 \plus{} 2a^2bc),$ which is obvious.
For a proof of the right inequality we can use Cauchy-Schwartz too. :wink:[/quote]

My solution:
$ \sum_{cyc}\frac {a^2}{3a^2 \plus{} (b \plus{} c)^2} \ge \sum \frac{a^2}{3(a^2\plus{}b^2\plus{}c^2)}\equal{}\frac{1}{3}$
P/s:Em thưa thầy thầy gửi cho em đề NK đc hem ạ ??

Very nice proof! Thank you Dimitris X.

VBQC, what do you think about this solution? Surpricing?

[quote="Dimitris X"][quote="Namdung"]What about right inequality. I think it is harder.

@caubetoanhoc94: Em xem ở http://www.mathscope.org nhé.[/quote]
Nice!!!
The right part:
$ 2a^2 \plus{} a^2 \plus{} (b \plus{} c)^2 \ge 2a^2 \plus{} 2a(b \plus{} c) \equal{} 2a(a \plus{} b \plus{} c)$
So $ \sum \frac {a^2}{3a^2 \plus{} (b \plus{} c)^2} \le \sum\frac {a^2}{2a(a \plus{} b \plus{} c)} \equal{} \frac {1}{2}$ :P[/quote]
Very nice your proof, Dimitri X. This solution is out of my thinking. I used Cauchy Schwarz to prove the right part (different from arqady's one), but your proof is nicer. Thank you very much. :)

[hide="left "]
Since $ (x\plus{}y)^2 <3x^2 \plus{}3y^2$ we have that

$ \sum_{cyc}\frac {a^2}{3a^2 \plus{} (b \plus{} c)^2} \geq \sum_{cyc}\frac {a^2}{3a^2 \plus{}3b^2 \plus{}3c^2} \equal{}\frac{1}{3}$

 [/hide]

Nice and neat enndb0x. Thank you.

One of my students also found this shortest way. Note that equality for (a, 0, 0) and permutations.

We have:
$ \sum\frac{a^2}{a^2\plus{}k(b\plus{}c)^2} \ge$min{$ 1,\frac{3}{4k\plus{}1}$}
This result can prove easily by AM-GM or cauchy. :)

I send herewith 16 problems I suggested for Vietnamese team in Training Camp for IMO 2009. Almost of them are well-known, but methods of their solutions are usefull.

Please find the attached for details and feel free to discuss about the solutions.

[quote="enndb0x"]Thank you Namdung ,but why you don't use LaTeX.I think you can use mathlinks, just preview and save as image.I just think it would be easier for you than to write in word,however it's your choice.[/quote]

Yes, friend. I will move to Latex soon. I have some psychological difficulty only: On my time, Latex was quite complicated and this imagination is still on my mind till now :).


function
inequalities
advanced fields
advanced fields open

I need to find a closed formula for the following function (desirable, or at least a very nice bound), where all the variable involved are comples (all, $\nu, \omega, \tau$ and $\zeta$).

\[\Lambda(\omega,\zeta) = \inf_{\nu\in\mathbb{C}}\left( \frac{\max_{|\tau|=1}|\tau^2-(\omega+\nu)\tau+\omega\nu|}{|\zeta^2-(\omega+\nu)\zeta+\omega\nu|}\right)\]

Does anyone has an idea? I was able to prove that $\Lambda(\omega,\zeta)\leq\frac{|\omega|+1}{|\zeta-\omega|}$, but I can't find my scratch paper (man!!, it is time to put in order this desk!!), but the bound may not be good enouth. It would be great if $\Lambda(\omega,\zeta)\leq\frac{|\omega|}{|\zeta-\omega|}$, but I don't think it is that way!

[quote="fedja"][quote="djimenez"] I was able to prove that $\Lambda(\omega,\zeta)\leq\frac{|\omega|+1}{|\zeta+\omega|}$... It would be great if $\Lambda(\omega,\zeta)\leq\frac{|\omega|}{|\zeta+\omega|}$, [/quote]
Some $+$ should be $-$ or vice versa because, as written, you have
\[
\Lambda(\omega,\zeta)=\frac1{|\zeta-\omega|}\inf_{\nu\in\mathbb C}\max_{|\tau|=1}\frac{|\tau-\omega|\cdot|\tau-\nu|}{|\zeta-\nu|},
\]
which clearly gives you $|\zeta-\omega|$ in the denominator instead of $|\zeta+\omega|$ mentioned...
Also, as written, it is clear that $\Lambda(0,\zeta)$>0, so an estimate with pure $|\omega|$ in the numerator is impossible. But, perhaps, there is a typo in your definition of $\Lambda$  :?[/quote]

Yes, there was a $+$ where it should be a $-$ in my bound, that has a very easy proof by the way. No, there is no mistake in the definition of $\Lambda$. Thanks for the remark!  ;)

[quote="djimenez"] Yes, there was a + where it should be a - in my bound, that has a very easy proof by the way.   [/quote]
Sure, just take $\nu=\infty$. Another useful value is $\nu=0$. So, you have the simple bound $\Lambda(\omega,\zeta)\leq \frac{1+|\omega|}{|\zeta-\omega|}\min\left\{1,\frac 1{|\zeta|}\right\}$. The problem is that this bound is sharp up to a factor not exceeding $2$, i.e., $\Lambda(\omega,\zeta)\geq \frac12\frac{1+|\omega|}{|\zeta-\omega|}\min\left\{1,\frac 1{|\zeta|}\right\}$.
So, could you, please, be more specific about what you need from your bound so that it would become clear which estimates are good for you and which aren't?

[quote="fedja"]Could you, please, be more specific about what you need from your bound so that it would become clear which estimates are good for you and which aren't?[/quote]

Well, as I said, the best for me would be to have a closed formula, what I think is not easy to get. I was specting a bound smaller than the one I had. But....

[quote="fedja"]The problem is that this bound is sharp up to a factor not exceeding $2$, i.e., \[\Lambda(\omega,\zeta)\geq \frac12\frac{1+|\omega|}{|\zeta-\omega|}\min\left\{1,\frac 1{|\zeta|}\right\}\][/quote]

Could you please tell me how did you found that $2$? It would be very appreciated!

Thanks!

[quote="djimenez"]Could you please tell me how did you found that 2? It would be very appreciated! [/quote]
As I mentioned before, $\Lambda(\omega,\zeta)=\frac1{|\zeta-\omega|}\inf_{\nu\in\mathbb C}\max_{|\tau|=1}\frac{|\tau-\omega|\cdot|\tau-\nu|}{|\zeta-\nu|}$
Now, for every $\xi\in\mathbb C$, we have $|\tau-\xi|\geq \sqrt{1+|\xi|^2}$ on at least half of the unit circumference. Therefore, for every $\nu\in\mathbb C$, we can always choose $\tau$ with $|\tau|=1$ such that  $|\tau-\omega|\cdot|\tau-\nu|\geq \sqrt{1+|\omega|^2}\sqrt{1+|\nu|^2}$.
Now, $\sqrt{1+|\omega|^2}\geq \frac{1+|\omega|}{\sqrt 2}$ and $\frac{\sqrt{1+|\nu|^2}}{|\zeta-\nu|}\geq \frac{\sqrt{1+|\nu|^2}}{|\zeta|+|\nu|}\geq \frac{\sqrt{1+|\nu|^2}}{\max\{|\zeta|,1\}(1+|\nu|)}\geq \frac1{\sqrt 2}\min\left\{1,\frac 1{|\zeta|}\right\}$. Multiplying these two inequalities, we get the result.
Actually, we get a slightly better estimate from below than $1/2$ of the upper estimate. I just wanted to emphasize that your simple upper estimate couldn't be improved more than twice, which for most analysis problems I can imagine is as good as an identity. But your problem may be a subtler one, so I really wondered where the boundary between "good" and "bad" estimates is (I doubt anyone will be able to provide an exact formula for $\Lambda$). :)

[quote="fedja"]I doubt anyone will be able to provide an exact formula for $\Lambda$[/quote]

I doubt it too. Nevertheless, your posts were very helpful (at least for me :D ). Thank you very much!  ;)


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You can join this group, the COMC Veterans Club, no matter what's your COMC experience.