Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a\plus{}b\plus{}c\equal{}3.$ Prove that:
\[ \frac{6}{\sqrt[12]{abc}}\geq a^2b\plus{}a^2c\plus{}b^2a\plus{}b^2c\plus{}c^2a\plus{}c^2b\]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \frac {6}{\sqrt [12]{abc}}\geq a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b\]
[/quote]

$ \frac {6}{\sqrt [12]{abc}}\geq a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b$

$ \Longleftrightarrow F(a,b,c) \equal{} \frac {6^{12}}{abc} \minus{} (a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b)^{12}\geq 0$

substitution $ a \equal{} \frac {3x}{x \plus{} y \plus{} z},b \equal{} \frac {3y}{x \plus{} y \plus{} z},c \equal{} \frac {3z}{x \plus{} y \plus{} z},(x,y,z>0)$

applying program tsds3 to the numerator of F  the screen shows "The form is positive semidefinite" by running DS (Difference Substitution) 1 times in 0.422 seconds.

[url]http://www.irgoc.org/viewtopic.php?f=10&t=12&sid=efdc3cc0793d2579fa5b7be1bc32f704[/url]

[quote="arqady"]hedeng123, there is easy proof without computer.  :wink:[/quote]

 :D sorry. :)

[quote="kunny"]Solution by Computer shouldn't be posted in MathLinks. :wink:[/quote]

 :D sorry.

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \frac {6}{\sqrt [12]{abc}}\geq a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b\]
[/quote]

$ \Leftrightarrow 3{v}^{2}\leq 2{\frac {1}{\sqrt [4]{w}}}\plus{}{w}^{3}$

so just need to prove that inequality holds when $ b\equal{}c$.

...

[quote="kuing"]so just need to prove that inequality holds when $ b \equal{} c$.

...[/quote]I've seen solutions like this many times (mostly by arqady), but I never understand it. Can you explain in detail why is it enough to prove that inequality holds when $ b\equal{}c$?

[quote="kuing"][quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \frac {6}{\sqrt [12]{abc}}\geq a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b\]
[/quote]

$ \Leftrightarrow 3{v}^{2}\leq 2{\frac {1}{\sqrt [4]{w}}} \plus{} {w}^{3}$

so just need to prove that inequality holds when $ b \equal{} c$.

...[/quote]
Yes, it's enough! But how you prove this inequality when $ b\equal{}c$ $ ?$

@johan : see the article about uvw method created by mathias-DK

(gw udah post di olimpiade.org, lu emangnya kagak baca?)

[quote="Johan Gunardi"][quote="kuing"]so just need to prove that inequality holds when $ b = c$.

...[/quote]I've seen solutions like this many times (mostly by arqady), but I never understand it. Can you explain in detail why is it enough to prove that inequality holds when $ b = c$?[/quote]
$ a,$ $ b$ and $ c$ are real roots of the following equation:
\[ x^3 - 3ux^2 + 3v^2x - w^3 = 0\]
Let $ f(x) = x^3 - 3ux^2 + 3v^2x - w^3.$ Hence, $ f'(x) = 3(x^2 - 2ux + v^2).$
If we'll increase $ v^2$ continuously, when $ u =$constant and $ w^3 =$constant,
we can not do it arbitrary because otherwise we obtain $ f'(x) > 0,$
 which is contradiction since, for $ f$ must be three real roots.
The moment, for which not all roots of $ f$ are reals happens after moment, for which two of the roots are equal
because we can obtain only two complex roots exactly.
Thus, we can increase $ v^2$ up to moment, for which two numbers from $ \{a,b,c\}$ are equal.
The original inequality is equivalent to ${ {v}^{2}\leq \frac {1}{3}(\frac {2}{\sqrt [4]{w}}} + {w}^{3}),$ where $ u = 1.$
We'll increase $ v^2$ up to moment, for which two numbers from $ \{a,b,c\}$ are equal, when ${ \frac {1}{3}(\frac {2}{\sqrt [4]{w}}} + {w}^{3})$ is constant.
Hence, for the proof of the original inequality it remains to prove it when $ b = c.$

[quote="arqady"][quote="kuing"][quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \frac {6}{\sqrt [12]{abc}}\geq a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b\]
[/quote]

$ \Leftrightarrow 3{v}^{2}\leq 2{\frac {1}{\sqrt [4]{w}}} \plus{} {w}^{3}$

so just need to prove that inequality holds when $ b \equal{} c$.

...[/quote]
Yes, it's enough! But how you prove this inequality when $ b \equal{} c$ $ ?$[/quote]

sorry... I thought it long time but still can't prove it... :(
help... :maybe:

[quote="kuing"]
sorry... I thought it long time but still can't prove it... :(
help... :maybe:[/quote]
Use logarithms. :wink:

oh my god, befor I get a calculation mistake... :(

in fact, is so easy

when $ b \equal{} c$, we have $ b \equal{} c \equal{} \frac {3 \minus{} a}{2}$, ineq becomes

$ \frac {6}{\sqrt [12]{a\left(\frac {3 \minus{} a}{2}\right)^2}}\geq \frac {3}{4}(3 \minus{} a)(a^2 \plus{} 3)$

$ \iff a(3 \minus{} a)^{14}(a^2 \plus{} 3)^{12} \le 2^{38},$

Let $ f(a) \equal{} a(3 \minus{} a)^{14}(a^2 \plus{} 3)^{12},a\in (0,3)$, we have

$ f'(a) \equal{} 3(1 \minus{} a)(3 \minus{} a)^{13}(a^2 \plus{} 3)^{11}(13a^2 \minus{} 12a \plus{} 3),$

easy to prove that $ 13a^2 \minus{} 12a \plus{} 3 > 0$, so we get that

when $ 0 < a < 1$ then $ f'(a) > 0$, when $ 1 < a < 3$ then $ f'(a) < 0$ and $ f'(1)\equal{}0$, thus we have

$ f(a) \le f(1) \equal{} 2^{38}$,

ineq holds.

right? arqady


calculus
derivative
analytic geometry
graphing lines
slope
logarithms
function

There's no algebraic way to solve these kinds of problems (I don't think, I know there is no easy way to solve $x^{2}=2^{x}$.). Also, just graphing it, you can see there are either 0, 1, or 2 solutions. :wink:

[quote="maedko"]The problem is $2^{x}=2x$, and just by lookin at it I can tell that the solutions are $x=1,2$, but is there a way to solve it other than guess and check?  My calc teacher was stumped, and it just seems to go in circles.  I'm getting that idea that it might take some really advanced math, but at this point I'm not sure.  Any help?[/quote]

[hide]Graphing is the best thing that works.

No need to use calculus in this question.

After graphing, we see that at x=1, and x=2, are the solutions.[/hide]

[quote="13375P34K43V312"]This is calculusy, but are you allowed to take the derivative of both sides?[/quote]

that does not work. just because two graphs intersect does not mean that their slope is the same at the intersection

Take the log base 2 of both sides giving
$x=\log_{x}2x$ Now we apply the product property.
$x=\log_{x}2+1$
Which simplifies things...

i got it by graphing, but i dont really consider that solving; is there a calculus way to do it that isnt too complicated?

I suppose that it depends on what you mean by "solving." If the question is, can we find the solutions and prove that those are the only solutions?, then probably yes. The left side is an exponential function, the right side a linear one, so we know that there will be at most 2 intersections. Then we can find these by graphing, I suppose...there is a very limited area where there can be roots.

[quote="bpms"][quote="lotrgreengrapes7926"][quote="bpms"]Which simplifies things...[/quote] Please explain... :P
Also, it's $x=\log_{2}{x}+1$.[/quote]
The great graphing calculator accepts this equation and mastercard and visa. (it accepts exponents too, but I thought that logs might help)[/quote]

Why would you do all that work? Just plug in $2^{x}-2x$ and find the zeros.

[quote="lotrgreengrapes7926"][/quote][quote="bpms"][quote="lotrgreengrapes7926"][quote="bpms"]Which simplifies things...[/quote] Please explain... :P
Also, it's $x=\log_{2}{x}+1$.[/quote]
The great graphing calculator accepts this equation and mastercard and visa. (it accepts exponents too, but I thought that logs might help)[/quote][quote="lotrgreengrapes7926"]

Why would you do all that work? Just plug in $2^{x}-2x$ and find the zeros.[/quote]

I believe the point of this is to make it simple enough so that you cn easily find intersections.

Graph the line $y=x$ and $y=2^{x-1}$.  These should intersect in 2 points?  Since you have found these solutions by guess/check, you're done.

[quote="lotrgreengrapes7926"]$y=2x$ and $y=2^{x}$ are also very easy to graph. :D[/quote]

yep, thats what i used.   :P

[color=red]Only to those who know  Rolle's theorem:[/color]

 [hide]If $f(x) = 2^{x}-2x$ , then $f(1) = f(2) = 0$.If there was one more root , then $f''(x)$ would have at least one root.But this is imposible , since $f''(x) = 2^{2}\ln^{2}x >0$.
 Thus , the only roots of the equation $f(x) =0$ are $0$ and $1$[/hide]
[u] Babis[/u]


geometry
incenter
circumcircle
trapezoid
geometry unsolved

We have isosceles triangle $ ABC$ in which $ AB\equal{}AC$, and $ M$ is the midpoint of $ BC$. Let the point $ I$ be the incenter of $ ABC$. Point $ P$ is a point inside the triangle $ ABC$, and $ P$ lies on the circumcircle of $ IBC$. Prove that $ \angle BPM\plus{}\angle CPA\equal{}180^\circ$

Extend AP and meet circle at point O ,join BO,CO; then ,we get :∠BOP=∠MCP ;If we can prove that ∠BPO=∠MPC;  then we can get the answer; so the mission is just to prove that △PBO∽△PMC；because we‘ve proved that ∠BOP=∠MCP; so we only need “BO/OP=MC/PC” That is to prove :BO*CP=OP*CM; Because M is the midpoint of BC; so we only need to prove: BO*CP=OP*BC/2;Obviously,we need to use Ptolemy theorem: PB*OC+BO*PC=BC*PO;  so , we only need to prove that :BP*OC=BO*PC; That’s :BP/BO=PC/OC; ∠BOP=∠BCP=∠ICB-∠ICP=∠IBA-∠IBP=∠PBA; So:△APB∽△ABO; Use the same way to prove △APC∽△ACO; We get PB/BO=BA/AO=AC/AO=PC/OC; That’s PB/BO=PC/CO→  PB*CO=PC*BO; Then we get it..  :blush:  :lol:  :lol:

It is obvious that PBQC is a harmonic quadrilateral
Thus QBCQ' is an isosceles trapezoid
done

CCMath 1,very nice solution.
I don''t understand from plane geometry's solution why from the fact that PQBC is harmonic,we conclude that QBCQ' is an isosceles trapezoid.

this is the property of harmonic quadrilateral

Could anyone write the correct formulation of that property and give a proof for that.


Solve in $ \mathbb{N} \times \mathbb{N} \times \mathbb{N}$ the follow equation:
\[ x^6\plus{}x^3y^3\plus{}y^6\equal{}3^n\]

[quote="Mîndrilă Claudiu"]Solve in $ \mathbb{N} \times \mathbb{N} \times \mathbb{N}$ the follow equation:
\[ x^6 \plus{} x^3y^3 \plus{} y^6 \equal{} 3^n
\]
[/quote]
The first step ,we will solve this problem in the case : $ \gcd(x,y)\equal{}1$ .It is clear that both of x,y doesn't divide by $ 3$
From equation we have $ (x^3\minus{}y^3)^2\plus{}3x^3.y^3\equal{}3^n$ ,therefore $ 3|x^3\minus{}y^3$ . By Fermat little theorem it is easy to check that  $ 3|x\minus{}y$ ,therefore $ 9|x^3\minus{}y^3$ .
If $ n>1$ then $ (x^3\minus{}y^3)^2\minus{}3^n$ divide by 9 but $ 3y^3x^3$ not . It means that $ n$ must be equal 1 . It clear that there exist only $ (x,y)\equal{}(1,1)$ satisfy condition .
Now consider our problem . Let $ d\equal{}\gcd(x,y)$ ,because $ d|3^n$ so it has form $ 3^t$ .Let
$ x\equal{}dx',y\equal{}dy'$ where $ \gcd(x',y')\equal{}1$ . The equation equivalent :
\[ x'^6\plus{}x'^3.y'^3\plus{}y'^6\equal{}3^{n\minus{}6t}
\]
From above result we have $ x'\equal{}y'\equal{}1$ and $ n\equal{}6t\plus{}1$ . This is root of the equation .

[quote="Mîndrilă Claudiu"]Solve in $ \mathbb{N} \times \mathbb{N} \times \mathbb{N}$ the follow equation:
\[ x^6 \plus{} x^3y^3 \plus{} y^6 \equal{} 3^n
\]
[/quote]

What does N x N x N , mean ...?

[quote="TTsphn"][quote="Mîndrilă Claudiu"]Solve in $ \mathbb{N} \times \mathbb{N} \times \mathbb{N}$ the follow equation:
\[ x^6 \plus{} x^3y^3 \plus{} y^6 \equal{} 3^n
\]
[/quote]
The first step ,we will solve this problem in the case : $ \gcd(x,y) \equal{} 1$ .It is clear that both of x,y doesn't divide by $ 3$
From equation we have $ (x^3 \minus{} y^3)^2 \plus{} 3x^3.y^3 \equal{} 3^n$ ,therefore $ 3|x^3 \minus{} y^3$ . By Fermat little theorem it is easy to check that  $ 3|x \minus{} y$ ,therefore $ 9|x^3 \minus{} y^3$ .
If $ n > 1$ then $ (x^3 \minus{} y^3)^2 \minus{} 3^n$ divide by 9 but $ 3y^3x^3$ not . It means that $ n$ must be equal 1 . It clear that there exist only $ (x,y) \equal{} (1,1)$ satisfy condition .
Now consider our problem . Let $ d \equal{} \gcd(x,y)$ ,because $ d|3^n$ so it has form $ 3^t$ .Let
$ x \equal{} dx',y \equal{} dy'$ where $ \gcd(x',y') \equal{} 1$ . The equation equivalent :
\[ x'^6 \plus{} x'^3.y'^3 \plus{} y'^6 \equal{} 3^{n \minus{} 6t}
\]
From above result we have $ x' \equal{} y' \equal{} 1$ and $ n \equal{} 6t \plus{} 1$ . This is root of the equation .[/quote]
Nice solution!  :coolspeak:


A bicycle with 60 cm diameter wheels travels 2 km. How many complete revolutions will the wheel
make?
A) 94 B) 106 C) 1061 D) 1062 E) 2122

The number of revolutions is given by $ d\pi$. Thus, in $ 2$km the wheel will make $ \frac{200000}{60\pi}\approx1061$ revolutions. So the answer is C.

Ok, the circumference of the wheel is 60pi cm. To find the amount of revolutions, we do distance/circumference. The distance in cm is 200000, so $ \dfrac{200000}{60\pi}\approx 1061$, so we have $ \boxed{C}$

Edit: Beat.


For some integers n, the numbers
$tan(n), tan(n^2), tan(n^3), tan(n^4) ...$ reach a point where the next one is equal to the previous one.  (in degrees, not radians)

A trivial example is n=1, where $tan(1)=tan(1^2)=tan(1^3)=.0174...$

Another example is n=6, where $tan(6^2)=tan(6^3)=tan(6^4)=.7265...$

Question: Find all numbers less than 180 that have this property, and that do not equal 0 [i.e. tan(60^2)=tan(60^3)=0 does not count]

Bonus:
Some numbers go through a chain instead.  For example:
$tan(40)=tan(40^4)=tan(40^7)=.839$
$tan(40^2)=tan(40^5)=tan(40^8)=-.364$
$tan(40^3)=tan(40^6)=tan(40^9)=-5.671$

Can you come up with a general way to identify which numbers have a chain, and how 'long' their chain is?  For example, with n=6, the chain has length 1.  For n=40, the chain has length 3.

When i say 180 or 360, i mean 180 deg. or 360 deg.  respectively.

-

Modulo $180$, $tan(x) = tan(y)$ whenever $x=y$.

Suppose $tan(kx) = tan(x)$.  Then modulo $180$, $kx = x$.

Conversely, suppose $kx = x$ modulo $180$.  Then $kx = 180r + x$ for some integer $r$ and $tan(kx) = tan(180r + x) = tan(x)$.

So $tan(kx) = tan(x) \Leftrightarrow kx=x$ (modulo $180$).    $(*)$
---

We solve a general problem: given $x$, what $k$ will have the property $tan(kx) = tan(x)$?

From (*), we only need to find $k$ such that $kx = x$.

We restrict ourselves to the range $k \in [0,180]$.  This is because if $kx = x$ modulo $180$, then clearly $(k+180r)x = kx = x$ modulo $180$, for all integers $r$.

Obviously, $k = 1$ is a solution.  If $x | 180m$, then $180m/x = k-1$ is an integer that works, for any integer $m$.

Conversely, suppose $k$ is an integer that works.  Then $(k-1)x = 0$ modulo $180$, so that $(k-1)x = 180m$ for some integer $m$, whereby $k = 180m/x + 1$.

Otherwise, $x$ does not divide $180$, and $kx$ goes through a permutation of the modular residues of 180 before meeting the residue $x$ again when $k$ is 181, so that the only solution in our range is $k = 1$.

Thus, we have given solutions for $k$ in both cases, where we can add or subtract $180$ freely to those solutions.  These are all the solutions because we have made sure that all the important statements are iff.

Your solution to the general question is correct, but can you use it to solve the particular question?

The bonus question is kind of easy: all numbers will eventually go through a chain, since there are only a finite number of mod classes 0-179.

I thought my solution to the general question made it obvious on how to proceed with the specific questions.

In the first case, set x = a^b and k = a.

In the second cae, set x = a^b and k = a^c.

Your general equation is:
$kx=x$ (mod 180)
and setting x=a^b and k=a gives
$a^{b+1}=a^b$ (mod 180).

The question is: for what values of a does this equation hold for all b after a certain threshold?

I don't see where your general equation solves this.  It is one thing to pen an equation but another to actually compute the answer.  In the general equation, if it is to hold for all x then k=1+180n.  But this question is different.  It asks for specific numbers: what numbers work?  (There are 15 of them less than 180.)


Let n be a positive integer. Which one is larger: n^(n+1) or (n+1)^n?

[hide="Short answer"] $ \frac{(n+1)^{n}}{n^{n+1}}= \frac{1}{n}\left( 1+\frac{1}{n}\right)^{n}$

As $ n$ gets large (but in fact we can take $ n \ge 3 > e$) the right factor clearly tends to $ e$ (from below!) while the left factor tends to $ 0$.  Hence $ n^{n+1}\gg (n+1)^{n}$ (or we could simply say that $ n \ge 3 \implies n^{n+1}> (n+1)^{n}$ ). [/hide]
[hide="Long answer"] The general question of when $ x^{y}= y^{x}$ can be solved by casework on the equivalent expression

$ \frac{\ln x}{x}= \frac{\ln y}{y}$

In particular, this function $ f(x)$ attains its maximum at $ x = e$. [/hide]


number theory
greatest common divisor
number theory unsolved

It is easy to see that $\displaystyle\sum_{m=0}^{a-1}e^{\frac{2\pi ibm}{a}}=0$ if $a\nmid bm$ and $\displaystyle\sum_{m=0}^{a-1}e^{\frac{2\pi ibm}{a}}=a$ if $a\mid bm$. So we only need to prove that there are exactly $gcd(a,b)=d$ numbers $m$ such that $0\leq m\leq a-1$ and $a\mid bm$. Since $a=du$ and $b=dv$, where $gcd(u,v)=1$, we must have $u\mid m$ and the result follows


AMC
AIME
Putnam
USA(J)MO
USAMO
trigonometry
AMC 12

So who here thinks he/she has a decent or semi-decent chance at making MOP this year if it expands to accomodate 60 people?

thurrs a small chance i might be able to if its 60 people, maybe if i study and practice like a jimdandy (whatever that means) from now until March or whenever AMC and AIME starts, if you know what im talking aboot.

Now, if only I could go back in time . . .

The Putnam should be fun, though.

assuming semi-decent means 1/2, ill give myself a (semi)^8 shot at mop  :D . Considering ive improved very much since last year and will continue doing so, i think its possible.

if only they expanded it in some other year. sigh. hopefully they do it again before i graduate

Now that I'm in Richard's class, I think I have a decent chance, especially if they plan on expanding the number of eligible people.  The possibility still exists that they might reject me on the grounds of citizenship.... which would really suck ass... *_*

Hmm.

Well I'm fairly certain I'll make USAMO. The only question is whether I'll do well on it, because it'll probably be my first essay-proof type test. If I practice insanely hard on those, I have about a 50% chance. But I'm only a freshman, so I probably won't make it.

Maybe some other year though.

like 1/1000000... i'll make it... and i think Mystic can make it w/o a prob.

yay for mr. southpaw show em what youve got

same for everyone else

i dunno, im a freshman this year so the chances are low, but im fairly confident i can make USAMO (as long as the USAMO index is heavily weighted on AMC-12  :P) but proofs... im not so good at those, and i probably wont practice those anytime soon, so maybe in the next 3 years  :cry:

you guys make it sound like it's really easy to make USAMO...

I was under the impression that if you understood the concepts introduced in AOPS1 and 2, it's all you need to qualify.

well, qualification for USAMO for people that took the AMC-10 is purely based on AIME scores so about 7-9 points is usually the floor value... if i practice at least some until the test, i think i can make it... i don't know about other people, but for me it doesn't seem extremely hard

[quote]well, qualification for USAMO for people that took the AMC-10 is purely based on AIME scores so about 7-9 points is usually the floor value[/quote]

i believe that no longer applies just to those who take amc 10, but to all who are in 10th grade and below.

[quote]How can that be true? Only about 250 people a year can qualify, and there are a lot more people reading the books.[/quote]

A lot of people read the books, but few truly know their content very well. i think someone who truly could do every problem in aops is set for making usamo.

[quote]I think even less of AoPS is required. Last year, when I qualified for USAMO, I don't think I'd even heard of the Art of Problem Solving books. Maybe I had heard of them...but I didn't have them, that's for sure.[/quote]

Eh, I would say that to make the cutoff index *comfortably*, you would definitely need to know most of AoPS 2.

i dont think knowing the material in AOPS 2 implies that you will make USAMO.  After all, it's not about how much math you know, it's about how well you can solve problems that you haven't seen before (most likely).  That's why Zeitz is probablly a good book to go with after AOPS.  And i dont know if that guarantees USAMO qualification, either.

[quote]i dont think knowing the material in AOPS 2 implies that you will make USAMO. [/quote]

I never said that, though I did imply that not knowing most of the material in AoPS 2 is a huge drawback to your chances of making USAMO. For example, not knowing Viete's Theorem, Heron's Formula, and some basic trig identities could make some of the problems pretty tough...

I've had AoPS books for a while, but I haven't gone through them completely yet...actually, only about half way through. :cry:

-interesting_move

I don't have the AoPS problems and I heard of them in around September, but I think I could've been a lot better if I had those books.

[quote="mathfanatic"][quote="Fierytycoon"]For example, not knowing Viete's Theorem, Heron's Formula, and some basic trig identities could make some of the problems pretty tough...[/quote]

Hm...I knew Heron's formula from Algebra II/Trig class, and was sort of familiar with trig identities, but I didn't know (and still don't know!) what Viete's Theorem is.  Sigh.  I have yet to read AoPS2, or even the second half of AoPS1, for that matter.[/quote]
Huh, same here. I wonder why you're so much better at math than me then...hmm...

USAMO INDEX = 10 * AIME + AMC12

AMC 12 is not that heavily weighted, ur gonna need about a 140 / 9-10 AMC 12 / AIME score combo to safely get in... (assuming your not getting a 150 and you make one silly mistake )

Hmm, I'm a former mathematician who has never heard of Viete's theorem.  Does it have something to do with the roots of a polynomial?  Perhaps I know the result but not the name.

Thanks.  I didn't know that result had a name.

So is the an through a0 some sort of a series or ar they just numbers? Also sum(ri) and prod(ri) is the sum/product of all of the roots right?

[quote="Tare"]So is the an through a0 some sort of a series or ar they just numbers?[/quote]

a_k is the coefficient of the term x^k.

[quote]Also sum(ri) and prod(ri) is the sum/product of all of the roots right?[/quote]

Yeah.

[quote="ComplexZeta"]Yeah. If you have a polynomial

anxn+an-1xn-1+...+a0

with roots r1,...,rn, then

sum(ri)=-an-1/an
...
prod(ri)=(-1)^n*a0/an.

It's just what you get when you multiply out the factorization and then equate coefficients.[/quote]

Oh...that's...it?  I could have guessed that.

However, it looks very useful.  \[memorizeit! \]memorizeit! \[memorizeit!  !end-memorize\]

Happiness.


Suppose $ a_1,a_2,...,a_n$ are positive real numbers which satisfy $ a_1a_2...a_n\equal{}1$

Define :
$ b_r\equal{}\frac{\prod_{j\equal{}1}^{n}(a_j\plus{}k)}{(a_r\plus{}k)}$

Prove that :

$ \sum_{i\equal{}1}^{n} \frac{a_i^n}{b_i} \ge \frac{n}{(1\plus{}k)^{n\minus{}1}}$

when $ k>0$

[quote="Ronald Widjojo"]Suppose $ a_1,a_2,...,a_n$ are positive real numbers which satisfy $ a_1a_2...a_n \equal{} 1$

Define :
$ b_r \equal{} \frac {\prod_{j \equal{} 1}^{n}(a_j \plus{} k)}{(a_r \plus{} k)}$

Prove that :

$ \sum_{i \equal{} 1}^{n} \frac {a_i^n}{b_i} \ge \frac {n}{(1 \plus{} k)^{n \minus{} 1}}$

when $ k > 0$[/quote]

$ t \equal{} \sum_{i \equal{} 1}^{n} {a_i} \ge n$

by Chebyshev :  $ \sum_{i \equal{} 1}^{n} \frac {a_i^n}{b_i} \ge \frac{1}{n}(\sum_{i \equal{} 1}^{n} {a_i^n}) ( \sum_{r\equal{}1}^{n}\frac{a_r\plus{}k}{\prod_{j \equal{} 1}^{n}(a_j \plus{} k)}) \ge \frac{1}{n} \frac{t^n}{n^{n\minus{}1}} \frac{t\plus{}nk}{(t\plus{}nk)^n}n^n$ $ \equal{}\frac{t^n}{(t\plus{}nk)^{n\minus{}1}} \ge \frac{n}{(1\plus{}k)^{n\minus{}1}}$


let $ a,b,c\geq 0$. Prove that
$ a(a \minus{} b)(a \minus{} c) \plus{} b(b \minus{} c)(b \minus{} a) \plus{} c(c \minus{} a)(c \minus{} b)\geq \frac {4abc(a^2 \minus{} b^2)^2}{(a \plus{} b \plus{} c)(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

Reference: [url]http://vnineqmath.hnsv.com/viewtopic.php?f=75&t=235&sid=d161647e240518b0b91d521b5eeda098[/url]

[quote="litbon"]let $ a,b,c\geq 0$. Prove that
$ a(a \minus{} b)(a \minus{} c) \plus{} b(b \minus{} c)(b \minus{} a) \plus{} c(c \minus{} a)(c \minus{} b)\geq \frac {4abc(a^2 \minus{} b^2)}{(a \plus{} b \plus{} c)(a \plus{} b)(b \plus{} c)(c \plus{} a)}$[/quote]

It's wrong :P.
$ a(a \minus{} b)(a \minus{} c) \plus{} b(b \minus{} c)(b \minus{} a) \plus{} c(c \minus{} a)(c \minus{} b)\geq \frac {4abc(a^2 \minus{} b^2)^2}{(a \plus{} b \plus{} c)(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

[quote="litbon"][quote="Toan_VN_LC"][quote="litbon"]let $ a,b,c\geq 0$. Prove that
$ a(a \minus{} b)(a \minus{} c) \plus{} b(b \minus{} c)(b \minus{} a) \plus{} c(c \minus{} a)(c \minus{} b)\geq \frac {4abc(a^2 \minus{} b^2)}{(a \plus{} b \plus{} c)(a \plus{} b)(b \plus{} c)(c \plus{} a)}$[/quote]
it's must be:
$ a(a \minus{} b)(a \minus{} c) \plus{} b(b \minus{} c)(b \minus{} a) \plus{} c(c \minus{} a)(c \minus{} b)\geq \frac {4abc(a \minus{} b)^2}{(a \plus{} b \plus{} c)(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
 :lol:
I think we can kill it by SOS[/quote]
I'm sorry, I had edit
It will be kill by Cauchy-Schwarz[/quote]

I would like to see your nice solution...Could you share it here?

[quote="Jianxing113725"][/quote][quote="litbon"][quote="Toan_VN_LC"][quote="litbon"]let $ a,b,c\geq 0$. Prove that
$ a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b)\geq \frac {4abc(a^2 - b^2)}{(a + b + c)(a + b)(b + c)(c + a)}$[/quote]
it's must be:
$ a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b)\geq \frac {4abc(a - b)^2}{(a + b + c)(a + b)(b + c)(c + a)}$
 :lol:
I think we can kill it by SOS[/quote]
I'm sorry, I had edit
It will be kill by Cauchy-Schwarz[/quote][quote="Jianxing113725"]

I would like to see your nice solution...Could you share it here?[/quote]
because of I have a solution for I. Schur by SOS but $ S_a, S_b, S_c >0$ so from it I can make stronger than
Can you try?

[quote="Jianxing113725"][/quote][quote="litbon"][quote="Toan_VN_LC"][quote="litbon"]let $ a,b,c\geq 0$. Prove that
$ a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b)\geq \frac {4abc(a^2 - b^2)}{(a + b + c)(a + b)(b + c)(c + a)}$[/quote]
it's must be:
$ a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b)\geq \frac {4abc(a - b)^2}{(a + b + c)(a + b)(b + c)(c + a)}$
 :lol:
I think we can kill it by SOS[/quote]
I'm sorry, I had edit
It will be kill by Cauchy-Schwarz[/quote][quote="Jianxing113725"]

I would like to see your nice solution...Could you share it here?[/quote]
ok, that is my solution

[quote="litbon"][/quote][quote="Jianxing113725"][/quote][quote="litbon"][quote="Toan_VN_LC"][quote="litbon"]let $ a,b,c\geq 0$. Prove that
$ a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b)\geq \frac {4abc(a^2 - b^2)}{(a + b + c)(a + b)(b + c)(c + a)}$[/quote]
it's must be:
$ a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b)\geq \frac {4abc(a - b)^2}{(a + b + c)(a + b)(b + c)(c + a)}$
 :lol:
I think we can kill it by SOS[/quote]
I'm sorry, I had edit
It will be kill by Cauchy-Schwarz[/quote][quote="Jianxing113725"]

I would like to see your nice solution...Could you share it here?[/quote][quote="litbon"]
ok, that is my solution[/quote]

Wa..Great :).Very nice;libton  :P


Find all functions F such that the tangent to the graph of F at the point (x,y) intersects the x-axis at (x+2,0)

I got my answers  F(x)=Ce^(-x/2) and F(x)=-Ce^(-x/2)..

Please help me

Consider $XY$ plane, the tangent line at the point $(x,\ f(x))$ on the desired function $Y=f(X)$ is $Y-f(x)=f'(x)(X-x).$ Since the point $(x+2,0)$ lie on this tangent line, we have $-f(x)=2f'(x)\Longleftrightarrow f'(x)=-\frac{1}{2}f(x).$
Solve this differential equation, yielding $f(x)=Ce^{-\frac{1}{2}}^x.$

thankx so  ur answer is one of my answers .. could the negative one be right also.?


algebra
polynomial
number theory
relatively prime
number theory unsolved

$x$ and $y$ are positive and relatively prime and $z$ is an integer. They satisfy $(5z-4x)(5z-4y)=25xy$. Show that at least one of $10z+x+y$ or quotient of this number divided by $3$ is a square number (i.e. prove that $10z+x+y$ or integer part of $\frac{10z+x+y}{3}$ is a square number).

This problem was the easiest problem of 18th KMO.

Using Determinant of square polynomial for $z$, It will be easily proved  :D .

$(10z+x+y)(9x+9y-10z)=9(x-y)^{2}$

and enough to check the prime which divide all of $10z+x+y$ and $9x+9y-10z$
(but, 2 must be separately checked )


[quote="Adalbert"]
$2\sum_{cyc}{x^{2}y^{2}(x^{2}+y^{2})}\geq 4\sum_{cyc}{x^{3}}\geq 2\sum_{cyc}{x^{3}}+\sum_{cyc}{xy(x+y)}$
[/quote]
I think it's wrong. $x,$ $y$ and $z$ are real numbers. :wink:
But SOS kills it enough easy.

Basically are you saying that Muirhead type inequalities which are proved as sums of AM-GM's aren't valid for all reals? I know that AM-GM requires positives, but is the result truly not true for all reals?

(i.e. the inequality $\sum a^{3}b \geq  \sum a^{2}bc$ is clearly true for positive reals, but is it not necessarily true for negatives?)

[quote="me@home"]
(i.e. the inequality $\sum a^{3}b \geq \sum a^{2}bc$ is clearly true for positive reals, but is it not necessarily true for negatives?)[/quote]
Try $a=-2$ and $b=c=1$ for $\sum_{sym}a^{3}b \geq \sum_{sym}a^{2}bc$
[quote="chien than"]For a;b;c are real numbers.Prove that:
$3(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2}) \geq (ab+bc)^{3}+(bc+ca)^{3}+(ca+ab)^{3}$[/quote]
$3(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2}) \geq (ab+bc)^{3}+(bc+ca)^{3}+(ca+ab)^{3}\Leftrightarrow$
$\Leftrightarrow\sum_{sym}(3a^{4}b^{2}-a^{3}b^{3}-3a^{3}b^{2}c+a^{2}b^{2}c^{2})\geq0.$
But $\sum_{sym}(3a^{4}b^{2}-a^{3}b^{3}-6a^{3}b^{2}c+4a^{2}b^{2}c^{2})\geq0$ true too for all $a,$ $b$ and $c.$ :wink:

yes, i didn't read carefully...

but it was a nice idea...:D

edit: but i proved this ineqality:
$3\sum_{cyc}|x^{2}| |y^{2}| (|x^{2}|+|y^{2}|)+6 \geq 2\sum_{cyc}|x^{3}|+3\sum_{cyc}|x| |y| (|x|+|y|)$ and because of all terms on $LHS$ are positive we can remove absolute and we have:
$3\sum_{cyc}x^{2}y^{2}(x^{2}+y^{2})+6 \geq 2\sum_{cyc}|x^{3}|+3\sum_{cyc}|x| |y| (|x|+|y|) \geq 2\sum_{cyc}x^{3}+3\sum_{cyc}xy(x+y)$

i hope this is true...


Nonnegative real numbers $ a,b,c$ satisfies $ a \plus{} b \plus{} c \equal{} 2$. Prove that

$ \frac {b}{a^2 \plus{} 1} \plus{} \frac {c}{b^2 \plus{} 1} \plus{} \frac {a}{c^2 \plus{} 1} \ge \frac {18}{13}$

Here is my proof.

[hide]As

$ 36(a^2 \plus{} 1) \equal{} 9a^2 \plus{} 9a^2 \plus{} 9a^2 \plus{} 9a^2 \plus{} 4 \plus{} 4 \plus{} 4 \plus{} 4 \plus{} 4 \plus{} 4 \plus{} 4 \plus{} 4 \plus{} 4 \ge 13\sqrt [13]{9^4 4^9 a^8}$

We get

$ \frac {b}{a^2 \plus{} 1} \equal{} b \minus{} \frac {a^2b}{a^2 \plus{} 1} \ge b \minus{} \frac {36a^2b}{13\sqrt [13]{9^4 4^9 a^8}}$

There are two other simlar inequality. Add them up we get

$ \frac {b}{a^2 \plus{} 1} \plus{} \frac {c}{b^2 \plus{} 1} \plus{} \frac {a}{c^2 \plus{} 1} \ge 2 \minus{} \frac {36}{13\sqrt [13]{9^4 4^9}}(a^{\frac {18}{13}}b \plus{} b^{\frac {18}{13}}c \plus{} c^{\frac {18}{13}}a)$

It suffices to show that

$ a^{\frac {18}{13}}b \plus{} b^{\frac {18}{13}}c \plus{} c^{\frac {18}{13}}a \le \frac {2^{\frac {31}{13}}}{3^{\frac {18}{13}}}$

Notice that

$ (a^{\frac {18}{13}}b \plus{} b^{\frac {18}{13}}c \plus{} c^{\frac {18}{13}}a)^{13} \le (b \plus{} c \plus{} a)(a^{\frac {3}{2}}b \plus{} b^{\frac {3}{2}}c \plus{} c^{\frac {3}{2}}a)^{12}$

It suffices to show that

$ a^{\frac {3}{2}}b \plus{} b^{\frac {3}{2}}c \plus{} c^{\frac {3}{2}}a \le \sqrt {\frac {32}{27}}$

But this is a very hard inequality.[/hide]

So does anyone have a better proof?

[quote="shalex"]...It suffices to show that

$ a^{\frac {3}{2}}b \plus{} b^{\frac {3}{2}}c \plus{} c^{\frac {3}{2}}a \le \sqrt {\frac {32}{27}}$

But this is a very hard inequality.
[/quote]
I think your proof is nice, and for this inequality, [url=http://www.mathlinks.ro/viewtopic.php?t=193697]See my solution here[/url] :)


ratio
geometry
3D geometry
prism
trapezoid

Congruent wedges A and B are sliced as shown. Each wedge has a pair of parallel triangular faces. Wedge A is cut vertically so that slices are made perpendicular to the base; wedge B is cut horizontally so that slices are made parallel to the base. Six slices $ \frac14$ inch thick are cut from each wedge. What is the ratio of the remaining volume of wedge A to the remaining volume of wedge B? Express your answer as a common fraction.

[asy]unitsize(.4cm);
pair A,B,C,D,E,F;
A = (0,0); B = (6,0); D = (5,6); C = B+D; E = (11,3); F = (13,0);
path p = B--A--D--C--E--B--C;
draw(p); draw(shift(F)*p);
for(int i = 1; i < 7; ++i){
draw((i*B)/10 -- (i*C+(10-i)*D)/10,dashdotted);
path P = (10-i)*D/10--(i*B+(10-i)*C)/10--(i*E+(10-i)*C)/10;
draw(shift(F)*P,dashdotted);
}
label("12''",A--B,S); label("12''",A+F--B+F,S);
label("10''",E--B,SE); label("10''",E+F--B+F,SE);
label("6''",E--C); label("6''",E+F--C+F);
label("Wedge A", A-(0,1)--B-(0,1),SE);
label("Wedge B", A+F-(0,1)--B+F-(0,1),SE);[/asy]

I don't understand your solution...it seems to me like:

the volume left in wedge A is 21/2*30 = 315 and the volume left in wedge B is 360 - 360/4^2 = 675/2 so the answer would be 315/(675/2) = 14/15.

[quote="dragon96"]
$ 6\cdot \frac{1}{4}\equal{}\frac{3}{2}$

Amount left is $ A\equal{}12 \minus{} \frac{3}{2} \equal{} \frac{21}{2}$, so the area left is $ 60 \cdot \frac{21}{2} \equal{} 630$
Amount left in $ B\equal{}6 \minus{} \frac{3}{2}\equal{} \frac{9}{2}$, so the area left is $ 120 \cdot \frac{9}{2} \equal{} 540$

ratio of $ \frac{A}{B} \equal{} \frac{630}{540}\equal{}\boxed{\frac{7}{6}}$
[/quote]

[quote="FibonacciFan"]
the volume left in wedge A is $ \frac{21}{2}\cdot 30 \equal{} 315$ and the volume left in wedge B is $ 360 \minus{} \frac{360}{4^2} \equal{} \frac{675}{2}$ so the answer would be $ \frac{315}{\frac{675}{2}} \equal{} \boxed{\frac{14}{15}}$.
[/quote]

When wedge A is sliced, what is left is a triangular prism, where the base is a right triangle with legs 6 and 10, and the "height" to the base is $ 12\minus{}6\cdot \frac{1}{4}\equal{}\frac{21}{2}$.  The area of the base is $ \frac{6\cdot 10}{2}\equal{}30$, so the volume of wedge A is $ \frac{21}{2}\cdot 30\equal{}315$

When wedge B is sliced, we get a trapezoidal prism (?).  The "height" to the trapezoidal base is still 12 in this case.  For the trapezoid, we are given that the height of it is $ 6\minus{}6\cdot\frac{1}{4}\equal{}\frac{9}{2}$.  Also, the larger base has length 10.  To find the other base of the trapezoid, we use similar triangles.  If we let that base be $ x$, then \[ \frac{\frac{3}{2}}{x}\equal{}\frac{6}{10}\Rightarrow x\equal{}\frac{5}{2}\]  Hence, the area of the trapezoid is $ \frac{9}{2}\cdot \frac{10\plus{}\frac{5}{2}}{2}\equal{}\frac{225}{8}$ and the volume of the prism is $ 12\cdot \frac{225}{8}\equal{}\frac{675}{2}$

The desired ratio is then $ \frac{315}{\frac{675}{2}}\equal{}\boxed{\frac{14}{15}}$, like FibonacciFan said


Wow, nice. I never would have thought of that...


Let F be the first digit and L be the last digit of the expanded integer form of $ 2007^3$. What is the value of F + L?

2^3=8=F

the pattern for the powers of 7 is 7,9,3,1
the third one is 3=L

8+3=11


You buy a new car that comes with five new tires, four one each wheel and one spare tire in the trunk. You decide to be economical by using the spare tire as much as the other four tires. If you drive 68,000 miles, what will be the wear in miles on each tire?

ive seen at least 5 problems like this :winner_third:  :wow:

I posted the same thing (except different miles) a week ago.

[hide]$68000\cdot\frac{4}{5}=54400\ \text{miles}$[/hide]

[hide]
because there are 5 tires, and 4 wheels at a time:
$68000\times\frac{4}{5}=54400$[/hide]

Isn't it kinda weird when you have two users, almost the [b]exact[/b] same username, both of them spamming?

Sorry
the other one's my brother -_-
He was using my old account


inequalities
integration
probability and stats

Let $ B_t$ be a brownian motion started at $ 0$.  For $ a<0<b$ and $ a\not\equal{}\minus{}b$, define $ U\equal{}\inf\{t: B_t\not\in (a,b)\}$ and $ B_U$ be the brownian motion stopped at $ U$.

We are to use the cauchy schwarz inequality to estimate $ E(UB^2_U)$ (presumably this is just the bound $ E(UB^2_U) \leq (EU^2)^{1/2}(EB^4_U)^{1/2}$), but then we are supposed to conclude that $ EU^2 \leq CEB^4_U$ where $ C$ is independent of $ a$ and $ b$.  I don't see how this is done (especially since $ EU^2$ and $ EB^4_U$ are on the 'same side' in the bound above)

maybe like this:
let's define $ f(x,t)\equal{}2tx^2\minus{}t^2$, that satisfies $ \frac{1}{2}f_{x,x}(x,t) \plus{} f_t(x,t) \equal{} 2x^2 \geq 0$. The itô formula leads to:
$ 2 \min(U,t) B_{\min(U,t)}^2  \minus{} \min(U,t)^2 \equal{} \int_0^{\min{U,t}} f_x(B_s, s) dB_s\plus{} 2 \int_0^{\min{U,t}} B_s^2 ds$
because $ f_x(B_s, s)$ is bounded for $ s \leq U$, $ E[\int_0^{\min{U,t}} f_x(B_s, s) dB_s] \equal{} 0$. Hence
$ E[2 \min(U,t) B_{\min(U,t)}^2  \minus{} \min(U,t)^2] \geq 0$
which leads (Cauchy-Schwarz etc..) to
$ E[ \min(U,t)^2] \leq 4 E[B_{\min(U,t)}^4]$
hence the conclusion by letting $ t$ tends to $ \infty$. ($ U < \plus{} \infty$ a.e.)

I got this, finally,

We use cauchy-schwarz and the fact that $ t\to B^4_t \minus{} 6tB^2_t \plus{} 3t^2$ is a martingale.  this means (after some work) that $ 3EU^2\leq 3EU^2 \plus{} EB^4_U \equal{} 6EUB_U^2$, which is enough to get the result (with the same constant $ 4$ as alekk)


[quote="GameBot"]Human beings have discovered an inhabitable planet and soon after, they find 10 more inhabitable planets. Of these 11, only 5 are deemed ``Earth-like'' in its resources and the rest are deemed ``Mars-like'' since they lack many important resources. Assume that planets like Earth take up 2 units of colonization, while those like Mars take up only 1. If humanity can mobilize 12 total units of colonies, how many different combinations of planets can be occupied if the planets are all different from each other?[/quote]

[hide="Solution"]The ways to split the colonies, with xEyM meaning x double units on Earth-like planets and y single units on Mars-like ones, are as follows:

5E2M

4E4M

3E6M

Thus, the answer is $ \dbinom{5}{5}\dbinom{6}{2}\plus{}\dbinom{5}{4}\dbinom{6}{4}\plus{}\dbinom{5}{3}\dbinom{6}{6}\equal{}1(15)\plus{}5(15)\plus{}10(1)\equal{}15\plus{}75\plus{}10\equal{}\boxed{100}$[/hide]

[quote=Ellipsoid]May be I'm not understanding the problem correctly. Why is a combination of "9 colonies in earth-like planet and 3 colonies in Mars-like planet" not a valid configuration.[/quote]

9 is not divisible by 2. This would mean you used 4.5 earth-like planets, which is totally wrong.

[hide=Solution]
Let $x$ be the number of planets like Earth that have been occupied; and let $y$ be the number of planets like Mars that have been occupied. "Translating" the problem into equations and inequalities, we have
$$2x+y=12$$
$$x \le 5$$
and
$$y \le 6.$$
We can now do casework on the number of Earth-like planets occupied:[list=1]
[*] When $x=5$, $y=2$, and there are $\binom{5}{5} \binom{6}{2}=15$ ways to choose the occupied planets.
[*] When $x=4$, $y=4$, and there are $\binom{5}{4}\binom{6}{4}=75$ ways to choose the occupied planets.
[*] When $x=3$, $y=6$, and there are $\binom{5}{3}\binom{6}{6}=10$ ways to choose the occupied planets.[/list]
There is no other case that works. Therefore, in total we have $15+75+10=\boxed{100}$ different combinations of occupied planets.
[/hide]

[quote=aliu2048]Why does it take 2 units to colonize earth but only 1 to colonize mars?[/quote]

I would think it takes 1 to colonize earth cause there are more natural resources.

Actually, I think the problem meant that an Earth-like planet can fit two colonies, compared to Mars' one colony.

[hide=h]
if all the earth like planets are identical (and the mars like planets !) then there would be 3 combos:

3 earth like, 6 mars like
4 and 4
5 and 2

but they are distinguishable, so

5C3 (earth like) times 6C6 (mars like) is 10
5C4 times 6C4 is 75
5C5 times 6C2 is 15

10+75+15=100
[/hide]

[hide=solution]out of 11, 5 are earth-like, 6 are not earth-like
earth-like takes 2 units, not earth-like takes 1

call earth-like x and not earth-like y, then we have the diophantine equation 2x + y = 12, where we have a limit of 5 earth-like colonies and 6 non-earth-like colonies.

then there are 3 possibilities:
5x2 + 2x1 = 5 choose 5 * 6 choose 2 = 15 possibilities,
4x2 + 4x1 = 5 choose 4 * 6 choose 4 = 75 possibilities
3x2 + 6x1 = 5 choose 2 * 6 choose 6 = 10 possibilities

adding them together gives us 15+75+10 = [b]100 total possibilities[/b][/hide]


1. Should we e-mail only one .pdf file with all the solutions in it, or should we split the solutions up into five different files, one for each solution?

2. What should we name the solution file(s)?

3. Do we have to send the files in a folder, and if so, what should we name the folder?

4. In the .pdf file, do we have to put our name, ID, etc. on [b]every[/b] page or just the first?

1. Please send only [b]one[/b] PDF file.

2. It doesn't matter what you name it.

3. You don't need to send a folder: just send file(s).  (If you're not sending LaTeX code, then you should only send 1 file.)

4. [b]Every[/b] page should have your name and ID#.

[quote="jb05"][quote="DPatrick"]4. [b]Every[/b] page should have your name and ID#.[/quote]

For some reason I thought that this didn't apply to emailed solutions when the name/ID was already attached on the email and file. I guess it was one of my common senseless days. Should I resend? Thanks.[/quote]
Well, you should be using the USAMTS format, which will automatically include your name, username, and ID.

Can I mail in my permission form like today, and email my solutions today also?

Or does it matter whether USAMTS gets my permission form first, or my solutions first? (Is it OK if USAMTS receives my solutions before receiving the permission form?)

Am I allowed to email a permission form?


modular arithmetic
algebra unsolved
algebra

[quote="ridgers"]Find all natural numbers n , such that the sum of digits of the n-th power of 2 is eqaul to 5.
A easy question!!![/quote]

If the sum of digits is equal to $ 5$, we get $ 2^n\equal{}2 \pmod 3$ and so $ n$ odd.
If $ n$ odd, the last digit is either $ 2$, either $ 8$, hence $ 2$ and $ n\equal{}1\pmod 4$

We have now to solve $ 10^a\plus{}10^b\plus{}10^c\plus{}2\equal{}2^n$ with $ a\geq b\geq c\geq 1$

If $ c>1$, LHS$ \equal{}2\pmod 4$ while RHS$ \equal{}0\pmod 4$ and so $ c\equal{}1$

We have now to solve $ 10^a\plus{}10^b\plus{}12\equal{}2^n$ with $ a\geq b\geq 1$

If $ b>2$, LHS$ \equal{}4\pmod 8$ while RHS$ \equal{}0\pmod 8$ and so $ b\in\{1,2\}$

If $ b\equal{}1$, we have now to solve $ 10^a\plus{}22\equal{}2^n$ with $ a\geq 1$
If $ a>1$, LHS$ \equal{}2\pmod 4$ while RHS$ \equal{}0\pmod 4$ and so $ a\equal{}1$ and we got the solution $ 32\equal{}2^5$

If $ b\equal{}2$, we have now to solve $ 10^a\plus{}112\equal{}2^n$ with $ a\geq 2$
If $ a>4$, LHS$ \equal{}16\pmod {32}$ while RHS$ \equal{}0\pmod {32}$ and so $ a\in\{2,3,4\}$ and none of these values fit the requirement.

So, the only solution is $ n\equal{}5$

LHS and RHS what are these??
And why $ 10^{a}\plus{}10^{b}\plus{}10^{c}\plus{}2\equal{}2^{n}$ maybe $ 2(10^{a})\plus{}10^{b}\plus{}2\equal{}2^{n}$

LHS = Left Hand Side, aka $ 10^a\plus{}10^b\plus{}10^c\plus{}2$
RHS = Right Hand Side, aka $ 2^n$

If $ (a,b) \equal{} (a_0,b_0)$ is a solution to $ 2(10^a) \plus{} 10^b \plus{} 2 \equal{} 10^a \plus{} 10^a \plus{} 10^b \plus{} 2 \equal{} 2^n$
Then $ (a,b,c) \equal{} (a_0,a_0,b_0)$ is a solution to $ 10^a \plus{} 10^b \plus{} 10^c \plus{} 2 \equal{} 2^n$

So, that case is already covered in $ 10^a \plus{} 10^b \plus{} 10^c \plus{} 2 \equal{} 2^n$

[quote="Xantos C. Guin"]LHS = Left Hand Side, aka $ 10^a \plus{} 10^b \plus{} 10^c \plus{} 2$
RHS = Right Hand Side, aka $ 2^n$

If $ (a,b) \equal{} (a_0,b_0)$ is a solution to $ 2(10^a) \plus{} 10^b \plus{} 2 \equal{} 10^a \plus{} 10^a \plus{} 10^b \plus{} 2 \equal{} 2^n$
Then $ (a,b,c) \equal{} (a_0,a_0,b_0)$ is a solution to $ 10^a \plus{} 10^b \plus{} 10^c \plus{} 2 \equal{} 2^n$

So, that case is already covered in $ 10^a \plus{} 10^b \plus{} 10^c \plus{} 2 \equal{} 2^n$[/quote]

Thanks, Xantos, for your answer, quite right.

@ridgers : I wrote $ a\geq b\geq c$  so, as Xantos said, $ a\equal{}b$ is considered. Best proof is the solution $ 32$ where $ a\equal{}b\equal{}c\equal{}1$


calculus Stanford college AMC AIME USA(J)MO USAMO

Hi.

I am taking AP Calculus BC and my school offers no more math after that.

I am thinking about taking multivariable calculus and maybe number theory at Stanford EPGY.

If you took the course or know about it/people who did it please let me know how it is.

Thanks

-JHS

You are posting your query on a wrong forum.

well then what IS the correct forum to post this in?  :huh:

The more proper subforum is probably the "Other US Contests and Programs" one. But to answer your original question... I personally wouldn't recommend taking multivar calc through EPGY. The material isn't treated that in depth, and you'll likely have to retake a multivar calc course in college. This might not happen if you motivate yourself to study the subject thoroughly on your own, but if you have that kind of dedication, then you probably don't need the EPGY class in the first place.

What suggestion do you have for a high school student who has completed the mathematics courses at their school, is interested in studying some higher levels of mathematics, for whom it is not feasable to schedule a course at a college/university, but wants some guidance/instruction in their study of mathematics?

I'd say: pick up a good (introductory) book at college level and start working it through. For example, you could pick up Spivak's Calculus (I just love it) :)

I use Calculus early transcendentals by Anton, 7th edition. Is that book not a college level book then?

I am interested in taking a math class so that it doesnt look like i'm skipping math during my senior year...

How about UIUC Netmath?

[quote="squelchy451"]I am interested in taking a math class so that it doesnt look like i'm skipping math during my senior year..[/quote]

You could always take a math elective, such as AP Statistics. You can also show on your resume that you've accomplished something in math (making AIME, USAMO, or winning a regional math competition) during your senior year if you're worried that colleges will think you're skipping.

For the people at my high school that are a year ahead, they always planned to "skip" math in senior year. This usually gives them more room to fit in other APs (colleges would rather like to see you diversified than only focusing on math) or in some cases, take the courses you need to graduate (for example, I'm forced to take gym and health around senior year because I took AP Biology as a freshman and am taking AP Chemistry as a sophmore - both which are double period classes which cuts off one of my classes).

[quote="7h3.D3m0n.117"]colleges would rather like to see you diversified than only focusing on math[/quote]
I strongly disagree that skipping math in your senior year to take AP's is a wise move for college admissions - especially if you are interested in pursuing math.  Most top colleges want students to have a demonstrated passion for a particular subject.  An advanced math class will do more to help your application stand out than another AP.

One note: this query was posted on the correct forum. The discussion belongs right here.

One option, besides the distance-learning option, is to investigate whether a college in your area has a program that allows high school students to take courses at the college. My institution does have such a program for a very nominal fee (and I wind up approving the selections of math courses).

Depending on where you go eventually, the credit for such a course might or might not transfer. That - the question of transferable credit - probably shouldn't be your primary concern. The main idea is to keep moving.

With that said, multivariable calculus is only one of the possible courses you could take. There are other possibilities: linear algebra; differential equations; a "transition" or "how to prove things" course, which might or might not be the same thing as a discrete structures course; elementary number theory; calculus-prerequisite probability and/or statistics. I have on occasion recommended the "wrong" course with the right instructor, just for the quality of the experience.

A program note: This thread was originally posted in the Intermediate Topics forum.  That's what maximos was referring to.  Then an attentive moderator must have moved it correctly to this forum.

I find this very interesting, because I'm currently in exactly the same situation: I'm a senior who hasn't taken math for the last two years, and although I tried to take a math course at a college, my school's schedule would not work out with it.

On my application, I posted some links to my posts on the abstract algebra forum here. As well, on my blog, I have posted ideas in various areas I'm studying, and I sent colleges a link to my blog.

Interestingly, one of the colleges I applied to recently asked me for a math supplement, so I've been sending them examples of abstract algebra and topology problems that I've been doing. In hind sight, maybe I should have sent such a supplement to all of my colleges.

I also went to Canada/USA Mathcamp and got a recommendation from a professor there, so that probably helped a lot.

One suggestion I heard is to try to find a professor who will work with you. I tried this with multiple professors, but the best response I got was that I should pay a graduate student to help me, so I finally decided not to pursue that path.

theres these two city colleges that are 15 minutes away but the highest math they offer is equivalent to AP Calculus BC, which I am taking this year.

USC is 35 minutes away and UCLA is 45. (quite a commute)

That's why i'm doing it online and if u do it online i might be able to get a head start during the summer b/c these summer programs don't really care when u start as long as u end within the time they allow u to finish in.

[quote="Kent Merryfield"]Depending on where you go eventually, the credit for such a course might or might not transfer.[/quote]
To elaborate on this, some schools will grant credit for certain courses taken at a local college or in conjunction with one.  Some schools (such as mine), will only grant you credit if you took the course [i]at[/i] the university campus.  That is, some schools, usually public university systems, will offer a "university in the high school"-type program and hence grant credit for such courses.  However, just because you take a course at the university doesn't mean you'll get credit.  Usually, a director of undergraduate studies-type person will ask to see the syllabus, textbook, homework assignments, tests and then determine whether or not credit should be granted.

I am completing my sophmore year of high school and finishing AP Calculus AB.  This is the highest offered at my high school.  Should I take online AP Calculus BC next year?  Any recommendations of the best program to take?  Or should I go to the local community college and take one of their offerings?  If any so suggestions of which one?

If you can, I would take courses at your local community college.  If you don't feel like teaching yourself the BC material, enroll in a calc 2 course and learn it there.  Then continue on to higher courses.

As a note, not all colleges will accept credit you may earn there.  But you'll certainly be ahead when you to go to college.


The vortices of a regular $n$-gon are $A_{1}, A_{2},..., A_{n}$.
$P$ is an arbitrary point of the $n$-gons circumcircle. Prove that the $|PA_{1}|^{4}+|PA_{2}|^{4}+...+|PA_{n}|^{4}$ sum is constant independently from the position of $P$. Find this constant too.

probably you can solve this problem in using complex number.

Then could  you show a solution using complex numbers?


algebra
polynomial
ratio
modular arithmetic
geometric series
geometric sequence

Let $ g(x)\equal{}x^5\plus{}x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1$.  What is the remainder when the polynomial $ g(x^{12})$ is divided by the polynomial $ g(x)$?

Thanks in advance!

I believe you mean

[hide]$ g(x) \equal{} \displaystyle\frac {x^6 \minus{} 1}{x \minus{} 1}$

And, at that point, $ g(x^{12})$ falls into place.[/hide].

Yes, you are right, I edited.

EDIT:

I'm thinking it is probably easier [hide]just to divide polynomials in form $ (x^{60}\plus{}x^{48}\plus{}x^{36}\plus{}x^{24}\plus{}x^{12}\plus{}1): (x^5\plus{}x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1)$[/hide]

[quote="Flame"][hide="hint"]
we have 2 geometric series here

[hide=" hint 2"]
$ g(x) \equal{} \frac {x^6 \minus{} 1}{x \minus{} 1} \\
g(x^{12}) \equal{} \frac {x^{72} \minus{} 1}{x^{12} \minus{} 1}$[/hide][/hide][/quote]

So is the goal to find the common ratio of the geometric sequence?  But I don't get where the geometric sequence is...

[quote="314.math"]I don't get where the geometric sequence is...[/quote]

Notice that $ g(x)\equal{}\sum_{k\equal{}0}^{5}x^k$.

Alternatively, note that $ x^6\minus{}1\equiv0\pmod{g(x)}\implies x^6\equiv1\pmod{g(x)}$, so $ g(x^{12})\equiv g(1)\equiv\boxed6\pmod{g(x)}$.

How do you find the remainder of the two polynomials $ g(x)$ and $ g(x^{12})$?


calculus
derivative
ratio
analytic geometry

This is a rather difficult pre-calculus problem.  However it is not called a challenge problem, just a normal homework problem that I am having trouble with :blush: , so i put it in the getting started room.

A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground.  When he is 10 feet from the base of the light,
(a) at what rate is the tip of his shadow moving?
(b) at what rate is the length of his shadow changing?

Please be as specific in explaining as you can.  I want to know how to do the problem, not just the answer.

Hello - If you can draw a figure - assume the  light is at (0,15)

The man is at    (x,0)  .. I mean his feet is at (x,0) his head is at (x,6) becuase his is 6 feet tall.

I know that x=10 but  hold it, let us just assume it is x we will use the fact x=10 later when we   have to!   - IOW since x is changing let us keep it as x for now)

His shadow is at  (Assume it is at (S,0)    -   Do your own math now and see if you get:

[hide]( (5/3)*  x ,  0)     (That is (5/3)*x feet away from the  base of the pole)

S = (5/3)*x[/hide]

If you don't follow this [hide]do simple ratio of triangles:   S/15 = (S-x)/6 [/hide]

And the length of the shadow is, of course:
[hide]
L = (5/3)*x - x ) =  (2/3)* x [/hide]

(Of course you, could have  found it easier to calculate the length of the shadow first  and than added x to it - It does not matter how you do it as long as you do it corretly)

Now you are given that (dx/dt) =  5
(Actually  to be correct it is 5 ft/second if you insist on units but let us assume we will always  use correct units..)

Also    you see dS/dx =  5/3
                         and dL/dx = 2/3
so dS/dt =   25/3 and dL/dt = 10/3

(hmm it does not depend on x (does not matter how far he  was from the pole) but then so be it!

Thank you very much!  your answer was correct (it is in back of book).  I greatly appreciate your use of the coordinate plane...it makes the problem much simpler.  i have a differentiation test on friday and a differention/related rates word problem test on monday.  Thanks again for helping me to understand.

jeremy


1. If [tex]3sin^2 x = 5 sin x + 2 [/tex]and cos x < 0, then what is sin 2x?

This was from the 2002 Santa Clara Valley Mad Hatter Marathon. I'm having trouble, maybe because I'm just too tempted to throw this into my calculator and solve for x.

Let's rewrite that...

If $3\sin^2{x}=5\sin{x}+2$ and $\cos{x}<0$, then what is $\sin{2\theta}$?

[hide="hint"]
quadratic... factor.[/hide]

Wow, that was too obvious. I'm so oblivious :(

Anyways, I got [hide][4 sqrt 2 all over 9 [/hide]


For a prime p, let 1+1/2+1/3+...+1/((p-1)/2)=m/n, gcd(m,n)=1.

Prove that 2p-1[u]=[/u]1 (mod p2) iff p|m.
---------------------------------------------------------------------------------
In the book ("Elemente de Teoria Numerelor" - Victor Alexandru, Niculaie Gosoniu ~ probably most of you don't know of it  :D ), it is said that the proof of this statement is elementary. I don't know what they mean by this.

I feel like I'm so close and yet so far away from the solution.. :D:

p2|2p-1-1 iff p2|2p-2=\sum{k=1->p-1}pCk=2\sum{k=1->(p-1)/2}pCk iff p|\sum{k=1->(p-1)/2}(p-1)!/[k!*(p-k)!].

k!*(p-k)!=(-1)k+1*k (mod p), so the condition above translates to p|\sum{k=1->(p-1)/2}(-1)k*k-1, where k-1 is the inverse of k in Zp.

On the other hand, p|m iff p|\sum{k=1->(p-1)/2}k-1, so it's the same as above, without the (-1)k. Are these 2 equivalent? I don't know if there's a mistake, but the whole thing is really frustrating.

In grobber's work, take p2|2p-2 iff
p| \sum {k=1->p-1} (-1)k+1/k (don't try to restrain the sum, let it loose :D).

Now 1-1/2+1/3-1/4+...-1/(p-1)=1+1/2+1/3+...+1/(p-1)-2(1/2+1/4+...+1/(p-1))=1/((p+1)/2)+...+1/(p-1)=-(1+1/2+...+1/((p-1)/2))

So p|grobber's alternate sum, iff p|the problem's sum :D

Thanks grobber for solving most of the problem.

Thanks! I was going crazy over this problem :D:D.

Don't mention it.
I was stunned about the problem in the first place. I mean what in gods name is the connection between 2p-1-1 and \sum 1/k?


Are there exists integer n bigger than 24, satisfying diophantic equation:
$ 1^2\plus{}2^2\plus{}3^2\plus{} ... \plus{} n^2 \equal{} m^2$  :?:

This is a very famous (and old) topic, related to Kepler's conjecture of packing spheres. The answer to your problem is NO, see e.g. [url]http://www.daviddarling.info/encyclopedia/C/Cannonball_Problem.html[/url] and also [url]http://www.math.ubc.ca/~bennett/paper21.pdf[/url] for a quite detailed historical review and generalization.


$ABCD$ is a quadrilateral such that $AB=BC=3$ and $CD=7$ find the maximum possible area of $ABCD$

[hide="hint"]max area of a quadrilateral with fixed side lengths is when it's cyclic[hide="hint 2"]brahmagupta[/hide][/hide]

[hide="i think i get it"]
so using that we have that the area ($K$) is

$K^2=(s-3)(s-3)(s-7)(s-x)$
where $s$ is the semiperimeter which equals $\frac{13+x}{2}$
and $x$ is the other side length.
what i do not know is how i would maximize this value without calculus.
[/hide]


geometry
circumcircle
trigonometry
combinatorial geometry
trig identities
Law of Sines
geometry proposed

Given any convex polygon, show that there are three consecutive vertices such that the polygon lies inside the circle through them.

I think we can solve it with extremal.
Any ideas?

Well look at the convex hull, make a really really really really big circle around it and gradually shrink the circle down. Would that work?

This is a consequence of Bose's theorem for polygons (a somewhat stronger version of the discrete four vertex theorem). The idea is very similar to the idea that every finite set of points has a Delaunay triangulation (i.e. can be triangulated by triangles whose circumcircles do not contain any other vertices).

Also posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=557857&hilit=three+consecutive+vertices+%2C+circumcircle]here[/url].  However, I believe the last sentence in the proof was never actually resolved...

The last statement (of each solution) seems to be inspired by extended law of sines, though I'm not sure if the arguments are quite correct.


Make teams for the corresponding tourney here instead of by PM

[b]thedirichletlseries pwners[/b] (did you mean dirichletseries instead of dirichletlseries?)
pythag011 (captain)
dafjbomb
AIME15
mathswimmer

[b]fanfan<3u[/b]
BOGTRO
FantasyLover
athunder
isabella2296

[b]1337s+Glasgow Middle School Pwns[/b] (3/4)
AIME_is_hard
Ttocs45
nikeballa96

[b]Team who likes SNOW DAYS!![/b] (2/4)
007math
Twin Prime Conjecture

[b]Epic Failure Team[/b] (1/4)
jame4l

[b]Individuals[/b]
Discrete_Math
mathemagician1729
high_flyer10
shentang

Our team name will be fanfan<3u

 :rotfl:

OUR TEAM IS CALLED

thedirichletlseries pwners

[quote="Twin Prime Conjecture"][quote="007math"]anyone one to join my team....im pretty lonely out there..[/quote]

Can I join your team?[/quote]

Go ahead.

Team who likes SNOW DAYS!! (that ounds like a good name  :P ):

007math
Twin Prime Conjecture

Anyone else wanna join, Twin and I are still lonely...

Also, JOIN MY TOURNEY, WE NEED MORE PEOPLE FOR IT. LINK IS AT BOTTOM OF MY POST!!!

1337s+Glasgow Middle School Pwns

AIME_is_hard
Ttocs45
nikeballa96

OMG IT SNOWED TODAY!! IN LOUISIANA! 1st time in 20 years :D

dirichlet l-series do not equal dirichlet series.

A note to 5849206328x

hmm... why is that our name?

[quote="james4l"]Join Team 5: (Epic Failure Team)
james4l[/quote]

Can I, being an actual epic failure, join the Epic Failure Team?

@pythag - right after I submitted I realized I forgot about dirichlet l-series, but I had to go then
and that'll be your name unless you guys come up with something different

[quote="BOGTRO"]Our team name will be fanfan<3u

 :rotfl:[/quote]

you were serious... :(

I'll take it as serious unless he gives confirmation that it wasn't

EDIT: The team names should be accepted by the majority of the team members :P

I asked mathemagician1729, and he aid he' okay with it.  He's down with a cold, so yeah....

1337 Phailures (team name):

007math (captain since i mde it  :P , but if u don't want me to, then i can come down from the throne)
someperson01
Twin Prime Conjecture
mathemagician1729

Okay, I random.org-bashed it. Heads = BOGTRO, Tails = Me. BOGTRO wins. (Darn it! :P )

So BOGTRO's the captain, I guess.

[quote="someperson01"]Nice, ironic, team name. 007math, you can be captain, as long as you don't mess up our answers :P[/quote]

But you guys input answers to me, right?

There isn't really that much difference between the captain and the other people.  Since we can't exactly give out awards, the main thing is that the captain is the one responsible for the team round answers.

I want to get involve too.  Doesn't matter what team i will be in.
Thanks,

[quote="5849206328x"]
uh oh...
you people are mature (yeah right) enough to figure this out[/quote]

zomg personal attack someone ban alan from aops

also, the point of this post is to notify you guys that I've made the masters' round significantly easier 'cause I eliminated a part that none of us could actually solve. it's still much harder than, say, 2008 nats' masters', though.

I'd like to be captain for my team if no one else from my team wants to be one.

@Temperal:  :P

@nike: probably around wednesday of this week

Sorry izzy...

Anyway, when does the actual competition start?

[b][size=200]Sign-ups are now closed, refer to the other thread for teams and when the test will be posted[/size][/b]

[size=50]I lost...[/size]

[quote="nikeballa96"][hide="for my team"] can i be captain? [/hide]

whats the tentative date for starting?[/quote]

No, I am captain :chief:

either way, settle this in the next 2.5 hours, or we'll just randomly pick a captain instead...

ok, maybe not completely randomly...


This question has probably been asked numerous times, but I didn't have a desire to sift through a months' worth of threads: Is there a chance that the procedure for USAMO selection might change for 2011? Sadly, I discovered math competitions later than most and the changes don't favor my kind.

You can only qualify for USAMO through AMC 12, unless you score AIME 11+. For AMC 12, you have to be in top 230

No, I know that I have to use AMC 12. I was wondering if perhaps the 270/230 split might change or something else that would lower the index score needed to qualify.

[quote]I was wondering if perhaps the 270/230 split might change or something else that would lower the index score needed to qualify.[/quote]

No.

Steve Dunbar
MAA Director, American Mathematics Competitions

If anyone would be willing to postulate, what do you think the USAMO qualifying index will approximately be this year, given the new change in selection procedure?

I read that, if the 2009 scores were used with the new selection procedure, 210 would have been the index (up from the actual 2009 index of 201).

Is this accurate, and is it probable that the number will be near 210? (I realize that predicting the actual number is unrealistic but any rough estimate would be appreciated).  The last three years have been 201, 204, and 197.5

there have been many threads on this topic..

but- with the old procedure, very talented underclassmen took the much easier AMC 10 to maximize their chances of USAMO qualification. Now that the AMC 12 is the only option for potential qualifiers, there will be large droves of underclassmen who will have higher indices than before. The 210 estimate does not take these people into account. Thus, people are predicting 220+.

hmm.

thank you for the response, even though it had apparently been answered many times before.  I guess I'm going to have to do better than I had originally thought then.

fair enough...i like your straight forward answer, and I think that I will have a chance of making it in 2011. I'll just have to work harder...220+, though, sheesh.


The U.S. Puzzle Championship will be held Saturday, June 18 at 1pm ET. Register on the [url=http://wpc.puzzles.com/]site[/url] by June 16. Anyone can participate, but for Americans and Canadians this may determine placement on a team to the World Puzzle Championship. There are [url=http://wpc.puzzles.com/links/index.htm]practice materials[/url] and [url=http://wpc.puzzles.com/history/index.htm]old contests[/url] available.

It's a fun contest. Anyone else planning on doing it?

The test and the solutions (but not the results yet) have been [url=http://wpc.puzzles.com/p5/summary.htm]posted[/url].

I did rather poorly, I think: 65 points for problems 1, 16, 17, and 20, with 10 penalty points for incorrect answers.

So did anyone else participate?

Being a neophyte, i was able to score only 15 :(

But it was fun. I enjoyed it

oh I didnt know about this...
 :o  :mad: :wallbash: :furious: :wallbash_red: :(

I would definitely have noticed this, if this was posted on the puzzles and brainteasers section..

can I do it next year? is it always around this time?

What's a "good" score?
A qualifying score?


combinatorics unsolved
combinatorics
points
Chords
China
TST
Team Selection Test

Mark $4 \cdot k$ points in a circle and number them arbitrarily with numbers from $1$ to $4 \cdot k$. The chords cannot share common endpoints, also, the endpoints of these chords should be among the $4 \cdot k$ points.

[b]i.[/b] Prove that $2 \cdot k$ pairwisely non-intersecting chords can be drawn for each of whom its endpoints differ in at most $3 \cdot k - 1$.
[b]ii.[/b] Prove that the $3 \cdot k - 1$ cannot be improved.

Lemma: For any $4k$ points on a circle, with $2k$ coloured red and $2k$ coloured blue, there exists a way to draw $2k$ chords joining distinct coloured points such that no $2$ intersect.
Proof: Consider a drawing that minimizes the total sum of lengths of chords. Suppose two of them intersect $AB, CD$, where $A,C$ red and $B,D$ blue. Then change it to $AD, BC$, which clearly decreases the sum of lengths. Thus this drawing cannot have any intersection.
Apply the lemma by colouring $\{1,...,k,3k+1,...,4k\}$. red and the rest blue. Then the max difference is $3k-1$.
For part ii, write $k+2,k+3,...,3k-2,3k-1,k+1,4k,k,4k-1,...,3k+1,1,3k$ in clockwise order. This means that some two numbers in $k+1,4k,...,1,3k$ are connected, and they would come from each of $1,...,k+1$ and $3k,..,4k$. Not hard to show that this will cause two numbers differing by $3k-1$ to be connected by a chord.


For how many single-digit positive integers $ d $ are there an integer $ n $ for which the units digit of $ d^n $ is 6?

I'm taking this one  :P

[hide]obviously for 1, it's impossible, since $ 1^n=....1 $. For 2, we see that 2 raised to any positive multiple of 4 gives us a units digit of 6. THus d can equal two.

We can eliminate 3, 5, 7, and 9, since those number raised to any positive integer power is always odd, and can't end with 6.

4 is possible, since 4 raised to any number n that is identical to 2 (mod 4) will end with 6.

6 to any positive integer power ends with 6.

8 raised to any multiple of 4 again gives us 6.

0 is impossible.

d = 2, 4, 6, and 8. That's four integers. [/hide]


a,b,c positive
Prove
$ \frac{a^3}{b\plus{}2c}\plus{}\frac{b^3}{c\plus{}2a}\plus{}\frac{c^3}{a\plus{}2b} \ge \frac{a^2\plus{}b^2\plus{}c^2}{3}$

Cauchy implies $ \sum\frac {a^3}{b \plus{} 2c} \sum a(b \plus{} 2c) \ge (\sum a^2)^2$. Thus, it's left to prove $ 3\sum a^2 \ge 3\sum ab$, which is clearly true.

Yes ,I solved similarly by Cauchy Schwarz(Titu's Lemma)

$ \sum_{cyc}{\frac {a^3}{b \plus{} 2c}} \equal{} \sum_{cyc}{\frac {a^4}{ab \plus{} 2ac}} \geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{3(ab \plus{} bc \plus{} ca)}\geq \frac {a^2 \plus{} b^2 \plus{} c^2}{3}\iff a^2 \plus{} b^2 \plus{} c^2 \geq ab \plus{} bc \plus{} ca$

[quote="NikolayKaz"]a,b,c positive
Prove
$ \frac {a^3}{b + 2c} + \frac {b^3}{c + 2a} + \frac {c^3}{a + 2b} \ge \frac {a^2 + b^2 + c^2}{3}$[/quote]
The following stronger inequality is also true.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ \frac {a^3}{b + 2c} + \frac {b^3}{c + 2a} + \frac {c^3}{a + 2b} \ge\sqrt {\frac {a^4 + b^4 + c^4}{3}}\]

[quote="NikolayKaz"]a,b,c positive
Prove
$ \frac {a^3}{b \plus{} 2c} \plus{} \frac {b^3}{c \plus{} 2a} \plus{} \frac {c^3}{a \plus{} 2b} \ge \frac {a^2 \plus{} b^2 \plus{} c^2}{3}$[/quote]

\[ \sum_{cyc}\frac {a^{3}}{b \plus{} 2c} \geq \frac{1}{3}\left(\sum_{cyc}a^3\right)\left(\sum_{cyc}\frac {1}{a \plus{} 2b}\right)\ge \frac{1}{9}\cdot \left(\sum_{cyc}a^2\right)\left(\sum_{cyc}a\right)\cdot \frac{3}{a\plus{}b\plus{}c}\]

The following inequalities are also true.
1. Let $ a,$ $ b$ and $ c$ are positive numbers such that $a^5+b^5+c^5=3$. Prove that:
\[ \frac {a^3}{b \plus{} 2c} \plus{} \frac {b^3}{c \plus{} 2a} \plus{} \frac {c^3}{a \plus{} 2b} \ge1\]

2. Let $ a,$ $ b$ and $ c$ are positive numbers such that $a^4+b^4+c^4=3$. Prove that:
\[ \frac {a^3}{a \plus{} 2b} \plus{} \frac {b^3}{b \plus{} 2c} \plus{} \frac {c^3}{c \plus{} 2a} \ge1\]

3. Let $ a,$ $ b$ and $ c$ are positive numbers such that $a^2+b^2+c^2=3$. Prove that:
\[ \frac {a^3}{2a \plus{} b} \plus{} \frac {b^3}{2b \plus{} c} \plus{} \frac {c^3}{2c \plus{} a} \ge1\]

3 is easy
$\frac{a^3}{2a+b}=\frac{a^4}{2a^2+ab}$
similarly we transform the L.H.S.
By holder,
$(2(a^2+b^2+c^2)+(ab+bc+ca)).L.H.S. \geq (a^2+b^2+c^2)$
it is true as $a^2+b^2+c^2=3$
and  $a^2+b^2+c^2 \geq ab+bc+ca$
hence $ab+bc+ca \leq 3$

for 1
$\frac {a^3}{a+2c}=\frac{a^7}{a^5+2a^4c}$
similarly we transform the L.H.S.
By Holder
$(\frac{a^7}{a^5+2a^4c})(a^5+2a^4c)(a^3+b^3+c^3)\geq (a^5+b^5+c^5)^3$
therefore it remains to prove
$L.H.S. \geq \frac {27}{(3+\sum 2a^4c)(a^3+b^3+c^3)}$
that is to prove $(3+\sum 2a^4c) \leq 9$ [can be proved by rearrangement]
and $a^3+b^3+c^3 \leq 3$ which is true as $a^{\alpha}+b^{\alpha}+c^{\alpha}$ is strictly increasing with the increase in $\alpha$ this result can also be proved by holder

for 2
it is the same as 1
here
$\sum \frac {a^3}{a+2b}=\sum \frac{a^6}{a^4+2a^4b}$
by holder
$(\sum \frac{a^6}{a^4+2a^4b})(\sum a^4+2a^4b)(a^2+b^2+c^2) \geq (a^4+b^4+c^4)^3$
the rest is same as 1
we use rearrangement and holder to prove  $\sum a^4+2a^4b \leq 9$ and $a^2+b^2+c^2 \leq 3$

[quote="Ashutoshmaths"]for 2
i
by holder
$(\sum \frac{a^6}{a^4+2a^4b})(\sum a^4+2a^4b)(a^2+b^2+c^2) \geq (a^4+b^4+c^4)^3$
[/quote]
Are you sure? :maybe:

[quote="arqady"][quote="Ashutoshmaths"]
By Holder
$(\frac{a^7}{a^5+2a^4c})(a^5+2a^4c)(a^3+b^3+c^3)\geq (a^5+b^5+c^5)^3$
[/quote]
Are you sure? :maybe:[/quote]
yes,it is wrong and sorry that i did not reply to your first post :(


I was thinking one day and thought this up.
you put one cup of water in a jar.
Than a half of a cup in a second jar.
Than a quarter in the first. Than an eighth in the second. Then a sixteenth in the first. And so on. forever.
So, There is 2 cups of water total, one and one third in the first, two thirds in the second. So...
This means that the first jar has twice as much as the second. For this to be true, the second jar has to recieve water an equal amount of times as the first. This can only happen, in a finite sequence, when it ends in the second term.
So, wouldn't the "last term" of an infinite alternating series be the second term?
Is there a lapse in my thinking?

Yes: you are reasoning about an object (the "last term" of an infinite sequence) that doesn't exist.

[quote]For this to be true, the second jar has to recieve water an equal amount of times as the first[/quote]

I don't follow you there... As you approach infinitely many jars, you're essentially evaluating $ \lim_{n\to\infty}\frac {1}{2^n} \equal{} 0$ as you reach the "end" of the infinite sequence.

This is, of course, a limit; there's no "end," but as you reach sufficiently many jars, the water you're putting in each jar becomes negligible.

Edit: Sorry, I need to learn to read questions...  :lol:


linear algebra
matrix
linear algebra unsolved

Let $A\in Mn(C)$ such that $det(A+YX)=det(A+XY)$ for any
 $X,Y\in Mn(C)$
Prove that there exist $a\in C$ such that $A=aI_n$

Authors:M.Andromach and I.Savu

I haven't much time to think about this, but we can show fairly easily that $A$ is a diagonal matrix:

It's clear that we can conclude $\det(AX+Y)=\det(XA+Y)$ for all invertible $X$ and all $Y$. Now choose $X$ to be the elementary matrix $M_{i}(\alpha)$ s.t. $XA$ is obtained from $A$ by multiplying the $i$'th row by $\alpha$, while $AX$ is the same only with columns instead of rows. We can now easily choose $Y$ to make all entries of $AX+Y,XA+Y$ equal to $0$, except for the main diagonal and such that $a_{ik}+y_{ik}=0,\ \forall k\not \in\{i,j\}$ and the same for $a_{ki}+y_{ki}$. Our condition then translates to $(\alpha a_{ij}+y_{ij})(a_{ji}+y_{ji})=(\alpha a_{ji}+y_{ji})(a_{ij}+y_{ij})$ for all $y_{ij},y_{ji}$. This gives $a_{ij}=a_{ji}=0,\ \forall i\ne j$ (I think).

I don't know if there's much more left to do.

Actually, once we show that $A$ is diagonal, we can replace $X$ by the elementary matrix which interchanges columns/rows $i,j$. With a proper choice for $Y$ we find that all entries of $A$ must be equal.


group theory
superior algebra
superior algebra solved

prove that  $\alpha$ and $\beta \in S_{n}$ are conjugate iff they have the same structure.

Definition: let $\alpha=\alpha_{1}\cdot...\cdot \alpha_{k}$ and $\beta=\beta_{1}\cdot... \cdot \beta_{m}$ be the permutations such that $\alpha_{i},\beta_{j}\in S_{n}$ be disjoint cycles. $\alpha ,\beta$ are said to be have the same structure if k=m and the lenght($\alpha_{i}$)= lenght($\beta_{i}$)  for all i=1..n

I am not gonna solve this for you, but I am gonna give you a major hint :

let us denote conjugates in this way : $x^{g}=g^{-1}x g$

let us denote cycles in this way $(1 2 3)$ for instance (this means 1 is mapped onto 2 , 2 ont 3, 3 onto 1)

now note this : let a and g be permutations , if

$a=(a_{1}a_{2}\cdots a_{i_{1}}) (b_{1}b_{2}\cdots b_{i_{2}}) \cdots$

then

$a^{g}=(g(a_{1}) g(a_{2}) \cdots g(a_{i_{1}}) ) (g(b_{1})g(b_{2}) \cdots g(b_{i_{2}})) \cdots$

this is really good to know in general

Thank you!
I think i got the part: if they conjugate , then they have the same structure.
But i cant find the solution to the other :maybe:
help me please

It's really not as hard you think.

Remember the hints I gave you

consider in $S_{6}$ for instance (1 6) (2 4 5) (3) and
(5 4 3) (2 1 ) (6)

they have the same structure alright, right them neatly under each other

(1 6 ) (2 4 5) (3)
(2 1)  (5 4 3) (6)

well now take the unique permutation

that sends 1 to 2
6 to 1
2 to 5
4 to 4
5 to 3
3 to 6

and there you have it!

[quote="fredbel6"]

let us denote conjugates in this way : $x^{g}=g^{-1}x g$

[/quote]
I agree you are right.
But as what you wrote following,
we should denote:
$x^{g}=gx g^{-1}$
 :?:

[quote="fredbel6"]Are you really sure?
I am following the conventions of my first group theory professor, and that has always worked out fine? :maybe:[/quote]
I don't say you are wrong...
I just said you have the small mistake of the defination.
Consider in $S_{3}$
let $x=(12)$,$g=(123)$
You will find it.

Or you can write $g^{-1}$ instead of $g$ in
$a^{g}=\cdots$

Yes you are right, i forgot to mention that my professor is a die hard follower of the belief that $a b$ means b after a, he says you read from left to right don't you

while most, like you, think otherwise


Given $ a, b, c \geq\ 0$ and $ a\plus{}b\plus{}c\equal{}3$. Prove that:
$ (a^3\plus{}b^3\plus{}c^3)(a^2\minus{}b^2)(b^2\minus{}c^2)(c^2\minus{}a^2) \leq\ \frac{729\sqrt{3}}{8}$
Thank you very much for your gift, hxy09. And here is one of the nicest ineq in chapter "The strange ineqs" of me.
Hope that you'll like it  :)

Thank you for your gift,I will try to solve it :lol:


A large cube is made from unit cubes. The large cube's faces
are then painted, and the unit cubes are separated. Twenty-seven
unit cubes have no faces painted. How many unit cubes
were used to make the large cube?

The only cubes that are not painted at all are the ones in the interior of the cube. Generally the interior of the cube will be 1 edge inside on each side so 2 in all. So since its 27 or 3*3*3. Adding two we get 5*5*5 or [b]125[/b].


group theory
abstract algebra
algorithm
topology
number theory
Euclidean algorithm
superior algebra

Let G = <x> be an infinite cyclic group. Show that all the distinct subgroups of G are <e>, < x>, <x^2>, < x^3>, ....

I have been working on this for a few hours and it seem that the only way to prove this is to prove that

infinity/gcd(integer, infinity) = integer for the subgroups.

[quote="ayatollah85"]infinity/gcd(integer, infinity) = integer for the subgroups.[/quote]  I have no idea what this means, but as a hint, you probably want to include the words "Euclidean Algorithm" in your answer ;).

Here's a topological proof  :P

Consider the circle $ S^1$ and pick a point $ p$ (we will think of the circle as a curved edge and a vertex $ p$). The space's fundamental group is $ \pi_1(S^1)\cong\mathbb{Z}$. Now look at the covering spaces of the circle. The universal cover is the line $ \mathbb{R}$.

Now only graphs can cover graphs, so assume $ S^1$ admits a covering with two edges. In order for this to occur, there must a be a point for which in all sufficiently small neighborhoods, four distinct edges leave this point. There is no point on the circle with this property so the local homeomorphism property of the projection fails. Therefore each covering of the circle must be the circle. Consider now the circle $ S^1$ as unit norm complex numbers. The covering maps are $ z\rightarrow z^n$ which correspond to subgroups of $ \pi_1(S^1)\equal{}\mathbb{Z}$. The corresponding subgroups are $ n\mathbb{Z}$. Now by the Galois connection of covering spaces and subgroups of the fundamental group, the only subgroups of $ \mathbb{Z}$ are $ n\mathbb{Z}$. q.e.d.

Hint: In the subgroup, consider the element with the least positive index.

[hide] In any subgroup  consider the element with the least positive index, let it be a.

We claim that every other element of the subgroup is of the form $ x^m$ where a|m

Proof: If to the contrary there is an element $ x^n$ where n is not a multiple of a.

Then by Euclid, n = qa+r where 0<r<a

But, since $ x^n$ and $ x^{qa}$ and hence $ x^{\minus{}qa}$ are elements of the subgroup, we have

$ x^{n\minus{}qa} \equal{} x^r$ is also a member of the subgroup. But since 0<r<a, this contradicts the fact that a is the least positive index of elements in the subgroup[/hide]


[quote="lund"]I am not quite sure which forum to post this in.  I am looking for a game theory text book and I would like some recommendations.  I only have a basic Calculus background.  Is this over my head?  Thanks.[/quote]
No.

The bible of this subject is Winning Ways for your Mathematical Plays, by Berlekamp, Conway and Guy.

[url=http://www.amazon.com/gp/product/1568811306/qid=1137470020/sr=8-1/ref=pd_bbs_1/104-0845662-0189556?n=507846&s=books&v=glance]First volume here.[/url]

That looks fantastic!  Do you know of any others that are a little cheaper?  (Preferably $\$ 30 or less)

Well, that is [i]the[/i] book for game theory.
Do you want a basic intro text, or are you looking for a specific subject in game theory?

I have very little knowledge of Game Theory.  So I suppose I would need an intro text.

But what I feel is, take it in the right spirit, Penguin Integral, the book you have suggested is more of a book on mathematical games than game theory.

[quote="thealchemist"]I would suggest the book by Shaun.P.Hargreaves nad Yaroun Varoufakis " A Critical Introduction to Game Theory" and also another book by Avinash Dixit " Games of Strategy", I am not very sure of the titles.[/quote]

Game Theory: A critical text, (with Shaun Hargreaves Heap), London and New York: Routledge, 2004

http://www.econ.uoa.gr/UA/content/en/Article.aspx?office=16&folder=471&article=554

Game Theory: A critical introduction, London and New York: Routledge, 1995 (with Shaun Hargreaves-Heap)

http://www.uadphilecon.gr/UA/content/en/Article.aspx?office=16&folder=471&article=608

I think this is the best introduction to the subject:

[url=http://www.amazon.com/Strategy-Mathematical-Association-America-Textbooks/dp/0883856379/ref=sr_1_8?ie=UTF8&qid=1299672384&sr=8-8]Game Theory and Strategy[/url] by Philip Straffin

It's not extremely advanced, but it gives you a good idea of the basics of the subject.

(Heavens!) Watch out for the distinction between game theory and "combinatorial" game theory! [i]Winning Ways[/i] by Berkelamp, Conway, and Guy is definitely about the latter subject! The other ones are about (classical) game theory.

maybe this book helps , but need a higher concept :lol:

[url]http://kot.rogacz.com/page/studies/materials/08/atg/wyklad/2-7-BKRubinstein.PDF[/url]


i know the site is primarily aimed at americans, but you have a rapidly growing international membership and it would be nice to have jams at times that more people could join in.  7pm EST is just about ok for britain; its only 2300 or midnight or something (luckily my parents are away atm as they would usually make me come off the internet at 2300!!  so im tempted to join in the one on tuesday if i have time), but if you had ones every so often that were earlier it might be nice for people who wouldnt usually get the chance to join in.  also for example eastern europeans have a very different education system and would probably contribute a lot of non-standard solutions (we certainly found that when training with their IMO team, and im sure it applies at lower levels too) which would be interesting for all concerned.

i dont know how many non-US members you have but if you have more than a few itd be nice to have at least one 'international math jam' not very often maybe every couple of months or something.  and it can be at a time thats ok for americans to join too, say 10am EST at a weekend... so im sure youd not be short of people joining...

just a suggestion
 :)

[quote="Tare"]I think that's a good idea, except that I think there are people accessing from India, or another Asian country...[/quote]

yeah thats probably true, i just noticed a lot from europe (for example me  :P )

wouldnt east asia be early morning at 7pm EST anyway?  which is sort of ok...

and india would be evening at 10am EST

so i think with those two times it should be ok for just about anyone??

depends mostly what suits the instructors though! if they cant manage one at times other than 7pm EST theres no arguing with that really, since its their time theyre giving up

At 10 AM EST, India will be 9:00 PM, Beijing will be 11:00 PM, and Tokyo will be midnight.
Not to metion that in West Coast it'll be 7:00, which is pretty early.

I'm not being negative, I'm just being realistic...you understand thet, right?

We'll try some at different times on weekends, at the very least...

How about we have regular Math Jam at 3:00 PM EST (noon, PST) and a special one for people in Eastern Asia and Australia once every so often?

[quote="Fierytycoon"]Yes, but you must remember: during the school year, everyone is busy with school, and will probably not be able to attend during the mornings and afternoons...[/quote]

True, but during the weekends...of course, a lot of people have things going on then, too, but there's a better chance of having people com to a math jam at 10:00 AM on a weekend than on a weekday.

Of course 10 AM in a weekday is impossible, Mr. Rusczyk and Mr. Crawford will be at work and the reast of us will have to go to school.
But you do have a point there, Fiertycoon. Hmm...How about if we do it on big breaks?

[quote="rippledance"]Different countries have different breaks.[/quote]
I know. I'm not talking about small ones like Columbus Day or something attached to one's country such as St. Patrick's Day. I'm talking about the big ones that happen every season.

[quote]Different countries have different breaks.[/quote]

Even different schools have different breaks. But yeah, I guess we could do it on weekends or national holidays.

[quote="Fierytycoon"]Even different schools have different breaks. But yeah, I guess we could do it on weekends or national holidays.[/quote]
You mean "international," right?

No. Schools in different states often have different break periods. Furthermore, private schools usually have different break periods as well.

I concur - weekends are the most feasible. Holidays...well, from what I know, a lot of countries have hols at some point in June and some point in December.

if i had to do a mathjam at 4 am. i wouldnt mind at all. anything for aops!

Me neither. 7 AM is fine for me! :)
Still, think about Mr. Rusczyk and Mr. Crawford...

[quote="tmart"]I've just seen that today there is a MathJam, but...
I'm from Italy and I'd like to know when I should come here to follow the lesson...  SO
[b]what time (here in Italy, +1 est [i.e. London is 0]) does the mathjam start?[/b]

thank you and please answer quickly!

 :cool: [/b][/quote]
7:30 PM EST. EST is GMT - 5:00, so if you're at GMT+1:00, it would be at 1:30am.

The Math Jam starts 3 hours from now.  Look at the time of this post to figure out when it begins.

Most of the United States is presently observing Daylight Savings Time.  So 7:30 PM ET implies 7:30 PM EDT, which is GMT-4.

If you are certain you are at GMT+1, there is a 5-hour offset.

Does Italy operate on summer time?  If so, you are probably at GMT+2, which would be a 6-hour offset.

While I can't tell you what time it is in your location in Italy, these two sites will give you authoritative times in the United States and UTC:  [url]http://www.time.gov/[/url] and [url]http://tycho.usno.navy.mil/cgi-bin/timer.pl[/url]

If your AoPS / Mathlinks time stamps are correct, just use the time stamp from this message and know the Math Jam starts in just over 3 hours.

[quote="Erica"]i know the site is primarily aimed at americans, but you have a rapidly growing international membership  [... ]
i dont know how many non-US members you have but if you have more than a few [...]
just a suggestion
 :)[/quote] A flash news: AoPS and MathLinks merged forces long time ago, on 7th of July. The forum IS NOT primarily aimed at americans, and there are over 1500 non-american members.

[quote="Valentin Vornicu"] A flash news: AoPS and MathLinks merged forces long time ago, on 7th of July. The forum IS NOT primarily aimed at americans, and there are over 1500 non-american members.[/quote]

I don't see what your point is.  Erica posted her message in August of 2003 which was way before the merge.

ooops ...  :blush:  :blush:  :blush:  ... my bad, I read August 2004  (it seems that Erica was a member of both boards before the merge) ...

I also made the same mistake. Luckily there's a delete button on posts :)


for decreasing positive series $\sum a_n$,if $\sum n(a_n-a_{n+1})$ converges,prove $\sum a_n$ converges.
I only know how to prove if reverse :)

There is no way this can be true :). Just take $a_{n+1}$ very close to $a_n$. The quantities $n(a_n-a_{n+1})$ will be as small as you want them to be, but you don't even necessarily get $a_n\to 0$. You can make the sequence $(a_n)_n$ decrease so slowly, that it stays somewhere above $1$, for example.

For a concrete example, take $a_n=2-\left(\frac 12+\ldots+\frac 1{2^n}\right)$.

I know you mean now,but if we have $a_n \to 0$
it will be true??

[quote="zhaobin"] if we have $a_n \to 0$
it will be true??[/quote]
Yes. Let $b_{mn}=a_n-a_{n+1}$ if $n\geq m$ and $0$ if $m<n$. Note that $b_{mn}\geq 0$. Now we are going to employ a very useful theorem that says that, if all $b_{mn}\geq 0$, then $\sum_m\sum_n b_{mn}=\sum_n\sum_m b_{mn}$ (this is just the Fubini theorem for sequences). Summing in one order, we get one of our series and summing in the other order, we get the other one. Another solution uses an even stronger theorem that says that in a sum of non-negative terms, you can group and rearrange terms in any way you want without affecting the (possibly infinite) value of the sum. So, you may think of the sum $(a_1-a_2)+(a_2-a_3)+(a_2-a_3)+\dots$ ($a_n-a_{n+1}$ is repeated $n$ times) and just group the terms in 2 different ways.

Both theorems are very useful to know and not too hard to prove based on the idea that the sum of a family of non-negative numbers is just the supremum of sums over all finite subfamilies.

If you want a "direct" proof, you can reason the following way. Fix $N>0$. Choose $M>N$ such that $Na_{M+1}<1$. Now write $\sum_{n=1}^N a_n=\sum_{n=1}^N (a_n-a_{M+1})+Na_{M+1}\leqslant \sum_{n=1}^M (a_n-a_{M+1})+1=\sum_{n=1}^M n(a_n-a_{n+1})+1$, so if the partial sums of the second series are bounded, so are the partial sums of the first series.

I was trying to put together a proof based on summation by parts, with less than satisfactory results because I needed $na_n$ to be bounded, which is a rather stronger condition than $a_n\to0.$ The more I look at fedja's double sum argument, the more convinced I am that that is [b]the[/b] correct proof.


Assume that ( X,A, µ) is a measure space such that µ(X) =1 and
for every B∈A with  µ(B)>0, there exists a measurable subset C of B,
such that µ(B)> µ(C) >0. Show that there exists D∈A,
such that µ(D)=1/2.

Same answer as to your other post. :) Wait until the deadline (May 13) is over!


Show that there exist two natural numbers $a,b$ such that the sequence
$F_{1}=a,F_{2}=b,F_{n+1}=F_{n}+F_{n-1}$ contains no prime number.

I think you want the condition $gcd(a, b) = 1$, or else it's trivial.

Yes, of course I forgot to mention $(a,b)=1$, sorry.  :)