function
number theory solved
number theory

Let sigma(n) denote the sum of the positive divisors of the positive integer n and tau(n) the number of such divisors. Prove that tau(n) is a prime whenever sigma(n) is.

Suppose that sigma(n) is prime. Since sigma is well-known to be a multiplicative function, it follows that n = p a for some prime number p and positive integer a.
Thus tau(n) = 1+a and sigma(n) = (p a+1 -1)/(p-1).

Suppose, for a contradiction that tau(n) is not prime, that is 1+a = x*y for some integers x,y > 1.
Then : sigma(n) = (p xy -1)/(p-1) = [((p x) y  -1)/(p x -1)]*[(p y -1)/(p-1)] contradicting that sigma(n) is prime.
Therefore tau(n) is prime and we are done.

Pierre.


By induction $ x_n\equal{}ctg\frac{\pi }{3*2^n},y_n\equal{}tg\frac{\pi}{3*2^{n\minus{}1}}.$


geometry
circumcircle
analytic geometry
trigonometry
geometry proposed

[color=indigo]Let $ABC$ is an acute angled triangle and let $A_{1},\, B_{1},\, C_{1}$ are points respectively on $BC,\,CA,\,AB$ such that $\triangle ABC$ is similar to $\triangle A_{1}B_{1}C_{1}.$
Prove that orthocenter of $A_{1}B_{1}C_{1}$ coincides with circumcenter of $ABC$.[/color]

at the first sight this problem can be killed by using baricentric coordinates, denote $\overrightarrow{r_{A_{1}}}= x \cdot \overrightarrow{r_{B}}+(1-x) \cdot \overrightarrow{r_{C}}$ etc, and now we have $\overrightarrow{r_{H_{1}}}= \cot{B}\cot{C}r_{A_{1}}...$ etc, because the triangles are similar, and expand $\overrightarrow{r_{H_{1}}}$ and put the condition to be equal with $\overrightarrow{r_{O}}= a \cos{A}\overrightarrow{r_{A}}$ etc, the conclusion follows.

I don't understand your solution. You say that if $\overrightarrow{r_{A_{1}}}= x \cdot \overrightarrow{r_{B}}+(1-x) \cdot \overrightarrow{r_{C}}$ , $\overrightarrow{r_{B_{1}}}= x \cdot \overrightarrow{r_{C}}+(1-x) \cdot \overrightarrow{r_{A}}$, $\overrightarrow{r_{C_{1}}}= x \cdot \overrightarrow{r_{A}}+(1-x) \cdot \overrightarrow{r_{B}}$ then $A_{1}B_{1}C_{1}$ is similar to $ABC$ ? This is not true.

no, i refered just to write $ r_{A_{1}}$ and the analogues as a sum of $ r_{b}, r_{c}$ with constanst $ x,y,z$ not $ x$ in the same place, and put the condition of similarity just when you write $ r_{H_{1}}$.

[img]8292[/img]

Here is my proof:
Let H be the hortocentre of $A_{1}B_{1}C_{1}$. Since $\angle B_{1}AC_{1}=-\angle B_{1}A_{1}C_{1}= B_{1}HC_{1}$ (using directed angles), it's easy to see that A,B,C are on the circles obtained from the circumcircle of $A_{1}B_{1}C_{1}$ with symmetries on the sides of $A_{1}B_{1}C_{1}$.

We need to prove that $HA = HB = HC$. But HB, HC are chords of circles with the same radius. We need only to check that their respective angles on the circumference are equal. But this is indeed very easy! $\angle HA_{1}C = \angle HA_{1}B$, $\angle HC_{1}B = \angle HC_{1}A$, $HB_{1}A = HB_{1}C$ using directed angles!

It's also true a nice statament: in the picture, choose  arbitrary C on the first circumference, find B as intersection of $CA_{1}$ and the second circumference, find A as intersection of $BC_{1}$ and the third circumference.
Then ABC is a triangle, and is a triangle similar to $A_{1}B_{1}C_{1}$.

[quote="April"] [color=darkred]Let $ABC$ is an acute angled triangle and let $A_{1},\, B_{1},\, C_{1}$ are points respectively on $BC,\,CA,\,AB$ such that $\triangle ABC$ is similar to $\triangle A_{1}B_{1}C_{1}.$
Prove that orthocenter of $A_{1}B_{1}C_{1}$ coincides with circumcenter of $ABC$.[/color] [/quote]
[color=indigo][b]Proof.[/b] Let $O$ be the orthocentre of triangle $A_{1}B_{1}C_{1}$ and let $H=B_{1}O\cap C_{1}A_{1},\,K=C_{1}O\cap A_{1}B_{1}$

We have: $A_{1},\,H,\,O,\,K$ are cyclic.
$\Longrightarrow \angle B_{1}OC_{1}=\angle HOK=180^\circ-\angle B_{1}A_{1}C_{1}=180^\circ-\angle BAC$
So the quadrilateral $B_{1}OC_{1}A$ is cyclic.
$\Longrightarrow \angle OAC=\angle OC_{1}B_{1}=90^\circ-\angle A_{1}B_{1}C_{1}=90^\circ-\angle ABC$

Similar: $\angle OCA=90^\circ-\angle ABC$
So triangle $OAC$ is isosceles or $OA=OC$

Similar: $OC=OB$
Hence: $O$ is the circumcenter of triangle $ABC$[/color]


inequalities calculus integration AMC USA(J)MO USAMO

Given that $ abc \equal{} 1$ and $ a, b, c > 0$, show that $ \frac{1}{a \plus{} b \plus{} 1} \plus{} \frac{1}{a \plus{} c \plus{} 1} \plus{} \frac{1}{b \plus{} c \plus{} 1} \leq 1$.

Well, it has the potential of being straightforward if you don't mind a bit of multiplication:

[hide="A Brainless Solution"]Let $ a \equal{} x^3$, $ b \equal{} y^3$, $ c \equal{} z^3$. Then, the constraint becomes $ xyz \equal{} 1$, $ x,y,z > 0$, and thus the following homogeneous inequality (1's were replaced by xyz):

$ \sum_{cyc} \frac {xyz}{x^3 \plus{} y^3 \plus{} xyz} \leq 1$

We can multiply out the denominator and then rearrange terms to get:

$ \sum_{sym} x^5 y^2 z^2 \leq \sum_{sym} x^6 y^3 z^0$

Which is Muirhead's.[/hide]

[quote="Zhero"]After that substitution, we could also just note that $ a^3 \plus{} b^3 \leq ab(a \plus{} b)$, from which the desired result follows easily. No Muirhead there. :P

How would we approach this without that substitution? How did you think of that substitution? Just curious; I'm not very good with inequalities.[/quote]

I don't see how you get that $ a^3 \plus{} b^3 \leq ab(a \plus{} b)$.

[quote="Zhero"] we could also just note that $ a^3 \plus{} b^3 \leq ab(a \plus{} b)$[/quote]

Well , that's false . Actually it is :

$ a^3\plus{}b^3\ge ab(a\plus{}b)\Leftrightarrow (a\plus{}b)(a^2\minus{}ab\plus{}b^2)\ge ab(a\plus{}b)\Leftrightarrow (a\minus{}b)^2\ge 0$

Well, my goal was to make everything have the same degree. Since $ abc\equal{}1 \implies a^k b^k c^k \equal{} 1$, we could replace the 1's in there with $ (abc)^{1/3}$, and every term would have degree 1. But it's nicer if we make a substitution that gives us all integral exponents.

I think it's posted already,
By Cauchy-Schwarz,
$ (1\plus{}a\plus{}b)(c\plus{}1\plus{}1)\geq(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^2$
$ \Leftrightarrow \frac{1}{1\plus{}a\plus{}b}\leq \frac{c\plus{}2}{(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^2}$
then it's suffice to prove that,
$ \frac{a\plus{}b\plus{}c\plus{}6}{(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^2}\leq 1$
$ \Leftrightarrow \sum \sqrt{ab} \geq 3$ wich is true by AM-GM.

[quote="alex2008"][quote="Zhero"] we could also just note that $ a^3 \plus{} b^3 \leq ab(a \plus{} b)$[/quote]

Well , that's false . Actually it is :

$ a^3 \plus{} b^3\ge ab(a \plus{} b)\Leftrightarrow (a \plus{} b)(a^2 \minus{} ab \plus{} b^2)\ge ab(a \plus{} b)\Leftrightarrow (a \minus{} b)^2\ge 0$[/quote]

My bad, that's what I meant.

[quote="TZF"]Well, my goal was to make everything have the same degree. Since $ abc \equal{} 1 \implies a^k b^k c^k \equal{} 1$, we could replace the 1's in there with $ (abc)^{1/3}$, and every term would have degree 1. But it's nicer if we make a substitution that gives us all integral exponents.[/quote]

I see. Thanks for that; I'll use that technique later on.

[quote="Brut3Forc3"]It's equivalent to USAMO 1997/5:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2971#2971[/quote]
I know. :P That's where I originally got the problem from. I assumed it was homogeneous, and unintelligently performed the substitutions $ x \equal{} a^3$, $ b \equal{} y^3$, and $ z \equal{} c^3$. I was then stuck; I was just curious to see if there was a way to finish that inequality off without reversing my substitutions and doing it the slick way.

[quote="peine"]I think it's posted already,
By Cauchy-Schwarz,
$ (1 \plus{} a \plus{} b)(c \plus{} 1 \plus{} 1)\geq(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})^2$
$ \Leftrightarrow \frac {1}{1 \plus{} a \plus{} b}\leq \frac {c \plus{} 2}{(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})^2}$
then it's suffice to prove that,
$ \frac {a \plus{} b \plus{} c \plus{} 6}{(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})^2}\leq 1$
$ \Leftrightarrow \sum \sqrt {ab} \geq 3$ wich is true by AM-GM.[/quote]
Thanks.


Let $ a\neq0,$ $ b\neq0$ and $ c\neq0.$ Prove that:
\[ \frac{a^2}{b^2}\plus{}\frac{b^2}{c^2}\plus{}\frac{c^2}{a^2}\plus{}\frac{a}{c}\plus{}\frac{b}{a}\plus{}\frac{c}{b}\plus{}6\geq4\left(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\right)\]

zaya_yc, but $ x$ can be negative. :wink:

The inequality is equivalent to
$ 2(x\plus{}y\plus{}z\minus{}3)^2\plus{}x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}zx \ge 0$,
where $ xyz\equal{}1$.

My proof:
$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2} \plus{} \frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{b} \plus{} 6\minus{}4\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right)\equal{}\frac{1}{2}\sum_{cyc}\left(\frac{a}{b}\plus{}\frac{b}{c}\minus{}2\right)^2.$


trigonometry
vector
calculus
inequalities proposed
inequalities

A vector is in R3 such that the angles it make with the x,y,z axes are a,b,c which are always between 0 and 90. So we know that cos^2(a)+cos^2(b)+cos^2(c)=1.

Prove a+b+c<=180 without using lagrange multiplier.

[quote="weic"]A vector is in $ R_3$ such that the angles it make with the x,y,z axes are a,b,c which are always between $ 0^0$ and $ 90^0$. So we know that $ cos^2(a) \plus{} cos^2(b) \plus{} cos^2(c) \equal{} 1$.

Prove $ \angle a \plus{} \angle b \plus{} \angle c \leq 180^0$ without using lagrange multiplier.[/quote]

It isn't necessary to use Langrange multiplier; one can solve for one of the variables in the last equation and substitute that into the first equation and then partial differentiate.  As for a non-calculus approach, I haven't come up with any yet.


LaTeX
conics
parabola
AMC
USA(J)MO
USAMO
AoPS Books

Can someone tell me how to do the special P in Latex? It is used in AOPS books to represent parabolas and polygons and for an example, see No. 4 of the 2008 USAMO.

Thanks, (I did not find it to be in the Greek Symbols).

\mathcal{P} $ \mathcal{P}$
If you hover over the $ \mathcal{P}$ in [url=http://www.artofproblemsolving.com/Wiki/index.php/2008_USAMO_Problems/Problem_4]2008 USAMO Problems/Problem 4[/url] you'll see the LaTeX code.


geometry
ratio
combinatorics unsolved
combinatorics

There is a bird at each vertex of a regular n-gon. Suddenly the birds all fly away and then return a minute later. After their return there is again one bird at each vertex, but not necessarily the same bird as before. For which n can we always find three birds such that the triangle formed by their initial positions is of the same type (acute, obtuse or right-angled) as the triangle formed by their final position.

hint: for even $ n$ there are always 3 birds whose initial and final positions both form a right triangle, and this is pretty easy to show.

We can find it for any $ n$ except if $ n \equal{} 5$.
If we number the verticies of the polygon consecuantly $ 1,2,3,4,5$ and the birds that were on them at the beggining are now sitting in the order $ 1,4,2,5,3$ it is easy to check that every triangle changed form acute to obtuse and vice versa.

Now we show that the rest is possible:
[b]Case 1[/b] $ n \equal{} 2k$
Pick two birds in the final configuration that are sitting exactly opposite to each other and mark them $ B_i$ and $ B_j$, $ B_iB_j$ is the diameter of the circle that goes thorugh the vertecies of the polygon. Now let's look at the bird $ B_k$ that was sitting directly opposite to $ B_i$ at the beginning. If $ B_j \equal{} B_k$ then pick any other bird; it is easy to see that the triangle formed by this new bird and birds $ B_i$ and $ B_j$ was rightangled before and also is now in the final configuration (with right angle at this new bird). If $ B_j \neq B_k$ then tringle $ B_iB_jB_k$ was right angled at the beginning (with right angle at $ B_j$) and is also right angled now (with right angle at $ B_k$).

[b]Case 2[/b] $ n \equal{} 4k \plus{} 3$
Whenever $ n$ is odd, no triangle formed by the birds is rightangeled, therefore it suffices to prove that the number of
acute triangles doesn't equal the number of obtuse triangles.
The number of triangles formed by birds equals: $ \binom{4k \plus{} 3}{3} \equal{} \frac {(4k \plus{} 3)(4k \plus{} 2)(4k \plus{} 1)}{6}$
The only even term in the numerator is $ (4k \plus{} 2)$ and is not divisiable by $ 4$ therfore the number of triangles is odd, and therefore
can't be partitioned into two equal size sets, so the number of acute and obutse triangles can't be the same.

[b]Case 3[/b] $ n \equal{} 4k \plus{} 1$ and $ n > 5$
Let's count the number of obutse triangles(if $ n \equal{} 4k \plus{} 1$). Each diagonal that goes out of some vertex divides the polygon in two parts and the number of obutse triangles that have this diagonal as the longest side is equal to the number of vertecies on the shorter part. So the number of obutse triangles that have some vertex on their longer side equals to $ 2\frac {\frac {n \minus{} 3}{2}\frac {n \minus{} 1}{2}}{2}$ , so the number of obutse triangles equals to $ \frac {n(n \minus{} 1)(n \minus{} 3)}{8}$. If we now calculate ratio bettween the number of obutse triangles and all triangles,
which is after some division $ \frac {3(n \minus{} 3)}{4(n \minus{} 2)}$ which equals one half when $ n \equal{} 5$ and is not equal to that value for any other $ n$. So the number of triangles is different for any other $ n$ which finishes the problem.

I hope there isn't any mistake in above reasoning.


What is this, is this some kind of a joke :rotfl:  :rotfl:  :rotfl:  :rotfl:  even though it shouldnt be here!   dont you know $LaTeX$, nilesh???????? I cant believe that????!!!! :rotfl:  :rotfl:  :rotfl:


Three circles $ (O_1), (O_2), (O_3)$ intersect in one point $ O$. Three points $ A_1, A_2, A_3$ lie on the circles $ (O_1), (O_2), (O_3)$ respectively such that $ OA_1, OA_2, OA_3$ are parallel to $ O_2O_3, O_3O_1, O_1O_2$ in that order.
Prove that $ O, A_1, A_2, A_3$ are concyclic.

O2O3∥OA1  OF∥AA1
∠A3CB=∠A3OF=∠BAA1
Analogously,∠CAA1=∠CBA2
We have ∠A3OA2=∠BAC
∠ABA2=∠A3CA=180-(90+∠BAC)=90-∠BAC
sin∠A3CB/ sin∠A3CA * sin∠CAA1/ sin∠BAA1 * sin∠ABA2/ sin∠CBA2=
1
So we have AA1,BA2,CA3 are concurrent , denote D is the intersection
it’s easy to prove O,A1,A2,A3,D are concyclic with OD the diameter

Dear Mathlinkers,
D is the orthocenter of ABC and O, D, A1, A2, A3 are on the Mannheim's circle of ABC wrt to (O1), (O2), (O3).
see: http://perso.orange.fr/jl.ayme  vol.2 article 1.
Sincerely
Jean-Louis


[quote="ali_ms2004"]can somebody introduce me some nice book for learning algebra from the beggining ??!!  :P[/quote]
excuse me from the beggining means from the easy part !! basic levels !!  :lol:

[quote="ehsan2004"][quote="ali_ms2004"]can somebody introduce me some nice book for learning algebra from the beggining ??!!  :P[/quote]

ali_ms2004, at first ,do you have Mehdi safa's book ?  ;)[/quote]
unfortunately there is no more safa's book in the bazaar !!  :(

but in near future it will be available !! :)


quadratics
algebra
polynomial
number theory unsolved
number theory

$ p$ be a prime number
$ x^3\plus{}y^3\minus{}3xy\equal{}p\minus{}1$
find all $ x, y$ be positive integer

[quote="Theorem1"]You can use cardan's method! Let's see before posting the solution the other solutions.[/quote]
What is cardan's method?

Hello. We are going to use and identifity in order to solve the equation. I think it is well-known. x^3 + y^3 + z^3 - 3xyz =(x+y+z)(x^2 + y^2 + z^2 - xy - yz - xz). Now in our equation we add 1 to both of the sides. We write it as (x+y+1)(x^2 + y^2 + 1- xy - x - y). Now we have 2 alternatives. x+y+1=1 or x+y+1=p. The first alternative is wrong because we get x+y=0 which is false because x and y are both positive integers. Now x+y=p-1. x^2 + y^2 + 1 - xy - x -y =1. Here if we put x+y=p-1 we get that x^2 - xy +y^2 = p-1. So
x+y=x^2 - xy + y^2). Now we can write this equation as x^2 - (y+1)x + y^2 -y =0. Now think about the discriminant of this equation. This is a quadratic equation belongs to x. The discriminant is Y^2 + 2y + 1 - 4y^2 + 4y= -3y^2 + 6y + 1= -3y^2 + 6y -3 +4 = 4-3(y-1)^2. This is more than equal 0 so either y-1 = 1 or y-1 = 0. Now if you put them in the quadratic equation you will get the desired result.  :P  :P

\[ {x^3} \plus{} {y^3} \minus{} 3xy \equal{} p \minus{} 1 \Leftrightarrow {x^3} \minus{} 3xy \plus{} {y^3} \plus{} 1 \minus{} p \equal{} 0\]
Let : \[ x\equal{}a\plus{}b\]
so \[ {(a \plus{} b)^3} \minus{} 3(a \plus{} b)y \plus{} {y^3} \plus{} 1 \minus{} p \equal{} 0\]
\[ \Leftrightarrow {a^3} \plus{} {b^3} \plus{} 3(a \plus{} b)(ab \minus{} y) \plus{} {y^3} \plus{} 1 \minus{} p \equal{} 0\]
\[ ab \equal{} y \Rightarrow {a^3} \plus{} {b^3} \plus{} {y^3} \plus{} 1 \minus{} p \equal{} 0\]
So \[ ({a^3} \plus{} 1)({b^3} \plus{} 1) \equal{} p\]
Suppose that
\[ b \geqslant a\]
\[ ({a^3} \plus{} 1)({b^3} \plus{} 1) \equal{} p \Rightarrow {a^3} \plus{} 1 \equal{} 1\]
\[ \Rightarrow \,\,a \equal{} 0,\,\,p \equal{} (b \plus{} 1)({b^2} \minus{} b \plus{} 1)\]
\[ \, \Rightarrow x \equal{} 1,\,p \equal{} 2,y \equal{} 0\]
So:
\[ \left( {x,y,p} \right) \in \left\{ {(1,0,2),(0,1,2)} \right\}\]
Cardan's method is used to solve equations 3deg.

Cardano's method is used to solve degree $ 3$ polynomials in [b]one[/b] variable.  Unlike the quadratic equation, it tells you a lot less about the conditions necessary to solve them in more than one variable.

Yes i know but i consider here x the only variable!  so why i put x=a+b? it's clair!

My point is that if you tried that it would either not lead you anywhere or lead you to the factorization.  And if all you were looking for is a factorization, you should be trying to compute singularities instead.

Letting $ x\equal{}a\plus{}b$ and $ y\equal{}ab$ means that $ a$ and $ b$ are the solutions of $ u^2\minus{}xu\plus{}y\equal{}0$; $ a$ and $ b$ are not necessarily integers. That's the problem with your solution.


Find all $ n\ge 2$ ($ n\in\mathbb{N}$) for which the following holds: $ \forall (n,a)\equal{}1\text{ and }(n,b)\equal{}1\,\,\, n|(a\minus{}b)$ iff $ n|(ab\minus{}1)$.


Are there any Washington residents here that participated in Math is Cool on Saturday?

yep, my names also jeffrey, i got 3rd for 9/10 division I... im from bellevue high. my team also placed 3rd... we did pretty bad, but oh well.

I'm Jeffrey from Lewis and Clark and I got 6th in 9/10 divison. My team got 8th or something like that, but we don't practice or anything so we were fine with that.

The format for the individual competition is 40 questions in 35 minutes. The first twenty are pretty easy, the next 10 are medium, but some of the last 10 can be almost AIME level. At least it seems like that because you have less than 1 minute per question...

ya, that would be why 31/40 was the highest score... heh. i got 16 from the first 20, 8 from the next 9, and 3 from the last 11 (30-40 are all challenge so that would be 11 problems). 27 total.

besides the individual, there are:

Individual Multiple Choice (goes only to team score): 9 Q's, 15 min, +2 for correct, -1 for wrong, 0 for blank... i did horribly, 5 pts.

Pressure (ugh): 5 Q 10 min, turn in 1 Q every 2 min. 1st = 1, 2nd = 2, etc.

Team: 10 Q 15 min (i think)

College Bowl (vs other schools): 3 rounds of 7 Q

our Ind. MC was horrible... pressure was ok, team was not great, college bowl was ok.

btw, jeffrey, are you 9th or 10th grade?

[quote="s0mp"]hehehe *nasal laugh* what a lame name for a competition... :P

what are the questions like? around here we have competitions like that...[/quote]

Meh.  At least you have a competition.  All we have 'round here are national ones - AMC/AIME/USAMO, Mandelbrot, and the individual take-home-work-on-these-for-a-month ones like the USAMTS.

At least you have Mandelbrot, instead of that we have MAML Olympiad, not that interesting. I could also (try and) compete for MATHCOUNTS but not likely anymore :(
I could always pursuade but I am only one person and there's a whole school system to convince, trying itself is tiring me so much I can only do AoPS, eat, and sleep :)

Ya, but once you get to high school, you can do all the east-coast competitions.  Here in Nevada, there is nothing.  No, really, nothing.

For those of you looking for local math contests here is a nice website that lists them by region. I'm sure there are even more than this. But at least these are most of the ones with websites.

http://www.mualphatheta.org/Contests/MAO_Contests_Regional.htm

I'm not sure how far you are from the ones in Southern California mathfan, but maybe your school could get involved with some of those.

I was there this weekend, but I was leading one of the scoring rooms -- not nearly as fun as when I used to compete in such things.  :-)  I think Math is Cool is one of the best contests around.  The tests are no calculator, and are almost all free response (as opposed to multiple choice), so I think the results depend a lot more on mathematical abilities than application of technology or testing skills.  The Pressure Round is a really neat format, too.

I know that Math is Cool is interested in expanding beyond Washington state, and that hosting the contest is a great fundraising opportunity for math teams.  If your coach might be interested in doing something like that, have him/her get in touch with the organizers via the website:  http://www.academicsarecool.com.

I don't want to be pushy, but for people who don't have a lot of local contest options, my company administers a number of mail-in contests throughout the year.  More information is available on our website, http://www.natassessment.com, or you can send me a private message.

yea im from thomas jefferson
and i won the 9/10th individual on state.

DIE!! PERSON FROM EXCELLENT SCHOOL...unlike mine...lol, jk, welcome to AoPS ;)


If the reciprocal of $ (x\minus{}2)$ is $ (x\plus{}2)$, what is the greatest possible value of $ x$?  Express your answer as a common fraction.

$ \frac{1}{x\minus{}2}\equal{}x\plus{}2 \implies 1\equal{}x^2\minus{}4 \implies x^2\equal{}5.$

answer : $ \sqrt{5}$.

express your answer as a common fraction??


Note that the 1's all add up, giving us $ \boxed{123454321}$.

Yes, super's pattern works only up to 9 ones (I think). When it reaches ten ones, there is carrying and 0's, which would completely ruin the pattern.


For $ S\equal{}{x_{1},x_{2},...,x_{n}}$ a set of $ n$ real numbers, all at least 1, we count the number of reals of the form $ \sum_{i\equal{}1}^{n} \epsilon_{i} x_{i}$, (where $ \epsilon_{i} \in {0,1})$ lying in an open interval I of length 1.  Find the maximum value of this count over all I and S.

First, can someone actually explain what the problem is asking?

Second, if someone gets a soltuion, could they explain how to do it?

Thanks

This is more a combinatoric problem rather than an algebric one.
The result is $ \binom {n} {\frac n 2}$ and follows from Sperner's theorem about antichains.

Pierre.

can you explain your solution please pbornsztein?


hi !!

let $ a,b,c$ be nonnegatives reals , with $ abc(a \plus{} b \plus{} c) \equal{} 4$

prove that :

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

[hide]ya yahoudi ya himar , tfou 3lik [/hide]

[quote="mehdi cherif"]hi !!

let $ a,b,c$ be nonnegatives reals , with $ abc(a \plus{} b \plus{} c) \equal{} 4$

prove that :

$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq 1 \plus{} \frac {1}{2\sqrt {2}}$
[/quote]
Hello!
Your inequality is wrong. Try $ a \equal{} b \equal{} 10$ and $ c \equal{} \sqrt {100.04} \minus{} 10.$ :wink:

[hide="to mehdi cherif"] Tizdayen besavlanut. :) [/hide]

are u sure hhhhhhhh  :rotfl:   $ (40 ??!!)$

[hide]sir tkawad alyhodi mata3rfoch hta t7asbo. tghalbo mmatkom machi tghalbo zagha , and i dont speak hebrew  :rotfl: [/hide]

[quote="mehdi cherif"]are u sure hhhhhhhh  :rotfl:   $ (40 ??!!)$

[/quote]
I have fixed it. :P

maybe you need more lessons in calculation  :P

[hide="@ Arcady"]lech tizdayen  :) [/hide]

[quote="mehdi cherif"]maybe you need more lessons in calculation  :P [/quote]
It is not interesting already since, your inequality is wrong. :wink:
[quote="mehdi cherif"]
[hide="@ Arcady"]lech tizdayen  :) [/hide][/quote]
[hide="to mehdi cherif"] Thx! I like to do it.   :lol:  [/hide]

may be my problem is complicated for u, but not enought as your ;) (i know u inderstand)

[hide="@arcady"] :) אה, שמח לשמוע את זה! אז אמא שלך יהיה ניק יהודי מלוכלך
אתה איש מת![/hide]

[hide="to mehdi cherif"]
[quote="mehdi cherif"]
אתה איש מת![/quote]

לא אסכים איתך! חייתי אלפי שנים לפניך ואחיה אלפי שנים אחרי שתיעלם.
אתה חי שנאה. זאת בחירתך החופשית אך מכאן בלעדיי
[/hide]

[hide="@arcady"]לפי נגדנו, אנחנו שונאים אותך, אבל עכשיו היא מי אתה תשלם על זה, כי אתה שלנו האחים להתנקשות![/hide]

so now try to solve it  :rotfl: שוטה

Dear mathlinks ,I'm sorry for all this , just because somebody forgot  his calculator for hiding and here is my correct problem :)

let $ a,b,c$ be nonegatives reals , with $ ab\leq2$ , $ bc\leq 2$ and $ ca\leq 2$

prove the following inequality :

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

:)

[hide="Remark"]this inequality is equivalent to the hard one :

$ x,y,z$ are nonegatives such that $ xy\plus{}yz\plus{}zx\equal{}1$ then :

$ \frac{1}{\sqrt{x\plus{}y}}\plus{}\frac{1}{\sqrt{z\plus{}x}}\plus{}\frac{1}{\sqrt{y\plus{}z}}\geq 2\plus{}\frac{\sqrt{2}}{2}$[/hide]


I need state and nationals problems to practice before my state competition. Please PM them to me or post them here, thanks.

(I want sprint and target most, but team and countdown would help too.)

states can be found here

[url]http://www.yashar.biz/mathcounts[/url]

My website: http://www.freewebs.com/totallyzealous
NOTE: THIS IS A NEW URL.  :)  :)

Dude, that like ur 4th url or wat?

Check the "National problems for mathcounts" thread.

can u do something on MATHCOUNTS geometry?

thx
jorian

[quote="#H34N1"]http://www.mathprob.com/math/?page_id=39

There are lessons posted a few times a month that I make about MC level.[/quote]

Wow, that website is actally useful to me!

Okay. I'm writing one on Geometry right now :)

ok, thank you

i might comment on how good it is later

you should post this in ur blog ani! (comments)

jorian


In how many ways can 4 on/off switches in a row be set so that no two adjacent switches are on?

8 ??? I counted so i could've miscounted.

[quote="biffanddoc"][hide]
First, call the switches 1,2,3,4
You can have all of them off. (1 way)
You can switch any of them on (4 ways)
You can have 1,3 or 2,4 or 1,4 on. (3 ways)
You can't have 3 on or 4.
Add them up to get 8[/hide][/quote]
I forgot about 1,4.  :D  :D  :D :D :D


Which format do you use for AoPS?

Mathlinks, or AoPS?

I prefer the AoPS layout.
Though I can I always help make a new one. :evilgrin:

I use mathlinks because it works better with my crappy computer.

I prefer AoPS because it's fancy. Bleh. Just my taste.

I prefer MathLinks because it is wider. I like to switch every now and then, though.

Mathlinks of course. It was the reverse, I didn't know about AoPS...
But ML is prettier and wider.

[quote="andrewjjiang97"][quote="bluecarneal"][quote="jjfun1"]I prefer AoPS because I didn't find out about mathlinks until later, and by that time I was used to AoPS. W/e.[/quote]
Me too. :!:[/quote]

Me three.[/quote]

Me four.

[quote="spaceguy524"][/quote][quote="andrewjjiang97"][quote="bluecarneal"][quote="jjfun1"]I prefer AoPS because I didn't find out about mathlinks until later, and by that time I was used to AoPS. W/e.[/quote]
Me too. :!:[/quote]

Me three.[/quote][quote="spaceguy524"]

Me four.[/quote]
\Join the group!!! :D

[quote="eqjj168"][/quote][quote="spaceguy524"][/quote][quote="andrewjjiang97"][quote="bluecarneal"][quote="jjfun1"]I prefer AoPS because I didn't find out about mathlinks until later, and by that time I was used to AoPS. W/e.[/quote]
Me too. :!:[/quote]

Me three.[/quote][quote="spaceguy524"]

Me four.[/quote][quote="eqjj168"]
\Join the group!!! :D[/quote]

Me five.

But imo, AOPS is way better.  I like the color scheme.

Plus, it looks really good with google chrome!

I'm using Mathlinks right now, but normally I use AoPS.  I might switch, though, because the MathLinks layout is more prettifulz.  :P But then I might not because I'll miss FTW....

So I would say that I use both board layouts.

how do you make boxes for what other people said?

ME 8

Me 9. :lol:

[quote="piegraph"]how do you make boxes for what other people said?[/quote]

Use the quote tags.

[quote="piegraph"]how do you make boxes for what other people said?[/quote]
like this: [code][quote="piegraph"]how do you make boxes for what other people said?[/quote][/code]

[quote="Maybach"][/quote][quote="eqjj168"][/quote][quote="spaceguy524"][/quote][quote="andrewjjiang97"][quote="bluecarneal"][quote="jjfun1"]I prefer AoPS because I didn't find out about mathlinks until later, and by that time I was used to AoPS. W/e.[/quote]
Me too. :!:[/quote]

Me three.[/quote][quote="spaceguy524"]

Me four.[/quote][quote="eqjj168"]
\Join the group!!! :D[/quote][quote="Maybach"]

Me five.
[/quote]

Me six!

[quote="piegraph"]like this?

[quote]Maybach wrote:
eqjj168 wrote:
spaceguy524 wrote:
andrewjjiang97 wrote:
bluecarneal wrote:
jjfun1 wrote:
I prefer AoPS because I didn't find out about mathlinks until later, and by that time I was used to AoPS. W/e.[/quote]
me 7 or 8 or 9 or something like that[/quote]
[size=150][b]Notice  at the bottom right of each post, there is a "Quote" button that automatically quotes the post for you.[/b][/size]

[hide="FINAL RESULT"][quote="cyberspace"][/quote][quote="Maybach"][/quote][quote="eqjj168"][/quote][quote="spaceguy524"][/quote][quote="andrewjjiang97"][quote="bluecarneal"][quote="jjfun1"]I prefer AoPS because I didn't find out about mathlinks until later, and by that time I was used to AoPS. W/e.[/quote]
Me too. :!:[/quote]

Me three.[/quote][quote="spaceguy524"]

Me four.[/quote][quote="eqjj168"]
\Join the group!!! :D[/quote][quote="Maybach"]

Me five.
[/quote][quote="cyberspace"]

Me six![/quote]
[code][quote="cyberspace"][/quote][quote="Maybach"][/quote][quote="eqjj168"][/quote][quote="spaceguy524"][/quote][quote="andrewjjiang97"][quote="bluecarneal"][quote="jjfun1"]I prefer AoPS because I didn't find out about mathlinks until later, and by that time I was used to AoPS. W/e.[/quote]
Me too. :!:[/quote]

Me three.[/quote][quote="spaceguy524"]

Me four.[/quote][quote="eqjj168"]
\Join the group!!! :D[/quote][quote="Maybach"]

Me five.
[/quote][quote="cyberspace"]

Me six![/quote][/code][/hide]

[quote="eqjj168"][/quote][quote="piegraph"]like this?

[quote]Maybach wrote:
eqjj168 wrote:
spaceguy524 wrote:
andrewjjiang97 wrote:
bluecarneal wrote:
jjfun1 wrote:
I prefer AoPS because I didn't find out about mathlinks until later, and by that time I was used to AoPS. W/e.[/quote]
me 7 or 8 or 9 or something like that[/quote]
[size=150][b]Notice  at the bottom right of each post, there is a "Quote" button that automatically quotes the post for you.[/b][/size]

[hide="FINAL RESULT"][quote="cyberspace"][/quote][quote="Maybach"][/quote][quote="eqjj168"][/quote][quote="spaceguy524"][/quote][quote="andrewjjiang97"][quote="bluecarneal"][quote="jjfun1"]I prefer AoPS because I didn't find out about mathlinks until later, and by that time I was used to AoPS. W/e.[/quote]
Me too. :!:[/quote]

Me three.[/quote][quote="spaceguy524"]

Me four.[/quote][quote="eqjj168"]
\Join the group!!! :D[/quote][quote="Maybach"]

Me five.
[/quote][quote="cyberspace"]

Me six![/quote]
[code][quote="cyberspace"][/quote][quote="Maybach"][/quote][quote="eqjj168"][/quote][quote="spaceguy524"][/quote][quote="andrewjjiang97"][quote="bluecarneal"][quote="jjfun1"]I prefer AoPS because I didn't find out about mathlinks until later, and by that time I was used to AoPS. W/e.[/quote]
Me too. :!:[/quote]

Me three.[/quote][quote="spaceguy524"]

Me four.[/quote][quote="eqjj168"]
\Join the group!!! :D[/quote][quote="Maybach"]

Me five.
[/quote][quote="cyberspace"]

Me six![/quote][quote="eqjj168"][/code][/hide][/quote]
like this?

YEAH, I DIDN'T EPIC FAIL.

I LIKE AOPS BECAUSE IT'S SIDE IS COOL


algorithm
number theory
Euclidean algorithm

Prove that you can color positive rational numbers with two colors such that for each positive rational number $q$ is color of $q$ same as color of $\frac{1}{q}$ and different to color of $q+1$.

Let $a,b$ be two positive integers.
Now, use the euclidean algorithm to define two sequences of integers, namely $(q_i)$ and $(r_i)$ such that $a=r_0$
$b=q_1a+r_1$, (1)
$a = r_0 = q_2r_1 + r_2$,
...
$r_{i-1} = q_{i+1}r_i + r_{i+1}$ and so on.

Since $0 \leq r_i < r_{i-1}$ the algorithm will stop, and since the only reason to stop is to reach $r_i = 0$ for some $i$, we may consider the integer $n$ such that $r_n=0$.
Now let $f(\frac a b) = q_1+q_2 + \cdots + q_n$.

Now assume that $b>a$. Then $a = 0 \cdot b + a$ and then the algorithm associated to $\frac a b$ follows the one associated to $\frac b a$. Therefore $f( \frac a b ) = f( \frac b a).$ (2)
In another hand $\frac b a + 1 = \frac {a+b} a$, and $(b+a) = (q_1+1)a+r_1$ where $q_1,r_1$ are the same as in (1). Thus, from that step, the algorithm associated to $\frac {a+b} a$ is the same as the one associated to $\frac b a$. Therefore $f( \frac b a +1) = 1 + f( \frac b a)$. (3)

Thus, it suffices to define the coloring as follows :
For each positive rational $q$ let $q = \frac b a$ where $\gcd ( a,b) = 1$. Then color $q$ in red if $f( \frac b a )$ is even, and in blue otherwise.
From (2) and (3), we deduce that this coloring as the desired property.

Remark.
We may slightly simplify the above solution by defining $f( \frac b a)$ to be the first non-zero term of the sequence $(q_i)$

Pierre.

This was exactly my solution :)
One more thing to note, that there are only two such colorings (just put one color to 1 and you can determine all colors of all other rational numbers).


symmetry
superior algebra
superior algebra solved

the relation $R$ on the class of all groups defined by $GRG^{\prime}$ if and only if $G$ and $G^{\prime}$ are isomorphic. Show that $R$ has the properties of reflexivity, symmetry, and transitivity.  thanks for the help

Reflexivity: $f: x\mapsto x$  is an isomorphism from $G$ to $G$. Hence $GRG$.
Symmetry: If $GRG'$, then there is an isomorphism $f$ from $G$ to $G'$. $f$ is a bijection, hence it has inverse. It is easy to verify that $f^{-1}$ is an isomorphism from $G'$ to $G$.
Transitivity: Suppose $G_1RG_2$ and $G_2RG_3$. $f,g$ are isomorphisms from $G_1$ to $G_2$ and $G_2$ to $G_3$ respectively. Then $g\circ f$ is an isomorphism from $G_1$ to $G_3$.


Find all positive integers $n,x,y,k$ with $(x,y)=1$ and $k>1$ such that $3^n=x^k+y^k$

The only two solutions are $x=2,y=1,n=2,k=3$ and $x=1,y=2,n=2,k=3.$

First, note that none of $x$ and $y$ can be divisible by $3$ since $k>1$ and $x,y$ are supposed to be coprime.
It follows that $x^2=y^2=1 \mod [3]$ so that $k$ has to be odd.
Moreover we clearly must have $x+y \geq 3.$

Let $k=pq$ where $p \geq 1$
Then $3^n = (x^p+y^p)(x^{p(q-1)}-x^{p(q-2)}y^p+ \cdots +y^{p(q-1)}).$
It follows that $y^p = -x^p \mod [3]$ thus the parenthesis gives $qx^{p(q-1)} = 0 \mod [3]$ and then $q=0 \mod [3].$
Therfore $k$ is a perfect power of $3$, say $k=3^p.$

The same reasoning as above shows that $x+y = 3^a$ and $x^3+y^3 = 3^b$ for some positive integers $a,b.$
Thus $x^2-xy+y^2 = 3^c$ for some positive integer $c$, which means that $3xy = (x+y)^2-3^c = 3^{2a}-3^c$, and then $xy = 3^{2a-1} - 3^{c-1}.$
Since $xy \neq 0 \mod [3]$ we deduce that $c=1.$
Thus $x^2-xy + y^2 = 3.$
It follows that $4xy \leq (x+y)^2 = 3 + 3xy$ so that $xy \leq 3$, from which we deduce that $x=1,y=2$ or $x=2,y=1$ (remember that $x,y$ are coprime and add up to a multiple of $3$).
Now, the given equation reduces to $2^k+1=3^n$, or $3^n-2^k = 1.$
It is a classical result (even non using Catalan's equation) that the only solutions are $n=2,k=3$, and we are done.

Pierre.

Assume that $3^n=x^k+y^k$ $(*)$ with $(n,x,y)\in\mathbb{N}_*^3$ and $k>1$.
--- case $k$ even :
$3^n=X^2+Y^2$ is a sum of two squares and $3\not\equiv1[4]$ so $n$ is even and $X=0$ or $Y=0$ $\Rightarrow$ impossible.
--- case $k=3$ :
$*$ Assume that $x \not \equiv 0[3]$.
$3^n=(x+y)(x^2-xy+y^2)$ so $x+y=3^m$ and $x^2-xy+y^2=3^r$.
Thus $x^2-xy+y^2=3(x^2-3^mx+3^{2m})=3^r$ $\Rightarrow r=1 \Rightarrow (x,y,n)=(1,2,2)$ or $(x,y,n)=(2,1,2)$.
$*$ Assume that $3 \mid x$. Then $3 \mid y$ and $m=ord_3x=ord_3y$.
$\Rightarrow (x,y,n)=(2 \times 3^m,3^m,2+3m)$ or $(x,y,n)=(3^m,2 \times 3^m,2+3m)$.
--- general case $k$ odd :
If $3 \mid x$. Then $3 \mid y$ and $m=ord_3x=ord_3y$. Dividing both side $(*)$ by $3^{k \times ord_3x}$, we can assume $x\not \equiv 0[3]$.
We have $3 \mid s=x+y$ and $3 \mid \frac{x^k+y^k}{x+y}=\sum_{j=0}^{k-1}(-1)^jx^{k-1-j}(s-x)^j \equiv kx^{k-1}[3]$. So $3 \mid k$ and we can can use the case $k=3$.

------ $Consequence$:  the only solutions are $(n,x,y,k)=(2+3m,3^m,2 \times 3^m,3)$ or $(n,x,y,k)=(2+3m,2 \times 3^m,3^m,3)$

Pierre, you are too fast for me :juggle:
And I have forgotten the hypothesis $(x,y)=1$  :wallbash:


[quote="sks119"]A swimmer heads directly across a river, swimming at 3.70 m/s relative to still water. He arrives at a point 30.0 m downstream from the point directly across the river, which is 54.0 m wide. What is the speed of the river current?
What is the swimmer's speed relative to the shore? In what direction should the swimmer head so as to arrive at the point directly opposite his starting point? Express your answer in terms of the angle with respect to a line drawn directly across the river.[/quote]
[hide]
$\frac{v_{\text{current}}}{3.70}= \frac{30.0}{54.0}$

Swimmer's speed = ${v_{\text{current}}}$

Swimmer's direction = $-\tan^{-1}\left(\frac{30.0}{54.0}\right)$.[/hide]


algebra
polynomial
real analysis
combinatorics proposed
combinatorics

If $ i$, $ j$, $ n$ are nonnegative integers such that $ i\plus{}j\leq n$, then prove that

$ 2^{n \minus{} \left(i \plus{} j\right)}\mid\sum_{p \equal{} 0}^n\binom{n}{p}\binom{p}{i}\binom{p}{j}$.

I am wondering whether the Lucas theorem can provide a simple proof. Meanwhile, here is the sketch of a rather unusual proof:

The polynomial $ \binom{X}{i}\binom{X}{j}$ (with $ X$ being the polynomial variable) is a polynomial of degree $ i \plus{} j$ which takes integer values for all integers $ X$. It is known that any polynomial of degree $ \lambda$ which takes integer values for all integers $ X$ must be a linear combination of the polynomials $ \binom{X}{0}$, $ \binom{X}{1}$, ..., $ \binom{X}{\lambda}$ with integer coefficients. Thus, in order to prove that

$ 2^{n \minus{} \left(i \plus{} j\right)}\mid\sum_{p \equal{} 0}^n\binom{n}{p}\binom{p}{i}\binom{p}{j}$,

it will be enough to show that

$ 2^{n \minus{} \left(i \plus{} j\right)}\mid\sum_{p \equal{} 0}^n\binom{n}{p}\binom{p}{\kappa}$

for all $ \kappa < i \plus{} j$. But in fact, the stronger relation

$ 2^{n \minus{} \kappa}\mid\sum_{p \equal{} 0}^n\binom{n}{p}\binom{p}{\kappa}$

holds, and in fact it is obvious since $ \sum_{p \equal{} 0}^n\binom{n}{p}\binom{p}{\kappa} \equal{} 2^{n \minus{} \kappa}\binom{n}{\kappa}$ (as you can prove by noticing that $ \binom{n}{p}\binom{p}{\kappa} \equal{} \binom{n}{\kappa}\binom{n \minus{} \kappa}{n \minus{} p}$).

  darij

I found this one:

notice that the coefficient of $ x^iy^j$ in $ (1\plus{}x)^p(1\plus{}y)^p$ is $ \binom{p}{i}\binom{p}{j}$ so our number is the coefficient of $ x^iy^j$ in $ \sum_{p\equal{}0}^{n}\binom{n}{p}(1\plus{}x\plus{}y\plus{}xy)^p\equal{}(2\plus{}x\plus{}y\plus{}xy)^n$ but if we get i x's and j y's from the product then we must take at least n-i-j 2's therefore the conclusion!

A little easy generalization is:
If $ n$, $ i_1$, $ i_2$, ..., $ i_k$, $ i$ are nonnegative integers such that $ n\geq i \equal{} i_1 \plus{} i_2 \plus{} \cdots \plus{} i_k$, then
$ 2^{n \minus{} i}| \sum_{p \equal{} 0}^n \left(\binom{n}{p}\prod_{r \equal{} 1}^k \binom{p}{i_r}\right)$

[quote="Albanian Eagle"]A little easy generalization is:
If $ n$, $ i_1$, $ i_2$, ..., $ i_k$, $ i$ are nonnegative integers such that $ n\geq i = i_1 + i_2 + \cdots + i_k$, then
$ 2^{n - i}| \sum_{p = 0}^n (\binom{n}{p}\prod_{r = 1}^k \binom{p}{i_r})$[/quote]

Nice one! A combinatorial proof:

We begin with some obvious observations.

[color=green][b](1)[/b][/color] If $ B$ is a set, and $ \mathcal{A}(X)$ is an assertion for every subset $ X$ of $ B$, then

$ \sum_{X\subseteq B;\;\mathcal{A}(X)\text{ holds}}1 = |\{X\subseteq B\mid \mathcal{A}(X)\text{ holds}\}|$

is the number of all subsets $ X$ of $ B$ satisfying the assertion $ \mathcal{A}(X)$.

Hence, in particular:

[color=green][b](2)[/b][/color] If $ B$ is a set with $ |B| = u$, and $ v$ is an integer, then

$ \sum_{X\subseteq B;\;|X| = v}1 = |\{X\subseteq B\mid|X| = v\}| = \binom{u}{v}$.

Next triviality:

[color=green][b](3)[/b][/color] If $ N$ is a set with $ |N| = n$, and $ I$ is a subset of $ N$, then $ |\{P\subseteq N\mid I\subseteq P\}| = 2^{n - |I|}$.

This is because the subsets $ P\subseteq N$ satisfying $ I\subseteq P$ are in a one-to-one correspondence with the subsets $ Q\subseteq N\setminus I$ (this correspondence is given by $ Q = P\setminus I$ and, the other way round, $ P = Q\cup I$), and therefore,

$ |\{P\subseteq N\mid I\subseteq P\}| = |\{Q\subseteq N\setminus I\}| = 2^{|N\setminus I|} = 2^{|N| - |I|} = 2^{n - |I|}$.

Now, let $ N$ be a set with $ |N| = n$. Then,

[color=green][b](4)[/b][/color] $ \sum_{p = 0}^n (\binom{n}{p}\prod_{r = 1}^k \binom{p}{i_r}) = \sum_{p = 0}^n (\sum_{P\subseteq N;\;|P| = p}1\cdot \prod_{r = 1}^k \binom{p}{i_r}))$  (by [color=green][b](2)[/b][/color])
$ = \sum_{p = 0}^n (\sum_{P\subseteq N;\;|P| = p}\prod_{r = 1}^k \binom{p}{i_r}))$
$ = \sum_{p = 0}^n (\sum_{P\subseteq N;\;|P| = p}\prod_{r = 1}^k(\sum_{I_r\subseteq P;\;|I_r| = i_r}1))$  (by [color=green][b](2)[/b][/color], since $ |P| = p$ for all $ P$ in our sum)
$ = \sum_{P\subseteq N}\prod_{r = 1}^k(\sum_{I_r\subseteq P;\;|I_r| = i_r}1)$
$ = \sum_{P\subseteq N}\sum_{\begin{array}{c}k\text{ - tuples }(I_1,I_2,...,I_k) \\
\text{ such that }I_r\subseteq P\text{ and} \\
|I_r| = i_r\text{ for} \\
\text{every }r\in\{1,2,...,k\}\end{array}}1$
$ = \sum_{\begin{array}{c}k\text{ - tuples }(I_1,I_2,...,I_k) \\
\text{ such that }I_r\subseteq N\text{ and} \\
|I_r| = i_r\text{ for} \\
\text{every }r\in\{1,2,...,k\}\end{array}}\sum_{\begin{array}{c}P\subseteq N\text{ such that} \\
I_r\subseteq P\text{ for every} \\
r\in\{1,2,...,k\}\end{array}}1$.

Now, for any fixed $ k$-tuple $ (I_1,I_2,...,I_k)$ such that $ I_r\subseteq N$ and $ |I_r| = i_r$ for every $ r\in\{1,2,...,k\}$, we have

[color=green][b](5)[/b][/color] $ \sum_{\begin{array}{c}P\subseteq N\text{ such that} \\
I_r\subseteq P\text{ for every} \\
r\in\{1,2,...,k\}\end{array}}1 = \sum_{\begin{array}{c}P\subseteq N\text{ such that} \\
I_1\cup I_2\cup ...\cup I_k\subseteq P\end{array}}1$
$ = |\{P\subseteq N\mid I_1\cup I_2\cup ...\cup I_k\subseteq P\}|$  (by [color=green][b](1)[/b][/color])
$ = 2^{n - |I_1\cup I_2\cup ...\cup I_k|}$  (by [color=green][b](3)[/b][/color]).

Since $ |I_1\cup I_2\cup ...\cup I_k|\leq|I_1| + |I_2| + ... + |I_k| = i_1 + i_2 + ... + i_k = i$, we have $ n - |I_1\cup I_2\cup ...\cup I_k|\geq n - i$, and thus $ 2^{n - i}\mid 2^{n - |I_1\cup I_2\cup ...\cup I_k|}$, so that, by [color=green][b](5)[/b][/color], we get

$ 2^{n - i}\mid \sum_{\begin{array}{c}P\subseteq N\text{ such that} \\
I_r\subseteq P\text{ for every} \\
r\in\{1,2,...,k\}\end{array}}1$. Thus,

$ 2^{n - i}\mid \sum_{\begin{array}{c}k\text{ - tuples }(I_1,I_2,...,I_k) \\
\text{ such that }I_r\subseteq N\text{ and} \\
|I_r| = i_r\text{ for} \\
\text{every }r\in\{1,2,...,k\}\end{array}}\sum_{\begin{array}{c}P\subseteq N\text{ such that} \\
I_r\subseteq P\text{ for every} \\
r\in\{1,2,...,k\}\end{array}}1$.

According to [color=green][b](4)[/b][/color], this becomes

$ 2^{n - i}\mid\sum_{p = 0}^n (\binom{n}{p}\prod_{r = 1}^k \binom{p}{i_r})$,

and we are done.

[b]PS.[/b] This proof can be easily translated into a completely algebraical one (using the Vandermonde convolution), at least for my original problem (that is, the case $ k = 2$). As a result, we then obtain the identity

$ \sum_{p = 0}^n\binom{n}{p}\binom{p}{i}\binom{p}{j} = \sum_{q = 0}^n\binom{q}{i}\binom{i}{q - j}\binom{n}{q}2^{n - q}$.

  darij

Let $ q\equal{}a^2\plus{}b^2$
then
$ q^{n\minus{}i\minus{}j}|\sum_{p\equal{}0}^n \binom{n}{p} \binom{p}{i} \binom{p}{j} \left(\frac{a}{b}\right)^{2p}$ (as in the numerator)

:)

[quote="Albanian Eagle"]Let $ q \equal{} a^2 \plus{} b^2$
then
$ q^{n \minus{} i \minus{} j}|\sum_{p \equal{} 0}^n \binom{n}{p} \binom{p}{i} \binom{p}{j} \left(\frac {a}{b}\right)^{2p}$ (as in the numerator)[/quote]

Please be more precise: Is $ \left(\frac {a}{b}\right)$ a fraction or a Jacobi symbol? And is $ q$ supposed to be prime?

If $ \left(\frac {a}{b}\right)$ means a fraction, I don't really understand why you consider $ a^2$ and $ b^2$ rather than just $ a$ and $ b$. In fact, if I am not mistaken, if $ a$ and $ b$ are two [i]coprime[/i] integers, then

$ \left(a \plus{} b\right)^{n \minus{} \left(i \plus{} j\right)}\mid\sum_{p \equal{} 0}^n \binom{n}{p} \binom{p}{i} \binom{p}{j} \left(\frac {a}{b}\right)^p$.

This is meant numerator-wise, so if you want to stick to integers, this becomes

$ \left(a \plus{} b\right)^{n \minus{} \left(i \plus{} j\right)}\mid\sum_{p \equal{} 0}^n \binom{n}{p} \binom{p}{i} \binom{p}{j} a^pb^{n \minus{} p}$.

  darij

Yup, that's exactly what i meant sorry :)


[quote="wavy78"]This is an inequality that I found in one of the Inequality documents on web(Samin Riasat)
-------------------------
Given that a,b,c are positive real numbers such that abc=1.

PT  (a/b +b/c +c/a)>= (a+b+c)

-------------------------

I homogenized the equation, but couldn't find an appropriate AM-GM to solve it.[/quote]
By [b]AM-GM Inequality[/b] ,we have

       $ \frac {a}{b} \plus{} \frac {a}{b} \plus{} \frac {b}{c} \ge 3 \sqrt [3]{\frac {a^2}{bc}} \equal{} 3 \sqrt [3]{\frac {a^3}{abc}} \equal{} 3a$   ( note $ abc \equal{} 1$)   (1)

Similarly,we have
$ \frac {b}{c} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3b$   (2)

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

(1),(2),(3) ->Q.E.D

Equality holds if only if $ a\equal{}b\equal{}c\equal{}1$

[quote="wavy78"]This is an inequality that I found in one of the Inequality documents on web(Samin Riasat)
-------------------------
Given that a,b,c are positive real numbers such that abc=1.

PT  (a/b +b/c +c/a)>= (a+b+c)

-------------------------

I homogenized the equation, but couldn't find an appropriate AM-GM to solve it.[/quote]

With $ k$ be a positive interger and $ a,b,c$ be positive real numbers such that $ abc \le 1\#$.
We have

                  $ \frac{a}{b^k}+\frac{b}{c^k}+\frac{c}{a^k} \ge a+b+c$


Do there exist $ f: (0,+\infty)\to\(0,+\infty)$ so that :

$ 1,f(x) < f(y)$ if $ x < y$.

$ 2,f(x+\frac{1}{f(x)})\ge 2f(x)$ .

 :)

[code]edited [/code]

[quote="quangpbc"]Do there exist $ f: (0,+\infty)\to\(0,+\infty)$ so that :

$ 1,f(x) < f(y)$ if $ x < y$.

$ 2,f(x+\frac{1}{f(y)})\ge 2f(y)$ .

 :)[/quote]

It does not.

$ f(x)$ is an increasing function, an so we have two possibilities :

1. $ \lim_{x\rightarrow+\infty}f(x)=+\infty$
Then, when $ y$ grows towards $ +\infty$ in $ f(x+\frac{1}{f(y)})\ge 2f(y)$, LHS is $ <f(x+\frac{1}{f(1)})$ and RHS is as great as we want $ \implies$ impossible.

2. $ \lim_{x\rightarrow+\infty}f(x)=L$
Then, when $ y$ grows towards $ +\infty$ in $ f(x+\frac{1}{f(y)})\ge 2f(y)$, LHS is $ <L$ and RHS may be as near as we want of $ 2L$ $ \implies$ impossible.

[quote="quangpbc"]$ \lim_{x\to\infty}\equal{}\text{L}$

I think, the function isn't a continuous function , so maybe it doesn't have a limit .  :wink:[/quote]

It's an increasing one.
So , either $ \lim_{x\rightarrow\plus{}\infty}f(x)\equal{}\plus{}\infty$, either $ \lim_{x\rightarrow\plus{}\infty}f(x)\equal{}L$, even if the function is not continuous.

It would look like $ x$ may take any value in condition 2., without any kind of limitation.  Is that correct?

Anyway, if no mistake in condition 2., it is clear that, taking $ x \equal{} y\minus{}\frac{1}{f(y)}$, we would have $ f(y)\geq2f(y)$ for any $ y$, which is absurd since $ f(y) > 0$.  If $ x$ must take on values from the starting set, then this means that $ f$ cannot exist unless $ y\leq\frac{1}{f(y)}$ for all $ y$, or $ f(y)\leq\frac{1}{y}$ for all $ y$.  Then, $ 2f(y)\leq\frac{f(y)}{xf(y)\plus{}1}\leq f(y)$ because $ x$, $ f(y)$ are positive.  Absurd, and we are done.

[quote="pco"]It's an increasing one.
So , either $ \lim_{x\rightarrow+\infty}f(x) =\infty$, either $ \lim_{x\rightarrow+\infty}f(x) = L$, even if the function is not continuous.[/quote]

Oh i see   . Thanks for your solution.

Sorry all members. I mistyped   :blush: .I have edited it  :)
 My solution :

Consider a real number $ a$ .Define the sequence :$ (x_{n})$: $ x_{0}= a,x_{n}= x_{n-1}+\frac{1}{f(x_{n-1})}$

We have $ f(x_{n})\ge 2^{n}f(a)$ so $ f(x_{n})\to\infty$.

But $ x_{n}= x_{0}+\sum_{i = 1}^{n-1}\frac{1}{f(x_{i})}\le x_{0}+\sum_{i = 1}^{n-1}\frac{1}{2^{i+1}f(a)}< x_{0}+\frac{1}{f(a)}$

Because $ f$ is an increasing function so $ f(x_{n}) < f(x_{0}+\frac{1}{f(a)}) =\text{constant}$  Contracdition because $ f(x_{n})\to\infty$ when $ n\to\infty$  :)


Where do I get the "Latex". pls sombody tell me


Which book would be most helpful for the current AMC 12s? I already have #5 and pretty much anything after that can be found on the internet. So which one should i get (1,2,3, or 4)?

I personally prefer that one. It has good problems in it.


Find the greatest constant $c>0$ with the following property. For every (not necessarily convex) polygon $F$ of area $A$, there is a chord which splits $F$ into two parts, each with area at least $cA$.

I think something like the following can be used to show that the maximal $c$ is at most $\frac 13$ (see the picture below).

We make the "center" very small compared to the three branches. There are a few details to work on, but I think basicly, it's Ok. I know for a fact that the maximal $c$ is at least $\frac 13$ because i've seen the problem on Kalva before :). I'll try to show that :).

Edit: sorry about the quality of the picture. What do you guys think I should use to convert .wmf to .jpg?

It looked to me earlier today that I could show $\max c\ge \frac 13$, but I don't know if it's correct. Anyway, I'll post it here, but please give me some feedback here, guys (vess :)).

Triangulate the polygon (it wasn't mentioned, but it should be a simple polygon, i.e. it shouldn't intersect itself). Now look at its dual tree (to each triangle in the triangulation associate a vertex, and connect two vertices iff they correspond to triangles which share a side). In this tree, each vertex has degree at most $3$. If all the vertices have degree at most $2$, I think we can even bisect the area of the polygon, so there's no problem there. In what follows, let's assume that there is a vertex with degree $3$.

Let $\Delta$ be the area of the triangle representing the vertex of degree $3$ (if there are more such vertices, just pick one of them). Let $S_1,S_2,S_3$ be the areas of the parts of the polygon "glued" to the three sides of that triangle. We have $\min(S_1,S_2,S_3)\le\frac 13\cdot S=\frac 13\cdot(S_1+S_2+S_3+\Delta)$. Assume $S_1\ge S_2\ge S_3$. We then have $\frac 13\cdot S-S_1\le\frac 23\cdot S-S_1-S_3\ (*)$ and $\frac{2\Delta}3\ge 0\ge \frac {S_2-S_1+S_3-S_1}3\ (**)$. Add $\frac{\Delta}3$ to $(**)$ to turn it into $\Delta\ge \frac 13\cdot S-S_1\Rightarrow \Delta+S_1\ge\frac 13\cdot S\ (\#)$. Combining $(*),(\#)$, we find that there is $\lambda\in[0,1]$ s.t. $\frac 13\cdot S\le S_1+\lambda\cdot \Delta\le S_1+\lambda\cdot\Delta+S_3\le\frac 23\cdot S\ (!)$.

From here it should be easy: assume the triangle with area $\Delta$ we have been talking about is called $ABC$, and $S_1$ is "glued" to $AB$, while $S_3$ is "glued" to $BC$. Now consider a chord which passes through $A$ and $D\in[BC]$ s.t. $[ABD]=\lambda\cdot \Delta$. From $(!)$ we should be able to draw the wanted conclusion.

Comments? Please? :) Is it Ok?

Randomly stumbled upon this.

I don't think grobber's argument works when $S_1>2S/3$, but triangulations seem to be pretty effective at describing simple polygons, and we can patch the solution as follows: For any triangle $T$ in the triangulation, let $M(T)$ be the area of the largest region "glued" to $T$ (there are at most $3$ such regions for fixed $T$), and let $N(T)$ be the triangle (sharing a side with $T$) corresponding to $M(T)$. Go by contradiction; then $M(T)>2S/3$ for all $T$, or else we can apply grobber's argument (treating $S_3=0$ if $\deg{T}=2$, it's obvious when $M(T)\ge S/3$, and otherwise we can take $\lambda\in(0,1]$ such that $S_1+\lambda M(T)=1/3$; clearly $S_3\le S/3$). Now consider any $T$ with $M(T)$ minimal; then $M(N(T))\ge M(T)>2S/3$, so $N(N(T))=T$. But then $S = M(N(T))+M(T) > 4S/3$, contradiction.

---

Alternatively, suppose $F=P_1P_2\ldots P_n$ (indices mod $n$); go by contradiction again. Suppose out of all chords with endpoints as vertices of $F$, $P_iP_{i+j-1}$ ($j\ge2$) has the largest smaller half (WLOG $P_iP_{i+1}\ldots P_{i+j-1}$ is the smaller half with area $M<1/3$); such a chord exists because we can simply take a side of $F$. Now take $k\in\{i+j,\ldots,i+n-1\}$ with $d(P_k,P_iP_{i+j-1})$ minimal. Clearly $P_iP_k$ and $P_{i+j-1}P_k$ must be chords of $F$, so by the maximality of (the smaller half of) $P_iP_{i+j-1}$, polygons $P_k\ldots P_{i-1}P_i$ and $P_{i+j-1}P_{i+j}\ldots P_k$ both have area at most $1/2$ and thus strictly less than $1/3$. But then $[P_iP_kP_{i+j-1}]+M>1/3$, so "sliding" up $P_iP_{i+j-1}$ towards $P_k$ we'll get a working chord before it hits vertex $P_k$, contradiction.


function
integration
calculus
linear algebra
matrix
algebra
polynomial

Here is a very interesting result :
      Consider $f$ a continuous function with period $2\cdot\pi$ and which is positive. Next, put $c_k$ its Fourier coefficients and consider the sequence of matrices $A_n$ that have $n+1$ lines and columns and elements $c_{j-i}$. Is it true that $\sqrt[n+1]{detA_n}$ converges towards $\frac{1}{2\cdot\pi}\cdot\int_{0}^{2\cdot\pi}{lnf(x)dx}$? For functions of the form $f(x)=a+bsinx+ccosx$ it works. Is it true in general?

Yes, it is true for all continuous functions. (with $\exp$ of the integral instead of the integral, of course) Actually, the condition of continuity can even be relaxed though I do not remember exactly what is the most general class of functions for which it holds.

As to the proof (for continuous case), note that $A_n$ is the matrix of the positive self-ajoint operator $T_n$ on the space $\mathcal P_n$ of algebraic polynomials of degree $\le n$ defined by $(T_n p,q)=\int_{\mathbb T}fp\overline q\,dm$ (where $\mathbb T$ is the unit circumference and $m$ is the normalized Lebesgue measure on $\mathbb T$) in the standard basis $\{z^j\}_{0\le j\le n}$. Recall that the determinant is pretty much the same as the inverse volume of the ellipsoid $(T_n p,p)\le 1$ (up to a constant and power $\frac 12$). From here it follows immediately that $\det A_n$ is a monotone function of $f$. Since each continuous function can be approximated by smooth functions from both above and below, it suffices to prove the statement for $f\in C^1$. The trick is to change basis to $p_j=\frac 1{\sqrt{n+1}}\frac{z^{n+1}-1}{z-\zeta_j}$ where $\zeta_j=e^{2\pi i\frac j{n+1}}$ (I leave it to you to check that $p_j$ is an orthonormal basis in $\mathcal P_n$). The matrix $B$ of $T_n$ in this new basis is very close to a diagonal one. More precisely, we have
\[ \begin{aligned} b_{ij}&=\int_{\mathbb T}p_i\overline{p_j}\,dm= f(\zeta_i)\delta_{ij}+O(\|(f-f(\zeta_i)p_i\|_\infty\|p_j\|_1) \\ &=f(\zeta_i)\delta_{ij}+O\left(\frac{\ln n}{n+1}\right) \end{aligned}  \]
where $\delta_{ij}=1$ if $i=j$ and $0$ otherwise. Thus, $B_n=D_n+R_n$ where $D_n$ is the diagonal matrix with entries $f(\zeta_j)$ on the diagonal and $r_{ij}=O\left(\frac{\ln n}{n+1}\right)$. Now write $B_n=D_n^{1/2}(I_n+R_n')D_n^{1/2}$ where $R_n'=D_n^{-1/2}R_n D_n^{-1/2}$. Note that the entries of $R_n'$ are of about the same size as the entries of $R_n$ and $I+R_n'$ is a bounded positive definite matrix, whose positive definiteness constant is bounded away from $0$. Let $\lambda_j$ be the eigenvalues of $R_n'$. Note that $1+\lambda_j\ge c>0$ regardless of $n$ and $\sum_j\lambda_j^2=\sum_{i,j}{r_{ij}'}^2=O(\ln^2 n)$. Thus
\[ \det(I+R_n')=\prod_j(1+\lambda_j)=e^{O(\sum_j|\lambda j|)}=e^{O(\sqrt n\ln n)}\,.  \]
The rest should be clear.

For those who are curious to see the full problem

Ecole Polytechnique 1987 Option M', Première Composition

ftp://ftp.drimm.u-bordeaux1.fr/UPS/files/corriges-1987/index.html
see Polytechnique M/énoncé Maths-1 by Quercia


im really confused over this problem.
Can anybody please help or advise please?

=========================
Mark each statement true or false
1. { x | x = A } = {A}
2. { x | x $\in$ A } = A
3. { x | {x} $\subset$ A } = {A}
4. { x | {x} $\subset$ $\varnothing$ } = $\varnothing$
5. { x | {x} = A } = A
6. { x | {x} $\in$ A } = {A}
=========================

could anyone give me some suggestions

well for number one?

[hide]
true because x=a
[/hide]

but then again i am in alg 2 and my teachers never taught me $\in$, etc.


Hello!

For the Easter Holidays I have a very old inequality and .. I couldn't solve it! I would really hope for some help!

Let x1,x2,x3 be positive real numbers. Prove that the following inequality holds:
sum(x1^2)+sum(x1*sqrt(x2*x3))>=2sum(x1*x2).
The sums are cyclic.

Thank you very much!
Happy Easter to everyone!

[hide]setting a = x1, b = x2, ...

By Schur inequality with k=2(http://planetmath.org/encyclopedia/SchursInequality.html) :

 :Sigma: a^4 +  :Sigma: a^2bc >=  :Sigma: a^3b

Now :Sigma: a^3b >= 2*:Sigma: ab  can be proved (for instance) by rearrangement.[/hide]

[quote]Now  :Sigma:  a^3b >= 2* :Sigma:  ab can be proved (for instance) by rearrangement.[/quote]

I dont think this is true... :Sigma:  a^3b >=  :Sigma:  ab is true by rearrangement, tho

oh, nevermind... the  :Sigma: a^3b was a symmetric sum, not a cyclic sum... then ya, ur proof works.


Nice problem: There are 50 cards and integers from 1 to 100 are writen in them (on both sides) each number appearing once. Now I put the cards on a table sou you can only see the top number. Now you are allowed to select some cards and turn them over, and  calculate the sum of the numbers on top. What is the biggest sum you can be sure to get?

Ok, I think the number is $S=\frac{n(n+1)}{4}$, that is, half of the sum on both sides of the cards. To achieve it from any situation, just look at first at the sum on the sides you see. If it is $\leq S$, then flip all the cards, otherwise, don't do anything. To prove this is tight, just take a combination of cards so that the sum on the sides you see equals the sum you don't see.
Also, for our problem, $n=100$ :D
I  hope I am not wrong!! :?

I also believe this is the answer, but i don't think it's as simple as that. You have shown that it's always possible to achieve $5050$, but what if there are strategies which help us be sure that we can get more than that? Not every combination of cards with $5050$ will do.

Assume, for example, that $n=8$, and assume you have $1,4,5,8$. If you turn the $1$, you can be sure you get at least a $2$ instead, so you'll get more than $18$. A combination which does show that we can't get more, I think (I haven't looked at it very carefully), would be the following one: assume you see the cards $26,27,\ldots,75$ at the beginning.

Actually, I think it is quite simple to assure such an answer.
Consider the cards with sides $a_i$ and $b_i$. We see only the set $a_1,\ldots , a_n$, so our choice of what cards to switch is determined only by this set, and we also have $\sum a_i = \sum b_i$. I think we can arrange the set $b_i$ so that when switching any particular set of cards, we will not obtain a greater number.......
[Edit: I just read more carefully grobber's above post and I see that the necessary and I hope that the necessary and sufficient condition is that $a_i$ majorizes $b_i$. I hope it is ok ] :?

Yes, that is how I found that arrangement: the initial configuration has to majorize the arrangement on the back of the cards.

yes, my arrengement is numbers of the form $4k+2$ and $4k+3$ on top, and the rest down.

(i edited my previous post because i didnt read you arrengement)

Was the arrangement I mentioned not correct? $26,27,\ldots,75$ on the up-faces of the cards.


Pretty trivial...

A circular sector of radius $ r$ and angle $ \alpha$ has twice the area as a sector of radius $ R$ and angle $ \beta$. Find the ratio $ \frac{\alpha}{\beta}$ if $ R \equal{} 2r$.

oooh this is fun!!!any answer choices

Ok so first i would substitute the value given to us and we get 2(4$ r^2\times \frac{\beta}{360}$) and we know that that equals $ r^2 \times \frac{\alpha}{360}$  so solving this we have

Now we have $ 8r^2 \times \frac{\beta}{360} \equal{} r^2 \times \frac{\alpha}{360}$

THen we get $ 8r^2 \times \beta \equal{} r^2 \times \alpha$ solving for $ \alpha$ in terms of $ \beta$ yeilds $ 8\beta \equal{} \alpha$ so $ \frac{\alpha}{\beta} \equal{} \frac{8\beta}{\beta} \equal{} \boxed{8}$

Hopefully that was the right answer!!!!Good problem!!!!

i think looking at the angle A and B as radian measures makes it a little easier.  :)
Area[sector]=(1/2)(R^2)(the angle)

yes it might but there is one enormous reason why i didnt do the problem that way!!!!I HATE RADIANS!!!!!!!


If you love inequalities , try this :
Prove that for every acute triangle ABC , we have :
  (cos A)^3 + (cos B)^3 >= 2 (cos ((A+B)/2))^3 [/tex]

This problem seems to be difficult , so noboby can prove it . :D  :blush:

I had told you it's easy.
It is second time you post it so you can see the solution at your last post reply. :D  :D


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ H$ be a subgroup of a finite group $ G$ and $ M$ an abelian group. Let  $ G \equal{} \bigcup \limits_{i \equal{} 1}^nHg_i$ be the decomposition of $ G$ over $ H$.

 $ \mathbb{Z}[G] \equal{} \bigoplus \limits_{i \equal{} 1}^n\mathbb{Z}[H]g_i$ and $ \mathbb{Z}[G] \simeq \mathbb{Z}[H]^{(n)}$. I ought to prove that a free resolution of $ \mathbb{Z}[G]$-modules is also a free resolution of $ \mathbb{Z}[H]$-modules.
 I proved that a free $ \mathbb{Z}[G]$-module is also a $ \mathbb{Z}[H]$-module :
 Let $ M$ be a free $ \mathbb{Z}[G]$-module and $ (a_i)_{1 \leq i \leq k}$ a basis of $ M$. Since $ \mathbb{Z}[G] \equal{} \bigoplus \limits_{i \equal{} 1}^n\mathbb{Z}[H]g_i$, $ (y_{ik})_{1 \leq i \leq n, 1 \leq j \leq k}$ with $ y_{ij} \equal{} g_ia_j$ is a basis of the free $ \mathbb{Z}[H]$-module $ M$.
 I fail to prove that an exact sequence of $ \mathbb{Z}[G]$-modules is also an exact sequence of $ \mathbb{Z}[H]$-modules.
 If somebody has an idea, thanks for advance!

[quote="Bob75"]I fail to prove that an exact sequence of $ \mathbb{Z}[G]$-modules is also an exact sequence of $ \mathbb{Z}[H]$-modules.[/quote]
There is nothing to prove!

More specifically, exactness for modules is just the same as exactness for abelian groups. The additive structure is the same either way, and the multiplicative structure doesn't matter.

Hmm, it may be that we also want some lemma like "a $ \mathbb{Z}[G]$-module homomorphism is also a $ \mathbb{Z}[H]$-module homomorphism."  One of Bob75's comments leads me to believe that this level of detail may be what he's after.  Anyway, this follows quite directly from the definitions.


On the last math league, #6 was a problem with probability, something like, "Alternately flip a coin and a dice, if you get a head or a (some number, i think it was 2), you stop.  if you start with the coin, what is the probability that you stop on a rolling of the dice (by rolling the 2)?"

when i took it i did it like the solution said, with infinite sequence, but i remembered from a math jam that you could do some types of probability with algebra.  how do you do it w/algebra in this case?

Just to let people know, the solution that they gave was:

\[\left(\frac 12\right)\left(\frac 16\right) + \left(\frac 12\right)^2\left(\frac 56\right)\left(\frac 16\right)+\left(\frac 12\right)^3\left(\frac 56\right)^2\left(\frac 16\right)+\cdots=\frac 16\cdot \frac 12 \cdot\frac{1}{1-\frac{5}{12}}=\frac 17.\]

The way which I think you are talking about is as follows:

Call the desired probability $p$.

Thus we can say that

\begin{eqnarray*}
p &=& \left(\frac 12\right) \left(\frac 16\right) +\left(\frac 12\right)  \left(\frac 56\right) p\\
  &&\Rightarrow p=\frac 17.
\end{eqnarray*}

We are basically saying that we can either end it on the first time around or any of the subsequent times around.  Even though I used the first method during the test the second method is pretty slick :D . The reason this works is because the process can go on ad infinitum.  It's kinda the same concept as evaluating:

\[{{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}}}.\]

thanks joml88, that's what i was thinking of, but wasn't sure how.

[quote]Couldnt you get 1/7 using an easier meathod?? The possible outcomes are H, T1, T2, T3, T4, T5, T6. T2 is the only one that works, so therefore its 1/7?? This is how i solved it. Did I get lucky or do i have a correct solution?[/quote]

i remember a while ago on a problem with infinite sequence (similar to this) i assumed that way and got it right as well.  but i was told that to do it right, you need the geometric sequence.
i dunno why it worked

Here's why you would get the right answer if you had a regular two sided coin and a  $k$-sided die:

If we list out all the possible outcomes for first round (like Braves did) we get one of them as just an H and $k$ others that have a T followed by a number 1 to $k$.  Only one of these is favorable thus the probability is $\frac{1}{k+1}.$

Doing it the other way we would get:

\[p=\left(\frac 12\right)\left(\frac 1k\right) + \left(\frac 12\right)\left(\frac{k-1}{k}\right)p.\]

And solving we get $p=\frac{1}{k+1}$.  So it works for a 2 sided coin and any sided die.

[quote="joml88"]That's interesting.  It does give the correct answer.  And if you used any $k$-sided die it would still work.  However, I can't find any sound reason why you could assume that at first.  The reason I say this is that {H, T1, T2, T3, T4, T5, T6} is not the complete set of possible outcomes.   The set is infinite.  There might be a way to go from here to the answer with sound reason (someone post if you know one).

Be more careful in the future however.  For example, suppose we had a coin that could give 3 equally likely outcomes (say Heads, Tails or Neither) and a 6-sided die.  Same rules as last time, i.e. flip the coin then the die...games stops if Heads is flipped or 2 is tossed.  Using your method we would have the outcomes as {H, T1, T2, T3, T4, T5, T6, N1, N2, N3, N4, N5, N6}.  There are 2 favorable outcomes and 13 total.  So this gives us 2/13.  Now using my method we have:

\[p=\left(\frac 23\right)\left(\frac 16\right) + \left(\frac 23\right)\left(\frac 56\right) p.\]

So $p=\frac 14.$  So I think you got a little lucky but I'm open for ideas  :) .[/quote]

I dont understand...how is that not the total possible outcomes.

because you can have, for example, T3T4T2
and this goes on infinitely

You stop when you get a head or a 2.  So if you got T3 you wouldn't stop...the sequence would continue until you got a head or a 2.  The possible outcomes would look like:

{H, T2, T1H, T1T2, T3H, T3T2,...}.

[hide="interesting solution..."]this problem is basically a geometric sequence...

the probability of stopping after a toss of each:
$\frac{1}{6} * \frac{1}{2}$
the probability of stopping after Two tosses of each:
$\frac{5}{6} * \frac{1}{2} * \frac{1}{6} * \frac{1}{2}$
the probability of stopping after threetosses of each:
$\frac{5}{6} * \frac{1}{2} * \frac{5}{6} * \frac{1}{2} * \frac{1}{6} * \frac{1}{2}$

so basically it becomes the sum of infinite geometric sequence with $F_1 = \frac{1}{12}, r = \frac{5}{12}$, thus by the formula $ \text{ sum } = \frac{F_1}{1-r}$, we get $\frac{\frac {1}{12}}{1 - \frac{5}{12}} = \frac{1}{7}$[/hide]


function
limit
trigonometry
calculus
derivative
real analysis
real analysis unsolved

Hello

I need to give an example of a function f(x) such that limf(x) as x goes to infinity exists but the limf'(x) does not. Any help?

Thx

$ f(x)\equal{}x\plus{}\sin(x)$,  $ f^{\prime}(x)\equal{}1\plus{}\cos(x)$.

$ \mathop{\lim}\limits_{x\to\infty}{f(x)}\equal{}\plus{}\infty$ and $ \mathop{\lim}\limits_{x\to\infty}{f^{\prime}(x)}$ does not exists.

If you want a finite limit, you need something oscillating with very high frequency even as its amplitude shrinks -- this suggests things of the form $ \frac{\sin x^2}{x}$ which has limit $ 0$ but which has derivative $ 2\cos x^2 \minus{} \frac{\sin x^2}{x^2}$.


Solve the equation   $ [\tan x]\{\tan x \}\equal{}\tan x$   :?:
______________________
Azerbaijan Land of the Fire   :)

If $ {\rm [tgx] \equal{} 0} \Rightarrow {\rm tgx \equal{} 0} \Rightarrow {\rm x \equal{} k}\pi {\rm ;}$
If $ {\rm \{ tgx\}  \equal{} 0} \Rightarrow {\rm tgx \equal{} 0} \Rightarrow {\rm x \equal{} k}\pi {\rm ;}$
Now we suppose that $ {\rm \{ tgx\} }{\rm ,[tgx]} \ne {\rm 0}$ and $ \frac{{\rm 1}}{{{\rm [tgx]}}} \plus{} \frac{1}{{\{ tgx\} }} \equal{} 1 \Rightarrow \frac{1}{{[tgx]}} < 0 \Rightarrow [tgx] < 0$
But $ {\rm ([tgx] \minus{} 1)(\{ tgx\}  \minus{} 1) \equal{} 1} \Rightarrow \frac{{\rm 1}}{{{\rm [tgx] \minus{} 1}}} \equal{} \{ tgx\}  \minus{} 1 \equal{} \{ tgx \minus{} 1\}  > 0 \Rightarrow [tgx] > 1,$false.

For [b]omegatheo[/b]: $ \{x\}\in[0;1)\Rightarrow 0\ge \{x\} < 1\Rightarrow \{x\} \minus{} 1 < 0; \ (\forall)\ x\in\mathbb{R}\Rightarrow \{tg x\}\minus{}1<0.$


[quote="xxxyyyy"]Prove that for all real numbers $a, b, c > 0$:

$(a^{2}+2)(b^{2}+2)(c^{2}+2) \geq 9(ab+bc+ca)$[/quote]

too tough for this level? Can someone move it to appropriate thread?

Uh... here's my first attempt at a proof... Im new to proofs so I'm not sure how valid this is...

[hide]
$(a^{2}+2)(b^{2}+2)(c^{2}+2) \geq 9(ab+bc+ca)$

If you plug in a, b, c = 0, then the equation becomes (2)(2)(2) => 9 (0 + 0 + 0), which is true.

$(a^{2}+2)(b^{2}+2)(c^{2}+2) \geq 9(ab+bc+ca)$

=$(a^{2}+2)(b^{2}+2)(c^{2}+2) \geq 9(1/a+1/b+1/c)$

Remember that for fraction 1/x, as x increases in value, the fraction as a whole decreases in value.

Therefore as a, b, or c increase, the RHS will decrease in value, while the LHS will increase. This is sort of an induction type proof, where all we need to do is prove that

$(a^{2}+2)(b^{2}+2)(c^{2}+2) \geq 9(ab+bc+ca)$

is true for a very small number of (a, b, c), and if that is true then all values (a, b, c) greater than the previous value will be true (since the RHS will decrease and LHS will increase, yeilding the LHS to be always greater than or equal to the RHS).

We have shown in the beginning that the inequality holds true for a, b, c = 0, therefore any value of a, b, c, > 0 will also hold true.
[/hide]

Choose A;B;C in $(0;\frac{\pi}{2})$ with $a = \sqrt{2}tanA$ , $b = \sqrt{2}tanB$, and $c = \sqrt{2}tanC$
Then we have
$\frac{4}{9}\geq cosAcosB cosC (cosAsinB sinC+sinAcosB sinC+sinAsinB cosC)$
$\frac{4}{9}\geq cosAcosB cosC (cosAcosB cosC-cos(A+B+C))$
Let $\alpha = \frac{A+B+C}{3}$
We have $cosAcosB cosC \leq (cos\alpha)^{3}$
The last thing is:
$\frac{4}{9}\geq (cos\alpha)^{3}((cos\alpha)^{3}-cos 3\alpha)$
$\frac{4}{27}\geq (cos\alpha)^{4}(1-(cos\alpha)^{2})$
Now it's easy with AM-GM inequality  :P
The inequality holds when $tanA=tanB=tanC=\frac{1}{\sqrt{2}}$ which means $a=b=c=1$

P/S: This is what I learned in my school (I remember that this was APMO 2004/5)


linear algebra
matrix
vector
linear algebra unsolved

Hello,

let $ A$ and $ B$ be square hermitian $ n\times n$ matrices.

I will say that $ A$ is "bigger than B" if and only if $ \forall$ vectors $ v : v^H A v\geq v^H B v$

Now which of these are true :

- $ A$ is bigger than $ B$ if and only if $ A\minus{}B$ is a sum of terms of form $ C^H C$

- $ A$ is bigger than $ B$ if and only if $ A\minus{}B$ is a sum of squares of hermitian matrices

- when $ A$ and $ B$ are simultaneously diagonizable with an orthonormal basis, $ A$ is bigger than $ B$ if and only if $ A\minus{}B$ is a sum of terms of form $ C^H C$

-when $ A$ and $ B$ are simultaneously diagonizable with an orthonormal basis, $ A$ is bigger than $ B$ if and only if $ A\minus{}B$ is a sum of squares of hermitian matrices

I only managed to prove the third and fourth statement, what do you think of the first and second? :ninja:

Note: reading the definition carefully, there is no difference between saying that $ A$ is bigger than $ B$ (which we might as well write $ A\ge B$) and saying that $ A - B$ is positive semidefinite.

Let $ X = A - B$ be Hermitian positive semidefinite. It is unitarily diagonalizable. That is, $ \exists\,U$ with $ U^*U = I$ such that $ U*XU = D = \text{diag}(\lambda_1,\lambda_2,\dots,\lambda_n).$

Since each $ \lambda_j\ge0$ because $ X$ is positive semidefinite, we can define $ D^{1/2} = \text{\ddag}\left(\sqrt {\lambda_1}, \dots,\sqrt {\lambda_n}\right)$ If $ C = UD^{1/2}U^*,$ then $ C^2 = C^*C = X.$

That pretty much takes care of everything. I don't see why we need to talk about simultaneous diagonability of $ A$ and $ B$ when we always have that $ A - B,$ being itself Hermitian, is diagonable.

The essential argument:

$ v^*Av\ge v^*Bv\,\implies\,v^*Av-v^*Bv\ge0\,\implies\,
v^*(A-B)v\ge0\,\forall\,v.$

Write $ A\minus{}B\equal{}U^{\dagger}DU$ with diagonal, real matrix $ D$ and unitary matrix $ U$, then

$ A\geq B\Longleftrightarrow D$ has only non-negative eigenvalues.

This should allow you to answer the questions

Of course! :o  Thanks. :wink:

So all statements are true, and to make it even stronger, it's a sum of several such squares if and only if it is the sum of exactly one square as well. :!:

But in order to defend myself a bit...  I was thinking of quantum mechanics and projections, which correspond to observables that can only be true or false: one or zero.
I had to consider one projection being bigger than another.

So that means that $ A$ and $ B$ are not only hermitian but also idempotent matrices.
The link should be that the space on which we project for $ B$ is contained in the one for $ A$.

I think I see that now, but maybe someone has a nicer proof (my proof has a$ \parallel{}A v \parallel{}\equal{}\parallel{}v\parallel{}$ and thus $ A v \equal{} v$ argument which I don't really like  :| )


Prove that If $a,b,c$ are positive real and $a+b+c=3$ then :

$(a+b^2)(b+c^2)(c+a^2) \le 13+abc$

 :)
________

Pham Kim Hung

Hi!
   It is a very hard equality. And unfortunately, I can't solve it. So...Hung can u give me some hints? :D
    Thanks u!

Az Gibbenergy said; I think it is too hard... :mad:

Oh... It seem very simple. Isn't it?

Try in one week. I will post solution.

Hungkhtn, when does the equal occur ? I think it should be "12 " , not "13 "  :?:

Oh... At first, I still prove for k=12, but finall, I see it's false...  :)

Why there isn't any solution here? It's nice but not too difficult. Try again and good luck!

Hungkhtn , when does the equality occur ?

Ok. If no one has solution for It, I need to prove more before it's missing...

The solution devices in two part :

(i) If $a \ge b \ge c$  then :

$(a+b^2)(b+c^2)(c+a^2) \ge (a^2+b)(b^2+c)(c^2+a)$

(ii) If $a \ge b \ge c$ then $f(a,b,c)= (a+b^2)(b+c^2)(c+a^2)-abc \le f(a+c,b,0)$

I think It's enough for everyone solve it.!

Hi Hung. Can u write detail your sols please? I am not good at math and also I don't understand what you mean:
Because$f(a,b,c)=(a+b^2)(b+c^2)(c+a^2)-abc$:
$f(a+c,b,0) -f(a,b,c)=-c(-3ba^2-3bca-c^2b-2b^3a-b^3c+ac^2+a^3c+b^3+b^2c^2+b^2a^2c)$.
How can u prove that $f(a+c,b,0) \geq f(a,b,c)$.
And then what do you do next?

 Moreover, you say that if $a \geq b \geq c$ then $(a+b^2)(b+c^2)(c+a^2) \le (a^2+b)(b^2+c)(c^2+a)$ but if I put $a=2$ $b=0.9$  $c=0.1$ it is wrong! I think it probably vice verse!

   By the way, I can prove that $f(a+d,b+d,c+d) \geq f(a,b,c)$ if $d>0$. But it is still useless in here.
   I hope you can write it detail . That would be nice for me to understand it.
Thanks!

For the first one, if $a \ge b \ge c$ then:

$(a+b^2)(b+c^2)(c+a^2)-(a^2+b)(b^2+c)(c^2+a)$

$=(a+b+c-ab-bc-ca)(a-b)(b-c)(a-c)$

$=\frac{1}{6}((a-b)^2+(b-c)^2+(c-a)^2)(a-b)(b-c)(a-c) \ge 0$

This help is enough to prove in one case $a \ge b \ge c$.

hi,anh Hùng
the problem is harder if

a,c,c>0. a+b+c=3 .
prove that:
                   (a+b^2)(b+c^2)(c+a^2) =<13

can you post a solution?

Are U crazy ? This year is 2008  :rotfl:

If $ x,y,z$ are nonnegative numbers such that $ x \plus{} y \plus{} z \equal{} 3,$ then

$ \left(x \plus{} y^2\right)\left(y \plus{} z^2\right)\left(z \plus{} x^2\right)\leq13 \minus{} 5xyz,$

with equality if and only if $ x \equal{} y \equal{} z \equal{} 1.$
[hide="Proof."]$ 13\left(\frac {x \plus{} y \plus{} z}{3}\right)^6 \minus{} 5xyz\left(\frac {x \plus{} y \plus{} z}{3}\right)^3 \minus{} \left[x\left(\frac {x \plus{} y \plus{} z}{3}\right) \plus{} y^2\right]\left[y\left(\frac {x \plus{} y \plus{} z}{3}\right) \plus{} z^2\right]\left[z\left(\frac {x \plus{} y \plus{} z}{3}\right) \plus{} x^2\right]$

$ \equiv\frac {F(x,y,z)}{729}.$

$ F(x,y,z) \equal{} F(x,x \plus{} s,x \plus{} t)$

$ \equal{} 729\left(s^2 \minus{} st \plus{} t^2\right)x^4 \plus{} 108(s \plus{} t)\left(11s^2 \minus{} 14st \plus{} 11t^2\right)x^3$

$ \plus{} 27\left(23s^4 \plus{} 20s^3t \minus{} 33s^2t^2 \plus{} 20st^3 \plus{} 23t^4\right)x^2$

$ \plus{} 9\left[(s \minus{} t)^2\left(17s^3 \plus{} 47s^2t \plus{} 74st^2 \plus{} 17t^3\right) \plus{} 11s^3t^2 \plus{} 38s^2t^3\right]x$

$ \plus{} \frac {s \plus{} t}{3125}\left[(4s \minus{} 5t)^2\left(2453s^3 \plus{} 1995s^2t \plus{} 10725st^2 \plus{} 1625t^3\right) \plus{} 1377s^5 \plus{} 16200s^4t\right]$

$ \geq 0,$

which is clearly true for $ x \equal{} \min\{x,y,z\}.$[/hide]
See also here : http://www.mathlinks.ro/viewtopic.php?t=133930

[quote=Ji Chen]If $ x,y,z$ are nonnegative numbers such that $ x \plus{} y \plus{} z \equal{} 3,$ then

$ \left(x \plus{} y^2\right)\left(y \plus{} z^2\right)\left(z \plus{} x^2\right)\leq13 \minus{} 5xyz,$

with equality if and only if $ x \equal{} y \equal{} z \equal{} 1.$
[/quote]
We can prove it without computer.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$
Since $x^2y^2z^2\leq xyz$, it remains to prove that $7u^2w^3+\sum_{cyc}(ux^3y+x^3z^2)\leq13u^5$, which is quadratic inequality of $w^3$
and the rest is smooth.


calculus
linear algebra
calculus computations

I am taking multivariable calculus, and I have two books at hand: Rudin's "Principles of Mathematical Analysis" and Hardy's "A Course of Pure Mathematics."  Which one would be better to start off with if I am only familiar with single-variable calculus?  Which one would be more useful for multivariable calculus and linear algebra?

They're probably both completely useless at the level you're talking about. Stick with textbooks on the subjects you're taking.


Reggie rents a bicycle for a fee of $ \$6$ plus $ \$5$ for each hour or fraction of an hour. What is the maximum number of hours for which he can rent the bicycle for a total cost of $ \$41$?

Reggie can rent the bike for $ \frac{41\minus{}6}{5}\equal{}\boxed{7}$ hours maximum.


If $ a,\ b,\ c$ are the sides of a triangle $ (a\ne b\ne c)$, then:

$ \left| \frac{a}{a\minus{}b}\plus{}\frac{b}{b\minus{}c}\plus{}\frac{c}{c\minus{}a}\right| \ > \sqrt{6}\minus{}1.$

[quote="Mateescu Constantin"]If $ a,\ b,\ c$ are the sides of a triangle $ (a\ne b\ne c)$, then:

$ \left| \frac {a}{a \minus{} b} \plus{} \frac {b}{b \minus{} c} \plus{} \frac {c}{c \minus{} a}\right| \ > \sqrt {6} \minus{} 1.$[/quote]

Let $ a\equal{}max(a,b,c)$

$ a \equal{} y\plus{}z$  , $ b\equal{} z\plus{}x$ , $ c\equal{} x\plus{}y$

1. $ a \ge b \ge c \Longrightarrow  z \ge y \ge x$

$ \left| \frac {a}{a \minus{} b} \plus{} \frac {b}{b \minus{} c} \plus{} \frac {c}{c \minus{} a}\right| \equal{} \frac {a}{a \minus{} b} \plus{} \frac {b}{b \minus{} c} \minus{} \frac {c}{a \minus{} c}\equal{} f(x,y,z)\equal{}$ $ \frac{y\plus{}z}{y\minus{}x}\plus{}\frac{x\plus{}y}{z\minus{}x}\minus{}\frac{z\plus{}y}{y\minus{}x}\equal{}$ $ \frac{(z^2\minus{}y^2)}{(y\minus{}x)(z\minus{}x)}\plus{}\frac{x\plus{}z}{z\minus{}y}\plus{}1$

$ min(f(x,y,z))$ will be for x = 0 . We can assume $ z\equal{}1:$

$ min(f(x,y,z)) \equal{}min( \frac{1}{y}\plus{}\frac{1}{1\minus{}y}\minus{}y\plus{}1)$ for $ 0<y< 1$; $ (*)$

2. $ a \ge c \ge b \Longrightarrow  y \ge z \ge x$

$ \left| \frac {a}{a \minus{} b} \plus{} \frac {b}{b \minus{} c} \plus{} \frac {c}{c \minus{} a}\right| \equal{} \frac {b}{c \minus{} b} \plus{} \frac {c}{a \minus{} c} \minus{} \frac {a}{a \minus{} b}\equal{} f(x,y,z)\equal{}$ $ \frac{x\plus{}z}{y\minus{}z}\plus{}\frac{x\plus{}y}{z\minus{}x}\minus{}\frac{z\plus{}y}{y\minus{}x}\equal{}$ $ \frac{(y^2\minus{}z^2)}{(y\minus{}x)(z\minus{}x)}\plus{}\frac{x\plus{}z}{y\minus{}z}\minus{}1$

$ min(f(x,y,z))$ will be for x = 0 . We can assume $ y\equal{}1:$

$ min(f(x,y,z)) \equal{}min( \frac{1}{z}\plus{}\frac{1}{1\minus{}z}\minus{}z\minus{}2)$ for $ 0<z< 1$; $ (**)$

From $ (*)$ and $ (**)\Longrightarrow$ $ min\left| \frac {a}{a \minus{} b} \plus{} \frac {b}{b \minus{} c} \plus{} \frac {c}{c \minus{} a}\right|\equal{} min( \frac{1}{t}\plus{}\frac{1}{1\minus{}t}\minus{}t\minus{}2)$ for $ 0<t< 1$;

$ g(t)\equal{}\frac{1}{t}\plus{}\frac{1}{1\minus{}t}\minus{}t\minus{}2$ , $ 0<t< 1$;

$ min(g(t))$ will be for $ t_0 \equal{} \frac{1\plus{}\sqrt 2}{2}\minus{}\frac{\sqrt{ 2 \sqrt{2}\minus{}1 }}{2}$

So  : $ min\left| \frac {a}{a \minus{} b} \plus{} \frac {b}{b \minus{} c} \plus{} \frac {c}{c \minus{} a}\right|\equal{} g(t_0)\equal{} \boxed{\frac{3\plus{}\sqrt{2}}{2}\sqrt{2 \sqrt{2}\minus{}1}\minus{}\frac{3}{2}} > \sqrt{6}\minus{}1$


Two people (X and Y) play a game. They have a $ \triangle ABC$ in front of them - not necessarily an equilateral triangle, but an acute
angled triangle. X goes first - he chooses a point $ P$ on $ AB$.
Y then chooses a point $ Q$ on $ BC$ and finally X chooses a
point $ R$ on $ CA$.

The objective of the game? X intends to push the $ \triangle PQR$
to be as large as possible, and Y intends to pull it as small
as possible.

What are their best strategies for choosing the points?

[size=150][b]Step 3[/b][/size]
Let $ P' \in BC$ such that $ PP' \parallel AC$
X wants to maximize the area of $ \triangle PQR$, so he maximizes the height which corresponds to $ PQ$
$ a)\ Q$ between $ C,P' \Longrightarrow$ he chooses the point $ A$
$ b)\ Q$ between $ B,P' \Longrightarrow$ he chooses the point $ C$
$ c)\ Q\equiv P'\Longrightarrow$ he chooses any point on $ AC$

[asy]pair A=(0,8), B=(4,0), C=(5,8); real p = 0.6, p2=0.6; real q=0.4; pair P,P2,Q; draw(A--B--C--cycle); dot(A^^B^^C); P = A+p*(B-A); dot(P); P2 = C+p2*(B-C); dot(P2); Q = C+q*(B-C); dot(Q); draw(P--P2); draw(P--Q,red); label("A",A,W); label("B",B,SW); label("C",C,E); label("P",P,W); label("P'",P2,E); label("Q",Q,E);[/asy]

[size=150][b]Step 2[/b][/size]
Y wants to minimize the area of $ \triangle PQR$

$ a)$ For any $ Q\in (P'B]$ we can see that $ (PQC)>(PP'C)$

[asy]pair A=(0,8), B=(4,0), C=(5,8); real p = 0.6, p2=0.6; real q=0.8; pair P,P2,Q; draw(A--B--C--cycle); dot(A^^B^^C); P = A+p*(B-A); dot(P); draw(P--C,red); P2 = C+p2*(B-C); dot(P2); Q = C+q*(B-C); dot(Q); draw(P--P2); draw(P--Q,red); label("A",A,SW); label("B",B,SW); label("C",C,SE); label("P",P,W); label("P'",P2,E); label("Q",Q,E);[/asy]

$ b)$ For any $ Q\in (P'C]$ we can see that $ (PQA)>(PP'A)$ because the height corresponding to the side $ AP$ is bigger in the first triangle than the second

[asy]pair A=(0,8), B=(4,0), C=(5,8); real p = 0.6, p2=0.6; real q=0.3; pair P,P2,Q; pair R1,R2; draw(A--B--C--cycle); dot(A^^B^^C); P = A+p*(B-A); dot(P); P2 = C+p2*(B-C); dot(P2); Q = C+q*(B-C); dot(Q); draw(Q--A,red); draw(Q--P,red); R1 = P2+(A-B); R2 =intersectionpoint (P2--R1, A--C); draw(P--P2); draw(P2--R2,dashed); draw(P2--A,blue); label("A",A,SW); label("B",B,SW); label("C",C,SE); label("P",P,W); label("P'",P2,E); label("Q",Q,E);[/asy]

So Y can achieve the smaller possible area for $ \triangle PQR$, if he choose the point $ Q$ in the position of $ P'$

[size=150][b]Step 1[/b][/size]
So, X has to solve the following problem:
Find $ P\in AB$ and $ P'\in BC$ such that $ PP' \parallel AC$ and the area of $ \triangle APP'$ is maximum

We can prove that the best position for $ P$ is the midpoint of $ AB$

[size=125][hide="Proof"][/size]
We have maximum $ (APP')$ when we have minimum $ (PBP')\plus{}(ACP')$
Let $ h_a$ be the height of $ \triangle ABC$ corresponding to the side $ BC$ and $ h$ the height of $ \triangle PP'B$ corresponding to the side $ BP'$
Let $ \lambda \equal{} \frac{BP}{BA} \equal{} \frac{BP'}{BC}$

$ (ACP') \equal{} \frac{1}{2} h_a\cdot CP' \equal{} \frac{1}{2} h_a\cdot(1\minus{}\lambda)BC$

$ (PBP') \equal{} \frac{1}{2} h\cdot BP' \equal{}  \frac{1}{2} (\lambda h_a)\cdot (\lambda BC)$

Then $ (PBP')\plus{}(ACP') \equal{} \frac{h_a \cdot BC}{2}(\lambda^2\minus{}\lambda\plus{}1)$

..and we know that the trinomial $ \lambda^2\minus{}\lambda\plus{}1$ has minimum for $ \lambda \equal{} \frac{1}{2}$[/hide]


A man builds a garden with dimensions $ x$ by $ y$ feet. He creates a two-inch wide border of pansies along the inside of the edges of the garden. He then builds a rectangular-prism greenhouse around the entire garden, with a height of 9 feet. Dissatisfied by this, he decides to create another greenhouse with one twenty-seventh the volume, but in the shape of a cube. Prove that the area of the border of pansies is always greater than or equal to the measure of one of the edges of the cube.

The likely reason why noone's answered this question yet is that it's kinda confusing... it seems like you're trying to compare a 1-dimensional measure with a 2-dimensional measure. Maybe if you forget the gardener and just use points and lines and shapes, it will be clearer what you mean  :)

Sorry... yeah, it's hard to explain, but the solution basically boils down to writing everything algebraically, rearranging, and using AMGM. I'm afraid it doesn't really get any better when you make the problem more abstract.

I understood the description of the problem, just not the last sentence.  What exactly do you mean by comparing area to length?

Was the inequality that you were driving at something like
[hide]
$ 4(x\plus{}y\minus{}4) \ge \sqrt[3]{4xy}$?
[/hide]

I mean comparing their measures.

@Andy: Yeah, that's basically the inequality.


Let ABCDEFG be a regular heptagon and take any H point in heptagon such that angle <DHG=x and |DH|=|DE|.If <HGF=270/7 than find x angle.


Please write a solution with details for this
With $4$ different numbers from the numbers $1,2,...,9$ we make $24$ three-digit numbers. Prove that the sum of the $24$ numbers is $6666$ times bigger than the sum of the digits of these numbers.

I think the numbers should be 4-digit. Than if there's a number x among the 4 chosen ones, it appears
in 3!=6 numbers on the position of 1s, in six numbers on the position of 10s and so on. The sum of multiples in which it appears is 6(1+10+100+1000)=6666, from wherere the answer follows.

[quote="Kondr"]I think the numbers should be 4-digit. [/quote]
You are sure?

If they were 3-digit and the greatest of them was x, the sum would be less then
111.24.x<6666x<6666(x+y+z).


Is there an elementary way to find all integer points of $ y^2 \equal{} x^4 \plus{} x^3 \plus{} 1$?

The thing to keep in mind about Diophantine equations of this form is that in general one should not [b]expect[/b] an elementary solution, even if one exists.  The general strategy is to write the Diophantine as an elliptic curve and compute its [url=http://planetmath.org/encyclopedia/RankOfAnEllipticCurve.html]rank[/url].  If it has rank zero, the integer points can be isolated using [url=http://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem]Nagell-Lutz[/url].

Okay. It would be nice to know how to find an integer point of an elliptic curve if rank is positive and is torsion group is known. I haven't found any material. Furthermore, what kind of transformations I am allowed to do when I transform curve to an elliptic curve?


geometry
geometric transformation
\/closed

All the links in my post
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198223
have been changed to some bullshiit by you righteous great Moderator. And I am not allow to edit it now.

Have you no shame?

Is this what you called freedom of speech?

I must thank you for moving my first post to
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198129

Seems like you can not edit the post there.  :rotfl:

[b]Freedom of speech does not exist in an online community.[/b]  The people who run an online community reserve the right to place any limitations on it that they wish.  You accepted those conditions when you signed up, whether you bothered to read the Terms of Service or not.

The moderator has already given his reasons:

[quote="Buffalo"][color=green]Moderator says: This post is
1) a double-post
2) on a subject for which a thread exists (and which was locked due to the misbehavior of the O.P.)
3) in violation of the general norms of this forum, i.e. that posts be attempts at discussion or requests for information rather than advertising, political propaganda, etc.[/color][/quote]

Pay a little more attention to the third reason, even if you aren't willing to accept the first two.

Please compare these two posts(the second one has been edited by JBL again, fortunately, I have a backup)
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198129

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198223
[quote="Buffalo"][url=http://www.tibet.org/]http://rfi.fr/actucn/articles/100/article_6734.asp[/url]

Click "收听 9 分钟 1 秒" 02/04/2008 to listen.

Part of the English translation(Which I think to be the most astonishing part). The bold text in the parenthesis was added for clarifying part of his lies.

[quote]Interview of Dawa Tsering, China Affairs Officer of the Tibetan Government-in-Exile

Q: The Chinese government on numerous occasions has questioned Dalai Lama's position on non-violence, and asked that if Dalai Lama truly advocates non-violence, why has he not condemned the cruel acts of murdering the Han Chinese committed by the Tibetan rioters during the March 14, 2008 riots in Tibet?

A: First of all, I must make it clear that the Tibetan ([b]rioters[/b]) has been non-violent throughout ([b]the incident[/b]).  From Tibetans'([b]Government-in-Exile[/b]) perspective, violence means harming life.  From the video recordings you can see that ([b]the Tibetans rioters[/b]) were beating Han Chinese, but only beating took place.  After the beating the Han Chinese were free to flee.  Therefore ([b]there were[/b]) only beating, no life was harmed.  Those who were killed were all results of accidents.  From recordings shown by the Chinese Communist government, we can clearly see that when Tibetan ([b]rioters[/b]) were beating on their doors, the Han Chinese all went into hiding upstairs.  When the Tibetan ([b]rioters[/b]) set fire to the buildings, the Han Chinese remained in hiding instead of escaping, the result is that these Han Chinese were all accidentally burnt to death.  Those ([b]Tibetan rioters[/b]) who set and spread the fire, on the other hand, had no idea whatsoever that there were Han Chinese hiding upstairs.  Therefore not only were Han Chinese burnt to death, some Tibetans were burnt to death too.  Therefore all these incidents were accidents, not murder.[/quote]

[url=http://www.tibet.com/]A full English translation on Youtube[/url]

Is he lying? Maybe, maybe not. Nevertheless, the fact is that they have such kind of glory tradition. For example, they call the 1959 rebellion "peaceful uprising" and claim that [url=http://www.freetibet.net/]peace and happiness prevailed in Tibet[/url] under Dalai lama's rule of theocracy and serf-ism.

Two more links
[url=http://www.savetibet.org/]Three dead in Indian 'Tibet' bomb[/url]
[quote]large quantity of explosives and detonators. timers[/quote]
They prepared all these for some peaceful purposes? They are not terrists?

[url=http://www.studentsforafreetibet.org/]Friendly Feudalism: The Tibet Myth[/url]
[quote]Since it was against Buddhist teachings to take human life, some offenders were severely lashed and then “left to God” in the freezing night to die.[/quote]

[color=green]Moderator says: This post is
1) a double-post
2) on a subject for which a thread exists (and which was locked due to the misbehavior of the O.P.)
3) in violation of the general norms of this forum, i.e. that posts be attempts at discussion or requests for information rather than advertising, political propaganda, etc.[/color][/quote]

What part of the mod-edit, or for that matter, common sense, do you not understand?

1. These topics do not invite discussion; they are heavily opinionated and overly long.
2. These topics have little relevancy to the forum.
3. These topics have a unattractively personal nature.

If you can tell me a successful topic that parallels this one's nature, I would be surprised.

Notice how I so boldly pointed out that your first link's topic was doomed.

[quote]Freedom of speech does not exist in an online community. [/quote]
[b]does not[/b] mean [b]other ones[/b] can speak with [b]my[/b] tongue.

[quote="distracted523"]:P The mod edit is in red, and [b]most[/b] people know it's a mod edit. Stop arguing.[/quote]
Please read before reply, OK?
JBL changed the links in my post, DO YOU UNDERSTAND?

I'm not really sure what you're trying to accomplish here. It's kinda obvious that your posts are in violation of some rule around here, and for all intents and purposes, the mods and admins are on the same team. Moreover, this is a math forum anyway, so why bother stressing over this so much?

[quote="moogra"][quote="ProtestanT"]so...why were the links changed?[/quote]
They were political propaganda.[/quote]
1. I do not agree that those are political propaganda.
2. Even if they were, delete them, ban me, that is fine. But why change them to something completely against the original post and pretend it to be my post? Is this what a moderator supposed to do?

The point here is not what I said, it is what and how JBL did, is it clear enough for you?

For example, suppose I were the infamous Saddam Hussin, I claimed that "There were no WMD in Iraq."  Dear president George W. Bush did not agree with me, he claimed that there were plenty of WMD in Iraq and sent GLA to Iraq. This is [b]fine[/b]. But what if Bush tell the people that "Saddam Hussin said that 'There are plenty of WMD in Iraq.'"?

Please compare

my links
http://rfi.fr/actucn/articles/100/article_6734.asp
[url=http://www.youtube.com/watch?v=3cp6l6yTi9M]A full English translation on Youtube[/url]
[url=http://www.tibet.net/en/ohhdl/future.html]peace and happiness prevailed in Tibet[/url]
[url=http://news.bbc.co.uk/2/hi/south_asia/7329040.stm]Three dead in Indian 'Tibet' bomb[/url]
[url=http://www.michaelparenti.org/Tibet.html]Friendly Feudalism: The Tibet Myth[/url]

After JBL's edit
[url=http://www.tibet.org/]http://rfi.fr/actucn/articles/100/article_6734.asp[/url]
[url=http://www.tibet.com/]A full English translation on Youtube[/url]
[url=http://www.freetibet.net/]peace and happiness prevailed in Tibet[/url]
[url=http://www.savetibet.org/]Three dead in Indian 'Tibet' bomb[/url]
[url=http://www.studentsforafreetibet.org/]Friendly Feudalism: The Tibet Myth[/url]

[quote="Buffalo"]
After JBL's edit
[url=http://www.tibet.org/]http://rfi.fr/actucn/articles/100/article_6734.asp[/url]
[url=http://www.tibet.com/]A full English translation on Youtube[/url]
[url=http://www.freetibet.net/]peace and happiness prevailed in Tibet[/url]
[url=http://www.savetibet.org/]Three dead in Indian 'Tibet' bomb[/url]
[url=http://www.studentsforafreetibet.org/]Friendly Feudalism: The Tibet Myth[/url][/quote]

Sounds to me like he has a good sense of humour, he must do to have not banned you already, but anyway....

Buffalo you are definitely posting the wrong thing in the wrong forum.
But I would also say that it is rather inappropriate to change the content without the poster's consent, in a way that would offense the poster and some members of the forum.
(I personally dislike the site http://www.tibet.org. It is as biased and propaganda-like as Buffalo's post)

I wonder why this topic isn't locked yet...

All right, that's it. Buffalo you have received your fair warning from me, it's time to say bye-bye. There are thousands and thousands of forums out there where you can militate your positions on international politics and I suggest you use them to do your flaming. This discussion is over.


conics
parabola
function
calculus
calculus computations

Determine the shortest distance from the point (0,5) to the parabola $16y=x^{2}$

$d(x) = \sqrt{(x-0)^{2}+\left(\frac{x^{2}}{16}-5 \right)^{2}}$

$D(x)=x^{2}+\left(\frac{x^{2}}{16}-5 \right)^{2}$
minimizing this:

$D'(x)=2x+\frac{1}{64}x^{3}-\frac{5}{4}x$

$\therefore \frac{3}{4}x+\frac{1}{64}x^{3}=0 \Longleftrightarrow \boxed{x=0}$ the only real solution.

Now, the minimum distance is $d(0)=5 \,$ units

We must minimize  $f(y)=x^{2}+(y-5)^{2}=16y+(y-5)^{2}$.Since f'(y)=2y+6,the only critical value of f is y=-3,which corresponds to an imaginary value of x.Hence the minumum distance does not exit.
Since , the only critical value of f is y=-3,which correspond to an imaginary value of x.Hence the minumum distance does not exit.

[quote="matholympicman"]Since $f'(y)=2y+6$, the only critical value of f is y=-3,which corresponds to an imaginary value of x.Hence the minumum distance does not exist.
[/quote]
You make a standard mistake here forgetting that, when minimizing a function over an interval with finite endpoints (in your case, the ray $[0,+\infty)$), the minimum may be attained at an endpoint, which is exactly what happens here.

I apologize for my solution. I did not see the word [b]parabola[/b]. I thought it is a curve $y=\frac{x^{2}}{16}$

But your solution was [b]correct[/b] (though, maybe, not the most elegant one) and the parabola in question is exactly the curve $y=\frac{x^{2}}{16}$! So, there is no need to apologize for anything  :) .