1)  Prove by contradiction that there do not exist integers m and n such that 14m + 21n = 100.

I know I need to use an if/then statement.  Still having trouble.

2) Using the fact that an integer n, either n >= 14 or n <= 15, or otherwise prove 7 does not divide 100.

I think if the problem was stated differently, it would be easier for me to comprehend.

Solution so far:  7*14 = 98 and 7*15 = 105

I was told I need one more step to complete the problem, but I am not sure what it is?

Thank you in advance!

[hide="1"]
Assume that there exists integers $m$ and $n$ such that $14m+21n = 100$.
$7$ divides the LHS, so it must also divide the RHS.
But $7$ does not divide $100$, so we have a contradiction.
Therefore, our initial assumption is incorrect, and no $m,n$ exists.
[/hide]

Thank you rjt.

So,
If m and n are integers, then 14m + 21n = 100
Factor out a 7 giving us 7(2m + 3n) = 100

From here can you clarify for me what you mean by 7 does not divide 100?  Thanks again.

He means that 100 divided by 7 is not an integer.  $2m+3n$ will always be an integer if $m$ and $n$ are integers, so by that assumption $100/7$ must also be an integer, so it is obviously false.

[quote="MathStudent2"]Got it!  Thank you.  Any suggestions on th e second problem below:

2) Using the fact that an integer n, either n >= 14 or n <= 15, or otherwise prove 7 does not divide 100.

I think if the problem were stated differently, it would be easier for me to comprehend.

Solution so far: 7*14 = 98 and 7*15 = 105

I was told I need one more step to complete the problem, but I am not sure what it is?

Thanks again![/quote]

I'm not sure what's being asked, but I think this is what you're looking for:

If $7$ does divide $100$ evenly, then there is some whole number number $k$ that when multiplied by $7$ equals $100$.
So, $7k=100$
As you said, $(7)(14)=98$.
So $k>14$.
Also, $(7)(15)=105$.
So $k<15$.
But there are no whole numbers in between $14$ and $15$, so $k$ does not exist. So $7$ cannot divide $100$ evenly.
:)

[quote="MathStudent2"]Thank you rjt.  One question after you multiply 7(14) = 98 should it be k > 14 or k >= 14?  Similiarly, 7(15) = 105 should it be k <15 or k <=15?  I think your right as far as what they are asking for and the solution to the problem.  The wording is not clear.  Thank you![/quote]
No problem :)

I'm not sure why the question refers to $k\geq 14$ and $k\leq 15$. As $(7)(14)<100$, we have to increase $k$, and so we should have $k>14$. Similarly with $k>15$.

[quote="MathStudent2"]rjt.  I made a typo!!  I went back and looked at the orginal problem and it shoud read:

Using the fact that for an integer n, either n <= 14 or n >= 15, or otherwise prove 7 does not divide 100.

Does this change anything?  Thanks![/quote]

That makes a bit more sense. The "$n \leq  14$ or $n \geq  15$" bit is just a hint so that when you get to the point where you have $14<n<15$, you realise that there is no possible solution when $n$ is a whole number.

[quote="MathStudent2"]So, is my solution correct below?

If 7 divides 100 evenly, then there is some whole number n that when multiplied by 7 equals 100.

So, Then 7n = 100
7(14) = 98
So, n <= 14
Also, 7(15) = 105
So, n >= 15

Therefore, there are no whole numbers for which 14 <= n <= 15 that when multiplied by 7 equals 100.  Therefore, 7 cannot divide 100 evenly.

Thank You![/quote]

Almost, but because $98<100$, we want to make $k$ larger, and so $n > 14$, not $n \leq  14$. And similarly it should also be $n < 15$.

Other than that, you've got it exactly.

Since in the orginal problem was n <= 14 and n >=15, to me that means n is less than or equal to 14 or greater than or equal to 15.  Not between 14 and 15.  Also, should the solution below be <= and >= and not < and >?  I am getting confused on the inequalities.

Made correction below.  Not sure if it is correct?

If 7 divides 100 evenly, then there is some whole number n that when multiplied by 7 equals 100.

So, Then 7n = 100
7(14) = 98
So, n >= 14
Also, 7(15) = 105
So, n <= 15

Therefore, there are no whole numbers for which 14 <= n <= 15 that when multiplied by 7 equals 100. Therefore, 7 cannot divide 100 evenly.

Thank You!

[quote="MathStudent2"]Also, should the solution below be <= and >= and not < and >?...
Therefore, there are no whole numbers for which 14 <= n <= 15 that when multiplied by 7 equals 100...[/quote]The correct inequalities should be $14<n<15$, they have to be strict so that you can get a contradiction. There is no contradiction if you write $14\le n\le15$ because this statement means $n$ can be 14 or 15.


trigonometry
conics
parabola
calculus
quadratics
geometry
similar triangles

Given a circle with radius 6 and centered at $ O$. A point $ A$ is such that $ AO\equal{}20$. A line passes through $ A$ and intersects with the circle at $ P$ and $ Q$. Find the max value of the difference of the distances from line $ OA$ to $ P$ and to $ Q$.

Maybe this is a start?

I'm not sure I did it right, but:

[hide]
Let P be the point closer to A.  Let $ \overline{AP} \equal{} x$ and $ \overline{PQ} \equal{} y$.  Let D be the point where a perpendicular from Q hits OA.  Let $ \overline{OD} \equal{} d$

Now Power of a Point gives us $ x(x\plus{}y)\equal{}(14)(26)\equal{}364$.  We want to maximize the quantity $ \sqrt{(x\plus{}y)^2\minus{}(20\plus{}d)^2} \minus{} \sqrt{x^2\minus{}196}$

Substitute from the previous PoP: $ \sqrt{\left(\frac{364}{x}\right)^2\minus{}(20\plus{}d)^2} \minus{} \sqrt{x^2\minus{}196}$

Now we can use similar triangles to see that $ 20\plus{}d\equal{}\frac{14(x\plus{}y)}{x}\equal{}\frac{14(364)}{x^2}$.

Now we want to minimize $ \sqrt{\left(\frac{364}{x}\right)^2\minus{}\left(\frac{14(364)}{x^2}\right)^2} \minus{} \sqrt{x^2\minus{}196}$

And I don't see an easy way to do that without a calculator...
[/hide]

[hide="Solution"]
Let $ \angle QAO \equal{} x$. We are looking  to maximize (AQ-AP)sinx. By law of cosines on triangles QAO and PAO, we have AP and AQ are the roots of $ 36 \equal{} 400 \plus{} n^2 \minus{} 40ncosx$, or $ n^2 \minus{} (40cosx)n \plus{} 364 \equal{} 0$. The difference of the roots is $ \sqrt {1600cos^2x \minus{} 1456}$, so we are looking for the maximum value of $ sinx\sqrt {1600cos^2x \minus{} 1456}$=$ 4sinx\sqrt {9 \minus{} 100sin^2x}$.
Let $ sin^2x\equal{}m$. We want to maximize $ 4\sqrt {9m \minus{} 100m^2}$. The vertex of the parabola is at m=9/200. Subbing in gives 9/5 as the answer.[/hide]

[b]dgreenb801[/b], my method is similar to yours, except that I didn't use calculus to maximize that quartic. It's just a quadratic in disguise, and we can surely maximize that by taking half of the sum of roots without calculus.

Timwu, I changed the solution before you posted that :lol:.
Does anyone have an easier (possibly purely geometric) proof?

Try this one.

[hide]Let $ H$ and $ K$ be the perpendicular feets from $ P$ and $ Q$.  Denote the area of triangle $ AOQ$ by $ [AOQ]$. Thus,
$ [AOQ]\minus{}[AOP]\equal{}[POQ]$
$ \frac{1}{2}(\overline{AO})(\overline{QK})\minus{}\frac{1}{2}(\overline{AO})(\overline{PH})\equal{}\frac{1}{2}(\overline{OP})(\overline{OQ})\sin{\theta}$
Simplify this we got
$ \overline{QK}\minus{}\overline{PH}\equal{}\frac{9}{5}\sin{\theta}\leq\frac95$[/hide]


algebra
polynomial
calculus
derivative
algorithm
Rational Root Theorem
calculus computations

HI

i need some help solving the following polynomial;

5x^7 - 21x^6 - 51x^5 + 141x^4 +264x^3 + 342x^2 - 560x + 96 = 0

i have to find all roots of the equation.
I tried horner's method but I am still confused.

Mathematica says the roots are

$-1\pm\sqrt{3}$, $-1\pm i\sqrt{2}$, $\frac{1}{5}$ and $4$ (double root)

[quote="frt"][quote="numero1"]And also what method you think would work best on the following problem:

1 + x + (x^2/2) + (x^3/6) = 0.

It says to show that the equation has exactly one root and then find all its roots.
Tossup between bisection method and newton's method.

thanks[/quote]
I think you can just take the derivitive of $f(x)= \frac{x^3}{6}+\frac{x^2}{2}+x+1$ and show that it's positive for all $x$.
If you can show that, then that means $f(x)$ is increasing at any point, so the graph $y=f(x)$ crosses the x-axis exactly once.

After that, you can use whichever method you want to find (or estimate) the root.[/quote]

I took the derivative and showed it was increasing and then used the bisection method.
here is what I got:

left endpoint      right endpoint      midpoint      p(midpoint)

-2                      -1                        -1.5              .375
-2                      -1.5                      -1.75           -3.17
-1.75                 -1                         -1.375         -1.75
-1.75                 -1                         -1.375         -1.75

and it got no closer than this to the root....it just kept repeating the last two lines.
What can I do now?

[quote="TheBB"][quote="numero1"]is there any way to do it by hand?[/quote]

Well, you can impress everyone by seeing in a split second that it factors into

$(x+1+\sqrt{3})(x+1-\sqrt{3})(x+1+i\sqrt{2})(x+1-i\sqrt{2})(x-\frac{1}{5})(x-4)^2=0$

Otherwise I'm afraid you're out of luck.[/quote]

haha
this was clever

[quote="jmerry"][quote="numero1"]and it got no closer than this to the root....it just kept repeating the last two lines.
What can I do now?[/quote]
You've programmed the algorithm wrong, obviously. The bisection method [b]always[/b] works (but it's slow).

Why does your list of steps go back to using -1 as an endpoint, after already establishing that we should be looking at the interval between -2 and -1.5? The third line should have -1.75 and -1.5 as the endpoints.[/quote]

OK thanks I saw my mistake.
A question though how do you find other roots using the bisection method. I found -1.50... so far.
but I am not sure how to find the others???

[quote="jmerry"]As the problem says, there's only one root. Don't go chasing ghosts.[/quote]

ok but how would you find the other roots if they existed with the bisection method.

[quote="jmerry"]As the problem says, there's only one root. Don't go chasing ghosts.[/quote]

How would I solve the following and what method do you reccomend?

5x^7 - 21x^6 - 51x^5 + 141x^4 +264x^3 + 342x^2 - 560x + 96 = 0

[quote="jmerry"]Step 1: Find the rational roots, and factor the appropriate terms out. (This one has three rational roots counting multiplicity)
I just did this myself with a spreadsheet- it's really quite easy.

By inspection of the values we calculated looking for rational roots, there are two more real roots and two complex roots. We can guess the approximate locations of the real roots easily. Anything works to find them, and they were rigged to be nice.

It's not as hard as it sounds- this one was rigged to come out nicely.

I'm not recommending any particular method except "Do [i]something[/i]". Just try things until the work.

With all the root-finding methods you know, the way to find different roots is to start from different places.[/quote]

Um..I have no clue how to find the rational roots.

The Rational Root Theorem:

If $a_nx^n+a_{n-1}x^{n-1}+\cdots a_1x+a_0=0,$ where the coefficients are integers and $a_n\ne0,$ has any rational roots, they are of the form $x=\frac pq$ where $p$ divides $a_0$ and $q$ divides $a_n.$

[This is essentially a theorem of number theory. It does not promise that any rational roots exists; it just limits the possibilities if they do.]

In your case, $a_n=5$ and $a_0=96.$ This gives the following list of possible rational roots:

$\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm16,\pm24,\pm32,\pm48,\pm96,$
$\pm\frac15,\pm\frac25,\pm\frac35,\pm\frac45,\pm\frac65,\pm\frac85, \pm\frac{12}5,\pm\frac{16}5,\pm\frac{24}5,\pm\frac{32}5,\pm\frac{48}5, \pm\frac{96}5.$

That's a long list, of course. Once you get into it, some of them can be ruled out on size grounds (watch the signs of the coefficients of the quotients.) We test them by synthetic division. Note jmerry's reference to having used a spreadsheet - clearly, he simply set up the synthetic division algorithm in that spreadsheet.

If you do find a factor, continue looking for factors not of the original polynomial but of the new quotient.

There's also a neat little theorem concerning real roots of polynomial.
[b]Theorem :[/b]
All real roots ,if there are any of them,of polynomial
$f(x)=a_{o}+a_{1}x+...+a_{n}x^n$
are located within interval:
$[-(\frac{a}{|a_{n}|}+1),\frac{a}{|a_{n}|}+1]$,where $a=max ( |a_{o}|,|a_{1}|,...,|a_{n}| )$
I find it useful sometimes.


There are $ 5$ different lines in the plane, any two of them intersecting in a point. Prove that among these lines there are two lines which form an angle with the measure at most $ 36^{\circ}$.

[hide="Key idea"]Translate the lines so that they all pass through a common point[/hide]


Two angles of a triangle measure $ 26^\circ $ and $ 67^\circ $.  How many degrees are in the measure of the third angle?

[hide="Answer"]$ \text{Total amount of degrees in a triangle: } $

$ 180^\circ $

$ 26^{\circ} + 67^{\circ} + x^{\circ} = 180^{\circ} $

$ 93^\circ + x^\circ = 180 $

$ x^\circ = 87^\circ $

${ \boxed{\boxed{\text{The third angle is 87}^{\circ}}}} $[/hide]

[hide]
The sum of degrees in a triangle is 180, 26+67=93, so the answer is 180-93=87[/hide]

[hide]we know that sum of the triangle angles are 180. $\rightarrow$
the third angle is $180-67-29=87$ but there is something that can you prove that sum of the triangle angles is 180(it is easy). [/hide]


trigonometry
geometry
inequalities
circumcircle
power of a point
triangle inequality
trig identities

Adding all the equations together, we have:
$2x^2+2y^2+2z^2+(xy+xz+yz)=50$

Now let $a=(x+y+z)^2, b=x^2+y^2+z^2, c=xy+xz+yz$
$a=b+2c$
$2b+c=50$

I'm stumped on how to solve for $c$, though. Any help? :huh:

this has been posted a while ago, but the idea is you have a point P inside a 3-4-5 triangle such that you form 120 degree angles between the lines connecting the verticies of the triangle, but i don't know if that helps...you might use power of a point or something...? but i remember yetti writing a huge post about it, i don't see a reasonable solution yet...i will look into it though

anybody want to generalize,

given the system
$x^2+xy+y^2=a^2$
$y^2+yz+z^2=b^2$
$z^2+zx+x^2=c^2$

where $a,b,c>0$ and $a<b+c$, $b<a+c$, $c<a+b$

find the value of $xy+yz+zx$

(it does become obvious that we have to have it be sides of a triangle with a generalization)

[quote="shobber"][quote="shyong"][quote="shobber"]Use the area equation to get the exact value of $xy+yz+xz$.[/quote]
shobber , what if x,y,z can be negative ? Then the method you suggested wont work ....[/quote]
Yes, that is what I've been thinking about. I think that there exists a way to prove that $x, y, z$ are positive..[/quote]

Wouldn't the geometric argument show $x, y, z$ positive as long as $a, b, c$ satisfy Triangle Inequality?

one thing to notice that if (x,y,z) is a solution, then (-x,-y,-z) is a solution also...

but just say you have the 3-4-5 triangle, and you use the law of cosines, and you get the system, so the triangle generates the same system, i think that is enough

If $x,y$ or $z$ is negative, couldn't you say you are using directed segments?  I don't really know much about directed segments, but I think the problem should still work... :?

Well, disregarding the parity argument, it's not that hard to generalize... Although I'm assuming $a^2+b^2=c^2$ as it is in this problem.
We make a point P with angles of 120 to each side, then law of cosines to get those equations, where x, y, and z are lengths to the vertices.  Then the area is $\frac12 ab = \frac{\sin 120}{2}(xy+yz+zx) \implies xy+yz+zx = \frac{2ab\sqrt 3 }{3}$

Outside of the triangle $ABC$ we construct the equilateral triangles $BCM$, $CAN$, $ABP$. The intersection $F$ of the circumcircles of these last triangles is the [u]Fermat's point[/u] and $m(\widehat {BMC})=m(\widehat {CNA})=m(\widehat {APB})=120^{\circ}$. The following relations are well-known (eventually you can prove easily):
$1.\ \ (FA+FB+FC)^2=\frac 12 (a^2+b^2+c^2)+2\sqrt 3 S$;
$2.\ \ FB^2+FC^2+FB\cdot FC=a^2$ a.s.o. ;
$3.\ \ FB\cdot FC+FC\cdot FA+FA\cdot FB=\frac{4\sqrt 3}{3}S$.

[u]Answer:[/u] $x+y+z=\sqrt{25+12\sqrt 3}\ ;\ xy+yz+zx=8\sqrt 3$.

[quote="Virgil Nicula"]Outside of the triangle $ABC$ we construct the equilateral triangles $BCM$, $CAN$, $ABP$. The intersection $F$ of the circumcircles of these last triangles is the [u]Fermat's point[/u] and $m(\widehat {BMC})=m(\widehat {CNA})=m(\widehat {APB})=120^{\circ}$. [/quote]

I'm not sure, but you meant $m(B\widehat {F}C)=m(C\widehat {F}A)=m(A\widehat {F}B)=120^{\circ}$?

virgil, so essentially, you are saying that the point $P$ in the center of the triangle $ABC$ such that $\angle BPC =\angle CPA =\angle APB=120$ is defined as the fermat point? i don't really know anything about the fermat points, eventhough i have seen them mentioned at this forum occasionally,

but this is really interesting! also more generally, we see that if you construct 3 circles with $AB$, $BC$, and $CA$ as chords of each one, and the sum of the arcs of those 3 circles that are on the opposide side of the chord as the triangle is $720$, then the circumcircles are concurrent

[quote="Altheman"] i don't really know anything about the fermat points[/quote]

Check [url=http://mathworld.wolfram.com/FermatPoints.html]here[/url].

$\{\begin{array}{lllllllll}x^2 &+& xy& +& y^2& =& 9 \\y^2& + &yz &+& z^2 &=& 16 \\ z^2\,& +&\, zx &+& x^2 &=& 25\end{array}$

Subtract Eq.2-Eq.1: $z^2-x^2+y(z-x)=7\Longleftrightarrow (z-x)(x+y+z)=7.$
Similarly we get $(x-y)(x+y+z)=9,\ (y-z)(x+y+z)=-16.$ Since both $x\neq y$ and $y\neq z$ and $z\neq x$ and $x+y+z\neq 0$, we can set $x-y=9k,\ y-z=-16k,\ z-x=7k\ (k\neq 0).$ On the other hand adding the given three equations,we have
$2(x^2+y^2+z^2)+xy+yz+zx=50$
$\Longleftrightarrow (x-y)^2+(y-z)^2+(z-x)^2+3(xy+yz+zx)=50,$ thus

we get $3(xy+xz+zx)=50-386k^2\Longleftrightarrow xy+yz+zx=\frac{50-386k^2}{3}.$

Good job [b]Virgil[/b] and [b]Kunny[/b]. Only one correction about Virgil answer. $S=6$, hence $xy+yz+zx=8\sqrt{3}$.


I have heard a lot of people refer to it, but I'm not quite sure what it is. Anyone want to help me out? thanks

Its really the USAMO Index

AMC 12 (or 10) score + 10* AIME score.

For juniors and seniors, used for USAMO qualification.

why for juniors and seniors only? what about soph/frosh?

Soph/frosh qualify for USAMO from getting the "floor value" on the AIME. Go to the AMC website to read about it. I don't feel like explaining it; it's quite complicated.

I'm not sure why underclassmen get to do the "AIME cutoff" business.  The AMC10 is much easier than the AMC12 anyways.

I think they're just trying to promote younger involvement in higher level math aka the USAMO. They need to train kids for the future, no?

Unless the AMC website is outdated, as it often is:

If you take the AMC 12, your USAMO index is computed by (AMC 12 score)+10*(AIME Score)

Then the people with the 160 highest indices go to USAMO.

The lowest AIME score of anyone going to USAMO so far determines a floor value.

Anyone who took AMC 10 and got an AIME score equal to or greater than the floor value also goes to USAMO.

Then there are some other exceptions.

[quote="qweretyq"]so basically for anyone <soph
its 150 + 10*AIME right?[/quote]

No... read RC-7th's post. It explains how sophomores and below qualify for USAMO. The lowest score of anyone in the first 160 does not imply that the person got 150 AMC12 and the lowest AIME. That would be quite interesting. It's merely the lowest AIME of the first 160 qualifiers which determines the floor.


simplify and remove roots:

$\sqrt{(a+b+x+y)(a+b+x-y)(a+b-x+y)(x+y-a-b)}\[=\]\sqrt{(a+x+z)(a+x-z)(a-x+z)(x+z-a)}+\sqrt{(b+z+y)(b+z-y)(b-z+y)(z+y-b)}$

$a,b,x,y,z>0$

Consider the triangle $ABC$ and point $D\in [BC]$ such that $|AB|=x$ $|AC|=y$ $|AD|=z$ $|BD|=a$ $|CD|=b$ then your identity becomes $S_{ABC}=S_{ABD}+S_{BCD}$ :)


geometry
3D geometry
pyramid
real analysis
real analysis unsolved

If the base of a pyramid is fixed in a plane and its height is given, every time a parallel plane to the base cuts the pyramid, does the area embedded in this plane depend only on the position of the plane?

yes.

that's how you show that two pyramids with equal bases and altitudes have same volume from Cavalieri's principle.

Ok, thanks, I was indeed checking the solution of that problem


algebra
polynomial
superior algebra
superior algebra theorems

Prove that in a commutative ring $ A$, if a nonzero polynomial $ f$ satisfies
\[ f(ut_{1},\cdots,ut_{n})=u^{d}f(t_{1},\cdots,t_{n})\]
for every $ u,t_{1},\cdots,t_{n}\in A$ and some $ d\in \mathbb{N}$, then $ f$ is homogeneous of degree $ d$.

Take $ A=\mathbb F_{2}$ and $ d>0$. In this case, your condition just says that $ f(0)=0$. owk

I need a general proof, not just a special case.  :mad:

And I want some friendly users that read what other (here owk) tell them.

Sorry. I wrote the wrong problem.  :blush: The correct one should be
Prove that in a commutative ring $ A$, if a nonzero polynomial $ f$ satisfies
\[ f(ut_{1},\cdots,ut_{n})=u^{d}f(t_{1},\cdots,t_{n}) \]
for every set of $ n+1$ algebraically independent elements $ u,t_{1},\cdots,t_{n}$ over $ A$ and some $ d\in \mathbb{N}$, then $ f$ is homogeneous of degree $ d$.

Original formulation: Lang, Algebra, page 103.

Please read again (p.102) what "algebraically independent [i]over[/i] $ A$" means and understand why this is not the same as "$ \in A$". owk

No. Your hypothesis tells you in particular that the polynomials $ f(UT_{1},\ldots,UT_{n})$ and $ U^{d}f(T_{1},\ldots,T_{n})$ in the indeterminates $ U,T_{1},\ldots,T_{n}$ are equal. Note that in $ f(UT)$ every monomial of $ f(T_{1},\ldots,T_{n})$ is multiplied by $ U^{k}$ where $ k$ is the total degree of the monomial. owk


Let $E$ be a normed vectorial space of infinite dimension, $F$-it's subspace of finite dimension and $a \in E\F$.
a). Show that there exists $b \in F$, s.t. $||a-b||=d(a,F)$
b).Show that there exists $c\in E$, s.t. $||c||=1=d(c,F)$.
c).Show that the unit ball in $E$ isn't compact.

It's basically all about Riesz's theorem. For 1) use also the fact that the map "distance" is continuous and $F$ is closed (whence if you restrict yourself to an appropriate bounded portion of $F$, this portion will be compact).


function
induction
combinatorics solved
combinatorics

Let S be a finite set with cardinality n. Let f: P(S) -> R be a function which satisfy:
 f(S/A)=f(A)
and f(a \cup B) <= max { f(A),f(B) } for all subsets A,B of S.
Prove that f can have at most n distinct value.
( Here P(S) denotes the set of subsets of S).

The result is obvious for n = 1 and n = 2. Now, suppose that it holds for n-1  \geq  2.

Let x_1,...,x_n be the elements of S.
Let M = max{f(A) / A in P(S)}.
Let A such that f(A) = M. Wlog, we may assume that A =/=  \emptyset  (if not choose S\A), and that A = {x_1,...,x_p} for some p.
Thus M = f(A)  \leq max{f(x_i) / i = 1,...,p}  \leq  M.
Thus, there exists i such that f({x_i}) = M. Wlog, we may assume that f({x_n}) = M.

Let S' = S\{x_n}. Clearly |S'| = n-1.
Now, let g : P(S') - > |R defined by:
g(A) = min{f(A) , f(A \cup {x_n}) for each A in P(S').  (1)
Since, for such a A, we have f(A \cup {x_n}) = f(S'\A), we deduce that
g(A) = min(f(A), f(S'\A)). (2)

Moreover, for each A in P(S'), we have :
M = f({x_n}) = f(S\{x_n}) = f(S') = f(A \cup (S'\A))  \leq  max{f(A),f(S'\A)}  max{f(A), f(S\(S'\A))} = max{f(A), f(A \cup {x_n})}  \leq  M.
Thus max{f(A) , f(A \cup {x_n}) = M. (3)

From (3) and (2), we deduce that for each A in P(S'), we have :
g(A) + M = min{f(A), f(A \cup {x_n})} + max{f(A), f(A \cup {x_n})}
that is g(A) + M = f(A) + f(A \cup {x_n}).  (4)

Now, we prove that g has the desired properties :

- For each A in P(S'), from (4) and the properties of f, we have :
g(S'\A) + M = f(S'\A) + f((S'\A) \cup {x_n})
= f(S\(S'\A)) + f(S\( (S'\A) \cup  {x_n}) )
= f(A \cup {x_n}) + f(A)
= g(A) + M,
thus g(A) = g(S'\A).

- Let A,B be in P(S') :
If g(A) = M or g(B) = M, clearly g(A \cup B)  \leq max{g(A),g(B)}. Thus, assume that g(A), g(B) < M.
From (1), it follows that one of the numbers f(A) and f(A \cup {x_n}) is less than M, and the same is true for B.
Let X be one of the sets A and A \cup {x_n} such that f(X) = g(A) < M, and defined Y similarly for B.
Note that X \cup Y is A \cup B or A \cup B \ {x_n}.

Thus :
g(A \cup B) = min{f(A \cup B), f(A \cup B \cup {x_n})}
 \leq  f(X \cup Y)
 \leq  max{f(X), f(Y)}
 =  max{g(A), g(B)}

From the induction hypothesis, it follows that g cannot take mare than n-1 distinct values.
But, for each A in P(S), one of the numbers f(A) and f(A \cup {x_n}) is g(A) and the other is M, so that f(P(S)) = g(P(S')) \cup  {M}, from which we deduce that |f(P(S))|  \leq  n, and we are done.

I wonder if there would not exist a solution using linear algebra....

Pierre.


[quote="e.lopes"]very similar to the second problem of first brazilian TST  :)

find that f(2) = 2 or -2 and find two ways to represent f(x+2) as a function of f(x).  :)[/quote]
Wouldn't that be $f(2)=2$ or $f(2)=0$?  ;)

i got $f(x)=0, f(x)=2$ and $f(x)=x$

my solution is non-ellegant so I won't post it unless nobody does, in few days.....

[quote="tranthanhnam"]find all function$f: R\to R$ such that :
 $f(x)f(y)+ f(x+y) = f(xy)+ f(x) +f(y)$ for all $x,y$ in real numbers
 i think that f(x)=0 ; f(x)=x, what else?[/quote]
Dear $tranthanhnam$ are you sure that this problem can be solved only with this conditions?
I think ther should be also one condition,that $f$ is continius or monice :? .
With this conditions I proved that,if $f$ isn't constant then $f(r)=r$ wher $r\in{Q}$.

i'm sure
  because this problem have in http://diendantoanhoc.net   of forum mathVietnam


Find the number of ways to select three distinct digits from the set (0,1,2,...9) if no two consecutive digits can be selected.

no idea where u got that but it's wrong..


modular arithmetic
number theory unsolved
number theory

Let $ \mathbb{N}^*$ be the set of positive integers. Find every $ p,q\in\mathbb{N}^*$ such that $ (p^2 \minus{} q^2 \plus{}4pq)^2 \minus{} 12p^2q^2$ is a perfect square.

[quote="Mashimaru"]Let $ \mathbb{N}^*$ be the set of positive integers. Find every $ p,q\in\mathbb{N}^*$ such that $ (p^2 \minus{} q^2 \plus{} 4pq)^2 \minus{} 12p^2q^2$ is a perfect square.[/quote]
hi all
(p^2 -q^2+4pq)^2 - 12p^2q^2
is prfect square for all p=+-q    belonges to N* :)

Sorry for reviving...

I claim that all of the solutions satisfy $(p,q)\in\{(k,0),(0,k),(k,k),(k,-k)\},\;\;k\in\mathbb{Z}$. To prove that there are no other solutions, proceed by contradiction; assume that there exists such an ordered pair $(p,q)$. Let $a=p^2-q^2$, $b=2pq$, and $c=p^2+q^2$, so that $a^2+b^2=c^2$ with $ab\ne0$. Then for some integer $m$,
\[m^2=(a+2b)^2-3b^2=a^2+b^2+4ab=c^2+4ab\implies ab=\left(\frac{m-c}{2}\right)\left(\frac{m+c}{2}\right),\]where $\frac{m-c}{2}$ and $\frac{m+c}{2}$ are integers because $m^2-c^2=4ab\equiv0\pmod2$ so $m,c$ have the same parity. By the [url=http://reflections.awesomemath.org/2007_6/factoring_lemma.pdf]Four Numbers Theorem[/url], there exist integers such that $a=xy$, $b=zw$, $\frac{m-c}{2}=xz$, and $\frac{m+c}{2}=yw$. Then from $a^2+b^2=c^2$,
\[(xy)^2+(zw)^2=(yw-xz)^2\implies x^2(y^2-z^2)+2xyzw-w^2(y^2-z^2)=0,\qquad(*)\]so the discriminant
\[\Delta_x = (2yzw)^2+4w^2(y^2-z^2)^2=(2w)^2(y^2z^2+(y^2-z^2)^2)=(2w)^2(y^4-y^2z^2+z^4)\]must be a perfect square. It is well-known that if $y^4-y^2z^2+z^4$ is a perfect square, then either $yz=0$ or $y=z$. Now, if $yz=0$, then $ab=(xy)(zw)=0$, and if $y=z$, then $(*)$ yields $0=xyzw=ab$ as well. But this is a clear contradiction.


function
Functional Analysis
advanced fields
advanced fields unsolved

Hello!

I'm rewiewing for my functional analysis exam and came across following kind of problem: Let $y=(\eta_{1},\eta_{2},\cdots),n_{1}\in \mathbb{C}$ be s.t. $\sum_{i}\xi_{i}\eta_{i}$ converges for every $x=(\xi_{1},\xi_{2},\cdots) \in c_{o}$, where $c_{0}\subset l^{\infty}$ is the subspace of all complex sequences converging to zero. Prove that $\sum_{i}|\eta_{i}| < \infty$.

I guess you're supposed to use uniform boundedness theorem by considering bounded linear operators $T_{n}x = \sum_{i}^{n}\xi_{i}\eta_{i}$, for which $\parallel T_{n}\parallel = \sum_{i}^{n}|\eta_{i}|$, and show that the norm of the limit function is bounded (it's known that $c_{0}$ is complete). The problem is to show that $|T_{n}x| \leq C_{x}$ for all $n$ and $x \in c_{0}$, so that the theorem could be used. I only get
\[|T_{n}x| \leq \sum_{i}^{n}|\xi_{i}\eta_{i}| \leq \textrm{sup}\ |\xi_{i}| \sum_{i}^{n}|\eta_{i}| = \parallel x\parallel \sum_{i}^{n}|\eta_{i}|,\]
but this depends on $n$ and I can't get forward. What am I doing wrong?

maybe you can try without the uniform boundedness principle. You want to show that if $(\eta_{i})_{\mathbb{N}}$ is a sequence such that $\sum |\eta_{i}| =+\infty$, then there exists a sequence $(\xi_{i})_{\mathbb{N}}\in c_{0}$ such that $\sum \eta_{i}\xi_{i}$ diverges. Clearly one can suppose that $\eta_{i}\geq 0$. One can find integers $n_{1}< n_{2}< \ldots < n_{k}< \ldots$ such that $\sum_{n_{k}+1}^{n_{k}}\eta_{i}\geq 1$ because $\sum \eta_{i}=+\infty$. Then it suffices to define $\xi_{i}=\frac{1}{k}$ for $n_{k}+1 \leq i \leq n_{k+1}$.

[quote="alekk"]One can find integers $n_{1}< n_{2}< \ldots < n_{k}< \ldots$ such that $\sum_{n_{k}+1}^{n_{k}}\eta_{i}\geq 1$ because $\sum \eta_{i}=+\infty$. Then it suffices to define $\xi_{i}=\frac{1}{k}$ for $n_{k}+1 \leq i \leq n_{k+1}$.[/quote]
Thank you for your proof. Although, it should probably read "$n_{k}< \ldots$ such that $\sum_{n_{k}+1}^{n_{k+1}}\eta_{i}\geq 1$", I mean $n_{k+1}$ in the upper bound? But still, I'd like to know how you can prove that with uniform boundedness :)


Can someone post nice, hard equilibrium problems (preferably ones involving multiple equilibria) which illustrate the application of approximations? Or just equilibrium problems in general (compleximetric, acid/base, redox, etc.)

Thanks.

I recommend problems 7 and 12 on the 2006 IChO Preparatory Problems, which you can access at [url=http://icho2006.kcsnet.or.kr/main/i_problems/i_p_prepexams.htm?qpage=i_p_en_prepexams]this page.[/url]

Thanks gregx007. Any more problems for me to practice with?

Thanks.

I have posted some about acid-base, which remain unanswered. However, let me post one or two about precipitation.


dei paavingala, :P  :P

help me answer my question. i have a test the day-after tommorrow...
i know jee is over and all that but you can spare some time to answer a little boy's doubts about chemistry... :wink:

[quote="hell_ever"]1. How will you decide which is more stable - o-methoxy toluene carbanion or m-methoxy toluene carbanion???
the negative charge is on the CH2 of toluene.
In general, between any meta and ortho/para compound???

2. Which factors affect nucleophilicity order??
Given any 5 compunds, how will yo decide the nucleophilicty order??

3. Im not sure about this question. If you dissolve sodium flouride in each of the foll. solvents, in which will F- be most reactive - DMSO, Me-CN, Me-OH[/quote]

1) Whichever is a better electron withdrawing group, that will reduce the charge on C and stablize the compound. Forget Resonance. :)

2) Its actually difficult predicting the order unless you know it byheart. However, generally size plays a role. By Nucleophilic nature, we compare the rate at which it can donate electrons. OS- is a better nucleophile than OH-, though the latter is a better base.

But 1) we are comparing an ortho and meta compound.....how can you talk of e- withdrawing groups....


Prove that for each prime $ p\geq 7$ there exists a positive integer $ n$ and $ x_1,x_2,...,x_n,y_1,y_2,...,y_n$ not divisible by p such that : $ x_1^2\plus{}y_1^2 \equiv  x_2^2$
                   $ x_2^2\plus{}y_2^2 \equiv  x_3^2$
                    ................................
                  $ x_n^2\plus{}y_n^2 \equiv  x_1^2$

[quote="srinath.r"]er....have you made any typo ?
Add all the equations we get .

$ \displaystyle\sum_{i \equal{} 1}^{n}\ x_{i}^{2} \plus{} \displaystyle\sum_{i \equal{} 1}^{n}\ y_{i}^{2} \equal{} \displaystyle\sum_{i \equal{} 1}^{n}\ x_{i}^{2} \implies \displaystyle\sum_{i \equal{} 1}^{n}\ y_{i}^{2} \equal{} 0$???????....which is not possible right ? :?:[/quote]
these are not equalities ,they are congruencies $ \sum_{i \equal{} 1}^{n}\ y_{i}^{2} \equiv  0 \pmod {p}$


A number, written on the desk, we can to double or to subtract from it $ 1$. In the beginning, on the desk is written number $ 1$. Is it possible to get number $ 800002$, not writing on the desk a number, which contains in its decimal system number $ 3$?

No.

Assume that you get $ 800002$ by the procedure, without having used a number with digit $ 3$.
Since $ 800003$ is forbidden, the last number before $ 800002$ was $ 400001$. It is an odd number, thus the last number before $ 400001$ was $ 400002$. Since $ 400003$ is forbidden, the last number before $ 400002$ was $ 200001$.
And so on, using these two type of arguments (odd number and forbidden number), you see that you reach $ 800002$ by the following path :
$ 800002$
$ 400001$
$ 400002$
$ 200001$
$ 200002$
$ 100001$
$ 100002$
$ 50001$
$ 50002$
$ 25001$
$ 25002$
$ 12501$
$ 12502$
$ 6251$
$ 6252$
But $ 6252$ can be reach only from $ 6253$ and $ 3126$, and both are forbidden, a contradiction.

Pierre.


inequalities
function
logarithms
real analysis
real analysis theorems

Which is bigger, $e^\pi$ or $\pi^e$?
[hide="solution"]
Let's consider the maximum of $x^{1/x}$. First we take a natural logarithm. This gives
$\frac{\ln(x)}{x}$. The maximum of this function is the maximum of the first function since $\ln(x)$ is increasing.  The maximum is easily found to be at $x=e$. So now we have:
$e^{1/e}>\pi^{1/\pi}$
Raise both sides to the $e\pi$ power.
[/hide]

This is probably stupid, but whenever I see a problem like this I'm strongly tempted to approximate it by hand.

Intuition tells us that $e^\pi$ is probably bigger, so we need to do a low estimate. Call $e = 27/10$ and $\pi = 22/7$. That should still be a small estimate. So we have $3^{66/7} / 10^{22/7}$. That's about $729 * 27 * 1.6$ in the numerator and $1000 * 1.4$ in the denominator. That reduces to $729 * 27 / (5^3 * 7) \approx 104 * 27 / 125 \approx 112 / 5 = 22.4$.

After the amount of time that took and the fact that I lost track of the magnitude of my error, I decided not to do the other one. But considering $e^\pi \approx 23.131$ and all of my calculations were mental, 22.4 isn't bad for an intentionally low estimate. :D

But since $\pi^e \approx 22.459$, I guess it isn't good enough...  :blush:

Define : $x\ s.s\ y\Longleftrightarrow sgn(x)=sgn(y)$, i.e. $x=y=0$ or $xy>0\ .$
Here are some examples :
$1.\blacktriangleright\frac ab\ s.s\ ab$ for $b\ne 0\ .$
$2.\blacktriangleright 0<a\ne 1\ ,\ \{x,y\}\subset R\Longrightarrow (a^x-a^y)\ s.s\ (a-1)(x-y)\ .$
$3.\blacktriangleright 0<a,b\ne 1\ ,\ x\in R\Longrightarrow (a^x-b^x)\ s.s\ x(a-b)\ .$
$4.\blacktriangleright 0<a\ne 1\ ,\ \{x,y\}\subset (0,\infty )\Longrightarrow \log_a x-\log_a y\ s.s\ (a-1)(x-y)\ .$
$5.\blacktriangleright 0<a\ne 1\ ,\ b>0\Longrightarrow \log_a b\ s.s\ (a-1)(b-1)\ .$
$6.\blacktriangleright (|x|-|y|)\ s.s\ (x^2-y^2)\ ;\ |x|\ s.s\ x^2\ .$
$7.\blacktriangleright (\arcsin x-\arcsin y)\ s.s\ (x-y)\ ;\ \arcsin x\ s.s\ x\ .$
$8.\blacktriangleright (\arccos x-\arccos y)\ s.s\ (y-x)\ ;\ \arccos x\ s.s\ (1-x)\ .$
$9.\blacktriangleright (\arctan x-\arctan y)\ s.s\ (x-y)\ ;\ \arctan x\ s.s\ x\ .$
$10.\blacktriangleright \{x,y\}\subset [0,\infty )\Longrightarrow (\sqrt [n] x-\sqrt [n] y)\ s.s\ (x-y)\ .$

$F.1\blacktriangleright f: D\rightarrow R\ ,\ f\nearrow\Longrightarrow f(x)-f(y)\ s.s\ (x-y)$ for any $\{x,y\}\subset D\ .$
$F.2\blacktriangleright f: D\rightarrow R\ ,\ f\searrow\Longrightarrow f(x)-f(y)\ s.s\ (y-x)$ for any $\{x,y\}\subset D\ .$

[b]Proof of the proposed problem.[/b]
$(e^{\pi}-\pi^e)\ s.s\ (\ln e^{\pi}-\ln \pi^e)=(\pi\ln e-e\ln \pi)\ s.s\ (\frac{\ln e}{e}-\frac{\ln\pi}{\pi})=$
$f(e)-f(\pi )$, where $f: (0,\infty )\rightarrow R\ ,\ f(x)=\frac{\ln x}{x}\ .$
But $f'(x)\ s.s\ (1-\ln x)=(\ln e-\ln x)\ s.s\ (e-x)\ .$ Therefore,
for any $0<x\ne e\ ,$ $f(x)< f(e)\Longrightarrow f(\pi)<f(e)\Longrightarrow\boxed {e^{\pi}>\pi ^e}\ .$

[b][u]A similar exercise :[/u][/b] $\{a,b\}\subset (0,e^2]\Longrightarrow (a^{\sqrt b}-b^{\sqrt a})\ s.s\ (a-b)\ .$
For example, $3^{\sqrt 5}<5^{\sqrt 3}\ .$
$\{a,b\}\subset [e^2,\infty )\Longrightarrow(a^{\sqrt b}-b^{\sqrt a})\ s.s\ (b-a)\ .$
For example, $3^{\sqrt 5}<5^{\sqrt 3}\ ;\ 512>81^{\sqrt 2}\ .$
[b]Remark.[/b] For any $0<x\ne e^2\ ,\ x^{\sqrt e}<e^{\sqrt x}\ .$


limit
trigonometry
real analysis
real analysis unsolved

Find the limit of: [sin(a+x).sin(a+2x)-sin^2a]/x when x goes to zero.

hello, applying the L'Hospital rule we have
$ \lim_{x \to 0}\frac{\sin(a\plus{}x)\sin(a\plus{}2x)\minus{}\sin^2(a)}{x}\equal{}\newline
\lim_{x \to 0}\cos(a\plus{}x)\sin(a\plus{}2x)\plus{}\sin(a\plus{}x)\cos(a\plus{}2x)2\equal{}\newline
\cos(a)\sin(a)\plus{}2\sin(a)\cos(a)\equal{}3\sin(a)\cos(a)$.
Sonnhard.

[hide]$ \lim_{x \to 0} \frac{\sin(a\plus{}x) \cdot \sin(a\plus{}2x) \minus{} \sin^2a}{x} \equal{} [\sin(a\plus{}x) \cdot \sin(a\plus{}2x)]'|_{x\equal{}0}$

$ \equal{} [\frac12(\cos x \minus{} \cos(2a\plus{}3x))]'|_{x\equal{}0} \equal{} [\minus{} \frac12 \sin x \plus{} \frac32 \sin(2a\plus{}3x)]|_{x\equal{}0} \equal{} \frac32 \sin(2a)$.[/hide]

My solution with only elementary method:
We have: sin(a+x).sin(a+2x)-sin^2a=sin^2a(cosx.cos2x-1)+sina.cosx.cosa.sin2x+sinx.cosa.(sina.cos2x)=sin^2a.sin(x/2)[-sin3x/2.cosx-2sin(x/2).cos^2(x/2) +sina.cosx.cosa.sin2x+sinx.cosa.(sina.cos2x)
Therefore, we easily infer that: lim [sin(a+x).sin(a+2x)-sin^2a]/x(x goes to zero)=0+2sina.cosa+0+sina.cosa=3sina.cosa=3/2.sin2a(by applying the formula: lim sin(x)/x=1 when x goes to zero and plugging x=0 into the rest of our equation)

$ 2\{\sin (a \plus{} x)\sin (a \plus{} 2x) \minus{} \sin ^ 2 a\}$
$ \equal{} \cos x \minus{} \cos (2a \plus{} 3x) \minus{} (1 \minus{} \cos 2a)$
$ \equal{} \cos 2a \minus{} \cos (2a \plus{} 3x) \minus{} (1 \minus{} \cos x)$
$ \equal{} 2\sin \left(2a \plus{} \frac {3}{2}x\right)\sin \frac {3}{2}x \minus{} (1 \minus{} \cos x)$.
$ \therefore \lim_{x\rightarrow 0} \frac {\sin (a \plus{} x)\sin (a \plus{} 2x) \minus{} \sin ^ 2 a}{x}$

$ \equal{} \lim_{x\rightarrow 0} \sin \left(2a \plus{} \frac {3}{2}x\right)\cdot \frac {\sin \frac {3}{2}x}{\frac {3}{2}x}\cdot \frac {3}{2} \minus{} \frac {1}{2}\lim_{x\rightarrow 0} \frac {1 \minus{} \cos x}{x}$

$ \equal{} \sin 2a \cdot 1\cdot \frac {3}{2} \minus{} \frac {1}{2}\lim_{x\rightarrow 0} \frac {\sin x}{x}\cdot \frac {\sin x}{1 \plus{} \cos x}$

$ \equal{} \frac {3}{2}\sin 2a \minus{} \frac {1}{2}\cdot 1\ \cdot \frac {0}{1 \plus{} 1} \equal{} \boxed{\frac {3}{2}\sin 2a}$.


inequalities
search
inequalities unsolved

[quote="bellahsayn"]show for all  $a_1,a_2,..........a_n$ in $R+$  then
$\frac{a_1}{a_1+a_2}+.........................+\frac{a_n}{a_n+a_1} >\frac{n}{4} $[/quote]
[quote="harazi"]I guess someone should indicate the recent (in fact old) inequality given this year in Moldova TST 2005 or national olympiad. I think it gives a much better estimation.[/quote]
Really, a better estimation :?:    What about the case $a_k=x^k$ and $x\longrightarrow +\infty $ ?? :D
 :cool: What's the correct estimation ? :cool:

But your initial inequality is not true either. I think the original USSR problem was something like $\frac{x_1}{x_2+x_3}+...>n/4$. If Alzheimer has not finished me till now, then in the moldavian olympiad this year (and in Kvant some 14 years ago) they showed we can replace $1/4$ by $\sqrt{2}-1$. Anyway, nobody is interested in the topic, so...

[quote="harazi"] Anyway, nobody is interested in the topic, so...[/quote]

Nice provocation  :P Though I'm one of those who are interested. Harazi, can you give the link or show the proof on the forum about this improvement?

Wait 5 minutes. I will find it even if it's the last thing I do.  ;)

It's simple. Click search and put as author Sasha, for topic "hard for me" and search in inequalities forum. It's the 4-th topic that will appear. Please, make a link, I don't know how.

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hard&t=30538

Thank you, I've must have missed when it was posted


So I saw that on the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]Classes Enroll[/url] page, that the Algebra classes are now split into Algebra 1, Algebra 2, and Algebra 3.

Looking at textbook choices, Algebra 1 + Algebra 2 = Introduction to Algebra, and Algebra 3 = Intermediate Algebra?  I would like this confirmed, please.

Thanks!

That is correct.  Algebra 1 + Algebra 2 = the old Intro Algebra class.  Algebra 3 = the old Intermediate Algebra class.

Why did you split it up like this?  Was it to add in

[quote]graphing, functions, [...] exponents and logarithms[/quote]

Sounds like algebra to me.  :D

It wasn't to add anything -- it was to make a clear decision for students who have taken algebra I in school.  Before we split, those students who had taken algebra I but not algebra II in school were simply not ready for Intermediate Algebra, but would already know a solid half of the Intro Algebra class.  So, we split it -- students with prealgebra, but no algebra I, should take our Algebra 1.  Those with algebra 1 but no algebra 2 should take our Algebra 2.  Everyone should take our Algebra 3 :)

Awesome; I know a lot of kids that definitely should be interested in taking your Algebra 2 class.  Time to advertise at MATHCOUNTS practice tomorrow :D

They like borrowing the AoPS books from our coach; he's missing about 4 of them right now :)

[quote="AIME15"]

They like borrowing the AoPS books from our coach; he's missing about 4 of them right now :)[/quote]

I believe our books are the most "borrowed and not returned" math books ever.

We even had someone steal some of our display copies at an NCTM Convention.  I wonder if McGraw-Hill or Saxon have this problem :)

just a question, will there be each a textbook for Algebra 1, Algebra 2, and Algebra 3 ???

Intro Algebra = Algebra 1 + Algebra 2 (one text for both)

Interm Algebra = Algebra 3 (course covers essentially the whole text)


Let $ a,b,c>0$ . Prove that $ \sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\leq \frac{3}{2}\sqrt{\frac{(a+b)(b+c)(c+a)}{abc}}.$

[quote="N.T.TUAN"]Let $ a,b,c>0$ . Prove that $ \sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\leq \frac{3}{2}\sqrt{\frac{(a+b)(b+c)(c+a)}{abc}}$[/quote]
$ \sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\leq \frac{3}{2}\sqrt{\frac{(a+b)(b+c)(c+a)}{abc}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt{\frac{ab}{(a+c)(b+c)}}\leq\frac{3}{2}.$
But $ 2\cdot\sum_{cyc}\sqrt{\frac{ab}{(a+c)(b+c)}}\leq\sum_{cyc}\left(\frac{a}{a+c}+\frac{b}{b+c}\right)=3.$ :)

to arqady: Good! :thumbup: This is solution of a my student:

We have $ 4abc\cdot (\text{LHS})^{2}\leq 4abc\cdot (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(2(a+b+c))=8(ab+bc+ca)(a+b+c)=8(ab^{2}+a^{2}b+b^{2}c+bc^{2}+c^{2}a+ca^{2}+3abc)\leq 9(ab^{2}+a^{2}b+b^{2}c+bc^{2}+c^{2}a+ca^{2}+2abc)=9(a+b)(b+c)(c+a)$ and we are done!


algorithm
function
geometry
geometric transformation
reflection
rotation
inequalities

[quote]Ive asked around but can anybody think of a fast way to prime factor a number? [/quote]

There is no easy way to do so. There are some algorithms (Pollard-Rho, Pollard-Strassen, Brent's, etc...), but none of them are manageable as the number you are trying to factorize increases. It is particularly difficult with numbers that only have a few very large primes in their factorizations. This is what makes the RSA algorithm secure.

Fierytycoon

Quote:If f(n+2)=n/f(n) for all positive integers n and f(2)=1, find f(8).

Answer below (highlight to view):

[hide]f(4)=f(2+2)=2/f(2)=2/1=2

f(6)=f(4+2)=4/f(4)=4/2=2

f(8)=f(6+2)=6/f(6)=6/2=3[/hide]

Fierytycoon

[quote="Rep123max"]
Can you think of a formula for the number of integer sided isoscels triangles with perimeter p?
[/quote]

If p is even it's (p-2)/2, and if p is odd it's (p-1)/2. Or you can write it as one formula using the Floor Function as [(p-1)/2].

[quote="Rep123max"]
Ive asked around but can anybody think of a fast way to prime factor a number?
[/quote]

It depends on what you mean by "fast", and also on the general size and type of the numbers you want to factor. There are lots of advanced methods for trying to factor large numbers or numbers of a special type, but the simplest method for small numbers is just trial division - Checking division by all the odd numbers up til the sqrt of the number you're trying to factor. You can write a simple program on a TI calculator to do this. But for very large numbers this because unwieldy.

[quote="Rep123max"]
And finally: Seven points are equally distributed on a circle. How many non-congruent quadrilaterals can be drawn with vertices chosen from among tthese seven points? I started using combinations but didnt know how to tell whether a quadrilateral was congruent or not!
[/quote]

Are you counting congruency by reflection too? Or just rotation. If it is just by rotation then the answer is 7C4/7 = 5. Take the number of ways of choosing 4 points from 7 and divide by 7 to account for the 7 possible rotations of each one. And if you are also counting congruency by reflection then the answer is 4 (just by inspection).

triangle inequality always satisfied:

p even:

p can range from 2 to p-2 --> p/2 - 1 ways.

p odd:

p can range from 1 to p-2 --> (p-1)/2 ways.


number theory
least common multiple
abstract algebra
group theory
superior algebra
superior algebra solved

Hi,

   Is any one can give me some hits on this problem:

     a).   Prove that an n-cycle has order n.

     b).   Prove that in general the order of a permutation is the lcm of the lengths of its cycles.

    Thank you!

    Lilly

Hi,

    I have another question on group homomorphism:

   Let f : G-->H be a group homomorphism.

   Show that f is one-to-one if and only if ker(f) = {1}.

    Thank you!

     Lilly

For the questions in your first post: look at the sequence $x_n=px_{n-1}$ where $p$ is our permutation and $x_0$ is an arbitrary element. $x_n=p^nx_0$ in general, and a permutation is the identity exactly when it fixes every element.

For your second post: It is clear that $f$ is not 1-1 if it maps two different elements to the identity. Conversely, if $f$ is not 1-1, we have $f(x)=f(y)$, so $e=f(x)(f(y))^{-1}=f(xy^{-1})$ and the kernel is nontrivial.


algebra
function
domain
superior algebra
superior algebra unsolved

Let $ R$ be a domain and let $ K$ be its fractions field.
Suppose $ I$ is a non zero ideal.
Prove that $ I \otimes K \equal{} K$
Hint: do not go into the construction of the tensor product. find a bilinear map $ f: I\times K \rightarrow K$ such that $ K$ equipped with this map satisfies the universal property.

Then take $ f(a,b)\equal{}ab$ and prove that it induces an isomorphism.

such a silly exercise!

tensoring with $ S^\minus{}1 A$ means localizing at $ S$ and for submodules of a field of fractions this means just writing down the fractions.


geometry
3D geometry
sphere
function
inequalities proposed
inequalities

Let $ x,y,z$ be the real numbers and $ a>0$.Prove that $ 4(axy\plus{}yz\plus{}zx)$≤$ (a\plus{}\sqrt{a^2\plus{}8})(x^2\plus{}y^2\plus{}z^2)$

[quote="SUPERMAN2"]Let $ x,y,z$ be the real numbers and $ a > 0$.Prove that $ 4(axy \plus{} yz \plus{} zx)$≤$ (a \plus{} \sqrt {a^2 \plus{} 8})(x^2 \plus{} y^2 \plus{} z^2)$[/quote]
Anyone has a solution?

[quote="SUPERMAN2"]Let $ x,y,z$ be the real numbers and $ a > 0$.Prove that $ 4(axy \plus{} yz \plus{} zx)$≤$ (a \plus{} \sqrt {a^2 \plus{} 8})(x^2 \plus{} y^2 \plus{} z^2)$[/quote]
Using Lagrange multipliers we could prove, that $ 4(axy \plus{} yz \plus{} zx)\le(\frac{8}{\sqrt{a^2\plus{}8}\minus{}a})(x^2\plus{}y^2\plus{}z^2)$
(just take $ x^2\plus{}y^2\plus{}z^2\equal{}3$, its sphere, so buy Weyerstrasse theorem function has min and max on this compact, and we must only decide system, one of the solutions will be $ (x;x;\frac{\sqrt{a^2\plus{}8}\minus{}a}{2}x)$,where $ x\equal{}\sqrt{\frac{6}{a^2\plus{}8\minus{}a\sqrt{a^2\plus{}8}}}$ :wink: )
But $ \frac{8}{\sqrt{a^2\plus{}8}\minus{}a}\equal{}a \plus{} \sqrt {a^2 \plus{} 8}$ so we are done :lol:  But I am really interested in "normal" solution :blush:

[quote="dima ukraine"][quote="SUPERMAN2"]Let $ x,y,z$ be the real numbers and $ a > 0$.Prove that $ 4(axy \plus{} yz \plus{} zx)$≤$ (a \plus{} \sqrt {a^2 \plus{} 8})(x^2 \plus{} y^2 \plus{} z^2)$[/quote]
Using Lagrange multipliers we could prove, that $ 4(axy \plus{} yz \plus{} zx)\le(\frac {8}{\sqrt {a^2 \plus{} 8} \minus{} a})(x^2 \plus{} y^2 \plus{} z^2)$
(just take $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$, its sphere, so buy Weyerstrasse theorem function has min and max on this compact, and we must only decide system, one of the solutions will be $ (x;x;\frac {\sqrt {a^2 \plus{} 8} \minus{} a}{2}x)$,where $ x \equal{} \sqrt {\frac {6}{a^2 \plus{} 8 \minus{} a\sqrt {a^2 \plus{} 8}}}$ :wink: )
But $ \frac {8}{\sqrt {a^2 \plus{} 8} \minus{} a} \equal{} a \plus{} \sqrt {a^2 \plus{} 8}$ so we are done :lol:  But I am really interested in "normal" solution :blush:[/quote]
I don't understand [quote]Using Lagrange multipliers we could prove, that $ 4(axy \plus{} yz \plus{} zx)\le(\frac {8}{\sqrt {a^2 \plus{} 8} \minus{} a})(x^2 \plus{} y^2 \plus{} z^2)$
(just take $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$, its sphere, so buy Weyerstrasse theorem function has min and max on this compact, and we must only decide system, one of the solutions will be $ (x;x;\frac {\sqrt {a^2 \plus{} 8} \minus{} a}{2}x)$,where $ x \equal{} \sqrt {\frac {6}{a^2 \plus{} 8 \minus{} a\sqrt {a^2 \plus{} 8}}}$ :wink: )[/quote]
Can anyone explain this to me please?
PS:I have an elementary solution to this problem.WLOG, suppose $ x^2\plus{}y^2\plus{}z^2\equal{}1$.Let $ z\equal{}cosa$,$ x\equal{}sinacosb$,$ y\equal{}sinasinb$.

Let $ k\equal{}a\plus{}\sqrt{a^2\plus{}8}$. Note that $ (k\minus{}a)^2\equal{}a^2\plus{}8\equal{}k^2\minus{}2ak\plus{}a^2$ or $ k^2\minus{}2ak\equal{}8$

$ [k]z^2\plus{}[\minus{}4a(x\plus{}y)]z\plus{}[\minus{}4axy\plus{}k(x^2\plus{}y^2)]$

To prove that this is non-negative, we must consider the discriminant $ \le 0$.

$ 4(k)(\minus{}4axy\plus{}k(x^2\plus{}y^2))\ge (4a(x\plus{}y))^2$

Dividing by $ 4$ and expanding,

$ \minus{}4akxy\plus{}k^2(x^2\plus{}y^2)\ge 4a^2(x^2\plus{}2xy\plus{}y^2)$

Collecting coefficients on $ (x^2\plus{}y^2)$ and $ (xy)$ and factoring,

$ (x^2\plus{}y^2)(k\minus{}2a)(k\plus{}2a))\ge4(xy)(a)(k\plus{}2a)$

This can be rewritten as

$ (k\plus{}2a)( (x\minus{}y)^2(k\minus{}2a)\plus{}2k(xy))\ge 0$

It is easy to see that this is positive, our steps are reversible, so we're done.


[hide="Hint"]
Find $ 10S$, and note that $ 9S \equal{} 10S \minus{} S$. Do you see a familiar sequence?
[/hide]

[hide="Solution"]Let $ S$ denote the sum, such that:
\[ S\equal{}\dfrac{1}{10^1} \plus{} \dfrac{1}{10^2} \plus{} \dfrac{2}{10^3} \plus{} \dfrac{3}{10^4} \plus{} \dfrac{5}{10^5} \plus{} \dfrac{8}{10^6}\plus{}\cdots\]Then:
\[ 10S\equal{}\dfrac{1}{10^0} \plus{} \dfrac{1}{10^1} \plus{} \dfrac{2}{10^2} \plus{} \dfrac{3}{10^3} \plus{} \dfrac{5}{10^4} \plus{} \dfrac{8}{10^5}\plus{}\cdots\]Then:
\[ 10S\minus{}S\equal{}1\plus{}0\plus{}\dfrac{1}{10^2} \plus{} \dfrac{1}{10^3} \plus{} \dfrac{2}{10^4} \plus{} \dfrac{3}{10^5} \plus{} \dfrac{5}{10^6}\plus{}\cdots\]Note that the RHS is just one-tenth of the sum $ S$. Then, we can write:
\[ 10S\minus{}S\equal{}1\plus{}\dfrac{S}{10} \\
\Rightarrow 9S\equal{}\dfrac{10\plus{}S}{10} \\
\Rightarrow 90S\equal{}10\plus{}S \\
\Rightarrow 89S\equal{}10 \\
\Rightarrow \boxed{S\equal{}\dfrac{10}{89}}\](Did you notice that this manipulation worked because the numerators were terms of the Fibonacci Sequence?)[/hide]


logarithms
real analysis
real analysis unsolved

Suppose a_n>0, s_n = a_1 + ... + a_n, and sum a_n diverges.

Prove that sum a_n/(1+ a_n) diverges.

What about the followings:
1)sum a_n/(1+ n * a_n)

2)sum a_n/(1+ n^2 * a_n)

If $a_n$ doesn't converge towards $0$, there exists $r, R$ such as $R > a_n > r > 0$ for infinitly many $n$
This proves $a_n/(1+a_n) > r/(1+R)$ for those $n$, and thus $\sum a_n$ diverges.

If $a_n$ converges towards $0$, then $0 < a_n \sim a_n/(1+a_n)$ and thus converges.

The other ones :

$a_{n^2} = 1$ and $a_n = 0$ elsewhere. Of course $\sum a_n$ diverges.

But $\sum a_n/(1+n*a_n)$ and $\sum a_n/(1+n^2 * a_n)$ both converge

So result doesn't applies in case 1) and 2)

Hm, I think these problems come from Rudin.

For 1), it can either converge or diverge; see what happens if you take $a_n=1/n$ and $a_n=1/(n\log^2 n)$; for 2), try comparing to a $p$-series.

[quote="mikeynot"][quote="tµtµ"]
The other ones :
$a_{n^2} = 1$ and $a_n = 0$ elsewhere. Of course $\sum a_n$ diverges.
But $\sum a_n/(1+n*a_n)$ and $\sum a_n/(1+n^2 * a_n)$ both converge
So result doesn't applies in case 1) and 2)[/quote]
Can you give some proof or reasoning?
Thanks.[/quote]

$\sum a_n$ diverges because $a_n$ does not converge toward $0$

$\sum_1^N a_n/(1+n*a_n) \leq \sum_1^\infty 1/(1+n^2)$ and $a_n \geq 0$

No ?


Ervina made 62.5$ \%$ of the shots she took. If she made 25 shots,
then how many shots did she take?

First, you convert 62.5% into fraction form.
62.5% is equivalent to 5/8.
Now, you set up a ratio.
5 shots/8 shots taken=25 shots/x shots taken.
Cross multiply and get 5x=200, or x=40.


On a scale drawing of a new school playground, a triangular
area has sides with lengths of 8 centimeters, 15 centimeters,
and 17 centimeters. If the triangular area located on the playground
has a perimeter of 120 meters, what is the length of its longest side?

(1) 24 m     (3) 45 m
(2) 40 m     (4) 51 m

[hide="Why do we bother hiding answers?"]The perimeter of the drawing is 40cm. To scale this up to 120m, we need a scale factor of 300. Scaling up  the longest side which has length 17cm by this much gives it length

$\boxed{51\textrm{m}}$[/hide]

Also, methinks this exercise would be better off in the [url=http://www.mathlinks.ro/Forum/index.php?f=300]Classroom Math[/url] forum.

If you scale up a figure, you multiply all its lengths by a constant (which is in this case 300)- it is like zooming in on or magnifying the figure.

Thanks for the definition of scaling up.


Le $ a, b, c \geq 0$ and $ ab\plus{}bc\plus{}ca\equal{}1$. Prove that: $ \frac{a^3}{1\plus{}b^2}\plus{}\frac{b^3}{1\plus{}c^2}\plus{}\frac{c^3}{1\plus{}a^2} \geq \frac{\sqrt{3}}{4}$

[quote="D.I.Culianop"]Le $ a, b, c \geq 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that: $ \frac {a^3}{1 \plus{} b^2} \plus{} \frac {b^3}{1 \plus{} c^2} \plus{} \frac {c^3}{1 \plus{} a^2} \geq \frac {\sqrt {3}}{4}$[/quote]
\[ \frac {{a^3 }}{{1 \plus{} b^2 }} \plus{} \frac {{b^3 }}{{1 \plus{} c^2 }} \plus{} \frac {{c^3 }}{{1 \plus{} a^2 }} \equal{} \frac {{a^3 }}{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)}} \plus{} \frac {{b^3 }}{{\left( {a \plus{} c} \right)\left( {c \plus{} b} \right)}} \plus{} \frac {{c^3 }}{{\left( {c \plus{} a} \right)\left( {a \plus{} b} \right)}}
\]

\[ \equal{} \frac {{a^3 \left( {c \plus{} a} \right) \plus{} b^3 \left( {a \plus{} b} \right) \plus{} c^3 \left( {b \plus{} c} \right)}}{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}} \ge \frac {{ab\left( {a^2 \plus{} b^2 } \right) \plus{} bc\left( {b^2 \plus{} c^2 } \right) \plus{} ca\left( {c^2 \plus{} a^2 } \right)}}{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}}
\]
\[ \equal{} \frac{{a^2  \plus{} b^2  \plus{} c^2  \minus{} abc\left( {a \plus{} b \plus{} c} \right)}}{{a \plus{} b \plus{} c \minus{} abc}} \ge \frac{{\sqrt 3 \left( {a \plus{} b \plus{} c} \right)^2  \minus{} \left( {a \plus{} b \plus{} c} \right)}}{{3\sqrt 3 \left( {a \plus{} b \plus{} c} \right) \minus{} 1}} \ge \frac{{\sqrt 3 }}{4}
\]

Let $ a, b, c \geq 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that: $ \frac {a^3}{1 \plus{} b^2} \plus{} \frac {b^3}{1 \plus{} c^2} \plus{} \frac {c^3}{1 \plus{} a^2} \geq \frac {\sqrt {3}}{4}$
\[ \frac {{a^3 }}{{1 \plus{} b^2 }} \plus{} \frac {{b^3 }}{{1 \plus{} c^2 }} \plus{} \frac {{c^3 }}{{1 \plus{} a^2 }} \equal{} \frac {{a^3 }}{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)}} \plus{} \frac {{b^3 }}{{\left( {a \plus{} c} \right)\left( {c \plus{} b} \right)}} \plus{} \frac {{c^3 }}{{\left( {c \plus{} a} \right)\left( {a \plus{} b} \right)}}
\]
We have:

$ LHS\equal{} \sum_{a,b,c} ( \frac{a^{3}}{(b\plus{}c)(b\plus{}a)}\plus{} \frac{b\plus{}c}{8}\plus{}\frac{b\plus{}a}{8}) \minus{} \frac{a\plus{}b\plus{}c}{2} \geq \frac{3(a\plus{}b\plus{}c)}{4} \minus{} \frac{a\plus{}b\plus{}c}{2}\equal{} \frac{a\plus{}b\plus{}c}{4} \geq \frac{\sqrt{3(ab\plus{}bc\plus{}ca)}}{4}\equal{}RHS$

^^! Thank you for your solution Honey_S

$ \sum\frac{a^3}{1\plus{}b^2}\geq\sum\frac{a^3}{1\plus{}a^2}\equal{}\sum a\minus{}\sum\frac{a}{a^2\plus{}1}\equal{}\sum a\minus{}\sum\frac{a}{a^2\plus{}3.\frac{1}{3}}\geq\sum a\minus{}\frac{3}{4\sqrt[4]{3}}\sum\sqrt{a}\geq\sum a\minus{}\frac{3\sqrt[4]{3}}{4}\sqrt{\sum a}$
that $ \sqrt{\sum a}\geq\sqrt[4]{3}$
$ \Rightarrow LHS\geq\frac{\sqrt{3}}{4}$

Let $ a, b, c > 0$ and $ a\plus{}b\plus{}c \leq \sqrt{3}$. Prove that: $ \sum_{a,b,c}\frac{a}{\sqrt{1\plus{}a^2}} \leq \frac{\sqrt{3}}{2}$

Well, the first thing I want to consider is whether hungctqb's problem is true or false. I think the exact problem should be
For a, b, c be positive real numbers such that $ a+b+c \leq \sqrt {3}$ then$ \sum_{cyc}{\frac{a}{\sqrt{a^2+1}} \leq \frac{3}{2}}$
and of course, there exists a very cauchy- schwartz proof for this problem.

[quote="hungctqb"]Let $ a, b, c > 0$ and $ a \plus{} b \plus{} c \leq \sqrt {3}$. Prove that: $ \sum_{a,b,c}\frac {a}{\sqrt {1 \plus{} a^2}} \leq \frac {\sqrt {3}}{2}$[/quote]
The author of this inequality is Cao Minh Quang! Please help me to solve this problem!  :huh:

[quote="hungctqb"][quote="hungctqb"]Let $ a, b, c > 0$ and $ a + b + c \leq \sqrt {3}$. Prove that: $ \sum_{a,b,c}\frac {a}{\sqrt {1 + a^2}} \leq \frac {\sqrt {3}}{2}$[/quote]
The author of this inequality is Cao Minh Quang! Please help me to solve this problem!  :huh:[/quote]
Notice that it suffices for us to consider the case $ a+b+c=\sqrt{3}$, then we can transform our problem into (for a better looking)
Let $ a,b,c$ be positive real numbers such that $ a+b+c=3$, then
\[ {\frac{a}{\sqrt{a^2+3}}+\frac{b}{\sqrt{b^2+3}}+\frac{c}{\sqrt{c^2+3}} \le \frac{3}{2}}\]
By Cauchy Schwarz and AM-GM, we obtain
\[ LHS^2 \le (a+b+c)\left( \frac{a}{a^2+3}+\frac{b}{b^2+3}+\frac{c}{c^2+3}\right) \le (a+b+c)\left( \frac{a}{2a+2}+\frac{b}{2b+2}+\frac{c}{2c+2}\right) =\frac{3}{2} \left( \frac{a}{a+1} +\frac{b}{b+1} +\frac{c}{c+1}\right) =\frac{3}{2} \left( 3-\frac{1}{a+1}-\frac{1}{b+1} -\frac{1}{c+1}\right) \le \frac{3}{2} \left( 3-\frac{9}{a+b+c+3}\right) =\frac{9}{4}\]
Done! :)

Sorry, but I don't understand in this proof

[quote="Honey_S"][quote="D.I.Culianop"]Le $ a, b, c \geq 0$ and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that: $ \frac {a^3}{1 \plus{} b^2} \plus{} \frac {b^3}{1 \plus{} c^2} \plus{} \frac {c^3}{1 \plus{} a^2} \geq \frac {\sqrt {3}}{4}$[/quote]
\[ \frac {{a^3 }}{{1 \plus{} b^2 }} \plus{} \frac {{b^3 }}{{1 \plus{} c^2 }} \plus{} \frac {{c^3 }}{{1 \plus{} a^2 }} \equal{} \frac {{a^3 }}{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)}} \plus{} \frac {{b^3 }}{{\left( {a \plus{} c} \right)\left( {c \plus{} b} \right)}} \plus{} \frac {{c^3 }}{{\left( {c \plus{} a} \right)\left( {a \plus{} b} \right)}}
\]

\[ \equal{} \frac {{a^3 \left( {c \plus{} a} \right) \plus{} b^3 \left( {a \plus{} b} \right) \plus{} c^3 \left( {b \plus{} c} \right)}}{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}} \ge \frac {{ab\left( {a^2 \plus{} b^2 } \right) \plus{} bc\left( {b^2 \plus{} c^2 } \right) \plus{} ca\left( {c^2 \plus{} a^2 } \right)}}{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}}
\]

\[ \equal{} \frac {{a^2 \plus{} b^2 \plus{} c^2 \minus{} abc\left( {a \plus{} b \plus{} c} \right)}}{{a \plus{} b \plus{} c \minus{} abc}} \ge \frac {{\sqrt 3 \left( {a \plus{} b \plus{} c} \right)^2 \minus{} \left( {a \plus{} b \plus{} c} \right)}}{{3\sqrt 3 \left( {a \plus{} b \plus{} c} \right) \minus{} 1}} \ge \frac {{\sqrt 3 }}{4}
\]
[/quote]

the step

$ \frac {{a^2 \plus{} b^2 \plus{} c^2 \minus{} abc\left( {a \plus{} b \plus{} c} \right)}}{{a \plus{} b \plus{} c \minus{} abc}} \ge \frac {{\sqrt 3 \left( {a \plus{} b \plus{} c} \right)^2 \minus{} \left( {a \plus{} b \plus{} c} \right)}}{{3\sqrt 3 \left( {a \plus{} b \plus{} c} \right) \minus{} 1}}$

 :blush:

I have a question connected with solution of

can_hang2007

 Could You explain why we can consider only case when $ a\plus{}b\plus{}c\equal{} 3^{1/2}$?
Probably the case when $ a\plus{}b\plus{}c<3^{1/2}$ holds but I can't see why.

Thanks for Your reply.

[quote="hungctqb"][quote="great math"]Well, the first thing I want to consider is whether hungctqb's problem is true or false. I think the exact problem should be
For a, b, c be positive real numbers such that $ a + b + c \leq \sqrt {3}$ then$ \sum_{cyc}{\frac {a}{\sqrt {a^2 + 1}} \leq \frac {3}{2}}$
and of course, there exists a very cauchy- schwartz proof for this problem.[/quote]
It's from Duong Duc Lam! I think it's true![/quote]

Dear Hung,
You can use the technique in my article , give us a very simple proof, as follows

We have $ ab+bc+ca\le \frac1{3}(a+b+c)^2\le 1$, hence
$ \sum_{cyc}{\frac {a}{\sqrt {a^2 + 1}} \le\sum_{cyc}{\frac {a}{\sqrt {a^2 + ab+bc+ca}}=\sum_{cyc}\sqrt {\frac {a}{a+b}\frac a {a+c}} \le\sum_{cyc}\frac1{2}(\frac {a}{a+b}+\frac a {a+c})= \frac {3}{2}}}$, done!  :)

[quote="matex (L)(L)(L)"]I have a question connected with solution of

can_hang2007

 Could You explain why we can consider only case when $ a \plus{} b \plus{} c \equal{} 3^{1/2}$?
Probably the case when $ a \plus{} b \plus{} c < 3^{1/2}$ holds but I can't see why.

Thanks for Your reply.[/quote]

Let $ a\equal{}kx,b\equal{}ky,c\equal{}kz$ such that $ x\plus{}y\plus{}z\equal{}\sqrt3$, then $ k\le1$, hence
$ \sum\frac {a}{\sqrt {1 \plus{} a^2}}\equal{} \sum\frac {kx}{\sqrt {1 \plus{} k^2x^2}}\le\sum\frac {kx}{\sqrt {k^2 \plus{} k^2x^2}} \equal{}\sum\frac {x}{\sqrt {1 \plus{} x^2}}$
So, we only need prove that $ \sum\frac {x}{\sqrt {1 \plus{} x^2}} \leq \frac {3}{2}$, with the consider $ x\plus{}y\plus{}z\equal{}\sqrt3$. That's all !  :)

[quote="hungctqb"]Let $ a, b, c > 0$ and $ a \plus{} b \plus{} c \leq \sqrt {3}$. Prove that: $ \sum_{a,b,c}\frac {a}{\sqrt {1 \plus{} a^2}} \leq \frac {\sqrt {3}}{2}$[/quote]

$ f(x) \equal{} \frac {x}{\sqrt {1 \plus{} x^2}}$  $ f'(x) > 0,f''(x) < 0$
with Jensen inequlity
easy know that
$ \sum_{cyc} f(a)\leq3 f(\frac {a \plus{} b \plus{} c}{3})\leq3 f(\frac {\sqrt {3}}{3})) \equal{} \frac {\sqrt {3}}{2}$


trigonometry
geometry
circumcircle
Euler
geometry unsolved

Let $ \triangle ABC$  for excircles $ (I_B);(I_C)$.Prove that:
 $ (p \minus{} b)ac.cosB \plus{} (p \minus{} c)ab.cosC \equal{} I_BI_C.\cos{\frac {B \plus{} C}{2}}(c(p \minus{} b) \plus{} b(p \minus{} c))$

Let, $ s \equal{} \frac {a \plus{} b \plus{} c}{2}$.

So,
$ \cos A \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc}$
$ r_A \equal{} \frac {S}{s \minus{} a} \equal{} \sqrt {\frac {s(s \minus{} b)(s \minus{} c)}{s \minus{} a}}$
$ \sin \frac {A}{2} \equal{} \sqrt {\frac {(s \minus{} b)(s \minus{} c)}{bc}}$

(The radius $ r_A$ of the excircle $ A$ is called its exradius.)

I also went the way of you, but I think change will be complex.
I really hope a solution nice,simply

[quote="Virgil Nicula"][quote="math10"]Let $ \triangle ABC$ with  excincenters $ (I_b)$ , $ (I_c)$. Prove that:
 $ (p \minus{} b)ac.cosB \plus{} (p \minus{} c)ab.cosC \equal{} I_bI_c.\cos{\frac {B \plus{} C}{2}}[c(p \minus{} b) \plus{} b(p \minus{} c)]$ .[/quote]

$ I_bI_c\cdot\cos\frac {B \plus{} C}{2} \equal{} \boxed {I_bI_c\cdot\sin\frac A2 \equal{} a}$ and the your identity is wrong !

Is correctly (evidently !) thus :  $ (p \minus{} b)ac \plus{} (p \minus{} c)ab \equal{} I_bI_c\cos{\frac {B \plus{} C}{2}}\cdot [c(p \minus{} b) \plus{} b(p \minus{} c)]$ .[/quote]
Why,I don't know equall:$ I_bI_c\cdot\sin\frac A2 \equal{} a$
It is true :?:

Please, if you don't know this relation, then you can try to prove it, i.e. $ \boxed {I_bI_c\cdot\sin\frac A2 \equal{} a}$ before to put question "Is it truly ?!".

it is not correct.
if it is correct, your proof?

[quote="math10"]it is not correct.[/quote]Why ?! I promise you, after your disproof, I"ll present my proof, O.K. ?!

Oh,If you have solution,can you post them?. I wish  that it is correct

I have tried, but I can not achieve results

[quote="Virgil Nicula"]I don't. You made a mistake when you said free "isn't correct".
The proof is simply, why try you no ?!
I don't understand you.
Here is 12:09 and I go to sleep. Good night ![/quote]

It's 1:39 p.m. in Korea here. :D

$ I_bI_c\sin\frac A2\equal{}\left(AI_b\plus{}AI_c\right)\sin \frac A2\equal{}AI_b\sin\frac A2\plus{}AI_c\sin\frac A2\equal{}(p\minus{}c)\plus{}(p\minus{}b)\equal{}a$ .

Let, $ s \equal{} \frac {a \plus{} b \plus{} c}{2}$.

So,
$ |AI_A| \equal{} 4R \cos \frac{B}{2} \cos \frac{C}{2}$
$ |BI_A| \equal{} 4R \sin \frac{A}{2} \cos \frac{C}{2}$

$ \sin \frac {A}{2} \equal{} \sqrt {\frac {(s \minus{} b)(s \minus{} c)}{bc}}$
$ \cos \frac {A}{2} \equal{} \sqrt {\frac {s(s \minus{} a)}{bc}}$

[b]Otherwise.[/b] The circumcircle of $ \triangle ABC$ is the [u][b]Euler[/b]'s circle[/u] of $ \triangle I_aI_bI_c\implies$ the length of the circumradius of $ \triangle I_aI_bI_c$ is equally to $ 2R$ .Thus,

$ I_bI_c\equal{}2\cdot 2R\cdot \sin\widehat {I_bI_aI_c}\equal{}$ $ 4R\sin \frac {B\plus{}C}{2}\equal{}4R\cos\frac A2$ $ \implies$ $ I_bI_c\sin \frac A2\equal{}2R\cdot 2\sin \frac A2\cos\frac A2\equal{}2R\sin A\equal{}a$ $ \implies$ $ I_bI_c\sin\frac A2\equal{}a$ .

I took it as follows at the start.
$ I_AI_B\equal{}r_A \cdot r_B$,
$ p$ is any real number.

But, at last a problem was able to understand it.
Is it as follows?
$ I_AI_B$ is $ \overline{I_AI_B}$? (segment?),
$ p\equal{}\frac{a\plus{}b\plus{}c}{2}$?

[quote="Virgil Nicula"][b]Otherwise.[/b] The circumcircle of $ \triangle ABC$ is the [u][b]Euler[/b]'s circle[/u] of $ \triangle I_aI_bI_c\implies$ the length of the circumradius of $ \triangle I_aI_bI_c$ is equally to $ 2R$ .Thus,

$ I_bI_c = 2\cdot 2R\cdot \sin\widehat {I_bI_aI_c} =$ $ 4R\sin \frac {B + C}{2} = 4R\cos\frac A2$ $ \implies$ $ I_bI_c\sin \frac A2 = 2R\cdot 2\sin \frac A2\cos\frac A2 = 2R\sin A = a$ $ \implies$ $ I_bI_c\sin\frac A2 = a$ .[/quote]
Sorry.I do not think my results are wrong because when I slove problem Viet Nam TST (link  http://www.mathlinks.ro/viewtopic.php?p=808480#p808480) I had have this problem(I had using product no direction,I don't defined,exam: $ \vec a.\verb=abcos(a,b)$).
So I did not try it. :blush:
I will check it.


[hide="Hint"]
You can use complementary counting.  You have $ 3$ cases, where you have $ 2$ $ 1$'s, $ 1$ $ 1$, and $ 0$ $ 1$'s.  Examine these cases, and subtract this probability from $ 1$.[/hide]

[hide="Start You Off"]
You have $ 6^4$ number of choices.

You can have $ 4$ $ 1$'s in a row which can be chosen $ \binom{4}{4}\equal{}1$ way.

You can have $ 3$ $ 1$'s in a row which can be chosen $ \binom{2}{1}\equal{}2$ ways. (A 'stick' of $ 3$ $ 1$'s and one other die being chosen) If you have $ 3$ ones in a row, that last number can be anything except a $ 1$, which leaves $ 5$ choices. Same applies when you have $ 3\;1$'s at the end. $ 2 \cdot 5\equal{}10$ different outcomes.
[/hide]

[quote="lenny"]Suppose you through a die 4 times. What is the probability that you are going to see 1 appears consecutively?

In other words, what is the probability of the events where two or more 1's are next to each other?[/quote]
[hide]
The throws represent Bernoulli trials -- experiments in which the only possible outcomes are success or failure. As such there are $ 2^4 = 16$ possible outcome patterns. Eight of these outcome patterns, have at least one pair of consecutive successes (a success is having a one showing; a failure is anything that is not a success).

The total probability of at least one pair of consecutive one's is then
\begin{align*}P & = \left(\frac16\right)^4 + 4\left(\frac16\right)^3\left(\frac56\right) + 3\left(\frac16\right)^2\left(\frac56\right)^2 \\
 \\
& = \frac {1 + 4\cdot 5 + 3\cdot 25}{1296} \\
& = \frac {96}{1296} \\
 \\
& = \frac {2}{27}\end{align*}
[/hide]


If $p,q$ and $M$ are positive numbers and $q < 100$, then the number obtained by increasing $M$ by $p \%$ and decreasing the result by $q \%$ exceeds $M$ if and only if

$\text{(A)} \ p > q \qquad \text{(B)} \ p > \frac{q}{100-q} \qquad \text{(C)} \ p > \frac{q}{1-q} \qquad \text{(D)} \ p > \frac{100q}{100+q}$
$\text{(E)} \ p > \frac{100q}{100-q}$


i just learn basic of real analysis...so there are basic problem...i need help...
does the sequence converge or diverge? if it convers, what is the limit(with simple solution please :) )
$ a_n \equal{} 1 \plus{} \frac{(\minus{}1)^n}{n}$
$ a_n \equal{} \frac {1 \plus{} (\minus{}1)^n}{n}$
$ a_n \equal{} (\minus{}1)^n(1\minus{}\frac{1}{n})$
$ a_n \equal{} 1 \plus{} (\minus{}1)^n$
$ a_n \equal{} \frac{e^n}{n^5 \plus{} n \minus{} 5}$

in this case, prove with epsilon-delta perhaps...
prove that $ a_n\equal{}\frac{1}{n}$ converges to $ 0$ and $ a_n\equal{}\frac{1000}{n^3}$
prove that if $ a_n \le b_n \le c_n$ and $ lim a_n\equal{}lim c_n\equal{}L$ then the $ lim b_n\equal{}L$
prove or give a counterexample. Let $ a_n$ be a sequence such that $ a_{n\plus{}1} \minus{} a_n \minus{}\minus{}>>0$. does $ a_n$ have to converge?

i know there are too easy problems, but i think i must have a good basic for this topic..
thanks before, sorry if my english is bad :D

In your textbook, look up the relation of the limit of a sequence with that of a subsequence. You will also need the uniqueness of the limit. If you have these facts under your control, then for the sequences for which you have $ \left(\minus{}1\right)^n$, consider evaluating the limits of the two subsequences: one of odd indices and the other of even indices

If you have learned the definition of limitation,just use the definition to decide whether it converges.$ (\minus{}1)^n$ may make the series diverge.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Let [i]R[/i] be an integral domain, [i]F[/i] its field of fractions. Let [i]S[/i] be a subset of [i]R[/i]. Assume that 0 is not in S, that 1 is in S, and that every product [i]st[/i] with [i]s[/i] and [i]t[/i] in [i]S[/i] is itself in [i]S[/i]. Prove that the fractions that can be written in the form [i]b/s[/i] with [i]s[/i] ∈ [i]S[/i] is a subring of [i]F[/i] containing [i]R[/i].

You are speaking about the concept of localization $ S^{ \minus{} 1}R$ which could be found in every textbook.
But it does not need for $ R$ to be an integral domain, for $ S^{\minus{}}R$ to be defined.


factorial
number theory
Diophantine equation
number theory proposed

Find all triplets (x,y,z) of positive integers that satisfy the equation
\[ x! y!\equal{}z!\]

Actually there is an infinite number of these.

Take $ x\equal{}n$, $ y\equal{}n!\minus{}1$ and $ z\equal{}n!$

also $ (x,y,z)\equal{}\{(1,s,s),(s,1,s),(s,s,1), s \in \mathbb{N}$ is also a solution.

[quote="doicu"]also $ (x,y,z) \equal{} \{(1,s,s),(s,1,s),(s,s,1), s \in \mathbb{N}$ is also a solution.[/quote]

$ (s,s,1)$ is not a solution, also, he said find [i]all[/i], not just an infinite number

Do you solve this problem? It may be hard!  :wink:

The  solutions are given by the triple
$ (x,y,z) \equal{} (n! \minus{} 1, n , n ! )$ , for all $ n\in N$ and $ n > 2$.( like said Nemion)
Hence,  there are infinitely  many   solutions in natural numbers greater than $ 1$.
Example for $ n \equal{}3$:
$ x! y! \equal{} ( 3! \minus{}1)! 3! \equal{} 5! 3! \equal{} 6!$.

But there is another example, which is not in the family $ (n! \minus{} 1, n , n ! )$ i.e.:

$ 6! 7 ! \equal{} 720 \cdot 5040 \equal{} 36 28 800 \equal{} 10!$.
So, these are the classic known solutions.
In fact I don't know if there are any other solutions.

So the problem is :
Prove or disprove that the only solutions of the diophantine equation

$ x!y!\equal{}z!$

are:
1° $ (x,y,z)\equal{} (n! \minus{} 1, n , n ! )$ , for all $ n\in N$ and $ n > 2$,
and
2° $ 6! 7 ! \equal{}10!$.


Asymptote LaTeX HMMT USAMTS search ARML email

If you want to get the problems you post put into the resources section:
1. Avoid any extraneous comments in the post that contains the information; reply with comments later.
2. Use the following format for answer choices:
[code]
$\textbf{(A)}\ AnswerA \qquad
\textbf{(B)}\ AnswerB \qquad
\textbf{(C)}\ AnswerC \qquad
\textbf{(D)}\ AnswerD \qquad
\textbf{(E)}\ AnswerE$
[/code]
3. Use LaTeX.
4. When making diagrams, use LaTeX's picture environment or (preferably) Asymptote. Don't link to an external site which will inevitably go down, and attachments won't show up in the resources section.

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

And use the following format for problem sources:

Contest Year Day/Grade Round Problem X

So for example:

AMC 10 2008A Problem 1
HMMT 2008 Algebra Problem 1
USAMTS 2008 Round 3 Problem 1

This makes it much, much easier to use Search to collect all the posts from a certain contest/year/day/grade in a single page, and thus much easier to put the contest into the Contests section.

[quote="Silverfalcon"]I'm not sure if we're allowed to do for ARML contest. I sent an email last year asking them to give us permission and I haven't gotten anything back. Unless you got something from them, I say don't do for ARML.[/quote]
Edited. That was unintentional, I was just coming up with contests off the top of my ahead. Good catch. :)

[quote="Binomial-theorem"][quote="worthawholebean"]And use the following format for problem sources:

Contest Year Day/Grade Round Problem X
[/quote]

Now that this forum has sources, please use the above format in the source tag, and [b] use a descriptive title of the problem [/b] as the title of the topic.  Thanks.[/quote]

Would these count as descriptive titles?

-2012 AMC 8 #1

-Sum of two numbers

I believe they want something like

Subject: "Summation of trigonometric expansion"
Source: "1976 AHSME #30"
(I don't actually know what problem that is - don't quote me :P)

Just give a rough description of the problem so someone browsing the forum knows what to expect. Treat it as if you were posting a problem for help - what would you put in the subject line then?

[quote="worthawholebean"]
2. (As always) hide your solution; the parser automatically strips out the hide tag, so it isn't an issue. The parser won't have trouble with unhidden solutions, but you should hide just for common courtesy.
[/quote]
Just as a reminder to everyone to please hide your solutions.  Others that have questions on the problem don't always want solutions directly in front of them.  Thanks ;)!


Must be

Prove that :
$\frac{a_{1}}{C^{1}_{n}}\ge \sqrt{\frac{a_{2}}{C^{2}_{n}}}\ge ... \ge \sqrt[n-1]{\frac{a_{n-1}}{C^{n-1}_{n}}}\ge \sqrt[n]{\frac{a_{n}}{C^{n}_{n}}}$
Where $C^{k}_{n}=\frac{n!}{k!(n-k)!}$
Because $a_{i}=\sigma_{i}$ positive, when all roots positive.


[quote="Bicameral"]Solve for $ x$:

$ 2^(4x) \minus{} 2^(2x \plus{} log2 5) \plus{} 4 \equal{} 0$[/quote]

You made a mistake in the latex, I believe you meant:

$ 2^{4x} \minus{} 2^{2x \plus{} \log_2 5} \plus{} 4 \equal{} 0$

and my approach would be to let $ 2^{2x} \equal{} a$ and then try to solve the problem from there.

[quote="Brut3Forc3"][hide="Solution?"]
\begin{eqnarray*} 2^{4x} - 2^{2x + \log_2 5} + 4 & = & 0 \\
2^{4x} - 2^{2x}2^{log_25} + 4 & = & 0 \\
2^{4x} - 5(2^{2x}) + 4 & = & 0 \end{eqnarray*}
Then, like BlackSwordsman said, let $ a = 2^{2x}$.
\begin{eqnarray*} a^2 - 5a + 4 & = & 0 \\
(a - 1)(a - 4) & = & 0 \\
a & = & 1,4 \\
\end{eqnarray*}
\begin{eqnarray*} \text{So: } 2^{2x} & = & 1 \\
2^{2x} & = & 2^0 \\
2x & = & 0 \\
x & = & 0 \end{eqnarray*}
\begin{eqnarray*} \text{Or: } 2^{2x} & = & 4 \\
2^{2x} & = & 2^2 \\
2x & = & 2 \\
x & = & 1 \end{eqnarray*}
Our answers are $ \boxed{x = 0,1}$.
[/hide][/quote]

On the 3rd line, I don't see how you suddenly get 5(2^{2x}).  I don't see how you just took out the log completely.

Well, what does $ \log_ba$ really signify? It's the power you must raise b to in order to get a. So, $ \log_25$ means the power you must raise 2 to to get 5. Thus, when you raise 2 to it, you will get 5$ \Leftrightarrow2^{\log_25}\equal{}5$.

On the 3rd line Bicameral:

$ 2^{2x+\log_{2}5}=2^{2x}2^{\log
_{2}5}$
Notice that $ 2^{\log_{2}5}=5^{\log_{2}2} \rightarrow 5$
Therefore, it equals $ 5(2^{2x})$


santa is giving out presents to $7$ children. He has $7$ different presents prepared for each child. But Santa is getting old and has forgotten which is for who. Only $3$ children received the right presents. How many ways could this happen?

[hide]

Isn't it just $\binom{7}{3}$?

[/hide]

I think it's this:
[hide]
3 children receive the right present. You can choose three right kids out of seven: $\binom{7}{3}$
The remaining 4 presents go to the remaining 4 children. That means, that you have to search the number of derangements of ${1,2,3,4}$. A derangement is a permutation where none of the elements comes back to the right place.
See http://mathworld.wolfram.com/Derangement.html for further explication.
this number is $9$, in total $\binom{7}{3}*9 = 315$

[/hide]

Actually its not its......7 choose 3 multiplied by  (4! - 7) = 455

Hmmm...I thought it would be $\binom{7}{3} \cdot 4!$ because you can arrange the $4$ remaining presents $4!$ ways after choosing $3$ of the $7$ presents :?:

EDIT: On the other hand, it could be $\binom{10}{3}$ because a child can get more than one present... $A+B+C\leq 7$

you cannot multiply by $4!$, because then you count also the situations where the other four children get their right present.
And I don't think Santa would give a child more than one present and no present to another child...

[quote="ndb"]
And I don't think Santa would give a child more than one present and no present to another child...[/quote]

True :P

[quote="maokid7"]is it  seven choose 3 times 3!. since of the remaining kids the first can get the wrong gift in 3 ways the second in 2 and the third and fourth each in 1. tell me if i am missing something.[/quote]

I don't think so.  Because:
Suppose the first remaining kid gets the present of the second remaining child. Then the second remaining child can get the presents of the first remaining, the third remaining and the fourth remaining child.
But these are three possibilities, not two.

[hide]I guess this might be called be brute force, but here are my thoughts..

For the first part, out of 7 kids, 3 can get their presents in $\binom{7}{3}$ ways.

OK, then we have 4 kids and 4 presents and no kid can get their designated present. Choose an arbitrary kid. That kid has 3 presents that he can receive. Say that kid receives the present intended for kid $\alpha$. Now, kid $\alpha$ has again 3 presents he can receive (his is already taken...). Say s/he receives the present for kid $\beta$. Now kid $\beta$ has 2 options for presents.....

Thus, I say we have $\binom{7}{3} * 3 * 3 * 2 * 1$ ways.[/hide]

Bumping an old topic.

We can use PIE for the 4 remaining children.

Let's say there are children A, B, C, and D.

Let a,b,c, and d denote the respective gifts that they are supposed to receive.

Let $C(n)$ denote the number of combinations.

C(1 person gets his own gift) = $(4C1)(3!)=24$

C(2 people getting their own gifts) = $(4C2)(2!)=24$

C(3 people getting their own gifts) = $(4C3)(1!)=4$

C(4 people getting their own gifts) = $(4C4)(0!)=1$

Total number of combinations for the 4 people : $24$.

Use PIE. $24-(24-24+4-1) = 21$

So....$(7C3)(21) = (35)(21) = 735$.


Show that every natural number $n\leq2^{1\;000\;000}$ can be obtained first with 1 doing less than $1\;100\;000$ sums; more precisely, there is a finite sequence of natural numbers $x_0,\ x_1,\dots,\ x_k\mbox{ with }k\leq1\;100\;000,\ x_0=1,\ x_k=n$ such that for all $i=1,\ 2,\dots,\ k$ there exist $r,\ s$ with $0\leq{r}\leq{s}<i$ such that $x_i=x_r+x_s$.

Generate first all integers from $1$ to $2^{11}-1$, it takes $2^{11}-2=2046$ sums, with $x_k=x_{k-1}+x_0$ for $k=1,2,\dots,2^{11}-2$, where by trivial induction $x_k=k+1$.

Then, write any number $n$ in base $2^{11}$ as $n=a_0+a_12^{11}+a_22^{22}+\dots$, where since $1000000=11\cdot90909+1$, there are $90910$ such $a_i$ (already generated since they are all less than $2^{11}$.  Construct $n$ as follows: sum $a_{90909}$ with itself, then sum the result with itself, and so on, do so $11$ times, then add $a_{90908}$, thus obtaining $a_{90909}2^{11}+a_{90908}$.  Proceed likewise to obtain $a_{90909}2^{22}+a_{90908}2^{11}+a_{90907}$, and continue doing so until obtaining $n$.  Each time that we add a new $a_i$ costs us $12$ sums ($11$ to multiply by $2^{11}$, one more to add the $a_i$), since there are $90909$ $a_i$ appart from $a_{90909}$, it takes us $90909\cdot12=1090908<1100000-2048$ sums, and we are done.


what exactly is it?  can someone give me a few examples of its use?  i keep seeing it but cannot find a good explanation or a clear example.  thanks.

here's the AoPSwiki article on it
[url]http://www.artofproblemsolving.com/Wiki/index.php/Simon%27s_Favorite_Factoring_Trick[/url]
and the thread it originated:
[url]http://www.artofproblemsolving.com/Wiki/index.php/Simon%27s_Favorite_Factoring_Trick[/url]


linear algebra
matrix
inequalities
linear algebra unsolved

Let $A\in M_n(R)$ and $X\in M_n(R)$ with all its elements equal to $x$.Prove that:
  $\det(A+X)\cdot\det(A-X)\leq \det(A^2)$.

[quote="spix"]... $X\in M_n(C)$ with all its elements equal to $x$..[/quote]

Do you mean $X=(m(ij))$ with $m(ij)=x$ ?

[quote="spix"]Let $A\in M_n(R)$ and $X\in M_n(C)$ with all its elements equal to $x$.Prove that:
  $\det(A+X)\cdot\det(A-X)\leq \det(A^2)$.[/quote]

Are you sure that $X$ is a complex matrix ?
because $\det(A+X)\cdot\det(A-X)$ is a complex number you cannot have inequality

I am sorry, of course it cannot be $C$... it is $R$. :fool:

How I did it.
Consider $f(x)=\det(A+xE)$ where $E=(e_{ij}),e_{ij}=1$.
Using the method of derivating a determinant I found that $f''(x)=0$ which means that $f(x)=ax+\det(A)$ .
So: $\det(A+X)\det(A-X)=f(x)f(-x)=\det(A^2)-a^2x^2\leq \det(A^2)$.


The degree of $(x^2+1)^4 (x^3+1)^3$ as polynomial has degree of what?

This is very basic problem for degree. I won't give the choices to this one. :)

[hide]

It is not necessary to do this but \[(x^2+1)^2(x^3+1)^3\] expands to

\[x^{13}+2x^{11}+3x^{10}+x^9+6x^8+3x^7+3x^6+6x^5+x^4+3x^3+2x^2+1\]

Hence, the degree of the polynomial is $13$.

[/hide]

EDIT: This is correct for the degree of $(x^2+1)^2(x^3+1)^3$. Sorry I misread the question!!!

Here is the easy way...i just felt like expanding

[hide]

The largest degree of $x$ from $(x^2+1)^4$ is $8$ and the largest degree of $x$ in $(x^3+1)^3$ is $9$. Hence when you multiply the two expressions the largest degree of $x$ will come from $x^8 \cdot x^9=x^{17}$. Hence, the polynomial has degree $17$.

[/hide]


Euler
geometry
circumcircle
geometric transformation
reflection
geometry proposed

Let ABC be a triangle with euler triangle DEF.
if the vertices D, E, F of euler triangle  are reflected wrt. euler line of ABC
prove that the images lie on the nine point circle of ABC.

yes , this is easy  :D because the ninepoint center is the circumcenter of DEF , then the reflection of D onto a line that passes through its circumcenter lies on the same circle DEF , here the ninepoint circle


I've given myself a problem that I'm kind of confused on:

K=a*b=c*d
What is the number a that yields the largest value for a^a^a^a^a...^a where the # of a's = b?
I intended for K to be virtually   :inf: so that you can compare it with any other number, like c, so like the  :rt5: 's in  :rt5: ^ :rt5: ^ :rt5: ^ :rt5: ^...^ :rt5:  add up to the same thing as the threes in 3^3^3^3...^3.  I tried this problem with integers as in 2+2+2=3+3, and found 2^2^2<3^3, but I was wondering what number from all positive reals works like this the best.  I think from experimenting that it is e, but I'm not sure.  One way to evalute these numbers is to use logs.  Can someone work with this a bit?

In general, you are correct that e will make such a thing the largest, I believe.  I do know that e is the number such that, for any positive number x, e^x >= x^e, with equality holding if and only if x = e.  I suspect that your fact and my fact are essentially the same fact.  Do you see how you might prove it from there?  (I don't, but I haven't worked on it.)

Relatedly -- if you have two numbers, x and y,  x > y > e or e > y > x > 0, then y^x > x^y.  Or in other words, that e is in some sense a maximum of this continuous function.  (I say in some sense because I didn't exactly define a function, but I hope you know what I mean.)

I don't quite buy it. I think the maximum will be at some number less than e. Take a look at K=12. 2^2^2^2^2^2 is way bigger than 3^3^3^3. That makes me think that the maximum will be at some number less than 2 (or at least not much above 2). Actually, a quick glance at K=10 with 2 and 2.5 yields that it's greater with 2 than with 2.5. Of course, these numbers soon start to become far too large for calculators. I suppose we can take logarithms several times, but I can see no clear way to proceed. I believe it's less than 2. In fact the limit may be 1, as counterintuitive as that might seem. All I can say is that it certainly doesn't seem like e to me.

Actually, that makes a lot of sense -- the limit being 1.  I'll think about it.

The limit being one makes a lot of sense, but I'm slightly skeptical on 2^2^2^2^2^2>3^3^3^3 just because both have more than a million digits, so calculating tools can't evaluate them well (my 89 says  :inf: ).  How did ComplexZeta know that it is larger without calculating it?  It  obviously appears to be larger, but actually proving it makes my head hurt.

If the problem had been about iterated multiplication, then I think JBL would have been right about e.  But since the problem is about iterated exponentiation, I think ComplexZeta is right, the limit is 1.

Proof that 2^2^2^2^2^2 > 3^3^3^3:

Let K = 2^2^2^2^2^2. Then log K = (log 2)*2^2^2^2^2.
So log (log K) = log (log 2) + (log 2)*2^2^2^2.

Then let L = 3^3^3^3. log L = (log 3)*3^3^3
And log (log L) = log (log 3) + (log 3)*3^3.

From there it's easy to see that K is much larger than L.


what type of music do you like
there are many choices this poll lists some of them .

hmm i don't like all rock but i love the song "Blow Me Away" by Breaking Benjamin

what about showtunes?  :lol:
im obsessed with the musical Rent

[quote="EUCLA"][quote="bpms"]Why isn't rap up here?[/quote]
yeah what about rap??  :huh:[/quote]

cuz rap sux
just kidding, I listen to it every now and then... but I'd go with rock more

personaly I like classic though sometimes I listen to rock. Does any one like brahms beethovhon or bach I do.

I like alternative as well.

What is your favorite alternative type song?

i like a lot arctic monkeys and panic at the disco, but my favourite song is "Welcome to the Black parade"( my chemical romance).

Rap-Only by Eminem
Popular-Hindi soundtracks
Rock-Avril,Hillary,Simple Plan,Guns N Roses, 1 or 2 by Greenday,
Classical-Carnatic by MS Subbalakshmi and Madurai Mani Iyer only...
And a lot of alternative...
how come I can't vote for all 5?  :rotfl:  :lol:

Hello,
    Great to know that you like Madurai Mani Iyer.My favourite is Semmangudi Srinivasa Iyer.However,I like to listen to all masters and a few contemporaries in Carnatic Music.

I like progressive rock music.
King Crimson, Camel, Genesis, Dream Theater, IQ, Banco del Mutuo Soccorso, Le Orme, La Locanda delle Fate, Premiata Forneria Marconi, ...!
And Pink Floyd, too, of course.

I'd like you to write some of your favourite groups/albums!

[quote="Padmavathi"]
Rock-Avril,Hillary,Simple Plan,Guns N Roses, 1 or 2 by Greenday,
[/quote]
Avril and Hillary are POP omg and Simple Plan and Green Day are alternative, I believe.

I like mostly alternative and metal.
Fav band right now: Bullet For My Valentine.
A new Welsh metalcore band, i think.
The songs Hit the Floor and Suffocating Under Words of Sorrow are the most played songs on my iPod, by far.
After that, my next fav band is either Green day or blink-182.
(currently looking for the demo-album Flyswatter-blink182.....can someone find me a link to places that give free downloads??? :( )

[quote="Padmavathi"]Popular-Hindi soundtracks[/quote]

I'm not sure if anything Hindi would be anything [b]close[/b] to popular.

[b]Junggi-[/b]Whenever I go to the I Tunes Music Store(which is quite rarely coz it's very dumb...)Avril always comes under rock, and Simple Plan comes under Punk Rock!! !!:huh:  :?: I really don't care about Greenday as i like only 1 song of theirs. Hilary is POP but I really don't bother about genres..more about artists...
[b]Treething-[/b]When u come 2 India and ask for popular-all u'll get is Hindi soundtracks...that's the only thing popular here.., (apart from a braying jackass called Himesh Reshammiya. Sorry if any of u have been unlucky enough to hear him...or like him!!) :lol:
[b]Sathej-[/b]Semmangudi's also nice...Have u heard Kurai Ondrum Illai by M S Subbalakshmi?? It's AMAZING!!! Its one of her best songs till today.

[quote="junggi"][quote="Padmavathi"]
Rock-Avril,Hillary,Simple Plan,Guns N Roses, 1 or 2 by Greenday,
[/quote]
[color=red][b]Avril and Hillary are POP omg and Simple Plan and Green Day are alternative, I believe.[/b][/color]I like mostly alternative and metal.
Fav band right now: Bullet For My Valentine.
A new Welsh metalcore band, i think.
The songs Hit the Floor and Suffocating Under Words of Sorrow are the most played songs on my iPod, by far.
After that, my next fav band is either Green day or blink-182.
(currently looking for the demo-album Flyswatter-blink182.....can someone find me a link to places that give free downloads??? :( )[/quote]

I think GD is punk, not altermative :P

[quote="Padmavathi"][b]Junggi-[/b]Whenever I go to the I Tunes Music Store(which is quite rarely coz it's very dumb...)Avril always comes under rock, and Simple Plan comes under Punk Rock!! !!:huh:  :?: I really don't care about Greenday as i like only 1 song of theirs. Hilary is POP but I really don't bother about genres..more about artists...
[/quote]
1)Avril IS pop
2)iTunes doesn't have punk rock under it's genres....But I think they put punk rock under alternative? I don't really know about simple plan since I never listened to them
3)Does that one green day song happen to be boulevard....?
[quote]I think GD is punk, not altermative[/quote]
Not ONLY punk...they're pretty much punk rock, pop punk, alternative rock.... which all kind of comes under alternative(i think).

[quote="Padmavathi"]I seriously DON'T classify my muzic on the basis of genres and so it really doesnt matter to me....but sorry for the rotten classifying!!   :lol:
Junggi-YES!!! that one song IS Boulevard....an AMAZING one...but how did u know??
 :o[/quote]
ERGHH....
that song is SOOOO overrated..... :(  :(
I don't think that song deserved that "Record of the Year" that it won. I mean, it's a great song, but it's not THAT great. Compare it to other Green day songs like Basket Case, American Idiot, Holiday, Jesus of Suburbia, When I come around, Longview, Time of Your Life, Minority, and many of the lesser known songs from albums like Insomniac or Nimrod or Warning or whatever...

[color=red][b]What is the best tune of 2006?
This is a very interesting question. I heard about 2006's awards but i want to hear from u (just ur opinion)[/b][/color]

Errr, I don't know, it'll depend on the type/genre of music you like.
I personally like alternative and metal, and my favorite song from this year would be "Suffocating Under Words of Sorrow"-Bullet for My Valentine.
It's from Bullet's first album, and it's a very nice metalcore song. Very nice(or, horrible, if you can't tolerate any kind of screaming. Their screams actually a lot of tolerable than screams of other bands, say Slayer).
"Hit the Floor"-Bullet is also a very nice song from that same album(and less screams, too).
If you like metal, I'd definitely suggest trying out those two songs.
"Diary of Jane"-Breaking Benjamin and "Welcome to the Black Parade"-My Chemical Romance are some amazing alternative songs from this year.
If you like alternative-punk, then +44's top songs are really good("When your heart stops beating," "Baby come on," "Lycanthrope") as well as some of AVA's songs("The Adventure," "The War").

I like metal. Something like Tool, Moonspell, Katatonia, Anathema,...
Some movie soundtracks like The Lord of the Rings, Dracula...
Madredeus though I don't know the genre they belong to.

[quote="bpms"]Why isn't rap up here?[/quote]

rap isn't music! it's just a bunch of ppl screaming about how not to live life...i love rock! have any of u heard hellogoodbye or the all-american rejects? i used to like GD before they got stale (like avril and a lot of others also)

there are only four choices up there. and theres a whole variety of music.

classical music?? :what?: ummm...ok? i didn't even think it was popular with our generation...

EDIT: 100th post!!!

isnt there another topic like this in this forum?  :ninja:
anyways, yeah... classical/rock/pop

i listen to:
alternative, emo, indie, punk, post-hardcore, and a tiny bit of screamo lol.

Alternative & Rap*
*- does not include like Soulja Boy or lame things like that. I only like GOOD rap

I play classical, occasionally listen to it, but I usually don't listen to music at all, yes, I know, I'm weird  :lol:

no, you're NOT wierd. i dont usually listen to music either, but when i do it's classical. 'cause i play it. on 2 instruments.

I play classical on the piano but i like...
[hide="I like this type of music"]
[hide="I really like it, it's pwns"]
[hide="OK, this is the last hide tag"]

RAP!

APPLE BOTTOM JEANS
[/hide][/hide][/hide]

DANG I USED TO ONLY LISTEN TO CLASSICAL

AND THEN I QUIT

SO NOW I BASICALLY LIKE EVERYTHING EXCEPT CLASSICAL KINDA

ROCK FTW
basically it alternative, metal, anything much . . .

current fav band: foo fighters