Let x; y; z be positive real numbers satisfying

1/3 <= xy + yz + zx <= 3:
Determine the range of values for (i) xyz, and (ii) x+y +z.

Applying AM-GM on $xy$, $xz$, and $yz$ yields

\[xy+xz+yz \geq 3\sqrt[3]{x^2y^2z^2}\]
\[(xy+yz+xz)^3 \geq 27x^2y^2z^2\]
\[(xy+yz+sz)^3 \geq 27(xyz)^2\]

Plugging in the maximum value for $xy+yz+xz$ we obtain $xyz \leq 1$. Now using the fact that

\[\frac{xy+xz+yz}{3x^2y^2z^2} \leq \sqrt[3]{x^2y^2z^2}\]
\[\Leftrightarrow (xy+yz+xz)^3 \leq 27(x^2y^2z^2)^4\]
\[\Leftrightarrow (xy+yz+xz)^3 \leq 27(xyz)^8\]

Plugging in the minimum value for $xy+yz+xz$ we obtain $xyz \geq \frac{1}{\sqrt[8]{729}}$. Combining both inequalities we see that $\frac{1}{\sqrt[8]{729}} \leq xyz \leq 1$.

[quote="seamusoboyle"]$\frac{x+y+z}{3} \geq \sqrt[3]{xyz}$

You can use this to get the minimum bound[/quote]
So $x+y+z \in [3, +\infty)$?

[quote="towersfreak2006"]$xyz=-1$ when $x=y=z=-1$ so something's weird here. I know those inequalities don't work when used with negatives numbers, but at the very least $xyz=0$ when $x=0, y=z=1$. What are we missing?[/quote]
im already falling a sleep and see that you are tired too? original statement says positive numbers... what? what is that little star........ :sleeping:  :sleep2:

:oops_sign: Heh, I guess I am tired. I should probably go to sleep, but I am too stubborn. What about $x=y=1, z=\frac{1}{3}$, or in general $x=y=1, z\rightarrow0$? In the meantime... :sleeping:

sorry, but i don't quite get why $\displaystyle \frac{xy+xz+yz}{3x^2y^2z^2} \leq \sqrt[3]{x^2y^2z^2}$ ? How does this follow from AM-GM?

I am lazy, but:

$(x+y+z)^2\geq 3(xy+yz+zx)$

$xy+yz+zx\geq 3\sqrt[3]{x^2y^2z^2}$

It's clear that we can have $xyz\to 0$.  It's similarly clear that we can have $x+y+z\to\infty$.  Use the earlier remarks to obtain the last two bounds, and we are done.

We get $x+y+z\in [1,\infty]$ and $xyz\in (0,1]$


there are $ (x\minus{}2)(y\minus{}2)(z\minus{}2)$ unpainted cubes when you paint the surfaces of cube with side lengths x, y, z.

since there are at least 2 cubes per side, 45=3*3*5.

so its $ (3\plus{}2)(3\plus{}2)(5\plus{}2)\equal{}5*5*7\equal{}175$

answer : 175

The answer is actually 125.  It can't be 175, because that's not a perfect cube :D

We know it is less than 216, because there would be at least 64 unpainted cubes, and greater than 64, because there would only be 4 unpainted cubes.  So it's 125 :P.

When I had that problem on FTW, I guessed right :P

[quote]When the cube is taken apart, there are exactly 45 unit cubes without any paint.[/quote]

there cant be 45 unit cubes without any paint when you take apart a cube.

if you take apart 5*5*5 cube, there are 3*3*3=27 unit cubes without any paint.

Also, we know that it must be either:

a $ 5 \times 5 \times 5$ cube

OR

a $ 4 \times 4 \times 4$ cube

This is because a $ 3 \times 3 \times 3$ cube cannot have $ 45$ unpainted cubes.

[quote="GameBot"]Unit cubes are glued together to make a cube several units on each side. Some of the faces of this large cube are painted. When the cube is taken apart, there are exactly 45 unit cubes without any paint. How many unit cubes were used to create the larger cube?[/quote]

Well, the official answer is 125...

Can anybody use asymptote to make it clearer?  Particularly, a 4x4x4 and 5x5x5 cube?

[quote="vallon22"][quote="GameBot"]Unit cubes are glued together to make a cube several units on each side. [b]Some[/b] of the faces of this large cube are painted. When the cube is taken apart, there are exactly 45 unit cubes without any paint. How many unit cubes were used to create the larger cube?[/quote]

maybe only 5 of the 6 faces are painted
maybe only 4 of the 6 are painted

with that information, we know that since a 6 by 6 by 6 cube is taken apart, there will be $ 4\times4\times4$ cubes unpainted, or 64 cubes
that's too much

in a $ 5\times 5\times5$ cube, when taken apart, there will be $ 3\times3\times3$ pieces unpainted, or 27 cubes
adding on some faces that aren't painted, there might be a total of 45 faces unpainted[/quote]

according to you, then the problem's asking for the value of x when $ 6x^3 \minus{} 6x^2\equal{}45$?

umm...
i'm not sure
what do you mean by that equation?

BTW, this sentence
[quote]adding on some faces that aren't painted, there might be a total of 45 faces unpainted[/quote]

it should say: adding on some faces that aren't painted, there might be a total of 45 [b]cubes[/b] unpainted
my mistake

my equation was based on '45 faces'

so the number of unpainted faces would be number of total faces of cubes (if one side of the cube is x, then there are $ x^3$ cubes. each cube has 6 faces, so its $ 6x^3$) minus painted faces (there would be $ 6x^2$ painted faces obviously), which is $ 6x^3\minus{}6x^2$.

this problem is weird

By guess and check, the inner cube is $ 3\cdot3\cdot3$. Now the problem states that all sides are not painted. So that means that some of the interior of one side of the cube is not painted. $ 45\equal{}3^2\cdot5$ which implies that the length of the side of the square is $ 5$, thus the volume of the cube is $ \boxed{125}$

OHH SOME OF THE FACES

not all faces are painted...

now i understand it...

[quote="vallon22"][quote="GameBot"]Unit cubes are glued together to make a cube several units on each side. [b]Some[/b] of the faces of this large cube are painted. When the cube is taken apart, there are exactly 45 unit cubes without any paint. How many unit cubes were used to create the larger cube?[/quote]

maybe only 5 of the 6 faces are painted
maybe only 4 of the 6 are painted

with that information, we know that since a 6 by 6 by 6 cube is taken apart, there will be $ 4\times4\times4$ cubes unpainted, or 64 cubes
that's too much

in a $ 5\times 5\times5$ cube, when taken apart, there will be $ 3\times3\times3$ pieces unpainted, or 27 cubes
adding on some faces that aren't painted, there might be a total of 45 faces unpainted[/quote]

Er... tsk tsk.


[quote="knickmid"]The perimeter of a right triangle equals 2[i]a[/i] and the length of the hypotenuse equals [i]c[/i]. Find the area of the incircle of the triangle.[/quote]

$ r\equal{}\frac{a_{1}\plus{}b_{1}\minus{}c_{1}}{2}\equal{}\frac{(2a\minus{}c)\minus{}c}{2}\equal{}a\minus{}c$


geometry
circumcircle
rhombus
trapezoid
parallelogram
geometry proposed

[quote="edriv"][b]Problem 3[/b]
Let ABC be a triangle, G its centroid, M the midpoint of AB, D the point on the line $AG$ such that $AG = GD, A \neq D$, E the point on the line $BG$ such that $BG = GE, B \neq E$. Show that the quadrilateral BDCM is cyclic if and only if $AD = BE$.[/quote]

This cannot possibly be true.

If $\triangle ABC$ is equilateral, then AD = BE, D is on the triangle circumcircle, but M is not.

Next, consider an isosceles $\triangle ABC$ with a = BC, b = AB = AC, such that $\frac{b^{2}}{a^{2}}= \frac{8}{5}.$ BDCG is a rhombus and $CD^{2}= CG^{2}= \frac{2a^{2}+b^{2}}{9}= \frac{b^{2}}{4}= BM^{2},$ CD = BM. Since in addition $BD \parallel GC \equiv MC,$ BDCM is an isosceles trapezoid, cyclic. But $AD \neq BE,$ because $a \neq b.$

I found the original problem and it stated that $BDCM$ is cyclic if and only if $AB=BE$

BGCD is a parallelogram, its diagonals BC, GD cutting each other in half. Hence BG = DC and $BD \parallel GC \equiv MC$ and BDCM is always a trapezoid.

Assume the trapezoid BDCM is cyclic, it is isosceles, BM = DC = BG, BA = 2BM = 2BG = BE.

Assume BA = BE, then BM = BA/2 = BE/2 = BG = DC, the trapezoid BDCM is isosceles, hence cyclic.

Why am i always too late to post my solution?

Like yetti's, but more detailed. Easy problem :)

For $AB=BE$ we have $BM=BG$ (1) since $AM=MB$ and $BG=GE$. Denote $P \in [BC]$ such that $BP=PC$ (2) . Then we have $AG=GD$, and from $GP=\frac{1}{2}AG$ we get $GP=PD$. From (2) and some triangle congruences we get $BDPG$ parallelogram, so $BG=DC$ and $BD\parallel CG$ (3). But using (1) we get $BM=DC$, and by (3) we find that $BDCM$ is an iscoscelles trapezoid. We have $\angle D+\angle M=\angle B+\angle C=\pi$ blah blah, it is cyclic.

I won't post the proof for $AB=BE$, i assume it is as simple as this one.

Observe that $GM \parallel AE$, and that $BGCD$ is a rhombus.
Now,
$BA = BE \iff \angle BAE = \angle BEA \iff \angle BMG = \angle BGM \iff \angle BMG + \angle BGC = 180^\circ \iff \angle BMC + \angle BDC = 180^\circ$
$\iff BMDC$ is cyclic. $\square$


linear algebra
matrix
vector
induction
strong induction
combinatorics unsolved
combinatorics

let $G$ be a graph with $2n$ vertix and $G$ has edge connectivity. prove that there exsist a spanning subgraph of $G$,that the degree of every vertix be odd.

What if your graph has no edges? Or, more generally, an isolated vertex?

Here's an algebraic proof.

Let $A$ be the adjacency matrix of our graph $G$: it's a $2n\times e$ matrix ($e$ being the number of edges of $G$). Its rows represent the vertices, and its columns represent the edges. The $(i,j)$ entry equals $1$ if vertex $i$ and edge $j$ are incident, and $0$ otherwise. All computations are carried over $\mathbb F_2$, the field with two elements. Also, all vectors are considered to be column vectors.

Let ${\bf 1}$ be the vector of size $2n$ with all entries equal to $1$. What we want to prove here is that there is a vector $x$ of size $e$ such that $Ax={\bf 1}$. This is equivalent to proving that there are no vectors $u$ of size $2n$ such that $u^tA=0,\ u^t{\bf 1}=1\ (*)$. $(*)$ means finding a set of vertices of $G$ consisting of an odd number of vertices, such that the edges which have at least one endpoint in this set have both endpoints here. Since the graph is connected, the only such set can be the set of all the vertices, but this set has an even number of elements, so we're done: $(*)$ can never happen.

[quote="amir2"]oh sorry the graph has edge connectivity[/quote]
if we show that every tree with $2n$ vertics has the condition then our proof is completed
because every graph has a spaning tree
now use strong induction on the number of vertics of tree $T$
assume the longest path ($P$) in this tree
it`s end points ($x,y$) are leaves
now let edge $xv \in P$ now we know that $v$`s neighbours (exept in $P-{x}$)are all leaves
we show these leaves by $N'(v)$
ok we have two conditions here:
$i)N'(v) \equiv 0 (\bmod 2)$
in this case remove all the members of $N'(v)$ from the graph now we have removed even number of vertics there for there are even number of vertics remaining
now according to the induction hypothesis the remaining tree ($T'$)has a sub graph which ....
now add $N'(v)$ to $T'$ and connect them to vertex $v$
every member of $N'(v)$ has odd degree
and also $d_{T'}(v)=odd$
now $d_{T}(v)=d_{T'}(v)+N'(v)=odd+even=odd$
$ii)N'(v) \equiv 1 (\bmod 2)$
in this case remove all the members of $N'(v)$ exept $u$ from the graph now we have removed even number of vertics there for there are even number of vertics remaining
now according to the induction hypothesis the remaininb tree ($T'$) has a sub graph which...
now add the removed vertics and conect them to $v$
every member of $N'(v)$ has odd degree
and also $d_{T'}(v)=odd$
now $d_{T}(v)=d_{T'}(v)+N'(v)=odd+even=odd$

our proof is completed
sorry if I didnt translate it well


1)Let $ A \in M_n{(\mathbb{C})}$,$ I_n$ is identity matrix.
Prove that :
        $ rank(A\bigotimes I_n) \equal{} rank(I_n \bigotimes A) \equal{} n.rank(A)$
From here we have : roots number of equation $ AX \equal{} O$ is similar roots number of equation $ XA \equal{} O$ here unknown $ X \in M_n{(\mathbb{C})}$
2) Let $ A \in M_{nxm}{(\mathbb{C})},rank(A) \equal{} r$ and put $ C(B) \equal{} \{B \in M_{mxn}{(\mathbb{C})}| AB \equal{} O,BA \equal{} O\}$
Find $ \dim(C(B))$

For 1) see      http://www.artofproblemsolving.com/Forum/viewtopic.php?t=270737
 Here $ U\otimes{V}: X\rightarrow{UXV^T}$.
For 2) Let $ m\equal{}n$ If $ A$ is invertible and $ X$ is a $ n\times{n}$ matrix then $ AX\equal{}0$ implies $ X\equal{}0$. Thus we may assume $ A\equal{}diag(J_{k_1},\cdots,J_{k_p})$ where $ J_k$ is the nilpotent Jordan matrix of dimension $ k$. $ \{X;AX\equal{}XA\equal{}0\}$ is of dimension $ p^2$
The idea is as follows for the case $ p\equal{}2$:  $ AX\equal{}0$ implies $ X$ has two full rows and the other are $ 0$. $ XA\equal{}0$ implies that the elements of the two rows are $ 0$ except two among them. Therefore the elements of $ X$ are $ 0$ except four terms.
Conclusion: If $ A$ is a $ n\times{n}$ matrix then  $ dim(\{X;AX\equal{}XA\equal{}0\})\equal{}dim(ker(A))^2$.


[quote="lotrgreengrapes7926"][quote="inom"]$tan72^\circ>\frac{1}{2}$[/quote]
How did you get $\frac{\tan36^{\circ}}{\tan54^{\circ}}=tan72^{\circ}$?[/quote]

He may have used some trigonometric identities, and then he figured it out.

But clearly $\frac{\tan36^{\circ}}{\tan54^{\circ}} \neq tan72^{\circ}$.  You can figure out why yourself.

$\frac{\tan36^{\circ}}{\tan54^{\circ}} \ge \frac{1}{2}$

$\frac{\sin36^{\circ}\cos54^{\circ}}{\cos36^{\circ}\sin54^{\circ}}\ge \frac{1}{2}$

$\frac{\sin36^{\circ}\sin36^{\circ}}{\cos36^{\circ}\cos36^{\circ}} \ge \frac{1}{2}$

$\tan36^{\circ}\ge \frac{1}{4}$

(Because $\tan36^{\circ}$ is inbetween $\tan30^{\circ}$ and $\tan45^{\circ}$. So it's inbetween $\frac{1}{\sqrt{3}}$ and $1$ which is greater then $\frac{1}{4}$)

[hide]But isn't $\frac{\cos54^\circ}{\sin54^\circ}$ less than one so your inequality may have a flaw in it somewhere right?[/hide]

May have been pointed out before, but one has to be careful ....with things like:
[quote](Because $\tan36^{\circ}$ is inbetween $\tan30^{\circ}$ and $\tan45^{\circ}$. So it's inbetween $\frac{1}{\sqrt{3}}$ and $1$ which is greater then $\frac{1}{4}$[/quote]

From this all you can say, that is $\tan (36^{o})> (1/4) \implies \tan^{2}(36^{o}) > 1/16$

Which, of course, is not very helpful, as you want it to be "> (1/2".

What you need is something like..
[hide]The quantity  on LHS in the original question is equal to $5-2 \sqrt 5$  and one can show this by showing $(5-1/2)^{2}> (2 \sqrt 5)^{2}$ or  by showing $81/4 > 20$[/hide]

[quote="kunny"]$\frac{\tan 30^\circ}{\tan 54^\circ}>\frac{\tan 30^\circ}{\tan 45^\circ}$[/quote]
Ummm, no.

[quote="10000th User"][quote="Chocoboom"]$\tan36^{\circ}\ge \frac{1}{4}$[/quote]The mistake is here. It should be $\tan^{2}36^\circ\ge\frac12$. A one step conversion is to use cotangent-tangent pair $\cot\theta=\tan(90^\circ-\theta)$ and this immediately gives you

$\frac{\tan36^\circ}{\tan54^\circ}=\tan36^\circ\cot54^\circ=\tan^{2}36^\circ$

You can use a 36-72-72 triangle and fiddle around and find that

$\tan36=\sqrt[4]{5}\left(\frac{\sqrt5-1}2\right)^\frac32$[/quote]
how did you found that $\tan36=\sqrt[4]{5}\left(\frac{\sqrt5-1}2\right)^\frac32$?

[quote="kunny"]$\frac{\tan 36^\circ}{\tan 54^\circ}>\frac{\tan 30^\circ}{\tan 54^\circ}>\frac{\tan 30^\circ}{\tan 45^\circ}=\frac{1}{\sqrt{3}}>\frac{1}{2}\Longleftrightarrow 2>\sqrt{3}.$[/quote]

Sorry kunny, but...

$\frac{\tan 30^\circ}{\tan 54^\circ}>\frac{\tan 30^\circ}{\tan 45^\circ}= \tan45^\circ > \tan54^\circ$, what is not true.  :(

Edit: Sorry, it was noticed already. Just see it right now  :( .

This is my mistake as I have mentioned in the last post. :wink:

Since $36^\circ+54^\circ = 90^\circ$, then $\cos36^\circ = \sin52^\circ$ and $\sin36^\circ = \cos52^\circ$

And $\tan\alpha = \frac{sen\alpha}{cos\alpha}$

$\frac{\tan36^{\circ}}{\tan54^{\circ}}= \frac{\frac{\sin36^\circ}{\cos36^\circ}}{\frac{\cos36^\circ}{\sin36^\circ}}= \frac{\sin^{2}36^\circ}{\cos^{2}36^\circ}$
...
Since $\sin^{2}36^\circ = 1-\cos^{2}36^\circ$, it will give:

$\frac{1-\cos^{2}36^\circ}{\cos^{2}36^\circ}$

We want to proof that:

$\frac{1-\cos^{2}36^\circ}{\cos^{2}36^\circ}> \frac{1}{2}$

$\cos36^\circ < \frac{\sqrt{6}}{3}$, what is true since $\cos36^\circ < \cos45^\circ < \frac{\sqrt{6}}{3}$

Sorry to point out again but there are  errors in some of the proofs above. Some errors have been pointed out  but I am pointing it out again ... In any case one has to be careful in making errors like the following...
[quote="Astronum"].....

$\frac{1-\cos^{2}36^\circ}{\cos^{2}36^\circ}$

We want to proof that:

$\frac{1-\cos^{2}36^\circ}{\cos^{2}36^\circ}> \frac{1}{2}$

$\cos36^\circ < \frac{\sqrt{6}}{3}$, what is true since $\cos36^\circ < \cos45^\circ < \frac{\sqrt{6}}{3}$[/quote]

Point is,  $\cos36^\circ < \cos45^\circ$ is not true.  In fact $\cos36^\circ$ is  greater than $\cos45^\circ$...

[color=darkblue][b]Lemma. [/b]$\boxed{\cos 36^{\circ}=\frac{1+\sqrt 5}{4}}\ .$
[b]The first proof.[/b] Let $ABC$ be a triangle for which $AB=AC=1$, $m(\widehat{BAC})=36^{\circ}$. Denote the middlepoint $E$ of the side $[AB]$ and the point $D\in (AC)$ for which $DA=DB$. Prove easily that $DA=DB=BC=x$, $DC=1-x$ and $ABC\sim BDC$. Therefore, $\frac{AB}{BD}=\frac{BC}{DC}$ $\Longrightarrow$ $\frac{1}{x}=\frac{x}{1-x}$ $\Longrightarrow$ $x^{2}+x-1=0\ ,\ x>0$ $\Longrightarrow$ $x=\frac{-1+\sqrt 5}{2}$. But $AB=2\cdot AE=2\cdot AD\cos 36^{\circ}$ $\Longrightarrow$ $1=2x\cos 36^{\circ}$ $\Longrightarrow$ $\cos 36^{\circ}=\frac{1}{2x}$ $\Longrightarrow$ $\cos 36^{\circ}=\frac{1}{\sqrt 5-1}$ $\Longrightarrow$ $\cos 36^{\circ}=\frac{1+\sqrt 5}{4}\ .$
[b]The second proof.[/b] Denote $u=36^{\circ}\ (5u=180^{\circ})$. Then $\sin 3u=\sin 2u$ $\Longleftrightarrow$ $\sin u(3-4\sin^{2}u)=2\sin u\cos u$ $\Longleftrightarrow$ $3-4(1-\cos^{2}u)=2\cos u$ $\Longleftrightarrow$ $4\cos^{2}u-2\cos u-1=0$ $\Longleftrightarrow$ $\cos u=\frac{1+\sqrt 5}{4}\ .$
=================================================================================================================
=================================================================================================================
$2\tan 36^{\circ}>\tan 54^{\circ}$ $\Longleftrightarrow$ $2\tan 36^{\circ}>\frac{1}{\tan 36^{\circ}}$ $\Longleftrightarrow$ $\tan^{2}36^{\circ}>\frac{1}{2}$ $\Longleftrightarrow$ $\frac{1-\cos 72^{\circ}}{1+\cos 72^{\circ}}>\frac{1}{2}$ $\Longleftrightarrow$ $3\cos 72^{\circ}<1$ $\Longleftrightarrow$ $3(2\cos^{2}36^{\circ}-1)<1$ $\Longleftrightarrow$ $\cos 36^{\circ}<\sqrt{\frac{2}{3}}$ $\Longleftrightarrow$ $\left(\frac{1+\sqrt 5}{4}\right)^{2}<\frac{2}{3}$ $\Longleftrightarrow$ $\frac{3+\sqrt 5}{8}<\frac{2}{3}$ $\Longleftrightarrow$ $9+3\sqrt 5<16$ $\Longleftrightarrow$ $3\sqrt 5<7$ $\Longleftrightarrow$ $45<49$, what is truly. [b]Very nice and strong inequality ![/b][/color]

[quote="srulikbd"]wow nice method! why does $\sin 3u=\sin 2u$?
I thought for a long time if we can convert sin x and cos x to numbers, and now you show me how  :)[/quote]

$u=36^{\circ}$, so $\sin 3u = \sin 108^{\circ}= \sin\left(180^{\circ}-108^{\circ}\right) = \sin 72^{\circ}= \sin 2u$.

May be of interest,
[hide]$(\cos (5a)+i \sin (5a)) = (\cos a+i \sin a)^{5}$ gives
$tan (5a) = \frac{5 \tan a-10 \tan^{3}a+\tan^{5}a}{1-10\tan^{2}a+5 \tan^{4}a}$
When $a=36^{o}$ $\tan (5a) = \tan (180^{o}) = 0$  this gvies [/hide]
(Calling $tan (36^{o}) = t$
$5t-10t^{3}+t^{5}= 0$
since $t$ is not zero, we can divide and get:
$t^{4}-10 t^{2}+5 = 0$
if $x=t^{2}$ we know that LHS is nothing but $x=t^{2}$ , x is the root of  (positive value only as we know  it is positive)
$x^{2}-10x+5$
One can easily find out the value of $x$ if one needs to, but here we don't even have to do that, putting$x=1/2$ we see $-10x+5$ is zero, and LHS is positive, so root  we want (near 1/2) must be greater than (1/2).


calculus
integration
calculus computations

$If$     $\int_a^b f(x) dx = a +2b$   $then$        $\int_a^b (f(x) +5) dx = ?$

  How do you work this one out?  The "+ 5" part is really screwing me up.

Hint: Integration is a linear operation. Dealing with sums is easy.

Solution:

[hide]$\int_a^b(f(x)+5)dx=\int_a^bf(x)dx+\int_a^b5dx=a+2b+[5x]_a^b=a+2b+5b-5a=7b-4a$[/hide]

yup thats right. just split up that second integral, and since youve already defined $\int^{b}_{a} f(x)dx$, you just do $\int^{b}_{a} 5 dx$ then use fundamental theorem. pretty straightforward.

Ahh... I see it.  Thanks so much guys.

adidasty - I like your avatar!  and your sig.  Biology and math do rule!

Thanks man! Yeah definitely:

$\text{Stones, Math, and Biology} > \text{Any other science...and english too!}$


algebra
polynomial
superior algebra
superior algebra unsolved

show that $ K$ is a splitting field over $ F$ for a set $ \{f_1,...,f_n\}$ of polynomials in $ F[x]$ if and only if $ K$ is a a splitting field over $ F$ for the single polynomial $ f_1...f_n$.

There is nothing to show.
The splitting field of a set of polynomials $ S\subseteq F[X]$ is simply the field over $ F$
generated by the set of roots of the polynomials in $ S$ in some algebraic closure of $ F$.
And clearly $ \{f_1,\ldots, f_n\}$ and $ f_1\cdots f_n$ have the same sets of roots.


so i was in python and they said the maxint is 32 bits which is equivalent to two million numbers with 16 million positive numbers.
now my problem is does 16 bits mean that the maximum number represention is 16 digits,cus when i did the sum i got $ {10^16\minus{}1}$

Generally in programming a 32 bit integer goes up to $ 2^{n/2}\minus{}1\Rightarrow 2^{16}\minus{}1 \text{ for 32 and } 2^8\minus{}1 \text{ for 16 bit }$, though i'm not sure specifically for python.


[quote="mdk"]Find all primes of the form n^3+1[/quote]

[hide]$ n^{3}+1 = (n+1)(n^{2}-n+1)$

Hence, it's going to have a factor other than $ 1$ (namely, $ n+1$) unless $ n+1 = 2$. This happens when $ n = 1$, so $ 2$ is the only such prime.

$ \boxed{2 \textrm{ is the only such prime}}$.[/hide]

Specify positive primes; otherwise, $ (-2)^{3}+1 =-7$ is also valid over $ \mathbb{Z}$.

quick Q:
are negatives considered primes?

Ultimately, it depends on the context.

When you're only dealing with positive integers, then the easiest way to define a prime is as a number with no (positive) divisors other than $ 1$ and itself.

When you're dealing with negative integers, however, $ -1$ also divides everything, so this definition doesn't work.  One way to fix this is the following:

[b]Definition:[/b]  $ p$ is a prime if for any integers $ a, b$,

$ p | ab \implies p | a \vee p | b$

[b]Exercise:[/b]  Verify that the two definitions of primality are equivalent for positive integers.


What percentage of four-digit numbers have a repeated digit? Express your answer as a decimal to the nearest tenth.

For it to NOT have a repeated digit, the first digit must be non-zero, the second cannot be the first, the third cannot be the first or second, and the last cannot be the first, second, or third. Thus the number of possibilities without a repeated digit is $ 9 \times 9 \times 8 \times 7\equal{}4536$.  There are $ 9000$ total, so $ 9000\minus{}4536\equal{}4464$ with repeated digit.  Then the percent is 4464/9000=0.496=49.6 percent.

Why must the first digit be a nonzero number???

Because if the first digit was 0, then it would be a 3-digit number.

Oh ok sorry about that, i misunderstood the answer because I thought by the first digit, you meant the units digit.


I was feeling pretty wasted so I just weighed my math books :D   and they weighed one-fourth of my weight  :o  wanna [hide="guess?"]12k.g[/hide]

What about you?

I don't use any math books right now...
If I weighed all of the math books in my house, they'd weigh a LOT...
But if you're wondering, my weight is around 60-65 kg...

My mathbook right now weighs about .5

[hide="want to know why?"]because we use a packet for the last month of school lol[/hide]

We just handed in our math books (or I lost it), and I don't have all that many other math books, so a random guess of their total weight is ... 6 kg  (1/9 my weight).


$ \sin \frac{\theta}2 \equal{} \pm \sqrt{\frac{1\minus{}\cos \theta}2}$
$ \cos \frac{\theta}2 \equal{} \pm \sqrt{\frac{1\plus{}\cos \theta}2}$

where $ \pm$ depends upon the quadrant of $ \frac{\theta}{2}$
but I also read that

$ \tan \frac{\theta}{2} \equal{} \frac{\sin \theta}{1\plus{}\cos \theta}$

without a $ \pm$.
Why doesn't there need to be a $ \pm$?

If $ \frac{\theta}{2}$ is in quadrant II:

Then
$ \sin \frac{\theta}2 \equal{}  \sqrt{\frac{1\minus{}\cos \theta}2}$
$ \cos \frac{\theta}2 \equal{} \minus{} \sqrt{\frac{1\plus{}\cos \theta}2}$

$ \tan \frac{\theta}2 \equal{} \minus{} \sqrt{\frac{1\minus{}\cos \theta}{1\plus{}\cos \theta}}$
$ \implies \tan \frac{\theta}2 \equal{} \minus{}\frac{\sin \theta}{1\plus{}\cos\theta}$
That's what the formula seems to be, but:
I took the example case $ \theta\equal{}270^\circ$

But going through the calculations,
$ \tan 135^\circ \equal{} \minus{}\frac{\sin 270^\circ}{1\plus{}\cos 270^\circ}\equal{}\frac{1}{1}\equal{}1$

but $ \tan 135^\circ \equal{} \minus{}1 \not\equal{} 1$. So is there a flaw somewhere in here?

The $ \pm$ signs are essentially because of the square root.  On the other hand, the third formula you gave is exact and does not involve square roots.  You took the wrong square root in your calculation.

Let $ t \equal{} \tan \frac{\theta}{2}$.  Try proving that

$ \sin \theta \equal{} \frac{2t}{1\plus{}t^2}$
$ \cos \theta \equal{} \frac{1\minus{}t^2}{1\plus{}t^2}$

from which your third identity immediately follows.

[quote="RisingStar"]$ \sin \frac {\theta}2 \equal{} \pm \sqrt {\frac {1 \minus{} \cos \theta}2}$
$ \cos \frac {\theta}2 \equal{} \pm \sqrt {\frac {1 \plus{} \cos \theta}2}$

where $ \pm$ depends upon the quadrant of $ \frac {\theta}{2}$
but I also read that

$ \tan \frac {\theta}{2} \equal{} \frac {\sin \theta}{1 \plus{} \cos \theta}$

without a $ \pm$.
Why doesn't there need to be a $ \pm$?

If $ \frac {\theta}{2}$ is in quadrant II:

Then
$ \sin \frac {\theta}2 \equal{} \sqrt {\frac {1 \minus{} \cos \theta}2}$
$ \cos \frac {\theta}2 \equal{} \minus{} \sqrt {\frac {1 \plus{} \cos \theta}2}$

$ \tan \frac {\theta}2 \equal{} \minus{} \sqrt {\frac {1 \minus{} \cos \theta}{1 \plus{} \cos \theta}}$
$ \implies \tan \frac {\theta}2 \equal{} \minus{} \frac {\sin \theta}{1 \plus{} \cos\theta}$
That's what the formula seems to be, but:
I took the example case $ \theta \equal{} 270^\circ$

But going through the calculations,
$ \tan 135^\circ \equal{} \minus{} \frac {\sin 270^\circ}{1 \plus{} \cos 270^\circ} \equal{} \frac {1}{1} \equal{} 1$

but $ \tan 135^\circ \equal{} \minus{} 1 \not \equal{} 1$. So is there a flaw somewhere in here?[/quote]
Drawing the unit circle, and seeing that 270 degrees is the same point as $ (0,\minus{}1)$. Thus $ cos x\equal{}0$ and $ sin x \equal{}\minus{}1$. So $ tan 135$ should equal $ \minus{}1$.

There shouldn't be a negative sign. There doesn't need to be a $ \pm$ sign because the range of $ \cos \alpha$ is $ [ \minus{} 1, 1]$, so both $ 1 \minus{} \cos \alpha$ and $ 1 \plus{} \cos \alpha$ are both nonnegative for any angle $ \alpha$.

[quote]The $ \pm$ signs are essentially because of the square root.  On the other hand, the third formula you gave is exact and does not involve square roots.  You took the wrong square root in your calculation.[/quote]

It's true that square roots are only needed because I took a square root but the sign is chosen based on the quadrant the angle is in. I'm just confused why this:

[quote="RisingStar"]
If $ \frac {\theta}{2}$ is in quadrant II:

$ \sin \frac {\theta}2 \equal{} \sqrt {\frac {1 \minus{} \cos \theta}2}$
$ \cos \frac {\theta}2 \equal{} \minus{} \sqrt {\frac {1 \plus{} \cos \theta}2}$

$ \tan \frac {\theta}2 \equal{} \minus{} \sqrt {\frac {1 \minus{} \cos \theta}{1 \plus{} \cos \theta}}$
$ \implies \tan \frac {\theta}2 \equal{} \minus{} \frac {\sin \theta}{1 \plus{} \cos\theta}$
[/quote]

is invalid. I can't find where I took the wrong square root. I just divided through to find the tangent.

The error is when you write $ \sqrt{ \frac{1 \minus{} \cos \theta}{1 \plus{} \cos \theta} } \equal{} \frac{\sin \theta}{1 \plus{} \cos \theta}$.  The correct square root to take in this quadrant is negative.


Determine $ k$ so that the equation $ x^4 \minus{} 14x^2 \plus{} 24x \minus{} k \equal{} 0$ has $ 4$ distinct roots.

[hide="Hint"]Let $ f(x)\equal{}x^4\minus{}14x^2\plus{}24x$, $ g(x)\equal{}k$.
So $ \boxed{\max \{\text{local minimum of }f(x)\}<k<\text{local maximum of }f(x)}$[/hide]


I was wondering if it was too late to sign up for 2008-2009 USAMTS.  If it is not, when is the deadline?

Thanks to all.

it's not too late; there is no deadline to register. if you wish to participate in round 1, however, you should have registered and turned in the permissions form/under 13 form, as well as your solutions, by october 14

so we all have 2 months then right? phew  :D

It's not too late, and it's never too late!  Even if you miss Round 1, you can still sign up later in the fall and participate in Rounds 2-4.

I only started round 2 last year (and that was last second...terrible score...)


probability
function
probability and stats

We roll a regular 6-sided dice $n$ times. What is the probabilty that the total number of eyes rolled is a multiple of 5?

Consider the generating function
\[ f(x) \equal{} (x \plus{} x^2 \plus{} ... \plus{} x^6)(x \plus{} x^2 \plus{} ... \plus{} x^6)...(x \plus{} x^2 \plus{} ... \plus{} x^6) \equal{} (x \plus{} x^2 \plus{} ... \plus{} x^6)^n \equal{} a_{6n}x^{6n} \plus{} a_{6n \minus{} 1}x^{6n \minus{} 1} ... \plus{} \plus{} a_1x \plus{} a_0.
\]
It is clear that if $ S \equal{} \sum_{k \equal{} 0}^{n} a_{5k}$ then $ \frac {S}{6^n}$ is the desired probability. Let $ \omega$ be the primitive root of $ 1$ of degree $ 5$. We have $ 1 \plus{} \omega \plus{} ... \plus{} \omega^4 \equal{} 0$ and therefore if $ 5 \not | k$ then $ a_k(1 \plus{} \omega^k \plus{} ... \plus{} \omega^{4k}) \equal{} 0$ and if $ 5|k$ then $ a_k(1 \plus{} \omega^k \plus{} ... \plus{} \omega^{4k}) \equal{} 5a_k$. From this observation we conclude that $ f(1) \plus{} f(\omega) \plus{} ... \plus{} f(\omega^4) \equal{} 5S$.But $ f(1) \equal{} 6^n$ and $ f(\omega^k) \equal{} \omega^{kn}$, hence $ S \equal{} \frac {6^n \minus{} 1}{5}$ if $ 5 \not | n$ and $ S \equal{} \frac {6^n \plus{} 4}{5}$ if $ 5|n$.

[just a small fix]
I think your $ k$ should be an $ n$, and you still have to divide by $ 6^n$ to obtain the probability.
[/just a small fix]

Yes, I've changed $ k$ to $ n$, thanks. But I've written before that the probability is $ \frac{S}{6^n}$ so I've only calculated $ S$ :).


If $ p$ is any prime number , and $ n$ is any integer number :

Prove or Disprove :

if $ p|n^2$ then $ p|n$.

 :D

[hide="The relevant lemma"] $ p \text{ is prime} \Leftrightarrow \forall a, b : p | ab \implies p | a \text{ or } p | b$. [/hide]


ARML
geometry
rectangle
probability
3D geometry
octahedron
email

[hide="Set 1"]$(1)$ How many integers between $100$ and $999$ inclusive contain atleast $2$ identical digits?

$(2)$ The number of different $8$ letter "words" that can be made from $\text{SHOBHITV}$ are equal to the number of $n$ letter "words" of $\text{MOCKARML}$
Find $n$.[/hide]

[hide="Set 2"]$(3)$ Consider each row in Pascal's Triangle. Each row contains certain terms. (For example, the first row contains $[1]$, the second row contains $[1,1]$, third row $[1,2,1]$, etc. What is the sum of the first $107$ terms? The answer should be expressed as $a^b+c$, where $a$ is a  prime positive integer.

$(4)$ If $a,b,c$ are the roots of $x^3-x+1=0$, what is $a^4+b^4+c^4$?[/hide]

[hide="Set 3"]$(5)$ How many rectangles with positive area are on an $8 \times 8$ chessboard?

$(6)$ Find the probability that $10!$ is divisible by a number chosen from ${1,2,3,....,99,100}$.[/hide]
[hide="Set 4"]

$(7)$ Find the sum of $log_{10}$ of the proper divisors of $10^6$.

$(8)$ A point in space is $2,6,7,8,9$ units away from all but one of the vertices of a regular octahedron, find the distance from the remaining vertex. [/hide]

Here is the PDF.  Once again, email me if you want the latex source file.

[b][size=200]THE PDF IS INCORRECT DUE TO THE FACT THAT SHOBHIT KEEPS CHANGING THE MOCK ARML.  DO NOT USE THE PDF.[/size][/b]
I will post a new one as soon as AoPS lets me.  It's going REALLY slowly for me.

Since the more natural interpretation $log_{10}\sum_{d|10^6}d$ is not an integer, I will assume that number 7 is asking for $\sum_{d|10^6}log_{10}d$. Please clarify or correct me if I'm wrong.

[quote="ShoeFactory"]No wonder i kept getting messed up answers on #2 and #7 was worded oddly[/quote]

Number 2 was screwed up, but it now has EXACTLY ONE CORRECT ANSWER.

OK, it's been a while.  When are you announcing the winners?

Ok, only three people actually took the mock arml this time...and right now, i think we're busy with the MIMC, so I don't feel like talking about it.

Anyways, MysticTerminator got an 8 (no surprise) and mysmartmouth got a 6.

The third person that took it (jb05) got a 6.

So I guess the average score on this mock arml is about 7 :P

Anyways, I'm going to hold off on these until after the MIMC

all problems have been posted in separate threads in HSM  , I just created the related post collections

[url=https://artofproblemsolving.com/community/c3734455_2006_mock_arml_ii_individual]here[/url]

enjoy  / start solving

more USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more ARML mocks [url=https://artofproblemsolving.com/community/c2442143_arml_mocks]here[/url],
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


Find the real solution(s) to the equation $ (x \plus{} y)^2 \equal{} (x \plus{} 1)(y \minus{} 1)$.

[hide="Solution"] Let $ a \equal{} x \plus{} 1, b \equal{} y \minus{} 1$ and write as $ (a \plus{} b)^2 \equal{} ab \implies a^2 \plus{} ab \plus{} b^2 \equal{} 0 \implies (a \minus{} \omega b)(a \minus{} \omega^2 b) \equal{} 0$.  The only real solution occurs when $ a \equal{} b \equal{} 0 \implies (x, y) \equal{} \boxed{(\minus{}1, 1)}$. [/hide]

I wrote is as a quadratic:$ x^2\plus{}x(y\plus{}1)\plus{}(y^2\minus{}y\plus{}1)\equal{}0$
Discriminant is $ \minus{}3(y\plus{}1)^2$, since it must be nonnegative, $ y\equal{}\minus{}1$, $ x\equal{}1$.

With t0rajir0u's solution, you could also multiply $ a^2\plus{}ab\plus{}b^2$ by $ a\minus{}b$ to get $ a^3\equal{}b^3$, or $ a\equal{}b$ (since it's real), and then it's easy.

Or, you can note that $ a^2\plus{}ab\plus{}b^2\equal{}\frac{1}{2}(a^2\plus{}b^2\plus{}(a\plus{}b)^2)\geq0$ with equality if and only if $ a\equal{}b\equal{}0$.


[quote="tuantam1lan"][quote="alex2008"]For all non-negative $ a,b,c,d$ and $ abcd \equal{} 1$ prove that

$ \frac {(a^2 \plus{} b^2)(a^2 \plus{} b^2 \plus{} 2)}{(a \plus{} b)^4} \plus{} \frac {(c^2 \plus{} d^2)(c^2 \plus{} d^2 \plus{} 2)}{(c \plus{} d)^4}\geq 1$[/quote]
$ \frac {(a^2 \plus{} b^2)(a^2 \plus{} b^2 \plus{} 2)}{(a \plus{} b)^4} \ge \frac {1}{ab} < \equal{} > (a^2 \plus{} b^2)(a^2 \plus{} b^2 \plus{} 2)(ab \plus{} 1) \ge (a^2 \plus{} b^2 \plus{} 2ab)^2 < \equal{} > ab(a^4 \plus{} 2a^2b^2 \plus{} b^4) \plus{} 2(a^2 \plus{} b^2) \ge 2ab(a^2 \plus{} b^2) \plus{} 4a^2b^2$
uses cauchuy inequality:
$ 2a^3b^3 \plus{} 2ab \ge 4a^2b^2$
$ ab(a^4 \plus{} b^4) \plus{} 2ab \ge 2ab(a^2 \plus{} b^2)$
=>$ \frac {(a^2 \plus{} b^2)(a^2 \plus{} b^2 \plus{} 2)}{(a \plus{} b)^4} \plus{} \frac {(c^2 \plus{} d^2)(c^2 \plus{} d^2 \plus{} 2)}{(c \plus{} d)^4}\geq \frac {1}{ab} \plus{} \frac {1}{cd} \equal{} 1$[/quote]
But $ \frac {1}{ab} \plus{} \frac {1}{cd}\geq 2\sqrt {\frac {1}{abcd}} \equal{} 2 \left(\because abcd \equal{} 1; a,b,c,d\geq0\right)$.......

......... :huh:

[quote="tuantam1lan"]
$ \frac {(a^2 \plus{} b^2)(a^2 \plus{} b^2 \plus{} 2)}{(a \plus{} b)^4} \ge \frac {1}{ab}$[/quote]

Not true. For $ 1 \equal{} a \equal{} b( \equal{} c \equal{} d)$ we have $ \frac {2\cdot4}{16} \geq \frac {1}{1}$, i.e. $ \frac {1}{2}\geq 1$.

Actually, it looks to me like you were proving

$ \frac {(a^2 \plus{} b^2)(a^2 \plus{} b^2 \plus{} 2)}{(a \plus{} b)^4} \ge \frac {1}{ab \plus{} 1}$

and that should be correct, so the final conclusion is $ LHS \geq \frac {1}{ab \plus{} 1} \plus{} \frac {1}{cd \plus{} 1}\equal{}\frac{ab\plus{}cd\plus{}2}{ab\plus{}cd\plus{}2}\equal{} 1$, right?

[quote="alex2008"]For all non-negative $ a,b,c,d$ and $ abcd \equal{} 1$ prove that

$ \frac {(a^2 \plus{} b^2)(a^2 \plus{} b^2 \plus{} 2)}{(a \plus{} b)^4} \plus{} \frac {(c^2 \plus{} d^2)(c^2 \plus{} d^2 \plus{} 2)}{(c \plus{} d)^4}\geq 1$[/quote]
we have lemma
$ \frac {(a^{2} \plus{} b^{2})(a^{2} \plus{} b^{2} \plus{} 2)}{(a \plus{} b)^{4}} \geq \frac {1}{ab \plus{} 1} \Leftrightarrow [(a^{2} \plus{} b^{2})^{2} \plus{} 2(a^{2} \plus{} b^{2})](ab \plus{} 1)\geq (a^{2} \plus{} b^{2} \plus{} 2ab)$
it is true because :
$ [(a^{2} \plus{} b^{2})^{2} \plus{} 4ab)](ab \plus{} 1)\geq (a^{2} \plus{} b^{2} \plus{} 2ab)$
combine with :
$ \frac {1}{1 \plus{} ab} \plus{} \frac {1}{1 \plus{} cd} \equal{} 1$ (abcd=1 )
Done !


Find the [b]increasing[/b] interval of:

$y=5 | \tan{(-3x+\frac{\pi}{5})}|$

No one solving it? I think this is a very nice problem.


geometry
circumcircle
geometric transformation
reflection
inequalities
triangle inequality
geometric inequality

Prove that in a triangle $R \geq r \sqrt{2}$

When is equality maintained?

i think you mean $R\geq 2r$ because this one actually has equality.

[hide="geometric solution"]
Let $\triangle ABC$ have circumcenter $O$. Reflect $A$ over $\overline{BC}$ to $X$. Ptolemy's inequality on quadrilateral $OBXC$ gives
\[OB\cdot XC+OC\cdot XB\geq OX\cdot BC\]
\[(b+c)R\geq a\cdot OX\]
\[(a+b+c)R\geq a(OA+OX)\geq a\cdot AX=2ah_{a}=4[ABC]\]
(where the triangle inequality used in the last part). Now just use the fact that $2[ABC]=r(a+b+c)$ to finish.

Equality occurs in the triangle inequality only when $A,O,X$ collinear, which means $\overline{AO}\perp \overline{BC}$ and $AB=AC$.

Equality occurs in the Ptolemy's Inequality only when $O,B,X,C$ lie on a circle in that order, which means $\angle BOC=2\angle A$ and $\angle BXC=\angle A$ are supplementary, so $\angle A=60^\circ$. Therefore, equality holds only in $\boxed{\text{equilateral triangles}}$.
[/hide]

or we could be like $OI^{2}=R(R-2r)$, equality when $OI=0$...$\angle OAB=90-C=\angle IAB=\frac{A}{2}$...and symmetrical equations...$(A,B,C)=(60,60,60)$ is clearly the only solution

[quote="scorpius119"][quote="diophantient"][quote="scorpius119"]i think you mean $R\geq 2r$ because this one actually has equality.

[hide="geometric solution"]
Let $\triangle ABC$ have circumcenter $O$. Reflect $A$ over $\overline{BC}$ to $X$. Ptolemy's inequality on quadrilateral $OBXC$ gives
\[OB\cdot XC+OC\cdot XB\geq OX\cdot BC \]

\[(b+c)R\geq a\cdot OX \]

\[(a+b+c)R\geq a(OA+OX)\geq a\cdot AX=2ah_{a}=4[ABC] \]
(where the triangle inequality used in the last part). Now just use the fact that $2[ABC]=r(a+b+c)$ to finish.

Equality occurs in the triangle inequality only when $A,O,X$ collinear, which means $\overline{AO}\perp \overline{BC}$ and $AB=AC$.

Equality occurs in the Ptolemy's Inequality only when $O,B,X,C$ lie on a circle in that order, which means $\angle BOC=2\angle A$ and $\angle BXC=\angle A$ are supplementary, so $\angle A=60^\circ$. Therefore, equality holds only in $\boxed{\text{equilateral triangles}}$.
[/hide][/quote]
What inspired you to do the reflection and use Ptolomey's Inequality?
That's a very nice solution by the way :)[/quote]

Well, Ptolemy's Inequality typically leads to sharp geometric inequalities, and it makes sense to have $O$ and two of the vertices be part of the quadrilateral (to pull the circumradius out), so we can have $O,B,C$ as vertices. To deal with the equality case, the fourth point should be some point $X$ such that if $\triangle ABC$ is equilateral, $\angle BXC=180^\circ-\angle BOC=60^\circ$ and $X$ is outside $\triangle ABC$.

There are a couple of candidates for $X$ (like constructing an equilateral triangle with $\overline{BC}$ as a side and outside $\triangle ABC$ and letting $X$ be the third vertex), but the reflection turned out to be the best because it gave the inequality immediately.[/quote]
So Ptolomey's is a very powerful geometric inequality?
I thought that the triangle inequality is more powerful, is it?


group theory
abstract algebra
vector
linear algebra
matrix
superior algebra
superior algebra theorems

hi all,

here a rather nice result:
let GL(n,C) the group of invertible n*n matrices. Let G a subgroup of
GL(n,C). Prove that if there exists an integer q such that g^q=Id for all
g in G then G is finite.

bye.

In kholle (oral training of math) the kholleur (the teacher or mathematician)
should give hints for a problem like this one.

As far I remenber, this was an Oral Examination for Ecole Normale Suprieure (Ulm) there were more questions.

1/ E a $C$ vector space dimension n, $u\in L(E)$ we suppose
$tr(u^k)=0,  \forall k \in [1,n] , integer$
Prove that $u$ is nilpotent

2/ G a sub-group of $GL_n(C)$, (A_1,...,A_m) basis of $Vect(G)$ consider
$w:G\rightarrow C^m$, $w(g)=tr(A_ig)$

i)Prove that if $w(g)=w(h)$ then $gh^{-1}-I_n$ is nilpotent
ii)Suppose all elements of $G$ are diagonalisable
Prove that $w$ is injective
iii) Suppose G is finite exponent there exist a positive integer $q$ s.t $g^q=I_n$
for all $g \in G$.
Prove $G$ is finite

Some hints:
1/ prove all eigenvalues of $u$ are 0 using Cayley-Hamilton and
 $tr(u^k)=a_1q_1^k+...+a_pq_p^k$ where $q_i$ eigenvalue with
multiplicity $a_i$

2/

First prove the
Lemma: If  $u \in L(E)$ s.t $tr(u^k)=n, 1\leq k\leq n$ then $u-id$ is nilpotent.

Hint Expand $tr((u-id)^k)$ and use 1/

i)Hint: See $tr(ag)=tr(ah)$ for all $a\in G$ and
$tr(g^{-1}h)^k=tr(g^{-1}h)^{k-1}=...=tr(id)=n$ use the lemma to get
$g^{-1}h-id$ is nilpotent.

ii) Hints matrix nilpotent and diagonalisable is $O_n$

iii) $x^q-1$ simple roots in $C$ and  vanishes $g$ is diagonalisable,
its eigenvalues are the q-roots of unity.
$Tr(g)$ take only a finite number of values,
$w$ is injective + $Im(w)$ is finite => $G$ is finite

So Alekk does your solution  is  shorter ?


[quote="vaibhav2903"]find all possible pols such that a,b,c are the roots of x^3+ax^2 + bx +c=0[/quote]
So $ (x\minus{}a)(x\minus{}b)(x\minus{}c)\equal{}x^3\plus{}ax^2\plus{}bx\plus{}c$ and so the system :

$ a\plus{}b\plus{}c\equal{}\minus{}a$
$ ab\plus{}ac\plus{}bc\equal{}b$
$ abc\equal{}\minus{}c$

1) $ c\equal{}0$
The system becomes :
$ b\equal{}\minus{}2a$
$ ab\equal{}b$

And so two solutions : $ (a,b,c)\equal{}(0,0,0)$ and $ (a,b,c)\equal{}(1,\minus{}2,0)$

2) $ c\ne 0$
and so $ ab\equal{}\minus{}1$ and $ b\equal{}\minus{}\frac 1a$
$ c\equal{}\minus{}b\minus{}2a\equal{}\frac 1a\minus{}2a$
And the last equation $ \minus{}1\plus{}1\minus{}2a^2\minus{}\frac 1a(\frac 1a\minus{}2a)\equal{}\minus{}\frac 1a$ and so $ 2a^4\minus{}2a^2\minus{}a\plus{}1\equal{}0$ $ \iff$ $ (a\minus{}1)(2a^3\plus{}2a^2\minus{}1)\equal{}0$

And so (using Cardan), two values for $ a$ :
$ a\equal{}1$

$ a\equal{}\frac{1}{\sqrt[3]{1\plus{}\sqrt{\frac{19}{27}}}\plus{}\sqrt[3]{1\minus{}\sqrt{\frac{19}{27}}}}$

Hence 4 solutions for this problem :
$ (a,b,c)\equal{}(0,0,0)$

$ (a,b,c)\equal{}(1,\minus{}2,0)$

$ (a,b,c)\equal{}(1,\minus{}1,\minus{}1)$

$ (a,b,c)\equal{}(u,\minus{}\frac 1u,\frac 1u\minus{}2u)$ where $ u\equal{}\frac{1}{\sqrt[3]{1\plus{}\sqrt{\frac{19}{27}}}\plus{}\sqrt[3]{1\minus{}\sqrt{\frac{19}{27}}}}$

And I'm very surprised that such a stupid (no theoretical difficulty and crazy calculus) problem is given in olympiad training in your country


Let $S_{n}$ be the number of permutations $(a_{1}, a_{2}, ... , a_{n})$ of $(1, 2, ... , n)$ such that $1 \leq |a_{k}-k | \leq 2$ for all  $k$. Show that $\frac{7}{4}S_{n-1}< S_{n}< 2 S_{n-1}$ for $n > 6.$

Easy to see that $S_n=S_{n-2}+S_{n-4}+2\times \sum_{i=0}^{n-3}{S_i}$ for all $n\geqslant 4$ where $S_0=1$.
This gives us $S_{n+1}=S_n+S_{n-1}+S_{n-2}+S_{n-3}-S_{n-4}$ for all $n\geqslant 4$ and we easily have the following first few values of $S_n$: $$S_0=1,S_1=0,S_2=1,S_3=2,S_4=4,S_5=8,S_6=15,S_7=28$$
For each $n\in \mathbb{Z}^+$, we denote certain statements as follows:
\begin{align*}
T(n) & : S_{n-1}<S_n\\
P_1(n) & : \frac{7}{4}S_{n-1}< S_n \\
P_2(n) & : S_n< 2 S_{n-1}\\
Q(n) & : S_n>S_{n-1}+S_{n-2}+S_{n-3}-S_{n-4}\\
R(n) & : S_{n-1}+S_{n-2}+S_{n-3}-S_{n-4}>\frac{3S_{n}}{4}
\end{align*}
Note that $T(n-3)\implies T(n+1)$ and $T(2),T(3),T(4),T(5)$ is true. So $T(n)$ is true for all $n\geqslant 2$.
Easy to see that $Q(n)\implies P_2(n+1)$ and $R(n)\implies P_1(n+1)$ for all $n\in \mathbb{Z}^+$.

For $n\geqslant 7$ we have
\begin{align*}
R(n+1)& \iff S_n+S_{n-1}+S_{n-2}+3S_{n-4}>7S_{n-3}\\
& \iff 2S_{n-1}+2S_{n-2}+4S_{n-4}>6S_{n-3}+S_{n-5}\\
& \iff 4S_{n-2}+6S_{n-4}+S_{n-5}>4S_{n-3}+2S_{n-6}.
\end{align*}
If $P_2(n-3)$ is true, combining it with $4S_{n-2}+S_{n-5}>S_{n-3}+2S_{n-6}$ which follows from $T(n-5),T(n-3)$ yields the last inequality. Hence we conclude that $P_2(n-4)\implies R(n)$ for all $n\geqslant 8$.

Also, we have
\begin{align*}
T(n-3) & \implies 2S_{n-3}>S_{n-4}\\
& \iff S_{n+1}>S_n+S_{n-1}+S_{n-2}-S_{n-3}\\
& \iff Q(n+1).
\end{align*}

Hence $Q(n)$ is true for all $n\geqslant 6\implies P_2(n)$ is true for all $n\geqslant 7\implies R(n)$ is true for all $n\geqslant 8\implies P_1(n)$ is true for all $n\geqslant 9$. This gives $\frac{7}{4}S_{n-1}< S_{n}< 2 S_{n-1}$ for all $n\geqslant 9$. The rest is just to manually confirm the case $n\in \{ 7,8\}$ which is easy.

@ThE-dArK-lOrD So from $S_{n+1} = S_n + S_{n-1}+S_{n-2}+S_{n-3}-S_{n-4}$ we can get close form of $S_n$.

@above Yes, but it will look complicated and seems harder to prove that relations.

No, I think the relations is :$$S_{n+1}=S_n+2S_{n-1}$$.


Ptolemy's theorem states that

ABxCD + ADxBC >= ACxBD,
iff ABCD is cyclic.

Prove Ptolemy's theorem.

Pls explain to me with steps.. THZ A LOT!!

It is Ptolemy's inequality, this is one of well-known theorems in Geometry, please see here  //http://planetmath.org/encyclopedia/ProofOfPtolemysInequality.html  :wink:

[quote="wailam010"]Ptolemy's theorem states that

ABxCD + ADxBC >= ACxBD,
iff ABCD is cyclic.[/quote]

Not quite! It states that $ AB\cdot CD+BC\cdot DA\geq AC\cdot BD$, and equality holds if and only if ABCD is cyclic or degenerate cyclic.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=109578 for details and proofs, but googling also helps, since it is a well-known theorem.

  Darij


Let a, b, c be positive numbers. Prove that

$ (a^3\plus{}b^3\plus{}c^3)^2\geq\frac{3}{8}(a\plus{}b)(b\plus{}c)(c\plus{}a)(a^2b\plus{}b^2c\plus{}c^2a)$

--------------------
 :wink:  Ligouras Panagiote

[quote="Ligouras"]Let a, b, c be positive numbers. Prove that

$ (a^3 \plus{} b^3 \plus{} c^3)^2\geq\frac {3}{8}(a \plus{} b)(b \plus{} c)(c \plus{} a)(a^2b \plus{} b^2c \plus{} c^2a)$

--------------------
 :wink:  Ligouras Panagiote[/quote]

By Horder and AM-GM we have :

$ 9(a^3\plus{}b^3\plus{}c^3) \geq (a\plus{}b\plus{}c)^3 \geq \frac{27(a\plus{}b)(b\plus{}c)(c\plus{}a)}{8}$

By AM-GM we have : $ a^3\plus{}b^3\plus{}c^3 \geq a^2b\plus{}b^2c\plus{}c^2a$

Done !

:)


symmetry
combinatorics unsolved
combinatorics

A table $ 2008\times 2008$ is given. Two players alternately make moves. A move consist of choosing a white or black pawn and placing it on a chosen field. A player wins if after his move there is a sequence of five pawn in the same color in a line (horizontal, vertical or diagonal). Determine if the first player has a winning strategy.

No. The second player can always reply a first player's movement using a pawn of different color and in the symmetric position with relation to a line of symmetry of the table. The first player, before win, must have at his disposal an "unblocked" sequence of four pawns, i.e., four pawns  followed by an empty space. But, if the second player is about to allow such a position, the first player necessarily allowed a similar sequence before (in a symmetric position and with the opposite color). Then the second player forgets his strategy and wins with the line made by the first player.


A face is bounded by at least $ 3$ edges, and every edge is the boundary of at most $ 2$ faces.  Therefore...?

Edit:  You also need the condition that the graph is simple.  Otherwise, the graph on two vertices with $ n$ edges has $ n$ faces.

[quote="t0rajir0u"]A face is bounded by at least $ 3$ edges, and every edge is the boundary of at most $ 2$ faces.  Therefore...?[/quote]

But are we talking about the region out side of the graph ?

If we are, then we simply rephrase the hint to read; it hardly makes a difference...

[quote]
A face is bounded by at least $ 3$ edges, and every edge is the boundary of exactly $ 2$ faces. Therefore...?[/quote]

(The boundary of any region/face is the set of adjacent edges regardless of whether the region/face is infinite or not.)


Hi marcopolo.

I'm not sure about this one as I don't think I understand the question.

Are there $ 7$ beads altogether on the bracelet, one of each colour and two more beads, each of which could be of any of the five colours, (red, red), (red, blue), ...(black, black), that is $ 5\times5 \equal{} 25$ possibilities?

[quote="marcopolo"]I have red, blue, yellow, purple and black beads. In how many ways can 7 different colored beads be made into a bracelet if the red ,. blue and yellow must be together and the purple and black mus not be adjacent? Please explain with solution :)[/quote]
You have only 5 colors of beads, how can you have a bracelet of 7 different colors?!

how is this possible? 7 different colors. reword and rewrite the problems. The formula, anyways, is :
 n!/k!(n-k)! i think. If you want to limit adjacency, divide by 2 for each limitation. anyhoo, if u reword the problem i can solve it in a :rotfl:

by 7 different colours i think he means having all colours differnt on a bracelet. i mean  it should not have any 2 colours adjacent


Determincati in multimea numerelor reale pozitive toate solutiile sistemului:

    a+b+c=1
    (x+0,2003)/(0,92003-a)+(y+0,2004)/(0,92004-b)+(z+0,2005)/(0,92005-c)=0,42005
unde a^2=x, b^2=y si c^2=z.

Mama Flavian m-ai omorat cu sistemul asta. Nici macar nu-mi dau seama cerinta problemei. :P
Poti sa scrii problema in Latex, please? ;)

Cred ca a vrut sa spuna:
[quote="flavian"]
Determinati in multimea numerelor reale pozitive toate solutiile sistemului:

 $a+b+c=1$
                     $\frac{a^2+0,2003}{0,92003-a}+ \frac{b^2+0,2004}{0,92004-b}+\frac{b^2+0,2005}{0,92005-c}=0,42005$
[/quote]

DA. Sistemul e cel scris de Valica. Asa e corect.

la ultima fractie e o greseala (la scrierea lui valica). La numarator e c la patrat, ci nu b la patrat


function
complex analysis
complex analysis unsolved

Let $ D$ be the unit disk and let $ \varphi: D\rightarrow \mathbb{C}$  be an analytic function with $ |\varphi(z)| < 1$ for all $ z\in D$.
 If $ \varphi(z) \equal{} \sum\limits_{n \equal{} 0}^{\infty}a_{n}z^{n}$, prove that

for $ |z| < \frac {1}{3}$ then $ |\left(\sum\limits_{n \equal{} 0}^{\infty}|a_{n}|z^{n}\right)| < 1$

mm.. it's follows from here
http://www.mathlinks.ro/viewtopic.php?t=198448
 :rotfl:


function
inequalities
inequalities proposed

Let $\ a,b,c$ be positive real numbers .
prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ $\geq$ $\((1+\sqrt{2}+\sqrt{3})(\frac{1}{a+\sqrt{2}b+\sqrt{3}c}+\frac{1}{b+\sqrt{2}c+\sqrt{3}a}+\frac{1}{c+\sqrt{2}a+\sqrt{3}b})$

[quote="Rzeszut"]It is in fact very easy.
Let $p=\frac{1}{1+\sqrt{2}+\sqrt{3}}$, $q=\frac{\sqrt{2}}{1+\sqrt{2}+\sqrt{3}}$ and $r=\frac{\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}$. Then we have to prove
$\sum\frac1a\geqslant \sum\frac{1}{pa+qb+rc}$.
But since the function $f(t)=\frac1t$ is convex, by Jensen:
$\sum\frac{1}{pa+qb+rc}\leqslant \sum\frac{p}{a}+\frac{q}{b}+\frac{r}{c}= (p+q+r)\cdot\sum\frac{1}{a}= \sum \frac{1}{a}$
which ends the proof.[/quote]
This proves a (very) special case of my [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83869]cyclic fractional sum conjecture[/url], viz. $\{1,0,0\}\ge\{p,q,r\}$ for positive $p,q,r$ such that $p+q+r=1$. ;)
I wonder if by a similar method we can prove something like $\{1,1,0\}\ge\{p,q,r\}$ for $1\ge p\ge q\ge r\ge0$ such that $p+q+r=2$, or even more general cases. :)  :?:

Using the inequality Chauchy-Shwarz
$\sum(a+\sqrt{2}a+\sqrt{3}c)\sum(\frac{1}{a}+\frac{1}{b}+)\frac{1}{c})\geq(1+\sqrt{2}+\sqrt{3})$
    then..........
   the inequality is stric !
 Which is true ?? :blush:  :D


Let $n  \in N^{*}$ be a given number and $a,b,c \in R$ so that $a^{n} = b-c$, $b^{n}=c-a$  and $c^{n}=a-b$. Prove that $a=b=c=0$.

This is true for even n since

a^n + b^n + c^n =0.

[quote="fourierseries"]This is true for even n since
$a^n + b^n + c^n =0$.[/quote]
For odd $n$: Assume one of the numbers is negative, let's say $a$. And $b$, $c$ are positive. But from equation $c^n=a-b$ we know that $c$ is also negative. Contrary.

Then assume two of the numbers are negative, let's say $a$ and $c$. But from $a^n=b-c$ we know that $a$ is positive, also imposibble.

It is trivial that $a$, $b$ and $c$ can't all be negative.

Therefore we know $a=b=c=0$.


There are 20 people in my club.  8 of them are left-handed.  15 of them like jazz music.  2 of them are right-handed and dislike jazz music.  How many club members are left-handed and like jazz music?

We know that there are $ 18$ who either are left handed and/or like jazz music.  8 are left handed and 15 like jazz.  Thus, there are 8+15-18=5 who are left handed AND like jazz music.

[hide]Every person is either right-handed or left-handed and either likes jazz or dislikes jazz.

Since $8$ are left-handed, $20-8=12$ are right-handed.  Since $15$ like jazz, $20-15=5$ dislike jazz.

One way:  Since $2$ are right-handed and dislike jazz, $12-2=10$ are right-handed and like jazz.  Since $10$ are right-handed and like jazz, $15-10=\boxed{5}$ are left-handed and like jazz.
Another way:  Since $2$ are right-handed and dislike jazz, $5-2=3$ are left-handed and dislike jazz.  Since $3$ are left-handed and dislike jazz, $8-3=\boxed{5}$ are left-handed and like jazz.[/hide]

Whenever I come across a problem like this, I like to make a table to organize my solution. Since I can't seem to get a LaTeX table to work here, I'll do it in Geogebra:

[geogebra]3a53233e71d3bd7d3b0b75f4086564886f0c5b1c[/geogebra]

[hide="Explanation of 2 ways to solve the table to find the answer"]
It turns out that I used the exact same logic that sray070 used in his (or her) post above.

Now it becomes a matter of filling in the missing values of the table. There are 20 people in the club, of which 8 are left-handed, so we know that the number of righties in the club is $20-8=12$. Of the 12 right-handers in the club, we know 2 of them don't like jazz, so there must be $12-2=10$ who do. Finally, of the 15 people in the club who like jazz, 10 are right-handed, which means the number of lefties who like jazz in the club is $15-10=\boxed{5}$.

Alternatively, I can conclude that of the 20 people in the club, 15 like jazz, so there are $20-15=5$ who don't like jazz. Of the 5 people who don't like jazz, 2 of them are right-handed, which means that $5-2=3$ of them are left-handed. Finally, of the 8 people in the club who are left-handed, 3 of them don't like jazz, therefore there are $8-3=\boxed{5}$ left-handed people in the club who like jazz.[/hide]

For the record, the problem probably should state that everyone in the club is either left-handed or right-handed--that there are no ambidextrous members of the club. I would [i]think[/i] that you could assume that, but everyone knows what can happen when you [i]assume[/i]... (I'll leave it at that since I don't want to trip any profanity alarms)

Tim

This one took quite a bit of insight, honestly...


LaTeX
number theory unsolved
number theory

Hello all,
Can you help me to solve this problem:
If the order of a mod(p^2) is d, what is the order of a mod(p).
Note that a may be non  primitive root.
Thank you

I undestand so. Given x: x^d=1(mod p^2) (for r<d x^r is not equal 1(mod p^2)). Find m: x^m=1 (mod p).
  It is so?
 Solution if p|d then m=p/d, else m=d.

Hello,
My problem exactly as you understood, can you give me an explanation of your solution, please?

Let $m=ord_p (x), \ d=ord_{p^2} (x)$. then $x^k =1 (mod \ p) \ only \ if \ m|k$
If $x^m=1 (mod \ p^2 ) \ then \ m=d \ and \ d\leq p-1 , \ else \ d=m p$.

Rust: I'm so happy you use $\LaTeX$   :)  :lol:  !!!

Also congrats for using it ;)

Is there anyone that hasn't gratulated till now¿  :lol:


Let a f be a real function satisfying
f(x+a) = 1/2 +  sqrt(f(x) - f(x)^2)

a) Prove f is periodic
b) In the case a = 1, give an example of such a function.

hint for part a): [hide]whats f(x+2a)?[/hide]

I don't get the problem. Do you mean that f(x+c) = f(x) = f(x + 3942598247359823)?

If so, I've solved for f(x) = (6- :sqrt:7) / 4, a constant value in the case a=0. I'm pretty sure I don't understand the question correctly though O.O.

a is a given constant, its not a function for all a.


We have $x$ different types of lollies, namely, types $a_{1}, a_{2}, a_{3}\cdots, a_{x}$.

There are $\alpha_{1}$ type $a_{1}$ lollies.
There are $\alpha_{2}$ type $a_{2}$ lollies.
...
There are $\alpha_{x}$ type of $a_{x}$ lollies.

Find a formula for determining the number of ways there are to choose $n$ lollies such that $0 \leq n \leq \alpha_{1}+\alpha_{2}+\cdots+\alpha_{x}$

no i dont think that is correct, because consider $n=1$ there are $x$ ways to do this.

M4RIO's formula only works when $a_{1}+a_{2}+...a_{x}=n$.

We divide into cases based on how many different types of lollipops we decide to pick (I can't bring myself to use the word "lollies" for some reason).

If we pick only 1 type of lollipop, there's only 1*x ways to do it (1 type*n combinations of types)

If we pick 2 types, there's $\binom{n-1}{1}\binom{x}{2}$, where we get the first term by the object/dividers analogy.

Continuing in this fashion, we find the total number of ways to be $\sum_{k=0}^{k = min(n-1,x-1)}\binom{n-1}{k}\binom{x}{k+1}$

[quote="justdudxit"]M4RIO's formula only works when $a_{1}+a_{2}+...a_{x}=n$.[/quote]

No, because in that case the correct answer is 1.  Mario's formula is the answer to an entirely different (but similar-sounding) question.

In fact, your argument is certainly wrong because your function also fails to give 1 at that point (and, I suppose, 0 thereafter).

I think that this problem is very hard.  Try the following three special cases: all the $a_{i}$ are equal to some fixed value, $a_{i}= 2^{i-1}$ or $a_{i}= i$.

D'ohhh....I was under the assumption that we were picking n lollipops out of an infinite number of each type.

This is my reasoning(which is probably wrong).

There are $x$ types of lollies and $\alpha_{1}+\alpha_{2}+\alpha_{3}+\cdots+\alpha_{x}$ total lollies.

Use the balls in urn formula:
Let $\beta_{n}=$ number that of lollies we choose of type $a_{n}$.

$\beta_{1}+\beta_{2}+\beta_{3}+\cdots+\beta_{x}= n$
Since $\beta_{k}\le \alpha_{k}$, replace $\beta_{k}$ with $\alpha_{k}-\beta_{k}$ so we only get numbers less than or equal to $\alpha_{k}$.

$\alpha_{1}-\beta_{1}+\alpha_{2}-\beta_{2}+\cdots+\alpha_{x}-\beta_{x}= n$
$\beta_{1}+\beta_{2}+\beta_{3}+\cdots+\beta_{x}= \alpha_{1}+\alpha_{2}+\alpha_{3}+\cdots+\alpha_{x}-n$

The number of combinations now is:

$\binom{\alpha_{1}+\alpha_{2}+\alpha_{3}+\cdots+\alpha_{x}-n+x-1}{x-1}$

I probably interpreted the problem wrong, bu t it was a good latex exercise on cdots.

The problem is in your second equation the $B_{k}$ can still exceeds $a_{k}$, so when we go back you have negative number of lollies.

But the main idea is correct, just that you have to use painful inclusion-exclusion principle; the transformation doesn't work.

Also, let's say $a_{1}= a_{2}$.  Does that mean the type $a_{1}$ lollies and type $a_{2}$ lollies become indistingushable???!!! :o

Now someone try solve this enhanced hard version...

[quote="JBL"]I am not sure at all what you were doing with the $\alpha_{k}-\beta_{k}$ bit.  Your answer is also clearly wrong, this time by setting $n = 0$.

[quote="pkerichang"]Now someone try solve this enhanced hard version...[/quote]
I think someone should try and solve any single case of the first problem first, don't you?[/quote]

Seems like I didn't make it clear.  Sorry.

so let $x_{1}, x_{2}, \dots x_{k}$ represent the number of the $a_{k}$ types of lollies.  Then, we have:

$\sum_{i=1}^{k}{x_{i}}= n$

with each term less than or equal to the number $a_{i}$.  Now, we can count the total number of solutions where at least one of the $x_{i}$ exceeds $a_{i}$ and subtract from the total solution.

The total will be $\binom{n+k-1}{k-1}$ by simple combinatoric argument which I believe many of you know.  Then:

if $x_{i}> a_{i}$, we replace $x_{i}$ by $y_{i}+a_{i}+1$ to create a bijection - each $y_{i}$ corresponds to a $x_{i}$ greater than $a_{i}$.  Solving we get $\binom{n-a_{i}+k-2}{k-1}$

if $x_{i}> a_{i}$ and $x_{j}> a_{j}$, we again create a bijection and replace them by $y_{i}+a_{i}+1$ and $y_{j}+a_{j}+1$.  Solving get $\binom{n+a_{i}-a_{j}+k-3}{k-1}$.

And so on and so forth.  Now, by inclusion-exclusion principle, we get:

$\binom{n+k-1}{k-1}-\sum_{i \le k}{\binom{n-a_{i}+k-2}{k-1}}+\sum_{i < j \le k}{\binom{n-a_{i}-a_{j}+k-3}{k-1}}-\dots$

as our nasty answer.  I don't believe there's an easy way around, because this type of problem is dealt with in mathematics of choice and this is the solution they gave (I just read it 30 min ago :rotfl: )

edit note: fix algebra error, and also choosing negative stuff in this case is defined to be 0.


Let $a, b \in \mathbb{N}$.

Prove that $a^{2}-b^{2}$ does not divide $a^{2}+b^{2}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=bezout&t=109997]here[/url] is a slightly more general version!


induction
analytic geometry
real analysis
real analysis unsolved

Let $ \mathbb{B}^n$ is Borel sigma algebra of $ \mathbb{R}^n$.

Prove $ \mathbb{B}^m$X$ \mathbb{B}^n\equal{}\mathbb{B}^{m\plus{}n}$

Is there any book that contains the proof of this problem?

or give some hints or outline... of course, full solution is better! PLEASE HELP ME!!

Sorry for the notation, but I'm going to denote the Borel $ \sigma$-algebra by $ \mathcal{B}\left(\mathbb{R}\right)$.  I think that you may prove this by induction, with the hard case being the base case.  That is, $ \mathcal{B}\left(\mathbb{R}^{2}\right) \equal{} \mathcal{B}\left(\mathbb{R}\right)\otimes\mathcal{B}\left(\mathbb{R}\right)$.  Here is an outline:

That $ \mathcal{B}\left(\mathbb{R}^{2}\right)\subset\mathcal{B}\left(\mathbb{R}\right)\otimes\mathcal{B}\left(\mathbb{R}\right)$ is easy since open sets in $ \mathbb{R}^{2}$ may be generated by countable unions of sets of the form $ \left(a,b\right)\times\left(c,d\right)$.

To prove $ \mathcal{B}\left(\mathbb{R}\right)\otimes\mathcal{B}\left(\mathbb{R}\right)\subset\mathcal{B}\left(\mathbb{R}^{2}\right)$, consider

\[ \mathcal{C} \equal{} \left\{\textrm{E}\in\mathcal{B}\left(\mathbb{R}\right) \ : \ \pi_{\alpha}^{\minus{}1}\left(\textrm{E}\right)\in\mathcal{B}\left(\mathbb{R}\right)\otimes\mathcal{B}\left(\mathbb{R}\right)\right\} ,\]
where $ \pi_{\alpha}$ is the projection of $ \left(x,y\right)$ onto either the first or second coordinate for $ \alpha \equal{}1$ and $ \alpha \equal{}2$, respectively.  You can show that $ \mathcal{B}\left(\mathbb{R}\right)\otimes\mathcal{B}\left(\mathbb{R}\right)$ is generated by $ \pi_{\alpha}^{\minus{}1}\left(\textrm{E}\right)$ for $ \alpha \equal{}1,2$ and $ \textrm{E}\in\mathcal{B}\left(\textrm{E}\right)$.  Then, show that $ \mathcal{C} \equal{} \mathcal{B}\left(\mathbb{R}\right)$.


I know this is an old problem but i'm looking for the solution

let $ a,b,c$ be real numbers , prove that

\[ (a^2\plus{}b^2\plus{}c^2)^2 \geq 3(a^3b\plus{}b^3c\plus{}c^3a)\]

[quote="shahriarta"]I know this is an old problem but i'm looking for the solution

let $ a,b,c$ be real numbers , prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)^2 \geq 3(a^3b \plus{} b^3c \plus{} c^3a)\]
[/quote]
$ (a^2 \plus{} b^2 \plus{} c^2)^2 \geq 3(a^3b \plus{} b^3c \plus{} c^3a) \Leftrightarrow \frac{1}{2}\sum\limits_{cyc} {(a^2  \minus{} b^2  \minus{} ab \minus{} ac \plus{} 2bc)^2 }  \ge 0$

thanks,
but i'm really interested in more solutions ,

It's all here
http://www.mathlinks.ro/Forum/viewtopic.php?t=6026

[quote="shahriarta"]I know this is an old problem but i'm looking for the solution

let $ a,b,c$ be real numbers , prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)^2 \geq 3(a^3b \plus{} b^3c \plus{} c^3a)\]
[/quote]

why don't use the fact that $ (a_1\plus{}a_2\plus{}...\plus{}a_n)(b_1\plus{}b_2\plus{}...\plus{}b_n) \geq n(a_1b_1\plus{}a_2b_2\plus{}...\plus{}a_nb_n)$
then $ (a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)\geq 3(a^3b\plus{}b^3c\plus{}c^3a)$
and then use the fact that $ a^2\plus{}b^2\plus{}c^2 \geq ab\plus{}bc\plus{}ca$
(u can easily prove it using AM-GM)
THEN it is done.

[quote="voong"]

why don't use the fact that $ (a_1 \plus{} a_2 \plus{} ... \plus{} a_n)(b_1 \plus{} b_2 \plus{} ... \plus{} b_n) \geq n(a_1b_1 \plus{} a_2b_2 \plus{} ... \plus{} a_nb_n)$
then $ (a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)\geq 3(a^3b \plus{} b^3c \plus{} c^3a)$
[/quote]

because this is only valid if the arrays are oppositely sorted. You can only get $ (a^2 \plus{} b^2 \plus{} c^2)(bc \plus{} ca \plus{} ab)\geq 3 (a^2bc \plus{} b^2ac \plus{} c^2ab)$ by this.  :!:


geometry
3D geometry
topology
real analysis
real analysis unsolved

Let $ (R^{n},d_{2})$ be the metric space of real n-tuples under the metric $ d_{2}(x,y)=(\sum_{i=1}^{n}|x_{i}-y_{i}|^{2})^{1/2}$.
Let $ Y$ be a subset of $ R^{n}$. A $ \epsilon-net$ of $ Y$ is a set of points $ S_{\epsilon}$ such that for every $ y\in Y$ there is at least one $ s\in S_{\epsilon}$ such that $ d_{2}(y,s)< \epsilon$
We say $ Y$ is totally bounded if for all $ \epsilon$ there is a finite $ S_{\epsilon}$.

Let $ Y={y\in R^{n}: \sum_{i=1}^{n}y_{i}^{2}\leq 1}$
Prove $ Y$ is totally bounded.

There is something wrong with your definition of $ Y$.
Every bounded subset of $ \mathbb R^{n}$ is totally bounded. Indeed, $ T_{\epsilon}=(\epsilon/n)\mathbb Z^{n}$ is an $ \epsilon$-net in $ \mathbb R^{n}$. A bounded set $ Y$ is contained in a cube, and the intersection of $ T_{\epsilon}$ with a slightly larger cube is a finite $ \epsilon$-net for $ Y$.

oops.
i already edited it.

anyway, i realized that it is not hard at all.


function
inequalities
algebra
polynomial
real analysis
real analysis unsolved

Let $K=[-\pi,\pi]$,$\alpha \in K$ be fixed.Suppose $f$ and $g$ are continious functions on $K$, differentiable at a fixed $\alpha \in K$, and suppose $f$ and $g$ have the properties
(1) $||f||_K=1,||g||_K<1$.
(2) there are $-\pi\leq x_1 < x_2 < ... < x_{2n}<\pi$ so that $f(x_j)=(-1)^j$.
(3) $f(\alpha)=g(\alpha),|g'(\alpha)|>|f'(\alpha)|$, and $sign(g'(\alpha))=sign(f'(\alpha))$.
Prove that then $f-g$ has at least $2n+2$ distinct zeros on $K$

For what theorem is this lemma used? I knew it in some special cases (which can be adapted to solve the more general case, I hope  :D ) as oral problem for X, having as purpose probably the easiest proof of Bernstein inequality.

Yes,this lemma is used for proving the Bernstein-Szego inequality $t'(\alpha)^2+n^2t^2(\alpha)\leq n^2||t||^2_{K}$, $\alpha \in K$, ($t$-trigonometric polynomial) ,from where the Bernstein inequality follows.

Are you sure that $2n+2$ isn't actually $2n+1$, eugene? I think I can easily draw examples with $2n+1$ zeros. :?

Actually $2n+1$ if sufficient to prove the main result based on this lemma, but what is confusing is that in the book it is stated exactly with $2n+2$
If you are interested in the book it is "Polynomials and polynomial inequalities" by Peter Borwein and Tomas Erdelyi.
Can you post you proof for $2n+1$,grobber?

If $\|\cdot\|_K$ is what I think it is, namely the maximum value of $|\cdot|$ on $K$, then I don't think it works for $2n+1$ either. Again, I can draw two such functions with $f-g$ having only $2n$ zeros. The graph, however, is getting too complicated for me to have the patience to actually construct the functions. Sorry, and I hope I'm wrong :).


What is the probability that the last three digits of a randomly selected phone number are all prime?  Express your answer as a common fraction.

There are four prime numbers from 1-10; 2,3,5 and 7. THus the probability that all three are prime is (2/5)^3, or 8/125

[hide]there's 4 prime numbers between 0-9: 2, 3, 5, 7... so it's gonna be 4/10 or 2/5... then you do (2/5)^3, which is 8/125[/hide]

[hide]1,2,3,4,5,6,7,8,9,0
2,3,5, and 7 are prime
p(unit digit is prime)=2/5
p(ten digit is prime)=2/5
p(hundreth digit is prime)=2/5
2/5*2/5*2/5=8/125
Since the probabilities are independent.

8/125  :rotfl:[/hide]


Let a,b,c are positive number. Prove that:
$\frac{a^{k+1}}{b}+\frac{b^{k+1}}{c}+\frac{c^{k+1}}{a}\ge \sqrt{3(a^{2k}+b^{2k}+c^{2k}})$
With k =1,2,3.
Is the inequality still true if k=4?


The number 100 can be written as the sum of a 1-digit prime number and a 2-digit prime number. What is the product of these prime numbers?

[hide="solution"]
In order for the numbers to be two digit and one digit, one of the numbers must be in the ninety's. The only prime number in the ninety's is $ 97,$ and its accompanying one digit prime number is $ 3.$ The product is \[ 3\times97\equal{}\boxed{291}.\][/hide]


algebra
binomial theorem
algebra unsolved

ive got this problem in math, it goes like this:

"(1+ax)^n = 1 + 30x + 375x^2...
- find the values of the two constants a and n "

How do you solve this problem???

I tried to do it.
this is what i did, i first compared the binomial at the top with the binomial theorem

(a+x)^n = a^n + (n nCr 1)*a^(n-1)*x + (n nCr 2)*a^(n-2)*x^2

comparing this to my binomial problem, the binomial at the top will look like this:

(1+ax)^n = 1 + (n nCr 1) * 1^(n-1)*ax + (n nCr 2) * 1^(n-2)*a^2*x^2

This is what I know:

1^(whatever) = 1

(n nCr 1) = n

so from this i can find that: (n nCr 1) * 1 * ax = 30x   <=> n * a = 30  (this i did by combining the problem i wrote at the top with the problem i made in the binomial theorem form)

ok, so, so far so good.
Now I need one more expression, so i can solve the two expressions simultaneously to find n and a.

Here comes the tricky part

I can of course find that    (n nCr 2) * a^2 * x^2= 375x^2 <=> (n nCr 2) * a^2 = 375

So with what I have done I end up with these two equations myself:

n*a = 30.
(n nCr 2) * a^2 = 375

any possible way to progress from this, or should the problem at the top be solved in a completely different way?

It is easy $na=30,n(n-1)a^{2}=750\Longrightarrow (n-1)a=25$, therefore a=5,n=6.

so you are basically saying that (n nCr 2) = n(n - 1) ??

can you please explain why its like this??

$^{n}C_{2}=\frac{n!}{(n-2)! \cdot 2!}=\frac{n(n-1)}{2}$ by very virtue of the formula!

He multiplied up the 2- that's why the 375 is replaced by 750.


logarithms
real analysis
real analysis unsolved
Miklos Schweitzer
inequalities

Let $ a_1,a_2,...,a_N$ be positive real numbers whose sum equals $ 1$. For a natural number $ i$, let $ n_i$ denote the number
of $ a_k$ for which $ 2^{1-i} \geq a_k \geq 2^{-i}$ holds. Prove that
\[  \sum_{i=1}^{\infty} \sqrt{n_i2^{-i}} \leq 4+\sqrt{\log_2 N}.\]

[i]L. Leinder[/i]

We have $ \sum_{i=1}^N a_i= 1 $ and $ \sum_{i=1}^{\infty} n_i = N $
Observe that $ 1 = \sum_{i=1}^{N} a_i \geq \sum_{i=1}^{\infty} \left( \mbox{ Number of } a_k \mbox{ such that } 2^{1-i} \leq a_k \leq 2^{-i} \right) \cdot 2^{-i} $
$ \Rightarrow 1 = \sum_{i=1}^N a_i \geq \sum_{i=1}^{\infty} n_i 2^{-i} $
Now $ \sum_{i=1}^{\infty} \sqrt{ n_i 2^{-i} } = \sum_{i=1}^{\lfloor \log_2{N} \rfloor } \sqrt{ n_i2^{-i} } + \sum_{i= \lfloor \log_2{N} \rfloor + 1}^{\infty} \sqrt{n_i 2^{-i} } $
By Cauchy Schwarz Inequality we have
$$ \sum_{i=1}^{\infty} \sqrt{ n_i 2^{-i} } \leq \left( \sum_{i=1}^{\lfloor \log_2{N} \rfloor } 1 \right)^{\frac{1}{2}} \left( \sum_{i=1}^{\lfloor \log_2{N} \rfloor } n_i2^{-i} \right)^{\frac{1}{2} } + \left( \sum_{i= \lfloor \log_2{N} \rfloor + 1}^{\infty}  n_i \right)^{\frac{1}{2} } \left( \sum_{i= \lfloor \log_2{N} \rfloor + 1}^{\infty}  2^{-i} \right)^{\frac{1}{2}} $$
$$ \Rightarrow \sum_{i=1}^{\infty} \sqrt{ n_i 2^{-i} } \leq \sqrt{ \lfloor \log_2{N} \rfloor} \left( \sum_{i=1}^{\infty} n_i 2^{-i} \right)^{\frac{1}{2}} + \left( \sum_{i=1}^{\infty} n_i \right)^{\frac{1}{2}} \left( 2^{ -\frac{ \lfloor \log_2{N} \rfloor }{2} } \right)  $$
$$ \Rightarrow \sum_{i=1}^{\infty} \sqrt{ n_i 2^{-i} } \leq \sqrt{ \log_2{N} } \cdot 1 + \sqrt{N} \left( 2^{ - \frac{ \lfloor \log_2{N} \rfloor }{2} } \right) $$
$$ \Rightarrow \sum_{i=1}^{\infty} \sqrt{ n_i 2^{-i} } \leq \sqrt{ \log_2{N} } + \sqrt{N} \left( 2^{ - \frac{ \log_2{N} - 1 }{2} } \right) $$
$$ \Rightarrow \sum_{i=1}^{\infty} \sqrt{ n_i 2^{-i} } \leq \sqrt{ \log_2{N} } + \sqrt{2} < \sqrt{ \log_2{N} } + 4 $$


How do we do a longdivision sign in latex? I tried combining lciel and overline, but it didn't look too good...

It depends what you are dividing. The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=longdiv]longdiv package[/url] will show long division of integers (using \include(longdiv)) and the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=polynom]polynom package[/url] (using \usepackage{polynom}) will show long division of polynomials. In both cases they will also carry out the calculations. With polynom you can tell it how many steps you want it to show (see the demo pdf included).

You could also play around with tables. A simple example is [code]12\,\begin{tabular}{|l}
\hline \hspace{-0.5em}
12345
\end{tabular}[/code]which gives $12\,\begin{tabular}{|l}\hline \hspace{-0.5em}12345 \end{tabular}$
or better still
[code]\setlength{\tabcolsep}{1pt}
12\,\begin{tabular}{|lllll}
\hline
1 & 2 & 3 & 4 & 5 \\
\end{tabular}[/code] which gives $\setlength{\tabcolsep}{1pt}12\,\begin{tabular}{|lllll}\hline 1 & 2 & 3 & 4 & 5 \\ \end{tabular}$


conics ellipse parabola hyperbola geometry ratio

[quote="probability1.01"]Hmmm, I think you need to give us some variables to work with.  Maybe tell us the initial position and velocity?  And also, I don't think there is a hyperbola case...  Anti-matter?[/quote]
Ok, let's say the equations are:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, and: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, and: $y=Ax^2$

I can understand Kepler's first law about the Ellipse motion, but i can't find examples and proofs of the other two...

[quote="probability1.01"]A parabola is possible if a body is moving fast enough.[/quote]
what is the lower bound of the velocity??
I asked my teacher about the Kepler's Law, but he only told me Kepler's is suitable for the sun, but if we consider those other than sun, we may recieve different results......
Can you give me an example of parabola motion??

Ooh... Haven't done any physics since I left high school. Not going to do anything substantial till college. So forgive my ignorance. I can't remember what Kepler's is.

Is it the one where you consider the angular momentum and derive something from it? Something about sweeping out the same area per time?

Certainly, you can find the energy and hence speed by considering conservation of energy. Then for velocity (or the direction rather) you can consider angular momentum. Or maybe you can just find the equation with a bit of calculus.

Actually, there are loads of calculation....
These problems are more like combinations of physics and conics. Maybe you feel bored when try to get the answer, but the conclutions are very useful for hard Olympiad problems. (which i shall post later)

Heh. Actually, I don't have internet access at home, so I'm usually stuck at my aunt's place when I post. And somehow I don't feel up to doing calculations here. No convenient table.

And I've forgotten most of my physics after more than two years... But since I'm planning to do some physics in college, I guess my memory needs a bit of refreshing! I must remember to bring paper here next time.

Um... What's Kepler's law again? Is it the one where the object sweeps out the same area per unit time?

Kepler has 3 laws.  One is the sweeping out area, one is the ratio between "radius" and period, and the last if I remember correctly is the fact that all orbits are ellipses.

Kepler 1: All motion around sun is Ellipse.

Kepler 2: $\frac{T^2}{a^3}=k$ where $k$ is a constant.

Kepler 3: Area equal. Look at my post of momentum in this forum.

[quote="shobber"]Subject: Planetary Motion
(Notice this one do not have a challenge icon. This is the warming up.)

Problem: If $M$ is a fixed planet and $m$ is moving under the gravitition.
(1) Ellipse case: Find the max and min of $m$ and the total energy of $m$ at the points: $(a,0), (0,b)$.
(2) Parabola case: Find the energy, velocity of $m$ at the vertex of the parabola.
(3) Hyperbola case: Find the Energy of $m$ and the velocity at the vertex and infinity.

I apologize for my poor English....... The problems must be very hard to understand, so just tell me where my mistake is!  :blush:  :P[/quote]

shobber,

The results for all three cases depend upon the Virial Theorem:

$2\langle T \rangle + \langle V \rangle = 0$

[quote="shobber"]Dr. No, i don't know what Virial Theorem is...

Ok, let's get back to the problems.

For 1, we have two equations:
\[v_1(a-c)=v_2(a+c)\]
(angular momentum)

\[\frac{1}{2} m v_1^2 - G\frac{mM}{a-c}= \frac{1}{2} - G\frac{mM}{a+c}\]

(Conservation of energy)

I'd like to see how solve these two equations...[/quote]

The Virial Theorem is as I stated in my previous post:

$2\langle \text{kinetic energy}\rangle + \langle \text{potential energy}\rangle =0$

The angle brackets mean the average over time.

The implication is that ellipses (circles are a special case) have total energy less than zero. (They are bound orbits.)

Parabolas have total energy equal to zero. (These are cometary "orbits.")

Hyperbolas have total energy greater than zero. (These are also cometary orbits.)

Go forth and solve, my friend, go forth and solve.


videos
analytic geometry
graphing lines
slope
parameterization

Hi, I am in desperate need of help. I have come across a dilemma which I am unable to solve.

I'm doing an experiment to see if there is a correlation between # of hours of video games played per week and spatial ability. Whether or not you agree or like my experiment is all right with me, but my real question is how to measure correlation.

Obviously using any two sets of random data you could get a correlation, but the real question is what kind of correlation is [i]enough[/i] to be significant. For example, if I got results that showed that each extra hour of video games increased spatial ability by an extra 3 correct answers, would that be [i]significant[/i] enough to show real correlation?

I guess the way to determine significance is based off the chances of this occurance happening by [i]chance[/i]. But, how much [i]significantly[/i] less than 50% (of this occurance happening by chance) does the data have to be for it to be significant? As you see, the problem remains.

Thanks in advance  :)

It sounds like you are trying to figure out how to interpret the data before you have recorded it. Have you collected this data? It seems logical that once you have a good sample of data, the data should answer your question for you (or it may not). You just have to wait and see.

For instance, if you see a linear relationship between hours and ability, then there you go.

However, there are other more important questions that should arise, and others SHOULD question about your study. Such as, did you collect enough data? Was it reliable? Was it from a random sample (as opposed to biased)? Was the test an adequate test of spacial ability? Etc.

I do not know how old you are, or how serious you are trying to make this, so what I've said is almost blind. Is this just for fun? Are you looking to publish the result? Some more information would be very helpful.

A personal comment: I think it is an interesting question and I'm curious as to what the correlation is :).

[quote="Spoon"]It sounds like you are trying to figure out how to interpret the data before you have recorded it. Have you collected this data? It seems logical that once you have a good sample of data, the data should answer your question for you (or it may not). You just have to wait and see.

For instance, if you see a linear relationship between hours and ability, then there you go.

I do not know how old you are, or how serious you are trying to make this, so what I've said is almost blind. Is this just for fun? Are you looking to publish the result? Some more information would be very helpful.[/quote]

My question though is, what kind of a slope is needed for the (for example) linear relationship you refer to is [i]enough[/i] of a slope to show correlation?

No publication, but I want to make this as valid is possible.

That depends on any number of error calculations... What is the range, and what is the standard deviation? What is the relative error? What is the chi^2? Then, it's a big matter of interpretation. This is where a lot of people may disagree.

Again, this isn't something you should worry about until after you've collected the data.

[quote="banblink182"]Yea, that's what I'm trying to figure out how to calculate, given a set of data: Relative error

What's chi^2?[/quote]

chi^2 (aka $\chi^2$, aka chi square) is a test of statistical significance.  It's not really hard to do (especially with a calculator :D ) so you could probably just post your data here and someone could compute it and interpret it for you.

[quote="banblink182"]I'm doing an experiment to see if there is a correlation between # of hours of video games played per week and spatial ability. (...) For example, if I got results that showed that each extra hour of video games increased spatial ability by an extra 3 correct answers, would that be [i]significant[/i] enough to show real correlation? [/quote]

There were several recent studies of this that got some publicity. One was with US Army soldiers playing battle games.  The general conclusion was that, surprisingly to some, the games improved visualization and more general cognitive performance.  There is also a recent book with title something like "Everything Bad is Good for You", that cites video games as one of the examples.

What is meaningful correlation depends on how strong the alternative explanations are for whatever you see.  People with more spatial ability may be more inclined to play video games (or more spatially demanding video games), for example.  Men play more videogames than women and tend to be better at spatial visualization.  It is probably tied to athletic ability as well (hand-eye coordination), which is also gender-skewed.

Also, it is almost certainly wrong to interpret spatial ability as simply proportional to  whatever parameter the study measures, such as number of correct answers on a test. A reliable interpretation would have to give some thought to the distribution of spatial ability in its own right and what is the correct measuring scale, or correct renormalization of whatever plays the role of the scale for the experiment.

So the construction of the experiment and common-sense questions about proposed interpretations are likely to be more important than Chi-square tests.

[quote="fleeting_guest"]
Also, it is almost certainly wrong to interpret spatial ability as simply proportional to whatever parameter the study measures, such as number of correct answers on a test. A reliable interpretation would have to give some thought to the distribution of spatial ability in its own right and what is the correct measuring scale, or correct renormalization of whatever plays the role of the scale for the experiment.[/quote]

Yes, I agree with everything that you've said, but I'm still not really understanding the procedure.

Let's say that I figure out the distribution of spatial ability in the general populace. Then I conduct this "experiment" and find that like 80% of those who play video games excessively (like greater than 35 hours a week or something, let's say) have an spatial ability that only 1/250 people have. Would that be significant?

If you are quick to say yes, then why? If your answer is, "I cannot make a conclusion because you have not given me enough information", then how [i]should[/i] I proceed if I [i]had[/i] more information?


probability
LaTeX
complementary counting
special factorizations
combinatorics unsolved
combinatorics

In a drawer, there are at most $ 2009$ balls, some of them are white, the rest are blue, which are randomly distributed. If two balls were taken at the same time, then the probability that the balls are both blue or both white is $ \frac12$. Determine the maximum amount of white balls in the drawer, such that the probability statement is true?

[hide="Solution"]Let there are $ a$ white balls and $ b$ blue balls. WLOG, $ a > b$.
The probability to get both of them white or blue, is:
\[ \frac {{a\choose 2} \plus{} {b\choose 2}}{{a \plus{} b\choose 2}} \equal{} \frac12\]

\[ \frac {a(a \minus{} 1) \plus{} b(b \minus{} 1)}{(a \plus{} b)(a \plus{} b \minus{} 1)} \equal{} \frac12\]

\[ 2a^2 \minus{} 2a \plus{} 2b^2 \minus{} 2b \equal{} a^2 \plus{} 2ab \plus{} b^2 \minus{} a \minus{} b\]

\[ a^2 \plus{} b^2 \minus{} 2ab \equal{} a \plus{} b\]

\[ (a \minus{} b)^2 \equal{} a \plus{} b\leq 2009\]
Let $ (a \minus{} b)^2 \equal{} a \plus{} b \equal{} k^2\leq 2009$, for a positive integer $ k$. Then, we get:
$ a \minus{} b \equal{} k$, $ a \plus{} b \equal{} k^2$.
Which leaves us to $ a \equal{} \frac {k^2 \plus{} k}{2}$. Since we wanted to maximize $ a$, then we must maximize $ k$. The highest possible value of $ k$ such that $ k^2\leq 2009$ is $ k \equal{} 44$.
Hence, the answer is:
\[ a \equal{} \frac {k^2 \plus{} k}{2} \equal{} \frac {44^2 \plus{} 4}{2} \equal{} \frac {1980}{2} \equal{} 990\]
[/hide]

Sorry for reviving, but there's a slightly more elegant method with complementary counting.

[hide]Let there be x white balls and n total balls. Complementary counting gives us

x (n - x) / (n (n - 1) / 2) = 1 / 2.

Rewriting, we obtain

4x^2 - 4nx = n - n^2.

Completing the square, we obtain

(2x - n)^2 = n,

so

x = (n +- sqrt n) / 2.

Since x is an integer, we must have perfect square n. Also, since x is being maximized, we take the positive square root and not the negative. The greatest perfect square less than 2009 is 1936 = 44^2, giving

x = (1936 + 44) / 2 = 990.[/hide]

How is it possible that you have more than 1000 posts and don't use LaTeX?!

Most people figure out that it's rude to write in a difficult-to-read way when a better alternative is available within say 50 or 100 posts.  It's remarkable to find someone who has spent so much time here and yet refuses to use the local dialect, if you will.

I don't see how it's rude since the solution is a favor. Besides, it takes too long to learn.

"too long to learn" is not true. It takes about 10 minutes to learn how to use LaTeX.

[color=#008000]Moderator says: this is right: if the user had devoted 10 seconds for every post he has made, he would be an accomplished LaTeX user by now.  (I mean, the instructions run to [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:LaTeX_on_AoPS]half a page[/url]!)  Not to mention the fact that he's already using mostly correct notation already ....[/color]


Prove that $ 641|2^{32}\plus{}1$. No calculator allowed ;p

You don't really need a calculator to figure out that $ 2^{32} \equal{} 4294967296$ and then do long division. This is basically how Euler did it, and moreover, he needed to find the $ 641$ in the first place  :) Of course he first proved that any prime divisor must be congruent to $ 1$ mod $ 64$.

A well-known shortcut is to observe that $ 641 \equal{} 5^4 \plus{} 2^4$ obviously divides $ A \equal{} 5^4\cdot 2^{28} \plus{} 2^{32}$, and also $ 641 \equal{} 5\cdot 2^7 \plus{} 1$ obviously divides $ B \equal{} (5 \cdot 2^7)^4 \minus{} 1 \equal{} 5^4 \cdot 2^{28} \minus{} 1$, so $ 641 | (A \minus{} B) \equal{} 2^{32} \plus{} 1$.


[hide]The radius at a height of 9 is 3.  $V=\frac{1}{3}\pi 3^{2}\cdot 9=\boxed{27\pi}$[/hide]

[quote="mr. math"]A conical tank with height 12 and radius 4 is filled with liquid. How much liquid has poured from the tip of the cone if the water level is 9 feet from the tip?[/quote]
[hide]Use similar triangles to set up a proportion and find that the radius of the cone of water is 3.

$V=\frac{3^{2}\cdot9\cdot\pi}{3}=\boxed{27\pi}$[/hide]

Wait, isnt 27pi, the amount of water left in the cone, not the amoujnt that has been poured out


I wonder if there's an interesting non-induction way without using difference equations .....

You can use the formula for the sum of third powers to find $S=1^3+2^2+\cdots+(2n)^3.$  Then break that into $S=(1^3+3^3+\cdots+(2n-1)^3)+(2^3+4^3+\cdots+(2n)^3)=(1^3+3^3+\cdots+(2n-1)^3)+8(1^3+2^3+\cdots+n^3).$

As you can see, you would be able to solve for $1^3+3^3+\cdots+(2n-1)^3.$


[quote="le00327146"]Is there any particular order in which papers are graded? And are all the scores released at once, or do they grade everyone's solutions, then release them all?[/quote]
The scores are released all at once, not in the order they are graded.

[quote="le00327146"]Is there any particular order in which papers are graded? And are all the scores released at once, or do they grade everyone's solutions, then release them all?[/quote]
The tests are graded problem by problem in two or three separate locations. Then they are shipped back to the Art of Problem Solving for summing and recording the scores.

We like to have the same team grade all submissions for the same problem for consistency. Grading all problems for Round 1 with 25 graders would take 8 solid hours, which is more than we can ask from a volunteer. So we rely on more than one group of volunteer graders.

The grading session that I will participate in is scheduled November 2. If the grading groups are the same as last year, we next ship the tests to Cambridge, Massachusetts, for another grading session. So you will have to wait a few weeks for your score.

Erin Schram
USAMTS grader


trigonometry
algebra
polynomial
complex numbers

Given $ \cos \theta \equal{} \dfrac 23$, is there a fast way to evaluate \[ \sum_{n\equal{}1}^\infty \dfrac{\cos(n \theta)}{2^n}?\]

I think you can do this using Chebychev polynomials and solving the linear recurrence, but is there a way to use $ \cos(n\theta) \equal{} \Re(e^{i\theta})$ for a faster solution?

You're on the right track, substituting $ \cos(n\theta) \equal{} \Re \left(e^{in\theta}\right)$ and using the fact that $ \Re(a) \plus{} \Re(b) \equal{} \Re(a \plus{} b)$ for all $ a,b\in \mathbb{C}$ yields,
\[ \sum_{n \equal{} 1}^\infty \frac {\cos (n\theta)}{2^n} \equal{} \Re \left\lbrace \sum_{n \equal{} 1}^\infty\left( \frac {e^{i\theta}}{2}\right)^n\right\rbrace\]
The right hand side is just the real part of a (complex valued) geometric series.  Though it's complex, it can be evaluated in the usual way(*), so the final answer is $ \Re \left\lbrace \frac {e^{i\theta}}{2 \minus{} e^{i\theta}}\right\rbrace \equal{} \frac {1}{5 \minus{} 4\cos \theta}\Re \left\lbrace 2e^{i\theta} \minus{} 1\right\rbrace \equal{} \boxed{\frac {2\cos \theta \minus{} 1}{5 \minus{} 4\cos \theta}}$.

(*)More generally, it is true that $ \sum_{n \equal{} 1}^\infty a^n \equal{} \frac {a}{1 \minus{} a}$ for all $ a\in \mathbb{C}$ satisfying $ \left\vert a\right \vert < 1$, and so $ \Re \left\lbrace\sum_{n \equal{} 1}^\infty a^n\right \rbrace \equal{} \frac {\Re (a) \minus{} |a|^2}{|1 \minus{} a|^2}$.  In our case $ a \equal{} \frac {e^{i\theta}}{2}$ and $ |a| \equal{} \frac {1}{2}$ for any $ \theta$.


Let $ M \equal{} n(n \minus{} 1)...(n \minus{} k \plus{} 1)$ where $ n\ge 2k$. Prove that there exist a prime number $ p > k$ such that $ p|M$.

This is a result due to Sylvester (and proved again by Erdös when he was young), but the proof it really too long to appear here.

Pierre.

The proof for this is in [i]Aigner, Ziegler "Proofs from the Book"[/i], you may find it there if you're interested.


[quote="Hong Quy"]$ 2^{81} < 5^{35}$
$ 2^9 < 5^4.41/10$.
hence : $ 2^{81} < 5^{18}41^9/10^9$.
 another: $ 41^3 < 5^6.9/2.$
   we have complete solution/[/quote]

Could you explain a little bit more your demo ?

[quote="Hong Quy"]$ 2^{81} < 5^{35}$[/quote]
is what we want to show

[quote="Hong Quy"]$ 2^9 < 5^4.41/10$.[/quote]
seems obviously true but useless (according to me) : $ 512<625\times4.1$

[quote="Hong Quy"]hence : $ 2^{81} < 5^{18}41^9/10^9$.[/quote]
I don't understand the "hence". You seem to say this statement is a consequence of the previous one. But a consequence, according to me, would be :

$ 2^{81} < 5^{36}41^9/10^9$ and not $ 2^{81} < 5^{18}41^9/10^9$

...
So I would be interested in some explanations.

[quote="Hong Quy"]prove that : $ \pi^e > e^{\pi}$[/quote]

You're wrong :

The function $ f(x)\equal{}\frac{\ln(x)}{x}$, for $ x>1$ has a maximum for $ x\equal{}e$ $ \left(f'(x)\equal{}\frac{1\minus{}\ln(x)}{x^2}\right)$.

So $ f(x)<f(e)$ $ \forall x>1,x\neq e$

So $ \frac{\ln(x)}{x}<\frac{\ln(e)}{e}$ $ \forall x>1,x\neq e$

So $ e\ln(x)<x\ln(e)$ $ \forall x>1,x\neq e$

So $ x^e<e^x$ $ \forall x>1,x\neq e$

So $ \pi^e<e^{\pi}$


geometry
circumcircle
vector
geometric transformation
rotation
symmetry
geometry unsolved

Let $ABC$ be an acute-angled triangle with orthocenter $H$. Points $A_1,B_1,C_1$ are chosen on the circumcircles of $BCH,CAH,ABH$, respectively, in such a way that none of these points coincides with a vertex of $ABC$.  Let $H_1,H_2,H_3$ be the orthocenters of $BCA_1,CAB_1,ABC_1$, respectively. Prove that the orthocenters of $H_1H_2H_3$ and $A_1B_1C_1$ coincide.

The first, easy observation is that the points $H_1$, $H_2$ and $H_3$ lie on the circumcircle of triangle ABC, but I don't know how to go further. (Maybe, it is better to use vectors and their scalar products.)

  Darij

I wonder what I'm doing wrong. I drew something on my computer but I find this to be untrue. I know there's something I'm missing, but I have no idea what it is. This is so frustrating :(...

I also get a contradiction.
Let S be the circumcircle of ABC. The circumcenters of BHC etc are symmetrics of S with respect to BC etc. Let their centers be O1,O2,O3=O's symmetrics with respect to ABC's sides.
How we pass to vectors. We use the identity OH=OA+OB+OC for O,H circumcenter, orthocenter of ABC.
Then O1H1=O1B+O1C+O1A1 and analogous.
SO OH1=OO1+O1H1=OB+OC+OA1 and analogous.
Now H1,H2,H3 lie on S. So if H' is the orthocenter of H1H2H3 the OH'=OH1+..
2(OB+OB+OA)+OA1+OB1+OC1=2OH+OA1+OB1+OC1(1)
Now let O' be the circumcenter of A1A2A3. The if H' is its orthocenter we get
O'H'=OA1+OB1+OC1.
So OH'=OO'+OB1+OC1+OA1.
Comparing with (1) we get O' is a fixed point.
But this is impossible!

Now I have also made a dynamic sketch and the theorem doesn't hold there, too. There obviously must be some typo or error in the formulation.

  Darij

Only A1  is chosen randomly.  B1 and C1 are on corresponding circumcircles and are
   collinear with A1,C    and  A1,B  .  Now the orthocenters of H1H2H3 and A1A2A3  coincide.

   Maj.  Pestich

Now it's mostly an angle chase :), so I'll just sketch it.

Let $H'$ be the orthocenter of $A_1B_1C_1$, and $O$ the circumcenter of $ABC$. We can show first that $OH=OH'$ (exercise :)). Then we compute the angle $\angle OHH'$, which must be equal to the angle made by $BC,B_1C_1$, in terms of $\angle A,\angle B,\angle C$ and, say $\angle A_1BC$, which allows us to find $\angle HOH'$  in terms of the same angles, because $OHH'$ is isosceles. After this, we compute $\angle AOH_1$ as being the double of $\angle ACH_1$ (or $\angle ABH_1$; it depends on the direction in which you have moved $A$ after starting from the midpoint of $B_1C_1$), which can, again, be computed in terms of those angles.

In the end, we find $\angle AOH_1=\angle HOH'$, and, since $OH=OH',OA=OH_1$, it means that the rotation centered at $O$ which maps $H$ onto $H'$ also maps $A$ onto $H_1$, and, due to symmetry, $B$ onto $H_2$, and $C$ onto $H_3$. $H_1H_2H_3$ is thus the image of $ABC$ through this rotation, meaning that its orthocenter must be the image of $H$. However, this image is $H'$, and we get the desired result.

[quote="grobber"]We can show first that $OH=OH'$ (exercise :)). [/quote]

How to do that??

[quote="vess"]Let $ ABC$ be an acute-angled triangle with orthocenter $ H$. Points $ A_1,B_1,C_1$ are chosen on the circumcircles of $ BCH,CAH,ABH$, respectively, in such a way that none of these points coincides with a vertex of $ ABC$.  Let $ H_1,H_2,H_3$ be the orthocenters of $ BCA_1,CAB_1,ABC_1$, respectively. Prove that the orthocenters of $ H_1H_2H_3$ and $ A_1B_1C_1$ coincide.[/quote]
Orthocenters of $ H_1H_2H_3$ and $ A_1B_1C_1$ coincide if and only if $ A_1B_1C_1$ is respectively similar to $ ABC$.