In how many ways can 8 people sit around a round table if 3 of the people -- Pierre, Rosa, and Thomas -- all want to sit together?

This is the same as arranging $ 6$ people around a table, and then multiplying by the $ 3!\equal{}6$ permutations of Pierre, Rosa, and Thomas, so our answer is $ \frac{6!}6\cdot6\equal{}\boxed{720}$.

Keep in mind that we can rotate the group of 3 anywhere on the circle after placing them. Now we only need to account for the arrangement of the Pierre, Rosa, and Thomas in that group of 3, which is just $3!$. Now multiply by the number of arrangements of the other 5 people ($5!$), yielding $720$ as the answer.

I get where we get $3!*5!$, but why don't we divide by $8$ to account for rotations. Can someone explain? Thanks!

$3!$ is the number of arrangements for the three that want to be together and $5!$ is the number arrangements the other five that sit at the table.

[hide]$3!\cdot5!=720$[/hide]

I get that, but aren't there 8 rotations you got to account for, since you can rotate the table 8 times?

You want the rotations in the total.

Edit: this is wrong oops

[quote=jkliu][quote=rubixsolver]You want the rotations in the total.[/quote]
But two rotations of the same seating are the same so wouldn't you divide?[/quote]

Yes, why????

I have no clue what I was thinking when I said that, but that is wrong (lol).

So suppose you treat Pierre, Rosa, and Thomas as a superhuman. Once we place the superhuman, we place the remaining people relative to the superhuman. So we get $5!$ placements. Since you can permutate the humans inside the superhuman $3!$ ways, the answer is $5!\times3!.$

[quote=MathNerd555]I'm pretty sure the idea here is that you fix the table around the "superhuman" block[/quote]

That's EXACTLY what I did! There are 3! for the "superhuman" block, and 5! ways for the rest of the people, so our answer is 3!*5!.

Suppose we have A, B, C, D sitting around a table should we not consider both clockwise and anticlockwise arrangments as rotations .

For example  arrangement ABCD shown below
              A
       D          B
             C
around a circle would also account for ADCB if we go anti clockwise from A but in circular counting this is considered a different arrangement. I am not able to get this.

For 4 people around a circle I get only

4!/ (4 * 2) = 3 unique placments
                A
         D         B
                C

                A
         C         B
                D

                A
         D         C
                B

All the other 21 permutations are rotations of these arrangments.

Based on this answer to the question should be 360 and not 720

Can someone please explain clockwise and counterclockwise are considered different arrangements when both are just rotations


trigonometry
geometry
geometric transformation
reflection
geometry proposed

I see that nobody tryed it, so I will post my solution.

Denote $\angle{MBC}=\angle{MDC}=\alpha$, $\angle{MCD}=\angle{MBA}=\beta$ and $\angle{BAC}=\angle{CAD}=t$.

Thus, $\frac{MC}{\sin{\alpha}}=\frac{BC}{\sin{BMC}}$, and $\frac{MC}{\sin{\alpha}}=\frac{CD}{\sin{(\alpha+\beta)}}$ by the sin law in $\triangle{MBC}$, and $\triangle{MCD}$.

So $\frac{BC}{CD}=\frac{\sin{BMC}}{\sin{(\alpha+\beta)}}$. $(1)$

Now from the sin law in $\triangle{ABC}$ and $\triangle{ADC}$, we have that:
$\frac{BC}{\sin{t}}=\frac{AC}{\sin{(\alpha+\beta)}}$ and $\frac{CD}{\sin{t}}=\frac{AC}{\sin{ADC}}$

So $\frac{BC}{CD}=\frac{\sin{ADC}}{\sin{(\alpha+\beta)}}$. $(2)$

By $(1)$ and $(2)$ we have that $\sin{ADC}=\sin{BMC}$, thus $\angle{ADC}=\angle{BMC}$, or $\angle{ADC}=180-\angle{BMC}$ etc.

[quote="pohoatza"]I see that nobody tryed it, so I will post my solution.

Denote $ \angle{MBC} \equal{} \angle{MDC} \equal{} \alpha$, $ \angle{MCD} \equal{} \angle{MBA} \equal{} \beta$ and $ \angle{BAC} \equal{} \angle{CAD} \equal{} t$.

Thus, $ \frac {MC}{\sin{\alpha}} \equal{} \frac {BC}{\sin{BMC}}$, and $ \frac {MC}{\sin{\alpha}} \equal{} \frac {CD}{\sin{(\alpha \plus{} \beta)}}$ by the sin law in $ \triangle{MBC}$, and $ \triangle{MCD}$.

So $ \frac {BC}{CD} \equal{} \frac {\sin{BMC}}{\sin{(\alpha \plus{} \beta)}}$. $ (1)$

Now from the sin law in $ \triangle{ABC}$ and $ \triangle{ADC}$, we have that:
$ \frac {BC}{\sin{t}} \equal{} \frac {AC}{\sin{(\alpha \plus{} \beta)}}$ and $ \frac {CD}{\sin{t}} \equal{} \frac {AC}{\sin{ADC}}$

So $ \frac {BC}{CD} \equal{} \frac {\sin{ADC}}{\sin{(\alpha \plus{} \beta)}}$. $ (2)$

By $ (1)$ and $ (2)$ we have that $ \sin{ADC} \equal{} \sin{BMC}$, thus $ \angle{ADC} \equal{} \angle{BMC}$, or $ \angle{ADC} \equal{} 180 \minus{} \angle{BMC}$ etc.[/quote]
you did'nt prove that A,M,C are collinear ???

If $A,M,C$ are collinear ,then we have easily $\angle{BMA}=\angle{CDA}$. Assume that $A,M,C$ aren't collinear. Let $N$ $K$ be reflections of $D$ and $B$ with $MC$. Then $NBMC$ and $KMCD$ are cyclic. Since  $\angle{MBA}=\angle{MCD}=\angle{MCN}$ We have: $N$ lies on the line $AB$. Let $E$ be intersection of $NB$,$DK$ and $CM$. Since $EM$ is bisector of $\angle{AED}$, $C$ is excircle center of $\triangle{EDA}$. (that touching to the side $AD$) hence $CD$ is external bisector of $\angle{ADE}$ Then: $\angle{BMC}=180^{\circ}-\angle{BNC}=180^{\circ}-\angle{CDK}=\angle{ADC}$. So we are done! (U.J.B.O.)

[quote=pohoatza]Let $ABCD$ be a convex quadrilateral, with $\angle BAC=\angle DAC$ and $M$ a point inside such that $\angle MBA=\angle MCD$ and $\angle MBC=\angle MDC$. Show that the angle $\angle ADC$ is equal to $\angle BMC$ or $\angle AMB$.[/quote]
Remark:
Actually this configuration same as Miquel's Theorem
Let points P,T,S are intersection of
MD,MC,MB with BC,AB,CD respectively
Using above facts ,it can be shown that
PBMT,TMDS,TSBC,TPCD,PBDS is cyclic
And this happen if and only if
T is Miquel Point of complete quadrilateral PBCDSM.

Denote $\angle BAC = \angle DAC = \alpha$, $\angle ABM = \angle MCD = x$ and $\angle MBC = \angle MDC = y$. By the Sine Law we have
\[ \frac{AC}{\sin\angle ADC} = \frac{CD}{\sin \alpha} = \frac{\frac{CM}{\sin y}\sin(x+y)}{\sin \alpha} = \frac{BC}{\sin \angle BMC}\frac{\sin \angle ABC}{\sin \angle BAC} = \frac{AC}{\sin \angle BMC} \]
thus $\sin \angle ADC = \sin \angle BMC$.

Since both angles have measures strictly between $0^{\circ}$ and $180^{\circ}$, we either have $\angle ADC = \angle BMC$ (and we are done), or $\angle ADC + \angle BMC = 180^{\circ}$.

In the latter case with $\angle BMC = z$, $\angle ADC = 180^{\circ} - z$ we obtain $\angle BCM = 180^{\circ} - y - z$ and now summing all angles in the quadrilateral $ABCD$ leads to $z = \alpha + x$. Now if $BM \cap AC = M_1$, then $\angle BM_1C = \alpha + x = z = \angle BMC$, so $M \equiv M_1$, i.e. $M$ lies on $AC$. Hence we conclude $\angle ADC = 180^{\circ} - \angle BMC = \angle AMB$, as desired.


algebra
polynomial
trigonometry
algebra proposed
Fixed point

Suppose that  $a,b,c$ are real numbers and let $f(x)=ax^{2}+bx+c .$
 [b] If $|f(10)| > 721$   show that there exists  at least a point $x_0$ in $[0,1]$   such that  $|f(x_0)| > 1$ . [/b]

I like the problem! can u post your solution? :)

[quote="flavian"]I like the problem! ...  your solution ? :)[/quote]
[color=green] [b] Suppose that $a,b,c$  are real numbers and let  $f(x)=ax^{2}+bx+c .$  If $|f(10)| > 721$  show that there exists
 at least  a point $x_0$ in $[0,1]$ such that $|f(x_0)| > 1 .$[/b] [/color]
[b]Solution.[/b]  Assume $| f(r) | \le 1$ for all real  (rational it's enough)  points $r$ from $[0,1] .$ Because
\[ f(x)=(2x-1)(x-1)f(0)-4x(x-1)f\left(\frac{1}{2}\right)+x(2x-1)f(1) ,  \]
for $x\in (-\infty,0)\cup (1,\infty)$ we find  $\displaystyle |f(x)|\le \displaystyle (2x-1)(x-1)+4x(x-1)+x(2x-1)=\displaystyle 8x^2-8x+1=: Q(x) .$
 Therefore   $|f(10)| \le Q(10)= 721$ [b] which is false ! [/b]  [size=75] Remark: observe that $|Q(t)|=|1-8t(1-t)|\le 1\; ,\; \forall t\in [0,1] .$ [/size]

[hide="Flip2004 and Flavian, see here !"]
$f\in R[X],\ gr f=p\in N,\ n>1\Longrightarrow$

$f=\frac{1}{p!}\prod\limits_{k=0}^p (pX-k)\cdot \sum\limits_{k=0}^p {p\choose k} (-1)^{p-k} f\left(\frac kp\right)\frac{1}{pX-k}$.

$\left( \forall \right) x\in [0,1],\ |{f(x)}|\le 1\Longrightarrow \left(\forall\right)x\in[0,1],\ |f(x)|\le g(x)$ where

$g=\frac{1}{p!}\prod\limits_{k=0}^p (pX-k)\cdot \sum\limits_{k=0}^p {p\choose k}\frac{1}{pX-k}$.

$721=6!+1$. But $10=\ ?$ a.s.o. $\Longrightarrow$ [b]Generalization ...[/b] [u]I wish you every success ![/u] [/hide]

[quote="Virgil Nicula"][hide=Flip2004 and Flavian, see here !] [/quote]
Thank you very much ! The generalization is the following :
[color=green][b] Let   $f(x)=\sum\limits_{k=0}^{n}a_kx^k$ ,  $a$ fixed in $(1,\infty),$  and
\[ \begin{array}{|c|} \hline \\ K_n(a)= \displaystyle \frac{\left(2a-1+2\sqrt{a(a-1)}\right)^n +\left(2a-1-2\sqrt{a(a-1)}\right)^n }{2} \\ \\ \hline \end{array}  \]
If we   assume  $\begin{array}{|c|} \hline \\ \displaystyle |f(a)|>\displaystyle K_n(a) \\ \\ \hline \end{array}\; ,\;$  then  there exists a point  $x_0$ in $[0,1]$  such that   $|f(x_0)| >1 \; .$   [/b][/color]
[size=75]
Remarks:
1)   $K_n(a) = T_n(2a-1)$ where $T_n(z)$ is Chebychev polynomial of first kind, defined for $|z| \le 1$ by  $T_n(z)=\cos\left(n\cdot\arccos{z}\right) .$
2) The constant $K_n(a)$ cannot be improved:  to see this it's sufficient to select \[ f(x): =T_n(2x-1)=\sum\limits_{k=0}^{n}(-1)^{n-k}\frac{n}{n+k} {n+k\choose 2k}2^{2k}x^k\; , \; n\ge 1, T_0(x): =1 .  \]
It's known that  $| T_n(2x-1) |\le 1 \; ,\; x\in{\mathbb R} ,$  iff   $x\in [0,1] \; .$
3)  $K_2(10) = \frac{(19+6\sqrt{10})^2+(19-6\sqrt{10})^2 }{2}=721\; .$
[/size]

Thanks for interest,Alex[=flip2004]

What is this Alex collection and why are the problems in it so difficult  :surrender:

[quote="t0rajir0u"]What is this Alex collection and why are the problems in it so difficult  ?[/quote]
I try to answer you by means of a private  e-mail message . / flip2004


[img]http://i706.photobucket.com/albums/ww68/jellymuffins2009/triangke.jpg[/img]

Angle <BDE = <BAC. So you can see that ED is parallel AC but then how do you know that DC = AE?

Thanks in advance!

I'm not quite sure if this should go here or not...  :huh:  if it doesn't can a mod please move it?

Is any other information given?

Otherwise, you can set $ B$ such that $ AE \perp AC$ and then clearly $ DC > AE$

The question I have comes from 5.18 in Intro to Geo.  It says, "Since <ABE = <ACD we know BE is parallel to CD.  Therefore the altituted from C and D to BE must be the same."

Would that make DC longer?

In $ \triangle{BDE}$ ,$ \angle BDE$=$ \angle BED$.
Hence, $ BD$=$ BE$.
Now, $ DE\|AC$
So $ \angle BDE$=$ \angle BAC$ and $ \angle BED$ =$ \angle ACB$.
Therefore, in $ \triangle ABC$, AB=BC.
AB-BD=BC-BE
or AD=CE.
In $ \triangle DAE$ and $ \triangle DCE$,
DE is common,
AD= CE
and $ \angle ADE$=$ \angle DEC$.
So $ \triangle DAE$  $ \cong$  $ \triangle DEC$.
So AE=DC.

oh, your diagram is labeled incorrectly then

[asy]draw((0,0)--(6,8)--(10,0)--cycle);
draw((3,4)--(8,4));
draw((3,4)--(10,0));
draw((8,4)--(0,0));
label("$A$",(6,8),N);
label("$B$",(3,4),W);
label("$C$",(0,0),SW);
label("$D$",(10,0),NE);
label("$E$",(8,4),E);[/asy]

What the book says is that the perpendicular from $ C$ to $ BE$ has the same length as the perpendicular from $ D$ to $ BE$

Because they are parallel, we know that the altitude to $ BE$ from both points $ C$ and $ D$ are equal since at any point on $ BE$, a point $ Q$ on $ CD$ is equidistant from $ BE$ in reference to a point $ P$ also on $ CD$ to $ BE$.  Because $ Q$ and $ P$ are arbitrary points, we can set them equal to $ C$ and $ D$ and our conclusion follows...much more wordier than it has to be.

[quote="muks"]In $ \triangle{BDE}$ ,$ \angle BDE$=$ \angle BED$.
[/quote]

How did you figure this out?..  I don't think it necessarily has to be..

@ 5849.. : Oh you're right!  My bad.. :blush:

@ Z-Coli : It's basically because they are parallel right? :P

Thanks everyone!


If $ a,b,c>0$ so that $ a\plus{}b\plus{}c\equal{}1$ then prove that:

\[ \frac{1\plus{}a}{1\minus{}a} \cdot \frac{1\plus{}b}{1\minus{}b} \cdot \frac{1\plus{}c}{1\minus{}c} \ge 8\]

[quote="moldovan"]If $ a,b,c > 0$ so that $ a \plus{} b \plus{} c \equal{} 1$ then prove that:
\[ \frac {1 \plus{} a}{1 \minus{} a} \cdot \frac {1 \plus{} b}{1 \minus{} b} \cdot \frac {1 \plus{} c}{1 \minus{} c} \ge 8\]
[/quote]

Take $ x \equal{} a \plus{} b$ and use $ (x \plus{} y)(y \plus{} z)(z \plus{} x) \ge 8xyz$

Very nice problem!
My solution:
The problem is equivalent to:
$ a,b,c>0,a+b+c=3\Rightarrow\prod\frac{3+a}{3-a}\ge 8\Leftrightarrow $
$ \Leftrightarrow\prod(1+\frac{2a}{b+c})\ge 8 $
With substitutions $ x=\frac{2a}{b+c},y=\frac{2b}{c+a},z=\frac{2c}{a+b} $ we obtain
1)
$ x,y,z>0,\sum\frac{1}{x+2}=1\Leftrightarrow xy+yz+zx+xyz=4 $
You must show that:
2)
${ x,y,z>0,xy+yz+zx+xyz=4\Rightarrow (1+x)(1+y}(1+z)\ge 8 $
After calculations, (2) is equivalent to:
3)
$ x,y,z>0,xy+yz+zx+xyz=4\Rightarrow x+y+z\ge 3 $
which is known! (or not!)
We conclude this proof! Remains as possibly to post proof of (3)!
____________
Sandu Marin


Substitute each of the capital letters in bold by a different digit from 1 to 9 to satisfy this cryptarithmetic equation.  It is known that [b]A[/b] > [b]D[/b] > [b]G[/b], [b]B [/b]> [b]C[/b], [b]E [/b]> [b]F[/b],  [b]H [/b]> [b]I[/b], and [b]C[/b] is neither 1 nor 5.

[b]A[/b]/([b]B[/b]+[b]C)[/b] + [b]D[/b]/([b]E[/b]+[b]F[/b]) + [b]G[/b]/([b]H[/b]+[b]I[/b]) = 1

[quote="U§ername"]so E could still be F and H could still be I[/quote]

Edit: I understand your query now, and I have edited the problem text to address the said concern.

[quote] Substitute each of the capital letters in bold by a different digit from 1 to 9 to satisfy this cryptarithmetic equation. It is known that A > D > G, B > C, E > F, H > I, and C is neither 1 nor 5.

A/(B+C) + D/(E+F) + G/(H+I) = 1 [/quote]

Well, I had almost solved the above problem, but I had committed an error somewhere which leads to a 'misplaced digit' (I apparently 'miss' the answer by a single digit).

Here's my near-complete solution.

[b] 8       +   2       +   1

    7+4         6 + 5       9 + 3

where

A = 8; D =2 ; G =1; B = 7; C = 4; E = 6; F = 5; H = 9, and I = 3 (this is incorrect)

[/b]


Let f(x)=$ ax^2\plus{}bx\plus{}c$ be any quadratic with real coefficients such that $ |f(x)|\leq 1$ for $ 0\leq x \leq 1$.  Find the smallest M such that $ |a|\plus{}|b|\plus{}|c|\leq M$.

[quote="kyyuanmathcount"]Let f(x)=$ ax^2 \plus{} bx \plus{} c$ be any quadratic with real coefficients such that $ |f(x)|\leq 1$ for $ 0\leq x \leq 1$.  Find the smallest M such that $ |a| \plus{} |b| \plus{} |c|\leq M$.[/quote]
I'm not sure how rigorous this is...
[hide]
If $ |a|\plus{}|b|\plus{}|c|>1$, then we have that $ f(x)\equal{}|a|\plus{}|b|\plus{}|c|$ (for x=1), but this violates $ |f(x)| \leq 1$, so M=1?[/hide]

[hide]Consider the function 8x^2-8x+1[/hide]

[quote="kyyuanmathcount"]Let f(x)=$ ax^2 \plus{} bx \plus{} c$ be any quadratic with real coefficients such that $ |f(x)|\leq 1$ for $ 0\leq x \leq 1$.  Find the smallest M such that $ |a| \plus{} |b| \plus{} |c|\leq M$.[/quote]
My answer is $ M\equal{}17.$
[hide]If this is correct, I will post the detailed answer.[/hide]


What is the maximum number of parts into which a plane can be divided by n circles?

Can the same be generalised to convex  figures as such??
(Recursion seems to be the key).


$ x$ and $ y$ are two numbres from $ \mathbb{N}^2$

prove that the smallest numbre $ t$ from $ \mathbb{N}$ ($ t>0$) such as $ y$ divides $ xt$ is a divisor for$ y$.

;)

[quote="sami.d"]$ x$ and $ y$ are two numbres from $ \mathbb{N}^2$

prove that the smallest numbre $ t$ from $ \mathbb{N}$ ($ t > 0$) such as $ y$ divides $ xt$ is a divisor for$ y$.

;)[/quote]Let me get this straight... so $ x,y\in\mathbb{N}^2$? How is that a number if it is the Cartesian product $ \mathbb{N}\times\mathbb{N}$? I guess $ \mathbb{N}\equal{}\{0,1,2,\dots\}$?
Assuming $ x,y\in\mathbb{N}$, prove that the smallest number $ t\in\mathbb{N}$ ($ t>0$) such that $ y$ divides $ xt$, is a divisor of $ y$, ie $ t|y$.  :ninja:

I'm guessing he means $ (x, y) \in \mathbb{N}^2$.


The first two names are caucasian, I believe.  Beyond that I'm not sure what you mean by real american.

[quote="pavel kozlov"][quote="MysticTerminator"]Zach Abel
Zarathustra Brady
Ryan Ko
Yi Sun
Arnav Tripathy
Alex Zhai[/quote]
Are there here any real american, except maybe Ryan Ko?[/quote]

Ryan Ko is of Asian descent.  The name they have listed on the AMC website is Taehyeon which would have been more of a giveaway :P

[quote="s1ck_psych0"][quote="amir2"]
[quote="Orl"]
It is a pretty smart trick to increase one's post count by asking questions all the time which can be figured out easily by searching them on the web[/quote]

There are smarter tricks too. Mr. Green[/quote]

I guess you're right! Wink[/quote]

lol, this could start an infinite chain of posts :rotfl:

[quote="pavel kozlov"][quote="MysticTerminator"]Zach Abel
Zarathustra Brady
Ryan Ko
Yi Sun
Arnav Tripathy
Alex Zhai[/quote]
Are there here any real american, except maybe Ryan Ko?[/quote]

Zach is American
I know because he graduated from the school I go to
whose math team will definitely not be as succesful next year without him
 :(

Maybe, this is somehow off topic but Zarathustra was an Iranian Prophet with the relegion of Zarathustrian (In farsi(Iranian) we pronounce it like Zartosht)
It's not a very popular relegion in Iran (But I think it's the second most relegion after Islam in Iran), in the world too (No offence to Zarathustrians)
I'm very interested about Zarathustra Brady, any one knows wether he is Zarathustrian or his parents just picked this name for him.
Does any one know about it? (Maybe Zarathustra himself is in this Forum)

[quote="Nima Ahmadi Pour"]the relegion of Zarathustrian (In farsi(Iranian) we pronounce it like Zartosht)[/quote]

In English, "Zoroastrian."

The question, "Are there any real Americans?" is a meaningless question.  A better question might be, "How many of those team members were born in America?"

[quote="pavel kozlov"][quote="MysticTerminator"]Zach Abel
Zarathustra Brady
Ryan Ko
Yi Sun
Arnav Tripathy
Alex Zhai[/quote]
Are there here any real american, except maybe Ryan Ko?[/quote]

Pavel, the situation is similar to that of USSR olympiad teams: the
general population and the IMO population of the country look very different.   In the USSR, a huge proportion of olympiad winners were Jewish (not all were permitted to travel to the IMO) and the same for scientists, mathematicians, engineers, chess-masters, and intelligentsia in general.    In the USA, immigrants from Asia (and ex-USSR), children of such immigrants, children of white American scientists and intelligentsia, and Jews form a large proportion of the IMO population.  Of the 6 names listed above for the USA IMO team, at least one sounds Jewish, three are obviously East Asian (whether US-born or not is unclear), and one Indian.

[quote="Nima Ahmadi Pour"]I'm very interested about Zarathustra Brady, any one knows wether he is Zarathustrian or his parents just picked this name for him.[/quote]
The latter. He goes by Zeb (his initials), and his AOPS username is also zeb.
I think the name is a reference to Nietzsche's work, possibly indirectly. Zarathustra is the German form.

Being named after a Nietzsche work is pretty awesome, actually.

Wow, I'm famous :P

The name is just there to throw you off... I'm German/Russian/Irish/Jewish.

The name was selected because of a paternal fascination with "Thus Spoke Zarathustra", and because of a paternal dislike for common names, such as Kevin, John, Daniel, etc.

[quote="JBL"] The question, "Are there any real Americans?" is a meaningless question. [/quote]

The questioner is from a country where "Politically Correct" is no longer an official phrase and where, unlike the Real America, one can ask and answer such questions openly.   There is, of course, a variety of obvious and non-trivial interpretations of the question.   By the way, it is a near-universal presumption by mathematically educated immigrants to the USA, that American IMO teams are dominated by immigrants (which is partially correct, and becomes more correct if one adds US-born children of the immigrant intelligentsia).

[quote="JBL"]I'm not sure what the purpose of your first sentence is.

[quote="fleeting_guest"]There is, of course, a variety of obvious and non-trivial interpretations of the question.[/quote]

Which makes the question effectively meaningless, lacking further elaboration.[/quote]

The question and its answers are clear and obviously meaningful, notwithstanding politically expedient deconstructions.  It is simply a demographic fact that in the US (and Canada, Australia, and elsewhere) the IMO population differs dramatically, and racially, from the general population.   Some racial groups have, so far, been nonexistent within that population, whereas others have been been hugely overrepresented compared to their share of national population.
It is of course possible to try and further deconstruct these unpleasant facts by "controlling" for various factors, but the discrepancies remain enormous.
I do not think any service is done by Political Correction of phrases such as "real American", particularly as used by non-US natives.

Added:
let me mention, e.g. for those reading this outside USA, what type of political considerations I refer to.

There remains in the US an ideology of "immigrant melting pot", as opposed to multiculturalism.    It has broken down somewhat in recent times, but differentiating between which citizens are more American than the others is still considered in many social situations as making an invidious and politically incorrect distinction.  In theory, Yanks are supposed to be one big family, just don't ask about the color of the garbage collectors or what language they speak.    Because of its more overtly competitive, zero-sum ethos as compared to other countries and also because of its troubled racial history (oppression of blacks), in the US when obvious patterns appear such as Asians/Jews/immigrants dominating science, this is not necessarily treated as good news ("look at those talented Chinese, isn't it nice they are here!") but rather as something to be covered over by either avoiding the subject or by invoking the We Are All Americans ideology.

If a question has a variety of obvious interpretations (as you wrote), then it is a poor question.  A good question has only one obvious interpretations (and hopefully no non-obvious interpretations).

You think the question is clear and well-posed?  Is a third-generation student of Japanese descent (say) a "real American?"  What about a second-generation Indian whose parents spoke English from childhood (in India, before emigrating)?  What about a twelfth-generation African-American?

Perhaps more importantly, the intent of the question was almost certainly closer to my interpretation (with immigration the relevant category) than yours (race).  I cannot be certain of this, because the question is not a good one.  But your last two posts are, most likely, much further off the point than the post of mine you are criticizing.

[quote="JBL"]If a question has a variety of obvious interpretations (as you wrote), then it is a poor question.  A good question has only one obvious interpretations (and hopefully no non-obvious interpretations). [/quote] Most questions, and most reasonable questions, contain some ambiguity. That is not a problem, much less something that renders the question "meaningless".

[quote] You think the question is clear and well-posed?  [/quote]
I think it was a reasonably clear question and that it was reasonable to ask it.   A PhD thesis on the definition of "real American" is not needed in order to grasp that someone was asking about an apparent difference between the composition of IMO team(s) from the USA and the composition of the USA national population.   Whether the question was specifically pointing to differences in birthplace, citizenship, immigrant status, racial group, social stratum, or other such paramaters was not assumed in the answers.

[quote] Is a third-generation student of Japanese descent (say) a "real American?"  What about a second-generation Indian whose parents spoke English from childhood (in India, before emigrating)?  What about a twelfth-generation African-American?   [/quote]We don't need to answer those questions in order to answer Pavel's question.  If, as you claim, I have misunderstood his question, I am sure that Pavel can clarify what was meant, or whether he had such a specific interpretation in mind.

This is absurd.  We agree the question contained ambiguity, yet you are criticizing me for attempting to disambiguate.  It is not necessary to answer my questions to answer Pavel's question, but it is most certainly necessary to answer them to support a claim that the question of "real Americanness" is clear and well-posed.  I did have not called Pavel's question "unreasonable," so I don't see why you think it necessary to note that Pavel's question was a reasonable one.
Moreover, Pavel had ample opportunity to suggest that I was giving a false interpretation of his question -- something he neglected to do.

[quote="JBL"] the question contained ambiguity, yet you are criticizing me for attempting to disambiguate.  [/quote]

I disputed the idea that ambiguous questions (either in general or this particular question) are meaningless or poor.   There was, obviously, no criticism of question-disambiguation as such.

[quote]  It is not necessary to answer my questions to answer Pavel's question, but it is most certainly necessary to answer them to support a claim that the question of "real Americanness" is clear and well-posed. [/quote]

You concede that Pavel's question (are there any real Americans on the American IMO team) was answerable, so presumably it was not meaningless, after all.   Some other question that you introduced and that only you have discussed here ("the question of real Americanness") may or may not be meaningful, clear or well-posed, but that is, literally, a different question.

[quote] Moreover, Pavel had ample opportunity to suggest that I was giving a false interpretation of his question -- something he neglected to do.[/quote]

Pavel did not falsify any of the interpretations of the question by at least five different respondents, so his "neglect" does not privilege the JBL interpretation over the others, but it does work against your statement that some non-JBL interpretations are worse than the JBL interpretation.

I think the purpose of this topic has been fulfilled.  Please stop arguing before it needs to be locked.

Uh, JBL is the moderator of this forum - I don't think you can tell him when the topics should be locked.

When you say Real American, do you refer to those whose family has lived in American ever since 16th century?

So that means no americans(whose parents are americans),who have no other country connections n stuff?

I thought the IMO rules made pretty clear that anyone who is a member of a national team to the IMO has to have some connection to the country sending out the national team. It's my understanding that the United States team to the IMO consists of citizens or permanent residents of the United States (as contrasted with persons who live in the United States with temporary visitor or diplomatic visas). The United States team might be glad to pick up some of the Canadians who score well on the USAMO   :D   but it's my understanding that those math competitors can only compete for Canada.

tokenadult

(a "real American" born in the United States and descended from pre-revolutionary settlers in New England and married to a "real American" first-generation immigrant who is now a naturalized citizen)

With any luck, this will be my final post on this topic.

Thank you, eryaman, shobber, shady and tokenadult for your various responses, all of which I like and all of which I will ignore from here onwards, as the rest of this post is addressed to fleeting_guest.

The initial question of Pavel's was, paraphrased, "Are there any real Americans on the American IMO team?"  I said, "The question, "Are there any real Americans?" is a meaningless question. A better question might be, "How many of those team members were born in America?""
The next day, you gave a perfectly reasonable response to Pavel's question, noting the discernable ethnicities/nations of origin of several of the USIMO team members.  More than 2 weeks later, you responded to my initial post, beginning with a sentence whose purpose I still don't understand (and which you did not explain when I requested)*.  You also allowed that the question "Are there real Americans?" has "a variety of obvious and non-trivial interpretations," one of which (presumably) is the question I had suggested in its place, "Are there any non-immigrant members of the US team?"  You argued that "the IMO population differs dramatically, and racially, from the general population," a proposition which is true (if perhaps unfortunate) and which has nothing to do with what I said.  Nor, I think, does it have very much to do with the question Pavel asked.  If it did, your initial (June 20) response to Pavel's question would seem to be sufficient to cover that territory.

Now, the position I have been arguing is, more or less, that Pavel's original question was so ambiguous so as to be effectively meaningless.  I therefore assume, since you have continued to joust with me, that you disagree with this contention.  It is unlikely that we will come to an agreement over this point.  I would note that in your first (June 20) response to Pavel's question, you noted that one of the team members was Indian.  That same team member, two days earlier, had posted a one-line response to Pavel's question, saying "I'm American!"  You could make it your business to tell him that he is not a "real American."  I suppose also that if the question were truly unambiguous, you would be able to tell me, in each of the three hypothetical cases I gave you, whether those individuals were "real Americans."  While it may not be necessary to answer my three hypothetical questions to answer Pavel's question, it is certainly necessary that [i]the ability to answer my question[/i] exists in order for Pavel's question to be meaningful.  The only way Pavel's question can mean anything is if, in the case of specific individuals with specific racial and national backgrounds, you can decide who is and who is not a "real American."  Now, if you think you can create a satisfactory delineation between those of us who really are Americans and the rest ("fake Americans?"  "Imported Americans?"  "racial minorities?"), go ahead.

Now, to nit-pick at your most recent post, I had (and have) no intention of conceding that Pavel's question is meaningful -- I was merely following your own construction.
"The question of real Americanness," that is whether or not such a thing is meaningful, is obviously fundamental to answering Pavel's question.  If the notion is meaningless, his question is as well.  It is indeed a different question, but it is a question whose answer is fundamental to the meaning of Pavel's question.

The question I offered in place of Pavel's question, that of immigrant status, has a key advantage over Pavel's: it is not ambiguous at all.  Moreover, I chose that question because I believed that it captured the issue that Pavel was actually interested in.  I may be wrong about that -- we don't know because Pavel hasn't re-entered the conversation.  On this point, we had the following exchange:[quote="fleeting_guest"]If, as you claim, I have misunderstood his question, I am sure that Pavel can clarify what was meant, or whether he had such a specific interpretation in mind.[/quote][quote="JBL"]Pavel had ample opportunity to suggest that I was giving a false interpretation of his question -- something he neglected to do.[/quote][quote="fleeting_guest"]Pavel did not falsify any of the interpretations of the question by at least five different respondents, so his "neglect" does not privilege the JBL interpretation over the others, but it does work against your statement that some non-JBL interpretations are worse than the JBL interpretation.[/quote]  I don't think I really need to say much about this -- you took my remark to imply that Pavel's silence supports my view, immediately after making an identical implication (that Pavel's silence would support your view, by not correcting it) and then criticized me for that implication.  I fully agree that Pavel's silence indicates nothing, and it certainly isn't the reason that I believe myself to be correct.

You may consider this my honest attempt to establish what we've been arguing about and what my position is.  If I have erred, please feel free to point out where.

* You also later implicitly accused me of "politically expedient deconstructions."  Miriam-Webster's Collegiate Dictionary tells me that "expedient" means 1: "suitable for achieving a particular end in a given circumstance" and 2: "characterized by concern with what is opportune; especially : governed by self-interest" with the additional note that "EXPEDIENT usually implies what is immediately advantageous without regard for ethics or consistent principles <a politically expedient decision>."  This also puzzles me, because I don't know what political agenda I am being accused of pursuing (at the expense of my ethics and consistent principles, no less) by attempting to disambiguate a clearly ambiguous question.

This has turned into a topic from the twilight zone or something...

Eryaman: mysmartmouth never told JBL to lock the topic.  He attempted to put the conversation back on track and said that it [i]might[/i] be locked (therefore implying that it would be up to JBL's discretion) if the conversation continued in that manner.

Anyway, the thread has gone far off topic...

re: tokenadult comment on nationality requirements for the IMO, I don't think the rules were as specific as permanent residency, but most countries impose such requirements in their selection procedures.  I think there was some requirement (or assumption within the language of the regulations) of affiliation with the school system of a given country.  This may have been rationalized and specified more in recent years in light of home-schooling and IMO contestants who exploit multiple citizenships to qualify in a less selective country than the one where they live or study.
re: JBL,

[quote="JBL"] [60-line made-for-PM metadiscussion] [/quote]It would take a 200-line dissertation to analyze it in full, which is unlikely to happen.  Some things need correction on the record, though:

[quote] More than 2 weeks later, you responded to my initial post, [/quote]My posting appeared less than 5 days after its predecessor; longer gaps of 5 and 13 days had already occurred in the thread and are apparently OK with you, so it is unclear what nefarious behavior you point to here.

[quote] beginning with a sentence whose purpose I still don't understand (and which you did not explain when I requested)*.  [/quote]

Please retract the falsehood.  You did not post any request for explanation.  As it happens, I did post a long explanatory paragraph about the political correctness of "real Americans" discussions in the United States, which is exactly the subject matter of the sentence you call unexplained.   It is outrageous to allege evasions of questions that were never asked but happen to have been answered anyway.

[quote]  You argued that "the IMO population differs dramatically, and racially, from the general population," a proposition which is true (if perhaps unfortunate) and which has nothing to do with what I said.  Nor, I think, does it have very much to do with the question Pavel asked.  [/quote]

The fact that 2-3 additional posters independently provided ethnic/racial/religious data on the American IMOers in response to Pavel's question contradicts both your points.

[quote] in your first (June 20) response ... you noted that one of the team members was Indian.  That same team member, two days earlier, had posted a one-line response to Pavel's question, saying "I'm American!" You could make it your business to tell him that he is not a "real American." [/quote]

An outrageous fabrication.  Please retract the falsehood and quote my actual words rather than rhetorically convenient straw-men.   I did not suggest that any team member "was Indian" or speculate on his level of Americanism.

[quote] [I suppose also that if the question were truly unambiguous, you would be able to tell me, in each of the three hypothetical cases I gave you, whether those individuals were "real Americans."  While it may not be necessary to answer my three hypothetical questions to answer Pavel's question, it is certainly necessary that [i]the ability to answer my question[/i] exists in order for Pavel's question to be meaningful.  The only way Pavel's question can mean anything is if, in the case of specific individuals with specific racial and national backgrounds, you can decide who is and who is not a "real American."  [/quote]

Pavel's question about the IMO, it is not itself an IMO question and need not reach that level of clarity.  Certainly it is possible to discuss the "average American" with his 2.2 children even though no particular American has that number of kids, or whether this year's World Cup 2nd place team consisted of "typical Frenchmen" or whether there are any "real Frenchmen" on the team.  What the question means in all its subtleties, ramifications, vagaries (or even its offensiveness) are interesting subjects but they are not grounds to simply deconstruct it as meaningless.

[quote] you took my remark to imply that Pavel's silence supports my view, immediately after making an identical implication (that Pavel's silence would support your view, by not correcting it) [/quote]

That was not my implication.  My implication was that he could speak for himself without your interposition, not that one view or another was preferred.

This thread has certainly gone off topic and has spun into a debate that currently offers no closure for the original poster..

[quote]this will be my final post on this topic. [/quote]

Why didn't you make it [i]the[/i] final post?  :lol:

There is no implicit expectation that all threads should conclude with a result satisfactory to the original poster (in particular, we don't know exactly what would constitute a satisfactory response, see the thread), nor should there be any problem with threads that wander off of their original topic.  The discussion is relevant to the question; that it fails to actually answer the question is mostly immaterial.

What is annoying is the useless and mostly contentless posts along the line of 'stop arguing', 'this thread should be locked', 'this thread is off topic'.  PM a moderator (lol), or better yet say/do nothing about it.

I refuse to stop posting reviews of the entire discussion that may or may not contain responses to other people's written statements.

What is annoying is people who reply to posts which are "useless and mostly contentless" with a request to employ the PM system (which they may or may not fail to use themselves) and recommend a course of inaction for any further doubts..

You're probably the kind of guy who thinks that backseat moderating and being passive aggressive is cool, but really, it isn't.  Go away.

I'm (still) American!

[i]Moderator's note:  What passed for discussion here has moved to PM.  This thread is locked.[/i]


Funny Strings

Let's consider a string of non-negative integers, containing N elements. Suppose these elements are $S_1, S_2, .. ,S_N$, in the order in which they are placed inside the string. Such a string is called 'funny' if the string $S_1+1, S_2, S_3, .., S_{N-1}, S_N -1$ can be obtained by rotating the first string (to the left or to the right) several times. For instance, the strings 2 2 2 3 and 1 2 1 2 2 are funny, but the string 1 2 1 2 is not. Your task is to find a funny string having N elements, for which the sum of its elements $S_1+S_2+..+S_N$ is equal to K.

Prove: GCD(N,K)=1 is a necessary condition for a string to be funny.

Is there anyone who has some ideas? I doubt its truth and is there anyone who can give a counterexample?


Is the level of the problem is so low that no one wants to post the solution?

It probably has more to do with the time at which you posted it.
$log(ax)+log(ax^2)=4$
$log(ax+ax^2)=4$
$ax+ax^2=10^4$
[if a≠0]
$x^2+x-\frac{10^4}{a}=0$
$x=\frac{-1\pm\sqrt{1+\frac{10^4}{a}}}{2}$
Two requirements must be met: $ax>0$ and ${1+\frac{10^4}{a}}\geq0$
If a>0 then the second condition is met.  Then needed is
$\frac{-1+\sqrt{1+\frac{10^4}{a}}}{2}>0$
$\sqrt{1+\frac{10^4}{a}}>1$
${1+\frac{10^4}{a}}>1$
${\frac{10^4}{a}}>0$ which is true.

If a<0 then ax^2 will be negative and log(ax^2) will not exist.
So the answer is:
(0,infinity)

[quote="Wumbate"]It probably has more to do with the time at which you posted it.
$log(ax)+log(ax^2)=4$
$log(ax+ax^2)=4$
[/quote]

I don't know how you got that.  I think there is no such rule like that in logarithms.  I believe you have mistaken it with $\lg a+\lg b=\lg ab\neq\lg (a+b)$.  Btw, the problem is $\lg ax\lg ax^2$.  Please tell me if I'm mistaking.

Masoud Zargar

I think 'Wumbate'  made some mistakes,its supposed to be product of logs,not [b]sum[/b] of logs...anyway...

[hide]$(\lg a + \lg x)(\lg a + 2.\lg x)=4$ for:$a,x>0$ and doing $\lg a=s$ and $\lg x=l$ we get:

$2l^2+3.ls+s^2-4=0$ solving for the positive real numbers:

$l=\frac{\sqrt{s^2+32}-3s}{4}$,its was given that $l>1$ consequently:$\frac{\sqrt{s^2+32}-3s}{4}>1 \implies 8s^2+24s-16<0$ solving again the inequation for the positive real numbers:

$1<s<\frac{\sqrt{17}-3}{2} \implies 1<\lg a<\frac{\sqrt{17}-3}{2} \implies 0<a<10^{\frac{\sqrt{17}-3}{2}}$

...problaby i missunderstand something if yes correct me.[/hide]

[quote="Gabriel(fr)"]
$l=\frac{\sqrt{s^2+32}-3s}{4}$,its was given that $l>1$ [/quote]
$x>1$ is NOT the same as $l=\lg x >1$.
Moreover, $s=\lg a$ may be less than or equal to 0.

[quote="ifai"]Given that the equation $\lg (ax) \lg(ax^2)=4$ has real solution in x, and all
the solutions are greater than 1. Find the range of a.

ps: $\lg x=\log_{10} x$[/quote]

The right solution
[hide]$(\lg a + \lg x)(\lg a + 2.\lg x)-4=0$ for:$a>0, x>1$

we need to have both roots positive for this equation.

so, the sum and product of the roots must be positive(discriminant is always positive for this equation)

hence we should have
$- 3\lg a > 0 \Rightarrow \lg a < 0$

and

$(\lg a)^2 - 4 > 0 \Rightarrow \log a \in ( - \infty ,\, - 2) \cup (2,\,\infty )$

implying  $\lg a < \, - 2 \Rightarrow 0 < a < \frac{1}{{100}}$[/hide]

Correcting Gabriel's solution:
$l = \frac{\pm\sqrt{s^2+32}-3s}{4} > 0$
$\pm\sqrt{s^2 + 32} > 3s$.

Case 1: $\sqrt{s^2 + 32} > 3s$
If $s \le 0$ this is obviously true. Otherwise we square both sides and get that $0 < s < 2$. Thus, $s < 2$ which means that $0 < a < 100$.

Case 2: $-\sqrt{s^2 + 32} > 3s$
If $s \ge 0$ this has no solutions. Otherwise we must have that $s^2 + 32 < 9s^2$. This is equivalent to $s < -2$, but this doesn't really add any solutions that weren't in case 1.

Answer: $0 < a < 100$.

It can be seen that the solution above mine is incorrect. For example, there are solutions for a = 1.

[quote="Kalle"]Correcting Gabriel's solution:
$l = \frac{\pm\sqrt{s^2+32}-3s}{4} > 0$
$\pm\sqrt{s^2 + 32} > 3s$.

Case 1: $\sqrt{s^2 + 32} > 3s$
If $s \le 0$ this is obviously true. Otherwise we square both sides and get that $0 < s < 2$. Thus, $s < 2$ which means that $0 < a < 100$.

Case 2: $-\sqrt{s^2 + 32} > 3s$
If $s \ge 0$ this has no solutions. Otherwise we must have that $s^2 + 32 < 9s^2$.

everything is right till here.
This is equivalent to $s<-2$,
Answer: $0 < a < 1/100$.

Think whether case1 is necessary.

Yeah, I turned the inequality the wrong way around in Case 2, but the solution is still right. You don't need two real solutions, only one, so it is Case 1 which decides the answer. As I said, there is a solution for $x > 1$ for the equation if $a = 1$, so your answer is wrong.

edit: oh, "all the solutions are greater than 1". Yeah, ok. I didn't read the question properly.

Kalle your answer is incorrect, and when a=1 the solutions are $x_1=10^{\sqrt 2}>1$, but $x_2=10^{-\sqrt 2}<1$.

Yeah, as I said, I thought the question only required one root to be greater than 1.


calculus
integration
inequalities
calculus computations

[hide="My solution"]Consider $ F(t) \equal{} \int_{1 \minus{} t}^{1 \plus{} t}x^{\frac {3}{x}}\ dx \minus{} 2t$. We know that $ F(t) \equal{} 0$ at $ t \equal{} 0$. Thus, we need only to show that $ F(t)$ is increasing on $ [0,1)$. Differentiating we get $ F'(t) \equal{} (1 \plus{} t)^{\frac {3}{1 \plus{} t}} \plus{} (1 \minus{} t)^{\frac {3}{1 \minus{} t}} \minus{} 2.$
However, by [url=http://www.mathlinks.ro/viewtopic.php?t=302774]an inequality of Vasc[/url], $ F'(t) \ge 0$ on $ [0,1)$. Hence $ F(t)$ is increasing, and the inequality has been proven.

Note: The inequality states that for positive reals $ a,b$ such that $ a \plus{} b \equal{} 2$, $ a^{\frac {3}{a}} \plus{} b^{\frac {3}{b}}\ge 2$. The proof of that inequality is hideous, though, and I'm not sure you had it in mind.[/hide]

For $ 0\leq t\leq 1$

$ \int_{1\minus{}t}^{1\plus{}t}x^{3/x}dx\geq \int_{1\minus{}t}^{1\plus{}t}x^{3/2}dx\equal{}\frac{2}{5}\left( \left( 1\plus{}t\right) ^{5/2}\minus{}\left( 1\minus{}t\right) ^{5/2}\right) \equal{}\frac{2}{5}\left( \frac{\allowbreak 10t\plus{}20t^{3}\plus{}2t^{5}}{\left( 1\plus{}t\right) ^{5/2}\plus{}\left( 1\minus{}t\right) ^{5/2}}\right) \allowbreak \geq \frac{2}{5}\left( \frac{\allowbreak 10t}{2}\right) \equal{}\allowbreak 2t$


group theory
superior algebra
superior algebra solved

I wonder if anyone in this forum knows/ has come across the following. It may be trivial.

Let $G$ be a finite group, and let $k$ be the order of an arbitrary element $g \in G$. Is it necessarily true that $G$ has at least $k$ equivalence classes? (in other words, does it follow that the number of irreducible complex characters of $G$ is at least $k$?)

i think this is trivial because you may send k to any k'th root of unity, so you get k different characters...i think one may even prove that the number of complex characters is divisible by k...but i may have misunderstood something.

:D  :D  :D
Trivial, indeed --- there are always at least $k$ explicitly given linear characters, and these are irreducible. And what an incredible proof had I found to this triviality!!  :cool: It's always a shame when you've thought you've developed an ingenious proof to something as simple as $1+1=2$.  :lol:  :lol:  :lol: Cool-cool.  :cool:

By the way, no Peter, $k$ does not necessarily divide the number of equivalence classes. Take, for example, the dihedral group of order $2p$, where $p$ is an odd prime (for the example, only $S_3$ would suffice, but anyway  :) ). This group has both elements of order $2$ and $p$, but the number of the equivalence classes is less than $2p$, and thus not divisible by both $2$ and $p$.

Also note that, if $G$ is nilpotent, then $k$ may be replaced with the exponent of the group.


geometry
3D geometry
sphere
induction
combinatorics solved
combinatorics

(1) What is the maximum number of regions that n plane divide the 3-dim. space into ?

(2) What is the maximum number of regions that n spheres divide the 3-dim. space into ?

for [1] I got to this answer 1/6(n^3+5n+6)!

cheers!!

and for the 2nd one I got : n/3(n^2-3n+8)

cheers! :D :D

The problems are similar to the ones posted before with lines and circles in the plane. I didn't check them out, but I assume the idea is the same? (induction; we see how many new regions are formed when we add a new sphere/plane)

Why don't you provide some arguments ? It would quite nice to get a more complete solution. But I guess it's too time-consuming...

Full solution for number one: :D

[1]First we have to compute the maximum dn of regions that a line is divided by n points.

We have that dn=n+1. Indeed, because a point divides the line d1 in 2 regions. After k points have divided the line in dk regions, then the "k+1"th is also in region which divides it in to 2 so d_(k+1)=dk+1 and induction.

[2]Now we must compute the maximum pn of regions which the plane is divided by n lines!

We have first the formula: p_(n+1)=pn+dn. Indeed, because if n lines have divided the plane pn in regions Ri (1<=i<=pn) then we consider the a_(n+1) line d, which intersects all the other points in n distinct points. From [1] we get that the n points divide d in a number of regions rj which can be equal to dn: (1<=j<=n). Each of these regions rj separetes an "old" region Ri in two : Ri=Ri'+Ri". So the number pn of the old regions Ri adds to the number dn to obtain p_(n+1). WE have the following equalities p_(i+1)-pi=di, and we add all these relations from 2 to n-1 and get that pn=(n^2+n+2)/2.

Now we get to our problem :D :D and we see that the number of regions that can be added by the (n+1) plane, this means that s_(n+1)-sn equals the number dn  of the regions in which the n+1 plane is divided by the intersection lines of the "old" planes! So we have that s_(n+1)-si=di. we sum all these form i=2 to n-1 and we get that sn=1/6(n^3+5n+6). This maximum is obtained if there are no paralel planes or there are 3 secantes!
hope I made my self clear! :)
cheers! :D :D :D

for the 2nd go and check:

http://www.mathlinks.ro/phpBB/viewtopic.php?t=1205

cheers! :D :D

Could someone elaborate lagrangia's solution? ..
thank you.


inequalities
linear algebra
matrix
blogs

Let $ A_1,A_2,...,A_n \in H^{ \plus{} \plus{} }_n(\mathbb{C})$ Such that $ \sum_{k \equal{} 1}^{n}{A_k} \equal{} nI_n$
1)Prove that : $ \det(\sum_{k \equal{} 1}^{n}{(A_k \plus{} A^{ \minus{} 1}_k)^2}) \geq \frac {(n^2 \plus{} 1)^{2n}}{n^n}$
2) Prove that : $ \det(A_1^k \plus{} A_2^k \plus{} ... \plus{} A_n^k) \geq \det(A_1^{k \minus{} 1} \plus{} A_2^{k \minus{} 1} \plus{} ... \plus{} A_n^{k \minus{} 1})$ , $ \forall k \in \mathbb{N^{*}}$
 $ I_n$ is identity matrix.
 $ H^{ \plus{} \plus{} }_n(\mathbb{C})$  are set $ n$x$ n$ Hermitian matrices and  positive definite.
[hide="Answer of me "]Very interesting ! :P [/hide]

Here I prove for (1) , for (2) don't dificult
Lemma1) Let $ A_1^{ \minus{} 1} \plus{} A_2^{ \minus{} 1} \plus{} ... \plus{} A_m^{ \minus{} 1} \geq m^2(A_1 \plus{} A_2 \plus{} ... \plus{} A_m)^{ \minus{} 1}$ for all $ A_i ,\forall i \in \{1,2,..,m\}$ are Hermitian matrices and positive define. you can read proof in http://www.mathlinks.ro/viewtopic.php?t=275033
Lemma2) $ m^2(A_1^2 \plus{} A_2^2 \plus{} ... \plus{} A_m^2) \geq (A_1 \plus{} A_2 \plus{} .. \plus{} A_m)^2$ , for all $ A_i ,\forall i \in \{1,2,..,m\}$ are Hermitian matrices . You can read  small proof in blog of me .
and a some lemma here .....
 Use lemma1) we have :
  $ n^2\sum_{k \equal{} 1}^{n}{(A_{k} \plus{} A^{ \minus{} 1}_{k})^{2}}\geq (\sum_{k \equal{} 1}^{n}{(A_{k} \plus{} A^{ \minus{} 1}_{k})})^2$
  by $ \sum_{k \equal{} 1}^{n}{(A_{k} \plus{} A^{ \minus{} 1}_{k})} \equal{} nI_n \plus{} \sum_{k \equal{} 1}^{n}{A^{ \minus{} 1}_{k}} \geq nI_n \plus{} n^2(A_1 \plus{} A_2 \plus{} .. \plus{} A_n)^{ \minus{} 1} \equal{} 2nI_n$
 then $ n^2\sum_{k \equal{} 1}^{n}{(A_{k} \plus{} A^{ \minus{} 1}_{k})^{2}}\geq 4n^2I_n$
 from here $ \sum_{k \equal{} 1}^{n}{(A_{k} \plus{} A^{ \minus{} 1}_{k})^{2}}\geq 4I_n$ or  $ \det(\sum_{k \equal{} 1}^{n}{(A_{k} \plus{} A^{ \minus{} 1}_{k})^{2}})\geq 4^n$
 Sorry  :blush:  for 1) is true if $ A_1 \plus{} A_2 \plus{} ... \plus{} A_n \equal{} I_n$ then
$ \det(\sum_{k \equal{} 1}^{n}{(A_{k} \plus{} A^{ \minus{} 1}_{k})^{2}})\geq\frac {(n^{2} \plus{} 1)^{2n}}{n^{2n}}$


first of all $x\neq \frac{k\pi}{2},k\in Z$ ..... [b](*)[/b]

equation is equivalent with:

$\sin x+\cos x=\cos 2x$

$\sin x+\cos x=-(\sin x+\cos x)(\sin x-\cos x)$

$(\sin x+\cos x)(\sin x-\cos x+1)=0$

$(\sin x+\cos x)\left(\cos\left(x+\frac{\pi}{4}\right)-\frac{\sqrt{2}}{2}\right)=0$

second factor of LHS is $\neq 0$ (condition [b](*)[/b]), so solution is

$\boxed{x=m\pi-\frac{\pi}{4}\ ,\ m\in Z}$

[hide=good one]mark  $\sin{x}=a$ and $\cos{x}=b$\\
the equation is equivalent to $a^3\left(1+\frac{b}{a}\right)+b^3\left(1+\frac{a}{b}\right)=b^2-a^2$\\
 we consider two cases $a+b=0$ and $a+b\neq 0$\\
for $a+b \neq 0$ we get:\\
$a^2+b^2=a-b \implies \sin2x=0$\\
clearly this don't works\\
so we have $a+b=0$\\
which gives $\tan{x}=-1$\\
now this gives $x=\varphi\pi-\frac{\pi}{4}  \qquad \forall \qquad \varphi \in \mathbb{Z}$[/hide]

[hide]$\sin x+\cos x=\cos 2x\Rightarrow {{(\sin x+\cos x)}^{2}}={{\cos }^{2}}2x\Leftrightarrow 1+\sin 2x=1-{{\sin }^{2}}2x$thus $\sin 2x(1+\sin 2x)=0$[/hide]


$a_1,...,a_n$ are non-negative reals. Let $x_i=\frac{a_i+a_{i+1}}{2}$ and $y_i=\frac{a_i+a_{i+1}+a_{i+2}}{3}$. Prove the following inequality:

\[\sum _{i=1}^n x_i ^{x_{i+1}} \geq \sum _{i=1}^n y_i ^{y_{i+1}}\]

where $a_{n+1}=a_1$, $a_{n+2}=a_2$

P.S. It may be easy though I'm not sure about my solution so I'm posting.

[quote="Megus"]

where $a_{n+1}=a_1$, $a_{n+2}=a_2$

[/quote]

Doesn't that make all the a's the same?

No, I just put it to reassure that indices are taken modulo n which should be obvious in such inequalities

Somebody, any solution?


geometry MATHCOUNTS AMC AIME ARML USA(J)MO USAMO \/closed

The Art of Problem Solving [b][i]Introduction to Geometry[/i][/b] textbook is now available.

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=9]Click here for excerpts and to order.[/url]

[b][i]Introduction to Geometry[/i][/b] is a complete introductory geometry text, covering all topics of a standard geometry curriculum, plus most of the problem solving tactics required for success in programs such as MATHCOUNTS and the AMC.  The textbook is 496 pages, and the solutions manual is 192 pages.  Full solutions to all problems are in either the solutions manual or the text itself.

Wow! it looks awesome, I can't wait to get it!!! :jump:  :jump:

It's certainly not too late for national MATHCOUNTS.

The books are an extention on the AoPS book 1 material, so as long as you know that, you should be fine for states  :D .

[quote="joml88"][quote="robinhe"]I guess it's too late to buy now  :(  :(

My states are on March 11...[/quote]

It's never too late to buy!  Even if you have hardly any time left for Mathcounts, there is still AMC, AIME, ARML, USAMO and more to prepare for :)[/quote]

Yeah, I was planning to buy it after states... :D  :D

My states are on March 17-18 so it might be too late to buy.

Nope, I win :winner_first::winner_first:!

Swipes it back  :ninja: .

These books are also great review, not only for MATHCOUNTS, but I feel they can help AIME's on some problems.

Yeah, like the book is just going ot randomly appear on your front doorstep tonight so you can study for like 5 minutes before going to bed...

I will buy it sometime in my lifetime (most likely in a few weeks) :D

Or you could make it your lifelong goal to someday own it.

This book is going to be awesome, like the counting one.

[quote="biffanddoc"]Im  buying it tonight!
(would I be the first order) :?:[/quote]

Not the first order, but close!  First order was the same as last time - AAST.

[quote="bubala"]Don't they own like 600 copies (or some other tremendous number) of your originals??[/quote]

We're big fans of AAST.  :)

Anyone has red the book? What you think about it? It can help also a future physic's student? What existensial questions!  :P

I placed the order 20 min after seeing the announcement, and paid extra for expedited shipping. Hope it arrives tomorrow :D

who can buy it and give it to me??  :lol:  :lol:

[quote="G-UNIT"]When will the Intro to Geometry course next be offered?[/quote]

We have not yet set the summer schedule, but we will probably not offer that class before the Fall at earliest.

[quote="easyas3.14159..."]I believe this is their website:
[url=http://www.bergen.org/AAST/]www.bergen.org/AAST/[/url]I googled it and just didnt think it was american association for surgery of trauma.
All hail aops for the geometry book! :omighty:[/quote][b] :roll:  :roll: [/b]


Let $ p $ be a prime number , $ x\in Z $ . {$ a_n $} :
 $ a_n = 1 +\frac{x}{n+1} + \frac{x^2}{2n+1} + ... + \frac{x^p}{pn+1} $
 Prove that : $ a_n $ doesn't equal to 0 for all $ n=1,2,..,p-1 $

Suppose the contrary.
Let $P=(n+1)(2n+1)...(pn+1)$. We see that $p\mid P$, more exactly: one of the numbers $kn+1$ is divisible by $p$ and all other numbers are not. Consider $a_nP$. It is a sum of $p+1$ summands and we know that $p$ of them are divisible by $p$, so $p\mid \frac{P}{kn+1}x^k$. It follows that $p\mid x$. It is clear $x<0$, so $x\leq p-1$. But then \[0=a_n \leq 1+\frac{|x|}{n+1}+\frac{|x|^2}{2n+1}+...+\frac{|x|^{p-1}}{n(p-1)+1}-\frac{|x|^p}{pn+1}=S.\] We have $\frac{|x^s|}{ns+1}<\frac{|x|^{s+1}}{n(s+1)+1}$ for $|x|\geq p$, therefore $0<S<1+\frac{(p-1)|x|^{p-1}}{n(p-1)+1}-\frac{x^p}{np+1}<0$ (easy task, for $p\geq 3$). Contradiction.

Oh, maybe you meant "$a_n$ is not an integer"?


Let $O$ be a point in the interior of a quadrilateral of area $S$, and suppose that
$2S = |OA |^2 + |OB |^2 + |OC |^2+ |OD |^2$ .

Prove that $ABCD$ is a square with centre $O$.

 :)  :)

Note that $[AOB] = \frac{1}{2} OA \cdot OB \cdot \sin \angle AOB \le \frac{OA \cdot OB}{2}$.  Summing with the other three analogous triangles yields

$S \le \frac{OA \cdot OB + OB \cdot OC + OC \cdot OD + OD \cdot OA}{2}$

with equality iff A, O, C collinear and B, O, D collinear, with BD perpendicular to AC.  Then

$S \le \frac{OA \cdot OB + OB \cdot OC + OC \cdot OD + OD \cdot OA}{2} \le \frac{OA^2 + OB^2 + OC^2 + OD^2}{2}$

with equality when all the lengths are equal.  Hence ABCD is a square of center O.

Source : 18th  British  MO  1982. The  same  problem appeared again  at  14th  Balkan  MO  1997


analytic geometry
graphing lines
slope
trigonometry
calculus
calculus computations

Suppose that you are climbing a hill whose shape is given bythe equation $z=1000-0.01x^{2}-0.02y^{2}$ and you are standing at a point with coordinates (60, 100, 764).
a) In which direction should you proceed initially in order to reach the top of the hill fastest?
b) If you climb in that direction, at what angle above the horizontal will you be climbing?

a) is easy. Its the $\nabla z$ at (60, 100, 764). $\nabla z=\left<-0.02x, -0.04y\right>$
$\nabla z(60,100)=\left<-1.2, -4\right>$

b) should be easy too, but I am drawing a complete blank. I thinks it's something related to the $tan(\theta)$ and the $\vec{r}'$

The slope in the direction of the most rapid increase is the largest possible directional derivative.

That is, $\max_{||u||=1}D_uz=\max_{||u||=1}\nabla z\cdot u=||\nabla z||.$

In this particular instance, $||\nabla z||=\sqrt{(-1.2)^2+(-4)^2}=\frac{2\sqrt{109}}5\approx 4.176.$

That's the slope. The relationship of a slope to an elevation angle is that $m=\tan \theta$ or $\theta=\arctan m\approx \arctan 4.176\approx 76.5^{\circ}.$

Very steep hill - probably best not to go there without ropes.

lol,

I was right about the $tan(\theta)$ part...heh.


Euler
function
LaTeX
number theory
number theory proposed

What are some questions I could use to test fundamentals in topics of Euler (0) - function, formulae for d(n) and funky symol (n) (means sum of divisors of n), multiplicative arithmetic functions, and the mobius inversion formula?

That's the Euler $\phi$ function. To write that in LaTeX, try (dollar) \phi(n) (dollar). The sum-of-divisors function is $\sigma(n)$ (that's (dollar) \sigma(n) (dollar) - the thing you called "funky symbol" is sigma. Some other useful multiplicative functions include $\tau(n)$, the number of divisors (that's (dollar) \tau(n) (dollar)), $\delta(n)$, which is 1 at 1 and zero everywere else, there's 1 as a function, and there's the Möbius function $\mu(n)$ (\mu(n)), which is zero if $n$ is divisible by any square and $(-1)^r$ if $n$ has $r$ prime factors taken one at a time.

I'm looking at a copy of Rosen's [i]Elementary Number Theory[/i]; the multiplicative functions are in chapter 7. He's got lots of exercises. Here are a few:

1. For which positive integers $n$ is $\phi(n)$ a power of 2?

2. Show that the equation $\sigma(n)=k$ has at most a finite number of solutions when $k$ is a positive integer.

3. For how many consecutive integers can the Möbius function $\mu(n)$ take the value $0$?

Thank you for questions. I think I'll look up the book too if it has alot of questions for number theory.


inequalities
function
logarithms
inequalities unsolved

[quote="April"]Prove the inequality
\[ \sqrt {a^{1 \minus{} a}b^{1 \minus{} b}c^{1 \minus{} c}} \le \frac {1}{3}
\]
holds for all positive real numbers $ a$, $ b$ and $ c$ with $ a \plus{} b \plus{} c \equal{} 1$.[/quote]
It is equivalent : $ (1\minus{}a)ln(a)\plus{}(1\minus{}b)ln(b)\plus{}(1\minus{}c)ln(c) \le \minus{}2ln(3)$
$ f(x) \equal{} (1 \minus{} x)ln(x)$ -  concave function for $ x > 0$ and by Jensen it is true

[hide="Alternative Solution"]

We have $ a^a b^b > \equal{} a^b b^a$ for all positive real numbers a and b (trivial),
By multiplying this inequality for (a,b) (b,c) (c,a), $ a^{1 \minus{} a} b^{1 \minus{} b} c^{1 \minus{} c} \equal{} a^{b \plus{} c} b^{c \plus{} a} c^{a \plus{} b} \leq a^{2a} b^{2b} c^{2c}$
Multiplying $ a^{2\minus{}2a} b^{2\minus{}2b} c^{2\minus{}2c}$ on both sides yields
$ (a^{1 \minus{} a} b^{1 \minus{} b} c^{1 \minus{} c} )^3 \leq a^2 b^2 c^2 \leq \frac {1}{64}$ (AM-GM)
Taking the sixth roots of both side yields the desired result.

[/hide]

By using the weighted AM-GM-inequality we get \[ a^\frac{1\minus{}a}2b^\frac{1\minus{}b}2c^\frac{1\minus{}c}2 \leq \frac{(1\minus{}a)a\plus{}(1\minus{}b)b\plus{}(1\minus{}c)c}2\equal{}\frac{1\minus{}(a^2\plus{}b^2\plus{}c^2)}2\]
Applying Cauchy-Schwartz-inequality yields $ 1\equal{}(a\plus{}b\plus{}c)^2 \leq 3(a^2\plus{}b^2\plus{}c^2)$
So we get \[ \sqrt {a^{1 \minus{} a}b^{1 \minus{} b}c^{1 \minus{} c}} \leq \frac{1\minus{}\frac1{3}}2\equal{}\frac1{3}\]

[quote="April"]Prove the inequality
\[ \sqrt {a^{1 \minus{} a}b^{1 \minus{} b}c^{1 \minus{} c}} \le \frac {1}{3}
\]
holds for all positive real numbers $ a$, $ b$ and $ c$ with $ a \plus{} b \plus{} c \equal{} 1$.[/quote]
It's equivalent to:
$ \sum_{cyc} (1\minus{}x) \cdot \ln(x) \le \ln \left ( \frac{1}{9} \right )$
$ \ln(x)$ is concave so applying Weighted Jensen gives us:
$ \frac{\sum_{cyc} (1\minus{}x) \cdot \ln(x)}{\sum_{cyc} (1\minus{}x)} \le \ln \left ( \frac{\sum_{cyc} x \cdot (1\minus{}x)}{\sum_{cyc} (1\minus{}x)} \right )$ $ \iff$
$ \sum_{cyc} (1\minus{}x) \cdot \ln(x) \le 2 \cdot \ln \left ( \frac{1 \minus{} \sum_{cyc} x^2}{2} \right ) \le 2 \cdot \ln \left ( \frac{1 \minus{} \frac{1}{3}}{2} \right ) \equal{} \ln \left ( \frac{1}{9} \right )$. QED

[quote="qwerty414"]
$ a^2 b^2 c^2 \leq \frac {1}{64}$ (AM-GM)
[/quote]
I think you have a mistake.
It will be modified $ a^2b^2c^2 \le \frac {1}{729}$, I think.

Sorry about my bad english.

$ a\log bc \plus{} b\log ca \plus{} c\log ab\leq \log (ab \plus{} bc \plus{} ca)\ \because a \plus{} b \plus{} c \equal{} 1$

$ \leq \log \frac {(a \plus{} b \plus{} c)^2}{3} \equal{} \log \frac {1}{9}$

$ \Longleftrightarrow (b \plus{} c)\log a \plus{} (c \plus{} a)\log b \plus{} (a \plus{} b)\log c\leq \log \frac {1}{9}$

$ \Longleftrightarrow \log a^{b \plus{} c}\log b^{c \plus{} a}\log c^{a \plus{} b}\leq \log \frac {1}{9}$

$ \Longleftrightarrow a^{b \plus{} c}b^{c \plus{} a}c^{a \plus{} b}\leq \frac {1}{9}$

$ \Longleftrightarrow \sqrt {a^{b \plus{} c}b^{c \plus{} a}c^{a \plus{} b}}\leq \frac {1}{3}$ Q.E.D.


The points on a circle are colored in three different colors.

Prove that there exist infinitely many isosceles triangles with vertices on the circle and of the same color.


function
Functional Analysis
topology
real analysis
real analysis unsolved

Let $ X \equal{} C([0,1])$ using the sup norm. Prove that the closure of every non-empty open subset of $ X$ is not compact.

I've been fighting with similar type theorems and problems since almost all references to Arzela-Ascoli assume that you are quite good with functional analysis already.

What I've picked up is that (since the set is closed) we either need to show not bounded or not equicontinuous by Arzela-Ascoli. My guess would be the latter.

Non-empty open: If I interpret this correctly I think it means if $ A\subset X$ is nonempty and open then for any $ f\in A$, there exists $ \varepsilon > 0$ such that whenever $ \sup_x |f \minus{} g| < \varepsilon$ we have $ g\in A$.

Oh darn. As I was typing this, it clarified where my misunderstanding was. I'll still post this, but I'll let other people try to answer it. (At the very least it will provide a good reference if someone is searching for this topic).

$ C[0,1]$ with the sup-norm is an infinite dimensional Banach space, hence the unit ball is not compact. Note that any closed subset of a compact set is also compact to finish the proof.

That was far easier than I was thinking.

I was trying to explicitly construct something along the lines of, if $ f$ is in some open set then there is some $ \epsilon$ neighborhood of continuous functions. So $ f_n(x) \equal{} f(x) \plus{} \frac{\epsilon}{2} \sin(nx)$ (or some variant, the idea that I thought of while typing is to have some oscillating function in the "wiggle room" such that no subsequence converges uniformly) are all in that open set. Thus they are all in the closure of that open set, but no subsequence converges uniformly, so the set is not compact.

Does this also work (in principle, I'm not certain that particular class of functions works)?


An antique dealer buys a chair for $ \$20$ and sells it for $ \$85$. The dealer later buys back the same chair for $ \$110$ and sells it for $ \$135$. How many dollars total profit did the dealer make on this chair?

First, the dealer gets $ \minus{}20$ dollars.  Then, he gets $ 85$ dollars. Then he gets $ \minus{}110$ dollars, and then he gets $ 135$.  Adding, the dealer made $ \minus{}20\plus{}85\minus{}110\plus{}135\equal{} \boxed{90}$ dollars.


geometry
rectangle
number theory proposed
number theory

Given a rectangle in plane. Prove that for every integer n one can part the origin rectangle into some sub-rectangles such that any line parallel to the side of the original rectangle that does not contain any side of any sub-retangles cuts exactly n sub-rectangles

The partition I described [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26382]here[/url], in post #2, should do the trick.