Hi everyone,

I would like to display the character '~' in Latex.
Ex: http://www.abc.com/~myname

I tried to type like this:
http://www.abc.com/\symbol{126}myname

but the resulting character '~' appeared a bit higher than normal.

Can you please tell me how to solve this problem?
Thank you very much

Best to use \textasciitilde rather than \symbol{126} which tends to rely on the font chosen. But anyway, you can change the height of the tilde by using \raisebox as in
[color=red]www.abc.com/\raisebox{-1ex}{\textasciitilde}myname[/color]

If you need to use ~ a lot then
[color=red]\newcommand{\mytilde}{\raisebox{-1ex}{\textasciitilde}}[/color]
will allow you to write
[color=red]www.abc.com/\mytilde myname[/color]

However, if you use the url package you can't use either method as \url assumes the text doesn't contain any commands and is literal text.

$\text{http: //studwww.ugent.be/\raisebox{-1ex}{\textasciitilde}pwvdendr/}$

--> doesn't really look good? maybe I should try another font?


geometry
3D geometry
sphere
calculus
calculus computations

Find the maximum and minimum values of $f(x,y,z)= x^{2}-2yz$ on  the curve of intersection of the sphere $x^{2}+y^{2}+z^{2}=4$ and the plane $x+y+z=0$.

Lagrange multipliers with two constraints ...

Look in the appropriate chapter of a calculus textbook, using those key words.


Find the smallest integer $ n$ such that among any $ n$ integers ,with repetition allowed ,there exists  $ 18$ integers whose sum is divisible by $ 18$.

general fact: given any natural number $ n$, among any $ 2n\minus{}1$ integers there exist $ n$ of them with sum divisible by $ n$.

And the bound is tight. To see that one needs to take $ n\minus{}1$ integer divisible by $ n$ and $ n\minus{}1$ integer congruent to $ 1$ mod $ n$.


[hide="Hmm?"] Since $f(n+1) = f(n)+\frac{1}{3}$ it follows by induction that that

$f(n) = f(1)+\frac{n-1}{3}$

So that

$f(1000) = 3+\frac{999}{3}= \boxed{ 336 }$. [/hide]

Problemsets, are you related to interval??

 :|

[hide]
I found a pattern
n=1 then f(n) equals 3
n=2 then f(n) equals 10/3
n=3 then f(n) equals 11/3
and so on
adds 1/3 each time n increases by 1
so
3+(1000-1)*(1/3)=3+333=336

[/hide]


calculus
integration
inequalities
real analysis
real analysis solved

Let $f:[0,1]\to\mathbb{R}$ be continious on $[0,1]$ and differentable  on $(0,1)$,$f(0)=0$ and $0 \leq f'(x) \leq 1$ for all $x\in(0,1)$. Prove the inequality $(\int_0^1 f(x)dx)^2 \geq \int_0^1 f^3(x)dx$

let $F(x) = \int_0^x f(y)dy$ and $G(x) = \int_0^x f^3(y)dy$.  for $x=0$, obviously $F(x)^2 = G(x)$.  so it suffices to prove that $(F(x)^2-G(x))' \geq 0$ for $x \in (0, 1)$.  now,

$(F(x)^2)' - G'(x) = 2F(x)F'(x) - f^3(x) = 2F(x)f(x) - f^3(x) = f(x)(2F(x)-f^2(x))$,

so, since $f(x) \geq 0$ for $x \in (0, 1)$ (because $f(0) = 0$ and $f'(x) \geq 0$), it suffices to prove that $2F(x) - f^2(x) \geq 0$ for $x \in (0, 1)$.

since $2F(0) - f^2(0) = 0$, it then suffices to prove that $(2F(x) - f^2(x))' \geq 0$.  and indeed,

$2F'(x) - 2f(x)f'(x) = 2f(x) - 2f(x)f'(x) = 2f(x)(1-f'(x)) \geq 0$

This is a very old problem, but slightly different.  I am sure a similar one has appeared before 1980, which, at least,  was used as a contest problem in my school before 1980. The only difference is that $f'(x)>0$, which enables us to use Cauchy's mean-value theorem.

We are going to prove that $\frac{\left(\int_0^1 f(x)dx\right)^2}{\int_0^1 f(x)^3 dx}\geq 1$
Let $F(x) = \left(\int_0^x f(t)dt \right)^2$ and $G(x) = \int_0^x f^3(t)dt $
Apply Cauchy's Mean-value theorem,
$\frac{\left(\int_0^1 f(x)dx\right)^2}{\int_0^1 f(x)^3 dx} = \frac{F(1)-F(0)}{G(1)-G(0)} = \frac{F'(\xi)}{G'(\xi)} = \frac{2f(\xi)\int_0^\xi f(t)dt}{f^3(\xi)} = \frac{2\int_0^\xi f(t)dt}{f^2(\xi)} = \frac{2\int_0^\xi f(t)dt - 2\int_0^0 f(t)dt}{f^2(\xi)-f^2(0)} = \frac{2f(\eta)}{2f(\eta)f'(\eta)} = \frac{1}{f'(\eta)} \geq 1$

[quote="liyi"]This is a very old problem, but slightly different.  I am sure a similar one has appeared before 1980, which, at least,  was used as a contest problem in my school before 1980[/quote]

Yes old, I have seen it 1982  in a old book at "l'île de la Réunion"


Let ${a_k}$ be a sequence of integers such that $a_1 = 1$ and $a_{m + n} = a_m + a_n + mn$, for all positive integers $m$ and $n$. Then $a_{12}$ is?
(a) 45
(b) 56
(c) 67
(d) 78
(e) 89

[hide]$a_{m+1}=a_m+a_1+m=a_m+m+1$.
$a_2=3$
$a_3=6$

From here, we can take two routes to the solution:
$a_{6}=2a_3+9=21$
$a_{12}=2a_6+36=78$.

OR

$1, 3, 6, \ldots$ What does that remind you of :?: Triangular numbers maybe?
We can prove through induction that $\{ a_n\}$ is the sequence of triangular numbers, so $a_{12}=\frac{12\cdot 13}{2}=78$.[/hide]


estas son las 7 nuevas maravillas del mundo  :!:

-Gran muralla china, China (entre el Río Amarillo y el Mar Bohai).
-Ciudad de Petra, Jordania.
-Cristo Redentor, Río de Janeiro, Brasil.
-Machu Picchu, Cusco, Perú.
-Coliseo, Roma, Italia.
-Templo de Kukulcán en Chichen Itzá (Yucatán, México).
-Taj Mahal, Agra, India.

 tres maravillas son de iberoamerica
 felicidades  iberoamerica  :first:
http://es.wikipedia.org/wiki/Nuevas_Siete_Maravillas_del_Mundo
http://www.new7wonders.com/index.php?id=633

Si!!!! aca en peru..estamos celebrando...por nuestro Macchu Picchu...!!!q bueno...es una gran noticia para nosotros!!! :lol:

Me dijeron que esto no tenia validez..  la unesco no lo ampara.

saludos


[quote="Own"]2 coins are, and throw, so 1 coin is T, other F -> Answer the probability[/quote]

the probability for one coin to be T(or F) is 1/2
both coins have the probabilty 1/2,
and it have to be happened at once, so you should multiply 1/2 by 1/2
1/2 × 1/2 = 1/4

but if the two coins are the same, u have to multiply 2 again,
and the answer is 1/2

another way is to count all cases.

(T F) ○
(T T) ×
(F F) ×
(F T) ○

∴1/2

The question is very poorly phrased, and one possible other interpretation is the following:
"Two coins have been tossed.  Given that at least one of the two landed heads, what is the probability that both landed heads?"
In this case, the answer is $ \frac{1}{3}$.

Thank you for your interpretation.
I have bad english.

But..
"Two coins have been tossed. Given that at least one of the two landed heads, what is the probability that both landed heads?"
In this case, the answer is 1/3? Why?

And..
my phrase means following   //hexa answered
(T F) ○
(T T) ×
(F F) ×
(F T) ○
∴1/2

Anyway..
Thank you

[quote="Own"]I appreciate your help.
I feel good to be understood by you.  //my phrase to be understood
This question is a test for my bad english to be understood or not.
I do not know A from B.

Next, I will post up more-meaning-qustions.
Thank you very much.

Remark : I think your answering is perfect.[/quote]

thanks! :)


inequalities
geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations

Let $x_{0},x_{1},...,x_{669}$ be distinct real numbers and everyone satisfies $0<x_{i}<1$. Prove that we can find always a pair of such numbers $x_{i},x_{j}$ such that we have $0 < x_{i}x_{j}(x_{j}-x_{i}) < \frac{1}{2007}$

If my English is so bad that you don't understand the problem, please tell me :)
Only one of the competitors got 8 out of 8, unfortunately I got 2 points (altough my solution was good enough for 5 points)

The cubes of the $x_{k}$ are also between 0 and 1.  Thus two of the cubes are less than $\frac{1}{669}$ away from each other.  In other words, there exist $i$ and $j$ such that
\[0 < x_{j}^{3}-x_{i}^{3}< \frac{1}{669}\, . \]
By factoring the difference of cubes, we have
\[(x_{j}-x_{i}) (x_{j}^{2}+x_{i}x_{j}+x_{i}^{2}) < \frac{1}{669}\, . \]
But $x_{j}^{2}+x_{i}x_{j}+x_{i}^{2}\ge 3 x_{i}x_{j}$, so we get
\[(x_{j}-x_{i}) 3 x_{i}x_{j}< \frac{1}{669}\, . \]
Dividing by 3 gives the desired inequality.


linear algebra
matrix
geometry
3D geometry
linear algebra unsolved

$A=(a_{i,j}) \in \mbox{GL}_{3}(\mathbb{Q})$, s.t. $A^{T}A=I_{3}$.
Denote $a_{i,j}= p_{i,j}/q_{i,j}$, with $p_{i,j},q_{i,j}$ coprime integers.
Show that for all $i,j$, $q_{i,j}$ is odd.

What are the positive integers $n$ s.t. the above is true is you change $\mbox{GL}_{3}(\mathbb{Q})$ by $\mbox{GL}_{n}(\mathbb{Q})$?

The first part is rather trivial, right? All you need is that if $a^{2}+b^{2}+c^{2}=1$ and $a,b,c$ are rational, then their denominators are odd. This is immediate with congruences mod $8$. I guess that there is some counterexample for $n=4$ and this would show that all $n>3$ fail to satisfy this condition.

Nice!
The following also shows that we cannot extend this to 3 cubes instead of 3 squares:
$(1/2)^{3}+(2/3)^{3}+(5/6)^{3}=1$


There is a problem that I am just bummed on. Please help :(

Problem:

How many times does the paper wind around the cardboard roller on a 300-sheet roll of bathroom tissue? Each sheet is 4.4 inches long, the diameter of the full roll is approximately 4.2 inches, and the diameter of the cardboard roller is approximately 1.6 inches.

If you know please help me. I need the answer and how you do it as soon as possible. Thank you.

The answer is less than 300.

My teacher already told me that.

Please people try. I need your help!!!

how can it be less than 300? even with the roll full the sheet is longer than the roll is wide so it would go all the way around and have extra, and when the roll is empty it can go around >2 times.

is anyone else nice enough to help?


dorosti hokme zir ra barresi konid:
har majmoe ke karan bala darad hatman kochektrain karan bala ham darad!

pas chi shod!!
kasi nemikhad ino hal kone!?

doroste:ba estefade az asl baze haye to dar to mishe halesh kard +farz kholf+halat bandi.kasi bishtar tozih khast bege

در واقع بازه های تو در تو اصل نیست. گزاره ای هستش که از اصل تمامیت نتیجه میشه.
و میگه که اگر تعدادی بازه ی ناتهی داشته باشیم که از هر دو تایی از آنها یکی زیرمجموعه ی دیگری باشد، همه ی این بازه ها اشتراک دارند.
اصل تمامیت هم میگه هر مجموعه ی ناتهی از اعداد حقیقی نامنفی یک کوچکترین کران پایین دارد.
اصل تمامیت به عنوان یکی از خواص اعداد حقیقی پذیرفته میشه.


geometry
Euler
circumcircle
vector
geometry proposed

In a scalene triangle $ ABC, H$ and $ M$ are the orthocenter an centroid respectively. Consider the triangle formed by the lines through $ A,B$ and $ C$ perpendicular to $ AM,BM$ and $ CM$ respectively. Prove that the centroid of this triangle lies on the line $ MH$.

Let $ A_{1}B_{1}C_{1}$ bounded by the perpendiculars in $ A$, $ B$, $ C$ to $ AM$, $ BM$ and $ CM$, respectively. Since $ MH$ is the Euler line of triangle $ ABC$, it is suffice to prove that the centroid of $ A_{1}B_{1}C_{1}$ lies on $ MO$, where $ O$ is the circumcenter of $ ABC$. According to [url=http://forumgeom.fau.edu/FG2008volume8/FG200812index.html]Lemoine's theorem[/url] that the symmedian (Lemoine) point of a triangle is the unique point which is the centroid of its own pedal triangle, we deduce that the centroid $ M$ of $ ABC$ is the symmedian point of $ A_{1}B_{1}C_{1}$. In this case, by rewriting our conclusion with respect to triangle $ A_{1}B_{1}C_{1}$, we have to prove the following problem:

[color=blue]Let $ ABC$ be a triangle with centroid $ G$ and symmedian point $ K$. Prove that the circumcenter of the pedal triangle of $ K$ lies on the line $ GK$.[/color]

Since $ G$ and $ K$ are isogonal conjugates with respect to $ ABC$, this is a particular case of the following theorem:

[color=darkblue][b]Theorem.[/b][/color] [color=blue]Let $ P^{\prime}$ be the isogonal conjugate of a point $ P$ in the plane of a given triangle $ ABC$. Then, the orthogonal projections of $ P$ and $ P^{\prime}$ on the sidelines of $ ABC$ lie on a same circle centered at the midpoint of the segment $ PP^{\prime}$.[/color]

For a synthetic proof, see Theorem 7 of [url=http://www.cip.ifi.lmu.de/~grinberg/]Darij Grinberg, Isogonal conjugation with respect to a triangle[/url].

Considering that this is a 9th-grade problem, I wonder whether there is an easier solution than this.

Denotes Hc is the orthocenter of ⊿BAC’ , Ha,Hb are similarly defined
It is easy to see HcAMB is a parallegrom , the same for the other two
We know M is the symmedian (Lemoine) point of M’ w.r.t to ⊿A’B’C’
(The Lemoine point of a triangle is the centroid of its pedal triangle)
Thus AM is the symmedian through C’  =>
∠MC’A=∠MBA=∠M’C’B=90-∠C’BA      =>
C’M’⊥AB  =>  C’,Hc,M’ are collinear =>  M’ is the orthocenter of ⊿HaHbHc
⊿HaHbHc and ⊿ABC are harmonious w.r.t M
DONE!

In the configuration of the triangle $ \bigtriangleup GBC,$ where $ G$ ( instead of $ M$ ) is the centroid  of $ \bigtriangleup ABC$ and based on the below [b][size=100]Lemma[/size][/b], we have that the line through the point $ A'$ as the problem states, and perpendicular to $ BC,$ passes through the point $ G',$ as the symmetric point of $ G,$ with respect to the circumcenter $ O$ of $ \bigtriangleup ABC$ ( Because this line is symmetrical of the $ G$-altitude of $ \bigtriangleup GBC,$ with respect to the midperpendicular of $ BC$ ).

Also, because this line bisects the segment $ PQ,$ where $ P,\ Q$ are the points of intersections of the line segments $ A'B,\ A'C$ respectively, from the line through $ G$ and perpendicular to $ GM$ ( $ M$ is here the midpoint of the side-segment $ BC$ ), applying the [b][size=100]Thales theorem[/size][/b], bisects also the segment $ B'C'$ and then, this line is the $ A'$-median of the triangle $ \bigtriangleup A'B'C'.$

Similarly, the other two medians of $ \bigtriangleup A'B'C',$ also pass through the fixed point $ G'.$

Hence, the centroid $ G'$ of the triangle $ \bigtriangleup A'B'C'$ as the problem states, lies on the [b][size=100]Euler Line[/size][/b] of the given triangle $ \bigtriangleup ABC$ and the proof is completed.

[b][size=100][color=DarkBlue]LEMMA. - A triangle $ \bigtriangleup ABC$ is given with circumcircle $ (O)$ and let $ A'$ be, the antidiametric ( = diametric opposite ) point of $ A.$ The line through the midpoint $ M$ of the side-segment $ BC$ and perpendicular to the $ AM,$ intersects the line segments $ A'B,\ A'C,$ at points $ P,\ Q,$ respectively. Prove that the line through the point $ A'$ and perpendicular to $ BC,$ bisects the segment $ PQ.$[/color][/size][/b]

Kostas Vittas.

PS. I will post here next time, the proof of the simple above [b][size=100]Lemma[/size][/b] I have in mind.

This can also be proved using vectors.  Let M be the origin, and $a,b,c$ be the vectors to the points of the original triangle and $A,B,C$ be the vectors to the newly created triangle.  Also let $v$ be the vector to the orthocenter of the original triangle.

Then,
$a+b+c=0$,
$(a-v)(b-c)=0$, or $v(b-c)=a(b-c)=-(b+c)(b-c)=c^2-b^2$ and similar
$(C-a)a=0$, or $Ca=a^2$, and similar.

So,
$A(a+b+c)=0$, so $Aa=-Ab-Ac=-b^2-c^2$, and similar.
$(A+B+C)(b-c)=Ab+Bb+Cb-Ac-Bc-Cc=b^2-(a^2+c^2)+b^2-c^2-c^2+(a^2+b^2)=3(b^2-c^2).$

Thus,
$(A+B+C)(b-c)= -3v(b-c)$, and similarly,
$(A+B+C)(c-a)= -3v(c-a)$,

which means that

$(A+B+C)/3 = -v.$

Any homothety solution for this problem???

Through the solution, $G$ will be the centroid of $\triangle{ABC}$.Let $A'B'C'$ be the triangle. If you have an accurate diagram, it is not hard to see that the median $A'M_A$ of $\triangle A'B'C'$ is perpendicular to $BC$. That is what we are going to prove.

Let $H_A$ be the orthocenter of $\triangle A'BC$. I am not going to detaily explain why all of the equalities are true, they are easy by angle and sine chasing
$$\dfrac{\sin H_AA'C}{\sin H_AA'B}=\dfrac{\cos A'CB}{\cos A'BC}=\dfrac{\sin BCG}{\sin CBG}=\dfrac{BG}{CG}$$
Now let $AA_1$ be the median of $\triangle ABC$. We also have:
$$\dfrac{\sin M_AA'B}{\sin M_AA'C}=\dfrac{A'C'}{A'B'}=\dfrac{\sin C'B'A'}{\sin B'C'A'}=\dfrac{\sin AGC}{\sin AGB}=\dfrac{\sin A_1GC}{\sin A_1GB}=\dfrac{BG}{CG}$$

Meaning that lines $A'H_A$ and $A'M_A$ cut the angle $B'A'C'$ into two angles, ratio of sines of which are equal. This implies, that lines $A'H_A$ and $A'M_A$ are one. So the median $A'M_A$ of $\triangle A'B'C'$ is perpendicular to $BC$.

Also, both of lines $CH_A$ and $BG$ are perpendicular to $BA'$, meaning $CH_A\parallel BG$. Analogously, $BH_A\parallel CG$. This means that quadrilateral $BGCA'$ is a parallelogram implying $GH_A$ passes through $A_1$, or points $A, G, H_A$ are collinear.

Finally, consider homothethy centered at $G$ with coefficient $-1$. Point $A$ maps to $H_A$, since $AG=2GA_1=GH_A$. Furthermore, the $A$- altitude of $\triangle ABC$ and line $A'M_A$ are parallel. These two facts are enough to prove that $A$ - altitude of $\triangle ABC$ maps to $A'$ - median of $\triangle A'B'C'$. Analogously, $B$ - altitude of $\triangle ABC$ maps to $B'$ median of $\triangle A'B'C'$. So, with this homotethy, the orthocenter of $\triangle ABC$ maps to the centroid of $\triangle A'B'C'$. This means that the needed points are collinear, furthermore $G$ is the midpoint of $HM$, where $M$ is centroid of $\triangle A'B'C'$.


trigonometry
integration
calculus
derivative
calculus computations

Find, for $m,n\in\mathbb{Z}$, $m=n$ and $m\neq n$,\[\int_0^L\sin\left(\frac{n\pi x}{L}\right)\sin\left(\frac{m\pi x}{L}\right)\,dx.\]

Suppose \[f(x)=\sum_{n=1}^\infty B_n\sin\left(\frac{n\pi x}{L}\right).\] How does one solve for $B_k$ using the above property?
I don't understand the textbook explanation.

There are several ways to do the first part. I'm partial to this one (something about the second derivative being a self-adjoint operator, but you don't have to know what that means):

For $m\ne n$, integrate by parts twice. Note that the boundary terms should all be zero, so you get one constant times the unknown integral equals a different constant times the same integral. Draw the appropriate conclusion.

For $m=n,$ use the fact that the integral of $\sin^2$ or $\cos^2$ over an interval that starts and ends at the boundaries of quadrants is "half the box."

As for the second part:

Multiply both sides of the equation by $\sin\left(\frac{m\pi x}L\right)$ and integrate $dx$ from $0$ to $L.$ If you've done the first part correctly, the right hand side should simplify nicely.

An alternative method for the first part: Use the identity
\[
  \sin \alpha \sin \beta
    = \frac{1}{2} \cos(\alpha - \beta) - \frac{1}{2} \cos(\alpha + \beta) .
\]


inequalities
geometry
inequalities unsolved

Let $m_{a},m_{b},m_{c}$ be the median of triangle $ABC$. Prove
$\frac{m_{a}m_{b}m_{c}}{m_{a}^{2}+m_{b}^{2}+m_{c}^{2}}\ge r$

I don't think it is hard :D. but this solution isn't of me :D
we consider the triangle $\triangle_{m}$ base on medians $m_{a},m_{b},m_{c}$ area $S_{m}=\frac{3}{4}S$
it is equivalent
$\frac{4R_{m}S_{m}}{m_{a}^{2}+m_{b}^{2}+m_{c}^{2}}\ge r \Leftrightarrow R_{m}\ge \frac{a^{2}+b^{2}+c^{2}}{2(a+b+c)}(*)$
now using the inequality in $ABC$
$MAGA+MBGB+MCGC\ge GA^{2}+GB^{2}+GC^{2}\forall M$ give $M\equiv O\Rightarrow R\ge \frac{GA^{2}+GB^{2}+GC^{2}}{GA+GB+GC}$ change it into $\triangle_{m}$
$\Rightarrow R_{m}\ge \frac{a^{2}+b^{2}+c^{2}}{2(a+b+c)}$ thus $(*)$ is true :D


Let there be an acute angle O. Let a point P lie between the sides of angle O. Let A and B be the intersection of a line through P and the sides of angle O. Prove that the area of AOB is smallest when P bisects AB.

[hide]
Okay, so we will take XY so that P bisects XY.  We will also take MN so that MN goes through P, but MP is not equal to NP (O, X, and M are collinear as are O, N, and Y.  Also, without loss of generality, let us say that ON<OY and thus OX<OM)  So, we wish to prove that the area of triangle OXY is greater than the area of triangle OXY is smaller than the area of triangle OMN.  So, we wish to prove that
\[ \frac {(OX)(OY)\sin O}{2} < \frac {(OM)(ON)\sin O}{2}
\]

\[ (OX)(OY) < (OM)(ON)
\]

\[ (OX)(ON \plus{} NY) < (ON)(OX \plus{} XM)
\]

\[ (OX)(ON) \plus{} (OX)(NY) < (ON)(OX) \plus{} (ON)(XM)
\]

\[ (OX)(NY) < (ON)(XM)
\]

\[ (\frac {OX}{XM})(\frac {NY}{ON}) < 1
\]
Extend OP to hit MY at Z.  By Ceva's Theorem, we have that $ (\frac {OX}{XM})(\frac {MZ}{ZY})(\frac {NY}{ON}) \equal{} 1$.  So, if we show that $ \frac {MZ}{ZY} > 1$, then our proof is complete.  Now, draw PW parallel to MY.  So, we have that $ MW \equal{} WY$.  So, we wish to show that $ MZ > MY$.  Notice that we can show this by showing that
\[ \angle YXM > \angle ZPY
\]

\[ 180 \minus{} \angle YXM < 180 \minus{} \angle ZPY
\]

\[ \angle OXY < \angle OPY
\]

\[ \angle OPY \equal{} \angle OXY \plus{} \angle POX \rightarrow \angle OXY < \angle OPY
\]
Thus, our proof is complete.  [/hide]


probability
geometry
circumcircle
incenter
trigonometry
algorithm
area of a triangle

I just made a new one because the other one is getting too long.  And since nationals is 1 1/2 weeks away.

So [size=200][color=red]problem 1[/color][/size]
[size=200]team[/size]

How many differnet ways are there to write 100 as the sum of two or more consecutive integers.

[hide]Find how many odd factors 100 has basically.  So that gives us 5 and 25.  So 2.  i think...[/hide]

I agree if the consecutive integers must be nonnegative.
[hide]
But the sum of the integers from -8 to 16 adds to 100 as well.[/hide]

ahh ... but it's included with that i think.  25 in a row is exactly that way.

you're overlooking a case

9 10 11 12 13 14 15 16

[hide]Set x to be the lower of the added numbers.
If there are 2 numbers, 2x + 1 = 100. This is impossible because x is an integer.
If there are 3 numbers, 3x + 3 = 100. Still impossible.
If there are 4 numbers, 4x + 6 = 100. Still impossible.
If there are 5 numbers, 5x + 10 = 100. x = 18.
If there are 8 numbers, 8x + 28 = 100. x = 9.

[b]Only 2 ways, 18 + 19 + 20....+22, and 9 + 10 + 11 +....+ 16 add up to 100.[/b]

....somehow find a formula for the pattern.
oops, forgot to consider negative integers.[/hide]

uh #1 hasn't been solved

[hide="hint for 1"]
consider what happens when there are an odd number of numbers, and an even numbers. Where is the arithmetic mean if there is an odd number of numbers? What if there is an even number of numbers?[/hide]

Is the 1st 10 in the problem in base 9?
Either way, I count 11 sticks.

[hide="2"]
The only configuration that doesn't work is $5_{9},6_{9},12_{9}$.  In base 10, those would be $5_{10}, 6_{10}, \text{and}~ 11_{10}$.  Since there is 1 way to choose the $5_{9}$ stick, 2 ways to choose the $6_{9}$ stick, and 1 way to choose the $12_{9}$ stick, there are only 2 ways that don't work.  There are $\binom{11_{10}}{3_{10}}$, or $165_{10}$ total ways.  So the answer is $\frac{2_{10}}{165_{10}}$[/hide]

[quote="alanchou"][size=200]Problem 2 [/size]

All numbers in this problem are in base 9. There are 10 sticks on a table. The lengths of the sticks are 5,6,6,7,7,7,8,8,10,10, and 12. Find the probability, expressed as a common fraction, that if you pick 3 random sticks, you can not form a triangle. Express your answer in base 10.  Assume that degenerate triangles do not count as triangles.[/quote]

But when it says ALL NUMBERS IN THIS PROBLEM ARE IN BASE 9, doesn't that mean, express your answer in base $10_{9}$? or Base $9_{10}$
[hide]
So it should be $\frac{2_{9}}{203_{9}}$ I think.[/hide]

However, like CircleSquared said, there are 11 sticks, and 10 in base 9 is 9
.9 does not equal 11

Yeah, lots of mistakes on my part.

Shame, can't edit it anymore.

So let's just forget it now.

[b][color=blue][size=150]Problem 3[/size][/color]
[/b]

How many integers between 1 and 44100 exclusive are not divisible by the first 4 primes?

btw. Why couldn't you edit it?

[quote="CircleSquared"]@magixter: You hear a loud *BANG*.  Then another. And another.  :wallbash:

@Anaxerzia: Where did you get $4^{6}$?[/quote]

What do you mean? Anyway,

[size=150][color=red]Question 10[/color][/size]

How many 3-digit integers not divisible by 5 have 5 or 7 as any of their digits?

328, I have no idea what I did with complementary counting.

Well, to answer your question mathcrazed, I think that it's any of the digits in any of the places.

[quote="mathcrazed"]wait, only 5 or 7 as their digits? can you explain[/quote]

is 328 correct? If it is I can explain..

328 is correct. Let me restate the problem.

How many 3 digit integers satisfy these 2 conditions simultaneously?

1. It is not divisible by 5
2. Let's say the 3-digit integer is $xyz$ as $x$ being the hundredth digit, $y$ tens digit, $z$ unit's digit.

    $x=5$ or $y=5$ or $z=5$ or $x=7$ or $y=7$ or $z=7$

Adapted from the other marathon...

[size=200][b]Problem 11[/b]
[/size]
A number is considered odd-continous if all odd digits in the number have another odd digit immediately following them, unless the odd digit is the last digit. For example, 6111, 5531, 2413, and 2420 are odd-continous integers. How many 4-digit odd-continous integers are there?

is it [hide]2625?
I got this because the possible configurations are :
EEEE, EEEO, EEOO, EOOO, and OOOO, = 4*5*5*5 + 4*5*5*5+4*5*5*5+4*5*5*5+5*5*5*5[/hide]

Correct. Can somebody post the next problem?

another countdownish question:
PROBLEM 12:

 What is the sum of the finite series
 1+2-3+4+5-6+7+8-9+ ... +151+152-153 ?

i believe that is correct, can you post the next problem?

[color=blue]Problem 13[/color]

Find 1/(1*2*3*4)+1/(2*3*4*5)+.....+1/(10*11*12*13) and express it as a common fraction.


Because of parallel sides and summed lengths, it's just $ 8\plus{}8\plus{}7\plus{}7\plus{}4\plus{}4\equal{}\boxed{38}$.

(Note how the 3 vertical sides on the right have lengths summing to 7, the bottom horizontal side has length 8, and the two interior horizontal sides both have length 4)

Hi;

Sama as a square:

2 * 8 + 2 * 7 but with additionally 2 * 4:

16 + 14 + 8 = 38

[quote="markoboy"]same as square 8+8 each side 7+7 each side 4+4 same side add all and get 38

[hide="the answer is here"]38[/hide]

[size=50][hide="im as small as an ant and you wont see anything beyond this he he : )"][color=#FFFFFF]YOU CANT SEE ME HA HA HA !! : )[/color][/hide][/size]

[hide="at least you can see me now"][color=#008000]PLEASE RATE ME POST THANK YOU SO MUCH!!![/color][/hide]

 :showoff:+ :play_ball: = :stretcher: --> :wacko:[/quote]

"YOU CANT SEE ME HA HA HA !! : )"

I see what you did there!

That makes so much more sense now! Thanks gauss1181!

[quote=markoboy]same as square 8+8 each side 7+7 each side 4+4 same side add all and get 38

[hide="the answer is here"]38[/hide]

[size=50][hide="im as small as an ant and you wont see anything beyond this he he : )"][color=#FFFFFF]YOU CANT SEE ME HA HA HA !! : )[/color][/hide][/size]

[hide="at least you can see me now"][color=#008000]PLEASE RATE ME POST THANK YOU SO MUCH!!![/color][/hide]

 :showoff:+ :play_ball: = :stretcher: --> :wacko:[/quote]

Yes I can

[quote=markoboy]same as square 8+8 each side 7+7 each side 4+4 same side add all and get 38

[hide="the answer is here"]38[/hide]

[size=50][hide="im as small as an ant and you wont see anything beyond this he he : )"][color=#FFFFFF]YOU CANT SEE ME HA HA HA !! : )[/color][/hide][/size]

[hide="at least you can see me now"][color=#008000]PLEASE RATE ME POST THANK YOU SO MUCH!!![/color][/hide]

 :showoff:+ :play_ball: = :stretcher: --> :wacko:[/quote]

lol, just open the second post and highlight


Έστω $ a,b,c$ θετικοί πραγματικοί. Να αποδείξετε ότι $ \frac{a\plus{}2b}{c\plus{}2b}\plus{}\frac{b\plus{}2c}{a\plus{}2c}\plus{}\frac{c\plus{}2a}{b\plus{}2a}\geq 3$

Ισοδυναμα,
$ \sum \frac{a\minus{}c}{c\plus{}2b} \geq 0 \leftrightarrow 2 \sum a^3\plus{}2\sum a^2c \geq 9abc \plus{} \sum b^2c$
που ισχυει αφου ειναι γνωστες οι ανισοτητες,
$ \sum a^3 \geq \sum a^2b$
και
$ \sum a^3\plus{} 2\sum a^2b \geq 9abc$

[quote="Athinaios"]Ισοδυναμα,
$ \sum \frac {a \minus{} c}{c \plus{} 2b} \geq 0 \leftrightarrow 2 \sum a^3 \plus{} 2\sum a^2c \geq 9abc \plus{} \sum b^2c$
που ισχυει αφου ειναι γνωστες οι ανισοτητες,
$ \sum a^3 \geq \sum a^2b$
και
$ \sum a^3 \plus{} 2\sum a^2b \geq 9abc$[/quote]

gia kane kai ena expand thn anisotita opos einai na deis ti prokuptei :)


I was sitting in economics class today reading this girls mathbook, and at the end it had some series studies... At one point it gave the nth fibinacci number as an explicit function. This was something I had never seen before, and the book did provide a proof.

nth fibinacci number: F(n) = 1/:rt5:*(((1+:rt5:)/2)^n-((1-:rt5:)/2)^n)

Can someone provide me with a full proof on how to go from

F_n = F_n-1 + F_n-2 to that function... I just can't even see where to start.

If it's to much to post and you could provide a link for me to read it on the web somewhere, please do that.

Thanks.

I don't know if you learned about linear recurrence ...

From the equation
F(n + 2) - F(n + 1) - F(n) = 0
we derive the caracteristic equation
x - x - 1 = 0.
This equation has roots
A = (1 + :sqrt:5)/2, B = (1 - :sqrt:5)/2.
Hence a general formula for the n - th term is given by
F(n) = p(A^n) + q(B^n)
with p and q constants. To find p and q we plug in F(1) = F(2) = 1.
Then we find that p = 1/:sqrt:5 and q = -1/:sqrt:5.
We're done !

If you can get your hands on an AoPS2, you can read a more detailed explanation of what Arne did starting at pg 244.

(We'll also be going into this method & its usefulness in problem solving in the Intermediate Counting class).

I knew how to find the roots and everything up to that point, i did not know however that:

"Hence a general formula for the n - th term is given by
F(n) = p(A^n) + q(B^n) "

is this true for all sequences?

I hope to get those books soon.

Yep - just solve the caracteristic equation and plug in two given values to determine your constant coefficients.

This can be extended of course ... for example, recursions of the form u(n + 3) = a u(n + 2) + b u(n + 1) + c u(n) ... :)

If you don't want to be relying on stuff you can't prove easily and probably have to take on faith, you can use generating functions. What you do is to write F_(n+2)=F_(n+1)+F_n and F_0=0, F_1=1. Then we multiply the first equation by x^n to get F_(n+2)*x^n=F_(n+1)*x^n+F_n*x^n and then sum everything from n=0 to infinity. Let us then write sum(F_n*x^n) from to infinity as S(x). Then the equation transforms into

(S(x)-x)/x^2=S(x)/x+S(x),

and then we can solve for S(x) and decompose it into partial fractions to get the formula. By the way, the formula is called Binet's formula, and it's sometimes useful to know that name.

(If you're worried about convergence issues when we sum infinite series, which we usually ignore while manipulating them, pick some small value for x and convince yourself that the sums will converge. For example, let x=.1, and then it must converge.)

[color=green]Once you have the answer, by the way, you can throw a bit of reasoning to explain why that form should cover everything.  First, the powers of the two roots of the quadratic will satisfy the recursion.  From this, we can show without much difficulty that any linear combination of the two roots will work.  Then, since we can make a linear combination of the two roots to satisfy any starting conditions (that is, if you give me any values for the first 2 numbers in the Fibonacci sequence, I can give you a linear combination of the powers of the two roots that gives those two numbers as consecutive terms) I can use it to make any similar recursion.  So every linear combination of the powers of the two roots will give you the fibonacci recursion, and any fibonacci recursion can be expressed in such a form.  [/color]

basically
Arne's technique can be used to find a closed form expression for any recursively defined sequence

it works for fibonacci, Lucas, and Catalan

Catalan requires a bit more ingenuity than Fibonacci and Lucas I think. I don't know of any very simple proof for the explicit formula for Catalan numbers.

hmm
yea
that's true, but it's still rather doable


How will u [b]show that $A$ is a triangular matrix, then $A$ is nonsingular if and only if $a_{ii} \neq 0$[/b] by using the conditions of $(1)$ the columns of $A$ are linearly independent and $(2)$ the rows of $A$ are linearly independent ($A$ is a square matrix)

Please help, thx

I guess I shall ask how to show that if the columns of $A$ are linearly independent then $A^{-1}$ exists

This is a well-known result.  If the columns of a matrix are linearly independent, then you can row reduce it into upper triangular form so that all of the entries on the diagonal are nonzero.  This implies that the determinant is nonzero, and hence the matrix is invertible.

There are about a million different ways you can establish those equivalences, but that one seems fairly natural to me.

For the other part, you can use the fact that row rank equals column rank.


Let $k$ be a positive integer.
$S = \{(a,b)|1\leq a<b\leq 2k, a,b\in N \}$

Prove that $S$ can be partitioned into $2k-1$ sets such that the ordered pairs in each set cover all the numbers from $1$ to $2k$.

Did I miss something or it is quite trivial?  ;)

Just consider the pairs as integer points, and consider the $2k-1$ sets $S_i$ , for $i=1,2, \cdots , 2k-1$, where $S_i$ is formed by the points on the lines $y=x+i$ and $y=x+i-2k$.

Why do we have to suppose that we use $2k$ instead of any positive integer?

Pierre.

[quote="pbornsztein"]$y=x+i-2k$.
[/quote]

I think the $x$ co-ordinate should be less than the $y$ co-ordinate


In how many ways can the integers from $1$ to $2006$ be divided into three nonempty disjoint sets so that none of these sets contains a pair of consecutive integers?

Haven't been here for a long time now. When people post questions, am I supposed to try to answer them?

[hide="A guess?"]
Do this by recursion:
Suppose we want to divide 1, 2, ... , [i]n[/i] into three sets. Let there be $a_{n}$ ways to do this.

Case 1:[i]n[/i] is in it's own set
Therefore 1, 2, ... , [i]n[/i]-1 are divided into two sets with no consecutive integers. There is one way to do this; one set has odds and the other sets has the evens.

Case 2:[i]n[/i] is not in it's own set.
Take the three sets, and remove [i]n[/i] from the set that it is in. We then have the case for [i]n[/i]-1. We can map each case in [i]n[/i]-1 to two cases of [i]n[/i], because we can either put  [i]n[/i] in either set that does not contain [i]n[/i]-1. There are $2 a_{n-1}$ ways to do this.

Therefore we have $a_{n}= 2 a_{n-1}+1$. It's easy to see that $a_{3}= 1$. Then using the formula $a_{4}= 3$, $a_{5}= 7$, $a_{6}= 15$, $a_{7}= 31$. So it's a safe guess that $a_{n}= 2^{n-2}-1$.

(I have an inelegant proof of this, expand out $a_{n}= 2 a_{n-1}+1 = 2 (2 a_{n-2}+1)+1 = 2^{2}a_{n-2}+2+1$ which will eventually expand to $2^{n-3}a_{3}+2^{n-3}+2^{n-1}+\cdots+1 = 2^{n-3}+2^{n-3}-1 = 2^{n-4}-1$).

So the final answer is [b]2^2004 - 1[/b].[/hide]

I had another way to solve it. Assume that the three sets are three boxes in which you put the integers from $1$ to $2006$. We start with number $1$, so we can choose between $3$ boxes. For number $2$ we only have $2$ possibilities, because none of the boxes may contain a pair of two consecutive integers, this is the same for number $3$ and so on. So we have $3.2^{2005}$ possibilities. But now we are forgotten that the boxes can't be empty. It could be, that one box is empty, two is impossible, because all consecutive integers would be in the same box then. In the case we have one empty box, another contains all even numbers and another all odd numbers. We can order these three boxes in $3!=6$ ways, so there are $6$ cases in which one box is empty.
Finally, we must realize that the boxes can be moved in another combination, so we have $\frac{3.2^{2005}-6}{6}$ ways to create three sets. And this is equal to $2^{2004}-1$.
Correct me if I made a mistake.

Even without the 6: $1$ and $2$ take a category each, for the other $2004$ integers we have each time 2 choices (for $k$: the two sets $k-1$ is not in). And there's only 1 such way that leaves the third set empty, so $2^{2004}-1$.

[quote="FieryHydra"][quote="Myself"]Haven't been here for a long time now. When people post questions, am I supposed to try to answer them?
[/quote]
Well, if you know the solution in advance, then providing a hint or two to others may be a good idea. If you solved it yourself, then I guess you could post the solution.[/quote]

Ah, thanks! As a matter of fact, I did do this one by myself.

And I like your way, Peter VDD. Very nice :)

Thanks, but my method is the same as Aralc's, just removing the unnecessary ordering part.

My solution is the same as Aralc's, because at first, during the competition, I forgot about the ordering part. So my original answer was $3\cdot 2^{2005}-6$, but then I had 3 hours left (yes, I finished in 1.5 hours - it was that easy) to check my answers and saw the mistake.


In the $ 6\times4$ grid shown, $ 12$ of the $ 24$ squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column.  Let $ N$ be the number of shadings with this property.  Find the remainder when $ N$ is divided by $ 1000$.
[asy]size(100);
defaultpen(linewidth(0.7));
int i;
for(i=0; i<5; ++i) {
draw((i,0)--(i,6));
}
for(i=0; i<7; ++i) {
draw((0,i)--(4,i));
}[/asy]

Brute force?
[hide]I represented the shaded boxes in each row as A,B,C,D
so we must have 3 sets of paired rows such as AB/CD or AC/BD.
Pattern one is AB and CD
Pattern two is AC and BD
Pattern three is AD and BC

For the 4 by 6 box, we must have 3 pairs of those things for a total of 12 shades boxes.

We have 3 scenarios:

1) We have 3 of the same pattern:
Ways: 6C3 times 3 because there are 3 patterns.
= 60

2) We have 2 of the same pattern, so we have to :
Ways: 3C1 * 2C1 * 6!/(2!2!)
=3*2*180 = 1080

3) We have one of each pattern:
6! = 720

Total = 60 + 1080 + 720 = 1860.

Answer: 860[/hide]

[hide="Here is how I saw it"]
Consider the first column.  There are $ {6\choose3} \equal{} 20$ ways that the balls could be chosen, but WLOG let them be the first three rows.  (Change the order of the rows to make this true.)  We will multiply whatever answer we get by 20 to get our final answer.

Now consider the 3x3 that is next to the 3 boxes we have filled in.  We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each).  We split into three cases:

*All three balls are in the same column.  In this case, there are 3 choices for which column that is.  From here, the bottom half of the board is fixed.

*Two balls are in one column, and one is in the other.  In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball.  Then, there are 3 ways to choose which row the lone ball is in.  Now, what happens in the bottom half of the board?  Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here.  In the column with two balls already, we can choose any of the 3 boxes for the third ball.  This forces the location for the last two balls.  So we have $ 3 \cdot 2 \cdot 3 \cdot 3 \equal{} 54$.

*All three balls are in different columns.  Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes.  (The location of the ball in column 4 is forced.)  Again, we think about what happens in the bottom half of the board.  There are 2 balls in each row and column now, so in the 3x3 where we still have choices, each row and column has one square that is '''not''' filled in.  But there are 6 ways to do this.  So in all there are 36 ways.

So there are $ 20(3 \plus{} 54 \plus{} 36) \equal{} 1860$ different shadings, and the solution is [b]860[/b].
[/hide]

[quote="besttate"][hide]
Here's how I thought of it:

First, pick the first column. There are $6C3=20$ ways to do this. Now we break it into cases based on the second column:
In the second column (with respect to the first column)...

[b]Case 1: All of the balls line up[/b]
There is one way to chose the second column, then 1 more way to arrange the next two. So this gives us $20$ ways.

[b]Case 2: Exactly two pairs of balls line up[/b]
There are $3C2$ ways to pick the balls that line up, then $3C1$ ways to pick where the last ball goes. Then, we need to place one ball in the third column, and there are $2C1$ ways to do this. $20 \cdot 3 \cdot 3 \cdot 2 = 360$.

[b]Case 2: Exactly one pair of balls line up[/b]
There are $3C1$ ways to pick the ball that lines up, then $3C2$ ways to pick where the last ball goes. Then, we need to place two balls in the third column, and there are $4C2$ ways to do this. $20 \cdot 3 \cdot 3 \cdot 6 = 1080$.

[b]Case 3: No balls line up[/b]
There is one way to chose the second column. Then, we need to place three balls in the third column, and there are $6C3$ ways to do this. $20 \cdot 20 = 400$.

[b]Answer: $20+360+1080+400=1860$[/b]
[/hide][/quote]

Hey, that’s exactly how I did it, except I beat you to it on the AoPSWiki [url=http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_I_Problems/Problem_10]2007 AIME I Problems/Problem 10[/url].

[quote="besttate"][hide]
Here's how I thought of it:

First, pick the first column. There are $6C3=20$ ways to do this. Now we break it into cases based on the second column:
In the second column (with respect to the first column)...

[b]Case 1: All of the balls line up[/b]
There is one way to chose the second column, then 1 more way to arrange the next two. So this gives us $20$ ways.

[b]Case 2: Exactly two pairs of balls line up[/b]
There are $3C2$ ways to pick the balls that line up, then $3C1$ ways to pick where the last ball goes. Then, we need to place one ball in the third column, and there are $2C1$ ways to do this. $20 \cdot 3 \cdot 3 \cdot 2 = 360$.

[b]Case 2: Exactly one pair of balls line up[/b]
There are $3C1$ ways to pick the ball that lines up, then $3C2$ ways to pick where the last ball goes. Then, we need to place two balls in the third column, and there are $4C2$ ways to do this. $20 \cdot 3 \cdot 3 \cdot 6 = 1080$.

[b]Case 3: No balls line up[/b]
There is one way to chose the second column. Then, we need to place three balls in the third column, and there are $6C3$ ways to do this. $20 \cdot 20 = 400$.

[b]Answer: $20+360+1080+400=1860$[/b]
[/hide][/quote]
Why are there two case 2's?

Reviving thread!

In Beststate's solution--> Case 2,
At the end, he has 2 choose 1 and in the second case 2, he has 4 choose 2. Why?
Isn't it 4 choose 3 and 5 choose 3 respectively?

We start with the first column and there are ${6\choose 3}$ ways we can shade 3 squares.

Now we have 4 cases.

Case 1: In the next column, the next three shaded squares are in the same row as the previous 3 shaded squares

For this case we have ${6\choose 3}$ ways.

Case 2: In the next column, two of the next three shaded squares are in the same row as two of the previous three squares

For this case we have ${6\choose 3}{3\choose 2}{3\choose 1}{2\choose 1}$ ways.

Case 3: In the next column, one of the three shaded squares is in the same row as another one of the previous three squares

For this case we have ${6\choose 3}{3\choose 1}{3\choose 2}{4\choose 2}$ ways.

Case 4: In the next column, none of the three shaded squares is in the same row as either of the previous three squares

For this case we have ${6\choose 3}{6\choose 3}$ ways.

Adding all 4 of these cases we get $20 + 360 + 1080 + 400 = 1860$ mod $1000 = \boxed{860}$.

When I frst saw this problem, I thought I had to use casework from top to bottom.

But then I realized that I could go from left to right! And then I got my answer :D

I really liked this problem, anyone else?   :love:

The trick is actually common:
Think column-wise.First column can be coloured in $\binom{6}{3} = 20$ ways .Now note that there are 4 types of colouring for the 2nd column
Case $I$
 The colour of the second columnis exactly same to the first columns and then note that the remaining two columns can be coloured uniquely and done!
Case $(ii)$
The colour of the second column is actually complementary with the first row and that makes the whole cycle repeated .The third column can be coloured in $20$ ways and the fourth can be coloured uniquely with the complementary colour of the third row
Case$(III)$
The colour of second column coincides with the first column in exactly $1$ square.Then this colouring can be done in $9$ ways and the Third column can be coloured in $6$ ways then
$Case(IV)$
The colour of second column coincides with the first column in exactly $2$ squares.Then this colouring can be done in $9$ ways and the Third column can be coloured in $2$ ways then
So total colouring is $20[1+20\times 1 +3\times 3\times 6 + 3 \times 3 \times 2 ]$


Let a quadrilateral ABCD, AC meets BD at K, $ \angle ABC \equal{} \angle ADC \equal{} 90^o$, AC=AB+AD. Prove that the radii of incircles of triangles ABK,ADK are equal.

Wow! This problem is very nice, but I don't have a solution in my mind  :(  Who solved?

Let $ r_1,r_2$ be radii of incircles of triangle $ ABK,ADK$ respectively.
$ r_1 \equal{} r_2$
<=>$ \frac {KB}{KD} \equal{} \frac { AB \plus{} AK \plus{} KB}{AD \plus{} AK \plus{} KB}$
<=>$ \frac {KB}{KD} \equal{} \frac {AB \plus{} AK}{AD \plus{} AK }$
<=>$ \frac {KB^2}{KD^2} \equal{} \frac { (AB \plus{} AK)^2 \minus{} KB^2}{(AD \plus{} AK)^2 \minus{} KD^2}$
<=>$ \frac {KB^2}{KD^2} \equal{} \frac {(1 \plus{} cos\angle BAC)AB}{(1 \plus{} cos\angle DAC).AD}$
<=>$ \frac {AB^2}{AD^2}.\frac {sin^2 \angle BAC}{sin^2 \angle DAC} \equal{} \frac {(1 \plus{} cos\angle BAC)AB}{(1 \plus{} cos\angle DAC).AD}$
<=>$ \frac {AB}{AD} \equal{} \frac {cos \angle {BAC} }{cos \angle DAC}$  (Since $ cos \angle BAC \plus{} cos \angle DAC \equal{} 1$ )
Done :D
[hide]
Chào anh thuận :P .Lâu hem gặp ^^
[/hide][/hide]

For a nice problem, you had better solve it without trigonometry. I have a proof which uses Carnot's theorem and I'd like to have Mathlinkers post synthetic proofs.


vector
linear algebra
matrix
linear algebra unsolved

I am hoping someone can explain more with the question about vector space as following. Thanks alot.

Let $V=k^{2}$ Define coordinatewise addition on $V$, and scalar multiplication: $\alpha \begin{bmatrix}x\\y \end{bmatrix}= \begin{bmatrix}\alpha x \\ 0 \end{bmatrix}$ ; $\alpha \in K$. $\begin{bmatrix}x \\y \end{bmatrix}\in K^{2}$

Is $V$ a vector space to these operations?

No it is not. Since we have  $1\begin{bmatrix}x\\y \end{bmatrix}= \begin{bmatrix}1 x \\ 0 \end{bmatrix}$, we don't even have the multiplicative identity element property in $V$.

Hi, thx for your explaination.

I thought I had to show all the steps of vector space (I think it's 8 steps). But what you mentioned about the " we don't even have the multiplicative identity element property in $V$ "

Could you explain more? Thx in advanced.  :)

One of the steps you mention is the property $1.x = x$ for all $x\in V$, and this is not valid here.. :)


With a is a constant,calculate
$ \sum\limits_{i \equal{} 1}^{\infty }\frac {1}{1 \plus{} a^{i}}$

hello, for $ 0<a\le1$ your sum diverges and for $ a>1$ Maple 12 gives $ \sum_{i\equal{}1}^{\infty}\frac{1}{1\plus{}a^i}\equal{}\minus{}\frac{\psi _{\frac{1}{a}}^{(0)}\left(\frac{i \pi \minus{}\log (a)}{\log \left(\frac{1}{a}\right)}\right)}{\log \left(\frac{1}{a}\right)}\plus{}\frac{\log (a\minus{}1)}{\log (a)}\minus{}1$.
Sonnhard.

hello, for $ \minus{} 1 \le a\le 0$ the sum does also not converge.
Sonnhard.


Let be $ a,b,c\in \mathbb{R}_\plus{}^*$ such that $ a\plus{}b\plus{}c\equal{}1$ . Show that :

$ 3\plus{}\sum_{cyc}\frac{(a\minus{}b)^2\plus{}(a\minus{}c)^2}{1\plus{}a}\le 4(a^2\plus{}b^2\plus{}c^2)\left(\sum_{cyc}\frac{1}{a\plus{}1}\right)$

[quote="alex2008"]Let be $ a,b,c\in \mathbb{R}_ \plus{} ^*$ such that $ a \plus{} b \plus{} c \equal{} 1$ . Show that :

$ 3 \plus{} \sum_{cyc}\frac {(a \minus{} b)^2 \plus{} (a \minus{} c)^2}{1 \plus{} a}\le 4(a^2 \plus{} b^2 \plus{} c^2)\left(\sum_{cyc}\frac {1}{a \plus{} 1}\right)$[/quote]
The inequality is equivalent to
$ \sum\limits_{cyc} {\frac{{1 \plus{} a \plus{} 2a^2  \plus{} b^2  \plus{} c^2  \minus{} 2a\left( {b \plus{} c} \right)}}{{1 \plus{} a}}}  \le 4\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)\left( {\sum\limits_{cyc} {\frac{1}{{a \plus{} 1}}} } \right) $
  $ \Leftrightarrow \sum\limits_{cyc} {\frac{{4\left( {a^2  \plus{} b^2  \plus{} c^2 } \right) \minus{} \left( {1 \plus{} a \plus{} 2a^2  \plus{} b^2  \plus{} c^2  \minus{} 2a\left( {b \plus{} c} \right)} \right)}}{{a \plus{} 1}}}  \ge 0 $
  $ \Leftrightarrow \sum\limits_{cyc} {\frac{{a \minus{} 1 \plus{} 3\left( {b^2  \plus{} c^2 } \right)}}{{1 \plus{} a}}}  \ge 0 $
Used that $ b^2  \plus{} c^2  \ge \frac{{\left( {b \plus{} c} \right)^2 }}{2}$ and $ a\plus{}b\plus{}c\equal{}1$
It sufficies to prove that:
$ \sum\limits_{cyc} {\frac{{\left( {b \plus{} c} \right)^2 }}{{1 \plus{} a}} \ge \sum\limits_{cyc} {\frac{{2a\left( {b \plus{} c} \right)}}{{1 \plus{} a}}} }$
But it's true because $ \sum\limits_{cyc} {\frac{{\left( {b \plus{} c} \right)^2 }}{{1 \plus{} a}}}  \ge \frac{{\left( {2\left( {a \plus{} b \plus{} c} \right)} \right)^2 }}{{3 \plus{} a \plus{} b \plus{} c}} \equal{} 1 $
 $ \sum\limits_{cyc} {\frac{{2a\left( {b \plus{} c} \right)}}{{1 \plus{} a}} \le \sum\limits_{cyc} {\frac{{\left( {2a \plus{} b \plus{} c} \right)^2 }}{{4\left( {1 \plus{} a} \right)}} \equal{} } } \sum\limits_{cyc} {\frac{{1 \plus{} a}}{4} \equal{} 1}  $ :)  :rotfl:

Let be $ a,b,c\in \mathbb{R}_ \plus{} ^*$ such that $ a \plus{} b \plus{} c \equal{} 1$ . Find the greated coefficient $ k$ such at :

$ 3 \plus{} k.\sum_{cyc}\frac {(a \minus{} b)^2 \plus{} (a \minus{} c)^2}{1 \plus{} a}\le 4(a^2 \plus{} b^2 \plus{} c^2)\left(\sum_{cyc}\frac {1}{a \plus{} 1}\right)$


Does there exist $k\in\mathbb{Q}_+\setminus\{1,2\}$ such that $\sqrt[k-1]k\in\mathbb{Q}$?

[quote="ifai"]Does there exist $k\in\mathbb{Q}_+\setminus\{1,2\}$ such that $\sqrt[k-1]k\in\mathbb{N}$?[/quote]

For K≥3, we have that (K-1)√K became closer and closer to 1 when K-->∞. So, the answer is NO.

it will probabilly less trivial to ask if $k^{\frac{1}{k-1}}$ is a rational number.

i had made a mistake and i corrected the question.


vector
linear algebra
linear algebra unsolved

let $E$ be a $K$ vectorial space and $f(x,y)$ a bilinear form form.
A vectorial space $F$ is called a SETI if $f(x,x)=0$ for all $x \in F$.
A vectorial space $F$ is called a SETIM if it is a SETI and if there is no other SETI that contains $F$ strictly.

Prove that all SETIM have the same dimension

Ok, here it goes. Denote $U^{*}$ the orthogonal of the subspace $U$ and $ i(V_1,V_2)$ the intersection of $V_1$ and $V_2$.
  Lemma 1
    If $U,V$ are SETI and $x\in i(U, V^{*})$, then $V+Kx$ is also a SETI.
   Proof:
    Simple computation.

   Second lemma
      Let $U,V$ two SETI and choose $F,G$ supplementaries of $i(U,V)$  in $U$ and $V$. Then $i(F, G^{*})$ is included in $i(U,V^{*})$.
    Proof:
     Suppose $x\in i(F,G^{*})$. Of course, $x\in U$. Now, let $v\in V$ and write $v=s+g$ with $s\in i(U,V), g\in G$. Since of course $ f(x,y)=0$ for any $x,y$ belonging to the same SETI, it follows that $ f(x,v)=f(x,s)+f(x,g)=0$. This shows that $ f(x,v)=0$ and thus $x\in V^{*}$.
    The lemma is proved.

    Now, observe that $dim i(F, G^{*})\geq dim F-dim G$.
    It is not difficult to prove that any SETI is in fact included in a SETIM. Now, suppose we have $U,V$ two SETIM of different dimension, WLOG $dimU>dimV$. Choose $F,G$ two supplementaries of $i(U,V)$ in $U,V$. From the previous observation about the dimension we find that $i(F, G^{*})$ is not reduced to ${0}$. Thus we can find some non-zero vector $x$ in it. From the second lemma we have that this $x$ is in $i(U,V^{*})$ and thus from lemma 1 $V+Kx$ is a SETI, which strictly contains $V$ (since $x$ is not in $V$), contradiction.

Continuing this nice  problem: prove that  that the same dimension of all the SETIMs is $\leq \frac{n}{2}$, where $n=dimE$

That's easy, since the sum of dimensions of any subspace and its orthogonal is at most n and for any SETI, the orthogonal contains the space.

I think the following result would fit here:
$p+q+n=\dim E$
Where $p$ - the highest dimension of the space where $f$ is positive definite, $q$ for negative definite, and $n$ for null....

harazi , why we have dim(i(F,G*)>=dim(F)-dim(G) , i can't see this !! i dont think what you have told eugene is true ...it's a least n not at most !!


I desperately need a 2-month Xbox Live trial card.  They come inside XBOX Live-enabled games, such as Halo 2.  If anyone has one that they don't want, can you please PM the scratch-off code to me.  I can give you something in return.  Just ask.

Thanks!

You can play XBOX over the internet, on your tv.

talk about stealing... :diablo:
go buy xbox live if you like it so much.

No... I need the code for a promotion in my clan...

[quote="DarkKnight"]I think I might have a code if I can find it. I bought halo 2 and i'll probably never play on Xbox live. I'll look for it.[/quote]

Thanks!


Is it true that for all $ a_1,\ldots,a_n \geq 0$ we have
\[ (a_1\plus{}2a_2\plus{}\cdots\plus{}na_n)^n \geq a_1 \cdot (a_1\plus{}4a_2) \cdots (a_1\plus{}\cdots \plus{}(n\minus{}1)a_{n\minus{}1}\plus{}n^2a_n) \ ?\]

No
Take a_1,...a_n to be integers and a_1 to be prime larger that
2a_2+...+na_n. Then (a_1+...+na_n) n is not divisible by a_1, and the right side of your identity is.

yeah this would be a great  and neat solution if only ... harazi would have posted an identity. Unfortunately he posted an inequality (>=). :D

Well, in fact  it is " extremely difficult" to see that it cannot be an identity.

Nobody has any answer? Unfortunately, I have made this problem long time ago and I don't know anymore the idea. But I remember the answer was yes!

Does any one have a solution iam eager to know it!
i think the problem is
\[ (a_1  \plus{} 2a_2  \plus{} ... \plus{} na_n )^n  \ge a_1 *(a_1  \plus{} 4a_2 )*...(a_1  \plus{} ... \plus{} (n \minus{} 1)a_{n \minus{} 1}  \plus{} n^2 a_n )
\]

\[ \forall {\rm{ }}a_1 ,...,a_n  \ge 0{\rm{ }}
\]

Maybe I am missing something, but this seems like a simple application of AM-GM,
so the answer is yes.


topology
abstract algebra
advanced fields
advanced fields unsolved

Hello,

I am trying to prove that I can write every nonempty closed subset of a Noetherian topological space as a finite union of distinct irreducible closed sets (without one containing the other), in a unique way.

It is that uniqueness that I have a problem with. :huh: It seems to me that they do this :

The closure of $(A \cup B)\backslash B)$ with $A,B$ closed, and $B$ irreducible, is $A$.

Is this true? Is the noetherian property necessary?

[quote="owk"]No, take $A=B$. owk[/quote]
 :D [quote="fredbel6"] (without one containing the other)[/quote]

So my problem should have been:

The closure of $(A \cup B)\backslash B)$ with $A,B$ closed and $A$ not contained in $B$, and $B$ irreducible, is $A$.

What do you think now? :blush:

Certainly $\overline{A\setminus B}\cup(A\cap B)=A$, so if $A$ is irreducible and $A\nsubseteq B$, then $\overline{A\setminus B}=A$. For your statement, take $B$ something irreducible not consisting of a single point and containing a closed point $b$. Take the "big" space to be the disjoint union of $B$ and some nonempty space $Y$, and $A=Y\cup\{b\}$. owk


For positive reals $x_{1},x_{2},\cdots,x_{n}$ such that $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n}}=1$. Prove that:

$x_{1}x_{2}\cdots x_{n}\geq\left(n-1\right)^{n}$

WooH :D That was Fast,  Nice Quick Solution...

the other is:

Let $y_{i}=\frac{1}{1+x_{i}}$ for all $1\leq i\leq n$ then:

$\cdots\cdots\cdots$

it'll be completely like your solution but in other Form... :lol:


Let F=(6x^2+16x+3m)/6 be the square of an expression which is linear in x. Then m has a particular value between...

A)3 and 4
B)4 and 5
C)5 and 6
D)-4 and -3
E)-5 and -6

[color=cyan]Where did this question come from?

And, no calculators!  (I think).[/color]

teratomato wrote:A.

Agreed. Still, please pleeeeaaaaase use spoilers! :evil:

Here's how I got it:[hide]Given equation   :sqrt: x:^2:+(8/3)x+(1/2)m  (x+4/3) :^2:  (1/2)m=(4/3) :^2:   m=32/9=3.555... :approx: 3.6

So the answer is A.[/hide]


For what values of the variable $x$ does the following inequality hold: \[ \dfrac{4x^2}{(1-\sqrt{2x+1})^2}<2x+9 \ ?  \]

We also have
$2x+1\geq 0$
$x\geq -\frac12$
So the solution is $-\frac12\leq x<0$ or $0<x<\frac{45}{8}$

Note that the function $f(x)=(1+\sqrt{1+2x})^2-2x-9=2\sqrt{1+2x}-7$ is increasing function. Also $f(\frac{45}{8})=0$

[hide=Solution]We have $4x^2 < (2x+9)(2x-2\sqrt{2x+1}+2)$. This expands and simplifies to $\sqrt{2x+1}(4x+18)<22x+18$. Squaring both sides gives $(2x+1)(4x+18)^2<(22x+18)^2$, which expands to $32x^3-180x^2<0$. Solving gives $\boxed{-\frac{1}{2} \leq x < 0, 0 < x < \frac{45}{8}}$ after noting that $2x-1 \geq 0$ and that $x$ cannot be $0$.[/hide]

We know that $2x+1\ge0$ so $x\ge-\frac12$.

We also know that $1-\sqrt{2x+1}\ne0$ so $x\ne0$.

We can multiply both sides of $\frac{4x^2}{(1-\sqrt{2x+1})^2}<2x+9$ by $\frac{(1+\sqrt{2x+1})^2}{(1+\sqrt{2x+1})^2}$ for a difference of squares to get $(1+\sqrt{2x+1})^2<2x+9$.

We can now solve the inequality to get
\begin{align*}
1+2\sqrt{2x+1}+2x+1 &< 2x+9 \\
2\sqrt{2x+1} &< 7 \\
\sqrt{2x+1} &< \frac72 \\
2x+1 &< \frac{49}4 \\
2x &< \frac{45}4 \\
x &< \frac{45}8
\end{align*}
so our answer is $-\frac12\le x<0$ and $0<x<\frac{45}8$.

[quote=OlympusHero][hide=Solution]We have $4x^2 < (2x+9)(2x-2\sqrt{2x+1}+2)$. This expands and simplifies to $\sqrt{2x+1}(4x+18)<22x+18$. Squaring both sides gives $(2x+1)(4x+18)^2<(22x+18)^2$, which expands to $32x^3-180x^2<0$. Solving gives $\boxed{-\frac{1}{2} \leq x < \frac{45}{8}}$ after noting that $2x-1 \geq 0$.[/hide][/quote]

x cannot be 0 :o

Henceforth assume $x\geq-\tfrac{1}{2}$ and $x\neq0$ to have $\sqrt{1+2x}\in\mathbb{R}$ and $(1-\sqrt{1+2x})^2\neq0$, respectively. As $(1-\sqrt{1+2x})^2$ is positive, we can multiply by it to get
$$4x^2<(1-\sqrt{1+2x})^2(2x+9)=(2+2x-2\sqrt{1+2x})(2x+9)=4x^2+22x+18-(4x+18)\sqrt{1+2x},$$
or, equivalently, $2x^2<2x^2+11x+9-(2x+9)\sqrt{1+2x}\iff(2x+9)\sqrt{1+2x}<11x+9$. Since all terms are positive as $x\geq-\tfrac{1}{2}$, we can square both sides to obtain an equivalent formulation: $(2x+9)^2(1+2x)<(11x+9)^2$. Now, the LHS equals
$$(2x+9)^2(2x+1)=(4x^2+36x+81)(2x+1)=8x^3+76x^2+198x+81,$$
while the RHS equals $121x^2+198x+81$, so we have
$$8x^3+76x^2+198x+81<121x^2+198x+81\iff8x^3-45x^2<0.$$
Dividing by $x^2$ since it is positive (recall the assumption $x\neq0$) gives $8x-45<0\iff x<\tfrac{45}{8}$. Therefore, the solutions to the inequality are given by $\boxed{x\in[-\tfrac{1}{2},\tfrac{45}{8})/\{0\}}$. $\blacksquare$

Rearranging we have that $(2x+9)(2x-2\sqrt{2x+1}+2)>4x^2.$ If we expand and simplify we get $\sqrt{2x+1}(4x+18)<22x+18$, now we square both sides to get that $(2x+1)(4x+18)^2<(22x+18)^2 \implies 32x^3-180x^2<0$, so $\boxed{-\frac{1}{2} \leq x < 0, 0 < x < \frac{45}{8}}.$


A local sandwich shop has three cash registers. In how many possible ways can four customers line up at the registers?

Yeah, yeah, yeah, this may sound embarrasing, but I kept on getting 3^4=81 ways.  But that's not the answer. Could someone explain the solution to me?

Hint:
[hide]Balls and Urns[/hide]
Solution:
[hide]I think this is right...
We have 3 registers, so there are 2 dividers, and four people.  In all, 4 people and 2 divisors.  There are 6C2 or 15 ways to order these, 15 ways for them to line up.  [/hide]

yeah... i see your logic, but the answer is incorrect.

The real answer is 360.

You forgot to multiply by 4!=24 because there are 24 ways to arrange the people....

But thanks!

Whoa.. I totally forgot about the people... :(

[quote="Quickster94"]Hint:
[hide]Balls and Urns[/hide]
[/quote] How exactly does it work again?


a student has a three liter unmarked container a five liter unmarked container and a unlimited supply of lemonade. how can she use them to measure out exactly four liters of lemonade? ( determine 2 methods of solving)

[hide][b]First method:[/b]
[u]3 liter container[/u][color=white]_____[/color][u]5 Liter container[/u]
0[color=white]________________[/color]0
0[color=white]________________[/color]5
Pour 3 liters of lemonade from the 5 liter container to the other:
3[color=white]________________[/color]2
Pour out the lemonade from the 3 liter container:
0[color=white]________________[/color]2
2[color=white]________________[/color]0
2[color=white]________________[/color]5
Pour 1 liter of lemonade from the 5 liter container to the other:
3[color=white]________________[/color][b]4[/b]

[b]Second method:[/b]
[u]3 liter container[/u][color=white]_____[/color][u]5 Liter container[/u]
0[color=white]________________[/color]0
3[color=white]________________[/color]0
0[color=white]________________[/color]3
3[color=white]________________[/color]3
Pour 2 liters of lemonade from the 3 liter container to the other:
1[color=white]________________[/color]5
Pour out the lemonade from the 5 liter container:
1[color=white]________________[/color]0
0[color=white]________________[/color]1
3[color=white]________________[/color]1
0[color=white]________________[/color][b]4[/b]
[/hide]


Well I think this complex is square-planar, so there are two isomers, the one where Cl and Br are next to each other (cis) and the one where Cl and Br are across (trans).  I am not sure if the trans isomer would be very stable as there would definitely be some angle strain, but you still should denote which isomer you are talking about.

Ok and for naming, you have to just apply the rules, which I have attached at the bottom of this post.  I found them at [url]http://wwwchem.uwimona.edu.jm:1104/courses/inorgnom.html[/url].  Anyway, first you have to determine what your ligands are.  You have a chloro, bromo, and ethylene diamine (en).  Arranging these in alphabetical order gives chlorobromo(en) as your prefix.  For the base/suffix, you need to consider the oxidation state on Pt.  The chloro and bromo ligands are both negative one, (en) is neutral, so Pt must be positive two.  The overall charge on the complex is zero.  Hence the total complex is chlorobromo(en)platinum(II).

1. In naming the entire complex, the name of the cation is given first and the anion second (just as for sodium chloride), no matter whether the cation or the anion is the complex species.

2. In the complex ion, the name of the ligand or ligands precedes that of the central metal atom. (This procedure is reversed from writing formulae.)

3. Ligand names generally end with 'o' if the ligand is negative ('chloro' for Cl-, 'cyano' for CN-, 'hydro' for H-) and unmodified if the ligand is neutral ('methanamine' for MeNH2).
Special ligand names are 'aqua' for water, 'ammine' for ammonia, 'carbonyl' for CO, 'nitrosyl' for NO.

4. A Greek prefix (mono, di, tri, tetra, penta, hexa, etc.) indicates the number of each ligand (mono is usually omitted for a single ligand of a given type). If the name of the ligand itself contains the terms mono, di, tri, eg triphenylphosphine, then the ligand name is enclosed in parentheses and its number is given with the alternate prefixes bis, tris, tetrakis instead.
For example, NiCl2(PPh3)2 is named dichloridobis(triphenylphosphane)nickel(II).

Again, one would use diammine, for (NH3)2, but bis(methylamine), for (NH2Me)2, to make a distinction from dimethylamine. (Note that this ambiguity does not arise if the preferred IUPAC name, methanamine, is used instead of methylamine).
There is no elision of vowels or use of a hyphen, e.g. in tetraammine and similar names. Some texts suggest that if a ligand is "complicated" then use the bis, tris multipliers. What constitutes "complicated" is not spelled out however, so a simpler approach is to use them if the name of the ligand is three or more syllables long!

5. A Roman numeral or a zero in parentheses is used to indicate the oxidation state of the central metal atom.

6. If the complex ion is negative, the name of the metal ends in 'ATE' for example, ferrate, cuprate, nickelate, cobaltate etc.

7. If more than one ligand is present in the species, then the ligands are named in alphabetical order regardless of the number of each. For example, NH3 (ammine) would be considered an 'a' ligand and come before Cl- (chloro). (This is where the 1971 rules differed from the 1959 rules. Some texts still say that ligands are named in the order: neutral then anionic).

Hello,
   I must say gregx007, has given an extremely detailed and informative answer to the question.Great work.
Sathej

Thanks... although I must say that I made a mistake in alphabetizing the ligands.  The correct name should be bromochloro(en)platinum(II).


I just copy-pasted from test forum....WHat are you congratulating me about? :rotfl:

awesome I didn't know you could use for loops on asymptote


algebra
polynomial
Galois Theory
group theory
superior algebra
superior algebra theorems

Suppose I have a polynomial of degree n over a field F (char F = 0).  What can be inferred about the polynomials or its roots if we say that its Galois group is S(n) ?  Similarly, what if the Galois group is not isomorphic to S(n) - then must some of the roots lie in the original field F ?

When the galois group is $S_n$ then the polynomial is irreducible for sure (and for $n\geq 5$ the equation isn't solvable by radicals).
But the other direction is wrong:
The polynomial $x^4+x^3+x^2+x+1$ over $\mathbb{Q}$ is irreducible, but it's galois group is just $C_4$, the cyclic group on $4$ elements.

Hmm :? . I think we cannot say anything about those roots of the polynomial that belong to $F$. In fact, if two polynomials have the same roots except for the roots that belong to $F$ then the corresponding field extensions are isomorphic. So the Galois groups are isomorphic as well.

thanks for the responses,

But i was wondering more about relationships between the roots themselves (ie. the automorphisms) in the case when the Galois group is S(n).  For example, can any of the roots be the negative of each other? or are only certain terms of the roots being affected by the automophisms.
Knowing this case may help, say we have a root 'alpha' in the splitting field, and we consider the extension field F('alpha') - what automorphisms can occured in that extension?

Well, it's pretty easy to show that any $n-1$ of the roots and $1$ are linearly independent over the base field; if not, we could take a dependence relation and apply an automorphism to change one of the roots to the missing root, fixing everything else.


Determine all real numbers $a$ such that \[4\lfloor an\rfloor =n+\lfloor a\lfloor an\rfloor \rfloor \; \text{for all}\; n \in \mathbb{N}.\]

Divide all sides by $ n.$ We will have:
\[ \frac{4\lfloor an\rfloor}{n}\equal{}1\plus{}\frac{\lfloor a\lfloor an\rfloor \rfloor}{n}\]
\[ \lfloor an\rfloor \equal{} an \minus{} \{an\}\]
\[ \lfloor a\lfloor an\rfloor \rfloor \equal{} a^2 n \minus{} a\{an\} \minus{} \{\lfloor a\lfloor an\rfloor \rfloor \}\]
We know, that $ 0\leq\{x\}<1$ always. If we take $ n \to \infty$ we will get equation with $ a$ to solve it:
\[ 4a\equal{}1\plus{}a^2\]
The answer is $ a\equal{}2\plus{}\sqrt 3$
Am i right?

[quote="Yuriy Solovyov"]
The answer is $ a \equal{} 2 \plus{} \sqrt 3$
Am i right?[/quote]

I don't think so $ \lfloor 2 \plus{} \sqrt {3} \rfloor \equal{}3$

Using $ n \equal{} 1$ equality doesn't hold.

[quote="SimonM"][quote="Yuriy Solovyov"]
The answer is $ a \equal{} 2 \plus{} \sqrt 3$
Am i right?[/quote]

I don't think so $ \lfloor 2 \plus{} \sqrt {3} \rfloor \equal{} 3$

Using $ n \equal{} 1$ equality doesn't hold.[/quote]

 :mad:   :D
$ 2 \plus{} \sqrt 3 \equal{} 3.732... etc$
$ \lfloor 2 \plus{} \sqrt {3} \rfloor \equal{} 3$  :wink:
If $ n \equal{} 1$ we get:
$ 4\cdot 3 \equal{} 1 \plus{} \lfloor (2 \plus{} \sqrt 3)\cdot 3 \rfloor$
$ 5 \equal{} \lfloor 3\sqrt 3 \rfloor$
$ 3\sqrt 3 \equal{} 5.169...etc$
Last is correct.

a=2+sqrt(3),prove  that  4[an]=n+[a[an]]  .

To  prove  that  4[an]=n+[a[an]], where a^2+1=4a ,a=2+3^(1/2),
It  is  enough  to   prove  that   4[an]-n<=a[an]<4[an]-n+1  ,
To  prove  the  left  part : [an](4-a)<=an(4-a)=n(4a-a^2)=n.
The  right  part:4[an]-a[an]=(4-a)[an]>(4-a)(na-1)=n(4a-a^2)-(4-a)=n-(4-a)>n-1<=>a>3  ,which  is  true  because  a=2+3^(1/2)>3  .

I came up with the following nice generalization. Perhaps further generalizations are possible.

[b]Theorem.[/b] Let $p,q \in \mathbb{N}$ be such that $1 \le q < p-1$. Then $a =\frac{1}{2} \left(p + \sqrt{p^2 - 4q}\right)$ is the only real number $a$ such that $p \lfloor an \rfloor = qn + \lfloor a \lfloor an \rfloor \rfloor$ for all $n \in \mathbb{N}$. \\

[b]Proof.[/b] To prove this, first note that we always have $pan \ge p \lfloor an \rfloor > pan - p$. On the other hand, it's fairly clear that $a < p$, otherwise the right-hand side of our equation would always be bigger. This in turn implies that we also have bounds on the right-hand side: $qn + a^2n \ge qn + \lfloor a \lfloor an \rfloor \rfloor > qn + a\lfloor an \rfloor  - 1 > qn + a(an - 1) - 1 > qn + a^2n - (p+1)$.\\

We claim that $a$ must be such that $a^2 - pa + q = 0$. If not, then there is an $n$ such that $|a^2n - pan + qn| > p+1$. So then either $pan > qn + a^2n + p+1$, in which case $p \lfloor an \rfloor > pan - p > qn + a^2n + 1 >  qn + \lfloor a \lfloor an \rfloor \rfloor$; contradiction. Or $pan < qn + a^2n - (p+1)$, in which case $p \lfloor an \rfloor \le pan < qn + a^2n - (p+1) < qn + \lfloor a \lfloor an \rfloor \rfloor$; contradiction. Conclusion: $a$ must be such that $a^2 - pa + q = 0$. \\

The two solutions to that quadratic equation are $a_1 = \frac{1}{2} \left(p - \sqrt{p^2 - 4q}\right)$ and $a_2 = \frac{1}{2} \left(p + \sqrt{p^2 - 4q}\right)$. Since $p > \sqrt{p^2 - 4q} > \sqrt{p^2 - 4p + 4} = p-2$, we get $0 < a_1 < 1$ and $p-1 < a_2 < p$. The first option can be ruled out by checking $n = 1$. From now on assume $a = a_2 = \frac{1}{2} \left(p + \sqrt{p^2 - 4q}\right)$. We will prove that it works.  \\

Define $\epsilon_n$ to be $\epsilon_n = an - \lfloor an \rfloor = \{an\}$ and define $\delta_n$ to be $\delta_n = a \lfloor an \rfloor - \lfloor a \lfloor an \rfloor \rfloor = \{a \lfloor an \rfloor\}$. Then:

\begin{align*}
\delta_n &= \{a \lfloor an \rfloor\} \\
&= \{a (an - \epsilon_n)\}  \\
&= \{a^2n - a\epsilon_n\}  \\
&= \{(pa - q)n - a\epsilon_n\}  \\
&= \{pan - qn - a\epsilon_n\}  \\
&= \{p \lfloor an \rfloor + p\epsilon_n -  qn - a\epsilon_n \} \\
&= \{(p - a)\epsilon_n\} \\
&= (p - a)\epsilon_n
\end{align*}

With this in mind we can finish the proof.

\begin{align*}
qn + \lfloor a \lfloor an \rfloor \rfloor &= qn + a \lfloor an \rfloor - \delta_n \\
&= qn + a (an - \epsilon_n) - \delta_n \\
&= qn + a^2n - a\epsilon_n - (p-a)\epsilon_n \\
&= qn + a^2n - p \epsilon_n \\
&= pan - p\epsilon_n \\
&= p \lfloor an \rfloor
\end{align*}


Hello people.
I've been searching through some books of my friend's about the maths Olympiad.  Here are two problems I completely have no idea about.

a)There are more chess masters in New York City than in the rest of U.S. combined.  A chess tournament is planned to which all American masters are expected to come.  It is agreed that the tournament should be held at the site which minimizes the total intercity traveling done by the contestants.  The New York masters claim that, by this criterion, the site chosen should be their city.  The West Coast masters argue that a city at or near the center of gravity of the players would be better.  Where should the tournament be held?

b)The nine councillors of a certain local body are not all on speaking terms.  The exact score is set out below, "1"meaning "is on speaking terms with", "0"meaning "is not on speaking terms with", and the letters standing for the councillors' names:

   A B C D E F G H I
 A - 0 0 1 0 0 1 0 0
 B 0 - 1 1 1 1 1 1 1
 C 0 1 - 0 0 0 1 1 0
 D 1 1 0 - 1 0 1 0 1
 E 0 1 0 1 - 0 1 0 1
 F 0 1 0 0 0 - 0 0 1
 G 1 1 1 1 1 0 - 0 0
 H 0 1 1 0 0 0 0 - 0
 I 0 1 0 1 1 1 0 0 -

Councillor A recently started a rumour which, surprisingly enough, came back after being heard by each councillor once and once only.  Each councillor heard it from and passed it to a councillor with whom (s)he was on speaking terms.  Councillor E was one of the last to hear it.  In what order did the rumour pass?

For question a, I do not understand if the answer is New York City or West Coast or completely different city.  We are not given where all the players are.

For question b, is a person supposed to be telling one other person?  Also, if each councillor hears the rumour only once, how does A hear it back? As that would be his second time.  Also, only D and G can tell him.  Therefore, surely if the rumour is heard by each person only ONCE, it would be heard by either D or G?

[quote="max1546"]
a)There are more chess masters in New York City than in the rest of U.S. combined.  A chess tournament is planned to which all American masters are expected to come.  It is agreed that the tournament should be held at the site which minimizes the total intercity traveling done by the contestants.  The New York masters claim that, by this criterion, the site chosen should be their city.  The West Coast masters argue that a city at or near the center of gravity of the players would be better.  Where should the tournament be held?
 [/quote]
Both the West Coast and New York city are considered as points on a 2-D coordinate plane. Now, draw a straight line between the two locations, and label the West Coast and New York by the coordinates $(0,0)$ and $(a,0),$ where $a > 0, a \in \mathbb{R}.$ Also, let the number of chess players in West Coast be $n_{1}$ and that in New York be $n_{2}$ ($n_{2}> n_{1}.$)

In the solution below, we will use the fact that if $k$ points have masses $m_{1},m_{2},...,m_{k}$ and coordinates $(x_{1},y_{1}), (x_{2},y_{2}),...,(x_{k},y_{k}),$ respectively, then the center of mass of these $k$ points is
$(\frac{x_{1}m_{1}+x_{2}m_{2}+...+x_{k}m_{k}}{m_{1}+m_{2}+...+m_{k}},\frac{y_{1}m_{1}+y_{2}m_{2}+...+y_{k}m_{k}}{m_{1}+m_{2}+...+m_{k}}).$

Lastly, we assume that each chess master has the same mass $m$ (a fair assumption for our purposes.)

So, the center of mass (we only need calculate the $x$ coordinate) of all the chess players from the West Coast and New York $= \frac{n_{1}(m \cdot 0)+n_{2}(m \cdot a)}{(n_{1}+n_{2})m}= \frac{n_{2}a}{n_{1}+n_{2}}.$

Now, if we heed the claim of the West Coast players, then the amount of intercity travel needed to be done by the chess players will be $D_{1}= \frac{n_{2}a}{n_{1}+n_{2}}\cdot n_{1}+(a-\frac{n_{2}a}{n_{1}+n_{2}}) \cdot n_{2}= \frac{2n_{1}n_{2}a}{n_{1}+n_{2}}.$

And, if we heed the claim of the New York masters, then the amount of intercity travel needed to be done by the chess players will be $D_{2}= an_{1}.$

Therefore,
$D_{1}> D_{2}$

$\Leftrightarrow \frac{2n_{1}n_{2}a}{n_{1}+n_{2}}> an_{1}$
$\Leftrightarrow 2n_{2}> n_{1}+n_{2}$
$\Leftrightarrow n_{2}> n_{1},$ which is true from our assumptions.

Hence, the tournament should be held in New York.


For a triangle ABC, let w,v be the circles with diameter AB, AC respectively. Let D=w∩AC, E=v∩AB. If P=BD∩v,Q=CE∩w  then prove that AP=AQ.They tell that this problem is easy, but I feel difficult. :(

$ \triangle{ADP}\sim \triangle{APC}\Rightarrow AP^2\equal{}AD\cdot AC$ analogously $ AQ^2\equal{}AE\cdot AB$ but $ \triangle{AED}\sim \triangle{ACB}\Rightarrow AE\cdot AB\equal{}AD\cdot AC$ $ \Rightarrow$ $ AP^2\equal{}AQ^2$

Thanks.I am ashamed of myself to see that is easy. :blush:

[quote="GioMott"][quote="Nemion"]$ \triangle{AED}\sim \triangle{ACB}$[/quote]
Maybe I'm just too tired to see the obvious demonstration but... how do you prove this? :blush:

Good night
GM[/quote]
notice they are right-angled triangle


calculus
integration
induction
number theory solved
number theory

$a_0 = a_1 = 1$ and ${a_{n+1} . a_{n-1}} = a_n . (a_n + 1)$ for all positive integers n.

prove that $a_n$ is one integer for all positive integers n.

A nice problem, but the method is canonical: Find an integral recurrence.

[color=blue][b]Problem.[/b] Let $\mathbb Z^+$ denote the set of all nonnegative integers.

Define a recurrent sequence $\left(a_n\right)_{n\in\mathbb{Z}^+}$ by $a_0=a_1=1$ and the recurrence equation $a_{n+1}a_{n-1}=a_n\left(a_n+1\right)$ for every positive integer n.

Prove that $a_n$ is an integer for every $n\in\mathbb{Z}^+$.[/color]

[i]Solution.[/i] Applying the recurrence relation $a_{n+1}a_{n-1}=a_n\left(a_n+1\right)$ to n = 1, we get $a_2a_0=a_1\left(a_1+1\right)$, what, in view of $a_0=a_1=1$, becomes $a_2\cdot 1=1\left(1+1\right)=2$, so that $a_2=2$.

Now, we will show that the equation $a_{n+1}+a_{n-1}+1=4a_n$ holds for every positive integer n. This will be proved by induction over n:

For n = 1, the equation $a_{n+1}+a_{n-1}+1=4a_n$ can be directly verified: $a_2+a_0+1=2+1+1=4=4\cdot 1=4a_1$.

Now assume that the equation $a_{n+1}+a_{n-1}+1=4a_n$ holds for n = k, where k is some positive integer; in other words, assume that we have $a_{k+1}+a_{k-1}+1=4a_k$. Then, we have to show that the equation $a_{n+1}+a_{n-1}+1=4a_n$ also holds for n = k + 1; in other words, we have to prove the equation $a_{k+2}+a_k+1=4a_{k+1}$.

Well, by the recurrent definition of the sequence $\left(a_n\right)_{n\in\mathbb{Z}^+}$, we have $a_{k+2}a_k=a_{k+1}\left(a_{k+1}+1\right)$ and $a_{k+1}a_{k-1}=a_k\left(a_k+1\right)$. Thus,

$a_{k+2}a_k+a_k\left(a_k+1\right)=a_{k+1}\left(a_{k+1}+1\right)+a_{k+1}a_{k-1}$.

In other words, $a_k\left(a_{k+2}+a_k+1\right)=a_{k+1}\left(a_{k+1}+a_{k-1}+1\right)$. Now, since $a_{k+1}+a_{k-1}+1=4a_k$, this becomes $a_k\left(a_{k+2}+a_k+1\right)=a_{k+1}\cdot 4a_k$, and thus, upon division by the number $a_k$ (which is obviously positive), we obtain $a_{k+2}+a_k+1=4a_{k+1}$. Thus, the induction step is completed, and the equation $a_{n+1}+a_{n-1}+1=4a_n$ is proven for every positive integer n.

Now, using this equation, we can recurrently calculate all terms of the sequence $\left(a_n\right)_{n\in\mathbb{Z}^+}$ given the first two terms $a_0$ and $a_1$. Moreover, since these first two terms $a_0$ and $a_1$ are integers ($a_0=a_1=1$), it follows that all terms $a_n$ of the sequence must be integers as well, and thus the problem is solved.

  Darij

My solution is the same.  :)

this problem was proposed in Brazilian Olympic Revenge 2004.


geometry
3D geometry
sphere
Gauss
integration
function
calculus

A volume $V_{f}$ of fluid with uniform charge density $\rho$ is sprayed into a room, forming spherical drops.  As they float around the room, the drops may break apart into smaller drops or coalesce into larger ones.  Suppose that all of the drops have radius $R$.  Ignore inter-drop forces and assume that $V_{f}\gg R^{3}$.

(10 pts)  a.  Calculate the electrostatic potential energy of a single drop. (Hint: suppose the sphere has radius $r$.  How much work is required to increase the radius by $dr$?).

(4 pts)  b.  What is the total electrostatic energy of the drops?

Your answer to (b) should indicate that the total energy increases with $R$.  In the absence of surface tension, then the fluid would break apart into infinitesimally small drops.  Suppose, however, that the fluid has a surface tension $\gamma$.  (This value is the potential energy per unit surface area, and is positive.)

(4 pts)  c.  What is the total energy of the drops due to surface tension?

(7 pts)  d.  What is the equillibrium radius of the drops?

a) Actually I don't know how to do it using the hint (I'm a bit rusty on electrostatics, as I have already mentioned :blush:), so I'll find the total energy by integrating the energies of the interactions of elementary volumes with the entire sphere.

By using Gauss' law, we first find the electric field at distance $x$ from the center:
\[E(x) = \frac{\rho x}{3\varepsilon_{0}}. \]
Then the absolute value of the potential is
\[\varphi(x) = \int E \text dx = \frac{\rho x^{2}}{6\varepsilon_{0}}, \]
where we've chosen the potential to be zero at $r = 0$.

Then,
\[W = \frac{1}{2}\int_{V}\varphi \text dq = \frac{1}{2}\int_{0}^{r}\frac{\rho x^{2}}{6\varepsilon_{0}}4\rho\pi x^{2}\text dx = \frac{\pi \rho^{2}r^{5}}{15 \varepsilon_{0}}. \]
The one half is because we've counted the interaction between two elementary parts exactly twice.

b) Since the drops do not interact,
\[W_{0}= N W = \frac{V_{f}}{4r^{3}\pi/3}\frac{\pi \rho^{2}r^{5}}{15 \varepsilon_{0}}= \frac{\rho^{2}r^{2}V_{f}}{20 \varepsilon_{0}}. \]
c) The energy due to surface tension of one drop is $-4\gamma r^{2}\pi$ (minus because it tends to contract the drop), the total s. t. energy is
\[-4 \frac{V_{f}}{4r^{3}\pi/3}\gamma r^{2}\pi =-\frac{3\gamma V_{f}}{r}. \]
The total energy is
\[E = \frac{\rho^{2}r^{2}V_{f}}{20 \varepsilon_{0}}-\frac{3\gamma V_{f}}{r}. \]
d) The energy of a drop has to be minimized; the total energy of a single drop is
\[E_{0}(r) = \frac{\pi \rho^{2}r^{5}}{15 \varepsilon_{0}}-4\gamma r^{2}\pi; \]

\[E'_{0}(r) = \frac{\pi \rho^{2}r^{4}}{3 \varepsilon_{0}}-8\gamma r\pi. \]
Equating this with zero we get $r = 0$ (not what we want) and
\[r = \sqrt[3]{\frac{24 \gamma \varepsilon_{0}}{\rho^{2}}}. \]
I'm probably rambling, but whatever. Is this even close to the correct answer? Compared to the previous problem, this one is way too hard :maybe:

[quote="durt"]a. I think the idea is to find the energy needed to construct a sphere by adding thin layers of fluid to smaller spheres. To do this we must find the work required to surround a sphere of radius $r$ with a layer of thickness $dr$. The magnitude of the potential at the surface is:

$V=\frac{\rho r^{2}}{3 \epsilon_{0}}$

So:

$dW = V dq = \frac{4 \pi \rho^{2}r^{4}dr}{3 \epsilon_{0}}$

Integrate this to find the total work required to form a sphere, which is also its potential energy.

$U = \int dW = \int_{0}^{R}\frac{4 \pi \rho^{2}r^{4}dr}{3 \epsilon_{0}}= \frac{4 \pi \rho^{2}r^{5}}{15 \epsilon_{0}}$

And for part d., I figured you would minimize the total energy. I know its counterintuitive to have the radius of a drop depend on the total volume, but if the system were in a state with total energy minimized you'd have to input energy to get different radii.[/quote]

Agreed.

Nice idea! However, I don't see why $V = \rho r^{2}/3\varepsilon_{0}$. How did you get this?

I used the Green functions as a solution for the poisson equation  for a homogenous shpere, and got the same as Djole, but with much longer calculation.  :D

Did you get the same result for $V$ or for $W$?

For $W$ I got the same answer as you  :D
If I didnt screw anything for there is a big chance I did, because there were lots of integrals  in my solution :maybe:

To find $V$, treat the sphere as a point charge. So you have:
$V = \frac{1}{4 \pi \epsilon_{0}}\frac{Q}{r}$, where
${Q = \frac{4}{3}\pi r^{2}\rho}$.

Can you do that? I mean, $V = kQ/r$ means that $V = 0$ for $r \rightarrow \infty$, and that's clearly not the case once you substitute $Q$...

Ok, I see what you mean. Assume that the sphere has a radius $r$. Then the potential at a distance $x>r$ from the center of the sphere is $\frac{kQ}{x}$. So $V=0$ for $x \rightarrow \infty$. Then plug in $x = r$ to get the potential at the surface.

I'm still not convinced, but let's leave that for now. I'm rather confused and I'm not sure I can think of that any more today...

However, my new question is this: have you included the potential energy of each infinitesimaly thin sphere in your calculation?

Furthermore, here's another approach: let's imagine a ball of radius $R$ made out of some dielectric. Let's take a charge $\text dq$ and bring it to the ball from infinity; once near the edge of the ball, the energy of interaction between the ball and the charge is $\text dW = kq\text dq/R$. Integrating from $0$ to $Q = 4\rho R^{3}\pi$ we get
\[W = \frac{2R^{5}\pi\rho^{2}}{9\varepsilon_{0}}. \]
Now which one is correct?

Now the last one is not right for sure, because the energy of interacting between the charge is not the same distance from all of the small parts of the ball, so it cant be $\frac{kqdq}{R}$

Good point. Does this rebuke [b]durt[/b]'s solution as well?

I dont understand Durt's solution.
From where does he get thet:
$dW=Vdq$?
once we put an extra layer on the sphere then the potential at the surface of the sphere will change because the "older" surface will be in the sphere. So $V$ is not constant in his solution. I think thats the mistake. but I may be wrong cause I dont understand what he did.  :huh:

This is the formula for the potential energy of a point charge $\text dq$ in a potential $V$. I think :maybe:

Ok, so I've read Feynman's lectures on electromagnetism and there it's solved in [b]durt[/b]'s way, so that's correct. I've made a mistake with the potential, it is $\varphi = \rho r^{2}/ 3\varepsilon_{0}$, but I still don't seem to get the right thing :(


If $ a,b,c$ (distinct) are the side lengths of a triangle, then

$ \sum\frac {a(b \plus{} c)}{(b \minus{} c)^2}\geq2$

$ \sum\frac{a(b\plus{}c)}{(b\minus{}c)^{2}} \ge \sum\frac{a(b\plus{}c)}{a^{2}} \equal{} \sum\frac{(b\plus{}c)}{a} \ge 6$
this is ture because $ a\ge b\minus{}c$,
i think u meant 6 instead of 2.(sry if wrong)


what is the discriminant?

something to do with the quadratic formula...
_________________
Don't Even Think About IT :rotfl:

[url=http://www.math123xyz.com]Interactive Math Resources[/url]


search
number theory proposed
number theory

If p is a prime show $C(pa,pb) \equiv C(a,b) (mod p)$ .

If I could find it I didnt post it here. :)

http://ecademy.agnesscott.edu/~lriddle/ifs/siertri/LucasProof.htm

P.S. Have you tried to search for it?  ;)


calculus
integration
geometry
function
derivative
real analysis
real analysis unsolved

Let $I_{n}=\int_{0}^{1} (1-x^{n})^{\frac{1}{n}} dx$. Prove that the serie $\sum_{n=1}^{\infty} (1-I_{n})$ is convergent. :)  ;)

$1-I_n=\int_0^1 dx - \int_0^1 (1-x^n)^{\frac{1}{n}} dx = \int_0^1 \left[ 1 - (1-x^n)^{\frac{1}{n}} \right] dx$

When $n$ becomes very large, $x^n$ becomes very small ($0<|x|<1$). We use that $(1-x)^\alpha = 1-\alpha x$ for small $x$. Then,

$1-(1-x^n)^{\frac{1}{n}}=1-\left(1-\frac{x^n}{n}\right) = \frac{x^n}{n}$

So for large $n$,

$1-I_n = \int_0^1 \frac{x^n}{n} dx = \frac{1}{n(n+1)}$

So the tail of the sequence clearly converges, being $O(n^{-2})$. Since the integral is proper, the entire series must also converge.

[quote="TheBB"]When $n$ becomes very large, $x^n$ becomes very small ($0<|x|<1$). We use that $(1-x)^\alpha = 1-\alpha x$ for small $x$. [/quote]

This can be made just a bit more precise:

By Taylor's Theorem, for $0 < \epsilon < 1$, $(1+t)^\epsilon \ge 1 + \epsilon t - \frac{\epsilon - \epsilon^2}{2}t^2$ for $t > -1$. Therefore,
$1 - (1-x^n)^{1/n} \le \frac{x^n}{n} + \frac{n-1}{2n^2}x^{2n}$ for $0 < x < 1$ and thus
$\int_0^1 \left(1 - (1-x^n)^{1/n}\right) dx \le \frac{c}{n^2}$ for some $c > 0$.

The integral $\int_0^1 \left(1 - (1-x^n)^{1/n}\right) dx$ is the area outside the curve $x^n + y^n = 1$, inside $[0,1]^2$.

Here a semi-explicit calculation, using the beta integral identity: for $\alpha,\beta>0,$

$\int_0^1(1-u)^{\alpha-1}u^{\beta-1}\,du=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma (\alpha+\beta)}.$

With the substitution $u=x^n,$ we have

$I_n=\frac1n\int_0^1(1-u)^{\frac1n}u^{\frac1/n-1}\,du= \frac{\Gamma (1+\frac1n)\Gamma(\frac1n)} {n\Gamma(1+\frac2n)}$

 $=\frac{(\Gamma(1+\frac1n))^2}{\Gamma(1+\frac2n )}.$

What can we say about the Gamma function? It is certainly a smooth function (and we can write integral formulas for its derivatives) and $\Gamma(1)=1.$

$\Gamma(1+\frac1n)=1+\frac1n\Gamma'(1)+\frac1{2n^2}\Gamma''(1)+ O(n^{-3})$

$\Gamma(1+\frac1n))^2=1+\frac2n\Gamma'(1)+ \frac1{n^2}(2(\Gamma'(1))^2+\Gamma''(1))+O(n^{-3})$

$\Gamma(1+\frac2n)=1+\frac2n\Gamma'(1)+\frac2{n^2}\Gamma''(1)+ O(n^{-3})$

Dividing, we get

$I_n=1+\frac1{n^2}(2(\Gamma'(1))^2-\Gamma''(1))+O(n^{-3}).$

So $\sum_{n=1}^{\infty}(1-I_n)$ converges by limit comparison to $\sum\frac1{n^2}.$


Consider following game on table n times n: On every cell is a bulb (which can be turn on or off). At the begining some bulbs are turned on. When we "click" on on bulb we interchange state of that bulb and all bulbs around this one (cells that share an edge). Prove that no matter what is the configuration of turned on bulbs, we can always do some moves, so that all bulbs are turned on or all bulbs are turned off.

That's all.

For n = 3 this was on SCG national competition.

I don't think the statement is true for every value of $n$.  For example, suppose $n=5$.  The problem is equivalent to showing that every configuration is reachable from either the all off or all on board, i.e. that we can label each configuration with an ordered pair $(x, S)$, where $x$ is either the all empty or all full board and $S$ is the set of clicks made.

There are $2^{25}$ possible board positions, and $2^{26}$ possible ordered pairs.  However, some ordered pairs lead to the same board position.  Note that if we start with the empty board and press the buttons marked in either of the below configurations:

01110
10101
11011    (call this one A)
10101
01110

or

10101
10101
00000  (call this one B)
10101
10101

leads to the empty board again.  What this means is that $(x, S), (x, S+A), (x, S+B), $and $(x, S+A+B)$ all lead to the same final configuration (by "+" I mean pushing one set of switches, then the other), so there can only be at most $2^{24}$ configurations reachable from either the all on or the all off position.

[quote="alexilic"]..we can always do some moves, so that all bulbs are turned on or all bulbs are turned off.
[/quote]

Just an observation:

It's a consequence of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=graph+AND+odd&t=5976]this problem[/url] that there is a sequence of moves turning the all-bulbs-on situation into the all-bulbs-off one, so there's no need for the distinction "all bulbs are turned on or all bulbs are turned off".

alexilic, where did you find the problem? Did you just generalize the case $n=3$, or did you actually see it used somewhere?

I'm really surprised to find this problem here on Mathlinks. The case [tex]n=6[/tex] was a kind of subgame in an old PC adventure game called "The Feeble Files"  :!:  Luckily it wasn't necessary to solve this problem in order to proceed in the game  :P


let r is positive rational
pro there exist a,b,c,m,n,p distinct satisfying:
r=$\frac{a^2+b^3+c^5}{m^7+n^{11}+p^{13}}$
goodluck

It is very similar, to Polish problem (LI Polish math olimpiad-2000).
Prove that every positive rational number can be writen in the form: $\frac{a^2+b^3}{c^5+d^7}$, and $a,b,c,d\in \mathbb{Z}_+$

It's really easy.

If $p$ is prime, $v_p$ denote the $p$-adic valuation.

[b]lemma:[/b] Let $(t,s,m,n)\in(\mathbb{N}^*)^4$ such that $\gcd(m,n)=1$.
We can find $(a,b)\in(\mathbb{N}^*)^2$ such that $a^m t=b^n s$.

[b]proof[/b]: If $p$ is prime then, by Bezout, we can find $(\alpha_p,\beta_p)\in\mathbb{N}^2$ such that $\alpha_p m- \beta_p n=v_p(s)-v_p(t)$. It's now easy to show that we can take $a=\prod p^{\alpha_p}$ and $b=\prod p^{\beta_p}$.
_ _ _

Thus if $r\in\mathbb{Q}^+_*$, we can choose $(a,b,c,m,n,p)\in(\mathbb{N}^*)^6$ s.t. $a^2=rm^7$, $b^3=rn^{11}$ and $c^5=rp^{13}$ $\Rightarrow\frac{a^2+b^3+c^5}{m^7+n^{11}+p^{13}}=r$.

okie
this pro is similar  SL-99: show that for every positive rational r, there exist a,b,c,d positive integer such that:
[tex]r=\frac{a^3+b^3}{c^3+d^3} [/tex]
and we tend to general pro: r is positive rational,m,n,p,q positive integer.Is there exist a,b,c,d positive integer such that
[tex]r=\frac{a^m+b^n}{c^p+d^q} [/tex]
goodluck


Suppose that after I wrote this problem, Richard Rusczyk thought he could be more clever than I could, so he wrote his own problem. Suppose that both of our problems are in the set of 12 problems you are currently working on. If you sat down at your computer this morning and randomly loaded 4 of the 12 problems, what is the probability that both this problem and Richard Rusczyk's problem were among the four you loaded?

There are 10 choose 2 ways you can choose the other two problems in the four with RR's and this one. There are a total of 12 choose 4 ways to choose the problems, so the answer is:

$ \frac{\binom{10}{2}}{\binom{12}{4}}\equal{}\frac{\frac{10 \cdot 9}{2 \cdot 1}}{\frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1}} \equal{} \boxed{\frac{1}{11}}$

How did you get 10 choose 2? I get the other part, though.

We count the total number of ways there can be the two wanted problems and 2 other problems in a set of 12. This is just how many ways we can select 2 problems out of 10 problems. We don't care about the order, so it is $ \binom {10}{2}$.

[quote=pickten]This problem ought to get a secret achievement. I mean, honestly, its a shame not to have something special for a problem about Richard.[/quote]

Sorry for the bump, but agreed.

Question: Does Alcumus actually give problems in sets of 12?

Sorry for the bump, but I don't understand why the answer is:
$\frac{\binom{4}{2}}{\binom{12}{4}}$

I got this answer, since there is $\binom{12}{4}$, or $495$, ways to choose 4 problems. There is $\binom{4}{2}$, or 6 ways, to choose 2 problems, out of 4. $\frac{6}{495}$ equals $\frac{2}{165}$.

Sorry again, for the bump.

[quote=wizwilzo]its always ok to bump gamebot topics[/quote]

Yes, it's ok if you have a question like pycops did, or an alternative solution or pointing out that the only person who posted there was wrong... But bumping isn't ok for posting, say, the answer only when it's been posted multiple times.

This problem was interesting, especially the story :P

I like self-referring problems like this! They're very interesting to think about. [hide][size=50]Am I allowed to bump this way?[/size][/hide]

[quote=debasishmohanta]The fact that you are currently seeing this problem has no influence on the outcome; one way to know this is that the problem asks you what the probability was this morning, before you saw this problem. There are $\binom{12}{4} = 11\cdot 10 \cdot 9 / 2$ ways to pick 4 of the 12 problems. If you pick both my problem and Richard Rusczyk's, there are $\binom{10}{2} = 10 \cdot 9 / 2$ ways to pick the other 2 problems of the 4, out of 10 choices. So the probability that you picked both my problem and Richard Rusczyk's is $\frac{\text{successful outcomes}}{\text{total outcomes}} = \frac{10 \cdot 9 / 2}{11\cdot 10 \cdot 9 / 2} = \boxed{\frac{1}{11}}$.[/quote]

You shouldn't do that, because you are just copying and pasting from the solution, making it sound as if you wrote it. :spam:

P.S. Sorry for the bump-just cause that's what everyone else said. [size=50]I'm not really sorry LOL[/size]

[font=Times new roman][hide=Easier than it looks]We simply have to take the number of combinations in which the 2 problems are given to be chosen over the total combinations. This amounts to $$\cfrac{\dbinom{10}{2}}{\dbinom{12}{4}}=\boxed{\dfrac1{11}}.$$ $$\definecolor{color}{rgb}{0.5, 0.5, 0.5} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$

[quote=ben7]You can learn about him [url=http://www.artofproblemsolving.com/School/index.php?page=school.instructors]here.[/url][/quote]

Paul joined AoPS in 2009. He works on Alcumus with much pride and teaches with much joy in the Art of Problem Solving Online School. Paul got an early start on math and programming when he discovered QBasic on his computer when he was six years old and remarked that sine functions looked very pretty, while tangent functions looked very angry. Before attending Harvard, he was a third-place winner at National MATHCOUNTS, a two-time perfect scorer on the American Mathematics Contest (AMC) 12, a participant at the Research Science Institute, and a prize winner at the Intel International Science and Engineering Fair. In his free time, he enjoys using emacs, waterbending, collecting and playing musical instruments, and talking about himself in the third person. Paul lives with four cats, including a Siamese kitten named after Elijah Wood.

Lol my friend said the same thing, it's not reffering to this problem, it's talking about another one.

[hide=Answer]$1/11$[/hide]

I remember getting this problem.

[quote=acornfirst]Ya'll realize this problem isn't possible? If Richard wrote it after he wrote the problem, how does the problem include the fact that Richard wrote it?[/quote]

there are more than one person writing questions

[quote=Critical][quote=CR7SC30]Did you guys realize that this thread is over 11 years old? IT IS AS OLD AS ME!!! OMG[/quote]

There are some SS threads from 2006...[/quote]

2006= Start of Alcumus. 2003=Start of AoPS

[quote=CR7SC30][quote=ilovepizza2020][quote=CR7SC30]Did you guys realize that this thread is over 11 years old? IT IS AS OLD AS ME!!! OMG[/quote]

Now we know your age...[/quote]

idc tho[/quote]

LOL

lol how is this lvl 23 :thinking:
so misplaced

[hide=simple sol]Just place Richard's and my problems by default. The answer is then just 10c2/12c4=45/495=1/11, because we need to choose 2 more out of the 10 left, and there are 12c4 total outcomes.[/hide]

[hide=simpler sol]Lets view necessary problems as one[/hide]

[hide=even simpler sol]4/12 probability that richard got in, 3/11 probability that other guy got in, you get 1/11[/hide]

The answer is 3/11 because you are already viewing this problem and now that you take away one problem from the total you get a 1/11 + 1/11 + 1/11 chance to see RR's problem ;)


Find all positive integer a such that there exists infinitely many positive odd integers n such that $ a^n\plus{}2009^n$ is divisible by $ n$

No one can solve it. I think it must be posted on Olympiad forum?


I'm re-posting this thread: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314195[/url]
I had no replies. Others solutions to (i) are welcome but specially to (ii).  :roll:

$ (O)$ is a semicircle and $ AB$ is a diameter. $ (O')$ is the semicircle with diameter $ AO$ (in the same half-plane than $ (O)$).
$ C\in AO$, $ C$ variable. The perpendicular to $ AB$ at $ C$ cuts $ (O')$ at $ D$ and $ (O)$ at $ F$. $ AD$ intercepts $ (O)$ at $ E$. $ (I)$ is the circumcircle of triangle $ DEF$.
[b](i)[/b] Prove that $ AF$ is tangent to $ (I)$.

[b](iii)[/b] Prove that $ I$ belongs to the circle with center $ O$ and radius $ \sqrt {2}/2\cdot AO$.

I made part (i). I'm missing (ii).

[hide]I first proved using similarity that $ AF/AE \equal{} \sqrt {2}$ . Then $ AD\cdot AE \equal{} 1/2\cdot AE^2 \equal{} AF^2$.
Therefore, power of $ A$ with respect to $ (I)$ implies $ AF$ is tangent to it.[/hide]

1) It's obvious that $ D$ is the midpoint of $ AE$ ( 1 )
2) If $ F'$ is the intersection of $ DF$ with the circle $ O$, we have $ \angle AFF' \equal{} \angle AEF$ ( 2 ), so $ AF$ is tangent to the circle $ (DEF)$, consequently its center lies on $ BF$ and, with observation (1), we get $ AE \equal{} AF \sqrt {2}$.
3) $ EF$ is the common chord of the circles $ O$ and $ O"$, so $ \angle EOO" \equal{} \angle EBF$, hence $ EO"OB$ is cyclic. From all the above mentioned we get that $ \triangle AFE \sim  \triangle OO"E$, hence $ OE \equal{} OO" \sqrt {2}$, done.

Best regards,
sunken rock