The hexagons in the second column to the left must be yellow and green, but either color can be on top: 2 possibilities. With either possibility, the rest of the figure is easily colored in, with the color of every other hexagon being forced. (ex. In the third column, the middle hexagon must be red, and the top and bottom must be the opposite of the top and bottom in the second column) Thus, there are only 2 possibilities.

[quote="GameBot"]The hexagon with the R is colored red.  Each hexagon is colored either red, yellow or green, such that no two hexagons with a common side are
colored the same color.  In how many
different ways can the figure be colored?

[/quote]
For some reason, I got $ 3!\equal{}6$ ways.

[quote=GameBot]The hexagon with the R is colored red.  Each hexagon is colored either red, yellow or green, such that no two hexagons with a common side are
colored the same color.  In how many
different ways can the figure be colored?

[asy]path a=(0,0)--(10,0)--(15,8.7)--(10,17.3)--(0,17.3)--(-5,8.7)--cycle;
draw(a);
draw(shift((15,8.7))*a);
draw(shift((15,-8.7))*a);
draw(shift((30,17.3))*a);
draw(shift((30,0))*a);
draw(shift((30,-17.3))*a);
draw(shift((45,26))*a);
draw(shift((45,8.7))*a);
draw(shift((45,-8.7))*a);
draw(shift((60,17.3))*a);
draw(shift((60,0))*a);
draw(shift((60,-17.3))*a);
draw(shift((75,8.7))*a);
draw(shift((75,-8.7))*a);
label("$R$",(5,10),S);[/asy][/quote]

Why is it two? I guessed but I want to know why. I didn't get the alcumus colution.

Um, they didn't make it very clear in the question, I also got $6$ at first. The first hexagon must always be red, so that only leaves us with $2!$ possibilities and everything else would be predetermined based off of those colors, so you would end up with $2!$, or $\boxed{2}$ ways.

[hide = ez solution]The second row of hexagons, there are two choices. for all the other rows, it's forced from the previous hexagons. So the answer is $\boxed{2}$.[/hide]

is this the 14th problem ever on alcumus??? does anyone know what the first one was

if you looked deep enough into alcums forum you would get it

thanks, but it would take a long time to scroll

Actually, that's inaccurate. The Alcumus forum does not display all the threads ever created. It displays a few threads to prevent cheating.


geometry
trigonometry
quadratics
trig identities
Law of Cosines

Source: 360 PMC
Let $ABC$ be a triangle such that $(8AB-7BC-3CA)^{2}=6(AB^{2}-BC^{2}-CA^{2})$.
Prove that $\angle A=60^\circ$.

[quote="gotztahbeazn"]errr i did some stuff with law of cosines, which seems to be the only thing you can do for this problem? but unfortunately things got messy and it didn't work out right

anyone have any ideas?[/quote]

[hide="hint"]
that's what I tried first time.
but for this problem, just expand out the given equation and use the fact that the equation is a quadratic eqaution (i.e. discriminant). i agree that it is messy...[/hide]

[quote="computer_nuke"]Ok, I got it, just set AB = x, BC = y, AC = z, and expand the equation given. Then solve for x using the quadratic formula. The descriminant becomes -6(6y-14z)squared which means 6y-14z = 0 or else when you square root it, you have a square root of -6 which would creat immaginary numbers equalling to a side length of a triangle. So 6y=14z, then you continue with the quadratic and substitute the equation you already have (3y = 7z) into it to find a common relationship between x, y, and z and then you see that when x = 8, then z = 3 and y = 7, now use the law of cosine and you see than cos(A) = 0.5 which has to mean that it is 60 degrees.[/quote]
nice! :D

Thanks, squaring 112 was a pain though lol

[quote="computer_nuke"]Ok, I got it, just set AB = x, BC = y, AC = z, and expand the equation given. Then solve for x using the quadratic formula. The descriminant becomes -6(6y-14z)squared which means 6y-14z = 0 or else when you square root it, you have a square root of -6 which would creat immaginary numbers equalling to a side length of a triangle. So 6y=14z, then you continue with the quadratic and substitute the equation you already have (3y = 7z) into it to find a common relationship between x, y, and z and then you see that when x = 8, then z = 3 and y = 7, now use the law of cosine and you see than cos(A) = 0.5 which has to mean that it is 60 degrees.[/quote]
Thats what i did, except maybe a bit longer. Nice solution Computer nuke :D


calculus
integration
function
real analysis
real analysis unsolved

Allow p and k to be real numbers.  For which values of p and k does the following double series converge:

$\sum_{n,m=1}^\infty \frac{1}{n^{p}+m^{k}}$

That is not the correct answer. For example, if $p=k=2,$ the series diverges.

It seems that I've got the answer: $k>2$,$p>\frac{k}{k-1}$ and $p>2$,$k>\frac{p}{p-1}$.But I'm not quite sure.Maybe I should post my proof and let you guys check up on it.

That could be written more symmetrically as $\frac1p+\frac1k<1$, which is related to Holder's inequality.

I think your calculations are correct. I also solved  the problem, I think completely, and I got that the sum is, of course we consider that $p$ and $k$ are positive real numbers:

$=\infty$ if either $p\leq 1$ or $k\leq 1$.

$=\infty$ if $p=2$ and $1<k\leq 2$.

$=\infty$ if $p>2$ and $1<k\leq \frac{p}{p-1}$.

$<\infty$ if $p>2$ and $k>2$.

$<\infty$ if $p=2$ and $k>2$.

$<\infty$ if $p>2$ and $\frac{p}{p-1}<k<2$.
Hence all the cases are discussed. The sum is symmetric so only some cases are to be discussed.

Thank you!More concisely,I employed a method of decomposition,which further gives an estimate of the total sum.

It seems that our methods are different.

Does your method imply a generalization ?

Here's my argument.

The first step is to claim that the sum converges if and only if the following integral converges:

$\int_{0}^{\infty}\int_{0}^{\infty}\frac1{x^{p}+y^{k}+1}\,dx\,dy$

Think of this as an argument with Riemann sums on a partition using squares, or as a two-dimensional verision of the integral test. The big advantage of doing this, of switching from sums to integrals, is that I'm now free to do with the integrals the one thing I can't do with sums: change variables.

The first set of changes of variables is to let $u=x^{p}$ and $v=y^{k}.$ The integral now becomes

$\frac1{pk}\int_{0}^{\infty}\int_{0}^{\infty}\frac{u^{\frac1p-1}v^{\frac1k-1}}{u+v+1}\,du\,dv$

This potentially introduces extra singularities at the axes. Note, however, that we must have $p>0,k>0$ or else we have obvious divergence of the sum. That means that $\frac1p-1>-1$ and $\frac1k-1>-1,$ so those singularities shouldn't cause any trouble with the integrability.

It's now time to make a second change of variables. This time, we set $u=ts$ and $v=(1-t)s$ for $0\le t\le 1,0\le s<\infty.$ The Jacobian is $\frac{\partial(u,v)}{\partial(s,t)}=s$ so that $\,du\,dv=s\,ds\,dt.$ Make this change of variables and sort out the integrand to get that our integral is now:

$\frac1{pk}\left(\int_{0}^{1}t^{\frac1p-1}(1-t)^{\frac1k-1}\,dt\right)\left(\int_{0}^{\infty}\frac{s^{\frac1p+\frac1k-1}}{s+1}\,ds\right)$

Assuming that $p,k>0,$ the first integral, the one in $t,$ always converges. (It can even be computed exactly, as a Beta function, but that's not important.) The second integral, the one in $s,$ is integrable for $s$ near zero, so the issue is what happens for large $s.$ The integrand looks like $s$ to the power $\frac1p+\frac1k-2,$ and we need that power to be less than $-1.$

$\frac1p+\frac1k-2<-1$ if and only if $\frac1p+\frac1k<1.$

This is exactly ffrog's result, expressed in the symmetrical language jmerry used.

Here is the generalization of this nice problem.

Generalization. An n dimensional harmonic series. Let $p_{i}$, $i=1\cdots n$ be positive numbers. Then the series

$\sum\frac{1}{k_{1}^{p_{1}}+\cdots+k_{n}^{p_{n}}}$ converges exactly when $\frac{1}{p_{1}}+\frac{1}{p_{2}}+\cdots+\frac{1}{p_{n}}<1$.

Solution.

The behavior of the series is the same as the behavior of the n dimensional integral:

$I=\int_{0}^{\infty}\cdots\int_{0}^{\infty}\frac{dx_{1}\cdots dx_{n}}{x_{1}^{p_{1}}+\cdots+x_{n}^{p_{n}}+1}$.

We use the substitution $x_{i}^{p_{i}}=y_{i}$, and we get that

$I=\prod \frac{1}{p_{i}}\int_{0}^{\infty}\cdots\int_{0}^{\infty}\frac{y_{1}^{1/p_{1}-1}\cdots y_{n}^{1/p_{n}-1}}{y_{1}+\cdots+y_{n}+1}dy_{1}\cdots dy_{n}$.

However the behavior of $I$ is the same as that of

$J=\int\cdots\int_{y_{1}+\cdots+y_{n}\geq 1, y_{i}\geq 0}\frac{y_{1}^{1/p_{1}-1}\cdots y_{n}^{1/p_{n}-1}}{y_{1}+\cdots+y_{n}+1}dy_{1}\cdots dy_{n}$.

Using Louville's formula we get that

$J=\frac{\Gamma(1/p_{1})\cdots \Gamma(1/p_{n})}{\Gamma(1/p_{1}+\cdots+1/p_{n})}\int_{1}^{\infty}\frac{x^{1/p_{1}+1/p_{2}+\cdots+1/p_{n}-1}}{x+1}dx$,

which converges exactly when $\frac{1}{p_{1}}+\cdots+\frac{1}{p_{n}}<1$


induction
real analysis
real analysis theorems

Hi I want to show that for x>3

(x^5 - 1)/(x-1) ≠ y^2

where y is any integer

basically the expression is not a perfect square

I've tried proof by induction but can't seem to get anywhere

So are you both cheating off the same exam?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276719


Prove that there are an infinite number of functions $f: N\to N$ such that:
(a) $f(2)=2$
(b) $f(xy)=f(x)f(y)$

[hide="Uhhh..."] The multiplicative functions $\in \mathbb{N}$ have values that are uniquely determined by the values of the function at the primes, since $\mathbb{N}$ has unique prime factorization.  Obviously fixing one prime value has no effect on the others, so clearly infinite such functions exist. [/hide]


invariant
analytic geometry
geometry
rectangle
combinatorics unsolved
combinatorics

For an 8x8 chessboard, if you remove two squares of opposite colors (one black and one white), is it always possible to tile the remaining squares with 1x2 dominoes with no overlap? How would you prove this?

Yes, it is always possible.  To prove it, first find a Hamiltonian cycle of the chessboard.

hum...i believe this is a (very) classic invariant problem, and the answer is yes. There is a diagram that outlines the pattern of solution, it looks much like a snake, i can't draw it on computer, err...this is frustrating  :(

If your squares have coordinates $(i,j),(k,l)$ (with $i,j,k,l\in\overline{1,8}$), first isolate the rectangle having $(i,j),(k,l)$ as two opposite vertices: one of the sides of this rectangle is even, and $8$ (the length of any row or column of the board) is also even, so you can easily fill the outside of the rectangle with dominoes.

We're only left with the rectangle now, and two of its vertices must not be covered by dominoes. Now you can do that "snake thing" al.M.V. was talking about: just place dominoes one row at a time (assuming the rows have even length; if the columns have even length, do it one column at a time), each time placing a domino so that you cover as many squares as possible of the current row (if the rows have length $2k$, for instance, on the first row we will have $k-1$ horizontal pieces at first, and a vertical one to cover the last square of the row, assuming that the first square of the first row is one of the forbidden corners).

Thanks grobber! :D I didn't feel like typing all that... :lol:


conics
geometry
trapezoid
projective geometry
geometry solved

Let be given two parallel lines $k$ and $l$, and a circle not intersecting $k$. Consider a variable point $A$ on the line $k$. The two tangents from this point $A$ to the circle intersect the line $l$ at $B$ and $C$. Let $m$ be the line through the point $A$ and the midpoint of the segment $BC$. Prove that all the lines $m$ (as $A$ varies) have a common point.

Another solution:

[quote="Megus"]Let be given two parallel lines $k$ and $l$, and a circle not intersecting $k$. Two tangents from a variable point $A \in k$ to the circle intersect the line $l$ at $B$ and $C$. Let $m$ be the line through $A$ and the midpoint of $BC$. Prove that all the lines $m$ (as $A$ varies) have a common point.[/quote]

In fact, we will show that all the lines m pass through the pole N of the line k with respect to our circle.

We know that the lines AB and AC are tangents to our circle; let them touch the circle at the points M and K. Also, let O be the center of the circle.

If a point lies outside a circle, then its polar with respect to the circle can be constructed by joining the points where the two tangents from the point to the circle touch the circle. Hence, the polar of the point A with respect to our circle is the line MK. Since the line k passes through the point A, the pole N of the line k with respect to our circle lies on the polar MK of the point A. On the other hand, since N is the pole of the line k with respect to our circle, whose center is O, we have $ON\perp k$; since $k\parallel l$, this becomes $ON\perp l$, or, equivalently, $ON\perp BC$.

So our point N is a point on the line MK satisfying $ON\perp BC$. Thus, by http://www.mathlinks.ro/Forum/viewtopic.php?t=5830 post #4 Theorem 2, it follows that, if the lines AN and BC intersect at a point Q, then this point Q is the midpoint of the segment BC. In other words, the line AN passes through the midpoint of the segment BC, i. e. through the point M. Hence, the line m through the point A and the midpoint of the segment BC coincides with the line AN and thus passes through the point N. This completes the proof.

  darij

Isn't it a genetically modified butterfly?

  T.Y.

  M.T.

[quote="grobber"]Let $AB,AC$ touch the circle in $U,V$ respectively. Also, let $M$ be the midpoint of $BC$, $Q=UV\cap k$, and $P=AM\cap UV$, and $X,Y$ the intersections between the circle and $AM$.

$UXVY$ is a harmonic quadrilateral, so the pole of $XY$ lies on $UV$. At the same time, we have $(AU,AV;XY,AQ)=-1$, so the same pole of $XY$ lies on $AQ=k$. This means that the respective pole is $Q$. Since $A$  is the pole of $UV$, it must be that $P=XY\cap UV$ is the pole of $AQ=k$. This shows that $AM$ always passes through $P$, the pole of $k$ wrt the circle.[/quote]
Sorry,I can't understand this solution.Who can help me ?
I have known what is the pole and polar .but I can't understand this setence"At the same time, we have $(AU,AV;XY,AQ)=-1,$so the same pole of $XY$ lies on $AQ=k$,"

[color=darkblue][b]An analytical method.[/b] The equation of the fixed circle $w$ is $x^{2}+y^{2}-1=0$ ; the equations of the fixed lines $k$, $l$ are $y=a$ with $|a|>1$ and $y=b$ respectively ; the mobile point is $A(m,a)\in k$. The conic $h$ of the tangents to the circle $w$ from the point $M$ is $(x^{2}+y^{2}-1)(m^{2}+a^{2}-1)=(mx+ay-1)^{2}$. Thus, the abscises  $x_{B}$, $x_{C}$ of the points $\{B,C\}\subset h\cap l$ are the roots of the equation $(x^{2}+b^{2}-1)(m^{2}+a^{2}-1)=(mx+ab-1)^{2}$, i.e. $(a^{2}-1)x^{2}-2m(ab-1)x+\ldots =0$. Therefore, the middlepoint $M$ of the segment $[BC]$ has the co-ordinates $x_{M}=\frac{x_{B}+x_{C}}{2}=\frac{m(ab-1)}{a^{2}-1}$ and $y_{M}=b$. The equation of the line $AM$ is : $\left|\begin{array}{ccc}x & y & 1\\\ m & a & 1\\ m(ab-1) & b(a^{2}-1) & a^{2}-1\end{array}\right|=0$ $\Longleftrightarrow$ $m(b-a)(ay-1)-(a^{2}-1)x=0$, i.e. the demand fixed point is $F\left(0,\frac{1}{a}\right)$. The polar line of the point $F$ w.r.t. the circle $w$ is $0\cdot x+\frac{1}{a}\cdot y-1=0$, i.e. $y=a$- the equation of the line $k$. Therefore the fixed point $F$ is the pol of the line $k$ w.r.t. the circle $w$.

[b]Remark (memorize ! ).[/b] Let $f(x,y)\equiv x^{2}+y^{2}+2mx+2ny+p=0$ be the equation of the circle $w=C(O,R)$, where $O(-m,-n)$ and $R^{2}=m^{2}+n^{2}-p\ .$
$1.\blacktriangleright$ The equation of the polar of the point $P(a,b)$ w.r.t. the circle $w$ is : $\boxed{\ d(x,y)\equiv ax+by+m(x+a)+n(y+b)+p=0\ }\ .$
If the point $P\in w$, then the its polar line w.r.t. the circle $w$ becomes the tangent $PP$ to the circle $w$.
The co-ordinates of the pol $S$ for the line $Ax+By+C=0$ w.r.t. the circle $w$ you can obtain from the relations : $\boxed{\ \frac{m+x_{S}}{A}=\frac{n+y_{S}}{B}=\frac{mx_{S}+ny_{S}+p}{C}\ }\ .$
 $2.\blacktriangleright$ The equation of the conic $h$ of the tangents to the circle $w$ from the point $P(a,b)$ is : $\boxed{\ f(x,y)\cdot f(a,b)=d^{\ 2\ }(x,y)\ }\ .$[/color]

[color=darkblue][b]The synthetical method.[/b] The tangents $AB$, $AC$ touch the circle $w=C(O)$ in the points $D$, $E$ respectively. Denote : the fixed point $P\in DE$
(the pol of the line $k$ w.r.t. the circle $w$) for which $OP\perp k$ ; the points $F\in AC$, $G\in AB$ for which $P\in FG$ and $FG\parallel k$.
From the [u]lower lemma[/u] $(\searrow )$ results that $PF=PG$. Therefore, in the trapezoid $BCGF$ the fixed point $P$ belongs to the line $AM$.

[b]Lemma.[/b] Let $ADE$ be an isosceles triangle $(AD=AE)$. Denote the point $O$ for which $OD\perp AD$ and $OE\perp AE$.
For a point $P\in (DE)$ define the points $F\in AE$, $G\in AD$ for which $P\in (FG)$ and $FG\perp OP$. Then $PF=PG$.
[u]Proof.[/u] The quadrilaterals $FEPO$ and $GDOP$ are inscribed $\Longrightarrow$ $\widehat{FOP}\equiv\widehat{AED}\equiv\widehat{ADE}\equiv\widehat{GOP}$ $\Longrightarrow$
$\widehat{FOP}\equiv\widehat{GOP}$, i.e. the triangle $FOG$ is isosceles and $OP\perp FG$ $\Longrightarrow$ $PF=PG$.

[b]Remark.[/b] This lemma offers a solution for the following remarkable construct problem :[/color]

[color=darkred][b]Construct problem.[/b] Given are two rays $[AX$, $[AY$ and a point $B$ which belongs to the interior of the angle $\widehat{XAY}$.
Construct the points $M\in [AX$ and $N\in [AY$ so that $B\in MN$ and $BM=BN$.[/color]

[color=darkblue]Indeed, let $C\in [AX$, $D\in [AY$ be the points for which $B\in CD$ and $AC=AD$. Denote the point $O$ for which $OC\perp AX$ and $OD\perp AY$.
Then the points $M$, $N$ are the intersections of the perpendicular line in the point $B$ to the line $OB$ with the rays $[AX$ and $[AY$.[/color]

[quote="Virgil Nicula"][color=darkblue][b]An analytical method.[/b] The equation of the fixed circle $w$ is $x^{2}+y^{2}-1=0$ ; the equations of the fixed lines $k$, $l$ are $y=a$ with $|a|>1$ and $y=b$ respectively ; the mobile point is $A(m,a)\in k$. The conic $h$ of the tangents to the circle $w$ from the point $M$ is $(x^{2}+y^{2}-1)(m^{2}+a^{2}-1)=(mx+ay-1)^{2}$. Thus, the abscises  $x_{B}$, $x_{C}$ of the points $\{B,C\}\subset h\cap l$ are the roots of the equation $(x^{2}+b^{2}-1)(m^{2}+a^{2}-1)=(mx+ab-1)^{2}$, i.e. $(a^{2}-1)x^{2}-2m(ab-1)x+\ldots =0$. Therefore, the middlepoint $M$ of the segment $[BC]$ has the co-ordinates $x_{M}=\frac{x_{B}+x_{C}}{2}=\frac{m(ab-1)}{a^{2}-1}$ and $y_{M}=b$. The equation of the line $AM$ is : $\left|\begin{array}{ccc}x & y & 1\\\ m & a & 1\\ m(ab-1) & b(a^{2}-1) & a^{2}-1\end{array}\right|=0$ $\Longleftrightarrow$ $m(b-a)(ay-1)-(a^{2}-1)x=0$, i.e. the demand fixed point is $F\left(0,\frac{1}{a}\right)$. The polar line of the point $F$ w.r.t. the circle $w$ is $0\cdot x+\frac{1}{a}\cdot y-1=0$, i.e. $y=a$- the equation of the line $k$. Therefore the fixed point $F$ is the pol of the line $k$.

[b]Remark (memorize ! ).[/b] Let $f(x,y)\equiv x^{2}+y^{2}+2mx+2ny+p=0$ be the equation of the circle $w$ and let $P(a,b)$ be a point. Then :
$1.\blacktriangleright$ The equation of the polar of the point $P$ w.r.t. the circle $w$ is : $\boxed{\ d(x,y)\equiv ax+by+m(x+a)+n(y+b)+p=0\ }\ .$
$2.\blacktriangleright$ The equation of the conic $h$ of the tangents to the circle $w$ from the point $P$ is : $\boxed{\ f(x,y)\cdot f(a,b)=d^{\ 2\ }(x,y)\ }\ .$
[/color][/quote]
Thank you for the nice solution.But who can explain the grobber's post ?

In the previous my message I added a synthetical method for this problem.


[quote="#H34N1"]If $ 60^{a}\equal{} 3$ and $ 60^{b}\equal{} 5$, then find $ 12^{[(1\minus{}a\minus{}b)/2(1\minus{}b)]}$.[/quote]

[hide]Note that $ a\equal{}\log_{60}3$, $ b\equal{}\log_{60}5$. So $ a\plus{}b \equal{}\log_{60}15$. Hence,

$ 1\minus{}(a\plus{}b) \equal{}\log_{60}4 \equal{}\log_{60}2^{2}$.

Also, note that:

$ 2(1\minus{}b) \equal{} 2\cdot\log_{60}12 \equal{}\log_{60}12^{2}$.

Therefore,

$ \frac{1\minus{}a\minus{}b}{2(1\minus{}b)}\equal{}\frac{\log_{60}2^{2}}{\log_{60}12^{2}}\equal{}\log_{12^{2}}2^{2}\equal{}\log_{12}2$

So, raising both sides to the 12-th power gets us the answer of $ \boxed{2}$.[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ H\cong \mathbb{Z}_2$ be a normal subgroup of $ G$ and $ G/H \cong \mathbb{Z}$

Show that $ G\cong \mathbb{Z}\times  \mathbb{Z}_2$

$ \mathbb{Z}$ is free. Therefore $ 1\to H\to G \to \mathbb{Z} \to 1$ splits, i.e. $ G\cong H\ltimes\mathbb{Z}$. The only automorphism of $ H$ is the identity so that this semidirect product is in fact a direct one.

[quote="SD5225272"]Hint: Show $ H$ is in $ Z(G)$.

Steve[/quote]
Well, $ H \equal{} \{e,a\}$, so $ gHg^{ \minus{} 1} \equal{} \{e,a\} \Rightarrow ag \equal{} ga$ witch proves that $ H$ is in $ Z(G)$

But I don't really see how to move further on.

There is a theorem saying if $ H$ and $ K$ are subgroups of $ G$ st $ H \cap K \equal{} \{e\}$, $ \forall k\in K, h \in H, hk \equal{} kh$ and $ HK \equal{} G$

Then $ G \cong H\times K$

don't know if that can be useful


So in aA + bB --> cC + dD

$R =-\frac{\delta [A]}{a \delta t}=-\frac{\delta [B]}{b \delta t}=\frac{\delta [C]}{c \delta t}= \frac{\delta [D]}{d \delta t}$

When solving a differential rate reaction, is the value of R obtained equal to the generalized R described above?

If so, would the rate of an individual species be equal to the species' coefficient multiplied by the R from the rate equation?

In general, you solve differential rate equations with respect to a certain species.  For example, it would be common to see the (first order) differential rate equation

- d[x]/dt = k [x].

Here, we are talking about the concentration of a specific species, x.  If you were to use the unique instantaneous reaction rate (defined as R above) then you would have to rewrite this equation as    - R * (stoichiometric coefficient of x) = k [x].  As you can see, this gets rather messy (and conveys no more information, as k is an arbitrary constant anyway).  If you try to integrate the R equation, you get something that is, in general, no more useful than a normal integrated rate law and likely a bit more confusing (i think you end up with an equation describing equivalents remaining, or something along those lines).

in answer to your second question, you are correct in that given R, you just need to multiply by the stoichiometric coefficient to get the rate for a given species.

hope that helps.

Thanks.

Is the overall rate of reaction equal to R?

Again, generally that's not the form that is asked for.  Technically, that is what you should report if a problem does not ask for rate with respect to a certain species.  Just be sure, in any case, that you report what species you are taking the rate with respect to.


any free book for projective geometry on the net...

Yeah, plenty.

http://www.geometer.org/mathcircles/projective.pdf
http://www.maths.ox.ac.uk/~hitchin/hitchinnotes/Projective_geometry/Chapter_1_Projective_geometry.pdf
http://ai.stanford.edu/~birch/projective/projective.pdf
http://www.galaxy.gmu.edu/stats/syllabi/csi877/Lecture8.pdf
http://modelingnts.la.asu.edu/pdf/PGwithCA.pdf
http://com2mac.postech.ac.kr/Lecture/Lec-9.pdf

It just depends on what level you're reading/working on.  If you want more, let me know.  Or you can type [b][code]projective geometry filetype:pdf[/code][/b] in Google and search!

sorry no projective geometry, but projective transformation and their aplication to elementary geometry.

For now, try:

http://www.robots.ox.ac.uk/~ian/Teaching/CompGeom/lec2.pdf
http://www.fen.bilkent.edu.tr/~franz/ag05/ag-07.pdf

Tell me if you need more.

EDIT: Also try:
http://www.umiacs.umd.edu/~ramani/cmsc828d/ProjectiveGeometry.pdf


PRove or disprove :
Supposed that $ A$ and $ B$ are two matrices with the same size, such that for every complex number $ l$ and natural number $ n$, $ rank((lI\minus{}A)^n)\equal{}rank((lI\minus{}B)^n)$, then $ A$ is similiar with $ B$.
Thanks in advance.

Clearly, we can assume $ A,B\in M(n,\mathbb{C})$ already in Jordan normal block form, i.e.

$ \begin{array}{lcl} A & = & J_{\alpha_1}(a_1)\oplus\ldots\oplus J_{\alpha_r}(a_r)\\
B & = & J_{\beta_1}(b_1)\oplus\ldots\oplus J_{\beta_s}(b_s)\end{array}$

in particular, $ n=\sum_{i=1}^r\alpha_i=\sum_{i=1}^s\beta_i$ holds. Define for $ l\in\mathbb{C},m\in\mathbb{N}: t_A(l,m)=|\{i|a_i=l,\alpha_i=m\}|$ and similarly define $ t_B(l,m)$.

Now, for arbitrary $ l\in\mathbb{C},n\in\mathbb{N}$ direct calculation gives:

$ \begin{array}{lcl}0 & = & rk[(lI-A)^n]-rk[(lI-B)^n]\\[0.1cm] & = & \sum_{i: a_i\not=l}\alpha_i+\sum_{i: a_i=l}\max\{\alpha_i-n,0\}-\left[\sum_{i: b_i\not=l}\beta_i+\sum_{i: b_i=l}\max\{\beta_i-n,0\}\right]\\[0.2cm]
& = & n+\sum_{i: a_i=l}[\max\{\alpha_i-n,0\}-\alpha_i]-\left[n+\sum_{i: b_i=l}[\max\{\beta_i-n,0\}-\beta_i]\right]\\[0.1cm]
 & = & \sum_{i: b_i=l}\min\{\beta_i,n\}-\sum_{i: a_i=l}\min\{\alpha_i,n\}\\[0.1cm]
 & = & \sum_{r=1}^{n-1}r\,[t_B(l,r)-t_A(l,r)]+n\sum_{r\geq n}[t_B(l,r)-t_A(l,r)]\end{array}$

This implies $ J: =\{a_1,\ldots,a_r\}=\{b_1,\ldots,b_s\}$ and for $ l\in J: \,\sum_{r\geq 1}t_A(l,r)=\sum_{r\geq 1}t_B(l,r)$. Replacing this in the last line of the previous equation gives: $ 0=\sum_{r=1}^{n-1}(n-r)\,[t_A(l,r)-t_B(l,r)]$, i.e. iterating through $ n=2,3,\ldots$ we get $ \forall r\in\mathbb{N}: t_A(l,r)=t_B(l,r)$.

Summarising the observations, we have that $ A,B$ have the same set of eigenvalues and the number of corresponding Jordan blocks for an arbitrary eigenvalue and block size is the same, i.e. $ A,B$ are similar.


On a frictionless pool table, a triangle of 15 balls is hit with the cue ball. Since there is no friction, all of the balls will fall into one of 6 pockets, including the cue ball. Suppose each ball is equally likely to fall into any one pocket.

1. What is the probability that at least one pocket is empty?

2. Disregard the cue ball. The "value" of a pocket is the sum of the numbers of the ball(s) in that pocket. If there is a 14, 6, and 5, the pocket's value is 25. How many ways can all the balls be pocketed such that all non-empty pockets have equal value? (Don't count permutations of pockets; all balls in pocket 1 is the same as all balls in pocket 4.)
One way is: 15+5, 14+6, 13+7, 12+8, 11+9, 10+4+3+2+1

3. Same as problem 2, but include the cue ball, and give it a value of 16. How many ways?

tarquin--I think you need to add a ${n \choose k}$ in front of some of those.

1.

[hide]
To find the probability of there being at least one pocket empty, consider the probability of the single case where NO pocket is empty.  We will simply subtract this from 1.

First give each pocket one ball, and then distribute the remaining 10 balls (treated as indistinguishable) to all of the pockets.  There are ${10+6-1} \choose {10}$ ways to do this.  The total number of ways to distribute all of the balls to the pockets is thus ${{16+6-1} \choose {10}}$.  So the probability of there being at least one pocket empty is $1- {10+6-1 \choose 10} / {16+6-1 \choose 10}$.[/hide]

[quote="tarquin"]1.
[hide]$(\frac 56)^{16}*6 + (\frac 23)^{16}*15 + (\frac 12)^{16}*20 + (\frac 13)^{16}*15 + (\frac 16)^{16}*6 = \frac {5^{16}*6 + 4^{16}*15 + 3^{16}*20 + 2^{16}*15 + 1^{16}*6}{6^{16}} = \frac {10216810111}{29386561536}$.[/hide]
For 2 and 3 all I know is that.. It's a lot![/quote]

it doesn't really work well like that since if you a probability of 5/6 that it's going to go into one of the 5 cups and not the 6th one, it may all go into 4 of the 5 cups. not all 5 cups are filled. and same for the other probabilities.


[b] PROBLEM[/b]

It is known of a polynomial over $Z$ that $p(n)>n$ for every positive integer n. Consider $x_{1}=1$, $x_{2}=p(x1)...$ . We know that, for any positive integer $N$, there exist a term of the sequence divisible by $N$. Prove that $p(x)=x+1$.

P.S. There is a lemma about some properties of polynomials in $Z$ that may help a lot for this problem and especially for problem 5 (IMO).


How can I prove that if:

$ x\equiv a (mod  p)$

and

$ x\equiv a ( \hspace mod q)$

then:

$ x\equiv a (mod \hspace p*q)$

only when $ gcd(p,q) = 1$.

Thanks.

if $ GCD(p,q)\equal{}1$ then there exist integers $ v,u : pu\plus{}qv\equal{}1 \leftrightarrow pu(x\minus{}a)\plus{}qv(x\minus{}a)\equal{}x\minus{}a$
we have $ p|x\minus{}a \equal{}> pq|qv(x\minus{}a)$ and $ q|x\minus{}a \equal{}> qp|pu(x\minus{}a)$ so $ qp|x\minus{}a \equal{}> x \equiv a (mod qp)$
if $ GCD (p,q)\equal{}d >1$ then there is a possibility as u see (if u try the same way) that when $ qp | d(x\minus{}a)$, $ qp$  does not divide $ x\minus{}a$
so a counterexample for the last case is
$ (p\equal{}3, q\equal{}6,x\equal{}18, a\equal{}0)$

$ x\equal{}pk\plus{}a\equal{}qm\plus{}a$, implying $ pk\equal{}qm$ $ (*)$. If $ (p;q)\equal{}1$, than from $ (*)$ we deduce $ k\equal{}qr$, where $ r$ is an integer number. So, $ x\equal{}pqr\plus{}a$, or $ x\equiv a (mod pq)$, q.e.d. But it doen't mean that if $ (p,q)>1$, than $ pq$ can't divide $ x\minus{}a$, as for example when $ p,q$ are random positive integers with $ (p,q)>1$ and $ x\equal{}pq\plus{}a$, the given conditions hold but $ (p,q)$ is not $ 1$.

$ \begin{cases} p|(x\minus{}a) \\
q|(x\minus{}a) \end{cases} \Leftrightarrow pq|(x\minus{}a)$ as $ p, q$ coprime .

please post this in the pre olympiad section


integration
limit
real analysis
real analysis unsolved

Let $ w: [0,\infty) \to \mathbb{R}$ be bounded and non-negative such that $ \int_{0}^\infty \frac{w(x)}{x+1}dx < \infty$. Show that $ \lim_{T \to \infty}\frac{1}{T}\int_{0}^{T}w(x) dx = 0$.

Very classical stuff :). We fix $ A>0$ and write
\[ \frac{1}{T}\int_{0}^{T}w=\frac{1}{T}\int_{0}^{A}w+\frac{1}{T}\int_{A}^{T}w \le \frac{1}{T}\int_{0}^{A}w+\int_{A}^\infty\frac{w(x)}x\,. \]
The first term in the last sum clearly tends to $ 0$ as $ T\to\infty$ and the second can be made arbitrarily small by choosing sufficiently large $ A$.


Yale
college
MIT
AMC
AMC 12
AMC 12 B
Stanford

I was told that the universities that did ask whether one is an AMC10/12 veteran would ask his/her best score in that contest because they feel that AMC contests cover a deeper level of math than usual high school currculum and that these institutions feel that one's AMC10/12 score would help determine one's readiness to follow a rigorous undergraduate program.

What universities were reported to actually ask for it?

Yale asks for them, and supposedly MIT does too. I don't know of any list of others that do. You could always mention them in an essay, of course.

But do Washington State University ask for it? Because I may advise a friend of mine (though we both live in Quebec) to take part in AMC12B so she may have better odds of being accepted there, apart from her rowing background (my friend is a Canamex champion; both WSU and Texas A&M courted her).

Stanford does as well IIRC.

Edit: Okay, I don't remember correctly.

[quote="MCTP"]Yale asks for them, and supposedly MIT does too.[/quote]

MIT does have a spot on their application for AMC 10/12 scores and AIME scores, however, they don't require it.

"If you have taken the AMC 10, AMC 12, or AIME exams, list dates and scores. (If you haven’t taken these exams or don’t know what they are, don’t worry about it.)" -- MIT application part 2.

MIT, Caltech, and Brown were the schools I applied to that specifically asked for AMC scores.  Stanford did not ask for them.

I highly doubt a school like Washington State would even know what the AMC was, much less ask for scores.  But I could be wrong.

I can also confirm that Stanford doesn't ask for them.


function
algebra
domain
functional equation

$ f : \mathbb{N} \to \mathbb{N}$ satisfies $ f(m^2 \plus{} n^2) \equal{} f(m)^2 \plus{} f(n)^2$ and $ f(1) \equal{} 1$.  Show $ f(n) \equal{} n$.

Let $ m\equal{}1,n\equal{}0$. Then we have $ f(0)\equal{}0$.
Now let $ n\equal{}0$, then $ f(m^2)\equal{}f(m)^2$
So $ f(m^2\plus{}n^2)\equal{}f(m^2)\plus{}f(n^2)$ and this satisfies the cauchy equation, so $ f(x)\equal{}xf(1)$ for all $ x$. But $ f(1)\equal{}1$, so $ f(n)\equal{}n$ for all $ n$.

Incorrect. $ f(m^2 \plus{} n^2) \equal{} f(m^2) \plus{} f(n^2)$ only means the function satisfies the Cauchy equation for perfect squares.

Oops. But we have $ f(m)^2\equal{}f(m^2)$ so we have $ f(m)^2\equal{}m^2$ so $ f(m)\equal{}m$ for all $ m$

[quote="malcolm"]Oops. But we have $ f(m)^2 \equal{} f(m^2)$ so we have $ f(m)^2 \equal{} m^2$ so $ f(m) \equal{} m$ for all $ m$[/quote]

I don't see how this follows. Would you explaining?

[quote]Is $ 0\in \mathbb{N}$?[/quote]

Depends what country you're in. But I would guess yes in the context of this problem, though I don't think it actually matters except in the method of solving it.

From the cauchy bit, we have that $ f(m^2)\equal{}m^2$ for all perfect squares.
From letting $ n\equal{}0$, we have $ f(m^2)\equal{}f(m)^2$. Combining these, we have $ f(m)^2\equal{}m^2$ so $ f(m)\equal{}m$ ( we can exclude $ f(m)\equal{}\minus{}m$ since it is naturals)

[quote="malcolm"]From the cauchy bit, we have that $ f(m^2) \equal{} m^2$ for all perfect squares.[/quote]
No we don't.  The squares aren't closed under addition.

By the way, in the original problem $ 0 \not \in \mathbb{N}$, which makes the problem harder.  There's a pretty nice thread about this problem somewhere, although I can't find it, which also considers the generalization to cubes.

In the variant I have, $ 0 \in \mathbb{N}$.

[quote="t0rajir0u"][quote="malcolm"]From the cauchy bit, we have that $ f(m^2) \equal{} m^2$ for all perfect squares.[/quote]
No we don't.  The squares aren't closed under addition.
[/quote]

Can we "extend" it to a function of the reals which satisfies the cauchy equation, and use the fact that $ f(x^2)\equal{}f(x)^2$ implies it is positive on the positives, and thus must be a linear function since bounded solutions to the Cauchy equation must be linear?

[quote="malcolm"]Can we "extend" it to a function of the reals[/quote]
No; there's no reason that any extension of the function to a larger domain should satisfy the same functional equation as the original function.  For example, it's easy to come up with a function that satisfies $ f(n \plus{} m) \equal{} f(n) \plus{} f(m), n, m \in \mathbb{Z}$ by making it zero on the integers, but that doesn't imply that it's linear on $ \mathbb{R}$.


abstract algebra
superior algebra
superior algebra unsolved

Let a and b be members of a non-abelian (non-communative) group. Prove that if (a^-1)bba=bbb and (b^-1)aab=aaa then a=b=1


analytic geometry
quadratics
algebra unsolved
algebra

Show that the system
$ x\plus{}y\plus{}z \equal{} 4$
$ x^3\plus{}y^3\plus{}z^3 \equal{} 10$
has not integers solutions with some coordinate less or equal than zero.

[quote="Lousin Garckz"]Show that the system
$ x \plus{} y \plus{} z \equal{} 4$
$ x^3 \plus{} y^3 \plus{} z^3 \equal{} 10$
has not integers solutions with some coordinate less or equal than zero.[/quote]

The first equation implies $ z\equal{}4\minus{}x\minus{}y$ and the the second becomes $ x^3\plus{}y^3\plus{}(4\minus{}x\minus{}y)^3\equal{}10$ which may be written $ xy(x\plus{}y)\minus{}4(x\plus{}y)^2\plus{}16(x\plus{}y)\minus{}18\equal{}0$

So $ x\plus{}y|18$ and $ (x\plus{}y)\in\{\minus{}18,\minus{}9,\minus{}6,\minus{}3,\minus{}2,\minus{}1,\plus{}1,\plus{}2,\plus{}3,\plus{}6,\plus{}9,\plus{}18\}$ and $ xy\equal{}4(x\plus{}y)\minus{}16\plus{}\frac{18}{x\plus{}y}$

$ x\plus{}y\equal{}\minus{}18$ $ \implies$ $ xy\equal{}\minus{}89$ $ \implies$ $ (x,y)$ solutions of the quadratic $ X^2\plus{}18X\minus{}89$ $ \implies$ no integer solution
$ x\plus{}y\equal{}\minus{}9$ $ \implies$ $ xy\equal{}\minus{}54$ $ \implies$ $ (x,y)$ solutions of the quadratic $ X^2\plus{}9X\minus{}54$ $ \implies$ no integer solution
$ x\plus{}y\equal{}\minus{}6$ $ \implies$ $ xy\equal{}\minus{}43$ $ \implies$ $ (x,y)$ solutions of the quadratic $ X^2\plus{}6X\minus{}43$ $ \implies$ no integer solution
$ x\plus{}y\equal{}\minus{}3$ $ \implies$ $ xy\equal{}\minus{}34$ $ \implies$ $ (x,y)$ solutions of the quadratic $ X^2\plus{}3X\minus{}34$ $ \implies$ no integer solution
$ x\plus{}y\equal{}\minus{}2$ $ \implies$ $ xy\equal{}\minus{}33$ $ \implies$ $ (x,y)$ solutions of the quadratic $ X^2\plus{}2X\minus{}33$ $ \implies$ no integer solution
$ x\plus{}y\equal{}\minus{}1$ $ \implies$ $ xy\equal{}\minus{}38$ $ \implies$ $ (x,y)$ solutions of the quadratic $ X^2\plus{}X\minus{}38$ $ \implies$ no integer solution
$ x\plus{}y\equal{}1$ $ \implies$ $ xy\equal{}6$ $ \implies$ $ (x,y)$ solutions of the quadratic $ X^2\minus{}X\plus{}6$ $ \implies$ no real solution
$ x\plus{}y\equal{}2$ $ \implies$ $ xy\equal{}1$ $ \implies$ $ (x,y)$ solutions of the quadratic $ X^2\minus{}2X\plus{}1$ $ \implies$ $ (x,y)\equal{}(1,1)$
$ x\plus{}y\equal{}3$ $ \implies$ $ xy\equal{}2$ $ \implies$ $ (x,y)$ solutions of the quadratic $ X^2\minus{}3X\plus{}2$ $ \implies$ $ (x,y)\equal{}(1,2)$ or $ (2,1)$
$ x\plus{}y\equal{}6$ $ \implies$ $ xy\equal{}11$ $ \implies$ $ (x,y)$ solutions of the quadratic $ X^2\minus{}6X\plus{}11$ $ \implies$ no real solution
$ x\plus{}y\equal{}9$ $ \implies$ $ xy\equal{}22$ $ \implies$ $ (x,y)$ solutions of the quadratic $ X^2\minus{}9X\plus{}22$ $ \implies$ no real solution
$ x\plus{}y\equal{}18$ $ \implies$ $ xy\equal{}57$ $ \implies$ $ (x,y)$ solutions of the quadratic $ X^2\minus{}18X\plus{}57$ $ \implies$ no integer solution

And the only integer solutions of the requested equation are $ (1,1,2),(1,2,1),(2,1,1)$
Hence the result.

You're welcome.

Glad to have helped you.

thanks pco, you are solving the problems i'm posting and helping me a lot!  :D


Here are my two cents... I personally have only taken the AIMEs once, this year but I found with the AMC tests I've taken, doing past tests helps immensely (AMC 8, 10, AIME [altho I didn't study for the AIMEs]). I'm sure there is some better advice out there, so sorry if this is poor advice, but I assume there is some correlation

This is my first year going... but the way I'm going to prepare is mostly by doing as many old USAMO problems that I can (which isn't very many, but that's ok). I also have some books for reference as well as my transcripts from my AoPS classes =).

Man I don't know how I could sit there for nine hours cumulatively staring at six proof-oriented problems... it would drive me nuts by the fourth hour. Anyways good luck to anyone who qualified (:( not me)

[quote="MithsApprentice"]Simply put, there's no way you could start studying for USAMO now and expect to cover a significant portion of the material that could be covered. Just work some olympiad problems on this site, pray hard, sleep well, and eat a lot during the test.[/quote]

That's great advice.

[quote="tekno10m"]How do [i]you[/i] prepare for the USAMO? (Or, what is the best way to prepare for the USAMo?)

Do you read and work through problem solving books like Engel's "Problem-Solving Strategies", doing all the exercises/problems?

Do you simply work on past USAMO's?

Do you read and work on math journals like CRUX?

Do you simply work on [i]any[/i] olympiad problem, e.g. working on the book "Mathematical Olympiads from Around the World 199_ to 199_"?

Do you post regularly on the Olympiad section of AOPS/Mathlinks?

Do you do something else?

Or do you do a combination of the above?

Personally, I think I should simply work on past olympiad problems from the USAMO or other olympiads, because most USAMO problems are so distinct that working on books wouldn't really make you learn how to do well on the USAMO.[/quote]

I haven't qualified for the USAMO yet (just 2 points away...), but I'd probably say All of the Above except probably the Crux Mathemawhatever (you won't get to many issues between now and the USAMO :) ).

[quote="DPopov"]For all past USAMOers:

Does analytical goemetry come in handy when dealing with USAMO geometry questions? I am just wondering because I am not very strong in that area and if it is helpful I will probably try to improve by April 19th. :)[/quote]

I've tried to reply to this in the best way possible in another thread. :) gl dimitar

Yeah, sorry about that, thanx for the reply. I don't think I'm really going to waste a week trying to learn how to APPLY analytical geometry to complex USAMO problems which probably won't require it. I think I'll stick to the old "take lots of old tests" strategy.


MATHCOUNTS
AMC
AIME
trigonometry
geometry
AIME I

[quote="tiger21"]I often make stupid mistakes.
So I do double checks to reduce my mistakes.

If I do double check right after I finish each problem, I often find myself cannot finish the test.
If I do double check after I finish all problems, I found I have to start my processes over which takes longer time to double check my work.

When is the best time do double check?

Any good ways to double check your work?

Thanks.[/quote]

I usually eyeball the problem and try and come up with a reasonable estimate in 10 seconds or less. Then I do it and compare it to my estimate. If its close, that comprises my double-check of the problem. If its far away, I quickly skim my work and reread the problem to be sure of what it is asking.

You should try to check for computational errors after the problem and check for other things at the end (correct units, common fraction,...). Of course, this is my opinion and it may not be the best.

Make sure you write the same answer you came up with.

[quote="tiger21"]I often make stupid mistakes.
So I do double checks to reduce my mistakes.

If I do double check right after I finish each problem, I often find myself cannot finish the test.
If I do double check after I finish all problems, I found I have to start my processes over which takes longer time to double check my work.

When is the best time do double check?

Any good ways to double check your work?

Thanks.[/quote]

Practice Practice Practice!! is one way to do it. You time yourself and continually do problems and time yourself.

Another way is to look at the question and just mutter under your breath what you did really fast. Saying it quietly sometimes helps you catch stuff. The best time to look over something is after you are finished. Finishing all the problems with effort gives you more of a chance of getting higher points than you getting like half of them done by checking them.

Of course, whispering what you did might give people around you a clue as to how to solve the problem.

I find it sometimes helps to jot down a few notes on how you solved each one. They don't have to be anything fancy, just short, abbreviated ones to make it easier to check and find mistakes.

[quote="PowerOfPi"]You don't have to say it loud. Anyway, I whisper to myself even when I am solving the problem.[/quote]

Isn't whispering "out loud"? :wink:

I really hope this isn't going to be a topic of how loud you whisper, so I'm going to clarify a few things.

What I meant by whispering was saying a few numbers and words once in a while. Like this:

So you~~~~~5*6~~~carry the one~~~and etc.....

But as you can see in my last post, I said muttering.

Original Topic:
Actually, I found a new way to practice. Since you mess up a lot on questions, check only the easy ones such as 1-15 or how ever many you think is easy to you. People make more mistakes on easy questions because they underestimate it and just go over it too quickly. And don't check the last 5 if you don't have time. Just give it your best shot and finish. You'll have a better chance of getting more questions right.

Also, if you write numbers down or put them into a calculator, make sure you're writing the same number as it says in the problem. Make sure you read the whole problem too, don't skim.

[quote="tiger21"]I often make stupid mistakes.
So I do double checks to reduce my mistakes.

If I do double check right after I finish each problem, I often find myself cannot finish the test.
If I do double check after I finish all problems, I found I have to start my processes over which takes longer time to double check my work.

When is the best time do double check?

Any good ways to double check your work?

Thanks.[/quote]
I normally don't check the problems until I'm all done since I'd rather get a wrong answer and get partial credit than not answer at all. Also, if I don't have a lot of time, I check the ones that I find easy since that's when I make the most mistakes.

If it's multiple choice, I consider my answer checked if it's one of the options. This usually works, but since the incorrect choices are supposed to be common mistakes, it doesn't always work and is better for school math tests.

Read each problem carefully and keep in mind what it's asking for (there isn't always enough time for this in MATHCOUNTS countdown round and similar contests). Many of the mistakes I make are ones I could catch by reading the whole question.

If you don't know how to do a problem, consider skipping it because chances are you won't figure it out in the amount of time you have, time that could be better spent on other problems.

Don't be overconfident on easy problems, but also don't get the wrong answer by trying to make them harder.

I often double check as I am going through the problem, as there is no time for me to check after I finish the test... Muttering to yourself works to some extent, but people might get a hint or 2...

Good Luck!

[quote="AIME15"]Er, there's no partial credit on MATHCOUNTS :wink:[/quote]
I know. I'm taking about school math.

[quote="jonathanchou711"]Wait. Tiger21. Are you talking about mathcounts or just math?[/quote]

I am talking about MathCounts.

I posted this in the "MathCounts" category.
Now they merged it with many other categories.

MathCounts gives no partial credits. No matter how much do you know a problem, if you make a stupid mistake.
You get nothing -- the same as someone who has no clue on how to solving it and didn't work on it at all.

I am shooting for going to National.
I cannot afford stupid mistakes. 1 or 2 of them will kill your whole year's effort.

For me, making silly mistakes really hurt in the state competition.  In the sprint round I got 13 out of thirty(OUCH!).  That made me 63rd in Indiana.  However, I recently made new strategies(thanks to looking over the 2009 State rounds) to help.  1: First, do all of the problems that you can finish in 35 minutes.  Be careful not to rush(so you don't throw away points) and take about 5-15 seconds per problem to check for reasonableness.  If you're not sure whether its right but want to keep going, put a tick mark to give it special thought in your last 5 minutes. On the hard problems, make a reasonable guess(you may be right).  Then, on your last 5 minutes, check the "tick marked problems" as well as finishing up any other problems you want to complete.  If you came up with a solution easily on a problem late in the Sprint Round, it is likely a trap, so be extra cautious.  Finally, READ THE PROBLEM CAREFULLY!  It seems obvious but I can't stress this enough.

Mewto, what I meant was you can RARELY simply find a solution without thinking through truely what the problem is asking.  However, many times near the end, the problems may try to trip you up through subtleties.  Take 2009 Sprint Round Problem 25, for instance.

For example, MATHCOUNTS gives you little time for some problems.... so you need to effectivdly check your work quickly. What I do is I create a lot of space for each problem that needs work so that it will be much more easier to check and get my work right the first time... :) That's what I do.. :arrow:

[quote="Mewto55555"][quote="maybach"] If you came up with a solution easily on a problem late in the Sprint Round, it is likely a trap, so be extra cautious.  [/quote]

I don't know what you mean by this, almost ALL non-arithmetic intense MATHCOUNTS problems have a really quick nice solution. See like #29 on 2008 state. It was basically doing 7^2-9.[/quote]

This technically could "apply" to anything. Take AIME I 2009, #10. It was "basically" doing 1^3+4^3+6^3+4^3+1^3.

Wat he means is that there is a hidden solution that is nice and easy under the words

[quote="maybach"] If you came up with a solution easily on a problem late in the Sprint Round, it is likely a trap, so be extra cautious.  [/quote]
That's not always true. Sometimes it just comes easily.

To take an EXTREME case, this years national sprint round's last page could be done under 5 minutes using a combination of assumption, basic trig, some basic arithmetic, and a little bit of thinking outside the box.

When I double check, I start from the end and go to the beginning, since I usually get the end ones wrong. Also, I won't forget the end problems when I double check them, since I did them recently.

[quote="PowerOfPi"][quote="maybach"]Take 2009 Sprint Round Problem 25, for instance.[/quote]
School, Chapter, State, or Nats?[/quote]
State.  Sorry.

[quote="MathWhiz2009"]Back on topic:

You should make your scratch paper neat so that you can go back and look at it while double checking, so that you don't have to re-do the problem over again. For me, I often catch my mistakes on the scratch paper without writing the steps again.[/quote]
I know a lot of people recommend that, but wouldnt it waste time? After all, you have to have blank pieces of paper when you start the test.

[quote="tennis123"]At the MathCounts competitions, at least in my area, they provide a lot of blank paper, so that wouldn't be a problem.[/quote]
Yeah, but you cant make the boxes before the time starts, right?

Now that i think about it, making the boxes would probably take around 30 seconds. Never mind then.

[quote="MathWhiz2009"]Back on topic:

You should make your scratch paper neat so that you can go back and look at it while double checking, so that you don't have to re-do the problem over again. For me, I often catch my mistakes on the scratch paper without writing the steps again.[/quote]

I never ever do this. For one thing, the effort required to do so takes too much time (imo). For another, when I see my work when checking, its harder to catch mistakes. Its better to just rememebr what I did, rework the problem, and compare my answers.

Well, what I do is first do the whole test and mark down some answers I'm not really sure of, but [b]still do them[/b]!!!! Then if I have time later I come back and go over these. Then, if I still have time I go through the easier ones.


Let $\triangle ABC$ have $\overline{AC}=5$, $\overline{AB}=13$, and $\angle C=90$
point $E$ is on the angle biscector of $\angle ABC$, inside $\triangle ABC$
point $F$ is on $\overline{AB}$ such that $\angle FEB$ is a right angle
let $M$ and $N$ be the midpoints of $\overline{AC}$ and $\overline{FE}$, respectively
if $\overline{MN}=3$, find $\overline{BE}$


For this problem I spent 2 hours without any concludent result. I've lost my head around the indices :mad: It doesn't look a hard problem I think...  :roll:

Let $A\in\mathcal M_{n}(C)$ and $^{t}A$ it's transpose. If the sum of elements of $A\cdot ^{t}A$ is $0$, then $\det(A)=0$.

let $X=(1,1,...,1)$ $XA^tA^tX=\sum (A^tA)_{ij}$
$A^tA$ is positive and if $\sum (A^tA)_{ij}=0$ it won't be definie.
so if $\sum (A^tA)_{ij}=0$ then $det(A)=det(A^tA)=0$


[quote="spanferkel"][quote="fjwxcsl"]We have found  an  interesting  proof  for  this inequality  ...[/quote]May you share it with us?  :)[/quote]

This is a very useful inequality, we will be in an paper

  to prove it , dear spanferkel!


Is there a list of the quotes from the box on the bottom of AoPS pages?

No, we do not have such a file listed on the site.

[quote="mysmartmouth"][quote="Valentin Vornicu"][quote="mysmartmouth"]How are the quotes generated?  It appears to be 100% random.[/quote]Exactly.[/quote]

But how does it work?  I would like to know.  Where are the quotes "plucked" from?[/quote]Google "how to get random quote on your web page" and you'll easily find the answer ;)


It seems to take forever to load an AoPS page these days...well not forever, but 2-3 minutes perhaps....and for a cable modem connection that's pretty ridiculous. It started just about 2 weeks ago...and it's never occured before.

Anyone happen to know the problem?

I don't know the solution, but I have the same problem. ;) I also have cable.

No idea - have you tried deleting your cookies (though I have no idea why this would be the problem)?

I tried- nothing happened... it still takes a minute to view aops webpages, and sometimes it doesn't load at all.

Have you tried changing your style to Mathlinks? I also have cable but mathlinks loads at <0.1 sec as usual...

Please let us know if it's a style issue - that will help us diagnose and fix the problem.

I have Cable, I use AOPS style, and it loads just fine. It might help if you could post your ISPs so we can isolate the problem, MithsApprentice and hatsup123.

This thing happened to me a couple of weeks ago, the site was running very very slow from my computer, however Richard was browsing perfectly, and so were other users on the site. I accessed the site with my mobile (which has obviously a different ISP :P) and it worked perfectly.

There is a big chance that this is actually an ISP / Local setting problem, rather than a general server side one.

Isn't it a bit wide-spread to be a local settings problem? I haven't ever heard of an alike problem on any other site... at least not as consistent and wide-spread...

2 people out of 4500 (registred) users complain about this. I would not call this wide-spread.

hm...using mathlinks style does help slightly...not a lot though...

[quote="Peter VDD"]comon Vali, don't do like it's the first time people tell about having this problem ;-)[/quote]I had this problem for a couple of days a couple of weeks ago. It was caused by my ISP (some other folks from Astral in Bucharest could not access the site, but others from RDS could do it easily). There is a 90% chance that this is an ISP problem.

hmm...well i use roadrunner (houston)...

We've had quite a few problems with roadrunner.  Every once in a while, for example, they randomly start denying all emails we send to roadrunner accounts.  We complain, then they say 'oops sorry about that'.

This may be the best organized and most informative site concerning mathematical competitions (much better than [url]http://192.154.43.167/goebel/statecon/State.htm[/url], the NC State Competition site) but it is still bloated and slow.  The flash proofs, logos, and quotes are nice, but should be optional.  What are the chances of AoPS loading on a multiuser system based on misadministered 200 Mhz Win-95 boxes like the ones at my high school?  Stick to HTML, except for the most important images for the forum, and allow registered users to turn on the gadgets, please.

I have cable and it takes a long time to load the page when I try to log in......

a lo-fi version of the site is on the tasks list ... It will be done at some point in the future.

[i]Posted from a Sony Ericsson K700i[/i]


geometry
angle bisector
geometry proposed

If $ AD$ is the altitude, $ BE$ the angle bisector, and $ CF$ the median of a triangle $ ABC$, prove that $ AD,BE,$ and $ CF$ are concurrent if and only if:

$ a^2(a\minus{}c)\equal{}(b^2\minus{}c^2)(a\plus{}c),$

where $ a,b,c$ are the lengths of the sides $ BC,CA,AB$, respectively.

hello, i think your theorem is wrong,
setting
$ c\equal{}\sqrt{a^2\plus{}b^2}$
we get
$ a^2(a\minus{}\sqrt{a^2\plus{}b^2})\minus{}(b^2\minus{}a^2\minus{}b^2))(a\plus{}\sqrt{a^2\plus{}b^2})\equal{}$
$ a^3\minus{}a^2\sqrt{a^2\plus{}b^2}\minus{}(\minus{}a^2)(a\plus{}\sqrt{a^2\plus{}b^2})\equal{}$
$ a^3\minus{}a^2\sqrt{a^2\plus{}b^2}\plus{}a^3\plus{}a^2\sqrt{a^2\plus{}b^2}\equal{}2a^3\equal{}0$?
Sonnhard.

[quote]we get ... $ \equal{}2a^3\equal{}0$?
Sonnhard.[/quote]

... which only shows that a right-angled triangle indeed is never suitable (no concurrency there)!

[quote="mavropnevma"][quote]we get ... $ \equal{} 2a^3 \equal{} 0$?
Sonnhard.[/quote]

... which only shows that a right-angled triangle indeed is never suitable (no concurrency there)![/quote]

^^^^

In my opinion, is corectly thus :

[quote]Let $ AD$ - altitude, $ BE$ - angle bisector and $ CF$ - median of $ \triangle ABC$ .

 Prove that $ AD\cap BE\cap CF\ne\emptyset\ \Longleftrightarrow\ a^2(a\mp c) \equal{} (b^2 \minus{} c^2)(a \pm c)$ .[/quote]

i.e. pairs $ ( \minus{} , \plus{} )$ or $ ( \plus{} , \minus{} )$ .


Functional Analysis
advanced fields
advanced fields unsolved

First some definitions:
1. A tree on a set $X$ is a partially ordered subset $T \subseteq \bigcup_{n=1}^{\infty} X^{n}$ such that $(x_{1},\ldots,x_{m}) \leq (y_{1},\ldots,y_{n})$ implies $m \leq n$ and $x_{i}=y_{i}$ for $i=1,\ldots,m$.
2. The terminal nodes of a tree are the maximal elements.
3. An $\ell_{1}$-$K$ (for $K \geq 1$) tree on a Banach space $X$ is a tree on $X$ such that $T \subseteq \bigcup_{n=1}^{\infty} S(X)^{n}$ and $\frac{1}{K} \sum_{1}^{m} |a_{i}| \leq ||\sum_{1}^{m} a_{i}x_{i}||$ for all $(x_{i})_{1}^{m} \in T$ and $(a_{i})_{1}^{m} \subset \mathbb{R}$.
4. Let $T$ be a tree on $S(X)$. $S$ is a block tree of $T$ if $S$ is a tree on $S(X)$ such that there exists a subtree $T' \subset T$ and an isomorphism (bijective order isomorphism) $f: T' \rightarrow S$ satisfying $f((x_{i})_{1}^{m}) = (y_{i})_{1}^{n}$ is a normalized block basis of $(x_{i})_{1}^{m}$ for each $(x_{i})_{1}^{m} \in T'$ and if $(x_{i})_{1}^{k} \in T'$ for some $k < m$, then there exists $l < n$ such that $(y_{i})_{1}^{l} = f((x_{i})_{1}^{k})$ and $(y_{i})_{l+1}^{n}$ is a normalized block basis of $(x_{i})_{k+1}^{m}$.

Now for the question:
Let $T$ be an $\ell_{1}$-$K$ tree for some $K \geq 1$. If $(x_{i})_{1}^{n} \in T$, let $F = F((x_{i})_{1}^{n}) \subset S(X^{*})$ be a finite set which 1-norms a (1/2)-net in $S([x_{i}]_{1}^{n})$. There exists a block tree $T_{(x_{i})_{1}^{n},F}$ of $T$ contained in $X_{F}= \{ x \in X: x^{*}(x)=0 \: \forall x^{*} \in F \}$ (by some result). Let $(a_{k})$ be the collection of terminal nodes of $T$ and if $a_{k}=(x_{i}^{k})_{i=1}^{n}$, $F_{k}=F(a_{k})$, let $T(k)=\{ a_{k} \cup x: x \in T_{a_{k},F_{k}} \}$. Then the claim is that $T' = T \cup \bigcup_{k} T(k)$ is an $\ell_{1}$-$6K$ tree.


The letters of the word `SIXTEEN' are randomly arranged.  What is the probability that the two E's are not next to each other?

Although I got this problem correct on the second attempt by considering the complement ($ 1\minus{}2/7\equal{}5/7$), I'm still puzzled why my first method didn't work (without using complement):

First place a gap between the two E's. We have to place 5 letters into 3 boxes (the leftmost box, middle box, and the rightmost box, with the gap being the middle box). The total number of ways is $ 3^5$. However, we have to elminate the one's where the middle box is empty, and this happens in $ 2^5$ ways. So the total number of desired ways is $ 3^5\minus{}2^5$. This is out of $ 7!/2$. So the probability is $ 2(3^5\minus{}2^5)/7!\equal{}211/2520$, but this does not agree with the answer I obtained using the complement approach...  :huh:

The second part you are doing without accounting for permutations, but the weighting is not equal.

Sorry I don't understand what you meant... could you please elaborate?

SIXTNEE and SIXTEEN are both only counted once, but the first should be counted $ 5!\equal{}120$ times and the second only $ 4!\equal{}24$ (or any equivalent ratio of the two, which is what i meant by weighting).

I see... thanks (haha I always make these type of dumb mistakes... :()
Is there a way that does not use the complement approach (and could my first one be modified?)?

Also, what is the proper way of counting what i was trying to count in my first incorrect method (i.e., place 5 distinct objects into 3 distinct urns with the order of the objects in each urn matters)?

There's not any efficient method that I know of.

... but is there any possible way of solving it (not very efficient is ok)...? of course without enumerating all the possibilities...

You could find all the ways that the Es can be at least one apart without considering the other letters, but really, there's no motivation at all to find a constructive solution instead of complementary counting.

mathwizarddude's solution is the best.  (quick one about complements)  period.

The complementary probability method is more helpful.
The probability that two E's are next to each other is $2/7$. Subtract this from $1$ in order to get $5/7$, which is our final answer.


any free book for inequalities  on the net...

http://www.olimpiadematematika.net , and then you search in "gudang arsip" . click forum "Buku - Buku Yang Sangat Berguna" and download directly.  :D

are you talking about mildorf, hoojo lee, etc.'s notes? or is there something else?


We should have written "1. CH3MgBr and 2. H3O+" (why?). Mechanism:

1) Attack of Grignard reagent's carbon on the carbonyl carbon;

2) Recovery of the C=O bond by eliminating the alcoxide group;

3) Attack of another Grignard reagent molecule on the carbonyl carbon;

4) Acid workup yields the neutral diol.

Final product: $ HO\minus{}(CH_2)_3\minus{}C(OH)(CH_3)_2$.

That is because, I believe, if we think back to one of the restrictions of Grignard Reagent:
[hide] We cannot use the reaction is the presence of acidic protons because the reagent is itself a base which will cause acid base reaction.

$ \ce{CH3MgBr} \plus{} \ce{H2O} \rightarrow \ce{CH4} \plus{} \ce{Mg(OH)Br}$

Which is why we produce Grignard Reagent in situ with ether solvent rather than aqueous medium.

We write the process in two steps because experimentally we do the acidic work-up afterwards to prevent it from reacting with the reagent[/hide]

Yes, that's right. What are other restrictions on the use of Grignard reagents?

we can't restrict the reaction to the aldehydic or ketonic stage
but the alcohol is obtained . I think this is another disadvantage of Grignard's


In rhombus AMCD, angle A = 60. A circle passes through vertices A, B, and D and intersects AC at P. If PC = 4, find AC.

[hide]
Answer:12

we know that by PoP, $BC^2 = PC \cdot AC$. Now, let's denote $AC=x$. Then $BC^2=4x$. If we draw a line $BD$, $\triangle ABD$ becomes an equalateral triangle. so,  $BC=AB=\frac{\sqrt{3}}{{3}}AC= \frac{\sqrt{3}x}{3}$. then $\frac{x^2}{3}=4x$, $x=12$.
[/hide]

check my work please.

I get the same answer as you through a slightly different method:
[hide]
answer: 12
I set $AC \cdot PC = BC^2$
then $BC = x$. Then $AC = x\sqrt{3}$
so $x^2 = 4x\sqrt{3}$
making $x=4\sqrt{3}$
substitute back in for AC $\sqrt{3} \cdot 4 \cdot \sqrt{3} = 12$
used law of cosines.
[/hide]


Given three circles in the plane such that no circle is inside any other, no two of the circles intersect, and the radii of the circles are all of different lengths.  We join the center of each circle to the point formed at the intersection of the common internal tangents of the other two.  Prove that these three lines are concurrent.

[hide]We just use the reciprocal of Ceva's thm in the triangle formed by the centers of the circles.[/hide]

Wait...isn't a common internal tangent just a line that is tangent to two circles? So if the circles don't intersect at a single point, how can they have a common internal tangent?

Nope. The common internal tangents are the two lines that are tangent to both circles and intersect between the two circles.

Hope this helps.

As grobber said before, the most straightforward approach seems to be to use the inverse of Ceva's Theorem (I've never used the inverse before though, so I'm not sure it exists).

Anyway, the diagram is color-coordinated for your convenience. Circles are in black, intersecting tangents are in red, etc.

Also, a question. Does the property still hold when you align the three circles vertically like a disjointed snowman?

Osiris:

Ah, I see. Thanks. So what would be the common external tangents of those two circles? Would they be the tangents to the two circles that do not intersect (as opposed to the two common external tangents, which would intersect in your diagram)?

The external common tangents intersect unless the 2 circles have equal radii. In that case they're parallel.

Well, they don't intersect *between* the circles anyway.


MIT
college
Putnam
topology
real analysis

My family is far from poor; but they're also far from being able to afford a $\$$38k-45k/year education. Assuming lack of a stellar financial aid package, how do people who aren't going to receive a great deal of help from their parents finance their undergraduate educations?

Is your family committed to spending all of what the college calculates as the expected family contribution (EFC), or not? If not, you need to look at colleges with unusually low list prices (so that you can work your way through college, which is what I happened to do) or colleges with "merit scholarships" (reductions in list price for desired students) that don't leave a gap between what they offer and what you can afford.

The first step is to fill out financial aid applications to the schools you are considering, and see what is on offer. Not all colleges offer the same aid package (that is, reduction of list price for price-sensitive customers), but the only way to know what's on offer is to apply.

Good luck. Be sure to expose your parents to the idea that many parents think it is worthwhile to THEM to pay their calculated EFC, and that helps you have a broader range of possibilities.

Most good graduate schools offer stipends to the majority (if not all) of the students they admit.  This may be contingent upon performing a certain amount of teaching or something similar.  There are also external fellowships available.

Yes, usually being a graduate student is mostly like a job, so you get paid to be there. Your pay can come from a portable fellowship (like the National Science Foundation fellowships in the United States) or from the department funds available at the graduate school to which you apply. It's not usually a good idea to accept an offer of admission to a graduate school unless that offer comes with an attached offer of employment as a research assistant or teaching assistant.

As for the basic question about whether or not it is possible to get into a good graduate program after going to a so-so undergraduate program, yes, that is possible, but it is difficult. And that's especially difficult if the student has to work to pay for tuition while going to the undergraduate college. Not impossible, but difficult.

It's really important to compare cost of undergraduate programs by looking at actual financial aid offers. Sometimes the "big schools" with high list prices also offer the best need-based financial aid over a broad range of family incomes. If your state university and [insert name of famous college here] had an equal out-of-pocket cost, after financial aid is considered, it would likely be worthwhile to attend [famous college]. And the only way to know what the financial aid offer is from each college is to apply for financial aid from all the colleges. If your parents are concerned about the cost of your undergraduate education, chances are that you are eligible for financial aid from such colleges, so it is worthwhile to apply.

[quote="nigma"]Thanks for the response tokenadult. Do you know how plausible it would be to go to a lesser-quality, but cheap in-state school, and then try to get into a better university for graduate studies? I don't usually hear about people paying for grad school (w/ exceptions), and I assume that's because it's much more typical to work something out where you promise x years of work for someone and in return they pay for your school.[/quote]

There is a decreasing scale of funding at grad school that goes from engineering down to philosophy, and this determines the number of positions and the difficulty of the competition.   For the cold hard subjects such as math, science and engineering there is not only more money, there are more ways to give objective documentation of your qualifications and the faculty at MIT will easily overcome their contempt for Bucktooth State University if you ace the Putnam competition or do serious research.  They will be less impressed by courses with fancy titles that don't mean anything at standardless colleges, such as Topology, Real Analysis, and Quantum Mechanics, if those are even offered.  It doesn't matter all that much where (or whether) you go if your intellectual maturation and specialization can be demonstrated to be at the admissible level.   References can be a problem, or rather, you will need to take many more actions in order to either generate references that have value, or to circumvent the need for a lot of references.


Find all positive integers $ n$ such that acute-angled $ \triangle ABC$ with $ \angle BAC<\frac{\pi}{4}$ could be divided into $ n$ quadrilateral. Every quadrilateral is inscribed in circle and radiuses of circles are in geometric progression.
[hide] be carefull !  :lol: [/hide]


find the minimum of $\lambda\in\mathbb{R}$ such that forall $x,y,z\in\mathbb{R}$ there is
$\lambda(x^{2}-x+1)(y^{2}-y+1)(z^{2}-z+1)\geqslant x^{2}y^{2}z^{2}-xyz+1$
I just know
$\frac{12(7\sqrt[4]{108}-13\sqrt{3}+12\sqrt[4]{12}-23)}{(\sqrt[4]{108}-\sqrt{3}-3)^{3}}\leqslant\lambda_{\min}\leqslant9/4$
but don't know the exact value of $\lambda_{\min}$.
If we set that not all of $x,y,z$ are positive, we'll get $\lambda=16/9$(take $x=y=1/2,\ z=0$)

[quote="fjwxcsl"]lmin=1+2/3*3^(1/2).[/quote]
Could you please show me why?


I found this problem in Bona's combinatorics book. I would like to know whether my solution to it is right.

$ 2n$ balls are put into $ n$ boxes so that no box contains more than $ n$ balls and no box is empty. Prove that we can choose a set of boxes so that sum of the balls in the boxes of that set equals to $ n$

elegant official solution
[hide]
Let $ a_1,a_2,\dots,a_{100}$ be the number of balls in boxes numbered $ b_1,b_2,\dots,b_{100}$ respectively arranged in such a way that $ a_1\ne a_2$ ( Impossibility of such an arrangement implies that all boxes have equal number of balls ($ \equal{}2$) and the problem statement holds in that case ). Consider the sums,
\[ a_1,a_1\plus{}a_2,\dots,a_1\plus{}\dots \plus{} a_{100}\]
In case some two leave same remainder when divided by $ 100$, we are done. Else consider the sums,
\[ a_1,a_2,a_1\plus{}a_2,\dots,a_1\plus{}\dots \plus{} a_{100}\]
and we are done as $ a_1$ and $ a_2$ cannot leave same remainder.[/hide]

my idea
[hide]When every box contais exactly $ 2$ balls, the conclusion follows. If there is a box that contains $ 2<k\le 99$ balls, then there exist $ k\minus{}2$ boxes which contain exactly one ball each. Hence if there exists a box with more than $ 51$ balls, the solution is straight forward. In the other case, arrange boxes (b's) in non increasing order,
\[ b_i\ge b_j\ge \dots\]
such that $ 3\le b_i\le 51$. we shall prove that every such a distribution can be obtained from $ 99,1,1,\dots,1$ distribution. Just partition it appropriately ! This can be done, back and forth.[/hide]

I am sorry, there is a question about the problem: consider the case $ n$ is odd and we put $ 2$ balls in every box. Now whatever boxes you take, the sum of the number of balls in them is an even number, which can not be $ n$.

Sorry for the mistake in my previous post.

The problem is : $ 4n$ balls are put into $ 2n$ boxes so that no box contains more than $ 2n$ balls and no box is empty. Prove that we can choose a set of boxes so that sum of the balls in the boxes of that set equals to $ 2n$.

The original problem in the book was asked with $ n\equal{}50$

A other way to see the problem

[img]http://img526.imageshack.us/img526/4215/tarte.jpg[/img]

The disc represent the 2n ball ( one sector= one ball ) and boxes are between two rayons. If no diameter is draw , you only have to change two neighbours boxes wich don't have the same size to finish .

Domi

PS : sorry for my poor english  :blush:


For $ a,b,c > 0$ and $ abc\equal{}1$ prove that:

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

$ (2)$ $ \frac {ab} {b \plus{} c} \plus{} \frac {bc} {c \plus{} a} \plus{} \frac {ca} {a \plus{} b} \ge \frac {9} {2(a^2 \plus{} b^2 \plus{} c^2)}$

$ (3)$ $ \frac {a^2} {b \plus{} c} \plus{} \frac {b^2} {a \plus{} c} \plus{} \frac {c^2} {a \plus{} b} \ge \frac {ab} {b \plus{} c} \plus{} \frac {bc} {c \plus{} a} \plus{} \frac {ca} {a \plus{} b}$

[quote="moldovan"]For $ a,b,c > 0$ prove that:

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

$ (2)$ $ \frac {ab} {b \plus{} c} \plus{} \frac {bc} {c \plus{} a} \plus{} \frac {ca} {a \plus{} b} \ge \frac {9} {2(a^2 \plus{} b^2 \plus{} c^2)}$

$ (3)$ $ \frac {a^2} {b \plus{} c} \plus{} \frac {b^2} {a \plus{} c} \plus{} \frac {c^2} {a \plus{} b} \ge \frac {ab} {b \plus{} c} \plus{} \frac {bc} {c \plus{} a} \plus{} \frac {ca} {a \plus{} b}$[/quote]
1) $ (\frac {a^2} {b \plus{} c} \plus{} \frac {b^2} {a \plus{} c} \plus{} \frac {c^2} {a \plus{} b})((b\plus{}c)\plus{}(a\plus{}c)\plus{}(a\plus{}b))\ge(a\plus{}b\plus{}c)^2$ by CS so $ \frac {a^2} {b \plus{} c} \plus{} \frac {b^2} {a \plus{} c} \plus{} \frac {c^2} {a \plus{} b} \ge \frac {a \plus{} b \plus{} c} {2}$
2) Try a=b=c=1/2 :wink:
3) After using 1) we must prove  $ (a\plus{}b)\plus{}(b\plus{}c)\plus{}(c\plus{}a)\ge 4(\frac {ab} {b \plus{} c} \plus{} \frac {bc} {c \plus{} a} \plus{} \frac {ca} {a \plus{} b})$ but by Rearrangement ineq $ (a\plus{}b)\plus{}(b\plus{}c)\plus{}(c\plus{}a)\ge \frac{(a\plus{}b)^2}{b\plus{}c}\plus{}\frac{(b\plus{}c)^2}{c\plus{}a}\plus{}\frac{(c\plus{}a)^2}{a\plus{}b}\ge 4(\frac {ab} {b \plus{} c} \plus{} \frac {bc} {c \plus{} a} \plus{} \frac {ca} {a \plus{} b})$ last because $ (a\plus{}b)^2\ge 4ab$

Thanks for the observation. For $ (2)$ put the condition $ abc\equal{}1$. I will correct the terms of the problem.

dima ukraine, I think $ \sum\frac {(a \plus{} b)^2}{b \plus{} c}\leq\sum(a \plus{} b)$ is wrong because ineq's conversions position
Here is my solution:
$ \sum\frac{a^2}{b\plus{}c}\geq\sum\frac{ab}{b\plus{}c}\Leftrightarrow\sum\frac{a}{b\plus{}c}(a\minus{}b)\geq 0$
Case $ a\geq b\geq c:$
with$ a \plus{} c\geq 2b: \Rightarrow a \minus{} b\geq b \minus{} c\geq c \minus{} a\Rightarrow\sum\frac {a}{b \plus{} c}(a \minus{} b)\geq\frac {1}{3}\sum\frac {a}{b \plus{} c}\sum(a \minus{} b) \equal{} 0$
with$ a \plus{} c\leq 2b: \Rightarrow b \minus{} c\geq a \minus{} b\geq c \minus{} a$ and $ 2(a\minus{}b)\geq(b\minus{}c)\plus{}(c\minus{}a)\equal{}b\minus{}a,Right$
$ \Rightarrow\sum\frac {a}{b \plus{} c}(a \minus{} b)\geq\sum\frac {1}{3}\sum\frac {a}{b \plus{} c}\sum(a \minus{} b) \equal{} 0$
the similar for case$ a\geq c\geq b$

[quote="moldovan"]For $ a,b,c > 0$ and $ abc \equal{} 1$ prove that:

$ (3)$ $ \frac {a^2} {b \plus{} c} \plus{} \frac {b^2} {a \plus{} c} \plus{} \frac {c^2} {a \plus{} b} \ge \frac {ab} {b \plus{} c} \plus{} \frac {bc} {c \plus{} a} \plus{} \frac {ca} {a \plus{} b}$[/quote]

It's trues with all $ a, b, c \geq\ 0.$
After expand,it'll become:
$ \sum\ a^4\plus{}\sum\ ab(a^2\plus{}b^2) \geq\ \sum\ a^3b\plus{}\sum\ a^2b^2\plus{}\sum\ abc(a\plus{}b\plus{}c)$
(Obviously trues )

For 2
$ \sum \frac{ab}{b\plus{}c} \equal{}\sum \frac{1}{bc\plus{}c^2} \ge \frac{9}{a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ac} \ge \frac{9}{2(a^2\plus{}b^2\plus{}c^2)}$

[quote="moldovan"]For $ a,b,c > 0$ prove that:

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

[/quote]
The following inequality is true too.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that \[ \frac {a^2} {b \plus{} c} \plus{} \frac {b^2} {a \plus{} c} \plus{} \frac {c^2} {a \plus{} b} \ge \frac {3} {2}\cdot\sqrt[5]{\frac{a^5\plus{}b^5\plus{}c^5}{3}}\]

[hide="I liked them, for the entire 30 seconds they lasted :)"]
The first two are just Cauchy Engel form
the last one is fun:
\[ \sum\frac {a^2}{b \plus{} c} \minus{} \sum\frac {ab}{b \plus{} c}\geq0\iff
\]

\[ \sum \frac {a}{b \plus{} c}(a \minus{} b) \geq 0\iff
\]

\[ (a \plus{} b \plus{} c)\sum\frac {a \minus{} b}{b \plus{} c}\geq 0 \iff
\]

\[ \sum \frac {a \plus{} c}{b \plus{} c}\geq 3
\]
Which is evidently true by Am-Gm :)

[/hide]


calculus
probability
real analysis
integration
linear algebra
number theory
complex analysis

what prerequisites do you need to do real analysis?

depends what level of real analysis. Many colleges offer an advanced calculus course designed to be taken following the calculus/differential equation sequence. its somewhat rigorous and teaches how to communicate with proofs. if you refer to graduate real analysis you'll need a solid basis in writing proofs. such prerequisites for classes like this is an undergraduate analysis course.

Like what Poincare said, if it's an undergraduate real analysis course, the standard calculus sequence and maybe an Intro to Proof/Bridge to Higher Math course would be required.  If it's graduate real analysis, a course in the undergraduate form of analysis is usually the requirement :].

At my institution (CSULB), I've produced an advising document which classifies all of our courses by level and has arrows showing the important prerequisites. This chart has six levels: precalculus, the calculus sequence, intermediate level courses, early advanced courses, later advanced courses, and graduate level courses (we have an M.S. program but not a Ph.D. program.)

On that list, I've got the first semester of our two semester analysis sequence as an "early advanced course" and the second semester as a "later advanced course." The "intermediate level course" band - the things to be taken immediately after the calculus sequence - has five courses on it. The "bridge" or "discrete structures/how to prove things" course, the intro linear algebra course, the differential equations course, the (calculus prerequisite) probability course, and (for teaching credential candidates only) the early history of mathematics course. I've got the analysis course on the next band, the early advanced courses with a prerequisite of either the "bridge" course or the linear algebra course. Examples of other courses on that early advanced level include the second linear algebra course, the number theory course (which for us comes before abstract algebra), the numerical analysis course, and the mathematical statistics course.

This chart was made for typical students and our typical students have very little experience with proofs and no exposure to "our" kind of problem-solving. There are a few students who can jump over prerequisites, and I can usually recognize them.

We have several graduate level analysis courses, including both real analysis (meaning Lebesgue measure and integration) and complex analysis; they have both semesters of that undergraduate analysis sequence as prerequisites.


how many ways can 12 different coins, 3 quarters, 4 dimes and 5 nickels, be put on a book shelf if:
2 particular coins must be in the middle;
two particular coins cannot be beside each other;
if each type of coin must be beside each other;

Treat the coins as if they are distinguishable.

[hide="solution"]The middle consists of the 6th and 7th coins. If the nickels are to be put on the outside, the middle coins are both quarters or both dimes. If the nickels are to be put in the inside, the middle coins are two nickels.

If two particular coins are to not be beside each other, they must be of different types, and those types must be on the outside, giving the third type of coin the middle.

Since there are restrictions, we must only consider one case: a particular coin is in the middle. That gives the other two coin types the second restriction. Now, since a particular coin type must be in the middle, the other two coin types are interchangeable to the two sides of the bookshelf. Therefore, there are only [b]2[/b] ways.[/hide]


See http://www.mathlinks.ro/Forum/viewtopic.php?t=19244 (mainly post #22 remark [b]2.[/b]). Also, http://www.kalva.demon.co.uk/usa/usoln/usol775.html .

  darij


logarithms
function
number theory
relatively prime
complex analysis

Solve for all complex $ x$ such that

$ x^{2/4} \equal{} 1$.

How many distinct solutions should there be?

How about $ x^{1/2} \equal{} 1$?

[quote="ZetaX"]Giving your definitions would be helpful ;)[/quote]

What definition?  The exponenet follows the usual definition of exponent.

Which definition do you consider "usual"?  In particular, how are you defining the square root of a complex number?  There are two of them.

^

How do you mean? Taking square root. Picking branch of the logarithm and setting $ z^{1/2}\equal{}exp(1/2 log z)$. What's the other one?

It's not clear which branch to take, and additionally $ x^{\frac 24}$ seems to be different from $ x^{\frac 12}$ in his definition.

I guess by definition (of [i]fractional exponent[/i]) of $ x^{m/n}$ for relatively prime integers $ m$ and $ n$, you mean to whether consider $ x^{m/n}$ as $ (x^m)^{1/n}$ or $ (x^{1/n})^m$.  Well, I would take both cases.  What would the difference be?

I'd like to try again please (I deleted an earlier post which I felt was not accurate):  I think there is only one solution.  Need to distinguish between the value of a multifunction and the multifunction at a value.  The expression $ M(z)=z_0$ is a particular value of the multifunction at the value $ z$.  However, $ M(z_0)$ is multivalued.  In your case $ M(z)=z^{1/2}=z^{2/4}=r^{1/2}e^{i/2(\Theta+2k\pi)}=1$ and that occurs when $ z=1$ and $ \Theta=0$.  Had I made the conversion: $ z=1^2$, no extraneous  roots would have entered into the equation.  That's not the case if the inverse of the exponent is non-integer.  Take for example, $ z^{4/5}=1$.
That's:
\begin{align*}
z^{4/5}&=e^{4/5\log(z)}\\
&=e^{4/5}\big[\ln(r)+i(\Theta+2k\pi)\big] \\
&=e^{4i/5(0+2k\pi)}
\end{align*}
and $ z^{4/5}=1$ occurs when $ k=0$.  That is, $ z^{4/5}=1$ has one solution.  However, when I make the algebraic manipulation: $ z=1^{5/4}$, I'm now equating the solution to a multivalued function $ f(z)=z^{5/4}$ at the value $ z=1$ and that function has $ 4$ values, one of which is the solution to the problem $ z^{4/5}=1$.

That's how I see it anyway and like the other one, that's what I'm turning in and hoping for the best.


calculus
integration
Diophantine Equations

Show that the equation \[x^{3}+y^{3}+z^{3}+t^{3}=1999\] has infinitely many integral solutions.

to answer to this kind of qestion we have to find a general formua and that is one of as formulas
$(10+60n^{3})^{3}+(10-60n^{3})^{3}+(-60n^{2})^{3}+(-1)^{3}=1999$. :wink:
Driss Dahbani.

Well the idea is that variable terms have to cancel out in the expansion, and at the same time we must get an infinite sequence, so we try the easiest way:
$ (a\plus{}bk)^3\plus{}(a\minus{}bk)^3\equal{} 2a^3 \plus{} 6ab^2k^2$, the next term should cancel that $ k^2$ out, so put $ k\equal{}n^3$ to get $ (a\plus{}bn^3)^3\plus{}(a\minus{}bn^3)^3\plus{}(\minus{}bn^2)^3\equal{}2a^3\plus{}(6a\minus{}b)b^2n^6$.
If we put $ b\equal{}6a$, we're nearly there, since we have a constant. Now, $ 2a^3\plus{}c^3\equal{}1999$ should have a solution to be done. And indeed: $ a\equal{}10,c\equal{}\minus{}1$ is such a solution.

Hope that explains the reasoning. :)

[quote="mbucko]Nevertheless, I would like to point out one thing, i.e. generalization, e.g. $ A^n \plus{} B^n \plus{} C^n \plus{} D^n \equal{} 1999$. Haven't gone through this either[quote]

Be careful, $ n$ has to be even otherwise all the terms are positive and limited by $ 1999$ so finite.

In the trivial case $ n\equal{}1$ with two term is enough $ (1999\plus{}n)^1\plus{}(\minus{}n)^1\equal{}1999$
so probably $ a_1^{2k\minus{}1}\plus{}\cdots\plus{}a_{2k}^{2k\minus{}1}\equal{}m$ has infinite solutions.


What percent of the integers from 1 to 100, inclusive, leave
remainder 1 when divided by 5?

Numbers leave mod 0, 1, 2,3, or 4 when divided by 5. Since each has the same probability of occuring, it is 1/5 or 20%.

mod 1 occurs in numbers ending with $1$ or $6$, which occurs 2 times out of every set of 10, so $20\%$


geometry
perimeter
combinatorial geometry
geometry unsolved

Let $ABC$ and $A'B'C'$ be triangles such that $AB > A'B', BC>B'C', CA>C'A'$. Consider an arbitrary point $M$ in the interior of $ABC$. Prove that there exists a point $M'$ in the interior of $A'B'C'$ such that $A'M' < AM, B'M' < BM, C'M' < CM$.

Well, actually there exist more than just one point $M'$. The locus of such points is an open convex subset of $A'B'C'$, whose perimeter is made of segments and arcs of circles.

Ok, I'll give an intuitive proof: let's draw three circles, with centers in $A$, $B$, $C$ and radii $AM$, $BM$ and $CM$ respectively. Obviously, they share just a common point, that is $M$.
Now, draw three circles with the same radii as before, but with centers in $A'$, $B'$, $C'$ respectively, and see what happens. The distance between any two centers is lower than before, so the intersection of the three circles and the triangle is more than one point, and is the locus described above.

I think that a more formal proof can be given by considering the triangle as the convex hull of its vertices. So, for example you can express any point inside $ABC$ as $\vec M=x\vec A + y\vec B + z\vec C$, where $x+y+z=1$. Then, you may try to find your proof in a purely algebrical way, for example trying to prove that $\vec{M'}=x\vec{A'} + y\vec{B'} + z\vec {C'}$ is one of the points you're looking for (just guessing, I'm not sure it works!)... But I'm already satisfied with my first proof, so why bothering with this one?


The AIME statistics show that there is only one 15 on both tests, scored by a male 12th grader on AIME I.

who is it? :)

They send it out to USAMO qualifiers for free?  If so, AWESOME.  But does it have the names, too?  Are they even allowed to disclose people's names?

They do disclose names, or at least last name, first initial and school.

The person you're looking for doesn't live in California. The highest AIME score in California is a 13 (shared by 3 people.)

[quote="Kent Merryfield"]They do disclose names, or at least last name, first initial and school.[/quote]

If all you're looking for is the person who made the 15, this should definitely be enough information. People don't just pop out of the ground and make 15's. You'll probably be able to trace the identity to USAMO/MOP qualifier from previous years.

No 15's from Washington, either. WE had couple of 14s, but thats it.

well, i suck at math, i didn't qualify for USAMO ;)

i wanna say Thomas Mildorf from my school (TJ ;)), just to be loyal, but there are many other talented people in this world.

It was I who scored 15[size=42]/3[/size] on the AIME.

Yeah I did really bad.

[quote="Derek"]It was I who scored 15/[size=200]3[/size] on the AIME.

Yeah I did really bad.[/quote]

 :D

no 15s in FL either
there was a 13 and two 12s

I've been told (unofficial but reliable source as usual) that the 15 on AIME was scored by Natee Pitiwan, who was also a winner on the USAMO. But, the kicker is, he's gone to IMO for Thailand before and is 20 years old?

how is a 20yr old eligible for the USAMO???

[quote="darktreb"]You'll find out in the little summary of results booklet they send out to USAMO qualifiers every summer.[/quote]

does that booklet also have the USAMO results???

You can't do USAMO if you're 20 can you? Why do I seem to remember the max age being 19.5?


If I were to take two X-chromosomes from two of my sperms, and put them in an egg, would I end up with a female embryo? :huh: Is there a woman lurking inside my DNA?  :o

Remember each sperm has only one copy of the sex chromosome, either X or Y. To answer your question, if you fused two sperm that each had the X chromosome and tried to make a woman, I do not think it would result in the intended outcome. I'm trying not to laugh right now; what an obscure question! lol

[quote="adidasty"]if you fused two sperm that each had the X chromosome and tried to make a woman, I do not think it would result in the intended outcome[/quote]

Why not? Sure, she'd probably have a lot of recessive traits, but why couldn't that work in principle?

Recessive traits would have nothing to do with it. Each gamete receives one allele for each gene, be it dominant or recessive. The problem with your question is that the fusion of sperm cells is particularly difficult and I don't think it's been done. The egg is much easier to work with. Also, sperm are not arranged in the same way as eggs. The cell wouldn't have the correct proportions of mitochondria and other organelles. Theoretically, it's funny to entertain the idea, but pragmatically the idea is off.

Ok, you could physically fuse them, but nothing would happen on a chemical level. The anatomical shape and physiology of the egg and the sperm are designed so that one sperm and one egg is a gamete... that's just how we work.

If you had a host egg cell so you had the microcondra and the rest of the cell (including neculus) structure, you should be able to use two copies of your X to make a woman, sorta clone.

Note that you need a woman to provide the egg.  A woman couldn't make a man without a man providing the X, so we need both genders, even on the theoritical sub-cellular level to reproduce.

None of this has been done.


There is a 15 x 16 grid of lights, and each light is on or off. You may choose a 2x2 square of lights from the grid, and if three of these are on, turn the fourth light on as well. Initially 211 lights are off. Prove that you cannot turn the entire grid on.

There is a really nice 2-line solution for this one.

[quote="lingomaniac88"][hide]You increase the number of lights on by 4, 2, 0, -2, -4.  So the number of lights that are on is always odd, so it can never eventually be 0.[/hide][/quote]

um, where did you come with this?

(Note that 211 is the minimum, because with only 210 lights off, you can start with the boundary lit and eventually, you can light them all.)

[hide="proof"]
$(15-1)(16-1)=210$ copies of the $2\times 2$ square appears in the grid. Since you can turn on at most one light per square, you can turn on at most 210 lights!
[/hide]


function
vector
integration
calculus
geometry
3D geometry
sphere

All vector fields of the form $ \textbf{F} \equal{} \bigtriangledown g$ satisfy the equation curl F = 0 and that all vector fields of the form F = curl G satisfy the equation div F = 0 (assuming continuity of the appropriate partial derivatives). Are there any equations that all functions of the form $ f \equal{} \rm div \textbf{G}$ must satisfy? Show that the answer in this question is "No" by proving that every continuous function $ f$ on $ R^{3}$ is the divergence of some vector field.

Here is my work,

Let $ \textbf{G}(x, y, z) \equal{} <g(x, y, z), 0, 0>$, where $ g(x, y, z) \equal{} \int^{x}_{0} f(t, y, z) dt$. We now take $ \rm div \textbf{G}$ and using the Fundamental Theorem of Calculus we see that $ \rm div \textbf{G}$ is $ f(x, y, z)$. So every continuous function on $ R^{3}$ is the divergence of some vector field.

Is this correct?

That looks good. There's a little work to show that what you get is continuous, and you don't get differentiability in all directions unless you assume more smoothness for $ f$- that may be a problem, depending on your definitions.

This is, at least in part, a topological fact about $ \mathbb{R}^3$. We could define a divergence for vector fields* on (tangent to) the $ 3$-sphere $ \{(x,y,z,w): x^2\plus{}y^2\plus{}z^2\plus{}w^2\equal{}1\}$; only functions that integrate to zero are divergences in that case.

*The standard definition uses 2-forms, but with both a metric and an orientation, we have an isomorphism.

Stewart's Calculus: Early Vectors is somewhat inconsistent at this point.
[quote]If $ \vec F \equal{} P\vec \imath \plus{} Q\vec \jmath \plus{} R\vec k$ is a vector field on $ \mathbb R^3$ and the partial derivatives of $ P$, $ Q$, and $ R$ all exist, then the [b]curl[/b] of $ \vec F$ is ...[/quote]
This requires $ \partial P/\partial x$ to exist, even though it does not appear in the formula for curl.
[quote]If $ \vec F \equal{} P\vec \imath \plus{} Q\vec \jmath \plus{} R\vec k$ is a vector field on $ \mathbb R^3$ and $ \partial P/\partial x$, $ \partial P/\partial y$, and $ \partial P/\partial z$ exist, then the [b]divergence of[/b] $ \vec F$ is ...[/quote]
This only requires the existence of derivatives that actually appear in the formula for divergence. Another important difference: "of" is typed in bold font in the second definition, but not in the first. :roll:

What should we do if $ \vec F$ is required to be differentiable as in the definition of curl?

Here's another approach, which doesn't have the same problems with differentiability.

Suppose $ f$ is nice enough*. Let $ F$ be the field defined by the inverse-square law $ F(\vec{x})\equal{}\frac1{4\pi}\int_{\mathbb{R}^3}f(\vec{y}) \frac{\vec{x}\minus{}\vec{y}}{|\vec{x}\minus{}\vec{y}|^3}\,dV(\vec{y})$.
That doesn't mean differentiability is trivial, but it doesn't stumble where the other does.

*What is "nice enough" here? The integral has to converge. One way to ensure that is the overkill condition of being compactly supported. Another way is if $ |f(x)|$ decays at least as fast as some integrable function of $ |x|$.

One can improve convergence a little bit: $ \vec F(\vec{x})\equal{}\frac1{4\pi}\int_{\mathbb{R}^3}f(\vec{y}) \left(\frac{\vec{x}\minus{}\vec{y}}{|\vec{x}\minus{}\vec{y}|^3}\plus{}\frac{\vec{y}}{|\vec{y}|^3}\right)\,dV(\vec{y})$ has kernel with decay $ O(|y|^{\minus{}3})$ at infinity. This allows for $ f$ such that $ \int_{\mathbb R^3}|f(\vec y)|\frac{1}{|\vec y|^3\plus{}1}\,dV(\vec y)<\infty$.

But I still want a $ C^1$ vector field $ \vec F$ with $ \mathrm{div}\vec F\equal{}f$, for any $ f\in C(\mathbb R^3)$.  :wallbash:

I think this works: use the construction of the original poster, but average it over all appropriate coordinate systems. In a formula, $ \vec F(\vec x) \equal{} \int_{S^{2}}I(\vec x,\vec\nu)\vec\nu \,d\sigma (\vec \nu)$ (the vector-valued integral is taken with respect to the normalized surface measure $ \sigma)$, where $ I(\vec x,\vec\nu) \equal{} \int_0^{\vec x\cdot \vec\nu} f(\vec x \minus{} t\vec\nu)\,dt$ (as above, $ \vec\nu$ is a unit vector). It's probably not easy to prove that $ \vec F\in C^1$, but it should be true.

@jmerry Hmmm... I tend to end up with singular integrals when trying to differentiate your formula. Are you sure that it, indeed, gives a $ C^1$-field for every continuous $ f$?

@mlok Same question basically. How do you show $ C^1$ for your last construction? (since it was posted just a few minutes ago I had no time to think it over yet but I suspect that everything is not that simple...)

It's not obvious from the formula, and I don't have a proof. In practice- it seems to work.
An example: Suppose $ f$ is a radial function decaying fast enough at $ \infty$: $ f(\vec{x}) \equal{} g(|\vec{x}|)$. Then I get $ F(\vec{x}) \equal{} h(|\vec{x}|)\frac {\vec{x}}{|\vec{x}|}$, where $ h(r) \equal{} \frac1{r^2}\int_0^r s^2g(s)\,ds$

That's $ C^1$ as long as $ g$ is continuous.

The physics inspiration is strong here.

[b]jmerry[/b]: in an attempt to direct [b]fedja[/b]'s attention away from the flaws of my construction :) , I'd like to point out that the field you construct is conservative. Basically, you solve the Poisson equation $ \Delta u \equal{} f$ and then let $ \vec F \equal{} \nabla u$. But as far as I remember, the equation $ \Delta u \equal{} f$ does not have a $ C^2$ solution for every continuous $ f$, even compactly supported. (However, if $ f\in C^{\alpha}$, then you get $ u\in C^{2,\alpha}$.) It's somewhere in Gilbarg-Trudinger, but I don't have the book here.

Sure. Of course, that's a side effect of the physical inspiration- it's a gradient because gravitational/electrical potential energy works.

[hide="Wrong"]For my radial example, the potential $ P(r) \equal{} \frac1r\int_0^r s^2g(s)\,ds$ is $ C^1$ and not necessarily $ C^2$- but its gradient is still $ C^1$.[/hide]

I don't see how a function can have a $ C^1$ gradient without being $ C^2$ itself. If $ \nabla u\equal{}\langle u_x,u_y.u_z\rangle$ is a vector field in which all components have continuous 1st order partial derivatives, then all 2nd order partial derivatives of $ u$ are continuous.

Yeah, you're right there. That's not the right expression for the potential; it should be $ \int_0^\infty \min(\frac1r,\frac1s)\cdot s^2g(s)\,ds$ instead, which is $ C^2$.

This is from the book of Gilbarg and Trudinger, p.71: choose $ \eta\in C^{\infty}$ such that $ \eta\equiv 1$ in the unit ball $ B$ and $ \eta\equiv 0$ outside of $ 2B$. Let $ \phi(x)\equal{}\Delta(x_1x_2\eta(x))$: this function is supported in $ 2B\setminus B$. Finally, let $ f(x)\equal{}\sum_{k\equal{}0}^{\infty}\frac{1}{k}\phi(2^k x)$.
Although $ f$ is continuous, the Poisson equation $ \Delta u\equal{}f$ does not have any $ C^2$ solutions near the origin. Indeed, any such $ u$ is a harmonic function plus $ x_1x_2\sum_{k\equal{}0}^{\infty}\frac{1}{k}\eta(2^kx)$, and the latter function has a problem with $ \frac{\partial^2}{\partial x_1\partial x_2}$ at the origin.

This goes to show that we need non-conservative fields here, if the problem has a positive solution at all.

[quote="mlok"]...non-conservative fields here,...[/quote]

http://farside.ph.utexas.edu/teaching/301/lectures/node59.html

After a few hours of rather fruitless thinking, I decided to ask a few friends of mine whether they know the answer. It turned out that it is negative. Probably, the first one to observe it was Preiss: [i] Preiss, David; Additional regularity for Lipschitz solutions of PDE. J. Reine Angew. Math. 485 (1997), 197–207[/i]. There were a couple of similar papers after that by other authors.


conics
hyperbola
analytic geometry
geometry unsolved
geometry

Given three circles on a plane, try to construct a circle that is externally tangent to the all three circles.

We tried to solve such a problem  in the past (the problem was asking the radius of the 4th circle)

This is a very intersesting problem and I would like to see a solution  :)

Also, the problem with the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=44942]Soddy Circles[/url] is a particular case of this problem, hence it's easier

By the way, shobber, if you remember, we did discuss in my problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=66246]Equation of an hyperbola with given focuses[/url] (post #4), how we can find the equation of an hyperbola in cartesian coordinates

If you read all my post [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64502]here[/url], you will see for which reason I posted this query

Given are three circles $w_k=C(O_k,r_k),\ k\in \overline {1,3}$. Suppose w.l.o.g. $r_3\le r_1,\ r_3\le r_2$. Denote the circles $w'_k=C(O_k,r_k-r_1),\ k\in \overline {1,2}$. Then the proposed problem $(C.C.C.)$ is equivalently with the problem $(P.C.C.)$ to construct a circle which passes through the center $O_1$ and which is exterior tangent to the circles $w'_1,\ w'_2\ .$

[b]Remark.[/b] This problem is a particular case of a well-known Appollonios' construct problem and which was posted in this section no many time ago.