geometry
incenter
circumcircle
IMO Shortlist
IMO 2006
httpbit.ly3vjUcdg
IMO

Let $ABC$ be triangle with incenter $I$. A point $P$ in the interior of the triangle satisfies \[\angle PBA+\angle PCA = \angle PBC+\angle PCB.\] Show that $AP \geq AI$, and that equality holds if and only if $P=I$.

Let CP meet the circumcircle of triangle ABC at P', then <P'BP=<P'BA+<PBA=<PCA+<PBA=<PBC+<PCB=<P'PB. Thus P'P=P'B. Since <P'=<A, so we have $\angle{BPC}=90^{o}+\frac{A}{2}=\angle{BIC}$. Thus B, P, I, C lie on the same circle (Let's say with centre $D$). But it is a well-known result that $D$ lies on line $AI$, since we have $\angle{DIP}\leq 90^{o}$, so $\angle{AIP}\geq 90^{o}$. Hence $AP \geq AI$ with equality holds iff $I=P$.

(Almost) same solution as mine.
This was very easy.....
Seems as tough this year will be a year with many many honorable mentions.....

Ok………………………………………………………………….

We have: $ \angle PBA \plus{} \angle PCA \equal{} \angle PBC \plus{} \angle PCB$

If we add $ \angle PBC \plus{} \angle PCB$, to both side of the equality we have:

$ 2(\angle PBC \plus{} \angle PCB) \equal{} \angle PBA \plus{} \angle PCA \plus{} \angle PBC \plus{} \angle PCB \equal{} \angle B \plus{} \angle C$

then, it means that $ \angle PBC \plus{} \angle PCB$, is fixed for each $ P$.

then we conclude that $ \angle BPC$ is also fixed, and it is clear that, the angle is equal to$ \angle BIC$

then the locus of $ P$ is arc$ BIC$.

We know that this arc is tangent to $ AB,AC$ at $ B,C$ respectively.

And also know that $ AI$ goes through the center of circle $ BIC$.

Then $ I$ is the nearest point of circle to $ A$.

And for each point $ P$ on the circle we have:

$ AI < AP$

and if $ P$ is $ I$, it is clear that $ AP \equal{} AI$.

How ever it was nice and simple.

[spam]Finally a geo problem from IMO which I can solve.

It is a very nice restate of the fact that the intersection $J$ of the perpendicular bisector of $BC$ with $AI$ satisfies $JB=JI=JC$.[/spam]

You don't need to construct the point P', Shobber, but of course that works.

I like the idea of having an easy question for problem 1, but isn't this too simple?

just prove that BIPC is cyclic than let intersection of AI and circle aroun BIPC be V and AP and that circle be W.
 Than using angles we can prove that IBV is 90 from which we conclude IV is diameter.(2 radius).Than IV>=PW and we use potention of point A,we have AI*AV=AP*AW and if we assume that AI>AP we easy get a contradiction.From that we conclude AP>=AI and easy solve the cases of equality :)
 I hope this is enough for 7  :P

The problem is too much easy at IMO stage.

My solution is:

Actually P is a point on the circum circle of BIC and of course inside ABC. Let P is on arc BI.

To prove, $AP \geq AI$ it is sufficient to prove $\angle{AIP}\geq \angle{API}$.

$\angle{AIP}=360-\angle{AIC}-\angle{PIC}=360-(90+B/2)-(180-\angle{PBC})=90-B/2+\angle{PBC}$

$\angle{API}=360-\angle{APB}-\angle{BPI}=360-\angle{APB}-(180-\angle{BCI})=180-\angle{APB}+C/2$

$\angle{AIP}\geq \angle{API}$
implies, $90-B/2+\angle{PBC}\geq 180-\angle{APB}+C/2$
implies, $90-B/2+\angle{PBI}+\angle{IBC}\geq 180-\angle{APB}+C/2$
implies, $\angle{PBI}+\angle{APB}\geq 90+C/2$
implies, $\angle{PBI}+\angle{APB}\geq \angle{AIB}$

But, $\angle{APB}\geq \angle{AIB}$ Since P is inside AIB.

So the above inequality is true. Equality when P is I and then $\angle{PBI}=0$

What happened to my attached picture? It was there when posted the proof but now it is gone!

As I read "triangle triviality" I thought that it just can't happen at the IMO;
but in a while I realized it was true - the only facts I used to solve this problem are very, very basic things like sum of angles in a triangle :huh:

My solution was as shobbers solution but at the end i have used inequalities defferently. I hope my solutin is also true

  Abdurashid

I had a much longer solution to this, not sure if that's a good thing though.. probably isn't. Most people here seem to have got this one right, so the limits for medals are likely to be higher this year than last.

:o i couldn't believe that this is an imo problem when i saw it.it only took me 10 minutes to solve it(and i'm not good at geometry).i imagine all of you guys did it in....3 min :o

It was a $\le$10 minute problem for me, too.

My take on the problem:

First, we note that $P$ must be a point on circle $BIC$; $\angle PBA+\angle PCA-\angle PBC-\angle PCB=\angle CBA+\angle BCA-2\angle PBC-2\angle PCB$
$=(180^\circ-\angle BAC)-2(180^\circ-\angle BPC)=2\angle BPC-\angle BAC-180^\circ$. This is zero if $P=I$ (by the original form), so $\angle BPC=\angle BIC$ and $BPIC$ is cyclic.

It took a lot less time to notice the fact than to write this out; anyone that has done olympiad angle chasing should see the cyclic quadrilateral immediately.

Now, we want to show that $I$ is the closest point to $A$ on circle $BIC$. Equivalently, the line $AI$ passes through the center of circle $BIC$. The center must lie on the perpendicular bisector of $BC$, which meets the bisector $AI$ of angle $BAC$ at $A'$ on the circumcircle of $ABC$ (the midpoint of arc $BC$). We want to show that $A'$ is the center of circle $BIC$.
Chasing some angles, $\angle IA'C=\angle AA'C=\angle ABC=2\angle IBC$. Similarly, $\angle IA'B=2\angle ICB$. The only point other than $I$ making these (signed) angles is the center of circle $BIC$, so $A'$ is that center and we are done.

ooooooooooooops! I can't beleive it was IMO problem,really easy.(It took me 30 seconds to solve)

we have $\angle ABP+\angle PCA=\angle PBC+\angle PCB=90^{\circ}-\frac{A}{2}\rightarrow \angle BPC=90^{\circ}+\frac{A}{2}$
hence points $B,P,I,C$ are on a circle,we know that $AI$ pass through circumcenter of $BIC$ therefore $AI\le AP$

Since $\angle PBA+\angle PCA = \angle PBC+\angle PCB,$ $\angle PBC+\angle PCB = \frac{1}{2}(\angle ABC + \angle ACB) = \frac{1}{2}(180^\circ - \angle BAC).$ Hence, $$\angle BPC = 180^\circ - \angle PBC-\angle PCB = 90^\circ + \frac{\angle BAC}{2}.$$ Since $\angle BPC=90^\circ + \frac{\angle BAC}{2}$ and $\angle BI_AC = 90^\circ - \frac{\angle BAC}{2},$ $BPCI_A$ is cyclic with incenter $I$ being a point on the circumcircle of $BPCI_A$ and $M_A$ being the center of the circle. Thus, $PM_A = IM_A.$ Since $A,$ $I,$ and $M_A$ are collinear, applying triangle inequality on $\triangle APM_A$ gives $AP + PM_A \geq MA = AI + IM_A \implies AP \geq AI.$ When $P=I,$ $AP=AI$ since $PM_A=IM_A$ and vice versa. Hence we are done. $\blacksquare$

Well this was my 5000th post  :P

Note that this is equivalent to
$\measuredangle CBP + \measuredangle PCB = \measuredangle PBA + \measuredangle ACP$
so $\measuredangle CBP + \measuredangle PCB = \measuredangle CBI + \measuredangle ICB$
and thus quadrilateral $PIBC$ is cyclic, So
by incenter-excenter lemma implies that
$(PIBC)$ has center $L$ where is the
intersection of $AI$ and the circumcenter.

Since $A, I, L$ are collinear,
the result follows.

[hide=motivation]
The equality motivates us to wonder whether or not the sum is constant, and it is! This leads us to think if other angles are constant, and after some thought, it happens that the sum of $\angle{APB}$ and $\angle{APC}$ is also constant. This also means that $\angle{BPC}$ is constant, meaning that $P$ lies on $(BIC)$. Denote the center of the $(BIC)$ as $O'$. Now we need to show that $AIO'$ lie on a line because that would mean a circle of radius $AI$ centered at $A$ would only touch $(BIC)$ at one point, and that $AP \ge AI$ with the equality condition iff. $P=I$. By the incenter-excenter lemma, $O'$ lies on $(ABC)$ and on the angle bisector of $A$. Since $I$ also lies on the angle bisector of $A$, we are done.
[/hide]
[hide=solution]
Literally the motivation lmao
[/hide]

[hide=Solution]
$$\angle PBA+\angle PCA = \angle PBC+\angle PCB=180-\angle BPC$$
$$\angle ABC+\angle ACB=180-\angle BAC=2(\angle PBC+\angle BPC)$$
$$180-\angle BAC=2 \cdot \angle BPC $$
$$\angle BPC=180-\frac{\angle ABC+\angle ACB}{2}=\angle BIC$$
Since $\angle BPC=\angle BIC$, $P$ is on the circumcircle of $\triangle BIC$.\\
$ $\\
Let $D$ be the $A$-excenter of $\triangle ABC$ and let $E$ be the circumcenter of $BIC$.\\
Since $\angle IBD=\angle ICD=90^\circ$, the quadrilateral $IBDC$ is cyclic and it's circumcenter,$E$, lies on the angle bisector of $BAC$.\\
$ $\\
The length of $AE$ is constant as $P$ varies along the circle and $AE=AP+PE$. The radius is also constant, therefore $PD$ is constant, so the lentgh of $AP$ is the least when $A,P,E$ are collinear, which is equivalent to $P=I$. If the points aren't collinear, from the triangle inequality $AE<AP+PE$, therefore $AP>AI$.
[/hide]

Define $D = AI \cap (ABC)$. It's well known that $DI = BD = CD$. Thus, the circle, $\omega$ centered at D, going through B and C, also passes through I. We claim that all $P$ lie on $\omega$. Note that $\angle{BPC}$ is constant because $\angle{PBC}+\angle{PCB}$ is also constant, proving that all $P$ lie on $\omega$. By triangle inequality with $ADP$, we have that $AP \geq AI$.

Let $D$ be the intersection of $AI$ and $(ABC)$, so $DI=BD=CD$, so the circle centered at $D$, that passes through $B$ and $C$, also passes through $I$, so $\angle BPC$, is constant. So all $P$ lie on the circle. So using triangle inequality on $\triangle ADP$, finishes.

Observe that $2(\angle PBC + \angle PCB) = \angle B + \angle C$. Hence we have that the sum of these two angles is entirely independent of the choice of $P$. In particular, we find \[\angle BPC = 180^{\circ} - (\angle B - \angle C)/2 = 90^{\circ} + \angle A/2 = \angle BIC.\]

Therefore points $B, I, P, C$ lie on a circle with center $D$, midpoint of minor arc $\widehat{BC}$. Now by the triangle inequality, \[AP + PD \geq AD = AI + ID \iff AP \geq AI,\] with equivalence provided only when $P = I$, as desired. $\blacksquare$

Note that
\[\angle PBC+\angle PCB=\frac12\left(\angle B+\angle C\right),\]
so $180^{\circ}-\angle PBC-\angle PCB=\angle BPC$ is constant. Since $P=I$ satisfies the requirement, $\angle BPC=\angle BIC$, so quadrilateral $BPIC$ is cyclic. Now by fact 5, the circumcenter of triangle $BIC$, say $L$, is collinear with $A$ and $I$, so $I$ is the closer intersection of $AL$ and $(BPIC)$, completing our proof. $\blacksquare$

Solved this a while ago while doing EGMO. The angle condition implies $$\angle BPC = \angle A + \angle PBC + \angle PCB = \angle A + 180 - \angle BPC \implies \angle BPC = 90 + \frac{\angle A}{2}$$ Hence $P \in (IBC)$. Now the conclusion follows by Incenter-Excenter lemma. Namely $M$, the midpoint of minor arc $\widehat{BC}$ is the center of $(PIB)$ and $M-I-A$, so by Triangle Inequality we are done.

$\angle BPC=\angle ABP+\angle ACP+\angle A=\angle CBP+\angle PCB+\angle A=180-\angle BPC+\angle A$
$2\angle BPC=180+\angle A$
$\angle BPC=90+\frac{\angle A}{2}=\angle BIC$
so $BPIC$ is cyclic.
Let $E$ be the point where the line $AI$ intersects the circumcircle of $ABC$
By the incenter excenter lemma, $E$ is the center of the circumcircle of $BIPC$, so $EI=EP$
By the triangle inequality, $AP+PE \ge AE$
$AP+PE \ge AI+EI$
$AP \ge AI$

Notice that $\angle BIC = 90^{\circ} + \angle A \implies \angle IBC + \angle ICB = \frac{1}{2} (\angle B + \angle C).$

Due to the equality, we have both sides of the equation equaling $\frac{\angle B + \angle C}{2}$, so it follows that $\angle BIC = \angle BPC$, so $BIPC$ is cyclic, with the circumcenter being the midpoint of arc $BC$(by Incenter-Excenter Lemma.)

Then by the Incenter-Excenter Lemma, the circle with center $A$ passing through $I$ is tangent to $(IBC)$, so there is only one point where $AP \leq AI$, and that is when $AP = AI$ or $P = I$(due to the circles only intersecting once.)

Observe that from angle chasing, $P$ lies on the circle passing through $B, I, C$ . Also from the incentre-excentre lemma, this circle is externally tangent to that with centre $A$ passing through $I$ at $I$, giving us the required relation. $\square$

The LHS and RHS sum to $180^\circ-\angle BAC$ so the LHS and RHS are both $90^\circ-\frac{\angle BAC}{2}$ so $$\angle BPC=90^\circ+\frac{\angle BAC}{2}=180^\circ-\frac{\angle ABC+\angle ACB}{2}=\angle BIC$$ so $P$ must lie on $(BIC)$. Let $O$ be the center of $(BIC)$. Then, we just need to prove that $I$ lies on $AO$. We see that $BO=OC$ and $$\angle BOC=2(180^\circ-\angle BIC)=\angle ABC+\angle ACB=180^\circ-\angle BAC$$ so $O$ is the midpoint of arc $BC$ of $(ABC)$. Then, $\angle BAO = \angle BAI = \frac{\angle BAC}{2}$, as desired.

I haven't seen a solution with this approach yet (comparing angle sizes to deduce side lengths), so I'm wondering if this would get me full points (or any at all).

First we claim that $P$ does not lie in the interior of $\Delta IBC$. For the sake of contradiction, suppose otherwise. Then, we have $$\begin{cases} \angle PBA > \angle IBA \\ \angle PCA > \angle ICA \end{cases}$$ It follows that $\angle PBA + \angle PCA > \dfrac{1}{2} (\angle ABC + \angle ACB)$. However, we have $$\begin{cases} \angle PBC < \angle IBC \\ \angle PCB < \angle ICB \end{cases}$$ which leads to $\angle PBC + \angle PCB < \dfrac 12 (\angle ABC + \angle ACB)$, contradiction.

Now, WLOG suppose that $P \notin \overline{AI}$ lies on the half-plane of $AI$ that contains $B$, or $\angle PBC > \angle IBC$ and $\angle PCB < \angle ICB$. The given condition is equivalent to \begin{align*}
\angle ABI - \angle PBI + \angle ACI + \angle PCI &= \angle IBC + \angle PBI + \angle ICB - \angle PCI \\
\dfrac 12 \angle B + \dfrac 12 \angle C - \angle PBI + \angle PCI = \dfrac 12 \angle B + \dfrac 12 \angle C + \angle PBI - \angle PCI\\
\angle PCI = \angle PBI
\end{align*}
It follows that $BPIC$ is a cyclic quadrilateral. Consider
\begin{align*}
\angle AIP &= \angle AIB - \angle PIB
&= 90^{\circ} + \dfrac 12 \angle ACB - \angle PIB
&= 90^{\circ} + \angle ICB - \angle PCB
&= 90^{\circ} + \angle ICP
\end{align*}
Since we are considering the case $A,I,P$ are not collinear, $\Delta AIP$ is a nondegenerate triangle, and $\angle ICP > 0^{\circ}$. It follows that $\angle AIP > 90^{\circ}$, and it is well known that the side opposite to the largest angle of an obtuse triangle has greatest length, so it must be true that $AP > AI \, \forall P \notin \overline{AI}$.

The only remaining case is $P \in \overline{AI}$. If $AP = AI$, then we must have $P=I$. Moreover, it is not hard to see that if $P \neq I$, then the left side of the given condition is strictly less than the right side, yielding a contradiction. Thus, $AP = AI$ only if $P=I$, and checking we see that the given condition holds. $\blacksquare$

We have:
$\angle PBA + \angle PBC = \angle ABC$
$\angle PCA + \angle PCB = \angle ACB$
$\Rightarrow \angle PBA + \angle PCA + \angle PBC + \angle PCB = \angle ABC + \angle ACB$
$\Rightarrow \angle PBC + \angle PCB = \frac{\angle ABC + \angle ACB}{2}$
$\Rightarrow \angle BPC = 180 - \frac{\angle ABC + \angle ACB}{2} = \angle BIC$
Therefore, $BIPC$ is cyclic. By the incenter-excenter lemma, if $O$ is the center of circle $(BIPC)$, then $A$, $I$, $O$ are collinear. Since $OI = OP$, by the triangle inequality we know $AP \geq AI$. Equality occurs when $P = I$. $\blacksquare$


Diophantine equation
number theory unsolved
number theory

I want a solution integer with equation
   [tex]2(x(4x^2+1))^2-1=y^2[/tex] (I think it's no solution )
can help me? thanhks!!!!!!


(a) Find an expression for 3/5 as a finite sum of distinct reciprocals of
positive integers.

(b) Prove that any positive rational number can be so expressed.

a) 3/5 = 1/2 + 1/10

b) Let the rational number be p/q.

Then choose m such that 1/m </= p/q < 1/(m-1)

This is the first term of the sum.

The rest of the terms must sum to p/q - 1/m = (pm - q)/(qm).

However, since p/q < 1/(m-1), pm - p < q, or pm - q < p.

Thus our numerator has decreased by at least one. When we repeat the procedure with (pm - q)/(qm), the numerator will decrease again. Eventually, the numerator will become 1, and we will be done.


inequalities
combinatorics unsolved
combinatorics

each integer from 1 to 100 (inclusive) is colored red ,blue,green or yellow. prove that there are two numbers of the same color whose difference is of the same color.

its pretty hard :D

Well, I think the bound $100$ is too large because it is a classical result to prove that the Schur number $s(n)$ (i.e. the greatest integer $k$ such that $\{1, \cdots, k \}$ can be split into $k$ sum-free subsets) satisfies
\[\frac {3^n-1} 2 \leq s(n) \leq [e \cdot n! ] - 1\]

Thus, for $n=4$, and for $k \geq 65$, we are sure (ah, ah  :blush:  ) that in any $4$-colouring of $\{1, \cdots , k \}$ we may find $a,b,c$ (not necessarly distinct) with the same color such that $a+b = c$, thus $(a,c,c-a)$ gives the desired result.

Pierre.

whoa¡¡¡ your solution looks very advanced and useful¡¡¡ :D  where did you learn this?? the solution i have to this problem is highly dificult yours looks very easy¡¡ so do you have any links to these shur numbers and maybe some other useful things?? :blush:

You may start from here
http://mathworld.wolfram.com/SchurNumber.html

Pierre.


[quote="jimchill"]Just wondering:
Suppose x+e<y, for all e>0, would that give me x<=y?[/quote]

Actually, if $e>0$, it actually means that $x<y$, not $x\leq y$. Think about it, it makes sense; if you and I both have money, and I find MORE money on the ground (a positive amount), and I *still* have less money than YOU, then obviously I had even less money before I found the money on the ground.

[quote="10000th User"][quote="jimchill"]sorry...i wasn't clear about the question.  it does seem like a dumb one, but i was thinking in a "limit"-kinda way, where e (epsilon) denotes a small number.

i was told a "cool" way to prove two numbers are equal, that is suppose x, y are numbers, and |x-y|<e for every e>0, implies x=y, which is why i think of this in the first place.[/quote]Even if you were referring to "limit"-kinda way, which I'm sure you mean delta-epsilon definitions, $|x-y|<e$ does NOT imply that $x=y$. You have something missing, you need to put $0<|x-y|<e$ because you can have something undefined and yet have a limit. For example, take the limit $\lim_{h\to 0}\frac{h}{h}$. It is clear that if $h=0$ it is undefined but the limit is equal to 1. This example explains to you why it says that $h$ [b]approaches[/b] to 0 instead of saying $h$ [b]equals[/b] to 0 because you can be reaching it very close and yet it is not equal.

[EDIT]PS: the technique that uses $e$ works only on continuous functions, and cannot be used in general to prove two numbers are equal, [u]unless[/u] either you assume first that $e>0$ and are trying to get a contradiction or that you initially introduce $e$ and then able to conclude that $e$ must be equal to 0.[/quote]

10000th User, I have no idea what you are talking about.  If $|x-y|<\epsilon$ for every $\epsilon>0$, then $x=y$.  This trick, if you can call it such, is used often in analysis.

Jimchill's first statement is correct.

Nowhere do we need to, nor should we, use limits here.

[quote="blahblahblah"]

10000th User, I have no idea what you are talking about.  If $|x-y|<\epsilon$ for every $\epsilon>0$, then $x=y$.  This trick, if you can call it such, is used often in analysis.

Jimchill's first statement is correct.

Nowhere do we need to, nor should we, use limits here.[/quote]
I guess you are right about that implication, I have been thinking about limits all along :blush:

But jimchill's first statement, I have doubts. He says:
[quote="jimchill"]Suppose x+e<y, for all e>0, would that give me x<=y?[/quote]
If $x+e<y$ for all $e>0$, then $0<e<y-x$ for all $e>0$. So, $y-x$ is unbounded? Yes, I know $0\le y-x$ but I'm just wondering on why is 'for all $e>0$' in the statement? Because $e$ can be any particular positive real number and his statement is correct as well...

[quote="blahblahblah"]My mistake, I read into it what I expected to be there.

It should read 'if $x<y+\epsilon$ for every $\epsilon>0$, then $x\leq y$.[/quote]

My mistake in the first post... It should read as you said.  Thank you guys for helping me clarifying that.

[quote="0714446459923"]Suppose that y-x>0
then if you make e=y-x then
y<y
so y-x=<0
so x>=y[/quote]I see you want to prove by contradiction but in fact you only did a circular reasoning because $x+e<y$ is the same as $e<y-x$. How can you assume $e=y-x$?  :? . And you forgot that $x,y$ are both not fixed, they are variables...

[quote="0714446459923"][quote="Hriundel"]eeeer.....where did you get [b][i]y<y[/i][/b] from?
 :huh:[/quote]

I got it by making e = y-x and substituting and simplifying.[/quote]

$y<y$ ????????????

[quote="0714446459923"][quote="Hriundel"]eeeer.....where did you get [b][i]y<y[/i][/b] from?
 :huh:[/quote]

I got it by making e = y-x and substituting and simplifying.[/quote]

by doing that gives you y<=y, which is ok, but not y<y.

Unless if I'm mistaken, Jimchill's question comes from a special case of the delta-epsilon definition of a limit. So we can't just treat E as any integer.

Also, are limits and stuff allowed to be posted here? It's calculus, but because it's earlier in the curriculum I think we can be more lenient about it.


Prove that the three inequalities can't hold good simultaneously, if   $ a_1 , a_2 , a_3 , b_1, b_2, b_3$  are positive reals.

$ \frac{a_1b_2}{a_1 \plus{} b_2} < \frac{a_2b_1}{a_2 \plus{} b_1}$

$ \frac{a_2b_3}{a_2 \plus{} b_3} < \frac{a_3b_2}{a_3 \plus{} b_2}$

$ \frac{a_3b_1}{a_3 \plus{} b_1} < \frac{a_1b_3}{a_1 \plus{} b_3}$

$ \frac {a_1b_2}{a_1 \plus{} b_2} < \frac {a_2b_1}{a_2 \plus{} b_1}$

$ \implies \frac 1 {a_1} \plus{} \frac 1 {b_2} > \frac 1 {a_2} \plus{} \frac 1 {b_1}$

similarly, rewrite the other 2 inequalities and add all 3 to get some expression to be greater than itself.

hence, all 3 inequalities cannot hold good simultaneously.


Let $k\geq 2$ be integers and let $C$ be the curve $y=\sqrt{2x+1}\ (x>0)$. Given the tangent line at the point $(a_{k},\ \sqrt{2a_{k}+1})$ from the point $(0,\ k)$ to $C.$

Evaluate \[\lim_{n\to\infty}\frac{1}{n^{3}}\sum_{k=2}^{n}a_{k}.\]

[b]Kunny[/b], I think we should evaluate $\lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=2}^{n}a_{k}$ instead of $\lim_{n\to\infty}\frac{1}{n^{3}}\sum_{k=2}^{n}a_{k}.$ Otherwise, I guess the answer to the original limit is just $0.$

Could you please recheck your question?

No, kunny meant it the way he had it. I have that $a_{k}=O(k^{2})$ and hence the sum is $O(n^{3}),$ leading to an answer that is neither zero nor infinity.

Prof Merryfield, indeed, you are correct. I made a "small" error in computation earlier, which gave me $a_{k}= O(k).$ But, now I have rectified that. Thanks!

Here is the complete solution.

$y = \sqrt{2x+1}\Rightarrow \frac{dy}{dx}= \frac{1}{\sqrt{2x+1}}.$

So, the equation of the tangent line at the point $(a_{k}, \sqrt{2a_{k}+1})$ is given by $y-\sqrt{2a_{k}+1}= \frac{1}{\sqrt{2a_{k}+1}}(x-a_{k}).$

Since the point $(0,k)$ lies on this line, we must have
$k-\sqrt{2a_{k}+1}= \frac{1}{\sqrt{2a_{k}+1}}(-a_{k})$

The above, after squaring and some algebraic manipulation yields
$a_{k}^{2}+2(1-k^{2})a_{k}+1-k^{2}= 0$
$\Rightarrow a_{k}= (k^{2}-1)+k(\sqrt{k^{2}-1})$ ... (We drop the negative root because $a_{k}> 0$ for all $k\ge 2$)

Now, let $\{b_{n}\}$ and $\{c_{n}\}$ be two sequences such that $b_{n}= \sum_{k=2}^{n}a_{k}$ and $c_{n}= n^{3}.$
Note that $\{c_{n}\}$ is a positive, increasing and unbounded sequence.

$\therefore \lim_{n\rightarrow \infty}\frac{b_{n+1}-b_{n}}{c_{n+1}-c_{n}}= \lim_{n\rightarrow \infty}\frac{\sum_{k=2}^{n+1}a_{k}-\sum_{k=2}^{n}a_{k}}{(n+1)^{3}-n^{3}}= \lim_{n\rightarrow \infty}\frac{a_{n+1}}{3n^{2}+3n+1}= \lim_{n\rightarrow \infty}\frac{(n+1)^{2}-1+(n+1)\sqrt{(n+1)^{2}-1}}{3n^{2}+3n+1}$

$= \lim_{n\rightarrow \infty}\frac{(1+\frac1{n})^{2}-\frac1{n^{2}}+(1+\frac1{n})\sqrt{(1+\frac1{n})^{2}-\frac1{n^{2}}}}{3+\frac{3}{n}+\frac{1}{n^{2}}}$ ... (Or we could just drop the non-leading terms)

$=\frac23.$

$\therefore$ By the Stolz- Cesaro theorem, we have

$\lim_{n\rightarrow \infty}\frac{b_{n}}{c_{n}}= \frac23$

Or, $\lim_{n\rightarrow \infty}\frac{\sum_{k=2}^{n}a_{k}}{n^{3}}= \boxed{\frac23}$


Define a set $ S$ as $ S \equal{} \{n\in\mathbb{N}|n^3 < 65\}$. What is the supremum of $ S$?

I am thinking that the supremum should be $ 4$ as $ 4$ is an upperbound and also the least upperbound of the set. Could it be wrong or other possibility? like $ \sqrt[3] {65}$?

Thanks

The only possible values for the supremum (aka least upper bound) of a set of integers are integers and $ \infty$. Try to prove that.

Equivalently, the positive integers are well-ordered, so the supremum of any subset is either its maximum element or $ \infty$.


I think most people on Aops will know more than one language.
Let's see..

For some reason the poll disappeared so i fixed it and put it back in.

EDIT: OK, the poll kept dissappearing so I kept adding options and now it is a big mess!

Can a mod somehow fix the poll?!?!

I'm trilingual. English, urdu, some/most punjabi. I also understand a bit of hindi, but its too little to count. :)

[quote="Xevarion"]I only know one. :( (I assume you mean fluent. I can read Latin decently well, but that doesn't really count...)[/quote]
What about LaTeX? I'm bilingual: English and Telugu

english cantonese spanish

java latex c++

Dutch as native, English as second (but from an educational perspective third), French as third (but from an educational perspective as second), and some German (aber nicht so gut  :oops: ).

I also studied Latin (but I stopped doing that six years ago).

3 + 1 = 4

english, hindi, some spanish LaTeX :P

If you are including foreign languages taken at school, then 4 in all.  :)

Edit: that's not including $\LaTeX$. If you include that then 5 :P

english
american
$\LaTeX$
c++
and german
 :D

[quote="mtms5467"]What foreign languages does everyone here take?[/quote]
Lol, I don't mean to be offending but from a question like that you can immediately deduce that a person is from the USA.
More than half of the languages I can say a few words in are also native languages somewhere in my country.  The term "foreign" is not that uniform.
And we don't really take the languages, we simply have to learn them.  :|

[quote="mtms5467"]What foreign languages does everyone here take?[/quote]
Yo tomo espanol.

[quote="LadyKn1ght"]You should put an option for only one language :) . Although most will probably know more than one...[/quote]

sorry I forgot...I was in a hurry. I'll try to change it. Don't want the poll to be biased.

Since we are including computer languages too, i know 5 in total:
add: Latex and VB!!! :D

[quote="Kurt Gödel"]That's not so evident, akalra! For example: \[french+dutch = belgium \not\longrightarrow french(founders)+dutch(founders)\] I don't think every Canadian guy can speak both English and French. I guess it's like in Belgium  :D[/quote]
Well Kurt, read my last post (you posted in the same time). That's where I am getting at. (It's not like in Flanders in any case Kurt, we are trilingual remember, and I'm not talking about German)

Yeah, I can't speak a full German sentence without saying "euhm". But take Wallonia for example. They are supposed to speak Dutch, but they can't (not all of them, in any case) So maybe it's the same in Canada, where a part of the population can speak French and English and another part can't? I doubt that everyone speaks both.

(Oh, Frédéric, I've seen that Yves Leterme said that those who are pleading for a united Belgium should better learn Dutch before doing that. Elections are coming, so I guess he wants some votes from Vlaams Belang? I know someone who's losing many votes *Onckelinckx* :D )

[quote="fredbel6"][quote="rem"]English, Russian, a bit of French, Latex, Turing(programming language), Q-basic, C++[/quote]
I think it's kinda confusing everyone persistently adds those computer languages.... :|

Russian, that is quite something else though.  Are you of Russian ancestry?  I don't get it, are all Canadians supposed to be bilingual?? :o[/quote]
Yes, I was born in Russia and lived there for $12$ years.

[quote="Kurt Gödel"] So maybe it's the same in Canada, where a part of the population can speak French and English and another part can't? I doubt that everyone speaks both.[/quote]
Go check out the Canadian mathlinks community :roll:

I think we were in the end blessed by not being taught a "world language" at birth, we learnt one and didn't stop there... I once translated between the Canadians in their community (nobody replied though)

[quote="Kurt Gödel"]](Oh, Frédéric, I've seen that Yves Leterme said that those who are pleading for a united Belgium should better learn Dutch before doing that. Elections are coming, so I guess he wants some votes from Vlaams Belang? I know someone who's losing many votes *Onckelinckx* :D )[/quote]
Well I am currently trying to understand our own Belgian system better.  [hide]I actually made a post on Wetenschapsforum (Wiskunde) about that.  It's really pathetic, they explain how the distribution parliamentary seats is in accordance to the ratio of votes for a party... but the only really questionable thing behind that system (=those ratios won't give nice numbers) are ignored as their examples work out neatly.(If you want I can give you the official governmental mathematical explanation(in Dutch))[/hide]
Politicians often have a hidden agenda, Leterme is now our strong man, Di Rupo is the strong man on the other side now.  Our own state might be a good idea. For now, I wouldn't hold my breath though.
Onckelinkx : as I said I don't understand the system enough to conclude that this is all her own fault.  It's like firing ministers when Dutroux escaped.

English, a little Japanese (1 year), if computer languages count then $\LaTeX$, Java and a little C++.

English, Bengali, Hindi, piglatin, BASIC, c++, a little java, a little $\LaTeX$a little chtspk, a little leetspeak - my gosh, that's a lot!

[quote="fredbel6"I think it's kinda confusing everyone persistently adds those computer languages.... :| [/quote]They are languages just like any other kind. They are perhaps more exact and more rigorous, but they are languages...

[quote="Peter VDD"][quote="fredbel6"]I think it's kinda confusing everyone persistently adds those computer languages.... :| [/quote]They are languages just like any other kind. They are perhaps more exact and more rigorous, but they are languages...[/quote]
 :huuh:  O really, times have changed  :(

It doesn't make sense, can you communicate with someone else by writing code on a screen?

Yes. You'd have to define some extra commands though (like greeting, asking, ... but that is purely a matter of new vocabulary, the structure of the language is not a problem).

me witout computer lang.

chatspeak, english, hindi, pig latin, some spanish

english latin and some greek/french

i can also speak indo-european!

and pig-latin

and chat speak but that is so hard to i mean what kind of word is peep?!? i sent a fake email with a ghost peep(easter food) i hope nobody caught on...

lotsa of languages I invented...but only i can understand, and are useless in communicating with other people in anyways...

Also, English, Japanese (not fluently, but been learning 3 years now), tex

i speak 5 languages some fluent and some not.

[quote="Peter VDD"]Eum... programming languages, ok... but TeX is not a language. Then you can also call HTML a language. ;)[/quote]

how are u feelin today?


analytic geometry
real analysis
real analysis unsolved

Is it possible that we can draw uncountable circles such that any circle doesn't intersect another circle and any circle doesn't contain another circle inside of this?

no , it's not possible, since in every cirlce $ R$ we can choose  number $ h$ with rational coordinate  and such  map  $ R \to h$ is surjective since any circle doesn't intersect another circle and any circle doesn't contain another circle inside of this.
And as you already  know (maybe) $ \mathbb{Q}*\mathbb{Q}$ is countable.


Let $ a,b,c\ge0,k\ge1$ such that $ abc\equal{}1$. Prove that

$ a \plus{} b \plus{} c\ge \frac {k \plus{} a}{k \plus{} b} \plus{} \frac {k \plus{} b}{k \plus{} c} \plus{} \frac {k \plus{} c}{k \plus{} a}$

are you sure there isn't a mistake? when you put $ a\equal{}b\equal{}c\equal{}1/2$ we get that $ 3/2\geq3$ which is false.

[quote="dduclam"]Let $ a,b,c\ge0,k\ge1$ such that $ abc \equal{} 1$. Prove that

$ a \plus{} b \plus{} c\ge \frac {k \plus{} a}{k \plus{} b} \plus{} \frac {k \plus{} b}{k \plus{} c} \plus{} \frac {k \plus{} c}{k \plus{} a}$[/quote]
is's old and it's killed by S.O.S ineq

[quote="Hong Quy"]
is's old and it's killed by S.O.S ineq[/quote]
Really? I want to see your solution,Hong Quy  :)

Can you post your solution,Hong Quy?
Thank you  :roll:

sori ,   with k =1 .
  we have :
 $ \frac {(a \minus{} b)(a \minus{} c)}{(a \plus{} c)(c \plus{} b)} \plus{} \frac {(b \minus{} c)^2}{(a \plus{} b)(b \plus{} c)} \geq 0$

 suppose  $ a \equal{} max (a,b,c)$
  with k >1  i promise  i'll solve  later
    notice we let  $ x\equal{}a/b, y\equal{}b/c$ and$ z\equal{}c/a$
$ xyz\equal{}1.$
   only  change  varies in this problem

[quote="Hong Quy"]sori ,   with k =1 .
  we have :
 $ \frac {(a \minus{} b)(a \minus{} c)}{(a \plus{} c)(c \plus{} b)} \plus{} \frac {(b \minus{} c)^2}{(a \plus{} b)(b \plus{} c)} \geq 0$

 suppose  $ a \equal{} max (a,b,c)$
  with k >1  i promise  i'll solve  later[/quote]

Are you sure for your solution?

With $ k \equal{} 1$,this problem is very easy  :wink:

$ \sum (a \plus{} 1) \ge \sum \frac {a \plus{} 1}{b \plus{} 1} \plus{} 3$

<=> $ \sum \frac {(a \plus{} 1)b}{b \plus{} 1} \ge 3$  which true by AM-GM !


In one step you are allowed to substitute the set of three numbers $ (a,b,c)$ (the order of the number does not matter) with the other three numbers $ (a_1, b_1, c_1)$ according to the following rule:
$ a_1 = \sqrt{2b^2+2c^2-a^2}$
$ b_1 = \sqrt{2c^2+2a^2-b^2}$
$ c_1 = \sqrt{2b^2 +2a^2 - c^2}$.
Is it possible to obtain
a) ${ (\sqrt{3^{2007}.2}, \sqrt{3^2007}.4}, \sqrt{3^{2007}.6})$
b) $ (\sqrt{3^{2008}.4}, \sqrt{3^{2008}.13}, \sqrt{3^{2008}.33})$
from the three numbers $ (3, \sqrt{19}, \sqrt{22})$ in finite number of steps?

In every step sum of squares of 3 numbers is geting larger 3 times. At the begining it's 50, so case a) is impossible.
b) I don't know :oops:

b) impossible too. See what happens applying twice the rule


[quote="anhnguyen"]I am going away so I cannot take any class this summer. I know that the AMC12 problem series will be in October,2005. So will there also be an AIME Problem Series this fall?[/quote]

Yes.  We'll probably alternate the two different AIME Problem Series between Summer and Winter each year.


Just wanted to see if i was right or wrong
11253 base 7 is 2880 base 10


[quote="Flakky"]Prove that $ \left(p \plus{} q \plus{} r\right)\left(\frac {1}{p} \plus{} \frac {1}{q} \plus{} \frac {1}{r}\right)\geq 9 \plus{} 3\left(1 \minus{} \frac {p}{q}\right)^{\frac {2}{3}}\left(1 \minus{} \frac {q}{r}\right)^{\frac {2}{3}}\left(1 \minus{} \frac {r}{p}\right)^{\frac {2}{3}}$  $ p, q, r > 0$[/quote]
use AM-GM inequality,we have:
$ (p \plus{} q \plus{} r)(\frac {1}{p} \plus{} \frac {1}{q} \plus{} \frac {1}{r}) \minus{} 9 \equal{} \frac {(p \minus{} q)^2}{pq} \plus{} \frac {(q \minus{} r)^2}{qr} \plus{} \frac {(r \minus{} p)^2}{rp} \ge 3 \sqrt [3]{\frac {(p \minus{} q)^2(q \minus{} r)^2(r \minus{} p)^2}{p^2q^2r^2}} \equal{} RHS$    :wink:  :)


function
algebra
domain
real analysis
real analysis unsolved

Let $ f_{n}: \mathbb{R}\to\mathbb{R}$ be continuous for each $ n$ and suppose $ f_{n}$ converges uniformly to $ f$ on each closed, bounded interval $ [a,b]$.  Prove $ f$ is continuous on $ \mathbb{R}$.

Obviously $ f$ is continuous on each $ [a,b]$.  How do I make this all of $ \mathbb{R}$?

If there's a point $ p \in \mathbb{R}$ at which $ f$ is not continuous, can $ f$ be continuous in a neighborhood of $ p$?

Obviously not, that just seems too simple.

It's not too simple; it's exactly right.  Alternately, you can show that if two functions $ f_1, f_2$ are continuous in domains $ D_1, D_2$ and agree on $ D_1 \cap D_2$ (non-empty), then the "pasted" function $ f : D_1 \cup D_2$ which is equal to $ f_1$ on $ D_1$ and $ f_2$ on $ D_2$ is also continuous.

The "pasting lemma".  Yep, I have a proof of that.

A function $ f$ continuous on each $ [a,b]$ must be continuous on $ \mathbb{R}$. Is there any problem?

[quote="JRav"]Obviously $ f$ is continuous on each $ [a,b]$.  How do I make this all of $ \mathbb{R}$?[/quote]

Given $ x\in\mathbb{R}$, $ f$ is continuous on $ [x\minus{}\varepsilon,x\plus{}\varepsilon]$.  In particular, $ f$ is continuous at $ x$.  Since $ x$ was arbitrary, this is true for all $ x\in\mathbb{R}$.  But this is just the definition of continuity on $ \mathbb{R}$.

Yeah, exactly what I thought.  As I said before, it just seemed too easy.


Solve for $ x$: $ \sqrt{3x \plus{} 1} \equal{} 12 \minus{} 2\sqrt{3x \plus{} 1}$

[hide] We quickly get $ \sqrt{3x\plus{}1}\equal{}4$ so $ 3x\plus{}1\equal{}16$ so $ x\equal{}5$.[/hide]


assume $n$ is a natural number. then the set $(a_1,a_2,a_3,...,a_n)$ with $n$ numbers, is know to be "good" if we have
$a_1+a_2+a_3+...+a_n=2n$ [u]and[/u] that if you choos any number of $a_i$, then $\sum a_i\not=n$
find all of the "good" sets with $n$ number.
for example the set of $(1,1,4)$ is "good", but the set $(1,2,1,2,4)$ is not good because the sum of first, second and forth term is equal to 5.

The sets (2, 2, 2) (2, 2, 2, 2, 2) and so on, all sets with an odd number of twos, all work.

if any subset of the numbers has sum n, its bad.

The mean of the numbers is 2.  If a number is greater than 2, by x, we must have x 1's to "balance" it.

I have an idea to seperate it so that we can write it like this, for example:  (4,1,1), (6,1,1,1,1) , etc. where we group together the ones and the other number that 'supports' the ones, and keep the 2's for the end.

now let y be the largest element: y = n+1 is clearly the upper bound; n+1 > y > n/2 and we have enough ones to make a sum of n, so its bad.

If y < n/2, then we have y-2 ones from this y.  Now take the sequence to be decreasing order; we consider a_1 + a_2 + ....   and we stop if we would go over the target of n.  Then we definetly have enough ones, since we must include a_2, a_2 upholds a_2 - 2 ones, and if we are overboard, we are less than a_1 from the target, but we have a_1 - 2 + a_2 - 2 ones.

This only fails if a_1 = a_2 = 2.  So a necessary condition to satisfy is that it is (n+1,1...1) or (2,...2).

It is clear that (n+1, 1...1) and (2,..2) are the only working ones, for odd.


ratio
geometry
area of a triangle
Heron\u0027s formula

Prove that if two polygons are similar, the ratio of their areas is the square of the ratio of their sides.

Scientifically, we could use dimensional analysis. If that's too lame...

[hide]Prove it for triangles with e.g. Heron's formula, then use [url=http://en.wikipedia.org/wiki/Triangulation_(advanced_geometry)]triangulation[/url] to extend it to arbitrary polygons.[/hide]


How many even perfect square factors does $ 2^4 \cdot 7^9$ have?

We have $ 4/2\plus{}1$ times $ 5\equal{}15$.

Since you want an square number, the exponent should be even.  However, we cannot have $ 2^0 \times 7^n$, as that would be odd.

So we have either $ 2^2$ or $ 2^4$ and either $ 7^0$, $ 7^2$, $ 7^4$, $ 7^6$, or $ 7^8$.

Thus, we have $ 2$ possibilities for the $ 2^n$ and $ 5$ for $ 7^n$.  Multiplying yields $ \boxed{10}$.

AIME counted the evens as well as the odds.


Although i`m in my senior year ( for american standard, in Italy i have my last year still to do ) i `ve never had the opportunity to study math deeply and to try math contest. I spent this year in a canadian school and i caught up with the math curriculum that we dont do in my school back home ( i took math 11 , math 12 and calculus Ap all in one year achieving grades of 95/96% in each class) . I feel though that is time now to learn something more useful than just the academic math and i was looking at the books this website offers( the art of solving problem). Which one should i buy? first , second volume or both?

tommy

Start with the first.  It's a great resource, and you don't necessarily need the second to carry you far until you become a serious competitor.  Studying the first well will take you far, I've never gone beyond the third chapter in Volume 2 myself :)

[quote="cooljoe"]t0rajir0u, I'm a freshman (in precalc), and I have a question about the AoPS books. I've looked at the table of contents of the first volume, and all the stuff seems to be stuff I learned in algebra and geometry already. I got a 2 on the AIME and a 104.5 on the AMC 12A, and I just wanted to know how much would it really help me? How much did it help you?[/quote]

Studying AoPS Vol. 1 and working on problems on this forum got me from a $5$ to an $11$ on AIME.  You'd be surprised how much you can learn from it :)

[quote="cooljoe"]Wow! That is so awesome! Do you think I should buy the solution manuals to go with it? How long do you think I should study just volume 1? This whole summer? Thanks.[/quote]

Go ahead and get the solutions, but they generally tend to be a little fast and hard to understand, so on the problems you actually want solutions for you won't get much help from them.  You'd be better off asking clarification about concepts on the forum than looking at the solution.  Once you get the concepts you should be able to solve it yourself.  :)

Study it as long as you think you need to.  I think after you get important basic concepts down, working on problems on the forum is more helpful.


calculus
integration
calculus computations

...but I don't see how to show that $\int^{y}_{0}\frac{1}{\sqrt{uy-u^{2}}}du = \pi$. :?

Something tells me the integrated form will be some awkward arctangent, but I fail to see how to deduce it in a simple way.

It's an arcsine, not an arctangent.
Complete the square: $yu-u^{2}=\left(\frac{y}{2}\right)^{2}-\left(u-\frac{y}{2}\right)^{2}$.

Of course, that's why it didn't work out! :lol: Stupid me, thanks!


Let $ \triangle{XOY}$ be a right-angled triangle with $ m\angle{XOY}\equal{}90^\circ$. Let $ M$ and $ N$ be the midpoints of legs $ OX$ and $ OY$, respectively. Given that $ XN\equal{}19$ and $ YM\equal{}22$, find $ XY$.

$ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 26 \qquad
\textbf{(C)}\ 28 \qquad
\textbf{(D)}\ 30 \qquad
\textbf{(E)}\ 32$

[hide]Answer: $ (B)$ $ 26$

We are given $ 2OM \equal{} OX$ and $ 2ON \equal{} OY$ so we can write the equations

$ (2OM)^2 \plus{} ON^2 \equal{} 19^2$ and $ (2ON)^2 \plus{} OM^2 \equal{} 22^2$

We then solve for $ ON$ and substitute to get:

$ (2\sqrt {361 \minus{} (2OM)^2})^2 \plus{} OM^2 \equal{} 484$

Simplifying, we get        $ 1444 \minus{} 150M^2 \equal{} 484$

                                 $ OM^2 \equal{} 64$

We substitute this value into the equation $ OM^2 \plus{} OY^2 \equal{} 484$

We solve this to get $ OY^2 \equal{} 420$

Since $ OX \equal{} 2OM$, we have $ 16^2 \plus{} 420 \equal{} XY^2 \equal{} 676$ and $ XY \equal{} 26$[/hide]

[quote="Illos"][hide]Answer: $ (B)$ $ 26$

We are given $ 2OM \equal{} OX$ and $ 2ON \equal{} OY$ so we can write the equations

$ (2OM)^2 \plus{} ON^2 \equal{} 19^2$ and $ (2ON)^2 \plus{} OM^2 \equal{} 22^2$

We then solve for $ ON$ and substitute to get:

$ (2\sqrt {361 \minus{} (2OM)^2})^2 \plus{} OM^2 \equal{} 484$

Simplifying, we get        $ 1444 \minus{} 15OM^2 \equal{} 484$

                                 $ OM^2 \equal{} 64$

We substitute this value into the equation $ OM^2 \plus{} OY^2 \equal{} 484$

We solve this to get $ OY^2 \equal{} 420$

Since $ OX \equal{} 2OM$, we have $ 16^2 \plus{} 420 \equal{} XY^2 XY^2 \equal{} 676$ and $ XY \equal{} 26$[/hide][/quote]

[hide="same idea, but faster"]
call ON b and OM a.
pythagorean on OMY gives $ a^2\plus{}4b^2\equal{}484$
pythag on ONX gives $ 4a^2\plus{}b^2\equal{}361$
add to get $ 5a^2\plus{}5b^2\equal{}845$
after dividing a bit $ 4a^2\plus{}4b^2\equal{}676$
we know $ XY^2\equal{}4a^2\plus{}4b^2$ by pythag on OXY
so $ xy\equal{}26$

[/hide]

[b]Solution:[/b] We have that $(\frac{x}{2})^2 + y^2 = 361$ and $(\frac{y}{2})^2 + x^2 = 484$. Solving, we find that $x^2 + y^2 = 676$. So our answer is B.


These are challenging reactions...

Excess concentrated hydrochloric acid is added to a 1.0M solution of cobalt(2_ chloride.
Acidified solutions of potassium permanganate and iron(2) nitrate are mixed together.
Equal volumes of 1 M sodium phosphate and 1 M hydrochloric acid are mixed

1. Just a guess, is it a complexation?
2. Looks like oxidation of Fe(II) to Fe(III) and reduction of permanganate to Mn(II)
3. Are you sure thats the only data given?


Solve the equation $y^{3}= x^{3}+8x^{2}-6x+8$ in positive integers.

Posted and solved [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105950]here[/url].


Split it into two terms. The substitution $ t\equal{}x\minus{}3$ may help with this.


For nonnegative integers $n$ and $k$, define $Q(n, k)$ to be the coefficient of $x^{k}$ in the expansion $(1+x+x^{2}+x^{3})^{n}$. Prove that

$Q(n, k) = \sum_{j=0}^{k}\binom{n}{j}\binom{n}{k-2j}$.

[hide="hint"]
Think of $\binom{n}{j}$ as the number of ways you can pick the $x^{2}$ term in the expansion.[/hide]

The way you usually define it:

$\frac{(n-1)(n-2)\cdots(n-k+1)}{k!}$

[quote="Farenhajt"]How do you define ${n\choose k}$ when $k>n$?  :huh:[/quote]

If $k > n$, then $\binom{n}{k}= 0$. That's standard. How many ways can YOU choose $20$ things out of $5$?

And the definition otherwise is usually $\binom{n}{k}= \frac{n!}{(n-k)! \cdot k!}$.

[quote="Farenhajt"]Stupid :roll: Yup, the latter is what I had in mind, and the negative factorial got me.

And we can solve it thus:

[hide]The expression is equal to $(1+x)^{n}(1+x^{2})^{n}$, hence the general term is ${n\choose r}{n\choose j}x^{2j+r}={n\choose j}{n\choose k-2j}x^{k}$[/hide][/quote]

Nice. I thought of it this way: $\binom{n}{j}$ is the number of ways we can choose $j$ of the $n$ terms to be $x^{2}$. Then $\binom{n}{k-2j}$ is the number of ways we can choose $k-2j$ of the terms terms to increment by one power (i.e. $x^{2}\rightarrow x^{3}$ or $1 \rightarrow x$). The total power is then $2j+(k-2j) = k$. Summing over all $j$ gives us the desired exponent.

[quote="amirhtlusa"]i think its a putnam problem. :)[/quote]

Probably. I saw it in a problem set that Stanford people use to practice for the Putnam.


This year, [url=http://www.arml.com]ARML[/url] is happening May 29-30. Connecticut generally brings three teams, chosen by performance in a runoff held in early March. We go to Penn State in State College, PA, leaving on the 28th and returning on the 30th; the competition is during the day on the 29th, a Saturday. It's tons of fun, and I highly recommend going to your local runoff to try out for the team. It's not terribly selective. All the runoff locations are attached; you should probably get in touch with the contact at the closest school instead of just showing up. I don't know what time, specifically, any of this is happening.

Note: I'm just a student interested in getting more people interested, so some of my information may be wrong/out of date.

[quote="Ihatepie"]My teacher says we're taking the arml at my school, and it's not on the list. Are those all of the schools?[/quote]Your teacher has better information than I do.

I don't know much about the CT team so hopefully you can answer a few of my questions.

How hard are the ARML runoff questions (compared to individual ARML/AMC/AIME)?
How many questions are there on the runoff?
What is the time limit?
How well (on average) does one need to perform to make it?
How many people compete? How many make it?

There are generally 20-24 questions, comparable to, say, 10-15 on the AMC 12. I don't know how well you need to do, but generally about 100 people take the test and 45 make it. The results are announced at the state competition.

So how'd you guys do? I think I got around a 15. Is that enough to make it?

Not well :( I made too many stupid mistakes (a few arithmetic and overcounting on one). Which location did you take it at?

Are we allowed to compare answers? I think I only missed one or two, but I want to be sure.

Don't; some people may be taking it later for whatever reason.

Ok, my school didn't do it this year, how can my school get in on this next year? It sounds pretty cool.


geometry
geometric transformation
reflection
projective geometry
circumcircle
power of a point
cyclic quadrilateral

Let $A$ be point in the exterior of the circle $\mathcal C$. Two lines passing through $A$ intersect the circle $\mathcal C$ in points $B$ and $C$ (with $B$ between $A$ and $C$) respectively in $D$ and $E$ (with $D$ between $A$ and $E$). The parallel from $D$ to $BC$ intersects the second time the circle $\mathcal C$ in $F$. Let $G$ be the second point of intersection between the circle $\mathcal C$ and the line $AF$ and $M$ the point in which the lines $AB$ and $EG$ intersect. Prove that
\[ \frac 1{AM} = \frac 1{AB} + \frac 1{AC}. \]

$MG \times ME = MB \times MC = (AB-AM) \times (AC-AM)$

$MG \times ME = AB \times AC - AB \times AM - AC \times AM + AM^2$

Now if we show that $MG \times MF = AM^2$ then the problem is finished, and it is true since

the triangles $AMG$ and $AME$ are similar, thus $\frac{MG}{AM} = \frac{AM}{ME}$

equivalent to $AM^2 = MG \times MF$. So we have desired $AB \times AC= AB \times AM + AC \times AM$ which is

$\frac{1}{AM} = \frac{1}{AB} + \frac{1}{AC}$.

the problem is finished.

Since the lines $DF$ and $AC$ are parallel, it follows that $\angle DFA=\angle CAF.$ On the other hand, $\angle DFA=\angle DEG,$ because both angles subtend
the arc $DG.$ We deduce that $\angle CAF=\angle DEG,$ hence triangles $AMG$
and $EMA$ are similar. This implies
\[ \frac{AM}{MG}=\frac{EM}{AM},  \]
that is, $AM^{2}=MG\cdot ME.$

From the power of a point theorem we have
\[ MG\cdot ME=MB\cdot MC,  \]
hence
\[ AM^{2}=MB\cdot MC.  \]
Writing the latter as
\[ AM^{2}=\left( AB-AM\right) \left( AC-AM\right) =AB\cdot AC-AM\left( AB+AC\right) +AM^{2},  \]
gives
\[ AM\left( AB+AC\right) =AB\cdot AC,  \]
or, equivalently,
\[ \frac{1}{AM}=\frac{1}{AB}+\frac{1}{AC},  \]
as desired.

[color=darkblue][b]Proof.[/b] $DF\parallel AC$ $\Longleftrightarrow$ $(\widehat {AGM}\equiv\widehat {EGF}\equiv\widehat {EDF}\equiv\widehat {EAM})$ $\Longleftrightarrow$
$\widehat {AGM}\equiv\widehat {EAM}$ $\Longleftrightarrow$ $\triangle AGM\sim\triangle EAM$ $\Longleftrightarrow$ $MA^2=ME\cdot MG$ $\Longleftrightarrow$ $MA^2=MB\cdot MC$ $\Longleftrightarrow$ $\boxed {\ \frac{1}{AM}=\frac{1}{AB}+\frac{1}{AC}\ }\ .$

[b]Remark 1.[/b] If the point $N$ is the reflection of the point $A$ w.r.t. the point $M$ then $\frac{1}{AM}=\frac{1}{AB}+\frac{1}{AC}$ $\Longleftrightarrow$ $\frac{2}{AN}=\frac{1}{AB}+\frac{1}{AC}$ $\Longleftrightarrow$
the division $(A,B,N,C)$ is harmonically, i.e. $\frac{NB}{NC}=\frac{AB}{AC}\ .$[/color]

[color=darkred][b]Remark 2.[/b] See the following proposed problems:
[b]nr. 1.9.[/b], [u]the page 9.[/u], about the characterization of a harmonical division;
[b]nr. 5.103.[/b], [u]the page 58.[/u], with three extensions in the my book
[u]Geometrie plana (sintetica, vectoriala, analitica).
Culegere de probleme., Editura GIL, Zalau, 2002, Romania.[/u][/color]

Consider the inversion of pole $A$ wich invariates the circle.
If $M'$ is the inverse of $M$ then $M'FDA$ is an isosceles trapezoid.
Thus $AM'=AB+AC$ and the conclusion follows.

[color=darkblue][b][u]A generalization.[/u][/b] Consider two circles $C_k=C(O_k)\ ,\ k\in \overline {1,2}\ ,$ a fixed point $A\in C_1$ and a mobile point $M\in C_2\ .$ Denote: the tangent $d\equiv MM$ in the point $M$ to the circle $C_ 2\ ;$ the second intersection $N$ of the line $AM$ with the circle $C_1\ ;$ the point $P$ of the circle $C_1$ for which $NP\parallel d\ ;$ the second intersection $R$ of the line $MP$ with the circle $C_1\ ;$ the intersection $L\in AR\cap d\ .$ Prove that the relation $LA\cdot LR=LM^2$ and ascertain the geometrical locus of the point $L$ when the point $M$ covers the circle $C_2\ .$[/color]

This probelm have a simple projective solution.

From Desargue th for the cyclic quadrilateral $DEFG$ and the line $AM$ we have that this line cuts the pairs of opposite sides, the diagonals and the circumcircle of $DEFG$ in point in the same involution.

Denote the ecuation of this involution with $axx'+b(x+x')+c=0$ with origin in $A$. Since $(A,A)$ is a pair of points in involution we have that, $c=0$, hence the involution is $\frac 1 x +\frac 1 {x'}=-\frac a b$ which is const.

Hence, $-\frac a b=\frac 1 {AM}=\frac 1 {AM}+\frac 1 {A \infty}=\frac 1 {AB}+\frac 1 {BC}$ q.e.d.

We consider the following:
Perspectivity from E gives $(A,B;M,C)=(D,B;G,C)$.
Perspectivity from F gives $(D,B;G,C)=(A,C;B,\infty)$
Now, compare the cross ratios so obtained.
We get the conclusion.

Let $P$ be the reflection of $A$ in $M$ and let $N$ be the midpoint of $\overline{BC}.$ Since $CFDB$ is an isoceles trapezoid, we have $\widehat{CF} = \widehat{DB}.$ Therefore, \[\measuredangle MAG = \frac{1}{2}\left(\widehat{CF} - \widehat{GB}\right) = \frac{1}{2}\left(\widehat{DB} - \widehat{GB}\right) = \frac{1}{2}\widehat{DG} = \measuredangle DEG = \measuredangle AEG.\] Thus, by the Alternate Segment Theorem, $MA$ is tangent to $\odot (AGE)$, and in particular, $MA^2 = MG \cdot ME = MB \cdot MC.$ Therefore, it is [url=http://olympiads.hbcse.tifr.res.in/uploads/notes-on-harmonic-pencils-imotc-2013]well-known[/url] (Lemma 1.5) that $(B, C; A, P)$ is harmonic. By the same Lemma 1.5, we have $AB \cdot AC = AP \cdot AN = 2 \cdot AM \cdot AN.$ Hence, \[\frac{1}{AM} = \frac{2 \cdot AN}{AB \cdot AC} = \frac{AB + AC}{AB \cdot AC} = \frac{1}{AB} + \frac{1}{AC}. \; \square\]


Let $ b_k\equal{}\frac{a_k}{k}\le 1$. Then

$ b_1\ge b_1$
$ b_1\plus{}b_2\plus{}b_2\ge 3\sqrt[3]{b_1b_2^2}$
$ b_1\plus{}b_2\plus{}b_2\plus{}b_3\plus{}b_3\plus{}b_3\ge 6\sqrt[6]{b_1b_2^2b_3^3}$
$ b_1\plus{}b_2\plus{}b_2\plus{}b_3\plus{}b_3\plus{}b_3\plus{}b_4\plus{}b_4\plus{}b_4\plus{}b_4\ge 10\sqrt[10]{b_1b_2^2b_3^3b_4^4}$

Multiplying the relations above, we get the LHS of the original inequality on the left and on the RHS we get:

$ 3*6*10*b_1^{1\plus{}\frac{1}{3}\plus{}\frac{1}{6}\plus{}\frac{1}{10}}*b_2^{\frac{2}{3}\plus{}\frac{2}{6}}*b_3^{\frac{3}{6}\plus{}\frac{3}{10}}*b_4^{\frac{4}{10}}$

Now observe that the exponent of each $ b_i$ is less than $ 2$. So we can use the fact that $ 0<b_i\le 1\implies 0<b_i^e\le 1$ where $ e>0$.

Thus we get LHS $ \ge 3*6*10 (b_1b_2b_3b_4)^2$, and replacing the $ b_i$'s with the $ a_i$'s gives the result.

This is a nice problem :)


logarithms
integration
probability
function
real analysis
limit
algebra

$ X_{1},X_{2},...$ is the sequence of independent random variables uniformly distributed on interval $ [0,e]$. Check whether sequence
\[ X_{1}X_{2}^{\frac{1}{2}}X_{3}^{\frac{1}{3}}...X_{n}^{\frac{1}{n}}\]
converges almost surely

I assume that you're talking about the sequence $ Y_n$ where $ Y_n\equal{}\prod_{j\equal{}1}^nX_j^{1/j}.$

Note that $ \ln Y_n\equal{}\sum_{j\equal{}1}^n\frac1j\ln X_j .$

Since we can compute that $ E(\ln X_j)\equal{}0$ and $ \text{Var}\,(\ln X_j)\equal{}1,$ we see that $ E(\ln Y_n)\equal{}0$ and $ \text{Var}\,(\ln Y_n)\equal{}\sum_{j\equal{}1}^n\frac1{j^2}$

I would expect that $ \ln Y_n$ converges [i]in distribution[/i] to a normal distribution with mean zero and variance $ \frac{\pi^2}6.$

There is no reason to think that this would be almost sure convergence to anything.

[quote="Kent Merryfield"]
Since we can compute that $ E(\ln X_j) \equal{} 0$ and $ \text{Var}\,(\ln X_j) \equal{} 1,$ we see that $ E(\ln Y_n) \equal{} 0$ and $ \text{Var}\,(\ln Y_n) \equal{} \sum_{j \equal{} 1}^n\frac1{j^2}$
[/quote]

$ E(\ln X_j) \equal{} \frac {1}{2}$ We know that $ X_j\sim U([0,e])$  :maybe: and rest isn't also true.

$ E(\ln X)\equal{}\frac1e\int_0^e\ln x\,dx\equal{}\left.\frac1e(x\ln x\minus{}x)\right|_0^e\equal{}0.$

Where did $ \frac12$ come from?

My wrong :| .  But i still don't see why $ Y_{n}$ doesn't have to converges almost surely. Can you elaborate it little more ?

Why doesn't $ Y_n$ converge almost surely? Because the moment we realized that the limit of the variances tended to anything other than zero - that is, the moment we realized that any kind of limit would have to involve some non-constant distribution - we should have realized that the question of almost sure convergence (or convergence in probability) didn't even make sense as a question.

Let's elaborate. A probability space is simply a measure space, and a random variable is a measurable function on that space. Saying that something happens almost surely is saying that it happens for each specific point (outside a negligible set) in that space - hence saying something like that about a random variable or variables requires us to know something about those function at each point of that measure space.

And you didn't tell me that about the $ X_j.$ You told me how they were distributed, and you told me that they were independent - but you didn't tell me what they were as functions.

Here's a modest illustration. Let $ [0,1]$ with Lebesgue measure be a probability space. (It does meet all the definitions.) Consider the functions $ f(t)\equal{}t$ and $ g(t)\equal{}1\minus{}t.$ Both are measurable functions, hence random variables. Both have the same distribution - they're both $ U([0,1]).$ And they're not the same function, not even close. The sequence $ (f,f,f,f,\dots)$ does converge almost surely to $ f,$ but the sequence $ (f,g,f,g,\dots),$ while it converges in distribution to $ U([0,1]),$ does not converge [i]a.s.[/i] or in probability to anything.

Now if the proposed limit were a constant, it would make sense to ask about possible [i]a.s.[/i] convergence, since it no longer matters what point in the probability space you're talking about. But for the non-constant limit, the question makes no sense precisely because you only told me the distributions of the $ X_j.$

[quote="Kent Merryfield"]Why doesn't $ Y_n$ converge almost surely? Because the moment we realized that the limit of the variances tended to anything other than zero - that is, the moment we realized that any kind of limit would have to involve some non-constant distribution - we should have realized that the question of almost sure convergence (or convergence in probability) didn't even make sense as a question.

Let's elaborate. A probability space is simply a measure space, and a random variable is a measurable function on that space. Saying that something happens almost surely is saying that it happens for each specific point (outside a negligible set) in that space - hence saying something like that about a random variable or variables requires us to know something about those function at each point of that measure space.

And you didn't tell me that about the $ X_j.$ You told me how they were distributed, and you told me that they were independent - but you didn't tell me what they were as functions.

Here's a modest illustration. Let $ [0,1]$ with Lebesgue measure be a probability space. (It does meet all the definitions.) Consider the functions $ f(t) \equal{} t$ and $ g(t) \equal{} 1 \minus{} t.$ Both are measurable functions, hence random variables. Both have the same distribution - they're both $ U([0,1]).$ And they're not the same function, not even close. The sequence $ (f,f,f,f,\dots)$ does converge almost surely to $ f,$ but the sequence $ (f,g,f,g,\dots),$ while it converges in distribution to $ U([0,1]),$ does not converge [i]a.s.[/i] or in probability to anything.

Now if the proposed limit were a constant, it would make sense to ask about possible [i]a.s.[/i] convergence, since it no longer matters what point in the probability space you're talking about. But for the non-constant limit, the question makes no sense precisely because you only told me the distributions of the $ X_j.$[/quote]

I disagree!

Statements in probability theory are (almost) never statements about random variables as functions, but statements about the joint distribution of random variables. Thus it is never required to know the definition of the random variables as functions on the sample space, but it is sufficient to know their joint distribution. In the case of a.s. convergence we have to determine whether $ 1\equal{}P(X_n converges) \equal{} P( (X_n)_n \in A)$, where $ A$ is the set of sequences of reals numbers that converge. So clearly this is a property of the distribution of $ (X_n)_n$, i.e. the joint distribution of the RVs.

In the given problem we are given the distributions of the random variables and their independence, so we know their joint distribution, and thus the question about a.s. conergence makes sense! Furthermore the convergence in distribution to something nonconstant alone doesn't tell you anything about a.s. convergence. (Take any a.s. convergent sequence with a nontrivial limit, then the sequence also converges in probability and in distribution.) Note that in the above example  $ (f,f,f,f,\dots)$ and $ (f,g,f,g,\dots)$ have different joint distributions, so it is not surprising that their behaviour concerning a.s. convergence is different.

As Kent Merryfield pointed out the problem is equivalent to the investigation of the a.s. convergence of a random series. There are very powerful theorems that give sufficient and necessary conditions for the convergence of a random series with independent increments. Are you familiar with this kind of stuff?

[quote="solyaris"]

As Kent Merryfield pointed out the problem is equivalent to the investigation of the a.s. convergence of a random series. There are very powerful theorems that give sufficient and necessary conditions for the convergence of a random series with independent increments. Are you familiar with this kind of stuff?[/quote]

You mean Kolomogorow theorem about three sequences? (That states that  necessary condition for almost surely convergence of  $ \sum_{n\equal{}1}^{\infty}X_{n}$ is the convergence (for all $ c>0$ )  of sequences:
\[ \sum_{n\equal{}1}^{\infty}E(X_{n}^{c}) \ \ \ \sum_{n\equal{}1}^{\infty}Var(X_{n}^{c}) \ \ \ \sum_{n\equal{}1}^{\infty}P(|X_{n}|>c)\]
and sufficient condition is convergence of these sequences for some $ c>0$. )

Thanks for your answers. They were very helpful

Yes, the kolmogorov 3-series theorem is one of the ones I meant (and the most general one at that).
There are other theorems of the same type for nonnegative increments and increments satisfying
certain integrability conditions if I remember correctly.

[quote="solyaris"]
Statements in probability theory are (almost) never statements about random variables as functions, but statements about the joint distribution of random variables. Thus it is never required to know the definition of the random variables as functions on the sample space, but it is sufficient to know their joint distribution. In the case of a.s. convergence we have to determine whether $ 1 \equal{} P(X_n converges) \equal{} P( (X_n)_n \in A)$, where $ A$ is the set of sequences of reals numbers that converge. So clearly this is a property of the distribution of $ (X_n)_n$, i.e. the joint distribution of the RVs.
[/quote]
But your interpretation of the problem is something completely different from a.e. convergence. Do you see this type of convergence called almost sure convergence in literature?

I am not sure I understand what you mean. I am using the usual definition of a.s. convergence:
\[ Y_n \to Y \text{ a.s.} \quad \Leftrightarrow \quad P(Y_n \to Y) \equal{} 1\]
so
 \[ Y_n \text{ converges a.s.} \quad \Leftrightarrow \quad P(\lim Y_n \text{ exists})
\equal{} P(\liminf Y_n \equal{} \limsup Y_n ) \equal{}1\]
In measure theory this is sometimes called a.e. convergence, but it is exactly the same thing.

In any case, it is pretty clear that "$ Y_n$ converges a.s." is a statement about the joint
distribution of the $ Y_n$, or do I make some stupid mistake here?

[quote="solyaris"] or do I make some stupid mistake here?[/quote]
Absolutely.
The explanation which Kent Merryfield gave is the precise one.
Please refer some classical Prob. theory book . You are saying that .
\[ Y_n \to Y \text{ a.s.} \quad \Leftrightarrow \quad P(Y_n \to Y) \equal{} 1
\]
But what does that mean? It means that the event $ \{\omega: Y_{n}(\omega)\to Y(\omega)\}\in \mathcal{F}$ has probability $ 1$ where $ (\Omega, \mathcal{F}, \mathbb{P})$ is the probability space.
This is what exactly Kent Merryfield is saying.

Since we seem to agree on the definition of a.s. convergence, I think we can also agree on that a.s. convergence is something which only depends on the joint distribution of the RVs involved.  (And this of course was the point of my post.) Note that:
\[ P(Y_n \to Y)  \equal{} 1 \quad \Leftrightarrow \quad P((Y,Y_1,Y_2,....) \in A)\equal{} 1\]
\[ \quad  \Leftrightarrow \quad P\circ (Y,Y_1,Y_2,...)^{\minus{}1}(A) \equal{} 1,\]
where $ A \equal{} \{(y,y_1y_2,...) \in \mathbb{R}^{\mathbb{N}}: y_n \to y\}$ is  a set of sequences of real numbers.  In the last formulation of a.s. convergence above there is no need to know $ \Omega$ or $ P$ or how the $ Y_n$s are defined as functions on $ \Omega$, the only thing you need to know is $ P\circ (Y,Y_1,Y_2,...)^{\minus{}1}$, which is the joint distribution of all RVs involved.
Similarly you can reformulate "$ Y_n$ conveges a.s." in terms of the joint distribution of the $ Y_n$.

As a more basic example: If you are given a standard normal RV $ X$ and want to compute some $ P(X \ge 1)$ you don't care about what $ P$ is and what $ \Omega$ is and how $ X$ is defined  as a function on $ \Omega$, you only need the distribution of $ X$, which you are given to be normal. Right?

Whether or not $ Y_n$ converges pointwise almost everywhere indeed seems to be a property of the joint distribution, but you of course cannot tell what the limit function is just based on the joint distribution! I know next to nil about probability when presented in the "probability-way", but considering convergence in this fashion has a very strange feel to it to me. I have a hard time pointing out what is strange, but it is related to the fact that if we only care about distributions and not functions then we probably only care about the limit as a probability distribution also. But the convergence question then depends on how we choose functions to induce the distributions in a complicated way.

The a.e. convergence just seems to have the character of a technical device in this setting of distributions, and so the question in the original post comes across as weird when presented in a vacuum, and I think this is the reason for the confusion of several posters. But yes, you are right, the question does make sense - sorry we messed the thread up!

Hi, it is sufficient to show that $ ln(Y_n)$ converges almost surely to a random variable almost surely finite.

\[ ln(Y_n)\equal{}\sum_{k\equal{}0}^{n} \frac{1}{k} ln(X_k)\] which is clearly a martingale due to the fact that $ \mathbb{E}ln(X_k)\equal{}0$

But, this martingale is bounded in $ L^{2}(\Omega)$ , so the martingale converge almost surely.

[quote="limsup"][quote="solyaris"] or do I make some stupid mistake here?[/quote]
Absolutely.[/quote]

ugh no, solyaris's posts in this thread are really good; you should read them carefully.

Yes blahblahblah, you are right. Though I am not convinced completely, I could realize solyaris's arguments.
To solyaris:
When you talk about $ (Y,Y_{1},Y_{2},...)^{\minus{}1}$ what is the domain of $ (Y,Y_{1},Y_{2},...)$?

Sorry solyaris, if I m mistaken!

Blahblahblah, thanks for confirming my posts.

Limsup, the domain of $ (Y,Y_1,Y_2, \ldots)$ can be any measurable space  $ (\Omega,\mathcal{F},P)$. However, for asking and answering the question in the original post, the knowledge of the domain is irrelevant if the joint distribution is known.
(If you are given a standard normal RV $ X$, you can do all sorts of computations of probabilities, expectations, and so on, without knowing the domain of $ X$ and the definition of $ X$ as a function on the domain. So of course $ X$ has to be defined on some domain, but the knowledge of the domain is irrelevant for the calculations.)


A traffic light runs repeatedly through the following cycle: green for $ 30$ seconds, then yellow for $ 3$ seconds, and then red for $ 30$ seconds.  Leah picks a random three-second time interval to watch the light.  What is the probability that the color changes while she is watching?

$ \textbf{(A)}\ \frac {1}{63}\qquad \textbf{(B)}\ \frac {1}{21}\qquad \textbf{(C)}\ \frac {1}{10}\qquad \textbf{(D)}\ \frac {1}{7}\qquad \textbf{(E)}\ \frac {1}{3}$

This is a good place to use geometric probability, if you are becoming accustomed to doing these sorts of questions the "normal way". I like to make some solutions very visual, and others not so much.

LOL, I love  how my friend solved this problem using process of elimination. Here it is..

Since each cycle is 63 minutes, u must have an answer $\frac{n}{63}$, eliminate C.
A, B are way too small and E is too big, leaves us D

 :rotfl: , he bubbled in D because of that

[hide=What is wrong with this?]I viewed this cycle as a numberline with length 63. Leah can begin watching at [27,33] or at 60. But how can Leah start watching from after 60 seconds when it says Leah watches for 3 seconds? I thought Leah can only start watching from [0,60] and only [27,33] work + very small fraction of 60 seconds. So $\frac{33-27}{60}=\frac{1}{10}$.[/hide]

Why does this not give the same answer?

@create
The traffic light cycle doesn't stop after 63 seconds (it goes back to the green light and the cycle restarts). Therefore, Leah can watch any point on the interval (60, 63) as well and the entire interval length is 63.

[quote=junggi][hide]So Leah can start watching 3 seconds before the green-to-yellow change, or start watching right at the yellow-to-red change. She can also start watching any time during that interval, and still see a color change. That interval is 9 seconds long, while the entire cycle is 63 seconds long.
$P=\frac{9}{63}=\frac{1}{7}=D$
[/hide][/quote]

Why isn't the transition from red to green accounted for? I have the 9 seconds from green to yellow and yellow to red, but also an additional 6 seconds when it transitions from red to green. I feel like this explanation should make sense if you view the cycle as a circle with circumference of 63.

Because that's wrong. It's actually a 6-second interval at the green to yellow to red change, and a 3-second interval at the red to green change, in terms of the start times.

Sorry for the bump, but i got $5/21$
consider the interval $[0,63]$
for green to yellow, it can be $[27,30]$
for yellow to red, it can be $[33,36]$
for both green to yellow and yellow to red, it is $[30,33]$
for red to green, it is $[60, 63]$ and $[0,3]$
adding these up, we get $(3+3+3+3+3)/63=5/21$??

I'm not the best at probability so I'm not sure
@create also the length of the interval [27,33] is [b]7[/b] not 6 :P you have to do $33-27$[b]+$1$[/b]

[quote=AOPS12142015]@above, that's assuming the times are integers, but they need not be[/quote]

But that includes the end points of the range so integers can be part of the set.
It need not be integers but it can be


What is the balanced net ionic equation for [i]the reaction between ammonia and hydrogen chloride, both in the gas phase.
[/i]

In a net ionic equation, should I include electrons?  For instance, in the balanced equation:

8H+ + MnO4- + 5e- $ \rightarrow$ Mn2+ + 4H2O

Do I need the 5e-?

Thanks!

*Bump* I changed the question.  (I solved the one previously there.)

$ NH_3 (g) \plus{} HCl (g) \longrightarrow NH_4Cl(s)$.

In a net equation, there should not be electrons. The one you gave for MnO4- ---> Mn2+ is simply a redution half reaction, which is certainly accompanied by an oxidation half reaction in such a way that the number of electrons transferred in both is the same.


Predict the positions of the following aromatic compounds that are expected to be nitrated with the mixture $HNO_{3}/H_{2}SO_{4}$:

(1) p-Chlorotoluene
(2) Ethyl p-etoxibenzoate
(3) m-Difenoxibenzene
(4) p-tert-Butyltoluene
(5) Thioanisole
(6) Furan.

[hide="Answer"]
(1) Position ortho to methyl group.
(2) Position ortho to ethoxy group.
(3) Position ortho to fenoxi group, but not between both fenoxi groups.
(4) Position ortho to methyl group.
(5) Ortho and para positions.
[/hide]


vector
analytic geometry
advanced fields
advanced fields unsolved

Show that there is a smooth vector field on S^2 that vanishes at exactly one point.

...hmm...I am thinking about two ways

1) Let $ X: {\mathbb S}^2 \minus{} > T{\mathbb S}^2$ be such that $ X(p) \equal{} s_1 d/dx \plus{} s_2 d/dy$ where $ s_1, s_2$ are stereographic coordinates of $ p$ with respect to the chart $ {\mathbb S}^2 \minus{}
\{ N \}$, where $ N$ is the north pole.

....now, this is clearly zero at $ S$ (south pole)...however I am not sure this map is smooth. I think it's not smooth at infinity (north pole).

2) $ X(p) \equal{} d(N, p) d/dx$, here $ d(N ,p)$ is the distance between the north pole and a point $ p$. this is smooth (?) and zero at the north pole.

Is any of these approaches correct or should I do something else?

Thanks

[quote="HilbertThm90"]I think you are making it much harder than it needs to be. Construct it as a pushforward under stereographic charts of a constant vector field. That way you get smooth for free, and except at infinity it is non-vanishing. All you'll need to do is check that it actually vanishes at infinity. This is pretty interesting conceptually, since under the pushforward, the constant length vectors actually become shorter.

I'll leave the details out there unless you get really stuck.[/quote]

thanks a lot.

Could you give me a bit more details of how to get a vector field on $ {\mathbb S}^2$ as a pushforward?

I understand that if $ F: M \minus{}> N$ is a diffeomorphism then we get a smooth vector field on $ N$ as a pushforward of a smooth vector field on $ M$.

Are you saying that we construct smooth vector field on $ {\mathbb S}^2$ locally on each chart via a coordinate diffeomorphism? Will we get a smooth vector field on the whole $ {\mathbb S}^2$ then?

Also, by a constant vector field on $ {\mathbb R}^2$ you mean a field $ X(p) \equal{} d/dx \plus{} d/dy$?


limit
real analysis
real analysis unsolved

[b]Assume that   $\left(a_n\right)_{n=1}^{\infty}$  and  $\left(b_n\right)_{n=1}^{\infty}$  are real sequences such that  for some  $\lambda_0 >0 ,\lambda_1> 0 ,$  the limits
1)   $L_1: = \lim\limits_{n\to \infty}\left( \lambda_0 a_{n+1}+(1-\lambda_0) \frac{1}{n}\sum\limits_{k=1}^{n}a_k \right)$ ,
2)  $L_2: = \lim\limits_{n\to \infty}\left( \lambda_1\left(b_{n+1}\sqrt[n+1]{(n+1)!} - b_n\sqrt[n]{n!} \right)+(1-\lambda_1) b_n\frac{\displaystyle\sqrt[n]{n!}}{n} \right)$
exist.[/b] [color=green][b]   Prove that
\[ \begin{array}{ll} 1')\; \; \; \; & \displaystyle \; \lim\limits_{n\to \infty} a_n = \displaystyle\lim\limits_{n\to \infty} \frac{1}{n}\sum\limits_{k=1}^{n}a_k = L_1 \\ \\ 2') \; \; \; \; & \displaystyle \; \lim\limits_{n\to \infty} \left(b_{n+1}\sqrt[n+1]{(n+1)!} - b_{n}\sqrt[n]{n ! }\right) = L_2 \; \; \; \; \mbox{and}\; \; \; \lim\limits_{n\to \infty} b_n = e\cdot L_2 \\ \\ \end{array}\; .  \]
[/b][/color]
[size=75] [color=blue][b]Consequences of  a theorem by James Mercer (1893--1932) [/b][/color] [/size]

Denote by $l=\liminf x_n$ and $L=\limsup x_n$ taking a subsequence which tends to $l$ we get that $\lambda l+(1-\lambda)L\ge L_1$ similarly $\lambda L+(1-\lambda)l\le L_1$. Hence $\lambda(L-l)\le (1-\lambda)(L-l)$. Assume that $\lambda>\frac 1 2$ . Then, we will get $L-l\le 0$ so the sequence converges. And the problem became a simple exercise of Cesaro-Stolz theorem.

I don't know how to solve for $\lambda\le \frac 1 2$. Anybody else?

[quote="xirti"] ...... the problem became a simple exercise of Cesaro-Stolz theorem...... [/quote]
[color=blue][b]Please take care ! It's possible to apply Cesaro-Stolz ? Please justify ! [/b] [/color] My impression is that   Cesaro-Stolz is a consequence !

In my case, I proved that $x_n$ is convergent. Then I can apply Cesaro-Stoltz theorem for $\lim \frac{\sum x_n} n=\lim x_n=\ell$. So writing this in $1)$ we get $\lambda x_n+(1-\lambda)\frac{\sum x_n}{n}\to \ell$. So, the conclusion follows? I'm sure that it works. But only for $\lambda \ge \frac 1 2$


Select numbers $ a$ and $ b$ between $ 0$ and $ 1$ independently and at random, and let $ c$ be their sum. Let $ A, B$ and $ C$ be the results when $ a, b$ and $ c$, respectively, are rounded to the nearest integer. What is the probability that $ A \plus{} B \equal{} C$?

$ \textbf{(A)}\  \frac14 \qquad \textbf{(B)}\  \frac13 \qquad \textbf{(C)}\  \frac12 \qquad \textbf{(D)}\  \frac23 \qquad \textbf{(E)}\ \frac34$

I'm not sure if this works, as I did have to guess a bit. ;) :(

[hide="Possible Solution"]First, notice that there are 4 unique possibilities for $A$ and $B$: $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$. We can now examine the cases.

When $a$ and $b$ are less than $\frac 12$, it doesn't work because $A+B=0$ and $C=1$.

When one of $a$ or $b$ is less than $\frac 12$, then things do work. We get $1+0=1$.

When both $a$ and $b$ are greater than $\frac 12$, things sometimes work. for example, if we got $a=.6$ and $b=.7$, $A+B\ne C$, but if we got $a=.9$ and $b=.7$, we'd get $A+B=C$.

Thus, the answer is between $\frac 12$ and $\frac 34$, and should be $\text{(D)}$. [/hide]

[quote="scorpius119"]Not quite what I get...

[hide]Use areas to deal with this continuous probability problem. Set up a unit square with values of $a$ on x-axis and $b$ on y-axis.

If $a+b<1/2$ then this will work because $A=B=C=0$. Similarly if $a+b>3/2$ then this will work because in order for this to happen, $a$ and $b$ are each greater than $1/2$ making $A=B=1$, and $C=2$. Each of these triangles in the unit square has area of 1/8.

The only case left is when $C=1$. Then each of $A$ and $B$ must be 1 and 0, in any order. These cut off squares of area 1/2 from the upper left and lower right corners of the unit square.

Then the area producing the desired result is 3/4. Since the area of the unit square is 1, the probability is $\frac 34 \Rightarrow \boxed{E}$.[/hide][/quote]Yeah, that's right.  :)

[hide]Draw two graphs, ranging from 0 to 1.

The first graph, will represent $A+B$.  The x axis and y axis represent $a$ and $b$.  Split this 1 by 1 square (the coordinate plane) fourths.

The upper right quadrant should have $A+B=2$.  Its corners are at .5 and .5, which both round to 1 to make 1+1=2.

The upper left and lower right quadrants should have $A+B=1$.

The lower left quadrant should have $A+B=0$, because both $a$ and $b$ round to 0.

The second graph still has $a$ and $b$ as the values for the axis.  However, this time we are trying to partition the graph based on values of $C$, instead of $A+B$.
this one is split into eight squares.
By testing values, the stuff under the line connecting (0,.5) and (.5,0) should have $C=0$.  The area above the line connecting (.5,1) and (1,.5) should have $C=2$.  The area in between should have $C=1$.

Combe these two graphs by finding the areas where the $A+B$ from the first graph has the same value as $C$ in the second graph, which is what we want to find.

We see that the upper right corner has the upper right part shaded, $\frac{1}{8}$

The lower left corner has the lower left corner has the lower left part shaded, $\frac{1}{8}$

The upper left and lower right sections are both entirely shaded, $\frac{1}{2}$.

So the answer is $\frac{1}{8}+\frac{1}{8}+\frac{1}{2}=\frac{3}{4}$.  The answer is $\boxed{E}$

[/hide]

[hide]You could start with the sections that don't work. Clearly, if $\frac{1}{2} \le a+b \le 1$, then $C$ rounds to $1$. Now if both $a$ and $b$ are less than $\frac{1}{2}$, then both $A$ and $B$ round to $0$.

This area is bound by $x + y = \frac{1}{2}$, $x = \frac{1}{2}$, and $y = \frac{1}{2}$.

The same goes if $1 \le a+b \le \frac{3}{2}$, but both $a$ and $b$ are both greater than $\frac{1}{2}$.

This area is bound by again $x = \frac{1}{2}$, $y = \frac{1}{2}$, and $x + y = \frac{3}{2}$.

Both these areas have areas of $\frac{1}{8}$.

So the answer is $\frac{3}{4}$.[/hide]

[b]Solution:[/b] Casework! If both $a, b \le 0.5$, we must have that $a+b \le 0.5$. It is easy to see that the probability of this is $\frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}$. A similar logic can be applied if both $a, b \ge 0.5$ (so the probability is $\frac{1}{8}$ too). However, notice that if, WLOG, $a \ge 0.5$ and $b \le 0.5$, the condition is always satisfied, giving us a probability of $2 \cdot \frac{1}{4} = \frac{1}{2}$ (we multiply by $2$ since the variables can be switched and it is also possible that $b \ge 0.5$ and $a \le 0.5$). So, adding up all our separate probabilities, our answer is $\frac{1}{8} + \frac{1}{8} + \frac{1}{2} = \boxed{E}$. NOTE: The $\ge$ and $\le$ signs are used very loosely here as that has an infinitesimal impact on our final value's probability.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Prove that RXS,  R and S are rings, RXS is the Cartestian product of R and S, that is the set of all odered paris (r,s) with r in R adn s in S, is an intergral domain iff one of R or S is an integral domain and the other contains only a zero element.

this is a very basic result
I don't think it is good for you if you get it completely written out, and it is a lot of work for people here

Write down the conditions for a ring, check them, then tell us what you find difficult.

what are the conditions to prove something is a ring... the book i am using is very unclear about it...
do i have to prove all of the following...

zero element of R is unique
each element of R has a  unique negative
left and right cancellation law
a+x=b and x+a=b have unique soltuions
-(-a)=a and -(a+b) = (-a) + (-b)
and distribute property?

please help...

for the definition of ring : wikipedia usually has the answer http://en.wikipedia.org/wiki/Ring_%28mathematics%29#Formal_definition
now your question is weird too, you ask : prove R , S $R\times S$ are all rings
now don't you mean, if R and S are rings, so is $R\times S$?


pl solve q no 6,9, 10 of this assignment

Rather than just telling us to solve three problems, why don't you show some of your own work and show us where you are having problems.

The second and third problems in particular just require you to follow a specific technique.


geometry
geometric transformation
rotation
linear algebra
matrix
function
linear algebra unsolved

Let $A_k=\left(\begin{array}{cc}cos(k2pi/n)&-sin(k*2*pi/n)\\sin(k2pi/n)&cos(k2pi/n)\end{array}\right).$ where k,n in N
 $S_p=(A_0)^p+(A_1)^p+...+(A(n-1))^p= ?$.

notice that there are all rotations, and that is you know the matrix of a rotation $R$, it is easy to compute $R^k$...

cute ... the matrices are all the n-th roots of one so this can be viewed as a question about complex numbers.
so express $S_p$ (which is symmetric) as a function of the elementary symmetric functions of the n-th roots of 1 (and these are mostly zero.


function
inequalities
calculus
calculus computations

Hello yet again!

Okay, this time I'm expanding a function using Taylor's formula, and I think I'm messing up the remainder portion (Lagrange form).

Here's what I've got (about $a=0$):

$f(x) = (1-x)^{\frac{-1}{2}}$

$f'(x) = \frac{1}{2}(1-x)^{\frac{-3}{2}}$

etc.

$f^{(n)}(x) = \frac{1 \cdot 3 \cdot \ldots \cdot (2n-1)}{2^n}(1-x)^{\frac{-(2n+1)}{2}}$

Plugging it all in:

$(1-x)^{\frac{-1}{2}} = 1 + \frac{x}{2} + \frac{1 \cdot 3}{2 \cdot 4}x^2 + \ldots + \frac{1 \cdot 3 \cdot \ldots \cdot (2n-1)}{2 \cdot 4 \cdot \ldots \cdot 2n}x^n + R_n(x)$

Now, for Lagrange's remainder, I get the following:

$R_n(x) = \frac{1 \cdot 3 \cdot \ldots \cdot (2n+1)}{(n+1)! \cdot 2^{n+1}}(1-X)^{\frac{-(2n+3)}{2}}x^{n+1}$

where $0\leq X \leq x$ or $x\leq X \leq 0$

Assuming I've not mucked it up at this point, I'm to show the following:

a) $\left|R_n\right|<\frac{1 \cdot 3 \cdot \ldots \cdot (2n+1)}{2 \cdot 4 \cdot \ldots \cdot (2n+2)}\left|x\right|^{n+1}$ for $-1<x<0$

b) $\left|R_n\right|<\frac{1 \cdot 3 \cdot \ldots \cdot (2n+1)}{2^{n+1} \cdot n!}\frac{x^{n+1}}{(1-x)^\frac{3}{2}}$ for $0<x<1$

For part a), I can see this works by letting, say, $x=-0.5$ and $n=4$, for example.

But for part b), the inequality doesn't seem to work.

I can't figure out where I went wrong with the remainder.

I'd greatly appreciate your help.

Best wishes,
Bailey

You may need to use a different form of the remainder for that side; notice the missing factor of $n+1$, among other things.


[quote="bodan"]Suppose $a,b,c$ are sidelengths of a triangle and $m_{a},m_{b},m_{c}$ are its medians. Prove the inequality \[\frac{m_{a}}{a^{2}}+\frac{m_{b}}{b^{2}}+\frac{m_{c}}{c^{2}}\geq \frac{\sqrt{3}(a^{2}+b^{2}+c^{2})}{2abc}.\] [/quote]

$m_{a}^{2}=\frac{b^{2}+c^{2}}{2}-\frac{a^{2}}{4}$
After transfer we have:
$\sum_{cyc}\ a^{4}b^{4}(2a^{2}+2b^{2}-c^{2})\geq \frac{(a^{2}+b^{2}+c^{2})^{2}(abc)^{2}}{3}$

Now here i have problems!
If put $a=x+y, b=x+z, c=z+y$  :roll:

[quote="gemath"]Fitim if you want to square both $LHS$ and $RHS$ you shouldn't forget
$2\sum\frac{m_{b}m_{c}}{b^{2}c^{2}}$ of $LHS$[/quote]
You're right!!! But you knows solutions?

[quote="bodan"]Suppose $a,b,c$ are sidelengths of a triangle and $m_{a},m_{b},m_{c}$ are its medians. Prove the inequality \[\frac{m_{a}}{a^{2}}+\frac{m_{b}}{b^{2}}+\frac{m_{c}}{c^{2}}\geq \frac{\sqrt{3}(a^{2}+b^{2}+c^{2})}{2abc}.\][/quote]
$\sqrt{3a^2(2b^2+2c^2-a^2)}\leq\frac{3a^2+2b^2+2c^2-a^2}{2}=a^2+b^2+c^2$.
Hence, $\sum_{cyc}\frac{m_{a}}{a^{2}}\geq\sum_{cyc}\frac{\sqrt3(2b^2+2c^2-a^2)}{2a(a^2+b^2+c^2)}$.
Thus, it remains to prove that $\sum_{cyc}bc(2b^2+2c^2-a^2)\geq(a^2+b^2+c^2)^2$, which after
substitution $a=y+z$, $b=x+z$ and $c=x+y$ gives $\sum_{cyc}xy(x-y)^2\geq0$. Done! :)


let a,b,c be positive numbers. such that abc=1
prove that $ \sqrt{\frac{a}{a\plus{}b}}\plus{}\sqrt{\frac{b}{b\plus{}c}}\plus{}\sqrt{\frac{c}{c\plus{}a}}\leq \frac{3}{\sqrt{2}}$

[quote="lian109"]let a,b,c be positive numbers. such that abc=1
prove that $ \sqrt {\frac {a}{a \plus{} b}} \plus{} \sqrt {\frac {b}{b \plus{} c}} \plus{} \sqrt {\frac {c}{c \plus{} a}}\leq \frac {3}{\sqrt {2}}$[/quote]
It's true for all positive $ a,$ $ b$ and $ c.$ See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17549


Let the particles horizontal velocity be $v$. On some approximately linear tiny segment of the hill, the horizontal component of the normal force is $m g \sin{\theta}\cos{\theta}$, where $\theta$ is the elevation of the hill. Substituting:

$\sin{\theta}= \frac{-dy}{\sqrt{{dx}^{2}+{dy}^{2}}}$
$\cos{\theta}= \frac{dx}{\sqrt{{dx}^{2}+{dy}^{2}}}$

And assuming there is no retarding force:

\begin{eqnarray*}\frac{d^{2}x}{dt^{2}}&=&-g  \frac{\frac{dy}{dx}}{\left(\frac{dy}{dx}\right)^{2}+1}\nonumber \\ & = & \frac{2 g x e^{-x^{2}}}{1+4 x^{2}e^{-2 x^{2}}}\nonumber \end{eqnarray*}

... And that's as far as I can get. If the retarding force is proportional to the velocity, then add a $-k v$ to the RHS.

Yes, I meant the retarding force is $-k \vec v$.

[quote="PTynan89"]Say we have a particle of mass $m$, constrained to move along a hill in the shape of the function $y = e^{-x^{2}}$.

At time $t = 0$, the particle starts from rest at the point $(0, 1)$.

Gravity points in the $-y$ direction.

Now, a retarding force $-k \vector v$ acts on the particle as it travels.

How far does the particle travel? ($x$ value and/or arc length distance traveled)[/quote]

PTynan,

Considering the constraint $y = e^{-x^{2}}$

If you then write the Lagrangian, taking care not to flog any previously decreased equine, as:

$L = \frac12 m\dot{x}^{2}+\frac12 m\dot{y}^{2}-mg y+\lambda\left(y-e^{-x^{2}}\right)$

then the resulting equations of motion are

$m\ddot{x}-\lambda\left(2x e^{-x^{2}}\right) =-k\dot{x}$

$m\ddot{y}+mg-\lambda =-k\dot{y}$

(One is free to collect any forces not due to a potential gradient on the right hand side of the equals sign, e.g. $-k\vec{v}.$)

Rearranging these equations some what,

$\ddot{x}+\frac{k}{m}\dot{x}-\frac{\lambda}{m}\left(2x e^{-x^{2}}\right) = 0$

$\ddot{y}+\frac{k}{m}\dot{y}= \frac{\lambda-mg}{m}$

If we then multiply the x-EOM by $\dot{x},$ we obtain:

$\frac{1}{2}\frac{d \dot{x}^{2}}{d t}+\frac{k}{m}\dot{x}^{2}-\frac{\lambda}{m}\dot{y}= 0$

The y-EOM is a first order DE in $\dot{y}$ that is easily solved:

$\dot{y}= \frac{mg-\lambda}{k}\left[e^{-\frac{k}{m}t}-1\right]$

Substituting this into the x-EOM gives:

$\dot{x}^{2}= \frac{\lambda\left(\lambda-mg\right)}{k^{2}}\left(e^{-\frac{2k}{m}t}-2 e^{-\frac{k}{m}t}+1\right)$

I'm sure you can do the rest.


Hi,
following problem:

vector  r(t)=(vector)af(t)+(vector)bg(t)

which relation has to exist between f(t) and g(t) , that r(t) x d/dt r(t) = 0

can somebody help me please ?

$\frac{d \vec r}{dt}= \vec a f'(t)+\vec b g'(t)$.

Therefore, $\vec r (t) \times \frac{d \vec r (t)}{dt}= (\vec a f(t)+\vec b g(t)) \times (\vec a f'(t)+\vec b g'(t))$
$= (\vec a \times \vec b)(f(t)g'(t)-f'(t)g(t)) = 0$.

Therefore, either $\vec a \times \vec b = 0$, implying that either $\vec a$ and $\vec b$ are parallel, or at least one of them is $0$, in which case the use of both vectors would be unnecessary to begin with; or $f(t)g'(t)-f'(t)g(t)) = 0$.

From the latter case, $\frac{f'(t)}{f(t)}= \frac{g'(t)}{g(t)}$.
Thus, upon integrating, $\ln f(t) = \ln g(t)+C$.
Therefore, $f(t) = k g(t)$, meaning that $f(t)$ and $g(t)$ are [i]not[/i] linearly independent.

In either case, this implies that both vectors ($\vec r(t)$ and $\vec r'(t)$) have constant [u]direction[/u], although the magnitude may change (or become negative, representing a possible reversal, but still along the same line).


Let $n$ be a positive integer. Consider the
alphabet ${1,2,..,n}$ with the natural order on letters. Denote by $b_{n}$ the number of words in this alphabet which
satisfy the following condition: Between any two occurrences of the same letter there should be an occurence
of a  smaller letter and an occurence of a bigger latter (for example, the word $2132$ satisfies this condition
since we have $1<2$ and $3>2$ occuring between the two $2$'s, whereas the word $2432$ does not satisfy  this
condition as there is no letter smaller than $2$ occurring between the two $2$'s). It is very easy to show  that $b_{n}$
is finite. Question: find a closed  formula for b_n.

So: $B_{0}= \{\emptyset\}$, $B_{1}= \{1, \emptyset\}$, $B_{2}= \{\emptyset, 1, 2, 12, 21\}$, $B_{3}= \{\emptyset, 1, 2, 3, 12, 21, 13, 31, 23, 32, 123, 132, 213, 231, 312, 321, 2132, 2312\}$ thus $b_{0}= 1, b_{1}= 2, b_{2}= 5, b_{3}= 18$.  Also, I believe $b_{4}= 115$.  The only sequence in the OEIS which matches is [url=http://www.research.att.com/~njas/sequences/A125625]A125625[/url], defined there as the number of elements of the semigroup of type $K_{n}$; I don't know what this means, but it could be counting the same thing, in which case a closed form seems highly unlikely.  (Indeed, a closed form for this problem also seems unlikely -- I think just an asymptotic would be interesting.)

Sorry to bump this up, but I wanted to add that the next term is $ b_5 \equal{} 1710$, making it almost certainly the same as that sequence from the OEIS.


Mike has five boxes. The first contains two squares and eight triangles, the second three squares and two triangles, the third three squares and four triangles, the fourth four squares and three triangles and the fifth five squares and four triangles. The side lengths of all the squares and triangles in all the boxes are the same. Mike wishes to make a few polyhedra by glueing together squares and triangles along their edges. Using all the pieces from one box for each polyhedron, for how many boxes can he succeed?

[hide]would f-e+v = 2 (Euler) be useful?

The number of edges in the polyhedron will be the total number of edges in the raw materials over two.

Then count the number of faces (given). Use f-e+v=2, then:[/hide]

1

[hide]See if the total number of edges will be integral.[/hide]

2

[hide]See if v is positive.[/hide]

Then

[hide]Using the above procedure, this will be true only for the boxes 1,2,3, and 5, thus the answer is 4.[/hide]

Could you please explain the part 1 and 2?

I don't quite understand the phrase that "See if the total number of edges will be integral"  and "See if v is positive".

The total nmber of edges in the polyhedron will be half the total number of edges on the raw materials. Suppose that in 4, there are 4 squares and 3 triangle, thus e = [4(4) + 3(3)]/2 = 12.5, not integral. So Setup 4 is invalid.


Let N be the number of positive integers that are less than or equal to 2003, and whose base-2 representation has more 1's than 0's. Find the remainder when N is divided by 1000.

I figured for example, I could find all integers whose base-2 representation were less than 11 digits.

So 1 _ _ _ _ _ _ _ _ _ for integers n, where 512  <=  n < 1024.

And so on.

For that example. I just had 2^9 - C(9,6) - C(9,7) - C(9,8) - C(9,9). To exclude numbers whose arrangements of 1's and 0's had 6 or more 0's. And so on. But it led me to a wrong answer... A quicker solution anyone?

whitehorseking88, what about the numbers less than 512 and between 1024 and 2003?

[hide]well, the casework isnt so bad...i divided pascals triangle in half and added up the first 11 lines of one side (the last one starts with $1,10,45...$) including the numbers on the divide.  this yields 1199, which is for all numbers up to and excluding $2^{11}$ however, this is greater than 2003.  any number greater than 11111000000 (1984) will be wrongly included if it is also greater than 2003.  so we subtract the numbers from 2004-2047.  1199-44=1155
155 is the answer.

though i do admit there is enormous potential for arithmetical error[/hide]

Well, I just gave that as an example.

I said so on as to say I did include all those cases in my "casing" which obviously had many arithmetic errors or just case errors in general. Though thanks for the help. Much appreciated.

This may or may not lead to a slightly faster solution:
[hide]
Like alan did, just consider 1-2048 and then subtract 44.

If the number starts with 1 and then has an odd number of digits following it, exactly half of them fit the condition (since for any good case, swapping the 0s and 1s results in a bad case.)

So for this case, we have simply $\frac{2^1+2^3+2^5+2^7+2^9}/{2}$.

If the number starts with 1 and then has an even number of digits, again we can take half of them but then we must adjust the value for the case where those even digits are split exactly half 0s and half 1s (because this works both ways whether we swap 0s and 1s or not.)

So we have (grouping the powers of 2 together) $\frac{(2^0+2^2+2^4+2^6+2^8) + (1 + 2C1 + 4C2 + 6C3 + 8C4 + 10C5)}{2}$

Final total = $\frac{2^0 + 2^1 + 2^2 \ldots + 2^10 + 351}{2} - 44 = \frac{2^11 - 1 + 351}{2} - 44 = 1155$.

Now to try to get you to understand what I wrote about the halves thing with some examples:
1 _ _ _
exactly half work: 111 (not 000), 110 (not 001), 101 (not 010) etc.

1 _ _ _ _
If we again split them into halves:
1111 not 0000, 1110 not 0001, ...
However we are counting 1100 not 0011, 1010 not 0101 etc. We want to reinclude these values we aren't including, which is $\frac{4C2}{2}$.
[/hide]

I remember doing this question on the AIME a few years back, so I looked at Kalva's solution, which is essentially the same as TripleM's.  It's quite a nice approach, IMHO.


On Kedlaya's Inequality notes, it is printed 2 pages on a single page. However, since I couldnt get the pdf for his geometry notes, I used LaTex to generate pdf, which just had 1 single page on a page. Since that was 83 pages (yikes), how to i make it like his inequality notes??? Is there any specific command for LaTex?

The best way I know to print 2 pages on 1 is within your printer.  When you select "print", look around in the propeties.  You should have options like "landscape" and "pages by sheet".

This is the way I do it.

Ledurt

On windows---Go to my computer>control panel>printers and faxes>go2pdf or adobe printer. right click and select set printer preferences(also found on the sidebar). On the main page, just press on 2 in multiple pages and press ok. then you are set!


how many re.al solut.ions

$ x^{3}= 2y-1$
$ y^{3}= 2z-1$
$ z^{3}= 2x-1$

x=y=z=1 is obvious,but how to show that is the only one (if it is)

Let $ f(x)=\frac{x^{3}+1}{2}$. Equation $ f(x)=x$ had 3 solutions $ x_{0}=1>x_{1}=\frac{\sqrt 5-1}{2}>\frac{1}{2}>-1.5>x_{2}=-\frac{1+\sqrt 5 }{2}>-2.$
It is easy to sheck, that if $ x>x_{0}$, then $ x_{0}<x<y<z<x$ - contradition.
If $ x_{1}<x<x_{0}$, then $ x_{0}>x>y>z>x>x_{1}$ - contradition.
If $ x_{2}<x<x_{1}$, then $ x_{2}<x<y<z<x<x_{1}$ - contradition.
If $ x<x_{2}$, then $ x_{2}>x>y>z>x$ - contradition.
Therefore only 3 solutions $ x=y=z=x_{i}, i=0,1,2$.

[quote="anonymous1173"][quote="Virgil Nicula"][color=darkred][b]Extension.[/b] For any $ \{X,Y\}\subset\mathbb R$ define the following equivalence relation ("[b]S[/b]ame [b]S[/b]ign") : $ X\ .s.s.\ Y\Longleftrightarrow XY>0\ \vee\ X=Y=0$.

Consider two strict increasing functions $ f,g: \mathbb R\rightarrow\mathbb R$. Prove that the following chain of equivalencies :

$ (x-y)\ .s.s.\ (y-z)\ .s.s.\ (z-x)\Longleftrightarrow$ $ \left\|\begin{array}{c}f(x)=g(y)\\\ f(y)=g(z)\\\ f(z)=g(x)\end{array}\right\|$ $ \Longleftrightarrow x=y=z$.[/color][/quote]
i can't understand the last part with 2 "if and only if" signs.the first one can be "and" i think.[/quote]
[color=darkblue]You are right ! Correctly is $ \boxed{\left\|\begin{array}{c}f(x)=g(y)\\\ f(y)=g(z)\\\ f(z)=g(x)\end{array}\right\|\implies (x-y)\ .s.s.\ (y-z)\ .s.s.\ (z-x)\Longleftrightarrow x=y=z}$. Indeed, $ \left\|\begin{array}{c}f(x)=g(y)\\\ f(y)=g(z)\\\ f(z)=g(x)\end{array}\right\|$ $ \implies$

$ \boxed{x-y}\ .s.s.\ f(x)-f(y)=$ $ g(y)-g(z)\ .s.s.\ \boxed{y-z}\ .s.s.\ f(y)-f(z)=$ $ g(z)-g(x)\ .s.s.\ \boxed{z-x}$ $ \implies$ $ (x-y)\ .s.s.\ (y-z)\ .s.s.\ (z-x)$.

$ \left\{\begin{array}{c}(x-y)+(y-z)+(z-x)=0\\\ (x-y)\ .s.s.\ (y-z)\ .s.s.\ (z-x)\end{array}\right\|$ $ \implies$ $ x=y=z$. Reciprocally is evidently.[/color]


What is the sum of all positive integer cubes that are less than 100?

Also note that $ 1 \plus{} 2 \plus{} 3 \plus{} \dots \plus{} n$ is just $ \frac{n(n\plus{}1)}{2}$.


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real analysis
real analysis unsolved

Partition the real line $ \mathbb{R}$ into a countable union of compact subsets.

Let $ E_{0}=\{0\}$ and $ E_{n}=\{x\in\mathbb R\colon 1/n \le \mathrm{dist}(x,E_{n-1})\le n\}$ for $ n\ge 1$.

Actually, such a partition is not possible. :)

[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=385579269&t=25908[/url]

All I did was define a bunch of compact sets... I never said they covered $ \mathbb R$.  :)  :blush:

A natural question: how small can we make $ \mathbb R\setminus\bigcup E_{n}$? It must be uncountable, of course. But it seems that it can have Hausdorff dimension $ 0$.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=62703]here[/url] for another proof; the link at the end is to the thread grobber already linked to.
The remaining set is large in the Baire sense; it can certainly be small with respect to measures.