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The circles $ C_{1}$ and $ C_{2}$ touch externally at $ M$ and the radius of $ C_{2}$ is larger than that of $ C_{1}$. $ A$ is any point on $ C_{2}$ which does not lie on the line joining the centers of the circles. $ B$ and $ C$ are points on $ C_{1}$ such that $ AB$ and $ AC$ are tangent to $ C_{1}$. The lines $ BM$, $ CM$ intersect $ C_{2}$ again at $ E$, $ F$ respectively. $ D$ is the intersection of the tangent at $ A$ and the line $ EF$. Show that the locus of $ D$ as $ A$ varies is a straight line.

And that line is the perpendicular to the line of the centers from $M$. $EF$ is the image of $BC$ through the homothety of center $M$ turning $C_1$ into $C_2$, which means that the pole of $EF$ wrt $C_2$ is the image of $A$ through this homothety. Call this point $A'$. Since $A,M,A'$ are collinear, we get that $AA'$ passes through $M$, so the intersection of the polars of $A,A'$ wrt $C_2$ must lie on a line, which is the polar of $M$ wrt $C_2$ (duality principle), and this line is the tangnt from $M$ to $C_2$.

Okay, I have the same proof but in a different form and a bit more detailed, so I'm posting it...

In fact, the locus of the point D is the common tangent of the circles $C_1$ and $C_2$ at the point M. In order to prove this, I will show that the point D always lies on the common tangent of the circles $C_1$ and $C_2$ at the point M.

Here is how I prove it (sorry, the proof uses pole-polar relationship): Let m be the common tangent of the circles $C_1$ and $C_2$ at the point M. Then we have to show that the point D lies on the line m.

The polar of the point A with respect to the circle $C_1$ is the line BC, since the points B and C are the points where the tangents from A to $C_1$ touch $C_1$. Now, since the line MA passes through the point A, the pole of the line MA with respect to the circle $C_1$ must lie on the polar of the point A with respect to $C_1$. Hence we have obtained the fact that the pole of the line MA with respect to the circle $C_1$ lies on the line BC.

Now, since M is the point of tangency of the circles $C_1$ and $C_2$, there exists a homothety h with center M mapping the circle $C_1$ to the circle $C_2$. This homothety h must take the points B and C to the points E and F, respectively (since the points E and F lie on the circle $C_2$ and on the lines BM and CM, respectively). Hence, this homothety h takes the line BC to the line EF. Also, this homothety h leaves the line MA invariant (since this line passes through the center M of the homothety). Finally, polar relation is clearly invariant under homotheties. Hence, from the fact that the pole of the line MA with respect to the circle $C_1$ lies on the line BC, we can derive using our homothety h that the pole of the line MA with respect to the circle $C_2$ lies on the line EF. But the pole of the line MA with respect to the circle $C_2$ is the point of intersection of the tangents to $C_2$ at the points M and A. The tangent to $C_2$ at M is the line m that we have met before. Hence, we see that the point of intersection of the line m with the tangent to $C_2$ at A lies on the line EF. In other words, the point of intersection D of the tangent to $C_2$ at A with the line EF must lie on the line m. And this is exactly what we wanted to prove.

  Darij

The condition:...the radius of  $C_2$ is larger than that of $C_1$...
It is not necessary.
$AFME$ performed in this way is harmonic quadrilateral, also: $AE.MF=AF.ME$
From the another view point: $AB,AC$ cut at $J,K$ to circle $C_2$, we know:
$MB$ and $MC$ are external bisector of $AMJ$ and $AMK$ triangles.
Here the proof:
The line $AM$ cut at $L$ to circle $C_1$.
Draw the line $T$ common tangent at $M$ for two circles $C_1$ and $C_2$.
The tangent line $T$ cut at $H,I$ to $AB,AC$ ,then:
$<HMJ=<MAJ=\gamma$,$<LMB=<MAB+<MBA=\gamma+\delta$,
$<MBA=<MBH=<HMB=\delta$, so $<BMJ=<BML=\gamma+\delta$ then $MB$ is external bisector of $AMJ$triangle.
$<KMI=<KAM=\alpha$,$<CML=<CAM+<ACM=\alpha+\beta$,
$<ACM=<ICM=<IMC=\beta$, so $MC$ is external bisector of $AMK$ triangle.
The another angular relations in the figure.
It is not hard prove that the tangent line $T$ at $M$ and the tangent at $A$ with $EF$ are concur at $D$.
Therefore the locus is the tangent line $T$ at $M$.

Let $O_{1},\ O_{2}$ be centers and $r_{1},\ r_{2}$ radii of the circles $\mathcal C_{1},\ \mathcal C_{1}$ touching at M. These 2 circles can touch either externally or internally. Only in the case of internal tangency we need $r_{2}> r_{1},$ otherwise the tangents to $\mathcal C_{1}$ from $A \in \mathcal C_{2}$ would not exist. Assuming external tangency, M is the internal similarity center of the circles $\mathcal C_{1},\ \mathcal C_{2},$ hence $\mathcal C_{1}\sim \mathcal C_{2}$ are centrally similar with similarity center M and similarity coefficient $-\frac{r_{1}}{r_{2}}.$ The triangles $\triangle MBC \sim \triangle MEF$ are centrally similar with the same similarity center and coefficient, hence their corresponding sides $BC \parallel EF$ are parallel. Since BC is a polar of A with respect to $\mathcal C_{1},$ $BC \perp AO_{1}$ and consequently, $EF \perp AO_{1}$ as well. Let AM meet $\mathcal C_{1}$ at Z. Since AB is a tangent of $\mathcal C_{1}$ at B, $\angle ABM = \angle MZB = \angle MAE.$ Using the sine theorem for the triangles $\triangle ABM,\ \triangle AEM$ then yields

$\frac{EA}{EM}= \frac{\sin \widehat{AME}}{\sin \widehat{MAE}}= \frac{\sin \widehat{AMB}}{\sin \widehat{ABM}}= \frac{AB}{AM},\ \ \ EA^{2}= EM^{2}\cdot \frac{AB^{2}}{AM^{2}}$

Because of the central similarity of $\mathcal C_{1},\ \mathcal C_{2}$ with center M, $AZ = AM+MZ = AM \left(1+\frac{r_{1}}{r_{2}}\right)$ and $EB = EM+MB = EM \left(1+\frac{r_{1}}{r_{2}}\right).$ Power of A to $\mathcal C_{1}$ is then $AB^{2}= AM \cdot AZ = AM^{2}\left(1+\frac{r_{1}}{r_{2}}\right).$ Substituting this to the above equation yields $EA^{2}= EM \cdot EB,$ which means that E lies on the radical axis of the point A and the circle $\mathcal C_{1}.$ Since $EF \perp AO_{1},$ EF is their radical axis. On the other hand, the tangent $t_{A}$ of $\mathcal C_{2}$ at A is the radical axis of the point A and the circle $\mathcal C_{2}.$ The radical axes $EF,\ t_{A}$ meet at the radical center D of the point A and the circles $\mathcal C_{1},\ \mathcal C_{2},$ which lies on the radical axis $t_{M}$ of the circles $\mathcal C_{1},\ \mathcal C_{2},$ their single common internal tangent at M.

And I could have made this simpler, too:

... Since AB is a tangent of $\mathcal C_{1}$ at B, $\angle ABM = \angle MZB = \angle MAE.$ The triangles $\triangle ABE \sim \triangle MAE$ with a common angle at the vertex E are then similar, having equal angles, hence

$\frac{EA}{AB}= \frac{EM}{AM},\ \ \ EA^{2}= EM^{2}\cdot \frac{AB^{2}}{AM^{2}}$

etc.

[color=darkblue][b]Very nice both the proposed problem and the [u]Yetti proof ![/u][/b]

[b]Lemma.[/b] If the circles $c\ ,\ c'$ are tangent in the point $T$ and are given
the points $\{A,B\}\subset c$ , $\{A',B'\}\subset c'$ so that $T\in AA'\cap BB'\ ,$ then $AB\parallel A'B'\ .$

[b][u]Another proof of the proposed problem[/u] (similarly with the Yetti's).[/b]
Denote the second intersection $L$ of the line $AM$ with the circle $C_{1}\ .$ Therefore,
$\begin{array}{ccccccc}BL\parallel AE & \Longrightarrow & \widehat{ABE}\equiv\widehat{BLA}\equiv\widehat{EAM}& \Longrightarrow & EAM\sim EBA & \Longrightarrow & EA^{2}=EM\cdot EB\\\ CL\parallel AF & \Longrightarrow & \widehat{ACF}\equiv\widehat{CLA}\equiv\widehat{FAM}& \Longrightarrow & FAM\sim FCA & \Longrightarrow & FA^{2}=FM\cdot FC\end{array}\|$
$\implies$ the line $EF$ is the radical axis between the circle $C_{1}$ and the point $A$ (null circle).

$D\in EF\cap AA$ and $AA$ is the radical axis of the circles $C_{2}$ and $A$ $\implies$ $D\in MM\ .$[/color]

Thank you very much. Very nice solution mr Virgil .
I remember I friend Nick Rapanos solved it by homothety ,but I don't remember the full solution .Could anyone find such one ?

We denote as $L,$ the intersection point of the circle $(O_{1}),$ from the segment line $AM.$ Also we denote as $D,$ $H,$ the intersection point of the common tangent line at $M$ $($ radical axis $),$ of $(O_{1}),$ $(O_{2}),$ at points $A,$ $L,$ respectively.

By applying the polar theory, because of the segment line $LM,$ as the polar of $H,$ with respect to the circle $(O_{1}),$ passes through the point $A,$ we have that the segment line $BC,$ as the polar of $A,$ with respect also to $(O_{1}),$ passes through the point $H.$

$($ We can also prove this result, without polar theory. If we denote as $K,$ $N,$ the intersection points of the segment line $LH,$ from the tangent lines $AB,$ $AC$ respectively, we have the configuration of the triangle $\bigtriangleup AKN,$ taken the circle $(O_{1})$ as it’s incircle and so, based on a well known $($ at least to me $)$ Lemma, we have that the segment lines $KN,$ $BC$ and the tangent line of $(O_{1})$ at $M,$ are concurrent. An elementary proof can be found, by Newton’s, Menelaus’s and Ceva’s theorems $).$

We will prove now, that the segment line $EF,$ passes through the point $D.$ From similar isosceles triangles $\bigtriangleup DAM\sim \bigtriangleup HLM,$ $\Longrightarrow$ $\frac{MD}{MH}= \frac{MA}{ML}= \frac{R_{2}}{R_{1}}$ $,(1)$ $($ because of $\bigtriangleup O_{1}LM\sim \bigtriangleup O_{2}AM$ $).$

It is easy to prove that $BC\parallel EF$ $($ from $\angle CBM = \angle CMH = \angle FMD = \angle FEM$ $).$

We denote as $D',$ the intersection point of $EF,$ $HM.$ From similar triangles $\bigtriangleup CMH\sim \bigtriangleup FMD'$ $\Longrightarrow$ $\frac{MD'}{MH}= \frac{MF}{MC}= \frac{R_{2}}{R_{1}}$ $,(2)$ $($ because of $\bigtriangleup O_{1}MC\sim \bigtriangleup O_{2}MF$ $).$

From $(1),$ $(2)$ $\Longrightarrow$ $MD' = MD$ $\Longrightarrow$ $D'\equiv D$ $,(3)$

From $(3),$ we conclude that the concurrency point of the segment line $EF,$ with the tangent line of $(O_{1}),$ at point $A,$ lies on the radical axis of $(O_{1}),$ $(O_{2})$ and the proof is completed.

Kostas Vittas.

Let me restate the problem so that I can fit my solution:
"The circles $ (C_1), (C_2)$ externally touches at $ A$ ($ (C_2)>(C_1)$). Let $ B$ be an any point on $ (C_2)$. From $ B$, let $ BM,BN$ be the tangents wrt $ (C_1)$. $ AM$, $ AN$, respectively, intersects $ (C_2)$ at $ P,Q$. $ PQ$ intersects the tangents from $ B$ of $ (C_2)$ by the point $ K$. Prove that $ K$ blongs to a fixed line."
[i]Proof:[/i]
Consider the homothety with center $ A$, ratio $ \frac {\minus{}r_1}{r_2}\equal{}k$,  we get:
$ H(A,k): P\mapsto M$,  $ Q\mapsto N$.  Hence,  $ H(A,k): PQ\mapsto MN$.   Also, the tagents from  $ B$  also maps to the tangents from  $ C$. Let  $ H$  be the intersection of  $ MN$  to the tangent from  $ C$. It is easy to see that  $ H(A,k): H\mapsto K$.  Now, we need to prove that  $ H$  belongs to the fixed line. But this is obviously true because the fact that  $ AMCN$  is the harmonic quadrilateral, therefore,  $ H$  belongs the the common internal tangent of $ (C_1),(C_2)$. The homothety center  $ A$, ratio  $ k$, which turns the tangent at  $ A$  of  $ (C_1),(C_2)$  into itself. Therefore, K is also a member of the common internal tangent of  $ (C_1),(C_2)$,  which is fixed.
Our proof is completed :D
p/s: Sorry if my solution is the same to anyone's :lol:

Another solution using properties of symmedian:
Easy to show that $LO_{1} // AO_{2}$ and $CO_{1} // FO_{2}$ and $LC // FA$, also $LB // AE$, therefore $BC//EF$.
$\Delta BLC \sim \Delta EAF$. Since $LC // AF$, point $K$ in $\Delta BLC$ corresponds to point $H$ in $\Delta EAF$.
As $AB$ and $AC$ are two tangents to the circle $C_{1}$, $AL$, resp. $KL$ is the symmedian in $\Delta BLC$, so $AH$ is the symmedian in $\Delta EAF$.
Furthermore, $HE$, resp. $DE$ is the symmedian in $\Delta MEA$, so $D$ must be the intersection of the tangents at $A$ and $M$ to the circle $C_{2}$.
With the solution, other interesting facts can be seen, e.g. the two tangents at $E$ and $F$ to the circle $C_{2}$ must intersect at a point on line $AL$.

Hello, can anyone please check the above proof?

[quote=Ankoganit]Sorry to revive an old thread, but can someone please tell me if the following proof is correct or not? :maybe:

Suppose line $AM$ meets $C_1$ again at $G$. Evidently, $MCGB$ is a harmonic quadrilateral, and so $M(MCGB)$ is a harmonic pencil. Intersecting this with $C_2$, we conclude that $MFAE$ is a harmonic quadrilateral. If the tangent to $C_2$ from $D$ touches $C_2$ at $D'$, then $D'FAE$ is harmonic as well. Thus $D'\equiv M\implies D\in $ the common tangent of $C_1$ and $C_2$, which is a straight line.

Please let me know if it's wrong, since it appears to be too short to be true. :help:[/quote]

It seems okay. Also, you should note that for a specific $\triangle ABC$ inscribed in $(O)$ (unrelated to our problem) there are three distinct points on the circle, each completing $\triangle ABC$ to a harmonic quadrilateral. What makes $D' \equiv M$ in your case is that they are both on the same arc determined by $E$ and $F$.

@bonciocatciprian Thank you for your comments.  :)  :clap:

Indeed,the locus of $D$ is the internally common tangent of $C_1$ and $C_2$ except $M$.

Let $AM \cap C_1=J$. Since BC is the polar of A w.r.t C1 the lines BM,BJ,BC,BA form a harmonic pencil $\implies (M,J;C,B)=-1$.Seeing this from M we get that MM,MJ,MC,MB form a harmonic pencil.Projecting this onto C2 we get $(M,A;F,E)=-1$.Thus AM,AD,AF,AE form a harmonic pencil $\implies AM$ is the polar of $D$ w.r.t C2.Therefore $D$ lies on the polar of M w.r.t C2 (duality principle) which is a straight line.

Symmedians!

Let $AM$ intersect $C_2$ again at $A'$. Then, since $A'A$ is a symmedian, we have $$-1=(M,A';B,C)\stackrel{M}{=}(M,A;E,F)$$
Thus quadrilateral $MEAF$ is harmonic. Since $D$ lies on the intersection of tangent of $A$, and the $EF$, $DM$ must be tangent to $C_1$, which means it's also tangent to $C_2$. As, $A$ varies, $D$ traces the radical axis of $C_1$ and $C_2$, which is a line, as needed.

It is sufficient to prove $MD$ is tangent.

Let $N=AM\cap (MBC)$. Since $AB, AC$ is tangent so $MBNC$ is a harmonic quadrilateral. A homothety centered at $M$ takes $MBNC$ to $MEAF$. So $MEAF$ is also harmonic quadrilateral. Meaning $MD$ is tangent because $AD$ is tangent.


[quote="joml88"]You've been posted a lot on rate problems.  This sounds a lot like homework  :alien: .  Set up the equations with $d=rt$ and try going from there.

If you show some work that you tried, then it would be easier for people to help you.[/quote]

The point is that I don't know where to start. That's why I'm asking for help, and if you can't or don't want to help me, no need to give me attitude.
Anyways, I do know the formula s=d/t, but how can you do it if you only have the time for each of them...?

lol, I didn't mean to offend you.   I understand having trouble with problems.  I'll show you how to do this one and then hopefully you can make a little progress on the other problems.

So Frances can go some distance (let's call it $d$) in 1.5 hours.  Assuming she always goes at the same rate (call it $r_f$) we have:

\[d=1.5r_f.\]

So $r_f=\frac 23d.$

For Enrique we let his pace be $r_e$ and we get:

\[d=r_e.\]

So now we want to find how long it takes for them to meet if they start a distance of $d$ away from each other.  Now, here's the tricky part.  We consider the [i]rate at which they come closer[/i].  Since they are coming toward each other, we can say that the rate at which they are getting closer is equal to $r_f+r_e$.  Using $d=rt$ we can express the desired time as $t=\frac dr$ where the rate that we want is $r_f+r_e$.  Now, we can express this in terms of $d$:

\[r_f+r_e=\frac 23 d+d=\frac 53 d.\]

So, $t=\frac{d}{\frac 53 d}=\frac 35 \ \text{hours}.$  So our answer is 36 minutes.

Think of the meeting as a job.

Frances can finish the job in 1.5 hrs; i.e, his rate is 2/3.
Iglesias ( :P ) in 1 hr; his rate is 1.

x is the time for them to meet

2/3 * x + x = 1

x = 3/5 hr = 36 minutes.

[quote="Ani"]Frances can walk a certain distance in 1.5 hours, and Enrique can walk the same distance in 1 hour. if they started walking toward each other at the same time, from opposite ends of the route, how long would it take them to meet? (give answer in hours and minutes, and assume all speeds are constant.[/quote]Here's where you might start. Several references are made to the length of the route, so make a name for this length, such as $d$. From the conditions of the problem, $r_1\cdot1.5=d$ and $r_2\cdot1=d$, where $r_1,r_2$ are the two guys' speeds and the $1.5$ and $1$ are the times needed for each to travel the distance $d$. Now you can say $r_1=\frac{d}{1.5}=\frac{2}{3}d$ and $r_2=\frac{d}{1}=d$.

At the instant that they meet you can say $r_1t+r_2t=d$, where $t$ is the time it took for them to meet. This is because $r_1t+r_2t$ is the distance the first guy travelled plus the distance the second guy travelled, which is equal to the starting distance between them (that should be easy to visualize). So \[r_1t+r_2t=d\Longrightarrow\frac{2}{3}dt+dt=d\Longrightarrow\left(\frac{2}{3}+1\right)t=1\Longrightarrow{t}=\frac{3}{5}.\]


we could do something like a march madness bracket for who will win nationals. i think that would be pretty awesome. something like 1 point for every person you guess right making cd. 3 points for every person you guess right for their place

nationals is unpredictable. Probably half the people that make cd will be people you've never heard of.

it doesn't matter, it's just for fun.

I'm gonna go for Justin Ahmann and Bobby shen making cd  :D

I predict that at least 12 people will make the countdown round. :maybe:
And also that at least 228 people will make the top 228. :maybe:
Oh yeah, and that I will be somewhere in the top 228 also. :maybe:

I have a feeling Justin Ahmann won't do as well as last year...cd's gonna be really competitive...

[quote="Ihatepie"]
/me predicts that he will get in the top 50.[/quote]

Wow, that's very nice of you!

Oh wait, that means more pressure on me... uh-oh... :|

Unless you're not talking about me.

I'm talking about myself, (and you kind of)

usually when you say /me, it means your name, so I was saying,

Ihatepie thinks that he will get in the top 50.

Judging from how your doing in the forums, you'll probably get some where there also.


quadratics
integration
function
real analysis
real analysis unsolved

I'm wondering what the solution is for h(t) to this differential equation \[ \frac{d^2h}{dt^2} = \frac{a}{(h+b)^2}  \]
(a and b are constants)
I've always seen the simple quadratic model of an object under gravity, but I'd like to know what it would be taking into account the fact that gravity lessens as an object goes farther from Earth.

Begin with the substitution
 $v = \frac{dh}{dt}, \ \frac{dv}{dt} = \frac{dv}{dh} \frac{dh}{dt} \\ = v \frac{dv}{dh}$

Good luck!

mathisfun's suggestion leads to a very general solution, with some physical insight behind it. Our particle is subject to a force $F(h)$ that depends only on position; calc rulz was making that the inverse-square gravitational force. (You might want a negative sign in there somewhere.)

So $m\frac{d^2h}{dt^2}=F(h).$

Employing mathisfun's suggestion gives us

$mv\frac{dv}{dh}=F(h).$ Integrating that:

$\frac12mv^2=\int F(h)\,dh +C$

And right there is the interplay of kinetic energy and potential energy.

In one dimension, use energy- potential energy in this model is $\frac{-a}{h+b}$. That gets you speed as a function of height, and you can integrate the reciprocal to get time. In two dimensions, look up an explanation of Kepler's laws. Everything is easier if you put the origin at the center of gravitational attraction

Physical insights can be useful on strictly mathematical problems.


Find all $ f: \mathbb{R}\to\mathbb{R}$ such that for all $x,y \in \mathbb R$,
\[ f(x\plus{}y) \equal{} f(x)f(y) \plus{} \gamma\sin{(x)}\sin{(y)}\]
where $ \gamma$ is a real number.

[quote="reason"]hi!!
find all $ f: \mathbb{R}\to\mathbb{R}$ such that:

   $ f(x \plus{} y) \equal{} f(x)f(y) \plus{} \gamma\sin{x}\sin{y}$

  with $ \gamma\in{\mathbb{R}}$.

thanx.[/quote]

Let $ P(x,y)$ be the assertion $ f(x \plus{} y) \equal{} f(x)f(y) \plus{} a\sin(x)\sin(y)$

Let $ f(\frac {\pi}2) \equal{} b$

1) $ a \equal{} 0$
========
We get $ f(x \plus{} y) \equal{} f(x)f(y)$, classical equation with two solutions :
1.1) $ f(x) \equal{} 0$ $ \forall x$
1.2) $ f(x) \equal{} e^{h(x)}$ where $ h(x)$ is any solution of Cauchy equation $ h(x \plus{} y) \equal{} h(x) \plus{} h(y)$

2) $ a\ne 0$
===========
Then $ f(x)$ is not equal to zero for every $ x$ and so exists $ w$ such that $ f(w)\ne 0$
$ P(w,0)$ $ \implies$ $ f(w) \equal{} f(w)f(0)$ and so $ f(0) \equal{} 1$

$ P(\frac {\pi}2,\frac {\pi}2)$ $ \implies$ $ f(\pi) \equal{} b^2 \plus{} a$

$ P(\pi,\frac {\pi}2)$ $ \implies$ $ f(\frac {3\pi}2) \equal{} b(b^2 \plus{} a)$

$ P(\frac {3\pi}2,\frac {\pi}2)$ $ \implies$ $ f(2\pi) \equal{} b^2(b^2 \plus{} a) \minus{} a$

But $ P(\pi,\pi)$ $ \implies$ $ f(2\pi) \equal{} f(\pi)^2$ and so $ b^2(b^2 \plus{} a) \minus{} a \equal{} (b^2 \plus{} a)^2$ and so $ a(a \plus{} b^2 \plus{} 1) \equal{} 0$ and so $ a \plus{} b^2 \plus{} 1 \equal{} 0$

This implies $ a\le \minus{} 1$ and so $ 1\ge \minus{} \frac 1a > 0$ and $ \exists x_0$ such that $ \sin(x_0)^2 \equal{} \minus{} \frac 1a$

$ P(x_0, \minus{} x_0)$ $ \implies$ $ 1 \equal{} f(x_0)f( \minus{} x_0) \minus{} a\sin(x_0)^2$ and so $ f(x_0)f( \minus{} x_0) \equal{} 0$ and so the equation $ f(x) \equal{} 0$ has real roots.

Let then $ u$ any root of $ f(x) \equal{} 0$ :

$ P(u, \minus{} u)$ $ \implies$ $ 1 \equal{} \minus{} a\sin(u)^2$
$ P(x \minus{} u,u)$ $ \implies$ $ f(x) \equal{} a\sin(x \minus{} u)\sin(u) \equal{} \minus{} \frac {\sin(x \minus{} u)}{\sin(u)}$

And it's easy to check back that this indeed is a solution.

3) Synthesis of solutions :
==========================
If $ a \equal{} 0$ : $ f(x) \equal{} 0$ $ \forall x$ or $ f(x) \equal{} e^{h(x)}$ where $ h(x)$ is any solution of Cauchy equation $ h(x \plus{} y) \equal{} h(x) \plus{} h(y)$

If $ a > \minus{} 1$ and $ a\ne 0$ : No solution

If $ a\le \minus{} 1$ : For each $ u$ such that $ \sin(u)^2 \equal{} \minus{} \frac 1a$, we get a solution $ f(x) \equal{} \minus{} \frac {\sin(x \minus{} u)}{\sin(u)}$


Are you allowed to take both the AMC 10 and the 12? I know sirorange did, but I'm not sure if they were both official.

Yes, you are allowed to.

Yes, they are both going to be official if you were eligible for both(mainly the AMC 10)


sorry for all these questions...but one more

at constant temperature, the behavior of a sample of a real gas more closely approximates that of an ideal gas as its volume is increased because the

a. collisions with the walls of teh container become less frequent

b. average molecular speed decreases

c. molecules have expanded

d. average distance between molecules becomes greater

e. average molecular kinetic energy decreases

thanks

An ideal gas is one in that the molecules are assumed to be infinitely small particles with no intermolecular forces that move randomly. The closest choice to that is C.

I think you probably mean (D) and not (C).
 (C) sounds a bit of rather active imagination doesn't it?

Oops, you're right.  :blush:
Somehow I mixed up the choices.


Express $ \frac {719}{770}$ as a sum of fractions of the form $ \frac 1 x$.

[i]Hint: Try to figure out a strategy of expressing a fraction into the given form first.[/i]

[quote="DonkeyKong"][quote="abacadaea"]take the recipricols of both sides 770/719[/quote]

How did you see that?!! You are smart... :lol:

But that doesn't lead us to the answer.. does it????

Because then x= 553630....[/quote]
ummm
$ 719/770\equal{}1/x$ multiply both sides by 770x to get rid of the fractions
$ 719x\equal{}770$
$ x\equal{}770/719$
in short we ARE taking the recipricols but be careful because you cant always do this

[quote="abacadaea"][quote="DonkeyKong"][quote="abacadaea"]take the recipricols of both sides 770/719[/quote]

How did you see that?!! You are smart... :lol:

But that doesn't lead us to the answer.. does it????

Because then x= 553630....[/quote]
ummm
$ 719/770 \equal{} 1/x$ multiply both sides by 770x to get rid of the fractions
$ 719x \equal{} 770$
$ x \equal{} 770/719$
in short we ARE taking the recipricols but be careful because you cant always do this[/quote]
Oh..! I see.. Thanks

Is this possible for any fraction $ \frac{p}{q}$ ?

[quote="Fraenkel"]Is this possible for any fraction $ \frac {p}{q}$ ?[/quote]

I think so..

The website in the following link will provide the answer to the question in the last second post above:

http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fractions/egyptian.html#intro

[quote="luimichael"]Maybe I didn't express the question clearly.
Try to express the given fraction as $ \frac 1 a \plus{} \frac 1 b \plus{} \frac 1 c \plus{} ...$.

And the given hint is not clear also...
So you may ignore the hint.  Sorry to tell you about that... :oops:

[b]Hint(revised):[/b]
[hide]You may assume that the given expression can be expressed as a sum of fractions of the required form.
Then apply the greedy algorithm:
Note that 719/770 = 0.9337..... , the fraction of the form 1/x  which is just smaller that 0.9337..... is 1/2.
Now, subtract 719/770 by 1/2 and get 167/385.
Next, apply the same idea to 167/385:
167 /385- 1/3 = 116/ 1155;
Then apply the same idea again to to 116/1155:
116/1155 - 1/11 = 1/105.
Then add up these equations... and result follows.[/hide][/quote]

I'm not sure if that qualifies as a hint if it essentially gives us the answer...

how do you add a horrifically long harmonic series?

The harmonic series is unbounded above. As we know the result in advance, we don't really need to add the harmonic series.

I guess you are talking about how to evaluate the sum of many fractions?
The answer is.... using a computer or a calculator, or else it would be a great pain to add up so many fractions.  :)


algebra polynomial trigonometry function

Polynomial $P$ is such that for all real x we have $P(sin x) + P(cos x) = 1$. What can the degree of this polynomial be?

So... anybody wanna struggle with this one?

[quote="t0rajir0u"][quote="lotrgreengrapes7926"][quote="harazi"]I don't understand something: am I the only idiot who thinks that some of the problems posted in the begginer corner are quite tough? As far as I remember, I met some very tough people who couldn't solve this problem. I think I have an explanation: we're getting old and idiot.  :(  :D[/quote]
No, you are not the only one. I can only get about one half of them.

[hide="1 solution"]
$P(x)=x^2$ because we know the Pythagorean identity
$\sin^2{\theta}+\cos^2{\theta}=1$ :D [/hide][/quote]

I think he's trying to say that this problem shouldn't belong here because it is too hard.   ;)[/quote]

Oh. :blush:

[quote="Hokkage"][hide="Possibly?"]Well, $\sin x = \sqrt{1- \cos ^2 x}$ so $P(x)+P(\sqrt{1-x^2})=1$. So the polynomial has only even powers (or else the $\sqrt{1-x^2}$ term will have a fractional exponent and can't cancel with the other terms). For degree 0, $P(x)=.5$. For degree 2, the solution $P(x)=x^2$ works. For degree 4 or higher, say $2n$ for $n \ge 2$, there can't be a solution because the term $\sqrt{1-x^2}$ will give a positive $x^{2n}$ to be added to another positive $x^{2n}$, and thus because the term isn't negative, the terms don't cancel, so there are no other solutions. So degree 0 or 2.[/hide][/quote]

Not true.  Consider $n$ odd ;)

[hide="Solution"]
[b]Lemma:[/b]  $P(x)$ is an even polynomial.

Proof:  $P(\sin x) + P(\cos x) = 1$
$P(\sin (-x)) + P(\cos (-x)) = 1$
$P(- \sin x) + P(\cos x) = 1$
$P(\sin x) = P(- \sin x)$, or, phrased in another way,
$P(y) = P(-y) \, \forall y \in [-1, 1]$

So $P(x)$ must be an even function, and have no odd terms.  :)

Let $P(x) = Q(x^2)$ for some polynomial $Q(x)$.  The problem condition becomes

$Q(\sin^2 x) + Q(1 - \sin^2 x) = 1 \forall x \in \mathbb{R}$

Let us substitute $u = \sin^2 x - \frac{1}{2}$, and we can write

$Q(u + \frac{1}{2}) + Q(-u + \frac{1}{2}) = 1 \forall u \in [-\frac{1}{2}, \frac{1}{2}]$

Finally, let us substitute $R(x) = Q(x + \frac{1}{2}) - \frac{1}{2}$.  The problem becomes

$R(x) + R(-x) = 0$

And so $R(x)$ can be any odd polynomial or $R(x) = 0$.  Plugging all the way back in, $P(x)$ can have degree $0, 2(2k-1), k \in \mathbb{N}$.  :)
[/hide]


$a$ and $b$ are integers such that $2^na + b$ is a square for all non-negative integers $n$. Show that $a = 0$.

I know it was posted before (more than once I suppose) but I was unable to find it. I'll be thankful if anybody can find it (or possibly solving would be quicker  :) )


Hey y'all,

I am sorry if this question has already been posted. But if a 12th grader and a 11th grader score the same on the USAMO, are 11th graders given preference? And can 12th graders make it to blue MOSP only if they are awarded Honorable Mention??

A 12th grader has to win (top 12) or at least get within a tiebreaker of winning in order to qualify, except in rare cases when enough winners decline that they start pulling more in. I've never seen MOP qualification rely on tiebreakers - a score cutoff is usually set and everyone who gets that makes it.


algebra
polynomial
function
inequalities
quadratics
quadratic formula
algebra unsolved

Let $P(x) = a_nx^n + \ldots + a_1x + a_0$ be a polynomial with real coeficients,
where $n = deg P$ is even. Suppose that
[b](i)[/b] $a_0 > 0, a_n > 0$.
[b](ii)[/b] ${a_1}^2+...+{a_{n-1}}^2\le\frac{4min({a_0}^2,{a_n}^2)}{n-1}$.
Prove that $P(x) \ge0$ for all real $x$.

Oh, this problem is very hard. Has anyone any solutions ?

[quote="draperp"]I believe you when you say this is hard to prove (especially since it's false).

Consider  P(x) when n=2, a0=1, a1=-3, and a2=1.
All constraints are satisfied.

P(1)=1(1^2)-3(1)+1=1-3+1=-1

Therefore, the attempted statement is not always true.

(Am I right in assuming that min function returns the least of the listed values?)[/quote]
Oh , you are right . I'm really sorry  :blush:
The second condition is :
[b](ii)[/b] ${a_1}^2+...+{a_{n-1}}^2\le\frac{4min({a_0}^2,{a_n}^2)}{n-1}$.

Wow, you weren't kidding when you said this is hard. All I have managed to do is to prove that for the first two values of n (n=0 and n=2).

When n=0:
P(x)=a0
P(x)>0 for all real x since a0>0.

When n=1:
From (ii) we get:
a1^2<=4(a0^2)
a1^2<=4(a2^2)
Both sides of eqach inequality are positive, so we take the sqroot(|x|) (which is a strictly increasing function for its postive domain).
|a1|<=2(a0)
|a1|<=2(a2)
By multiplying the left sides (two lesser positive values) and the right sides (two greater positive values).
(a1^2)<=4(a0)(a2)

We shall use the descriminant of the quadratic formula to determine if there are any real zeros of P.
If it is greater than 0, there are multiple real zeros.
Iff there are multiple real zeros:
(a1^2)-4(a0)(a2)>0
(a1^2)>4(a0)(a2)
This directly contradicts our earlier inequality, so there are not two real zeros.

*Since P(x) is continuous and approaching +infinity on both the left and the right, and by the Intermediate Value Theorem, if P(x)<0 for some value(s) of x then P(x)=0 for multiple values of x. This latter part is not true, so
P(x)>=0.
*(I know this is more complicated than absolutely necessary (I could have appealed to intuition), but I wanted to do it right.)

(sorry I didn't use an equation editor)

Maybe there's a general pattern to determinants applicilable to this situation?

0th degree descriminant: a0?
1st degree descriminant: a1?
2nd degree descriminant: (a1^2)-4(a0)(a2)
3rd degree descriminant: I have it, but I'm not going to type it.
4th degree descriminant: ditto

Hooray Mathword!

[url=http://mathworld.wolfram.com/PolynomialDiscriminant.html]http://mathworld.wolfram.com/PolynomialDiscriminant.html[/url]

(This just seemed to make it much more complicated.  Maybe it's a step in the right direction?)

Well, the descriminant thing is useless, but here's (the first part of) how to prove the desired statement (the following proof is not valid for n=0, but that is a small matter):

By using the Property of Distribution of Multiplication/Addition,
(a_1+a_2+a_3...a_(n-1))^2 can be expanded to a polynomial comprising terms of form (a_I)(a_J) where I=1,2,3...n-1 and J=1,2,3...n-1.  There are (n-1)^2 of these terms.

Let a_M be value of the coefficient with the largest absolute value.  Then,
(a_1+a_2+a_3...a_(n-1))^2 <= ((n-1)^2)*(a_M^2)
which means
((a_1+a_2+a_3...a_(n-1))^2)/(n-1) <= (n-1)*(a_M^2)

Using a_M, the left side of (ii) can be written as (n-1)*(a_M^2).
By using our proven inequality and (ii) we can show that:
((a_1+a_2+a_3...a_(n-1))^2)/(n-1) <= 4min(a_0^2,a_n^2)/(n-1)
(a_1+a_2+a_3...a_(n-1))^2 <= 4min(a_0^2,a_n^2)

This can be split into two inequalites:
(a_1+a_2+a_3...a_(n-1))^2 <= 4(a_0^2)
(a_1+a_2+a_3...a_(n-1))^2 <= 4(a_n^2)

From this we get:
a_1+a_2+a_3...a_(n-1) >= -2(a_0)
a_1+a_2+a_3...a_(n-1) >= -2(a_n)

Continue from there (I'll put this in an equation editor when I have time).

I made a mistake.
I said "the left side of (ii) can be written as (n-1)*(a_M^2)," but that is false since (n-1)*(a_M^2) is greater or equal to the left side, so it can't be substituted in.
Sorry.

(I have now been trying to obtain the least possibility of P and then show that P(x)>=0 for all real x.)


Does $N$ denote the set of natural numbers? If that is true, then how can the third condition be valid?


superior algebra
superior algebra unsolved

This is a question related to Gröbner bases. Consider an admissible ordering $ <$ on the set $ [X]$ of monomials in some variables over some ring (i.e. and ordering such that $ 1 < t$ for all $ t$ and $ a < b \Longrightarrow at < bt$ for all $ a,t,b$). Now I want to show that there are no infinite decreasing sequences. How can I do that? I've constructed a proof using Hilberts basis theorem (I suppose there is an infinite sequence $ x_1 < x_2 < \dots$. I define $ A_k$ to be the ideal generated by $ x_1, \dots , x_k$. Then it follows from Hilberts that $ A_n$ for some $ n$ must be a basis for the union of all $ A_k$. Then I can look at $ x_{n\plus{}1}$, deduce that it's a linear combination of higher terms, which is a contradiction). But it feels rather complicated, I was under the impression that there was a much simpler proof, but I haven't been able to find it.

The proofs I have seen in literature also use Hilbert Basis Thm (or Dickson's lemma).


I was shock that I counldn´t solve the 1st easy geometry probelm which most people have done this question well, how abt all of ya? how many exercises have u done? :?:  I saw Chinese guy even has done an hour prior to the time, must b really nice of him!!! :maybe:

Please give the texts of the problems in English, Russian or Ukrainian.

Is this a place where all of us must post rumours about results?Or there is another place to do it.
That is our results:
1.+,+,-
2.+,-,-
3.+,+,-
4.+,+/-,-
5.+,+/-,-
6.+,-,-.
P.S:
I think it's gonna be a problem for us to know other results,as internet is placed far from our location,and most of us are tired and will not write up here.
I only can see some members of Romania team and some others unknown to me teams here...
Best regards...


geometry
incenter
circumcircle
trigonometry
geometry proposed

[quote="outback"]Let $ I$ and $ G$ be the incenter and centroid of triangle $ ABC$. Let $ r_A, r_B, r_C$ be the circumradius of triangles $ IBC, ICA$, and $ IAB$, respectively; let $ R_A, R_B, R_C$ be the circumradius of triangles $ GBC, GCA$, and $ GAB$. Prove that $ r_A \plus{} r_B \plus{} r_C \geq R_A \plus{} R_B \plus{} R_C$.[/quote]
Your post probably made a typo mistake. $ r_A \plus{} r_B \plus{} r_C \le R_A \plus{} R_B \plus{} R_C$, and I'd like to have you edit it.
$ R_A\equal{}\frac{4}{3}R\frac{m_bm_c}{bc}, r_A\equal{}2R\sin\frac{A}{2}$
$ \sum R_A\equal{}\frac{4}{3}R\sum\frac{m_bm_c}{bc}\ge \frac{4}{3}R\frac{9}{4}\equal{}3R\ge 2R\sum\sin\frac{A}{2}\equal{}\sum r_A$


geometry
circumcircle
symmetry
power of a point
radical axis
geometry unsolved

A convex quadrilateral $ABCD$ with $AC \neq BD$ is inscribed in a circle with center $O$. Let $E$ be the intersection of diagonals $AC$ and $BD$. If $P$ is a point inside $ABCD$ such that $\angle PAB+\angle PCB=\angle PBC+\angle PDC=90^\circ$, prove that $O$, $P$ and $E$ are collinear.

Let $\Sigma, \Sigma_{1}, \Sigma_{2}$ be respectively the circumcircles of $ABCD, \triangle PBD, \triangle PAC$. Let also $C_{1}, C_{2}$ be the centres of $\Sigma_{1}, \Sigma_{2}$ respectively.

Consider the radical axes of each two of them. Since the radical axis of $\Sigma, \Sigma_{1}$ is $BD$ and that of $\Sigma, \Sigma_{2}$ is $AC$, the radical axis of $\Sigma_{1}, \Sigma_{2}$ must be concurrent with $AC$ and $BD$, i.e. it passes through $E$. Therefore, the radical axis of $\Sigma_{1}, \Sigma_{2}$ is $PE$.

To prove $O$, $P$, $E$ are collinear, it suffices to prove $O$ has the same power with respect to $\Sigma_{1}$ and $\Sigma_{2}$.

Without loss of generality we may let $\angle A$ and $\angle B$ to be not smaller than a right angle. Then point $P$ lies in $\triangle CDE$ and it is outside $\Sigma_{1}$ and $\Sigma_{2}$. By some angle chasing we can prove that $\angle ODC_{1}=90^\circ$, and so the power of $O$ w.r.t $\Sigma_{1}$ is equal to $OC_{1}^{2}-CC_{1}^{2}=OC^{2}$. Similarly, the power of $O$ w.r.t. $\Sigma_{2}$ is also the square of the circumradius of $ABCD$. The result follows.

Consider inversion with center $P$ and any power. Then see that our hypothesis implies that $A^{*}B^{*}C^{*}D^{*}$ ($X^{*}$ denotes point $X$ after inversion) is a rectangle. Let $O'$ be center of circumcircle of $A^{*}B^{*}C^{*}D^{*}$. Point $E^{*}$ is the intersection of circumcircles of $PBD$ and $PCA$ and hence lies on line $PO'$ (as $O'$ is the intersection of $A^{*}C^{*}$ and $B^{*}D^{*}$). But well-known fact says that $P,O',O$ are collinear and hence $P,O',O^{*}$ are collinear, so $P,E^{*},O^{*}$ are collinear and hence $P,E,O$ are collinear.  :)  QED

Let $A', B', C', D'$ be the intersection point of the circle and $AP, BP, CP, DP$ respectively. By angle tracing, $A'B'C'D'$ is a rectangle. Therefore $O$ is the intersection point of $A'C'$ and $B'D'$. Use central projection to map $P$ to the centre of the circle. Then by symmetry, $P, E, O$ are collinear.

Diagram:
hope that helps a little (It helped me...)

Another solution:
Let $AP$ cut the circle at $F$, $BP$ cut the circle at $H$.
$\angle PAB + \angle PCB = \angle PCF = 90^{\circ}$,so $PC$ must cut $OF$ at a point $G$ on the circle.
Similarly, $DP$ must cut $OH$ at $K$ on the circle.
Let $GB$ cut $AK$ at $X$.
By Pascal theorem in $ACGBDK: E = AC \cap BD, P = CG \cap DK, X = AK \cap GB$ are collinear.
By Pascal theorem in $GFAKHB: O = GF \cap KH, P = FA \cap HB, X = GB \cap AK$ are collinear.
So $E, P, X, O$ are collinear

If $AP,BP,CP,DP$ cut to $\odot (ABCD)$ at $A',B',C',D'$ respectively. Consider $\mathcal{G}$ the composition of involutions with poles $P,E,P$ that fixed the conic $\odot (ABCD)$ is well-known that is a involution with pole on $EP$. Since $\mathcal{G}(A')=C'$ and $\mathcal{G}(B')=D'  \Rightarrow A'C'$ and $B'D'$ are secants at the pole of $\mathcal{G}$. Using the condition  $\angle PAB+\angle PCB=\angle PBC+\angle PDC=90^\circ$ we get $A'C'$ and $B'D'$, and are secants at $O$ so $E,P,O$ are collinear.

This was in Korean Postal Coaching..
Let $O_1$ be the pole of $AC$ with respect to $O$.
Draw a circle - with $O_1$ as its center and $O_1A=O_1C$ as its radius.
Then we have $2\angle APC = 2(90+\angle ADC)=180+\angle AOC = 360-\angle AO_1C$, so $P$ lies on the circle $O_1$.
Similarly, $P$ lies on the circle $O_2$. I claim that $O, P, E$ lie on the radical axis of $O_1, O_2$.
First, $O$ lies on the radical axis since $P_{O_1}(O)=OA^2=OB^2=P_{O_2}(O)$.
Also, $E$ lies on the radical axis since $AE \cdot EC = BE \cdot ED$. Therefore, we are done. $\blacksquare$


in the given figure $\frac{m_{1}}{m_{2}}=\nabla$ .The block $m_{2}$ is left from rest and it moves upwards find the maximum height reached by the block $m_{2}$.sorry i could not make the image better .both the pulleys are connected by the same rope that joins $m_{2}$.You can view thsi if u click on the image and expand it .

Just to clarify, the object of mass $m_{1}$ is a height $h$ above the ground, and the other one is initially on the ground. Neglect the masses of the pulleys as well as friction. Also, let's not get nambla involved here. Rather, let $\eta = \frac{m_{1}}{m_{2}}$. Note that $\eta>2$ when the lighter object rises.

[hide="Click to Reveal Hidden Solution"]
The tension $T$ in the rope is the same throughout. So a force of $2T$ is pulling up on $m_{1}$, and a force of $T$ is pulling up on $m_{2}$. The downward acceleration of $m_{1}$ is:
$a_{1}= g-\frac{2T}{m_{1}}$.
And the upward acceleration of $m_{2}$ is:
$a_{2}= \frac{T}{m_{2}}-g$.
Because moving the pulley holding $m_{1}$ a certain distance requires extending the rope on either side of the pulley, $a_{1}= 2 a_{2}$. Solving for $T$:
$T = \frac{3 m_{1}m_{2}g}{m_{1}+4 m_{2}}$.
Now solving for $a_{2}$:
$a_{2}= \left(\frac{2 m_{1}-4 m_{2}}{m_{1}+4 m_{2}}\right) g$.
The mass is accelerated until it reaches a height $2h$, at which point its kinetic energy carries it up an additional height $h_{m}$, where $h_{m}= \frac{2 a_{2}h}{g}$. So maximum height $H$ reached by block $m_{2}$ is:
$H = \left(\frac{6 \eta}{\eta+4}\right) h$.
[/hide]


induction
superior algebra
superior algebra solved

Prove that the sequence given by x_0=0
and
x_(n+1)=
sqrt(2+x_n), if n even
(6+x_n)^(1/3), if n odd.

cheers!

Well,  what do we have to prove Lagrangia  :?

Sorry for that! :D

a)converges;
b)compute it's limit! :D

. By induction x_n <= 2.

. Not difficult to see that sqrt(2+x_n) >= x_n, and
(6+x_n)^(1/3) >= x_n or x_n <= 2.

So x_n is upper bounded and increasing so it is convergent.

f(x) = sqrt(2+x). The only solution to f(f(x)) = x for 0<=x<=2 is x = 2 so x_n converges towards 2.


3)IN what triangle do we have: bsinB+csinC=a/sinA?

we know that sinA=a/2R and the other simmilar relations and the given relation becomes:

b^2+c^2=4R^2 so this takes place only in a right angled triangle! :D :D

cheers!

A little addon to the other solution!
We know that a=2RsinA and sin^2B=(1-cos2B)/2 and then we have that cos(B+C)*cos(B-C)=0.
So the triangle is  right angled in A or a triangle in which we have: B-C=90.
now the solution is complete! :D :D
cheers! :D :D


probability
combinatorics unsolved
combinatorics

In a graph with n nodes, each two nodes are connected by an edge with probability p. Let u and v be two distinct nodes. What is the probability that there is a path from u to v?

I think...
[hide]
$1-\prod_{i=0}^{n-2}i!\cdot{n-2\choose i}\cdot\frac{1}{2^{i+1}}$
which of course simplifies
[/hide]

Your answer will be negative for most if not all $n$ and does not contain $p$.  Also, full solutions are nice.


If $M$ is an infinite set, prove that $|M\times M|=|M|$. I know this is well well known, but I couldn't find a solution yet.

This is well-known, but not trivial.
It is in fact equivalent to the [url=http://en.wikipedia.org/wiki/Axiom_of_choice]Axiom of choice[/url] (AC).
The proof I know is an easy application of [url=http://en.wikipedia.org/wiki/Zorn%27s_lemma]Zorn's lemma[/url], which is also equivalent to AC.
It also uses $|\mathbb{N}|\leq |M|$ and $|N|\leq |M|\Rightarrow |M\cup N|=|M|$ for all infinite sets $M,N$,
which both need AC to prove.
Zorn's lemma is similar to induction in a way, and just as powerful.


Let $\, a$, and $b \,$ be odd positive integers.  Define the sequence $\{f_n\}_{n\ge 1}$ by putting $\, f_1 = a,$ $f_2 = b, \,$ and by letting $\, f_n \,$ for $\, n \geq 3 \,$ be the greatest odd divisor of $\, f_{n-1} + f_{n-2}$.  Show that $\, f_n \,$ is constant for sufficiently large $\, n \,$  and determine the eventual value as a function of $\, a \,$ and $\, b$.

$ f_{n \plus{} 2}\leq \max\{f_{n},f_{n \plus{} 1}\}$
So $ \{f_{n}\}$ is bound .
Exist $ k > l$ such that $ f_{l \plus{} 1} \equal{} f_{k \plus{} 1},f_l \equal{} f_k$
So $ f_{n \plus{} k \minus{} l} \equal{} f(n)$ for n large enough.

Are you sure it's a $ \ge$ there?
Also, you have to prove it becomes constant, while I think you only proved it is periodical.

$ f_{n \plus{} 2}\leq \max\{f_{n},f_{n \plus{} 1}\}$
So $ f(n)$ is bound.
If $ f(n),f(n \plus{} 1)$ is diferent then $ f(n \plus{} 2)\leq max\{f(n),f(n \plus{} 1)\} \minus{} 1$
$ f(n) > 0,\forall n$
So when n large enough we must has $ f(n) \equal{} f(n \plus{} 2) \equal{} f(n \plus{} 1)$
It mean that for n large enough then it constant.

Let $ gcd(a,b)\equal{}d*2^k$, were d is odd, then $ d|f_n$ and for $ n>2log_2(max(\frac{a}{d},\frac{b}{d}))$ we get $ f_n\equal{}d.$

Step 1:Show $ f_n$ is constant for sufficiently large $ n$(see post $ 4$)
Step 2:It is easy enough to show $ (f_n,f_{n\plus{}1})\equal{}(f_{n\plus{}1},f_{n\plus{}2})$
Then the constant is $ (a,b)$


Let $ 0\prec a\le b\le c$ and $ abc\equal{}1$,prove that:
$ a\plus{}b^2\plus{}c^3\ge\frac{1}{a}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^3}$.
Find the generalization.


Let $ABCDE$ be a convex pentagon such that $AB=AE=CD=1$, $\angle ABC=\angle DEA=90^\circ$ and $BC+DE=1$. Compute the area of the pentagon.

[i]Greece[/i]

Continue discussion at http://www.mathlinks.ro/Forum/viewtopic.php?t=21866 .

  darij


[quote="nr1337"]I have a Mac OS X computer, a Windows XP computer and I am currently installing Red Hat Linux 9 :).[/quote]
I've used (on my computers) DOS, Windows 3.1, 95, XP, Redhat 8, Redhat 9, Mandrake (7?), Knoppix 3.2, Knoppix 3.4, and FreeBSD, and perhaps a few other linux distros. Primarily, though, I use XP.

I used to have Mandrake Linux 9, and I liked it.  I'm installing Redhat 9 since it's more popular right now and it's supposed to have more command line  :D.  A good thing about Unix based computers like Macs and Linux computers is they include bunches and bunches of development tools (not like I can do much programming besides calculators and Hello World programs lol).

[quote="nr1337"]I'm installing Redhat 9 since it's more popular right now and it's supposed to have more command line  :D.  [/quote]
No.. [url=http://www.slackware.com/]slack[/url] has more command line, at the very least.

Right now I use [url=http://www.linux.org/]Gentoo[/url] and Windows XP. If it wasn't for my dad not liking linux, I'd probably dump Windows in favor of Gentoo. Gentoo rocks. I've tried FreeBSD and Knoppix before, and a little slack. I'm about to install slack on my computer.

For anyone who's interested, Mandrake's first CD is good for when you want to resize partitions, but I think qtparted from GNU also does that..

Can you actually use the command line, though? Do you use X?

Windows is bloated.

I use Windows XP and MacOS X (at least I think that's what it's called), but I really don't know how to open up applications except for the ones at the little thing at the right of the screen.

Windows XP, and an old computer with Windows ME.  The really scary part is that until recently we had one with 3.1 but it finally died after merely crashing every few months for the past ten years.

I use Windows 2K. I can't seem to shake loose the notion that with Windows XP, Microsoft tries to treat its customers as computer illiterates.

But even I may have to migrate one day soon when Microsoft stops officially supporting Windows 2K (which is scheduled for December of this year) and security flaws surface with no fixes in sight.

Haul it out until Longhorn? I think not. ;)

To Hatsup123,
Since your computer is totally different what is it like?

By the way Mac OS 10.1
I also have Windows XP but I prefer the previous.

[quote="tandjsnell"]
By the way Mac OS 10.1
I also have Windows XP but I prefer the previous.[/quote]

YAY! Another mac person!

[quote="tandjsnell"]That's what I thought when I saw your first post in the topic.
The only reason I have the windows machine is because the computing world is dominated by windows.[/quote]

Really if you look around - on commercials, in some businesses - you will see a lot of macs.

Yes I have noticed that (once saw a commercial for some microsoft product with a mac in it.thought it was kinda funny) But  the majority of software is windows only.

[quote="confuted"][quote="Duck-Billed Platypus"]For anyone who's interested, Mandrake's first CD is good for when you want to resize partitions, but I think qtparted from GNU also does that.[/quote]
I like the partitioner built into the Redhat install more than the Mandrake version. Overall, I found Mandrake difficult to use since the command line was not clearly accessible and I couldn't log into root directly. Without root priviledge on a fresh install of Linux... it's not worth it. If my sound card worked with Linux (Soundblaster Live!), I very likely would make the switch.[/quote]
I didn't actually install Mandrake, just used the first CD because someone recommended it back then..

I have a Soundblaster Live sound card, and it works really well with Linux.. But then, my speakers suck... Why doesn't your sound card not work? Did you use OSS or ALSA or something else? the 2.6 kernels now have ALSA support built in. I've heard that OSS is more ready-out-of-the-box, but it's deprecated in favor of ALSA.

What do you mean the command line wasn't clearly accessible? Do you mean you couldn't access a terminal emulator like xterm? Have you tried doing ctrl-alt-fx where fx is one of the function keys (f1, f2, f3, etc)? That should bring you to the command prompt..

You might want to try out Gentoo. It gives you more control over what you want to install what not to install, but for that you need to know more or you might be in for some trouble.. You could also try LFS just as an experiment, but for your main desktop.. It's not automated, so you might have some trouble keeping it up to date.

[quote="Osiris"]I was considering dual-booting W2K and Debian on my laptop before I got it; but the HDD isn't as large as I would've liked (only 20 GB), so that plan will have to wait until another day.[/quote]
I have Gentoo running on about 3-4GB.. You don't need that much space to run Linux.. Just make sure that you don't use one of those distros that come with all sorts of software bundled with it. Slackware is also pretty lean. Or you could try *BSD or something.. Windows on the other hand..

just the classical windows ME. Why install another thing? It's a good training to handle the worst. :maybe:

Slide rule.

Actually, I have XP.  I hate it, and I think I'm going to install some sort of Linux.

[quote="Duck-Billed Platypus"]I found Windows 95, 98, and XP more enjoyable than ME. ME just crashed too much.[/quote]

I know. But I manage to make any system crash so that doesn't matter. :D

[quote="takuanitromars36"]Windows 3.1, Windows 95, Windows 98, Windows 98 SE, Windows ME, Windows 2K & Windows 98 Chinese Edition (simultaneously), and Windows XP.

In that order too.[/quote]

Ascending or descending? :D

I like Windows 8.1. It costs like 250 bucks though, so it brought my build cost up to 1400 bucks instead of 1100 bucks. :(

I refuse to tell you my operating system because you will probably use it to stalk me.

then why post?

i have macos x raspian(rspberry pi)
and i HATE windows

windows 8 is not exactly the best.
still, windows ftw

My dad has Windows 8 (this one has 7) and it is SO confusing. I literally spent ten minutes trying to get to the internet. Then I couldn't go back to the desktop

[quote=mathleticguyyy]Windows 10 is good[/quote]

Do not bump if the bump isn’t 14 yr


logarithms
real analysis
real analysis unsolved

I don't quite know how to formulate my question but I will try:

It is known that ln2 is the sum of the Leibniz series.

What famous/non-trivial/interesting numerical series has  sum of ln3 or ln5 for example?

$ \ln 2$ is special because the Taylor series $ \ln (1 \plus{} x) \equal{} x \minus{} \frac{x^2}{2} \plus{} \frac{x^3}{3} \mp ...$ converges at $ x \equal{} 1$ but for no $ x > 1$.  If you want to evaluate the natural logs of larger integers you have to take advantage of some properties of the logarithm.  For example, you can write $ \ln 3 \equal{} 2 \ln \sqrt{3}$; then

$ \ln 3 \equal{} 2 \left( \sum_{i\equal{}1}^{\infty} \frac{(\minus{}1)^{i\plus{}1} (\sqrt{3} \minus{} 1)^i}{i} \right)$.

Of course, there are lots of other ways to find a convergent series that works.  For example, you can write $ \ln 3 \equal{} \ln 2 \plus{} \ln \frac{3}{2}$; then

$ \ln 3 \equal{} \ln 2 \plus{} \sum_{i\equal{}1}^{\infty} \frac{(\minus{}1)^{i\plus{}1}}{2^i \cdot i}$.

Here's a very general way to approach this.

$ \ln\left(\frac {1 \plus{} x}{1 \minus{} x}\right) \equal{} 2\left(x \plus{} \frac {x^3}3 \plus{} \frac {x^5}5 \plus{} \frac {x^7}7\plus{}\cdots\right)$

(If you think this looks like the arctangent with the signs altered - yes, it's the inverse hyperbolic tangent.)

So we want to find $ \ln y.$ We let $ \frac {1 \plus{} x}{1 \minus{} x} \equal{} y,$ solved to $ x \equal{} \frac {y \minus{} 1}{y \plus{} 1}$ and then use that $ x$ in the series.

So, for instance for $ \ln 3,$ we have $ y \equal{} 3$ and $ x \equal{} \frac12,$ so

$ \ln 3 \equal{} 1 \plus{} \frac1{12} \plus{} \frac1{80} \plus{} \frac1{2^6\cdot7} \plus{} \frac1{2^8\cdot9} \plus{} \cdots$

The series converges pretty quickly. In fact, go ahead and use this with $ x \equal{} \frac13$ to get $ \ln 2$ from a quickly converging series, not that conditionally converging nightmare.

I was looking at that and thought of a nice cheap estimate for $ e.$

$ e$ is the root of $ \ln x \minus{} 1 \equal{} 0.$ By that series, to a first-order approximation, we have that $ \ln 3 \minus{} 1\approxeq\frac1{12}.$ Start there and apply one round of Newton's method to get $ 3 \minus{} \frac {\frac 1{12}}{\frac13} \equal{} 2.75,$ and take that as an approximation for $ e.$ Not too bad for that little computation, and there are many means for improving on it. For instance, just including the $ \frac1{80}$ term in the series for $ \ln 3$ gets us $ e\approxeq 2.7125.$

I just rigged a spreadsheet to perform the calculation above: convert $ y$ to $ x$ and then add up 30 or so terms of the series. I reported the result to 14 digits.

I get $ \ln2 \equal{} 0.69314718055995$ in 15 terms.

I get $ \ln3 \equal{} 1.09861228866811$ in 23 terms.

I get $ \ln5 \equal{} 1.60943791243410$ in 37 terms.

It was an easy spreadsheet to build.