[hide="hint"]

Re-write every number using logs such that they all have the same base. Ex/ we can rewrite $ 8^{x}$ to the base $ 2$ to by doing the following:

$ \log^{8}_{2}= 3 \Longrightarrow 8^{x}= (2^{3})^{^{x}}= 2^{3x}$

[/hide]

[quote="diophantient"][quote="duy_nguyen"]Find the roots of the equation:
 $ \frac{3}{2}.(5^{x}+1)=2^{x}+3^{x}+4^{x}$[/quote][hide]
x=1, x=2 work.
After that the LHS grows exponentially much much faster than the RHS, consider
$ \frac{1.5 \cdot 5^{x}}{2^{x}+3^{x}+4^{x}}$ as x gets incredibly large.[/hide][/quote]

Er actually, only 0 and 1 works in $ \mathbb{Z}$...now my question to duy_nguyen is, did you intend for roots in $ \mathbb{Z}$ or $ \mathbb{R}$?  I would assume the latter...in which case, you can use calculus to show there are no other roots...


logarithms
number theory unsolved
number theory

Prove that there exist positive integer which is divisible by 1996 and the sum of his digits it is also 1996!

You mean that the sum of digits is $1996!$ and the number is divisible by $1996$ ?

Let's say we can generalize it for every $n$.

Let's construct a number $100...00100...00100...00100.......00100...00100...000$ where there are $k$ zeros at the end of the number - where $k$ is the $2max(\log_2[n],\log_5[n])$ and there are $n!$ $1's$ and between each $1's$ there are exactly $\varphi(n)$ zeros.

Sum of digits of $1996$ is $25$ and sum of digits of $2\cdot 1996=3992$ is
$23$, and $1996=78\cdot 25+2\cdot 23$ so there you go,
concretely:\[\underbrace{19961996...1996}_{78}39923992\] is divisable by 1996 and
its sum of digits is 1996.

Nice and elementary  :D  :D  :D  :D

Is sum of digits suposse to be $1996$ or $1996!$ but if is $1996!$ only write that my number $1995!$ time and there you go

ok, sorry that one above was cheating  :blush: ,
$111...111000...000$ where there are $n!$ $1's$ and $k$ zeros is divisible by $n$ when $gcd(n,3)=1$ so it's ok for $n=1996$.

I dont mean $1996!$ digits but 1996 digits. Sorry


How to write something like this?
\[(a_1+a_2+\cdots+a_m)^n=\sum_{\begin{array}{c}n_1,n_2,\dots,n_m\geqslant0\\n_1+n_2+\cdots+n_m=n\end{array}}\frac{n!}{n_1!n_2!\cdots n_m!}a_1^{n_1}a_2^{n_2}\cdots a_m^{n_m}\]

I used \begin{array} to display two rows below Sum, but the leters are too big, is there any more elegant way?

I won't write the whole thing again but to have multiline subscripts, use the amsmath package and then:

\begin{displaymath}
\sum_{\substack{0<i<n \\ 1<j<m}}
P(i,j) =
\sum_{\begin{subarray}{l} i\in I\\
1<j<m
\end{subarray}} Q(i,j)
\end{displaymath}

$
\sum_{\substack{0<i<n \\ 1<j<m}}
P(i,j) =
\sum_{\begin{subarray}{l} i\in I\\
1<j<m
\end{subarray}} Q(i,j)
$

the subarray command gives you more control on the alignement of the subscripts.

ledurt


calculus
integration
number theory solved
number theory

Determine all integeral solutions of a^2+b^2+c^2=a^2b^2.

Use modulo 4 and then use the infinite descendence of fermat!

cheers! :D :D


quadratics
function
logarithms
trigonometry
algebra proposed
algebra
Fixed point

[quote="ajorza"]Hi,

Problem: Let $(x_n)$ be a sequence such that $x_0=0$ and for every $n\geq 1$ we have \[x_n=\frac{x_{n-1}^2+1}{2}.\]

Find an "almost" closed form for $x_n$, i.e., find a nice estimate of $x_n$ with error $O(1/n^2)$. The estimate should be in closed form and should use only standard functions (eg, $\log,\cos,\sin,\zeta$).

Hint: not trivial. The first few terms of the sequence are: $0,1/2, 5/8, 89/128,\ldots$.

Andrei[/quote]

$\frac{x_n}{2}$ is the Mandelbrot set recurrence starting at $1/4$.  It increases toward a fixed point 1/2 and one is left with studying $1/2 - {x_n}/2 = y_n = y_{n-1} - y_{n-1}^2$. The associated differential equation $y' = -y^2$  predicts that $y_n$ is approximately $1/(n+C)$, so that we should consider $1/y$.   One finds that $1/{y_n}$ is $n + log(n) + c_n$ where $c_n$ converges to a limit.

1/4 is at the boundary of the Mandelbrot set.  The rate of divergence of the $1/4 + \epsilon$ iteration involves $\pi$; if I remember, the number of iterations to escape some fixed region grows like $\pi/{\epsilon^2}$.


You and five friends need to raise $ \$1500$ in donations for a charity, dividing the fundraising equally. How many dollars will each of you need to raise?

$ \textbf{(A)}\ 250\qquad
\textbf{(B)}\ 300\qquad
\textbf{(C)}\ 1500\qquad
\textbf{(D)}\ 7500\qquad
\textbf{(E)}\ 9000$

[quote="#H34N1"]You and five friends need to raise $\$$1500 in donations for a charity, dividing the fundraising equally. How many dollars will each of you need to raise?

A. 250
B. 300
C. 1500
D. 7500
E. 9000[/quote]
[hide] B =1500/5[/hide]

[quote="#H34N1"]You and five friends need to raise $\$$1500 in donations for a charity, dividing the fundraising equally. How many dollars will each of you need to raise?

A. 250
B. 300
C. 1500
D. 7500
E. 9000[/quote]

[hide="solution"]You need to split it into 6 people.  $\frac{1500}{6}=\boxed{\$250}$[/hide]

[quote="Quevvy"]Isn't this kinda easy for the AMC 10?[/quote]If the first few questions were as hard as USAMO level, I don't think a lot of students will get much encouragement to participate in the competition, and you probably wouldn't survive for the next level. Take it for granted that the first few are easy.

lol, it's not THAT easy. when i did it for practice i got B.300 and then realized my mistake.

The trickiness of the text does not mean that its hard -__-

this is pretty easy.  just watch out for the "you and 5 friends"  that means 5+1=6

1500/6=250


Dear everyone.

find the envelope of the following.

$ \frac {x}{a} \plus{} \frac {y}{b} \equal{} 1$, $ a \plus{} b \equal{} 1$

I know the method for finding envelope, but no why.
May I ask you a favor ? I want to know why.

sorry for my poverty in math. :(
thank you.

Check this link [url]http://en.wikipedia.org/wiki/Envelope_(mathematics)[/url]
The easiest way to find useful links is to enter Google with suitable keywords. I got this by using "mathematical envelope" (of course, you have to be careful with your key words, since if you just enter "envelope", guess what you'll get; try "convex envelope" for yet a different meaning).

Dear mavropnevma.

thank you for good advice. I don't know how wonderful google is  :blush:


Level Sets.
a. Let z = x + y. Putting x and z on the axes, graph the relationship
between z and x when i) y = 0 ii) y = 1.
b. Let z = x + y. Putting x and y on the axes, plot all the values of x
and y that make z = 2. This is a level set.
c. Let z = x + |y|. Putting x and y on the axes, plot all the values of x
and y that make z = 2. (This one is trickier).

Can anyone show me these graphically on a plane and any other necessary working so i can get a better idea of how to work these out in the future.

I won't give you the solutions, but think of it this way:

If one of the variables is a constant, why not first try graphing it on an x-y plane with the other two variables.

i.e. Your first problem says to find the relationship between x and z, and gives a predefined value for y. So why not make y a constant, giving you z=x, and then graph that in 2-d? (The line you would graph is y=x.)

On your third problem, just plug in z, leaving 2=x+|y|. You can graph that in 2d as well.

etc.

[url=http://imageshack.us][img]http://img255.imageshack.us/img255/6362/iqaq2fb1.png[/img][/url]

Is this what the graphs should look like?????

Check your last graph. It should be a line segment from (0,2) to (2,0) and another segment from (2,0) to (0,-2). But otherwise, they're okay. Now just imagine these extended on another axis.

It's helpful to relate level curves to the lines on an [url=http://upload.wikimedia.org/wikipedia/commons/7/79/Topographic_map_example.png]elevation map[/url]. Taking one particular level curve is basically taking a cross-section of the 3D manifold.

That's what I meant (the cross-section part), just forgot the proper vocab.  :oops:  :P


ARML
rotation
geometry
ratio
3D geometry
tetrahedron
geometric transformation

Three spiders and an invisible fly are placed randomly on the edges of a regular tetrahedron.  They are only allowed to walk on its edges.  The spiders are trying to catch the fly, but the fly is not only invisible, it knows what the spiders are doing and it knows what they will do in the future.  However, the spiders are slightly faster than the fly.  (That is, if a spider was chasing the fly in a straight line, the spider would eventually catch the fly.)  If any spider runs into the fly, it will feel the fly and eat it, and the spiders will win.

Demonstrate the strategy that allows the spiders to catch the fly, or prove that none exists.

(I must warn you, my friend gave this to our entire ARML team and no one could solve it, and he gave it to people at MOSP and only a few have solved it there  :P )

(Note:  Probabilistic or random strategies that will 'eventually' catch the fly do not count.)

How much faster is a spider than the fly?  Negligibly faster?  Can we put a bound on it, or do we just have to say it's infinitesimally more?

I haven't seen a good problem in a while...nice one!

Is your friend certain of the "solution" of this problem?  If Moppers could get it, then I feel slightly better, because I don't know your friend.  It just seems like not enough information was given.

Are the spiders even capable of knowing where they have been?

Do the spiders know how much faster they are?  As far as I can tell at this time, the speed is crucial to the solution of this problem, because it yields information about how much area the spiders can guarantee the fly does not exist it, assuming spiders and flies have finite thickness.

Do these spiders and flies have finite thickness?

Interesting problem, congratulations.

[quote="FMako"]Is your friend certain of the "solution" of this problem?  If Moppers could get it, then I feel slightly better, because I don't know your friend.  It just seems like not enough information was given.

Are the spiders even capable of knowing where they have been?

Do the spiders know how much faster they are?  As far as I can tell at this time, the speed is crucial to the solution of this problem, because it yields information about how much area the spiders can guarantee the fly does not exist it, assuming spiders and flies have finite thickness.

Do these spiders and flies have finite thickness?[/quote]

I'd say he's certain, he's a MOPper himself so he's pretty good at stuff like this, and other people have arrived at the same solution as him.

The spiders plan beforehand exactly how they will move, either in a pattern or according to a set of rules (i.e., "if you have not found the fly yet, move to vertex A", etc.).  They can be considered to know where the other spiders are, and where they have been.  The crucial point is that the fly also knows what your strategy is, and also that the fly can start anywhere.

I've asked my friend about the speed ratio, and he says it's negligibly faster - i.e. a spider chasing a fly in an infinite straight line will eventually catch the fly in a large finite (or infinite) amount of time.  He also says that the solution doesn't change regardless of how small you  make the speed ratio, as long as the spiders are faster.  Consider it on the order of $1000: 999$.

As far as thickness, if a spider or a fly is on a vertex, it can be considered to occupy the entire vertex - i.e., a spider passing the fly on the vertex will eat it, and a fly cannot pass a spider on a vertex without being eaten.  So I suppose the spider and the fly could be considered to have width equal to the width of the edges of the tetrahedron, which (if you consider them as lines) have zero width, so... it shouldn't matter.

[quote="zanttrang"]I feel like a promising way to start is to do the following: have the spiders all converge on one vertex.  Then have them spread out and move at the same time to the three remaining vertices, one to each vertex.  Because they haven't hit the fly, they know that the fly isn't on the three edges they just traversed or the vertex they left behind.  Thus the fly must be trapped on one of the segments of the triangle formed by the vertices the spiders occupy.

My initial thought was then to have the spiders rotate simultaneously to catch the fly, but because they're only slightly faster, the fly could be standing right next to a vertex, and then slip back onto the remaining portion of the tetrahedron.[/quote]

That's what I was thinking.  Right now, I'm trying to figure out a strategy by which the three spiders can end up at three vertices and be sure that the fly is not on the triangle formed by those three vertices.
I'm still thinking about it.  Suppose there's some repeated sort of motion which they can make, so that at some point, the fly will be caught?

I told my dad about this problem.  He thought for a few seconds, and the said "So you have to find the expected time for the fly to be caught?"  Supposedly, he can prove that the fly will be caught eventually...now for finding the expected time...that's an interesting problem.

[quote="K81o7"]
That's what I was thinking.  Right now, I'm trying to figure out a strategy by which the three spiders can end up at three vertices and be sure that the fly is not on the triangle formed by those three vertices.
I'm still thinking about it.  Suppose there's some repeated sort of motion which they can make, so that at some point, the fly will be caught?

I told my dad about this problem.  He thought for a few seconds, and the said "So you have to find the expected time for the fly to be caught?"  Supposedly, he can prove that the fly will be caught eventually...now for finding the expected time...that's an interesting problem.[/quote]
There are two configurations whereby the spiders can immediately catch the fly: the spiders are standing on the vertices of a triangle and they know that the fly is on one of two edges between them, or they know that the fly is on none of the edges.  However, given that the fly knows their strategy, I can't figure out a way how we wouldn't avoid being trapped in such a configuration.

[quote="K81o7"]Maybe he was using a probabalistic method...so he meant in an infinite amount of time...
Well I'm thinking about it, and I don't think it's possible...the spiders can't catch the fly.  Whenever a spider moves from one vertex, that opens up a pathway.  If we can show that when the spiders are on three vertices of the triangle, and the fly knows what they are going to do at any point, then it can get to safety, then we are done...  We need to show that for any sequence of moves which they make, the fly has a way to safety.
Maybe you could, as a hint, tell us at least whether it's possible.  I don't think it is...Ah, forget the hint...[/quote]
Well, it's evident that for some moves, the fly couldn't get away (i.e. if two spiders on opposite vertices converged, or if three spiders converged on one vertex, and the fly where between them).  What we'd have to prove would be that the fly could avoid ending up in a situation where the spiders would know where he was, so that they would know to converge on him.  Otherwise, he could be in those positions but the spiders wouldn't know that he was there, and so they wouldn't want to converge in case they then let the fly free from somewhere else.  So I don't think it would be easy to prove that the fly could always get away, either.

[hide="not much of a hint"]There's definitely a solution for catching the spider, but I don't think anyone here knows of it (myself included). The problem was told to me by Kamior (that's his screenname here), and he never told anyone the solution as far as I know. I believe he said two people at MOP solved it.[/hide]

hehehe
this problem...
a very nice problem, i have thought about it almost everyday since he gave it us...yet no conclusion :D

edit: i remmeber him saying that: "at least two of the spiders must be moving at all times

this is quite an old topic,

but i believe today i have found a solution to this problem :) (yet i might be wrong though)

[hide="hint"]
we want to force the fly in one and only one loop (rather obvious)
[/hide]
[hide="hint2"]
not all the spiders must move at their maximum speed[/hide]

[quote="t0rajir0u"](Note:  Probabilistic or random strategies that will 'eventually' catch the fly do not count.)[/quote]Is Schrödinger's fly dead yet?
Thoughts:  Spiders circle around one face each. Spiders avoid going on the same segment twice in one loop by switching with another spider that has not gone on that segment. Spiders converge on one point like zanttrang suggested. Think about it as 2d? It's the exact same as a triangle with a point inside that's connected to the vertices, where the segments are the possible paths. Just my two cents. Anyone (amirhtlusa) mind posting a solution?

i believe you have to specificly say where do the three spiders start, and where each one goes....

Ha, I kinda gave up on this problem, being engulfed by schoolwork...
Now with a day and a half left of vacation, I'll think some more...

[quote="t0rajir0u"]The spiders have to agree on a deterministic strategy beforehand, so each spider is aware of the location of the other spiders at any point in this strategy (i.e. they cannot make random decisions, even if the resulting strategy terminates in finite time with probability $ 1$.)[/quote]
But my question is, whatever the starting position is, do the spiders know where each other are?

Yes.  I don't think it matters much, though.

I heard this one more than three years ago, and I [b]finally[/b] have it. It would be a shame to spoil this problem, so I won't post the strategy here.

I do have a random strategy as well, which works with small finite expected value even if the fly is much faster than the spiders. The only way to beat it is with a fly that actively predicts the future/strategy, as in this setup.
[hide="Best random strategy"]Start with the spiders at the same vertex. Choose one of the unoccupied vertices at random. Diverge along the three edges, then converge to the chosen vertex, with the spider already there staying still. Repeat this process.
Each iteration has at least a $ \frac23$ chance of catching the fly and takes two spider travel units to execute, for an expected time to capture of at most 3 spider travel units.[/hide]

If the tetrahedron were small enough, it would be possible to trap the fly in a triangle wherein the entirety of the space were occupied by the spiders.

Sorta ccheap, but it works.

Okay, here is the setup, which can be obtained easily:
Two spiders begin on the same vertex. One spider crawls along the left face clockwise, and the other crawls along the right face anticlockwise. The third spider walks back and forth along the back edge. Because the third spider only has to back and forth, while the other two go along a triangle, the timing is not matched. This means that if n is the time it takes for a spider to walk an edge, the fly will be caught at most 7n after the spiders set up their formation. I gave it thought, and it is all about LCM and number theory.

That strategy doesn't catch the fly. Name the paths so that the spiders' cycles are $ ABDABD\dots$, $ ACDACD\dots$, and $ BCBCBC\dots$. The fly starts near $ B$, ducks onto edge $ BC$ as spider 1 passes, follows it to $ D$, goes to $ C$, ducks onto edge $ CB$ as spider 2 passes, follows it to $ D$, and goes back to $ B$. That's a little more than 4 fly moves in 6 spider time units, so the strategy works if the fly is $ >\frac23$ as fast as the spiders.

[quote="mathelete"]This means that if n is the time it takes for a spider to walk an edge, the fly will be caught at most 7n after the spiders set up their formation. I gave it thought, and it is all about LCM and number theory.[/quote]
If the number of steps required doesn't depend on how much faster the spiders are, your solution is wrong.

Think about it this way:  if the chase were happening on a flat line, then the amount of time required for the spiders to catch the fly from any given position is proportional to $ \frac {1}{\epsilon}$, where the spiders are $ 1 \plus{} \epsilon$ times faster than the fly, because $ \epsilon$ is the rate at which the distance between the spiders and the fly decreases.

Furthermore, in the "best" case that the spiders start out opposite to, but can see, the fly, it cannot possibly take less than $ \frac {3}{2} \epsilon$ steps.

[hide="It turns out that..."] The amount of time the optimal strategy takes is something like $ \frac {2}{\epsilon} \plus{} C$ if one edge takes one unit of time to travel; I've forgotten the constant.  I was given this more specific problem by Hesterberg; I still don't know the solution.  There is a strategy that works but takes time $ \frac {4}{\epsilon} \plus{} C$. [/hide]

I would be very interested if someone were willing to write up an applet that simulated this problem.

Take a look at the random strategy- that doesn't depend on the speeds, to the point that it works even if the fly is much faster than the spiders. It is a function of strategy- it's just that the conditions here are finely tuned enough for only a limited class of strategies to work.

I've got the $ \frac2{\epsilon} \plus{} 2$ strategy. PM sent.

Also, that's wrong about the case in which the spiders get to look. If they get to look all the time, they catch the fly in constant time even if it's much faster than them. If they start together and look once at that time, it's $ \frac1{2\epsilon}\plus{}C$.

Could someone with correct solution post it now? Or at least PM me?

I know this is a very old topic, but I thought I would post some of my own extentions to the problem. The solution posted to the case with the tetrahedron is nice, but the stratergy will possibly need to be revised to solve these :)

[b]1)[/b] Replace the three spiders on a tetrahedron with four spiders on the edge of a cube. Can they still catch the fly?

[b]2)[/b] Suppose that the spiders are more than twice as fast as the fly, show that three spiders on a cube are now sufficient to catch the fly.

I'd imagine that (1) is possible; as long as you can find a cycle that traverses all of the edges and such that the spiders can close out the other routes in time, which seems to be possible, then they should be able to catch the fly. I'm not sure how to do (2) though; the strategy from (1) doesn't work, and a simple modification of the solution for the original problem doesn't seem to work, though it does work if the spiders are more than three times as fast as the fly.

Some other possible questions, I suppose: How about $4$ spiders on an octahedron? Are $n-1$ spiders sufficient on a $K_n$ graph, where the distance between any two nodes are the same? [I'd imagine the answer to both is no.] If not, if we can change the maximum speed the spiders can move, is the problem possible (and what is the minimum possible such maximum speed)? For a given graph whose edges have the same length (or even different lengths), can we find a way to estimate the minimum number of spiders necessary?

Have not seen this before.. here (interesting problem)!
 Easy to see  $ \frac2{\epsilon} \plus{}4$ type strategy.. ([hide]Spiders do step 1:  AB,AC,DB, step2:  two spiders rest while third , moves to BC , repeat this (step 3: CD,BD,CA, step four, just one spider moves    CD - and repeat) .) (Basically one spider moves loop  ABCD(A)  while two  others move back/forth on AC and BD making sure, there are  always two spiders  at  all the corner passings.
[/hide]

In this strategy(or just an obvious change) will work, even if fly can move much faster for the first  step.  (or last step)

Guess, when $\epsilon$ is  infinity, 2 is obvious answer, so it should require a little more thinking to reduce the answer to
 $ \frac2{\epsilon} \plus{}2$

[quote="filetmignon821"]If all the spiders converge on one vertex, spread to the the other vertices, rotate around the triangle a few times to make sure the fly Is not there, and simultaneously go back to the other vertex, the fly will be trapped.[/quote]

How many times will the spiders go around the triangle, though? You don't know the exact ratio, so if they agree to go around finitely many times, the fly's speed could be close enough to the spiders' so that they don't catch the fly when going around the triangle.


First we prove that, for $0\leq x,y,z\leq 1$, we have
\[ \frac{x^2+y^2-z^2-4}{x+y-z+2} \leq x+y-2.\]
This is equivalent to \[ x^2+y^2-z^2-4\leq(x+y + 2)(x+y -2)-z(x+y -2)\]
or equivalent to \[0\leq z^2 + 2xy-z(x+y -2).\]
The last one is true since $x+y\leq 2$ and $z\ge 0$.

For convenience, we assume that $a_{n+1}=a_1$ and $a_{n+2}=a_2$. We make a variable substitution by letting $b_i=a_i-2$ so $0\leq b_i\leq 1$.

For each $i=1,\ldots,n$, we have
\[
\frac{a_i^2+a_{i+1}^2-a_{i+2}^2}{a_i+a_{i+1}-a_{i+2}} = 4 + \frac{b_i^2+b_{i+1}^2-b_{i+2}^2-4}{b_i+b_{i+1}-b_{i+2} + 2}
\leq 4 +(b_i+b_{i+1} - 2) = a_i+a_{i+1}- 2.
\]
We get the desired result by summing up over $i=1,\ldots,n$. QED.


I really don't get how to do this:

If you have drag and you want to calculate the velocity of an object in free fall, you have the equation:

bv-mg=ma=m (dv/dt).

I suck at differential equations. I hate them with a passion. And i don't get how it becomes something with e^(-bm/t)...

help? :D anyone know a site where i can see the derivation...?

thanks!

I think that equation of movement for free fall is as follows.

\[ m\frac{dv}{dt}=-bv+mg  \]

\[ m\frac{dv}{dt}=-b\left(v-\frac{mg}{b}\right)  \]

\[ \frac{d}{dt}\left(v-\frac{mg}{b}\right)=-\frac{b}{m}\left(v-\frac{mg}{b}\right)  \] thus we have

\[ v-\frac{mg}{b}=Ce^{-\frac{b}{m}t}  \]

From the initial condition: $v=0$ when $t=0$ we obtain $C=-\frac{mg}{b}$

$\therefore v=\frac{mg}{b}\left(1-e^{-\frac{b}{m}t}\right)$

The terminal velocity is $\lim_{t\rightarrow \infty}v=\frac{mg}{b}$

wait-- how did you get from

the C*e^(...) part?

Thanks :)

For $y'=ky\Longleftrightarrow y'-ky=0\Longleftrightarrow e^{-kx}y'-ke^{-kx}y=0\Longleftrightarrow (e^{-kx}y)'=Const.$ We have
 $\boxed{y'=ky \Longleftrightarrow y=Ce^{kx}}$

Thanks Kunny you are saving my life as I have a physics test on this tonight...!

just one more question. how did you get d(v-mg/b)/dt from dv/dt?  :oops:

thank youuu

I transformed  $\frac{d}{dt} v$ into $\frac{d}{dt} \left(v-\frac{mg}{b}\right)$ so that we can use the given formula which I showed.

Let's practice!

1. $y'=y+1$

2. $y'=2y+1$

3. Find $I$ such that $L\frac{dI}{dt}=E-IR$, the initial condition $I=0\ when\ t=0.$

aaaah yes I get it you rock Kunny!!!! because it's a constant-- you can add it without any problem.

I'll go practice if the drag force is bv^2... :)

with drag force=bv^2 it just becomes the square root of the original answer right? :)

:?  lol yeah... you're good. :O

but I think I did pretty well, teacher said what if F-net=cv^2 and told us to prove x-xi=m/c*ln(vi/v)....  or something like that. and I got it so I'm pretty happy.

thanks for your help. are you in high school? o__o

can it be proved that the drag force equals $bv$ or $bv^2$ or...

That's pretty interesting, I had never seen the v^2 case.
The choice of whether to use v^1 or v^2 is based on experimental evidence.  I don't know of it being derived from anything else.  I bet that if you really looked into it, you could find some assumptions that would lead to one or the other, i.e. assume the object is spherical, assume something about the fluid... but other than that it just comes from experiments.

[quote="kunny"]Regrettably no. In this case solving the equation of motion is more complicated than previous problem.

The equation of motion in case of the drag force is $bv^2$ is as follows.

$m\frac{dv}{dt}=-bv^2+mg$

$\frac{dv}{dt}=-\frac{b}{m} \left(v^2-\frac{mg}{b}\right)$

Let $k^2=\frac{mg}{b}$, we have $\frac{1}{v^2-k^2} \frac{dv}{dt}=-\frac{b}{m}$

$\left(\frac{1}{v-k}-\frac{1}{v+k}\right)\frac{dv}{dt}=-2\frac{b}{m}k$

Integrate both sides with respect to $t$,

$\int \left(\frac{1}{v-k}-\frac{1}{v+k}\right)\frac{dv}{dt}\cdot dt=-2\int \frac{b}{m}k\ dt$

$\ln \left|\frac{v-k}{v+k}\right|=-2\frac{b}{m}kt+A\ (A: Constant)$

$\frac{v-k}{v+k}=Ce^{-2\frac{b}{m}kt}$ where $C=\pm {e^A}$

The initial condition $v=0\ when\ t=0$ gives $C=-1$, thus we have

${v=\frac{1-e^{-2\sqrt{\frac{bg}{m}t}}}{1+e^{-2\sqrt{\frac{bg}{m}t}}}\cdot \sqrt{\frac{b}{m}g}}$[/quote]

I have the same question as above but finallythey want me to find the terminal velocity( the maximum speed that can be attained by the body) ?
GOTTA SUBMIT WITHIN 10 HOURS!! :?

what does it mean? and what does $bv$ represent?

It's air resistance which is proportional to velocity.

[quote="phucnv87"]How do you solve this equation
$\large (m_o+\mu t)\frac{dv}{dt}+v\mu = F$ ($F$ :const)
The initial condition $v=0$ when $t=0$[/quote]

The answer is $v=\frac{Ft}{\mu t+m_0},$ right?

$\text{Your answer is correct: }\large v=\frac{Ft}{m_0+\mu t}$


geometry
3D geometry
tetrahedron
analytic geometry
symmetry
octahedron
vector

How would one prove that a given polyhedron is of a specific type (e.g. regular tetrahedron) given nothing but its coordinates?
Suppose, for instance, I wanted to show that the polyhedron given by the 4-dimensional coordinates $ (1,1,0,0),(1,0,1,0),(1,0,0,1),(0,1,1,0),(0,1,0,1),(0,0,1,1)$ is a regular octahedron. How would I go about doing so?
Should I show that the symmetry group of this polyhedron is identical to that of a regular octahedron and then show that it has 6 vertices, 12 edges and 8 faces (or any combination of 2 of those facts)?
Or would it be better/easier to use a vector approach and to say that the center of the polyhedron is equidistant from all six vertices and that any two pairs of opposite vertices form a square?
Any input will be greatly appreciated. Thank you!

If the polyhedron is convex, you can use 3-d polar coordinates to prove it's shape. Just set the origin of the polar coordinate system at the center of your polyhedron and calculate the distance and direction  each of the points are from that origin.


let  F:Z=>Z /F(m-n+f(n))=F(m) +F(n).when this problem was posted all mathlinkers find only F(n)=2n or F(n)=0.you can see here http://www.mathlinks.ro/forum/viewtopic.php?p=521359#p=521359 .but  in fact we can also get more solutions.in fact let E= { F(a);a in Z}.we can prove that  E is  SG of Z.we have that o in E.let x;y in E => there existe a;b in Z such F(a)=x ;F(b)=y .from the eguation we get that x-y and x+y in E then E  is SG of Z.and we know that a SG of Z is in the forme nZ where n=min(E+).then we get that the solutions are in the forme F(na)=2na.then if n=o we find F(a)=o if n=1 we have F(a)=2a.if n is deffirent to {0.1} let a in Z.and let a=nq +r then F(a)=F(nq +r)=F(r-nq +F(nq))=F(nq)+F(r)=2nq+f(r) that is mean that every integer are detriminite by the images of{1,2,.............,n-1}.what do you think. sorry for my bad english and latex :arrow:

I wait your observations .another time what do you think? :?:


When I click "community" on the toolbar to the left, I get to something ugly... is the forum home being updated?

If so, then it must have finished now, as I don't see anything.


Let $a;b;c$ be positive real numbers satisfying $abc=1$.Prove that:
${3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c+\frac{1}{a}+\frac{1}[b}+\frac{1}{c}$

[hide]
If we use the substitution
\[a=\frac{x}{z},\ b=\frac{y}{x},\ c=\frac{z}{y}\]
the inequality becomes
\[3+\frac{x^{2}}{yz}+\frac{y^{2}}{xz}+\frac{z^{2}}{xy}\geq \frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}\]
After multiplying by $xyz$ and subtracting, it follows from Schur's Inequality
\[x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y)\geq 0\]
[/hide]


limit
integration
calculus
logarithms
real analysis
real analysis unsolved

We start by computing $ \ln(a_n) \equal{} {1\over{n^2}} (\sum_{i\equal{}1}^n (n\plus{}1\minus{}i)(\ln(n\plus{}i)\minus{}\ln(i))$, which is clear if you count the number of times each term appears in the product.  We break this up into two sums:

$ {1\over{n^2}} \sum_{i\equal{}1}^n \ln(1\plus{}{i\over n}) \plus{} {1\over n} \sum_{i\equal{}1}^n (1\minus{}{i\over n}) \ln(1\plus{}{i\over n})$.

The first term is $ 1\over n$ times a Riemann sum for $ \int_0^1 \ln(1\plus{}x)dx$.  Since this integral exists and is finite, the first term approaches $ 0$ as $ n\minus{}>\infty$.

The second term is a Riemann sum for $ \int_0^1 (1\minus{}x)\ln(1\plus{}x)dx$, hence it approaches the value of this integral as $ n\minus{}>\infty$, which is $ 2\ln{2} \minus{} {5\over4}$.

We conclude that $ \lim_{n\minus{}>\infty} a_n\equal{}e^{\lim_{n\minus{}>\infty}\ln(a_n)}\equal{}e^{2\ln{2} \minus{} {5\over4}}\equal{}4e^{\minus{}{5\over 4}}$.


articles
Support
Gauss
geometry
geometric transformation
MIT
college

[url=http://www.washingtonpost.com/wp-dyn/content/article/2006/05/19/AR2006051901769_pf.html]This is a Saudi textbook (after the intolerance was removed).[/url]

I haven't read the whole thing but it really is time for countries like saudi arabia to open their eyes and bring lots of reforms.They habve to become a little less religious in the way they think and live.Religion dominates logic there.The women over there livereally live a tough public life adn peopel are troubled for the smallest ILLOGOCAL reason which is considered logical there.They themselves are breeding terorists and stuff in that atmosphere.

Is this really what they are taught in schools?  In general??

Two comments :

Quite a contrast with what see in school.  I even heard of high school students who got in trouble with a teacher for pointing out that some Palestinians cheered on 9/11.

Basically I don't get why the USA in its tour of liberation never seems to point a finger at its 'ally' Saudi Arabia.  Iran is a baddie because of human rights and no democracy.  But all those human rights violations can also be found in Saudi Arabia and there doesn't seem to be much democracy going on there.

Wow...these days mainstream Muslims (especially Saudi ones) think that Jihad teaches that Muslims must kill all the non-Muslims. This is not true.

Jihad literally means struggle. There are two types of Jihad: Jihad-e-akbar and Jihad-e-ashgar.

Jihad-e-akbar (the greater jihad) is a struggle within yourself to curb sinful inclinations. This is the most difficult form of Jihad.

Jihad-e-asghar (the lesser jihad) is the Jihad of the sword. This can only take place in certain conditions, however. The Quran states that fighting is permitted only against those who attack Muslims first.

The Western media often, if not always, ignores these distinctions. Those who they call "Islamic fundamentalists" are neither Islamic nor fundamental. There is no instance in the life of the Prophet Muhammad where he offered the alternative of the sword or Islam to anyone.

Saudi Arabia is a cool country though(talking about the malls and food) :D  oh the airport is hugeeee

Just to be the lone dissenter, I believe that many of the quotes shown at the end of the article are quotes from the Koran - but given as they are, are taken out of context. For what it's worth, they're also similar in tone and content to many passages of the bible.

I don't see anything quoted that explicitly says that Muslims should hate or kill non-Muslims. In the context of what is going on in the world today, it is easy to interpret them that way - but that's not what the majority of the quotes are really saying.

I'm not saying I support the Saudi government or its policies, or even Islam as a religion, I'm just saying that you have to recognize that it IS a religious state and as such you should expect it to teach and promote its religion in schools (even at the exclusion of other religions). And I think those quotes, more or less, do that as opposed to teaching unwarranted violence against other religions (although that sort of teaching is prevalent in SA also).

My main beef with Saudi Arabia is not its religious teachings in the schools, its more to do with its corrupt and repressive government and egregious human rights offenses. For example Saudi Arabia executes more of its citizens per capita than any other country in the world, by far.

Gauss-  IMO, To put it [b]mildly[/b],   NO. These are actual text books, and reality  is much  worse. Also since SA funds lot of schools  (religious) in other countries (Eg Pakistan) the text books are used there too. (Again, I am not talking about a passage here or there, but much more than that)

I suggest you actually read (original or in translation) some of the text books, even in some of the "moderate" states (eg Pakistan).... Or talk with some one who has actually taught or went to school there. (I have spent some time there, and do have first hand knowledge)

Pervez Hoodbhoy  (A well known educator/scientist  in Pakistan - He got his PhD in nuclear Physics from MIT, and is a proffesor in QIAU)   has written a lot about this. Here is one link[url=http://www.chowk.com/show_article.cgi?aid=00000753&channel=university%20ave&start=0&end=9&chapter=1&page=1]What are they Teaching in Pakistani Schools Today?[/url]

But a  google  search  can easily help you find more accurate picture.

I am told that things are changing    (There are many educators in countries themslelves who were/are speaking up)

TokenAdult - There was a CNN report yesterday about this.

[quote]My main beef with Saudi Arabia is not its religious teachings in the schools, its more to do with its corrupt and repressive government and egregious human rights offenses. For example Saudi Arabia executes more of its citizens per capita than any other country in the world, by far. [/quote]

Just  a minor remark : I think Kuwait and Singapore have a higher number of executions per capita

Don't get me wrong. I'm not saying that there isn't hate being taught in Saudi schools (I've never been there, and don't know anything about what goes on in their schools).

I'm just saying that I'm not convinced of it by the article. The quotes that were selected don't seem that out of line to me.

These books are of government schools right?Because i didn't have anything like this in my text books......


search
algebra
polynomial
algebra unsolved

Find $ (x; y)$ if:

$ \begin{cases}1\plus{}x^3y^3\equal{}\minus{}19x^3\\y\plus{}xy^3\equal{}6x^3\end{cases}$

We can isolate $ x$ from the first equation:
$ x^{3}(\minus{}19\minus{}y^{3})\equal{}1\Rightarrow x\equal{}\sqrt[3]{\frac{1}{\minus{}19\minus{}y^{3}}}$
Now plug this into the second equation:
$ y\plus{}y\sqrt[3]{\frac{1}{\minus{}19\minus{}y^{3}}}\equal{}\frac{6}{\minus{}19\minus{}y^{3}}$
It's ugly, but managable.

[quote="JRav"]We can isolate $ x$ from the first equation:
$ x^{3}( \minus{} 19 \minus{} y^{3}) \equal{} 1\Rightarrow x \equal{} \sqrt [3]{\frac {1}{ \minus{} 19 \minus{} y^{3}}}$
Now plug this into the second equation:
$ y \plus{} y\sqrt [3]{\frac {1}{ \minus{} 19 \minus{} y^{3}}} \equal{} \frac {6}{ \minus{} 19 \minus{} y^{3}}$
It's ugly, but managable.[/quote]
i think we can't solve your last equation :lol: .

Thank! But no body can give me the solution of my problem? May you help me?

Try solving for $ y$ in terms of $ x$ from the first equation and plug that into the second.  It's ugly and I'm not sure whether or not it's solvable but it's worth a try.

There is no simple solution. You'd need to solve a 15'th degree polynomial. :) At least according to Mathematica.

We find $ p(y)\equal{}0$ where $ p$ is a polynomial of degree 15 which is irreducible over $ \mathbb{Q}$. Hopeless !

cf    http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1800808255&t=199338 for a much better exercise.

I think it's:
$ \begin{cases}1\plus{}x^3y^3\equal{}\minus{}19x^3\\y\plus{}xy^2\equal{}6x^3\end{cases}$


[quote="who"]let a,b,c be positive real nos.
 show (a^3+b^3+c^3)^2>=(a^4+b^4+c^4)(ab+bc+ca)[/quote]
I'm sorry but I'll type it in TeX for more convenience
$(a^{3}+b^{3}+c^{3})^{2}\ge (a^{4}+b^{4}+c^{4})(ab+bc+ca)$
[code][Tex](a^3+b^3+c^3)^2 \ge (a^4+b^4+c^4)(ab+bc+ca)[/Tex][/code]

[quote="who"]let $a,b,c$ be positive real nos.
 show $(a^{3}+b^{3}+c^{3})^{2}\geq(a^{4}+b^{4}+c^{4})(ab+bc+ca)$[/quote]
Very nice fact!
$(a^{3}+b^{3}+c^{3})^{2}\geq(a^{4}+b^{4}+c^{4})(ab+bc+ca)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}a^{4}(a-b)(a-c)+2\sum_{cyc}(a^{3}b^{3}-a^{4}bc)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}a^{4}(a-b)(a-c)+\sum_{cyc}(a-b)^{2}(c^{4}-a^{2}b^{2})\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}a^{4}(a-b)(a-c)+\sum_{cyc}(a-b)(a-c+c-b)(c^{4}-a^{2}b^{2})\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a-b)(a-c)(a^{4}+b^{4}+c^{4}-a^{2}b^{2}-a^{2}c^{2})\geq0.$
$S_{a},$ $S_{b}$ and $S_{c}$ are non-negative.
Let $a\geq b\geq c$. Hence, $S_{c}\geq S_{b}\Leftrightarrow a^{2}b^{2}\geq a^{2}c^{2},$ which is true. Done! :)


A lot of other states have at least one thread about AMC's, and i just though we should have one too... u can just like post problems or scores or something...anything :noo: . this forum has so much more potential, it's almost pathetic

135 on 10A. forgot i was compliment counting on #16. i wrote $\frac{3}{8}$ instead of $\frac{5}{8}$  :(  i was like kicking my locker when i came downstairs after the test (figuratively speaking). it was torture. But i guess it must be worse to misbubble something...

By the way, do u know why they locked the whole AMC forum? do they do this every year before the AIME?

they locked it because they didnt want people discussing the AIME I, since people take it at different times due to the differenent time zones. lucky for me i am taking AIME II. :D


algorithm
floor function
number theory unsolved
number theory

Let  n>1  be an integer, and let  S  be a subset of  {1,2,...,n} such that  S  neither contains two coprime elements, nor does it contain two elements, one of which divides the other. What is the maximum possible number of elements of  S  ?

We want the maximum number of members such that no two are coprime - this implies they all share a common factor, so by greedy algorithm let that factor $= 2$.

Now we must eliminate elements that divide each other.  Let $2k < n$ be the largest even number in the subset.  Then we must eliminate all even numbers $\le k$.  If we chose to keep any one of them, we would have to eliminate at least one other member that is at least twice as large as that member, so this must be the maximum.  This leaves us with $\lfloor \frac{n}{4} \rfloor$ elements.

... Right?

[quote="t0rajir0u"]We want the maximum number of members such that no two are coprime - this implies they all share a common factor, so by greedy algorithm let that factor $= 2$.
[/quote]

Err, not quite. For example, you can have $\{6, 10, 15\}$. None are coprime, but they don't all share a common factor either.

The coprime condition seems a bit hard to work with, but we can get $\left\lfloor \frac{n}{2} \right\rfloor$ elements with the dividing condition...

Oh, damn, I knew it looked too easy.  =/

I got 548 as an upper bound, but i cant show if this is attainable...

What do you mean by the upper bound?I think when $n\to\infty$,the maximum possible number of elements of S will $\to\infty$ too.

For example,$p_1,p_2,\ldots,p_k$ is k different primes,then for $n>m=p_1p_2\cdots p_k$,we can let $S=\{\frac{m}{p_1},\frac{m}{p_2},\ldots,\frac{m}{p_k}\}$,then $|S|=k$,and k can be infinite large.

The problem is not from Romania, but from balkan 2005. I'm sure you can find it now on the Forum.
Edit: it's [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=balkan+2005&t=35936]here[/url].

Sorry! Thought i was replying to a different problem that i never put the solution in for! Whoops! :blush:


Prove that  $ R^n$  is connected.  In  other words prove that  there dont exist  open A and B in $ R^n$ such that their intersections  is empty  and their union is  $ R^n$.

Heres what  im  getting at:
  Suppose $ R^n$ is not connected. Then there exist  A and B such  that  establish  the non-connection. Ok  so let  a be a point  in  A and  b be a point in B.  Define A_1 to be the set  of all  reals t such  that  (1-t)a + tb belongs to A and B_1 to be the set  of all reals t  such  that  (1-t)a  + tb belongs in B. Since the union of A and B  is $ R^n$ then  it is clear  that  the union of A_1 and B_1 is R. since the intersection  of A and B  is empty  then the intersection of A_1 and B_1  is empty also. Now here comes the part im not sure of.  I  have to  prove  A_1 and B_1 are open  to reach some contradiction (  im still  not sure what it is).  Now since A is open  this means  that  the line through a and b passes through exactly 2  boundary  points of A. Let  x and  y be those boundary  points. Let  t_1 and t_2 such  that  x= (1-t_1)a + t_1b and y= (1-t_2)a + t_2b then  if im not mistaken A_1 =(t_1 , t_2) which  is  open. Analagously we prove B_1 is open. Now 0  is in A_1 and 1 is in B_1 , so  we have that  the union of A_1 and B_1 covers [0,1] (since it covers R)   and we already  know that their intersection is empty  so we have that  [0,1] is not connected which  is a contradiction. Or maybe  the contradiction is that  on  one side we have that  the union  of A_1 and  B_1 is R  but on  the other  i  have (kind of)  proved A_1 and B_1 are bounded  open  intervals  hence their union cannot be R.

 So  is this correct? it justs  seems to  me  like there must be something wrong.

I'm quite sure it's sufficient to show for n=1


geometry
perimeter
cyclic quadrilateral
geometry unsolved

$ABCD$ is a cyclic quadrilateral and $M$ a point on the side $CD$ such that $ADM$ and $ABCM$ have the same area and the same perimeter. Show that two sides of $ABCD$ have the same length.

let $AB = a$, $BC = b$, $CD = c$, $DA = d$, $MC = x$, $\angle{ADC} = m$.
first, we have $a+b+x = c+d-x \Longrightarrow c-2x = a+b-d$. (*)
second, we have $[ABCD] = 2[ADM]$.
so, $2d(c-x).sin(m) = (dc+ab)sen(m) \Longrightarrow d(c-2x) = ab \Longrightarrow d(a+b-d) = ab$

$\Longrightarrow (d-a)(d-b) = 0$.

so, the problem is solved.  :)


inequalities
function
inequalities proposed

For $a\geq 1,\ b\geq 1,\ c\geq 1,$ prove the following inequality.
\[a^{3}+b^{3}+c^{3}\geq \frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}+3\left(abc-\frac{1}{abc}\right). \]

So, the inequality is equivalent to

$(a+b+c)((a-b)^{2}+(b-c)^{2}+(c-a)^{2}))\geq (u+v+w)((u-v)^{2}+(v-w)^{2}+(w-u)^{2})$

where $u=1/a$, $v=1/b$, $w=1/c$.

Clearly $a+b+c\geq u+v+w$

So if we could show (for example) that $(a-b)^{2}\geq (u-v)^{2}$, then we would be essentially done.

But
\[(a-b)^{2}\geq \frac{(a-b)^{2}}{a^{2}b^{2}}= (u-v)^{2}\]
as desired.

[color=blue]Thank you for your reply and taking approriate measures , blahblahblah. [/color]:)

A tranquil situation can see us.

After all, $a^{2}+b^{2}\geq \frac{1}{a^{2}}+\frac{1}{b^{2}}+2\left(ab-\frac{1}{ab}\right)$ for $a\geq 1,\ b\geq 1.$

Is the following inequality hold? :maybe:

For $a_{1},\ a_{2},\ \cdots a_{n},\ \geq 1\ \ (i=1,\ 2,\ \cdots\, n),$

$a_{1}^{n}+a_{2}^{n}+\cdots+a_{n}^{n}\geq \frac{1}{a_{1}^{n}}+\frac{1}{a_{2}^{n}}+\cdots+\frac{1}{a_{n}^{n}}+n\left(a_{1}a_{2}\cdots a_{n}-\frac{1}{a_{1}a_{2}\cdots a_{n}}\right).$

kunny

[quote="puuhikki"][quote="blahblahblah"]where $u=1/x$, $v=1/y$, $w=1/z$.[/quote]
I don't understand this line.[/quote]

Probably $u=\frac{1}{a},\ v=\frac{1}{b},\ w=\frac{1}{c}.$

Yes, that's correct.  I've been so busy deleting a certain forum member's posts that my thinking is a little scrambled.

[quote="Kouichi Nakagawa"]there is the thing which I made with a PDF file before, I attach it.[/quote]

I only had a look at the second one, but this doesn't seem to be correct. If I am not mistaken
Jensen would be the other way round, as your function $f$ is not convex but concave.

[quote="Kouichi Nakagawa"]there is the thing which I made with a PDF file before, I attach it.[/quote]

On the other hand I just realized that it is enough to prove the inequality for $n = 2^{k}$ (as you did
in the first attachment): For general $n$ just choose any $k$ such that $n \le 2^{k}$.
In the corresponding inequality for $2^{k}$ just set $a_{i}$ to be the geometric mean of
$a_{1},\ldots,a_{n}$ for $i > n$ and you get the inequality for $n$.  :)

[quote="Kouichi Nakagawa"]there is the thing which I made with a PDF file before, I attach it.[/quote]

Yes, my solution is exactly same as yours.

Jensen is strong!


geometry
incenter
ratio
circumcircle
angle bisector

In a triangle $ ABC$ suppose that $ \angle AIO\equal{}90^{\circ}$ and $ \angle CIO\equal{}45^{\circ}$. Find the ratio $ AB: BC: CA$.

[quote="KrazyFK"]In a triangle $ ABC$ suppose that $ \angle AIO \equal{} 90^{\circ}$ and $ \angle CIO \equal{} 45^{\circ}$. Find the ratio $ AB: BC: CA$.[/quote]
hi
could u explain what are I , O plz
thanks

$ I$ is the incentre of triangle $ ABC$ and $ O$ is the circumcentre.

It's clear that KrazyFK means the incenter I and circumcenter O of ABC .

Hint

It is not hard to see that $ \angle B \equal{} 90^o$ and then $ O$ is the midpoint of $ AC$ , then apply  in the triangle AIC the generalized angle bisector theorem , which produces

$ \frac {IA}{IC} \equal{} \frac {sin C/2}{sin (45^o \minus{} C/2)} \equal{} \frac {sin90^o}{sin45^0}$

From here we can compute the angle C, then the angle A and the ratio of sides, but im a little lazy to deal with trigometry equations. :oops:


Billy Bob goes and starts a company called Billy Bob's Company. On the first day, one person applies for a job. On the second day, three people apply for a job. On the third day, seven people apply for a job. On the fourth day, fifteen people apply for a job. How many people will apply for a job on the eleventh day?

[hide]On day n,$2^{n}-1$ people apply for a job. When n=11, 2047 people apply.[/hide]


calculus
integration
logarithms
algebra
partial fractions
calculus computations

Is this the integral?
$ \int\frac{100x\pi}{1\plus{}x}\ dx$
I left out the limits of integration because you can do that.  Since $ 100\pi$ is a constant, you can pull this out and the integral becomes
$ 100\pi\int\frac{x\ dx}{1\plus{}x}$
Now think substitutions.  Can you figure out which one?

It doesn't seem a rational solution for me...

Note that $ \dfrac{x}{1\plus{}x} \equal{} 1 \minus{} \dfrac{1}{1\plus{}x}$

Better: Let $ u\equal{}1\plus{}x$.  Then $ du\equal{}dx$ and $ x\equal{}u\minus{}1$.  Then we substitute in and get
$ \int\frac{u\minus{}1}{u}\ du\equal{}u\minus{}\ln{u}\equal{}1\plus{}x\minus{}\ln(1\plus{}x)$.

It's not better, it's the exact same thing, except you found the division to be easier when doing a change of variables first. In general for these integrals you first perform the division then do partial fractions.

I prefer splitting up the numerator and having an easier demoninator.  I often find that partial fractions can be annoying and long-winded.


[quote="K09"]Let Fibonaici sequence F(1)=1,F(2)=1,...,F(n+2)=F(n+1)+F(n). p is prime number .P>2.
n min such that F(n)is divisible p.PROVETHAT F(n) is not divisible  p^2[/quote]
hmm... this problem is hard;who have a solution?

But it's really obvious!
This is in the first chapter in the course I give in university ;)

________
RY TVCM (The Very Crazy Math)

[quote="RobertuY"]But it's really obvious!
This is in the first chapter in the course I give in university ;) [/quote]
 :D but...have you sloved this one?

[quote="RobertuY"][quote="Kapi"]have you sloved this one?[/quote]
sloved? :D  :D  :D[/quote]

RobertY: Try to be a little more polite. I know that the interaction between the members of the list is by definition informal, but that is not a reason to be rude. If you want to do some remark on someone's mistake, try not to be despective. For example, the mistake of Kapi is just a typo, probably just a problem of typing a little fast.

By the way, with respect to

[quote="RobertuY"]
This is in the first chapter in the course I give in university
[/quote]

in English you teach a course, you don't give a course.

Best,


function
complex analysis
complex analysis unsolved

[quote="hien"]Given an entire function $w(z)$. Show that function $M(r)=max|w(z)|$, with $z$ run on circle $|z|=r$, is a continuous function of $r$[/quote]

The image of the circle $|z|=r$ varies continuously in the space of curves (continuous functions on the unit circle, with the uniform metric for distance between functions) and $M$ is a continuous function on that space of curves.


Hi, do you know that website that have all the putnam, IMO's?  kalva.demon.co.uk.  It's gone!  Where did it go?

You're not the first to notice...  Check these topics for a link to a web archive.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=208562[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=207824[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=207963[/url]


conics
ellipse
analytic geometry
function
calculus
calculus computations

Find the orthogonal trajectory to the ellipse:-

[b]x^2/(a^2 + k) + y^2/(a^2 + k)[/b]

Please post the full solution.   [/code]

That's not an ellipse. That's an origin-centered circle, and the orthogonal curves are all lines through the origin.

Regarding curves orthogonal to ellipses, you might want to check out [url=http://en.wikipedia.org/wiki/Elliptic_coordinates]elliptic coordinates[/url].

...A circle is an ellipse

But yeah, it is a circle.
When you multiply by $ a^{2}\plus{}k$ you get that $ x^{2}\plus{}y^{2}\equal{}a^{2}\plus{}k$, which is the standard form of a circle.
Or, since you know the focii are each located at $ (\plus{}/\minus{}\sqrt{a^{2}\plus{}k\minus{}a^{2}\minus{}k},0)\equal{}(0,0)$, it's just a circle centered at the origin.

No, I meant that it looks like he has a typo, since the (a^2+k)'s down there are irrelevant. Note that this is not one curve but a family of curves (that's why it isn't blah = 1, it is the family of curves of the form blah = c for c>0).

In any case, we can rigorously show that our curves are lines.

Our curves take the form $ f(r, \theta) \equal{} r \minus{} d\equal{}0$. We want solutions $ g$ to $ \nabla f \cdot \nabla g \equal{} 0$.

In polar, the gradient is given by $ \nabla F \equal{} \left( \frac {\partial F}{\partial r}, \; \frac {1}{r}\frac {\partial F}{\partial \theta} \right)$

So, we have $ \nabla f \equal{} \left( 1 , \; 0 \right)$, and our equation becomes $ (1,0) \cdot \nabla g \equal{} 0 \implies \nabla g \equal{} (0,k)$

Then, $ \frac {\partial g}{\partial r} \equal{} 0$, meaning $ g$ is not a function of $ r$, and $ \frac {\partial g}{\partial \theta} \equal{} k$.

Thus $ g \equal{} k \theta \plus{} c_1$, and the equation of an orthogonal curve is $ k \theta \plus{} c_1 \equal{} 0$, or $ \theta \equal{} \frac {\minus{} c_1}{k} \equal{} C$ for some constant $ C$. This is a line.

Sorry, Guys the curve is:-

[b]x^2/(a^2 + k) + y^2/(b^2 + K) =1[/b]

Now solve the question.


Construct two square matrices A, B s.t. $ AB\equal{}I$ but A, B are not inverses to each other, if any.

See discussion [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=277173]here[/url]. Asking the same question as [i]the top topic in the forum[/i] is really a waste of time.

wow... i was being an idiot...  :blush:  :o
hmm, it's kinda weird my textbook proved $ AB \equal{} I$ AND $ BA \equal{} I$ to show $ A,B$ are inverses, but won't only one of them suffice (which made silly me think it's possible to have $ AB \equal{} I$ but $ BA\neq I$)?

edit: wait, don't we have to prove a right inverse for B EXISTS first?


geometry
circumcircle
symmetry
perpendicular bisector

Let $ KL$ and $ KN$ be tangents to a circle $ k$. Point M is arbitralily taken on the extension of $ KN$ past $ N$, and point $ N$ is the second intersection point of $ k$ with the circumcirle of triangle $ KLM$. The point $ Q$ is the perpendicular from $ N$ to $ ML$. Prove that $ \angle MPQ\equal{}2\cdot \angle KML$.

Please without inversion.

Please check your problem. What are P and N?

1.we can prove ⊿PLK∽⊿PHM  (easy)  =>  ∠KML=∠KPL=∠HPM
2.denote I is the midpoint of MN  we can calculate that KI/IM=LQ/QH=tan(∠KNL)/tan(∠MNQ)  =>  ⊿PKI∽⊿PLQ   =>  ⊿QPH∽⊿IPM  => M,I,Q,P are concyclically  => ∠KML=∠HPM=∠IQM=∠IPM   =>  I,H,P are collinear  => ∠HPQ=∠IPQ=∠IQM=∠IMQ=∠HPM
DONE!

The statement of the problem above is so confused, I think the problem should be:

[b]PROBLEM-[/b] [b][color=darkblue]Let $ KL$ and $ KN$ be tangents to a circle $ (O)$. Let $ M$ be an arbitry point taken on the extension of $ KN$ through $ N$ which does not include $ K$. Let $ P$ be the second intersection of circumcircle of $ \triangle KLM$ and $ (O)$. The point $ Q$ is the projection of $ N$ onto $ ML$. Prove that $ \angle MPQ \equal{} 2\cdot \angle KML$[/color][/b].

[i]Solution.[/i] Let $ P'\equiv MP\cap (O)$, $ P'\neq P$ and $ Q'\equiv ON\cap ML$. Then it is followed that $ MN^2 \equal{} \overline {MQ}\cdot \overline {MQ'}$ $ \equal{} \overline {MP}\cdot \overline {MP'}$ $ \Longrightarrow PQQ'P'$ is concyclic. Therefore $ \angle MPQ \equal{} \angle P'Q'M$. We claim that $ LP'\parallel{}MK$. Indeed, we have $ (PM,PN)$ $ \equiv$ $ (LP',LN)$ $ \equiv$ $ (PL,PN) \minus{} (PL,PM)$ $ \equiv$ $ (NL,NK) \plus{} (KN,KL)$ $ \equiv$ $ (LN,LK)$ $ \equiv$ $ (NK,NL)$ $ \equiv$ $ (MK,NL)$ $ \pmod \pi$ $ \Longrightarrow LP'\parallel{}MK$. Therefore $ ON\perp P'L$ $ \Longrightarrow OQ'\perp P'L$. But $ O\in$ the perpendicular bisector of $ LP'$, thus $ Q'\in$ the perpendicular bisector of $ LP'$, which implies that $ \angle P'Q'M \equal{} 2\cdot \angle Q'LP'\equiv 2\cdot \angle MLP' \equal{} 2\cdot \angle KML$ $ \left (LP'\parallel{}MK\right)$ $ \Longrightarrow \angle MPQ\equal{}2\cdot \angle KML$ $ \square$.

Yes, sorry for my mistakes, i wrote it in a hurry, next time i will double check my posts.

[hide="My solution"]Let $ LM \cap (O)\equal{}\{J\}$
$ \angle JMP\equal{}\angle JKP, \angle PJM\equal{}180^o\minus{}\angle LJP\equal{}\angle KLP$
Then $ \angle JPM\equal{}\angle LPK\equal{}\angle LMK$
Denote $ T$ the intersection of $ PJ$ and $ KM$. We have $ TJ.TP\equal{}TM^2$
But $ TJ.TP\equal{}TN^2$ then $ TN\equal{}TM\equal{}TQ$
$ \Rightarrow \angle TQM\equal{}\angle TMQ\equal{}\angle TPM$
Then $ QTMP$ is cyclic we get $ \angle QPT\equal{}\angle QMT$
Or $ \angle QPM\equal{}2\angle QMT$ (QED)[/hide]

[quote="CCMath1"]Could anyone tell me where this problem comes from ? Thanks a lot! :)[/quote]
Iran National Math Olympiad (Second Round) 1997

Let $S$ be the midpoint of $LN$ and let $T$ be the midpoint of $MN$. By circle angle relations,

\[\angle{NPM}=\angle{LPM}-\angle{LPN}=(180^\circ - \angle{MKL})-\angle{KMN}=\angle{KNM}=\angle{LPN}.\]
Since $MN$ is tangent to circle $k$, $\angle{PNM}=\angle{PLN}$ which together with the fact that $\angle{LPN}=\angle{NPM}$ implies that $\triangle{LPN} \sim \triangle{NPM}$ which implies that $\triangle{SPN} \sim \triangle{TPM}$. Since $KL$ and $KN$ are tangent to $k$, it follows that $PK$ is the symmedian of triangle $\triangle{LPN}$ and therefore $\angle{LPK}=\angle{SPN}$. Since $Q$, is the foot of the perpendicular from $N$ to $ML$, $\angle{TQM}=\angle{KML}$. Therefore

\[\angle{TQM}=\angle{KML}=\angle{LPK}=\angle{SPN}=\angle{TPM}\]
which implies that $PMTQ$ is cyclic since $Q$ lies on segment $(ML)$ and segments $(PT)$ and $(ML)$ intersect. Therefore $\angle{MPQ}=180^\circ - \angle{MTQ}=2\cdot \angle{KML}$.

My solution:

Let $ X $ be the midpoint of $ LN $ .
Let $ T $ be the intersection of $ PX $ and $ LM $ .
Let $ R $ be the intersection of $ KP $ and $ (LNP) $ .

Since $ RLPN $ is a harmonic quadrilateral ,
so we get $ \angle TXN=\angle NXR $ .
Since $ \angle TLX=\angle KNL-\angle NML=\angle NPL-\angle KPL=\angle XLR $ ,
so $ T $ and $ R $ are symmetry with respect to $ NL $ .
Since $ \angle XQL=\angle QLX=\angle XLR=\angle NPR=\angle XPL $ ,
so we get $ Q, P, L, X $ are concyclic .
Since $ XL=XQ $ ,
so $ PX $ is the bisector of $ \angle QPL $ ,
hence we get $ \angle NPQ=\angle RPX $ .

Since $ \angle NPT=\angle RPL=\angle NMT $ ,
so $ M, N, T, P $ are concyclic ,
hence $ \angle MPQ=\angle MPN+\angle NPQ=180^{\circ} -\angle NTL+\angle RPX $
$ =180^{\circ} -\angle LRN+\angle RPX=\angle NPL+\angle RPX=2\angle NPT=2\angle KML $ .

Q.E.D

We have $\angle LPM= \angle LON(=180^{\circ}- \angle LKM)$ and $\angle LPN= \tfrac12 \angle LON$ so $\angle LPN= \tfrac 12 \angle LPN$. Therefore $PN$ is the bisector of $\angle LPM$.

$ML$ intersects $(O)$ at $R$. Let $J$ be the point such that $JQ=JR$. Since $\angle NJR= \angle NRL= \angle NPL= \angle KNL$ so $\angle LJN= \angle LNM$. Hence $\triangle LJN \sim \triangle LNM \; ( \text{A.A})$. This follows $LN^2=LJ \cdot LM$ or $\tfrac{NL}{LQ-QR}= \tfrac{LM}{NL}$. It is easy to prove that $\triangle MNR \sim \triangle MLN \; ( \text{A.A})$ so $\tfrac{ML}{NL}= \tfrac{MN}{NR}$. Hence $\tfrac{NL}{LQ-QR}= \tfrac{MN}{NR}$ or $\tfrac{NR}{LQ-QR}= \tfrac{MN}{NL}$. We also can prove $\triangle PMN \sim \triangle PNL \; ( \text{A.A})$ so $\tfrac{MN}{NL}= \tfrac{PN}{PL}$. We obtain $PL \cdot RN = PN \cdot (LQ-QR) \qquad (1)$.

Applying [b]Ptolemy's theorem[/b] in $LNRP$ we get $PR \cdot LN+PL \cdot NR= PN \cdot LR \qquad (2)$.
From $(1)$ and $(2)$ we obtain $2PN \cdot QR=PR \cdot LN$ or $\tfrac{PN}{NE}= \tfrac{PR}{RQ}$ with $E$ is the midpoint of $LN$. Hence, $\triangle NPE \sim \triangle RPQ \; ( \text{S.A.S})$. This follows $\angle LEP= \angle LQP$ or $L,E,P,Q$ are concyclic. We also have $EL=EQ=EN$ so $PE$ is the bisector of $\angle LPQ$. Therefore $\angle LPQ= 2 \angle NLM$. Thus, we get \[\begin{aligned} \angle QPM & = \angle LPM- \angle LPQ \\ & = 180^{\circ}- \angle LKM-2 \angle NLM \\ &= 2 (\angle LNK- \angle NLM) \\ & =2 \angle LMK. \end{aligned}\]

This problem took me way too long to solve -.-

We first look to a standard Yufei configuration as a lemma:

[b]LEMMA:[/b] Let $ABC$ be a triangle with circumcircle $\Gamma$.  Suppose points $D$ and $E$ reside on $\overline{AB}$ and $\overline{AC}$ respectively, and let $P=(ADE)\cap\Gamma$.  Then $P$ is the center of spiral similarity sending $\overline{DE}$ to $\overline{BC}$.

[i]Proof.[/i] Angle chasing.

Let $X=(NLP)\cap ML$.  Note that by this lemma, $P$ is the center of spiral similarity sending $\overline{XL}$ to $\overline{MK}$; in particular, this implies $\triangle PMX\sim\triangle PKL$, so $\angle MPX=\angle LPK=\angle LMK$.

Remark that from the equality $\angle MPX=\angle XML$ we get that $(MPX)$ is tangent to $MN$ at $M$.  Let $Y=PX\cap MK$.  Then $PX$ is the radical axis of $(MPX)$ and $(NLP)$.  Furthermore, $MN$ is the common tangent to these two circles.  Combining this information yields that $Y$ is the midpoint of $\overline{NM}$, i.e. $MY=YN=YQ$.

Finally, note that $\tfrac{YQ}{\sin\angle YMQ}=\tfrac{YM}{\sin\angle YPM}$, so by Law of Sines the circumradii of $\triangle YQM$ and $\triangle YPM$ are equal.  As a result, $YMPQ$ must be cyclic, and so $\angle YPQ=\angle YMQ=\angle YPM$.  Thus $\angle MPQ=2\angle KML$, which is what we wanted.

The key is to take midpts of $MN$ and $LN$. Let them be $X,Y$. Then $$x=\angle{KML}=\angle{KPL}=\angle{NPY}$$ by symmedian lemma. But $\angle{NXY}=x$ too ,so $XNYP$ is cyclic. This implies $$\angle{NLM}=\angle{NYX}=\angle{NPX}$$. Which implies $LM,PX$ and circle $C$ are concurrent, say at $R$. Then $XP.XR=XN^2=XM^2$ which implies $\angle{XPM}=x$. Finally we notice that $\angle{MQX}=x$ and hence $MPQX$ is cyclic, which leads to the desired conclusion.

From tangent line conditions, we have that $\angle MNP=\angle NLP, \angle NMP=180^{\circ}-\angle KLP=\angle LNP.$
Thus $\triangle PLN$ and $\triangle PNM$ is triangle. Thus $PQ$ passes through the mid point of arc $LRM.$
This situation looks a lot similar to mixtilinear circles.
So we want to construct new triangle that has point $N$ as incenter.

Let $R$ be the point such that $N$ is incenter of $\triangle RLM.$
Then, $\angle LRM=180^{\circ}-\angle RLM-\angle RML=180^{\circ}-2\angle LNM-2\angle LMN=180^{\circ}-2\angle LNK=\angle LKM.$
Thus, $R$ lies on the circumcircle of $\triangle KLM.$

Now, since $PN$ passes through the mid point of arc$LRM,$ $P$ is the tangency point of the mixtilnear circle of $\triangle LRM.$
Let $R'$ be the reflection of point $R$ respect to perpendicular bisector of $LM.$
Then $PQ$ passes through the $R'$ since $Q$ is the tangency point of incircle to $LM.$
Thus the length of the arc $R'M$ and $RL$ is same.
Thus, $\angle MPQ=\angle MPR'=\angle RML=2\angle LMR.$

This solution is kinda hard without knowing the lemmas regarding mixtillinear circle.

An easy solution :

Let ML intersect circle k at B and PB intersect KM at C. ∠LKP = BMP and ∠KLP = ∠KLB+∠BLP = ∠BPL+∠BLP = ∠MBP so ∠MPB = ∠KPL = ∠KML.
By power of C w.r.t circle k we have CN^2 = CB.CP . By having  ∠CMB = ∠MPB we have MC^2 = CB.CP .
So NC = MC and we also have ∠NQM = 90 so CQ = CM. ∠CMQ = ∠CQM = ∠CPM so CQPM is cyclic and ∠CPQ = CMQ.
∠MPQ = 2∠MPB = 2∠KML


There are $7$ chairs on stage, arranged in a straight line parallel to the front of the stage. There are $7$ people to occupy the $7$ chairs. Two of these people cannot sit next to each other. How many seating arrangements are possible?

[hide]
The possibilities without restrictions is $7!$.  Now, we'll look at the ones that aren't possible.  Group the 2 people.  The # of possibilities is then $6!$.  Multiply this by 2, as there are 2 ways the 2 people can't sit together can sit together.  Thus, the # of arrangements is $7!-2\cdot6!=3600$
[/hide]

[hide]I think you mean $7! - 2 * 6!$ ?[/hide]

Yeah, he edited it...
It was $26!$ at first.

[hide]
Without restrictions, there is $7!$ possible ways to arrange the people. If the two people are sitting near each other, it can b $AB$ or $BA$ and in any 6 possible places. The 5 other people have $5!$ ways to sit. So, the answer is $7!-12\cdot5!=3600$[/hide]

[hide]3600 I think but not sure. I'm never good at arrangements

[color=blue]Moderator kyyuanmathcount's note: If you're not sure, can you give reason so we can correct, then you ARE sure?  Giving an answer, then stating that it's probably wrong really doesn't help...[/color][/hide]

[hide]It is [b]3600[/b] because there is 7! ways to arrange 7 people in seven seats.  See how many ways you can arrange the two people together which is 5! * 2 * 6.  Subtract  that from 7! which is [b]3600[/b].[/hide]


calculus
analytic geometry
geometry
3D geometry
sphere
calculus computations

Can someone please show me the solution to this:

Use polar coordinates to find the volume of the solid above the cone $z=\sqrt{x^2+y^2}$ and below the sphere $x^2+y^2+z^2=1$.

This is what I have so far:

Their intersection is the circle x^2 + y^2 = 1/2 which has a radius of sqrt2/2.

I know you have to integrate two different volumes and subtract them...

Uck, help?

This isn't the sort of problem we usually deal with on this forum, nor should we. Go back and read through the examples in that calculus textbook, then think about this one. Visualize what's going on. Consider what was done in appropriate lower-dimensional examples (e.g., the area between two curves.) What you've said so far is correct, but you should call on your own resources to figure out how to finish it.


Hint:
Only $\frac{1}{m_{i}}$ of the integers are $\equiv a_{i}\mod m_{i}$.
Now add the $\frac{1}{m_{i}}$, and voila.

[quote="Rust"]It is not true. Must be $a_{i}=a_{j}(mod(gcd(m_{i},m_{j}))$.[/quote]
¿¿¿  :huh:

I am sorry I say about x, satisfy all congruences.
[quote="xeroxia"]Let $m_{1},m_{2},\ldots,m_{k}$ be integers with $2\leq m_{1}$ and $2m_{1}\leq m_{i+1}$ for all $i$. Show that for any integers
$a_{1},a_{2},\ldots,a_{k}$ there are infinitely many integers $x$ which do
not satisfy any of the congruences \[x\equiv a_{i}\ (\bmod \ m_{i}),\ i=1,2,\ldots k.\] [/quote]
In this case problem is trivial. Let $X_{i}=\{x|x\not =a_{i}(mod \ m_{i})\}$. Because $m_{i}\ge 2$ set $X_{i}$ have infinetely many elements.
All $x\in X_{i}$ for one i don't satisfyed any congruence.

rust i can't understand what are u talking about the problem asks you to prove your solution not say it again

i think it was a good problem.
first we will find what is the number of elements from ${1,...,[m_{1},...,m_{k}]}$ that satisfies at least one of the congruences. say that number $T$. every congruence has at most $[m_{1},... ,m_{k}]\cdot\frac{1}{m_{i}}$ different solution. so we have
$T=(\frac{1}{m_{1}}+...+\frac{1}{m_{k}})\cdot[m_{1},... ,m_{k}] \le(\frac{1}{m_{1}}+...+\frac{1}{m_{1}\cdot\ 2^{k-1}})\cdot[m_{1},...,m_{k}]\le\frac{2^{k}}{m_{1}\cdot\ 2^{k-1}}\cdot[m_{1},...,m_{k}]<[m_{1},...,m_{k}]$
hence we have at least one element that doesn't satisfy any of the congruences. moreover if $x$ is a solution $[m_{1},...,m_{k}]+x$ is a solution so there are infinitely many solutions.

It's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=480&t=150519]Pen D 19[/url]


vector
inequalities
geometric inequality
geometry
IMO
IMO 1973

Prove that the sum of an odd number of vectors of length 1, of common origin $O$ and all situated in the same semi-plane determined by a straight line which goes through $O,$ is at least 1.

We prove it by induction on the number $2n+1$ of vectors. The base step (when we have one vector) is clear, and for the induction step we use the hypothesis for the $2n-1$ vectors obtained by disregarding the outermost two vectors. We thus get a vector with norm $\ge 1$ betwen two with norm $1$. The sum of the two vectors of norm $1$ makes an angle of $\le\frac\pi 2$ with the vector of norm $\ge 1$, so their sum has norm $\ge 1$, and we're done.

I found the same solution as grobber... I don't see how we could solve it using a different and beautiful way.

what does 'norm' mean? Does it mean the length of a vector?

Well, you can also optimize gradually. Choose $O=(0,0)$ and let all vectors lie in the region $y\geq 0$.
Fix the first $2n$ vectors and let $w$ be their sum. We want to minimize $|v+w|$, where $|v|=1$ and $v$ has positive $y$-coordinate.

The possible locus $v+w$ is a semicircle $\Gamma$ of unit radius located above the origin. It's easy to see that the closest point of $\Gamma$ to $O$ is one of the endpoints. Thus, we by repeating the above procedure, we can assume all vectors are $(\pm 1,0)$, which trivialises the problem.


[quote="D.I.Culianop"][quote="Dormy"]Suppose that a,b,c are positive real numbers, prove that
$ \frac {xy \minus{} z^2}{4x \plus{} 5y \plus{} 7z} \plus{} \frac {yz \minus{} x^2}{4y \plus{} 5z \plus{} 7x} \plus{} \frac {xz \minus{} y^2}{4z \plus{} 5x \plus{} 7y}\leq0$[/quote]
^^ a, b, c or x, y, z???[/quote]

sorry, i have corrected that spelling mistake

[quote="Dormy"]Suppose that x,y,z are positive real numbers, prove that
$ \frac {xy \minus{} z^2}{4x \plus{} 5y \plus{} 7z} \plus{} \frac {yz \minus{} x^2}{4y \plus{} 5z \plus{} 7x} \plus{} \frac {xz \minus{} y^2}{4z \plus{} 5x \plus{} 7y}\leq0$[/quote]
[hide="hint"]Chebysev ineq works here.Note $ \sum (xy\minus{}z^2)(x\plus{}y)\equal{}0$[/hide]

[quote="Allnames"][quote="Dormy"]Suppose that x,y,z are positive real numbers, prove that
$ \frac {xy \minus{} z^2}{4x \plus{} 5y \plus{} 7z} \plus{} \frac {yz \minus{} x^2}{4y \plus{} 5z \plus{} 7x} \plus{} \frac {xz \minus{} y^2}{4z \plus{} 5x \plus{} 7y}\leq0$[/quote]
[hide="hint"]Chebysev ineq works here.Note $ \sum (xy \minus{} z^2)(x \plus{} y) \equal{} 0$[/hide][/quote]

what about the denominator?


geometry
circumcircle
incenter
trigonometry
vector
power of a point

Angle A=30 of triangle ABC. S and I are circumcenter and incenter, respectively. D and E are on BA and CA, respectively, such that: BD=CE=BC.
Prove that SI equal and perpendicular to DE.

Sorry for my poor english.

[quote="BenasKikutis"]Angle A=30 of triangle ABC. S and I are circumcenter and incenter, respectively. D and E are on BA and CA, respectively, such that: BD=CE=BC.
Prove what SI equal and perpendicular to DE.

Sorry for my poor english.[/quote]
Applying porcupine theorem we have:
$ BC.\vec{IH} \plus{} EC.\vec{IK} \plus{} DE.\vec{u} \plus{} DB.\vec{IJ} \equal{} \vec{0}$
$ \Rightarrow \vec{u} \equal{} \minus{} \frac {BC}{DE}.(\vec{IH} \plus{} \vec{IK} \plus{} \vec{IJ})$
To prove $ SI\perp DE$ we will show that $ \vec{SI}// \vec{u}$ or $ \vec{SI}//\vec{IH} \plus{} \vec{IK} \plus{} \vec{IJ} (*)$
But this result is well-known: SI is the [b]Euler's line [/b]of triangle $ JKH$ so (*) is true.
Therefore $ SI\perp DE$
From [url=http://www.mathlinks.ro/viewtopic.php?t=277079]here[/url] and $ BC\equal{}R$ we have:
$ \frac {DE^2}{R^2} \equal{} 3 \minus{} 2(cos A \plus{} cos B \plus{} cos C)$
On the other hand, $ OI^2 \equal{} R^2 \minus{} \frac {abc}{a \plus{} b \plus{} c} \equal{} R^2(1 \minus{} \frac {abc}{R^2(a \plus{} b \plus{} c)})$
So $ DE \equal{} R \Leftrightarrow 2(cos A \plus{} cos B \plus{} cos C) \minus{} 2 \equal{} \frac {abc}{R^2(a \plus{} b \plus{} c)}$
$ \equal{} \frac {16S^2}{abc(a \plus{} b \plus{} c)} \equal{} \frac {(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}{abc}$
$ \Leftrightarrow \frac {a(b^2 \plus{} c^2 \minus{} a^2) \plus{} b(a^2 \plus{} c^2 \minus{} b^2) \plus{} c(a^2 \plus{} b^2 \minus{} c^2) \minus{} 2abc}{abc} \equal{} \frac {(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}{abc}$
$ \Leftrightarrow (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) \equal{} a(b^2 \plus{} c^2 \minus{} a^2) \plus{} b(a^2 \plus{} c^2 \minus{} b^2) \plus{} c(a^2 \plus{} b^2 \minus{} c^2) \minus{} 2abc$ (True!)
$ \rightarrow$ QED

Since $ \angle A\equal{}30^\circ$, we get $ R\equal{}BC$. Hence from [url=http://www.mathlinks.ro/viewtopic.php?t=277079]this topic[/url] we have
\[ DE^2\equal{}BC^2\left(1\minus{}\frac{2r}{R}\right)\equal{}R^2\minus{}2Rr\equal{}OI^2.\]

DE is parallel to the line XYZ connecting feet of the external angle bisectors. This is radical axis of circumcircle (S') and 9-point circle (S) of the excentral triangle of the triangle ABC, with orthocenter I => DE is perpendicular to ISS'. If M is midpoint of AB and incircle touches AB at F, then EA = 2FM. By sine theorem for triangle ADE,
$ DE \equal{} EA \cdot \frac {\sin \angle EAD}{\sin \angle ADE} \equal{} \frac {FM} {\sin \angle ISM} \equal{} IS$

[quote="Mathias_DK"]
What is "porcupine theorem"? :)[/quote]
Porcupine theorem:
On the figure, $ |\vec{u_1}| \equal{} |\vec{u_2}| \equal{} ... \equal{} |\vec{u_n}|$ then $ A_1A_2\vec{u_1} \plus{} A_2A_3\vec{u_2} \plus{} ... \plus{} A_nA_1\vec{u_n} \equal{} \vec{0}$

May anyone prove that $ SI$ is radical axis of two circles with diameters $ ID$ and $ IE$ ?

Dear Mathlinkers,
an article concerning some results about “two equal segments on two sides of a triangle” has been put on my website
http://perso.orange.fr/jl.ayme    vol. 6 Deux segments égaux sur deux côtés d'un triangle  p. 23
You can use Google translator
Sincerely
Jean-Louis

Using vector makes my head dizzy @@

If $AB=BC$ or $AC=BC$ the result is easy to get.
If $AB \ne BC ; AC \ne BC$ .
WLOG :$AB \le AC$
Let $M$ be the midpoint of arc $BC$ having $A$
$K$ be apoint on $AB$ such that $MK=MD ; K \ne D$ .$F$ is midpoint of $KD$
It's well-known that :$FA=\frac{AC-AB}{2}$
$\Rightarrow BK=AB+AC-BC=2AI.cos\frac{A}{2}$
So that ,$\Rightarrow \frac{BK}{BM}=\frac{ 2AI.cos(\frac{A}{2})}{2Rcos(\frac{A}{2})}=\frac{ AI}{R}=\frac{AI}{AO}$
We have:
$ \frac{BK}{BM}=\frac{AI}{AO}$
$\angle KBM=\angle IAO \Rightarrow \triangle KBM\sim \triangle IAO$
$\Rightarrow \frac{OI}{R}=\frac{OI}{OA}=\frac{MK}{MB}=\frac{MD}{MB}$
And since $\triangle MDE \sim \triangle MBC \Rightarrow \frac{MD}{MB}=\frac{DE}{BC}$
$\Rightarrow \frac{OI}{R}=\frac{DE}{BC}$
So that , because $\angle A=30 \Rightarrow BC=R$
$\Rightarrow OI=DE$

Is the following a surprising proof? At least for me, yes!
Keeping original problems's notation:

See that $\triangle BID\cong\triangle BIC\cong\triangle EIC$; as $\angle BIC=105^\circ\implies\angle DIE=45^\circ\ (\ 1\ )$.
$\angle ABS=90^\circ-\angle C,\ BD=BS$ so $\angle BDS=45^\circ+\frac{\angle C}{2}\implies\angle IDS=45^\circ\ (\ 2\ )$; in a similar way $\angle SEI=45^\circ\ (\ 3\ )$, so we infer that $S$ is the orthocenter of $\triangle DIE$ and $SI\bot DE$.
With $\angle DIE=45^\circ$ we know that $SI=DE$ (the distance from a vertex to the orthocenter is double distance from circumcenter to midpoint of opposite side which is, for such triangle half of that side).

Best regards,
sunken rock

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14655]China TST 1988[/url] for more solutions.

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Highway%20to%20Geometry%204.pdf  p. 32...

Sincerely
Jean-Louis


probability
geometry
percent
number theory
divisibility tests
algebra
system of equations

1. Find a prime number with three digits, given that the product of the digits is 252.

2. The sum of three numbers is 846. Find the numbers given that the second is 2 larger than the first and twice as small as the third.

3. Can the fraction (4n+3)/(5n+4) be reduced, for some n?

Sources : Romanian 5th Grade Olympiads, 1. Local Round, '73 2. 2. (equivalent of state) Round '77 3. (equivalent of state) Round '94.

Solution to (1):

[hide]

Well I said let this number be ABC where A, B, and C are digits.  ABC is prime, and A*B*C = 252.

Now you want to eliminate all things that C could be.  Last digits are connected w/divisibility (tests), so C cannot be 0,5, 2,4,8.  6, 3, 9 are possible values b/c the divisibility tests have to do w/sums of digits.  7 i dunno but I dont think its conncected to the last digit.  THen I used casework:

C = 1 , A*B = 252 = 2*2*3*3*7 no way to express this as the product of two 1 digit #s

C = 3, A*B = 252/3 = 2*2*3*7, ditto

C = 6, A*B = 252/6 = 2*3*7, so 6 and 7 are the digits.  ABC = 676, 766

but both are divisible by 2 and thus not primes

C = 7, A*B = 2*2*3*3, so the digits are 4 and 9 or 6 and 6.  ABC = 667, 497, 947.  667=23*29, so thats not it.  947 works.  497 is divisible by 7, so that dosent work.

ABC = 947

C = 9, A*B = 2*2*7, so the digits are 4 and 7.  ABC = 479, 749.  479 works.  749 is divisible by 7.

ABC = 479

Therefore the two solutions are 947, 479.

[/hide]

Solution to (2):

[hide]

Numbers are x,y,z.

x+y+z=846

y=x+2

z=2y

Solve the system.

[/hide]

Solution to (3):

[hide]

The obvious solution is:

0/something = 0

4n+3=0

n = -3/4

But I dunno if you count this.

Other than that, I would say no, because if you graph y = (4x+3)/(5x+4) then as x--> :inf: y --> 4/5.  So it will never reduce other than for -3/4. [/hide]

Actually I am going to change my answer for #3.

[hide]

I think that it simplifies for an infinite number of x values.  I started w/

(x=n)

y = (4x+3)/(5x+4)

Solve for x in terms of y:

5xy + 4y = 4x + 3

x(5y -4) = 3-4y

x = (3 - 4y)/(5y -4)

Now lets say that we wanted the fraction to simplify to 1, ie y=1

x = (3-4)/(5-4) = -1/1 = -1

Not lets say we want the fraction to simplify to 2, ie y=2

x = (3-8)/(10 - 4) = -5/6

So yea, it works for a lot of numbers.

[/hide]

Sorry, I need to clarify that last problem. n must be a positive integer.

Then no, I dont think it simplifies.  Are my other answers correct?

Am I the only one who is completely flabbergasted that these problems were given to [b]5th[/b] graders? And not even at the national level? O.O

I don't think I would have been able to solve #3 as a 5th grader, but I think I could have done the other two, although they probably would have been tough for me.

I agree with Simon.  I started math team training and was taking a reasonably advanced math class (I'd done probability and combinations and boolean logic) by 5th grade.  I mean, they might not have been [i]easy[/i] for me in fifth grade, but I had all the tools and everything.  Number 3 is harder, but it certainly doesn't look like it ever reduces, which is probably what I would have guessed :)

My little brother is in 5th grade and he is learning things like introduction geometry (types of shapes and their properties) and long division, multiplying decimals, percents, easy probability.  In those grades (at least in RI) all the kids are put in classes that are not based on ability -- ie everyone studies the same stuff.  On the other hand ,my other brother is a freshman in HS and would be able to solve them as well as I did.  It is suprising that they are for 5th graders, but these are probably either given to only the gifted 5th graders who participate in math olympiads, or that the Russians are ahead of us (by far).  JBL -- is Boolean operators you are talking about the same thing as the AND, ADJ, NEAR, etc. that you use when searching the internet?

[quote="JS1527"]Another thing.  The major problem w/fifth graders from US doing these is that they don't know algebra.  So I think #2 would be hard since the only way they know is guess and check.[/quote]

When I did basic math competitions in elementary school, my mom would always get mad, because there would always be problems that could be solved with a really simple system of equations, but none of us had learned algebra.

Anyway.. the first problem is definitely solvable for a fifth grader.  It might be a little difficult, but I'm pretty sure any fifth grader seriously taking math competitions knows about primes and divisibility rules.

[quote="Combinatoricsman"]Honestly, I don't think that the 3rd problem is really a very good one for 5th graders, since the proof is so rigorous that they certainly have to guess.[/quote]

Yeah, I know a lot of [intelligent] ninth graders who probably couldn't do that... I guess it goes to show the different math levels in Romania and the US >.<

IMO 1 #1:

Prove that (21n+4)/(14n+3) is irreducible for every natural number n.

It's the same problem as #3 with just different constants. So much for 5th grade math.

The problem, since it asks for no proof, probably assumes that the 5th graders will try a few values, see that they don't work, and then say that it never reduces.

Since this is the begginer's forum, I translated the problems so that no proof was required. But the original problem was like: "Prove or disprove....". Like all math contests in Romania (excluding the 8th grade high school entrance exam, which is 1/2 multiple choice 1/2 written) proofs are required.

Well I couldn't solve the 3rd problem in 5th grade, but I think I probably could've solved #1 maybe #2

#3 solution:  [hide]it is just same as (4(n+1)-1)/(5(n+1)-1). let n+1=k

so (4k-1)/(5k-1). Let 5k-1=m.  (m-k)/m can be reduced iff k/m can be reduced. k/(5k-1) can never be reduced for k>1 since they are always relatively prime because (5k-1) mod k =k-1, and k and k-1 is always relatively prime. [/hide]

Yes, this is very hard for 5th graders, but how can it be an IMO problem?

That was the first problem on the very first IMO.  The problems hadn't reached today's level of difficulty.

Prove that (21n+4)/(14n+3) is irreducible for every natural number n.

It's easy because its the first question on the first IMO.

Soln:

[hide]

(14n + 3)*3 - (21n + 4)*2 = 1, so

14n+3 and 21n+4 are relatively prime.

[/hide]

Back in the day when IMOs were easy


Find the smallest natural number n,such that there exist positive integrs $ x_1,x_2,...,x_n$, such that $ x_{1}^3 \plus{} x_{2}^3 \plus{} ... \plus{} x_{n}^3 \equal{} 2008$

[hide="Hallo"]
Let $ x_1$ be largest such $ x_1^3<2008$, it will be $ 12$
Then a largest $ x_2\equal{}6$ and finally $ x_3\equal{}4$
which makes a perfect 2008
[/hide]

No. He has not proven that two cubes won't work.

[hide="answer"]$ n\equal{}3$ numbers, being $ 4 6 12$[/hide]

I think that it must be different numbers,so $ 2^3\plus{}10^3\plus{}10^3\equal{}2008$ doesn't work.
PS:I still wait foe complete solution of the problem.

[quote="barcelona"]I think that it must be different numbers,so $ 2^3 \plus{} 10^3 \plus{} 10^3 \equal{} 2008$ doesn't work.
[/quote]

The original problem doesn't say that the numbers must be different (although it doesn't matter anyway, since we've already found values for n=3 which are different and there is no value for n=2 with the two numbers equal).

Im pretty sure, to show that 2 cubes doesnt work, you can just do teh arithmetic and show that 2008-a^3 for a=0,1,2,...12  is not a cube.

assume there are two positive integers $ a,b$ such that $ a^3 \plus{} b^3 \equal{} 2008$

Then $ 2008 \equal{} a^3 \plus{} b^3 \ge \frac {(a \plus{} b)^3}{4} \Longrightarrow a \plus{} b < 2\sqrt [3]{1004} < 2\cdot 11 \equal{} 22$

Since $ 2008 \equal{} 2^3\cdot 251$ we have $ a \plus{} b \equal{} 1,2,4$ or $ 8$

But $ a^3 \plus{} b^3 \equal{} (a \plus{} b)(a^2 \minus{} ab \plus{} b^2)$ so $ a^2 \minus{} ab \plus{} b^2 \ge 251$ but $ a^2 \minus{} ab \plus{} b^2 \equal{} (a \plus{} b)^2 \minus{} 3ab < 8^2 \equal{} 64$

contradiction


vector
integration
calculus
calculus computations

This occurred to me while doing a problem and thought if there is a way to do the inverse of a curl:

Suppose $ \mathbf{F}\equal{}x^2\hat{x}\plus{}3xz^2\hat{y}\minus{}2xz\hat{z}$, where $ \hat{x},\hat{y},\hat{z}$ are Cartesian unit vectors. Now $ \mathrm{div}\,\mathbf{F}\equal{}0$ so there must exists a vector $ \mathbf{A}$ such that $ \mathbf{F}\equal{}\nabla\times\mathbf{A}$ or the curl of $ \mathbf{A}$. Question is ... can we find $ \mathbf{A}$?

I know the answer is not unique up to a gradient but can a solution be characterized? What are some conditions if it fails to yield a solution? I know I'm asking a lot but any input will be greatly appreciated. Thanks :)

Suppose $ \bold{F}(x,y,z) \equal{} f_1(x,y,z)\hat{x} \plus{} f_2(x,y,z)\hat{y} \plus{} f_3(x,y,z)\hat{z}$ is of the class $ C^1$ on $ \mathbb{R}^3$ and $ \text{div} \bold{F} \equal{} 0$. Let

$ \bold{A}(x,y,z) \equal{} \left( \int_{0}^{z} f_2(x,y,t) \, dt \minus{} \int_{0}^{y} f_3(x,t,0) \, dt \right) \hat{x} \minus{} \left( \int_{0}^{z} f_1(x,y,t) \, dt  \right) \hat{y}.$

Then you can check that $ \nabla \times \bold{A} \equal{} \bold{F}$.

Thx for the prompt reply, sos440. This sort of symmetric-looking solution seems to suggest something about a general solution! :)

[url=http://en.wikipedia.org/wiki/Vector_potential]Wikipedia article on vector potential[/url].

The general solution is just the sum of any particular solution with a conservative field.


Given $ f(x) \equal{} \dfrac{x^2 \plus{} 2x \plus{} 1}{x^2 \minus{} 1}$, find $ \dfrac{f(2)}{f(0)}$

So to be absolutely rigorous, one has to note that $ x \equal{} \pm1$ will not work in $ f(x)$ and check for it, but since this is CD it's like SCREW THIS I DON'T CARE.

Well the simplification (into (x+1)/(x-1)) assumes that x=-1 is not the case. x can't be 1 even in the simplified case as it's undefined.

That's kind of my point... :wink: But I also noted that this is a CD so the answer has to be defined.


floor function
number theory unsolved
number theory

Prove that the sequence defined by $a_{1}=1$ and
$a_{n}=a_{n-1}+a_{\lfloor\frac{n}{2}\rfloor}$ contains infinitely many terms divisible by 7.

Since you posted it in the "Unsolved Problems" section - check a simple solution at [url=http://www.om.edu.pl]www.om.edu.pl[/url]


[hide]Add 1 to each of them, and they can be factored (x+1)(y+1), and similarly for the other two. So it boils down to ab=24, ac=42, bc=28. I just stared at it and got a=6, b=4, c=7, so x=5, y=3, z=6.[/hide]

Adding 1 to the three equations, (realize that xy + x + y + 1 = (x + 1)(y + 1)) gives:

(x + 1)(y + 1) = 24
(y + 1)(z + 1) = 42
(x + 1)(z + 1) = 28

Multiply the three equations and take the square root, giving

(x + 1)(y + 1)(z + 1) = + or - 168

Taking these values, divide by each of the above three equations to obtain x, y and z. The solutions are (5,3,6) and (-7,-5,-8).

funny how degree said it will take ages...i got it in literally less than 30 seconds! and i didn't even look at bictor's answers, i just looked at the equations and took a wild guess, lol...

i know they gonna be small numbers, so i tried 3 ,4 for x, y, and turned up to be 19, so i tried 3,5 boom, 23...then i look down, i saw them both have z in it, and ones 41, ones 27, so i set x to be the 5, cuz its bigger, and tried 6, boom, worked again got 41, then 3,6 just automatically fell in place

damn i pwned that...lol! but i admit it's a lucky guess...if the numbers were bigger, and all composite, it might take much longer...lol

wow... not as long as I thought

can you learn that kind of insight ?

[quote="Altheman"]you have to do many problems...[/quote]

Well Altheman, you are correct I too need to do a lot of problems because the method I used was very long.
I got the values of x and y in terms of z with the help of equations given and then put up this value in the third equation to obtain the answer.
Please give some problems for practice.

ok, what i mean is that to gain insight, you have to keep challenging yourself with problems that require you to think, many problems on these forums, and in the aops books will be helpful, often there is a very elegant solution that once you see, the problem is done

but here is one problem that i thought was particularly clever (i forgot who posted it, but it is AMC)

if $(3x-1)^7=a_7x^7+a_6x^6+...a_1x^1+a_0$, then find the sum $a_8+a_7+a_6+...+a_1+a_0$

if you see it, you can do it in your head...

this illustrates the idea of the elegant solution, sometimes elegant solutions are not possible, but many difficult problems can be simplified with clever insight, you just have to keep doing problems to develop you insight,

i remember when i first started doing extra math (like last year), many of the problems that i looked at some of the problems on the AMC/AIME website for their examples and i did not understand the more difficult ones, but now i can usually get most of them

[quote="Kalle"]Since the identity is true for all x couldn't you just insery say x = 1 ? :P[/quote]Assuming he meant to say "for all $x$, etc." that would be true.


Prove that if $ gcd(a\plus{}b,c)\equal{}1,gcd(a,c)\neq 1,gcd(b,c)\neq 1$ then $ gcd(a,b)\equal{}1$

It's not true. For example let $ a\equal{}20,b\equal{}15,c\equal{}6$. Then $ (a\plus{}b,c)\equal{}(35,6)\equal{}1$, $ (a,c)\equal{}(20,6)\neq 1$, $ (b,c)\equal{}(15,6)\neq 1$ but $ (a,b)\equal{}(15,6)\equal{}3\neq 1$.


algebra
polynomial
quadratics
superior algebra
superior algebra unsolved

Let $\displaystyle \mathcal K$ be a finite field such that the polynomial $\displaystyle X^2-5$ is irreducible over $\displaystyle \mathcal K$. Prove that:

(a) $1+1 \neq 0$;

(b) for all $\displaystyle a \in \mathcal K$, the polynomial $\displaystyle X^5+a$ is reducible over $\displaystyle \mathcal K$.

[i]Marian Andronache[/i]

[Edit $1^\circ$] I wanted to post it in "Superior Algebra - Groups, Fields, Rings, Ideals", but I accidentally put it here  :blush:  Can any mod move it? I'd be very grateful.

[Edit $2^\circ$] OK, thanks.

(b) Let $p$ be the characteristic of $\mathcal{K}$. Clearly, $5$ is not a quadratic residue mod $p$. By quadratic reciprocity, $p$ is not a quadratic residue mod $5$, and either $p\equiv 2\mod 5$ or $p\equiv 3\mod 5$.
Since a quadratic polynomial over $\mathbb{F}_p$ is irreducible in $\mathcal{K}$, $\mathcal{K}$ has order $p^n$ for some odd $n$. Now $5\nmid p^n-1$, and $x\mapsto x^5$ is bijective on $\mathcal{K}^*$. Therefore, $a$ has a fifth root $b$ in $\mathcal{K}$ for every $a$, and $x-b$ is a factor of $x^5-a$.

[quote="jmerry"]Since a quadratic polynomial over $\mathbb{F}_p$ is irreducible in $\mathcal{K}$,[/quote]why is this?

$X^2-5$ is an irreducible quadratic by hypothesis.
The next statement follows from the fact that all quadratic polynomials over $\mathbb{F}_p$ split in $\mathbb{F}_{p^2}$.


function
integration
calculus
real analysis
real analysis theorems

Let $M>0$ and $u,v: I\to\mathbb{R_{+}}$ two continuous functions  and let $t_{0}\in I$ such that

                                                      \[ u(t)\leq M+|\int_{t_{0}}^{t}v(s)ds|,\forall t\in I,  \]

    then \[ \displaystyle u(t)\leq M\cdot\exp\left({|\int_{t_{0}}^{t}u(s)v(s)ds|\right).  }\]

Are you sure it's the right statement? [url=http://planetmath.org/?op=getobj&from=objects&id=3901]PlanetMath[/url] says that you have exchanged the integral in the hypothesis with the one in the conclusion.

They also provide [url=http://planetmath.org/?op=getobj&from=objects&id=3902]a proof[/url] to their version of the Lemma.


[b]Problema 4441(Mvsc, nr.5/1999):[/b]
$  \begin{array}{c}
a^2-b^2=bc\\
b^2-c^2=ac
\end{array}\}\Rightarrow a^2-c^2=ab; (\forall)a,b,c\in\mathbb{R}^*.$ ([i]Mirotin[/i]-Bielorusia)

[b]Problema 4415(Mvsc, nr.3/1999):[/b]
$ \mbox{Demonstrati, ca fiecare numar natural poate fi scris sub forma diferentei a doua numere naturale,}$
$ \mbox{care au acelas numar de divizori primi.}$ ([i]Takarev[/i])

Prima e o manipulare algebrica nu foarte complicata (eu am inmultit prima relatie cu $ c$, a doua cu $ a$ si le-am scazut...)
A doua problema e foarte simpatica.
Hint: daca $ n$ e par, atunci scriem $ n\equal{}2n\minus{}n$.

Daca $ n$ e impar, scriem $ n\equal{}np\minus{}n(p\minus{}1)$, unde $ p$ e  [hide]cel mai mic numar prim care nu divide $ n$.[/hide]

Eu aveam solutiile acestor probleme, intrucat solutiile tuturor problemelor din "Matematika v skole" si din "Kvant" apar dupa 4-5 numere.
Am tradus enunturile problemelor din aceste reviste pentru elevii nostrii care nu mai invata rusa in scoala.
[i][b]Vad insa, ca elevii nu prea se inghesuie sa rezolve aceste probleme.[/b][/i]
Multe dintre numerele revistei "Matematika v skole" dintre anii: 1929-1986, le puteti descarca de aici:
http://www.mathedu.ru/journals-collections/
La aceasta adresa gasiti si niste carti bune [u]toate in RUSA[/u]; de exemplu la sectiunea: Matematica si metodica/Geometrie, pe pag. a 2-a,
gasiti cartea lui [i]Zetel[/i]: [b]NOUA GEOMETRIE A TRIUNGHIULUI [/b](1940); iar la sectiunea Superioara(Vashaya) cartea lui
[i]Buhstab[/i]: [b]TEORIA NUMERELOR [/b](1966)(384p) si a lui [i]Joung[/i]: [b]GEOMETRIE PROIECTIVA [/b](1949)(184p) si multe altele.
Problemele din "Kvant" le gasiti aici: http://kvant.info/zkm_tex/zkm_main.pdf
iar numerele complete ale acestei reviste dintre anii 1970-2003, aici:
http://kvant.mirror1.mccme.ru/oblozhka_djvu.htm
sau de aici (pagina cu pagina in format jpg, dintre 1970-2008): http://kvant.eduhmao.ru/


Source: LBJ 2005

In a certain double-elimination chess tournament, there were 218 players, and no player went undefeated. How many matches took place?

are you asking for how many ways can these 218 player, have matches together.(also by saying no one went undefeted, do mean all of the player went with one lost?)

No, the question asks for the number of total games played in the tournament.

By rules of double elimination, a player is out of the tournament bracket once he/she has two losses. So what the question is essentially is saying is that everybody, save one person, was eliminated from the tournament, and the champion had one loss.

No, that's not correct.

How did you get that answer?

$435$?
(i got for every $n$ there must be $2n-1$ games)

yeah, i think it should be more than that. but i get the same thing over and over!
$435$

We're assuming that each game produced one winner and one loser, of course. No draws in this tournament. :)

[hide="Good place to start"]
How many total losses are there?
[hide]
435 is indeed the correct answer.  :D
[/hide]
[/hide]

Derek wrote:[hide]217 people were eliminated because they lost 2 games. The champion lost 1. So 435? Not sure   [/hide]

the reason it is the right answer is that every game that is played has two players in it so one game is counted for both of them, not only for one of them


Let $ \frac{a}{b}\equal{}x$ and $ \frac{b}{c}\equal{}y$ so $ x$ and $ y$ are $ \geq1$.

Inequality equivalents to $ (y^3\minus{}y^2)x^3\plus{}x^2\minus{}(y^2\plus{}1)x\plus{}y \geq0$.

Let $ (y^3\minus{}y^2)x^3\plus{}x^2\minus{}(y^2\plus{}1)x\plus{}y\equal{}f(x)$ so $ f'(x)\equal{}3(y^3\minus{}y^2)x^2\plus{}2x\minus{}(y^2\plus{}1) \geq 3(y^3\minus{}y^2)\plus{}2\minus{}y^2\minus{}1\equal{}(y\minus{}1)(3y^2\minus{}y\minus{}1) \geq0$

$ \Longrightarrow$     $ f$ is increasing for all $ x \geq1$

$ f(x) \geq f(1)\equal{}y^3\minus{}y^2\plus{}1\minus{}(y^2\plus{}1)\plus{}y\equal{}y(y\minus{}1)^2 \geq0$ and solution is done.  :wink:

[quote="crazyfehmy"]Let $ \frac {a}{b} \equal{} x$ and $ \frac {b}{c} \equal{} y$ so $ x$ and $ y$ are $ \geq1$.

Inequality equivalents to $ (y^3 \minus{} y^2)x^3 \plus{} x^2 \minus{} (y^2 \plus{} 1)x \plus{} y \geq0$.

Let $ (y^3 \minus{} y^2)x^3 \plus{} x^2 \minus{} (y^2 \plus{} 1)x \plus{} y \equal{} f(x)$ so $ f'(x) \equal{} 3(y^3 \minus{} y^2)x^2 \plus{} 2x \minus{} (y^2 \plus{} 1) \geq 3(y^3 \minus{} y^2) \plus{} 2 \minus{} y^2 \minus{} 1 \equal{} (y \minus{} 1)(3y^2 \minus{} y \minus{} 1) \geq0$

$ \Longrightarrow$     $ f$ is increasing for all $ x \geq1$

$ f(x) \geq f(1) \equal{} y^3 \minus{} y^2 \plus{} 1 \minus{} (y^2 \plus{} 1) \plus{} y \equal{} y(y \minus{} 1)^2 \geq0$ and solution is done.  :wink:[/quote]
My solution only use AM-GM inequality for two numbers :)

[quote="Allnames"][quote="quykhtn-qa1"]Given $ a \ge b \ge c > 0$.Prove that:
$ a^2(\frac {b}{c} \minus{} 1) \plus{} b^2(\frac {c}{a} \minus{} 1) \plus{} c^2(\frac {a}{b} \minus{} 1) \ge 0$ :)[/quote]
You should write clearly your source,dear  :mad: Because It is in[b] VNMO 1991[/b]  group A problems.
One of proofs is here
\[ \leftrightarrow (b \minus{} c)(a \minus{} c)(a^2b \plus{} bc(a \minus{} b)) \plus{} c^3(a \minus{} b)^2 \ge 0
\]
Which is obvious since $ a \ge b \ge c > 0$[/quote]
Sorry,I don't know. :wink:
Here is my solution
 We have: $ a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2 \minus{} a^2b^3 \minus{} b^2c^3 \minus{} c^2a^3 \equal{} (a \minus{} b)(b \minus{} c)(a \minus{} c)(ab \plus{} bc \plus{} ca) \ge 0$
Then,$ \frac {LHS}{abc} \ge \sum \ \frac{1}{2} a^2(\frac {b}{c} \plus{} \frac {c}{b}) \ge \sum \ a^2$
Then, we have Q.E.D :oops:

[quote="secrets"]Thank you,but my solution is use cauchy and SS.[/quote]
YES,IT 'S TRUE :)


From a standard deck of cards, one card is drawn.  What is the probability that the card is black and a jack?

[hide]  There are 2 kinds of black jacks: one jack of clubs and one jack of spades.  Since there are 52 cards in a deck, the probability is simply 2 out of 52, which is $\frac{2}{52}=\frac{1}{26}$.  [/hide]

Jack can answer a jack question!

[hide="Jacks answer to the jack question"]Since there are 52 cards in a deck, and 2 black jacks, the answer would be 2/52, or 1/26[/hide]


If $f: \mathbb R\to \mathbb R$ and $f(x)+y=f(x+y),\forall x,y \in \mathbb R$ it means that f is periodic ?

Choose $x=0$ we have $f(x)=x+C$, where $C$ is constant. So $f$ NOT periodic.


limit
logarithms
calculus
calculus computations

[quote="Ramanujan"]You can use $\lim_{x \to \infty }\left(1-\frac{1}{x}\right)^{x}= e^{-1}$.

First write the limit as $\lim_{x \to \infty}(1-\frac{4}{3x+2})^{x}$

and make the substitution $3x+2 = w$.

Notice that $w$ goes to infinity when $x$ goes to infinity .[/quote]

But $w$ and $x$ go to infinity at different rates, so the answer should be

$e^{-\frac{4}{3}}$

I think.

paladin8 is the one who's correct.

Let $L=\lim_{x\to\infty}\left(1-\frac4{3x+2}\right)^{x}$

$\ln L=\lim_{x\to\infty}x\ln\left(1-\frac4{3x+2}\right)$

$=\lim_{x\to\infty}x\left(-\frac4{3x+2}+o(x^{-1})\right)$

$=\lim_{x\to\infty}-\frac{4x}{3x+2}+o(1)=-\frac{4}{3}.$