induction
graph theory
combinatorics unsolved
combinatorics

Let be given a $ 100\times 100$ chessboard whose all cells are empty.
(a) Is it possible to color an odd number of cells so that every colored cell is adjacent to an odd number of colored cells?
(b) Is it possible to color some cells so that an odd number of them are adjacent to exactly four, while the remaining colored cells are adjacent to exactly two colored cells?
(c) Is it possible to color some cells so that an odd number of them are adjacent to exactly two, and the remaining colored cells are adjacent to exactly four colored cells?

I don't have a proof for (b), but I think that it is a NO.

(a) is just a special case of the first theorem in every graph theory book: in any graph, the sum of all the degrees is twice the number of edges, so it's an even number. Corollary: a graph where all the degrees are odd must have an even number of vertices. This applies here by viewing the colored cells as vertices and joining them with an edge if they are adjacent. So the answer is No.

(b) is quite easy to do:
[code]
XXX
X X
XXXXX
  X X
  XXX
[/code]

Here there is a single cell adjacent to 4 others.

(c) is impossible. The short proof is that a 4-regular multigraph has an even number of edges, and moreover every cut (= set of edges connecting a vertex in $ X$ to a vertex in $ V \setminus X$) has an even number of edges. In a checkboard pattern, the paths consisting of an even number of 2-cells connect cells of opposite colors, so there's an even number of those, and therefore there's an even number of odd-length paths, so an even number of 2-cells overall.

Here is a more detailed proof for those unfamiliar with graph theory. Draw some pictures; this is much simpler than it seems.

Let $ F$ be the collection of 4-cells (i.e. colored cells adjacent to 4 others) and let $ T$ be the collection of 2-cells. A "walk" is a sequence of cells $ (f,t_2,t_3,\ldots,t_m, g)$ such that $ f,g \in F$ while $ t_i$ is in $ T$ for all $ i$, and every cell is adjacent to the next one. The [i]length[/i] of the walk is defined to be $ m$. Note that it is possible that $ f \equal{} g$. Also note that a walk may have length 0: if $ f$ and $ g$ happen to be adjacent, there is a walk $ (f,g)$. With these conventiona, from every $ f \in F$ we can embark on a walk in precisely 4 directions and keep going until we hit $ F$ again, so there are exactly $ 4n$ walks where $ n \equal{} |F|$.

For every walk there is an opposite walk $ g,t_m, \ldots, t_1, f$ obtained simply by going backwards. A pair of opposite walks is called a [i]path[/i]: it is simply a (possibly empty) collection of cells in $ T$ connecting $ f$ and $ g$. A path has a well-defined length $ m$. The important point about paths is that every cell in $ T$ belongs to a unique path. Therefore, $ |T|$ is just the total length of all paths. We need to show that $ |T|$ is even, so we must show that there is an even number of odd-length paths. Since there are $ 4n$ walks, there are $ 2n$ paths, so it's enough to show that there is an even number of even-length paths.

Suppose that the members of $ F$ are partitioned into two sets in any way, $ F \equal{} A \cup B$. Some paths are now $ AB$ paths, connecting a member of $ A$ with a member of $ B$. We claim that the number of such paths is always even. Let $ k \equal{} |B|$, and use induction on $ k$. If $ k \equal{} 0$, there are no $ AB$ paths at all. If we move a cell $ x$ from $ A$ to $ B$, we lose the paths that start in $ x$ and end in $ B$, but we gain the paths that start in $ A$ and end in $ x$. The paths leading from $ x$ to $ x$, if any, don't matter either way, and there is always an even number of paths connecting $ x$ to something else (there could be 0, 2 or 4 such paths). So the total change in number of $ AB$ paths is always even. This proves the claim.

Apply a checkerboard pattern to the board, so that the cells are now partitioned into black and white ones. Let $ A$ be the white cells in $ F$ and $ B$ the black ones. Observe that the paths of even length are exactly the $ AB$ paths, and we just proved that there's an even number of those, so we're done.


modular arithmetic
Diophantine Equations
pen

[quote="Peter"]Prove that there are unique positive integers $a$ and $n$ such that \[a^{n+1}-(a+1)^{n}= 2001.\][/quote]
$-(a+1)^{n}\equiv 2001 \mod a$
$-1 \equiv 2001 \mod a$
$0 \equiv 2002 \mod a$
a=1, 2, 7, 11, 13, 154, 182, 286, 1001, 2002
Don't all of these work for a?

All of them work for your equation, but it's not because they satisfy that mod a equation that they'll satisfy the original problem...

The only thing you now know is that $a\in\{1, 2, 7, 11, 13, 154, 182, 286, 1001, 2002 \}$, now you have to prove that exactly one of those $a$'s has exactly one $n$ for which the equation is true.

As shown earlier a|2002. This means that a can only be
1, 2, 7, 11, 13, 14, 22, 26, 77, 91, 143, 154, 182, 286, 1001, 2002.
Mod a+1 gives us -1^(n+1)=2001 mod a+1
so either a+1|2002 or a+1|2000
If a|2002 and a+1|2002, we have that a=1 or a=13. a=13 works (n=2).
Our other case is a+1|2000. If this is true, then the only factors of a+1 are 2 and 5. Checking our factors above (and adding 1), we see that none are powers of only 2 and 5.
So, our only solution is $ 13^{3}\minus{}14^{2}$

Hi. Could you explain why Mod a+1 gives us -1^(n+1)=2001 mod a+1? I do not seem to be able to get this part, maybe I do not get it. Mind explaining? thanks.

We are given $ a^{n\plus{}1} \minus{} (a\plus{}1)^n \equal{} 2001$, and you expressed no surprise that then $ 2001 \equiv \minus{}1 \pmod{a}$. Similarly, $ 2001 \equiv ((a\plus{}1)\minus{}1)^{n\plus{}1} \minus{} (a\plus{}1)^n \equiv (\minus{}1)^{n\plus{}1} \pmod{(a\plus{}1)}$.

Ah... I see it now. Haha. Really need to practice manipulating delicately, eh. Thanks for the explanation.

A quick way to prove that $a|2002$ is to expand the LHS, move 2001 to the LHS, and use the Rational Root Theorem.

works mod 3 gives n even, so only the case a+1|2002.


Is there any elementary way (without calculus) to prove the volume of a pyramid is 1/3 times as the "pillar" with the same height and bottom area?

Searching on the forum for "Cavalieri" will get you what you need.  (There's one exchange between me and Altheman and one post by Kent Merryfield that look particularly helpful.)


vector
function
calculus
derivative
real analysis
real analysis theorems

I'm informed that this is a second order partial differential equation.

Two forms are present:Laplace equation in scalar form and in vector form.
What is difference?Is that just matter of notation of the same thing,or more fundamental difference is involved?
Thanks,
E88

I'm no expert, so I can't give a nice and detailed explanation. This is what I know:

The Laplace equation says that $\frac{\partial^{2}F}{\partial x_{1}^{2}}+...+\frac{\partial^{2}F}{\partial x_{n}^{2}}= 0$ for a function $F$ of $n$ variables.

The vector form you're thinking of is that div*grad is the same as laplacian. In fact I think that's how it's defined for vector functions. ("Laplacian" refers to the differentiation operation you do to the function in Laplace's equation.)

The difference is between whether you're acting on a vector function or a scalar function.

By the way, the identity $\Delta F=\mathrm{div}\nabla F$ is useful when you have to compute the Laplacian of a composition. Both $\mathrm{div}$ and $\nabla$ are first-order differential operators with constant coefficients, so the Chain Rule applies to both.

For a vector-valued function $\vec F$, $\Delta\vec F=(\Delta F_{1},\dots,\Delta F_{n})$. Nothing really new here.


Which of the two numbers is greater: $ 100!$ or $ 10^{150}$?

[quote="Jane_Bennet"]$ \sum_{n\, \equal{} \,1}^{100}{\log_{10}{n}} > 150$[/quote]
How do you know this? A calculator?

The antiderivative of $ \ln x dx$ is $ x(\ln x \minus{}1)$, so
$ \int\limits_{1}^{100}\log_{10} x dx\equal{}(\log_{10}e)\int\limits_{1}^{100}\ln x dx\equal{}(\log_{10}e)(100(\ln 100 \minus{}1)\plus{}1)>100\log_{10}\frac{100}{e}>150$
This integral is a lower bound on the sum in question.

[hide="Alternately..."] We could also use [url=http://mathworld.wolfram.com/StirlingsApproximation.html]Stirling's Approximation[/url], which gives

$ 100! \approx \sqrt{200 \pi} \left( \frac{100}{e} \right)^{100} > 25 \left( 1000 \right)^{50} \equal{} 25 \cdot 10^{150}$

In fact, Stirling's approximation gives $ 9.32... \times 10^{157}$. [/hide]

lol i've got a solution that's clearly more fun

So let's take this: $ (1\cdot20)(21\cdot40)(41\cdot60)(61\cdot80)(81\cdot100)$

and that is definitely greater than $ (1\cdot20)(20\cdot40)(40\cdot60)(60\cdot80)(80\cdot100) \equal{} 147456\cdot10^{10} > 10^{15}$

Now take these nine values...

$ (2\cdot19)(22\cdot39)(42\cdot59)(62\cdot79)(82\cdot99)$
$ \cdots$
$ (10\cdot11)(30\cdot31)(50\cdot51)(70\cdot71)(90\cdot91)$

..all of which are greater than $ (1\cdot20)(21\cdot40)(41\cdot60)(61\cdot80)(81\cdot100)$.

Now notice that when you multiply these ten ugly expressions together, you get (whoa) $ 100!$. Since each of these are greater than $ 10^{15}$, it follows that $ 100! > 10^{150}$.

Wow, that's nice, Lazarus!  :)

Happy New Year 2008!


[quote="MithsApprentice"]Are we going to be able to see the archives of older classes we took before the archives system started?[/quote]

It's not likely.  Many aspects of our system have changed since them and it's not something we're set up to do automatically.  It would take more time than we have spare.


3. X, Y , and Z are points on the sides BC, CA, and AB of triangle
ABC such that AX, BY , and CZ concur at point P. Show that if AY PZ and BZPX
are cyclic, then so is CXPY .

[hide]So call angle A a and angle B b.

So angle XPZ is 180-b and angle YPZ is 180-a (definition of a cyclic quadrilateral)

Then because the three lines are concurrent, angle YPX=360-(180-a)-(180-b)=a+b.

Also, because all angles in a triangle add up to 180, angle C is 180-a-b.

Because angle C and angle YPX are opposite angles in a quadrilateral and they add up to a+b+180-a-b=180 degrees, CXPY must be cyclic.[/hide]

$ \angle AYP\equal{}\alpha$ $ \implies$ $ \angle BZP\equal{}\alpha$ $ \implies$ $ \angle PXC\equal{}\alpha$.
$ \angle AYB\equal{}\alpha \implies  \angle PYC\equal{}180\minus{}\alpha$
therefore CXPY cyclic

[quote="the future"]So angle XPZ is 180-b and angle YPZ is 180-a (definition of a cyclic quadrilateral)[/quote]

To nitpick, that's not the definition of a cyclic quadrilateral. The definition is simply that all four points lie on the same circle; what you are using is a property of cyclic quadrilaterals.


ok so can I say, looking at a PE (of the system) vs Position (of a particle in the system) curve, that a turning point for the particle would be at the point that meets the following: PE = mechanical energy of the system and -dU/dx has the opposite sign of the particle's direction of motion as it approaches the turning point

reasoning: at this point because PE = mechanical energy, the particle would have 0 KE and consequently 0 velocity and the force on it, and consequently its acceleration, would be in the opposite direction to which the particle was going so it would turn around

is this right?


function
trigonometry
real analysis
real analysis unsolved

Prove that $ p: \mathbb{R}\to \mathbb{R}$, with $ p(x)\equal{}\sum_{i\equal{}0}^{n}a_{i}x^{i}$ is uniformly continuous if and only if $ i\leq 1$

$ p'(x)$ is unbounded; apply the Mean Value Theorem.

Ah, but there exist uniformly continuous on the entire real line smooth functions $ f$ with unbounded $ f'$ (say, $ \frac{\sin (x^{10})}{1\plus{}x^2}$). ;)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=243091]The same question[/url], two weeks ago.


[quote="kunny"]Can anyone teach me how to type roman figure in Tex?

Thanks in advance[/quote]
Sorry, but do you mean Roman font?

[quote="kunny"]For example, Ⅰand ⅰ :)[/quote]Unfortunately that has come out as squares on the screen so not clear what you want.
Roman numerals use the Latin alphabet so can be typed as L, X, iv directly. But is that what you mean?


geometry
circumcircle
geometric transformation
homothety
ratio
geometry proposed

In a circle S we fix the chord BC and A is movable on circumference of S such that S is the circumcircle of triangle ABC. what is the locus of centroid of ABC?

I know this problem is discussed here before but i have not idea about homothety and translations so if somebody could please explain this in simple terms. i have the proof only if somebody could help me prove the following,

i don't understand your problem: A homothety with center the midpoint M of BC and ratio 3 take the centroid G on A so the locus of G when A is moving on S is the result of S uder a homothety with center M and ratio $ \frac{1}{3}$ (that is a circle).

i'm sorry i don't understand transformational geometry well as i have already written in my last post.... my question is what is the locus of centroid of triangle ABC if the the base BC of the triangle is fixed while A is movable on the circumcircle. the second part of my post (the one with the picture) basically ask how to prove that X and Y form a circle if MC and MD move like hands of a clock but in opposite directions. if this can be proved the original problem with the locus follows immediately.


Hey guys, I wanna know the difference between these two vector problems.

1- A swimmer crosses the river in a direction S 30 DEGREE W at 2.5 km/hr. A current of
1 km/hr towards the east changes the swimmer's direction. What is the angle of deviation from the
intended direction.

2- A ferry is used to transport vehicles between Cap-Rouge and Cap-Espoire, a distance of 8 km. Its
direction with respect to Cap Rouge is N 35 degree E. The current is 3.5 km/hr to the west. So as to
land at the Cap Espoir in one hour, at what speed and in what direction should the captain set out in the ferry?

How do I know that in the first one we're looking for the resultant, yet in the second problem the resultant is given? Both problems look pretty much the same to me?

Resultant implies total or sum.  If you're asked to find the total or sum, you're looking for the resultant.  If it's given to you already, then you're looking for some component.


geometry
geometric transformation
reflection

$ \Delta ABC$ is an equilateral triangle and $ D$ and internal point. $ D_1,D_2,D_3$ are the reflections of $ D$ on $ BC,CA,AB$ . Prove that $ AD_1,BD_2,CD_3$ have a common point

prob looks familiar..... :huh:
it's frm challenge and thrill?it has a similar prob....


Heres a simple strategy game based on warcraft's tower defense. The mechanisms of this game are explained in the beginning. How many levels can you get to? :wink:
Edit: highest i've gotten is 83!

http://www.jeannettevejarano.com/tower-defence.swf

i tried it and got 60
i just made fire dudes and nothing else :D

I made it to level 55 on my first try, but my attention span is way too small to play again.

[quote="Chocoboom"]got to 86! a shiny gold penny to anyone who figures out how to reach 90[/quote]

CHOCOBO!!!

anyways, isn't there an online guide to this? there are guides to every single game

whats your strategy guys? I got to level 16.... :oops:
EDIT: i figured out how to upgrade towers and got to lvl 64.  :D

I tried 5 times and got:
69, 68, 67, 70, 69

That's somewhat consistent.

Plan out a strategy beforehand.
Think what has the most damage for the least money.
Think where to place the towers so they fire the fastest.
Efficiency!

curse you for getting me addicted to this game...i got to lvl 77...then i was like wtf the green towers are getting owned

are teh super towers worth investing in? i have not upgraded, but it seems unlikely...

level: 50, 54, 59, 62, 65, 69

The green towers/nature towers aren't worth it...the game is glitched so that you pay an insane amount of money for only one-quarter of the upgrade in damage you're supposed to get.

Anyways my strategy is to use maxed out -1 ice towers for the entire map except for the middle one, which is maxed out, cause it only increases range(doubles it), but no damage...

last word: Range is worth a lot more than you think...use my strategy for fire towers...you fail ~lvl55

well i got a 63 =( because i didn't have enough cash to fully upload my def

[img]http://img2.freeimagehosting.net/uploads/32e5233e84.jpg[/img]

[quote="7h3.D3m0n.117"]well i got a 63 =( because i didn't have enough cash to fully upload my def

[/quote]
LOL my little sis pwned you! :rotfl: Sorry, just had to say that, but if it makes you feel better, she pwned me too.

[quote="Xantos C. Guin"][quote="daermon"]i got 72  :D
(without cheating.  i got 144 with the spacebar glitch :P )

bpms did u use the spacebar glitch or something?[/quote]
Only lvl 144 w/ the spacebar glitch? I got to lvl 1061 with the spacebar glitch.[/quote]

wow how did u get that? wats the spacebar glitch btw?  :D

this was my first time playing i dont think its very good my bro (7h3.d3m0n.117) beat me by one on this round....lol

[img]http://img3.freeimagehosting.net/uploads/af24fd33b8.jpg[/img]
see its not very good

My 2nd try :lol: (1st time I scored a 10 :P )

[img]http://img244.imageshack.us/img244/3221/towerdefenceiy1.jpg[/img]

*goes back to reading AoPS Volume One*

58 my best. No matter how hard I try... I give up :(

Desktop Tower Defense is much better IMO. I think I got to 79 on this one... with all nature towers. Put them on the end of long runs so they don't have to turn around.

I just got 71 by using 2 fully upgrade ice towers and a bunch of living towers... Though I died in the end, because of bad placement IMO.  I'm going to see if I can beat 80, but not now, I should finish my homework now.  lol.

[quote="7h3.D3m0n.117"][quote="Miss"]The best I have got is 52, how can you guys get such a high score?[/quote]

ever fully upgrading ice towers? =P[/quote]

Thank you for your tip, :P  I have now gone up to 68. :oops:

i tried it once (totally addictive) and got a 68.  Sweet.  It's almost like Flash Element TD on mofunzone.com

I got a 76, can anyone who got higher post a screenshot of their strategy?

Cool. I just got up to level 40 in the first one :(


limit integration function algorithm calculus algebra domain

Is this new? You can solve for f(0) given a zero of f.

$ f(c) \equal{} 0$
$ \lim_{x \rightarrow c} \frac {1}{f(x)}* \int^{f(x)}_0 f(t) dt \equal{} f(0)$

Now that I realize, it can be expanded into a more general formula:

$ f(c) \equal{} A$
$ \lim_{x \rightarrow c} \frac {1}{f(x) \minus{} A}* \int^{f(x)}_A f(t) dt \equal{} f(A)$

Here's an approximation function given an error interval epsilon:
$ f(c) \equal{} A$
$ f(A) \equal{} \lim_{(x, \epsilon) \rightarrow (c, 0)} \frac {2\epsilon}{f(x \plus{} \epsilon) \minus{} f(x \minus{} \epsilon)}*\int^{f(x \plus{} \epsilon)}_{f(x \minus{} \epsilon)} f(t) dt$

Computers could use this as an algorithm to scheme the entire structure of f given f(c), f'(c) and some error epsilon.

I'm not quite sure what you are saying.  The variable you integrate with respect to needs to be different than the variable in your limit.  Are you trying to say

$ \lim_{x \to c} \frac{1}{f(x)} \int_0^{f(x)} f(t) dt$?

I think that this is not true.

[quote="facis"]I'm not quite sure what you are saying.  The variable you integrate with respect to needs to be different than the variable in your limit.  Are you trying to say

$ \lim_{x \to c} \frac {1}{f(x)} \int_0^{f(x)} f(t) dt$?

I think that this is not true.[/quote]

Yes sorry, I forgot to change the variable; I've tried with multiple functions and found it to be true, iff f(c) = 0.

actually perhaps it is true... I'll attempt a proof.  But to answer your question, if this is true, it is not usually stated as a general theorem.  You really need to know the function before you can do the integral, so how useful could this be?

$ \lim_{x \to c} \frac{1}{f(x)} \int_0^{f(x)} f(t) dt$.  Plugging in $ x\equal{}c$ clearly yields $ 0/0$, so we can use l'Hopital's rule.

Before we do, let $ F$ be an antiderivative of $ f$.  Thus $ \int_0^{f(x)} f(t)dt \equal{} F(f(x))\minus{}F(f(0))$

Now apply l'Hopital:
$ \lim_{x \to c} \frac{1}{f(x)} \int_0^{f(x)} f(t) dt \equal{} \lim_{x \to c} \frac{f'(x)f(f(x))}{f'(x)} \equal{} \lim_{x \to c} f(f(x)) \equal{} f(f(c)) \equal{} f(0)$

It is true!

It's about as useful as a lot of other elementary calculus theorems  :maybe: . I suppose a geometric representation of it would be interesting.

What you gave was just a corollary of the Mean Value Theorem for Integrals, which states that if $ f$ is continuous on $ [a,b]$, then there exists a $ c \in [a,b]$ such that:

$ f(c) \equal{} \frac {1}{b \minus{} a} \int_{a}^{b} f(x) \; dx$

You have the special case where $ f(c) \equal{} 0$, $ b \equal{} f(x)$ and $ a \equal{} 0$, while adding in the limit as $ x \to c$ to avoid an indeterminate form.

The question is, does this corollary have any practical value? In order to integrate $ f$, you need the function rule for $ f$, but if that is given, you should be able to compute $ f(0)$ directly. Moreover, computing the indefinite integral of $ \int_{0}^{f(x)} f(t) \; dt$ (no less the whole limit) will probably be more tiresome than simply computing $ f(0)$. Suffice to say, this theorem seems rather useless.

Edit: sorry for the delete and re-post epkid. I didn't immediately recognize this as Mean Value Theorem.

Now for the general form: I think it should be $ \lim_{x \to c} \frac{1}{f(x)\minus{}A} \int_A^{f(x)}f(t) dt$ - otherwise we are not gauranteed the indeterminate form necessary for l'hopital.

Now $ \int_A^{f(x)} f(t)dt \equal{} F(f(x)) \minus{} F(A)$ so from l'Hopital we have
$ \lim_{x \to c} \frac{1}{f(x)\minus{}A} \int_A^{f(x)}f(t) dt \equal{} \lim_{x \to c} \frac{f'(x)f(f(x))}{f'(x)}$
$ \equal{} \lim_{x \to c} f(f(c)) \equal{} f(A)$

As claimed.

I don't think this fact is useful enough to be a theorem in its own right.  The biggest issue is that if you want to find $ f(A)$, you need $ c$ such that $ f(c)\equal{}A$ first. However, it strikes me as slightly similar to Cauchy's integral formula, which deals with complex integrals.  See [url]http://www.math.gatech.edu/~cain/winter99/ch6.pdf[/url].

The big difference is that with your theorem you really need to know part of the function to do the integral - in the reals, there is only one way to get from one point to another.  With Cauchy's formula, however, you really only need to know the values of a complex differentiable function (analytic, actually) on a closed path through its domain to find the value of the function at any point inside the path.

EDIT: I suppose the other interpretation of this theorem is the average value of a function on an interval as we squeeze the interval down to a point.  The result then should be no surprise.

facis: I think I found a problem with your proof.

$ \lim_{x \to c} \frac { [ F(f(x)) \minus{} F(A)]' }{ [f(x) \minus{} A]' } \equal{} \lim_{x \to c} \frac { f(f(x))f'(x) \minus{} f(A) }{ f'(x) }$

how can you infer that $ f(A) \equal{} 0$?

No but $ F(A)$ is a constant, so $ \frac{d}{dx}[F(A)] \equal{} 0 \neq f(A)$.

That would only apply if $ F(x)$ were a constant function. By your logic, any function evaluated at a specific point would have a derivative of 0.

Eg/ $ f(x) \equal{} x^{2}$ then $ f(3) \equal{} 9$ and so $ f'(3) \equal{} 0$ even though $ f'(3) \equal{} 2(3) \equal{} 6$.

Though, perhaps there is a specific reason we can treat $ F(A)$ as a constant that I'm overlooking.

look again.  When we compute the integral, we plug in the limits.  $ F(A)$ is then a constant, since we've already plugged in $ A$.  A is fixed, not a changing value - at least, not changing with respect to $ x$.  We can express this as $ \frac{d}{dA}[F(A)] \neq \frac{d}{dx}[F(A)] \equal{} 0$.

When talking about theorems and such, how would one even check if something he thinks is unique in fact actually is?! Is there some "theorem data-base" where one can look into? Or is there another way to find out.. And also, suppose it turns out it is unique, how would one "put it out there"?

See if it's equivalent to any well-known theorems, or a special case thereof.  In either of those cases, it's not really worth thinking about it as its own theorem.  Generally, we name particular theorems and use them all the time because they happen to be useful in solving many problems and proving many other theorems, so anything that has less generality than a well-known theorem doesn't really need to be called its own theorem unless it's a special case that you can justify as having a lot of applications.  For example, Schur's lemma is just about the easiest lemma ever, but it is so useful that it is given its own name.

In this case, I'd call it an exercise, not a theorem.

It doesn't look useful for direct calculations, but it seems plausible that you might be able to predict values of A, c, or f using this equation, I'll look into it more.

On a different note, what would one do to get a proposed theorem recognized as a legit theorem?


Regardless, this definitely lacks any reason to be posted in the round table.

Indeed, and so it has now been moved to FF.

[quote="BOGTRO"][quote="1/(ln x)"]So I was wondering, when you read "LOL", what do you guys say/hear in your head?[/quote]

Uh... its ELL-OH-ELL.[/quote]

Yep, this is what I hear.

I hear the sounds of each individual letter, el oh el

i hear 롤 or ろる or loll. depending on what homework i was doing and ended up on aops.

lol...

I hear HAHAHAHHA! or loll. ELL-OH-ELL. It depends on how the message is sent to me, Aops, email, or text message :D

i hear "lawl" because a bunch of kids in my school say lawl.

I usually do ha-ha-ha, or I do ell-oh-ell or lawl if I'm being sarcastic or dry.

i usually write lolz so i think in my head:
ell-low-ell-zah!!


Si $A$ y $C$ son tus puntos fijos y $B$ un punto de tu lugar geometrico, las bisectrices (interior y exterior) en $B$ del triangulo $ABC$ cortan a la recta $AC$ en puntos fijos (digamos $P$ y $Q$)En particular, el lugar geometrico que describe $B$ es una circunferencia (de diametro $PQ$).

[quote="R.G.A.M."][color=blue]Hace unos meses, he leído acerca del teorema de Apolonio. Intenté demostrarlo, pero no tuve éxito.

Si alguien sabe cómo hacerlo, estaría muy agradecido.

[b]Teorema de Apolonio:[/b] [i]El lugar geométrico de todos los puntos tales que la proporción de las distancias de cada punto con relación a dos puntos fijos es una circunferencia, la cual se denomina "Circunferencia de Apolonio".[/i][/color][/quote]
bueno creo que es una aplicacion directa al aplicar geometria analitica.
chau.


I used guess and check and got it in about 5 secs

[hide]The answer is 7[/hide]

[quote="JesusFreak197"]What number $n$ has the property where $n!=n(n+1)(n+2)(n+3)$?

I don't know how difficult this problem is, as I just happened to notice the relation yesterday, and I haven't tried working it out yet. However, it shouldn't be too difficult to solve by guess-and-test, even if you can't do it very easily algebraically. :lol:[/quote]
multiply both sides by $(n-1)!$ and you should get $(n-1)!n!=(n+3)!$ hmm, im not sure if this helped  :|

[hide]First of all, you can see that numbers less than 5 are obviously too small to work. Trying numbers 5 and up by guess and check, we see 7 works. No other numbers will work because to get (n+1)!, we need to multiply both sides of the formula ( n! = n(n+1)(n+2)(n+3) ) by (n+1). Instead, the left side gets multiplied by (n+1), but the right side will be multiplied by (n+4)/n, which is obviously too small. For numbers larger than n+1, the situation is basically the same. So [b]7[/b] is the only answer.[/hide]

Next problem: are there any other integers $n$ where $n!=n(n+1)(n+2)...(n+k)$, where $k$ is also an integer? If so, find the first one, and give $\frac{n}{k}$ in simplest form.

I don't know if there are yet, but I would like to see if anybody can find another. :P

[quote="JesusFreak197"]What number $n$ has the property where $n!=n(n+1)(n+2)(n+3)$?

I don't know how difficult this problem is, as I just happened to notice the relation yesterday, and I haven't tried working it out yet. However, it shouldn't be too difficult to solve by guess-and-test, even if you can't do it very easily algebraically. :lol:[/quote]
[hide]
i used guess and check and got 7 as my answer.
I have been trying to find a way to solve the formula but i don't have any answers yet.[/hide]


Digits a,b,c,d,e,f in base m satisfy:

ab/bc=de/ef (this notation is not product, it is positional).

Show that(now the notation xy means product between x and y):

(af)^2 + (dc)^2 + 4ac(e^2) + 4df(b^2) - 2afdc - 4aebf- 4bdec = k^2 for some integer k.

ab/bc = de/ef in positional notation, m basis means :
(a*m+b)/(b*m+c) = (d*m+e)/(e*m+f) or
(a*m+b)*(e*m+f) - (b*m+c)*(d*m+e) = 0
which is a trinom in m :
(b*f - c*e) + (a*f - c*d)*m + (a*e - b*d)*m^2 = 0
the discriminant is :
(a*f - c*d)^2  - 4 (a*e - b*d)*(b*f - c*e)
which should be a perfect square (k^2) to allow integer solutions.


geometry
circumcircle
trigonometry
geometry open

ABC-triangle \[H\] is ortho center

if tg<Atg<B=2

prove: \[CH = CH_{1}\]

Circumradius of triangle $BHC$ is $R$- circumradius of a triangle $ABC$, easy by sine theorem. Therefore $BH=2R\cos{\beta}$, and $C_{1}H=2R\cos{\beta}\cdot \cos{\alpha}$. Also $CH=2R\cos{\gamma}$.

Now we are suposed to prove that $\cos{\alpha}\cdot \cos{\beta}=\cos{\gamma}$ which follows from $tg{\alpha}\cdot tg{\beta}=2 \implies \sin{\alpha}\cdot \sin{\beta}=2\cos{\alpha}\cdot \cos{\beta}\implies-\cos{\alpha}\cdot \cos{\beta}= \cos{(\alpha+\beta)}\implies \cos{\alpha}\cdot \cos{\beta}= \cos{\gamma}$.

qed


[quote="mysmartmouth"]Holi is, im my humble opinion, the coolest Indian holiday in the world.  A brief explanation for any Americans would be nice.  I'd do it myself, but I'm falling asleep right now.[/quote]
Ok .. All Americans have to do is to look at a picture from their own campus..[img]http://www.nd.edu/~iand/IANDWebPage/Holi-03-1.jpg[/img]

[quote="fireemblem13"][quote="mysmartmouth"]Yeah!  It's one giant fight with colors!  The whole nation participates all day, I believe.[/quote]
i wouldn't say "fight" with colors. that's kinda the exact opposite of what it really is.[/quote]

Yes, it's more like throw the colored powder at each other and have a good time. Eat, sing and DANCE DANCE DANCE
lol, my community decided to bring Holi to the next level with water and super soaker  :D :clap:

Well this can be sent as a Holi greeting.
Or you could create your own.
[url=http://artpad.art.com/?ixlhqcqtrfs]Holi[/url]

[quote="mysmartmouth"]Wow. That sounds fun. I've gotta try that! But I don't think that's strictly legal....lol.[/quote]

Why illegal ?

Not illegal as in against the law, but illegal as in against the rules of Holi.  But I've never been to India and seen how it works on a large scale, so I could be mistaken.

Well you do throw coloured water on everyone.Now a days there's some weird silver paint like thing people put on your faces....

[quote] But I've never been to India and seen how it works on a large scale, so I could be mistaken.[/quote]
Another way is to type in "holi" and your  ollege  in google... For example  this from MIT
[img]http://web.mit.edu/hsc/www/Media/Holi/Pictures/holi%2004%2015.jpg[/img]
You can also look, for example [url=http://web.mit.edu/hsc/www/Events/Events%20-%20Holi.html]here[/url] to see an article and more photos.

sadly, they dont allow the water holi in my community because of fights or grafiti

after holi, which is by far the besst, i would say diwali comes next


I found a [b][i]new[/i][/b] Art and Craft of Problem Solving. It won't be released until May next year.

http://www.amazon.com/exec/obidos/tg/detail/-/0471789011/ref=wl_it_dp/002-6163902-6913628?%5Fencoding=UTF8&coliid=I37MLEL7KXAA0B&v=glance&colid=3SJPYQWETYNQN

Anybody know anything about this? I know I won't be using this for a while (I'm still working through AoPS vol. 1), but I wanted to bring it up...

Er...the title is the same as Zeitz's current book. Maybe he's revising something or just making a newer edition of his book.

There were a bunch of little typos in his first edition.  He's probably fixing those.  Maybe there will be some other additions.  Probably nothing to get hyped up over though.

That's so cool!!...... and I was about to buy the old one. :)

Should I buy this book now then? Or should I wait?

Hmm, it looks like the release date for the new paperback version has been changed to August.  (2 months from now!)

Does anyone know how significant the changes will be?  The page numbers between the old and new versions increased by about 30 pages, so it seems like there may be more content added than just minor typos/revisions.

I'm looking to buy this book soon, so I'm not sure whether to wait or not.

The fact that it's paperback might account for some of the difference in the number of pages. Though hopefully some (or all) of it will be because of new material since I also plan to buy the new version when it's available  :P

I really want to buy acops right now.

but new edition coming out soon?

what to do!

I heard somewhere that a geometry chapter is being added.  I'm not sure though.

[quote="deimos"]wait, my shoe, just wait.[/quote]

Are you sure? Because if it only has minor changes to the last one I will just buy the old version.

[quote="ngade"]Are any of you guys buying the New Introduction to Number Theory Book?  :?:[/quote]

I'm waiting for the intermediate series.

If there's a geometry chapter maybe i'll buy the new one too actually no i'll just printscreen the geometry pages off amazon

Is ACoPS harder then AoPS  v.2? About the same?

It's harder.

(BTW: could a mod edit the link on the first post so the page doesn't stretch out?)


function
inequalities
combinatorics unsolved
combinatorics

Could anyone help me with the following problem? I am not familar with it, this problem is formulated in my current project, so I am not sure if it is solvable or come up with any reasonable solution. Thank you in advance.

Contraints:
x1 + x2 + x3 = a
y1 + y2 + y3 = b
x1 <= m1, x2<= m2, x3 <= m3
y1 <= n1, y2<=n2, y3<=n3
a < m1+m2+m3
b < n1+n2+n3
x1, x2, x3, y1, y2, y3 positive

Objective function:
f(x1, x2, x3, y1, y2, y3) = n1*m2*m3 + n2*m1*m3 +
n3*m1*m2 - y1*x2*x3 - y2*x1*x3 - y3*x1*x2

Question: Is there any lower bound for the function f?
I.e., does there exist any E>0 such that f >= E for any
xi, yi satisfying the constraints. If yes, what is E?

[quote="tt930101"]Could anyone help me with the following problem? I am not familar with it, this problem is formulated in my current project, so I am not sure if it is solvable or come up with any reasonable solution. Thank you in advance.

Contraints:
x1 + x2 + x3 = a
y1 + y2 + y3 = b
x1 <= m1, x2<= m2, x3 <= m3
y1 <= n1, y2<=n2, y3<=n3
a <= m1+m2+m3
b <= n1+n2+n3
x1, x2, x3, y1, y2, y3 positive

Objective function:
f(x1, x2, x3, y1, y2, y3) = n1*m2*m3 + n2*m1*m3 +
n3*m1*m2 - y1*x2*x3 - y2*x1*x3 - y3*x1*x2

Question: Is there any lower bound for the function f?
I.e., does there exist any E such that f >= E for any
xi, yi satisfying the constraints. If yes, what is E?[/quote]Your problem is one of linear programming. (That's what the discipline is called that gives rise to the problem.)

From the constraints it is seen that $y_1x_2x_3 \le n_1m_2m_3$

and $y_2x_1x_3 \le n_2m_1m_3$

... and $y_3x_1x_2 \le n_3m_1m_2$

Minus signs will reverse the inequalities such that:

$f(x_1,x_2,x_3;y_1,y_2,y_3) = n_1m_2m_3 + n_2m_1m_3 + n_3m_1m_2 - y_1x_2x_3 - y_2x_1x_3 - y_3x_1x_2 \ge n_1m_2m_3 + n_2m_1m_3 + n_3m_1m_2 - n_1m_2m_3 - n_2m_1m_3 - n_3m_1m_2 = 0$

Therefore, $f(x_1,x_2,x_3;y_1,y_2,y_3) \ge 0$.

[quote="tt930101"]Thanks Dr. No for your quick reply. I did see that in the worst case, E = 0. however it is not an interesting result. I need the value of E such that E>0.[/quote]Oh, well certainly that is the point of most linear programs: To find an optimum other than a trivial one. Nevertheless, if your objective function is as you defined it then E = 0 is its minimum. (This case will only occur when $x_1=m_1, x_2 =m_2,\ldots,y_3=n_3$, however.)

We did find that $f$ must be positive if it is to satisfy the constraints (in any case other than the trivial one), which I guess is a good thing for cost function. Giving you any better answer than what I have already is a bit out of my league. I'm a physicist, not an economist. Sorry :(

[quote="Diarmuid"]Unfortunately, the condition for equality that you've given is not achievable.  If $x_1=m_1, x_2 =m_2,\cdots,y_3=n_3$, then $a=x_1+x_2+x_3=m_1+m_2+m_3>a$, a contradiction.  In other words, the inequality $f\ge0$ should in fact be strict, and there may be another number $E>0$ such that $f\ge E$.

I'm not sure how to solve this problem in the general case though.[/quote]Thanks for pointing that out. Though it's kind of the point of the problem: $f = 0$ is trivial (and from the standpoint of cost, what $f$ usually expresses, nonsense); so, indeed if there is an E, such that $f \ge E$, then $E > 0$. However, I do not know how to solve the system.


calculus
derivative
algebra unsolved
algebra

Let be given func.$f(x):[a,b]\rightarrow R$,having derivative on$[a,b]$,s.t.
        $f(a)=f(b)=c$
Prove that$\forall \alpha \in R,\exists x_0\in (a,b)$ satisfies:
  $f'(x_0) + \alpha f(x_0) =\alpha c$

Aplly Rolle theoreme to $g(x)=e^{x\alpha }f(x)-e^{x\alpha}c$
$g(a)=g(b)=0$
there exist w such that $g'(w)=0$


$a,b,c$ are three positive numbers and $abc^2$ has the greatest value $\frac{1}{64}$.Find $a,b,c$

 :)


[quote="tranthanhnam"]$\int \frac{arctan}{x^{2}(x^{2}+1)}dx$ :blush:[/quote]

sorry  i suppose you wrote: $\int \frac{arctanx}{x^{2}(x^{2}+1)}dx$

$\int \frac{arctan}{x^{2}(x^{2}+1)}dx=\int \frac{arctan x dx}{x^{2}}-\int \frac{arctan x \ dx}{x^{2}+1}=-\int arctan x \ dx-(arctan x)^{2}/2 =(arctan x)^{2}/2-\frac{arctan x}{x}+\int \frac{dx}{x(x^{2}+1)}...$


trigonometry
trig identities
Law of Cosines

Well, I got this problem today in AP Physics and am having a difficult time solving it.  We are asked to prove the following:

D^2=A^2+B^2+C^2-2ABcos(x)-2BCcos(y)+2ACcos(x+y)

Any suggestions would be greatly appreciated...

It would help if you stated what exactly are $A, B, C, D, x, y$

I'd guess that A,B,C,D are sides of a quadrilateral, and x and y are angles.  Is this a cyclic quad?  Cause then we'd only need to know two of the angles.

Sorry about that, I forgot to attach the image.  Also, we were not given angles.  We were just asked to prove the formula.  Thanks for the help so far...

[img]http://i75.photobucket.com/albums/i298/deakstr/1-2.jpg[/img]

In AP Phys?? Anyway that's easy Law of Cosines... just apply it a couple times I'll let you work out the details..  :D
[hide="hint"][b]Law of Cosines:[/b] \[C^{2}=A^{2}+B^{2}-2AB\cos \angle C^\circ\] [/hide]
where A,B,C are the sides and angle C is the angle across from side C...

See the attached diagram.

$d^{2}=c^{2}+p^{2}-2pc\cos z$
$p^{2}=a^{2}+b^{2}-2ab\cos x$

These two give

$d^{2}=a^{2}+b^{2}+c^{2}-2ab\cos x-2cp\cos z\qquad(*)$

\begin{eqnarray*}p\cos z=BD'=A'D'-A'B=a\cos\angle AOB-b\cos(\pi-y)=a\cos\angle AOB+b\cos y\end{eqnarray*}

$\angle AOB=\pi-(\pi-x)-(\pi-y)=x+y-\pi\implies\cos\angle AOB=-\cos(x+y)$

Hence

$p\cos z=b\cos y-a\cos(x+y)$

Plugging that into $(*)$ we get

$d^{2}=a^{2}+b^{2}+c^{2}-2ab\cos x-2bc\cos y+2ac\cos(x+y)$

Very nice, I still don't see how this came up in a physics class??!!

Thanks guys, this helped tremendously.  I had got stuck about halfway through.

BTW, this for extra credit...not an actual assignement for physics.

as far as I know, I am the only one to get it so far.  I know that two or three other people will get it though.  I really appreciate everyones help and I hope I can return the favor...

what school do you go to? that's pretty good if 2 or 3 others will get it... at my school nobody would get it in my calc 3/4 class!!

in Indianapolis, IN.  We have a total of about 4,000 students, around 1,000 per class (freshman, sophomore,etc.).


In a country, there are some cities linked together by roads. The roads just meet each other inside the cities. In each city, there is a board which showing the shortest length of the road originating in that city and going through all other cities (the way can go through some cities more than one times and is not necessary to turn back to the originated city). Prove that 2 random numbers in the boards can't be greater or lesser than 1.5 times than each other.

Consider the shortest path originating in some city and going through all others, and WLOG let this path have length 1. This will be the smallest number among all boards. We now show that no other city can have a board with a number greater than 1.5.

From any city, we may travel to one end of the shortest path of length 1 (we choose the end such that this distance will be $ \leq$.5), then travel across the entire path of length 1. This path is of length $ \leq$1.5 and passes through all cities, so no board can have number greater than 1.5.

I have not solved many Russian combinatorics problems. Do Russian combinatorics generally have solutions as nice as the solution to this one?

[b]Solution[/b]

For a city $X$, let $|X|$ denote the length of the shortest path passing through all of the cities of the country $C$ originating from $X$. Consider the path of shortest length originating from $B$ that passes through all of the cities in $C$. Partition this path into the two paths $P$ and $Q$ such that $P$ begins with $B$ and ends with $A$ and $Q$ begins with $A$ and ends with the last city in the shortest path originating from $B$ passing through all of the cities in $C$. We now have that the length of $P$ and the length of $Q$ sum to $|B|$. Therefore one of $P$ or $Q$ is less than or equal to half of $|B|$. Without the loss of generality, let this element be $P$. Now consider the path originating from $A$ that uses path $P$ to get to $B$ and then the path of shortest length from $B$ to reach all cities in the country. The length of this path is the sum of $|B|$ and the length of $P$ which is less than or equal to $\frac{3\cdot |B|}{2}$. This implies that that $|A| \le \frac{3\cdot |B|}{2}$ for any $A, B \in C$.

[hide=" Solution"]
Let $C_1$ and $C_2$ be two random cities in the country, and let $\ell$ be the length of the minimal path originating from $C_1$. Let this path end at $C_3.$ If the distance between $C_2$ and $C_3$ is no more than $\dfrac{\ell}{2}$, consider the path which starts from $C_2$, goes to $C_3$, and then returns along the same route to $C_1$- this path has length at most $\ell + \dfrac{\ell}{2} = \dfrac{3}{2} \ell$. Otherwise, consider the path which starts from $C_2$, goes to $C_1$, and then goes to $C_3$ along the same route. Either way, the length of this path is no more than $1.5 \times \ell$. $\blacksquare$ [/hide]

Let $l$ be the lenght shortest path among all paths.Now,consider city C that is different from the city from which is the sorthest path.Now,divide the path in two paths:one that are all roads before that we came first time in city C and one after that.Now,we can pick a path that has lenght less or equal half of $l$,so just go like in that path and when finish then come back and go through the second path,so we are finished.


vector
Ring Theory
superior algebra
superior algebra unsolved

Let $ R$ be a commutative ring satisfying $ R^2 \neq {0}$ with exactly 2 ideals. Prove that $ R$ is a field.

so you don't assume $ R$ is unital and $ R^2$ means the set $ \{ab : a,b \in R\}$?

Yes. If we had a unit $ R^2\neq {0}$ wouldn't be needed.
We could just consider $ 1\in < a >$ so $ a$ has an inverse.

$ R$ has necessarily a unital element: Show that
\[ \Phi: R\to K: \equal{} Hom_R(R,R),\,a\mapsto (L_a: x\mapsto ax)\]
is an injective ring homomorphism into the [b]field[/b] $ K$ which is not shown to be commutative yet.

$ R$ is a $ K$-vector space with respect to evaluation, i.e. $ \varphi\cdot a: \equal{} \varphi(a)$. Since $ \Phi$ is $ K$-linear and non-trivial, we get that $ \Phi$ is surjective counting dimensions over $ K$. Hence $ \Phi$ is a ring isomorphism. This implies that $ K$ is commutative, $ R$ has a unital element and is a field.

Since R^2 ≠ {0}, ∃a≠0 such that aR ≠ {0}, which implies that aR = R by assumption.

Let e be an element of R such that ae = a.

Take any b in R, ∃c such that b = ac, and therefore,

be = (ac)e = (ae)c = ac = b.

It follows that R is a commutative ring with 1.

Then with 1 in R, proving that every non-zero element in R has an inverse is trivial.


on beginning of the year, lots of teachers start out the class with a simple trick that makes people surprised...

1. Think of a number and multiply it by 5.

2. After that, add 20

3. Multiply it by 2

4. Subtract 30 from the number

5. Divide the number you got by 10 and subtract 1

6. That's the number you thought of....

Tell me how it works

Ah.

Old classic number joke you used to use on people and making "WOW! YOU"RE SO SMART!" :P

[quote]1. Think of a number and multiply it by 5.

2. After that, add 20

3. Multiply it by 2

4. Subtract 30 from the number

5. Divide the number you got by 10 and subtract 1

6. That's the number you thought of....
[/quote]

Call the number x.
1. 5x
2 5x+20
3. 2(5x+20) = 10x+40
4. 10x+10
5. (10x+10)/10 = x+1 - 1 = x
6. x

Which is the number you thought of. :)

[quote]Old classic number joke you used to use on people and making "WOW! YOU"RE SO SMART!"[/quote]

ha ha ha

a big laugh to Silverfalcon!! :D

It's not that hard...

You just have to state the number as any variable such as x...

then , like Silverfalcon did, just keep writing the sentece as an equation...

then u know how t works :D

sorry for reviving a topic...

Call that number you thought of as x.

[hide="Here are the steps."]
x
5x
5x+20
10x+40
10x+10
x[/hide]

I just couldn't help myself. :P


algorithm
combinatorics unsolved
combinatorics

hey ppl actually this is an algorithms problem but the concept surely belongs to combinatorics:)
help me with this one..some definitions first.

An n-ary Tree is called a Strictly n-ary Tree if every nonleaf node in an n-ary tree has n children each. A Strictly n-ary Tree of depth d can now be defined as a Strictly n-ary Tree that has at least one root to leaf path of length d, and no root to leaf path in that tree is longer than d.

now given numebr n,d is there a recurrence relation for the strictly n-ary trees with depth d??
help!!
best regards

[quote="galois"]An n-ary Tree is called a Strictly n-ary Tree if every nonleaf node in an n-ary tree has n children each. A Strictly n-ary Tree of depth d can now be defined as a Strictly n-ary Tree that has at least one root to leaf path of length d, and no root to leaf path in that tree is longer than d.

now given numebr n,d is there a recurrence relation for the strictly n-ary trees with depth d??
help!!
best regards[/quote]

$a(d)$ - number of $n$-ary trees with depth $d$.
A simple recurrence after $d$ is:
$a(d+1)=a(d) \sum_{j=1}^n C^j_n (a(0)+a(1)+...+a(d-1))^{n-j} a^{j-1} (d)$
$a(0)=a(1)=1$, $a(2)=2^n-1$, ...

cud u plz let me know how u came up with that??
cheers!!

$a(d)$ - number of $n$-ary trees with the depth exactly $d$.
$s(d)$ - number of $n$-ary trees with the depth at most $d$.

$s(d)=\sum_{i=0}^d a(i)$

Compute $a(d+1)$ by choosing 1st node $i$ among all $n$ found on the 1st level, that has the depth exactly $d+1$ in order to avoid multiple counting.
For the node(s) $1$ to $i-1$ the depth is at most $d$:  $s(d-1)$
For the node $i$ we have the depth exactly $d+1$: $a(d)$
For the node(s) $i+1$ to $n$ the depth is at most $d+1$: $s(d)$

Then:
$a(d+1)=a(d) \sum_{j=1}^{n} s^{j-1} (d-1) s^{n-j} (d)$

$a(d+1)=a(d) \frac{ (\sum_{i=0}^d a(i))^n-\sum_{i=0}^{d-1} a(i))^n }{\sum_{i=0}^d a(i) -\sum_{i=0}^{d-1} a(i)}$

$a(d+1)=s^n (d) - s^n (d-1)$

$a(d+1)=a(d) \sum_{j=1}^n C^j_n (s(d-1))^{n-j} a^{j-1} (d)$

$a(d+1)=a(d) \sum_{j=1}^n C^j_n (\sum_{i=0}^{d-1} a(i))^{n-j} a^{j-1} (d)$


geometry
circumcircle
search
combinatorics proposed
combinatorics

Each point of a plane is painted in one of three colors. Show that there exists a triangle such that:
$ (i)$ all three vertices of the triangle are of the same color;
$ (ii)$ the radius of the circumcircle of the triangle is $ 2008$;
$ (iii)$ one angle of the triangle is either two or three times greater than one of the other two angles.

(i) and (ii) are simple.

(i) If there existed no triangle with three vertices of same colour then the all the points coloured with a colour say $ c_1$ lie of a straight line. We may at max have three distinct lines for each of three colours that do not fill a plane.

(ii) We can get a monochromatic triangle with circumradius equal to any positive real number. Pick a arbitrary point in space and construct a circle of radius $ r$, a positive real number. Pick a point on the circumference and let it have the colour $ c_1$ wlog.

If we  are not able to find another point (or if we find just one more point $ c_1$) on the circumference that is coloured $ c_1$ then every other point on the circumference is coloured either $ c_2$ or $ c_3$. If we do not find a point with $ c_2$ then we are done as rest of the points are $ c_3$. If we find just either 1 or 2 points coloured $ c_2$, then too we have rest of the points coloured $ c_3$.

No no no no... You should prove that there exist a triangle which fullfils all 3 conditions! These are not three diferent problems, these are conditions for triangle :)


[quote="cats..."]we have $C(m+n, n)\cdot m^{m}\cdot n^{n}<(m+n)^{m+n}$ which is equivalent to your inequality.[/quote]

Thank you

[quote="cats..."]we have $C(m+n, n)\cdot m^{m}\cdot n^{n}<(m+n)^{m+n}$ which is equivalent to your inequality.[/quote]

how is this true?

Expand $(m+n)^{m+n}$ with the binomial theorem. Then all the terms are positive, there must be more than one term, and one of the terms is ${m+n\choose n}m^{m}n^{n}$, so this must be less than $(m+n)^{m+n}$.


does a particle in projectile motion experiences weightlessness during whole motion or at only highest point of trajectory?

at the highest point of trajectory the particle has only horizontal velocitty and no force acts on it in the vertical direction.

Obv force acts on it whole the time and acc. is g hence it experience weightlessness.

well technically the projectile itself is still being weighed down by gravity, so only objects inside the projectile will experience weightlessness
when the projectile is at its peak, the objects inside will experience no $ y$ force (aka weightlessness), but when the projectile begins to fall, the objects will rise (in relation to the projectile)

What are you guys saying? No y force at highest point? There is a force mg acting on it all the time, only the velocity becomes 0 at the peak.

conjurer is right.the projectile will experience weightlessness during its whole motion.

@aadil
at the highest point of trajectory only its vertical velocity will be zero.still mg force acts on it


function
trigonometry
calculus
calculus computations

Let $ f: \mathbb{R}\rightarrow\mathbb{R}$ be the function given by $ f(x)\equal{}x^2\sin{\frac{1}{x}}\plus{}x$ if $ x$ is not equal to $ 0$;
and $ f(0)\equal{}0$

Prove that there is no positive number $ c$ such that the function $ f: (\minus{}c,c)\rightarrow\mathbb{R}$ is increasing.

Yes,I have computed it.
But it is difficult to know it is not increasing from $ f'$.
Did anyone solve this problem?


In some company there are $n$ people and each one of them know exactly $m$ other. Beside that we know that If $A$ and $B$ know each other then they have excatly $k$ comon peolpe which they know, and if $A$ and $B$ do not know each other then they have excatly $l$ comon people which they know. Prove that $m(m-k-1)=l(n-m-1)$.

Consider a people $A$, let $S$ be the set of people $A$ knows, and $T$ be the set of people $A$ don't know. Let $f(a,b)$ mark that $a$ knows $b$ and for a fix $a$ and $b$, we write $f(a,b) = 1$

Consider $B \in S$, the number of friends of $B$ in $T$ is equal to the total number of friends of $B$ minus the the number of friends of $B$ outside $T$ which is $m-(k+1)$
Since $|S| = m$, we have $\sum_{a \in S, b \in T}f(a,b) = m \cdot (m-k-1)$.

Furthermore, $|T| = n-m-1$ and each person in $T$ knows $l$ people in $S$. So, $\sum_{a \in T, b \in S}f(a,b) = (n-m-1) \cdot l$

Because $\sum_{a \in S, b \in T}f(a,b) = \sum_{a \in T, b \in S}f(a,b)$, then $m \cdot (m-k-1) = (n-m-1) \cdot l$, as desired.  :)

Consider the number of triples $(A, B, C)$ where $A$ does not know $B$ and $C$ knows $A$ and $B$.

Fixing C and A, we will have exactly $m-k-1$ choices for $B$. But we have $m$ choices for A. It gives $m(m-k-1)$, where $(A, B, C)$ and $(B, A, C)$ are considered different.

Fixing A, we have $n-m-1$ choices for $B$, so we have $l$ choices for $C$.
And it gives $l(n-m-1)$. So $m(m-k-1)=l(n-m-1)$.


quadratics
number theory unsolved
number theory

Is it true that for any prime integers $p \equiv 7 (mod8)$
 $2^{\frac{p-1}{2}} \equiv 1 (mod p)$

On other words: $2$ is a quadratic residue $\mod p$.
Answer: yes.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=23681 , the methode there is independent from $p \equiv 1 \mod 8$ and gives the desired result here.


how do you think this amc compares to the  2007AMC 12A? and the 2006 AMC?

Snap, it was hard!! i was going to take both the amc 10 a and the amc 12b, but only got to take the 12b because we had a SNOWDAY the day of the A exam.

i took the amc 10a and amc12a at home, and it was relatively moderate, but the 12B was excruciating!

paladin8 thought it was hard too, but I thought it was relatively easy... maybe because #25 was such a letdown  :rotfl:

The first 19 questions were significantly easier on the B than the A IMHO; questions 20 - 24 were all much tougher than their counterparts on the A (I only got one of them on the test); and question 25 was incredibly easy for a 20+ (I was too intimidated to make a significant attempt at it on the test, but I solved it in like two minutes afterwards [and then kicked myself]).
Overall, I'd say that the A was easier than the B.

That notwithstanding, I went from a 105 on the A to a 127.5 on the B. (Which I explain with the fact that I spent a fair amount of time preparing for the B after getting my A score.)


I can't open the file LaTeXiT.

The directions told me to run it, but I can't.

Can someone PLEASE help me?

From [url=http://www.apple.com/downloads/macosx/math_science/latexit.html]LaTeXIT's page[/url]:
[quote]System Requirements
Mac OS X 10.4 or later[/quote]

stevem, so then I don't have to have LaTeXiT?

EDIT: In other words, should I forget about it and trash it?

I'm not sure what you mean by "have to have". It depends what you want to use it for and whether you want to use something else to produce an image of a formula. There are lots of online facilities you can try instead such as AoPS's [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php]TeXer[/url] or this [url=http://www.codecogs.com/components/equationeditor/equationeditor.php]Equation Editor[/url].

it came in the Mac-2008 zip package.

What should I do with it?

I just need to be able to write equations and variables, and tables/charts.

Is there a version of LaTeXiT that works with Mac OS X 10.3.2 Panther?

Or is it just for 10.4?

LaTeXiT is for producing images of individual formulae so you can copy and paste them into other things like presentations. It's not for complete LaTeX documents. If you are producing complete LaTeX documents then obviously you won't need it. You will however find Panther versions at http://ktd.club.fr/programmation/latexit_en.php

Don't forget to upgrade to 10.3.9 as I mentioned at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1586483#1586483

For the latex page, I tried to download one for the Panter Mac, but it said there was a 404 error....


Let us have a 5-digit positive integer $n$, such that $n=100p + q$. How many integers $n$ are such,  that $11|q+p$?

(i hope 8181?  :? )

oops, already found it: [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=8181&t=25016]Click[/url]


Given the true statement: The picnic on Sunday will not be held only if the weather is not fair.  We can then conclude that:

$ \textbf{(A)}\ \text{If the picnic is held, Sunday's weather is undoubtedly fair.}$
$ \textbf{(B)}\ \text{If the picnic is not held, Sunday's weather is possibly unfair.}$
$ \textbf{(C)}\ \text{If it is not fair Sunday, the picnic will not be held.}$
$ \textbf{(D)}\ \text{If it is fair Sunday, the picnic may be held.}$
$ \textbf{(E)}\ \text{If it is fair Sunday, the picnic must be held.}$

I am almost certain that it is E, just because if it is fair, the picnic on Sunday will NOT not be held, meaning that it MUST be held.

Yes, it is $ \boxed{E}$.

This can easily be seen by common sense (as someperson01 shows) but also because a statement has the same truth value as its contrapositive (if you're unfamiliar with or don't understand this fact, learn the very basics of logic and it will follow from a truth table).  Since the given statement is true, its contrapositive must also be true, which is $ E$.
Another way of looking at it is $ p\Rightarrow q\equiv\sim q\Rightarrow\sim p$, if I remember the notation of logic correctly (it's been a while).


Twenty people are seated around a circular table. Each person is either a knight, who always tells the truth, or a knave, who always lies. If each person at the table says, ``The person on my right is a knave,'' how many knaves are seated around the table?

There are $ 20$ people. Say we start with a knight. Then the person on his right is a knave. The person to the right of the knave, that always lies, is a knight, who is left of a knave, and so on. This way, knights and knaves alternate in the seating arrangements. So, half of the people are knaves; $ 20 \div 2\equal{} \boxed{10}$.


Let $A,B \in \mathcal M_n \left( \mathbb C \right)$ s.t. $A^2 B = A^2 - B$. Prove that $AB = BA$ and construct two invertible matrices which fulfill the above relation.

My proof:
With \[ (A^2B=A^2-B \]
We have\[ (A-I)(A+I)B=-I,(A+I)(A-I)B=-I \]
Thus:\[ (A+I)B(A-I)=-I,(A-I)B(A+I)=-I \]
Thus :\[ (A+I)B(A-I)=(A-I)B(A+I)=-I \]
thus:\[ AB=BA \]
for first part

zhaobin i don't see how u deduced these two equalities you used them at first ,well here is my solution:
as (A²+I) B=A² then if A is invertible B is invertible and same goes for A²+I  and this case is sort of easy as we're gonna have B=(A²+I)^-1 *A² we obviously have (A²+I)^-1*A=A*(A²+I)^-1 so BA=AB.
for each complex number x we have (A²+I)(B-xI)=(1-x)A²-xI then for x different from 1 we have :
(A²+I)(B-xI)=(1-x)(A²-(x/(x-1))I)   thus we just have to chose x such as B-xI ans A²-(x/(x-1))I are invertible ,which is possible as x/(x-1) might have any complex value a part from 1 ,then we have to notice that A²+I is invertible too which means that i and -i might never be eigenvalues of A .
Thus we have B-xI=(1-x) (A²+I)^-1*(A²-(x/(x-1))I)  and then (B-xI)A=A(B-xI) so BA=AB.
For the example relating to invertible matrixes we can have a trivial one by chosing diagonal matrixes (it's 2 am i ain't in a good mood to that lol ;))

He used the fact that $XY = I$ implies $YX=I$. (and also that $Y$ is unique)


[quote="sharkman"]The tangent line to a circle may be defined as the line that intersects the circle in a single point called the POINT OF TANGENCY.

If the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b, show that the tangent line is perpendicular to the line containing the center of the circle and the point of tangency.[/quote]
Stay fly dude, don't play analytic, try to just use synthey and sides and angles and congruency junk.

I don't understand your reply.

Can you help me with this question?


People who believe in the ontological argument are idiots.

Discuss.

[color=green]Moderator says: To all: please see my comment at the bottom of this thread re: feeding trolls.[/color]

There is not much to discuss here: the statement, as written, is wrong and the time is not appropriate. This claim would be interesting to make around 1478 somewhere slightly southwest from Pirenea mountains. Now the only sensible reaction to it is to shrug your shoulders and remind you that no person is a complete idiot or completely free from idiocy.

Generally speaking, this forum is not for challenging people to various fights and calling them names directly or indirectly. If you are irritated with how the modern world works and what stands behind people's "normal" behavior, be sure, you are not alone. But be also sure that if you want a change, to cry loudly "it is idiotic to believe in unobservable entities/consider the interests of one irregular patch of land far superior to those of other patches/like rectangular pieces of paper so much/create a third hungry mouth/play rather boring and dangerous for your health games with various elongated metallic objects/etc." is the [b]least[/b] efficient way to get anywere. ;)

Anyway, you are welcome to this forum but there are a few rules and facts you should know of:
Rules:
1) Be as polite as you possibly can under the circumstances.
2) Try to make a clear and well argumented point when posting.
There are more, but these 2 are enough for now.
Facts:
1) The people here have only one thing in common: they like math and are not too bad in it. In all other respects, there are almost no similarities between them.
2) They come from almost every imaginable country and culture with values and beliefs varying  tremendously.
3) English is just the AoPS "official language", not a native tongue for most of the participants.
3) The main purpose of the AoPS forum is to discuss mathematics.
Again, there are more, but these 4 will do for now. Since you, probably, do not consider yourself an idiot, I leave the logical conclusions to you :).

could a mod please change the title or lock/delete this?

[quote="TZF"]René Descartes believed in this argument, and he certainly was not an "idiot".

If you have an objection to the logic behind this argument, post your objection, and we can have a [u]civilized[/u] discussion.[/quote]

It's just such a moronic argument that it's pretty difficult to actually attack on an analytical level. Here's what the ontological argument basically boils down to:

1. God is perfect.
2. Something good that exists is better than something equally good that does not exist.
3. Therefore God exists.

There are so many bizarre assumptions in this argument that it's really difficult to see how any rational person can be held under its sway. First of all, why should we believe that God is perfect? Vaguely analytical religious nuts like to say "because that's how God is defined," but that's not actually true, because God is generally defined as "an omnipotent creator of the universe." If you want to call a hypothetical "perfect" entity "god," fine, but it won't resemble the traditional definition whatsoever.

Moreover, the argument also assumes that it is possible for God to exist, which is also untrue.

[quote="fedja"]There is not much to discuss here: the statement, as written, is wrong and the time is not appropriate. This claim would be interesting to make around 1478 somewhere slightly southwest from Pirenea mountains. Now the only sensible reaction to it is to shrug your shoulders and remind you that no person is a complete idiot or completely free from idiocy.

Generally speaking, this forum is not for challenging people to various fights and calling them names directly or indirectly. If you are irritated with how the modern world works and what stands behind people's "normal" behavior, be sure, you are not alone. But be also sure that if you want a change, to cry loudly "it is idiotic to believe in unobservable entities/consider the interests of one irregular patch of land far superior to those of other patches/like rectangular pieces of paper so much/create a third hungry mouth/play rather boring and dangerous for your health games with various elongated metallic objects/etc." is the [b]least[/b] efficient way to get anywere. ;)[/quote]

What the heck are you talking about? You make it sound like atheism is some vestige of the past and religion is the status quo. While this may be true for the masses, there is a strong correlation between education and atheism. Few good philosophers or scientists believe the really childish reasoning that is the ontological argument.

The danger of the argument is that someone who gives it a quick read may leave with the impression that it assumes that God exists and concludes that God exists, and is therefore circular. The actual argument is a bit more subtle than that, though. I personally don't believe in it. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=216553]Here is a thread[/url] that may be of interest to you.

[quote="Kel_Varnsen"]Moreover, the argument also assumes that it is possible for God to exist, which is also untrue.[/quote]

If you don't think it's possible for God to exist, then it'll be hard to convince you of anything substantial on this topic ;)

[quote="Kel_Varnsen"]What the heck are you talking about?[/quote]

[b]Fejda[/b] is basically telling you you that you don't need to make a sweeping statement and call people you don't even know idiots, when you should actually be doing one of the following:
1. Realize it's an old argument and forget about it
2. Trying to inspect the possible subtler meanings of the argument to see if they have any merit

He's also saying to be careful who you call an idiot. That's just about the nicest response you'll get from people around here; many other senior members have significantly less patience for flaming. You probably want to show him a little more respect.

[quote]many other senior members have significantly less patience for flaming.[/quote]

and you're about to meet one such member, kel.  to pre-empt: no, i am not religious.

1) it's worth noting that the phrase "ontological argument" is usually a generic term for any argument that, via a priori proof, tries to prove the existence of god -- not just the descartian approach.  this is not inherently meritless, as mysticterminator's thread in this forum demonstrates.  we've had a pretty interesting discussion about ontological argumentation for a couple weeks so far.  in this way, your awkward misuse of the term suggests that you don't really know what you're talking about, so stop being such an arrogant prick.

2) more generally, "ontology" as a field of philosophy has nothing to do with religion or god -- it's a branch of metaphysics that deals with the notion of existence itself, and it's been an incredibly active field through the 20th and 21st centuries (see deleuze, baudrillard, guattari, derrida, etc.).  from there, one finds that the objections you raise as to how "god" is defined, what it would mean for a god to exist, and what might distinguish "god"-qua-omnipotent-creator from "god"-qua-manifestation-of-perfection are surrounded by hundreds of years and thousands of pages of relevant, germane philosophical debate which you flippantly dismiss (or, more likely, pass over from sheer ignorance) -- again putting your incompetence on display.

3) if an argument is flawed to the point of absurdity, it should make it easier, not harder, to attack it analytically since you can just point out all the flaws inherent in the assumptions, reasoning, and conclusion.  your claim to the contrary is not only a non sequitur, but since you go on to disprove it anyway, you also performatively contradict yourself.

4) you have basically used this thread as a pulpit to espouse your atheism by straw-manning all theology as variations on the descartian theme.  this belies the purpose of the round table forum as a venue for debate and discussion, showing that you don't even nominally understand why this forum exists.  either that, or you couldn't be bothered to read the six word description given in the index for the round table.

5) your inability to respond cogently to anything we've said (e.g. you quote fedja and then proceed to say something that is completely irrelevant) makes me pessimistic that you will actually be able to respond to me in a way that makes sense.  either that, or you're an internet troll -- in which case, i heartily call for you to be banned from the forums.

[color=green]Moderator says: and that will be the last word on this particular thread.  Folks, please don't feed the trolls.  It's better to leave their threads untouched so that a moderator can just delete them.[/color]


[img]http://www.ipix.lt/images/56213385.jpg[/img]

ABCD - trapezoid, also $ \frac{PC}{PB}\equal{}\frac{DC}{AB}$

Prove that AB=EB.

 :maybe:

[hide]
[asy]size(200); pathpen = linewidth(0.7); pen f = fontsize(10);
pair A = MP("A",(0,0),f), B=MP("B",(6,0),f), C=MP("C",(4,5),N,f), D=MP("D",(1,5),N,f), A_r = MP("A'",2B-A,f), P=MP("P",IP(B--C,D--A_r),NE,f), E=MP("E",foot(A,P,D),(1,0.5),f);
D(A--B--C--D--cycle); D(A--E--B); D(D--P); D(P--A_r--B,dashed);D(rightanglemark(A,E,P));[/asy]

Let $ A'$ be on $ AB$ such that $ AB \equal{} A'B$; then by given, $ \triangle PBA' \sim \triangle PCD$, so $ A',P,D$ are collinear. Then $ \triangle AEA'$ is right, so its median is half its hypotenuse, and $ AB \equal{} BE$.
[/hide]


Can you prove
$ \sum_{cyc}\frac {1}{a \plus{} b \minus{} c}\geq \sum_{cyc}\frac {2}{\sqrt{ab}}$ for any triangle? (just use Am-Gm)
your inequality can be translated to this one by considering the triangle with sides $ a^2,b^2,c^2$  :wink:

But $ a,b,c$ are side lengths of a triangle does not necessarily imply that $ a^2,b^2,c^2$ are as well.. :maybe:

Take for example $ (a,b,c) \equal{} (2,4,5)$.

EDIT: OK Eagle's proof is correct, since the triangle is acute. :blush:

[quote=xzlbq]But who can do this:

In acute triangle,prove

\[\sum{\tan{A}\cos^6{\frac{B-C}{2}}}\geq 2\sum{\sin{A}}---(1)\][/quote]

Dear Mateescu Constantin:
You must do out (1)!!!

[quote=xzlbq]In acute triangle,prove

\[\sum{\tan{A}\cos{(B-C)}}\geq 2\sum{\sin{A}}\][/quote]

Who nice do it?

[quote=xzlbq]But who can do this:

In acute triangle,prove

\[\sum{\tan{A}\cos^6{\frac{B-C}{2}}}\geq 2\sum{\sin{A}}\][/quote]

In acute triangle,prove

\[\sum{\tan{A}\cos^4{\frac{B-C}{2}}}\geq 3\sqrt{3}\]

[quote=nayel]Prove that for any acute triangle $ ABC$, the following inequality holds:
\[ \frac{1}{\minus{}a^2\plus{}b^2\plus{}c^2}\plus{}\frac{1}{a^2\minus{}b^2\plus{}c^2}\plus{}\frac{1}{a^2\plus{}b^2\minus{}c^2}\ge\frac{1}{2Rr}\][/quote]
Mathematical Reflections 3 (2008)

[quote=nayel]Prove that for any acute triangle $ ABC$, the following inequality holds:
\[ \frac{1}{\minus{}a^2\plus{}b^2\plus{}c^2}\plus{}\frac{1}{a^2\minus{}b^2\plus{}c^2}\plus{}\frac{1}{a^2\plus{}b^2\minus{}c^2}\ge\frac{1}{2Rr}\][/quote]
Very stronger and nice is:
Prove that for any acute triangle $ ABC$, the following inequality holds:
\[{\frac {r}{\sqrt {{b}^{2}+{c}^{2}-{a}^{2}}}}+{\frac {r}{\sqrt {-{b}^{2
}+{a}^{2}+{c}^{2}}}}+{\frac {r}{\sqrt {{b}^{2}+{a}^{2}-{c}^{2}}}}\geq \frac{1}{2}\sum{\tan{\frac{A}{2}}}\]
<=>
\[\sum{\sqrt{\tan{A}}}\geq (\sum{\tan{\frac{A}{2}}})\sqrt{\sum{\cot{\frac{A}{2}}}}\]this is best result today!
这是今天最好的结果之一！

[quote=xzlbq]

In acute triangle,prove

\[\sum{\tan{A}\cos^4{\frac{B-C}{2}}}\geq 3\sqrt{3}\][/quote]

stronger
In acute triangle,prove

\[\sum{\tan{A}\cos^4{\frac{B-C}{2}}}\geq 3\sum{\tan{\frac{A}{2}}}\]


x1,x2,...,xn>0
for 1 \leq i<j \leq n
 \sum xi*xj(xi 2 +xj 2 )  \leq  (\sum xi)^4/8

Let x1\geq x2\geq ... \geq xn - positive numbers, S=x1+...+xn.

\sum xi*xj(xi2+xj2)=\sum xi^3(S-xi)=S\sum xi^3-\sum xi^4

Let pi=xi/S (denote: p1\geq p2 \geq ...).

So our inequlity has alternative form
1/8+\sum pi^4 - \sum pi^3 \geq 0, p1+p2+...+pn=0

If n=2 then initial inequality is (x1-x2)^4\geq 0 - true
Suppose n>2.

Let q1=p1,...,q{n-2}=p{n-2},q{n-1}=p{n-1}+pn, a=p{n-1},b=pn.
Denote a+b<2/3 !!!!

We shall show that
1/8+\sum pi^4 - \sum pi^3 \geq 1/8+\sum qi^4 - \sum qi^3

Actually, it is
3ab(a+b)\geq 4ab(a^2+b^2)+6a^2b^2
or
3(a+b)+2ab\geq 4(a+b)^2, but 4(a+b)^2 \leq 4(a+b)*2/3<3(a+b)

We made reduction from n to n-1. Since for n=2 inequlity is true then it is true for all n>=2.


The number of real values of $ x$ satisfying the equality $ (\log_2x)(\log_bx) \equal{} \log_ab$, where $ a > 0$, $ b > 0$, $ a \neq 1$, $ b \neq 1$, is:

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \text{a finite integer greater than 2} \qquad \textbf{(E)}\ \text{not finite}$

hello, rewriting your equation in the form $ (\ln(x))^2\equal{}\frac{(\ln(b))^2\ln(2)}{\ln(a)}$ we get the following solutions $ a>1,0<b<1$ and $ x\equal{}e^{\minus{}\sqrt{\frac{(\ln(b))^2}{\ln(a)}}\sqrt{\ln(2)}}$ or
$ x\equal{}e^{\sqrt{\frac{(\ln(b))^2}{\ln(a)}}\sqrt{\ln(2)}}$. In the case of $ a>1$ and $ b>1$ we have
$ x\equal{}e^{\minus{}\sqrt{\frac{(\ln(b))^2}{\ln(a)}}\sqrt{\ln(2)}}$ or $ x\equal{}e^{\sqrt{\frac{(\ln(b))^2}{\ln(a)}}\sqrt{\ln(2)}}$.
So the answer $ \boxed{D}$ is correct.
Sonnhard.

We have
$ \log_a{b} \equal{} \frac {\log_2{b}}{\log_2{a}}$
$ \log_b{x} \equal{} \frac {\log_2{x}}{\log_2{b}}$

Therefore

$ (\log_2{x})(\frac {\log_2{x}}{\log_2{b}}) \equal{} \frac {\log_2{b}}{\log_2{a}}$
$ \log_2^2{x} \equal{} \frac {\log_2^2{b}}{\log_2{a}}$

$ \frac {\log_2^2{b}}{\log_2{a}}$ is an arbitrary constant, name it $ c$. Then $ \log_2^2{x} \equal{} c \implies \log_2{x} \equal{} \pm \sqrt {c} \implies x \equal{} 2^{\pm \sqrt {c}}$

There are exactly 2 solutions, so the answer is $ \boxed{C}$.

Observation: I had originally written "$ c$ is an arbitrary [i]positive[/i] constant, but upon closer inspection, that is not necessarily the case. Wouldn't $ 0 < a < 1$ imply no solutions, as it would lead to $ c < 0$? Without the condition $ a > 1$, the problem has no solution. I see that, even though Sonnhard's "cases" have the same solutions, he also states that $ a > 1$.


[quote="Virgil Nicula"]Let $ a$, $ b$ be two positive numbers for which $ (a \minus{} 1)\cdot (b \minus{} 1)\ge 0$. Then :

$ \boxed {a^b\ \plus{} \ b^a\ \ge\ 1\ \plus{} \ ab\ \plus{} \ (1 \minus{} a)(1 \minus{} b)\ \cdot\ \min\ \{\ 1\ ,\ ab\ \}}\ .$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42452  :lol:

[quote="arqady"]See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42452  :lol:[/quote]
[color=darkblue]Yes, [b]Arqady[/b], I forgot it ! Thank you, [b]Arqady[/b] !  I think this problem was the first my topic on M.L.S. Indeed, here is :

[b]Posted: Sun Jun 26, 2005 11:34 am Last edited by Virgil Nicula on Thu Aug 04, 2005 5:03 PM[/b][/color]

[color=darkred]The [b]moderators[/b] can delete this topic.[/color]


modular arithmetic
calculus
integration
number theory

1) Prove that a positive integer is the sum of a sequence of at least two consecutive positive integers if and only if it is not a positive power of 2.

2) Find all non-negative integers x such that $x^4 + x^3 + x^2 + x + 1$ is a perfect square.

3) If $a, m, n$ are positive integers with $a>1$ show that $a^m + 1 | a^n + 1 \implies m|n$

4) Prove that there are infinitely many integers which cannot be expressed as the sum of three squares.

Fourth problem:
If $n\equiv{7}\pmod{8}$, then $n$ cannot be expressed as a sum of three squares.
Moreover, if $n=4^l(8k+7)$, then $n$ cannot be expressed as a sum of three squares.

How do you conclude that if $n\equiv 7\pmod{8}$, it cannot be written as the sum of three squares?

Well, we know that $n^2$ is equivalent to $0$, $1$, or $4 \pmod{8}$.

[quote="amcavoy"]OK that makes sense looking back at the original statement.  I can see how $n^{2}\equiv 0,1,4\pmod{8}$ by trying different values of $n$, but is there a proof that this works for all $n\in\mathbb{Z}$?[/quote]

Say $n \equiv r \bmod 8$ where $r$ is the smallest possible positive integer that satisfies this property (that is, $r$ is the remainder of the division by $8$).  Then $n^2 \equiv r^2 \bmod 8$ and since there are a maximum of $8$ possible values of $r$ (that is, $r = 0, 1, 2, ... 7$) there are a corresponding maximum of $8$ possible values of the remainder of the division of $n^2$ by $8$, and these values also repeat with maximum period $8$.

That's interesting.  Someone just told me to show that $n^{2}\equiv\left(n+4\right)^{2}\pmod{8}$ to show it. Thanks for the help!

Well, using what I just learned here, I would like to attempt the first problem:

[hide="Problem 1"]If you have two consecutive positive integers $n$ and $n+1$, then their sum $2n+1=k\equiv 1\pmod{2}$.  If the sum could be written as a positive power of 2, then we would have $k\equiv 0\pmod{2}$, which isn't true.

Can someone tell me if this is a sufficient argument?  Sorry, I just don't know much about number theory (at all :))[/hide]

Solution to problem 1

First if n is odd, then set  k+(k+1) = 2k+1 = n
   Therefore we get an integral value for k.
   Thus n can be written as k+(k+1).

If n is even and not a power of 2,
  {A}  Then k+(k+1)+(k+2)+(k+3) = 4k+6 = 2(2k+3)
          Set n = 2(2k+3) .
                If n/2 is odd we get an integral value for k and we
                have proved n = sum of 4 pos. integers.

      But if n/2 is even consider 8 consec. integers.
           Set n = 4(2k+7) and continue the process until we get
           integral value for k. Thus proving it.

But if negative values are produced for k, i.e if the no. is of the form (2^k)pq , where p,q are primes, then consider p consec. integers.
For eg. 24 = 8*3 can be written as 3(k+3).

Here note that for powers of 2 the process {A} goes on  producing even nos. and finally we get 1 on the RHS thus producing negative values for k


calculus
integration
algebra
polynomial
algebra proposed

Let $ p(x) \equal{} x^8 \plus{} a_7 x^7 \plus{} a_6 x^6 \plus{} . . . \plus{} a_1x \plus{} a_0$  be a polynomial with integer coefficients and let $ q(x) \equal{} (p(x))^2 \minus{} 25$. Prove that there not more than $ 8$ distinct integers $ m$ such that $ q(m) \equal{} 0$.

PS:Is it an original(new) problem or copied?

No one interested?
It is quite a nice problem I guess.....  :)


Support
email
AMC
AIME
AIME I
number theory

I don't know what you are talking about, but you might want to ask your question in the "Support" section of the forum, since the instructors peruse that section more often and can answer your question there.

I agree. I don't suppose the normal registered users can answer those questions.  You might need to ask instructor or if you can't contact one, you could ask one of moderator.  Try to ask tokenadult he'll give you neat answer if you want.

[quote="wheezy81"]I just joined this math jam i got a confirmation email. My dad also got one and it said that i needed to download these certain sheets from the sites. Where do i get them?!?!?![/quote]

I think you are referring to the upcoming online class. I don't recall that there is any email notification like that for math jams--the jams are just first come, first served on the day that the jam occurs. You should be able to click on the tab "My Classes" (I can see such a tab on my screen right now) to get the information about what classes you are signed up for. And with the information about the class should be links for assigned readings. The only class we have signed up for so far is the class that happened the weekend before last to prepare for the AIME I test. We would LOVE to sign up for the number theory class that starts soon, except that the time conflicts with my son's algebra 2 class, which he takes across town at the U. of Minnesota in the late afternoons. Perhaps that is the class you are signed up for.

Hope this helps! I welcome any suggestions to the original questioner that other participants have here, as I am still gaining experience with the online classes that AoPS offers.

Hmm...well personally I get reminders for Math Jams :-)


Let $ A$ and $ B$ be real square matrices of the same size such that there exists nonzero real numbers $ p,q$ and $ r$ satisfy
$ pAB\plus{}qBA\equal{}I$ and $ A^2\equal{}rB^2$. Prove that if $ AB\neq BA$ then $ p\equal{}q$.

[hide="Hint"]We have  $ pA(pAB\plus{}qBA)\equal{}pA$ or $ p^2rB^3\plus{}pqABA\equal{}pA$ (1)  and $ (pAB\plus{}qBA)qA\equal{}qA$ or $ pqABA\plus{}q^2rB^3\equal{}qA$ (2)
 From (1) and (2) easy $ (p\minus{}q)(p\plus{}q)rB^3\equal{}(p\minus{}q)A$ ....[/hide]


Find all m such that 7x+31y=m has exactly 5 positive integer solutions.

leonsotelo

[hide="Not sure if this is correct, I just sketched it out."]

7x + 31y = m
31y = -7x + m
y = (-7/31)x + (m/31)

Since gcd(7,31) = 1, if $ x, y \in \mathbb{Z}$ then $ 31 | x$.
Graphing this, we see that $ x \equal{} 31, 62, 93, 124, 155$ must be the x coordinates of lattice points on the line in Quadrant I.

Notice that $ x \equal{} 186$ will give us an integral solution, and since there are exactly 5 positive integer solutions, $ x \equal{} 186$ must give a negative y. $ y \equal{} \minus{} 42 \plus{} m < 0$

Using this, we can find a bound on m.

$ 31 \times 31 < m < 42 \times 31 \Rightarrow 961 < m < 1302$

methinks that's the answer?

[/hide]


induction
combinatorics unsolved
combinatorics

Prove that for every n>=3, we can fill a n*n magic square with number 1 to n^2(every grid should have different nums) that the sum of numbers in every row, column and diagnol are same.

It should be a result already proved.

I try to use induction, since n=3 or 4 is fine.
so I suppose n=k works and try to prove n=k+2 also works by extend it two more rows and columns. Then fill in the rest numbers in an arranged way.But I can not work that out.

Check this link [url]http://en.wikipedia.org/wiki/Magic_square[/url]


ok, i have posted 2 different posts about finding the area with given vertices...

so far...learned nothing other than the shoelace formula given at the following website does not work! (either i did it wrong or..)

http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf

please [u]solve [/u]the following which i posted earlier.. but did not ever figure out how to solve...

Find the area of a convex polygon created by connecting the following points: (-2,-3)(-1,4)(2,-4)(3,5)(6,1)..

Please show allllllll work!! Please!

so to find the order in the "code" you always have to graph it, and write down the points in either clock or counterclockwise order...because doing the same thing in the order the question is given for example does not work.... :huh:

Don't you have to put the starting point again?

dont know about that but putting them in clock or counter clock wise order works, but anything else doesnt.... at least i think so... :huh:

Yes all the points have to be in consecutive order with the first point you used repeated in order to work.

Can someone tell me why this method works? I see how it's done, but I don't really understand it...


inequalities
geometry
linear algebra
matrix
induction

Is there any elementary way to prove AM-GM ?  :?

Like could it be prove by using basic algebra ?

[quote="fourierseries"]Yes it can.  Use smoothing:

Replace an element above the mean and an element below the mean with their sum minus the mean.[/quote]

I dont quite understand , can you elaborate more ? :?

[quote="varun"]What do you mean by AM-GM. :?[/quote]

Arithmetic Mean and Geometric mean ...

Here is what I know about the proof of AM-GM
1. By induction. There are three kind:
    -WLOG,asuming $a_1a_2 ... a_n =1$
    -asumming $a_1=a_2=....=a_{n-1}$
    -prove by case i)$n=2^k$ ii)otherwise
2. By using Langrange Multiplier.
3. By using Jensen
4. By using Smoothing Principle

anything else?? I'm asking about the proof using geometry, Matrix Algebra, or other if there are.

You can find the proof by Smoothing principle in Kiran keldaya's note, someone has posted it in intermediate forum-inequality lesson.

[quote="varun"]I know it is Arithmetic Mean and Geometric mean.
Can you please tell me actually what we have to prove. Please tell the expresision. :D[/quote]

Well , what I want is just that how we could prove the inequality for AM-GM is true .

And for [b]Beginner[/b] , could you show the induction prove ?  :roll:

if $a_1$, $a_2$, $...$, $a_n$ is positive real number then

$ \frac{a_1+a_2+...+a_n}{n} \geq ({a_1a_2...a_n})^\frac{1}{n}$
with equality if and only if $a_1 ,a_2,...a_n$ are all equal.

well, as it is real number, there must be $k$ so that, $a_1a_2...a_n = k^n$ and then for all $a_i$ there are correspondence $b_i$ so that $a_i = k b_i$, clearly we have that $b_1b_2..b_n=1$so proving AM-GM is equivalent to prove that
$k \frac{b_1+b_2+...+b_n}{n} \geq k ({b_1b_2...b_n})^\frac{1}{n}$ with $b_1b_2 .... b_n=1$. So $WLOG$ we may assume that $a_1a_2 .... a_n=1$. I'll skip the induction hypothesis, since its easy to prove it for 2 variable case.
Now assume we have proof for $n-1$ variabel. By defenition of AM and GM as the "Mean" there are some values $a_i$ and $a_j$ so that $a_i \geq GM=1 $ and $ a_j \leq GM=1$ for convenient we say that $a_1$ and $a_2$, thus by induction hypothesis we have  $ \frac{a_1a_2+...+a_n}{n-1} \geq ({a_1a_2...a_n})^\frac{1}{n-1} =1$ or $a_1a_2+...+a_n \geq n-1$, to complete the proof we need to show that $a_1+a_2+...+a_n \geq n$,or equivalent to prove that $a_1+a_2 \geq a_1a_2+1$ but we have
$a_1+a_2-(a_1a_2+1) = (a_1 -1)(1-a_2) \geq 0 $
I hope its correct, for another proof by induction you can see in  "inequality lesson maybe marathon"


In a particular dice game a player rolls two dice.  The player can choose either the sum or the product of the numbers rolled for her score for that turn.  To win the game, a player must get a total score of exactly 102 points from the sum of her scores.  What is the least possible number of turns needed to win the game?

Assuming she gets 6 and 6 for 3 times. Then whe rolls something that adds to 6 or multiplys to 6.
Ans is 4.

I got a different answer. If she gets $ 6 \times 6 \equal{} 36$ twice, then she's left with 30. We can get this from $ 5 \times 6$, so we have $ \boxed{3}$.


Let $ f$ bounded and positive. Prove that $ \sup [f^2] \equal{} [\sup f]^2$

For every $ x$ we have $ f^2(x) \equal{} f(x)\cdot f(x) \le [\sup f]^2$
So $ \sup[f^2] \le [\sup f]^2$

$ \sup[f] < f(x) \plus{} \epsilon$ for given $ \epsilon$ and some $ x$. Then
$ [\sup f]^2 < f^2(x) \plus{} \epsilon^2 \plus{} 2\epsilon f(x) \le \sup[f^2] \plus{} \epsilon^2 \plus{} 2\epsilon \sup[f]$
This is true for every $ \epsilon > 0$,
so $ [\sup f]^2 \le \sup [f^2]$

Thus $ [\sup f]^2 \equal{} \sup [f^2]$

[quote="PeterP"]
$ [\sup f]^2 < f^2(x) \plus{} \epsilon^2 \plus{} 2\epsilon f(x) \le \sup[f^2] \plus{} \epsilon^2 \plus{} 2\epsilon \sup[f]$
This is true for every $ \epsilon > 0$,
[b][color=red]so[/color][/b] $ [\sup f]^2 \le \sup [f^2]$
[/quote]

in think that there is a wrong in this step

$ a\leq b \plus{} e \,\,\, \forall e > 0$ doesn't implies $ a\leq b$

This propotition is true: $ a\leq b \implies a\leq b \plus{} e$ , $ \forall e > 0$,  but not  in reverse way

[quote="jorgeston"]
$ a\leq b \plus{} e \,\,\, \forall e > 0$ doesn't implies $ a\leq b$
[/quote]

Yes it does.

[quote="jorgeston"]
in think that there is a wrong in this step

$ a\leq b \plus{} e \,\,\, \forall e > 0$ doesn't implies $ a\leq b$

This propotition is true: $ a\leq b \implies a\leq b \plus{} e$ , $ \forall e > 0$,  but not  in reverse way[/quote]

To expand on what Igor already said: suppose the contrary, that $ a\leq b \plus{} \epsilon$ for all $ \epsilon > 0$, but $ a > b$.  Can you come up with a contradiction?

[quote="JoeBlow"][quote="jorgeston"]
in think that there is a wrong in this step

$ a\leq b + e \,\,\, \forall e > 0$ doesn't implies $ a\leq b$

This propotition is true: $ a\leq b \implies a\leq b + e$ , $ \forall e > 0$,  but not  in reverse way[/quote]

To expand on what Igor already said: suppose the contrary, that $ a\leq b + \epsilon$ for all $ \epsilon > 0$, but $ a > b$.  Can you come up with a contradiction?[/quote]

If  $ a > b$, we have $ a - b > 0\iff a - b\geq e \,\,\, \forall e\geq 0$

So, $ a - b\geq e$ and $ a - b\leq e$ , $ \implies a - b = e\implies a\leq b$

i'm sorry and thanks to you!


A singles tournament had six players. Each player played every other
player only once, with no ties. If Helen won 4 games, Ines won 3
games, Janet won 2 games, Kendra won 2 games and Lara won 2 games,
how many games did Monica win?

There are a total of 6C2=15 games.

So far, we know there are 4+3+2+2+2=13 games, so Monica won 15-13=2 games.


[quote="Gen8"]The graph of $ x^2$ + $ y^2$ = $ 29$ and the curve $ xy$= $ 10$ intersect at certain number of points. All these points of an intersection are joined successively by line segments to form a polygon. Which of the following best describes the polygon thus obtained?

(a) Rectangle (b) Square (c) Hexagon (d) Rhombus (e) Pentagon[/quote]
[hide]$ x^{2}\plus{}y^{2}\equal{}(x\plus{}y)^{2}\minus{}2xy\equal{}29$
$ (x\plus{}y)^{2}\equal{}49$
$ x\plus{}y\equal{}\pm 7$
$ (x,y)\equal{}(\pm 5, \pm 2), (\pm 2, \pm 5)$
These 4 points form a rectangle.[/hide]

[quote="ckck"][quote="Gen8"]
$ (x,y) \equal{} (\pm 5, \pm 2), (\pm 2, \pm 5)$
These 4 points form a rectangle.[/hide][/quote][/quote]

You are showing eight points, not four points.
Your post shows twice the number of intersection points than there are.

What you have is all of the ways of combining positive and negative
values for five and two.

The points of intersection are:

$ (5, 2)$
$ (2, 5)$
$ (\minus{}5, \minus{}2)$
$ (\minus{}2, \minus{}5)$


[b][color=blue]if (cos0,sin0)=(1,0) find (cos1, sin1)   ;0,1 degree[/color][/b] :?:  :?:  :?:

Definitely not..
But, heh, I can't get the point of this 'exercise'(/question)...

Well.  Put your calculator on degree mode.  Type in cos 1.  Type in sin 1.  On your homework, write down a left parenthesis, then the number you received for cos 1, then a comma, then the number you got for sin 1, and then a right parenthesis.

In my memory, cos 36 can be found by hand. Then half angle formula, half angle formula gives the value of cos 9. Reverse the triple angle formula to get cos 3. Do that again to get cos 1.

[quote="siuhochung"]Reverse the triple angle formula to get cos 3[/quote]
Easier said than done- that's solving a cubic equation, in the three real roots case. There is no formula involving strictly real arithmetic and radicals.

Actually, you can get to $\cos 3^{\circ}$ by starting with a 15-gon; it's the step down to $1^{\circ}$ that runs into trouble.

Of course, it's very easy to accurately approximate this cosine.


Put the 4-digit numbers into two groups based on whether

they are the products of two 2-digit numbers or not.

Which group is greater?

sol : $9999 = 99 times 101 > 99^{2}= 99 times 99$

[quote="b-flat"]Do you mean any four digit numbers that can be expressed as the product of two 2 digit numbers and those that can't, or the numbers that have 2 two digit prime factors?[/quote]
The first one I think... He didn't mention prime...


That explains why you created an 11-15 choice. I seem to have gotten a perfect.

9 :(
only 1 point improvement from last year.
i am too slow...

score is 4 :furious:
hope to get 8 next year


algebra
function
domain
irrational number
calculus
calculus computations

[quote="jmerry"]Assume that the domain is an interval; say, $ [0,1]$. If $ f$ is continuous, its range is also an interval $ I$. If it's nonconstant, $ I$ contains uncountably many irrational numbers. Those can't all be images of the countably many rational numbers, so every nonconstant continuous function on $ [0,1]$ maps some irrational number to an irrational number.[/quote]

Thanks.
It make sence. :idea:

Why do you go back and delete your original question after you get your answer?  You've done this in at least two threads now.  It confuses other people who are following the thread.  Do you have something to hide?

The question was whether there is a nonconstant continuous function that maps irrational numbers to rational numbers.  I agree that this behavior is extremely irritating.

wonder if this is some take home final problem..


In any acute triangle $ ABC$, let $ H$ be it's orthocenter. Let $ O$ be the midpoint of $ \overline{BC}$. Let $ \Omega$ be a circle with center $ O$ passes throw $ B,C$. If $ \overline{AN}, \overline{AM}$ tangent $ \Omega$ in $ M,N$. Prove that $ M,N,H$ are collinear.

 [geogebra]2c95f774b2381df91d3589518099e0a607f310ee[/geogebra]

Let $D,E$ be the orthogonal projections of $B,C$ onto $AC,AB$ $\Longrightarrow$ $D,E \in \Omega.$ If $ED$ cuts $BC$ at $P,$ then the line pencil $A(B,C,H,P)$ is clearly harmonic, thus it follows that $AH \perp BC$ is the polar of $P$ with respect to $\Omega.$ Therefore, the polar $\tau$ of $A$ with respect to $\Omega$ passes through $H$ $\Longrightarrow$ $MN \equiv \tau,$ i.e. $H \in MN.$

Here is another approach:
Consider the inversion through pole $ A$, power $ \mathcal {P}_{A/(O)} \equal{} k^2$. Let $ D$ be the projection of $ H$ onto $ BC$. It is easy to note that $ \mathcal {I}(A,k^2): H\mapsto D$, $ M\mapsto M$, $ N\mapsto N$. But $ \angle OMA \equal{} \angle ONA \equal{} \angle ODA \equal{} \frac {\pi}{2}\Longrightarrow \{A,M,N,O,D\}$ is a set of concyclic, which implies that $ H,M,N$ are collinear.
Our proof is completed then.

$ AH\cap BC\equal{}H'$
$ A,M,H',O,N$lies on $ \Omega$,then an inversion with center $ A$,and radius of the power of $ A$ to $ \Omega$
tells us that the inverses,$ M,H,N$should be colinear!
(not return of Ashegh,yet :( )

[quote="Ashegh"]$ AH\cap BC \equal{} H'$
$ A,M,H',O,N$lies on $ \Omega$,then an inversion with center $ A$,and radius of the power of $ A$ to $ \Omega$
tells us that the inverses,$ M,H,N$should be colinear!
(not return of Ashegh,yet :( )[/quote]
This is exactly my solution above, you can see it, my friend :D

right,I usually don't see other solutions before posting my solution!
because I don't want to get other's Ideas about the problem,before thinking about it.
whish U luck :wink:

Let $ D,\ E,\ F$ be, as  the points of intersection of $ BC,\ AC,\ AB$ respectively, from the line segments $ AH,\ BH,\ CH,$ where $ H$ is the orthocenter of the given triangle $ \bigtriangleup ABC.$

The points $ E,\ F,$ lie on the circle $ (O)$ ( instead of $ \Omega$ ), with diameter $ BC,$ as well.

We denote as $ K,$ the point of intersection of $ AO,$ from the line segment $ MN$ and from the right triangle $ \bigtriangleup MAO$ with $ \angle AMO \equal{} 90^{o}$ and $ MK\perp AO,$ we have that $ (AK)\cdot (AO) \equal{} (AM)^{2}$ $ ,(1)$

Let $ K'$ be, the orthogonal projection of $ H,$ on the line segment $ AO$ and it is enough to prove that $ K'\equiv K.$

Frome the cyclic quadrilaterals $ K'ODH,\ HDBF,$ $ \Longrightarrow$ $ (AK')\cdot (AO) \equal{} (AH)\cdot (AD) \equal{} (AF)\cdot (AB) \equal{} (AM)^{2}$ $ ,(2)$

From $ (1),\ (2)$ $ \Longrightarrow$ $ (AK)\cdot (AO) \equal{} (AK')\cdot (AO)$ $ \Longrightarrow$ $ AK' \equal{} AK$ $ \Longrightarrow$ $ K'\equiv K$ and the proof is completed.

Kostas Vittas.

[quote="Luis González"]The polar of $ H$ WRT $ (O)$ goes through $ A$[/quote]
How do you prove this?


integration
trigonometry
calculus
calculus computations

howdee doods !  :D

can you sum

$ \dfrac{1}{1\cdot (\pi\plus{}0)}\minus{}\dfrac{1}{2\cdot (\pi\minus{}1)} \plus{} \dfrac{1}{3\cdot (\pi\plus{}2)}\minus{}\dfrac{1}{4\cdot (\pi\minus{}3)} \plus{} \dfrac{1}{5\cdot (\pi\plus{}4)} \minus{}\dfrac{1}{6\cdot (\pi\minus{}5)}\plus{}\dfrac{1}{7\cdot (\pi\plus{}6)}\minus{} \dfrac{1}{8\cdot (\pi\minus{}7)} \plus{}\dfrac{1}{9\cdot (\pi\plus{}8)} \minus{}\dfrac{1}{10\cdot (\pi\minus{}9)} \plus{}\dfrac{1}{11\cdot (\pi\plus{}10)}\minus{}\;\;\ldots \;\;\plus{}\;\;\minus{}\;\;\ldots$

try to consider:
$ \int_{C}\frac {dz}{(\pi \plus{} ze^{\pi i z })\sin \pi z}$ where $ C$ is big rectangle.


solve the system equation \[ \left\{ \begin{array}{l}
 2{x^3} \plus{} y \equal{} 37x \plus{} 70 \\
 {y^3} \plus{} z \equal{} 18y \plus{} 40 \\
 3{z^3} \plus{} x \equal{} 56z \plus{} 100 \\
 \end{array} \right.\]

[quote="imedeen"]solve the system equation
\[ \left\{\begin{array}{l} 2{x^3} \plus{} y \equal{} 37x \plus{} 70 \\
{y^3} \plus{} z \equal{} 18y \plus{} 40 \\
3{z^3} \plus{} x \equal{} 56z \plus{} 100 \\
\end{array} \right.\]
[/quote]

[hide="partial result"]
The first equation may be written $ y\minus{}x\equal{}\minus{}2x^3\plus{}36x\plus{}70\equal{}\minus{}2(x\minus{}5)(x^2\plus{}5x\plus{}7)$
The second equation may be written $ z\minus{}y\equal{}\minus{}y^3\plus{}17y\plus{}40\equal{}\minus{}(y\minus{}5)(y^2\plus{}5y\plus{}8)$
The third equation may be written $ x\minus{}z\equal{}\minus{}3z^3\plus{}55z\plus{}100\equal{}\minus{}(z\minus{}5)(3z^2\plus{}15z\plus{}20)$

These lines show immediately that we have a solution with $ x\equal{}y\equal{}z\equal{}5$

It remains the most difficult part : to show this is the unique solution ... :(
[/hide]

i have another solution for your problem
suppose$ y\ge5$ from the first equation we get $ x\le5$from the third we get$ z\ge5$finally from the second we get$ y\le5$ so $ y\equal{}x\equal{}z\equal{}5$
and we easily prove that is the unique root

[quote="hoangvn.fix"]i have another solution for your problem
suppose$ y\ge5$ from the first equation we get $ x\le5$from the third we get$ z\ge5$finally from the second we get$ y\le5$ so $ y \equal{} x \equal{} z \equal{} 5$
and we easily prove that is the unique root[/quote]

How do you get $ z\ge 5$ from the third if $ x\le 5$ ?

[quote="hoangvn.fix"]from the first we get $ y \equal{} \minus{} 2x^3 \plus{} 37x \plus{} 70\ge5$
$ \Leftrightarrow(x \minus{} 5)(2x^2 \plus{} 10x \plus{} 13)\le0$ so we have $ x\le5$
from the third $ x \equal{} \minus{} 3z^3 \plus{} 56z \plus{} 100\le5$
$ \Leftrightarrow(z \minus{} 5)(3z^2 \plus{} 15z \plus{} 19)\ge0$ so we have $ z\ge5$
and similarily we get $ y\le5$ so $ y \equal{} 5$[/quote]

Yes, quite clear now. Thanks.