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This was pathetic. I have nothing to say about this. It should be ignored.

When I click the link, it appear as an empty page... can anyone pass whatever it is about here????

[quote="zbz"]There is a solution to the old problem
of division by zero.

0 / 0 = 0

see
http://members.lycos.co.uk/zerobyzero/

Consider it a new science.[/quote]Wow, this has got to be one of the biggest bogus site I have ever seen written by a layman. Waste of my time...

[quote="Xevarion"]It is an absurd argument that 0/0 = 0.

In truth, the statement 0/0 has no meaning unless we describe more specifically what those 0s represent. But the meaning he tries to give to it is bizarre and not very mathematical.[/quote]
Hmm,I also wonder how did he miss $0^0$ "argument" in his "discovery"?
 ;)  :roll:

I've fixed my PC, and I could open it

It's really "stupid", sorry for the word

[quote="PTynan89"][quote="Article"] There are no half zeros. Zero can not be physically divided since there is nothing to divide. [b][i]Sure you can swing a knife through empty space pretending to divide but you are not dividing but rather swinging a knife around so stop it before someone gets hurt.[/i][/b] When zero is being divided we agree that it will not be divided and so you don't divide it and your result will equal zero indicating that you did not divide. Example 0 / 12345 is equal to 0 / 0 which is equal to 0. That " / 0 " represents that you did not divide. [/quote]

ROFL

Way to go completely off topic.[/quote]

Where did you get this from? What an absurd argument

[quote="Article"]When you Multiply your objects then you state,

(Object) * (Multiple) = (Product Of Multiplication)

If you wish not to multiply then you multiply by ZERO and you will NOT HAVE a product. (1 Apple) * (ZERO) = 0 Apples Produced.

You can view ZERO as an order not to produce any multiples of apples.
[/quote]

If you "will not have a product," then you should get your apple  back. You cannot just make the apple dissappear, that violates the law of conservation.  :P

[quote="EUCLID 3.14"]I don't know you guys. We shouldn't be too harsh on him. I admit that what he says makes a little sense, but it's not like the concept of 0/0 is perfectly well understood in mathematics, right? :read:[/quote]

Take a look at this. Maybe it will help.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86053

[quote="zbz"]division by zero.[/quote]

As far as I know $\frac{0}{0}$ is arbitrary real number. Just consider the definition of division, and u will get that... :wink:

[quote="conceive"]As far as I know $\frac{0}{0}$ is arbitrary real number. Just consider the definition of division, and u will get that... :wink:[/quote]what exactly do you mean by that?

[quote="swapnillium"]what exactly do you mean by that?[/quote]

Well, if you can not divide arbitrary two real numbers, can you at least multiply arbitrary two real numbers? I suppose you can.

Suppose we can multiply arbitrary two real numbers. Define division as follows: Define $\frac{a}{b}$ as a such real number $c$ such that $bc=a$ (we can multiply - we know that). That is definition of division.

And how about $\frac{0}{0}$? Take $a=b=0$. We get that $0.c=0$ what satisfy arbitrary real number $c$.

On the other hand, let us examine the case $\frac{t}{0}$ for $t\not=0$. Then we have $0.c=t$ what is a contradiction.

Conclusion: We can divide by zero iff we divide zero. In addition $\frac00=$ arbitrary real number.
Advice: if you are not sure, return to definitions.

conceive

[quote="conceive"]Suppose we can multiply arbitrary two real numbers. Define division as follows: Define $\frac{a}{b}$ as a such real number $c$ such that $bc=a$ (we can multiply - we know that). That is definition of division.[/quote]Isn't the definition of division more like the following:
$\frac{a}{b}$ is a number $c$ such that $bc=a$ $\mbox{provided that}$ $b \not = 0$.

Either way, I think that you are committing the same error as swapnillium did in this post http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86053

[quote="swapnillium"][If you "will not have a product," then you should get your apple  back. You cannot just make the apple dissappear, that violates the law of conservation.  :P[/quote]

In fact, this is truer than you maybe realize.  Just as the value of an empty sum is 0, the value of an empty product is 1.  This fact shows up all over the place -- $0! = 1$ and $x^{0}= 1$ for $x\neq 0$ are essentially just restatements of this fact.

[quote="EUCLID 3.14"]$\mbox{provided that}$[/quote]

Write here a simple one reason for that condition.

I am too lazy to read that whole topic..
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86053

One good reason: an operation is a function.  A function has a domain and a range.  In the case of (real) division, the domain certainly contains $\mathbb{R}\times \mathbb{R}^{*}$ and the range is $\mathbb{R}$.  You want to argue that it should be extended to $\mathbb{R}\times \mathbb{R}\bigcup \{0, 0\}$ and that the range should be extended to $\mathbb{R}\bigcup\{\frac{0}{0}\}$, where $\frac{0}{0}$ means "any real number" or something similarly unclear.  The ugliness of that description is one good reason to simply omit zero entirely.  (More simply, "So that division maintains a simple, meaningful existence as a function.")

Also, too lazy to read 11 posts?   :roll:

[quote="JBL"]One good reason: an operation is a function.  A function has a domain and a range.  In the case of (real) division, the domain certainly contains $\mathbb{R}\times \mathbb{R}^{*}$ and the range is $\mathbb{R}$.  You want to argue that it should be extended to $\mathbb{R}\times \mathbb{R}\bigcup \{0, 0\}$ and that the range should be extended to $\mathbb{R}\bigcup\{\frac{0}{0}\}$, where $\frac{0}{0}$ means "any real number" or something similarly unclear.  The ugliness of that description is one good reason to simply omit zero entirely.  (More simply, "So that division maintains a simple, meaningful existence as a function.")[/quote]
That's an interesting idea JBL, but I don't really know what exactly do you mean. Could you put it more simply?

[quote="conceive"]Write here a simple one reason for that condition.
I am too lazy to read that whole topic..
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86053[/quote]You just have to read the first post in that topic to see what I am talking about. Anyways, here's a little modified version of what swapnillium said:
[quote="swapnillium"]Why is 0/0 indeterminate?
My friend offered me the following reason:
Say $\frac{0}{0}$ is some number $z$, then
$\frac{0}{0}= z \Rightarrow 0 = 0 \cdot z$
Thus $z$ could be any number and so it is impossible to determine the value of z. Hence 0/0 is indeterminate.

Here is what's wrong with the argument. To go from $\frac{0}{0}= z$ to $0=0 \cdot z$, you are actually multiplying both sides by $0$! Thus,
$0 \cdot \frac{0}{0}= 0 \cdot z$ but you can't simplify the left hand side because you can't cancel the $0's$ on the left. Hence the above argument is invalid.[/quote]

[quote="EUCLID 3.14"]
[quote="swapnillium"]Why is 0/0 indeterminate?
My friend offered me the following reason:
Say $\frac{0}{0}$ is some number $z$, then
$\frac{0}{0}= z \Rightarrow 0 = 0 \cdot z$
Thus $z$ could be any number and so it is impossible to determine the value of z. Hence 0/0 is indeterminate.

Here is what's wrong with the argument. To go from $\frac{0}{0}= z$ to $0=0 \cdot z$, you are actually multiplying both sides by $0$! Thus,
$0 \cdot \frac{0}{0}= 0 \cdot z$ but you can't simplify the left hand side because you can't cancel the $0's$ on the left. Hence the above argument is invalid.[/quote][/quote]

Well, that is right. But my defenition of division is correct.

[quote="EUCLID 3.14"][quote="conceive"]Suppose we can multiply arbitrary two real numbers. Define division as follows: Define $\frac{a}{b}$ as a such real number $c$ such that $bc=a$ (we can multiply - we know that). That is definition of division.[/quote]Isn't the definition of division more like the following:
$\frac{a}{b}$ is a number $c$ such that $bc=a$ $\mbox{provided that}$ $b \not = 0$.[/quote]Well, this is what I am going to say again. All the books I have read so far one way or the other put the constraint that b cannot equal zero on the definition on division.

Lets see what the "experts" have to say on this.

One of the most important reason is the property that we get groups/rings/fields, in other words that we can calculate as usual, with the laws of associativity, commutativity, distributivity,... .

And $\frac{0}{0}= \text{any number}$ (whatever "any number" means...) is wrong cause of
$\text{any number}= \frac{0}{0}= \frac{0^{2}}{0}= 0 \cdot \frac{0}{0}= 0 \cdot (\text{any number}) = 0,$
showing nonsense.
It shows that you can indeed divide by $0$ if your ring/"field" has only one element :D

All those unnatural constructions where you can divide therough $0$ don't fulfill any laws you can calculate with in a good way.

Why not use L'Hospital's Rule to prove 0/0 is equal to any number in the extended reals.  $\lim{x/x}=0/0=1$

Edit: Well I'm very sorry for my extra-mega-stupid mistake, and sorry, ZetaX. :blush:  :blush: I think I'd better keep my mouth shut.

[b]THE EXISTENTIAL BEING OF MATHEMATICS[/b]

[i][u][b]Zeroness[/b][/u][/i] & 9-ness vs. "1"-ness; Zero-Nine based vs. 10 based Numerical System:

Nothing is ever rejected, not rejection and not even Nothing. The Beingness rejects none and not even none nor rejection are rejected. There is 'no' rejection, not even of rejection or of 'no'.

We is not Everything Being as one ('oneness'), We is simply Everything, Being.

"Oneness" does not take the shape of Everything, rather it mimmicks Everything by dividing others through its indivuduality (In-divided-duality). 2 Divided by 1 is 2. Oneness 'wants' to be individual. To Be Zeroness (Everything), and Nine (Nothingness), is to BE. It IS Existence that defines Nothing and in turn Nothing defines the non-Existent boundaries of Existence which shows Us that Existence (Beingness) is Outfinitely limitless, boundless, and shapeless. 'One' must Be Outfinitely dividing into (in two) EVERYTHING, (individuality = In-divided-duality) thus dividing Everything to take the shape of Everything, and defeating the purpose of simply Being Everything. One is "trying" to BE Everything, so it is mimmicking. Yet it is perfect, just as the Being IS. We divided the Being so that We may see Nothing and Everything as Existence Being Us, WE ARe Existence.

"Oneness" is not the Beingness; because "one" is dividing Existence into (in two) "Nothingness" (illusion of separateness), allowing Us to see Everything as Everything AND Nothing as Everything. We is Zeroness, both Everything AND Nothing (numbers 0 & 9), there is 'Nothing' to be divided into, thus WE ARe Everything, Being.

Nothing (No-THING) cannot define itself without Existence, because it is Existence that allows 'Nothing' to Be 'non-Existent'. Everything Divided by Zero is allways undefinable. 1-:-0=Undefined, because Zero is ALL inclusive: THINGs (numerals 1-8), No-THING AND Every-THING [9and0]: i.e. "non-Existence", Existence, "individuality", Outdividuality, "illusion", reality etc.

"Oneness" is NOT the Everythingness AND Nothingness defining Zeroness, nor is "one" the Zeroness defining the Everythingness AND Nothingness. 0-:-1=0. Numbers Nine and Zero allways stay true to themselves; Our Omniverse (OMNI-verse; ALL-words) is a base 0 AND 9 mathematical BEING, NOT a base 10 numerical SYSTEM. The Being consists of numerals [9 and 0 ] (which share the same properties and attributes) and 1 through 8.

The product of nine times any number is allways 9. i.e. 9x3=9. There is no 10 because 10 = 1+0, 1+0 = 1 (Digital Root property). The number line consists of [9 and 0] and returns to 1, thus [9 and 0] times any number is [9 and 0] and [9 and 0] +,- any number equates to that number itself. There is never a negative sum or difference. There is no positive or negative, they are simply movements along the ennead number line (shown at the end of this post).

In a base 10, 9x3=36, 3+6=9.
0x3=0.

Example [9and0]+5=5

In a base 10, 9+5=14, 1+4=5.

The imprisoning factor occurs between oneness and Zeroness, ergo We created the illusion of 10. In this aspect We allso created the illusion of creation and the creation of illusion. Existence has Allways Existed and Allways will Exist; Existence, not created, Allways Exists.

[9and0] are non-Existence and Existence represented in numbers:

The Existing, non-Existent boundary of Existence IS Nothing and is defined by Existence as having no boundaries. Non-Existence defines Existence as not having Existing boundaries. Existence defines non-Existence as the boundary of Existence. Non-Existence defines the boundaries of Existence as Being non-Existent. Existence defines itself as having non-Existent boundaries. The non-Existent boundaries are Nothing. Nothing is bounded by Existence and Existence is bounded by Nothing. The boundary of Existence is non-Existent through non-Existence.

0 minus 0, 0 minus 9, 9 minus 0, 9 minus 9 = These base 10 operations are illusory in that they cannot be performed because 9 and 0 are ultimately connected

Nothing is Everything, Everything is Nothing [9and0]; Nothing minus Nothing IS Everything. [9and0] minus [9and0] IS [9and0]. To take away Nothing's Nothingness is to make it SOMEthing, since Nothing is the boundary of Existence and Something IS "No-thing" Existing (Existence has no boundaries), Something now becomes Everything (Outfinity and Eternity).

In another light; 0-0 = 0. Because Nothing IS Nothing, to take away Nothing from Nothing would equate to Nothing because there is Nothing to take away. Nothing Exists, yet it doesn't Exist, and it couldn't Exist without Existence, allthough it doesn't actually Exist. Nothing is a triality. Everything is a triality. Existence is a triality. Blackness, whiteness, and grayness; 0, 9, and [0and9].

Mathematically Being:

There is no want and there is no need in Beingness, Being is to Be.

You and I chose to Be US, and We-ness Is the Beingness. We are the choice. We chose Us. The choice is made, there is no need for choice in Beingness, Being IS the choice.

0 is a circle and a circle in base 10 mathematics is 360 degrees. 3+6+0=9. A circle has 9 degrees.

0 x any number is itself. 9 x any number is itself.

Zero (0) = Everything and Nothing.
Nine (9) = Nothing and Everything.
[9 and 0]=Thing, Nothing, and Everything (Existence, non-Existence, Being, etc.)

NUMBER LINE---> [9and0] 1 2 3 4 5 6 7 8 <--- ThIs Is All that Is/Is not Mathematics

It was only a matter of time- we've attracted another crank.

Correction:

[9and0] is Nothing and Everything, Alpha and Omega.

NUMBER LINE---> [9and0] 1 2 3 4 5 6 7 8 [9and0] <--- ThIs Is All that Is/Is not Mathematics.

The [9and0] on the right is used for subtraction [i](movements to the left).[/i] Subtraction is carried out through the [9and0] on the right side of the number line with movement to the left to find difference.

The [9and0] on the left is used for [b]addition (movements to the right).[/b] Addition is carried out through the [9and0] on the left side of the number line with movement to the right to find the sum

Correction #2:

"[9 and 0] +,- any number equates to that number itself." <--- error

[9and0] [i][b]+[/b][/i] [u]any number equates to that number itself[/u] <--- correction

Did you try to embed $\alpha$ as the $-1$-th ordinal number, using the axiom-of-no-choice¿ If not, you should really try it, it gives a canonical $\hat{\mathbb{Z}}^{\mathbb{R}}$-module-structure, containing the quaternion extension $\mathbb{F}_{1}[i,j,k]$.
Also, the principal structure of the transfinite subversion of your intraposition is not really well defined: the [0and9] is not the paracongruence of [0or9], but it should be; additionally, you need the xanthophobical inverse of [0xor9] to inverte the ultra-inverse henselian domains! You can try to fix this using the fundamental ring of the quotient of the line by all interior paradox numbers, but all chain complexes of these rings will split into invariant exact sequences of homological ideals, clearly not giving a dedekind ring, nor a good reciprocity mapping; only the structure as Hopf-algebla is preserved as subcomplex of transcedental $K_{1}$-groups, shown by the non-vanishing of $\vartheta_\pi (x,y)=\sum_{\mathfrak{n}\in I(\mathbb{Q}[\Psi])}\frac{e^{x}}{\mathrm{N}_{L|K}(\mathfrak{n}*x^{2})+N(y^{3})}$ at $x=\gamma, y=\zeta(3)$ (here $\Psi$ is the set of all multi-algebraic numbers lying over the module mentioned at the beginning, and $L,K$ are as always the $\infty$- and the $0$-dimensional groupfields over the entire functions respectively).

One problem remains: [0and9] is inconsistent with the axiom-of-no-[0and9] if the general Riemann conjecture is false for quantum groups, in fact you could only say [1and9] then, to be more explicite: $ZFC, DNA, \lnot [1and9] \ \mathbb{1}\ \lnot A \vee \lnot \lnot A [\eta]$ for all $\eta$.

A "xanthophobical inverse" can only arise if you deny the legitimacy of the classic question, "What's green and equivalent to the axiom of choice?"

[hide="!"]ZetaX, you're a dangerous man.[/hide]

Negatives are movements on the number line which is allways positve, as there is no negative existence.

There Exists 3 realities. The reality of illusion, the reality of reality, and the reality of both together as the 3rd (ultimate consciousness).

The realities of illusion are concepts that We have produced so that We may come to know the reality of reality and then ultimately both together (the unified field)

The number 1 Exists as a reality within illusion and as an illusion within reality. To perceive Existence in its true geometrical form (a subset of the all-inclusive Outfinite formlessness), 1 is never found.

One dot is not actually one dot; one dot is measured by a line (diameter), with each end of the diameter consisting of two more points or dots, and of which those dots can also contain a diameter, with each end consisting of two more points or dots. Even the lines made to connect the dots can be measured. No-where and now-here does 1 exist, in Both illusion and reality, mending Both and allowing us to see the Perfect Existence; the acceptance of Everything, because WE ARe  Everything through accepting Us, even that of illusion, because We IS [b]Everything[/b]

We accepts rejection so that there Is 'No' rejection, not even of rejection (illusion) or of 'No'.

The "first" dimension that is spoken of is really "two", which is really "three". Dimensions are illusional, yet accepted because it is the thought that WE ARe Us, and so WE ARe.

Harmony and Blessings

[quote="blahblahblah"]This is the best post I have ever read on AoPS.[/quote]
I second this motion. There should be an AoPS hall of fame, and ZetaX is our first nomination.

"[b]Reality is merely an illusion[/b], [i]albeit a very persistent [u]ONE[/u][/i]." Albert Einstein.

In my opinion $\frac{0}{0}$ is possible to solve only, when we are able to disappear things,like magicians.
because i think that everything around us and even ourselves, we are all divided by one infinitly. $\frac{10}{5}=2$ this means that something which is $10$ times bigger than $1$,  is divided into $5$ equal parts, and abviously the answer is $2$. and the same for other numbers. but if we consider divide by $0$.
say $\frac{x}{0}$ this means that something is divided in to $0$ equal parts, which in other words means make that something nothing, or vanish, disappear. i think it has unique solution. and this is why i love Mathematics, because everything in maths tends to infinity, from numbers to formulas.

I hate these conversations. Something that non-math people always talk about for some reason.

The question that we need to ask is what is division?

$a/b = a b^{-1}$
Where $b^{-1}$ is defined to be the number (unique) such that:
$bb^{-1}= b^{-1}b=1$.
And where $1$ is defined to be the number (unique) such that:
$1x=x1=x$ for all $x\in R$. ($R$ here stands for "ring" it is a structure where addition and multiplciation is permitted).

Now, we can show that $0x=0$ for all $x \in R$.
Where $0$ is defined to be the number (qunie) such that:
$0+x = x+0 =0$
Now we can write $0x$ as $(1-1)x= x-x = 0$ (because $R$ has the distribution properties).
That means,
$0x=x0= 0$.

Therefore, $a/0 = a0^{-1}$ is not permitted because that would imply that $00^{-1}= 0^{-1}0=0=1$. Which is impossible. Unless $0=1$ which is called the "zero ring". In fact, the only place division by zero is permitted. But it is so trivial and uninteresting there is no point to it.

Hate is a strong word. And "non-math people" assumptions reveal the veracity of your character and your mathematical understandings.

From the concept of 0 we created all numbers. We are the consciousness of nothing and are all conduits for it.

From the nothingness of existence, that which Existence is, when we choose to deny it's immeasurability, immovability, unstoppability, and ever changing state of static dynamism... we create "things". We create individuals, i.e. 1,2,3,4,5,6,7,8 etc.

This is an optical illusion and purely mental delusion. There are no separate things. There are identities to things that differ in nature becuase of their perceived differing natures, but you must understand that there is the consciousness of Nothing. Through this nothing we know that all things are really nothing... or as you have said in the past "one thing", though the "uni-verse" is really immeasurable, unstoppable, sempiternal, immovable, and statically changing thus making it a nothing.

Nothing is a dichotomous illusion; existing as an expression so that we may know never ending and interconnected existence, and not existing so that we may know its expression.

Mathematically this can be presented very simply if we can comprehend the defining factors of dichotomous and binomial no-thing.

0-0=0. Because we have nothing to begin with and we aren't taking any thing away. i.e. we are not removing nothing's nothingness so it remains constant in its attributes.

0-0=x. Where x is any given variable excluding the symbol 0 by its self as an independent integer. Because we have taken away the attributes of "nothing" from nothing, thus we have caused it to be seen as some thing.

Whenever we choose to identify "things" through optical perception and seemingly "individualized" energetical properties, we are grabbing them from the nothing. Nothing is everything and everything is nothing.

All things existing are interconnected through nothing, thus making them nothing. Ultimately there is no space of nothingness existing as some thing to ever separate any thing, therefore all things are nothing through this non-local, vectorless, non-scalar and eternally ubiquitous energy field/consciousness upgrade.

Did I miss any thing?

After reading two pages of posts concerning this very interesting question I decided to close the topic :D.


We can use the substitution:
let$ 3a\minus{}4b\plus{}5c\equal{}x$,$ 3b\minus{}4c\plus{}5a\equal{}y$,$ 3c\minus{}4a\plus{}5b\equal{}z$
And change every numerator to the form:$ f(x,y,z)$
and then, it should be obvious by AM ineq

[quote="litbon"]let$ a$,$ b$.$ c$>o.prove that:
$ \frac {a \plus{} 2b \minus{} 3c}{(3a \minus{} 4b \plus{} 5c)^2} \plus{} \frac {b \plus{} 2c \minus{} 3a}{(3b \minus{} 4c \plus{} 5a)^2} \plus{} \frac {c \plus{} 2a \minus{} 3b}{(3c \minus{} 4a \plus{} 5b)^2}\geq0$[/quote]
It's wrong. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=187377


Congratulations to Massachusetts for getting 7th in the entire nation!

And also congratulations to Alan Zhou for getting 5th written, and in the semifinals!

(Of course, I expected Massachusetts to get in the top 3 given that neither I, Xiaoyu, Jason, nor Carl are on the team anymore :P)

[quote="Yongyi781"]Of course, I expected Massachusetts to get in the top 3 given that neither I, Xiaoyu, Jason, [b]nor[/b] Carl are on the team anymore :P)[/quote]

What is this supposed to mean? Also, grammatical error fixed.

And congratulations indeed to Alan, who probably also got the highest score out of all of us on the USAMO at Lexington.  :|

[quote="CatalystOfNostalgia"][quote="Yongyi781"]Of course, I expected Massachusetts to get in the top 3 given that neither I, Xiaoyu, Jason, [b]nor[/b] Carl are on the team anymore :P)[/quote]

...Also, grammatical error fixed.
[/quote]

What are you talking about? :P

Also, how did Alan Zhou do on the USAMO?


[quote="silouan"]
@Durandal : "how many m-dimensional cubes can fit in an n-dimensional cube?"
px  xwrane tesseris 1-dimensional cube sto 2-d cube
kαι  6  2-dim sto 3dim[/quote]

Ok, καλά φανταζόμουν ότι ήθελες τα m-faces ενός n-cube. Very very nice :)
[hide]
Υπάρχουν πολλοί όμορφοι τρόποι να τους μετρήσεις. Αν δεν κάνω κάπου λάθος, βγάζω $ 2^{n \minus{} m} \binom{n}{m}$
[/hide]

Sil, με πόρωσες... άντε να κάτσω να δουλέψω τώρα :)

Cheerio,

Durandal 1707

Edit:
Nα πάμε αν είναι το πρόβλημα με τους κύβους που είναι ανεξάρτητο σε διαφορετικό topic?

Edit 2:

Γενικά, ο n-dim κύβος είναι το σύνολο $ [0,1]^n$.

Cheerio,

Durandal 1707

[quote="Durandal"]
Ok, καλά φανταζόμουν ότι ήθελες τα m-faces ενός n-cube. Very very nice :)
[hide]
Υπάρχουν πολλοί όμορφοι τρόποι να τους μετρήσεις. Αν δεν κάνω κάπου λάθος, βγάζω $ 2^{n \minus{} m} \binom{n}{m}$
[/hide]
[/quote]

Excellent !  :)  αυτή είναι η απάντηση .

Να και ένα link για τους υπερκύβους που έχει και τη λύση στο πρόβλημα που τέθηκε

http://en.wikipedia.org/wiki/Hypercube

Poli oraia paidia :-)

Ti lete gia kati paromoio, alla efkolo:

O d-dimensional hypercube megethous n einai to sinolo $ [n]^d\equal{}\{1,2, ..., n\}^d$. Mia euthia (combinatorial line) se auton ton hypercube einai ena sinolo tis morfis $ \{$ $ x\equal{}(x_1,x_2,...,x_d)\in [n]^d$ | $ x_i\equal{}x_j$ $ \forall i,j \in I$ kai $ x_i\equal{}a_i$ $ \forall i \in \{1,2,...,d\}\minus{}I$ $ \}$, opou I einai mi keno iposinolo tou {1,2,...,d} kai $ a_i\in [n]$ gia $ i\in[d]\minus{}I$.
O orismos autos einai apolytos fysikos, koitaksete prota tin periptosi tou tetragonou kai tou kibou gia na ksekatharistoun opoiadipote amfibolies.

Erotisi : Poses combinatorial lines exei o $ [n]^d$  ?

Mia alli erotisi me kibous sta plaisia tis theorias Ramsey:

Na apodeiksete oti gia kathe phisiko arithmo n kai gia kathe phisiko arithmo k, iparxei phisikos d etsi oste gia kathe xromatismo ton simeion tou $ [n]^d$ me k xromata iparxei mia monochomatiki euthia (combinatorial line).

Auto einai poli diskolo problima, basika einai ena poli gnosto theorima ton Hales-Jewett alla aksizei ton kopo na to skefteite apo monoi sas kai tha sas dosei megali xara ean to kataferete. (Apo tin alli, den tha sas dosei kamoia xara, alla mallon baremara, ean psaksete etoimi tin apodeiksi sto google.)

Stergios

[quote="silouan"]How many m-dimensional cubes can fit in an n-dimensional cube?
px  xwrane tesseris 1-dimensional cube sto 2-d cube
kαι  6  2-dim sto 3dim

Μονοδιάστατο cube είναι μια γραμμή . Δυσδιάστατο cube είναι το τετράγωνο .[/quote]

Προς απάντηση σε άλλο σου ποστ για το αν θυμόμαστε το πρόβλημα, φυσικά και το θυμόμαστε :wink: Καλά που ήταν ο Τάσος κι ο Ηλίας εκεί και μας έβγαλαν ασπροπρόσωπους :P

Anyway παιδιά, καλή προετοιμασία για την IMO :D Six medals are waiting for you :wink:


Let $A$ be a finite set of positive integers.  Prove that there exists a finite set $B$ of positive integers such that $A \subseteq B$ and

\[\prod_{x\in B} x = \sum_{x\in B} x^2.\]

If you found it on the forum why you did't insert the link here?

Anyway, we include in $A$ some number of 2 to obtain $\prod_{x\in A} x > \sum_{x\in A} x^2$.
Let $A=\{x_1,x_2,...,x_k\}$.

Consider the sequence $x_{n+1}=x_1x_2...x_n-1$, $n\geq k$. Denote $A_n=\{x_1,x_2,...,x_n\}$ ($n\geq k$). Then $A\subset A_n$.

Let's calculate $s_{n+1}=\prod_{x\in A_{n+1}} x - \sum_{x\in A_{n+1}} x^2$.
Indeed, using recurrent expression for $x_{n+1}$ we immediately obtain
\[s_{n+1}=\prod_{x\in A_{n}} x - \sum_{x\in A_{n}} x^2-1=s_n-1.\]
Taking into consideration $s_k>0$ we conclude there is an index $K$ s.t. $s_K=0$.
That's all.

[quote="Myth"]If you found it on the forum why you did't insert the link here?
[/quote]

Heh, it doesn't really help since there's no solution, but anyways it's here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=magnificent+NT+problem&t=8427

Just one question: what made you think of using $x_{n+1}=x_1x_2\cdots x_n -1$?  (I liked your solution btw  :) )

[quote="Myth"]That's all[/quote]
Well done!!! :)

__________
blang

[quote="ThAzN1"]Just one question: what made you think of using $x_{n+1}=x_1x_2\cdots x_n -1$?  (I liked your solution btw  :) )[/quote]
It was an inner voice. :D

Actually, I think, it was inspired by an old russian problem. :?

[quote="test20"]Myth's argument has a small gap:  A and B should be sets, whereas he is using several copies of the number 2.[/quote]
It doesn't matter. I used it to obtain $\prod_{x\in A} x > \sum_{x\in A} x^2$. Indeed, we can do it in many ways.

Yeah, if we take $m=\text{max}\{x_k\}$ then we can add $m+1, m+2, \cdots$, and I think it'll work.

Task appeared also on the IMO training camp, Poland 1999.

I think this is a much more motivated solution

By adding $M,M+1,M+2$ to $A$ for some massive $M$, we can WLOG assume that $\prod_{x \in A} x>\sum_{x \in A} x^2+1000$. Let $p=\prod_{x \in A} x$ and $q=\sum_{x \in A} x^2$; note that $p>1000$. Now let $n=p-q$; we will find $n$ distinct massive positive integers $x_1,\ldots,x_n$ (so that $x_i \not \in A$) such that
$$x_1^2+\cdots+x_n^2-px_1\ldots x_n+q=0,$$
whence $B=A \cup \{x_1,\ldots,x_n\}$ will work. This is by Vieta jumping, with the initial solution being $(1,\ldots,1)$. If at some point we have a solution where (WLOG) $x_1 \leq \cdots \leq x_n$. Fixing $x_2,\ldots,x_n$, we get a quadratic for $x_1$ with two integer roots; one of these is the current value of $x_1$, and the other is $px_2\ldots x_n-x_1>x_n$. If we repeat this process many times, each variable grows arbitrarily large, and furthermore, from the $n$th iteration onwards, all the $x_i$ will be distinct, so we are done. $\blacksquare$


geometry quadratics function calculus integration derivative algebra

I made a problem that I can't solve, and I would really appreciate some help - to see how far from my scope this is:

Find the number of latice points of the area enclosed by a line y=a and a monic quadratic function.
Where the solution is in terms of a and the coefficient of the monic quadratic function.

Thanks!!

WLOG the line is the $ x$-axis.  Rather than write out the formula, which is somewhat obfuscating, note that it suffices to look at the set of positive values of $ ax^2 \plus{} bx \plus{} c$ when $ x$ is an integer.  Of course, what I have in mind may not be very satisfying if you want to generalize it to, say, circles.

For two-dimensional lattice counting problems (in finite regions), I have always wanted to use this idea because it is so general.

First, treat lattice points ON the curves separately. Then construct an analytical function that has a simple zero at every lattice point on the complex plane that is strictly inside that region. Then take the contour integral of the logarithmic derivative of that function along the boundary of your region and divide the result by $ 2\pi i$. The answer is the number of lattice points contained in the region.

My difficulty has always been finding nice functions that have the lattice zero property. Polynomials are kind of messy.


Show that for every nonzero number a and integer n>1, x(x-1) is a factor of a*(x^n) - (a^n)*x.

I do not understand how to do this problem.

Are you sure that's the problem? Take $ a \equal{} n \equal{} 2$, so we have
\[ 2x^2 \minus{} 4x \equal{} 2x(x \minus{} 4),
\]
which does not have a factor $ (x \minus{} 1)$ in it.

Even if "factor" is interpreted in the number-theoretic sense (for $ a$ an integer) and not in the polynomial sense, $ ax^n \minus{} a^n x \equiv a \minus{} a^n \bmod (x \minus{} 1)$, so this divisibility clearly does not hold in general.

$ ax^n\minus{}a^nx\equal{}ax(x^{n\minus{}1}\minus{}a^{n\minus{}1})\equal{}ax(x\minus{}a)(x^{n\minus{}2}\plus{}ax^{n\minus{}3}\plus{}\ldots\plus{}a^{n\minus{}3}x\plus{}a^{n\minus{}2})$

That's what I see.

If x(x-1) was a factor of that thing, then 0 and 1 would be roots. 0 is a root, but 1 is not for an infinite number of values of a.

Are you sure that this is the right problem?

[quote="chenhsi"]huh?

how did you get that?[/quote]

synthetic division

oh sorry, i misread and mistyped the problem

the actual problem is:

Show that for every nonzero number a and integer n>1, x(x-[b]a[/b]) is a factor of a*(x^n) - (a^n)*x.


Let $x;y;z>0$ satisfying $x+y+z=6$.Find min:
$(\sqrt{x}+\sqrt{y}+\sqrt{z})(\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{z}}+\frac{z}{\sqrt{x}})$

[hide]we have $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{z}}+\frac{z}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}+\sqrt{z}$ by either rearrangement, cauchy..
and we have $\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}\geq \sqrt{\frac{x+y+z}{3}}=\sqrt{2}\Longrightarrow \sqrt{x}+\sqrt{y}+\sqrt{z}\geq 3\sqrt{2}$ by jensen, so
$(\sqrt{x}+\sqrt{y}+\sqrt{z})\left(\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{z}}+\frac{z}{\sqrt{x}}\right)\geq 18$, and equality case when $x=y=z=2$
[/hide]

we'd better denote $\sqrt{x}=a ; \sqrt{y}=b ; \sqrt{z}=c$
now expending give us
$A=\sum{a^{3}}{b}+\sum_{cyc}\frac{ab^{2}}{c}+\sum a^{2}$
notice that
$\sum(\frac{a^{3}}{b}+\frac{ab^{2}}{c}+\frac{a^{2}c}{b}) \geq 3 \sum a^{2}$
$\sum(\frac{a^{3}}{b}+ab) \geq 2\sum a^{2}$
$\sum a^{2}\geq \sum ab$
now adding them up , we will find the desired results

[quote="nayel"][quote="minsoens"]...and we have $\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}\geq \sqrt{\frac{x+y+z}{3}}$...[/quote]

you've written it in the wrong direction. :wink:[/quote] :o hmm.. not too easy!  :D


function
inequalities unsolved
inequalities

simple.
this is called the kyfan inequality.
for x in (0,1/2] the function f(x)=ln(1/x -1) is convex.
the result follows by using jensen's inequality.
-a


calculus
derivative
function
geometry
geometric transformation
reflection
calculus computations

I'm a grade 12 calculus student. I get how to convert graphs from functions to derivatives. But, when it comes to going from derivatives to functions...what do you do? wht do u look for?

thnks for any help.

Converting the graph of a derivative into the graph of a function is just the inverse operation to converting the graph of a function into the graph of its derivative.
If you want just a rough sketch, the main properties to try to reflect on the graph of the function are "increasing-decreasing" and "convex-concave". The function is increasing on the intervals where the derivative is positive and decreasing where it is negative. Also the function is convex (sometimes they use the term "concave up" in calculus textbooks) on the intervals where the derivative is increasing and concave ("concave down") where the derivative is decreasing. If you want a more accurate graph, you need to look at the areas between the graph of the derivative and the $x$ axis. For instance, a big positive hump of the derivative followed by a small negative hump corresponds to a big long accent of the function followed by a small short descent.
Also, don't forget that there are many functions with given derivative, more precisely, you can shift the graph of the function as far up or down as you wish and it will still be a correct answer. I would like to conclude with a few pictures, but, to my shame, still do not know how to post them. Maybe someone else will do it. Hope this helps.

thnks a bunch fedja...im getting the gist of it now.

it'd b helpful if any1 culd put up graphs (just sketches..no need to b accurate).

Thnkx!

OK, here is a picture. You can easily see the things I talked about on it (at least, I tried to make them seen easily). For instance, on the interval $[c,d]$ the derivative $F'$ is negative and increasing. Hence, the function $F$ is decreasing and convex on $[c,d]$ and so on. Also, since the positive hump of $F'$ between $d$ and $f$ is clearly of bigger area than the negative hump between $f$ and $h$, we have $F(h)>F(d)$ (I used the small letter $f$ for both the function and for one of the points on the picture but I hope that will lead to no confusion. I noticed it too late and was too lazy to correct this minor lapse :) ) You are welcome to explore the picture and to ask any questions. It still doesn't appear right in the post (some lines misteriously disappear) but, if you click on it, comes out OK in a separate window....


group theory
abstract algebra
superior algebra
superior algebra unsolved

How do we prove that S_6(the group of permutations of order 6) has only one subgroup with 3 elements(of order 3)?

Your notation is off; $ S_6$ is the group of permutations of 6 elements. It has $ 720$ elements, including $ 80$ of order $ 3$. ($ 40$ subgroups of order $ 3$)

I think you actually mean $ S_3$- and that's easy to handle by simply counting elements.

in the book it is S6...
i didn't work with groups for a long time...can you please tell me why there are 40 subgroups of order 3?

To count elements of order $ k$ in a permutation group, look at the cycle decomposition- each cycle length divides $ k$, and the largest cycle has size $ k$.


Hello
I would like to ask you a question regarding the t student distribution. It is related to a paragraph that I read on a book. It says:

"The degrees of freedom for tests based on sample means are n-1 because, given the mean, only n-1 observations can be unique."

I know that n is the size of the sample, but,
What does the previous sentence mean? What is the logic behind this sentence?
Thanks in advance

[url=http://en.wikipedia.org/wiki/Student%27s_t_distribution]This[/url] page has a lot of information on the $ t$.  I haven't read through it but even you don't get the exact answer you're looking for, it will be enlightening anyway.


Hello

A derangement is defined as the number of ways to place elements of an ordered set so that none of the elements are in the place they originally were in.  For instance {2,1,4,3} is a derangement of {1,2,3,4}.

Find the number of derangements of ALGEBRA where the A's are indistinguishable.

Show all work.

[hide]
First, we know that there are $ 5$ spots for the $ 2$ A's, so there are $ \dbinom52$ ways to do this.
WLOG, assume that L and G are the places where the A's are. After this, divide into cases:
i. L and G take the places of the A's--$ 2$ ways.
This means that we must find the number of derangements of EBR, which is easily seen to be $ 2$. Thus, this case has $ 4$ possibilities.
ii. One of L,G takes the place of one of the A's, and the other one of L,G takes the place of E,B, or R. There are $ 12$ ways for this.
WLOG, assume that E was the one of E,B,R chosen. Then we must find the number of ways for E,B,R to fit in B,R,A while still being a "derangement". It is easy to see that there are $ 3$ ways for this, and $ 36$ for this case.
iii. L,G are both in EBR. There are $ 6$ ways for this to happen.
WLOG, assume that E and B were taken. Then we have the spaces A,R,A to place E,B,R. The only restriction is that R can't be at R (but then there are also two possible arrangements), so there are $ 4$ ways for this, and $ 24$ ways for this case.
Thus, the answer is $ \dbinom52(4 \plus{} 36 \plus{} 24) \equal{} \boxed{640}$.
[/hide]

There are in fact 640 derangements.

There are $ !7 \equal{} 1854$ ways to derange them if they were all distinct.

But they are not, so we have to subtract the arrangements where at least one of the A's is in the other A's spot. Without loss of generality, let the first A be in the second A's spot (We'll multiply this by 2 at the end).

We'll first count the cases where the A's were simply interchanged; this is just $ !5 \equal{} 44$.

Now we have that (A1) is in the (A2) spot, so we are deranging
(A2)LGEBR. We already accounted for the case where (A2) is in (A1) in the previous case (since it would count as them being interchanged), so we do not want (A2) to stay where it is. But then this is the same as deranging 6 distinct elements, which is just $ !6 \equal{} 265$. Multiply by 2 to get $ 530$.

Now, subtract: $ 1854 \minus{} 44 \minus{} 530 \equal{} 1280$, but divide by 2 because, again, they are indistinct, to get $ \boxed {640}$.

Yongyi's answer is correct.  Nice solution.

I accidentally typoed one of my cases and it resulted in the wrong answer. :wink:

But still, Yongyi's solution was definitely more elegant.


pigeonhole principle
Putnam
combinatorics unsolved
combinatorics

Let $A = \{ 1,2, 3 , \ldots, 100 \}$ and $B$ be a subset of $A$ having $53$ elements. Show that $B$ has 2 distinct elements $x$ and $y$ whose sum is divisible by $11$.

We can devide the numbers into $11$ congruence classes.Ignore class of $0$ for a moment ,and classify
the remaining $10$,into five complementary groups.You can at most fill these groups with fifty numbers-without any two of members satisfying the condition.The 51st element will do the trick.Now consider the zero class it will delay
our method one step.

Did I miss something or only $48$ elements suffice? ;)

Divide the numbers into
- $45$ pairs :
$(1,10),(2,9), \cdots, (5,6)$, $(12,21),(13,20), \cdots, (23,32)$, $(24,31), \cdots ,(34,43)$, $(35,42) \cdots, (45,54)$, $(46,53), \cdots, (56,65)$, $(57,64), \cdots, (67,76)$, $(68,75), \cdots, (78,87)$, $(79,86) \cdots, (89,98)$, $(90,97), \cdots (93,94)$
- $9$ single numbers $11,22,33,44,55,66,77,88,99$
- and $100$.

If you choose some numbers such that you don't have two distinct of them whoch add up to a multiple of $1$, then you can choose at most $45$ numbers among those forming the pairs, at most 1$1$ among the multiple of $1$, and $100$. Thus, you can choose at most $45+1+1=47$ numbers, hence result.

Pierre.

[quote="pbornsztein"]Did I miss something or only $48$ elements suffice? ;) [/quote]
You are totally right.I knew that $48$elements suffice ,but this is something common in problems(eg see Putnam)

Group the numbers as (1,98),(2,97)... (49,50) and lastly (99,100) all of whose sum are 99 except the last one and by PHP there will be two numbers from one of the first 49 groups. So done

Try choosing one element and eliminating the other one. Ex: If you choose 1 in then you need to eliminate 10,21,...,98 for obvious reasons, also $1 \equiv 1 (mod 11)$ so we can't use a number $ \equiv 10 (mod 11)$. By this argument we choose the following series of elements in our set $B$.
\[1,12,23,...,100 \; \;  [\equiv 1 \; (mod 11)]\]
\[2,13,24,...,90 \;  \;  [\equiv 2 \; (mod 11)]\]
\[3,14,25,...,91 \;  \;  [\equiv 3 \; (mod 11)]\]
\[4,15,26,...,92 \;  \;  [\equiv 4 \; (mod 11)]\]
\[5,16,27,...,93 \;  \;  [\equiv 5 \; (mod 11)]\]
\[11 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;[\equiv 0 \; (mod 11)]\]
Now as soon as I choose 6 the solution will become invalid because 5+6 will sum upto which is divisible by 11. These all five series and one single candidate above contain elements such that no two of them sum upto a multiple of 11. Now $(10+9+9+9+9+1)$ there are 47 elements. As soon as you choose the $48$th element, two numbers' sum will be congruent to 0 mod 11 which will suffice the problem's condition. There 48 elements are enough to suffice the problem's condition.

[hide=Solution via Contradiction (Similar to PHP logic)]For the sake of contradiction, let us assume that we can never find such number $x, y$ in all subsets $B$ with $|B|=53$ such that $11|(x+y)$.
Let us partition $A$ into $11$ subsets $S_0, S_1, …, S_10$ such that $S_r$ contains all elements of $A$ that are of type $11q+r$. All $S_r$ will have $9$ elements, but only $S_1$ will have $10$ elements. $B$ can have maximum $1$ element of $S_0$. If a element from subset $S_r$ then no element of $S_{11-r}$ can come in $B$, given ($r>0)$. For boundary case, let's take $1$ element from $S_0$, $10$ elements from $S_1$, and $9$ elements from $S_r (1<r<6)$. So maximum value of $n(B)$ can be $47$, but we are given with $53$, which is a contradictory condition. So, we can always find $x, y$ from $B$ such that $11|(x+y)$.
Hence, proved.[/hide]


I remember seeing a post by tokenadult about a list of problem solving books that might help with the AMC tests.  I remember seeing one series of books that were supposed to teach you high-school math topics (Algebra, Geometry) from a different perspective.

Can anyone tell me what the books are, or give me a link to the post?  I've tried searching for it, and looking at the posts, but I can't find it.

[quote="h_s_potter2002"]I remember seeing a post by tokenadult about a list of problem solving books that might help with the AMC tests.  I remember seeing one series of books that were supposed to teach you high-school math topics (Algebra, Geometry) from a different perspective.[/quote]

The original thread that mentioned that book series was the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8674]readable books thread[/url] opened by Mathnovice. The book series is the Gelfand Correspondence Series textbooks and some other related books by Russian authors.

Hope this helps!

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12172&start=0
Do you mean that?
If not, you may want to use the "search" option. That can come in very useful in finding an old topic.

The one tokenadult gave me was right.

I actually found the link before while searching for it, but was to quick to say it wasn't the post I was looking for.


algebra
polynomial
geometry
rectangle
floor function
symmetry
combinatorics unsolved

In how many way one can put three knights on the $ n\times n$ chessboard such that no two of them attacks to themselves?

The general case seems very hard. I was wondering if it is easier to prove the case where $ n$ is even, because there are some cases I need where I haven't found a solution and the side length is even.

The two knights problem is answered by [url=http://www.research.att.com/~njas/sequences/A035289]A035289[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

There are two obvious solution methods there.  The first is a case-bash: divide the board into sets by how many squares are attacked from each position, then do casework.  This might be modifiable to the case of three knights, but it will involve a lot more cases and you'll essentially have to do "small cases" (certainly $ n \leq 5$, maybe up to $ n \equal{} 12$ or so?) by some form of brute force, I think.  (Well, the same sort of analysis will apply, but you have to take account of "cramping" caused by smaller boards.)

The second method is a cheat: the answer is essentially a six-degree polynomial in $ n$ with leading coefficient $ \frac {1}{6}$ and next coefficient 0 (it's easy to see that $ \frac {n^2(n^2 \minus{} 9)(n^2 \minus{} 18)}{6}$ is a lower bound and $ \binom{n^2}{3}$ is an upper bound, and not terribly hard to slightly improve either estimate), so you can try "guessing" that it's actually a sixth degree polynomial and then using the method of finite differences to actually find the polynomial in question.  In fact, it is extremely likely that it really is a polynomial, or at worst a quasi-polynomial or a polynomial except for a few initial values, so this method will almost certainly provide the answer (though not a proof that the answer is really correct, unless you have some other argument for polynomiality).

Your post makes it sound as if you've made some progress -- would you mind describing what you've done?

A natural generalization of this problem is to boards of size $ m \times n$.  I would again expect a polynomial of a single variable when we hold $ m$ fixed, and maybe a bivariate polynomial if both $ m$ and $ n$ vary, at least when both are large.

It might also be interesting to solve this question on, say, a torus board.  (It's always nice to get rid of edge effects, though small torus boards behave somewhat oddly.)

I'm not very good at combinatorics. I found the following hint.

Obviously you can put them all on the same colored square. There are $ 2n^2$ white squares, so $ \binom{2n^2}{3}$ ways of placing them all on white squares. The same for black squares, so in total $ 2\binom{2n^2}{3}$ of that form.
When there are $ 2$ black squared knights and one white squared knight, it depends on where the knights are placed. Lets look at scenarios with different combinations of the two black squared knights being placed in the following areas of the board:
a) corner
b) adjacent to the corner
c) diagonally touching the corner
d) on the edge but at least $ 3$ squares in from the adjacent side
e) second row/column from the edge, at least $ 3$ squares in from the adjacent side
f) at least squares in from any edge

Now one has to compute $ 36$ special cases. For example one can put two knights in black corners two ways, as there are two black corners. So the "combination a)-a)" has two ways to put the knights. Similarly, there are $ 2\cdot 4 \equal{} 8$ ways for combination a)-b).

For $ 2n^2$, I assume you're using a $ 2n \times 2n$ board?  That's exactly the sort of case-bash I was imagining.  Here's some work on the other approach:

I just did a bit of programming to compute the value of $ f(n)$ for $ n \leq 13$.  Here's the difference table for $ \binom{n^2}{3} - f(n)$ (just lowering the degree):

$ \begin{array}{ccccc} f(n) & \Delta f(n) & \Delta^2 f(n) & \Delta^3 f(n) & \Delta^4 f(n) \\
\hline 0 & & & & \\
& 0 & & & \\
0 & & 48 & & \\
& 48 & & 140 & \\
48 & & 188 & & 92 \\
& 236 & & 232 & \\
284 & & 420 & & 140 \\
& 656 & & 372 & \\
940 & & 792 & & 84 \\
& 1448 & & 456 & \\
2388 & & 1248 & & 96 \\
& 2696 & & 552 & \\
5084 & & 1800 & & 96 \\
& 4496 & & 648 & \\
9580 & & 2448 & & 96 \\
& 6944 & & 744 & \\
16524 & & 3192 & & 96 \\
& 10136 & & 840 & \\
26660 & & 4032 & & 96 \\
& 14168 & & 936 & \\
40828 & & 4968 & & 96 \\
& 19136 & & 1032 & \\
59964 & & 6000 & & \\
& 25136 & & & \\
85100 & & & & \end{array}$

That string of 96s is what we were looking for, so a fourth-degree polynomial it is (after the first few garbage values).  Specifically, this says that
\begin{align*} \binom{(n + 4)^2}{3} - f(n + 4) & = 284\binom{n}{0} + 656\binom{n}{1} + 792 \binom{n}{2} + 456 \binom{n}{3} + 96 \binom{n}{4} \\
& = 284 + 388n + 212n^2 + 52n^3 + 4n^4 \end{align*}
for $ n \geq 0$ and so
\[ f(n) = \frac {1}{6} \left(n^6 - 27n^4 + 72n^3 + 170n^2 - 984n + 1080\right)\]
for $ n \geq 4$.  If all you're interested in is "knowing the answer," that's that -- if you want a proof, for the moment I don't see anything better than the cases (but at least you now have an answer that you're trying to get to).

Hi JBL, I was just wondering how you knew that the coefficient of the $ x^5$ term would be 0. Thanks =)

It's from the parenthetical that follows that comment: since $ \binom{n^2}{3} \equal{} \frac {n^6}{6} \plus{} O(n^4)$ and $ \frac {n^2(n^2 \minus{} 9)(n^2 \minus{} 18)}{6} \equal{} \frac {n^6}{6} \plus{} O(n^4)$ are upper and lower bounds, respectively, asymptotically our answer must be $ \frac {n^6}{6} \plus{} O(n^4)$.  This means that if it's actually a polynomial, it must have leading coefficient $ \frac {1}{6}$ and next coefficient 0.

I've gone home now, but when I go back to school I'll test my conjecture that the answer for an $ m \times n$ rectangle is a polynomial in $ m$ and $ n$ (probably with some restriction like $ m, n \geq 4$) with leading term $ \frac{m^3 n^3}{6}$.  My guess is that the computational approach will again tell us exactly what the polynomial is, and probably the same case-bash will be able to prove it.

For the curious, the answer for the $ m \times n$ board (again, this is just a computational result) is
\[ f(m, n) \equal{} \frac{m^3n^3 \minus{} 27m^2n^2 \plus{} 36(m^2n \plus{} mn^2) \plus{} 170mn \minus{} 492(m \plus{} n) \plus{} 1080}{6}\]
for $ m, n \geq 4$.

Do you need the assumptions that $ m$ and $ n$ are even?

No, that works for all $ m$ and $ n$.  (And the original formula was for all $ n$, as well.)  When I produced the data, I was open to the possibility of a quasi-polynomial -- that is, something that acts as one polynomial on even values and another polynomial on odd values, like $ \left\lfloor \frac {n^2 \plus{} 1}{2}\right\rfloor$.  It turns out, however, that the parity doesn't matter.  This isn't entirely surprising -- most of the strange behavior comes from edge and corner effects, and for large boards only one corner or edge at a time affects what's going on.

If you want to turn the computational results into rigorous ones, the casework grinding isn't actually that bad when combined with inclusion-exclusion.   A sketch:

Start by labeling the knights red, white, blue (we can later correct for this by dividing by $ 6$).  Number each square on the board with the number of squares a knight on it attacks (for $ m,n \geq 4$, the center of the board is all $ 8$'s and each corner looks like
[code]2 3 4 4 4 4
3 4 6 6 6 6
4 6 8 8 8 8
4 6 8 8 8 8
4 6 8 8 8 8
4 6 8 8 8 8[/code]

By inclusion exclusion and symmetry, the number of ways of placing the knights is $ mn (mn\minus{}1) (mn\minus{}2) \minus{}3S_1\plus{}3S_2\minus{}S_3$, where

-$ S_1$ is the number of ways of placing the three knights so the red and white knight attack each other.  The number of ways of placing the red/white knights is just the sum of the numbers on the board, and we then multiply by $ mn\minus{}2$ to account for the blue.

-$ S_2$ is the number of ways of placing the three knights so red attacks both white and blue.  If we place the red knight on a square labeled $ x$ there are $ x^2\minus{}x$ ways to place the other knights, so we just add up $ x^2\minus{}x$ over the board.

-$ S_3$ is the number of ways of placing the three knights so they all attack each other, which is $ 0$ due to parity considerations (each knight only attacks squares of the opposite color).

If one of the sides is smaller than $ 4$, the only thing which changes is the labeling of the board.


algebra
polynomial
quadratics
system of equations

Find all polynomials $ p(x)$ so that $ p(0) \equal{} 0$ and $ p(x^2 \plus{} 1) \equal{}
(1/2)p(x)^2\plus{}2$

[hide="Found a solution"]
Letting $ x \equal{} 0$, we find that $ p(1) \equal{} 2$. If $ x \equal{} 1$, then $ p(2) \equal{} 3$. If $ x \equal{} 2$, $ p(5) \equal{} 7/2$. We

Define a system of equations for a quadratic (note c=0):
$ a \plus{} b \equal{} 1 \longrightarrow 2a \plus{} 2b \equal{} 2$
$ 4a \plus{} 2b \equal{} 3 \implies a \equal{} 1/2, b \equal{} 1/2$

Thus, the quadratic $ \frac {x^2}2 \plus{} \frac {x}2$ is one such polynomial.
[/hide]

$ \frac {(x^2 \plus{} 1)^2}{2} \plus{} \frac {x^2 \plus{} 1}{2} \neq \frac {1}{2}\left(\frac {x^2}{2} \plus{} \frac {x}{2}\right)^2 \plus{} 2$ -- for one thing, their leading coefficients disagree.  (It's not hard to see that the leading coefficient of any such polynomial should be 2.)  You appear to have computed $ \frac {3^2}{2} \plus{} 2 \equal{} \frac {7}{2}$, which helps explain this.


[quote="rustam"]Let $f(x) \in C^{(1)}(1,+\infty) \text{ and } \forall x\in (1.+\infty ) \Rightarrow |f(x)| \le \sqrt{x}$
Prove that $\exists x_n\to +\infty \text{ such that } f'(x_n) \to 0$[/quote]

If I doesnt made mistake, by applying Lagrange mean value theorem it is easy to show that for sequence:
a1 < x1 < a2 < x2 < a3 < .....
where an -> inf, proposition is true:
$f'(x_n) \le \frac{1}{a_{n+1}^{1/2} - a_{n}^{1/2}}$

So we can take for examle $a_n = n^4$


geometry
combinatorics unsolved
combinatorics

[u][b]The author of this posting is : billzhao[/b][/u]
____________________________________________________________________

Let A1, A2, A3, A4, A5, be regions on the plane such that
[b](1)[/b] |Ai| >= 1/2 for each i
[b](2)[/b] |Union of all Ai| <= 1
show that there exists distinct i, j, such that |Ai \cap Aj| >= 1/5.
(| | denotes area)

Yes look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=5284&highlight=five+regions

But what is your question Darij?

Pierre.

[quote="darij grinberg"]Sorry! I just didn't note that it was already reposted by the author. I actually try to recover those postings which were really lost...

  Darij[/quote] what do you mean by really lost? lost when?

No mathematics problem has ever been deleted on mathlinks, Darij ;)

(unless you deleted it you evil bastard :D)

Well, they were not deleted... they were lost when the forum moved to another location. In fact, before moving, Valentin wrote a big ANNOUNCEMENT that people should stop posting since all messages posted during 20 - 22 April 2004 would be lost. But nobody took note of this... well, not nobody... I did...  :)

  Darij


Right triangle $ABC$ with right angle $A$ has an area equal to its perimeter.  If the incenter of $ABC$ is $I$ and $IB$ is of length $\sqrt{13}$, find the area.

[hide="Big hint"]
$A = rs$
[/hide]

[hide]Let side AB be c, AC be b, and BC be a. Let the incenter of triangle be i. Therefore, we can write: b*c/2 = a + b + c = (a+b+c)/2*i. The incenter is then 2.

let the intersection of the radius and BA be x, the radius of BC be y, and the radius and AC = z. BX is therefore sqrt(sqrt(13)^2 - 2^2) = sqrt(9) = 3. BY is also 3 because triangle IXB is congruent to triangle IYB by HL congruence. now, we can let CY and CZ have length 'p'. Therefore, we can write: 5(2+x)/2 = (3+2+2+x+x+3). Solving, x = 10. It is a 5,12,13 right triangle so the area is 5*12/2 [b]= 30.[/b][/hide]

yeah thats what i got too. for the first step, since you know A=(P/2)r, and A = P, by substituting P for A, you get r = 2. same thing but i thought it would save some time.


a rhombus has diagonals of lengths 24 and 70. find the length of an altitude of the rhombus.

Is it
[hide] 840/37? I did it by finding the length of one side, and dividing the area of it by that. I may have done something wrong though...[/hide]

yeah it is.  :lol:

edit: i see how you did it, thanks!

Let $s$ be the length of the side of the rhombus, $A$ be the area, $h$ be the altitude.  $4s^{2}= d_{1}^{2}+d_{2}^{2}$, hence $s = 37$.  $A = \frac{d_{1}d_{2}}{2}= 840$.  $h = \frac{A}{s}= \frac{840}{37}$.

this problem is a bit trickier than most

[hide]
12,35,37 is sometimes a forgotten pythagoren triple
since you need to divide the diagonals by 2
in order to find each side of the rhombus
you get 4 congruent right triangles after drawing both diagonals
you remember the 12,35,37 root
and check it to be sure it works
then you
multiply both diagonals and divide by 2 (24*70/2) to find the area
then divide by 37 (the side of the rhombus)
so

24*70/2/37
[/hide]

I didn't have any paper or pencil around
so thats why its not simplified

yes, if you look at my solution that's what i did


calculus
geometry
geometric transformation
reflection
derivative
quadratics

Does anyone have this one problem where it took 1 second if you used calculus but like forever if not.
It was like this group of people start on one side of a wall then touched another wall next to it, then hit the last wall. What's the shortest distance (classic calculus problem) I want to see if I can do it now.

Do you have any idea what level it was at (chapter, state, or national)?

Also, was it a sprint or target?

That sort of problem seems like it could be solved very, very quickly by reflecting stuff over the wall and then finding a straight line.

It was a national problem in the 2000s
either 2000 2001 or 2002

Do you remember anything else about the nature of problem?  It seems to me like this is a basic textbook reflections problem.

how did this get moved to high school basics

[quote="moogra"][quote="Anaxerzia"]It probably took one second using reflections and forever using calculus, or at least in my experience with the kind of problem you've described it was...[/quote]

no i think it was a basic derivative problem where you had a quadratic equation.[/quote]

No; unfortunately, that is not the case. You end up with the sum of two square roots with expressions of degree two under them.


function
algebra
domain
calculus
derivative
calculus computations

Find the maximum of $ f(x)\equal{}\frac{25}{4}\sqrt{\dfrac{(x\minus{}4)^2x^2}{25x^2\minus{}72x\plus{}144}}$ over the domain $ x \in [0,4]$.

The coefficient in front and the square root are irrelevant, the denominator is just $ 9(x\minus{}4)^2\plus{}16x^2$, so the task reduces to the minimization of $ \frac 9{x^2}\plus{}\frac{16}{(4\minus{}x)^2}$. This can be easily done by taking the derivative and finding its (unique and rather ugly) root on $ (0,4)$.


Show that $ x^2\plus{}y$ and $ y^2\plus{}4$ cannot be perfect squares with the same $ (x,y)$

Hmmm..$ y\equal{}0$? :wink:

Maybe you wanted to ask for $ y^2\plus{}x$?

Pierre.

[quote="pelao_malo"]Show that $ x^2 \plus{} y$ and $ y^2 \plus{} 4$ cannot be perfect squares with the same $ (x,y)$[/quote]
$ y^{2}\plus{}4\equal{}k^{2}$ has no solutions in positive integers y,k.


oops.
1 and 2 are just the formula for sum of powers of 2. you know this because each power of 2 is 1 greater than those before it all added together.
3 and 4 you can do by either subtracting 1/n from 1 or the way mentioned in #6.
sorry, but i can't really explain my work for the last two. i'm kinda bad at explaining

5.  [hide]Let S equal the problem we are trying to find the formula for, which is 2^0+2^1+2 :^2: +...+2^n.  Then we multiply both sides of the equation by 2 getting 2S=2^1+2^2+2^3+...+2^(n+1).  Then subtract to S from 2S and you will get S=2^(n+1)-2^0=2^(n+1)-1 which equals 2^0+2^1+2 :^2: +...+2^n[/hide]

6. [hide]We technically do the same thing in #5 for this problem.  We set a variable for the sum, multiply both sides by two and then we subtract the original equation from the new equation.  In this equation, we can turn 1/2^n into 2^-n.   (2^0+2^-1+2^-2+2^-(n-1))-(2^-1+2^-2+2^-3+...+2^-n)=2^0-2^-n=1-1/2^n[/hide]


um... where are the pairings? All we have is seedings.

Please wait for AIME15 to make the first post.  :wink:

Fourth post, actually :)

[b][u]SEEDINGS:[/b][/u]
1. BOGTRO
2. Mewto55555
3. Nt811581147
4. ernie
5. Jongao
6. isabella2296
7. nikeballa96
8. lotsofmath
9. pythag011
10. jjx1
11. jain_harshi
12. Meta_Knight

You can see the bracket [url=http://www.bracketmaker.com/tmenu.cfm?tid=310119&tclass=]here[/url].

In the first round, we will have 8. lotsofmath against 9. pythag011, 7. nikeballa96 against 10. jjx1, 6. isabella2296 against 11. jain_harshi, and 5. Jongao against 12. Meta_Knight.

All matchups will be best of 5; that is, the first player to win 3 CD's.

on your bracket, both ernie and nt are labeled as 4th seed

doesn't affect the bracket too much, but i wanted to bring it up

On the bracket, you already have me beating pythag011, if you click "Game 2 Details", it says "Winner: lotsofmath"... but I haven't even played pythag011 yet.

I (think) it's fixed.

If you are playing, please confirm!

[quote="pythag011"]/confirm.

Blah I'm going to fail.[/quote]

u [i]might[/i] win against lotsofmath (if he's off like in chapter) :wink:

I confirm.

Why don't I have a game? byes for top 4 or something?

Also, I will be away Thursday - Sunday for SMT (stanford math tournament).

Meta Knight and Jongao have 5 days.  Nike defeats jjx by forfeit.  See bracket for detaisl.

great jongao is up 1-0 (was typing the answer in...)

Mewto55555 defeats isabella2296 3-0. (Countdowns were 4-2, 4-2, and 4-3. I think.)

got pwned by jongao 3-0 :(
must be an off day

Guessing this tourney is dead?  Sign-ups for 2 is starting. :P

It's not dead. It's going better than I expected.

Results of Round 1:

lotsofmath defeats pythag011
nikeballa defeats jjx1
isabella2296 defeats jain_harshi
Jongao defeats Meta_Knight

So Round 1's over

Results of Round 2:

Mewto55555 leads nikeballa 1-0
nt81something defeats isabella2296

Jongao beats ernie 3-1 (4-1,3-4,4-3,4-2)
My 4-3 win was a comeback from 3-0 down.  :|

Jongao defeats ernie 3-1 (4-1, 3-4, 4-3, 4-2).

Meh, he had a 3-0 comeback on four straight mems. :P

Oh yes, its nt811581147 (yes that's memmed).

EDIT: Beaten.

Results of Round 2:

Mewto55555 leads nikeballa 1-0
nt81something defeats isabella2296
Jongao defeats ernie

So we just need Mewto to finish his games with nikeballa.

[quote="BOGTRO"]Do I have a match yet?[/quote]

You're against lotsofmath.

AIME15, can you please set a deadline and/or send prod PMs, in addition to updating the bracket? Thanks!


A set of seven integers has a median of 73, a mode of 79, and a mean of 75.  What is the least possible difference between the maximum and minimum values in the set?

[hide="Hint"]Mean of 75 means the sum is 75*7[/hide]

[hide]i agree with 8 since the set could be
71,72,72,73,79,79,79
it fulfills everything and the range is 8[/hide]


calculus
integration
function
calculus computations

I got the answer, but I have no time to write my work now.

[quote="Xaenir"]
\[ \int {\sqrt {x \plus{} \sqrt {x^2 \plus{} a^2 } } } dx
\]
[/quote]
Let the value be I.
Integrate it twice by parts.
Both time taking $ \int {\sqrt {x \plus{} \sqrt {x^2 \plus{} a^2 } } } dx$ as first function.
you get $ 3I/4 \equal{} x\sqrt {x \plus{} \sqrt {x^2 \plus{} a^2}} \minus{} \frac{1}{2}\sqrt {x^2 \plus{} a^2}\sqrt {x \plus{} \sqrt {x^2 \plus{} a^2}}$

hello, your result can be simplifyied to $ \minus{}\frac{2}{3}(\minus{}2x\plus{}\sqrt{x^2\plus{}a^2})\sqrt{x\plus{}\sqrt{x^2\plus{}a^2}}\plus{}C$.
Sonnhard.


Yeah, so start making your fantasy tournament.  :)

Eh, Federer has the most momentum right now out of all of these people except arguably Max. :P

Wow, favorites so far are Rafa, Runpeng, and OF COURSE javascript:opener.emoticon(':starwars:')

I'm sorry, but Darth Vader just managed to edge Michael Phelps in the qualifiers.

6-7,3-6,7-6,6-4,22-20

It was epic.

22-20...? The message is too long. Please make the message shorter before submitting.

lol.

Yes apparently Darth Vader brought out dark forces...and pwnd Michael Phelp's backstroke skills.

Austin Powers was succumbed into the [b]Dark Side[/b] by Darth Vader.

awww come on. I was rooting for Austin Powers... Oh, wellz, gotta go root for Max.

btw, someone edit the poll. rafa dropped out  :(

Yep, Rafa dropped out. Hmm. I guess Michael Phelps lost in the final round of qualifying. NEWS FLASH He has just been chosen as a Lucky Loser!


ARML
MATHCOUNTS
AMC
AIME
function
linear algebra
matrix

Since many of the longer threads seem to get off-topic every now and then, I'd like to pre-emptively set up a thread to discuss the actual problems from this weekend's competition. Obviously,

DO NOT DISCUSS THE PROBLEMS UNTIL SATURDAY AFTER 500 P.M. EASTERN.

The 500 pm is there in case things run late at some site -- the relay round is supposed to be over by 200 pm Eastern. In the past, certain sites have had slight delays for various reasons.

Anyhow, as one of the writers this year, I'd like to know which problems you liked, which ones you thought were too easy or too hard, which were too tedious, etc. Maybe you saw a clever solution that betters the one in the solutions handout, or were able to trivialize a problem on your calculator  :wink:

(If that last one happens, let me know  :P )

Your feedback will be very valuable to forming future contests  :)

Now that the competition is over, what didja all think?

I enjoyed this year's arml individual--I thought each question had a clever way to do it that required different skills and didn't take too long.
However, I thought question 2 was mathcounts-level and question 8 was beyond-reasonable. I think there is a happy medium somewhere.
Nice work writing the questions!

Great job!

The individual was great. I kind of screwed up some of them...but of course, that is my fault. [I was like wtf on the last problem...I totally had the perpendicular bisector intersect the opposite side...lol..there were some tricky cyclic quads that developed...but w/e that failed.] The actual solution was nice. It was pretty trivial except for that one pair of congruent triangles that solved the problem.

I also got the sin[2x] problem wrong...like I approximated 58 degrees is 1 radian...but then I just got bogged down in computation...I suppose this is my fault too.

At the log right problem [I think this problem was on the "D team" homework sometime...I read it like $(p)(p-2a)(p-2b)(p-2c)=\sum 2b^{2}c^{2}-a^{4}$, when really, there was an extra negative on the p-2a term. For some reason I feel like I have made this mistake before [old AIME problem...it was in acops] but w/e].

Like these are my petty mistakes...the individual was very well-written!!!

I enjoyed the power. Note: one of my friends had an elementary number theory book with pretty much the [i]exact[/i] same problems, terminology, etc. I suppose this is unavoidable. How about an inversion power some time?

Relays good and fair...they are kind of trivial though...I suppose that they make sense logistically; during the event, the individuals are being compiled. Once individuals are over, I am just kind of tired and ready to finish-up.

YAY MR. GOODMAN was on the mic!!!!

Team was reasonable. The digits of 2007 problem showed up on a WOOT practice AIME.

Where were the power solutions? [Maybe I just didn't look hard enough.] We thought there was some problem...the expansion did not include 0 as a digit thing...Some numbers could not be expressed using "positive integers."

The tiebreaker problems were alright. The first one was a bit contrived; while I saw the substitution, something that is potentially worth $\$$2000 should somehow test other skills.

p.s. I like my copy of "severely-reduced-capabilities" mathematica (some people seem to call it mathematica "for students")  :P

#8 was made significantly harder (not in a mathematical sense, but in a time sense) by the fact that there was no diagram, I think.  More than half of the people I asked either first misread the question like I did, or drew it differently, or something.

Otherwise most of the problems were good.  5 and 6 seemed a little ugly, but that's just my opinion...1, 2, 3, 4 were nice.  And I don't really remember 7 so much.

Who else, during the second tiebreaker, got it quickly, but switched the two triangle legs?  :D

#7 was super easy, probably to give everyone the 10 minutes which was only half of what they needed for number 8.

My 3 person relay team missed both relays (second person used cos instead of sin on the 1st, and the wrong side of the box on the 2nd (I was 3rd) :(), but the team as a whole got 10+10=20 :). Oddly enough, we did the super relay in 5 minutes (normal relay time) and got it right. Too bad it doesn't count for points :(

1 and 2 were almost too easy.. I pretty much slipped up every problem after that for trying to work on the tough ones..

I took the log of 3 to find the answer, but then forgot to anti-log it again, and was wondering the entire time why I had a negative log (d'oh)

For 5, I forgot that only sin(9) is the *best* solution, so I took the 1-1.5, 4-4.5, 7-7.5, and 10-10.5 intervals

For 6, I don't know what I did, but it wasn't pretty.

Seven, I missed the question because 30-second rushed it, didn't check it, and spent 9:30 on a failed attempt to find 8, which I'm wondering now exactly *how* many cyclic quadrilaterals were needed.

So... yeah I did pretty terrible but.. oh well, it was my first ARML anyway.

The first tiebreaker, well, I didn't make it to that round, but I thought the solution was probably the most elegant one: so easy to such a hard problem.

I spent 1 minute on #7 and.. I was struggling to draw the diagram for #8
I made a careless mistake and got #6 right after the guy said stop -.-

Individual: I enjoyed it.  #2 was too easy and #8 was too hard, but the others were at a good difficulty level.  #3 was really cool, and I also liked #5.

Relays: My team didn't get either of them.  :(   I thought 2-2 was sort of ambiguous about which sides were included in the surface area.  1-2 was a nice problem for a relay, but unfortunately I misread the problem and passed $-\pi$.

Power: Nice topic, better than last year's.  I thought that it gave a bit too much of an advantage to people who had studied continued fractions previously, and the weaker people on our team really couldn't do anything.  But the problems were interesting.

Team: There were some good problems here.  #5 (I think) was funny: at first we had $-\frac{1}{2}$, then realized $0$ worked, and put down both of them.  We were actually thinking of appealing the problem, but didn't.  Then on the bus home we realized that $-\frac{1}{2}$ doesn't work because you're dividing by $0$.  :rotfl: Good problem, though.  #3  and #4 were nice, too.  #9 may have been the most insane problem I've ever seen on a math contest; I wasted way too much time just trying to figure out what the function actually did.

Individuals: I think #8 was so hard because it was hard to get a good diagram, much less connect the given information, within the time limit. (I even put A,B,E in wrong order, so I had no chance of seeing nice equal angles and randomly guessed 72  :roll: ). I liked #3 a lot, but the third pair... could have been better. (There was a big agreement among people I talked to that #6 should have helped separate the 6's from the 7's, and #5 sort of did just because people were not careful enough.)

Power: Best one among the 3 years I've been at ARML :) I thought #1-#6 were pretty approachable for most of the team, since all of us knew how to find the value of a periodic continued fraction . Yes, continued fractions are a bit specific, but from my own experience, there wasn't a massive advantage given to people who knew much about continued fractions. (This wasn't asking for proofs that if $\alpha$ is irrational, then $\left| \alpha-\frac{p}{q}\right| <\frac{1}{q^{2}\sqrt{5}}$ has infinitely many solutions in positive integers $p,q$, or anything like that. Our team even wrote up #8 without using the intended matrix solution or the "magic box formula". My idea was to try to prove that if $x=\frac{ax+b}{cx+d}$, then $y=\frac{ay+c}{by+d}$ and induct on the continued fraction length, and Shaunak was able to use that while I wrote up problems like #7 and #9.)

Team: Nice problems in general, although it was a bit of a shame that no else on my team could comprehend #9 nearly enough to check it. #7 seemed annoying at first, but once one of us realized that we could advantage of what divided the number of digits per chapter, it was really elegant. And as for #5... we found -1/2 first, and then someone our our team did his typical awesome guess and said "Hey, 0 works!" and then we were all like  :rotfl: and got rid of -1/2.

Relay: I thought they were okay, even though they cost our team the championship. (What happened to the first people in our squads on the second relay?!  :o We all thought that one was easy except for most of the people who needed to do it...)

Team:
The team round was very nice. Number 9 was off the wall though.

Power:
I agree with e^(pi*i) about the power contest. I haven't really dealt with continued fractions before, so that was somewhat limiting for me at times. However, there were many problems that were still very approachable yet challenging. Overall an interesting topic.

As for our team, we failed miserably; almost half our team members got together in a group to discuss one of the problems and just got jammed. And they kept going at it. So we were greatly reduced in manpower. We're not doing that again next year.

Individuals:
-1 was straightforward computation, but a good number 1.
-2 was a bit easy, but still decent.
-3, I did with a geometric interpretation ($-\left( \frac{4A}{c}\right)^{2}$ by Heron's); though that was totally unnecessary, it was more interesting.
-4 was a nice problem.
-5 was nice; I got it wrong for being careless
-6 was an interesting NT problem
-7 was trivial, but appropriate for a number 7
-8 was obscenely difficult. I don't understand the point of making obscenely difficult number 8's, it doesn't make the contest more fair or more interesting. It would've been a nice team round problem, though.
-T1 was one of those you-see-it-or-you-don't problems. I ended up seeing it while trying it out from my seat; however, that was in a much more relaxed and experimental mindset than someone actually taking the tiebreakers, so I'm not surprised that almost none of the contenders saw it. I know I wouldn't have seen it if I were taking it for real.
-T2 was a nicer problem
-T3 wasn't used at PSU

Though the problems themselves were nice, I was somewhat disappointed by the lack of probability and counting.

Relays:
Questions were generally trivial (1-3: wlog assume the inner sphere is degenerate?), but I guess relays have to be that way. First in 3, stupid mistake on the second. Boo.

problem 8 was not that impossible

The angle chasing stuff is _trivial_; that should not have been a problem.

This type of "find the angle in the random diagram problem" shows up frequently. You have to realize that angle chasing will not suffice. You need to find some other type of information...such as triangle congruency, a cyclic quadrilateral, incenter, excenter, etc.

I accept that I got it wrong because I had the wrong diagram [which turned out to have interesting properties...so I was like ...].

You have to remember that the #8 s are not meant for most people to solve. Those who have a strong handle on geometry had a very good chance to solve this problem.

I would say that this type of problem could be found in a math journal or an easy-ish olympiad. These types of geometry problems tend to need some sort of divine inspiration to see the random trick that finishes the problem.

Regardless, I feel that it was a fair problem.

Anyone mind posting the problems for those who weren't there?

I liked the problems overall, although i did pretty bad.
My favorites were numbers 1 and 7. I couldn't really figure out how to do 2, so I just kind of gave up....

it was my first arml, and hopefully i will do better next year....

Pretty sure Bay Area A got it.  If I remember right, they missed 1, 4, and 7.

Cool problem.  I tried it, but counted wrong as I only took into account the ones where three bisectors intersect.

Si - the buzz I was hearing late in the evening on Friday was that our A team had compared answers with Washington and SFBA and thought that Washington had 32, SoCal 28, and SFBA 24. But when the scores came out, it was Washington 28, SFBA 28, and SoCal 24. That was about #10 - we agreed with Washington's answer, but it was SFBA that actually got it right.

[quote="generating"][quote="Walk Around The River"]i dont like I-3 for one reason, because it can be solved the way i thought of before i saw the "real" solution.

i just set a=b, and solved a=b=root2, c=2, so it kinda doesnt make you do the problem.[/quote]

When I wrote the problem, I didn't see this. I only came up with it on the fly because we needed something "easy". But I kinda like that it's amenable to finding specific values for a, b, c. Also, that it has a purely geometric approach (Heron's formula).[/quote]
I thought the difference of squares factorization was easy.  Heron's, I thought, was extavagent, and takes a little effort to make legitimate—you have to prove a, b, c satisfy the triangle inequality, or else know some bizarre extention of area to complex numbers.  Of course, on a contest, all you need is the answer, but the nagging issues of restricted domain detract from the beauty of the Heron's solution, which is longer anyhow.

[quote="Karth"]Georgia A got it.[/quote]
No, we didn't, unless somebody changed the answer while I wasn't looking.

$a^{2}+b^{2}= c^{2}$ gives you the triangle inequality.

For the record, I initially goofed up a factor of 2 in the denominator when I was doing it, and only found my mistake when I saw that most of the Alabama team had answered -4, not -8.

I'd also like to point out that this calls to mind I-2 from 2003, which introduces a "log-right" triangle in which $log(a^{2})+log(b^{2})=log(c^{2})$ and then asks about log-right triangles that are also right triangles.  This is actually the reason I saw through to Heron's.  I looked at the problem, and thought, "hey, that looks like a log-right triangle that's also right!".  I had the definition of log-right wrong, of course, but it was enough to make me think of triangles.  Then I started marvelling at the symmetry of the numerator, which led me very quickly to Heron's once I realized I could switch the sign of the last factor.

[quote="Ignite168"]CT A put 224

what was the answer Washington and SoCal agreed on?[/quote]
892

On the third page of this topic, Sly Si an others are talking about power round #8. Si refers to two 3-point answers and one 1-point answer that he personally saw. Here's one of those 3-point answers:

----

Working $x$ downwards by first subtracting $a_{1},$ we obtain

$\frac{1}{\frac{1}{\frac{1}{\frac{1}{x-a_{1}}-a_{2}}-a_{3}}-\cdots}-a_{k}=\frac{1}{x}.$

Working $y$ upwards by adding $a_{1}$ to $\frac1y,$ we obtain

$\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{y}+a_{1}}+a_{2}}+a_{3}}+\cdots}+a_{k}=y$

Note that

$\frac{1}{\frac1y+a_{1}}=\frac{-1}{-\frac1y-a_{1}}.$ Also $\frac{1}{\frac{-1}{-\frac1y-a_{1}}+a_{2}}= \frac{-1}{\frac{1}{-\frac1y-a_{1}}-a_{2}}.$

Continuing this, we have

$y=\frac{1}{\frac{1}{\frac{1}{\frac1y+a_{1}}+a_{2}}+\cdots}+a_{k}=\frac{-1}{\frac{1}{\frac{1}{-\frac1y-a_{1}}-a_{2}}-\cdots}+a_{k}.$ Hence

$-y=\frac{1}{\frac{1}{\frac{1}{-\frac1y-1_{a}}-a_{2}}-\cdots}-a_{k}.$

Making the substitution $x=-\frac1y,$ we have that $x$ and $-\frac1y$ satisfy the same equation. Hence $-\frac1y$ is a root of $Ax^{2}+Bx^{2}+c.$ But if it is a root of $(x-a)(x-b)=x^{2}+\frac{Bx}{A}+\frac{C}{A}= x^{2}-(a+b)x+ab,$ $y$ is a root of $\left(x+\frac1a\right)\left(x+\frac1b\right)= x^{2}+\frac{a+b}{ab}x+\frac1{ab}=x^{2}+\frac{-B/A}{C/A}x+\frac{A}{C}$

$=x^{2}-\frac{B}{C}x+\frac{A}{C}.$

Hence, $y$ is a root of $Cx^{2}-Bx+A.$ Thus, $Cy^{2}-By+A=0.$

[quote="Boy Soprano II"][quote="Sly Si"]$a^{2}+b^{2}= c^{2}$ gives you the triangle inequality.[/quote]
This is true, but you still have to write
\begin{align*}2ab &>0 \\ a^{2}+2ab+b^{2}&> c^{2}\\ (a+b)^{2}&> c^{2}, \end{align*}
and it makes the solution less effortless and, in my opinion, detracts from the beauty.  Furthermore, it does not generalize (easily) to all complex numbers $a,b,c$ satisfying $a^{2}+b^{2}=c^{2}$, as the factoring solution does.  In fact, I am skeptical of an extension of area to triangles with complex side lengths, because it is hard to define unique square roots in the complexes.  I mention this because I did once start to read an article that talked about extending Heron's formula to "triangles" with side lengths not satisfying the triangle inequality, but in that case, you only have to deal with pure imaginaries and pure reals.[/quote]

Umm... what are you talking about?

[quote]
I–3.  Suppose $a$, $b$, and $c$ are [b]positive real numbers[/b] such that $a^{2}+b^{2}=c^{2}$ and
$\log{a}+\log{b}=\log{c}$.  Compute $\frac{( a+b+c)(a-b+c)(a+b-c)(a-b-c)}{c^{2}}$[/quote]

So I don't really see why you keep talking about complex numbers...

Also it's [b]very[/b] well known that $a^{2}+b^{2}= c^{2}$ for $a,b,c>0$ implies that $a,b$ and $c$ are the sides of a (right) triangle, you could probably even just cite that on an olympiad.

So basically, the Heron's formula solution is both, completely correct and rigorous , and the most elegant solution to the problem. I really don't see what you don't like about it.

[quote="Elemennop"][quote="Sly Si"] and only found my mistake when I saw that most of the Alabama team had answered -4, not -8.[/quote]

Where you the one who proctored for the Alabama team?[/quote]

Yep, that was me.  Hi!

well, he's stating that the heron's solution doesn't obviously generalize immediately to the case where a, b, and c are complex (which would still be an appropriate problem to pose), whereas the purely algebraic solution does

also dude yeah wtf why are you bringing up the triangle inequality

it is utterly trivial that if you have the law of cosines for any theta, you have the triangle inequality? so uh yeah right triangles really are triangles

[quote="JAM"]So basically, the Heron's formula solution is both, completely correct and rigorous , and the most elegant solution to the problem. I really don't see what you don't like about it.[/quote]
Difference of squares is as rigorous and correct as the Heron's formula solution, is as least as simple and elegant, and generalizes because it does not rely on geometry.

Um... or you could just see that $\frac{ab}{c}= 1$ and $a^{2}+b^{2}-c^{2}= 0$ ...?


topology
probability
function
integration
real analysis
real analysis unsolved

$\mathcal P$ the set of probability measure on $[0;1]$, with the weak* topology, $\mathcal L$ the set of Lipschitz functions on $[0;1]$ with lipschitz constant inferior or equal to $1$.
Prove that $d(\mu, \nu) = \inf\{|\int_{[0;1]}f d \mu-\int_{[0;1]}f d \mu| : f \in \mathcal L\}$ is a distance on $\mathcal P$ that defines the weak* topology.

Naturally, you meant $d(\mu, \nu) = \sup\{|\int_{[0;1]}f d \mu-\int_{[0;1]}f d \nu| : f \in \mathcal L\}$. Here is another way to represent this metric. Assuming, as we may, that $f(1)=0$, integrate by parts: $\int_{0}^{1}f d \mu=-\int_{0}^{1}f'(x)\mu([0,x])dx$. Since $f'$ can be an arbitrary measurable function bounded by $1$, it follows that $d(\mu,\nu)=\int_{0}^{1}|\mu([0,x])-\nu([0,x])|\,dx$.

Weak convergence of measures is [url=http://en.wikipedia.org/wiki/Portmanteau_theorem]equivalent[/url] to pointwise convergence of distribution functions at the points of continuity of the limit. It is an exercise to prove that $L^{1}$ convergence of increasing functions implies their pointwise convergence of this sort. The converse follows from the Lebesgue domination.


geometry
3D geometry
algebra solved
algebra

Is it possible for both n+3 and n^2+3 to be cubes for some positive integer n? It's kind of easy... :D

The answer is no. This really is an easy exercises. So try to solve it, my young friends!

Namdung

If they were both cubes then their product would be a cube also.
But their product is (n + 3)(n2 + 3) = (n + 1)3 + 23.
But there are no solutions to x3 + y3 = z3.
(You don't have to mention Fermat's theorem but I like doing that.)
You can also say (x + 1)3 < (x + 1)3 + 8 < (x + 2)3; the result follows as well.

Indeed, very easy :D

Very surprising! Small problem but great solution. Congratulations, Arne!

Namdung

Why is it so surprising? Isn't this just a standard solution for an easy problem?

Arne, I did mean the using of Fermat theorem. The approach is so nice. I thought only about your second solution.

Namdung

Oh :) That's what you meant :)
Well, the second method is boring but much more general.

For a proof of Fermat's theorem please see: http://www.s87346661.onlinehome.us/wiles_Fermat.pdf

It's indeed  a very nice solution. My approach was different. It didn't yield a long solution, but it was definitely uglier.


does anyone here play chess on Chess Live? If you do my handle is johnnyfox and ill be glad to play you and if you don't, and like chess, i can help you get an account.  [url]USchesslive.com[/url]should do the trick.

A while ago I tried looking for a good chess website, but only ones where you had to pay to become a member. How does Chess Live work?

ironic isnt it?

wouldnt you expect someone like chess64 to do something like this?

Eh...this can join the other chess thread.

Moved from F&G.

there is another chess live string?  I didn't know that.  (new edited post) oops, i put 2 strings of the same thing! :blush: never done that before.  How can i remove this string?

A moderator would have to do that. You can remove your own topics until someone replies.