geometry
geometric transformation
reflection
calculus
integration
trigonometry
derivative

I was just wondering what people think about different teaching styles in different countries.

As an example: I am American. My impression of a lot of American (public) school math is that students are simply expected to memorize a lot, often without having been given a satisfactory explanation. Of course, there are exceptions.

At the moment I am living in Ukraine, and I see an interesting difference in results (though I don't really know what the teaching style is like). I go to an international school, but according to my Ukrainian friends there is in general a much larger focus on memorizing things in school in Ukrainian schools--poems, short stories, etc. However, from what I have seen they don't seem to put such an emphasis on memorizing things in math. They all carry officially sanctioned "cheat sheets" around with them all the time. It seems to me (particularly after working out of Ukrainian mathbooks) that the emphasis is on the methods, tricks, etc. used rather than memorization of identities, etc.

Now, since I go to an international school, I can bring in a couple other examples. I am easily the best math student in the school (though not nearly good enough to go to IMO or anything like that), but because it is such a small school I am in classes with all the other people in my grade--I just do different work (the poor teacher has to teach people doing three different levels of work at the same time...). There is a Hungarian guy in my class who seems to have a very good intuitive grasp of mathematics. I'm not sure whether it is because he is just naturally good at it or because of his Hungarian education, but he is always looking a step ahead, very good at bringing things to their conclusions. He's very good at proofs for this reason.

There is also a Canadian girl who, although she isn't very advanced in math, has a surprisingly good intuition when it comes to math puzzles and such. I'm always surprised, because she usually comes to a correct solution (or declares, quite certainly, that the puzzle is impossible--something that I have a lot of trouble doing until I can prove it) very quickly, though she usually can't give a rigorous proof or explanation why.

I'm not sure whether these are all just individual traits of these people, or whether they reflect trends in education in these different countries. Any thoughts?

I like the ideals of the Moore Method (teaching style in the states) but I haven't taken a course that was taught in this way yet.

I go to an international school also!  I'm a freshman who goes to ASIJ (American School In Japan) in Tokyo.  There are probably more people at my school.  I'm taking pre-calc this year and I hope to do well in math competitions in the future, at the moment, hoping to do well on the AIME.  At my school it seems like there is even more focus on applications instead of pure math.  Not to mention there's more memorization here for some reason.  I [b]may possibly[/b] be the best person at math in my school, I wouldn't doubt it.

In my opinion, memorization is the worst possible way in which one could learn maths, but I'm no expert :D.

What's the Moore Method?

Tsk tsk tsk. I'd have to agree with grober about that. ;) Memorization's alright for basic "math facts" and for history and English vocabulary, but not for most of math. I've learnt mostly by the book method, which is where you recieve a book and pretty much teach yourself. For some people, it's pretty effective, but for others, it's not.

[quote="Ashiaro"]I've learnt mostly by the book method, which is where you recieve a book and pretty much teach yourself.[/quote]
Ditto. Well at least that's how I learn most new math now. When I was littler of course I needed a lot of help from my grandmother.

*recites a couple dozen integration formulas*
*recites the Law of Cosines*
*recites the Law of Sines*
*recites the Double Angle formulas*
*recites the trig addition and subraction formulas*
*recites Heron's formula*
*recites some differentiation formulas*
*recites the formula for finding the area between two curves*
*recites the fomula for finding the area of any shape rotated around*

That should show how I was taught...

i think being able to apply things you know is better than memorizing the names or whatever. for example i always mix up associative and commutative but that doesnt mean i dont know how or when to use them.

[color=green]Yeah. My mom told me that each person has a different learning style, so what works for me might not work for you. :)[/color]


[quote="t0rajir0u"]Wouldn't that make $30$ people total?[/quote]i think it's a trick question :wink: [hide="hint"][img]http://www.georgehart.com/sculpture/72pencils.jpg[/img][/hide][hide="or"]you could make two equilateral triangles[/hide]

[quote="akech"]Order 24 people in 6 rows but each row has to have 5 people![/quote]
[hide]A hexagon. 3 people are exclusively on each side.
Then put one person on each of the vertices.[/hide]


algebra
functional equation
algebra proposed

Here is a functional equation that has an important role in physics (Well guess where.. :D ). I can solve it the "physicist way" but I am curious on how it can be solved the mathematician way, you know what I mean... casework and all, no assumptions from reality etc. So here it is:

Let $ f: R\to R$, then for all $ x,y \in R$ and for constant $ \phi\in R$ we have:
\[ f\left(\dfrac{x\plus{}y}{1\plus{}\phi\cdot xy}\right)\equal{}f(x)f(y)(1\plus{}\phi \cdot xy)\]

Well, I don't think it can be solved in "nice way" without any pre-assumptions. First put into the equation $ x\equal{}y\equal{}0$ We will have $ f(0)\equal{}0\vee 1$ If $ f(0)\equal{}0$ then it's obvious by putting x=0 that we have $ f(y)\equal{}0$. But there are many other problems, for example:
if we try $ \phi \equal{}0$ then it's all messed up, because it simplifies into $ f(x\plus{}y)\equal{}f(x)f(y)$ which is known that has infinitely many solutions (because we know nothing about continuity or anything else).  :maybe:  :|


superior algebra
superior algebra unsolved

Is the set of all real numbers of the form a+b*sqrt(2)+c*sqrt(3)

where a, b, c are rational numbers a subfield of the field of real numbers? Explain your answer.

[quote="ZetaX"]It's not closed under multiplication.[/quote]

By "it's not closed under multiplication", you mean that if I multiply 2 numbers of that form, I will not get one of the same form, because there is a sqrt(6) that does not comply with the form.
PS. Could you please simply break it down for me, if you meant something else, because these problems are out of my league. Thanks

Yes, that's what I meant. Now try to prove that $ \sqrt 6$ is not of that form.


When the n's dice are thrown at the one time, find the probability such that the sum of the cast is $ n+3$

[url=http://www.mathlinks.ro/viewtopic.php?t=79185]You can find the solution here[/url]

i still don't understand how progress is...
thanks 4 those who reanswer

[hide="Explanation"]
Since the sum of the cast must be $ n+3$, and each die must give at least $ 1$, we  must have one of three distinct situations:

$ (1):$ Three dice show $ 2$, the rest show one. The probability of this is $ \frac{ \binom{n}{3}}{6^{n}}$
$ (2):$ One die shows $ 3$, one die shows $ 2$, the rest show $ 1$. The probability of this is $ \frac{ \binom{n}{2}\binom{2}{1}}{6^{n}}$
$ (3):$ One die shows $ 4$, the rest show $ n$. The probability is $ \frac{ \binom{n}{1}}{6^{n}}$

These three cases are distinct and exhausting, so we can sum the probabilities and obtain:
\[ \boxed{ P= \frac{ \binom{n}{3}+2 \binom{n}{2}+\binom{n}{1}}{6^{n}}}\]
[/hide]


Anyone want to swap/sell review books?  I have the following:

AP Statistics, Kaplan
AP European, Barron's
AP European, Princeton Review
AP European, REA
AP Physics B/C, Princeton Review

Post here if you're interested in trading.

I am going to take European History and Physics C next year, so I am interested in swapping/buying review books with you.

I have P.R. (Princeton Review) World History, Macro and Micro Econ, Chem, and Calc. and Barrons World History

My teacher says that the AP Euro review books suck.

I have:

Stats (Barrons)
Comp Sci, Java, A/AB (Barrons)
World History (Barrons)
Chem (Princeton Review)
Physics B/C (Princeton Review)

Chem is pretty heavily written in.

The postage for swapping books could get expensive.

I think the best way to go is just to sell used books off Amazon or Half.com and use the money to buy new ones. You can fetch about $10 for "very good" books, which are basically "You've used them for a year but you haven't written in it or torn anything out".

I always donate mine to the library....

Whatever you do, don't recycle or throw them away....Libraries will usually take them in....

GOOD IDEA ANKUR! i have been trying to pawn mine off to underclassmen but no one wants them. but.. i didnt even think of giving them to our tiny little library. AWESOME!

thanks.

I have to agree -- that's an excellent idea.  Thanks :)

Let's see...The ones that are in mint condition (no writing, no folds, not even looked through...)

Kaplan:
AP Physics B
AP English Language and Composition
AP English Literature and Composition

Barron's:
AP US History
AP Calculus

Sparknotes Packs:
AP US History
AP English Language and Composition
Literature 101
SAT II Physics
and about half of the sparkcharts and flashcards currently in production

Princeton Review:
SAT II US and World History

Released Exams:
AP US History 2001
AP Physics B and C 1998
AP English Language and Composition

hehe...as you can see, I bought a lot of stuff I didn't use this year. *sigh* Going through these review books took a lot longer than I thought it would take.

[quote="MithsApprentice"]Let's see...The ones that are in mint condition (no writing, no folds, not even looked through...)

Kaplan:
AP Physics B
AP English Language and Composition
AP English Literature and Composition
[/quote]

Why 2 of the same book? :?

Whoops, language =\= literature.  :blush:


Hey to all the CHEM Mahans haunting this forum, here are a few questions from a paper which i either got wrong or dint get at all.. So please help.One or more options could be correct.Please dont jus give me the answer, i have the key. I want the method.
Problem 1:
The enthalpy of formation of $ UF (g)$ is $ 22 k cal mol^{\minus{}1}$. and that of $ U (g)$  is $ 128 k cal mol^{\minus{}1}$. Also the bond energy of $ F\minus{}F$ bond is $ 37 k cal mol^{\minus{}1}$. The bond dissociation energy of $ UF (g)$ is / are
a] 1245.5
b] 131.1
c]521
d]623
Ans. :a,c
Problem 2:
Ground state Valence shell  configuration of Nitrogen is:
a] $ \uparrow \uparrow \uparrow$    b] $ \uparrow \downarrow \uparrow$
c] $ \uparrow \downarrow \downarrow$  d] $ \downarrow \downarrow \downarrow$
Ans :a,d

Problem 3:How many litres of ozone at STP will be needed to oxidise completely 1o mL  of 0.4 M KI to iodine.

Sorry, but i am

[b]THIS BAD [/b]at chem.  :D

[quote="Euclidean Geometer"]Problem 2:
Ground state Valence shell configuration of Nitrogen is:
a]  b]
c]  d]
Ans :a,d
[/quote]

For problem2 obviously all the electrons are of same spin so it can be either +1/2 or -1/2. So answer a,d.

[quote="Euclidean Geometer"]Problem 3:How many litres of ozone at STP will be needed to oxidise completely 1o mL of 0.4 M KI to iodine.
[/quote]

The equation is:
$ \ce{O_3}$ + $ \ce{6I^{ - }}$ + $ \ce{3H_2O}$ --> $ \ce{3I_2}$ + $ \ce{6OH^ - }$

in reply to Q1:

are u sure abt the question. i mean, how can a bond have 2 bond disocciation energies??

and anyway, the answer i m getting is 135.5 kcal. which is not there in the options. u can solve it by forming chemical reactions and adding them together, and in the process adding their enthalpies.

At achu_182
This is wat the question said. it dint make sense to me

and sorry ... er but plz may i know who this is :oops:

[quote="achu_182"]i am achyuth sanjay, appearing for JEE this year
i live in ghaziabad, study in Noida and FIITJEE.[/quote]

And a regular quizzer?

how did u know i was a regular quizzer??

perhaps from your blog. You're the AC part of  [url=http://quizzerboys.blogspot.com/]VIAC[/url]  right? :D
ive been to it, man. It's awesome  :coolspeak:

And just for the record, shreyas and i are regular quizzers as well, which is probably why we've heard of you :D


[quote="damhieuchien"]Find the minimum value of the function:
$ f(x) \equal{} (32x^5 \minus{} 40x^3 \plus{} 10x \minus{} 1)^{2006} \plus{} (16x^3 \minus{} 12x \plus{} \sqrt {5} \minus{} 1)^{2008}$[/quote]

this is an ongoing problem which in M&Y magazine !!  :!:


parameterization
integration
logarithms
calculus
calculus computations

What would a phase curve look like for the following equation?

$ x'\equal{}x$
$ y'\equal{}2x\plus{}3y$

Additionally, what would be the formula for the solutions?

I'm not exactly sure what you mean by a phase curve, but here is how to get solutions:
[hide]
Let $ f(t)\equal{}xe^{\minus{}t}$ and $ g(t)\equal{}ye^{\minus{}3t}$, where $ t$ is the parameter that $ x,y$ are differentiated with respect to. We then have
\[ f^{\prime}(t)\equal{}x^{\prime}e^{\minus{}t}\minus{}xe^{\minus{}t}\equal{}0\Rightarrow f^{\prime}(t)\equal{}A\Rightarrow x\equal{}Ae^{t}\]
\[ g^{\prime}(t)\equal{}y^{\prime}e^{\minus{}3t}\minus{}3ye^{\minus{}3t}\equal{}2Ae^{\minus{}2t}\Rightarrow g(t)\equal{}B\minus{}Ae^{\minus{}2t}\Rightarrow y\equal{}Be^{3t}\minus{}Ae^{t}\]
For an equation of $ y$ in terms of $ x$, in the case $ A\equal{}0$, we can have either $ x\equal{}y\equal{}0$, $ x\equal{}0, y>0$, or $ x\equal{}0, y<0$, depending on whether $ B$ is zero, positive, or negative.

In the case $ A\neq 0$, let $ C\equal{}\frac{B}{A^{3}}$ so that we get either $ x>0, y\equal{}Cx^{3}\minus{}x$ or $ x<0, y\equal{}Cx^{3}\minus{}x$, depending on whether $ A$ is positive or negative.
[/hide]

I'm sorry but I'm not sure if I understand your solution.. where did you get $ f(t), g(t)$
and how did you arrive at the conclusion that $ f'(t) \equal{} A$ and $ x\equal{}Ae^{t}$?

I will try and show what the phase curves look like before finding their equations. (Usually it is very difficult or impossible to find the phase curve explicitly).

Solution

The point $ (0,0)$ is an equilibrium point, and we must find its nature.

If we take the coefficients of $ x$ and $ y$ in the two equations we get:

$ a \equal{} 1$, $ b \equal{} 0$, $ c \equal{} 2$, $ d \equal{} 3$

$ p \equal{} a\plus{}d \equal{} 4$, $ q \equal{} ad\minus{}bc \equal{} 3$, $ \Delta \equal{} p^{2}\minus{}4q \equal{} 4$

Since $ q > 0$ and $ \Delta > 0$, then $ (0,0)$ is a node. Since $ p > 0$ then it is an unstable node.

We wish to find any straight line solutions so let $ y \equal{} mx$

$ \Rightarrow\frac{dy}{dx}\equal{} m \equal{}\frac{2x\plus{}3y}{x}$

$ \Rightarrow m \equal{}\frac{2x\plus{}3mx}{x}\equal{} 2\plus{}3m$

$ \Rightarrow m \equal{}\minus{}1$

Hence all phase curves tangent the line $ y \equal{}\minus{}x$ at $ (0,0)$

Now solving them (using the integrating factor):

$ \frac{dy}{dx}\equal{}\frac{2x\plus{}3y}{x}\equal{} 2\plus{}\frac{3}{x}y$

$ \Rightarrow\frac{dy}{dx}\minus{}\frac{3}{x}y \equal{} 2$

$ \mu (x) \equal{} e^{\int\minus{}3/x\,dx}\equal{} e^{\minus{}3\ln{x}}\equal{} e^{\ln{x^{\minus{}3}}}\equal{} x^{\minus{}3}$

$ x^{\minus{}3}\left(\frac{dy}{dx}\minus{}\frac{3}{x}y\right) \equal{} 2 x^{\minus{}3}$

$ \Rightarrow\frac{d}{dx}(x^{\minus{}3}y) \equal{} 2 x^{\minus{}3}$

$ \Rightarrow x^{\minus{}3}y \equal{} 2\int x^{\minus{}3}\,dx \equal{} 2\frac{x^{\minus{}2}}{\minus{}2}\plus{}C \equal{}\minus{}x^{\minus{}2}\plus{}C$

$ \Rightarrow y \equal{}\minus{}x\plus{}Cx^{3}$

Hello countdownmath.  I believe you mean "phase portrait" and "following system".  For me, there is no better introductory text for that than Devaney's book on Differential equations.  I know, some people rag on his book.  "Dumb-down" they say.  What, everyone's expected to be a brilliant mathematician I suppose.  His book works for me and I think if you spend some time with it, others too, then one day you'll look outside your window and think, "I no longer wonder why about a lot of things out there". :)

[quote="countdownmath"]I'm sorry but I'm not sure if I understand your solution.. where did you get $ f(t), g(t)$
and how did you arrive at the conclusion that $ f'(t) \equal{} A$ and $ x \equal{} Ae^{t}$?[/quote]
Actually, the part where I wrote $ f^{\prime}(t)\equal{}A$ is a typo, that should be $ f(t)\equal{}A$ (where $ A$ is an arbitrary constant) which follows from $ f^{\prime}(t)\equal{}0$.

To actually come up with $ f(t)$ and $ g(t)$for equations like these, it is well known (and can be deduced by solving the separable differential equation) that an equation of the form $ x^{\prime}\equal{}kx$ has $ x\equal{}je^{kx}$ as a solution, where $ j$ is another constant. So for the equation $ y^{\prime}\equal{}2x\plus{}3y$, it would make sense to write $ y$ as $ e^{3t}$ times something. That something is what $ g(t)$ is defined to be.


topology
real analysis
real analysis unsolved

Let's say two sets $E_{1}$ and $E_{2}$ in $\mathbb{R}$ are equivalent iff $m( (E_{1}-E_{2}) \cup (E_{2}-E_{1})) = 0$.
a). Prove that the cardinality of set of equivalence classes of measurable sets is continuum.
b). What is the cardinality of set of equivalence classes of all sets ?

(a) any measurable set is equivalent to a Borel set. The cardinality of the set of all open sets is continuum (consider a countable base of topology), hence the conclusion.

(b) not sure.


hi people ...
i have the APMO 2006 exam..

we waited 10 days to post it........ ;)

PD: If any body like the spanish version
you can find this in the Spanish Communities:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79241]HERE[/url]

Carlos Bravo ;)
Lima Peru

See : http://www.mathlinks.ro/Forum/resources.php?c=1&cid=20&year=2006 . :)

[quote="ambros"]THANK  :)  :)  YOU PAL WHEN ARE YOU GOING TO POST THE SOLUTIONS ?[/quote]How about you stop using Caps Lock to write all your messages?  :|


graph theory
combinatorics proposed
combinatorics
spanning tree
square grid

A subset of $ n\times n$ table is called even if it contains even elements of each row and each column. Find the minimum $ k$ such that each subset of this table with $ k$ elements contains an even subset

The answer is $ 2n \minus{} 1$.

(it is assumed that every even set must contain at least 1 element otherwise $ k \equal{} 0$)

Let's first show that $ 2n \minus{} 2$ (or less) is not enough.We can put $ n \minus{} 1$(or the half of those we have) in the first colum so that the last element of the colum is left open.Now we can put the other $ n \minus{} 1$(or the other half) elements in the last colum so that the first element (from top to bottom) is left empty.It is easy to check that there is no even subset.

Now if I have $ 2n \minus{} 1$ elements we only have to show that there exist some elements $ (i,x)$, $ (y,j)$ , $ (i,j)$. That is yet to be done ... :)

Consider the bipartite graph $G$ which has the columns and the rows of the table as its vertices, and  the entries of the table at the intersection of a row and a column  correspond to the edge linking the corresponding vertices of the graph.

Since each cycle correspond to an even subset of the table, the problem reduces to finding the maximal subgraph of $G$ which has no cycles. Since the spanning tree of the graph has exactly $2n-1$ edges the answer is $k=2n$.

The example for $2n-1$ is the set of all entries of the first row and the first column of the table, it corresponds to  a spanning tree on the graph $G.$


This is just quadrant and basic stuffs. I got the answers but I just want to be sure, you know. I"m not totally new to these stuffs but trig isn't my best so here it is:

1.
a) 5 :pi: /4 --> Qudrant III
b) 7 :pi: /4 --> Quadrant IV
2.
a) -1 --> Quadrant IV
b) -2 --> Quadrant III
3.
a) 5.63 --> Quadrant IV
b) -2.25 --> Quadrant III

Um... There are two more questions that I needed to draw but I guess I'll just name the quadrant. It will help me to see whether my picture is right or not.  :lol:

4.
a) -7 :pi: /4 --> Quadrant I
b) 2 :pi: /3 --> Quadrant IV
5.
a) 4 --> Quadrant III
b) -3 --> Quadrant II

It's kind of confusing but I usually can do it.
The reason I'm asking this is because I'm now freshman and I have four classes with sophomores, juniors, and even seniors now and as you know, when you are freshman, your grades began to count for the college.
So, try to get good at it and for math, there's no better helper than AoPS.

Thanks.  ;)

[quote="Silverfalcon"]

Um... There are two more questions that I needed to draw but I guess I'll just name the quadrant. It will help me to see whether my picture is right or not.  :lol:

4.
a) -7 :pi: /4 --> Quadrant I
b) 2 :pi: /3 --> Quadrant IV
5.
a) 4 --> Quadrant III
b) -3 --> Quadrant II

It's kind of confusing but I usually can do it.
The reason I'm asking this is because I'm now freshman and I have four classes with sophomores, juniors, and even seniors now and as you know, when you are freshman, your grades began to count for the college.
So, try to get good at it and for math, there's no better helper than AoPS.

Thanks.  ;)[/quote]
um for 4b) 1(pi)>(2/3)pi>pi/2.  Since pi is the border of the second quadrant and pi/2 us the border for the first, isnt (2/3)pi just in the second quadrant?
pardon my ignorance(if im wrong)

I got enough help from MithsApprentice so done. And 4b is actually wrong question I should have posted something else.

[i]Thread closed.[/i]


[quote="v235711"]Consider $ 100$ positive integers such that their sum is equal to their product.What is the least quantity of numbers "$ 1$" that can exist among these $ 100$ numbers?[/quote]

$ 3 \plus{} 3 \plus{} 3 \plus{} 2 \plus{} 2 \plus{} 1 \plus{} 1... \plus{} 1 \equal{} 3^32^2 \equal{} 108$ and so we have a solution with $ 95$ numbers $ 1$.

Let $ n_1\geq n_2\geq n_3 ... \geq n_{100}\geq 1$ the $ 100$ numbers in decreasing order.
Let $ a \equal{}$ the number of $ n_i$ whose value is $ 1$
Let $ b \equal{} 100 \minus{} a$ the number of $ n_i$ whose value is $ \geq 2$

We have $ a \plus{} \sum_{i \equal{} 1}^bn_i \equal{} \prod_{i \equal{} 1}^bn_i$ and so $ a \equal{} \prod_{i \equal{} 1}^bn_i \minus{} \sum_{i \equal{} 1}^bn_i$

We obviously have $ \prod_{i \equal{} 1}^bn_i \minus{} \sum_{i \equal{} 1}^bn_i$ $ \geq 2^b \minus{} 2b$ (consider that replacing a non-$ 1$ number by $ 2$ decreases $ \prod_{i \equal{} 1}^bn_i \minus{} \sum_{i \equal{} 1}^bn_i$).
So $ a\geq 2^b \minus{} 2b$
So $ a \plus{} 2b\geq 2^b$
So $ 100 \plus{} b\geq 2^b$
So $ b\leq 6$

Since we have a solution with $ b \equal{} 5$ (and so $ a \equal{} 95$), it just remains to see if we have a solution with $ b \equal{} 6$ (and so $ a \equal{} 94$).

So $ n_1 \plus{} n_2 \plus{} n_3 \plus{} n_4 \plus{} n_5 \plus{} n_6 \plus{} 94 \equal{}$ $ n_1n_2n_3n_4n_5n_6$

LHS$ \leq 6n_1 \plus{} 94$ while RHS$ \geq 32n_1$ so we need $ 6n_1 \plus{} 94\geq 32n_1$ and so $ n_1\leq 3$ and so we have just 7 possibilities to check :

$ 3 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 1... \plus{} 1 \equal{} 112\neq 3^6 \equal{} 729$
$ 3 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 2 \plus{} 1... \plus{} 1 \equal{} 111\neq 2\cdot 3^5 \equal{} 486$
$ 3 \plus{} 3 \plus{} 3 \plus{} 3 \plus{} 2 \plus{} 2 \plus{} 1... \plus{} 1 \equal{} 110\neq 2^2\cdot 3^4 \equal{} 324$
$ 3 \plus{} 3 \plus{} 3 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 1... \plus{} 1 \equal{} 109\neq 2^3\cdot 3^3 \equal{} 216$
$ 3 \plus{} 3 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 1... \plus{} 1 \equal{} 108\neq 2^4\cdot 3^2 \equal{} 144$
$ 3 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 1... \plus{} 1 \equal{} 107\neq 2^5\cdot 3^1 \equal{} 96$
$ 2 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 2 \plus{} 1... \plus{} 1 \equal{} 106\neq 2^6 \equal{} 64$

So the answer is "at least $ 95$ numbers $ 1$"


Hi everybody  :D

 [b]Problem:[/b]
 [hide]Prove that there exist a real number $x$ such that
 For every possitive integer $n$ we have $\{x^n\} \in \Big[\frac{1}{3},\frac{2}{3}\Big]$[/hide]
 Good luck !

sorry, maybe i misunderstood the problem, but i can only missprove it:

set $n=1$ we get $\frac13\le x\le \frac23$
so $0<x<1$

now we prove a Lemma for every $0<x,y<1$:
$x\ge y\implies x^n \ge y^n$

which is easy to see, because $f(x)=x^n$ is always increasing for $x,n>0$

so we have $\frac23\ge x \implies \left(\frac23\right)^n \ge x^n$
setting $n=3$ implies $\left(\frac23\right)^3=\frac8{27}<\frac13$ and since $\frac13>\left(\frac23\right)^3 \ge x^3$ we get that $x^3 <\frac13$ hence not in that interval anymore.

Peeta: already your first conclusion is wrong:
$n=1$ doesn't give $\frac13 \leq x \leq \frac23$ !
Remark that $\{ x \}$ means the fractional part of $x$.

o i see, wondered what $\{x\}$ was supposed to mean, but i didnt care and just left it  :D
well then, the problem is still open for everybody :lol:

already posted and solved on the forum. Try to use the fact that if $I_1$, $I_2$, .., $I_k$,.. is  a sequence of nonempty compact intervalles, $I_{k+1}$ included in $I_k$, then their intersection is nonempty.


note: this is pretty darn messy

suppose a+b+c=k
then $ k^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2ac \plus{} 2bc$ and $ k^2 \equal{} 5 \plus{} 2ab \plus{} 2ac \plus{} 2bc$ (substitution using the last equation)
so $ \frac {k^2 \minus{} 5}{2} \equal{} ab \plus{} ac \plus{} bc$
now multiply the right side by k and the left side by a+b+c
$ \frac {k^3 \minus{} 5k}{2} \equal{} 3abc \plus{} a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b \implies \frac {k^3 \minus{} 5k}{2} \minus{} 3abc \equal{} a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b$ (1)
since a+b+c=k and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 5$ then $ (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \equal{} 5k \implies a^3 \plus{} b^3 \plus{} c^3 \plus{} a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b \equal{} 5k$
substituting the second equation and (1) gives
$ \frac {k^3 \minus{} 5k}{2} \plus{} 7 \minus{} 3abc \equal{} 5k \implies 3abc \equal{} \frac {k^3 \minus{} 15k \plus{} 14}{2} \implies abc \equal{} \frac {k^3 \minus{} 15k \plus{} 14}{6}$
uh oh time:
raise both sides of $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 7$ to the second power
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2a^2b^2 \plus{} 2a^2c^2 \plus{} 2b^2c^2 \equal{} 49$
substituting the first equation gives  $ 2a^2b^2 \plus{} 2a^2c^2 \plus{} 2b^2c^2 \equal{} 40 \implies a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2 \equal{} 20$ (2)
now square both sides of $ \frac {k^2 \minus{} 5}{2} \equal{} ab \plus{} ac \plus{} bc$ to get $ \frac {k^4 \minus{} 10k^2 \plus{} 25}{4} \equal{} a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2 \plus{} a^2bc \plus{} b^2ac \plus{} c^2ab$
now substitute  (2) to get $ \frac {k^4 \minus{} 10k^2 \plus{} 25}{4} \equal{} 20 \plus{} abc(a \plus{} b \plus{} c) \implies \frac {k^4 \minus{} 10k^2 \minus{} 55}{4} \equal{} abc(a \plus{} b \plus{} c)$
substitute $ abc \equal{} \frac {k^3 \minus{} 15k \plus{} 14}{6}$ and a+b+c=k imto this final equation to get
$ \frac {k^4 \minus{} 10k^2 \minus{} 55}{4} \equal{} \frac {k^4 \minus{} 15k^2 \plus{} 14k}{6}$
$ 3k^4 \minus{} 30k^2 \minus{} 165 \equal{} 2k^4 \minus{} 30k^2 \plus{} 28k$
$ k^4 \minus{} 28k \minus{} 165 \equal{} 0$
$ (k \minus{} 3)(k^3 \plus{} 3k^2 \plus{} 9k \plus{} 55)$

$ k \equal{} 3$ or $ k \equal{} \text{a root of}\, k^3 \plus{} 3k^2 \plus{} 9k \plus{} 55$
ill assume k=3 since teh cubic has no rational solutions

I think this problem is false!
By Cauchy-Schwarz inequality, we always have:
$ (a^4\plus{}b^4\plus{}c^4)(a^2\plus{}b^2\plus{}c^2)\geq (a^3\plus{}b^3\plus{}c^3)^2$
$ \Rightarrow 45\geq 49 ???$ :huh:

[quote="Kyoshiro"]I think this problem is false!
By Cauchy-Schwarz inequality, we always have:
$ (a^4 \plus{} b^4 \plus{} c^4)(a^2 \plus{} b^2 \plus{} c^2)\geq (a^3 \plus{} b^3 \plus{} c^3)^2$
$ \Rightarrow 45\geq 49 ???$ :huh:[/quote]
OK!  :rotfl: I think that, too!  :wink:

[quote="Kyoshiro"]I think this problem is false!
By Cauchy-Schwarz inequality, we always have:
$ (a^4 \plus{} b^4 \plus{} c^4)(a^2 \plus{} b^2 \plus{} c^2)\geq (a^3 \plus{} b^3 \plus{} c^3)^2$
$ \Rightarrow 45\geq 49 ???$ :huh:[/quote]

Aren't $ a,\ b,\ c$ be limited to real numbers? :wink:

Let $ a \plus{} b \plus{} c \equal{} p,\ ab \plus{} bc \plus{} ca \equal{} q,\ abc \equal{} r$,

$ 7 \equal{} 5p \minus{} pq \plus{} 3r$
$ 9 \equal{} 7p \minus{} 5q \plus{} rp$
$ 2q \equal{} p^2 \minus{} 5$

[quote="xxxyyyy"]If
$ a^4 \plus{} b^4 \plus{} c^4 \equal{} 9$
$ a^3 \plus{} b^3 \plus{} c^3 \equal{} 7$
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 5$

then
$ a \plus{} b \plus{} c \equal{} ?$[/quote]

Can $ a,b,c$ are complex numbers?  :?:

they are complex numbers... ignore the cauchy inequality for this problem

[quote="Kyoshiro"]I think this problem is false!
By Cauchy-Schwarz inequality, we always have:
$ (a^4 \plus{} b^4 \plus{} c^4)(a^2 \plus{} b^2 \plus{} c^2)\geq (a^3 \plus{} b^3 \plus{} c^3)^2$
$ \Rightarrow 45\geq 49 ???$ :huh:[/quote]
That problem was edited!!!

Sorry  :blush:  because I haven't learned about complex number :D

we aren't talking about complecs numbers  :rotfl: look at wath stevenmeow said:

[quote="stevenmeow"]
raise both sides of $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 7$ to the second power
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2a^2b^2 \plus{} 2a^2c^2 \plus{} 2b^2c^2 \equal{} 49$[/quote]

And that wasn't right because in the enunciate is $ a^2\plus{}b^2\plus{}c^2\equal{}5$ and that was the error

Continue your solution steven, because it could be "the solution" because $ 45\geq25$ obviously :P :rotfl:


inequalities
LaTeX
inequalities proposed

a,b,c are positive real numbers such that ab+bc+ac=abc.
Prove that 1/a*2+1/b*2+1/c*2<=1-(12abc)*1/2 :P
sorry !But i am too lazy to use latex :oops:

since your inequality does not make much sense, i think you made a confusion with operators. you used * insted of ^

so, correcting this and the <= for >=

${a b c=a b+c b+a c}$

${\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{a^{2}}\geq \alpha-\beta \sqrt{a b c}}$

${(a b c)^{2}\left(\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{a^{2}}\right)\geq (a b c)^{2}\left(\alpha-\beta \sqrt{a b c}\right)}$

${(a b c)^{2}=(a b+\text{bc}+a c)^{2}\geq (a b)^{2}+(c b)^{2}+(a c)^{2}\geq (a b c)^{2}\left(\alpha-\beta \sqrt{a b c}\right)}$

${(a b c)^{2}\geq (a b c)^{2}\left(\alpha-\beta \sqrt{a b c}\right)\text{ }\Rightarrow (1-\alpha )+\sqrt{a b c}\beta \geq 0}$

${\{a,b,c\}>0}$ ,${\beta \geq 0\land \alpha \leq 1}$

I'm not good with inequalities, so i dont know if it is rigth


linear algebra
matrix
abstract algebra
vector
geometry
geometric transformation
algebra

Sheldon Axler, Linear Algebra 2nd ed, ch3 q26:
Suppose $ n$ is a positive integer and $ a_{i,j}\in F$ for $ i,j \equal{} 1,...,n$. Prove that the following are equivalent:
(a) the trivial solution $ x_1\equal{}...\equal{}x_n\equal{}0$ is the only solution to the homogenous system of equations
\[ \displaystyle\sum_{k\equal{}1}^{n}a_{1,k}x_k\equal{}0\]
\[ \vdots\]
\[ \displaystyle\sum_{k\equal{}1}^{n}a_{n,k}x_k\equal{}0\]
(b) For ever $ c_1,...,c_n \in F$, there exists a solution to the system of equations
\[ \displaystyle\sum_{k\equal{}1}^{n}a_{1,k}x_k\equal{}c_1\]
\[ \vdots\]
\[ \displaystyle\sum_{k\equal{}1}^{n}a_{n,k}x_k\equal{}c_n\]

This is what I think the intended solution is:
the system of field equations can be expressed as a matrix equation (beacuse all the matrix operations are compatible with field algebra), now if we provide a basis to make this a matrix of a linear transform $ F^n \rightarrow F^n$ (insert reason why we can here) then we have that the map is a bijection and so we can construct the matrix equation corresponding to condition (b) under a basis of $ F^n$ and then this implies that the condition on the field equations in (b) holds . (insert reason why here)

So my question is; is this correct? if so then what are the explicit reasons we can convert a matrix equation over a field to a matrix equation with a basis attached (which are isomorphic to linear transformations). I have a sort of fuzzy answer in my mind, the algebra of matrices with bases attached is no different to the algebra with no basis attached, so if a condition is imposed on a system of field equations we may convert it to a condition to on a linear transformation over the field and vice versa? this seems to be what is happening. Also, what would the solution be if no linear transformations were allowed? just field algebra.
what I am trying to get at is a better understanding of how linear transformations and fields fit together, please explain this to me.

We have a linear map $ f(x_1,x_2,\cdots,x_n)\equal{}(y_1,y_2,\cdots,y_n)$ where $ y_i\equal{}\sum_{j\equal{}1}^n a_i^j x_j$.
Statement (a) is that the kernel of $ f$ is trivial, which is equivalent to $ f$ being invective by standard algebra.
Statement (b) is that $ f$ is surjective.

We want to show that these are equivalent for linear maps from $ F^n$ to $ F^n$; the proof must be specific to finite-dimensional vector spaces of the same dimension, as it isn't true anywhere else.

Now, the translation to matrix terms is completely trivial- we already have a basis on each side, and it's the standard basis of $ F^n$. There is no such thing as a matrix equation with no basis attached; writing one down automatically presumes a basis on each side.

The specific proof for this result is Gaussian elimination. If there are $ n$ pivots, solutions always exist and are unique. If there are fewer than $ n$ pivots, we can construct unsolvable systems and extra solutions to the homogeneous system.

[quote="jmerry"]There is no such thing as a matrix equation with no basis attached; writing one down automatically presumes a basis on each side.[/quote]

 :oops: ok, I think I get this now, I was confused beacuse in first year linear algebra we did matrix equations over a field but we never dealt with any vector spaces and such, so we did matrix algebra with no concept of a basis.

what I still struggle to understand is the step where we go from a set of field equations to a linear transformation (a.k.a matrix, from now on). So I want to know what justifies us turning a field equation into a linear map, which introduces vectors, please explain?

By the structure of matrix multiplication, we can write that system of linear equations $ y_i\equal{}\sum_{j\equal{}1}^n a_i^j x_j$ as a single explicit matrix equation $ \begin{pmatrix}a_1^1&a_1^2&\cdots&a_1^n\\ a_2^1&a_2^2&\cdots&a_2^n\\ \vdots&\vdots&\ddots&\vdots\\ a_n^1&a_n^2&\cdots&a_n^n\end{pmatrix}\cdot\begin{pmatrix}x_1\\ x_2\\ \vdots\\ x_n\end{pmatrix}\equal{}\begin{pmatrix}y_1\\ y_2\\ \vdots\\ y_n\end{pmatrix}$, which we abbreviate $ Ax\equal{}y$. Each linear equation comes from a row of the matrix, and it's just a matter of reading off the coefficients.


I have been processing incoming solutions and entry and permission forms for the USAMTS and I have some important suggestions for how to do this properly.

1.  You must register for the USAMTS in order to participate.  Being registered on the Forum is not the same thing or sending in an entry form is not enough.  Go here ([url]http://www.artofproblemsolving.com/USAMTS/AoPS_U_Register.php[/url]) to register.

2.  You must enter your Username and User ID on all forms where indicated.  The USAMTS section of the website has information about how to find your User ID number if you do not know it.

3. VERY IMPORTANT!  You absolutely must write your Username and USAMTS User ID number on each page of solutions you submit.  This is how we identify who has written the solutions.  If you don't write your username and id number on each page of the solutions, then you are trusting that I will do it for you.

4.  We highly encourage you to use the blank solutions form for submitting solutions.  Your solutions will be easier to grade and you will never forget to write in your username and id.

I encourage everybody to read over the USAMTS section of the website to make sure they are adhering to the rules and for suggestions on how to write and submit solutions.

It is very exciting to see all the submissions coming in and we are looking forward to seeing more.  Thanks for your participation.

continued...

5.  A solution may be more than one page, but each solution should start on a new page.  I have received papers that have solutions to problems 1-3 on one page and 4-5 on the next page.  Please start a new page for each new solution.

Please read the [url=http://www.artofproblemsolving.com/USAMTS/AoPS_U_ProblemsSoln.php]Submitting Solutions[/url] section for more information.

[quote="joml88"]Would it be recommended to rewrite the problem before each solution?[/quote]
No.  Just make sure your name, your USAMTS ID#, and the problem number is clearly indicated on [b]each[/b] page of your solution.

We strongly recommend that you either use the LaTeX templates, or print out our blank solutions pages and write your solutions on them.  Both can be found here:
[url]http://www.artofproblemsolving.com/USAMTS/AoPS_U_MyForms.php[/url]

[quote="ZennyK"]May I send in the entry and permission forms via snail mail and then submit my solutions electronically?[/quote]
Yes.  This is, in fact, the recommended method.

In the Profile section, what is the School CEEB code?  What does it stand for and what is it used for?

It's a number used by the College Board and by colleges to identify high schools, I believe.  I think you may be able to find your school's CEEB code at [url=http://apps.collegeboard.com/cbsearch_code/codeSearchHighschool.jsp]this website[/url], if your guidance counselor or college counselor can't tell  you.

I have just sent in through e-mail my solutions. I sent the PDF file, and also the .blg, .tex, .aux, .log, .bbl files just to be safe. I plan on sending the Entry form and Permission form by snail mail in a few days.

Is there anything I'm missing? Have I done anything wrong? Please help me out. Thanks.

All we need are the PDF and the TEX files and any images you included (if any).  You don't need to send the others (but it's ok that you did - no need to resend).

Sending the Entry and Permission Forms separately is fine.

When we send a pdf file with an image, we have to send the image too (in a separate pdf)?

I thought the point of LaTeX is to typeset one pdf onto another. and you can just send the final pdf.

I know this is a latex question, but its about USAMTS.

You need to send the image so that if they choose to use your solution on the site, thne they still have your image in a convenient location and can access it easily. That way they can simply copy and paste your $\text{\TeX}$ code into a new $\text{\TeX}$ file after putting your image in the correct directory.

that makes sense.

but how do they get the TeX code from a pdf?

You have to send both the $\text{\TeX}$ source and the pdf.

I got few questions.

None of the links for USAMTS given above work.

So, how do I basically register?

Do I go to http://www.usamts.org and click register and fill out my name and ID? Do ID correspond to my AoPS ID or what?

And I heard people talking about Entry Form. What's the difference between registering on http://www.usamts.org and sending Entry Form?

At last, do I send Entry Form by mail? Where to?

Sorry for a lot of questions but I want to register properly. I didn't register yet because of these reasons.

There are directions for registering at http://www.usamts.org/HowJoin/U_Join.php .

[quote="mathclass"]Latex always crashes my computer so I just handwrote my solutions (in pen of course :)). Is this okay?[/quote]

Indeed, that's okay.

[quote="4everwise"][quote="zoogies"]Uh...the instructions merely read [i]'do not [b]fax[/b] your solutions in pencil'[/i], right? Because if you send it via snail mail, it doesn't really matter whether it's in pencil or not.

Right?

:huh:

[size=84]Wow, I can't believe this is my first post.[/size][/quote]If you mail it, I believe you can write in pencil.

But why not just write in pen to be safe? It's not like pens cost a million dollars apiece. :P[/quote]
The problem with writing in pen is that if you make a mistake, it's a pain to correct - pencils allow you to make your solutions a bit neater, so I can see why he might be concerned.

If you are mailing your solutions, then [b]neat[/b] pencil is acceptable, however, we still recommend pen.

For examples of how to write and how not to write solutions, see [url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteWrite.php[/url]

[quote="ehehheehee"]If I already wrote all my solutions on lined paper, would it be better to copy solutions all unto the solutions forms?[/quote]

If you have time, I would copy them, but that's only because I'm a perfectionist.  Plus, it would make it neater (probably).

[quote="ehehheehee"]I write crooked without lined paper, but that is the only problem with neatness I would forsee...[/quote]

as long as it's readable, i don't think the grader will take off points unless they're like, super mean.

[quote="zanttrang"][quote="buzzer11"]i've registered for the USAMTS yesterday, but i dont think i'll have time to do it.

Do i need to 'unregister', or am i automatically out if i dont send in my solutions by Oct. fifth?

Thanks,
Brian[/quote]
Just don't send in any solutions - it's no extra work for the graders if you register and don't participate, and it gives you the option of starting during a later round if you find eventually that you have more time than you thought.  It's too bad you won't be able to do it - it's a great contest, even if you have very little time to devote to it.[/quote]

Thanks. I hope I'll be able to do it next year...

[quote="topaz"]If I did not start each solution on a new page in my submission, will I be penalized or disqualified? Can I resubmit my solutions on separate pages even though the deadline was yesterday?  :([/quote]You won't get penalized, but the graders will probably write you a note complaining about it.  There's a small likelihood that a grader won't be able to find a specific problem (a different grader grades each problems, and they're trained to look at the top of the page for the start of the problem).  In any event, do [b]not[/b] resubmit anything, and next round, please read the instructions carefully.

[quote="xsk13"]Would it be possible to still register for USAMTS and participate starting the second round?[/quote]

I think it's possible. I did this during Year 16 - I started participating in Round 3.

[quote="xsk13"]Would it be possible to still register for USAMTS and participate starting the second round?[/quote]

Absolutely.

questions:

If a person (me) does not use any TeXer program (i am using AoPS texer to import images into word, later converting the word document to PDF file), is it okay to send the PDF file, and some text file that contains the TeX codes for the solutions?

Do we need to use asymptote if we draw diagrams and send the solutions electronically? Or can I just draw the diagrams on paint (old school :D ) and send a file containing the images seperately?

Thanks

You can only send [b]one[/b] file, and it must be a PDF file.  It doesn't matter how you create the PDF file (whether via LaTex, via a virtual PDF printer, or some other means).  You cannot separately send text, image files, TeX source files, or anything else.

Please carefully read [url]http://www.usamts.org/Rules/U_RulesSubmitting.php[/url] for the details.

[quote="lightstar"][quote="vRusczyk"]I have been processing incoming solutions and entry and permission forms for the USAMTS and I have some important suggestions for how to do this properly.

1.  You must register for the USAMTS in order to participate.  Being registered on the Forum is not the same thing or sending in an entry form is not enough.  Go here ([url]http://www.artofproblemsolving.com/USAMTS/AoPS_U_Register.php[/url]) to register.

[/quote]

link above is broken  :([/quote]
Old link? Try http://usamts.org/Register/U_Register.php


group theory
abstract algebra
complex numbers
absolute value
superior algebra
superior algebra unsolved

Let G be the group of real numbers under addition and let N be the subgroup of G consisting of all the integers. Prove that G/N is isomorphic to the group of all complex numbers of absolute value 1 under multiplication.

Hint: consider the mapping f: R-->C given by f(x)=e^[2pi(ix)]

So this says that a subgroup of Z is normal in R.
G/N is the quotient group of left cosets of N in G.
And I want to prove that G/N is isomorphic to (a+bi)(c+di) <---not sure if this is what I want to prove...but if it is then...it equals ac+adi+bci-bd= +/- 1
Which implies
ac+adi+bci-bd=ac-bd=+/- 1

Am I thinking of this right so far?

I'm not sure how to use the hint.

That's not what you want to prove (a group can't be isomorphic to a number).  A complex number of absolute value $ 1$ is of the form $ e^{2 \pi i r}$ where $ 0 \le r < 1$.  Now look more carefully at the definition of an isomorphism and recall that $ e^a e^b \equal{} e^{a \plus{} b}$.


I need help creating headers and footers for future assignments. I read the document of this site, but it didn't really work...  :(

Thanks.

Did you use the fancyhdr package?  What exactly didn't work?

I remember using everything in the instructions on this site. The text didn't apprear though.

I assume that you read and tried to do what is described on [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#fancy]this page[/url].  Did you try to LaTeX the example on that page?  Did it work?

Without a more specific description of what you did and what went wrong, it's hard for anyone to give you specific advice.

Hmm, it works now. I guess I'll just copy and paste the text on that page for my next assignment. Either something in my last document was messed up or I typed the commands incorrectly.

Thanks for the help.  :)


Together, Larry and Lenny have $ \$$35. Larry has two-fifths of Lenny's
amount. How many more dollars than Larry does Lenny have?

ok let larrys money=x, and lennys money=y
x=2y/5
y+2y/5=35
7y/5=35
y=25
x=10

25-10=15

Do you mind if I just write this in $ \text{\LaTeX}$? It makes it easier to see.

Larry's Money = $ x$

Lenny's Money = $ y$

$ x = \frac{2y}{5}$

$ y + \frac{2y}{5} = 35$

$ \frac{7y}{5} = 35$

$ y = 25$

$ x = 10$

$ 25 - 10 = 15$

fast method:

Larry's money = 2x
Lenny's money = 5x
difference = 3x

therefore, $ 35\times\frac{3}{7}\equal{}15$

answer : 15


induction
AMC
USA(J)MO
USAMO
ARML
USAMTS
pigeonhole principle

Wow, I hope I do these right; my extent of learning in induction is basically my AlgII teacher lightly mentioning it in passing and giving one very very very rough example of how it works.

Most of these will be TeX heavy, so I'll skip on the spoiler tags.

[b]6.[/b]Given:

$F_1+F_3+...+F_{2n+1}=F_{2n+2}$

If that is true, this must also be true:

$F_1+F_3+...+F_{2n+1}+F_{2n+3}=F_{2n+4}$

Replacing all but the last term on the left side:

$F_{2n+2}+F_{2n+3}=F_{2n+4}$, which is equivelent to the basic Fibonnaci rule, $F_n+F_{n+1}=F_{n+2}$

[b]7.[/b] We are given:

$1 + F_2 + F_4 + ... + F_{2n} = F_{2n+1}$

Assuming it's true, then this must also be true:

$1+F_2+F_4+...+F_{2n}+F_{2n+2} = F_{2n+3}$

Replace the all but the last term on the left with what is already given originally:

$F_{2n+1}+F_{2n+2}=F_{2n+3}$  which is equivelent to a basic Fibonacci rule, which is $F_n+F_{n+1}=F_{n+2}$

[b]8.[/b] We are given equation A:

$F_nF_{n+1}-F_{n-2}F_{n-1}=F_{2n-1}$

In terms of the same $n$ used above, then we can also say equation B:

$F_{n+1}F_{n+2}-F_{n-1}F_n=F_{2n+1}$

Now, let's attempt to put both equations into simpler (and congruent) terms.

Starting with equation A, we know that based off of the basic Fibonnaci rule $F_n+F_{n+1}=F_{n+2}$, then $F_{n-2}=F_n-F_{n-1}$. Replacing that into  equation A and simplifying, we come out with:

$F_nF_{n+1}-F_nF_{n-1}+F^2_{n-1}=F_{2n-1}$

Again, we repeat the process with the equation $F_{n-1}=F_{n+1}-F_n$, and after we plug it into equation A and simplify again, we come out with:

$F_nF_{n+1}-F_nF_{n+1}+F^2_n+F^2_{n-1}=F_{2n-1}$

We get some terms to cancel out, and wind up with:

$F^2_n+F^2_{n-1}=F_{2n-1}$

Back to Equation B:

We shall continue to do the same thing to equation B as we did to equation A. First of all, we know $F_{n-1}=F_{n+1}-F_n$, so plugging it into the equation and simplifying we get:

$ F_{n+1}F_{n+2}-F_nF_{n+1}+F^2_n=F_{2n+1}$

Repeat the process using the equation $F_n=F_{n+2}-F_{n+1}$. Plugging it in and simplifying (noticing a pattern here?), we get:

$F_{n+1}F_{n+2}-F_{n+1}F_{n+2}+F^2_{n+1}+F^2_n=F_{2n+1}$

Terms cancel out, and we are left with:

$F^2_{n+1}+F^2_n=F_{2n+1}$

Now, going back to equation A, we left off with $F^2_n+F^2_{n-1}=F_{2n-1}$. Replace the original $n$ in that equation with $n+1$, and you will get:

$F^2_{n+1}+F^2_{n}=F_{2(n+1)-1} \rightarrow F^2_{n+1}+F^2_{n}=F_{2n+1}$. This equation is equal to equation B, and thus the formula is true.

It looks like you have the basic idea behind induction down, but there are a couple of things you need to keep in mind when writing induction proofs.

First, you need to start with the base case.  This step is usually trivial, but it's necessary.  If you wrote an inuction solution up for the USAMO, ARML, USAMTS, or the Mandelbrot and didn't include the base case you can be sure you would get points docked.

Here is the way I would rewrite your solution to number 6:

Base case:
We know that $F_1 = F_2 = 1$.

Inductive step:
Assume that $F_1 + F_3 + \cdots + F_{2n+1} = F_{2n+2}$ for some value of $n$.

Adding $F_{2n+3}$ to both sides of this equation we get

\[ F_1 + F_3 + \cdots + F_{2n+3} = F_{2n+2} + F_{2n+3}. \]

By the recursive rule for the Fibonacci sequence we know that $F_{2n+2} + F_{2n+3} = F_{2n+4}$, thus

\[ F_1 + F_3 + \cdots + F_{2n+3} = F_{2n+4}. \]

We can now conclude that for all $n = 1, 2, 3, \ldots$ that

\[ F_1 + F_3 + \cdots + F_{2n+1} = F_{2n+2}. \]

Notice how nice and easy it is to read a proof like this.  Not only have I made the base case and inductive clear, but I made the inductive step constructive using the assumption that the sum we were trying to prove was true for some value of $n$.

Those of you who are interested in learning to solve much harder induction problems and also problems involving invariance, the extremal principle, the pigeonhole principle, and other Olympiad problem solving techniques not usually tested on regional math team tests might consider enrolling in the Olympiad Problem Solving class here.  Learning these techniques is the first step toward achieving a nice score on the USAMO and will also help at ARML and on the Mandelbrot.

There's a cool combinatorial model you can use to prove Fibonacci identities sometimes.  Here's the model:

Let $F_n$ be the number of ways you can climb n steps if you can take strides of 1 or two steps.

Clearly $F_1 = 1$ since you can only take one step.

$F_2 = 2$ since you could go 1+1 or just 2.

$F_3=3$ since you could go 1+1+1 or 2+1 or 1+2.

We have $F_{n-1} + F_n = F_{n+1}$ because to climb n+1 steps, we have two choices: climb n-1 steps (which we can do in $F_{n-1}$ ways) then take a step of size 2 or climb n steps (which we can do in $F_n$ ways) then take a step of size 1.

Try doing 2, 5, and 3 with this method.

(This is pretty tough, but awfully cool when you start to see them.)

How are these?

1. We know that $F_0 = 0$ and $F_1 = F_2 = 1$.

Assume that

 \[ F_n = \frac{\alpha^n - \beta^n}{\sqrt{5}}, \alpha = \frac{1 + \sqrt{5}}{2}, \beta = \frac{1-\sqrt{5}}{2}. \]

By the recursive rule of the Fibonacci sequence, $F_{n} + F_{n+1} = F_{n+2}$. Plugging into the formula:

\[ \frac{\alpha^n - \beta^n}{\sqrt{5}} +  \frac{\alpha^{n+1} - \beta^{n+1}}{\sqrt{5}} =  \frac{\alpha^{n+2} - \beta^{n+2}}{\sqrt{5}}\]

Multiply through by $\sqrt{5}$ and break up the exponents

\[\alpha^n - \beta^n + \alpha^n\alpha - \beta^n\alpha = \alpha^n\alpha^2 - \beta^n\beta^2\]

Plug back the values for $\alpha$ and $\beta$ that aren't raised to $n$

\[\alpha^n - \beta^n + (\frac{1+\sqrt{5}}{2})\alpha^n - (\frac{1-\sqrt{5}}{2})\beta^n = (\frac{3+\sqrt{5}}{2})\alpha^n - (\frac{3-\sqrt{5}}{2})\beta^n\]

Summing the left side:

\[(\frac{3+\sqrt{5}}{2})\alpha^n - (\frac{3-\sqrt{5}}{2})\beta^n = (\frac{3+\sqrt{5}}{2})\alpha^n - (\frac{3-\sqrt{5}}{2})\beta^n\]

gives reflexive equality, so we conclude that for all non-negative integers $n$

\[ F_n = \frac{\alpha^n - \beta^n}{\sqrt{5}}, \alpha = \frac{1 + \sqrt{5}}{2}, \beta = \frac{1-\sqrt{5}}{2}. \]

3. We know that $F_1 = F_2 = 1$.

Assume

\[ \sum_{i=1}^n F_i^2 = F_n F_{n+1}. \]

Then

\[ F_1^2 + F_2^2  + F_3^2 + ... + F_n^2 = F_n F_{n+1}. \]

Increasing $n$ by 1 would result in

\[ F_1^2 + F_2^2 + F_3^2 + ... +  F_n^2 + F_{n+1}^2 = F_{n+1} F_{n+2}. \]

By the Fibonacci recursive rule $ F_{n+2} = F_{n+1} + F_n $, so $F_{n+1}  F_{n+2} = F_{n+1}(F_{n+1} + F_{n}) = F_{n+1}^2 + F_n F_{n+1}$. Subtracting $F_{n+1}^2$ from both sides results in the original equation. We conclude that for all natural numbers $n$

\[ \sum_{i=1}^n F_i^2 = F_n F_{n+1}. \]

Nice job.

Two comments:

On both parts, you should be explicit about the base case (i.e. show the formula working for the base case - note that in the first one you'll need 2 base cases).

Second, while you'll often do your work 'backwards', you'll want to write your solutions forwards, like so:

\begin{eqnarray*}
F_1^2 + F_2^2 + F_3^2 + ... +  F_n^2 + F_{n+1}^2&=& F_nF_{n+1} + F_{n+1}^2\\
&=&F_{n+1}(F_n + F_{n+1})\\
&=&F_{n+1}F_{n+2}
\end{eqnarray*}

Finally, a TeX tip - put \left and \right before your parentheses to make them come out the right size.  For example (\frac{1}{2})  is $(\frac{1}{2})$, but \left(\frac{1}{2}\right) is $\left(\frac{1}{2}\right)$.

So would the base case be just the first few terms of the sequence or a recurssive formula? At least for these Fibonnaci inductions.

Let's say I was using mathematical induction to prove $1+2+3...+n=\frac{(n)(n+1)}{2}$; would I start by saying the base case is $t_1=1, t_{n+1}=t_n+1$, and then continue to prove it normally?

The base case for that example would just be $1 = \frac{1 \cdot 2}{2}$.

After that you can induct on the sum by adding terms.

The base cases for an induction problem just need to be enough that when you jump off of your assumption, you create all possible cases.  Often that just means the very first case, but if your inductive step requires more than one pre-existing case, then you will need as many base cases as it takes to jump from the assumption to the next step.


calculus
integration
analytic geometry
conics
parabola
calculus computations

Find the volume formed by the revolution of the region satisfying $ 0\leq y\leq (x \minus{} p)(q \minus{} x)\ (0 < p < q)$ in the coordinate plane about the $ y$ -axis.

You are not allowed to use the formula: $ V \equal{} \boxed{\int_a^b 2\pi x|f(x)|\ dx\ (a < b)}$ here.

if not cylinders then washers!
$ z\equal{}\left(\frac{q\minus{}p}{2}\right)^2$ is the top height of the parabola at, $ c\equal{}\frac{p\plus{}q}{2}$

$ V\equal{}2\pi \int_0^z x_{out}^2(y)\minus{}x_{in}^2(y) dy$

$ \equal{}2\pi \int_0^z (c\plus{}r(y))^2\minus{}(c\minus{}r(y))^2 dy \equal{}8c\pi \int_0^z r(y)dy$

where $ r(y)$ is the distance from $ x\equal{}c$ to the edge of the parabola at height y

$ y\equal{}\minus{}x^2\plus{}2cx\minus{}pq$ so by completing squares

$ \sqrt{c^2\minus{}pq\minus{}y}\equal{}|x\minus{}c|$ and since $ c^2\minus{}pq\equal{}z$, $ r(y)\equal{}c\minus{}\sqrt{z\minus{}y}$

$ V\equal{}8c\pi \int_0^z c\minus{}\sqrt{z\minus{}y}dy \equal{} 8c\pi [cz\plus{}\frac{2}{3}z^{\frac{3}{2}}]\equal{}\frac{8cz\pi}{3}[3c\plus{}2\sqrt z]$

$ \equal{}\frac{(p\plus{}q)(q\minus{}p)^2\pi}{6}[3(p\plus{}q)\plus{}4(q\minus{}p)]\equal{}\frac{\pi}{6}(q\plus{}p)(q\minus{}p)^2(7q\minus{}p)$

am i off by a multiple of two? because i just realized is $ \pi r^2$ not $ 2\pi r^2$ =/ or is it something else

[quote="kenn4000"]if not cylinders then washers!
$ z \equal{} \left(\frac {q \minus{} p}{2}\right)^2$ is the top height of the parabola at, $ c \equal{} \frac {p \plus{} q}{2}$

$ V \equal{} 2\pi \int_0^z x_{out}^2(y) \minus{} x_{in}^2(y) dy$

$ \equal{} 2\pi \int_0^z (c \plus{} r(y))^2 \minus{} (c \minus{} r(y))^2 dy \equal{} 8c\pi \int_0^z r(y)dy$

where $ r(y)$ is the distance from $ x \equal{} c$ to the edge of the parabola at height y

$ y \equal{} \minus{} x^2 \plus{} 2cx \minus{} pq$ so by completing squares

$ \sqrt {c^2 \minus{} pq \minus{} y} \equal{} |x \minus{} c|$ and since $ c^2 \minus{} pq \equal{} z$, $ r(y) \equal{} c \minus{} \sqrt {z \minus{} y}$[/quote]

$ V\equal{}\pi \int_0^z \{x_{out}^2(y) \minus{} x_{in}^2(y)\} dy$ is correct and $ r(y)$ seems to be incorrect.


I don't know if this problem is correct.
  [i]Problem[/i]
  Let $\ x,y$ be integer numbers such that $\ x^2-xy=p$.Prove that $\ p$ is prime number only and only if we have $\ |y|=p-1$.

Well, cause of $x(x-y)=x^2-xy=p$, we have that $x=\pm 1$, giving the result, or that $x-y= \pm 1$, giving $x=\pm p$, also a solution.
When $p$ is not prime, just take $x$ as any nontrivial divisor and then choose $y$ such that it holds.
Thus no (after some stupid mistkae): its correct.

The problem seems to be correct (and easy  ;) ) if it is interpreted in the following way:

Let $p$ be a natural number.
Prove that $p$ is a prime number if and only if the following holds:
For all integers $x,y$ $x^2 - xy = p$ implies $|y| = p-1$.

If $p=x(x-y)$ where $x,y\in\mathbb{Z}$, then there are only 4 cases:
1) $x=+1$, $x-y=+p$ $\to$ $y=-p+1$.
2) $x=-1$, $x-y=-p$ $\to$ $y=+p-1$.
3) $x=+p$, $x-y=+1$ $\to$ $y=+p-1$.
4) $x=-p$, $x-y=-1$ $\to$ $y=-p+1$.
All have in common $|y|=p-1$.
[Edit: It seems everyone posted their solution while I was writing this :blush: ]

Answering stancuiou sorin's problem:
If $p$ is not prime, it has at least one non-trivial divisor $0<d<\sqrt{p}$ and $x=\frac{p}{d}$, $x-y=d$ (or vice versa) $\to$ $|y|=\frac{p}{d}-d<p-1$

[quote="solyaris"]To ZetaX:
I don't think $x = p$ is a problem, as this implies $p-y = 1$, i. e. $y = p -1$.
Similar for $x = -p$.[/quote]
Yes right, I somehow concluded that $y=1$ then  :blush:


calculus
real analysis
real analysis unsolved

there was recently posted problem like this:
if $ g(g(g(x)))\equal{}x^2\plus{}3x\plus{}4$
find $ g(g(2))$
i would appreciate if u could find me a link. thank you

I don't remember anything like that in the Calculus subforum but I suspect that the right hand side should be $ x^2\color{red}\minus{}\color{black}3x\plus{}4$, which makes the problem fairly simple.


function
algebra
domain
limit
trigonometry
irrational number
real analysis

We had an odd question on our math test the other day.  It was: find a function such that all points in its domain are discontinuous.  And I was thinking about this and couldn't really come up with anything, but the teacher said that this function would work:

$ f(x) \equal{}\begin{cases}0 & x\in\mathbb{Q}\\ 1 & x\not\in\mathbb{Q}\end{cases}$

But then I thought that this doesn't seem like it would work either.  Isn't the density of the irrational numbers greater than the density of the rational numbers?  If so, then it couldn't be discontinuous at every point because some points would have the same value as others...

Anyone care to take a stab at this problem?

[quote="Potato Theory"]Isn't the density of the irrational numbers greater than the density of the rational numbers?[/quote]  There are more irrational numbers than rational numbers is several meaningful senses, and none of them has anything to do with whether or not this function is continuous.  If you want to find out whether this function is continuous, [i]apply the definition[/i] of continuity and see if it holds.  (In this case, it doesn't at any point, i.e. the function is in fact discontinuous everywhere and your professor is correct.)

Could you elaborate a little bit?  I'm having some trouble understanding how to apply the definition of continuity.  I mean, I know that it is: $ f(c) \equal{}\lim_{x\to c}f(x)$, but how do I find the limit of the aforementioned function at a given $ x$?

The usual epsilon-delta definition of continuity is equivalent to a sequential definition:

$ f$ is continuous at $ a$ if and only if for all sequences $ (x_{n})$ in the domain of $ f$ such that $ \lim x_{n}\equal{}a,$ then $ \lim f(x_{n})\equal{}f(a).$

This version is particularly useful for proving that a function is discontinuous: for the given $ a,$ find a particular sequence $ x_{n}\to a$ such that $ f(x_{n})$ does not tend to $ f(a).$

As an example of this: let $ f(x)\equal{}\begin{cases}\sin\frac{1}x,&x\ne0\\ 0,&x\equal{}0\end{cases}.$

How to show that this is not continuous at zero? Try the sequence $ x_{n}\equal{}\frac{1}{2\pi n\plus{}\pi/2}.$ We chose this because $ f(x_{n})\equal{}1$ and hence $ f(x_{n})$ tends to $ 1,$ which is not $ f(0).$ (You'll notice a certain amount of arbitrariness in this choice of sequence; many other similar proofs are possible.)

Now, back to your example. Given a rational numbers, can you find a sequence of irrational numbers that tends to it? Given an irrational number, can you find a sequence of rational numbers that tends to it? If you can answer these questions, you can get what you want.


My school has vacation the week of the AMC 10/12 B, but me and several of my friends want to take it. The list of places that offers the B exam doesn't show any for my state (New Hampshire). Which of the following, if any, are viable options?

Contact a high school that will likely be taking it (Exeter immediately comes to mind) and see if we can do it there.

Contact one of the listed places in a close state, such as Massachusetts, and see if we can take it there.

Contact a local college and ask them to host it.

In my opinion, it would be best to first contact local high schools, then local colleges, then other relatively close high schools. I would assume that a school like Exeter would let you take it (but of course, I'm not sure).

Contacting a local college and asking them to host you is the best among the 3 alternatives.

See also Frequently Asked Question 4.
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions


algebra
polynomial
quadratics
number theory proposed
number theory

Find all polynomials with (let's say...) integer coefficients such that $P(n)P(n+1)$ is of the form $P(k)$ for some integer and this for all $n$.

So, this seems a very beautiful problem and it's been one year since it was posted without any replies.
Harazi could you please give a hint of your solution. I'm sort of confused. all i've been able to prove so far is that the leading coefficient must be a power of the same degree as the polynomial (trivial) which is clearly not enough... (just look at first degree polynomials satisfying this)

Ok, I'll try to remember my approach, it's been a long time from then. And don't worry, practically all problems I posted from quite a long time received no attention. I think however there are some deserving some attention...  :D

Those are my ideas.

Let 's suppose that the leading coefficient of P(x) is positive. It seems to me that we could prove

There exist a constant a>0 and b<0 such that

P(n^2+n +b) <P(n).P(n+1) < P(n^2 + n+ a) for n is sufficiently large

As P(n).P(n+1) has the form P(k) for all n, there must be a constant C such that

P(n).P(n+1)= P(n^2+n+C) for infinitely many n. That yields

P(x).P(x+1)=P(x^2+x+C)

.....

Are you sure?  $ P(x) \equal{} x^2 \minus{} x \plus{} 1$ gives $ P(n) P(n \plus{} 1) \equal{} P(n^2 \plus{} 1)$.

what do you mean. every monic quadratic works for that matter...

Any solution to this beautiful problem? :maybe:

$P:$ is
$\bullet$ any quadratic polynomial with integer coefficients and leading coefficient = $1$.
or
$P(x)=1 \forall x$
or,
$P(x) = cx+d$

Does anyone have a solution to this nice problem? Thanks!

@above, I think I can narrow it down to showing (assuming $P$ is Monic, the other case being woefully technical) $P(X^2+X+c)=P(X)\cdot P(X+1)$ for all $X$ and some integer $c$. If it helps, I would post a proof till here. Let's work together! :)

I think the Polynomial $ P(x)= (x^2 +ax+b)^k $ Satisfies the Statement Of the Problem too, But for What  anantmudgal09 received , i can tell that for every $d$ there is at most one Polynomials of degree $d$ which satisfies the above relation.

Let's solve the problem under the assumption that $P$ is monic. First, as Anant has pointed out, the condition of the problem implies a polynomial identity of the form $P(x)P(x+1) = P(\pm x^2 + ux + v)$ for $u, v$. To formally justify this, denote $d = \deg P$. By matching the coefficients of $x^{2d-1}$ and $x^{2d-2}$, we can find unique rational numbers $u, v$ such that $P(x)P(x+1) - P(x^2 + ux + v) = o(x^{2d-2})$. Let $Q$ be a positive integer with the property that $Qu, Qv \in \mathbb{Z}$, then  $|k - (n^2+un+v)| \geq \frac{1}{Q}$ whenever $k \neq n^2 + un + v$. For $n$ sufficiently large this means that if $k > 0$ and $k \neq n^2 + un + v$, then $P(k) - P(n^2 + un + v) \geq \Omega(n^{2d-2}/Q)$ so that $P(k) \neq P(n)P(n+1)$. Therefore for $n$ sufficiently large, if $P(n)P(n+1) = P(k)$ then either $k = n^2 + un + v$ or $k < 0$. In the former case we obtain the polynomial identity. If $k < 0$ for all large $n$, then $P(x)$ must have even degree. In that case can instead consider $P(x)P(x+1) - P(-x^2 + u'x + v') = o(x^{2d-2})$ and deduce again the polynomial identity.

Now we work with the polynomial identity. First assume $P(x)P(x+1) = P(x^2 + ux + v)$. By directly comparing coefficients, we see that for any degree $d$ the coefficients of $P$ are uniquely determined. Now observe that $Q(x) = (x^2 + (u-1)x + v)^d$ is a candidate solution. Since $P^2(x)$ and $Q(x)$ have the same degree, we must have $P^2(x) = Q(x)$, and so $P(x) = (x^2 + (u-1)x + v)^{\frac{d}{2}}$. If $(u-1)^2 \neq 4v$ then $d$ must be even, otherwise $d$ can be arbitrary. A similar argument shows if $P(x)P(x+1) = P(-x^2 + u'x + v')$, then $P(x) = (x^2 + (-u'-1)x + (u'+1-v'))^k$.

I suspect that if $P$ is not monic, then $P = (cx+d)^k$, but I don't know how to prove that.

Very Nice Proof, Based on what i told in the Previous Post, we can Prove the general statement as:
[b]Lemma[/b]: Let $P,Q,R$ be three polynomials of Positive degree, and Positive leading Coefficients, then  for all $k>0$. There is at most one polynomial f of degree $k$, such that :
                                                                              $ f(P(x) ).f(Q(x) )=f(R(x))$

This lemma can easily being proven, via assuming the contrary and checking degree of Polynomials,
 now back to the problem(i will try to shed more light on what angiland wrote )   let :
                                                                          $P(x)P(x+1) = P(x^2 + ux + v)$
By above lemma, we can easily find the Only even degree Polynomial satisfy the above equation is $P(x) = (x^2 + (u-1)x + v)^d$ for some positive integer $d$, now assume there is an odd degree polynomial $P(x)$ satisfies the above equation, thus, $ P(x)^2$ satisfies the equation too, now the degree of the later polynomial is indeed even thus $P(x)^2 =(x^2 + (u-1)x + v)^k$ for some positive integer $k$ and we find the polynomial. now if the polynomial $x^2 + (u-1)x + v$ has not double root, then $k$ must be even, otherwise ( i.e $(u-1)^2 = 4v$) , $k$  could be arbitrary.

[b]Comment[/b], the first part of the proof of angiland, could derived by deciding both sides of the equation  $P(n)P(n+1)=P(k_n)$ by  $ (n^2 +n)^d$,  and take $n$ sufficiently large, where $d$ is the degree of the Polynomial $P(x)$ . thus  we find that $|\frac{ k_n}{n^2+n}|$tends to  unit, and etc...


[quote="rogue"]For every $a,b,c\ge0$ prove the inequality
$\sqrt{a^4+\frac{b^4}{2}+\frac{c^4}{2}}+\sqrt{b^4+\frac{c^4}{2}+\frac{a^4}{2}}+\sqrt{c^4+\frac{a^4}{2}+\frac{b^4}{2}}\ge\sqrt{a^4+b^3c}+\sqrt{b^4+c^3a}+\sqrt{c^4+a^3b}.$[/quote]
I found this one easy:
allow me to use $\sum$ to make the solution shorter:
$\sum \sqrt{a^4+\frac12(b^4+c^4)}=\frac12\sum \sqrt{a^4+\frac12(b^4+c^4)}+\frac12\sum \sqrt{a^4+\frac12(b^4+c^4)}\geq \frac12\sum (\sqrt{\frac14(a^4+b^4)}+\sqrt{\frac14(a^4+c^4)})+\frac12\sum \sqrt{a^4+b^2c^2}=\sum \sqrt{\frac14(a^4+b^4)}+\frac12\sum \sqrt{a^4+b^2c^2}=\sum (\sqrt{\frac14(a^4+b^4)}+\sqrt{\frac14a^4+\frac14b^2c^2})\geq \sum\sqrt{a^4+(\frac{b^2+bc}{2})^2}$


Why is the critical temperature found when:

$ ( \frac {\partial P}{\partial V})_{T_c} \equal{} (\frac {\partial^{2} P}{\partial V^{2}})_{T_c} \equal{} 0$

Using Van der Waals Equation of State:

$ (P \plus{} \frac {n^{2}a}{V^{2}})(V \minus{} nb) \equal{} nRT$. [06Jan08 <Upon request, typo corrected by gregx007>]

Also is it true that Critical Phenomenon do not arise from the Ideal Gas Equation (hence the need for the modified form)?

Thanks.

There is a typo there: it should be (V - nb). Anyway, considering a PV diagram, the critical temperature corresponds to the lowest isotherm for which the Van der Waals cubic equation has a triple real root, and that point corresponds to an inflexion point.

Well maybe, if you don't mind, run me step-by-step through generating the $ P\minus{}V$ plot and explaining why the Critical temperature occur at the triple root (my Physical Chemistry book doesn't explain this very well).

For a start, let's look at the Van der Waals equation of state: $ \left(P \plus{} \frac{an^2}{V^2}\right)\left(V \minus{} nb\right) \equal{} nRT$. Multiply both sides by $ V^2$ to get $ \left(PV^2 \plus{} an^2\right)\left(V \minus{} nb\right) \equal{} nRTV^2$, and now expand the product on the left:

$ PV^3 \minus{} nbPV^2 \plus{} an^2V \minus{} abn^3 \equal{} nRTV^2$

$ \equal{}> V^3 \minus{} \left(nb \plus{} \frac{nRT}{P}\right)V^2 \plus{} \frac{an^2}{P}V \minus{} \frac{abn^3}{P} \equal{} 0$  (1).

This means that, for given values of n, T, and P, the volume is found by solving a cubic equation. On a PV diagram, with V on the x axis and P on the y axis, and drawing isotherms according to equation (1), one sees that, for many moderate pressures there are three volumes that satisfy (1) - a line drawn parallel to the x axis cuts the isotherm at three spots. This means that equation (1) has 3 real and distinct roots. For high pressures, there is only one real root of (1), but the corresponding isotherms do not have an inflexion point. However, there is exactly one single value of T (and so just one isotherm) for which equation (1) shows three equal real roots for V, and this triple volume corresponds exactly to an inflexion point. Therefore, one must have at that point

$ \left(\frac{\partial P}{\partial V}\right)_T \equal{} 0$ and $ \left(\frac{\partial^2 P}{\partial V^2}\right)_T \equal{} 0$.

Then using these two conditions one can find the relation between the critical properties (T, V, and P) and the Van der Waals constants.

how does one prove the  Van Der Waals equation??? Are you just supposed to believe that it is correct. Are there any lemmas we can start with and then finally get his equation

The van der Waals equation was proposed by a cientist with the same name from simple microscopic arguments. It can, however, be obtained from statistical mechanics.


Hello everyone! Can you guys help me with this quadratic formula, i Need to isolate t:
[code]

0 = ½gt² + vt + s

[/code]

This does not belong here. Use the quadratic formula or "complete the square."


[quote="vishalarul"]How do you solve an equation like $3^{x}= x^{3}$? Is there a general formula for solving:
$n^{x}= x^{n}$
P.S. I did some research on it. It had to do something with Lambert's W-function, but I didn't get it.  :([/quote]
For $3^{x}= x^{3}$, $3$ obviously works. For any equ $n^{x}= x^{n}$, I think the solution is always $n$.
[hide=" :idea: "]Or maybe one of the solutions, though i think n will be the only one.[/hide]
Hope that helps. :)

[quote="vishalarul"]How do you solve an equation like 3^x = x^3? Is there a general formula for solving:
n^x = x^n

P.S. I did some research on it. It had to do something with Lambert's W-function, but I didn't get it.  :([/quote]

this type of problems have been posted earlier several times here... try this...

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81882

and an external link...
http://www.qc.edu.hk/math/Advanced%20Level/y^x_x^y.htm

By the way, there are 2 solutions, x=n, and x=something else. I was looking for the other solution, Inspired By Nature.

[color=darkblue]For fixed $p>0$ the equation $x^{p}=p^{x}\Longleftrightarrow f(x)=f(p)$, where $f(x)=\frac{\ln x}{x}\ .$ This equation has a only solution $x_{1}=p$ iff $p=e$; in opposite case, the equation has two solutions $x_{1}=p$ and $x_{2}\ne p$ so that $p<e\Longleftrightarrow x_{2}>e$ because $0<x<y<e\Longleftrightarrow f(x)<f(y)<\frac{1}{e}$ and $e<x<y\Longleftrightarrow \frac{1}{e}>f(x)>f(y)\ .$[/color]

[quote="vishalarul"]That is NOT CORRECT, calculuslover800! There is also a solution somewhere between 2.4 and 2.5. :mad:[/quote]

Hehe, you're right, sorry  :oops:

The other solution is at about 2.47805

Anyways, is there a formula that allows us to compute the EXACT (not approximate) values of the solutions of $n^{x}= x^{n}$?

I've tried finding the solutions on this website:
http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=mathcom&s1=equations&s2=solve&s3=basic

However, it returned with extremely bizzare answers!

This is what I found for, $x^{3}=3^{x}$,
http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=mathcom&s1=equations&s2=solve&s3=basic

In general, most arbitrary problems don't have a closed-form solution.

Edit:  We can rewrite the equation as

$\frac{\ln n}{n}= \frac{\ln x}{x}$ (can you see why?) and then it is obvious why there are zero, one, or two solutions for any particular $n$.  Then the exact solution depends on finding a closed form for the inverse of $\frac{\ln x}{x}$, which I do not think can be expressed in terms of elementary functions.

I researched and found that Lambert's W function is the inverse of the function
$f(x)=xe^{x}$
 :idea:
I am not sure how that can be applied to finding an inverse of
$\frac{\ln x}{x}$

However, I did make some progress on solving the equation:
Let Lambert's W function be $W(x)$

According to this website:
http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=mathcom&s1=equations&s2=solve&s3=basic

The solution to the equation $n^{x}=x^{n}$ is:

$x=-\frac{nW(-\frac{log(n)}{n})}{log(n)}$

Can anyone prove (or disprove! :cool: ) this?

[quote="Altheman"]uugh...the lambert function is an infinite sum...so it is not an exact solution[/quote]

The second does not follow from the first.

I'm not sure exactly how to translate between $W$ and an inverse to $\frac{\ln x}x$, but I can "almost translate":

Let $H(x)$ be the inverse function to $f(x) = \frac{\ln x}x$.  Then if $H(x) = e^{G(x)}$ we have $x = f(H(x)) = \frac{\ln e^{G(x)}}{e^{G(x)}}= G(x) e^{-G(x)}$ and so $G$ is a function very closely related to $W$.  Probably, we just have to switch some signs in the power-series expansion.

[quote="vishalarul"]The solution to the equation $n^{x}=x^{n}$ is:

$x=-\frac{nW(-\frac{log(n)}{n})}{log(n)}$

Can anyone prove (or disprove! :cool: ) this?[/quote]

I can prove it! I just recently learned about the Lambert W function and I have to say it is pretty sweet!

$n^{x}= x^{n}$

$\Rightarrow e^{\log(n^{x})}= x^{n}$

$\Rightarrow e^{x\log(n)}= x^{n}$

$\Rightarrow e^{x\log(n)\cdot\frac{1}{n}}= x$

$\Rightarrow e^{x\frac{\log(n)}{n}}= x$

$\Rightarrow 1 = xe^{-x\frac{\log(n)}{n}}$

$\Rightarrow \frac{-\log(n)}{n}=-x\frac{\log(n)}{n}e^{-x\frac{\log(n)}{n}}$

We now have the function in the forum $C = f(x)\cdot e^{f(x)}$ which implies that $f(x) = W(C)$
So $-x\frac{\log(n)}{n}= W(\frac{-\log(n)}{n})$

$\Rightarrow x = \frac{W(\frac{-\log(n)}{n})}{\frac{-\log(n)}{n}}$

$\Rightarrow x = \frac{-n W(\frac{-\log(n)}{n}) }{\log(n)}$


function
real analysis
real analysis unsolved

Suppose that $ f: U\rightarrow V$, $ U\subseteq\mathbb{R}^n,\ V\subseteq\mathbb{R}^m$ is a diffeomorphism (with the condition that $ f\in\mathcal{C}^1$).  Prove that $ m\equal{}n$.

I suppose $ U$ is open or it's not true

f is an open map (inverse function theorem). Perhaps some proofs assume that the map already is from R^n to R^n, but the proof of the inverse function theorem goes through for maps between any two Banach spaces

Differentiability that I know of exists over open sets, so the diffeomorphism is supposed to imply that $ U,V$ are open.


conics ellipse FTW percent articles search inequalities

Anyone of you in the totality region? I'm dying to see the eclipse :D

Which eclipse?
And totality,whaz dat? :huh:
I never heard anything :(

@scary
Don't you see TV or read newspapers?There is going to be a total solar eclipse and it will be visible for about 6 minutes,the longest in this century.The next one will be in the year 2114 :P
Totality is the region where the eclipse will be total and not partial.The region is the belt from Gujarat,MP,East UP,Bihar,WB,Arunachal Pradesh.

[quote="_MM_"]Aren't you in the totality region..?

Down here in chennai I can only hope of seeing the eclipse on TV.. :([/quote]
No, I'm in 97-98 percent region I think. But just made an insane plan with my friends to go off to varanasi today itself to see the eclipse!!

How can I?I am havin exams right now,and I am hardly getting 5 mins to visit AoPS. :(
Me in totality!! :D
It's very^1/0 cloudy over here,so my luck's bad I guess!

[quote="Scary math"]Yo!
Saw the eclipse through an X-ray film.t'was cloudy,saw glimpses,t'was pretty good!!!!!! :D
U guys saw?[/quote]
Are you nuts!!! You're NOT supposed to do that!

Ummmmm..................
Why?Everyone advised me so,since I had nothing else near at hand :P
And besides,what nut am I?Groundnut,coconut,Brazil nut,pea nut................. :rotfl:

[quote="Scary math"]pea nut................. :rotfl:[/quote]
You're a pee nut.

@ scary - dude i think sagar was right about you ...  :mad:
acting as if you dint know about the eclipse
anyway , you arent suppsed to look at the sun with an x ray

The eclipse sucks, didn't wait for me to wake up.

@Franky
Man I really didnt know.
After seeing this thread,I asked dad,only then I came to know the timings and all
Now,just tell me,why should I be pretending?
Had I known all that,would have I watched through a X-ray film? :mad:
I just came to know in the nick of time :mad:

@ernie
U r a democrat,rite?(seems from ur avatar) :rotfl:
And more news,guess what willl be the first elimination test in Rakhi ka sywamvar 2?
[hide]All rakhi fan gadhas who can consume cyanide for the sake of Rakhi(as a symbol of noble sacrifice) will be promoted to next round :rotfl: [/hide]

[quote="ernie"]Heh, I missed the eclipse.  I live in the US, and went to Gujarat (Ahmedabad) for 2 weeks, only I arrived on the 25th. :P[/quote]
Thanks for reminding us about the original purpose of this thread :P

When is the next one??????Any idea?
Dis thread must lie dormant till then,if u go by rules............
[hide]or till the airing of Swayamvar 2 :rotfl: [/hide]

January 15th, 2010 is the next solar eclipse. :P  And I'm not Democrat.  I prefer to call myself an Independent. :P

[quote="nuclear_alchemist"][quote="Scary math"]pioneering a new party,eh?good idea
dat eclipse will be visible from here? :huh:[/quote]
not really..[/quote]

Actually, it will be visible from most of India.  Not as long as the last eclipse, but it will be visible. :)  Stop hogging all the eclipses, India, America wants to see some too! :(

It will be a very partial eclipse even in the short duration.

[quote="Pocha"]pioneering a new party,eh?good idea[/quote]

Not a new party, it's one that just votes for the best candidate. :P

Yeah, the way we're going with choosing presidents, we'll become a third-world country by 2020...

Take it easy friend,we'll be(s far as my poor Bengal is concerned :( ) 5th world countries by then,u'll still have the upperhand. :rotfl:

are u referring to the annular eclipse?
if so yes i saw it and photographed it too. but it wasn't very good from kolkata ,


Let $ ABCD$ be a circumscribed quadrilateral of $ \odot O$.   $ AB,DC$and$ AD,BC$meet at $ P$and$ Q$ respectively.   There are two points $ M$,$ N$ in the two segments $ OB, OD$ in order.   Show that $ \angle MPN\equal{}\frac {1}{2}\angle APD$ if and only if $ \angle MQN \equal{} \frac {1}{2}\angle AQB$.

Apparently the same thing with [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=512703#512703[/url].


I have real 2x2 invertible matrices (K), and L is a subset of symmetric matrices, such that L=

 \[ \begin{pmatrix}a & b \\ b & c\end{pmatrix}\]

$ \in K: 1\le a\le b \le c \le 10$

For A, B $ \in L$ I need to define the relation $ \sim$ on L by A $ \sim$ B if A*(B^-1) is diagonal and show that $ \sim$ is an equivalence on L.

I need to prove that it is reflexive, symmetric and transitive. That's where I am stuck. Any help would be appreciated! Thanx.

I have found the diagonal matrix A*B(^-1) in a, b and c terms and have shown that the relation is reflexive by multiplying A*B(^-1) by the identity matrix and identity inverse which = A*B(^-1). I now need to show that it's symmetric (it's clear from the matrix, but I need to prove it algebraically) and transitive which I am not sure how to do.

Reflexivity means $ AA^{\minus{}1} \equal{} I$ is diagonal, which is clear (you don't seem to have said quite the right thing).

Symmetry means $ AB^{ \minus{} 1} \text{ is diagonal} \Leftrightarrow BA^{ \minus{} 1} \text{ is diagonal}$.  But $ BA^{ \minus{} 1} \equal{} (AB^{ \minus{} 1})^{ \minus{} 1}$, so this is clear.

Transitivity means $ AB^{ \minus{} 1} \text{ is diagonal}, BC^{ \minus{} 1} \text{ is diagonal} \implies AC^{ \minus{} 1} \text{ is diagonal}$.  But $ AC^{ \minus{} 1} \equal{} (AB^{ \minus{} 1})(BC^{ \minus{} 1})$, so this is again clear.

Hi thanks!

I have one more question. I need to find all B matrices so that

B*A^-1 are diagonal

Both A and B are real 2x2 invertible symmetric matrices.

A =
\[ \left[ \begin{array}{ c c } 1 & 2 \\
2 & 6 \end{array} \right]
\]
so,

A^-1 =
\[ \left[ \begin{array}{ c c } 3 & \minus{} 1 \\
\minus{} 1 & 1/2 \end{array} \right]
\]
So, when I multiply

A^-1
\[ \left[ \begin{array}{ c c } 3 & \minus{} 1 \\
\minus{} 1 & 1/2 \end{array} \right]
\]
by B
\[ \left[ \begin{array}{ c c } a & b \\
b & c \end{array} \right]
\]
= diagonal matrix
\[ \left[ \begin{array}{ c c } a & 0 \\
0 & b \end{array} \right]
\]
I cannot solve it as all terms cancel each other.

It does not work either if I put
\[ \left[ \begin{array}{ c c } a & 0 \\
0 & c \end{array} \right]
\]
as a diagonal matrix.

Is there any other way of solving it?

There should be several B matrices in the solution.

The above restrictions for a,b,c apply.

Thank you!

Your diagonal matrix should be something like $ \left(
\begin{array}{cc}
x&0\\
0&y\\
\end{array}\right)$. There's no reason why the diagonal matrix should have entries matching those of $ B$.


A wooden plank is $ 1\frac 12$ inches thick. To make the plank
$ 1\frac 18$ inches thick, a carpenter removes the same thickness of wood from the top and bottom of the plank. How many inches
does the carpenter remove from the top of the plank? Express
your answer as a common fraction.

$ 3/2\minus{}9/8\equal{}3/8$

Dividing by 2 gives you $ 3/16$

The total fraction of inches of wood the carpenter removes is $ \frac{3}{8}$. Since the carpenter removed the same amount from the top and bottom, $ \frac{3}{16}$ inches of wood was removed from the top and bottom.

@Pi-Dimension

You can write fractions like this: [dollar sign]\frac{a}{b}[dollar sign], which will look like this:

$ \frac{a}{b}$

 :)


I have an unfair coin that lands as head with probability of $ \dfrac{2}{3}$. If I flip the coin 5 times, what is the probability that I get exactly two heads?


graph theory
combinatorics unsolved
combinatorics

Let G be a k-regular, k-edge-colored, bipartite graph without multiple edges or loops.
Is it true that the number of vertices of G is at least [i]2^k[/i]. Why?

Isn't the complete bipartite graph $K_{3,3}$ an obvious counterexample for $k=3$? :?


How does one get quotes to turn the right way? For instance, in $\text{\"this is some text\"}$ I want that first quote to turn the other way.

Use the backwards apostrophe (which on most keyboards is at the extreme upper-left):

$\text{``Correct quotes''}$


a rocketship is receding from the earth at a speed of 1/5th the velocity of light. A light in the rocket appears blue to the passengers in the rocket . What colour would it appear to an observer on earth?

not sure whether this is even physics or not!

It should be shifted. Wavelegth of the light as observed from Earth will be $\lambda=\lambda_o\sqrt{\frac{1+v/c}{1-v/c}}$. If the original wavelength was 440 nm (typical blue), new wavelength at $v=1/5\,c$ will be approx. 540 nm which is green.

the options are

a         yellow
b          blue
c          yellowish orange
d          orange

Bus,ho did ya come to that solution??Do u need a particular solution??

I'd say blue.. :blush: (I dont know why)

FOr god sakes eminem,why have u written missisipee spelling bee winner in ur signature!!!

Perhaps 440 nm wasn't such a good guess. I looked on the internet for some resources on color & wavelength and I think that 465 nm is about the bluest blue you can get. In that case, you come to a result of 570 nm which is a little greenish but mostly yellow color.

[quote="Rushil"]I think Bus's solutions is absolutely correct. It uses a formula which comes from Einsteins's Special relativity(Really easy to derive if you think logically!!) Anyways, the formula is correct( I haven't verified - may be I'll correct if wrong). If Bus's calc. is correct, the answer should be green.

Hey guys, where did you get this relativity ques. in IIT prep. It's not in syllabus unless it's a trick question[/quote]

Rushil,how about u tell us some nice sources to develop our basics in physics,cause ur way more intelligent than us .It would really help us. :lol:

[quote="eminem"]soorry  bus, ya solution is not even an option ...................[/quote]Nevertheless, bus's solution is correct.

xtremely sorry bus, if every one is saying your solution is correct then it must be . I just said that it might be wrong cos it wasnt even a option .....

Heeeeeeeeeeeeeeyyy    Rushil Do YOU REAlly mean that this question is IIT worth ??? Boy ,its from a really lousy source ,( ummm an  ok source ) i cant really believe its IIT standard...

i'll tell you the source whenever SHADYSAYSURSPAMMED isnt around..(hehehehehe)

No, don't be sorry. I think that the light would actually really look more yellow than green. And they can't put options like "green", "greenish yellow" and "yellow" in there because then it'd be really hard to tell which one is correct. Now you know that orange and yellow-orange are probably wrong and that the correct answer should be yellow.

[quote="eminem"]hey bus i think you got it now! The answer IS yellow!!! coooooooll soln!

but hey i cant get how such a question can be there , i mean its a stupid 10 th grade book!!! I mean is it possible that you can somehow get the answer logically without applying ne formula or something???[/quote]Hm, I don't think so. You definitely need to use that formula. It weird, really... Or perhaps it's an experimental problem :) .

no probs thx for the soln, ne way . I was stuck on it for weeks.........


combinatorial geometry
combinatorics solved
combinatorics

Given 2n+3 points in the plane, no three on a line and no four on a circle, prove that it is always possible to find a circle C that goes through three of the given points and splits the other 2n in half, that is, has n on the inside and n on the outside.

Hm ... just a drawing actually. Consider two adjacent points, say A and B, of the convex hull. Consider the circles through A, B and one of the remaining points. You will get 2001 circles. In the "first" circle there won't be any point ; in the second one there will be precisely one point ; ... in circle 1001 there will be 1000 points. The other 1000 points lie outside this circle (no four concyclic). We are done.  :)

hmm and where did you find out that  n = 1001 ? :D:D:D

Difficult to explain. Just imagine those 2001 circles and you'll see it.

no I ment ... orl was talking about 2n+3 points. but you showed up with a solution for 2001 points! :D

Let A and B be two consecutive vertices of the convex hull of the 2n+3 points. Since no three points are colinear, the 2n+1 other points are on the same side of the line (AB).
Since no four are concyclic, we may label the remaining points P_1,...,P_(2n+1), such that <AP_iB  \leq <AP_(i+1)B for each i.
Thus, the circle through A,B and P_(n+1) gives a solution of the problem.

Pierre.

Indeed, that's what I wanted to say ... I must be very confused yesterday.
Sorry :)


linear algebra
matrix
linear algebra unsolved

$ A=\left[ \begin{array}{cc}\frac{\sqrt3}/2+1&\frac{-5}2\\frac12&\frac{\sqrt3}2-1\end{array}\right]$
find $ A^{2002}$ :D

$ \varphi _{{A}} \left( t \right) = 1 - \sqrt {3}t + {t}^{2}$ (determinant of matrix (A-t))
or $ \varphi _{{A}} \left( t \right) = \left( t - 1/2\,\sqrt {3} - 1/2\,i \right) \left( t - 1/2\,\sqrt {3} + 1/2\,i \right)$

then $ \lambda_{{1}} = 1/2\,\sqrt {3} + 1/2\,i$ and $ \lambda_{{2}} = 1/2\,\sqrt {3} - 1/2\,i$

Jordan matrix : $ T = \left[ \begin {array}{cc} 1/2\,\sqrt {3} + 1/2\,i & 0 \\
\noalign{\medskip}0 & 1/2\,\sqrt {3} - 1/2\,i\end {array} \right]$
 we can easy find matrix $ \ B$ and ${ \ B}^{ - 1}$ such that
${ \ B}^{ - 1}TB = A$
so we get :
$ \ B = \left[ \begin {array}{cc} 1/5 + 2/5\,i & - i \\
\noalign{\medskip} - 1 - 1/2\, i & 5/2\end {array} \right]$  and $ \ {B}^{ - 1} = \left[ \begin {array}{cc} 5/2 & i \\
\noalign{\medskip}1 + 1/2\,i & 1/5 + 2/5\,i\end {array} \right]$

then
$ \ A^{2002} = B^{ - 1}(T^{2002})B$
now we simply find
$ \ {T}^{2002} = \left[ \begin {array}{cc} \left( 1/2\,\sqrt {3} + 1/2\,i \right) ^{2002} & 0 \\
\noalign{\medskip}0 & \left( 1/2\,\sqrt {3} - 1/2\,i \right) ^{2002}\end {array} \right] = \left[ \begin {array}{cc} 1/2 - 1/2\,i\sqrt {3} & 0 \\
\noalign{\medskip}0 & 1/2 + 1/2\,i\sqrt {3}\end {array} \right] \rm \: \Rightarrow \:$
and  we get answer
$ \ A^{2002} = \left[ \begin {array}{cc} 1/2 + \sqrt {3} & - 5/2\,\sqrt {3} \\
\noalign{\medskip}1/2\,\sqrt {3} & 1/2 - \sqrt {3}\end {array} \right]$
Sorry for my language  :blush:

we have $ e^{\frac{i \pi}6}$ and $ e^{\frac{-i \pi}6}$ are eigenvalues of A so i think about this:
$ A^6= \left[ \begin{array} {cc}-1 & 0\\0&-1 \end{array} \right]$
do you agree with me loup blanc?

We can write $ A$ as $ A\equal{}R\left(\frac{\pi}{6}\right)\plus{}B$ such that $ B^4\equal{}I$.

From $ A^2\equal{}\sqrt{3}A\minus{}I,\ A^4\equal{}\sqrt{3}A\minus{}2I$ we have $ A^6\equal{}\minus{}I$, which is the idea of loup blanc, yielding $ A^{12}\equal{}I$.

$ 2002\equal{}12\cdot 166\plus{}10\Longrightarrow A^{2002}\equal{}A^{10}\equal{}A^6\cdot A^4\equal{}\minus{}A^4\equal{}2I\minus{}\sqrt{3}A$.


Consider two equilateral triangles ABC and QPR . The smallest triangle QPR is inside the triangle ABC . The following conditions are satisfied:
1) QP is parallel to AB
2) QR is parallel to AC
3) PR is parallel to BC
4) AQ=CR=BP and they bissect their respective angles;
5) The four interior regions AQPB, AQRC, BPRC, and QPR all have the same area and AB= 12 cm.
What is the number of centimeters in BP ? Give your answer to the nearest hundredth.

This question makes no sense whatsoever to me, and my answer is completely different from that of the source from which I obtained this.  Can anyone provide insight to this problem?

I believe the triangles in question are arranged as follows:

[asy]pathpen = black; pointpen = black;
real r = 2*3^0.5;
pair O=(0,0),Q=r*expi(-pi/6),P=(0,r),R=r*expi(7*pi/6);
pair A=2*r*expi(-pi/6),B=(0,2*r),C=2*r*expi(7*pi/6);
D(Q--P--R--cycle); D(A--B--C--cycle); D(Q--A); D(P--B); D(C--R); D(P--O--Q,dashed);
MP("12",(A+B)/2,NE); MP("A",A);MP("B",B,NW);MP("C",C);MP("C",C);MP("P",P,NW);MP("Q",Q);MP("R",R);[/asy]

I agree though that the wording is unnecessarily lengthy though.

[hide]
Let $ O$ be the common centers of the triangles. Since $ \frac {[\triangle QOP]}{[\triangle AOB]} \equal{} \frac 14$ and $ \triangle QOP \sim \triangle AOB$, by side squared area ratios we have that $ OP \equal{} \frac 12 OB$. By $ 30 \minus{} 60 \minus{} 90$ triangles, $ BP \equal{} OP \equal{} 2\sqrt {3} \approx \boxed{3.46}$.
[/hide]

[hide="My solution (a bit longer)"]Draw an altitude from vertex $ B$ to the midpoint of $ \overline{AC}$.  This altitude has a length of $ 6\sqrt{3}$.  This makes the area of triangle $ ABC$ $ 36\sqrt{3}$, and the areas of the four interior regions is $ 9\sqrt{3}$.  Knowing that the area of triangle $ PQR$ is $ 9\sqrt{3}$, we can determine the height to be $ 3\sqrt{3}$, and the side length to be $ 6$.  From this information, we can determine the height of the trapezoidal regions to be $ \sqrt{3}$ by solving the equation $ .5\cdot(6+12)\cdot{h}=9\sqrt{3}$.  The length of $ \overline{BP}$ is then $ 6\sqrt{3}-3\sqrt{3}-\sqrt{3}=2\sqrt{3}$.[/hide]


$ (\forall)a,b,c\in (0,\infty)$ show that :

$ \sqrt{\frac{ab}{ab\plus{}c(a\plus{}b\plus{}c)}}\plus{}\sqrt{\frac{bc}{bc\plus{}a(a\plus{}b\plus{}c)}}\plus{}\sqrt{\frac{ca}{ca\plus{}b(a\plus{}b\plus{}c)}}\le\frac{3}{2}$

the original inequality can be rewritten as follows
$ \sum_{cyc}{\sqrt{\frac{ab}{(c+a)(c+b)}} \le \sum_{cyc}{\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{b+c} \right)}}$
Likewise, we also get the same evaluations. Then, summing them up altogether, we get the desired result!

Let $a,b,c>0 ,a+b+c=1.$ Prove that$$\frac{a}{\sqrt{(c+a)(a+b)}}+\frac{b}{\sqrt{(a+b)(b+c)}}+\frac{c}{\sqrt{(b+c)(c+a)}}\le \frac {3}{2}.$$
German 2001?
[hide=Where?]
$$\sum \frac{a}{\sqrt{(c+a)(a+b)}}\leq \frac{1}{2}\sum \left(\frac{a}{c+a}+\frac{a}{a+b}\right)= \frac {3}{2}.$$[/hide]

[hide=Just][quote=arqady][quote=sqing]Let $a,b,c>0 ,a+b+c=1.$ Prove that$$\frac{a}{\sqrt{(c+a)(a+b)}}+\frac{b}{\sqrt{(a+b)(b+c)}}+\frac{c}{\sqrt{(b+c)(c+a)}}\le \frac {3}{2}.$$
German 2001?
Where?[/quote]

sqing, it's just $R\geq2r$, which is known 800 years at least.[/quote]
[/hide]
Let $ a, b, c$ be positive real numbers so that $ abc = 1$. Prove that
\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]
...


Find all $ a,b,c \in \mathbb{N}, a \neq 0$ such that:  $ \overline {abc} \equal{} a^3 \plus{} b^3 \plus{} c^3$

abc=a^3+b^3+c^3
-2abc=a+b+c(a^2+b^2+c^2)
so the factors on d right can b -2,abc
-1,2abc
a,-2bc
b,-2ab
c,-2bc

like dat u cn work out and get ur ans

No I believe on the left-hand side is the 3-digit number abc. I know for a fact 153 works, but I can't think about how to go about a solution.

All possible $ a$ are $ 0,1,2,3,4,5,6,7,8,9$, whose $ \bmod9$ residues are

$ 0,1,2,3,4,5,6,7,8,0$, respectively.

All possible cubes $ a^3$ are $ 0,1,8,27,64,125,216,343,512,729$, whose $ \bmod9$ residues are

$ 0,1,8,0,1,8,0,1,8,0$, in order.

$ a^3 \plus{} b^3 \plus{} c^3 \equal{} \overline{abc} \equal{} 100a \plus{} 10b \plus{} c$

$ \implies a^3 \plus{} b^3 \plus{} c^3\equiv a \plus{} b \plus{} c\pmod9$.

Since $ a^3,b^3,c^3\equiv\{0,\pm1\}\pmod9$, we can have these residues be
[hide="{0,0,0}"]
This means that $ a,b,c$ must be from the set $ \{0,3,6,9\}$.

Testing, we see that there are no solutions for this case.
[/hide]
[hide="{0,0,1}"]
This means that one of $ a,b,c$ must be from the set $ \{1,4,7\}$ and the others must be from $ \{0,3,6,9\}$.

Testing, we find that only $ 370$ works.
[/hide]
[hide="{0,0,8}"]
This means that one of $ a,b,c$ must be from the set $ \{2,5,8\}$ and the others must be from $ \{0,3,6,9\}$.

Testing, we find that there are no solutions for this case.
[/hide]
[hide="{0,1,1}"]
This means that one of $ a,b,c$ must be from the set $ \{0,3,6,9\}$ and the others must be from $ \{1,4,7\}$.

Testing, we find that $ 371$, $ 407$ work.
[/hide]
[hide="{0,1,8}"]
This means that there must be one of $ a,b,c$ from each of $ \{0,3,6,9\}$, $ \{1,4,7\}$, and $ \{2,5,8\}$.

Testing, we find that $ 153$ works.
[/hide]
[hide="{0,8,8}"]
This means that one of $ a,b,c$ must be from the set $ \{0,3,6,9\}$ and the others must be from $ \{2,5,8\}$.

Testing, we find that there are no solutions in this case.
[/hide]
[hide="{1,1,1}"]
This means that $ a,b,c$ must be from the set $ \{1,4,7\}$.

Testing, we see that there are no solutions for this case.
[/hide]
[hide="{1,1,8}"]
This means that one of $ a,b,c$ must be from the set $ \{2,5,8\}$ and the others must be from $ \{1,4,7\}$.

Testing, we find that there are no solutions in this case.
[/hide]
[hide="{1,8,8}"]
This means that one of $ a,b,c$ must be from the set $ \{1,4,7\}$ and the others must be from $ \{2,5,8\}$.

Testing, we find that there are no solutions in this case.
[/hide]
[hide="{8,8,8}"]
This means that $ a,b,c$ must be from the set $ \{2,5,8\}$.

Testing, we see that there are no solutions for this case.
[/hide]

Thus, the only solutions are $ \boxed{(a,b,c) \equal{} \{(1,5,3),(3,7,0),(3,7,1),(4,0,7)\}}$.

How about $ \overline {abcd} \equal{} a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4$ ?


Let split a  triangle into $ 98000$ polygonals such that we can't draw any straight line that it  intersecs  more $ 31$ polygonals  in them,(intersecing is mean having a common point)

Every body,,it is a hard problem,i really need your help,but i dont think no one here can solve it
So.hungrry up.post your help for me.please!!!!!!!!!!!!!!

I see no one can help me,ny problem is too hard for every body :lol:

[quote="Pul de Algodoncito"]ok ok ok... relax

Split a triangle into 98000 polygons such that we can't draw any straight line without intersecting 31 of them... is that what you wanted to say?

if so:

- you want to show that any line that crosses the triangle must intersect 31 of them?
- is the problem about a possible construction (we can do it or we can't do it) of 98 000 polygons such that we can't draw any straight line without intersecting 31 of them?
- is there a condition missing? anything about the line or the polygons? are they convex? doesn't matter?

Please, could you try to make the problem clear?[/quote]

No,no,thank your care,(you are the first people reply my topic)
But,My problem is full,98 000 polygons can be a convex or not,I dont miss any condition ,Her is a problem in our test(to select student )
Can you help me :(


[size=200]s h o w + m e + t h e = m o n e y[/size]

I'm guessing that if two letters are the same, they have the same value, and the letters represent digits.

[hide="part way"]We can write it this way:

$1000s+100h+10o+w+10m+e+100t+10h+e=10000m+1000o+100n+10e+y$

$1000s+110h+10o+w+10m+2e+100t=10000m+1000o+100n+10e+y$

$100s+110h+w+100t=99990m+990o+100n+8e+y$

s, t, and m may not be equal to 0.[/hide]

I need more information than this.

I think this is more logic than mathematic.
Because of the carry values, I'd say $m=1$ and $s=9$.