geometry
area of a triangle
geometry unsolved

To each side a of a convex polygon we assign the maximum area of a triangle contained
in the polygon and having a as one of its sides. Show that the sum of the areas assigned to all
sides of the polygon is not less than twice the area of the polygon.

Actually, IMO 2006/6. See here: [url]http://www.mathlinks.ro/viewtopic.php?p=572824#p572824[/url]


Prove that

$\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2} \geq 2$

where x,y,z are real numbers > -1

[quote="darij grinberg"]http://www.mathlinks.ro/Forum/viewtopic.php?t=6150

  darij[/quote]
Oh, sorry for the repost.


I would be, but my skills/experience is kinda limited.  Does anybody have any chips?

[quote="pilot"]Is anyone going to Mathcamp interested in playing Texas Hold'em?[/quote]

Absolutely.

As far as chips go, we can just use spare decks of cards if nothing works out. Or pennies! Otherwise, they're probably about 10 dollars for 250. We can all chip in.

[quote="DPatrick"][quote="MithsApprentice"]...We can all chip in.[/quote]Pun intended?  :rotfl:[/quote]

Indeed.  :lol: I didn't know how many people would get it. :)

Last year there was a bunch of people who played alot. (and I mean alot  :D ). Of course, everything is differnt from year to year. I for one will play a normal game on occasion, but never anything for real money  :)

But serously, your better off spening your time playing mao  :D

[quote="salmononpi"]Last year there was a bunch of people who played alot[/quote]
I won alot, and ordered/ate pizza alot as well.

[quote="Geehoon"]ordered/ate pizza alot[/quote]

Oh yes, good food is of the utmost importance. Someone (i.e. everyone) should acquire some phone numbers for food places as soon as we get there.

we ordered pizza (from domino's) almost every night.
There was 5-5-5 deal just on the commercials.

Later, they even recognized my voice(because there arent that many people ordering pizza at 11PM there(Maine)) :rotfl:

Does anyone have chips they can bring?

Otherwise we can all pitch in and buy a set as Mith suggested, but I doubt 10 dollars is enough.

Actually I just saw this and I wanted to mention that I am bringing crappy plastic chips, of 3 colors.

It is enough to play about 12 people tournament, can be stretched to maybe 16.


geometry
quadratics
probability
algebra
polynomial
AMC 10
AMC

Although the 'math culture' idea did not work out as planned, I have not entirely given up in that I am talking to several non-trivial friends of mine and recommending them the AoPS books.  I figured volume 1 would be more useful to a high schooler than the introductory books since volume 1 seems to be a curricular supplement whereas the introductory books are curriculum replacements, and they have already endured much of the standard curriculum.  They will be taking the AMC 10 next year, and I was wondering which chapters of AoPS 1 would be most important (I could do this myself, but I don't own AoPS 1--only AoPS 2).

Would it be something like:

Most important: [b]25[/b]
Very important: 9-18 (the geometry chapters, excepting shifts, turns, flips, whatev)
Important: 1,3-5,7, 26
Not very important: everything else?

I would appreciate advice on this matter from not only the administrators but anyone who has knowledge as to the content of AoPS 1 (or the new books).

I think the most important chapters are 5, 24-26 (Integers, Sequences and Series, Counting, and Probability).  In my opinion, these are the subjects that are very important in math contests but are not tought much in school curriculum.

Chapters 9-18, 20 (Chapter 20 is just a bunch of geometry problems) are also very important, but your friends might find that they already know much of this material if they have already taken a geometry class in school (That's what happened to me :) ).

Chapters 1, 3-4, 6-7 are also important, but I'm sure your friends already know some, if not most of this material, especially the stuff about linear and quadratic equations.

So I would say

Most Important: 5, 24-26
Very Important: 9-18, 20
Important: 1, 3-4, 6-7

For me, the geometry chapters were the most important.  I had learned many of the theorems and such presented in those chapters but I had no idea how to apply them to the end of the chapter problems.  I don't own a copy of Volume 1 either, and that's about all I can remember.

Though it does greatly depend on the year, I would say the geometry chapters are VERY useful.  Some years, they have 3-4 tougher geometry problems.

Counting and probability is important, but not AS important as the algebra, try the word problems chapter.

[quote="Karth"]What about volume 2?[/quote]

What do you mean?  It depends on what you're preparing for.  Volume 2 wouldn't be very helpful in preparing for the AMC 10.

hi david aops is cool i am jumping through cool parts of aops2 the limits part is very nice. thx.

For volume two, definitely the stuff on Algebraic Expressions, Polynomials, Counting tricks, and Number Theory was amazing.


If i have to compare two equations that involve the addition of two fractions to see which is larger, how would I do that with out having to get a common denominator and putting it into 1 fraction. For example, such as 1/2 + 1/7 vs 1/3 + 6/19. Is there any way else?

Well, you could estimate.  $\frac{1}{2}= 0.5, \frac{1}{7}\approx 0.14$, so their sum is $\approx 0.64$ whereas $\frac{1}{3}\approx 0.33, \frac{6}{19}\approx \frac{6}{18}\approx 0.33$ so their sum is $\approx 0.66$.

This isn't totally accurate, though, so I wouldn't trust it.

Usually, the best way is to combine fractions... it is not so bad!
The only way to get around this is:
a. Prove an inequality about fractions or at least notice a pattern and generalize
b. Use a well known inequality that happens to apply to your fractions

Of course, this requires a lot of insight and doesn't work a good portion of the time, theres no issue with just combing

well, the thing I posted above was an example, but what if if the fractions were 4002/4001 + 3999/4000 vs. 3998/3999 + 3999/4000
At first I would have done the multiplying... but my bad handwriting  confused me so I just guessed and got it wrong... It seemed really annoying to just multiply it out with all the 0's.. So is that the only way?... I thought that there was going to be a more elegant method.. but i dunno..

[quote="turtlecloud"]well, the thing I posted above was an example, but what if if the fractions were 4002/4001 + 3999/4000 vs. 3998/3999 + 3999/4000[/quote]

You could've just posted this problem :P

This is how I would do it:  Let $x = 4000$ and rewrite the two expressions as

$1+\frac{1}{x+1}+1-\frac{1}{x}= 2-\frac{1}{x(x+1)}$

And

$1-\frac{1}{x-1}+1-\frac{1}{x}= 2-\frac{2x-1}{x(x-1)}$

Edit:  The problem's a little too easy this way.  Did you mean, say, $\frac{3998}{3999}+\frac{4001}{4000}$ on the other side?  That would be interesting.  In the same vein, this is

$1-\frac{1}{x-1}+1-\frac{1}{x}= 2-\frac{1}{x(x-1)}$

Which is [b]very[/b] close to (but slightly larger than) the other expression.


Determine all triples of positive real numbers $ (x,y,z)$ such that:

$ x\plus{}y\plus{}z\equal{}6,$
$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}2\minus{}\frac{4}{xyz}.$

$ x,y,z\not \equal{}0,\frac 12$. Let $ x\equal{}\frac 12$, then $ y\plus{}z\equal{}11,5$ and $ \frac 1y \plus{}\frac 1z \equal{}\minus{}\frac{8}{yz}\to y\plus{}z\equal{}\minus{}8$ - contradition.
Let $ xyz\equal{}a$, then $ xy\plus{}yz\plus{}zx\equal{}2a\minus{}4$, therefore $ x,y,z$ are roots of $ t^3\minus{}6t^2\plus{}(2a\minus{}4)t\minus{}a\equal{}0$.
We can chose one of roots $ x\not \equal{}0,\frac 12$, define $ a\equal{}\frac{x^3\minus{}6x^2\minus{}4x}{1\minus{}2x}$ and find $ y,z$ from $ y\plus{}z\equal{}6\minus{}x,yz\equal{}\frac ax\equal{}\frac{x^2\minus{}6x\equal{}4}{1\minus{}2x}\equal{}b$ as roots
$ t^2\minus{}(6\minus{}x)t\plus{}b\equal{}0$. For reality roots, must be $ D\equal{}(6\minus{}x)^2\minus{}4b\ge 0$.

[quote="moldovan"]Determine all triples of positive real numbers $ (x,y,z)$ such that:

$ x \plus{} y \plus{} z \equal{} 6,$
$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 2 \minus{} \frac {4}{xyz}.$[/quote]
Apply AMGM inequality we have $ xyz\le 8$.
The CS inequality in Engel form gives us $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\ge \frac {9}{6}$.
Thus $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \plus{} \frac {4}{xyz}\ge 2$.
The equality occurs if only if $ x \equal{} y \equal{} z\equal{}2$.It is the unique solution.

Yes, I forget, that x,y,z all are positive.


geometry
geometric transformation
reflection
incenter
symmetry
trigonometry
angle bisector

Hi!, How are you? ;)

I ask for its aid with the following problem:

In the $\bigtriangleup ABC$ triangle, $AB=AC$ and $<BAC=20$°. Let $D$ be a point on segment $AB$ such that $BD = BC$ and let $E$ be a point on segment $AC$ such that $<EBC=60$°. Let $F$ be a point of intersection of segment $BE$ with $DC$.

Show that $\bigtriangleup DEF\sim\bigtriangleup CED$

Beforehand I thank for its aid.

Greetings! :lol:

Reflect $D$ through $AC$ to $D'$. Since $\angle{ACD}=30^o$, so $\triangle{DD'C}$ is an equilateral triangle. Hence: $D'D=D'C$. But we also have: $BD=BC$, so $D'DBC$ is a kite, so $D'B$ bisects $\angle{ABC}$. Thus: $\angle{ABD'}=40^o$. Since $\angle{ABE}=20^o$, $\angle{EBD'}=40^o-20^o=20^o$, so $BE$ is the angle bisector of $\angle{ABD'}$. Remember we also have $AE$ bisects $BAD'$, so $E$ is the incentre of $\triangle{ABD'}$. Therefore $\angle{ADE}=\angle{AD'E}=\frac{1}{2}\angle{AD'B}=50^o$. Hence $\angle{BED}=50^o-20^o=30^o=\angle{DCE}$. Hence: $\triangle{DEF} \sim \triangle{DCE}$.

Thank you very much :lol: your solution is wonderfull!!

How it was happened to apply symmetry to you?
The truth I do not believe that it finds another solution, nevertheless geometry always is very flexible...

Thanks.

[quote="elchuy"]How it was happened to apply symmetry to you?
[/quote]
Because there is a $30^o$ angle. So symmetry will give an equilateral triangle.

The problem with trigonometry is straightforward, but this make you to make more
job.

Basically, we must prove that $\angle DEF = \angle ACD = 30^\circ$.
Take $P$ on $(AC)$ so that $BP = BC$, easily to see that $\triangle BDP$ is equilateral and $\triangle BEP$ isosceles, with $\angle BEP = \angle BEC = \angle PBE = 40^\circ$, hence $DP = PE$, so the triangle $\triangle DEP$ is isosceles, with $\angle DPE = 180^\circ - (80^\circ + 60^\circ) = 40^\circ$, hence $\angle DEP = \angle PDE = 70^\circ$, so $\angle BED = \angle PED - \angle PEB = 70^\circ - 40^\circ = 30^\circ$. Now, the triangles $\Delta CED$ and $\Delta DEF$ have $\angle CDE = \angle FDE$ and $\angle DEF=\angle DCE = 30^\circ$.

Best regards,
sunken rock


[hide="#1"]$ f(0) \equal{} \minus{} a | \minus{} a| \geq 1$

Case 1: $ a > 0$

$ \minus{} a | \minus{} a| \equal{} \minus{} a^2 \geq 1$

$ a^2 \leq \minus{} 1$
No real solutions.

Case 2: $ a < 0$

$ \minus{} a | \minus{} a| \equal{} a^2 \geq 1$

$ a \leq \minus{} 1$ or $ a \geq 1$ but $ a < 0$ so

$ \boxed{a \in ( \minus{} \infty , \minus{} 1] }$[/hide]

EDIT: Fixed it thanks

[quote="modularmarc101"][hide="#1"]$ f(0) = - a | - a| \geq 1$

Case 1: $ a > 0$

$ - a | - a| = - a^2 \geq 1$

$ a^2 \leq - 1$
No real solutions.

Case 2: $ a < 0$

$ - a | - a| = a^2 \geq 1$

$ a \leq - 1$ or $ a \geq 1$

$ \boxed{a %Error. "fork" is a bad command.
( - \infty , - 1] \cup [ 1 , \infty)}$[/hide][/quote]

If $ a < 0$, then $ a \not \ge 1$.

Thus, only $ a \in ( - \infty , - 1]$ work.


I am relatively new to LaTeX, but as a math teacher I know that there is no other way to go to get the results that I need as easily as I can in LaTeX.

My current problem is font sizes.  I need a font size that is very large, something around 40pt, maybe bigger.  The onyl wy I have been able to change the font size in LaTeX so far is with commands like \Large, \LARGE, \huge, and \Huge.  But this still isn't big enough.  Is there a way (hopefully an easy way) to set a specific font size?  I will also need to be able to do this in math mode.

Any suggestions or help would be appreciated.

Hi,
Try the scalefnt package - it provides a command called \scalefont{x}, which will scale your current font by a scale of x. Assuming your document has the default option of 10 pt, either:
[code]\normalsize
\scalefont{4}[/code]
or
[code]\huge
\scalefont{2}[/code]
should give you a 40 pt font.
SSN.

Sweet!   Works like a charm.  Thanks so much.

Cheers,
Ben


If I received 39/50 on an exam and the exam total was 30% of my final grade, how would I figure out what I receved out of 30. thank you.

I don't get what the question asked.

39/50 is 78/100=78%

You received 30*0.78=   23.4 out of 30.

How did you get 7.02%?

[quote="fantasylover"]39/50=78/100=78 %

$ 78\%\times\frac {3}{10} \equal{} 23.4\%$

$ 23.4\% \equal{} \frac {23.4}{100} \equal{} \frac {2.34}{10} \equal{} \frac {7.02}{30}$

answer : $ 7.02\%$[/quote]

fixed your latex

edit: now that i think about it, i think that you shoulda stopped at $ 23.4$
you got $ 78\%$ on the test, so your points out of 30 would be $ 0.78\times30 \equal{} 23.4$

Oh, I get it now.

Fantasy interpreted the total score, when I thought he meant it was just for the test.

ok...
i'm not sure what you mean by that, but i think the answer is supposed to be 23.4, not 7.02%

[quote="captainonion"]If I received 39/50 on an exam and the exam total was 30% of my final grade, how would I figure out [b]what I receved out of 30[/b]. thank you.[/quote]

adding some text of my own

Okay, but then wouldn't the answer be the number 23.4, rather than the percent 7.02?

[quote="captainonion"]If I received 39/50 on an exam and the exam total was 30% of my final grade, how would I figure out what [b]I receved out of 30[/b]. thank you.[/quote]

then what does bold part do in this problem?

i'm guessing "what i received out of 30" means what part of the 30% he would've gotten
that's how i understood the problem, but we should ask captainonion

I interpreted it as how many points I received out of 30.

So I thought that 23.4 was correct.

captainonion, care to clarify :?:

hey fishy, you are right. I was looking for my total out of thirty, which would be 23.4/30=78%


What three-digit integer is equal to the sum of the factorials of its digits?

I remembered the answer is 145 AS the same question was asked in my Computer science two years ago.
such type of nos are known by some specific term.I didn't remember the term.

These are called factorions. There are only 4 such numbers, 1,2,145, and 40585.

I had a question to the same problem — the solution says that "the hundreds digit cannot be greater than 1, because the digits 5 or 6 are needed to make a 3 digit integer but will not fit with the larger hundreds digit." I don't understand why it can't be 2 or 3, for example.

Is there a number theory solution to this other than brute force and limiting cases?

[quote=xue1]but 153 also works - I did it on a CALCULATOR[/quote]

$1! + 5! + 3! = 1+120 + 6 = 127$. $  127 \neq 153.$

but how do you prove it's the only answer? This one just says use brute force

[quote=AIME15][quote="ernie"]Obviously, none of the digits can be $ \ge 7$, because $ 7! > 999$.

You can deduce that this problem uses a $ 5$.

The $ 5$ probably won't be in the hundreds digit.

Then, I used brute force to find $ \boxed{145}$.[/quote]

Or, we can have a nice proof used by pythag011 and written by me:

[hide=":)"]
No digit can be $ 7$, $ 8$, or $ 9$, because their factorials are greater than $ 999$, the largest $ 3$-digit number.  Also, no digit can be $ 6$, because $ 6! \equal{} 720$, so you need a digit in the hundreds spot that it as least $ 7$, but that is impossible.  This narrows down our possibilities greatly.

The hundreds digit cannot be greater than $ 3$, because the largest possible now is $ 555$, and the sum of the factorials of their digits is $ 360$.  But if the hundreds digit is $ 3$, the largest possible is $ 355$, but the sum of the factorials of the digits of $ 355$ is $ 246$.  Thus the hundreds digit must be $ 1$ or $ 2$.

If the last two digits are $ 5$, the sum of the factorials of the digits is $ 240 \plus{} \text{first digit}$.  $ 155$ obviously doesn't work, and $ 255$ doesn't either.  Thus, one of the last two digits is at most $ 4$.  Thus, the greatest possible sum of the factorials now is $ 5! \plus{} 4! \plus{} 2! \equal{} 120 \plus{} 24 \plus{} 2 \equal{} 146$.  So the hundreds digit can only be $ 1$!  Also, because we know that the sum of the factorials is at least $ 100$, we must have one $ 5$, because $ 4! \cdot 3 \equal{} 72$, which is not a $ 3$-digit number!  So it must be of the form $ 1a5$ or $ 15a$.  If it was in the form $ 15a$, the largest possibility would be $ 154$, but summing the factorials of the digits yields $ 120 \plus{} 24 \plus{} 1 \equal{} 145$, so it can only be in the form $ 1a5$ where $ 0\le a \le 4$.

If we take this modulo $ 10$, we know the sum of the factorials is $ 1 \plus{} 120 \plus{} a! \equiv 5 \pmod{10}$.  Thus $ a! \equiv 4 \pmod {10}$.  Obviously, the only solution where $ 0\le a \le 4$ is $ 4$, so our number is $ \boxed{145}$.[/hide][/quote]

you might want to consider that 6! = 720, and since you can't use 7, you can't use 6 either


There are 8 red, 8 blue and 8 yellow marbles in a jar.  What is the fewest marbles you can remove from the jar so that the ratio of red to non-red marbles is 3 to 7 and so that the ratio of yellow to non-yellow marbles is also 3 to 7?

wait, do you mean "the fewest marbles you can remove from the jar so that the ratio of red to non-red marbles is 3 to 7 and so that the ratio of yellow to non-yellow marbles is also 3 to 7" inside or outside the jar?

[hide]well, whichever side you mean, there needs to be 10 marbles on that side because out of the 10 marbles, there can be 3 red and 3 yellow and 4 non-red-or-yellow marbles.[/hide]

[hide][color=indigo]You remove 4 (2 red and 2 yellow) :D  [/color][/hide]

[hide]it's 4. 2 of both yelow and red. since there would be 20 marbles.
6 to 14 is 3 to 7 both ways too.[/hide]

[hide]
You must remove 4 marbles from the jar.

3:7 = 6:14

Six marbles are red, and 14 are non-red.

Therefore, six marbles are yellow, and 14 are non-yellow.

So there are 6 red, 6, yellow, and 8 blue marbles remaining.

You took away 4 marbles.

 :)
[/hide]

[hide]
Remove 2 red and 2 yellow. (4)
[/hide]


Denote $ S(n)$ the sum of digits of $ n$
Does there exist an integer $ n$ such that $ S(n) \equal{} 2009$ and $ S(n^2) \equal{} 1996^2$?

Since the sum of the digits of $ n^2$ is $ 1996^2$, $ n^2$ has to have at least $ (1996^2)/9 \equal{} 44134$ digits. that means that n has to have at max $ 44134/2 \equal{} 22067$ digits. In those 22067 digits, even if the first $ 2009/9 \equal{} 223$ digits were 9's, the square of n would not have 44134 digits, so such a number n does not exist.

Sorry,but I think it is not true.Since $ ((10^{223} \minus{} 1)*10^{21844})^2 \ge 10^{44133}$
So the square of $ n$ may have $ 44134$ digits
Am I right? :blush:

Obviosly, if $ S(n)\equal{}2009$, then $ n\equal{}2\mod 9\to n^2\equal{}4\mod 9 \to S(n^2)\equal{}4\mod 9 \not \equal{}1996\equal{}7\mod 9$.

[quote="Rust"]Obviosly, if $ S(n) \equal{} 2009$, then $ n \equal{} 2\mod 9\to n^2 \equal{} 4\mod 9 \to S(n^2) \equal{} 4\mod 9 \not \equal{} 1996 \equal{} 7\mod 9$.[/quote]

No,$ 1996^2 \equal{}4 \mod 9$ :wink:

My solution is $ 110...011...0....,,110...0999996$
$ 11$ repeats $ 979$ times,and there are enough many $ 0$ between every consecutive $ 11$
Hope it is right :)

[quote="hxy09"]My solution is $ 110...011...0....,,110...0999996$
$ 11$ repeats $ 979$ times,and there are enough many $ 0$ between every consecutive $ 11$ [/quote]

My check did not quite come out right.  I guess the square (if there are plenty of zeros) will have
979 blocks of 121
478731 blocks of 242
one block of 99999216
979 blocks of 21999912

so I get for the digit sum: 3874936, while 1996^2 = 3984016

But maybe I counted wrong...

Sorry,I haven't checked it seriously :blush:
Another:$ 20..030...020...030...020...30......010...010...010...01$
There are $ 5785$ pairs of $ 2,3$ and the rest are $ 1$,$ 0$ are added between them as above.


algebra
polynomial
group theory
abstract algebra
vector
geometry
geometric transformation

Let $ a,b,c$ be roots of an irreducible polynomial $ f\in\mathbb{Z}[X]$ such that $ a^{2}=bc$. Prove that $ a/b$ is a root of unity.

Do you have a better solution?  I just crunched it.

Say deg $ f$ = $ n$ and $ a = r_{1}, b = r_{2}, c = r_{3}, r_{4}, \ldots, r_{n}$ are distinct $ n$ roots of $ f$.  We can assume that $ a,b,c$ are pairwise distinct since otherwise $ \frac{a}{b}$ is either 1 or -1, clearly a root of unity.  Since $ f$ is irreducible, for each $ r_{i}$, you can find an automorphism (of say, splitting field of $ f$) sending $ a \rightarrow r_{i}$.  That is, $ r_{i}^{2}= r_{f(i)}r_{g(i)}$ for some permutations $ f,g \in S_{n}$ with no fixed points such that $ f(i) \neq g(i)$ for every $ i \in \{1, 2, \ldots, n \}$.  To fix the idea, we have $ f(1) = 2$ and $ g(1) = 3$.

Take $ G = <f,g>$, a subgroup of $ S_{n}$ generated by $ f$ and $ g$.  Take $ X$ to be the orbit of $ 1 \in \{ 1, 2, \ldots, n \}$ under $ G$.  So $ X = \{1, 2, 3, \ldots \}$.  Renumbering if necessary, say $ X = \{1, 2, 3, \ldots, k \}$.  Then it's clear that $ \frac{a{}^{2{}^{r}}}{b{}^{2{}^{r}}}= \frac{r_{1}{}^{2{}^{r}}}{r_{2}{}^{2{}^{r}}}$ can be written as a quotient $ \frac{r_{1}^{s}_{1}r_{2}^{s}_{2}\ldots r_{k}^{s}_{k}}{r_{1}^{u}_{1}r_{2}^{u}_{2}\ldots r_{k}{}^{u}{}^{k}}$ with $ s_{1}+\ldots+s_{k}= u_{1}+\ldots+u_{k}= 2^{r}$.  Now if $ s_{1}+s_{2}+\ldots+s_{k}> 2kM$ where $ M$ is the maximal of $ j$ among the lengths of products of $ f,g$ required to take $ j$ to $ 1$, then some $ s_{i}> 2M$ and applying the corresponding product of $ f,g$ required to take $ i$ to $ 1$ using the relations $ r_{i}^{2}= r_{f(i)}r_{g(i)}$, we get at the quotient with $ s_{1}> 0$ and $ u_{1}> 0$.  So one can reduce the fraction $ \frac{r_{1}^{s}_{1}\ldots r_{k}^{s}_{k}}{r_{1}^{u}_{1}\ldots r_{k}^{u}_{k}}$ further if $ s_{1}+\ldots+s_{k}= u_{1}+\ldots+u_{k}> 2kM$.  This shows that $ \frac{a{}^{2{}^{r}}}{b{}^{2{}^{r}}}$ can be written as $ \frac{r_{1}^{s}_{1}\ldots r_{k}^{s}_{k}}{r_{1}^{u}_{1}\ldots r_{k}^{u}_{k}}$ with $ s_{1}+\ldots+s_{k}= u_{1}+\ldots+u_{k}\leq 2kM$ for every $ r$.  But there are only finitely many expressions of this form so for some $ r \neq s$, we have $ (a/b){}^{2{}^{r}}$ = $ (a/b){}^{2{}^{s}}$.  So $ (a/b){}^{2{}^{r}-2^{s}}= 1$.  So $ a/b$ is a root of unity.

Here's what I did.

We assume $ \frac{a}{b}$ is not a root of unity, and derive a contradiction.

First, let's make the following modification: notice that for all positive integers $ n$, the conjugates of $ a^{n}$ are precisely the distinct $ n$'th powers of the conjugates of $ a$. Moreover, if $ \frac{x}{y}$ is not a root of unity ($ x$ and $ y$ being two complex numbers), then $ x^{n}$ and $ y^{n}$ are different for all positive integers $ n$. Now, if instead of the algebraic numbers we initially had we work with their distinct $ n$'th powers for suitably chosen $ n$, then $ b^{n},a^{n},c^{n}$ will still be three distnict conjugates in arithmetic progression, and in addition, we can make sure that $ \frac{x}{y}$ is not a root of unity for [i]any[/i] distinct conjugates $ x,y$ of $ a^{n}$. In order to keep the notation simple, I'll still name the numbers $ a,b,c$ etc., but keep in mind that now we make the extra assumption that no two conjugates of $ a$ have a root of unity quotient.

Notice that [i]all[/i] roots $ a$ of our polynomial are geometric means of two other roots. This follows from the fact that the Galois group of the polynomial acts transitively on the roots. Now, if the root $ a$ has maximal absolute value, then $ b,c$ also have maximal absolute value whenever $ a^{2}=bc$. Now forget about all number-theoretic considerations. The situation is this: we have $ n\ge 3$ complex numbers $ a_{i},\ i=\overline{1,n}$, all with the same absolute value, such that no quotient $ \frac{a_{i}}{a_{j}},\ i\ne j$ is a root of unity. We want to reach a contradiction. Assume in addition that the set $ \{a_{i}\}$ is minimal with respect to these properties, and also that $ a_{n}=1$ (clearly, this does not affect the problem).

The complex numbers $ a_{i},\ i\le n$ lie on the unit circle; moreover, because of our assumption that $ \frac{a_{i}}{a_{j}}$ are not roots of unity, the $ a_{i}$ are mapped in a one-to-one fashion into $ V$, the quotient of the unit circle through the group of roots of unity. Notice that $ V$ is a vector space over $ \mathbb Q$, being a torsion-free divisible abelian group. Let $ E$ be the $ \mathbb Q$-vector space spanned by the images $ e_{i}$ of the $ a_{i}$ in $ V$, with the numbering arranged so that $ e_{1}, \ldots,\ e_{t},\ 1\le t\le n-1$ is a basis of $ E$. Among all the $ e_{i}$ there must be at least one ($ e_{s}$, say) which has the largest coefficient of $ e_{1}$ when written in the basis $ e_{1},\ \ldots,\ e_{t}$. Then, if $ a_{s}^{2}=a_{i}a_{j}$, the coefficients of $ e_{1}$ in $ e_{i}$ and $ e_{j}$ are equal to that same maximal value. Repeat the procedure with $ a_{i},a_{j}$ instead of $ a_{s}$, and so on. The set we obtain after the finitely many distinct steps we can make has all the properties of $ \left\{a_{i}\ |\ i=\overline{1,n}\right\}$ (that is, it consists of complex numbers on the unit circle, and each element is the geometric mean of two other elements), but is strictly smaller, because all $ a_{i}$ we looked at are such that the coefficient of $ e_{1}$ in $ e_{i}$ is non-zero, so, for example, the set in question does not contain $ a_{n}=1$ (which maps onto $ 0$ in $ V$). This contradicts the minimality of the set $ \{a_{i}\}$, and finishes the proof.

I think your proof is quite nice using | | of $ \mathbb{C}$.  I didn't think of doing this as, in the back of my mind, I replaced $ \mathbb{Z}$ with an abstract field $ \mathbb{F}$.

So in general, for any irreducible $ f \in \mathbb{F}[x]$ having some roots $ a_{0}, \ldots, a_{k}$ in geometric progression, i.e. $ a_{0}^{k}= a_{1}a_{2}\ldots a_{k}$ ($ k$ not necessarily the total number of roots of $ f$, i.e. $ k \leq$ deg $ f$), $ a_{i}/a_{j}$ is a root of unity for all $ i,j$.

1) Hi sheaf.
We can have $ f(i)=f(j)$ with $ i\not=j$ (cf. for example the polynomial $ x^{n}-\alpha$). Perhaps you must work by contradiction as Grobber did.
2) Hi Grobber.
I don't understand: "We assume  $ \frac{a}{b}$ is not a root of unity" implies "we can make sure that $ \frac{x}{y}$ is not a root of unity for any distinct conjugates  $ x,y$ of $ a^{n}$."
Moreover "assume that  $ a_{n}=1$ (clearly, this does not affect the problem)" is a detail but it isn't clear for me.
3) I don't see very well the polynomial $ f$:
 Can somebody construct a such polynomial $ f$ which have complex roots of different modules ?
 Can somebody construct a such polynomial $ f$ which have all its roots of same module and s.t.  $ f$ isn't a divisor of a polynomial of the form $ x^{n}-\alpha$ ?

[quote="loup blanc"]
I don't understand: "We assume  $ \frac{a}{b}$ is not a root of unity" implies "we can make sure that $ \frac{x}{y}$ is not a root of unity for any distinct conjugates  $ x,y$ of $ a^{n}$."
[/quote]

Let $ x_{1},\ldots,x_{s}$ be the roots of the minimal polynomial of $ a=x_{1}$ (so $ b$ and $ c$ are also among them, of course). For all $ n\ge 1$, the roots of the minimal polynomial of $ x_{1}^{n}$ are precisely the distinct values appearing in the multiset $ x_{1}^{n},\ldots,x_{s}^{n}$ (why?). Now, we assumed that $ \frac{a}{b}$ is not a root of unity. This means that $ a^{n}\ne b^{n}$ for all $ n$ (same for $ a^{n},c^{n}$ and $ b^{n},c^{n}$), so $ a^{n},b^{n},c^{n}$ will always be distinct conjugates of one another, satisfying $ (a^{n})^{2}=b^{n}c^{n}$. Simply take $ n$ which "eliminates" all roots of unity of the form $ \frac{x_{i}}{x_{j}}$, i.e. such that $ x_{i}^{n}=x_{j}^{n}$ whenever $ \frac{x_{i}}{x_{j}}$ is a root of unity. Now, instead of working with the original algebraic numbers $ x_{1},\ldots,x_{s}$, work with the conjugates of $ x_{1}^{n}$, which are precisely the distinct elements of the form $ x_{i}^{n}$.

[quote="loup blanc"]
Moreover "assume that  $ a_{n}=1$ (clearly, this does not affect the problem)" is a detail but it isn't clear for me.
[/quote]

The problem, as stated there, no longer depends on the $ a_{i}$ being roots of an integer polynomial and all that. We're concerned only with their positions on a circle centered at the origin in the complex plane, and it's only their positions relative to one another that matter. Simply apply a homothety and a rotation around the origin to make $ a_{n}=1$. If you prefer, look at it this way: replace each $ a_{i}$ with $ \frac{a_{i}}{a_{n}}$.

[quote="loup blanc"]
3) I don't see very well the polynomial $ f$:
 Can somebody construct a such polynomial $ f$ which have complex roots of different modules ?
 Can somebody construct a such polynomial $ f$ which have all its roots of same module and s.t.  $ f$ isn't a divisor of a polynomial of the form $ x^{n}-\alpha$ ?[/quote]

You mean integer irreducible polynomials with three different roots $ a,b,c$ in geometric progression? I don't know if those examples exist, and right now I'm not particularly interested in the issue :).


Prove for every irrational real number a, there are irrational numbers b and b' such that a+b and ab' are rational while a+b' and ab are irrational.

[hide="Solution"] Given our irrational real numbers $a$, let $a+b = p, ab' = q$ be two rational numbers we hope exist.  Then $b = p-a, b' = \frac{q}{a}$ and we want to show that there exist $p, q$ such that

$a+\frac{q}{a}, a(p-a)$

Are irrational.

If $a+\frac{q}{a}= x$ some irrational then $a$ satisfies the (irrational) quadratic $a^{2}-xa+q = 0$.  If $a$ is not the root of a rational quadratic then any rational $q$ produces an irrational $x$.  Otherwise $a$ is the root of some rational quadratic with some rational constant term, so pick $q$ not equal to that constant term and $x$ cannot then be rational.

Similarly, if $a(p-a) = y$ some rational then $a$ satisfies the (irrational) quadratic $a^{2}-pa+y = 0$.  Again, if $a$ is not the root of a rational quadratic then any rational $p$ produces an irrational $y$.  Otherwise $a$ is the root of some rational quadratic with some rational $a$ -coefficient, so pick $p$ not equal to that constant term and $y$ cannot be rational.  [/hide]
This seems a little too easy, so I'm not sure if I'm missing subtleties.

I don't understand the part about he irrational and rational quadratics. Surely I find a constant term not equal to the term by which x is multiplied and still have it have rational roots, e.g. $x^{2}+5x+6$.

[hide="Some relevant background"] [b]Definition:[/b]  Let the minimal polynomial $m(x)$ over $F[x]$ (in practice, over $\mathbb{Q}[x]$) of some $a$ be the unique monic polynomial of smallest degree such that $m(a) = 0$.

If $a$ is rational then $m(x) = x-a$.  Otherwise $m(x)$ is of some higher degree.

[b]Lemma:[/b]  Given a polynomial $p(x)$ over $F[x]$ such that $p(a) = 0$, the minimal polynomial $m(x) | p(x)$.

Proof:  Division algorithm.  The remainder $r(x)$ when $m(x)$ divides into $p(x)$ must either be $0$ or be a polynomial of degree lower than $m(x)$ such that $r(a) = 0$ - contradiction.

[b]Corollary (conjugate root theorem):[/b]  If $p+\sqrt{q}$ is a root of a rational quadratic, where $p, q$ are rationals and $q$ is not the square of a rational, then $p-\sqrt{q}$ is also a root. [/hide]
[hide="A further explanation of my logic"] If the monic polynomial of $a$ over the rationals is quadratic, then $a$ cannot be the root of any other monic rational quadratic (or else the difference between the two quadratics, which is both linear and rational, must have $a$ as a root, contradicting $a$ irrational).

Hence if we fix one of the coefficients to be a rational not equal to the corresponding coefficient in the minimal polynomial of $a$ then the other coefficient cannot be rational.

I am basically just using the idea of the uniqueness of the minimal polynomial, but the other results following from that idea are interesting too. [/hide]


Find all natural numbers $ a,b$ such that :
                                                   $ (a^2\minus{}1)^2\plus{}(a^2)^2\equal{}b^2$


Can someone remind me what the formula is for finding the day of the week of a given date? I can't remember it...

wouldn't it depend on the month and year?  :maybe:

The formula for the day of the week is called the Doomsday Formula ( or rule, like at [url]http://en.wikipedia.org/wiki/Calculating_the_day_of_the_week[/url])

[hide="ummm..."]Can non-Stanford math circle participants post on this forum?[/hide]


How would I get involved in the AMC competitions if my school does not participate?

[quote="Xantos C. Guin"]Find a nearby school that does participate in the AMC Contests and ask them if you could take the test along with their students. Most schools will gladly let you do so.[/quote]

What if you win something? Which school will your name go under?

See the FAQ thread, which is stickied.  You could also pay for the tests, in which case it would be really weird if your school didn't let you take them, especially as it gives other students at your school an opportunity to take the test.

That can get a little complicated, SplashD.  It [i]should[/i] go under your real school's name, but you have to be careful to do the book-keeping right.  Mistakes have occurred.


inequalities
rearrangement inequality
inequalities open

let a,b,c in R .prove that
$ \sum_{sym}(\frac {a}{b \plus{} c})^2\geq\sum_{cyc}(\frac {b}{b \plus{} c})^2$

[hide="hint"]Straightforward application of the rearrangement inequality[/hide][hide="solution"]
WLOG $ a^2 \leq b^2 \leq c^2 \Rightarrow (a\plus{}b)^2 \leq (a\plus{}c)^2 \leq (b\plus{}c)^2 \Rightarrow (\frac{1}{b\plus{}c})^2 \leq (\frac{1}{a\plus{}c})^2 \leq (\frac{1}{a\plus{}b})^2$

Because of this, we see that the LHS of the original inequality consists of the product of similarly ordered sequences and the RHS consists of differently ordered sequences, thus it is true by the rearrangement inequality.[/hide]

$ a$, $ b$, $ c$ must be positive if you want the inequality to hold (else, $ a\equal{}b\equal{}1$, $ c\equal{}\minus{}2$ is a counterexample).

  darij

Yes, you're right... I just realized that if a,b,c are not nonnegative then $ a \leq b \leq c$ does not imply that $ (a\plus{}b)^2 \leq (a\plus{}c)^2 \leq (b\plus{}c)^2$ but I believe my solution still works otherwise.


function
combinatorics unsolved
combinatorics

$ Problem$

Enumerate the number of $ 3$-regular simple graphs.

(i) Vertices are numbered.
(ii) Up to Isomorphism.

A graph $ G(v,e)$ with $ n$ vertices is called $ k$-regular if degree of every vertex is $ k$.

http://www.research.att.com/~njas/sequences/A005638 is unlabeled; http://www.research.att.com/~njas/sequences/A002851 is unlabeled connected; http://www.research.att.com/~njas/sequences/A004109 is labeled connected.  (If you can get the connected ones you can get all of them via the exponential formula.)

The original poster asked via PM:
[quote="I hate rating"]I did look at the sequences. But I could not get a closed form or a gf, do you know any standard text or a paper that enumerate this ?( I dont have JSTOR access)

You said : "If you can get the connected ones you can get all of them via the exponential formula" . I dont understand ?[/quote]
For the first, this is unfortunately everything I know about the question.  It's a natural enough question (and the sequence numbers are low enough) that I would guess not much is known about their enumeration that isn't in those sequences.  (There are also several sequences for different kinds of cubic multigraphs and graphs or multigraphs with loops allowed.)

For the second question, the [url=http://en.wikipedia.org/wiki/Exponential_formula]exponential formula[/url] is just the observation that if $ A(x)$ is the exponential generating function for the sequence $ a_n$ of connected, labeled objects (e.g., connected graphs, trees, connected cubic graphs, but also many other classes) then $ B(x) \equal{} e^{A(x)}$ is the generating function for $ b_n$, the number of not-necessarily-connected labeled objects of the same sort (e.g. graphs, forests, all cubic graphs).  This becomes clear when you write down the coefficients of the latter series (as Wikipedia does in the link above).


algebra
polynomial
function
functional equation

Find all polynomials $ f$ over $ \mathbb{C}$ such that
\[ f(x)f(\minus{}x) \equal{} f(x^2)
\]

If $ f(x)$ is constant, then it is equal to $ 0$ or $ 1$.

If $ f(x)$ is nonconstant, then it has a finite number of roots. Let $ r$ be the root for which $ |r|$ is minimal. If $ r$ is a root, then $ r^2$ is a root also $ \sqrt{r}$ is a root. Suppose for sake of contradiction that $ 1>|r|>0$. Then
\[ |r|\minus{}|r|^2\equal{}|r|*(1\minus{}|r|)>0\]
which is a contradiction to the minimality of $ r$. For sake of contradiction, suppose that $ |r|>1$, then
\[ |r|\minus{}|\sqrt{r}|\equal{}|\sqrt{r}|*(|\sqrt{r}|\minus{}1)> 0\]
which contradicts the minimality of $ r$.

Consider the cases:
i) $ |r|\equal{}0\implies r\equal{}0$
If $ 0$ is root, then it is a double root by considering the LHS of the functional equation. Hence $ x^2|f(x)$. At some point, $ f(x)\equal{}x^{2n}*g(x)$ where $ g(x)$ is a polynomial that satisfies the same functional equation.

ii) $ |r|\equal{}1$. Now let $ r'$ be the root where $ |r'|\equal{}1$ and $ |\arg (r')|>0$ is minimal. But if $ r$ is a root, then $ \sqrt{r}$ is a root, which contradicts the minimality of $ \arg(r')$. Hence $ \arg(r)\equal{}0$ so $ r\equal{}1$. If $ f(x)\equal{}(1\minus{}x)*g(x)$ where $ g(x)$ is a polynomial that satisfies the same functional equation. Eventually, $ f(x)\equal{}(1\minus{}x)^n*g_1(x)$.

If we look at the minimal root, then divide it out, and repeat, eventually we get a constant polynomial (since $ f(x)$ has a finite degree). It is clear that any solution must be either
\[ f(x)\equal{}(1\minus{}x)^n*(x)^{2m} \qquad\text{or}\qquad f(x)\equal{}0\]

where $ m$ and $ n$ are non-negative integers. It is clear that both of these functions satisfy the equation so we are done.

The function $ x^a \minus{} x^b$ where $ a$ is an even nonnegative integer and $ b$ is an odd nonnegative integer works, but it's not in your solution space.

In addition, the polynomial $ x^2\plus{}x\plus{}1$ is missing.

[hide]Regard $ f(x)$ as a power series.
Let $ f(x) \equal{} \sum_{k \equal{} 1}^{\infty} a_k x^k$.
Expand $ f(x)f( \minus{} x)$ and then compare coefficients with that of $ f(x^2)$.
We have a set of equalities:
$ a_0^2 \equal{} a_0$,
$ 2 a_0 a_2 \minus{} a_1^2 \equal{} a_1$,
$ 2 a_0 a_4\minus{} a_2^2 \equal{} a_2$,
$ 2 a_0 a_6\minus{} a_3^2 \equal{} a_3$,
$ 2 a_0 a_8 \minus{} a_4^2 \equal{} a_4$,
.........
Case 1
$ a_0 \equal{} 0$
......
Case 2
$ a_0 \equal{} 1$
......[/hide]

Well it is pretty clear that i screwed up the $ |r| \equal{} 1$ case...the mistake is that $ r$ is a root, implying that $ \sqrt {r}$ is a root is not always true. $ \pm\sqrt {r}$ is true.

But that screws up the proof a bit because if $ \arg(r) \equal{} \theta$, then $ \arg( \minus{} \sqrt {r}) \equal{} \pi \plus{} \frac {\theta}{2}$.

[quote="t0rajir0u"][hide="Observation"] The important case is when $ |r| \equal{} 1$.  Let $ r \equal{} e^{ix}$.  If $ x$ is not a rational multiple of $ \pi$, then $ e^{2^k ix}$ is also a solution for any $ x$, so there are infinitely many roots; contradiction.  Hence $ r$ is an $ n^{th}$ root of unity. [/hide][/quote]Hmm... so this means that the solution set is the set of all polynomials whose roots are roots of unity with their conjugates?

EDIT: $ f(x) \equal{} x^m(x\minus{}1)^n\Phi_k(x)$, cyclotomic polynomials $ \Phi_k(x)$?

That is not true. $ \Phi_{10}(x) \equal{} x^4 \minus{} x^3 \plus{} x^2 \minus{} x \plus{} 1$ does not work. Neither does $ \Phi_{12}(x)$...or for that matter, $ \Phi_2(x) \equal{} x \plus{} 1$

This is because $ 2$, $ 10$ and $ 12$ are even.

If we have $ r$, then we have $ r^{2k}$. But if $ r$ is a primitive $ 2n$th root of unity, then $ r^2$ is not; it is a primitive $ n$th root of unity.

I believe that cyclotomic polynomials of odd numbers all work.

[hide="Finishing up the solution"] We've established that all roots of $ f(x)$ are $ k^{th}$ roots, hence primitive $ k^{th}$ roots (for appropriate $ k$).  We've established that $ k$ is odd.  It's also not hard to verify that $ f(x) \equal{} x^2$ is a solution.

[b]Lemma:[/b]  If $ f(x), g(x)$ are solutions to the functional equation, then so is $ h(x) \equal{} f(x) g(x)$.

The solution set is therefore closed under multiplication; it remains to consider its generators.  We will do so by investigating the primitive $ k^{th}$ roots for a given $ k$ individually.

Let $ n \equal{} \text{deg}(f)$.  The leading coefficient of $ f(x)$ is either $ 1$ (when $ n$ is even) or $ \minus{} 1$ (when $ n$ is odd).  Therefore, let

$ f(x) \equal{} \prod_{i \equal{} 0}^{n \minus{} 1} (r_i \minus{} x)$.  Then

$ f(x) f( \minus{} x) \equal{} \prod_{i \equal{} 0}^{n \minus{} 1} (r_i^2 \minus{} x^2) \equal{} f(x^2) \equal{} \prod_{i \equal{} 0}^{n \minus{} 1} (r_i \minus{} x^2)$

so the roots of $ f(x)$ have the property that when squared, they are the same set of roots.

For a given odd $ k$, let $ \zeta_{m} \equal{} e^{2\pi i \frac {m}{k} }$.  Suppose that $ \zeta_{2j}$ were a root of $ f(x)$.  Then $ \zeta_{2^{\varphi(k) } j} \equal{} \zeta_j$ would also be a root.  Let $ j$ be the smallest odd natural number such that $ r_j$ is a root.  Then $ \zeta_{2^l j}$ is a root for $ 0 \le l < \text{ord}_{k}(2)$.

This set of roots satisfies the desired property.  There are $ \frac {\varphi(k) }{\text{ord}_{k}(2) }$ such polynomials for every $ k$ (their product is $ \Phi_k$), and they form the generators of our solution set.

In particular, for $ k \equal{} 1$ we have $ f(x) \equal{} 1 \minus{} x$.
For $ k \equal{} 3$ we have $ f(x) \equal{} x^2 \plus{} x \plus{} 1$.
For $ k \equal{} 5$ we have $ f(x) \equal{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$.
For $ k \equal{} 7$ we have $ f(x) \equal{} (\zeta_1 \minus{} x)(\zeta_2 \minus{} x)(\zeta_4 \minus{} x), (\zeta_3 \minus{} x)(\zeta_6 \minus{} x)(\zeta_5 \minus{} x)$ because $ \text{ord}_7(2) \equal{} 3$.  Note that the coefficients of these polynomials is [b]not purely real.[/b]

And so forth. [/hide]

Engel says that for p+q even we have the following solution

$ f(x)\equal{}x^p(x\minus{}1)^q(1\plus{}x\plus{}x^2)^r$  for p,q,r natural numbers

Strange.  Engel has missed at least the solutions $ f(x) \equal{} 1 \minus{} x$ and $ f(x) \equal{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$.  What is the solution in the text?

I have missed some solutions as well (for example, $ x(x \minus{} 1)$).  It seems that a discussion of multiplicities was more important than I had expected (particular, the multiplicity of the root $ 0$).

My Engel book gives the set of solutions $ f(x)\equal{}(\minus{}x)^p(1\minus{}x)^q(x^2\plus{}x\plus{}1)^r,\ p,q,r\in\mathbb{Z}$.  For emphasis, I will say that we can even take $ p,q,r<0$.  The solution is not as short as an average "Engel" solution size, that is, I will post the solution later (it is 2:10 a.m. now and I have to go to school tomorrow).

Is it so hard to verify that

$ (x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1)(x^4 \minus{} x^3 \plus{} x^2 \minus{} x \plus{} 1) \equal{} (x^8 \plus{} x^6 \plus{} x^4 \plus{} x^2 \plus{} 1)$?

While Engel is correct in identifying the generator $ \minus{}x$ (rather than the generator $ x^2$ which I identified), $ f(x) \equal{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$ is plainly in the solution set.  I admit that the full solution set I gave is strange and uncharacteristically unclean; however, I don't see anything in the problem statement that suggests that I am wrong.  Is there an extra condition?

Engel does not actually claim that his solution is complete. He notes that boxedexe's polynomials constitute a solution, then asks, "Are there any others?"

[quote="t0rajir0u"]Is it so hard to verify that

$ (x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1)(x^4 \minus{} x^3 \plus{} x^2 \minus{} x \plus{} 1) \equal{} (x^8 \plus{} x^6 \plus{} x^4 \plus{} x^2 \plus{} 1)$?

While Engel is correct in identifying the generator $ \minus{} x$ (rather than the generator $ x^2$ which I identified), $ f(x) \equal{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$ is plainly in the solution set.  I admit that the full solution set I gave is strange and uncharacteristically unclean; however, I don't see anything in the problem statement that suggests that I am wrong.  Is there an extra condition?[/quote]

Clarification: It appears that you are frustrated, thus I must clarify that Engel does not claim that his solution set is complete, as Phelpedo stated above; I beleive you are correct.

Does anyone know the actual source of this problem?


Hi,

I am new here. Hopefully share good things.

I think this question is simple but can not understand. And don't have the tools of experiment too. Any help is very good for me. Unbelieveble confused nowadays. Please don't laugh.

Let me explain the question, as you may see my attached picture, a hollowed and almost spherical conductor has (-3q) charge. Another charged conductor has (+2q). If i tie this (+2q) charged one to inside of hollowed sphere with a conductor cable what will be the charge of them? And how? I suppose the picture which i attached will explain better.

My problem is; is there a difference of directly touching to inside and tie with a cable again to inside? I suppose this question will be help to understand.

welcome fisher !!!
[hide="hint"]
the potential in both the sphere becomes same...
some charges get collected on the inner hollw some remain at the outer hollow.[/hide]

Thank you pardesi,

[quote="pardesi"]some charges get collected on the inner hollw some remain at the outer hollow.[/quote]
Maybe could not understand but there will be charges inside? As i know, they must be present only outsides of conductor.

Do you mean before, after, both cases? I suppose later. Then what about Z sphere? Will have charge and if it's, which kind of charge?

Also, if i would touch the Z sphere to inside directly, it would be lose all. Is this different? Couldn't understand enough.

charges can only reside on the boundary of a conductor...
so some charge resides on the inner boundary and some on the outer...
the charge distribution is such that both have same potential

Thank you again,

But if;

1) i touch the "Z" without cable(with insulated handle) to inside of "K"?

2) "K" was almost closed from upside like whole sphere?

More clear, in Faraday ice pail experiment, all charges were only outside of pail. Now some of them inner, some on the outer. How?

---------------
For example here:
http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ShellsConductors.xml

When a charged object is placed within a metal container, an equal charge of the same sign is FORCED to the outer surface of the container.

All of the charge on any metal object resides on its outer surface if a conducting path is provided so that the charge can move there. Remember that charges will flow between two positions as long as there is a potential difference between those positions. When the voltage has been equalized, all charges will cease to flow.
-----------------

[quote="fisher"]
---------------
For example here:
http://dev.physicslab.org/Document.aspx?doctype=3&filename=Electrostatics_ShellsConductors.xml

When a charged object is placed within a metal container, an equal charge of the same sign is FORCED to the outer surface of the container.

-----------------[/quote]

I really don't understand your question above but let me just explain the above statement. If a charged object is placed inside a metallic container, then an equal and OPPOSITE charge is forced on the inner surface of the container because even a hollow metallic container has some thickness and inside that net field must be zero. (Use gauss law).

And so by conservation of charge equal charge of same sign is forced outside.

Excuse me Cyber-Shreyas, the link and pasted sentences are not identing my upper question. Just noticed.

Yesterday, pardesi and sentence from another book make me think. I suppose i will explain different now. This is;

If a hollowed cylinder(or any "COMPLETELY" closed) conductor charged then the charges must sit on outer surface. And everywhere has only one potential. The thing which confused me is, the figure and pictures of Faraday ice pail experiment in net, books,.... Everywhere, the pail is uncovered. Not completely closed. But still, says that if a conductor touch inside of it, all charges travels to outside of pail. So the inside of faces(facing each other) in pail has no charge, anyway everywhere has same potential at last. I mean the figure of hollow in my first post.

In that figure, the Z(hollowed) has no charge in faces at first. Now i am not sure this. After pardesi, i think, the inner faces must has charges too since, it is not completely closed sphere(without a cover). Is this true?

Again, if i understood pardesi, everytime, before touch or after touch, inner surfaces has charges. Again, if it is not completely closed with a cover, inside will has charges too. IF THIS IS TRUE, this is my question; in another experiment, an electroscope sits inside of a Faraday Cage, knob and cage tied via cable. But electroscope has no charge anytime, even cage has. We may not assume cage is COMPLETELY CLOSED. But anyhow, electroscope is never has charge. How? When inside has charges?

------------
BTW, does my English understandable enough?

ok u partially followed me...
know these
1)charges on a conductor reside only at the boundary(the boundary may be an inner one or a outer one).
That is it is possible to have charges on the inner surface of a shell
Ex.U have a [b]thick[/b] shell with a point charge $ q$ at it's centre then the inner surface will have charge $ \minus{}q$ and the outer surface has charge $ q$
2)the conductor is equipotential
3)in ur case the charges will redistribute on inner as well as outer surfaces
4)both the surfaces will have same potential as they are connected by conducting wire

Thank you for your patience. I guess your boredom. But for long years i surmise some wrongs as true. I can not easily change my mind. Maybe for this reason, understand slow. You teach me new things, you clear my pessimism. I am thankful. But feel embarrassed too.

These are final 2 questions(certain, promise :oops: ),

1) About the conductors of above figure, if i would have dip and touch the Z(ball) into the hollow(without a cover) and put out again, some charges would have left on Z(ball)? yes and no sufficient. Please.

2) In my previous post, at last paragraph, mentioned about Faraday Cage experiment. The cage is not completely closed. There are plenty of holes on wire fence. But the electroscope in cage(connected to cage with a wire) don't has charges. When i think comperative, the electroscope would have had charge a bit(or more). Because the wire between cage and electroscope is contacted to outer face of cage. What did i omit now?

Regards


I accidentally posted a HUGE asymptote (10000 size) in my blog, and now it doesn't load no matter what. I meant to post it as size: 1000, but I accidentally typed another zero and clicked submit. Could someone help?

I don't think it can be fixed without going into the SQL or something. Its best to kill your blog, I guess.

I've fixed the bug. It's good that you've found that hole. That's what caused the interruption in our server.


symmetry
combinatorics proposed
combinatorics

Let $M$ be a finite set and $\Omega$ be a set of subsets from $M$ such that :
(i) For all $A,B \in \Omega$, if $|A \cap B | \geq 2$, then $A=B$
(ii) There are $A,B,C \in \Omega$, pairwise distinct, such that $|A \cap B \cap C | = 1$
(iii) For every $A \in \Omega$ and every $a \in M \setminus A$ there is a unique $B \in \Omega$ such that $a \in B$ and $A \cap B = \emptyset.$

Prove that there are numbers $p$ and $s$ such that :
(a) For every $a \in M$ the number of sets which contain $a$ is $p$.
(b) For every $A \in \Omega$, we have $|A| = s.$
(c) $s+1 \geq p.$

My solution sounds a bit strange for an olympiad problem. So what is yours? :P

Pierre.

Let's call the sets lines and the elements of $M$ points :). Furthermore, whenever $d,d'\in\Omega,d\cap d'=\emptyset$.

The third axiom (Euclid's $5$'th postulate, actually :)) says that $\|$ defined by $d,d'\in\Omega,d\|d'\iff$ either $d=d'$ or $d\cap d'=\emptyset$ is an equivalence relation (actually, it says that the relation is transitive, the reflexivity and symmetry being obvious).

We can thus partition $\Omega$ in $\|$ classes. It's also clear from the third axiom that each $\|$ class partitions $M$.

If $\|_1,\|_2$ are $2$ distinct $\|$ classes (we can find at least three distinct classes according to the second axiom), then every line in $\|_1$ has exactly $|\|_2|$ points, because every such line intersects each line in $\|_2$ in exactly one point, and those are all its points. This means that in each $\|$ class $\|_i$, all lines have the same number of points $s_i$. At the same time, all the other $\|$ classes contain $s_i$ lines each, since each line from another $\|$ class $\|_j,j\ne i$ intersects one of the lines in $\|_i$ in exactly one point among the $s_i$. Again, this means that for all $j\ne i$, $s_j$ is the same. This, together with the second axiom (which we fully use this time) shows that all the $s_i$ are equal, and we denote this common value by $s$.

If $p$ is the number of $\|$ classes, then every point in $M$ belongs to $p$ lines.

We have thus shown the existence of $p,s$, and the fact that $p$ is, in fact, the number of $\|$ classes. Take $d\|d',a\in d$. Each $\|$ class different from the one to which $d,d'$ belong contains one line (and only one) passing through $a$ and some point on $d'$. Since $d'$ has $s$ points, it means that except for the class of $d,d'$, we have at most $s$ other classes. In other words, $p\le s+1$, as desired.

Yes, that's what I did (even if I did not call the sets 'lines'). But, I thought that using an equivalence relation was a bit strange for an olympiad problem  ;)

Pierre.


In general how do you do those problems where it gives you $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{z}$ where x, y, and z are integers and it asks like what is the smallest possible value of x+y or how many total possiblilities there are.

What do you mean by how many total values?

Are you sure you didn't accidentally put an "equal" sign instead of an "inequalities" sign or forget to put absolute value?

I don't understand your question; make it clearer.

He probably means something like, how many ordered pairs $ (x,y)$ of positive integers satisfy $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{42}$ or whatever.

[url=http://www.artofproblemsolving.com/Wiki/index.php/SFFT]See here[/url]

Wait, in these types of problems, they usually give you a value of z. Multiply by xyz and factor using Simon's Favorite Factoring Trick.

what if you were asked the find the smallest possible value of x+y

$ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{x\plus{}y}{xy}$ so $ x\plus{}y\equal{}\frac{xy}{z}$. The smallest sum occurs when xy is the smallest possible, so perhaps brute force and find the smallest product?

For greater numbers of unit fractions where expanding and factoring may not help as much, one can use a technique similar to that illustrated [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1336088#1336088]here[/url].

[quote="leewongjao"]In general how do you do those problems where it gives you $ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {1}{z}$ where x, y, and z are integers and it asks like what is the smallest possible value of x+y or how many total possiblilities there are.[/quote]

Good question.


There are six 1 x 2 x  :pi: bricks stacked one on top of another to form a tower six bricks high. Each brick can be in any orientation as long as it rests flat atop the brick (or floor) below it. How many possible heights for the tower are possible?

Why isn't the answer just 3^6?

[hide]assume that the tallest bricks are always on the bottom.  Then all of the bricks on the bottom will have height pi, then all of the bricks in the middle will have height 2 and all the bricks on top will have height 1.

then we just have to divide 6 into 3 parts to get the number of possible heights, which is 28

the answer is not just 36 because if we have one arrangement of bricks and we rearrange them then we have the same height but a different tower[/hide]

this resembles an aime problem i did recently

For a moment I thought "What if they were 1x2x3 bricks ? Then some of the 28 arrangements would give the same height ... but which ones, and how many ?".

Then I realised that there is a much simpler method for 1x2x3 bricks  ....

Yes, 1x2x3 has a nice (simple) alternate method.

Not so nice, but still doable.  1x2x3 is particularly nice, though.

Isn't that an old AIME or AHSME question?


Hola amigos, segun veo no se ha presentado un problema en un buen tiempo, asi es que los animo a resolver este pequeño problema, y que de paso se animen a presentar sus problemas...

Durante el curso de una campaña de elecciones, p suertes de promesas diferentes han sido dadas por los diferentes partidos politicos (p>0). Mientras que varios partidos han hecho la misma promesa, toda pareja de partidos tienen la menos una promesa en comun, sin embargo ninguna pareja de partidos ha hecho lasmismas promesas. Mostrar que no existe mas de 2^(p-1) partidos.

En un conjunto $S$ de $n$ elementos no podemos encontrar mas de $2^{n-1}$ subconjuntos tal que para cualquier par $A,B \subset S$, la interseccion de $A$ y $B$ no es nula.
Esto es ya que $S$ tiene $2^n$ subconjuntos y estos forman $2^{n-1}$ parejas $X, S-X$ (donde $X \subset S$) de las cuales solo podemos contar un miembro.

[quote="manuel"]Un problema interesante (que esta un poco relacionado con lo anterior, quizas esta un poco incoherente pero me vino a la mente y esta bueno jaj) seria demostrar la siguiente generalización:

Dados $n$ subconjuntos convexos de $R^d$, si la interseccion de cualquiera $d+1$ subconjuntos no es nula, entonces la interseccion de todos los $n$ subconjuntos no es nula.[/quote]
¿No eso el teorema de Helly, o al menos una generalización? Lo divertido no es ese sino verlo coloreado y toda clase de generalizaciones horribles que le han hecho a ese pobre teorema

si eso es el de helly...
lo sabes demostrar dave?
Ah! y enseñeme la versión coloreada que nunca he oido de esa


Ya tenia rato que no posteaba nada x aki, debido a que andaba preparandome para la XX OMM, la cual fue la semana pasada y en la que sake ORO!!!!
jeje
Aki posteo el problema 1 del primer dia:

Sea ab un numero  de dos digitos. Un entero positivo n es "pariente" de ab si:
1)el digito de las unidades de n tambien es b
2)los otros digitos de n son distintos de cero y suman a
Por ejemplo, los parientes de 31 son 31, 121, 211, 1111.
Encuentra todos los numeros de dos digitos que dividen a todos sus parientes.

Espero sus soluciones!!!

Si $a=1$ el unico pariente de $ab$ es si mismo.
Si $a=2$ el unico pariente de $ab$ es $11b$ que manualmente nos damos cuenta que no es posible llenar.
Si $a\ge 3$ entonces hay dos parientes de $ab$ que son de la forma:
$n12b$ y $n21b$. La diferencia de esos dos parientes es $90$, por lo que si $ab$ los divide a los dos es porque $ab$ divide a $90$. Los divisores de 2 digitos de noventa (con $a>2$) son 30, 45, y 90. Es facil ver que los 3 funcionan.
QED.

Bueno, tu solucion fue practicamente igual k la mia, solo unas cuantas diferencias:

Para $a=1$ solo ab es su pariente, de manera que si se cumple.
Para a$\geq 2$ consideremos el numero $100+10(a-1)+b$ vemos que $10a+b$ debe dividir a dicho numoer, el cual es igual a $90+10a+b$. Entonces basta con que $10a+b$ divida a $90$, es decir, que sea un divisor de 90, y mayor o igual que 20 ($a\geq 2$). Entonces se cumple para 30, 45 y 90.
 :lol:


[hide]Assume there is such a square, call it $x$ then for some positive integer $k$, $k\geq 2$
\[\displaystyle 2^{k+2}5^k+9=x^2\]
\[\displaystyle \Longrightarrow 2^{k+2}5^k=(x+3)(x-3)\]
Then we have two equations:
\[2^a5^b=x+3\]
\[2^{k+2-a}5^{k-b}=x-3\]
\[\Longrightarrow 2^{a-1}5^b-2^{k+1-a}5^{k-b}=3\]
Consider the above equation mod 2 to prove no set of integers $a,b,k$ can satisfy the equation.[/hide]

[hide]

Suppose there exist a perfect square , there can be two such cases

(i) When the perfect square is in the form $(10k+3)^2$ for $k\in \mathbb{N}$
Then we have $(10k+3)^2=4\times 10^{n+1}+9$ for integer  $n>0$ .

$100k^2+60k=4\times 10^{n+1}$ $\Leftrightarrow$ $k(5k+3)=2^{n+1}5^n$

Since $5k+3$ cannot be divide by $5$ , it left only $k=5^n$ . So $5^{n+1}+3=2^{n+1}$ . But $5^{n+1}>2^{n+1}$ . Hence , no solution in this case.

(ii)When the perfect square is in the form $(10k+7)^2$ for $k\in \mathbb{N}$ similarly

$100k^2+140k+40=4\times 10^{n+1}$  $\Leftrightarrow$ $(5k+2)(k+1)=2^{n+1}5^n$ . So this implies $k=5^n-1$ and $5^{n+1}-3=2^{n+1}$  Again we see that there are no solution here .

From the two cases , there are no way to form a perfect square by the given number .[/hide]


3 nafar miran ye goldoon mikharan 300 toman(ahde bogh masalan) pas nafari 100 toman sahmeshon mishe bad maloom mishe ke 50 toman geroon kharidan sahebe maghaze 50 toman mide be shagerdesh onam 20 toman mizare to jibesh va nafari 10 toman beheshon mide yani nafari 90 toman dadan 3*90=270   ,  270+20=290 payda konid 10 tomane gom shode ra.mamnon misham rahnamaei konid

kafie yebar az dide forooshonade be ghazie negah koni yebar az dide kharidar yebaram az dide shagerde,oonvakh khodet motevajjeh mishi ke hich moshkeli nadare ghazie va hamcehi dorosteo faghat bazie ba kalamateo oon 20 tomane... :)


I'll put here problems that i can;t solve :D, hoping that u will help me :D

a,b - two natural numbers. proof that a^2 + b^2 is the difference of two square roots only if ab is even

Assume that $ ab$ is odd then $ a,b$ are odd.

So we have $ a^2 \plus{} b^2\equiv 2 (\mod 4)$, so if $ a^2 \plus{} b^2$ is the difference of two square roots $ x$ and $ y$, we have $ x^2\equiv y^2 \plus{} 2 (\mod 4)$, which is impossible

prove that
x²y + z ≥ 2x√yz , (v) x,y,z > 0
[/code]

[quote="brasuceava"]prove that
$ x^2y \plus{} z \ge 2x\sqrt{yz} \forall x,y,z > 0$[/quote]
You know what is AM-GM inequality? - If not take time to go through this [url=http://url]link[/url]
If yes then,
$ x^2y \plus{} z \ge 2\sqrt{x^2yz} \equal{}   2x\sqrt{yz}$
Which is the same as -
$ (x\sqrt{y})^2 \minus{} 2(x\sqrt{y}\cdot\sqrt{z} \plus{} \sqrt{z}^2 \ge 0 \implies (x\sqrt{y} \minus{} \sqrt{z})^2 \ge 0$
which is true

\:d/ there goes another one \:d/

Find out real, positive numbers a, b, c knowing that:

$ ( ab \plus{} bc \plus{} ca) \minus{} 1 \geq a^2 \plus{} b^2 \plus{} c^2 \geq 3 ( a^3 \plus{} b^3 \plus{} c^3)$

From QM-AM we get easily that if
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} S$ then $ a \plus{} b \plus{} c \le \sqrt {3S}$ which means $ S \plus{} 2(ab \plus{} ac \plus{} bc) \le 3S$ which means $ ab \plus{} ac \plus{} bc \le S$. Than, going back our inequality, we get:
$ S \minus{} 1 \ge ab \plus{} ac \plus{} bc \minus{} 1 \ge S$ which is impossible.
Hence there are no such numbers.

$ \sqrt{\frac{\sum_{k\equal{}1}^n a_k^2}{n}} \le \frac{\sum_{k\equal{}1}^n a_k}{n}$

correction:

$ \sqrt{\frac{\sum_{k\equal{}1}^{n}a_{k}^{2}}{n}} \geq \frac{\sum_{k\equal{}1}^{n}a_{k}}{n}$

This is the Quadratic Mean-Arithmetic Mean Inequality.

If that looks alien:

$ \sqrt{\frac{{a_1}^2 \plus{} {a_2}^2 \plus{} ... \plus{} {a_n}^2}{n}} \geq \frac{a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n}$

For real numbers $ a_1, a_2, ..., a_n$.


superior algebra
superior algebra theorems

I'm interested in Harari's book on Algebra in English(I don't exactly know it's name, i have Harari's book on groupes,fields,ring but they are in French)-
Does anybody have this book or books in electronic format, or maybe you know where they are available.
I'd be very grateful for someone who could help me. Thanks

if it exists in electronic format, I'd also be interested.
(preferrably english, but french may work too)


Express three-halves of one-half as a common fraction.

naturally, we analyze the problem, break it down.....

3 halves is 3/2

1 half is 1/2

3 halves of 1 half = 3/2*1/2   (the term "of"in math word problems is usually interpreted as "multiplied")

3/2*1/2= [b]3/4[/b]

[b]
Your answer is 3/4[/b]

We have $ \frac{3}{2}(\frac12)\equal{}\boxed{\frac34}$.


inequalities
trigonometry
inequalities proposed

dear mathlinker, i have a problem
let$ x,y,z$ such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$. Prove that $ xy \plus{} yz \plus{} 2xz \le \frac {{1 \plus{} \sqrt 3 }}{2}$

Here is my proof
$ \begin{array}{l} f(x,y,z) \equal{} xy \plus{} yz \plus{} zx \minus{} \frac {{1 \plus{} \sqrt 3 }}{2} \\
f(\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ;y,\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ) \equal{} 2y\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} \plus{} {x^2} \plus{} {y^2} \minus{} \frac {{1 \plus{} \sqrt 3 }}{2} \\
f(x,y,z) \minus{} f(\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ;y,\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ) \equal{} \minus{} {(x \minus{} z)^2}\left( {\frac {y}{{(x \plus{} z) \plus{} \sqrt {2({x^2} \plus{} {y^2})} }} \plus{} 1} \right) \\
\end{array}$
noitice that $ (x \plus{} y)^2 \le2(x^2 \plus{} y^2)$. so we can show that

if $ y > 0$ or if $ y < 0,x > 0,z > 0$ , then $ f(x,y,z) \minus{} f(\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ;y,\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ) \le 0$
if $ y < 0$.WLOG $ x > 0,z < 0$ it follow $ xy \plus{} yz \plus{} 2xz < \frac {{1 \plus{} \sqrt 3 }}{2}$

So, we have proved that $ f(x,y,z) \minus{} f(\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ;y,\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ) < 0$

we have to prove that if $ x \equal{} z \equal{} t$ and $ y^2 \plus{} 2t^2 \equal{} 1$ then $ 2ty \plus{} 2t^2 \le \frac {{1 \plus{} \sqrt 3 }}{2}$ [b] (*)[/b]

[b]but i have a long proof for this, use $ f'(t)$[/b]  :blush:  :blush:

who can help me prove (*)

Thanks

[quote="imedeen"]dear mathlinker, i have a problem
let$ x,y,z$ such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$. Prove that $ xy \plus{} yz \plus{} 2xz \le \frac {{1 \plus{} \sqrt 3 }}{2}$

Here is my proof
$ \begin{array}{l} f(x,y,z) \equal{} xy \plus{} yz \plus{} zx \minus{} \frac {{1 \plus{} \sqrt 3 }}{2} \\
f(\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ;y,\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ) \equal{} 2y\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} \plus{} {x^2} \plus{} {y^2} \minus{} \frac {{1 \plus{} \sqrt 3 }}{2} \\
f(x,y,z) \minus{} f(\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ;y,\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ) \equal{} \minus{} {(x \minus{} z)^2}\left( {\frac {y}{{(x \plus{} z) \plus{} \sqrt {2({x^2} \plus{} {y^2})} }} \plus{} 1} \right) \\
\end{array}$
noitice that $ (x \plus{} y)^2 \le2(x^2 \plus{} y^2)$. so we can show that

if $ y > 0$ or if $ y < 0,x > 0,z > 0$ , then $ f(x,y,z) \minus{} f(\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ;y,\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ) \le 0$
if $ y < 0$.WLOG $ x > 0,z < 0$ it follow $ xy \plus{} yz \plus{} 2xz < \frac {{1 \plus{} \sqrt 3 }}{2}$

So, we have proved that $ f(x,y,z) \minus{} f(\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ;y,\sqrt {\frac {{{x^2} \plus{} {y^2}}}{2}} ) < 0$

we have to prove that if $ x \equal{} z \equal{} t$ and $ 2y^2 \plus{} 2t^2 \equal{} 1$ then $ 2ty \plus{} 2t^2 \le \frac {{1 \plus{} \sqrt 3 }}{2}$ [b] (*)[/b]

[b]but i have a long proof for this, use $ f'(t)$[/b]  :blush:  :blush:

who can help me prove (*)

Thanks[/quote]
I know this is a bit off topic: Let $ \alpha \equal{} \frac{1\plus{}\sqrt{3}}{2}$. Then $ \frac{1}{\alpha} \frac{x^2 \plus{} (\alpha y)^2}{2} \ge xy$, $ \frac{1}{\alpha} \frac{z^2 \plus{} (\alpha y)^2}{2} \ge yz$, $ x^2\plus{}z^2 \ge 2xz$ which gives $ xy \plus{} yz  \plus{} 2zx \le \alpha(x^2\plus{}z^2\plus{}y^2) \equal{} \frac{1\plus{}\sqrt{3}}{2}$. Since $ 2\alpha^2\minus{}2\alpha\minus{}1 \equal{} 0 \iff \frac{1}{2\alpha}\plus{}1 \equal{} \alpha$.

You have a tiny mistake: You have to prove $ 2ty \plus{} 2t^2 \le \frac{1\plus{}\sqrt{3}}{2}$ when $ y^2 \plus{} 2t^2 \equal{} 1$

Let $ \alpha \equal{} \frac{1\plus{}\sqrt{3}}{2}$ again:
$ 2ty \le \frac{1}{\alpha} (t^2\plus{}(\alpha y)^2)$ Which gives $ 2ty\plus{}2t^2 \le \alpha(2t^2\plus{}y^2) \equal{} \frac{1\plus{}\sqrt{3}}{2}$

[quote="FelixD"][quote="Mathias_DK"]I know this is a bit off topic:[/quote]
Why is this off topic?[/quote]
Because he wanted a proof of (*) :P

[quote="Mathias_DK"][quote="imedeen"]i see that your solution is not really simple :([/quote]
Why don't you think so?[/quote]

because i have a proof for (*) use $ f'$ seem nicer :)  ( i think so  :( )


Let AC,AB are tangents from A to (O).OA meets  (O) at H ,I,D (O in segment ID,H in segment IA).Bisector of DCO meets DB,(O) at E,F;AE meets FH at K.Pover that IK perpendicular with DB

[quote="Herodatviet"]OA meets  (O) at H ,I,D (O in segment ID,H in segment IA)[/quote]

How on Earth does a line meet a circle in three points? :?


[quote="ksun"]Show that if $ \delta(G) \geq 3$ then $ G$ contains a circuit with a chord.[/quote]
I'm sorry,but what does $ \delta(G)$ mean?

[quote="ksun"]$ \delta (G)$ denotes, for a given graph $ G$, the minimal degree of the vertex; so if $ \delta (G) \geq 3$ then every vertex of the graph has a degree at least 3.[/quote]
[b]Proof:[/b]
-Consider a path of a maximum length:$ A \equal{} A_1A_2\dots A_n$,
Obviously $ A_1$ is connected only with vertexes from $ A$(otherwise we can find a path with longer length),but degree of $ A_1$ is at least $ 3$,so there exist $ A_i,A_j\in A$,such that $ A_1$ is connected with $ A_i$ and $ A_j$($ i,j\neq 2$).
WLOG suppose $ i > j$,but then $ A_1A_j$ is a chord in a cycle $ A_1A_2\dots A_i$.


Three numbers a, b, and c, none zero, form an arithmetic progression. Increasing a by 1 or increasing c by 2 results in a geometric progression. Find b.

You posted this problem before in another forum
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=48229]http://www.mathlinks.ro/Forum/viewtopic.php?t=48229[/url]

Oups. I don't remember these things. That was a month ago though.  ;)

I think that this is still a good practice problem and should stay.

OK ;)
So other people, DONT click on my previous link :D

This question has not yet been answered?  :huh:

Guess my posts are not good...  :(


Triangle $ ABC$ with $ AB \equal{} AC \equal{} a$, and altitude $ AH$. Construct a circle with center $ A$, radius $ R$, $ R < a$. From points $ B$ and $ C$, draw the tangents $ BM$ and $ CN$ to this circle ($ M$ and $ N$ are the points of tangency) so that they are not symmetric with respect to the altitude $ AH$ of triangle $ ABC$. Let $ I$ be the point of intersection of $ BM$ and $ CN$.
1. Find the locus of $ I$ when $ R$ varies;
2. Prove that $ IB\cdot IC \equal{} |a^2\minus{}d^2|$ where $ AI \equal{} d$.

Notice BM=BM’ =CN=CN’  =〉 △BMA=△CAN   =〉 ∠BAC=∠MAN
=〉 BACI are concyclic
IB*IC=BM^2-MI^2=a^2-R^2-( d^2- R^2)= a^2- d^2
Where  R is the radius of the circle A


search
inequalities
geometry
email
number theory
IMO

[quote="Davron"]People!

I think that last year this forum ( http://www.mathlinks.ro ) was much more cooler, there were many active users which are not using this website now ... really oops!!

Here is my question:
who can say what are the reasons??

davron latipov[/quote]
Hey Davron why have you disaded that now there are less active users than last year?For me I just can say that now I'm less active now than last year but I'm just have no time to spend on math,but I hope from Junuary of 2007 I will again become active in solving problems and posting solutions here:)!

Davron is right.

What I really miss for example are the discussion about really difficult (but solved/solvable) nice problems (like in the beginning of MathLinks, look up thread numbers under 20000). Now most problems posted are reposts by people that don't search before. And the rest is often "standard" olympiad stuff, often not trivial, but still not of the type above. It got a bit better the last days (e.g. Harazi posts again), but it still is not like in earlier times (I joined more at the end of this nice period, so older users can probably judge better, but this my oppinion).
That what drove Harazi away from Inequalities is also happing in the other forums, just a bit slower.
I'm not sure if one reason simply is that there are not many nice problems left in the world.

I think a part of what Davron means is just the current point of a slow but constant lowering of niveau here (as all the spam shows)...
But it's difficult to express the happaning in total.

[quote="silouan"]Do you believe that that there exist a way to help the forum ? :maybe:[/quote]

of course we can help the forum ...

davron latipov

Guys during the last months I wasn't so active in watching topics here and didn’t really notice that this problem has grown to such big sizes, but I think the roots of this problem were seeable long ago. Saying the truth I think that one of the reasons for this is the neutral position which in some cases take moderators. What I am actually trying to say is that the moderators should sometimes be severer about persons who tries to spam here, or tries to fill themselves like they can do everything they want in this forum. This kind of behavior in some cases confusing the true funs of math and they just didn’t posting here. This question is really needs to be solved and we all have to try to help moderators solving this big problem, if not in the near future our most favorite forum will notable increase in its quality and rank and soon will become like thousand other middle quality math forums in the net.
Thank you for attention!
Tigran

[quote="Tiks"] What I am actually trying to say is that the moderators should sometimes be severer about persons who tries to spam here, or tries to fill themselves like they can do everything they want in this forum. [/quote]
It is true. But I see that in many forums, the moderators listed for the forum are no longer active in that forum, or even on the site at all! So perhaps it is not just because our moderators are not strict enough, but we don't have enough -- the number of moderators decreases while the number of users increases. Maybe some of us should volunteer to moderate the forums that we care about.

I generally agree with Zetax about the Old mighty Mathlinks topics(I wasnt born in mathlinks at that time , but i have seen the posts)

But Xevarion , What do you expect from the moderators , Now that the summer is over , they have their lives to cope with.
Although, Mathlinks/AOPS site is a magnificent place to learn new facts and improve your skills, Its just an artificial world
built in the Memories of computers.

No, ZetaX, there are still infinitely more nice, difficult problems in the world that are still not discussed on mathlinks. This is not a reason for what is happening here. It's quite simple actually: at the beginning of mathlinks, there were much fewer users, practically all passioned about nice difficult problems and not about doing their homework or posting for 189020 times the same problem. Now, things have changed, every day there are more posts than we had in a whole week last year. Interesting enough, when you want to struggle with a hard problem, or even with an easy but nice one, you don't look through the problems posted recently (unless the topic is started by sam-n or vess or some other 2 or 3 people), but you search in the old topics (at least, this is what I'm doing). It's such a pity all the problems posted nowadays are just reformulation of old discussed problems. Why on earth do we have a search button, which works actually very well? Why on earth do we accept in the advanced area all kind of homework exercices and stuff with absolutely no value? I think I said on the forum at least 10 times that there should be created new sections, with the most valuable topics, in a few number, so that we shouldn't be obliged to search for hours for them. Of course, nothing happenned. I also asked at least 4 times to erase my name from the moderators list-name, because I have no longer the time and especially the patience to read all this boring stuff and to take actions. Nothing happened. I proposed that active users should be moderators: nothing again. Then, how can you expect the same level of discussions? Unfortunately, there are also many extremely good users who left us and this is again a pity. But how to convince them to discuss problems for homework or problems they have already seen 10 times? I really hope that mathlinks will not become just a place where each discussed topic has  4 links to previous discussions on that theme. As I said, there are still (too) many problems left!

Agreed!

What do you think about a subforum just for threads we are sure that were posted before¿
One who wants new stuff could ignore it. The difference would be that the moderators move the threads there, even when _not_ giving a link. It will be let to the proposer or friendly users to find them.
And I would also be for the section with really good problems, but who decides¿

The other is stronger actions against people reposting without any attempts of search (like even using the title gives the result).

And well, http://www.artofproblemsolving.com/Wiki/index.php/AoPS-Mathlinks_Rules_and_Tips should finally be forced on all (new) users to be read.

wouldn't it be more practical if the moderators just started "cleaning" the advanced section from all kind of rubbish which flourishes among the few interesting threads?
I mean taking threads as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107397]this[/url] to the preolympiad or HS basics etc. Now I don't know if that would really help solve our problem but it would at least bring some seriosity to the forum.
It is really embarracing to have such discussions in the olympiad (advanced) section. Now I understand the "old" users complaining for the situation at the moment :lol:

About $\frac{2}{3}$ of the posted problems this week are ones already posted I think.
We can't simply delete them, nor can we just move them to Solved Problems without giving the author anything. Thus best is to make a subforum for this I think.
And cause of the very good topic naming (combined with the lack of being able to search TeX-code) it's simply impossible to find anything  :roll:
When I started moderating, I regulary corrected titles, but if the user again and again post that way (some even changed them back!), you will give up or at least do it less regulary...

To the linked thread: well, at first one should delete the spam there ;)
Then moving is probably best.

[quote="ZetaX"]
We can't simply delete them, nor can we just move them to Solved Problems without giving the author anything. [/quote]
ZetaX I didn't get your point completely, what do you mean saying “we can’t”? You aren’t supposed to do that or you just unable? I think if someone can delete or replace the topics (or spams) from the places that they aren’t belong is the best way to get back the popularity of ML!

Well, if I want, I could, but it would not be nice to those persons who posted this (they often have no idea that it was posted before).
Another way of solving the repost-problem is deleting them and notifying the proposer that it was posted.

The problem is really caused in searching: as already said, most user don't think of it, and the way search works and how titles are named makes it hard to search.
For finding a topic I've no clue about I need more than five minutes, if I've an idea of what was posted there less (just a heuristic, I never stopped ;) ).
It would be so simple of all users would take deep searches for their stuff, it would only take them ten minutes per problem; but for the moderators, this five till ten minutes add up to hours, but if they could just move topics away without being "forced" to give a link, this hours will get minutes.

And it would be nice to help moderators by always linking to older threads if possible, even at/from older topics :)
A link is never spam if it has anything to do with the problem, but some people seem to see this different...

Well, I will probably try a (even) stricter rule in number theory, especially deleting anything spammy without notifying. But I already see people complaining  :roll:

[quote="harazi"] I also asked at least 4 times to erase my name from the moderators list-name, because I have no longer the time and especially the patience to read all this boring stuff and to take actions. Nothing happened. I proposed that active users should be moderators: nothing again. Then, how can you expect the same level of discussions? Unfortunately, there are also many extremely good users who left us and this is again a pity. But how to convince them to discuss problems for homework or problems they have already seehttp://www.artofproblemsolving.com/Forum/posting.php?mode=quote&p=646797n 10 times? I really hope that mathlinks will not become just a place where each discussed topic has  4 links to previous discussions on that theme. As I said, there are still (too) many problems left![/quote]

Who did you ask?  Sending an email to aops@artofproblemsolving.com is the best way to get the admins' attention.

Every society,empire, forum :)  etc.. has its good and bad days.
Let us hope that AoPS will prosper again!!!  :lol:

The reason for appearing easier and easier problems is simple: there are more people logging into forum and with growing popularity the level has to decrease, that's kind of natural to me. There are many people who after getting the problem have a strong urge to share it with the whole world.

Another subforum with extraordinary problems/solutions should be surely made. Posting new topics into it should be locked with just continuing what was already started available. Other questions is how to choose these problems as if it were only a work for moderators it would take HUGE amount of time. I'd imagine some system of notifying by users (other than PMs) would work well if made simple ,e.g.: you can just tick the problem you like if you have enough number of posts [that's the only barrier which could be made from people just doing it for fun I guess], and then moderators could see a database  of votes that would contain informations on who voted and on which problem.


hi
I have a table like this
\begin{tabular}[hb]{c|l|c|c|c|c|}

Objectifs & Sous objectifs & Paul & L-P & Kevin & P-A \\ \hline

Définir le projet & Compréhension du projet & R & R & R & R \\ \cline{2-6}
& Définir les objectifs & S & R & S & S \\ \cline{2-6}
& Conception de la matrice des responsabilités &  & R &   & S \\ \hline

Recherche d'informations & Recherche(Internet, périodiques, etc) & R &  & R &  \\ \cline{2-6}
& Analyse de l'information trouvée & R &  & R &  \\ \cline{2-6}
& Prise de mesures sur le moteur & A & R &  & S \\ \hline

\end{tabular}

and i want want to enumerate the items in the first colum and have subnumber in the second colum to have something like that

1. Bla        1.1 bla bla
                 1.2 blabla
2. Buz       2.1 buzbuz
                 2.2 ....

and so on...all in different cells

also... :blush:

is there a special command to write an exponential  else then \mathrm{e}^x  ?

and on a title page, how can I add more info like the location or the person to who the report is given?

thank you very much!!

ledurt

[quote]and on a title page, how can I add more info like the location or the person to who the report is given? [/quote]

I mean in the same way

\title{Analyse vibratoire}
\author{Paul Doyon (04 500 800)\and
Louis-Philippe Riel (04 480 225)\and
Kevin Tremblay (04 498 537)\and
Pierre-André Trudel (04 515 331)}
\date{\today}
\maketitle

is there a command to add the infos I talked about?

thanks again
ledurt


vector
geometry
geometric transformation
reflection
linear algebra
linear algebra unsolved

Let $ V$ be a finite dimensional complex vector space.
Let $ T_1$, $ T_2$, ... , $ T_m$ span the dual space of $ V$.

If there exists $ x$ in $ V$ such that, for any $ y$ in $ V$,
$ T_1 (x) T_1 (y) \plus{} T_2 (x) T_2 (y) \plus{} ... \plus{} T_m (x) T_m (y) \equal{} 0$

Show that $ x \equal{} 0$.

The idea is the following: consider the sum T2+...+Tm. It is easy to prove that its orthogonal is $ A_1: \equal{} \bigcap_{i \equal{} 2}^m T_i^{\perp}$, and it has to be $ \neq 0$. Moreover, there exists $ y_1$ in $ A_1$ such that $ T_1(y_1)\neq 0$. Using $ y \equal{} y_1$ you show that $ T_1(x) \equal{} 0$. So $ T_i(x) \equal{} 0$ for all $ i$ and then x=0.

We should be able to write this entirely as matrices.

We have that $ T$ is an invertible matrix. ($ T_1,T_2,\dots,T_n$ are the rows of that matrix.)

We are given that $ \exists\,x$ such that $ \forall\,y,\ (Tx)^T(Ty)\equal{}0.$ That is, $ x^TT^TTy\equal{}0$ for all $ y.$ That means that $ x^TT^TT$ must be the zero vector, since it generates the zero functional. We can take the transpose to write this as $ T^TTx\equal{}0.$ But since $ T$ was given to be invertible, we also have that $ T^TT$ is invertible, and hence $ x\equal{}0.$

Yet another proof:

The condition implies that the linear functional $ \sum_{i} T_{i}(x)T_i$ is identically zero, but since the $ T_i$ are a basis, this means that $ \forall i$, $ T_{i}(x) \equal{} 0$.  But that means that $ x$ is mapped to zero by every single element of the dual space.

The last statement implies that $ x \equal{} 0$, and there are a million ways to see that.  Pick an inner product and consider $ \cdot \minus{}> < x,\cdot >$, for example, or argue that the dual space would have dimension less than $ dim\ V$.

Edit: On further reflection, this is exactly the same as Kent's proof, though I think it still adds something to the discussion.


[img]http://i2.photobucket.com/albums/y8/XHellsFireX/alge.png[/img]

I'm not very good at Algebra. I have no idea how to go about solving both of these to figure out whether or not this is a true statement.

What is the bold equal sign in the middle supposed to be.

Anyways, interpreting these equations in any of the >4ish ways possible, it seems false.

This should be moved to High School Basics. This forum is olympiad level.


If 5 geometric means are inserted between 8 and 5832, what's the fifth term in the geometric sequence that starts with 8?

[hide="Solution"] 5832/8=729, which is 3^6, conveniently. Then the common ratio is 3, and the sequence is 8,24,72,216,648,1944,5832... The fifth term is 648.[/hide]


Could someone explain what an international film is, along with some examples?  Too bad I don't get chance to watch movies.  :(

Hmm...a film that comes from a country other than your own? I live in America and some of my favorites are Pan's Labyrinth, Delicatessen, The City of Lost Children, Amelie, Yuume Juya, Potemkin, The Bubble, The Edukators, The Seventh Seal, and Cha no Aji.


Does anybody know the area of a pentagon with all of its angles equaling 108 degrees in terms of the side length, x?

It is $\frac{5x^2}{4\tan{36}}$

For a regular n-gon, with side s, the area is:

$\frac{ns^2}{4\tan{\frac{180}{n}}}$

[quote="nat mc"]It is $\frac{5x^2}{4\tan{36}}$[/quote]
$\tan 36^\circ$ is a constructible number and therefore you can expressed it in terms of radicals. Challenge: find it  :D

Trigonometry is definitely above the level of MATHCOUNTS.  Please do not use trigonometry in this forum.  :police:

There are solutions to pgpatel's question that do not require trigonometry.  If nobody gives it a whirl, we'll have to move this topic up to a High School forum.

woah! slow down there. how did you get $A = \frac {a^2}4 \sqrt {25+10\sqrt{5}}$ without knowing $\tan 36$?

also, how would you get $\frac{5a^2}{4\tan{36}}=\frac{a^2}4\sqrt{25+10\sqrt 5}$ without knowing $\tan 36$ or the area formula? It seems to me that you need one or the other to get the other your way....

but anyway, I'll do the simplifying

$\tan 36 = (\frac{\sqrt{25+10\sqrt 5}}{5+\sqrt 20})(\frac{5-\sqrt 20}{5-\sqrt 20})$

$\tan 36 = \frac{5\sqrt{25+10\sqrt 5}-\sqrt{500+200\sqrt 5}}{25-20}$

$\tan 36 = \frac{5\sqrt{25+10\sqrt 5}-10\sqrt{5+2\sqrt 5}}5$

$\tan 36 = \sqrt{25+10\sqrt 5}-2\sqrt{5+2\sqrt 5}$

$\tan 36 = \sqrt 5\sqrt{5+2\sqrt 5}-2\sqrt{5+2\sqrt 5}$

$\tan 36 = (\sqrt 5 - 2)(\sqrt{5+2\sqrt 5})$

$\tan 36 = (\sqrt {9-4\sqrt 5})(\sqrt{5+2\sqrt 5})$

$\tan 36 = \sqrt {45 + 18\sqrt 5 - 20\sqrt 5 - 40}$

$\tan 36 = \sqrt {5 - 2\sqrt 5}$

If you can simplify that rad I don't know how...

I thought that I'd post where this came from. It comes from the equation to find the area of a regular n-gon, which is:

asn, or apothem x side-length x number of sides. This can also be shown as ap or apothem x perimeter.

(*the apothem is line from the center of the n-gon to one of its vertices)

By drawing out the apothems, you get several congruent triangles. The trigonometry that shouldn't be posted in this forum,  :P , is used to find the area of one of these triangles, and then you multiple by the side length to get the total area.

Although the original question by pgpatel is answered, the solution required methods beyond MATHCOUNTS level.

Mods: pls move to appropriate level

[quote="Hamster1800"]woah! slow down there. how did you get $A = \frac {a^2}4 \sqrt {25+10\sqrt{5}}$ without knowing $\tan 36$?
[/quote]

$A = \frac {a^2}4 \sqrt {25+10\sqrt{5}}$ is the formula for the area of a regular  pentagon. I got it from MathWorld. It has nothing to do with nat mc's formula, except for the fact that it gives the same result  :lol:

but how would you derive the formula $A = \frac{a^2}4\sqrt{25+10\sqrt5}$ without $\tan 36$? I'll check mathworld but i still don't get it...

also, I don't think any regular polygons except triangles, squares, and hexagons are in mathcounts. I could be wrong though.

no, I don't think you'll need it. If they DO involve pentagons on MC problems though, chances are that they'll give you the formula, that or finding the area will be unnessesary

[color=darkblue][size=75]Moved from the Middle School MATHCOUNTS Forum by moderator rcv[/size][/color]

Definitely is.

My class at school was just introduced to it, and we're in 10th grade, pre - IB (the highest level at the school).

Not saying I didn't learn it a bit before then, but that's how schools around here, and I assume most places, work.

Well, now that trig is legal, I'm still wondering how you would get the formula for area without trig. I'm watching the other post, and the golden ratio is there, but I don't quite get how you can find that that is the golden ratio (the pentagram inside the pentagon) or how you can find stuff using angle measures without using trig. I think trig is the easiest way but other ways work.

This stuff was in Mathcounts ??? Anyone telling as to how you can solve the stuff without using trignometry ??
P.S : Problem is , now that I do know trignometry , that's the first thing that comes into my mind when I see such problems.....  :(

well the way to get it without trig is in mathcounts now but I think trig is the easiest way :)


abstract algebra
superior algebra
superior algebra unsolved

In a finite group $G$, whose order is not divisible by 3, for any $a,b\in G$ we have $(ab)^{3}=a^{3}b^{3}$. Prove that $G$ is abelian.

[quote="finister"]... ${=(ba)^{3}=ba^{2}b^{2}a}$ ...[/quote]
No.
You can't prove it without using the second condition, there are counterexamples (even finite ones).

[quote="ZetaX"]You can't prove it without using the second condition, there are counterexamples (even finite ones).[/quote]

You are right, I've corrected the proof. Sorry for not paying enough attention.

let F endmorphisme in G such that F(x)=x^3. .we have ord(G)=3k+1or3k-1 then we can prove easly that F is surjective.and we know that if a endomorphisme is surjective in G.then G is abelian. :ninja:

[quote="fermat3"]and we know that if a endomorphisme is surjective in G.then G is abelian. :ninja:[/quote]
¿¿¿
You need to make some restrictions on it ;)


This problem requires logical thinking, so get ready.

3 Mathematicians (we call them this so that they can be considered able to figure out any logical condition) are standing in a line so that the last one can see two in front, the middle can see the one in front, and the first can see none.

A hat-seller comes along with 5 hats - 3 white, 2 black.  He shows them to the 3 mathematicians.  He places 1 on each head randomly and hides the remaining 2.  Then he says:

"I will give as much money as could possibly be wanted to the first mathematician that can name the color of the hat on his head.  If you cannot justify your solution, you will not get the money."

No mathematician talks, turns around, or communicates in any way with anybody, looks in any mirror, or anything.  [b]After a long pause[/b], one mathematician speaks up.  Who is he?  He states the color of the hat on his head.  What color is it?  He gives his justification.  What is it?

The solution to this problem is under the topic "Solution to 3 mathemeticians" which will be posted in a few minutes.  If you have a question, post it.  If you have a solution, do [i]not[/i] post it.  Check your answer.  If you want a hint, the biggest hint is in this problem in bold.

This problem belongs in the Puzzles and Brainteasers forum.

ok suppose the first peron has a black hat, if the second person has a black hat then the third person will know right away.  but if the third person doesn't know right away then the second person will know that he must have a white hat.  So thus, if the second and third people don't know fairly fast then the first person knows that he must have a white hat.


Hey guys  :)  I am in year 9 and was interested in studying for the Australian mathematics competition and the best books i have seen around are the AoPS volume 1 and 2. I was wondering are these books suitable for an Australian maths comp as they are written in the USA?
An example of some practice questions are here http://www.amt.edu.au/wuamcqi04.pdf or http://www.itute.com/mathline/2006_aus_maths_comp_int_sol.pdf.(the second link is only solutions which may give an idea of the difficulty )
Of what i have seen of volume 1 it seems to be all high school basics, but are there any problem solving tricks in there?
 :( Any help would be appreciated!

woops :blush:  sorry guys i was actually asking for a friend that is in year 9, i am actually in year 12........dam it now i look like an idiot.

Please don't curse.

It should be fine for him.

sorry. Are you offended because your christian?

The content should be fine. There aren't any specifically Australian problems, but math contests have more in common with each other than with the regular curriculum anywhere.
The question I don't know the answer to- does AOPS ship them outside North America?


Show that there are no positive integers $ a,b$ such that both $ a^2\plus{}b^2$ and $ a^2\minus{}b^2$ are perfect squares.

$ a^2 \minus{} b^2 \equal{} n^2$ and$ a^2 \plus{} b^2 \equal{} m^2$ so we have $ m^4 \minus{} n^4 \equal{} (2ab)^2$
now i prove that this equation dont have solution in $ N$:$ x^4 \minus{} y^4 \equal{} z^2$
suppose the inverse ,take on solution such that $ x$ is minimum($ x > 0$).
we have$ (x^2 \minus{} y^2)(x^2 \plus{} y^2) \equal{} z^2$ but $ (x^2 \minus{} y^2)$ and$ (x^2 \plus{} y^2)$ are coprime.
so we have :
$ (x^2 \minus{} y^2) \equal{} s^2$
$ (x^2 \plus{} y^2) \equal{} r^2$
this means that:$ r^4 \minus{} s^4 \equal{} (2xy)^2$ but we have $ r < x$.

Nice, short and simple. Thank you for the beautiful solution. :)

[quote="cfheolpiixn"][quote="sumita"]$ a^2 \minus{} b^2 \equal{} n^2$ and$ a^2 \plus{} b^2 \equal{} m^2$ so we have $ m^4 \minus{} n^4 \equal{} (2ab)^2$
$\boxed{\text{now I prove that this equation dont have solution in  }  \mathbb{N}: x^4 \minus{} y^4 \equal{} z^2}$
suppose the inverse ,take on solution such that $ x$ is minimum($ x > 0$).
we have$ (x^2 \minus{} y^2)(x^2 \plus{} y^2) \equal{} z^2$ but $ (x^2 \minus{} y^2)$ and$ (x^2 \plus{} y^2)$ are coprime.
so we have :
$ (x^2 \minus{} y^2) \equal{} s^2$
$ (x^2 \plus{} y^2) \equal{} r^2$
this means that:$ r^4 \minus{} s^4 \equal{} (2xy)^2$ but we have $ r < x$.[/quote]
I think the last line $ r<x $ is not correct,
since $ x^2+y^2=r^2 $ implies $r>x $, which makes the descending method fails.

Does anyone come up with other ways to prove it?[/quote]


in $$ mode, if you want to include text into it by \text{}, the spacing between the math stuff and the text comes out really weird.  is there a way to fix that?

Have you put spaces at the beginning and end of the text box ie \text{ this is text }? If that's not the problem could you show us a minimal example? It looks OK here: \[\sqrt{2}\text{ this is text }\frac{1}{2}\]