I practiced previous sprint rounds when I first started MathCounts, and I also studied from the AoPS book.

/shameless plug

AoPS volume 1, and the introduction books are good
Practice some sprint rounds, target and coun tdown rounds.
You can get some from this site:
http://freewebs.com/tzpublicsite

[quote="Arvind_sn"]Really, I am just starting with my own competitions, so what I am doing is hanging around this site a lot and doing AoPS 1... :D[/quote]
This website IS very helpful, but more to the experienced competitors. There is relatively little activity in the middle school forums compared to the high school forums. I think the MathCounts forum is a good place for random problems, though many topics there aren't actually about math.  :roll:

[quote="als2312"]I plan to participate in Mathcounts next year. This year I take algebra 1 for my school math course. It'll be my first year doing competition math. What should I do over the summer to prepare? What did you guys practice when you first started?[/quote]

Try to find good books to practise, and start little by little, not half a book in 1 or 2 days.and do not forget!! try to solve problems and more problems!!!!!

I took Mathcounts last year as a sixth grader.  I didn't prepare well enough, and that showed in the state competition, where I got 63rd place(not even close to a national qualifier!).  Practice sprint rounds and look at AoPs books(especially Introduction to Algebra!)

I would practice previous competitions and handbook problems. Also, in the handbook, there's a list of concepts , math vocabulary, and formulas. I'd read over that.

3 useful sites (other than aops)
mathcounts.org (obviously)
http://www.mathcountscentral.com/
http://www.mnspe.org/mathcounts/coach.htm

Well the amount you should study and what you should study honestly depends how far you want to advance and what state you live in. If you just want to make states, then dont go crazy (unless you live someplace like California or Texas). The reason I think a lot of beginners dont make nationals is that they only aim low like top 15 in chapters. If you want to make nationals, aim high and study your *tail* off.
As for what to study, I would say do the AOPS volume1 before you start loading on the practice tests (practice state and national tests, and make sure you really understand the answers). After a few (5 or so) practice tests, find out what you are weak in and then read that chapter in your AOPS book.
Last year was my first year in Mathcounts, and my only goal was making states. I did, but after that I really didnt have any goals. After going to the state competition and watching all the beastly people there (including my teammate who got 1st in state), I got really inspired and now I am really studying like crazy.

You could also get some books here:
[url]http://www.artofproblemsolving.com/Books/AoPS_B_CP_MC.php[/url]
They're really helpful for practice.

You may also be intereted in this:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276439[/url]

Apart from all of these, you should also be doing a lot of studying. Make sure you know
[b]all[/b] the stuff from your Algebra 1 Class, and you can also take AoPS Classes if you want to.

 :)

Guys, do you realize that this thread was created THREE YEARS ago??


How many times does the digit 1 appear when all positive two-digit numbers are written?

uh... i think the answer is 19:
10, 11, 12, 13, 14, 15, 16, 17, 18, 19
     21,
     31,
     41,
     51,
     61,
     71,
     81,
     91,

Counting all the ones gives us [b]19[/b].

Or you can remember there are 10 numbers with a tens digit 1, and 9 [u]two[/u] digit numbers with ones digit 1.
11 is counted twice, but contains 2 ones, so its negligible.   10+9= 19.[/u][/b]

Kouichi Nakagawa um your kinda wrong it is 19

Kouichi Nakagawa
It's asking for tens, which doesn't include 1, so it makes it 19 and not 20


quadratics
number theory proposed
number theory

If Q and q are quadratic residues of m and Q - q = r (mod m) for a given r, find Q and q. NB. r is odd and m is even or m is odd and r is even.

I don't quite understand what the content of this problem is.

e.g. Q - q = 12 (mod 43) where Q and q are quadratic residues of 43

It had many solutions. Easy way t o find one $r+sm=4kl\Longrightarrow r=(k+l)^{2}-(k-l)^{2}(mod \ m)$.

I'm not looking for r, I'm looking for Q and q. My solution is as follows-

Clearly r*m + r = r(m + 1) = r (mod m)

By Fermat's Factorisation Method we have-

x = (m + 1 + r)/2 and y = (m + 1 - r)/2 so that x and y are the roots of Q and q respectively.

x^2 - y^2 = Q - q = r (mod m)

From the above example we have-

x = (43 + 1 + 12)/2 = 28
y = (43 + 1 - 12)/2 = 16

28^2 - 16^2 = 10 - 41 = -31 = 12 (mod 43)


modular arithmetic
number theory unsolved
number theory

Sequence $ {a_n}$ satisfiess $ a_n\equal{}2^n\plus{}49$.Find all positive integer $ n$ such that $ a_n\equal{}pq$,$ a_{n\plus{}1}\equal{}rs$ where $ p,q,r,s$ are all primes and $ p<q,r<s$,$ q\minus{}p\equal{}s\minus{}r$

[quote="math10"]Sequence $ {a_n}$ satisfiess $ a_n \equal{} 2^n \plus{} 49$.Find all positive integer $ n$ such that $ a_n \equal{} pq$,$ a_{n \plus{} 1} \equal{} rs$ where $ p,q,r,s$ are all primes and $ p < q,r < s$,$ q \minus{} p \equal{} s \minus{} r$[/quote]

Nice !
[hide="My solution"]

$ 2^n\plus{}49\equal{}pq$ with $ n>0$ and so $ p,q$ odd.
Modulus $ 3$, we get $ pq\equal{}(\minus{}1)^n\plus{}1\pmod 3$ and $ rs\equal{}\minus{}(\minus{}1)^n \plus{}1\pmod 3$ and so one of the four primes is $ 3$. It cant be $ r$ or $ s$ since then $ p$ would be $ 2$ even. So $ p\equal{}3$

Let $ a\equal{}q\minus{}p\equal{}s\minus{}r$
We get $ 2^n\plus{}49\equal{}3q$ and $ 2^{n\plus{}1}\plus{}49\equal{}(3\plus{}a)(q\plus{}a)\equal{}3q\plus{}a(q\plus{}3)\plus{}a^2$ $ \equal{}2^n\plus{}49 \plus{}a(q\plus{}3)\plus{}a^2$ and so $ 2^n\equal{}a(q\plus{}3)\plus{}a^2$

So $ a|2^n$, and so $ a\equal{}2^u$ and $ 2^{n\minus{}u}\equal{}q\plus{}3\plus{}2^u$ which implies $ q\equal{}2^{n\minus{}u}\minus{}2^u\minus{}3$ and so $ 3q\equal{}3\cdot 2^{n\minus{}u}\minus{}3\cdot 2^u\minus{}9$

So $ 2^n\plus{}49\equal{}3\cdot 2^{n\minus{}u}\minus{}3\cdot 2^u\minus{}9$ $ \implies$ $ 2^n\plus{}58\equal{}3\cdot 2^{n\minus{}u}\minus{}3\cdot 2^u$ and so $ \min(n\minus{}u,u)<2$ (else $ 4|58$)

$ 2^{n\minus{}u}\equal{}q\plus{}3\plus{}2^u$ implies $ n\minus{}u>u$ and so either $ u\equal{}0$, either $ u\equal{}1$ but $ u\equal{}0$ $ \implies$ $ a\equal{}2^u\equal{}q\minus{}p\equal{}s\minus{}r\equal{}1$ impossible.

So $ u\equal{}1$ and $ a\equal{}2$
$ 2^n\plus{}58\equal{}3\cdot 2^{n\minus{}u}\minus{}3\cdot 2^u$ becomes $ 2^n\plus{}58\equal{}3\cdot 2^{n\minus{}1}\minus{}6$ and so $ 2^{n\minus{}1}\equal{}64$

So $ n\equal{}7$ and $ 3q\equal{}2^n\plus{}49$ gives $ q\equal{}59$

Hence the result :
$ 2^7\plus{}49\equal{}3\cdot 59$
$ 2^8\plus{}49\equal{}5\cdot 61$

[/hide]


What simplified common fraction is equivalent to $ 0.0\overline{57}$?

$ 0.0\overline{57}$ is the same as $ \frac{1}{10}$ of $ \frac{57}{99}$. ($ 0.\overline{ab}\equal{}\frac{ab}{99}$) Thus, the answer is $ \frac{57}{990}\equal{}\boxed{\frac{19}{330}}$.


floor function
pigeonhole principle
graph theory
Ramsey Theory
combinatorics unsolved
combinatorics

Six points are connected in pairs by lines, each of which is either red or blue. Every pair of points is joined. Determine whether there must be a closed path having four sides all of the same colour. (A path is closed if it begins and ends at the same point.)

Yes such a path must exist.
Let $A,B,C,D,E,F$ be the vertices. This is a well-known result that the graph contains a monochromatic triangle, say $BDF$ and this triangle is red.

Assume that there is no monochromatic path with length $4$.
Then, among the edges $AB,AF,AD$ at least two, say $AB,AF$ are blue.
Then, among the edges $EF$ and $EB$ one is red (otherwise $ABEF$ is blue) and the other is blue (otherwise $DBEF$ is red). Let say that $EB$ is red and $EF$ is blue. Thus $ED$ is blue (otherwise $DEBF$ is red).
As above, to avoid a red quadrilateral $BCDF$, one of the edge $CB$ and $CD$ is blue. This forces $CF$ to be red (to avoid a blue quadrilateral). In that case, both $BC$ and $CD$ are  blue.
Then $AD$ is red (otherwise $ABCD$ would be blue).
Then $AE$ is blue, otherwise $ADBE$ would be red.
But, if $EC$ is red then $ECFB$ is red. And if $EC$ is blue then $ABCE$ is blue. A contradiction.

Pierre.

See also the (much later) post [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=392171&hilit=six+points&[/url].

Note that the famous bound $||G|| \leq \left \lfloor \frac {n} {4} ( 1 + \sqrt{4n-3}) \right \rfloor$ given by Istvan Reiman for a graph with $|G| = n$ containing no $4$-cycle, only yields $||G|| \leq 8$. This is not enough for an immediate conclusion, since, by pigeonhole principle,we can only guarantee that out of the $\binom {6} {2} = 15$ edges of a bichromatic complete graph $K_6$ at least $8$ will bear a same color.

A short proof, not requiring the result that a bichromatic $K_6$ contains a monochromatic $3$-cycle (first non-trivial Ramsey number $R(3,3)$), can be given, by considering a vertex of maximal degree in a graph $G$ with $||G||=8$, and counting edges so that no $4$-cycle is made.

A clarification required:

May be I am wrong . But is not this result imply $R(4,4) \leq 6$ where actually $R(4,4)=18$ ?

Thanks for any answer.

no, since in $R(4,4)$, we are searching for a clique of size $4$ that all of it's edges have the same color, but here only existence of a cycle of length $4$ was discussed.


geometry trigonometry AMC AIME MATHCOUNTS USA(J)MO USAMO

lol, next year you should study everything.  :D
Geometry, Trig, Counting, Algebra, Number Theory...have fun.
But seriously, we can't tell what we'll have next year for the AIME.

[quote="simulacrum"]It felt like there were alot of counting questions on the AIME. does this mean for next yr i should study counting?[/quote]
The topic distribution each year is independent of previous years. Don't count on anything.

Also, I only see two problems on the AIME that are primarily counting. Yes, there were also a few problems that involved a small amount of counting, but most AIME problems involve more than one topic.

If you really just want to study a few topics, knowing geometry to an IMO level and nothing else would get you very close to the floor. Throw in a MathCounts level understanding of math and I think you could qualify for the USAMO. Geometry is just so heavily weighted.

Of course wizardly algebra and number theory skills are essential. Hard core counting practice is probably the last thing I would do...

Yeah you gotta know everything...
Seriously SO MANY COUNTING PROBLEMS THO...

The new Intermediate Counting text by AoPS should be coming this summer, and apparently it should be good for AIME/easy USAMO problems, so if you want to brush up on it... :P

[quote="randomdragoon"]Don't count on anything.[/quote] Was that supposed to be a pun? :lol:

What?

The only subject that I thought was overused in this AIME was geometry. Of which there was a hell of a lot.

Yeah, there were only about two counting problems, if I recall correctly; the frog problem and the grid. And the frog hardly counts as counting (pun not intended).

Is number 14 counting or algebra?
The only reason I'm thinking counting is because of the recurrence relation.

Nah, I'd say that's algebra. Or it might be a bit of both. But still, the only problems that [i]really[/i] needed counting skills were #6 and #10.

There were only 2 problems which were completely counting. But lots lots lots of others involved cases and annoying stuff like that. oh yes and if you are going to take PenguinIntegral's advice and are going to rely heavily on Geometry for next year, you should also brush up on computation.  :D


calculus
integration
algebra
polynomial
logarithms
real analysis
real analysis unsolved

I was given to calculate the integral

   $ I = \int {\frac {xe^x}{1+e^x} dx}$

 I have tried with elementary methods , but I have failt.  Can you help or is this imposible without expansion to  a series ?

 Thanks - Babis

[quote="stergiu"]I was given to calculate the integral

   $ I = \int {\frac {xe^x}{1 + e^x} dx}$

 I have tried with elementary methods , but I have failt.  Can you help or is this imposible without expansion to  a series ?

 Thanks - Babis[/quote]

$ I = \int {\frac {xe^x}{1 + e^x} dx = \int \left(x - \frac {x e^{ - x}}{1 + e^{ - x}}\right)\, dx}$

Now $ \int \frac {x e^{ - x}}{1 + e^{ - x}} \, dx = \int \frac {x e^{x}}{1 + e^{x}} \, dx = I$
So
$ I = \int {\frac {xe^x}{1 + e^x} dx = \int x\, dx - I \Rightarrow I = \frac {x^2}4}$

EDIT: Sorry... I cant replace u by x again :( Thanks to kevinatcausa.

Το ολοκλήρωμα ανάγεται στο \[ \int \frac{\ ln x}{x\minus{}1}dx\]  το οποίο δεν υπολογίζεται στοιχειωδώς.

Differentiate  $ x\log (1\plus{}e^x)$  to find
\[ I\equal{}x\log (1\plus{}e^x) \minus{}\int{\log (1\plus{}e^x)dx}\]

Making the substitution $ t\equal{}\minus{}e^x$, we can express this integral via the dilogarithm $ \text{Li}_2$: http://mathworld.wolfram.com/Dilogarithm.html

\[ I\equal{}x\log (1\plus{}e^x)\plus{}\text{Li}_2 (\minus{}e^x)\]

[quote="Ravi B"][quote="isomorphism"]Now $ \int \frac {x e^{ \minus{} x}}{1 \plus{} e^{ \minus{} x}} \, dx \equal{} \int \frac {x e^{x}}{1 \plus{} e^{x}} \, dx$[/quote]
This step seems faulty.  Besides, the integral of a non-polynomial can't be a polynomial.[/quote]

Substituting $ x \equal{} \minus{}u$,

$ \int \frac{x e^{ \minus{} x}}{1 \plus{} e^{ \minus{} x}} \, dx \equal{} \int \frac{\minus{}u e^{ u}}{1 \plus{} e^{u}} \, d(\minus{}u) \equal{} \int \frac{ue^{ u}}{1 \plus{} e^{u}} \, du \equal{} \int \frac{xe^{x}}{1 \plus{} e^{x}} \, dx$

What is wrong with that?  :maybe:

The problem's the same thing that's wrong with the following argument.

$ \int 2x^3 dx\equal{} \int (x^2)( 2x dx) \equal{} \int u du \equal{} \int x dx \equal{} \frac{x^2}{2}\plus{}c$.

The trouble is that $ \int u du \equal{} \frac{u^2}{2}\plus{}c$, while $ \int x dx\equal{} \frac{x^2}{2}\plus{}c$.  They may look the same, but they're not equal because $ u$ and $ x$ are different.

[quote="kevinatcausa"]The problem's the same thing that's wrong with the following argument.

$ \int 2x^3 dx \equal{} \int (x^2)( 2x dx) \equal{} \int u du \equal{} \int x dx \equal{} \frac {x^2}{2} \plus{} c$.

The trouble is that $ \int u du \equal{} \frac {u^2}{2} \plus{} c$, while $ \int x dx \equal{} \frac {x^2}{2} \plus{} c$.  They may look the same, but they're not equal because $ u$ and $ x$ are different.[/quote]

Oh my  :oops_sign:
What you said is perfectly true. Now I realize my mistake, sorry  :oops:

Very well !

 It is not possible to calculate it with elementrary method. That was my op;inion too.

 Thanks to all of you.

 Babis Stergiou


induction
number theory proposed
number theory

Prove that there are infinitely many positive integer $ n$ such that
$ n|2^ n \plus{}2$ and $ n\minus{}1|2^n \plus{}1$

$ a_{1}\equal{}2$
$ a_{n\plus{}1}\equal{}2^{a_n}\plus{}2$
Induction to prove this sequence satisfy conditon.


$\setlength{\unitlength}{3cm}\begin{picture}(2,2) \qbezier(2,1)(2,2)(1,2) \qbezier(1,2)(0,2)(0,1) \qbezier(0,1)(0,0)(1,0) \qbezier(1,0)(2,0)(2,1) \end{picture}$

that doesn't look exactly like a circle? can anyone help me?

[quote="anirudh"]$\setlength{\unitlength}{1mm}\begin{picture}(0, 0) \put(0,0){\circle{32}}\end{picture}$

You tried to use bezier curves to draw it. Circles are better

[code]
$\setlength{\unitlength}{1mm}
\begin{picture}(0, 0)
\put(0,0){\circle{32}}
\end{picture}$[/code][/quote]

k thanks lol i got this off of the LaTeX- show your cool ideas or something like that

Is it possible to change the diameter or radius?

[quote="stevem"]Yes. In anirudh's code \circle{32} means a circle of diameter 32mm so just change the number. However, not all diameters are allowed (unless you use a special package) so LaTeX chooses the closest acceptable value.[/quote]
$\setlength{\unitlength}{1mm}\begin{picture}(0, 0) \put(0,0){\circle{8}}\end{picture}$
Ok, Thanks


Let a subset of A={log x, log y, log(x+y/x)} be B={0,1}. Then the probable maximum element of A is
(A)  1
(B)  2log 3
(C)  log 11
(D)  log 15

Assuming the third term means log((x+y)/x)

[hide]We want x to be as small as possible, so
log x = 0 => x=1

So log y=1 => y=10 and log (x+y)/x=log 11
or  log (1+y)=1=>log y=log 9

of these, the largest is log 11 (C)[/hide]

[quote="prototypex24"]Assuming the third term means log((x+y)/x)[/quote]
The third term should be $\log\left(x+\frac yx\right)$.

[quote="prototypex24"]The answer (I think ) is the same following my previous reasoning.[/quote]
yes, your ans is correct.


[quote="encyclopedia"]Let $ a,b,c,d > 0$ prove that
\[ \frac {1}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca} \plus{} \frac {1}{b^2 \plus{} c^2 \plus{} d^2 \plus{} bc \plus{} cd \plus{} da} \plus{} \frac {1}{c^2 \plus{} d^2 \plus{} a^2 \plus{} cd \plus{} da \plus{} ac} \plus{} \frac {1}{d^2 \plus{} a^2 \plus{} b^2 \plus{} da \plus{} ab \plus{} bd}\ge\frac {32}{3(a \plus{} b \plus{} c \plus{} d)^2}
\]
[/quote]
It's wrong. Try $ a\equal{}b\equal{}c\equal{}1$ and $ d\rightarrow0^\plus{}.$ :wink:

[quote="encyclopedia"][quote="arqady"][quote="encyclopedia"]Let $ a,b,c,d > 0$ prove that
\[ \frac {1}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca} \plus{} \frac {1}{b^2 \plus{} c^2 \plus{} d^2 \plus{} bc \plus{} cd \plus{} da} \plus{} \frac {1}{c^2 \plus{} d^2 \plus{} a^2 \plus{} cd \plus{} da \plus{} ac} \plus{} \frac {1}{d^2 \plus{} a^2 \plus{} b^2 \plus{} da \plus{} ab \plus{} bd}\ge\frac {32}{3(a \plus{} b \plus{} c \plus{} d)^2}
\]
[/quote]
It's wrong. Try $ a \equal{} b \equal{} c \equal{} 1$ and $ d\rightarrow0^ \plus{} .$ :wink:[/quote]

Hi arqady, that's ok, but is the following wrong
\[ \frac {1}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca} \plus{} \frac {1}{b^2 \plus{} c^2 \plus{} d^2 \plus{} bc \plus{} cd \plus{} da} \plus{} \frac {1}{c^2 \plus{} d^2 \plus{} a^2 \plus{} cd \plus{} da \plus{} ac} \plus{} \frac {1}{d^2 \plus{} a^2 \plus{} b^2 \plus{} da \plus{} ab \plus{} bd}\ge\frac {21}{2(a \plus{} b \plus{} c \plus{} d)^2}
\]
[/quote]
I think it must be
\[ \frac {1}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca} \plus{} \frac {1}{b^2 \plus{} c^2 \plus{} d^2 \plus{} bc \plus{} cd \plus{} db} \plus{} \frac {1}{c^2 \plus{} d^2 \plus{} a^2 \plus{} cd \plus{} da \plus{} ac} \plus{} \frac {1}{d^2 \plus{} a^2 \plus{} b^2 \plus{} da \plus{} ab \plus{} bd}\ge\frac {21}{2(a \plus{} b \plus{} c \plus{} d)^2}
\]
This inequality is true and nice with Cauchy Schwarz. :)


Find the measure of the smallest angle formed by the lines
y=3.5x+3 and y=(2/3)x+1

49.6
50.8
51.9
52.3
none of these

[hide]The formula for the acute angle between two lines is $ \tan \theta \equal{} \frac{m_1\plus{}m_2}{1\plus{}m_1 m_2}$.
Letting $ m_1\equal{}\frac{7}{2}$ and $ m_2\equal{}\frac{2}{3}$, we have:
\[ \tan \theta \equal{} \frac{\frac{7}{2}\plus{}\frac{2}{3}}{1\plus{}\frac{7}{2} \cdot \frac{2}{3}}\equal{}\frac{5}{4}\]Solving for $ \theta$, we have:
\[ \theta \approx 51.3^{\circ} \Rightarrow \boxed{\text{none of these}}\][/hide]

123456789, you've got your formula wrong, unless i'm mistaken
it's  $ \tan \theta \equal{} \frac{m_1\minus{}m_2}{1\plus{}m_1 m_2}$
so, we have
           $ \tan \theta \equal{} \frac{\frac{7}{2}\minus{}\frac{2}{3}}{1\plus{}\frac{7}{2} \cdot \frac{2}{3}}\equal{}\frac{5}{4}$

$ \equal{}>\theta \approx 40.4^{\circ} \Rightarrow \boxed{\text{none of these}}$

answer remains the same, but just wanted to clarify the method :)


Find all positive integers N with at most 4 digits such that the number obtained by reversing the order of digits of N is divisible by N and differs from N.

1) Number with one digit is clearly solution.
2) Let $N=10a+b$ with $1\le a\le 9$ and $1\le b\le 9$
The problem give us that $10a+b|10b+a$
[b]Case 1 : [/b] $10b+a=2(10a+b)$ $\iff$ $8b=19a$ but $19$ can't divide the left hand side.
[b]Case 2 : [/b] $10b+a=3(10a+b)$ $\iff$ $7b=29a$ but $29$ can't divide the left hand side.
[b]Case 3 : [/b] $10b+a=4(10a+b)$ $\iff$ $2b=13a$ but $13$ can't divide the left hand side.
[b]Case 4 : [/b] $10b+a=k(10a+b)$ with $k\ge5$. In this case if $a\ge 2$ $10b+a\ge 5*10*2=100$ but $100$ doesn't have 2 digits. So $a=1$
                            $10b+1=10k+kb$ $\iff$ $b(10-k)=10k-1$ So $10-k|10k-1$ so $10-k\in\{1.3.9\}$ we compute that there are no solutions.

3) Let $N=100a+10b+c$ with $1\le a,c\le 9$ and $0\le b\le 9$
The problem give us that $100a+10b+c|100c+10b+a$
[b]Case 1 : [/b] $100c+10b+a=2(100a+10b+c)$ $\iff$ $98c=199a+10b$ so $c>2a$ and $c\ge3a$ so $a\in\{1.2.3\}$ but $a$ is even so $a=2$
                           $98c=398+10b$ we find that $c\equiv 1\pmod5$ so $c=6$ but that implies that $b=19$ this is clearly impossible.
[b]Case 2 : [/b] $100c+10b+a=3(100a+10b+c)$ $\iff$ $97c=299a+20b$ so $c>3a$ and $c\ge4a$ so $a\in\{1.2\}$ but if we look $\pmod4$ we find
                            that $a\equiv 3c\pmod4$ so if $a=2$ $c\equiv 2\pmod4$ but $c\ge 8$ and there are no digits which respect this conditions and
                            $a=1$ we find that $c\equiv3\pmod4$ but we have $c\ge4$ so $c=7$ but that give us $b=19$ impossible
[b]Case 3 : [/b] $100c+10b+a=k(100a+10b+c)$ with $k\ge4$ $\iff$ $(100-k)c=(10k-10)b+(100k-1)a$ so $c>ka$ so $10>c\ge a (k+1)\ge5a$ so
                           $a=1$ so $100c+10b+1=k(100+10b+c)$ so if we look the number of hundred we find that $k=c$ or $k+1=c$ but $k=c$ give us
                           $10b(c-1)+c^2-1=0$ but it's clear that $10b(c-1)+c^2-1>0$. If $k=c-1$ we have $20b=10bc+c^2-c$ but $k\ge4$ so $10bc\ge40b\ge20b$ and $c^2>c$ so it is impossible.


What type of the triangle ABC if we have
$ 4sin^2A\equal{}8cosB.cosC\plus{}1$

I can show that a certain triangle is the solution
[hide]
45,45,90
[/hide]

But I can't prove it's the only one (yet)

[quote] [color=darkred]What is the type of the triangle $ ABC$ if $ 4\sin^2A \equal{} 8\cos B\cos C \plus{} 1$ ?[/color] [/quote]
[quote="SimonM"] [color=darkred]I can show that a certain triangle is the solution $ 45,45,90$ .
But I can't prove it's the only one (yet) . [/color][/quote]
[color=darkblue][b]SimonM[/b], maybe $ 30,60,90$ ! The your solution doesn't verify the upper relation.

Another solution is $ A\equal{}90^{\circ}$ , $ B\equal{}\arctan\frac {4\plus{}\sqrt 7}{3}$ , $ C\equal{}\arctan\frac {4\minus{}\sqrt 7}{3}\ !$

In the my opinion, there is a mistake in this problem.

Exist many nonsimilar triangles which verify this relation.[/color]

[color=darkblue][b][u]Remark.[/u][/b] $ 4\sin^2A \equal{} 8\cos B\cos C \plus{} 1$ $ \Longleftrightarrow$ $ a^2 \minus{} R^2 \equal{} 2\cdot HB\cdot HC$ $ \Longleftrightarrow$

$ \boxed {a^2 \minus{} R^2 \equal{} 4R\cdot HD}$ , where $ H$ is the orthocenter and

$ R$ is the length of the circumradius ; $ D\in BC$ , $ AD\perp BC$ .[/color]

[quote] [color=darkred]What is the type of the triangle $ ABC$ if $ 4\sin^2A \equal{} 8\cos B\cos C \plus{} 1$ ?[/color] [/quote]
[quote="SimonM"] [color=darkred]I can show that a certain triangle is the solution $ 45,45,90$ .
But I can't prove it's the only one (yet) . [/color][/quote]
[color=darkblue][b]SimonM[/b], the your solution doesn't verify the upper relation. Some  solutions are

$ (A,B,C)$ : $ (60,60,60)$ ; $ (30,60,90)$ ; $ \left(\ 90\ ,\ \arctan\frac {4 \plus{} \sqrt 7}{3}\ ,\ \arctan\frac {4 \minus{} \sqrt 7}{3}\ \right)\ \ !$

In the my opinion, there is a mistake in this enunciation of this problem !

I think that exist many nonsimilar triangles which verify this relation.[/color]


Στη λύση ενός γενικότερου προβλήματος σε PDE μου προέκυψε η εξής (seemingly simple) ανισότητα. Η λύση που κατασκεύασα δε μου αρέσει γιατί φαίνεται ιδιαίτερα ad hoc (αν και όχι μεγάλη), οπότε παραθέτω το πρόβλημα σε εσάς που σίγουρα μπορείτε να βρείτε κάποια όμορφη λύση. Αν υπάρχει ενδιαφέρον, θα σας δώσω και τη δική μου λύση.

Έστω $ a_0, a_\ell \in \mathbb{R}$ (σταθερές). Να αποδειχθεί ότι υπάρχει σταθερά $ M$ τέτοια ώστε:
$ \int_{[0,\ell]} f'' f \leq M$
για οποιαδήποτε πραγματική συνάρτηση $ f \in \mathcal{C}^2$ στο $ [0,\ell]$ που να ικανοποιεί $ \int_{[0,\ell]} f^2 \equal{} 1$ και $ f'(0) \minus{} a_0 f(0) \equal{} f'(\ell) \plus{} a_\ell f(\ell) \equal{} 0$.

Για όποιον ενδιαφέρεται, η ανισότητα προέκυψε γιατί ήθελα να αποδείξω ότι ο τελεστής $ f \mapsto \left \langle \minus{}f'', f \right \rangle$ είναι κάτω-φραγμένος για τις $ \mathcal{C}^2$ συναρτήσεις $ f : [0,\ell] \to \mathbb{R}$, μοναδιαίες στο $ \mathcal{L}^2$, που να ικανοποιούν Robin boundary conditions, όπου $ \langle \cdot, \cdot \rangle$ το κλασικό inner product του $ \mathcal{L}^2$. (Έτσι με ένα variational approach μπορούμε εύκολα να κατασκευάσουμε eigenvalues/eigenfunctions.)

ΥΓ. Η άσκηση αυτή μπορεί να τεθεί για εξάσκηση και σε μαθητές λυκείου, καθώς τουλάχιστον η λύση μου είναι αρκετά λυκειακή.

[quote="Demetres"]Η λύση μου μου φαίνεται αρκετά απλή και δεν χρησιμοποιώ αρκετές από τις συνθήκες. Μάλλον κάτι δεν διάβασα καλά ή κάπου κάνω λάθος:

$ \int_0^{\ell} f''f \equal{} f(\ell)f'(\ell) \minus{} f(0)f'(0) \minus{} \int_0^{\ell} (f')^2$

Αλλά η f' είναι παραγωγίσιμη άρα και συνεχής στο $ [0,\ell]$, άρα είναι φραγμένη στο $ [0,\ell]$.[/quote]

Χωρίς να την έχω κοιτάξει, μάλλον εδώ θα χρειαστούν οι Robin (που πρώτη φορά τις βλέπω :P)... νομίζω ο cmad θέλει ένα uniform lower bound $ M$ του ολοκληρωτικού του τελεστή ώστε για κάθε $ f$ να έχουμε $ \int\ldots\leq M$, και όχι απλώς να δείξουμε ότι το ολ/μα είναι πεπερασμένο για κάθε $ f$.

My tuppence anyway...

Durandal 1707

Edit: π.χ. αν $ f_n(x) \equal{} x^n$, $ x\in[0,1]$, έχουμε $ \int f f'' \equal{} \frac {n^2 \minus{} n}{2n \minus{} 1}\to\infty$, οπότε σίγουρα χρειαζόμαστε και το $ \int f^2 \equal{} 1$.

Έχεις δίκιο. Μου υπεδείχθει και από τον cmad σε προσωπικό μήνυμα.

[quote="SorrowMan"]Tha sou itan kopos na eksigiseis ti einai kato fragmenos telestis??? Giati proswpika exw sinantisi mono tin enia tou fragmenou telesti[/quote]

Ummm, θα υπέθετα ότι ο $ T$ λέγεται κάτω φραγμένος όταν για όλες τις $ f$ που ανήκουν σε κάποιο προκαθορισμένο σύνολο (π.χ. $ \int f^2 \equal{} 1$ or whatever...) $ Tf\geq M$ για κάποιο $ M\in\mathbb{R}$ :D

Cheerio,

Durandal 1707


Simple. All you have to know is the fraction of the circumference of a circle the arc is part of and multiply that by 360.  If the arc is 1/2 of the circumference of the circle, the measure is 180 degrees.

You find the section of the circle. Let's say the arc is of length 2 and the circumference is of length 8. Thus, the arc is 1/4 the circle because it is 1/4 the circumference and thus it is 1/4 of 360 degrees. This also works the other way around.

[quote="Cliu0301"]How do you find the number of degrees in an arc?[/quote]
Draw the line segment that contains the "endpoints" of the arc and draw radii to the center from each one. The angle formed by the radii is the measure of the arc

Have the two points of the arc connect to the center so you have a portion of the circle. The measure of the angle formed by the two points of the arc and the center of the circle is the degree measure of the arc.

arc/circle geometry is basically covered in Chapters 11/12 in Intro to Geometry


linear algebra
matrix
vector
function
number theory
relatively prime
superior algebra

assume that $m\geq 1,n\geq 2$ are postive integer .find the number of invertible $m*m$ matrices modulo n.

This isn't the complete answer, but we might as well start with the easy case.

If $n=p$, a prime, then the number of invertible matrices is $(p^m-1)(p^m-p)(p^m-p^2)\cdots(p^m-p^{m-1}).$

The argument: build the matrix by columns. The first column can be any nonzero vector, and there are $p^m-1$ of those. After that, in order for the columns to be independent, each additional column must avoid the span of the previous $k$ columns. The vectors in the span of $k$ independent columns can be indexed by the $p^k$ coefficients of the linear combination.

The situation for $n$ composite is more complicated. Since the entries of the matrix belong to a ring that is not a field, we cannot use the language of vector spaces and independence. A matrix is invertible if and only if its determinant is invertible mod $n$.

As a brute force count for the smallest interesting case: for $2\times2$ matrices mod $4$, there are $96$ invertible matrices out of the $256$ possibilities.

Let $f(n)$ be the number of $m\times m$ invertible matrices in the ring of integers mod $n$.
We can build the full answer in three steps:

Step 1: If $n=p$ is prime, Kent's argument applies and
$f(n)=p^{m^2}(1-\frac1p)(1-\frac1{p^2})\cdots(1-\frac1{p^m})$

Step 2: If $n=p^k$ with $p$ prime, the matrix is invertible mod $n$ if and only if it is invertible mod $p$. Each invertible matrix mod $p$ pulls back to $(p^{k-1})^{m^2}$ matrices mod $p^k$, so
$f(n)=(p^{k-1})^{m^2}\cdot p^{m^2}(1-\frac1p) (1-\frac1{p^2})\cdots(1-\frac1{p^m})$
$f(n)=n^{m^2}(1-\frac1p)(1-\frac1{p^2})\cdots(1-\frac1{p^m})$

Step 3: $f$ is a multiplicative function.
If $a$ and $b$ are relatively prime, the Chinese Remainder Theorem gives us a 1-1 correspondence between pairs of matrices mod $a$ and mod $b$ and matrices mod $ab$. The matrix mod $ab$ is invertible in the integers mod $ab$ if and only if each of the matrices in the pair is; after all, the determinant follows the same transformation as the elements.

I won't write out the full expression, but it's obvious from the above. If $m=1$, this specializes to Euler's function $\phi(n)$


arithmetic sequence
Miscellaneous Problems

The set $S=\{ \frac{1}{n} \; \vert \; n \in \mathbb{N} \}$ contains arithmetic progressions of various lengths. For instance, $\frac{1}{20}$, $\frac{1}{8}$, $\frac{1}{5}$ is such a progression of length $3$ and common difference $\frac{3}{40}$. Moreover, this is a maximal progression in $S$ since it cannot be extended to the left or the right within $S$ ($\frac{11}{40}$ and $\frac{-1}{40}$ not being members of $S$). Prove that for all $n \in \mathbb{N}$, there exists a maximal arithmetic progression of length $n$ in $S$.

Consider the arithmetic progression with fixed $ n$, $ 1 \plus{} n \cdot k \forall k \in \mathbb N$
let $ p_{n_j} \in \mathbb P$ and $ p_{n_j} \equal{} 1 \plus{} n \cdot j$ (by Dirichlet's theorem on arithmetic progressions)
$ \frac {1}{1 \cdot (1 \plus{} j) \cdot (1 \plus{} 2j) \cdots(1 \plus{} (n \minus{} 1)j)}$
$ \frac {1 \plus{} j}{1 \cdot (1 \plus{} j) \cdot (1 \plus{} 2j) \cdots(1 \plus{} (n \minus{} 1)j)}$
$ \frac {1 \plus{} 2j}{1 \cdot (1 \plus{} j) \cdot (1 \plus{} 2j) \cdots(1 \plus{} (n \minus{} 1)j)}$
$ \cdots$
$ \frac {1 \plus{} (n \minus{} 1)j}{1 \cdot (1 \plus{} j) \cdot (1 \plus{} 2j) \cdots(1 \plus{} (n \minus{} 1)j)}$
is maximal arithmetic progression of length $ n$ in $ S$.

For Example: $ n\equal{}1996$
$ p_{1996_1}\equal{}1\plus{}1 \cdot 1996\equal{}1997$ is prime then:
$ \frac{1}{1 \cdot 2 \cdot 3 \cdots 1996}$
$ \frac{2}{1 \cdot 2 \cdot 3 \cdots 1996}$
$ \frac{3}{1 \cdot 2 \cdot 3 \cdots 1996}$
$ \cdots$
$ \frac{1996}{1 \cdot 2 \cdot 3 \cdots 1996}$
is a maximal arithmetic progression of length $ 1996$ in $ S$

Another example: $ n\equal{}1997$
Pick $ p\equal{}1\plus{}1997 \cdot 4$ prime


In the following frequency distribution, find the number of fours (that is, find the value of k) given that the mean is equal to one of the data elements.

$ \begin{tabular}[t]{|l|ccccc|c|} \multicolumn{7}{c}{} \\
Data Elements & Frequencies \\
0 & 1 \\
2 & 3 \\
3 & 2 \\
4 & k \\
5 & 1 \\
\end{tabular}$

[hide="taco"]We know that the mean will be the sum of the products of the data frequencies and their respective probabilities.  The total possible outcomes is $ 1 \plus{} 3 \plus{} 2 \plus{} k \plus{} 1 \equal{} 7 \plus{} k$, for some positive integer $ k$.  So we have $ 0\cdot\frac {1}{7 \plus{} k} \plus{} 2\cdot\frac {3}{7 \plus{} k} \plus{} 3\cdot\frac {2}{7 \plus{} k} \plus{} 4\cdot\frac {k}{7 \plus{} k} \plus{} 5\cdot\frac {1}{7 \plus{} k} \equal{} \frac {17 \plus{} 4k}{7 \plus{} k}$.  As stated, we want this to equal an integer, $ x$, $ [2,5]$ since it is clearly not equal to $ 0$ because $ 17$ is not divisible by $ 4$.  We have that $ 17 \plus{} 4k \equal{} 7x \plus{} kx \rightarrow17 \minus{} 7x \equal{} (x \minus{} 4)k$  For $ k$ to be an integer, $ 17 \minus{} 7x$ must be divisible by $ x \minus{} 4$.  So we want $ x \minus{} 4 \equal{} \pm{1}\rightarrow{x \equal{} 3,5}$.  For this to be valid (making $ k$ positive, since you can't have negative frequency, $ 17 \minus{} 7x$ must share the same sign as $ x \minus{} 4$.  You can see that both values of $ x$ make $ 17 \minus{} 7x < 0$, so $ x \minus{} 4$ must be negative, leaving the only solution as $ x \equal{} 3$.  Since $ k \equal{} \frac {17 \minus{} 7x}{x \minus{} 4} \equal{} \frac { \minus{} 4}{ \minus{} 1} \equal{} 4$,  the data element $ 4$ occurs $ 4$ times.[/hide]


Given the numbers $1,2,2^{2}, ...2^{n-1}$, for a specific permutation $\sigma = x_{1},x_{2},...,x_{n}$ of these numbers we define $S_{1}(\sigma) = x_{1}, S_{2}(\sigma)=x_{1}+x_{2}$, ... and $Q(\sigma) = S_{1}(\sigma)S_{2}(\sigma)...S_{n}(\sigma)$. Evaluate $\sum 1/Q(\sigma)$, where the sum is taken over all possible permutations.

[hide="Give Away Hint"]Induction :wink:[/hide]
Masoud Zargar

Alright, I'm digging this up from my blog, so if there are any questions, chances are I won't be able to answer. :)
[quote]Given the numbers $1,2,2^{2},\ldots,2^{n-1}$, for a specific permutation $\sigma=x_{1},x_{2},\ldots,x_{n}$ of these numbers we define $S_{1}(\sigma)=x_{1},S_{2}(\sigma)=x_{1}+x_{2},\ldots$ and $Q(\sigma)=S_{1}(\sigma)S_{2}(\sigma)\cdots S_{n}(\sigma)$. Evaluate $\sum\frac{1}{Q(\sigma)}$, where the sum is taken over all possible permutations.[/quote]
[hide="Solution"]Claim: $\sum\frac{1}{Q(\sigma)}=\prod_{i=1}^{n}\frac{1}{x_{i}}$.

Proof, by induction:
[hide="Base Case:"]$n=1$. Then $S_{1}(\sigma)=1$ and $\sum\frac{1}{Q(\sigma)}=\frac{1}{1}=\prod_{i=1}^{1}\frac{1}{x_{i}}$.
[/hide]
[hide="Inductive Step:"] Take an $n$ for which $\sum\frac{1}{Q(\sigma)}=\prod_{i=1}^{n}\frac{1}{x_{i}}$. We want to show this relationship is satisfed for $n+1$ as well.

Suppose we take the sum over all permutations of $\sigma=x_{1},\ldots,x_{n},x_{n+1}$. Set $S=\sum_{i=1}^{n+1}x_{i}$, $P=\prod_{i=1}^{n+1}x_{i}$ and define $\psi(k)=\sum\frac{1}{Q(\sigma)}$, where the sum is taken over all permutations of $\{x_{1},\ldots,x_{n},x_{n+1}\}\setminus\{x_{k}\}$. Thus,
\[\sum\frac{1}{Q(\sigma)}=\left(\frac{1}{S}\right)\cdot\left(\sum_{i=1}^{n+1}\psi(i)\right). \]
From the inductive hypothesis, the sum taken over all permutations of a set of $n$ elements is just the reciprocal of the product of the elements. That is,
\[\psi(k)=\frac{x_{k}}{x_{1}\cdot x_{2}\cdots x_{n}\cdot x_{n+1}}=\frac{x_{k}}{P}. \]
Hence,
\[\sum\frac{1}{Q(\sigma)}=\left(\frac{1}{S}\right)\cdot\left(\sum_{i=1}^{n+1}\frac{x_{i}}{P}\right). \]

\[\sum\frac{1}{Q(\sigma)}=\left(\frac{1}{S\cdot P}\right)\cdot\left(\sum_{i=1}^{n+1}x_{i}\right). \]

\[\sum\frac{1}{Q(\sigma)}=\left(\frac{1}{S\cdot P}\right)\cdot(S)=\frac{1}{P}\prod_{i=1}^{n+1}\frac{1}{x_{i}}, \]
and the prove is completed through induction.[/hide]
Therefore,
\[\sum\frac{1}{Q(\sigma)}=\prod_{i=0}^{n}\frac{1}{2^{i-1}}. \]
[/hide]

There's a slight typo at the end, but its a very nice proof.


For each $n\geq 2$, show that there exist $n$ different positive integers such that for any two $a,b$, $a-b$ divides $a+b$.

For $n=2$, take $x_{1}=1$ and $x_{2}=2$.
Now suppose that there are $n$ numbers $x_{1}<\dots <x_{n}$ satisfying the property, we will show that there are $n+1$ such numbers.
Let
\[P=\prod_{1\leq i<j\leq n}(x_{j}-x_{i}). \]
Take an integer $t$ such that $t P>x_{n},$ and let
\[k =-t+\prod_{i=1}^{n}(t P-x_{i}). \]
Let $y_{i}= x_{i}+k P$ for $i=1,\dots,n$ and $y_{n+1}= t P+k P.$ Then $y_{1},\dots,y_{n+1}$ satisfy the required property:
* for $1\leq i<j\leq n,$ $y_{j}-y_{i}=x_{j}-x_{i}$ divides $y_{j}+y_{i}= x_{j}+x_{i}+2k P$ since $x_{j}-x_{i}$ divides each of $x_{j}+x_{i}$ and $P$;
* for $1\leq i\leq n,$ $y_{n+1}-y_{i}= t P-x_{i}$ divides $y_{n+1}+y_{i}= 2tP+2kP-(tP-x_{i})$ since $tP-x_{i}$ divides $2 t P+2 k P = 2 P (t+k).$


calculus
integration
function
real analysis
real analysis solved

P4

Let f:R->R be a function which has primitives s.t. f(x)=f(f(x)-x). Show that f is continuous

Lets denote g:R-->R, g(x)=f(x)-x , for any x in R, we have that:

g(g(x))=x for any x in R.

So g is continuous, it's injective and it has Darboux property.

cheers! :D

Thx. I particularly like this one. Thx for the other ones too. But you have a tremendous talent for finding exactly the same solns as at the end of the book where I found them! :D

hehe.. like I said these one's with the integrals I have took them from the book.. I don't know integrals and stuff like that! :D :D

cheers! :D


There are 2000 points on a circle and each point is given a number which is equal to the average of the two numbers which are its nearest neighbors. Show that all the numbers must be equal.

[hide="Hint"]
Well, look at the largest number there... ;)[/hide]

[hide="Another solution with generalization, although a bit more complicated"]Call the numbers $ a_{1},a_{2},a_{3}\ldots a_{k}$, going around the circle in order.  In addition, let $ a_{k+n}=a_{n}$ for any positive integer $ n$.

Lemma 1: Any three adjacent terms form an arithmetic sequence.
Proof: Let $ n$ be an integer from 2 to $ k$ inclusive.  $ a_{n+1}=\frac{a_{n}+a_{n+2}}{2}\Rightarrow 2a_{n+1}=a_{n}+a_{n+2}\Rightarrow a_{n+2}-a_{n+1}=a_{n+1}-a_{n}$, which is true by the definition of an arithmetic sequence.
$ \mathbb{QED}$

Lemma 2: $ \{a_{1},a_{2},a_{3}\ldots a_{k+1}\}$ is an arithmetic sequence.
Proof: We prove it by induction.  Our base case is $ \{a_{1},a_{2},a_{3}\}$, which is arithmetic by lemma 1.
Now suppose $ \{a_{1},a_{2},\ldots a_{n}\}$ is arithmetic for an integer $ n\ge 3$.  Let $ d=a_{n}-a_{n-1}$.  So $ d=a_{n-1}-a_{n-2}=a_{n-2}-{n-3}=\ldots=a_{2}-a_{1}$.  $ \{a_{n-1},a_{n},a_{n+1}\}$ is arithmetic, so $ a_{n+1}-a_{n}=d$, hence $ \{a_{1},a_{2},\ldots a_{n+1}\}$ is also arithmetic, and the induction is complete.
$ \mathbb{QED}$

So $ \{a_{1},a_{2},\ldots a_{k+1}\}$ is arithmetic with common difference $ d$.  Therefore $ a_{k+1}=a_{1}+kd$.  Since $ a_{k+1}=a_{1}$, $ d=0$, hence $ a_{1}=a_{2}=\ldots=a_{k}$, as desired.
$ \mathbb{QED}$

For the specific case, let $ k=2000$.[/hide]

i dont think that u need to do these all.
just assume that $ M$ is the largest number here and all # arent equal.but its the average of neighboring two numbers.so if the numbers are bnot equal then one of them is greater than $ M$.
$ Contradiction.$
now if these $ 3$ are equal then inductively all are equal.


[quote="Ligouras"]Let $ a, b$ and $ c$ be the sides, and $ R$ circum-radius, of an triangle $ ABC$. Prove or disprove that

$ 4\sqrt {\frac {1}{a^4 \plus{} b^4 \plus{} c^4}}\le R$.

_________
 :([/quote]

It's not true for $ a\equal{}b\equal{}c\equal{}1$, $ R\equal{}\frac{1}{\sqrt{3}}$

[quote="Ligouras"]For every problem it always exists at least a POSITIVE SOLUTION... Dear [b]SaYaT[/b] retries

________
 :wink:  :blush:[/quote]

What do you mean? :wink:
I found counterexample for your problem with conditions that was set before.

[quote="Jan"]He means that you forgot the condition $ abc \equal{} 4$.[/quote]

It's not me who had forgot this condition :wink:
as you see before edition it was proposed so:

[quote="Ligouras"]Let $ a, b$ and $ c$ be the sides, and $ R$ circum-radius, of an triangle $ ABC$. Prove or disprove that

$ 4\sqrt {\frac {1}{a^4 \plus{} b^4 \plus{} c^4}}\le R$.

_________
 :([/quote]

[quote="Ligouras"]Let $ a, b$ and $ c$ be the sides, and $ R$ circum-radius, of an triangle $ ABC$ such that $ abc \equal{} 4$. Prove or disprove that

$ 4\sqrt {\frac {1}{a^4 \plus{} b^4 \plus{} c^4}}\le R$.

_________
 :([/quote]

[hide="Solution"]
Let's $ S$ - is area of $ \triangle ABC$
We have that $ R\equal{}\frac{abc}{4S}\equal{}\frac{1}{S}$
So this inequality equivolent to :
$ \sqrt{a^4\plus{}b^4\plus{}c^4} \geq 4S$
or
$ a^4\plus{}b^4\plus{}c^4 \geq 16S^2\equal{}(a\plus{}b\plus{}c)(b\plus{}c\minus{}a)(a\plus{}c\minus{}b)(a\plus{}b\minus{}c)$
We have that:
$ a^4\plus{}b^4\plus{}c^4 \geq \frac{(a^2\plus{}b^2\plus{}c^2)^2}{3} \geq \frac{(a\plus{}b\plus{}c)^4}{27}$
So it's left to prove:
$ \frac{(a\plus{}b\plus{}c)^4}{27} \geq (a\plus{}b\plus{}c)(b\plus{}c\minus{}a)(a\plus{}c\minus{}b)(a\plus{}b\minus{}c)$
$ <\equal{}>$
$ (\frac{a\plus{}b\plus{}c}{3})^3 \geq (b\plus{}c\minus{}a)(a\plus{}c\minus{}b)(a\plus{}b\minus{}c)$
Let's get $ x\equal{}b\plus{}c\minus{}a$, $ y\equal{}a\plus{}c\minus{}b$, $ z\equal{}a\plus{}b\minus{}c$, then our inequality will become :
$ (\frac{x\plus{}y\plus{}z}{3})^3 \geq xyz$ which is right(becouse of AM-GM inequality)[/hide]

[quote="Ligouras"]Dear JAN and dear SaYaT I have never said that SaYat has not found the solution (banal problem...). I have said to retry with the condition it joins (forgotten by me the first time). I apologize for the incomprehension.

Tak[/quote]
It's not necessary to apologize. I absolutely agree that it was incomprehension. Indeed, i didn't understand your post(may be becouse of my english=) :)

[quote="SaYaT"][quote="Ligouras"]Let $ a, b$ and $ c$ be the sides, and $ R$ circum-radius, of an triangle $ ABC$ such that $ abc \equal{} 4$. Prove or disprove that

$ 4\sqrt {\frac {1}{a^4 \plus{} b^4 \plus{} c^4}}\le R$.

_________
 :([/quote]

[hide="Solution"]
Let's $ S$ - is area of $ \triangle ABC$
We have that $ R \equal{} \frac {abc}{4S} \equal{} \frac {1}{S}$
So this inequality equivolent to :
$ \sqrt {a^4 \plus{} b^4 \plus{} c^4} \geq 4S$
or
$ a^4 \plus{} b^4 \plus{} c^4 \geq 16S^2 \equal{} (a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \minus{} c)$
We have that:
$ a^4 \plus{} b^4 \plus{} c^4 \geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{3} \geq \frac {(a \plus{} b \plus{} c)^4}{27}$
So it's left to prove:
$ \frac {(a \plus{} b \plus{} c)^4}{27} \geq (a \plus{} b \plus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \minus{} c)$
$ < \equal{} >$
$ (\frac {a \plus{} b \plus{} c}{3})^3 \geq (b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \minus{} c)$
Let's get $ x \equal{} b \plus{} c \minus{} a$, $ y \equal{} a \plus{} c \minus{} b$, $ z \equal{} a \plus{} b \minus{} c$, then our inequality will become :
$ (\frac {x \plus{} y \plus{} z}{3})^3 \geq xyz$ which is right(becouse of AM-GM inequality)[/hide]

[quote="Ligouras"]Dear JAN and dear SaYaT I have never said that SaYat has not found the solution (banal problem...). I have said to retry with the condition it joins (forgotten by me the first time). I apologize for the incomprehension.

Tak[/quote]
It's not necessary to apologize. I absolutely agree that it was incomprehension. Indeed, i didn't understand your post(may be becouse of my english=) :)[/quote]

Very nive proof. Thanks :wink:


AMC
AIME
ratio
geometry
email
angle bisector

Here is the problem.
In triangle ABC , AB=300, BC=507, and CA =780. Let M be the midpoint of CA and let D be the point on CA such that BD bisects angle ABC. Let F be the point on BC such that DF is perpendicular to BD.  Suppose that DF meets BM at E.  The ratio DE to EF can be written in the form m/n, where m and n are realtively prime positive integers. Find m+n.

I have another question
Was my proctor for the AIME supposed to give me the solutions?  Somehow she never did.
Thanks.
 [/img]

The proctor is not required to give out the solutions. You can ask her though.

Fierytycoon

Anyone figure this one out?

Great pure geometry question.  But very hard...

You can find the answers at <http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2003-aa/03answersaime1.html>, but they do not come with reasons :( .

From a MOP email:

Construct X on BC with BA = BX. Then (<BDC + <BDX)/2 = (<BDC + <BDA)/2 = 90
= <BDF, so DF bisects <CDX. Using the angle bisector theorem twice, we find
that XF/FC = XD/DC = AD/DC = AB/BC = c/a, so FC = XC * FC/(XF + FC) = (a -
c)a/(a + c).

Having gotten this far during the test, I can't believe failed to see the
crucial step: we can now use Menelaus' theorem on triangle DFC and line BM.
(Thanks Po-Ru!)

DE/EF * FB/BC * CM/MD = 1
DE/EF * (a - (a - c)a/(a + c)) / a * (b/2) / (ab/(a+c) - b/2) = 1
DE/EF * 2c/(a + c) * (a + c)/(a - c) = 1

So DE/EF = (a - c)/2c = (507 - 360)/(2*360) = 49/240.


how much circles pass through $n$ points?

[hide]Three points determines one circle. Then $n$ points determines $\binom{n}{3}$ circles. (Only if three of this points never lies on the same straight-line). Am I correct?  :? [/hide]

[quote="bobthebuilder"][quote="mathgeniuse^ln(x)"][hide]Actually, couldn't there be infinitely many circles, since any amount of circles can pass through any two points.[/hide][/quote]
[hide]You mean there [b]could[/b] be infinitely many circles. I believe that if n=1,2, then there are infinitely many circles. n=3, one circle. n>3 is indeterminite. Please correct me.[/hide][/quote]
he is right!  :)
and the proof? (I know it's logic, but try to prove it...)

[quote="seungjun"]What do you mean by indeterminate?[/quote]
You cannot determine how many circles pass through the points. It could be 0 circles or 1 circle (if the points are concyclic).

Well, for n>3 it has to be 1 or 0 if the question is one circle passing through all points.

There's the proof...


If I have 72 right now for the unofficial score, will I be able to make the Olympiad??
Or, would I have to write the Repechage??

I'm guessing that I might have to write the CMOR cause the cutoff for the CMO last year was 72... (which was a bit harder than this year..)

But, also, Many of the talented people seem to have graduated this year, so I'm sensing a feeling of Hope :P (j/k..)

anyways, can anyone tell me what the cutoff line for the CMO might be this year?
thx

[quote="un_o"]I got 68. . . got my mail for repechage today.
What's the usual cutoff for repechage?
If I had 68 on the COMC, would I need maybe 8 or 9 on the repechage to make it to CMO?

hopefully people with previous experience will reply XD
thanks[/quote]

hmm, what i heard is that around 100 people after the first 50 are invited, so the cutoffs would be around 63 to 64.. 66 probably this year since it was easy...

last year, i wrote the repechage, and received an email saying that my score was 30 - 40 ...

they only give you your score range, not the exact score..

As a result?? I didn't write the CMO... :P
 my friend wrote it with a repechage score of 40 - 50..

hope this helps :D

[quote="Nawahd Werdna"]Was the repechage invite sent by mail to your school or by email???[/quote]

I received an email :P
(last year and this year)

they take the next 200 after top 50 for cmor
and last year, the cmo cutoff from cmor was 42

When they choose the people who will be invited to take the CMO,
is there any other criterium besides COMC and Repechage (for those who didn't make the clear cut)?

I got a 73 and I made CMO directly. I don't know if this was the cutoff for the CMO though.
James

Yeah, I'm pretty sure 73 is the cutoff for making CMO directly

I'm pretty sure that the cutoff for repechage is 67 this year.


Two people take turns in drinking the contents of a glass of water. The first person drinks half the water in the glass while the second drinks half of what is remaining and the first drinks half of what is remaining and so on until an inconsequential drop is left behind.
Find the ratio of the water drunk by the first person to that of the second person.

Call the amount of water at the beginning $ x$.  The first person drinks $ \frac {1}{2}x$, then the 2nd person will drink $ \frac {1}{2}\cdot\frac {1}{2}x \equal{} \frac {1}{4}x$, then the first person will drink $ \frac {1}{4}\cdot\frac {1}{2}x \equal{} \frac {1}{8}x$, and so on.
The first person would drink a total of $ x\left(\frac {1}{2} \plus{} \frac {1}{8} \plus{} \frac {1}{32} \plus{} \dots\right)$.  This can be expressed as $ \frac {\frac {1}{2}}{1 \minus{} \frac {1}{4}}x \equal{} \frac {2}{3}x$.
The second person would simply drink the rest, or $ x \minus{} \frac {2}{3}x \equal{} \frac {1}{3}x$.  Therefore, the ratio of the water drunk by the first person to that of the second person is $ \frac {2}{3}x: \frac {1}{3}x$, or $ \boxed{2: 1}$

Or a more logical approach would be that the 2nd person always drinks half of whatever the first person drank before, so the ratio would clearly be 2:1.

How about a follow up problem?  What fraction of the water does the second person have to drink from the glass so that at the end each person drinks the same amount?

EDIT:  Maybe I didn't think my problem through.  That was a very simple answer.

100% They'll both have drunk half.

I bet I'm missing something. oO

Interesting...I'm not sure but I think it doesn't make sense for the second person to ever be completely able to drink the *same* amount of water as the first one as the glass would never be empty, and since the first person drank half of the whole glass first, he is going to be always leading, I think.

$ \frac{1}{2} \plus{} \frac{1}{8} \plus{} \frac{1}{32} \plus{} \frac{1}{128} \plus{} \dots > \frac{1}{4} \plus{} \frac{1}{16} \plus{}\frac{1}{64} \plus{} \frac{1}{256} \plus{} \dots$

If you evaluate the sum of the two infinite geometric series, the former is clearly bigger than the latter, even intuitively.

Am I going wrong anywhere?

unless the 2nd guy drinks all that's left, but otherwise you're right.

limac factor 1/2 from the right side of that  and you'll see the that $ x>\frac{x}{2}\Rightarrow 2x>x$. The ratio $ 2: 1$ follows immediately.


abstract algebra
superior algebra
superior algebra unsolved

[b]Theorem:[/b] Let $ M$ be a finite $ A$-module and $ I$ and ideal of $ A$. If $ IM \equal{} M$ then there is an $ a \in A$ such that $ aM \equal{} 0$ and $ a \equal{} 1 \ mod \ I$.

[b]Exercise:[/b] Let $ A$ be a ring and I a f.g. ideal satisfying $ I \equal{} I^2$, then $ I$ is generated by an idempotent $ e$.

[b]Solution:[/b] By NAK there is an $ e \in I$ such that $ (1 \minus{} e)I \equal{} 0$. One sees easily that then $ I \equal{} Ie \equal{} Ae$ and $ e^2 \equal{} e$

I don't understand the reasoning. Have we taken in the theorem $ M \equal{} I$ as $ A$-module. Then we can apply the theorem, since the exercise says that $ I \equal{} I^2 \equal{} I \cdot I$, but then there is some $ e \in I$ such that $ eI \equal{} 0$... but then I don't see the link

euhm, right. I seriously did not pay attention  :rotfl:


Could anyone help me solve this problem. I tried all my best but I can't still reach the solution.

Find the integral: $\int\frac{dx}{\sin^3x+\cos^3x}$.

This should go in the calculus forum.  I can get an answer from mathematica, but it's not pretty at all.  (Tan and arctanh are involved.)

Hint:[hide]
Try the universal substitution $t = tan\frac{x}2$[/hide]


Let a,b,c be three sides of triangle prove that
\[ a^{10}\plus{}b^{10}\plus{}c^{10}\plus{}\sum b^4c^4(b^2\plus{}c^2)\plus{}4a^2b^2c^2(a^4\plus{}b^4\plus{}c^4)\geq 2\sum b^2c^2(b^6\plus{}c^6)\plus{}3a^2b^2c^2(b^2c^2\plus{}c^2a^2\plus{}a^2b^2)\]

[quote="blackmouse"]Let a,b,c be three sides of triangle prove that
\[ a^{10} \plus{} b^{10} \plus{} c^{10} \plus{} \sum b^4c^4(b^2 \plus{} c^2) \plus{} 4a^2b^2c^2(a^4 \plus{} b^4 \plus{} c^4)\geq 2\sum b^2c^2(b^6 \plus{} c^6) \plus{} 3a^2b^2c^2(b^2c^2 \plus{} c^2a^2 \plus{} a^2b^2)
\]
[/quote]
It's equivalent to $ \sum_{cyc}(a^2 \minus{} b^2)^4(a^2 \plus{} b^2 \minus{} c^2)\geq0,$ which obviously true for all $ a,$ $ b$ and $ c.$

I think we need $ b^2\plus{}c^2\geq a^2$ e.t.c don't I?

[quote="blackmouse"]I think we need $ b^2 \plus{} c^2\geq a^2$ e.t.c don't I?[/quote]
No! If $ a^2\geq b^2\geq c^2$ then $ a^2\plus{}b^2\minus{}c^2\geq0,$ $ a^2\plus{}c^2\minus{}b^2\geq0$ and $ (a^2\minus{}c^2)^4\geq(b^2\minus{}c^2)^4.$
Thus, $ \sum_{cyc}(a^2 \minus{} b^2)^4(a^2 \plus{} b^2 \minus{} c^2)\geq(b^2\minus{}c^2)^4(a^2\plus{}c^2\minus{}b^2\plus{}b^2\plus{}c^2\minus{}a^2)\geq0.$ :wink:

Oh thanks it looks like sum of square!!! :)


[hide]x*1 = x+1+1x
x+1+1x = 7
2x = 6
x is [color=green]3[/color][/hide]

I'm going to overkill this problem.

[hide]

define / such that a*b/b = a. To find out /, solve for a in terms of b, because a*b = b*a. Say we have a*b = a+b+ab. Then, substitue x=a*b.

x = a + b + ab. Now we want to find a in terms of b and x.

x - b = a + ab
x - b = a(b+1)

$\frac{x-b}{b+1} = a$.

So: $a/b = \frac{a-b}{b+1}$.

Now, to find a, we do a*b/b. So: $7/1 = \frac{7-1}{1+1} = \frac{6}{2} = \boxed 3$

I felt like doing more than I had to :P

However, the inverse function is interesting. Say we had 1*2. 1*2 = 1+2+2 = 5. So, 5/2 = 1. But how do we find this?

$5/2 = \frac{5-2}{2+1} = \frac 33 = 1$
[/hide]

[quote="236factorial"]Suppose a*b=a+b+ab. If x*1=7, what is x?[/quote]

[hide="MY REALLY EASY ANSWER"]$x+1+x=7
2x+1=7
x=3$[/hide]

No... * is not multiplication, if you would pay attention to the first sentence. $a\cdot b\not= a+b+ab$

[quote="jli"][quote="236factorial"]Suppose a*b=a+b+ab. If x*1=7, what is x?[/quote]

why do you evan need the first sentence?

isnt it just x=7 because 7*1=7[/quote]

The star doesn't mean multiply in this case

[quote="jli"][quote="236factorial"]Suppose a*b=a+b+ab. If x*1=7, what is x?[/quote]

why do you evan need the first sentence?

isnt it just x=7 because 7*1=7[/quote]

C'mon guys, there's not so many keys on the keyboard  :P

Yeah, Hamster, you sort of overdid the problem.


Let $A_1A_2A_3\ldots A_n$ be a regular $n$-gon. Let $B_1$ and $B_{n-1}$ be the midpoints of its sides $A_1A_2$ and $A_{n-1}A_n$. Also, for every $i\in\left\{2,3,4,\ldots ,n-2\right\}$. Let $S$ be the point of intersection of the lines $A_1A_{i+1}$ and $A_nA_i$, and let $B_i$ be the point of intersection of the angle bisector bisector of the angle $\measuredangle A_iSA_{i+1}$ with the segment $A_iA_{i+1}$.

Prove that $\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n=180^{\circ}$.

[i]Proposed by Dusan Dukic, Serbia and Montenegro[/i]

Using lemma, This can be proved easily

$lemma$. Let $XAB$ be a triangle such that $XA=XB$.

$D, C$ be points on $XA, XB$ respectly such that $DC \parallel AB$.

and $S= AC \cap BD$ . and $N$ be the intersection of the bisector of $\angle CSB$ and $BC$.

Finally, $M$ be the mid point of $BC$. Then $D, N, M, A$ are on one circle.

$pf)$ Let $AN \cap CD = E$, $DM\cap AB = F$. And the length of $DC=p$.

It can be easily shown that $BF=CE=p$. so $EF\parallel XB$.

Therefore, $\angle EAF=\angle FDE$. and from $\angle CAB=\angle BDC$,

We get $\angle CAN = \angle MDB$. cuz $\angle CAD= \angle DBC$,

$\angle NAD = \angle DMN$. so lemma is proved.

Now, Let $C_i$ be the mid points of $A_i A_{i+1}$.

Using lemma, We get $\angle A_1 C_i A_n = \angle A_1 B_i A_n$.

With using similarity, We can concentrate angles into $B_1$.

So, the sum of angles is $180$.

$Q.E.D.$

[quote="Pascual2005"]Let $B_1$ and $B_n$ be the midpoints of its sides $A_1A_2$ and $A_{n-1}A_n$.
[/quote]

$B_n$ should be $B_{n-1}$.

[quote="Pascual2005"]. Also, for every $i\in\left\{2,3,4,\ldots ,n-1\right\}$.
[/quote]

$i\in\left\{2,3,4,\ldots ,n-1\right\}$ should be $i\in\left\{2,3,4,\ldots ,n-2\right\}$.

Let $C_i$ be the midpoint of $A_iA_{i+1}$. The central claim is that $\angle A_1B_iA_n=\angle A_1C_iA_n$. We only need to consider when $B_i\ne C_i$, so when $A_1A_n$ and $A_iA_{i+1}$ intersect at $X$ outside of the polygon. Note that $\angle XSB_i=90^\circ$, so $(XB_i;A_iA_{i+1})=-1$. Hence, $XA_i\cdot XA_{i+1}=XB_i\cdot XC_i$, from which we obtain $A_1A_nB_iC_i$ is cyclic by Power of a Point. The claim now follows.

To finish, note that $\angle A_1C_iA_n=\angle A_iMA_{i+1}$ by symmetry, where $M$ is the midpoint of $A_1A_n$, so the angles sum to $\angle A_1MA_n=180^\circ$, as desired.

The pith of this problem is this lemma:\\

[b]Lemma.[/b] Let $ABCD$ be an isosceles trapezoid, so that $\overline{AB}\parallel\overline{CD}$ and $BC=DA$. Let $S=\overline{AC}\cap\overline{BD}$, and suppose the angle bisector of $\angle BSC$ intersects $\overline{BC}$ at $N$. If $P$ is the midpoint of $\overline{BC}$, then $ADPN$ is cyclic.\\

[i]Proof.[/i] Let $T=\overline{AD}\cap\overline{BC}$ (if they are parallel the proof is trivial). Since $\overline{ST}$ bisects $\angle ASB$, by a well-known lemma, $-1=(B,C;T,N)$.  By the Midpoints of Harmonic Bundles Lemma, $TN\cdot TP=TB\cdot TC=TA\cdot TD$, and the desired result follows. $\blacksquare$
[asy]
size(4.5cm);
defaultpen(fontsize(10pt));
pair A, B, C, D, SS, NN, P, T;
A=(-3, 4);
B=(3, 4);
C=(6, 0);
D=(-6, 0);
SS=extension(A, C, B, D);
NN=extension(SS, SS+(1, 0), B, C);
P=(B+C)/2;
T=extension(A, D, B, C);
draw(B -- T -- A -- B -- C -- D -- A -- C, red);
draw(B -- D, red);
draw(T -- SS -- NN, orange);
draw(circumcircle(A, D, P), orange);
clip((-100, -1) -- (100, -1) -- (100, 100) -- (-100, 100) -- cycle);

dot("$A$", A, NW);
dot("$B$", B, NE);
dot("$C$", C, E);
dot("$D$", D, W);
dot("$S$", SS, S);
dot("$N$", NN, NE);
dot("$P$", P, E);
dot("$T$", T, N);
[/asy]
Let $M$ be the midpoint of $\overline{A_1A_n}$. Since $A_1A_n=A_iA_{i+1}$ and $A_1A_iA_{i+1}A_n$ is cyclic, $A_1A_iA_{i+1}A_n$ must be an isosceles trapezoid, whence by our lemma, $$\sum_{i=1}^{n-1}\angle A_1B_iA_n=\sum_{i=1}^{n-1}\angle A_1M_iA_n=\sum_{i=1}^{n-1}\angle A_iMA_{i+1}=\angle A_1MA_n=180^\circ,$$and we are done. $\square$
[asy]
size(8cm);
defaultpen(fontsize(10pt));
pair A1, A2, A3, A4, A5, A6, S2, S3, S4, B1, B2, B3, B4, B5, B6, M, M2, M4;
A1=dir(90); A2=dir(30); A3=dir(330); A4=dir(270); A5=dir(210); A6=dir(150);
S2=extension(A1, A3, A6, A2);
S3=extension(A1, A4, A6, A3);
S4=extension(A1, A5, A6, A4);
B1=(A1+A2)/2;
B2=extension(A1+S4-S3, S2, A2, A3);
B3=(A3+A4)/2;
B4=extension(A6+S2-S3, S4, A4, A5);
B5=(A5+A6)/2;
M=(A1+A6)/2;
M2=(A2+A3)/2;
M4=(A4+A5)/2;
pair X=A1+S4-S3;
pair O=(X+S3)/2;

draw(circle((0, 0), 1), red);
draw(circle(O, length(O)), orange);
draw(A1 -- A2 -- A3 -- A4 -- A5 -- A6 -- A1, red);
draw(A3 -- A1 -- A5 -- A1 -- A4 -- A6 -- A2 -- A6 -- A3, red);
draw(B2 -- X -- B3 -- X -- B4, orange+dashed);
//draw(B1 -- A6 -- B2 -- A1 -- B3 -- A6 -- B4 -- A1 -- B5, orange+dotted);
//draw(A6 -- M2 -- A1 -- M4 -- A6, fuchsia+dotted);
draw(B1 -- M -- B2 -- M -- B3 -- M -- B4 -- M -- B5, pink);
label("$A_1$", A1, A1);
label("$A_2$", A2, A2);
label("$A_3$", A3, A3);
label("$A_4$", A4, A4);
label("$A_5$", A5, A5);
label("$A_6$", A6, A6);
label("$S_2$", S2, NE);
label("$S_3$", S3, E);
label("$S_4$", S4, W);
label("$B_1$", B1, unit(B1));
label("$B_2$", B2, E);
label("$B_3$", B3, unit(B3));
label("$B_4$", B4, unit(M4));
label("$B_5$", B5, W);
label("$M$", M, dir(75));
label("$M_2$", M2, E);
label("$M_4$", M4, unit(M4));
label("$X$", X, unit(X));
dot(A1); dot(A2); dot(A3); dot(A4); dot(A5); dot(A6); dot(S2); dot(S3); dot(S4); dot(B1); dot(B2); dot(B3); dot(B4); dot(B5); dot(B6); dot(M); dot(M2); dot(M4); dot(X); dot(O);
[/asy]

We use complex numbers with $\omega$ a primitive $n$th root of unity, and $A_k = \omega^k$.

[asy] size(8cm);	pair A(int k) { return dir(90 + 180/11 - 360*k/11); } 	pair S = extension(A(1), A(5), A(4), A(0)); 	pair B = extension(S, incenter(S, A(4), A(5)), A(4), A(5));

dot("$A_1 = \omega^1$", A(1), A(1), blue); 	dot("$A_2 = \omega^2$", A(2), A(2), blue); 	dot(A(3), blue); 	dot("$A_k = \omega^k$", A(4), A(4), blue); 	dot("$A_{k+1} = \omega^{k+1}$", A(5), A(5), blue); 	dot(A(6), blue); 	dot(A(7), blue); 	dot(A(8), blue); 	dot(A(9), blue); 	dot(A(10), blue); 	dot("$A_n = \omega^n = 1$", A(11), A(11), blue); 	draw(S--B, heavygreen);

dot("$B_1$", midpoint(A(1)--A(2)), dir(A(1)+A(2))); 	dot("$B_{n-1}$", midpoint(A(10)--A(11)), dir(A(10)+A(11))); 	dot("$B_k$", B, dir(B), deepgreen); 	dot("$S_k$", S, dir(S), deepgreen);

draw(A(1)--A(5), deepgreen); 	draw(A(4)--A(0), deepgreen); 	draw(A(1)--A(4), lightred + dashed); 	draw(A(0)--A(5), lightred + dashed);

filldraw(A(1)--A(2)--A(3)--A(4)--A(5)--A(6)--A(7)--A(8)--A(9)--A(10)--A(0)--cycle, 		invisible, blue); [/asy]

[b][color=red]Claim:[/color][/b]  	For each $2 \le k \le n-2$ we have 	\[ b_k = \frac{2\omega^{2k+1} - \omega^{k+2} - \omega^k} 		{\omega^k - \omega + \omega^{k+1} - 1}.  \]

[i]Proof.[/i]  Note that $A_1 A_k A_{k+1} A_n$ is an isosceles trapezoid. Since the triangles $\triangle A_1 A_k S_k$ and $\triangle A_n A_{k+1} S_k$ are similar and isosceles, we have \[ \frac{|A_{k+1} B_k|}{|A_k B_k|} 	= \frac{|A_{k+1} S_k|}{|A_k S_k|} 	= \frac{|A_{k+1} A_n|}{|A_k S_1|} 	= \frac{\omega^{k+1}-1}{\omega^k-\omega} \] with the right hand side being a positive real number, owing to the fact that $\overline{A_1 A_k} \parallel \overline{A_n A_{k+1}}$ and point in the same direction. Thus, \[ b_k = \frac 	{(\omega^k-\omega) \cdot A_{k+1} + (\omega^{k+1}-1) \cdot A_k} 	{(\omega^k-\omega) + (\omega^{k+1}-1) } \] which is what we wanted to prove. $\blacksquare$

We can now compute the desired angle: \begin{align*} 	\frac{b_k-1}{b_k-\omega} 	&= \frac{ 		2\omega^{2k+1}-\omega^{k+2}-\omega^k 		- \omega^{k+1} - \omega^k + \omega + 1} 		{2\omega^{2k+1}-\omega^{k+2}-\omega^k 		- \omega^{k+2} - \omega^{k+1} + \omega^2 + \omega} \\ 	&= \frac 	{2\omega (\omega^k)^2 - (\omega^2+\omega+2) \omega^k + (\omega+1)} 	{2\omega (\omega^k)^2 - (2\omega^2+\omega+1) \omega^k + \omega(\omega+1)} \\ 	&= \frac 	{(\omega \cdot \omega^k - 1)(2 \omega^k - (\omega+1))} 	{(\omega^k - \omega)(2\omega \cdot \omega^k - (\omega+1))} \\ 	&= \frac{1}{\omega} \cdot \frac{\omega^{k+1}-1}{\omega^{k-1}-1} 	\cdot \frac{2\omega^k - (\omega+1)}{2\omega^{k+1} - (\omega+1)}. \end{align*} Unsurprisingly, this telescopes when multiplied: \begin{align*} 	\prod_{k=2}^{n-2} \frac{b_k-1}{b_k-\omega} 	&= \frac{1}{\omega^{n-3}} 	\cdot \frac{(\omega^{n-1}-1)(\omega^{n-2}-1)} 	{(\omega-1)(\omega^2-1)}  	\cdot \frac{2\omega^2-(\omega+1)}{2\omega^{n-1}-(\omega+1)} \\ 	&= \frac{1}{\omega^{-3}} 	\cdot \frac{\left( \frac{1}{\omega}-1 \right) 	\left( \frac{1}{\omega^2}-1 \right)} 	{(\omega-1)(\omega^2-1)} 	\cdot \frac{2\omega^2-(\omega+1)}{2\omega^{n-1}-(\omega+1)} \\ 	&= \frac{2\omega^2-(\omega+1)}{\frac{2}{\omega}-(\omega+1)} 	= -\frac{\omega(2\omega+1)(\omega-1)}{(\omega+2)(\omega-1)} 	= \frac{\omega(2\omega+1)}{\omega+2} \end{align*} On the other hand, the complex number with angle $\angle A_1 B_1 A_n = \angle A_1 B_{n-1} A_n$ is \[ \frac{1 - \frac{\omega + \omega^2}{2}}{\omega - \frac{\omega + \omega^2}{2}} 	= \frac{2-(\omega+\omega^2)}{\omega-\omega^2} 	= \frac{(2+\omega)(1-\omega)}{\omega(1-\omega)} 	= \frac{2+\omega}{\omega}.  \] So we finally arrive at \[ 	\frac{\omega(2\omega+1)}{\omega+2} 	\left( \frac{2+\omega}{\omega}\right)^2 	= \frac{(2+\omega)(2\omega+1)}{\omega} 	= 2(\omega+\omega^{-1}) + 4 \] which is real.

This proves that the requested sum $\sum_{i=1}^{n-1} \angle A_1B_iA_n$ is an integer multiple of $\pi$. It remains to do the annoying check that this is less than $2\pi$. One way is to note that for $k = 2, \dots, n-2$ we have \begin{align*} 	\angle A_1 B_k A_n 	&= \pi - (\angle A_n A_1 B_k + \angle A_1 A_n B_k) \\ 	&< \pi - (\angle A_n A_1 A_{k+1} + \angle A_1 A_n A_k) \\ 	&= \frac{2\pi}{n} \end{align*} and $\angle A_1 B_1 A_n < \pi - \angle A_2 A_1 A_n = \frac{2\pi}{n}$ as well. Thus the sum in question is less than $\frac{n-1}{n} \cdot 2\pi$ as required.

For each $i$, let $M_i$ be the midpoint of $\overline{A_iA_{i+1}}$, and let $T_i = \overline{A_1A_n} \cap \overline{A_iA_{i+1}}$. Note that $-1 = (T_iB_i;A_{i+1}A_i)$, so $T_iB_i \cdot T_iM_i = T_iA_{i+1} \cdot T_iA_i = T_iA_n \cdot A_iA_1$, so $A_nB_iM_iA_1$ is cyclic. Now $\angle A_1B_iA_n = \angle A_1M_iA_n = \angle A_iM_nA_{i+1}$, so summing cyclically we find $\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n = \angle A_1M_nA_n = 180^{\circ}$.

[font=MV Boli]Here is mine (may have typing issues I'll check soon)
Name midpoint of $A_1 A_n$ as $N$.
by symmetry $\angle A_1B_{n-1}A_n =\angle  A_{n-1}NA_{n}$ and Similarly $\angle A_1B_1A_n =\angle  A_2NA_1$
[rule]
[font=MV Boli]$\textbf{Claim:- }$  ($\forall i\neq 1,n,n-1$) $\angle A_1B_iA_n =\angle A_iNA_{i+1}$
Proof:
Consider $(A_1A_iA_{i+1}A_n)$ that is a isosceles trapizode.
Call midpoint of $A_iA_{i+1} = M_i$
by symmetry $\angle A_iNA_{i+1}=\angle A_1MA_n $
So it's equivalent to $\angle A_1B_iA_n = \angle A_1M_iA_n$
Enough to prove $A_1A_nB_iM_i$ is cyclic
Call $K_i=\overline{A_1A_n} \cap \overline{A_iA_{i+1}}$
Now notice $(K_i,B_i,A_i,A_{i+1})=-1 \implies K_iB_i \cdot K_iM_i = K_iA_i \cdot K_iA_{i+1}=K_iA_1 \cdot K_iA_n$
So PoP gives the cyclic result.
[rule]
[font=MV Boli]Now notice summing all these angle gives us $A_1NA_n = 180$
$\boxed{\mathcal{Q.E.D}}$

[img]https://cdn.discordapp.com/attachments/879997824564068375/932471678158663680/SPOILER_IMG_0685.jpg[/img]
This diagram gives a nudge for how to synthetically prove that $A_1A_nB_iM_i$ is cyclic. Using the point names in the diagram, we see that $OS_2B_2M_2$ is cyclic by easy angle chasing. We also know that $X_2$ is on $\overline{OS_2}$ by symmetry. Thus, we have \[X_2A_1 \cdot X_2A_7=X_2O \cdot X_2S_2=X_2B_2 \cdot X_2M_2\] by Power of a Point. This can be generalized.

Let $M_i$ denotes midpoint of $A_iA_{i+1}$ with cyclic numeration.

[b]Claim.[/b] $A_1,A_n,B_i,M_i$ are concyclic for $i\notin \left\{ 1,n-1,n\right\} .$
[b]Proof.[/b] Let $C_i=A_1A_n\cap A_iA_{i+1}.$ We easily conclude $$(A_iA_{i+1}B_iC_i)=-1\implies |C_iB_i|\cdot |C_iM_i|=|C_iA_i|\cdot |C_iA_{i+1}|=|C_iA_1|\cdot |C_iA_n|\text{ } \Box$$

Now we obtain $\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n=\sum_{i=1}^{n-1} \measuredangle A_1M_iA_n=\sum_{i=1}^{n-1} \measuredangle A_{i+1}M_nA_i=\measuredangle A_nM_nA_1=\pi \text{ } \blacksquare$

Let $M_i$ be the midpoint of $A_iA_{i+1}$ for $i\in \{1,2,...,n-1\}$, while $M$ is the midpoint of $A_1A_n$. We claim that the points $A_1,A_n,B_i$ and $M_i$ are concyclic:
[i]Proof:[/i] for simplicity, take an isosceles trapezoid $BDEC$ such that $BC\parallel DE$. Let $H$ be the midpoint of $EC$, $A=BD\cap EC$, $F=BE\cap EC$ and $G\in \overline{EC}$ such that $\angle GFE= \angle GFC$. By interior-exterior angle bisectors perpendicularity , we have $\angle AFG=90$, Applying [b]Right Angles and Bisectors[/b] theorem we get that $(A,G;E,C)=-1$, so by [b]Harmonic Bundles and Midpoint[/b] theorem we have
$$AG\cdot AH=AE\cdot AC=AD\cdot AB$$
Hence by power of the point $A$ we derive the needed result.$\square$

$$\sum_{i=1}^{n-1} \angle A_1B_iA_n = \sum_{i=1}^{n-1} \angle A_1M_iA_n = \sum_{i=1}^{n-1} \angle A_iMA_{i+1} = \angle A_1MA_n = 180$$
where the second equality holds by symmetry. $\blacksquare$


Among the numbers 1,2,3,.....,n prove that there is a number divisible by a higher power of 2 than any of the others. Hence, prove that the sum 1/1+1/2+1/3+.....+1/n cannot be an integer

[hide="Solution"]Assume that there are two numbers in this sequence - $ 2^mk$ and $ 2^ml$ and $ 2^m$ is the highest power of 2. Because k,l are odd there is an even number between them. Hence $ 2^mk<2^m\cdot 2h<2^ml$ - contradiction.[/hide]

will someone tell me how to derive the result ?

limes123 has prved this lemma :
if we denote by $ m \equal{} \sup${$ k ; \exists i \in${$ 1,..,n$} such that $ 2^k/i$} then such an $ i$ is unique .
so the $ M \equal{} lcm(1,..n)$ is in the forme $ M \equal{} 2^m(2p \plus{} 1)$ furthermore, if we wright $ \frac {1}{1} \plus{} .. \plus{} \frac {1}{n} \equal{} \frac {a}{M}$ then we have : $ a \equal{} 2l \plus{} 1$ because there is one odd number in the sequence $ (\frac {M}{j})_{1 \le j \le n}$ so:$ \frac {1}{1} \plus{} .. \plus{} \frac {1}{n}$ can not be an integer.


Sorry I just wanted to try something ... $ f(x)\equal{}x$, ...

[quote="mathnice1"]Show that there is not the function $ f: Z \rightarrow Z$ such that for all $ x,y \in Z :$
$ f(x \plus{} f(y)) \equal{} f(x) \minus{} y$[/quote]

$ x\equal{}0 \implies f(f(y))\equal{}f(0)\minus{}y$ as we know $ fof$ is increasing function , and $ T(y)\equal{}t\minus{}y$ is decreasing function (with $ t\equal{}f(0)$)    $ \implies$ increasing =decreasing $ \rightarrow$ contradiction .

[quote="javlik"][quote="mathnice1"]Show that there is not the function $ f: Z \rightarrow Z$ such that for all $ x,y \in Z :$
$ f(x\plus{}f(y))\equal{}f(x)\minus{}y$[/quote]
x=y=o
x=0      f(f(y))=f(0)-y   y=z   f(f(z))=f(0)-z     z=0 f(f(0))=f(0)
y=f(x)   f(x+f(f(x)))=0   x=z    f(z+f(f(z)))=f(z+f(0)-z)=f(f(0))=0   f(0)=0

x=0   f(f(y))=-y
x=-f(y)   f(0)=f(-f(y))-y=0   y=f(-f(y))=-f(f(y))  (1)  function   f(-x)=-f(x)
y=-f(x)   f(x+f(-f(x)))=2f(x)
     (1) ---       f(x-f(f(x)))=2f(x)
                     f(x-x)=2f(x)
                       2f(x)=0
                           f(x)=0  ---   0=0-y
not function
[/quote]


Salut,

Daca [i]A[/i] este o matrice patrata de ordinul 2 care admite 2 valori proprii reale, distincte, poate fi acest lucru utilizat la calculul puterii de ordinul [i]n[/i] a matricii? Exista vreo formula prin care $A^{n}$  sa se poata exprima cu ajutorul lui [i]A[/i]?

Multumesc

Dacă valorile proprii sunt distincte, atunci există $P$ inversabilă astfel încât $A = P \left( \begin{matrix}\lambda_{1}& 0 \\ 0 & \lambda_{2}\end{matrix}\right) P^{-1}$, unde coloanele lui $P$ sunt chiar eigenvectorii lui $A$ corespunzători celor două valorii proprii $\lambda_{1}, \, \lambda_{2}$.
Prin inducţie demonstrăm că $A^{n}= P \left( \begin{matrix}\lambda_{1}^{n}& 0 \\ 0 & \lambda_{2}^{n}\end{matrix}\right) P^{-1}$.

Teoretic, e destul de interesant. Practic, calculele sunt foarte urâte.

Ceea ce a zis perfect_radio e total adevarat.
Eu ma gandesc ca daca stii $\lambda_{1},\lambda_{2}$ valorile proprii atunci Hamilton-Cayley se scrie $A^{2}-(\lambda_{1}+\lambda_{2})A+\lambda_{1}\cdot\lambda_{2}=0$. De unde poti calcula $A^{n}$ dupa o banala relatie de recurenta.

[quote="jeanvaljean"][b]
Daca [i]A[/i] este o matrice patrata de ordinul 2 care admite 2 valori proprii reale, distincte, poate fi acest lucru utilizat la calculul puterii de ordinul [i]n[/i] a matricii? Exista vreo formula prin care $A^{n}$  sa se poata exprima cu ajutorul lui [i]A[/i]?
[/b][/quote]
[u][b] Raspuns afirmativ ![/b] [/u]Fie $A=\left(\begin{array}{cc}a & b\\ c & d\\ \end{array}\right)$ si sa notam $s\equiv$[b]tr(A)[/b]$: =a+d$ , $\Delta\equiv$[b]det(A)[/b]$: =ad-bc$. Polinomul (in $\lambda$ )
\[p_{A}(\lambda): =\begin{array}{|cc|}a-\lambda & b\\ c & d-\lambda \\ \end{array}=(\lambda-\lambda_{1})(\lambda-\lambda_{2})=\lambda^{2}-s\lambda+\Delta \]
se numeste "polinomul caracteristic" al matricii $A$, iar ecuatia $(*)\; \; \; \; p_{A}(\lambda)=\lambda^{2}-s\lambda+\Delta =0$ mai poarta denumirea de "ecuatia caracteristica a matricii $A$".  Numerele (reale sau complexe ) $\lambda_{1}, \lambda_{2}$ sunt  [b]valorile proprii [/b] ale  matricii $A$.
Prin intermediul unor calcule elementare se stabileste egalitatea
\[(**)\; \; \; A^{2}-sA+\Delta I_{2}=O\; , \]
care in realitate este teorema lui Hamilton- Cayley pentru matricile $2\times 2$, si care afirm'a ca orice matrice patrata isi verifica ecuatia caracteristica (compara cu  (*) ).

Algoritm:
Step 1. Notam  $w_{n}= \left\{\begin{array}{lcl}\frac{\lambda_{2}^{n}-\lambda_{1}^{n}}{\lambda_{2}-\lambda_{1}}&,&\lambda_{2}\ne \lambda_{1}\\ \\ n\lambda_{1}^{n-1}&, & \lambda_{2}=\lambda_{1}\end{array}\right.$.

Step 2. Se poate utiliza egalitatea (stabilita prin inductie matematica avand in vedere (**)) :
\[\begin{array}{|c|}\hline \\ A^{n}=w_{n}\cdot A-\Delta w_{n-1}\cdot I_{2}\\ \\ \hline \end{array}. \]
Exemplu "nostim": Daca $F: =\left(\begin{array}{cc}1 & 1 \\ 1 & 0\\ \end{array}\right)$ (matricea Fibonacci), gasiti $F^{n}$.
 Dupa aceea utilizati egalitatile $[$[b]det[/b]$(F)]^{n}=$ [b]det[/b]$(F^{n})$ si $F^{n}F^{m}=F^{n+m}$ , pentru a gasi anumite identitati verificate de numerele lui Fibonacci.


induction
inequalities unsolved
inequalities


geometry
circumcircle
exterior angle
angle bisector
geometry unsolved

[quote="Inequalities Master"]K is the intersection of the bisectors of interior angle B, and exterior angle C of the triangle ABC. L is the intersection of bisectors of the interior angle C, and exterior angle B. M is the midpoint of KL. Prove that M lies on the circumsircle of triangle ABC.[/quote]
 let X=(BL,CK) then the circumcircle of triangle ABC is the nine points circle of the triangle XLK so the problem solved  :wink:

sorry but i have a question
why did you say that the circuncircle of the triangle ABC
is the same that the 9 point circle of the triangle XKL???
i think that K and L lies in infinite circles.
 :blush:  :blush:  :blush:

it is clear dear :wink:
its obvoius that $ L,A,K$ are colinear.and $ KB,LC,XA$ are the altitudes of $ \triangle XLK$
then the circumcircle of triangle $ ABC$,is the nine_point circle of $ \triangle XLK$,and goes through the midpoint of $ KL$
if u have any problem,we can help U :D

[quote="pelao_malo"]i think im really bad, cuz i can't understand why L,A,K are collinear.
i think i have to practice more, please, help me  :blush:  :oops:
bye[/quote]

Both are on the exterior angle bisector of $ A$.


I've been using
\cup and \cap

for union and intersetion. Is there a better way?

Is there a command for the complement (instead of A^{c})

[quote="shake9991"]I've been using \cup and \cap for union and intersetion. Is there a better way?}[/quote]Those are the usual symbols. What do you mean by "better"? You can use \bigcap and \bigcup for things like $ \bigcup_{i=1}^\infty A_{i}$
[quote="shake9991"]Is there a command for the complement (instead of A^{c})[/quote]The amssymb package (which you should always load) provides \complement $ A^{\complement}$

You'll find lots of other symbols in [url=http://tug.ctan.org/tex-archive/info/symbols/comprehensive/symbols-letter.pdf]The Comprehensive LaTeX Symbol List[/url].


Hello,

I will be your moderator for Mafia Game 10. I hope everyone here is familiar with the rules; if not, you can find them on the Sticky at the top of the F&G forum.

The following people are signed up - please PM me to confirm or you may post a simple "here" below.

k2c901_1 +
Probability1.01
mwahahahaha +
anirudh
math92 +
agolsme +
RC-7th +
noneoftheabove +
xdl615x
G-UNIT +
blueywinkle +
jli +
zeb +
MithsApprentice +
PenguinIntegral +
bmsiv
whirligigs
slamchess
samath +
falconwing64 +
Super Naterul +
Fuzzeh_Panda
Drunker_Math +
mark +
Sly Si +
Peter VDD +
L_Li +

Edit note: If you have a + by your name you have confirmed as of 10:50 P.M. Thursday.

The roles:

Civilian: 7
Mason: 4
Miller: 1
Assassin: 1
Mafia: 4
Inspector: 2
Angel: 1
Werewolf: 4
Seer: 2
Sorcerer: 1

Some general rules:

From game 9:

Mafia may communicate during the day. Inspectors and Seers may not (you will know each other). Masons may not.

You may not falsely claim to be the assassin - this is stated in the rules, but I want to state it again. If you do, there will be fatal consequences.

Absolutely no discussion before the killings are announced the morning after the night. Those who are out of the game - please don't ruin it for others. I'm leaving much of this to your judgment of right and wrong.

For day lynchings: First you vote for who to kill, and the top 2 gets a defense speech. After that, everyone votes for only one of the two - you can also vote not guilty of course. Everybody - do not vote until defense speeches are given. After defense speeches are given, voting will proceed. You may change your vote exactly once before vote results and the lynchings are announced. So please vote wisely.

If you feel these rules need to be modified, PM me.

I will PM the roles today. Then, tomorrow, day 1 will commence.

Time Edit:

9:00 PM - Everybody should have received their roles by now, if you have a question please contact me. Other than that, I am waiting for more confirmations before starting.

Here.

What role am I? I don't think I've recieved a PM yet...  :?

I'm here! This is going to be fun! :D

EDIT: I think that we should start now!!

Here!  Werewolves should be interesting.

Er, I'm not sure how that would work potentially.

[quote="white_horse_king88"]You may not falsely claim to be the assassin - this is stated in the rules, but I want to state it again. If you do, there will be fatal consequences.[/quote]I have a question on this rule since the moment I read it: is it considered 'cheating' or are you just 100% sure you will get lynched by the community? :)

Besides that, I'm alive and playing. (unless someone suddenly stabs me in the back, that is)

[quote="Hexaditidom"]Ooh is it too late to sign up?
I didn't see a mafia 10 sign up.[/quote]

There's almost bound to be someone who doesn't confirm. You can have one of their parts. Otherwise, next game starts in about 2 weeks.

Peter VDD, it's not illegal but I'll assassinate you if you do. :D

Goodness gracious. 44 pages...

Anyhow, yeah I guessed he was werewolf *hmph*. Anyways, I'll give you guys a few days to comment on the game and whatnot before closing this topic...

Hi I'm back from the dead!

That was a really close game!

Wow.  Good game.  I was guessing that LadyKnight was the werewolf, until jli counter-claimed inspector, which was pretty clearly a last-ditch effort.

w00t! go citizens!

Ya i had inspected both of them in the past and i saw that jli was innocent and ladyknight was guilty. Good game people.

wow good job citezins!
i didnt get accused until dinoman inspected me w00t!!!!

Just curious...

Can someone get a list of all the players from the beginning and their roles? Looking back, I want to see what happened.

[quote="Peter VDD"][quote="MithsApprentice"]Anyways, I'll give you guys a few days to comment on the game and whatnot before closing this topic...[/quote]Why would you close it? :?[/quote]

That's generally how Mafia games go. When the game ends, I close the topic so that people don't come back in the future to continue the topic.

Go ahead, the game is over.

Why wouldn't we be aloud to claim roles?

[quote="slamchess"]but what if it is during the game?[/quote]often you must claim roles in order to win/survive. Great part of the fun in the game is exactly the roleclaiming. (like claiming good cards in poker)

Ah yes. Poker. :)

Anyhow, great game folks. And now uh...yeah let's put this (terribly) long and drawn-out match behind us. Moving on to 13...


I am using Microsoft Windows XP, IE 6.

http://www.artofproblemsolving.com/Forum/privmsg.php?folder=inbox

On the top-right corner, there is the "Your inbox is 49% full"

The bar is hilighted backwards from the right; the interval [49,100] is hilighted rather than [0,49].

I could not reproduce this error on Windows XP / IE6.  (My inbox is 85% full and the bar appears on the left as it should, in both the AoPS and Mathlinks views).

Is this still happening?

One trick that sometimes works with IE 6 is to slightly resize the window horizontally -- this sometimes causes rendering bugs to fix themselves.

Yes, the problem is still there. I tried resizing the window, but it turns out that it actually produces another bug - the text hangs over the right side out of the frame. What is even more odd is that I tried this with several other windows and the effect only happens on Private Messages.

I tried to attach a screenshot here, but there were some errors with that. So I will (attempt to) PM it to DPatrick.

I think Philip lied to me.   :mad:  (Not on purpose though -- I assume  :roll: )

He was really using IE7 (when he said he was using IE6).  There was indeed a bug on IE7 with the quota bar.  I think it should now be fixed.

Yes, it turns out that it actually was IE7. The bar now works fine, but there is still text hanging out over the right frame when it is restored down.


limit
integration
logarithms
real analysis
real analysis unsolved

Compute $\lim_{n\to \infty}\sum_{k=1}^{n}\frac{1}{k\cdot 2^{k}}$.

We are to prove that $0\leq \frac{1}{1-x}-\sum_{k=1}^{n}x^{k-1}\leq 2x^{n}\ (n=1,\ 2,\ \cdots)$ for $0\leq x\leq \frac{1}{2}.$
[hide="Answer"]$\ln 2$[/hide]


AMC
USA(J)MO
USAMO
inequalities proposed
inequalities

Obvious? One of the easiest USAMO problems ever! Anyway, nice problem...

[quote="Pascual2005"]Let $x_1,x_2...x_n$ be real numbers such that:
$x_1+x_2+...x_n>=n$ and $x_1^2+x_2^2+...x_n^2>=n^2$
prove some of the n numbers is at least as big as $2$.[/quote]
??? Do you ask to prove that there is $x_i\geq 2$?
It is really easy, though I considered two cases.
Maybe Harazi knows much more trivial proof?  :D

Kalva Site USAMO99 b/2 solution is by contraddiction.

anyway. my solution is nice.
by contradiction:
all number lie in the interval $(-n+2,2)$ so the parabola $x^2+(n-4)x-2(n+2)$ is <0 for all $x_i$ suming over all $x_i$, we get a contradiction.


If the date is March 1, 2007, in how many whole years will the number representing the year be the smallest perfect square greater than 2007?

Some quick arithmetic shows that $ 44^{2}<\sqrt{2007}<45^{2}\equal{}2025$.  So that's 18 years.


modular arithmetic
algebra
IMO Shortlist
equation
number theory

another way:
cheking by moduls $ 3$ and $ 4$ we conclude that $ x$ and $ z$ are even therefore $ x\equal{}2a$ and $ z\equal{}2b$ now we have
$ 4^y\equal{}(5^b)^2\minus{}(3^a)^2\equal{}(5^b\minus{}3^a)(5^b\plus{}3^a)$ and consequently $ (5^b\minus{}3^a)$ and $ (5^b\plus{}3^a)$ are obviously both even but not both devisable by $ 4$ we can conclude one of these two numbers equals $ 2$ and therefore the only possibility is $ 5^b\minus{}3^a\equal{}2$ leading to the unique solution of $ (x,y,z)\equal{}(2,2,2)$

Sorry to revive, but how can you prove the only solution to $5^a-3^b=2$ is $(a,b)=1$?

Once we get $ 4^y\equal{}(5^b)^2\minus{}(3^a)^2\equal{}(5^b\minus{}3^a)(5^b\plus{}3^a)$, it follows $5^b\minus{}3^a = 2^{\alpha}$ and $5^b\plus{}3^a = 2^{\beta}$, with $\alpha+\beta = 2y$. So $5^b = 2^{\beta-1} + 2^{\alpha-1}$, $3^a = 2^{\beta-1} - 2^{\alpha-1}$. Therefore we must have $\alpha=1$, and now the equation $3^a = 2^{\beta-1} - 1$ is known (and easy to prove) it has as only solution $\beta = 2$, $a=0$ or $\beta = 3$, $a=1$; the only working for $5^b = 2^{\beta-1} + 1$ is the latter, for which $b=1$.

[quote="me@home"][quote="snw"]Supose $a>1$ , it's easy to see which $b$ is odd then by hensel's lemma $v_2(a)=v_2(b)$ then $a$ is odd. Let $p$ the less prime number such that $p|5^a-1$ and let $d$ the order of 5 module $p$, then $d|\gcd(p,a-1)=1$ because $2\nmid a$. Contradiction, then $b=1$ and the map $x\mapsto 3^x$ in growing it's conclude that the only solution is $(a,b)=(1,1)$[/quote]
Where do you show that $p\not | (a-1)$?[/quote]

it's wrong, is $d|\gcd(p-1,a)$

[quote="me@home"][quote="snw"]Supose $a>1$ , it's easy to see which $b$ is odd then by hensel's lemma $v_2(a)=v_2(b)$ then $a$ is odd. Let $p$ the [b]less prime number[/b] such that $p|5^a-1$ and let $d$ the order of 5 module $p$, then $d|\gcd(p,a-1)=1$ because $2\nmid a$. Contradiction, then $b=1$ and the map $x\mapsto 3^x$ in growing it's conclude that the only solution is $(a,b)=(1,1)$[/quote]
Where do you show that $p\not | (a-1)$?[/quote]
 In fact it was a typo. It would be $\gcd(p-1,a)$ and he meant $p$ to be the smallest prime factor of $5^a-1$, as bolded  and so no primes less than $p$ can't be in $a$. Thus $\gcd(p-1,a)=1$. No fact of $p\not|a-1$

[quote="orl"]Find all positive integer solutions $ x, y, z$ of the equation $ 3^x \plus{} 4^y \equal{} 5^z.$[/quote]
Reducing modulo 3 gives $1\equiv 2^z$, so $z=2z_1$. Reducing modulo 4 gives $(-1)^x\equiv 1$, so $x=2x_1$. Hence by difference of squares $4^y=(5^{z_1}-3^{x_1})(5^{z_1}+3^{x_1})$. Now $LHS$ is a power of $2$ so $5^{z_1}-3^{x_1}=2^a$ and $5^{z_1}+3^{x_1}=2^b$ with obviously $a<b$. Adding we get $2\cdot 5^{z_1}=2^a(2^{b-a}+1)$, so $a=1$. Subtracting and using $a=1$ we obtain $5^{z_1}-1=2^{b-1}$, so $4(5^{z_1-1}+5^{z_1-2}+\cdots +5+1)=2^{b-1}$, hence $b=3$. So $5^{z_1}+3^{x_1}=8\Rightarrow (z_1,x_1)=(1,1)$, and $2y=a+b=4\Rightarrow y=2$. It follows after reverse substituting for $z=2z_1$ and $x=2x_1$ that $x=y=z=2$ is the only solution: $3^2+4^2=5^2$.

Looks like this problem was proposed long before 1991. Here it is at the second Romanian TST from 1979:

We can also use Fermat's theorem once we get X=y to conclude (X,y) belongs to(2,2)

[quote=cosinator][hide]Consider the equation modulo $ 3$.  We have $ 1\equiv 2^z\pmod{3}\Rightarrow z \equal{} 2v$ for positive integer $ v$.  Considering the equation modulo $ 4$ yields $ ( \minus{} 1)^x\equiv 1\pmod{4}\Rightarrow x \equal{} 2u$ for positive integer $ u$.  Thus, we have the equation

$ 3^{2u} \plus{} 2^{2y} \equal{} 5^{2v}$

That is, $ (3^u,2^y,5^v)$ is a pythagorean triple.  So, it is well known that we have $ 3^u \equal{} m^2 \minus{} n^2,2^y \equal{} 2mn,5^v \equal{} m^2 \plus{} n^2$ for integers $ m,n$.  From $ 2^y \equal{} 2mn$ we find that $ m,n$ are powers of $ 2$.  It is clear from $ 3^u \equal{} m^2 \minus{} n^2$ that $ n \equal{} 1$ or $ n \equal{} 0$.  If $ n \equal{} 0$ then $ m^2 \equal{} 3^u \equal{} 5^v\Rightarrow m \equal{} 1$ which gives has no positive integer solutions.  If $ n \equal{} 1$, let $ m \equal{} 2^t$.  Then we have $ 3^u \equal{} 2^{2t} \minus{} 1$.  Consider the equation modulo $ 4$.  We get $ ( \minus{} 1)^u\equiv \minus{} 1\pmod{4}$ and so $ u \equal{} 2k \plus{} 1$ for nonnegative $ k$.  If $ t > 1$, we reduce the equation to

$ 3^{2k \plus{} 1} \minus{} 3 \equal{} 4^t \minus{} 4$

$ \Rightarrow 3(3^k \minus{} 1)(3^k \plus{} 1) \equal{} 4(2^{t\minus{}1} \minus{} 1)(2^{t\minus{}1} \plus{} 1)$

It is clear that since $ t > 1$ that $ 4$ is the largest power of $ 2$ which divides the right hand side.  But, whether or not $ k$ is even or odd, we have that $ 8$ divides the left hand side.  Thus, we must have $ t\le1$ which gives the only solution $ t \equal{} 1,k \equal{} 0$.  This yields the unique solution $ (x,y,z) \equal{} (2,2,2)$.   [/hide][/quote]

How do you now that n is 0,or 1 please explain more i don't understand.

[quote=orl]Find all positive integer solutions $ x, y, z$ of the equation $ 3^x \plus{} 4^y \equal{} 5^z.$[/quote]

Cute diophane :3. Obviously $y>1$ becuase for $y=1$ there is no sol.
[color=#f00][b]Claim 1:[/b][/color] $x$ and $z$ are even.
[color=#f00][b]Proof:[/b][/color] Consider using the modulo 3 on the following diophane:
$$3^x+4^y \equiv 5^z \pmod 3 \implies (-1)^z \equiv 1 \pmod 3 \implies z \; \text{even}$$
Now consider using modulo 4 on the folllowing diophane:
$$3^x+4^y \equiv 5^z \pmod 4 \implies (-1)^x \equiv 1 \pmod 4 \implies x \; \text{even}$$
From here we will set $x=2m$ and $z=2n$ so the given diophane becomes to:
$$9^m+4^y=25^n$$
[color=#f00][b]Claim 2:[/b][/color] $5^n=3^m+2$ is true.
[b][color=#f00]Proof:[/color][/b] Factor $25^n-9^m$ and use that $y>1$, note that $4^y$ is a multiple of $16$ so:
$$4^y=(5^n+3^m)(5^n-3^m) \implies 5^n-3^m=2 \; \text{becuase} \; 5^n-3^m \not\equiv 0 \pmod 4$$
Replace $25^n=9^m+4 \cdot 3^m+4$ on the given diophane to get:
$$4^{y-1}=3^m+1$$
[color=#f00][b]Claim 3:[/b][/color] The only sol is $(2,2,2)$
[color=#f00][b]Proof:[/b][/color] Put on both sides of the equation $v_3$ and then by LTE:
$$v_3(4^{y-1}-1)=1+v_3(y-1)=m \implies m-1=v_3(y-1) \implies y=3^{m-1} \cdot a+1 \; \text{where} \; a \not\equiv 0 \pmod 3$$
[b]Case 1:[/b] $y=4$
From here we can see that $a=1$ and $m=2$ so $x=4$. Replacing on the diophane we get:
$$81+256=337=5^z \; \implies \; \text{contradiction!!}$$
[b]Case 2:[/b] $y \ne 4$
Since $y \ne 4$ we can use Zsigmondy to get that $4^{y-1}-1$ has a prime factor such that $p \ne 3$.
Assume that $m>1$. And now divide by $3$ both sides and use modulo p:
$$3^{m-1}=\frac{4^{y-1}-1}{3} \equiv 0 \pmod p \implies \; \text{contradiction!!}$$
So we have that $m=1$ and this implies $x=2$ and $y=2$. Replace this on the initial diophane:
$$3^2+4^2=5^z \implies 25=5^z \implies z=2$$
So the only solution is $(x,y,z)=(2,2,2)$. Thus we are done  :blush:
Note: I think there is a trivial way to do this, i think using many powerful theorems for an easy problem is bad.

[quote=cosinator][hide]Consider the equation modulo $ 3$.  We have $ 1\equiv 2^z\pmod{3}\Rightarrow z \equal{} 2v$ for positive integer $ v$.  Considering the equation modulo $ 4$ yields $ ( \minus{} 1)^x\equiv 1\pmod{4}\Rightarrow x \equal{} 2u$ for positive integer $ u$.  Thus, we have the equation

$ 3^{2u} \plus{} 2^{2y} \equal{} 5^{2v}$

That is, $ (3^u,2^y,5^v)$ is a pythagorean triple.  So, it is well known that we have $ 3^u \equal{} m^2 \minus{} n^2,2^y \equal{} 2mn,5^v \equal{} m^2 \plus{} n^2$ for integers $ m,n$.  From $ 2^y \equal{} 2mn$ we find that $ m,n$ are powers of $ 2$.  It is clear from $ 3^u \equal{} m^2 \minus{} n^2$ that $ n \equal{} 1$ or $ n \equal{} 0$.  If $ n \equal{} 0$ then $ m^2 \equal{} 3^u \equal{} 5^v\Rightarrow m \equal{} 1$ which gives has no positive integer solutions.  If $ n \equal{} 1$, let $ m \equal{} 2^t$.  Then we have $ 3^u \equal{} 2^{2t} \minus{} 1$.  Consider the equation modulo $ 4$.  We get $ ( \minus{} 1)^u\equiv \minus{} 1\pmod{4}$ and so $ u \equal{} 2k \plus{} 1$ for nonnegative $ k$.  If $ t > 1$, we reduce the equation to

$ 3^{2k \plus{} 1} \minus{} 3 \equal{} 4^t \minus{} 4$

$ \Rightarrow 3(3^k \minus{} 1)(3^k \plus{} 1) \equal{} 4(2^{t\minus{}1} \minus{} 1)(2^{t\minus{}1} \plus{} 1)$

It is clear that since $ t > 1$ that $ 4$ is the largest power of $ 2$ which divides the right hand side.  But, whether or not $ k$ is even or odd, we have that $ 8$ divides the left hand side.  Thus, we must have $ t\le1$ which gives the only solution $ t \equal{} 1,k \equal{} 0$.  This yields the unique solution $ (x,y,z) \equal{} (2,2,2)$.   [/hide][/quote]

how did you get $ n = 1$ or $ n = 0 $ as the $2$ only possibilities?

Since $m$ and $n$ are positive integers and the difference of their squares is odd, one of the numbers must be odd. If $n>1$ then since $m>n$ and both are powers of $2$, that would mean both numbers are even, which cannot happen.

\begin{align*}
    (-1)^x \equiv 1^y \pmod{4} \implies 2 \mid x \\
    1^y \equiv 2^z \pmod{3} \implies 2 \mid z
\end{align*}

Let $x = 2a, z = 2c$ and thus we have \[ 2^{2y} = 5^{2c} - 3^{2a} = (5^c - 3^a)(5^c + 3^a). \] Notice that $5^c \equiv 1 \pmod{4}$ and that $3^a \equiv \pm 1 \pmod{4}$. This means that exactly $1$ of $5^c - 3^a, 5^c + 3^a$ is divisible by $4$. Now this means that one of them is equal to 2.

$\textbf{Case 1:}$ $5^c + 3^a = 2$

This means that $a = c = 0$ which is a contradiction since $x, y, z > 0$.

$\textbf{Case 2:}$ $5^c - 3^a = 2$

It is easy to see that $(a, c) = (1, 1)$ is a solution. We claim that this is the only solution. For the rest of the solution we assume that $a, c > 0$.

Taking mod 5 we get \[ -3^a \equiv 2 \pmod{5} \implies 3^a \equiv 3 \pmod{5} \] The cycle is $(3, 4, 2, 1)$ which means that $a \equiv 1 \pmod{4}$.
Now let $a = 4\beta + 1$.

Taking mod 16 we get \[ 5^c-3^{4\beta} \cdot 3 \equiv 2 \pmod{16} \implies 5^c \equiv 5 \pmod{16} \] The cycle is $5, 9, 13, 1$ which means that \[ c \equiv 1 \pmod{4} \qquad (2) \]
Let $c = 4\alpha + 1$.

Taking mod 13 we get \[ 5^{4\alpha} \cdot 5 - 3^a \equiv 2 \pmod{13} \implies 3^a \equiv 3 \pmod{13} \] The cycle is $3, 9, 1$ which means that \[ a \equiv 1 \pmod{3} \qquad (3) \] Combining $(1)$ and $(3)$ we get \[ a \equiv 1 \pmod{12} \qquad (4) \]
Let $a = 12\lambda + 1$.

Taking mod 7 we get \[ 5^c - 3^{12\lambda} \cdot 3 \equiv 2 \pmod{7} \implies 5^{c} \equiv 5 \pmod{7} \] The cycle is $5, 4, 6, 2, 3, 1$ which means that \[ c \equiv 1 \pmod{6} \qquad (5) \] Combining this with $(2)$ we get \[ c \equiv 1 \pmod{12} \qquad (6) \]

Finally taking mod 9 we get \[ 5^c \equiv 2 \pmod{9} \] Note that since we assumed that $a, c > 0$ this means that from $(4)$ we have $\lambda > 0$ which means that $3^{12\lambda + 1} \equiv 0 \pmod{9}$. The cycle is $5, 7, 8, 4, 2, 1$ which means that $c \equiv 5 \pmod{6}$ which is a contradiction with $(5)$.

Since $(a, c) = (1, 1)$ is the only solution to the above diophantine equation, we get that $x=z=2$. Thus the only solution is $\boxed{(2, 2, 2)}$.

[quote=x3yukari][quote]$\textbf{Case 2:}$ $5^c - 3^a = 2$[/quote]

here's another way >.<
Taking $\mod 4$, $1-3^a \equiv 2 \pmod{4}$, so $a$ is odd.
Taking $\mod 3$, $2^c \equiv 2 \pmod{3}$ so $c$ is also odd.
Taking $\mod 7$, we can see that $5^c$ cycles through $5, 4, 6, 2, 3, 1 \pmod{7}$, and $3^a$ cycles through $3, 2, 6, 4, 5, 1 \pmod{7}$. But since both $a$ and $c$ are odd, the only possibility is for $5^c \equiv 5 \pmod{7}$ and $3^a \equiv 3 \pmod{7}$. So $c \equiv 1 \pmod{6}$ and $a \equiv 1 \pmod{6}$
Taking $\mod 9$, we can see that $5^c$ cycles through $5, 7, 8, 4, 2, 1 \pmod{9}$, and for $a \geq 2$, $3^a \equiv 0 \pmod{9}$. But if $3^a \equiv 0 \pmod{9}$, then $5^c \equiv 2 \pmod{9}$ which implies $c \equiv 5 \pmod{6}$, contradiction. So the only solution is when $3^a \equiv 3 \pmod{9}$, which is when $a=1$. So $(a,c)=(1,1)$.[/quote]

Cool solution! :D


linear algebra
matrix
linear algebra unsolved

Let $ P$ be a matrix in $ SO_3(\mathbb{C})$.
Prove that if $ X$ ia an eigenvector with eigenvalue $ 1$ and if $ P \ne I$, then $ X^t X\ne 0$.

Can you check again about your problem :
Supposed $ A\equal{}[{a}_{ij}]$, with the property : $ {a}_{11}\equal{}{a}_{22}\equal{}1$ and $ {a}_{33}\equal{}\minus{}1$, and $ {a}_{ij}\equal{}0$ besides, note that $ x\equal{}(1,i,0)^t$ is an eigenvector with eigenvalue $ 1$, but $ x^tx\equal{}0$.
I am sorry if I am wrong.

Qruni, that matrix is in $ O(3)$ but not $ SO(3).$

[quote="Stiger"]Let $ P$ be a matrix in $ SO_3(\mathbb{C})$.[/quote]Is $ SO_3(\mathbb{C})$ the same as the unitary group $ SU(3)$?

Also, does $ X^t$ refer to the conjugate tranpose or ordinary transpose?

Has to be transpose, otherwise the assertion would be trivial.


Everyone please vote your favorite number ... this will be a nice stat to view after a hundred votes :P

I voted 3.

Don't know why, but it is a great number.

The Ancient Egyptians would agree with me, right?

Or were they more of 6-type people?

i like 7 because its the highest score i've ever gotten on an AIME (practice, no real ones taken yet :( )

+, 7 is a lucky #. (not really, but i like to say that it is)

i like none of those........my fav. num. is 225

my favorite number is not an integer  :D

[quote="mcalderbank"]my favorite number is not an integer  :D[/quote]

Agreed.
However, my favorite natural number less than 10 is 3.
One of my friends has devised several 'proofs' of the fact that every number is equal to 3.

Please do not ask me why.   ;)

my favorite number is actually 21...but i really like the numbers 3&7 bcuz they're factors of 21. :) i voted for 7 cuz i write that one really cool so i like that one best :) i'm sucha  dork! :lol:

i voted for ten because:
i have 10 fingers
10 toes
so we work in base 10
and no one else voted for it

I think 3 is a very interesting number, very interesting!!!

[quote="Danbert"]
Agreed.
However, my favorite natural number less than 10 is 3.
One of my friends has devised several 'proofs' of the fact that every number is equal to 3.

Please do not ask me why.   ;)[/quote]

I would like to see some of these proofs. Sounds interesting....

Prove that $1$ is the largest number.

Proof: Compare the set $\{x\}$ with $\{-x\}$ where $x>0$. Obviously the former set is strictly larger than the latter. Thus no number in the latter set can be the largest.
Compare the set $\{x^2\}$ with $\{x\}$ where $x>1$. Obviously the former set is strictly larger than the latter. Thus no number in the latter set can be the largest.
Compare the set $\{x^2\}$ with $\{x\}$ where $0<x<1$. Obviously the former set is strictly larger than the latter. Thus no number in the latter set can be the largest.

The only numbers which are not in either of the three latter sets are $1$ and $0$. But $1>0$ QED

Well, 7 is a nice number, but i totally prefer 127.

[quote="mcalderbank"]what it makes it so special vs. 1, or 2?[/quote]

Eventually you will figure it out!  :D .

Best,

My favortie number between 1 and 10 isnt on the list. You need to change it to 'favorite integer'

Anyways I can make a case for all of them.

1 is nice and small. It is the only number with one divisor.
2 is even, which of course makes is the oddest prime of them all.
3 is half of 8, yet 3*2 = 6.
4 is the first sqaure. Gotta like it for that
5 is the only number on the list that I cant make a good case for. Of course, this in and of itself is a readon to vote for 5 :P
6 Is perfect. Whats not to like?
7 Is lucky. Also ive always liked to for some reason
8 Is round and pretty. Its fun to draw
9 is the first number you can spell without using any curves(in english).
10 Gives the smallest possible reult when you divide letters in its english name by digits in the number itself(1.5).

[quote]9 is the first number you can spell without using any curves(in english).[/quote]

What about 5? FIVE (i see no curves)  :P

8 is my lucky number. I can usually associate most of my math competition results to eight(eighth place, eight points, minus eight points, whatever). So great. That rhymes!

you're forgeting phi. What does the symbol for phi look like? And how is it pronounced?

If you mean the golden number (or the golden ratio) it looks like $\ro$ actually ... *at least in romanian*

why don'twehavea vote of the greek letters?

[quote="Valentin Vornicu"][quote="fluffybunny123"]lol, no pi or e?  :([/quote]Nope, just integers :D - it would have been unfair for the integers to compete with $\pi$ or $e$ :P[/quote]

By the way, happy $\pi$ day (Today it is 3.14)

Best,

the year 2015 will be a good year for pi.

How could anyone not vote for 6.  It is perfect.

The only other acceptable vote would have been for 10 which the ancient Greeks called tetrakus - the holy four-foldness - representing the ancient elements of earth, wind, water, and fire.  And no heart by the way, this isn't captain planet.

\[1 + 2 + 3 + 4 = 10\]

hmm i have always loved 4 and 104, don't know why, but they always turn up on everything. they have always loved me, too, i guess!


calculus
integration
inequalities
trigonometry
function
real analysis
real analysis unsolved

Prove the following inequality.

\[(1+\sqrt{2})\ \pi\leqq \int^\pi_{-\pi }\sqrt{1+\cos ^2 x}dx\leqq\sqrt{6}\ \pi\]

The function is even and periodic of period $\pi$. That means we can take any interval of length $\frac{\pi}2$ as representative of the whole. Rewrite the inequality as

$\frac{(1+\sqrt{2})\pi}4<\int_0^{\frac{\pi}2}\sqrt{1+\cos^2x}dx< \frac{\sqrt{6}\pi}4.$

The left hand side is the trapezoidal rule estimate (with one interval) for the integral on that interval and the right hand side is the midpoint rule (with one interval) for the same. The function is concave downward. For such functions, the trapezoidal rule estimate is always less than the integral and the midpoint rule estimate is always greater than the integral.

[quote="Kent Merryfield"] The function is concave downward. For such functions, the trapezoidal rule estimate is always less than the integral and the midpoint rule estimate is always greater than the integral.[/quote]

It does not preserve the convexity...

Oops, you're right. It's not concave down.