What are the advantages to using LaTeX vs. the equation editors that come with MS Word?

Clearly I'm a newbie, so go easy on me.  ;)

It is usually faster to just type $ \text{\LaTeX}$ code then click a bunch of buttons.  Of course, this is why in Office2007 Microsoft put in a sort of "$ \text{\LaTeX}$" parsing for its equation editor.

The most obvious advantages of LaTeX are
1. much better quality output of typesetting standard as used in books;
2. easier to type in equations than constantly picking symbols with a mouse;
3. some versions of word can't cope with too many euqations in a document; LaTeX can cope with hundreds of pages;
4. LaTeX is free;
5. you couldn't easily have equations on this forum using Word.

Some links describing the benefits of LaTeX
[url=http://www.andy-roberts.net/misc/latex/latexvsword.html]Latex vs. MS Word[/url]
[url=http://nitens.org/taraborelli/latex]The Beauty of LaTeX[/url]
[url=http://ricardo.ecn.wfu.edu/~cottrell/wp.html]Word Processors: Stupid and Inefficient[/url]
[url=http://www.sixthform.info/steve/Slideb.pdf]How Was I Made?[/url] presentation ([i]save on downloading then open it. ESC gets out of full screen, CTRL+L goes to full screen[/i])

It would be interesting to hear from other users why they use LaTeX.

The MS version is a bit more tedious to use. While learning the LaTeX code can also be long and tedious, you can accomplish many more things using LaTeX.

Thinking of a particular LaTeX outcome and it's code can take no time at all. However, using Word's confusing interface can cause misinformation (I heard some functions of the Word Equation do not allow certain types of math to be inputed?).


[hide]Well the first discount, 10%, is a multiplication by 0.9. The second, 20%, is a multiplication by 0.8. Multiplying these, this is an overall multiplication by 0.72, and thus a 28% total discount.[/hide]

Is it....[hide]35.2%

lets say it costs 100 dollars
-----------
at first:100
first discount (10%):90 (-10)
second discount (10%):81 (-9)
third discount (20%):64.8 (-16.2)

so together (100 - 64.8), we get a 35.2% discount
:)
[/hide]?
-jorian

i actually posted that earlier.  but no, and i know why.

in the question, it says THE 2 succesive...of 10 and 20.  so its just .9 times .8.

if it said 2 succesive followed by 20, then youd be right.

can u or someone else explain what 1337 said?
-jorian

sure.

[hide]all it is is a ten percent discount followed by a twenty percent.

so you have x, or 1x.  10% discount is 90% or .9. so x times .9 is .9x.  and a 20% discount is 80% or .8. so .9x times .8 is .72x.  so the final price is 72% of the original, which is a 28% discount
[/hide]
 :ninja:

oh, i thought there were 2 different discounts of 10% :(
-jorian

That's the same thing that happened to me.

Um, actually I think that there are two 10 % discounts. It says two succesive discounts of 10 [b]followed[/b] by one of 20.

I think it meant two successive discounts, the first being 10 and the second being 20.

[quote="davidli"]What is the single discount that is equivalent to the two successive dicounts of 10% off followed by 20% off the dicounted price?[/quote]

[hide]An easy way to do it is to assume the value is $\$$100. It becomes $\$ 100\longrightarrow \$90\longrightarrow\$ 72.\ 100-72=\mbox{28\% Discount}$

I think this problem should have been worded more clearly. It was worded badly.[/hide]

Yes. I agree. Most people thought it meant 3 discounts:wink:
-jorian

didnt i already clear this up???

it said THE 2 succesive discounts, meaning only two, of 10 and 20 respectively. if the "The" wasnt there, then itd be 2 10s and a 20.

The point is that the question was worded unclearly, and in my opinion, I would consider either way as correct.
There's no point arguing which way it was intended, since you could make a strong argument for both interpretations, so how you interpret it really just depends on your own opinion.

(By the way, in case you were wondering, I'm pretty sure the MathCounts Question Writing Committee is much more careful with ambiguities like this on competitions than on the handbook warm-ups, so you won't need to worry too much about something like this happening during a competition.)

[quote="bpms"][quote="humblemastermind"]No, the answer sheet's answer is as if it's just one of each.[/quote]
Which warmup is this from what year?[/quote]
Can someone just say the answer to my question please?


In the 1st answer.-1 is not in the root thing.same with -2 in the 2nd ans
sorry,i dont know latex :blush:

aFter rationalizing the denominator for the second one, I get $-2-\sqrt6 = \sqrt6+2$. It works on my calculator too.

[quote="shadysaysurspammed"]2nd one 6^1/2-2?

if both my answers r wrong, please laugh hard at me cause these questions r pretty simple  :oops:[/quote]
hahhaahhaahahhahahahahahahahhahahahahahahaha
please do not take it seriously :rotfl:

You guys got the right answers :)
These were pretty simple, weren't they?

What were the answers? I didn't quite catch it... :P

I just wanted to say PLEASE do NOT double post.

oh and i also wanted to say...

[hide]PLEASE do NOT triple post.[/hide]

actually, you can just press the edit button.

Similar to number 1:

[hide]If $x=\frac{\sqrt{33}-5}{2}$, the what is the value of $(x+1)(x+2)(x+3)(x+4)$? [/hide]

Let the chain continue(easy ones)
this question as given to me my eminem,so its really his question-

x^2+(x^2)/(x+1)^2=3
find x. :D

[quote="shadysaysurspammed"]is the answer 48?(to factorial's question)[/quote]
[hide="my solution"]$x=\frac{\sqrt{33}-5}{2}\Longleftrightarrow 2x+5=\sqrt{33}$
$\Longrightarrow 4x^2+20x+25=33\Longleftrightarrow x^2+5x-2=0$

Then,

$(x+1)(x+2)(x+3)(x+4)=(x+1)(x+4)\times (x+2)(x+3)$
$=(x^2+5x+4)(x^2+5x+6)=(2+4)(2+6)=\boxed{48}$[/hide]
Is this what you did?

[quote="shadysaysurspammed"]x^2+(x^2)/(x+1)^2=3
find x. :D [/quote]
Did you mean $x^2+\frac{x^2}{(x+1)^2}=3$?
Hmm, I don't know any cool way... :(

I hope u guys can understand my question.could someone please convert it into latex? :lol:
by the way frt,there's no cool way to do it,its just a sum to solve. :lol:
ps-no binomial theorem is required in it.only simple 9th grade formulae. ;)

I dont seem to be able to factorise this....
$x^4+2x^3-x^2-6x-3=0$

I've tried that, and I don't think that's the solution.


Let a circumscribed (tangential) quadrilateral ABCD. AC meets BD at P. $ h_1,h_2,h_3,h_4$ are repectively distances between P and sides AB,BC,CD,DA. Prove that $ \frac{1}{h_1}\plus{}\frac{1}{h_3}\equal{}\frac{1}{h_2}\plus{}\frac{1}{h_4}$

Not completely solved at http://www.mathlinks.ro/Forum/viewtopic.php?t=173533 but I gave hints...
[b]EDIT:[/b] now completely solved

  darij

We can make a use of the formula $ \frac{1}{r}\equal{}\frac{1}{h_A}\plus{}\frac{1}{h_B}\plus{}\frac{1}{h_C}$ in a triangle.


[hide="I think we have to prove that..."]$\left(x+\sqrt{1+y^{2}}\right)\left(y+\sqrt{1+x^{2}}\right)=1\ \Rightarrow\ x^{2}=y^{2}$[/hide]

[quote="cincodemayo5590"]For $x, \ y \in \mathbb{R}$ such that $(x+\sqrt{1+y^{2}})(y+\sqrt{1+x^{2}})=1$, prove $(x+\sqrt{1+x^{2}})(y+\sqrt{1+y^{2}})=1$.[/quote]

let $y=\tan\theta$ for $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$, then the first expression becomes: \[(x+\csc\theta)(\tan\theta+\sqrt{1+x^{2}})=1\] let $x=-\tan\phi$ \[(-\tan\phi+\csc\theta)(\tan\theta+\csc\phi)=1\] it can be shown that $\phi=\theta$, then the last part is trivial


inequalities
logarithms
inequalities proposed
Computer solutions
109

[quote="Ji Chen"]If $ a, b, c$ are positive numbers such that $ abc \equal{} 1,$ then

$ \frac {1}{a^a(b \plus{} c)} \plus{} \frac {1}{b^b(c \plus{} a)} \plus{} \frac {1}{c^c(a \plus{} b)}\leq\frac {3}{2},$

with equality if and only if $ a \equal{} b \equal{} c \equal{} 1.$[/quote]
Nice Inequality. Here is my proof:

Without loss of generality, assume that $ c\equal{}\min \{a,\;b,\;c\}.$ From $ abc\equal{}1,$ we get $ c\leq 1.$ Thus, by Bernoulli's Inequality, we have
\[ \frac{1}{c^{c}}\equal{}\left( 1\plus{}\frac{1}{c}\minus{}1\right) ^{c}\leq 1\plus{}c\left( \frac{1}{c}\minus{}1\right) \equal{}2\minus{}c.\]
On the other hand, it is known that $ x^{x}\geq x$ for any positive real number $ x.$ According these two inequalities, we see that it suffices to show that
\[ \frac{1}{a(b\plus{}c)}\plus{}\frac{1}{b(a\plus{}c)}\plus{}\frac{2\minus{}c}{a\plus{}b}\leq \frac{3}{2}.\]
Since
\[ \frac{1}{a(b\plus{}c)}\plus{}\frac{1}{b(a\plus{}c)}\equal{}\frac{bc}{b\plus{}c}\plus{}\frac{ac}{a\plus{}c}\equal{}2c\minus{}c^{2}\left( \frac{1}{a\plus{}c}\plus{}\frac{1}{b\plus{}c}\right)\]
and
\[ \frac{1}{a\plus{}c}\plus{}\frac{1}{b\plus{}c}\minus{}\frac{2}{\sqrt{ab}\plus{}c}\equal{}\frac{\left( \sqrt{a}\minus{}\sqrt{b}\right) ^{2}\left( \sqrt{ab}\minus{}c\right) }{(a\plus{}c)(b\plus{}c)\left( \sqrt{ab}\plus{}c\right) }\geq 0,\]
we get
\[ \frac{1}{a(b\plus{}c)}\plus{}\frac{1}{b(a\plus{}c)}\leq 2c\minus{}\frac{2c^{2}}{\sqrt{ab}\plus{}c}.\]
Besides, it is clear from the AM-GM Inequality that
\[ \frac{2\minus{}c}{a\plus{}b}\leq \frac{2\minus{}c}{2\sqrt{ab}}.\]
Therefore, it suffices to prove that
\[ 2c\minus{}\frac{2c^{2}}{\sqrt{ab}\plus{}c}\plus{}\frac{2\minus{}c}{2\sqrt{ab}}\leq \frac{3}{2}.\]
Setting $ t\equal{}\sqrt{ab},$ we get $ c\equal{}\dfrac{1}{t^{2}}$ and the inequality becomes
\[ \frac{2}{t^{2}}\minus{}\frac{\dfrac{2}{t^{4}}}{t\plus{}\dfrac{1}{t^{2}}}\plus{}\frac{2\minus{}\dfrac{1}{t^{2}}}{2t}\leq \frac{3}{2},\]
which is equivalent to the obvious inequality
\[ \frac{(3t^{4}\plus{}4t^{3}\plus{}t^{2}\plus{}2t\plus{}1)(t\minus{}1)^{2}}{2t^{3}(t^{3}\plus{}1)}\geq 0.\]
The proof is completed. Equality holds if and only if $ a\equal{}b\equal{}c\equal{}1.$

See also here: http://canhang2007.wordpress.com/2009/11/27/inequality-85-j-chen/

Thank you, can_hang2007 for your interest. I approached it as follows :

Let $ f(x) \equal{} x\ln{x} \plus{} \ln{2} \minus{} \ln{\left(x^2 \plus{} 1\right)}$ on $ (0, \plus{} \infty).$ Then

$ f'(x) \equal{} \ln{x} \plus{} \frac {(x \minus{} 1)^2}{x^2 \plus{} 1},f'(1) \equal{} 0,$

$ f''(x) \equal{} \frac {x^2(x \plus{} 1)^2 \plus{} (x \minus{} 1)^2}{x(x^2 \plus{} 1)^2} > 0\Longrightarrow$

$ f(x)\geq f(1) \equal{} 0 \Longrightarrow 2x^x\geq x^2 \plus{} 1.$

This comes down to an symmetric algebraic inequality :

$ \frac {1}{(a^2 \plus{} 1)(b \plus{} c)} \plus{} \frac {1}{(b^2 \plus{} 1)(c \plus{} a)} \plus{} \frac {1}{(c^2 \plus{} 1)(a \plus{} b)}\leq\frac {3}{4},$

which holds for positive numbers $ a,b,c$ with $ abc \geq 1.$

Arne's one see : http://www.mathlinks.ro/viewtopic.php?t=7854

what an nice inequality and nice sulotion!...
 :)
learning...


1) find all $ x \in \mathbb{R}$ such that:
$ x(x \minus{} 4)(x \minus{} 1)(x \plus{} 2)(x \plus{} 6) \equal{} \minus{} 100$

2) consider real numbers $ a,b,x,y$ such that :

$ ax \plus{} by \equal{} 3$
$ ax^2 \plus{} by^2 \equal{} 7$
$ ax^3 \plus{} by^3 \equal{} 16$
$ ax^4 \plus{} by^4 \equal{} 42$

find $ ax^5 \plus{} by^5$ ?

1. Had 3 real roots, one $ \minus{}7<x_1<\minus{}6$, second $ 1<x_2<2$, third $ 3<x_3<4.$
2. Let $ z_n\equal{}ax^n\plus{}by^n$, then $ z_{n\plus{}1}\equal{}pz_n\minus{}qz_{n\minus{}1}$. It is easy to find $ p\equal{}\minus{}14,q\equal{}\minus{}80$. Therefore $ z_5\equal{}\minus{}14*42\plus{}80*16\equal{}692.$

But the third is $ ax^3\plus{}by^2\equal{}16$

sorry,,, it's my mistake $ ax^3 \plus{} by^3 \equal{} 16$  :blush:

[quote="Rust"]1. Had 3 real roots, one $ \minus{} 7 < x_1 < \minus{} 6$, second $ 1 < x_2 < 2$, third $ 3 < x_3 < 4.$
2. Let $ z_n \equal{} ax^n \plus{} by^n$, then $ z_{n \plus{} 1} \equal{} pz_n \minus{} qz_{n \minus{} 1}$. It is easy to find $ p \equal{} \minus{} 14,q \equal{} \minus{} 80$. Therefore $ z_5 \equal{} \minus{} 14*42 \plus{} 80*16 \equal{} 692.$[/quote]

thank you Rust, but for the second, I think you miscalculated something -if I'm not wrong- :

if $ z_{n \plus{} 1} \equal{} pz_n \minus{} qz_{n \minus{} 1}$ where, $ (p,q) \equal{} ( \minus{} 14, \minus{} 80)$, then :

$ z_4 \equal{} \minus{} 14 \times 16 \plus{} 80 \times 7 \equal{} 336$ which is wrong!!!

I think : $ p \equal{} \minus{} 14$ and $ q \equal{} \minus{}38$, so $ z_5 \equal{} 20$.

ps. I changed $ q \equal{} 38$ to $ q \equal{} \minus{}38$ after Rust commented so.

thanx Rust,

btw, any Idea for special factorization solution for the first problem?


Please do not discuss any aspect of the AMC 8 contest until after November 21.  The AMC 8 rules allow some school to administer the contest up to 1 week after the "offical" date (which was today, November 14).  Please protect the integrity of the contest and do not discuss any aspect of it, including what you think you scored.

Oh, I'm very sorry for posting my possible score. Have they been deleted?
-jorian

Not discussing it is a good thing for the people who took the test. That means the people who take it on the 21st wont have a lot of detail. :D

Yes, please do not discuss the AMC 8 yet. It is still too early, as some people will not be taking it until November 21st. After that time, you will be able to discuss it.

If you make a topic discussing the AMC 8 before then, it will be either locked or deleted at the discretion of the moderators.

This announcement applies to discussing the questions, guessing your scores, etc.

For more information, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119172]this topic[/url].

Thanks.

yeah lock it, and delete mine if u think it is inappropriate, altho i hardly said anything  :maybe:

no, because that is where a good description of what not to say about the amcs 8 is
-jorian

Ugh, I am so going to be happy when that date comes.

Can we discuss the AMC 8 on the afternoon of the 14th, or on the 15th?

Edit: 21st instead of 14th, 22nd instead of 15th

I think you meant 21st and 22nd (instead of 14th and 15th).

I believe you should wait until November 22nd before discussing the AMC 8, since many people will be taking it on that day. Also, different schools may take the test at different times during the day, so that we cannot be sure that everyone has taken the exam until November 22nd (which is also the day this announcement ends).


Can someone attach the official solutions ? :) .
 I see ''Delegat'' posted the problems, so maybe he could post the official solutions too.

Ok I will scan and publish them in following days.

Scanned version of solution to problem 1. is attached.

Hi delegat.
You have problem JBTST??Can you post them?

Shortlist version of solution of problem 2 - just small modification was made (almost same as problem that appeared in competition).

I will post solutions to problems 3 and 4 later.

Scanned version of solution to Problem 3 is attached.

Scanned version of solution to Problem 4. is attached.


geometry
parallelogram
angle bisector
geometry proposed

Given points $ A,B$ on the circle ($ A\not\equal{} B$) and  point $ C$ inside of the circle. Construct the circle that tangents sides $ AC,BC$ and arc $ AB$.

[quote="¬[ƒ(Gabriel)³²¹º]¼"]Costruct the rombus with sides parallel to AC and CB and tangent to the circle. Call D the vertex of the rombus such that ACBD is a parallelogram and call E the opposit vertex of D. Then EC meet the circle on G such that G is in ACBD. CAll O the center of the circle. Then the angle bisector of $ \angle ACB$ meet GO at the center of the circle tangent to AC and BC.[/quote]

  :maybe: ACBD is not rombus. Because $ B,C$ is not tangent points of sides of rombus with circle. Can you explain?

[quote="Litlle 1000t"]:maybe: ACBD is not rombus. Because $ B,C$ is not tangent points of sides of rombus with circle. Can you explain?[/quote]

I have explayn bad but the rombus is the only rombus tangent to the circle with side parallel to CA and CB, it isn't necessary that B,C are the tangent point, for an exemple see this figure :wink:

[hide="figure"][img]http://img143.imageshack.us/img143/1048/costruzionecerchioka5.png[/img][/hide]


[quote=chien than]Let $ x;y$ be non-negative numbers satisfying $ x^2+y^2+(x-y)^2=2.$  Prove that
$$ \frac{x}{\sqrt{2(y+1)-x(x+1)}}+\frac{y}{\sqrt{2(x+1)-y(y+1)}} \geq \sqrt{2}$$[/quote]
Any one?

[hide=Very nice.][quote=anhduy98][quote=chien than]Let $ x;y$ be non-negative numbers satisfying $ x^2\plus{}y^2\plus{}(x\minus{}y)^2\equal{}2$
Prove that
$ \frac{x}{\sqrt{2(y\plus{}1)\minus{}x(x\plus{}1)}}\plus{}\frac{y}{\sqrt{2(x\plus{}1)\minus{}y(y\plus{}1)}} \geq \sqrt{2}$[/quote]

$\bullet x^2+y^2+(x-y)^2=2\rightarrow \left\{\begin{matrix}
\bullet 2(y+1)-x(x+1)=(y+1)(y+1-x)\\ \\ \\
\bullet 2(x+1)-y(y+1)=(x+1)(x+1-y)\\

\end{matrix}\right.$

$\bullet x^2+y^2+(x-y)^2=2  \rightarrow  \dfrac{x}{y+1}+\dfrac{y}{x+1}=1 . $

$\bullet a= \frac{x}{y+1},b=\frac{y}{x+1} \rightarrow  a+b=1  \rightarrow LHS=\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}} \ge \sqrt{2(a+b)}=RHS.$

.[/quote]

[/hide]
$$ x^2+y^2+(x-y)^2=2  \Rightarrow  \dfrac{x}{y+1}+\dfrac{y}{x+1}=1 . $$

Let $ x,y$ be non-negative numbers satisfying $ x^2+y^2+(x-y)^2=2.$ Prove that
$$\frac{x}{\sqrt{3(y+1)-x(x+1)}}+\frac{y}{\sqrt{3(x+1)-y(y+1)}}  \geq 1$$


Could you explain newton's bionmial? It's not in the wiki.

Well, it is in (some) wiki! See [url]http://en.wikipedia.org/wiki/Binomial_theorem[/url]

In general, try to use key words in Google ...

Newton's binomial..? Hmm... I guess there are other names for the *binomial theorem*.

You might find the *binomial theorem* on AoPS wiki. :wink:


[quote="tdl"]Let $ S(n)$ denotes the sum of all digits of the positive integer $ n$. Prove that there exist infinitely many $ n$, with $ S(n) > 1996S(3n)$.[/quote]

Let $ n\equal{}\frac{10^p\plus{}2}{3}$

$ S(n)\equal{}3p\plus{}1$
$ S(3n)\equal{}3$

$ p\geq 1996$ implies $ S(n)\equal{}3p\plus{}1>1996\times 3\equal{}1996.S(3n)$
Q.E.D.


geometry
rectangle
combinatorics proposed
combinatorics

Given a grid of mXn squares.
Find the total area of all the rectangles lying in the grid.

Example:
When m = 1 and n = 3, we have
1 *3 + 2 * 2 + 3 * 1 = 10.

When m = 2 and n = 2, we have
1*4 + 2*4 + 4*1 = 16.

Directly is easy enough.  The square that is $ i$th from the left and $ j$th from the bottom lies inside $ i(m \plus{} 1 \minus{} j) \plus{} j(n \plus{} 1 \minus{} i)$ rectangles (choose a rectangle by choosing an upper-left and lower-right corner).  Summing this expression over all $ i, j$ is straightforward.

(m*1 + (m-1)*2 + ... + 1*m)(n*1 + ... + 1*n)
When one expands this product, it corresponds to the area of the rectangle of length (second variable), and its position along the axis (first variable).  So the sum would be the above product.

Thanks for your explanations, Soarer and JBL.

I have applied recursion method to obtain the formula $ \binom {m\plus{}2} 3 * \binom {n\plus{}2} 3$ but I could not explain the result.

Now, everything is clear enough.
I prefer Soarer's solution as it is neat.  :idea:

Looking at the formula again, I feel that there could be another way to explain the occurrence of the binomial coefficients.
{Keep thinking......}


I am a sixth grade student from Suffolk County NY and I'd like to participate in MathCounts.  My school does not have a club.  They offered it several years ago but they dropped the program.  Is there any way for me to join without my school??

Thanks,
LMB

You can join as an individual but I'm not sure if the deadline has passed.

http://mathcounts.org/Document.Doc?id=229

It says: "Late registrations for the Competition may still be accepted at the discretion of the National Office."

I'm pretty sure the deadline has passed. There is a very small chance that it has still not passed. You have to check if your region has done regionals. If they have finished it, then probably not.  :maybe:

if you want to be part of a team, take a class at the local junior high and join their team.
I think that's the only way if you want to be on a team

Point out the pros of MATHCOUNTS to your school and hope they make a team.

This may seem nerdy/geeky/dorky but try to convince your fellow students the benefits of this program (this may be difficult) also.

If that fails, register as an individual and try to get separate coaching or coach yourself.

Er... I'm pretty sure it's kind of too late now seeing as many people have already had their chapters.

You can still try to register like the other people above you,  but if you can't try to convince your school to make a team next year.  That's what I did this year and everything worked out just great!  Make sure to talk about how having a mathcounts team will help the students [i]and[/i] the school.


Compute all $ x$ in the interval $ 0 \leq x \leq 2\pi$ such that $ f(x) \equal{} \sin \lfloor2x \rfloor$ takes on its largest value.

[hide]I got $ \left\{ x| \frac {\pi}{6}\leq x \leq \frac {5\pi}{6}, x\neq \frac {\pi}{4} \right\}$.  Is what I did wrong obvious?  Thanks![/hide]

$ \lfloor 2x \rfloor$ only takes on the values $ 0,1,2,3,...,10,11,12$ when $ x \in [0,2 \pi]$. So, you are essentially asking which of $ \sin(0), \sin(1), \sin(2),...,\sin(11), \sin(12)$ is the largest. Is that what you meant to ask?

Building on FibonacciFan's post, we only have to check how close Pi/2, 5Pi/2, and 7Pi/2 are to integers between 0 and 12 (there's no need to go any farther than that).  Respectively, they are approximately 1.57, 7.85, and 14.13.  It seems then that 8 would be the closest?  That would give an interval of x = 4 through 4.5.


topology
geometry
3D geometry
sphere
advanced fields
advanced fields theorems

Hi, I am trying to understand what is special about symplectic manifolds. I am trying to visualize how, say, a symplectic morphism from a sphere to an ellipsoid is different from just an ordinary homeo/diffeomorphism. Geometrically speaking, what are the extra constraints on such a deformation?

The constraints are given by the choice of symplectic forms on both surfaces. Any diffeomorphism $f\colon S\to E$ can be made into a symplectomorphism if you equip $S$ with the pullback form $f^{*}\omega$, where $\omega$ is a symplectic form on $E$. Compare this with the Riemannian case: if you have a diffeomorphism $f\colon S\to E$, you can make it into an isometry by changing the metric on one of the manifolds.


geometry
trigonometry
circumcircle
analytic geometry
geometry theorems

[color=darkblue][b]Usual notations.[/b] Given are a circle $w=C(O,R)$ and a point $X$.
$\blacktriangleright\ p_{w}(X)$- the power of the point $X$ w.r.t. the circle $w$, i.e. for any two points $\{U,V\}\subset w$ for which $X\in UV\ ,\ p_{w}(X)\equiv \overline{XU}\cdot\overline{XV}=XO^{2}-R^{2}$ (constant).
$\blacktriangleright\ XX$- the tangent to the circle $w$ in the point $X\in w\ .$
$\blacktriangleright\ \delta_{d}(X)$- the distance from the point $X$ to the line $d\ .$

[b]Lemma 1 (well-known).[/b] For a point $X$ which belongs to the sideline $BC$ of the triangle $ABC$ there is the relation $\boxed{\ \frac{XB}{XC}=\frac{AB}{AC}\cdot\frac{\sin\widehat{XAB}}{\sin\widehat{XAC}}\ }\ .$

[b]Lemma 2 (well-known).[/b] Given are a constant real value $0<k\ne 1$ and two concurrent lines $d_{1}$, $d_{2}$ for which $m\left(\widehat{d_{1},d_{2}}\right)=\phi <90^{\circ}\ .$
Then the geometrical locus (the set of the all positions) of the mobile point $L$ for which $\delta_{d_{1}}(L)=k\cdot \delta_{d_{2}}(L)$ is the reunion of the two lines $l_{1}$, $l_{2}$ for which ${m\left(\widehat{l_{1},l_{2}}\right)=\theta\ ,\ \boxed{\ \tan\theta=\frac{2k}{|k^{2}-1|}\cdot \sin \phi\ }=|\tan (2\cdot\arctan k)|\cdot\sin\phi\ }$ and [b]the concurrent fascicle $(d_{1},d_{2},l_{1},l_{2})$ is harmonically (!)[/b],
i.e. for any line $d$ which meets the lines $d_{1}$, $d_{2}$, $l_{1}$, $l_{2}$ in the points $X$, $Y$, $Z$, $T$ respectively there is the relation $\frac{ZX}{ZY}=\frac{TX}{TY}\ .$
[b]Particular case.[/b] Let $ABC$ be a triangle and let $M$ be the middlepoint of the side $[BC]$. Then $\tan\left(\widehat{AMB}\right)=\frac{2bc}{|b^{2}-c^{2}|}\cdot \sin A\ .$

[b]Definition.[/b] Denote the middlepoint $M$ of the side $[BC]$ of the triangle $ABC$. The [b]$A$- symmedian[/b] $[AS$, where $S\in (BC)$, is the ray for which $\widehat{BAS}\equiv\widehat{CAM}$.
I'll prove the relation $\frac{SB}{SC}=\left(\frac{c}{b}\right)^{2}$. Indeed, aplly the [b]lemma 1.[/b] in the triangle $ABC$ for the cevians $[AS$ and $[AM$, where I denoted
$x=m(\widehat{SAB})=m(\widehat{MAC})$ and $y=m(\widehat{SAM})\ :$ $\frac{SB}{SC}=\frac{c}{b}\cdot\frac{\sin x}{\sin (y+x)}\ \ ;\ \ 1=\frac{MB}{MC}=\frac{c}{b}\cdot\frac{\sin (x+y)}{\sin x}\Longrightarrow \frac{SB}{SC}=\left(\frac{c}{b}\right)^{2}\ .$

[b]Remark 1.[/b] Prove easily that $\frac{\delta_{AB}(S)}{\delta_{AC}(S)}=\frac{\delta_{AB}(T)}{\delta_{AC}(T)}=\frac{c}{b}$, where $T\in BC\cap AA$ and $\tan\left(\widehat{AA,AS}\right)=\frac{2bc}{|b^{2}-c^{2}|}\cdot \sin A$, from the [b]lemma 2.[/b] for $k=\frac{c}{b}\ .$

[b]Remark 2.[/b] The above mentioned property of the symmedian is a particular case of the well-known [b]Steiner's theorem[/b] :

[b]Steiner's theorem.[/b] Let $M,N$ be two points which belong to the sideline $BC$ of the triangle $ABC$
so that the angles $\widehat{MAN}$ and $\widehat{BAC}$ have the same bisector ray. Then exists the relation $\boxed{\ \frac{MB}{MC}\cdot\frac{NB}{NC}=\frac{AB}{AC}\cdot\frac{\sin\widehat{XAB}}{\sin\widehat{XAC}}\ }\ .$
[b]Proof.[/b] Suppose w.l.o.g. that the points $M$, $N$ belong to the side $(BC)$ (can prove analogously the another case).
Denote $x=m(\widehat{MAB})=(\widehat{NAC})$ and $y=m(\widehat{MAN})$. Apply the [b]lemma 1.[/b] to the cevians $[AM$ and $[AN$ in the triangle $ABC\ :$
$\frac{MB}{MC}=\frac{AB}{AC}\cdot\frac{\sin x}{\sin (y+x)}\ ,\ \frac{NB}{NC}=\frac{AB}{AC}\cdot\frac{\sin (x+y)}{\sin x}$ $\Longrightarrow$ $\frac{MB}{MC}\cdot\frac{NB}{NC}=\left(\frac{AB}{AC}\right)^{2}\ .$

[b]Theorem 1.[/b] Let $w$ be the circumcircle of the triangle $ABC$. Define the intersections $T\in AA\cap BC$, $X\in BB\cap CC$
and the $A$- symmedian $AS$, where $S\in (BC)$. Then there are the following properties :

$1.\blacktriangleright\ \boxed{\ \frac{TB}{TC}=\frac{SB}{SC}=\left(\frac{c}{b}\right)^{2}\ }$, i.e. the point $T$ is the conjugate of the point $S$ w.r.t. the points $B$ and $C\ \ ;\ \ \boxed{\ S\in AX\ }\ .$

$2.\blacktriangleright\ \boxed{\ TA=\frac{abc}{\left|b^{2}-c^{2}\right|}\ }\ \ \ ;\ \ \boxed{\ s_{a}=\frac{2bc}{b^{2}+c^{2}}\cdot m_{a}\ }$, where $m_{a}=AM$ and $s_{a}=AS\ \ ;\ \ \boxed{\ XA\cdot\left|\cos A\right|=m_{a}\ }\ .$

$3.\blacktriangleright$ The quadrilateral $ABPC$ is harmonically, i.e. $AC\cdot PB=AB\cdot PC\ .$

[b]Proof of the theorem.[/b] Denote: the intersection $R\in BC\cap AX$ and
the second intersection $P$ of the line $AX$ with the circle $w\ ;$ $u=m(\widehat{BAR})$ and $v=m(\widehat{CAR})$.
$1.\blacktriangleright$ Aplly the [b]lemma 1.[/b] in the triangle $ABC$ for the cevian $[AT\ : \ \frac{TB}{TC}=\frac{c}{b}\cdot\frac{\sin \widehat{TAB}}{\sin \widehat{TAC}}=\frac{c}{b}\cdot\frac{\sin C}{\sin B}$ $\Longrightarrow$ $\frac{TB}{TC}=\left(\frac{c}{b}\right)^{2}\ .$

$\triangle BXP$ $\Longrightarrow$ $\frac{XB}{XP}=\frac{\sin\widehat{BPX}}{\sin\widehat{PBX}}$ $\Longrightarrow$ $\boxed{\frac{XB}{XP}=\frac{\sin C}{\sin u}}\ \ ;\ \triangle CPX\Longrightarrow$ $\frac{XC}{XP}=\frac{\sin\widehat{CPX}}{\sin\widehat{PCX}}\Longrightarrow$ $\boxed{\frac{XC}{XP}=\frac{\sin B}{\sin v}}\ .$
$XB=XC$ $\Longrightarrow$ $\frac{\sin C}{\sin u}=\frac{\sin B}{\sin v}$ $\Longrightarrow$ $\frac{\sin u}{\sin v}=\frac{\sin C}{\sin B}$ $\Longrightarrow$ $\boxed{\ \frac{\sin u}{\sin v}=\frac{c}{b}\ }\ \ (1)\ .$
Apply the [b]lemma 1.[/b] in the triangle $ABC$ for the cevian $[AR\ :$ $\frac{RB}{RC}=\frac{c}{b}\cdot\frac{\sin u}{\sin v}$ $\Longrightarrow$ $\frac{RB}{RC}=\left(\frac{c}{b}\right)^{2}$ $\Longrightarrow$ $R\equiv S$ $\Longrightarrow$ $S\in AX\ .$

$\star\blacktriangleright$ Otherwise, because the point $T$ belongs to the polar $BC$ of the point $X$ w.r.t. the circle $w$, then
the point $X$ belongs to the polar $AS$ of the point $T$ w.r.t. the same circle, i.e. $S\in AX\ .$

$\star\blacktriangleright$ Otherwise, $\frac{\delta_{AB}(X)}{\delta_{AC}(X)}=\frac{XB\sin \widehat{ABX}}{XC\sin\widehat{ACX}}=\frac{\sin C}{\sin B}=\frac{c}{b}\Longrightarrow$ $\frac{\delta_{AB}(X)}{\delta_{AC}(X)}=\frac{c}{b}$ (apply the [b]lemma 2.[/b] and the [b]remark 1.[/b]) $\Longrightarrow$ $X\in AS$, i.e. $S\in AX\ .$

$2.\blacktriangleright\ p_{w}(T)\equiv\overline{TB}\cdot \overline{TC}=TA^{2}\ \ ,\ \frac{TB}{c^{2}}=\frac{TC}{b^{2}}=\frac{a}{|b^{2}-c^{2}|}$ $\Longrightarrow$ $TA^{2}=\frac{ac^{2}}{|b^{2}-a^{2}|}\cdot \frac{ab^{2}}{|b^{2}-c^{2}|}$ $\Longrightarrow$ $\boxed{\ TA=\frac{abc}{|b^{2}-c^{2}|}\ }\ .$
${\frac{SB}{c^{2}}=\frac{SC}{b^{2}}=\frac{a}{b^{2}+c^{2}}\ .}$ Apply the [b]Stewart's theorem[/b] in the triangle $ABC$ to the cevian $AS\ :$ $a\cdot s^{2}_{a}+a\cdot SB\cdot SC=c^{2}\cdot SC+b^{2}\cdot SB$ $\Longrightarrow$ $s^{2}_{a}+\frac{a^{2}b^{2}c^{2}}{(b^{2}+c^{2})^{2}}=\frac{2b^{2}c^{2}}{b^{2}+c^{2}}$ $\Longrightarrow$ $s^{2}_{a}=\frac{b^{2}c^{2}[2(b^{2}+c^{2})-a^{2}]}{(b^{2}+c^{2})^{2}}$ $\Longrightarrow$ $s_{a}=\frac{2bc}{b^{2}+c^{2}}\cdot m_{a}$ because $4m^{2}_{a}=2(b^{2}+c^{2})-a^{2}\ .$

Apply the [b]lemma 1.[/b] in the triangle $ABS$ for the cevian $[BX\ : \ \frac{XA}{XS}=\frac{BA}{BS}\cdot \frac{\sin\widehat{XBA}}{\sin\widehat{XBS}}=$ $\frac{c}{\frac{ac^{2}}{b^{2}+c^{2}}}\cdot \frac{\sin C}{\sin A}=$ $\frac{b^{2}+c^{2}}{ac}\cdot\frac{c}{a}=$ $\frac{b^{2}+c^{2}}{a^{2}}$ $\Longrightarrow$ $\frac{XA}{b^{2}+c^{2}}=$ $\frac{XS}{a^{2}}=$ $\frac{AS}{b^{2}+c^{2}-a^{2}}$ $\Longrightarrow$ $XA=\frac{b^{2}+c^{2}}{b^{2}+c^{2}-a^{2}}\cdot AS=$ $\frac{b^{2}+c^{2}}{b^{2}+c^{2}-a^{2}}\cdot \frac{2bc}{b^{2}+c^{2}}\cdot m_{a}$ $\Longrightarrow$ $XA=\frac{2bc}{b^{2}+c^{2}-a^{2}}\cdot m_{a}$ $\Longrightarrow$ $XA\cdot \left|\cos A\right|=m_{a}\ .$

$3.\blacktriangleright\ AC\cdot PB=AB\cdot PC\Longleftrightarrow b\cdot 2R\sin u=c\cdot 2R\sin v\Longleftrightarrow \frac{\sin u}{\sin v}=\frac{c}{b}$, i.e. the true relation $(1)\ .$

See :

http://www.mathlinks.ro/Forum/viewtopic.php?t=76906
http://www.mathlinks.ro/Forum/viewtopic.php?t=43652
http://www.mathlinks.ro/Forum/viewtopic.php?t=43202
http://www.mathlinks.ro/Forum/viewtopic.php?t=46146 [/color]

Some others:
Three symmedians are concurrent at the point $L$ that we called Lemoine point. Lemoine has many geometric properties! It has barycentric coordinates $(a^{2},b^{2},c^{2})$  :P
And $LA=\frac{2m_{a}.bc}{a^{2}+b^{2}+c^{2}}$
more references
http://mathworld.wolfram.com/SymmedianPoint.html


limit
real analysis
inequalities
calculus
calculus computations

$ \left\{  a_{n}\right\}$ and  $ \left\{  b_{n}\right\}$ are bounded, and $ a_{n}\leqslant b_{n}$ for nature  $ n$  ;  then $ \underset{n\rightarrow\infty}{\lim}\sup
a_{n}\leqslant\underset{n\rightarrow\infty}{\lim}\sup b_{n}$ and $ \underset{n\rightarrow\infty}{\lim}\inf a_{n}\leqslant\underset
{n\rightarrow\infty}{\lim}\inf b_{n}$

Is it true and  trivial?

I got it on rudin book.

[quote="§outh§tar"]Trivially true thanks to the analogue for convergent sequences. If you are dealing with unbounded sequences then you can decide for yourself what the inequality means.[/quote]

YES! it is ture fllowing from defines of limit inferior and limit superior.

 $ a_{n}\leqslant b_{n}$ for nature  $ n$  ;  then $ \underset{n\rightarrow\infty}{\lim}\sup a_{n}\leqslant\underset{n\rightarrow\infty}{\lim}\sup b_{n}$ and $ \underset{n\rightarrow\infty}{\lim}\inf a_{n}\leqslant\underset {n\rightarrow\infty}{\lim}\inf b_{n}$

-------------------------------------------------------------------------------------------------------------------------

I use this to prove Stolz-Cesaro theorem and other several questions .

it is a basic, well know and is quite obivious


First of all this is not hard and just a basic result from Quantum Mechanics. It follows from Schrodinger's Steady state eqn and a paricular definiton.

Schrodinger's eqn(steady state)

$ \bigtriangledown ^2 {\psi} + \frac{2m}{\hbar ^2} (E-U) \psi = 0 $
where $ \bigtriangledown ^2 $ is the laplacian
I hope you understand these symbols for without them you can't have asked this question and don't expect to understand the solution.Also you should know about vector operators.

Define the hamiltonian operator  $ \hat{H} = - \frac{\hbar^2}{2m} \bigtriangledown ^2 + U $

Now using the schrodinger steady state eqn. and by simple manuipulation as

$ (- \frac{\hbar^2}{2m} \bigtriangledown ^2 + U ) \psi = E \psi $
Here I got this by simple manipulation as you can see easily. Also I have written the LHS according to the rule of operators.( A constant in an operator is simply multiplied to the quantity it is operating on as in the case of U )
Comparing with the definition of the Hamiltonian

we get $ \hat{H} \psi = E \psi $

This implies that solving the Schrodinger eqn in the steady state is an eigenvalue problem i.e. it involves the eigenoperator$ \hat{H} $ (hope you know what eigenoperators are...)(The subscipts are immaterial since they only refer to different waves)

I am sure you know all the symbols involved and sth about operators, laplacian etc.

[quote="Rushil"]Hey do you understand the solution????

At least post to inform me that the solution is acceptable to you!!!![/quote]

yes i Understood,thanks for solution!!


If $ a\equal{}2,162$, $ b\equal{}7,389$, and $ c\equal{}25,412$, find the remainder when the product $ abc$ is divided by 10.

We have $ abc\equiv2162\cdot7389\cdot25412\equiv2\cdot\minus{}1\cdot2\equiv\minus{}4\equiv\boxed6\pmod{10}$.

you can also just multiply the last digits of each number to get 36. Ten goes into 36 3 times with a remainder of 6 so the answer is [b]6[/b]


Please help me solve the following problem, as i m a newbie to point Set topology.
Let $ f$ be a real valued function defined for each $ x \in [0,1]$. Suppose there exists a positive number $ M$ having the following property:
For every choice of finitely many points $ x_{1}, x_{2}, \cdots x_{n}$ in the interval $ 0 \le x \le 1$ the sum
$ |f(x_{1}) \plus{} f(x_2) \plus{} \cdots f(x_n)| \le M$
Let $ S$ be the set of those $ x$ in $ 0 \le x \le 1$ for which $ f(x) \neq 0$. Prove that $ S$ is countable.

Suppose that it is uncountable. So one of {x | f(x) > 0 }, {x | f(x) < 0} is uncountable.
Suppose the first from above is uncountable. It is the union of {x | f(x)>1/n} so for some n
set {x| f(x)>1/n} is uncountable. It is impossible.


Hi there. I suspect that between 10^(k-1) and 10^k there is at least one prime number p such that the length of it`s period in base 10 is p-1, for every natural number k. I would like to test that suspicion partially, for some values of k but I am not good at programming and I do not know how to write and execute a program that would solve that. So, for those that wanna help me I propose this problem: To test if between 10^(k-1) and 10^k there exist at least one prime number p such that his period length in base 10 is equal to p-1, for some values of k (for a first few values of k). Send your results here...

Thank you for your help...

The length of the repeating decimal of $ \frac{1}{p}$, i.e. the order of $ 10 \bmod p$.  I don't think too much is known about the distribution of primitive roots, so this may be open (if it isn't just false).

[quote="bus"]What is a "period" of prime number?[/quote]

Oh, I forgot to write that i mean the period of 1/p but thanks to [b]t0rajir0u[/b] for answering that...

[quote="bus"]Using a naive algorith I checked this up to k=10. The smallest corresponding primes were

7
17
109
1019
10007
100019
1000171
10000019
100000007
1000000007
10000000019

They all seem to be pretty small. (Or rather, close to 10^(k-1).) What is a good way to check the period of prime p other than iterating the *= 10 loop?[/quote]

Just as I have suspected...  You discovered a rather interesting fact, that they are really close to 10^(k-1). Can you check it for up to k= let`s say 50? Or that would take too much time to execute? Thank you for your help....


linear algebra
matrix
linear algebra unsolved

$p,q,r$ primes numbers such that $q=p+r$
Find one couple $(A,B)\in M_{3}(Z)\times M_{3}(Z)$ explicitly such that

$AB-BA=\left(\begin{array}{ccc}0&p&q\\-p&0&r\\-q&r&0\end {array}\right)$

Is there something missing in the problem statement?  Any matrix of trace = 0 can be represented as a commutator.

Of course, but I think he asked for an explicit representation of this matrix as a commutator.

yes as grobber said find an explicit matrices A,B


search
number theory
number theory unsolved

Hello I was unsuccesful in Math Olympiad and I won't enter this exam next year but I love Number Theory and I want to learn algebraic number theory,elementary number theory ,What should I do?

search amazon.com or your local library for "algebraic number theory." and just read. duh!

PS. this ain't no unsolved problem in number theory ;)


Euler
inequalities proposed
inequalities

[quote="Fermat -Euler"][quote="chien than"]Let $a;b;c$ be non-negative and satisfying $a+b+c=2$.Prove that:
$(a^{2}+ab+b^{2})(b^{2}+bc+c^{2})(c^{2}+a^{2}+ac)\leq\ 3$[/quote]
it's wrong
take
$a=2$
$b=c=0$[/quote]
No, Fermat -Euler, it's true. I look for nice proof.

As I am too lazy to write a complete solution, I just send my file because I have already add this famous problem of Murray Klamkin in my book

Down load here

As I understand that my solution is just theorical, as Arqady, I also wait for a better one without mixing variable.

[quote="chien than"]Let $ x,y,z$ be nonnegative and satisfying $ x \plus{} y \plus{} z \equal{} 2$. Prove that

$ \left(y^{2} \plus{} yz \plus{} z^{2}\right)\left(z^{2} \plus{} zx \plus{} x^{2}\right)\left(x^{2} \plus{}xy\plus{} y^{2}\right)\leq 3$.[/quote]$ 3(x\plus{}y\plus{}z)^6\minus{}64\left(y^{2} \plus{} yz \plus{} z^{2}\right)\left(z^{2} \plus{} zx \plus{} x^{2}\right)\left(x^{2} \plus{}xy\plus{} y^{2}\right)$

$ \equiv F(x,y,z)\equal{}F(x,x\plus{}s,x\plus{}t)$

$ \equal{}459x^6\plus{}918(s\plus{}t)x^5\plus{}765(s\plus{}t)^2x^4\plus{}36(s\plus{}t)\left(13s^2\plus{}10st\plus{}13t^2\right)x^3$

$ \plus{}(s\plus{}t)^2\left(71s^2\minus{}50st\plus{}71t^2\right)x^2\plus{}63(s\plus{}t)^3\left(9s^2\minus{}14st\plus{}9t^2\right)x$

$ \plus{}(s\minus{}t)^2\left(3s^4\plus{}24s^3t\plus{}26s^2t^2\plus{}24st^3\plus{}3t^4\right)\geq0,$

which is clearly true for $ x\equal{}\min\{x,y,z\}.$

A generalization see: http://www.mathlinks.ro/viewtopic.php?t=62544


find all functions $ \zeta$ : $ N$---->$ N$
such that :
$ \zeta$ is  surjective.
and $ \zeta (\zeta(k))$ divides $ (\zeta(k))^2$
(this problem is  very dificult and require many technic and  theormes) :roll:

It seems to me that $ \zeta (k) \equal{} k$ is the only solution, as functions of higher (or "lower") 'degree' are not surjective $ \mathbb{N} \longrightarrow \mathbb{N}$ (the same goes for linear functions with coefficients $ \neq 1$), and linear functions with constant terms do not meet the condition $ \zeta(\zeta(k)) \mid (\zeta(k))^{2}$:

$ \zeta (k) \equal{} k \plus{} m \Longrightarrow \zeta(\zeta(k)) \equal{} k \plus{} 2m \nmid (\zeta(k))^{2} \equal{} (k \plus{} m)^{2}$.

Perhaps there are some other functions I'm missing?

The problem is equivalent to finding surjective functions $ \zeta$ with $ \zeta(k) \mid k^2$. Since $ \zeta$ is surjective, for any $ m$ there is an $ n$ such that $ \zeta(n) \equal{} m$, so we can replace all instances of $ \zeta(k)$ with any number in the problem. This has loads of solutions, which probably can't be concisely described. For instance, $ \zeta(p^2) \equal{} p, \zeta(p) \equal{} p^2$ for all primes $ p$, $ \zeta(n) \equal{} n$ everywhere else.

True, but I don't like piecewise-defined functions.  XD


algebra
polynomial
combinatorics proposed
combinatorics

Compute the number of strings of length $ k$ with entries in $ [n] = \{1, 2, \ldots, n\}$ such that between any two occurrences of the letter $ i$ there is an occurrence of a letter strictly larger than $ i$.

For example, with $ n = 3$ and $ k = 3$ there are 9 such strings: 121, 131, 232 and the six permutations of 123.

I don't know a solution, so in particular I don't know whether there is a nice answer  :oops:.  So, if you can say anything else interesting about these sequences ....  (An easy statement: $ k \leq 2^n - 1$, with a single string when $ k = 2^n - 1$.)

Edit: Conjecture: this is the same as [url=http://www.research.att.com/~njas/sequences/A122888]A122888[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

Later edit: the conjecture is true.  Let $ f(n, k)$ be the number of such strings of length $ k$ on $ [n]$.  I claim that this is the coefficient of $ x^{k + 1}$ in the $ n$-fold composition of $ g(x) = x + x^2$ with itself.  For $ n = 1$ this is clear: there is one string of length 0 and one of length 1 and no others.  Now, we show that they obey the same recursion.  By polynomial arithmetic,
\begin{align*} [x^{k + 1}] g^{(n)}(x) & = [x^{k + 1}]g^{(n - 1)}(x) + [x^{k + 1}]\left(g^{(n - 1)}(x)\right)^2 \\
& = [x^{k + 1}]g^{(n - 1)}(x) + \sum_{i = 1}^{k} [x^{i}]g^{(n - 1)}(x) \cdot [x^{k + 1 - i}]g^{(n - 1)}(x) \end{align*} where $ [x^*]$ is a coefficient extractor.  On the other hand we have
\[ f(n, k) = f(n - 1, k) + \sum_{i = 0}^{k - 1}f(n - 1, i)\cdot f(n - 1, k - i)
\]
because the first term counts strings in which $ n$ does not appear while the second term counts strings in which $ n$ does appear.  (The latter is true because when $ n$ appears, exactly one copy is present and the substrings of our original string to the left and right of this copy of $ n$ can be any valid strings on $ [n - 1]$.)

Follow-up: how many strings are there of length $ k$ with entries in $ [n]$ such that between any two occurrences of the letter $ i$ there is an occurrence of $ i \plus{} 1$?

Or, as a special case: now we have that $ k \leq \frac {n(n \plus{} 1)}{2}$.  How many strings of length $ \frac{n(n \plus{} 1)}{2}$ are there with entries in $ [n]$ such that between any two occurrences of the letter $ i$ there is an occurrence of $ i \plus{} 1$?

[quote="JBL"]Or, as a special case: now we have that $ k \leq \frac {n(n \plus{} 1)}{2}$.  How many strings of length $ \frac {n(n \plus{} 1)}{2}$ are there with entries in $ [n]$ such that between any two occurrences of the letter $ i$ there is an occurrence of $ i \plus{} 1$?[/quote]

This appears, with the same description (among others), in http://www.research.att.com/~njas/sequences/A003121 .

Nice result!@first post

  darij

Thanks and thanks!  A reasonably nice formula with some other good combinatorial interpretations -- so it seems still possible that there is a nice formula of some sort.


I'm not sure where I should start off (or go on?). I'm currently doing well in math, and am the best in my class, but I'm not sure as to where to go. I've been trying to win math tournaments, and have succeeded in doing so in some, but I haven't been able to approach higher level stuff... I've ordered the Volume 1 and 2 of the Art of Problem Solving books... are they good? Do they help win tournaments? (*cough cough if anyone lives around the southeast US, the UGA Math tournament*) How would I start approaching these problems?

And would the AIME be considered as Olympiad level? How would I learn how to solve these problems?

If anyone could help me solve my issues, I would really appreciate it

Thanks!

 :)

Karth

The UGA Mathematical Tournament is particularly AMC style.  So honestly, if you learn all the material in volume 1 very well so that you can apply it to anything, you should be able to score quite high on the UGA Individual.  If you want to take yourself to the very top, try to learn as much as possible from Volume 2.  The Team round is quite similar to some ARML and AIME questions. (AIME Problems http://www.kalva.demon.co.uk) You may want to try to practice alot of ciphering.  You can make alot of points to add to your individual in that round.  Thats usually what seperates the good from the best at that competition.

Good luck trying to win that tournament.  I think you'll have quite a bit of competition.  :D
Although I'm not personally going.  :(


conics
parabola
geometry
analytic geometry
calculus
graphing lines
slope

Let $ A$, $ B$, and $ C$ be three distinct points on the graph of $ y\equal{}x^2$ such that line $ AB$ is parallel to the $ x$-axis and $ \triangle{ABC}$ is a right triangle with area $ 2008$. What is the sum of the digits of the $ y$-coordinate of $ C$?

$ \textbf{(A)}\ 16 \qquad
\textbf{(B)}\ 17 \qquad
\textbf{(C)}\ 18 \qquad
\textbf{(D)}\ 19 \qquad
\textbf{(E)}\ 20$

I don't think this is possible...The only case of a right triangle that satisfies this has an area of $ 1$.
($ ( \minus{} 1,1),(1,1), (0,0)$ for A B and C respectively)
Where did I go wrong?

(If AB is anywhere above or below, it's impossible to draw a right triangle)

[quote="randomdragoon"]C is not necessarily on the y-axis.[/quote]
I don't think it's possible for it to be anywhere else though...I tried working it out and got that either the $ y$ value of C has to be $ 0$ or the $ y$ value of C has to equal the $ y$ values of A and B.

edit- nvm i see my mistake--I squared something too many times
edit2-[hide="Solution"]
The point C has to be on the circumference of a circle with diameter AB. Draw a miniature right triangle to C from the diameter, and see that the hypoteneuse, which is the radius, is $ \sqrt {y_B}$. (I'll use $ y_B$ as the $ y$ value of A and B).
The part of the diameter that is on the miniature right triangle is the $ x$ value of C, which is $ \sqrt {y_C}$. The other leg of the miniature right triangle is $ y_B \minus{} y_C$
Using the pythagorean theorem, one can work out that $ (y_C \minus{} y_B)(y_C \minus{} y_B \plus{} 1) \equal{} 0$, which means $ y_C \equal{} y_B \minus{} 1$.
The $ y$ value of $ C$ is $ 1$ less than the $ y$ value of $ AB$.
The altitude to side $ AB$ of the right triangle in the problem is $ 1$, so the length of $ AB$ is twice $ 2008$, so the $ x$ value of $ B$ = $ 2008$.
To find the $ y$ value of $ C$, take the square of $ 2008$ and add $ 1$.
the square is $ 4032064$. Subtracting $ 1$ from the ending $ 4$, the sum of the digits is $ 18$, or $ (C)$
[/hide]

[hide="outline"]Let $ B \equal{} (x,y)$ so $ A \equal{} (\minus{}x,y)$.  Let $ C \equal{} (m,n)$.  Now we know that $ AC$ is perpendicular to $ BC$, so write an equation for that.  Then use the fact that $ x^2 \equal{} y$ and $ m^2 \equal{} n$.  Finally, express the area of the triangle in terms of the base $ AB\equal{}2x$ and the height $ y\minus{}n$.[/hide]

$ 4\plus{}3\plus{}2\plus{}6\plus{}3\equal{}18$, and the answer sheet says $ \boxed{C}$, but it's a good solution.

I do get 4032063, but I think the sum of the numbers is 18.
You guys are talking about the answer sheet. What is that?

OK maybe this is not very helpful but I forgot to subtract 1 on this one. Ugh.

Anyway, I believe I found the intersection of the circle with center M, where M is the midpoint of AB, with the parabola. Then angle C is right. Then I did some analytic geo stuff. (sorry if this has been posted before but I needed to make my post useful :P )

[quote="randomdragoon"][hide]Since $ A, B, C$ are on the parabola $ y \equal{} x^2$ and $ AB$ is parallel to the x-axis, let point $ A$ be $ (a, a^2)$, $ B$ be $ ( \minus{} a, a^2)$, and $ C$ be $ (c,c^2)$.

$ AB$ is the hypotenuse and therefore the midpoint of $ A$ and $ B$ is equidistant to $ A$, $ B$, and $ C$. The midpoint has coordinates $ (0, a^2)$, and has distance $ a$ from point $ A$.

Therefore, using distance formula, $ a^2 \equal{} c^2 \plus{} (a^2 \minus{} c^2)^2$. Solving for $ (a^2 \minus{} c^2)$, we have
$ a^2 \minus{} c^2 \equal{} \sqrt {a^2 \minus{} c^2}$, which gives us $ a^2 \minus{} c^2 \equal{} 1$ (it can't equal 0, or else the triangle will be degenerate)

Also, because the triangle has area $ 2008$, we know that $ 1/2(2a)(a^2 \minus{} c^2) \equal{} 2008$. From the above, we know that $ a^2 \minus{} c^2 \equal{} 1$, so $ a \equal{} 2008$. Then $ c^2 \equal{} 2008^2 \minus{} 1 \equal{} 4032063$, and the sum of its digits is $ \boxed{18}$. (C)[/hide][/quote]

How do you know that the midpoint of $ AB$ has to be equidistant to $ A$, $ B$, and $ C$?  Where is $ C$ on the parabola?

Thanks in advance!

The midpoint of the hypotenuse of a right triangle is equidistant from all three vertices. This is one of those facts you know. It's pretty easy to prove; just drop altitudes from that midpoint and connect it to the third vertex and you get four congruent triangles. It's also equivalent to saying that a right triangle is inscribed in a semicircle, which is very useful as well.

Oh, we also know that $ AB$ is the hypotenuse becuase otherwise two points would be at the same $ x$-coordinate, a contradiction.

[quote="314.math"][quote="randomdragoon"]Therefore, using distance formula, $ a^2 \equal{} c^2 \plus{} (a^2 \minus{} c^2)^2$. Solving for $ (a^2 \minus{} c^2)$, we have
$ a^2 \minus{} c^2 \equal{} \sqrt {a^2 \minus{} c^2}$, which gives us $ a^2 \minus{} c^2 \equal{} 1$ (it can't equal 0, or else the triangle will be degenerate)[/quote]

Why couldn't $ a^2 \minus{} c^2$ equal something other than $ 1$?  Thanks![/quote]

you have $ a^2\minus{}c^2\equal{}\sqrt{a^2\minus{}c^2}$

Then,
\[ (a^2\minus{}c^2)^2\minus{}(a^2\minus{}c^2)\equal{}0 \implies (a^2\minus{}c^2)(a^2\minus{}c^2\minus{}1)\equal{}0\]

So, $ a^2\minus{}c^2\equal{}1$ or $ a^2\minus{}c^2\equal{}0$

[quote="randomdragoon"][hide]Since $ A, B, C$ are on the parabola $ y \equal{} x^2$ and $ AB$ is parallel to the x-axis, let point $ A$ be $ (a, a^2)$, $ B$ be $ ( \minus{} a, a^2)$, and $ C$ be $ (c,c^2)$.

$ AB$ is the hypotenuse and therefore the midpoint of $ A$ and $ B$ is equidistant to $ A$, $ B$, and $ C$. The midpoint has coordinates $ (0, a^2)$, and has distance $ a$ from point $ A$.

Therefore, using distance formula, $ a^2 \equal{} c^2 \plus{} (a^2 \minus{} c^2)^2$. Solving for $ (a^2 \minus{} c^2)$, we have
$ a^2 \minus{} c^2 \equal{} \sqrt {a^2 \minus{} c^2}$, which gives us $ a^2 \minus{} c^2 \equal{} 1$ (it can't equal 0, or else the triangle will be degenerate)

Also, because the triangle has area $ 2008$, we know that $ 1/2(2a)(a^2 \minus{} c^2) \equal{} 2008$. From the above, we know that $ a^2 \minus{} c^2 \equal{} 1$, so $ a \equal{} 2008$. Then $ c^2 \equal{} 2008^2 \minus{} 1 \equal{} 4032063$, and the sum of its digits is $ \boxed{18}$. (C)[/hide][/quote]

I'm confused, what distance did you use to get $ a^2 \equal{} c^2 \plus{} (a^2 \minus{} c^2)^2$.?

[hide="Hopefully a more succinct explanation"]
Observation 1: From calculus, we have that the slope for the line between points $ P$ and $ Q$ is the slope at the midpoint of $ PQ$ for some functions, which happens to include the function $ y\equal{}x^2$. To be honest, the investigation of this fact is much more interesting than the rest of this problem.

Anyways, since $ AB$ is parallel to the x axis, we have that $ A\equal{}(a,a^2)$ and $ B\equal{}(\minus{}a,a^2)$ for some real $ a$. Let $ C\equal{}(c,c^2)$. Since the slopes of two perpendicular lines (we know this because ABC is right) fulfill the relation m_1m_2 = -1, we may apply observation 1 to show that $ (2(\frac{c\plus{}a}{2})) \cdot (2(\frac{c\minus{}a}{2})) \equal{} \minus{}1$ which reduces to $ c^2\minus{}a^2\equal{}\minus{}1$.

Calculating the area of the triangle as $ \frac{1}{2} \cdot (a\minus{}\minus{}a)(a^2\minus{}c^2) \equal{} a(a^2\minus{}c^2) \equal{} a \equal{} 2008$, we may plug $ a \equal{} 2008$ back into $ c^2\minus{}a^2\equal{}\minus{}1$ to get $ c^2 \equal{} 2008^2\minus{}1 \equal{} 4032063$. The sum of these digits is $ 18$, and thus our answer is $ \boxed{C}$.[/hide]

ummm easiest?
[hide=Solution]
Let the coordinates be $(-x,x^2)$, $(x,x^2)$, and $(c,c^2)$. WLOG $x>c$. From $bh/2$, we have $x(x+c)(x-c)=2008$. Since the triangle has a $90^{\circ}$ angle, we have $\frac{c^2-x^2}{c+x}=-\frac{c-x}{c^2-x^2}$, formed by the slopes. This simplifies to $(x+c)(x-c)=1$, so $x=2008$.Then $c={2008}^2-1$ and the rest is a matter of computation to get $\boxed{\textbf{(C)} \text{18}}$


Let r and s be the roots of 2x^2+xsqrt3-5=0. Find r^2+2rs+s^2.

Can you do this without finding the roots of the equation?

[hide]$ r^2 \plus{} 2rs \plus{} s^2 \equal{} (r\plus{}s)^2$
By Viete's, the sum of the roots of the equation
$ 2x^2 \plus{} \sqrt3 x \minus{} 5 \equal{} 0$ is $ \minus{}\frac{\sqrt3}{2}$
So $ (r\plus{}s)^2 \equal{} \frac{3}{4}$
[/hide]


A train traveling from Aytown to Beetown meets with an accident after 1 hours, The train is stopped for 30 minutes, after which it proceeds at four-fifths of its usual rate, arriving at Beetown 2 hours late. If the train had covered 80 miles more before the accident, it would have been just one hour late. What is the usual rate of the train?

Also;

An airplane flies at some number of miles per hour and arrives at it's destination 1 hour late.
If the airplane had flown at 850 miles per hour, it would've been 20 minutes late.
How fast was the airplane going? (yeah i just made this up in my head, lol)

The key is the formula $ \text{rate}=\frac{\text{distance}}{\text{time}}$.

[hide="#1"]
(from V1)

Let the train's normal rate be $ x$, where $ x$ is in miles per hour.

Notice that the only difference between the two trips (the one-hour-late trip and two-hour-late trip) is an 80 mile stretch where if the train goes $ x$ mph it will get there 1 hour faster than if it goes $ \frac{4}{5} x$ mph. Thus, the difference in the time it takes for that 80 miles is 1 hour.

Let the time it takes for the train to travel $ 80$ miles in $ x$ mph be $ a$. Using our rate definition from above, we have $ x = \frac{80}{a}$ and we wish to solve for $ a$ in terms of $ x$. This gives $ a=\frac{80}{x}$.

A similar thing can be done for the second time, $ b$. Our definition gives $ \frac{4}{5} x = \frac{80}{b}$, from which we get $ b=\frac{100}{x}$.

Thus, the difference between our longer time (with smaller rate) and shorter time (with faster rate) is 1 hour, or $ b-a=1$. Substituting from above we just solve for $ x$:
\begin{eqnarray*} b-a &=&1 \\ \\
\frac{100}{x}-\frac{80}{x} &=& 1 \\ \\
100-80&=&x \\ \\
x&=& \boxed{20} \end{eqnarray*}
So our answer is 20.
[/hide]

couldn't there be multiple answers for #2.... because if he went 850 mph for 1 hour, there would be an answer... but if he went 2 hours or 3 hours or etc.... there would be different answers

i remember doing this in volume 1 of aops book and in chapter 3,maybe 4


find all function $ f: R \to R$ such that
$ xf(y) - yf(x) = f(\frac {y}{x})$ with all $ x;y \in R;x \neq 0$

Put $ x \equal{} a, y \equal{} b$, we have $ af(b) \minus{} bf(a) \equal{} f(\frac {b}{a})$
Put $ x \equal{} b,y \equal{} a$, we have $ bf(a) \minus{} af(b) \equal{} f(\frac {a}{b})$
Adding them together , we have $ f(x) \plus{} f(\frac {1}{x}) \equal{} 0$ for all $ x$

Now put $ y \equal{} a, x \equal{} \frac {1}{b}$, we have $ \frac {1}{b}f(a) \plus{} af(b) \equal{} f(ab)....................1$
put $ a \equal{} b, b \equal{} a$  we have $ \frac {1}{a}f(b) \plus{} bf(a) \equal{} f(ab)..........................................2$

$ 1 \minus{} 2$, we have $ \frac {f(x)}{x \minus{} \frac {1}{x}} \equal{} k$ is a constant for all $ x \neq 1, \minus{} 1$

It's easy to see that $ f(x) \equal{} k(x \minus{} \frac {1}{x})$ , where $ k$ is a constant, can satisfy the conditions.

[quote="stephencheng"]Put $ x \equal{} a, y \equal{} b$, we have $ af(b) \minus{} bf(a) \equal{} f(\frac {b}{a})$
Put $ x \equal{} b,y \equal{} a$, we have $ bf(a) \minus{} af(b) \equal{} f(\frac {a}{b})$
Adding them together , we have $ f(x) \plus{} f(\frac {1}{x}) \equal{} 0$ for all $ x$

Now put $ y \equal{} a, x \equal{} \frac {1}{b}$, we have $ \frac {1}{b}f(a) \plus{} af(b) \equal{} f(ab)....................1$
put $ a \equal{} b, b \equal{} a$  we have $ \frac {1}{a}f(b) \plus{} bf(a) \equal{} f(ab)..........................................2$

$ 1 \minus{} 2$, we have $ \frac {f(x)}{x \minus{} \frac {1}{x}} \equal{} k$ is a constant for all $ x \neq 1, \minus{} 1$

It's easy to see that $ f(x) \equal{} k(x \plus{} \frac {1}{x})$ , where $ k$ is a constant, can satisfy the conditions.[/quote]
Do you mean $ f(x)\equal{}k(x\minus{}\frac{1}{x})$?

[quote="Brut3Forc3"][quote="stephencheng"]Put $ x \equal{} a, y \equal{} b$, we have $ af(b) \minus{} bf(a) \equal{} f(\frac {b}{a})$
Put $ x \equal{} b,y \equal{} a$, we have $ bf(a) \minus{} af(b) \equal{} f(\frac {a}{b})$
Adding them together , we have $ f(x) \plus{} f(\frac {1}{x}) \equal{} 0$ for all $ x$

Now put $ y \equal{} a, x \equal{} \frac {1}{b}$, we have $ \frac {1}{b}f(a) \plus{} af(b) \equal{} f(ab)....................1$
put $ a \equal{} b, b \equal{} a$  we have $ \frac {1}{a}f(b) \plus{} bf(a) \equal{} f(ab)..........................................2$

$ 1 \minus{} 2$, we have $ \frac {f(x)}{x \minus{} \frac {1}{x}} \equal{} k$ is a constant for all $ x \neq 1, \minus{} 1$

It's easy to see that $ f(x) \equal{} k(x \plus{} \frac {1}{x})$ , where $ k$ is a constant, can satisfy the conditions.[/quote]
Do you mean $ f(x) \equal{} k(x \minus{} \frac {1}{x})$?[/quote]

You are right .It's a typing mistake ,now corrected.


number theory
prime numbers
number theory proposed

Prove that there exist infinitely many composite natural numbers,such that
$ 3^{n\minus{}1}\minus{}2^{n\minus{}1}$ is divisible by $ n$.

Does there exist an infinity of sets of prime numbers $ \{p_{1}, p_{2}, \dots p_{k}\}$, such that $ (p_{1}\minus{}1)(p_2\minus{}1)\cdots (p_k\minus{}1)$ divides $ p_{1}p_{2}\cdots p_{k}\minus{}1$ ?

Sorry,this problem was posted at PEN.Here is the link:
http://www.mathlinks.ro/viewtopic.php?t=150409
Sorry again and i ask moderators to lock this topic pls.Thank you.

[Moderator: wish granted  :wink:  ]


Which compass should you get if:

- you don't want the lead to snap off alot
- want to be able to easily adjust the angle between the legs of the compass (none of those things that can only be adjusted by rotating a disk)
- you want the legs to stay the way they are after you moved them?

It isn't a big deal but it is sort of frustrating when the first of the above bullets happens and you have to replace the compass or when the angle between the "legs" of the compass increases as you are drawing your circle.

I recommend a Staedtler compass, the one with rigid but also adjustable legs and a circular rotating disc. I've used this compass since middle school (and thats more than 7 years of use). Pretty good I'll say.

Interesting: I have a Staedtler with circular disc (since I was 8: that's 12 years!)too: although aufha asked for a compass without the disc, I have to say that they're much more accurate then others (I have another, Rotring one without a disc-very stable, but it was quite expensive).

I have a Westcott compass, buy it and you won't regret it. Even though it looks like one of those cheap hole Academia compasses, it really rocks :)


logarithms
inequalities
floor function
ceiling function

Can anyone do this problem without actually multiplying it out?

How many digits are there in the expanded form of 2 to the 40th power?

[hide="WARNING: do the problem before you read this."]
[hide="hint (solution comes after)"]
don't multiply it out unless you want to waste an hour of your life.
[hide=" answer"]
13. SHHHH!! Don't tell anyone! it's top-secret.
 :wink: [/hide]
[/hide][/hide]

Thanks.
-LHM

I suppose that since this is on a sprint round, we can't use the log thing without already knowing what $ \log 2$ is.

You could take $ 2^{10}$, square that, then square the result, but it's still multiplying

I have that memorized since I've used it so much.
I guess we could do the wimpy pattern way, but there might not be a pattern...

There is a pattern, for every three times $ 2$ is multplied there is another digit place.  :)
[hide="EDIT So, now trivially wrong"]
$ \left\lfloor\frac {40}{3} \right\rfloor \equal{} 13$.
[/hide]

That pattern doesn't hold, $ 1024\times 2^3$ doesn't have 5 digits[/i]

[quote="AstroPhys"]Guess not  :oops: .
[hide="Revised Pattern"]
For every $ 13$ multiples of $ 2$ there are $ 4$ digit places so, $ \left\lceil \frac {4 \cdot 40}{13} \right\rceil \equal{} 13$ .
[/hide]

Find if there are more unevaluated cases.  :wink:[/quote]

nope

for any such patterns you find (every __ factors of 2 adds __ digits), it will eventually break down when you make the power very large, say, 2^1000.

There isn't a formula that works for all powers of two, because there is no power of 2 that is also a power of 10.

[quote="nebula42"]You can use some simple inequalities to bound $ 2^{40}$ between two numbers with the same number of digits.

[hide]$ 2^{40} \equal{} (2^{10})^4 \equal{} 1024^4 > 1000^4 \equal{} 10^{12}$.

$ 2^{40} \equal{} 1024^4 < 1100^4 \equal{} 1000^4 \cdot 1.1^4 < 1000^4 \cdot 2 \equal{} 2 \cdot 10^{12}$.

Thus, $ 2^{40}$ has $ \boxed{13}$ digits.[/hide][/quote]

I don't quite follow the inequalities.  Could someone explain more, please?  Thanks.

All of that is multipling

but i kno there is a way to do without multiplying , i did it before, but i forgot :P

Actually... you can take logs...

$ \log2^{40}=40\log2$

We want to find $ \log2$.

$ 2^3=8,2^4=16\rightarrow\frac{1}{4}<\log2<\frac{1}{3}$
The answer thus must lie between $ 10$ and $ 14$, either being $ 11,12,13$

We attempt to rule out $ 12$.

If $ 40\log2=12,\log2=\frac{3}{10}\rightarrow\10^{\frac{3}{10}}=2$

Now... $ \sqrt[10]{1000}<2\rightarrow\log2>12$.

So.... the answer must be $ 13$..

xD =). Not practical at all.


[quote="Superficial"]The circles $ w_1(O_1),w_2(O_2),w_3(O_3)$ are such that $ w_1,w_2$ are externally tangent at $ A$, $ w_2,w_3$ at $ B$, and $ w_3,w_1$ at $ C$. Let $ A_1,B_1$ be points on $ w_1$ diametrically opposite to $ A,B$ respectively. Let $ AB_1$ meet $ w_2$ again at $ M$, $ BA_1$ meet $ w_3$ again at $ N$, $ AA_1,BB_1$ meet at $ P$. Prove that $ M,N,P$ are collinear.[/quote]
I think that your statement is incorrect,how could $ B_1$ be on $ w_1$,if $ B$ doesn't lie on $ w_1$?

You are right, Erken.

If I let $ w_2,w_3$ be tangent at $ C$, and $ w_3,w_1$ at $ B$ instead, is the problem right now?


Hi all!

I hope that this is the right forum for this problem; I don't remember where I found it, sorry - but I think it was proposed in 1996 ;)

Let [tex]F:=\{A_1,...,A_n\}[/tex] be a set of subsets of the set  [tex]S:=\{1,2,...,n\}[/tex] such that the following conditions hold:
(a) Every two of the subsets [tex]A_i[/tex] have exactly one element in common,
(b) Every element from [tex]S[/tex] is contained in exactly [tex]k[/tex] of the sets [tex]A_i[/tex] (,where [tex]k[/tex] is a natural number with [tex]k=1,2,...,n[/tex])

Is this possible for [tex]n=1996[/tex]?

Hanno

no.
let's do some counting first;if we count the total number of pairs $(x,A),x\in A$ with $x\in S,A\in \mathcal{F}$ then we have \[nk=\sum_{A\in\mathcal{F}} |A|\].
fix a set $A\in\mathcal{F}$and say $A=\{x_1,x_2,\ldots,x_l\}$.since every other set meets this at precisely one point if we denote by $\mathcal{F}_x$ the subfamily of $\mathcal{F}$ consisting of all those sets containing $x$, it follows that $\mathcal{F}\setminus \{A\}=\bigcup_{x\in A}\mathcal{F}_x$ and since $|\mathcal{F}_x|=k\ \forall x$ it follows that $n-1=l(k-1)$ and so in particular all the sets have the same size as well.but coupling this with the earlier observation, we have $nl=nk\Rightarrow l=k$ and so we have in particular that $n-1=k(k-1)$ and this is not possible for $n=1996$ with $k$ being an integer.


Hey guys,

If any of you brought the team contest problems back, do you think you could scan it and post it as a PDF? I'd really appreciate it a lot.

Thanks.
Zhero

I would post my round, but I'm saving many of the problems to use in a mock AIME for early 2010.

[quote="stickfigure"]I have both rounds, but I'm not sure Basham/Jay would be too happy about me posting these.[/quote]

shouldn't be a problem, the '08 team contest is right here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218962


[quote="kunny"]Are there positive integers $ x,\ y$ such that $ x^2 \plus{} y^2 \equal{} 2009^n$ for a positive integer $ n$?[/quote][hide]Yes, let $ x\equal{}5\cdot7, y\equal{}4\cdot7, n\equal{}1$.[/hide]

No, it isn't the only solution ;)

[hide]We'll prove that for every $ n$ there are $ x,y$ which satisfy the condition.

First case: $ n$ is odd.

Then $ x \equal{} 5 \cdot 7 \cdot 2009^\frac{n \minus{} 1}{2}$ and $ 4 \cdot 7 \cdot 2009^\frac{n \minus{} 1}{2}$ are solutions.

Second case: $ n$ is even.

Then $ x \equal{} 40\cdot 7\cdot 2009^\frac{n \minus{} 2}{2}$ and $ 9\cdot 7 \cdot 2009^\frac{n \minus{} 2}{2}$ are solutions. (it's easy to spot $ 40$ and $ 9$ in $ 41^2 \equal{} 1681$).[/hide]

[hide]
Note that $ 2009 = 41\cdot 7^2 \equiv 0 \mod 7$

Quadratic residues $ \mod 7$ are $ 0,1,2,4$. Hence, $ x\equiv y \equiv 0 \mod 7$

So let $ x=7^na, y=7^nb \Longrightarrow a^2 + b^2 = 41^n$

Notice that $ 5^2 + 4^2 = 41$, So using Pythag triples $ (5^2-4^2,2\cdot 5\cdot 4,5^2+4^2) = (9, 40,41) \rightarrow 9^2+40^2 = 41^2$

From here we can write a general solution
$ (x,y) = (5\cdot 7^n\cdot 41^{\frac{n-1}{2}},4\cdot 7^n\cdot 41^{\frac{n-1}{2})}$ for odd $ n$
$ (x,y) = (9\cdot 7^n\cdot 41^{\frac{n}{2}-1}},40\cdot 7^n\cdot 41^{{\frac{n}{2}-1})$ for $ n$ even

Note that these are not the only solutions; e.g. we can take the pythag triple $ (4,5,41)$ and induct upward to get solutions for all $ n=2^k, k\in\mathbb{N_0}$[/hide]

Here's a way to prove that there are an infinite number of triples $ (x,y,n)\in\mathbb{N}$ that satisfy $ x^2\plus{}y^2\equal{}2009^n$:  Notice that $ 2009\equal{}7^2\cdot 41$ and $ 41\equal{}5^2\plus{}4^2\implies 2009\equal{}(7\cdot 5)^2\plus{}(7\cdot 4)^2\equal{}35^2\plus{}28^2$.  Consider $ z\equal{}35\plus{}28i$.  Then $ 2009^n\equal{}(|z|^2)^n\equal{}|z^n|^2\equal{}|x\plus{}iy|^2\equal{}x^2\plus{}y^2$ by expanding $ z^n$ using Newton's Binomial Theorem.  So for each $ n$, there are $ x,y$ that satisfy the equation, i.e., there are an infinite number of such triples. $ \Box$

Or you could multiply both $ x$ and $ y$ by an integral power of 2009.

Or you can do use the simple fact that

$ (a^2\plus{}b^2)(c^2\plus{}d^2)\equal{}(ac\plus{}bd)^2\plus{}(ad\minus{}bc)^2$.

And if you can show for n=1, which 2009=35^2+28^2, then this means we can extend this for any n=k, as we can just build from n=1.

So, we have infinitely many solutions (x,y,n).


function
inequalities unsolved
inequalities

[quote="duc 95"]Given 0< a,b,c
Prove that[/quote]

substitution    $ a\equal{}\frac{1}{1\plus{}x},b\equal{}\frac{1}{1\plus{}y},c\equal{}\frac{1}{1\plus{}z},(x,y,z>0)$.

Problem: Given $ 0<a,b,c\le1$. Prove that \[ \frac1{a\plus{}b\plus{}c}\ge\frac13\plus{}(1\minus{}a)(1\minus{}b)(1\minus{}c)\]

[hide]Let $ a\equal{}1\minus{}x,b\equal{}1\minus{}y,c\equal{}1\minus{}z$. So $ 0\le x,y,z<1$.

\[ \frac1{a\plus{}b\plus{}c}\ge\frac13\plus{}(1\minus{}a)(1\minus{}b)(1\minus{}c)\]

\[ \Leftrightarrow\frac1{3\minus{}x\minus{}y\minus{}z}\ge\frac13\plus{}xyz\]

\[ \Leftrightarrow3\ge3\minus{}x\minus{}y\minus{}z\plus{}9xyz\minus{}9xyz(x\plus{}y\plus{}z)\]

\[ \Leftrightarrow9xyz(x\plus{}y\plus{}z)\plus{}x\plus{}y\plus{}z\ge9xyz\]

\[ \Leftrightarrow x\plus{}y\plus{}z\ge9xyz(1\minus{}x\minus{}y\minus{}z)\]

Let $ x\plus{}y\plus{}z\equal{}a$. So $ xyz\le\left(\frac{x\plus{}y\plus{}z}3\right)^3\equal{}\frac{a^3}{27}$. Therefore it is enough to prove that

\[ a\ge\frac{a^3}3(1\minus{}a)\]

\[ \Leftrightarrow3a\ge a^3\minus{}a^4\]

\[ \Leftrightarrow a^4\minus{}a^3\plus{}3a\ge0\]

\[ \Leftrightarrow a^3\minus{}a^2\plus{}3\ge0\]

Consider the function $ f(x)\equal{}x^3\minus{}x^2\plus{}3$. We have $ f'(x)\equal{}3x^2\minus{}2x$. So the turning points are $ x\equal{}0,x\equal{}\frac23$. But $ f(0)\equal{}3,f(\frac23)>0$, so clearly $ a^3\minus{}a^2\plus{}3\ge0$. Q.E.D.[/hide]


Consider quaderilateral $ ABCD$ such that $ [ABC] \equal{} [ADC]$ (when I say $ [\triangle]$, I refer to the area of $ \triangle$).  Allow the diagonals of this quaderilateral to meet at $ E$.  Through $ E$, let the lines parallel to $ AD$, $ AB$, $ BC$, and $ CD$ meet $ AB$, $ AD$, $ CD$, and $ BC$ at $ K$, $ N$, $ M$, and $ L$ respectively.  Find $ [ABCD]: [KLMN]$.

$ [ABC]: [ACD] \equal{} [BEA]: [DEA] \equal{} BE: ED \equal{} 1$. It is clear that $ KENA$ and $ LEMC$ are parallelograms that each have half the area of $ ABD$ and $ BDC$, so the ratio is $ 2: 1$.


geometry
3D geometry
quadratics
similar triangles
algebra
quadratic formula

A right circular cone of height 2 is cut into two pieces by a plane parallel to its base. The lateral areas of the two pieces are equal. What is the distance from the cone's vertex to the plane?

[hide="Solution"]Let the distance be $ h$. Let the cone-shaped piece be $ A$ and the other one be $ B$.

Using similar triangles, the radius of $ A$ is $ r_a = \frac {hr}{2}$.

$ S_A = \pi r_a\sqrt {r_a^2 + h^2},\quad S_B = \pi r\sqrt {r^2 + 4} - \pi r_a\sqrt {r_a^2 + h^2}$

Set $ S_A = S_B$ and simplify:
\begin{align*}\pi r_a\sqrt {r_a^2 + h^2} & = \pi r\sqrt {r^2 + 4} - \pi r_a\sqrt {r_a^2 + h^2} \\
 \\
r_a\sqrt {r_a^2 + h^2} & = r\sqrt {r^2 + 4} - r_a\sqrt {r_a^2 + h^2} \\
 \\
2r_a\sqrt {r_a^2 + h^2} & = r\sqrt {r^2 + 4}\end{align*}
Substitute $ \frac {hr}{2}$ for $ r_a$ and simplify further:
\begin{align*}hr\sqrt {\left(\frac {hr}{2}\right)^2 + h^2} & = r\sqrt {r^2 + 4} \\
 \\
h\sqrt {\left(\frac {hr}{2}\right)^2 + h^2} & = \sqrt {r^2 + 4} \\
 \\
h^2\sqrt {\frac {r^2}{4} + 1} & = \sqrt {r^2 + 4} \\
 \\
h^2\sqrt {r^2 + 4} & = 2\sqrt {r^2 + 4} \\
 \\
h^2 & = 2 \\
 \\
h & = \boxed{\sqrt {2}}\end{align*}
[/hide]

[hide="Alternative method."]
Call the distance from the bottom of the cone to the plane $ x$ and from the plane to the top of the cone $ 2\minus{}x$.  Since they have the same volume, you know that:

$ 2^2 \minus{} (2\minus{}x)^2 \equal{} (2\minus{}x)^2$

This simplifies to:

$ x^2 \minus{} 4x \plus{} 2 \equal{} 0$

Using the quadratic formula, you get that $ x \equal{} 2 \minus{} \sqrt{2}$ (since $ 2 \plus{} \sqrt{2}$ is an extraneous root as it is greater than $ 2$).

However, you're interested in $ 2\minus{}x$, so your answer is $ \boxed{\sqrt{2}}$.[/hide]

[hide="Fast"]
[b]Proposition:[/b] The lateral area of similar cones is proportional to the square of their heights.

[b]Proof:[/b]Let's cut the cone by a segment connecting a point on the base to the vertex, and flatten it out. This results in a sector of a circle. Let the angle of this sector be $ \theta$, in radians.

Then, the area of such a sector is $ \frac{\theta}{2 \pi} \pi r^2 \equal{} \frac{\theta}{2} r^2$. Thus, the lateral area is proportional to $ r^2$. Since $ r$ and $ h$ are proportional by similar triangles, the lateral area is proportional to $ h^2$.

[b]Solution:[/b] We have two similar cones, one with height $ 2$ and one with height $ H$, and the lateral area of the first is twice the lateral area of the second. So, $ 2 ^2 \equal{} 2 H^2 \implies H \equal{} \sqrt{2}$[/hide]


Find the coefficient of $x^4$ in the expansion of $\left(1-\frac32x-x^2\right)^5$.

Solution only, just in case you can't find it
[hide=":)"] -515/16[/hide]

[quote="kunny"][hide]The answer is $\frac{1645}{16}$[/hide][/quote]

how did you get that?

[quote="deej21"][hide]Hmm... I completed the square on the inside, then used binomial expansion on that, then used binomial expansion on the terms to find the $x^4$ coefficient in those... and since it would take me a while I don't want to show all the work that that encompassed.  Anyway, my answer was $\frac{-515}{16}$, which looks wrong, and probably is because my solution was really bad.[/hide][/quote]

Seems your answer is not wrong.  I too got the same answer, using multinomial theorem.

[hide]
The general format is
$-\frac{5!}{p!q!r!}(x^2)^p\left(\frac{3}{2}x\right)^q(-1)^r$

where $p+q+r=5$ , $2p+q=4$ .Hence we have

$(p,q,r)=(0,1,4),(1,2,2),(2,3,0)$ . plugging this three value into the general equation and add up to get $\frac{-515}{16}$

 :)
[/hide]


If a1<=a2<=..<=an with n>=2 then prove that the following inequality holds:

1/(n+1)(a1+...+an)>=min(a1,an/2)
cheers! :D :D


superior algebra
superior algebra unsolved

How to prove that a union of c sets of cardinality c has cardinality c?

Let $\left(A_{i}\right)_{i\in I}$ a system of sets with $\forall i\in I: |A_{i}|=|I|=\kappa$.
Then $\left|\bigcup_{i\in I}{A_{i}}\right|\leq \left|\dot{\bigcup_{i\in I}}{A_{i}}\right|=\left|\bigcup_{i\in I}{\{i\}\times\kappa}\right|=|I\times\kappa|=\kappa^{2}=\kappa$. (provided of course, that $\kappa$ is an infinite cardinal)

maybe i'm wrong, but i think he's been asking for a proof of the same fact you use: it's not at all obvious that for every cardinal $\kappa^{2}= \kappa$...

sketches of two different proofs of this (well-known and widely used) fact have been posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=106016]here[/url].


conics
projective geometry
invariant
geometry proposed
geometry

Let's call a sextuple of points (A, B, C, D, E, F) in the plane a [b]Pascalian sextuple[/b] if and only if the points of intersection $AB\cap DE$, $BC\cap EF$ and $CD\cap FA$ are collinear.

Prove that if a sextuple of points is Pascalian, then each permutation of this sextuple is Pascalian.

This is, of course, trivial using the fact that a sextuple of points is Pascalian if and only if the six points lie on one conic. But how about a proof without using conics? There is one, and I find it quite nice.

[EDIT: I have lately found the above theorem as Theorem 2 at http://www.esil.univ-mrs.fr/~mdaniel/GTMG/AS-Contrainte/Publis-Presentations/detecting_incidences.pdf .]

  Darij

[quote="sprmnt21"]does this way make use of Desargue's theorem?[/quote]

My proof doesn't. Maybe some other proofs do...

  darij

the main idea is this: every permutation of ABCDEF is a sequence of swap of two close points.

for example (since my english, is better to provide an example  :) ) the permutation CFADEB is the limit of the following sequence of swapping.

starting from ABCDEF we need to move C at the firts place on the left, then, swapping BC we have ACBDEF, and swapping AC, we have CABDEF. Now we can apply recursively the metod to the sub-sets ABDEF and FADEB and so on.

Then the problem is reduced to that to prove the claim for the case where only a "simple" swapping of close points is considered.

Let's to prove, for example, that if G=AB.DE, I=BC.EF, H=CD.FA are collinears then  J=AC.DE, I=CB.EF, K=BD.FA are collinears too.

But this derive from the Pappus theorem, applied to the sextuple AJCDBK.

I hope at least the idea above is clear.

[quote="sprmnt21"]the main idea is this: every permutation of ABCDEF is a sequence of swap of two close points.[/quote]

This is the core of my proof, too. It's just the well-known theorem that every permutation of the set {1; 2; ...; n} is a composition of transpositions of adjacent elements.

[quote="sprmnt21"]for example (since my english, is better to provide an example  :) ) the permutation CFADEB is the limit of the following sequence of swapping.[/quote]

I think that, instead of "limit", you should have better said "result" (there is no limiting process here  ;) ).

[quote="sprmnt21"]starting from ABCDEF we need to move C at the firts place on the left, then, swapping BC we have ACBDEF, and swapping AC, we have CABDEF. Now we can apply recursively the metod to the sub-sets ABDEF and FADEB and so on.[/quote]

Nice proof; your idea is actually the same as I had in mind.

[quote="sprmnt21"]Then the problem is reduced to that to prove the claim for the case where only a "simple" swapping of close points is considered.

Let's to prove, for example, that if G=AB.DE, I=BC.EF, H=CD.FA are collinears then  J=AC.DE, I=CB.EF, K=BD.FA are collinears too.

But this derive from the Pappus theorem, applied to the sextuple AJCDBK.[/quote]

I'm sorry, but I don't see how to derive this from the Pappus theorem. Maybe I'm blind, but from some reasons I doubt it is [i]that[/i] simple. Maybe one should apply Pappus or Desargues not one time but several times.

Anyway, my solution of the original problem doesn't use Pappus or Desargues, but rather uses Menelaos:

I first prove the following lemma, which is, I believe, called "Carnot theorem":

[b]Lemma 1.[/b] Let ABC be a triangle, let $D_1$ and $D_2$ be two points on its sideline BC, let $E_1$ and $E_2$ be two points on its sideline CA, and let $F_1$ and $F_2$ be two points on its sideline AB. Then, the points $E_1F_2\cap BC$, $F_1D_2\cap CA$ and $D_1E_2\cap AB$ are collinear if and only if

$\frac{BD_1}{D_1C}\cdot\frac{BD_2}{D_2C}\cdot\frac{CE_1}{E_1A}\cdot\frac{CE_2}{E_2A}\cdot\frac{AF_1}{F_1B}\cdot\frac{AF_2}{F_2B}=1$.

[i]Proof of Lemma 1.[/i] Let $X=E_1F_2\cap BC$, $Y=F_1D_2\cap CA$ and $Z=D_1E_2\cap AB$. Then, we have to show that the points X, Y, Z are collinear if and only if

$\frac{BD_1}{D_1C}\cdot\frac{BD_2}{D_2C}\cdot\frac{CE_1}{E_1A}\cdot\frac{CE_2}{E_2A}\cdot\frac{AF_1}{F_1B}\cdot\frac{AF_2}{F_2B}=1$.

But since the points X, Y, Z lie on the sidelines BC, CA, AB of triangle ABC, the Menelaos theorem yields that they are actually collinear if and only if

$\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=-1$.

So all we have to do is to prove that the equations

$\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=-1$

and

$\frac{BD_1}{D_1C}\cdot\frac{BD_2}{D_2C}\cdot\frac{CE_1}{E_1A}\cdot\frac{CE_2}{E_2A}\cdot\frac{AF_1}{F_1B}\cdot\frac{AF_2}{F_2B}=1$

are equivalent. Indeed, this will be trivial once it will be shown that we always have the identity

$\left(\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}\right)\cdot\left(\frac{BD_1}{D_1C}\cdot\frac{BD_2}{D_2C}\cdot\frac{CE_1}{E_1A}\cdot\frac{CE_2}{E_2A}\cdot\frac{AF_1}{F_1B}\cdot\frac{AF_2}{F_2B}\right)=-1$.

Thus, in order to prove Lemma 1, it will be enough to establish this identity.

Well, since the points $E_1$, $F_2$ and X on the sidelines of triangle ABC are collinear, the Menelaos theorem yields

$\frac{BX}{XC}\cdot\frac{CE_1}{E_1A}\cdot\frac{AF_2}{F_2B}=-1$.

Similarly, the collinearity of the points $F_1$, $D_2$ and Y yields

$\frac{CY}{YA}\cdot\frac{AF_1}{F_1B}\cdot\frac{BD_2}{D_2C}=-1$,

and the collinearity of the points $D_1$, $E_2$ and Z yields

$\frac{AZ}{ZB}\cdot\frac{BD_1}{D_1C}\cdot\frac{CE_2}{E_2A}=-1$.

Multiplying these three equations, we get

$\left(\frac{BX}{XC}\cdot\frac{CE_1}{E_1A}\cdot\frac{AF_2}{F_2B}\right)\cdot\left(\frac{CY}{YA}\cdot\frac{AF_1}{F_1B}\cdot\frac{BD_2}{D_2C}\right)\cdot\left(\frac{AZ}{ZB}\cdot\frac{BD_1}{D_1C}\cdot\frac{CE_2}{E_2A}\right)=\left(-1\right)^3$,

or, equivalently,

$\left(\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}\right)\cdot\left(\frac{BD_1}{D_1C}\cdot\frac{BD_2}{D_2C}\cdot\frac{CE_1}{E_1A}\cdot\frac{CE_2}{E_2A}\cdot\frac{AF_1}{F_1B}\cdot\frac{AF_2}{F_2B}\right)=-1$.

But this is exactly the identity we have to prove, and thus the proof of Lemma 1 is complete.

Now, in order to solve the initial problem, I consider a Pascalian sextuple of points (A, B, C, D, E, F) in the plane. The problem requires me to prove that every permutation of this sextuple is Pascalian, but, noting - like you did - that any permutation can be written as a composition of transpositions of adjacent elements, I see that it is enough to show that every sextuple resulting from the sextuple (A, B, C, D, E, F) upon a transposition of adjacent elements is Pascalian. Since it doesn't make any difference whether we transpose A with B, or B with C, or C with D, or D with E, or E with F (in fact, the definition of a Pascalian sextuple is cyclically invariant), we will restrict ourselves to considering only the transposition of B with C. In other words, we have to show that the sextuple of points (A, C, B, D, E, F) is Pascalian.

But this is just a matter of applying Lemma 1: Let $U=DE\cap FA$, $V=FA\cap BC$ and $W=BC\cap DE$. Then, the points B and C lie on the sideline VW of triangle UVW, the points D and E lie on its sideline WU, and the points F and A lie on its sideline UV. Since the sextuple of points (A, B, C, D, E, F) is Pascalian, the points $AB\cap DE$, $BC\cap EF$ and $CD\cap FA$ are collinear; in other words, the points $AB\cap WU$, $EF\cap VW$ and $CD\cap UV$ are collinear. Thus, after Lemma 1, we have

$\frac{VC}{CW}\cdot\frac{VB}{BW}\cdot\frac{BE}{EU}\cdot\frac{BD}{DU}\cdot\frac{UA}{AV}\cdot\frac{UF}{FV}=1$.

But this can be rewritten in the form

$\frac{VB}{BW}\cdot\frac{VC}{CW}\cdot\frac{BE}{EU}\cdot\frac{BD}{DU}\cdot\frac{UA}{AV}\cdot\frac{UF}{FV}=1$.

And thus, using Lemma 1 again, we see that the points $AC\cap WU$, $EF\cap VW$ and $BD\cap UV$ are collinear. In other words, the points $AC\cap DE$, $CB\cap EF$ and $BD\cap FA$ are collinear. This yields that the sextuple of points (A, C, B, D, E, F) is Pascalian. And this completes the solution of the problem.

  Darij

[quote="sprmnt21"]the main idea is this: every permutation of ABCDEF is a sequence of swap of two close points.
...

Then the problem is reduced to that to prove the claim for the case where only a "simple" swapping of close points is considered.

Let's to prove, for example, that if G=AB.DE, I=BC.EF, H=CD.FA are collinears then  J=AC.DE, I=CB.EF, K=BD.FA are collinears too.

But this derive from the Pappus theorem, applied to the sextuple AJCDBK.

[/quote]

I will try to clarify this last stentence, with the help of the figure attacched.

The triples of points {A,J,C} and {B,D,K} are collinear. Then by Pappus G=AB.JD, I'=CB.JK and H=CD.KA are collinear points.  Then I' = GH.CB = I therefore {J,I,K} is a triple of collinear points.

[quote="sprmnt21"]The triples of points {A,J,C} and {B,D,K} are collinear. Then by Pappus G=AB.JD, I'=CB.JK and H=CD.KA are collinear points.  Then I' = GH.CB = I therefore {J,I,K} is a triple of collinear points.[/quote]

Thanks, now it is clear. You actually applied the Pappus theorem to the collinear point triples (A; J; C) and (D; B; K), with corresponding points on corresponding places.

Thank you for this nice proof!

  Darij


function
functional equation
algebra
IMO Shortlist

Let $ \mathbb{N}$ denote the set of all positive integers. Prove that there exists a unique function $ f: \mathbb{N} \mapsto \mathbb{N}$ satisfying
\[ f(m \plus{} f(n)) \equal{} n \plus{} f(m \plus{} 95)
\]
for all $ m$ and $ n$ in $ \mathbb{N}.$ What is the value of $ \sum^{19}_{k \equal{} 1} f(k)?$

I dont know if this can be taken as a totally correct solution but :
by observation I can see that a function satisfying the given condition is f(k) = k+95 (but i have to prove that it is unique...how?)Then the value of the sum is 1995

We have $ f(f(m) \plus{} f(n)) \equal{} n \plus{} f(f(m) \plus{} 95) \equal{} m \plus{} n \plus{} f(2.95)\forall m,n\in\mathbb{N}$ and if $ f(m) \equal{} f(n)$ then $ m \equal{} n$ therefore $ f(n \plus{} 1) \plus{} f(n \minus{} 1) \equal{} 2f(n)\forall n > 1$. Done! :P

Today i've known a nice proof for this problem:, i will replace $95$ by $K$:
- First, $1 + f(m + f(n)) = n + 1 + f(m + K) \implies f(1 + f(m + f(n))) = f(n + 1 + f(m + K)) \implies m + f(n) + f(K + 1) = m + K + f(n + K + 1) \implies f(n + K + 1) = f(n) + f(K + 1) - K$ (*)
- Second, $P(m, n) is f(m + f(n)) = n + f(m + K)$:
i) $P(m, K + f(n)): f(m + n + f(2K)) = f(m + f(K + f(n))) = f(n) + K + f(m + K)$
ii) Change $m, n$ in i), $f(m + n + f(2K)) = f(m) + f(n + K) + K$
Therefore, $f(m + K) + f(n) + K = f(m) + f(n + K) + K \implies f(m + K) - f(m) = f(n + K) - f(n) \implies f(n + K) - f(n) = f(K + 1) - f(1)$ (**)

[b]$$P(m, n + 1) in (**): f(n + K + 1) - f(n + 1) = f(K + 1) - f(1) \implies f(n) + f(K + 1) - K - f(n + 1) = f(K + 1) - f(1) \implies f(n + 1) - f(n) = f(1) - K$$[/b]
[b]$$\implies f(n) - f(1) = (n - 1)(f(1) - K) \implies f(n) = n(f(1) - K) + K $$[/b]
Use it in the first, we have: (Set $a = f(1)$)
$(m + f(n))(a - K) + K = n + (m + K)(a - K) + K$
$\implies f(n)(a - K) = n + K(a - K)$
$\implies (n(a - K) + K)(a - K) = n + Ka - K^2$
$\implies n(a - K)^2 = n$
So $a = K + 1$ or $a = K - 1$
Case $a = K + 1$ we get $f(n) = n + K$
Case $a = K - 1$ we get $f(n) = K - n$, let $n := 2K$ then f(2K) < 0, contradition. So there exist a unique function $f$.
Done.
P.s: Sorry for bump up this topic. I want to share nice idea :).


modular arithmetic
articles
algebra
polynomial
floor function
algorithm
quadratics

Find all solutions to
$ x^{10} \equiv 1 \pmod{14}$

Can someone teach (as in basics) me how to do the modulo stuff?
I can't get a hang of it?

[hide="fryeggs"]
it takes practice.  one of the harder things to deal with is the lack of division.  Anyway, see nsato's article on Number Theory: [url]http://www.artofproblemsolving.com/Resources/Papers/SatoNT.pdf[/url].  chapter 4 is all modular arithmetic, and many later chapters continue in that direction.  anyway, for problems like this one it is often useful to go back to definitions.

important ideas:
definition: $ a \equiv b \pmod{p}$ means $ p|a-b$ (read $ p$ divides (is a factor of) $ a-b$)
if $ a\equiv b \pmod{p}$ and $ c \equiv d \pmod{p}$ then $ a+c \equiv b+d \pmod{p}$ and $ ac \equiv bd \pmod{p}$.

Putting these together, we obtain the lemma that if $ P(x)$ is a polynomial with integer coefficients, $ P(a) \equiv P(b) \pmod{p}$.  Thus to find all the solutions of a polynomial equivalence you only need to check $ a=0,1,2 \ldots p-1$.  And if the polynomial is even, you only need to check $ a \equiv 0,1,2 \ldots \lfloor p/2 \rfloor$ since $ p-1 \equiv -1 \pmod{p}$, $ p-2, \equiv -2 \pmod{p}$, etc. and $ -k \pmod{p}$ will only be a solution if $ k$ is a solution.

also note $ ad \equiv bd \pmod{pd}$ implies $ a \equiv b \pmod{p}$ (since the first relation is equivalent to $ ad-bd = kpd$ for some $ k$ and the second is the same thing after $ d$ is divided out)

Other important ideas
modular inverses
extended euclidean algorithm

(you can wikipedia these):
about reducing exponents:
fermat's little theorem
euler's totient theorem
carmichael's theorem

about composite moduli:
chinese remainder theorem

equation solving - hensel's lemma for solving polynomials in mod $ p^k$ where $ p$ is prime

and there is tons of stuff about quadratic congruences too, but don't worry about that until you get through the other stuff.  see [url]http://www.imomath.com/tekstkut/quadcong_ddj.pdf[/url] when you are ready for that.
[/hide]

[hide="solution"]
First - apply carmichael's theorem to see if we can reduce the exponent.  $ \lambda(14)=\text{lcm}(\lambda(2), \lambda(7)) = \text{lcm}(1,6)=6$.  Therefore, since $ x$ clearly must be relatively prime to $ 14$, our given relation is the same as $ x^4 \equiv 1 \pmod{14}$.

Hence we have $ 14|x^4-1$.  breaking this up, $ 2|x^4-1$ implies $ 2|(x^2-1)(x^2+1)$, so $ 2|x^2-1$ or $ 2|x^2+1$.  In either case, $ x \equiv 1 \pmod{2}$.  duh.

Also, we have $ 7|x^4-1$ implies $ 7|x-1$, $ 7|x+1$, or $ 7|x^2+1$.  Hence $ x \equiv 1,6 \pmod{7}$ or $ x^2 \equiv -1 \pmod{7}$.  But the quadratic residues mod $ 7$ are $ 0,1,2,4$, so this second relation has no solutions.

So we need to put together $ x\equiv 1,6 \pmod{7}$ and $ x\equiv 1 \pmod{2}$.  By chinese remainder theorem, our solutions should be in mod $ 14$.  Now the first relation give $ x \equiv 1,6,8,13 \pmod{14}$, and taking only the odds, we have that $ x \equiv 1, 13 \pmod{14}$ are the only solutions.

btw, this problem would be easy to brute force since x^{10} = (-x)^10 you would only need to check $ 0,1, \ldots 7$.[/hide]

$ x^{10}\equal{}1(mod 2) \Leftrightarrow x$ is odd.
$ x^{10}\equal{}1(mod 7) \Leftrightarrow x^{10 \times 2\minus{} 3 \times (7\minus{}1)}\equal{}x^2\equal{}1(mod 7)$ by Fermat's little theorem. $ \Leftrightarrow x\equal{}1,\minus{}1(mod 7)$

So $ x\equal{}1,\minus{}13(mod 14)$ are the solutions.


[quote="cnyd"]$ f : \mathbb{Z}\mapsto\mathbb{Z}$      $ f(f(n) \minus{} 2n)) \equal{} 2f(n) \plus{} n$[/quote]

Let $ g(n)\equal{}f(n)\minus{}2n$. The equation becomes $ g(g(n))\equal{}5n$

And this is a very classical equation whose general solution is :

Let $ E$ be the set of all non zero integers which are not divisible by $ 5$.
Let $ A$ and $ B$ two equinumerous subsets of $ E$ such that $ A\cap B\equal{}\emptyset$ and $ A\cup B\equal{}E$ and let $ h(x)$ a bijection from $ A\to B$.

Then $ g(x)$ is defined as :

$ g(0)\equal{}0$
$ \forall x\neq 0$ :
If $ x5^{\minus{}v_5(|x|)}\in A$ : $ g(x)\equal{}5^{v_5(|x|)}h(x5^{\minus{}v_5(|x|)})$
If $ x5^{\minus{}v_5(|x|)}\in B$ : $ g(x)\equal{}5^{1\plus{}v_5(|x|)}h^{\minus{}1}(x5^{\minus{}v_5(|x|)})$


MIT college videos algebra function domain

What would you do if you would have one free year, without any duty at all? I mean if you would have enough money to live from.. What would you do?

Well if nothing happened to me, like no weight gaining or heart attacks or strokes, I would just play games all day and eat stuff.

Well I'll buy everything I want because I have infinite money...

Behold the philosophical nexus between prison and retirement.

-Watch every movie that comes up
-Spend time in University bookshop (and buy books)
-Go on Youtube a lot
-Very little sleep
-Eat food I normally won't it (because I normally won't have enough money)
-Facebook a lot

I don't know. Probably just all the things I love doing.

[quote="t0rajir0u"]Visit mathematicians at various universities.  :)[/quote]

That sounds neat too actually  :D . I would probably be visiting Chemists though (like Professor Schrock at MIT). Of course assuming that they won't mind  :P .

obtain all my technical needs regarding education n then do some research in math  :D

1 free year. Assuming my age remained constant it would be training for the IMO. I would love to go soo much!

If my age was affect then probably doing more and more advanced maths

maybe vacation or stay home playing games and watching movies :D

Personal math research, eat, sleep, basketball, gaming, youtube. Put it in any order you want, and you can't go wrong  :lol:

You make it sound like taking a free year is some kind of imaginary thing. A lot of people go abroad and do nothing useful for half a year.

Given unlimited funds, I'd travel, practice languages, practice instruments, and walk a lot. I'd probably hire a guide or find very learned friends to teach about wherever I am.

I would buy a tank and drive it on the street as my car(loaded of course). Also I would travel around the world, buy a Farrari and buy the biggest house in the world. After I did all this I would spend my time partying! WOOT

I'd have the ability to go where I'd want, and gain ahold of every possibility. For example, I would buy the books that are new and I'm interested in and read them in the airplane on a ride to wherever I want (probably England). Listening to music (while obtaining it from iTunes), using the laptop, on business class so things can run smoothly, etc. However, I wouldn't be pompous and rent a [i]mansion[/i] for a two-night stay. Something good enough that can withstand a kid, adult, whatever age I'd be by the time I get this free year.

study real hard and apply for MIT, study harder there so I can graduate in a month, buy a huge piece of land, hire a hundred workers to make 20 Earthships there, and 10 container homes. Then get all my relatives to those homes, and we can be a clan!

Assuming that I get to keep the infinite money, buy my way out of school after I'm done.  Read a textbook once or twice during the year so I don't make myself feel dumb, and go form my own new nuclear power nation with the money.  Hopefully it won't become the next North Korea...hopefully.

I'd probably be really happy, compared to now.
Well, (video games, computer, TV) ^10000000000000
I'd also buy an Xbox360 and play Halo all day after beating Brawl
Probably wouldn't have to buy out of school, just ditch...every day
Buy my way into college sounds fun
Buy a car
Buy my way into a driver's license
Become the next President
Blow up Iraq and North Korea....just kidding, lol
THen, I'd flame out after 1 year
oh well

An awful lot of computers. A lot of AoPS, games, and programming. No, really. Enter some singing contests, maybe... Junk food. I can't remember the last time I ate a french fry... oh yeah, the day before yesterday I got two.

Maybe when I get bored of all that, I'll try to accomplish my goal for the year.

Oh yeah, I forgot... buy myself a web domain and begin learning CGI. Learn how to make Firefox addons and themes. Buy a laptop and go back to Paint.NET. Buy Deluxe version of every Popcap game available, write some interesting software, read a lot of books, write a lot of books, and so on. Wonder how that might fit into one year...


Trying to solve a somehow related problem - to find chains of length larger than 6 containig only primes such that the next prime is the product of the digits of the previous prime plus one - I put myself the following question:

Start from any positive integer and insert as many digits of 1 as you want between the digits of the original number.

After checking some of 2 or 3 digits composite number I conjecture that this is possible every time, but I have no other ideas - maybe I'm not the first that thought about this.

Does anyone have some ideas?

[quote="ZetaX"]What's the question¿
Sorry, I don't undestand what you ask for.[/quote]

I mean that if you insert between any two original digits of any integer as many digits of 1 as you want - as well as none - at some time you will get a prime number. You are allowed to put 1's at the beginning and as well at the end.
Trivial this is true for all the primes.
For example if you take 25 you will add an 1 at the end and 251 is such a prime.
For 80 you have 1801 as a prime.
For 339 one may choose 31319 and so one - I hope it is more understoodable now!


[quote="Treething"]two circles, circles A and B, are drawn. If circle A is 69% larger than circle B, how many times longer is circle A's circumference than circle B's circumference?[/quote]

[hide]$\sqrt{\frac{100+69}{100}}=\boxed{1.3}$[/hide]

[quote="moogra"][quote="Treething"]two circles, circles A and B, are drawn. If circle A is 69% larger than circle B, how many times longer is circle A's circumference than circle B's circumference?[/quote]

[hide]
root 169/10 = 1.3[/hide][/quote]
Did you mean [hide]169/100[/hide]?


How many of you are applying to Ross?  Currently, I'm working on the application problems with 7/11 of them successfully solved.

Any comments on Ross would be appreciated if you are an alumnus etc.

I was there last year, and will most likely be returning.  It is a great program, and if you have any questions feel free to ask me.

zscool, how many of the 11 problems do you think you have to solve to get into the program?

I do not think that the point of the questions is to determine who is better at problem solving, but to see how one approaches different types of problems.

try your best, and do as many as you can, and for ones that you aren't sure about, write something (as in what you tried to do, or how you think it may be solved, etc.).  Just remember that the questions are not there to sieve out the weaker people, only to see how you think, and get inside your head.


Υπολογίστε το

$\displaystyle {\int_{0}^{\frac{\pi}{2}}} \frac{1+\cos^{n}x}{2+sin^{n}x+cos^{n}x} dx$ όπου $n>0$

 :)

Νομίζω θέτοντας

      $x = \frac{\pi}{2} - u$ .
 προσθέτουμε τα αρχικό και το τελικό ολοκλήρωμα. Έτσι 2Ι = .... π/2 κλπ

Θα το ξαναδώ !

 Μπάμπης

Mου  φαίνεται  ότι  είναι   γνωστή  μέθοδος  για  τον  υπολογισμό  ορισμένων  ολοκληρωμάτων  με  τριγωνομετρικούς  όρους .  Όμως  ισχύει  μονάχα  για ν  περιττό

   $I= \displaystyle {\int_{0}^{\frac{\pi}{2}}} \frac{1+\cos^{n}x}{2+sin^{n}x+cos^{n}x} dx$  (1)

   τότε  θέτοντας   $u =\frac{\pi}{2} - x$  έχουμε  $du = - dx$   ενώ  αλλάζουμε  τα  άκρα  $u_1 = 0$
    και  $u_2 = \frac{\pi}{2}$

        οπότε  ${J = \displaystyle {\int_{0}^{\frac{\pi}{2}}} \frac{1+\cos^{n}({\frac{\pi}{2}-x})}{2+sin^{n}(\frac{\pi}{2}-x)+cos^{n}(\frac{\pi}{2}-x)}}dx$

   $\displaystyle {\int_{0}^{\frac{\pi}{2}}} \frac{1+\sin^{n}x}{2+sin^{n}x+cos^{n}x} dx$  (2)

  Προσθέτοντας  τις  (1)  και (2)  έχουμε  ότι  ${\int_{0}^{\frac{\pi}{2}}} 1 dx = \frac{\pi}{2}$   αν όμως  ο   n είναι  άρτιος  δεν  εφαρμόζεται  η  μέθοδος ότι  δηλαδή  I = J

George,

 Τι νομίζεις ότι αλλάζει αν βάλουμε τον $n$ περιττό ;

   Μάλλον τίποτα. Αλλά ξαναδέστο πάλι και γράψε. Θα το ψάξουμε αν χρειαστεί.

Μπάμπης

 PS. Οι υποψήφιοι της Γ΄Λυκείου να προσέξουν τη μέθοδο που περιάγραψε πιο αναλυτικά ο Gdeskos. Αν δεν την ξέρεις από πριν , πάει το θέμα , χάαθηκε !!!!!!!!!!!!!!!!

Λοιπόν  κύριε  Στεργίου  έχετε  δίκιο μάλλον  εγω έγραψα  βιαστικά  από  το  σχολίο ( μάλλον  για  αυτό  είπα  ότι μου  φαίνεται ότι  βρήκα  κάτι )  . Οντως  Ι = J  για  κάθε  ν φυσικό  αριθμό  . Λοιπόν τι  λέτε  δεν  συνεχίζουμε  στο  ελληνικό  Φόρουμ  και  με  ανάλυση ........Λέω  να  κάνω  την  αρχή

Καλό!!!!. Μη βάζετε όμως σε φροντηστηριακό στυλ προβλήματα . Μου τη σπάει...

Σιλουανέ  όντως  είναι  σε  φροντιστηριακό στύλ, αλλά  αν  δεν  την  ξέρει  κάποιος την  μέθοδο  αυτή  δύσκολα  βγαίνει  η άσκηση.


linear algebra
matrix
algorithm
number theory
relatively prime
linear algebra unsolved

Let R be an Euclidean domain.  If the Elements $a_{11}, a_{12}, \ldots, a_{1n} \in R$ of an $n\times n$-matrix over $R$ are relatively prime, then you can choose $a_{ij} \in R, 2 \leq i \leq n, 1 \leq j \leq n$ such that the matrix is invertible.

already posted. One idea is to apply Euclid algorithm on the first line to obtain $(1,0,..,0)$, and then to work backward...

Is it this thread: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=euclidean&t=44914
or is there another one?


Edited  Later: -- Catching up on  my sleep :)  ... (Some how while  one is   sleeping, one may think "sum" and "average" means the same ...:) ..

[quote="Gyan"][quote="socrates"]The sum of two positive reals is less than their product.
Show that their sum exceeds 4.[/quote]

Hmm... $1+1 < 1 \cdot 1$  but    "sum of 1 1 does NOT exceed 4" .    ;)
(Now if you said 2 then, that would  make sense...( :!:[/quote]

did you have enough sleep last night? :)

so you're saying $2<1$!!!??


geometry
combinatorics proposed
combinatorics

A square of side 1 is decomposed into 9 equal squares of sides 1/3 and the one in the center is painted in black. The remaining eight squares are analogously divided into nine squares each and the square in the center is painted in black. Prove that after 1000 steps the total area of black region exceeds 0.999[/b]

After each step 1/9 of the white area is painted black so the total area of the black part after n steps is 1-(8/9)^n

(8/9)^1000=7.038*10^-52 so 1-(8/9)^1000 > 0.999

So we need white area <0.0001, or 1/10,000. (8/9)^1000=((8/9)^250)^4 ? (1/10)^4, or (8/9)^250<1/10 to prove. The closest would be (8/9)^20 which is less than 1/10, so we are done (reasoning by @above for how the (8/9)^k was obtained), but I don’t know how to do this plausibly in a contest.


logarithms
combinatorics proposed
combinatorics

Let $f(n)$ be the smallest number of 1s needed to represent the positive integer $n$ using only 1s, $+$ signs, $\times$ signs and brackets $(,)$. For example, you could represent 80 with 13 1s as follows: $(1+1+1+1+1)(1+1+1+1)(1+1+1+1)$. Show that $3 \log(n) \leq \log(3)f(n) \leq 5 \log(n)$ for $n > 1$.

so is something like $(1+1+1)(1+1)+(1+1)(1)$ allowed?

[hide="hint for the upper bound (which can probably be greatly improved)"]Let $a_0=n, a_{k+1}=[\frac {a_k}3]$ for $k=1,...,[\frac{\log n}{\log 3}]$ and $b_k:=a_k-3a_{k+1}$.[/hide]

Actually, this is called The (Mahler-Popken) complexity of $n$.