MATHCOUNTS AMC AIME ARML calculus USA(J)MO USAMO

Since these are problem solving resources, I figure I might as well post it here:

[size=200]THE LIST OF ALL PROBLEM SOLVING MARATHONS ON AoPS!![/size]

Here they are, sorted by forum:

[size=150]Middle School:[/size]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38673]Basics Forum Marathon[/url] - Posts: 85 , Status: still going on
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39305]MOEMS Marathon[/url] - Posts: 43 , Status: still going on

[size=150]Mathcounts:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=11174]Difficult Question Marathon[/url] - Posts: 4 , Status: ended Mar 16, 2004
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=11084]Marathon #1[/url] - Posts: 162 , Status: ended May 05, 2004
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36158]Marathon #2[/url] - Posts: 277, Status: still going on

[size=150]Getting Started:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=11091]Marathon #1[/url] - Posts: 100 , Status: ended Dec 22, 2004
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=33505]Marathon #2[/url] - Posts: 965 :first: , Status: still going on

[size=150]Intermediate Topics:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36316]Proof Marathon[/url] - Posts: 52, Status: still going on
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36254]Intermediate Marathon[/url] - Posts: 146, Status: still going on

[size=150]Other Problem Solving Topics:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=37312]Puzzle/Brainteaser Marathon[/url] - Posts: 160  Status: still going on
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39010]Criptarithm Marathon[/url] - Posts: 26 Status: still going on
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40762]Chess Marathon[/url] - Posts: 17 Status: still going on

[size=150]Contests and Programs:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=11655]AMC Marathon[/url] - Posts: 2 , Status: ended Oct 30, 2004
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=21418]AIME Marathon[/url] - Posts: 101 , Status: still going on
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=40531]ARML Marathon[/url] - Posts: 7 , Status: still going on

[size=150]College:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36321]Calculus Marathon #1[/url] - Posts: 19 , Status: still going on
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=38866]Calculus Marathon #2[/url] - Posts: 1 , Status: still going on

[size=150]Arts and Sciences:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36546]Chemistry Marathon[/url] - Posts: 110 , Status: still going on
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=37184]Physics Marathon[/url] - Posts: 73 , Status: still going on

actually, you should get this stickied on a more general forum, like Round Table or Site & General Informatoin Forum

Yeah, that's true, but I have a very good reason for posting it here:

I already posted it here and only a mod can move threads.

[size=150]Mods, I'd appreciate it if you move this thread...[/size]

I can move it, but I'm not sure of where would be the best place for it. What do people think?

I'm not sure actually. This thread doesn't seem to fit anywhere. Maybe one of the Site and General Information forums?

This is probably best in "Other Problem Solving Topics", though I'm not sure I have a good reason why.  I'll stick it there.

Many of the marathon has been posted by this guy I know.

[hide="Yay"]
Me[/hide]

Haha.. :D

Me too, me too   :clap2:  :gleam:

Only one and I hope it will grow big and bigger and huge...:harhar:

There's a Middle School Classroom marathon now, too :P

The [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42194]Classroom Marathon[/url] in the middle school forum has 222 posts as of 8/2/05 and is still going.

There's even a games and fun factory marathon, but I doubt you'd want to include that, because it's all spam (I regret starting it).

Marathon #2 in MATHCOUNTS is now officially dead. Also, there is a MATHCOUNTS Target Round Marathon :)

Yeah, this should be updated every few weeks.

[size=200]THE [b] UPDATED[/b] LIST OF ALL PROBLEM SOLVING MARATHONS ON AoPS!![/size]
[hide="(this version by me@home)"]

MORE GROUNDRULES
1) [b][color=red]red indicates section I am working on,[/color] [color=blue]blue indicates completed section[/color][/b]

2) Any necessitated addition to a topic title enclosed in [brackets]

3) Topics with less than 20 posts (less than one page) with the title "marathon" do not count as marathons

4) Topics ranked:
- firstly by group, that is: status: active, status: questionable, status: ended
- secondly by number of posts
- in case of a tie in the above two, they are thirdly ranked alphabetically

[/hide]

___________________________

Here they are, sorted by forum >> subfora >> status:

[color=blue][size=184]Middle School:[/size]

[size=150]Classroom Math:[/size]
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=42194]Classroom Marathon[/url] - Posts: 781 Status: ended Tue May 30, 2006

[size=150]MOEMS:[/size]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39305]MOEMS Marathon[/url] - Posts: 524, Status: ended Sat Oct 07, 2006

[size=150]Mathcounts:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=104081]MathCounts Problem Solving Marathon...2![/url] - Posts: 1010, Status: still going

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36158]MC Problem Solving Marathon! [Locked][/url] - Posts: 1814, Status: ended Sat Jul 29, 2006
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38673]Basics Forum Marathon[/url] - Posts: 553, Status: ended May 5, 2006
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=107008]MATHCOUNTS Target Round Marathon[/url] - Posts: 168, Status: ended Sat Sep 23, 2006
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=11084]MATHCOUNTS marathon...[/url] - Posts: 163, Status: ended May 05, 2004
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=46626]MathThonI--READ THIS[/url] - Posts: 73, Status: ended Fri Aug 05, 2005

[size=184]High School:[/size]

[size=150]High School Basics (aka Getting Started):[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=47851]Problem Solving Marathon 3[/url] - Posts: 2021, Status: still going

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=11091]Problem Solving Marathon[/url] | [url=http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=2274]PDF archive by Palytoxin[/url] - Posts: 101, Status: ended Wed Dec 22, 2004

[size=150]Intermediate Topics:[/size]
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=33505]Marathon #2[/url] - Posts: 2009 , Status: still going on
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70276]The Intermediate Geometry Marathon[/url] - Posts: 233 , Status: still going on
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=126345]The Intermediate Complex Numbers and Trigo Marathon[/url] - Posts: 59, Status: still going on
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=126370]Intermediate Combinatorics/Probability Marathon[/url] - Posts: 34 Status: still going on

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84848]Proof Contest/Marathon[/url] - Posts: 71, Status: coming to an end (?) last post Fri Jan 12, 2007
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=127316]Geometric Constructions[/url] - Posts: 29 Status: coming to an end (?) last post Fri Jan 12, 2007

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36254]Intermediate Problem Solving Marathon[/url] - Posts: 225, Status: ended Sep 5, 2005
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=46950]Intermediate Forum AIME Marathon[/url] - Posts: 199 Status: ended Wed Feb 08, 2006
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=73967]AIME Marathon   [II][/url] - Posts: 187 Status: ended Thu Apr 13, 2006
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=85929]Which is bigger? - the marathon[/url] - Posts: 140 Status: ended Fri Jun 09, 2006
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=70984]Intermidiate chess marathon (started!)[/url] - Posts: 71 Status: ended Sat Jan 28, 2006
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36316]Proof Marathon[/url] - Posts: 53, Status: ended May 29, 2005
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=91494]Intermediate marathon[/url] - Posts: 41 Status: ended Fri Jul 07, 2006
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=91306]Root finding marathon[/url] - Posts: 33 Status: ended Fri Jun 16, 2006
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=126461]AIME Marathon[/url] - Posts: 26 Status: ended Sun Dec 31, 2006
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=70338]Algebra, nt, and polynomials marathon #2[/url] - Posts: 23 Status: ended Fri Jan 27, 2006[/color]

[color=red][size=150]Other Problem Solving Topics:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=37312]Puzzle/Brainteaser Marathon[/url] - Posts: 422 Status: still going on??
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39010]Criptarithm Marathon[/url] - Posts: 53 Status: ended May 9, 2006
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40762]Chess Marathon[/url] - Posts: 169 Status: ended August 17, 2005
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85261]Lateral Thinking Puzzle Marathon[/url] - Posts: 63 Status: still going on

[size=184]Contests and Programs:[/size]

[size=150]American Mathematics Competitions:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=11655]AMC Marathon[/url] - Posts: 3 , Status: ended Oct 30, 2004
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=21418]AIME Marathon[/url] - Posts: 13, Status: ended November 9, 2005
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=40531]ARML Marathon[/url] - Posts: 52 , Status: ended April 14, 2006

[size=150]College:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36321]Calculus Marathon #1[/url] - Posts: 20 , Status: ended May 28, 2005
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=38866]Calculus Marathon #2[/url] - Posts: 2 , Status: ended May 25, 2005

[size=150]Arts and Sciences:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36546]Chemistry Marathon[/url] - Posts: 393 , Status: still going on
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=37184]Physics Marathon[/url] - Posts: 112 , Status: ended October 13, 2005
[/color]

[size=200][hide="Version Before Me"]
[/size][size=100](for records purposes in case of edits etc.)

_____________________________

[/size]
[quote="lotrgreengrapes7926"][size=200]THE [b] UPDATED[/b] LIST OF ALL PROBLEM SOLVING MARATHONS ON AoPS!![/size]

Here they are, sorted by forum:

[size=150]Middle School:[/size]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39305]MOEMS Marathon[/url] - Posts: 443, Status: still going on

[size=150]Mathcounts:[/size]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38673]Basics Forum Marathon[/url] - Posts: 553, Status: ended May 5, 2006
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=11174]Difficult Question Marathon[/url] - Posts: 5 , Status: ended Mar 16, 2004
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=11084]Marathon #1[/url] - Posts: 163 , Status: ended May 05, 2004
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36158]Marathon #2[/url] - Posts: 1802, Status: still going on

[size=150]Getting Started:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=11091]Marathon #1[/url] - Posts: 101 , Status: ended Dec 22, 2004
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47851]Marathon #3[/url] - Posts: 1714, Status: still going on

[size=150]Intermediate Topics:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36316]Proof Marathon[/url] - Posts: 53, Status: ended May 29, 2005
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36254]Intermediate Marathon[/url] - Posts: 225, Status: ended Sep 5, 2005
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70276]The Intermediate Geometry Marathon[/url] - Posts: 196 , Status: still going on
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84848]Proof Contest/Marathon[/url] - Posts: 52, Status: still going on
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=33505]Marathon #2[/url] - Posts: 1892 , Status: still going on

[size=150]Other Problem Solving Topics:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=37312]Puzzle/Brainteaser Marathon[/url] - Posts: 422 Status: still going on??
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39010]Criptarithm Marathon[/url] - Posts: 53 Status: ended May 9, 2006
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40762]Chess Marathon[/url] - Posts: 169 Status: ended August 17, 2005
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85261]Lateral Thinking Puzzle Marathon[/url] - Posts: 63 Status: still going on

[size=150]Contests and Programs:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=11655]AMC Marathon[/url] - Posts: 3 , Status: ended Oct 30, 2004
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=21418]AIME Marathon[/url] - Posts: 13, Status: ended November 9, 2005
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=40531]ARML Marathon[/url] - Posts: 52 , Status: ended April 14, 2006

[size=150]College:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36321]Calculus Marathon #1[/url] - Posts: 20 , Status: ended May 28, 2005
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=38866]Calculus Marathon #2[/url] - Posts: 2 , Status: ended May 25, 2005

[size=150]Arts and Sciences:[/size]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=36546]Chemistry Marathon[/url] - Posts: 393 , Status: still going on
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=37184]Physics Marathon[/url] - Posts: 112 , Status: ended October 13, 2005[/quote]
[/hide]
[size=117]

I hope others take up this organizational "quest" to clear up the boards...[/size]

erm. there was (and still is, hopefully) a bay area marathon
in the bay area forum

i still have hope, but it looks on the verge of death.

Update:
The Bay Area Marathon is alive and well, although it has since its inception outgrown its original "Tough AMC/ Medium AIME" title.

Umm, I am the author of a marathon i did not see there. It's a mathcounts marathon and it has like 100,000 (no exaggeration) views. Do i win first place?


logarithms
real analysis
real analysis solved

compute the sum of teh following series:
                    n+1
:Sigma:  _________
               n* (2^n)

i think the result is 1+ln 2
(the sum of the first 200 terms is about 1+ln 2. i found it using a pascal program to compute the sum), but i don't know how to prove the result

Do you mean $\sum_{n\geq 1}\frac{n+1}{n2^n}$?

well, this is indeed trivial. we know by the power series of the logarithm that $\ln 2=-\ln \frac 12=\sum_{n\geq 1} \frac 1{n2^n}$. on the other hand, $1=\sum_{n\geq 1}2^{-n}=\sum_{n\geq 1} \frac{n}{n2^n}$. add both to get your result.

It was so trivial that I asked whether I correctly understood what he meant.  :lol:  :lol:  :lol:


Let $ x$ be an [i]integer[/i]. What is the smallest possible value of $ |2x \minus{} 7| \plus{} |2x \minus{} 9| \plus{} |2x \minus{} 11| \plus{} |2x \minus{} 13| \plus{} |2x \minus{} 15|$.

I really have now forml way of solving but after using common sense x can either be 4 or 7 to get a minimum sum of 17

[hide="Solution?"]This expression is symmetrical about $ |2x-11|,$ so the best sum is when $ 2x=11.$ This would give an overall sum of $ 0.$ However $ x$ must equal $ 5.5$ for that to happen, and $ x$ must be an integer. We want to make the middle term as small as possible, which happens when $ x=5,6.$ Plugging each in gives the one answer \begin{align*}|2\cdot5-7|+|2\cdot5-9|+|2\cdot5-11|+|2\cdot5-13|+|2\cdot5-15|&=3+1+1+3+5\\&=\boxed{13}.\end{align*}[/hide]

[hide]the first thing i noticed is that 2x must be as close to 11 as possible since 2x-11 is the middle number. as c can not be 5.5, it must be 5, or 6. it does not mattewr which one we choose because there are 2 lesser values, and 2 greater values (by lesser i mean (2x-13), and (2x-15) and by greater i mean (2x-7) abd (2x-9))
but if that doesn't make sense, just try both cases
choosing 5 gives us 3+1+1+3+5= 13
choosing 6 gives us 5+3+1+1+3= 13
therefore the answer is 13[/hide]

well we can see that 2x has to be close to 11

because 2x-11 is  the median

so we can try 5 and 6 will hav the same answer let us try 5 because it is smaller.

l10-7l+l10-9l+l10-11l+l10-13l+l10-15l
3+1+1+3+5=13


superior algebra
superior algebra unsolved

Please show that free group there is a unique homomorphism
and
Please proof  , thank you very much

[b]theorem 1[/b]   Let (G,a) and (G',a' ) be free groups on the set S. Then there is a unique isomorphism ( in the category of groups) from G to G'
[b]
theorem 2 [/b]  For any set S , ther exists a free group on S

We need it as soon as possible
THANK YOU VERY MUCH

1. the first sentence makes no sense.
2. @theorem 1: let $G$ be the free group on $2$ generators $x,y$. Then there are two essentially different automorphisms given by $x \to x , y\to y$ and $x \to y , y\to x$.
3. we are not there to make your homework! Post your attempts etc., but don't expect us to do your stuff.

Thank you for let me know that.
But there's something i want you to know too

1.thank you for your answer but it's not help too much

2.we first study this subject with Italian teacher and he gave us a paper of theorems and said "next week you have to teach these theorems to your friends" so we have to study by ourselves , [b][color=red]but English is not my MOTHER LANGAUGE[/color][/b]  it's difficult to understand,OK?

we didn't want you to do our homework for us, we know you guys are not here to do somebody's homework, just an advice that we can use to understand the way to proof or if someone can proof them for us it will be helpful

we didn't post it, waiting for the answer and go sleep , we try to proof it too (Now it's 3.38 am in our country)

3.now we can proof theorem 1, if somebody want to help us , just  2nd theorem

I still don't understand how theorem 1 can be correct, see my post above.
For the second one: this is the group of finite words with letters from $S$ (and their inverses), connected by simple concatenation. And this exists.


geometry
circumcircle
parallelogram
geometry proposed

Let ABCD be a cyclic quadrilateral.Let the orthocenters of BCD,CDA,DAB,ABC be A1,B1,C1,D1,
respectively.prove that quadrilateral A1B1C1D1 is congruent to ABCD. :D

Using the fact that in any triangle the distance from the orthocenter to a vertex is twice the distance from the circumcenter to the opposite side, it follows that $\overline{BA_1}$ and $\overline{B_1A}$ are parallel and equal to twice the distance from $O$ to $ DC$ (O is the circumcenter of ABCD). Hence, $B_1BA_1A$ is a parallelogram $\Longrightarrow$ $\overline{ A_1B_1} = \overline{AB}.$ By similar reasoning, we have that $\overline{ B_1C_1} =\overline{ BC},$ $ \overline{C_1D_1} = \overline{CD}$ and $ \overline{D_1A_1} = \overline{DA}.$ Likewise, $ C_1ACA_1$ is parallelogram $\Longrightarrow$ $\overline{AC} =\overline{ A_1C_1}$ and  $\overline{ BD }= \overline{B_1C_1}$ $\Longrightarrow$ Quadrilaterals $ ABCD$ and $ A_1B_1C_1D_1$ have congruent sides and diagonals.


A metric space $ X$ is called totally bounded if given $ \epsilon > 0$ there exist $ x_{1},...,x_{n}\in X$ st : $ X \equal{} \bigcup_{i \equal{} 1}^{n}B(x_{i},\epsilon)$

for $ k \in\mathbb{N},  (X_{k},p_{k})$ is a metric space

let $ Z \equal{} \prod_{k \equal{} 1}^{\infty} X_{k}$

 for $ x \equal{} (x_{k}), y \equal{} (y_{k})\in Z$, define

$ p(x,y) \equal{} \sum_{k \equal{} 1}^{\infty}\frac {1}{2^{k}}[\frac {p_{k}(x_{k},y_{k})}{1 \plus{} p_{k}(x_{k},y_{k})}]$

then we know that $ (Z,p)$ is a metric space.

prove $ Z$ is totally bounded if $ X_{k}$ is totally bounded for all $ k\in\mathbb{N}$


I just learned Newtons sums today and I tried all the practice problems in the AoPs book. Does anyone have anymore problems or sources that are good practice problems for Newton's Sums? I'm still quite slow at solving them, so I need more practice. If anyone has such problems, please tell me and that would be great!

look at this problem, and use newton's sums

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=401739#401739[/url]

my solution at the bottom uses them...


quadratics
invariant
symmetry
number theory unsolved
number theory

Find all pair of positive integers $(x, y)$ satisfying the equation

\[x^2 + y^2 - 5 \cdot x \cdot y + 5 = 0.\]

$x^2 - 5xy + y^2$ is invariant under $(x,y) \to (y,5y-x)$, and one can use this to reduce to a minimal solution, then reverse the process by iterating the inverse transformation.  Of course this gives the same recurrence as for solutions of the Pell equation.

Is this method you state an easier one, fleeting_guest? If it is, could you give a detailed explanation or any site which could explain this? Thanks.

To write down the transformation I gave, consider the equation as a quadratic in $x$ and ask what is the formula for the second root in terms of the one ($x$) that is already known.  Then use the symmetry of $x$ and $y$.

Suppose  $x \geq y > 0$.  Then, for $x > 5$.
(a) $5y-x > 0$ (our transformation conserves positivity of the solutions). Indeed, $x(5y-x) = y^2 + 5 > 0$.
(b) $y \geq 5y-x$ (the transformation reduces solutions).  Indeed, $ yx >= y^2 = (5y-x)x - 5$ so that $y > 5y-x - 1$ whenever $x > 5$.

So the minimal solution, if one exists, has $5 \geq x_1 \geq y_1> 0$. Then, one can reverse the above transformation to generate the sequence of positive solutions.  The result will be a sequence $(a_0, a_1, a_2, a_3, a_4, ... )$ where $a_0 = y_1$, $a_1 = x_1$, and $a_n$ satisfies a second-order linear recurrence, and the positive solutions are $(x,y) = (a_{n+1}, a_n)$.   Of course, one has to check whether a minimal solution exists, in this case $(2,1)$, and then $a_{n+2} = 5a_{n+1} - a_n$ producing $(2,9)$, $(9,43)$, $(43,206)$, etc  as the solutions.

[quote="ak007"]
(1) \[ s^{2} - 21y^{2} = -20 \]
... To get the other solutions we solve the equation
(2) \[ \[s^{2} - 21y^{2} = 1 \]

Thus the solutions can be found by
$s_{1} = 55 + 12 \times 21 = 307$
$y_{1} = 12 + 55 = 67$

$s_{p} = 55s_{p-1} + 252y_{p-1}$
$y_{p} = 12s_{p-1}+55y_{p-1}$
$x_{p} = \frac{5y \pm \sqrt{21y_{p}^{2}-20}}{2}$[/quote]

I don't see the connection between (1) and (2) and I don't know how can we get the solutions of (1) from (2) because clearly what ak007 shows next is the recurrence solution of this Pell's equations rather than (1)- can somebody explain it to me ?

[quote="borz"][quote="fleeting_guest"] in this case $(2,1)$, and then $a_{n+2} = 5a_{n+1} - a_n$ producing $(2,9)$, $(9,43)$, $(43,206)$, etc  as the solutions.[/quote]
What about (3,1) , (14,3)...?[/quote]

It is the same sequence $a_n$, extended to $n < 0$.

You Can Also See The Answer In The Following Link:

[code]http://www.irysc.com/forum/t6886-post90843/#post90843[/code]
Post #49 Is An Answer And Also Post #50 Is Another Answer ;)


calculus
integration
number theory
relatively prime
prime numbers

Show that if m is a positive rational number then [tex]m + {{1}\over{m}}[/tex] is an integer only if m = 1.

Source: Five Hundred Mathematical Challenges (E.J. Barbeau, M.S. Klamkin, W.O.J. Moser), problem 23

Good Idea, but you can make it look a bit more elegant

Hint: [hide]Let m=a/b,where a and b are relatively prime,  you must show a/b+b/a can not be an integer[/hide]

EDIT- I misread the expression as 1+1/m before.

JS1527 wrote:I don't know how to write proofs, but here goes:
[hide]
(1) If m>1, then it would always be a mixed number, with 1/m being the non-integral part.
(2) If m=1, then it would be 1 + 1/1 = 2
(3) If 0<m<1 then it would be (fraction)+ 1/(fraction).  If it was a unit fraction then 1/(fraction) would equal an integer, but the (fraction) term would always be there, making m + 1/m not be an integer.
[/hide]

Good start, but there are two things to note.

1. m can be any rationtal number

2. The sum of two fractions can be an integer (e.g., 4/5+1/5).  If m is a rational number, then m+1/m is the sum of two fractions except in special cases.  Therefore, it is not sufficient to prove that at least one of the two terms is a fraction.

[hide]

First, we know that m + (1/m) = (m^2 + 1)/m.  Thus, m must divide m^2 + 1.  m always divides m^2, so we need m to divide one.  Thus m must be one.

[/hide]

I think nr1337's idea does work, but it's a little shaky and will require some casework.  I'm not sure what beta is trying to hint at, but here's a different, hopefully beginner-friendly hint:

[hide]Since integers except 1 will obviously not work, let m=p/q, where p and q are integers, and the fraction is fully reduced.  WLOG (Without loss of generalization), p > q, since switching p and q around does not change the sum.  Let p :equiv: x (mod q), meaning (essentially) that when we divide p by q, our remainder is x.[/hide]

Edit: some of this is the same as beta's hint, now that he has edited his post.

I was trying to hint at nr1337's idea before, then I realized that m doesn't have to be an integer.

New try:

[hide]

Let m=a/b, where a and b are positive integral values.  If a/b > 1 then m + 1/m = a/b + b/a.  This means that you are adding a number and its recripocal.  Since a/b + b/a = (a^2 + b^2)/ab, this irreducible since you can't factor a^2 + b^2 over the reals.  Only when ab=1 and a=b=1 does this reduce.  If a does not equal b, it can't reduce.  Therefore it will not be an integer (the expression m + 1/m). [/hide]

JS1527 wrote:New try:
[hide]
Since a/b + b/a = (a^2 + b^2)/ab, this irreducible since you can't factor a^2 + b^2 over the reals.  [/hide]

This argument doesn't quite work, because we can have, for example,

[hide]a+2b | a^2 + b^2 if a=3, b=1, and a+2b is indeed a "factor" of a^2 + b^2.  Hence, a^2 + b^2 doesn't need to have a generalized factorization.  ab might be a factor of a^2 + b^2 in certain special cases.  You're definitely on the right track though.[/hide]

Yet another try:

[hide]

Let m = a/b.  We want a/b + b/a = (a^2 + b^2)/ab to be an integer.  Doing some long division gives us remainders of (b-1)a^2 + b^2, (b-1)a^2 + (a-1)b^2, (b-2)a^2 + (a-1)b^2, (b-2)a^2 + (a-2)b^2, etc.  In order for one of these to be zero, a and b must either differ by one or be equal.  a/(a+1) + (a+1)/a = 2 + (1/(a^2 + a)), which is not an integer, so a must equal b, so m must be 1.

[/hide]

nr1337 wrote:Yet another try:
[hide]
(b-1)a^2 + b^2
[/hide]

Getting closer, but not quite true:

[hide]

Using my favorite example

a = 1, b = 3

(a^2+b^2)/ab = 3 remainder 1

((b-1)a^2 + b^2)/ab = (2*1^2 + 3^2)/3 = 3 remainder 2

[/hide]

Here's another hint, but it might give away a cool solution that requires a little knowledge about relatively prime numbers:

[hide](a-b)^2 or (a+b)^2[/hide]

Reviving this because no one has solved it yet, and it's good practice.

I guess I'll post a solution, but try to do it yourself:

[hide]The case where m is an integer is trivial.  Let m = p/q, where p and q are relatively prime.  WLOG, p > q.  Let p :equiv: x (mod q).  p/q + q/p is an integer iff the fractional component of p/q plus q/p is equal to 1.

We thus have

x/q + q/p = 1

x/q = 1 - q/p

x = q - q^2/p

px = pq - q^2

px = q(p - q)

px/(p - q) = q

Since p - q does not divide p, p - q must divide x.

For a certain natural number n, p = nq + x, and so p - q = (n-1)q + x.

(n-1)q + x < or = x iff (if and only if) n = 1; we check this case by plugging in:

px/(p - q) = q

px/(q + x - q) = q

px/x = q

p = q

That's ridiculous!

[/hide]

There's a significantly shorter way, but I'll leave that up to other people.

Hint: [hide]Use my previous hint [/hide]

Since people were so close to solving this, I think someone should try this again (or for the first time).  It's really a pretty good problem that's good practice for beginners.  If people really don't want me to keep reviving this, though, I'll stop.


How does conduction of heat is different from flow of current?

Both those properties of metals - electrical and thermal conductivity - are a direct consequence of the high mobility of their electrons. An electrical current is an ordered flow of charged particles - electrons, protons, or ions. In a metal, an electrical current of electrons can be easily set up because the electrons are loosely held by the metal cations, and so they readily respond to an applied potential difference. In thermal conduction, there is no ordered motion of electrons: an electron collide with a vibrating cation, pick up its energy, and tranfer it to another cation elsewhere in the metal.

i would rather support for an analogy but not too much of it between heat flow and current (resiatance equations,kirchoff's laws..)
but as carcul rightly pointed out the most imp thing here is orderly & disorderly.


Un problema para q piensen  :D
Hallen el valor de:
$ \sum_{i\equal{}1}^{n\plus{}1}(\minus{}1)^{n\plus{}1\minus{}i} i^n{n \choose i\minus{}1}$
nota: eviten operar, usen combinatoria :mad:

[quote="feeurl1"]Un problema para q piensen  :D
Hallen el valor de:
$ \sum_{i \equal{} 1}^{n \plus{} 1}( \minus{} 1)^{n \plus{} 1 \minus{} i} i^n{n \choose i \minus{} 1}$
nota: eviten operar, usen combinatoria :mad:[/quote]
$ \sum_{i \equal{} 1}^{n \plus{} 1}( \minus{} 1)^{n \plus{} 1 \minus{} i} i^n{n \choose i \minus{} 1}\equal{} n!$

No se si este es el tipo de demostración que querías; al principio se modifica ligeramente la expresión:
$ \sum_{i \equal{} 1}^{n \plus{} 1}( \minus{} 1)^{n \plus{} 1 \minus{} i} i^n{n \choose i \minus{} 1} \equal{} \sum_{i \equal{} 0}^{n }( \minus{} 1)^{i} (n \plus{} 1 \minus{} i)^n{n \choose i}$
Luego se consideran los conjuntos:
$ X \equal{} \{\, f: \{1 \dots n \}\to \{ 0 \dots n \}\, \}$
$ X_{k} \equal{} \{\, f\in X\,|\, k\notin f(\{1\dots n \})\,\},\;\;k \equal{} 1\dots n$
Entonces:
$ S_{n} \equal{} \{\, f: \{1 \dots n \}\to \{ 1 \dots n \}\,|\,f(\{ 1 \dots n \}) \equal{} \{ 1 \dots n \}\, \} \equal{} X \minus{} \bigcup_{k \equal{} 1}^{n}X_{k}$
Además, dado $ Y\subset X$ con $ |Y| \equal{} r$ se tiene:
$ |\bigcap_{k\in Y}X_{k}| \equal{} |\{\, f: \{1 \dots n \}\to \{ 0 \dots n \} \minus{} Y\,\}| \equal{} (n \plus{} 1 \minus{} r)^{n}$
Aplicando el principio de inclusión-exclusión obtenemos:
$ n! \equal{} |S_{n}| \equal{} \sum_{i \equal{} 0}^{n}[( \minus{} 1)^{i}\sum_{\substack{Y\subset X \\
|Y| \equal{} i}}|\bigcap_{k\in Y}X_{k}|] \equal{} \sum_{i \equal{} 0}^{n}( \minus{} 1)^{i}(n \plus{} 1 \minus{} i)^{n} { n \choose i}$


algebra
polynomial
Ring Theory
function
calculus
derivative
real analysis

Let $ f\in \mathbb{R}[X_{1},\ldots, X_{n}], \ n \ge 1.$
Explain why $ \# \{x\in \mathbb{R}^{n} \ : \ f(x) \equal{} 0\} \neq \aleph_{0}.$

[b]Remark:[/b]
Please, try to avoid using such 'heavy' results as Tarski-Seidenberg theorem.

Here some alternative proof:

For $ n\equal{}1$, $ f$ decomposes into a finite product of irreducible polynomials in the UFD $ \mathbb R[X]$ which are of the form $ X\minus{}\alpha$ ($ \alpha\in\mathbb R$) and polynomials of degree 2 without any real zero. This shows that the number of zeros of $ f$ is finite (or zero).

For $ n>1$ assume that $ f$ is not constant. (If $ f$ is constant, the claim is trivial to see). Then the gradient of $ f$ does not vanish identically. Pick some $ x_0\in\mathbb R^n$ with $ grad_{x_0} f\neq 0$ and apply the implicit function theorem to see that the set of zeros of $ f$ contains a graph of a differentiable function in a small open neighbourhood of $ x_0$. So $ \#\equal{}card(\mathbb R)>\aleph_0$.

That argument doesn't work. In the multidimensional case, it's possible for the gradient to be zero at the roots, and thus to have finitely many roots; $ x^2\plus{}y^2\equal{}0$ in $ \mathbb{R}^2$ has exactly one root.

You are right of course.  :(  So we are left with the case that $ f$ is irreducible and that every partial derivative vanishes on the set of zeros of $ f$, since the approach I gave works otherwise.

Finally I've taken some rest after exam, and I can go back to this problem.

[quote="Third Edition"]What about using the Noether normalization lemma?[/quote]

Please, could You give some details how this lemma may be helpful to solving this problem? I don't have much experience in using it  :maybe:
Precisely - I learned what is Noether normalization a few days ago.
I know how to prove claim for uncountable algebraically closed field using Noether normalization: in case of algebraically closed field normalization implies that there is a surjective map with finite fibers from set of zeros of polynomial to some affine space; but similarly constructed map over reals doesn't have to be surjective.
I'm especially interested whether this way leads to general result for arbitrary uncountable field instead $ \mathbb{R}$ (I'm not sure that this result holds in general) or must we use somewhere special properties of $ \mathbb{R}$  ?

As to the second approach, I've tried this idea before and I met the same problems. I thought of proving that if $ f$ is irreducible with countably infinite set of zeros, and $ f\nmid g$ then the set of common zeros of $ f,g$ cannot be countably infinite, but I have no idea how to do it.

And some observation, rather inessential to the main problem - in fact we don't need the implicit function theorem to find out, that $ \text{grad}\ f$ is zero in the zeros of $ f$. It suffices to observe that otherwise $ f(x) < 0 < f(y)$ for some $ x,y\in \mathbb{R}^{n}$ and then use Darboux property along curves connecting $ x,y$ to obtain uncountable many zeros of $ f.$


An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:

(a) the selection of four red marbles;
(b) the selection of one white and three red marbles;
(c) the selection of one white, one blue, and two red marbles; and
(d) the selection of one marble of each color

What is the smallest number of marbles satisfying the given condition?

A. 19
B. 21
C. 46
D. 69
E. More than 69

[quote="probability1.01"]Isn't the probability of a) the selection of four marbles equal to one? :P Maybe you mean 4 whites or something.[/quote]

Hehe.. I meant four red marbles. Fixed.  :D

[hide="Answer"]The denominator in every probability is always the same, so we can disregard that.
$r(r-1)(r-2)(r-3)=4(w(r)(r-1)(r-2))=6(w(b)(r)(r-1))=4(rwgb)$
$(r-1)(r-2)(r-3)=4(w(r-1)(r-2))=6(w(b)(r-1))=4(wgb)$
$4(w(r-1)(r-2))=4(wgb)\Rightarrow (r-1)(r-2)=gb$, so$(r-1)(r-2)(r-3)=4(wgb)\Rightarrow 4w=r-3\Rightarrow r=4w+3$
$(r-1)(r-2)(r-3)=6(w(b)(r-1))\Rightarrow (r-2)(r-3)=6(w)(b)$, so $(r-2)(4w)=6wb\Rightarrow \frac{3b}{2}=r-2\Rightarrow r=\frac{3b+4}{2}$
$(r-1)(r-2)(r-3)=(r-1)(\frac{3b}{2})(4w)=4(wgb)\Rightarrow \frac{3}{2}(r-1)=g\Rightarrow r=\frac{2g+3}{3}$
$r=\frac{3b+4}{2}=\frac{2g+3}{3}=4w+3$
$w=2$, $g=15$, $b=6$, $r=11$, $w+g+b+r=34$[/hide]

[b]EDIT:[/b] Okay... no matter how much time I spend looking at it, I can not figure out what I did wrong this time. :huh:

order matters, JesusFreak197

[hide]$\frac{4!}{4!}r(r-1)(r-2)(r-3)=\frac{4!}{3!}r(r-1)(r-2)w=\frac{4!}{2!}r(r-1)wb =\frac{4!}{1!}rwbg$
solving adjacent equations for $r$, we get $r=4w+3=3b+2=2g+1$.  plugging in small values for $w$, we stop at $w=2, r=11, b=3, g=5$ for a total of $21$ or $\boxed{B}$.[/hide]

My solution proceeds similarly to Alan's:

[hide]
We have
$r(r - 1)(r - 2)(r - 3) = 4r(r - 1)(r - 2)w = (4)(3)wbr(r - 1) = 4! wbrg$
Canceling the $r$ from the simultaneous equations and equating the first and second, first and third, and third and fourth parts of the equation, respectively, we have

$r - 3 = 4w$
$(r - 2)(r - 3) = 12wb \Rightarrow r - 2 = 3b$
$r - 1 = 2g$

This leads to $(r - 3)/4 + (r-2)/3 + (r-1)/2 = n - r$, where n is the total number of marbles.  The smallest $r$ that gives an integral value of $n$ is $r = 11$, leading to $n = 21$.

[/hide]


AMC USA(J)MO USAMO search function AoPSwiki


A rectangular prism $p \times q \times r$ where $p,q,r$ are pairwise coprime. This solid is cut into cubes of $1 \times 1 \times 1$. An internal diagonal pierces how many of these cubes?

[quote="worthawholebean"]I don't see how this is solvable. Aren't there many different possibilities for the dimensions of the prism?[/quote]
He wants the formula I think.
[hide="hint"]PIE[/hide]

[quote="bpms"][hide="hint"]he formula I think.
[hide="hint"]PIE[/hide][/hide][/quote]

I don't quite follow. Can you elaborate please?


geometry
geometric transformation
homothety
circumcircle
geometry solved

dears, its a very very well known problem (apollonius problem).
of course i know there are many proofs for it. but i want to know if my solution is true or false.
if there is any new solution i will be glad to see it.
i need u ideas.

[i]the problem:[/i]

we have an angle, and a circle (C) (with the center O) wich is betwin the sides of the angle.
draw a circle wich is tangent to the sides of the circle and circle (C).

[i]the solution:[/i]

first i draw a circle (C') (with the center O') wich the sides of the angle are tangent to it at points A and B.
we can find the angle AO'B. because if we suppose that the angle is w degree, the angle AO'B will be 180-w.
then, we draw a line wich is parallel to AB and intersects circle (C) at points C and D. in wich
the angle AO'B=COD=180-w.
now we can find the center of homothecy betwin the shapes (circle (C') with its chord AB, and circle (C) with its chord CD)
wich is the point that AD and CB meet each other at it. lets name it K.
now i claim : if we draw the other circles like (C'), the locus of the points like K will be a line :!: .
then we draw it and let it meet the circle (C) at x.
then we draw XD. and let it meet one of the sides of the angle at Y.
then we draw XC. and let it meet the other side of the angle at Z.
the circumcircle of triangle XYZ is the circle that we wanted to draw. :D

[color=red][Moderator edit: Combined multiple posts and put spaces between different words. Yes, the solution is right.][/color]

[quote="ashegh"]we draw a line wich is parallel to AB and intersects circle (C) at points C and D. in wich
the angle AO'B=COD=180-w.[/quote]

How would you find $C$ and $D$ satisfying $\angle{AO'B} = \angle{COD}$?
You are talking about a ruler and compass construction right?

Hm.

dear arne , its not hard to draw line CD.

Of course you can draw any line $CD$ parallel to $AB$, but how would you construct the line $CD$ which satisfies $\angle{COD} = \angle{AO'B}$ with ruler and compass?

If it is that easy, it cannot be that hard to explain either. Or am I missing something obvious maybe?

we have the angle O'AB = w/2.
we draw a line from O, wich is perpendicular to line AB. and intersects it at R
and then we draw a line wich made angle 90-(w/2) with line OR.

i dont know: does it injer  the solution if we draw the line by ruller and compass? :?:  :?:

we can find the center of homothecy betwin the circles (C) and (C'). wich its name is K.
and then let BK and CK meet circle (C) at C and D respectivly.

Yes, I figured that out myself, but I just want to show that you should be more careful and rigorous in the write-up of your solution. By the way, it's not a good idea to make so many short posts in a row, rather make one longer post.

Clear enough? Well, I don't know.

[quote="ashegh"]now i claim : if we draw the other circles like (C'), the locus of the points like K will be a line :!: .[/quote]

For example, you haven't even proven this.

i think u here this theorem that:

MON and KO'M are two angles that meet each other at points K,S,N,M.
we draw aline from O, wich intersects O'K and O'M at points A and B respectivly.
then SB and AN meet each other at point X. and again AM and KB meet each other at point Y.
we have:
X,Y,O' in a line.

after u read all the solution,plz think about the things that u think they arent clear .
i know my solution may not be clear but ,i also know ure brain will help u to find them so easily :D .


Yeah, I dont post a lot so i'm not sure if this question has already been asked.

A hundred prisoners are placed in a single file line. They are each to be given a blue or red hat, and starting from the back of the line, each prisoner must state the color of his hat. A prisoner can see all the hats in front of him, but none of those behind him. If the prisoner answers wrong, he will be sentenced to death. Given that the prisoners are allowed to strategized prior to being lined up and hatted, how many prisoners can be saved?

[hide="Hint"]Let a red hat be $0$ and a blue hat be $1.$ Use a parity argument.

[hide="Answer"]99 can be saved.[/hide][/hide]

you AoPSers are too smart

why do I even bother trying to challenge you guys?

actually I think thaat it has been asked at the brainteaser marathon

Really? I don't think I saw it. But yeah, that's the right answer.[hide="what they do"]The person in back counts up all the number of red hats and blue hats. Then, for example, if there are and odd number of red hats, he will say red, and all the prisoners in front of him will know there is an odd number of red hats. The person second to last will be able to use what the last person said to find his hat color, and similarly everyone else will be able to find out their hat color using what the people before him/her said. Thus, everyone except for the last person will be saved.

Those are some smart prisoners. :) [/hide]

the last prisoner could get it right, though...


well, I know first 15 (to $ 2^0$ from $ 2^{14}\equal{}16384$) without thinking. I don't remember $ 2^{15}$, but I know $ 2^{16}\equal{}65536$.
How many do you know?

Eh I know up to 2^19=524288. Spent a few days calculating them out in 5th grade :D I actually got them all up to 2^46, where I made a computation error, invalidating all my values up to 2^66, but couldn't remember more than 2^19.

for me, it's different. If you ask me for example, what 2^18 is, I wouldn't get it right away, since I don't have it memorized, but I could get it within 2-3 seconds, as I have this sequence down:

1,2,4,8,16,32,64,128,256,512,1024.... I've done it many times in my head and I could say it nearly as fast as I would counting regularly (1,2,3,4.....) I've  gone well over the 60th power this way.

I can go up to 8192. Of course I can double it in my head but I can't remember off the top of it

2^30 = 1073741824, mental. Useful for Gigabyte => byte conversion...

one day i was bored in ARML practices (indiv round) and calculated 2^64 = 18446744073709551616.

I have only memorized 2^6 and 2^10, but with those, I can get 2^0-2^5,2^7-2^9,2^11-2^12 in one millesecond.  My math club teacher told me that.

[quote="davidyko"]Um, to the OP, if you know $ 2^{14}$, you can easily double that in your head to get $ 2^{15}$.[/quote]

The trouble is because I CAN'T!!!!!!!!!!!!!!!!!!

MY MEMORY SUCKS!!!!! :wallbash_red:  :wallbash_red:  :wallbash_red:

I know up to $ 2^{13}\equal{}8192$ by heart.  I could easily double it in my head and get the next one (although it may take a few seconds), and I also know $ 2^{16}\equal{}65536$.

I know all of them in binary!
$ 10$
$ 100$
$ 1000$
$ 10000$
$ 100000$
etc.
In decimal, I know up to $ 2^{16}$ and $ 2^{20}$; I can calculate the ones in between pretty quickly.

If this is for speed stuff like countdown, then knowing the powers of 2 above $ 2^{13}$ is pretty useless since using the numbers would probably just slow you down.

And I have up to $ 2^{13}$

I know $ 2^{\minus{}14}$ through $ 2^{14}$ as common fractions, as well as $ 2^{\pm 16}$.

I guess that means I know 31 of them... owned  :D  :rotfl:

Is it just me, or is one of my favorite numbers $ 2^{100} \equal{} 1267650600228229401496703205376$? (Yes, I did that all in my head without a calculator!) My other favorite number is 57.

I know $ 2^{50} \equal{} 1125899906842624$ because it contains a string of three 9's!
I know $ 2^{60} \equal{} 1152921504606846976$ because it's special!

As for consecutively, I know... um... up to the first 50 or so powers of 2, I think.
All the ones up to 10 digits are obvious to me because they fit on a scientific calculator. Then comes 17179869184, 34359738368, 68719076736, ..., (2^40) 1099511627776, 2199023255552 (that one has a string of four 5's in a row!), 4398046511104, ..., (2^45) 35184372088832, 70368744177664, 140737488355328, 281474976710656, 562949953421312, 1125899906842624, 2251799813685248, 450something, 90071something, er...

Next goal: memorize powers of 3! (Unlikely to be obtained in a year) :D

I think 2^3 or 2^4.  :| I never really tried.

this is making me feel bad *mostlyongyi* like the time when ppl were memorizing 100+ digits of pi
...
i only know like 33 ish
1
2
4
8
16
32
64
128...
1048576
2097152
4194304
8388608
16777216
33445532
67108864
134217728
268435456
...
i guess not...
im like the other ppl, i have to do like 1-3 secs to do like 21+

Wow... this make me feel relieved due to my epic fail memorizing pi digits... I suck at it :mad:

You do realize that one one question in the WHOLE Nationals countdown round had anything to do with powers or 2, right?

Yes, this is weird coming from somebody who can memorize, uhh... some powers of 2. I'll just say this: during the summer before 6th grade, I went back to China and had a LOT of free time. :)


A person pushes horizontally with a force of 220 N on a 55 kg crate.The coefficent of kinetic friction is 0.35 What is the magnitude of the frictional force?

[color=blue]The magnitude of the friction force

$F=\mu mg=0.35\times 55\times 10=192.5 N$

If the force of the man has its magnitude fewer than $192.5N$, then the magnitude of the friction force is equal to the man's force[/color]


Let be $ a,b,x,y\in \mathbb{R}^*$ such that $ |ax \plus{} by|\le xy$ . Prove that :

$ |ax \plus{} by|\ge 4ab$

Proof
[hide]
If $ ab\le 0$ it is evident.

If $ ab > 0$:
  If $ x\equal{}y\equal{}0$ is false so suppose that $ x$ and $ y$ can't be simultaneously $ 0$.

  If $ xy\equal{}0$:
Suppose $ x\equal{}0$ then $ by\equal{}0$ then $ b\equal{}0$ since $ y\ne 0$. From here the conclusion is evident.

  If $ xy>0$:
$ xy\ge |ax \plus{} by|\Rightarrow x^2y^2\ge (ax \plus{} by)^2\Rightarrow xy(xy \minus{} 4ab)\ge (ax \minus{} by)^2\ge 0\Rightarrow$

$ 4ab \minus{} xy\le 0\Rightarrow 4ab(4ab \minus{} xy)\le 0\Rightarrow 4ab(4ab \minus{} xy)\le (ax \minus{} by)^2\Rightarrow$

$ 16a^2b^2\le (ax \plus{} by)^2\Rightarrow 4ab\le |ax \plus{} by|$
[/hide]

:)

[quote="alex2008"]Let be $ a,b,x,y\in \mathbb{R}$ such that $ |ax \plus{} by|\le xy$ . Prove that :

$ |ax \plus{} by|\ge 4ab$[/quote]

When $ x\equal{}y\equal{}0$, it is wrong.

You are right stephencheng.

My earlier proof didn't work for $ xy\equal{}0$ since we can't conclude that $ 4ab\minus{}xy\le 0$ from $ xy(xy\minus{}4ab)\ge 0$. I have edited my proof. The conclusion is false only if $ x\equal{}y\equal{}0$.

:)


This is for misan:
$ \prod_{n=0}^{\infty}\left( \frac{\pi^{6n}+\pi^{3n}+1}{\pi^{3n}}\right)$

It diverges to $ \infty$, of course. What do ypu expect when the terms don't go to 1?


geometry
geometric transformation
rotation

Does anyone have a value of the pKa of p-methylanisole? I think that it will be slightly greater than that of toluene... is this reasonable to assume?

Cheers

Ehh... it will be slightly higher indeed.  However, most of the numbers you get for such weak acids vary from source to source anyway.  Might I ask what you need the number for?

Just for a project I'm doing regarding acid guest interactions in calixarenes-looking at the rotation of methyl groups of the guests and seeing if theres a correlation to pKa. Stronger interactions = slower methyl rotation. Seems that the weaker the acid, the stronger the interaction and slower the methyl rotation...
Just needed to establish whether PMA was less acidic than toluene.

Cheers


Two play the following game. The first takes a 10-digit number $\overline{a_9a_8\dots a_0}$. The second, on each move, choses a set $S\subset \{0,\,1,\,\dots,9\}$ and asks about the multy-set $\{a_s|s\in S\}$ (i.e. he may ask, which digits are on coressponding positions, without exact correspondence between digits and places). Find the minimal number of questions, necessary to guess a number. Sorry for terrible English.

Sorry, it was not a multy-set, but a set (i.e. if the answer is (2,3), then we only know that there are some 2-s and 3-s and no other digits on respective places, we do not know how many 2-s are there). And I know, how to solve a problem with this condition. But however, is not a question with multy-set challenging?

I posted a false proof of the original problem some time ago. Hoping that the proposition in post number $7$ from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=283877#p283877]this topic[/url] is correct, I'll try to solve the original question (i.e. the one with the sets, not the multisets).

I think the minimal number of questions needed is $n$, if we try to guess a number with $n$ digits, even if it only has digits in the set $\{0,1\}$ (of course, no matter how many variants there are for a digit, $n$ questions always suffice, since we can ask a question for each digit).

Suppose we can find any number in $n-1$ questions $q_1,\ldots,q_{n-1}$. Suppose, now, that we can find $0<p\le n-1$ questions $q_1,\ldots,q_p$ among these which are only concerned with $\le p$ digits $d_1,\ldots,d_p$. We can change these questions s.t. $q_i,\ i\in\overline{1,p}$ asks what digit $d_i$ is. Then, these new questions, together with $\bar q_{p+1},\ldots,\bar q_{n-1}$ obtained from $q_{p+1},\ldots,q_{n-1}$ by deleting any information about the digits $d_1,\ldots,d_p$ contained in them, can still determine any number, so we can find any number with $n-p$ digits in $n-p-1$ questions, and we can use induction.

The above means that we can restrict our attention to sets of questions which satisfy the following condition: for any $0<p\le n-1$ of them, these $p$ questions ask for information about at least $p+1$ digits. This is the condition mentioned [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=283877#p283877]here[/url], so we can conclude (assuming that proposition is correct, which I think it is :)) that we can assign values $0,1$ to the digits s.t. each question covers a bicolor set of digits. In this situation, however, the one who asks the questions cannot distinguish between two different numbers, one of them obtained by negating all the digits of the other one (i.e. turning every $0$ in a $1$ and vice-versa), so we have a contradiction.


MATHCOUNTS probability geometry rectangle ratio inequalities

1.
Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4.  If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?

2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can.  If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls.  Express your answer in terms if $\pi$.

3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass.  The rectangle has dimensions of 4 feet by 3.5 feet.  How many feet of weather-stripping will be needed to go around the glass?  Express your answer to the nearest integer.

4.
The numbers from 1 through 8 are separated into two sets, A and B.  The numbers in A are multiplied together to get $a$.  The numbers in B are multiplied together to get $b$.  The larger of the two numbers $a$ and $b$ is written down.  What is the smallest possible number that can be written down using this procedure?

5.
Eric's Florist Shop sells roses for 20 dollars per dozen, daises for 5 dollars per dozen, and chrysanthemums for 10 dollars per dozen.  At these rates how much would an arrangement of 9 roses, 6 daises, and 18 chrysanthemums cost in dollars?

6.
Given the equation $\displaystyle \frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ such that $m>n$ and $m$ is not a multiple of $n$.

7.
Use the pattern below to express $100^2$ in the form $a^2+b^2-c^2$.  Then find $a+b+c$.
\begin{eqnarray*}
12^2 = 8^2 + 9^2 - 1^2\\
14^2 = 10^2 + 10^2 - 2^2\\
16^2 = 12^2 + 11^2 - 3^2\\
18^2 = 14^2 + 12^2 - 4^2
\end{eqnarray*}

8.
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

9.
Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?

10.
What is the difference between the largest and the smallest values of $x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*}
2x-6 \le 6\\
4x+3 \ge 13
\end{eqnarray*}

Any middle schoolers care to try some of these cool MathCounts problems?

[hide=Problem 2][font=Century Schoolbook]SUBWAY SOLUTION: GEOMETRY PROBLEM 2

Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can. If the radius of each ball is $4$ cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls. Express your answer in terms of $\pi$.

On this geometry problem, we can take advantage of our knowledge of the cylinder volume and sphere volume formulas.

We will start by finding the volume of the cylindrical container. We are told that the radius of each tightly-sealed ball is $4$cm, so we can conclude that the radius of the cylindrical container is the same. We also know that the height of the cylinder is three times the diameter (or six times the radius) of the balls $(24)$, since there are exactly three tightly sealed balls inside. We can now use the cylinder volume formula: V=πr²h, where $V=\text{volume}$, $r=\text{radius}$, and $h=\text{height}$.

After inputting the values, we get:

$V=\pi (4)^2\times 24=384\pi cm^3$.

The volume of the cylinder is $384\pi$ cubic centimeters.

We must now find the volume of a ball, multiply it by $3$, and subtract from the volume of the cylinder to get the correct answer.

To do this, we use the formula $V={4\pi r^3}{3}$. However, we see that, since we are going to be multiplying by $3$, we do not need to divide by $3$ to find the volume of all three balls. Our new $3$-ball volume formula is: $V=4\pi r^3$, where $V=\text{volume}$ and $r=\text{radius}$.

After inputting the radius, we get:

$V=4\pi (4)^3=256\pi cm^3$.

We now subtract the volume of the balls from the volume of the cylinder to get the answer:

$384\pi cm^3-256\pi cm^3=128\pi cm^3$.

After rounding to the nearest integer, we find that the volume of the space within the cylinder (excluding that taken up by the balls) is approximately $402cm^3$.
[/font][/hide]

[hide=Problem 1][font=Century Schoolbook]Darts are thrown randomly at a board showing three concentric circles of radii $2$, $3$, and $4$. If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius $3$ but not in the interior of the circle of radius $2$?
We will start by drawing the board for visual purposes.

[asy]
filldraw(circle((0,0),4),red,red+linewidth(1));
filldraw(circle((0,0),3),yellow,yellow+linewidth(1));
filldraw(circle((0,0),2),green,green+linewidth(1));
[/asy]

In geometric probability problems,

$P(\text{success})=\dfrac{\text{Size of Successful Region}}{\text{Size of Possible Region}}$.

That means that we must find the size of the successful region, “The interior of the circle of radius 3 but not the interior of the circle of radius 2,” and the possible region, “A board with radius 4.”

To find the size of the successful region, we use the formula for finding the area of a circle: $A=\pi r^2$. This formula tells us that the area of a circle is equal to the product of $\pi$ and the square of the radius.

We first find the area of the circle with a radius of 3:
$A=\pi(3)^2=9\pi$.

We then find the area of the unwanted circle with a radius of 2:
$A=\pi(2)^2=4\pi$.

After that, we subtract the area of the circle with radius 2 from that of the circle with radius 3 to find the size of the successful region:
$9\pi-4\pi=5\pi$.

To find the size of the possible region, we simply input r=4 into the formula for the area of a circle:
$A=\pi(4)^2=16\pi$

We can now determine the probability of success:
$P(\text{success})=\dfrac{\text{Size of Successful Region}}{\text{Size of Possible Region}}=\dfrac{5\pi}{16\pi}=\dfrac{5}{16}$
[/font][/hide]

You can submit solutions to these problems, but I would appreciate it if you could do so as text within the post.  (LaTeX is available on this forum.)  For security reasons, I prefer that readers not have to open attachments or click on links to external web sites.  Thanks.

[font=Century Schoolbook]I will convert them to text and $\text{\LaTeX}$ with hide boxes, if that works.[/font]

[hide=Problem 7][font=Century Schoolbook]Use the pattern below to express $100^2$ in the form $a^2+b^2-c^2$. Then find $a+b+c$.

\begin{eqnarray*}
12^2 = 8^2 + 9^2 - 1^2\\
14^2 = 10^2 + 10^2 - 2^2\\
16^2 = 12^2 + 11^2 - 3^2\\
18^2 = 14^2 + 12^2 - 4^2
\end{eqnarray*}

We start by determining the relationship between the values in $x^2=a^2+b^2-c^2$, where $x$, $a$, $b$, and $c$ are all positive integers.

We see that the value of n increases by 2 in each equation.

We can also see that $a=x-4$, because $8=12-4$, $10=14-4$, $12=16-4$, etc.

In addition, we notice that c represents the term number of the arithmetic sequence ($12, 14, 16, 18$).

(Note: We can define this sequence of values of $x$ recursively as $a_{1}=12, a_{n}=a_{n-1+2}$ for $n>1$.)

Finally, we notice that the values of $b$ are exactly $8$ greater than those of $c$.

Now, to solve for the required $b$ and $c$, we must determine how many terms there are in the sequence: $12, 14, 16, \cdots , 96, 98, 100$.

We see that there is a common difference of $2$, so we divide all terms by $2$. We get:

$6, 7, 8, \cdots , 48, 49, 50$.

We now have a common difference of one, and we want to make the first term $1$ for our convenience, so we subtract $5$ from each term to do this. (Note that $5$ is the difference between the first term, $6$, and $1$.) When we do this, we get:

$1, 2, 3, \cdots , 43, 44, 45$.

We now know that the appropriate value of $c$ will be $45$ and the appropriate value of $b$ will be $45+8=53$ for $x=100$. We also can determine that the appropriate value of $a$ will be $100-4=96$ for $x=100$.

We can now express $100^2$ in the form $a^2+b^2-c^2$:

$100^2=96^2+53^2-45^2$.

We can also take the sum of $a$, $b$, and $c$:

$96+53+45=194$.
[/font][/hide]

[hide=Problem 3][font=Century Schoolbook]A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass. The rectangle has dimensions of $4$ feet by $3.5$ feet. How many feet of weather-stripping will be needed to go around the glass? Express your answer to the nearest integer.

We can start this problem by using the given information to draw the window and get a more visual understanding of the problem. We are told that the rectangle has dimensions of $4$ feet by $3.5$ feet, so we draw a scale drawing of the rectangle with those dimensions.

[asy]
unitsize(1.1cm);
path SUBWAYSDIAGRAM,SEMICIRCLE;
SUBWAYSDIAGRAM=(0,0)--(0,4)--(3.5,4)--(3.5,0)--cycle;
draw(SUBWAYSDIAGRAM);
label("3.5 Ft.",((3.5)/2,0),S);
label("4 Ft.",(3.5,2),E);
[/asy]

We are also told that a portion of glass attached to the side of length $3.5$ feet is semicircular, so we add it to our diagram, as well. (Note: We include a line in the diagram to show where the rectangular and semicircular portions of the window meet.)

[asy]
unitsize(1.1cm);
path SUBWAYSDIAGRAM,SEMICIRCLE;
SUBWAYSDIAGRAM=(0,0)--(0,4)--(3.5,4)--(3.5,0)--cycle;
draw(SUBWAYSDIAGRAM);
label("3.5 Ft.",((3.5)/2,0),S);
label("4 Ft.",(3.5,2),E);
draw(arc((1.75,4),(3.5,4),(0,4)));
draw((1.75,4)--(1.75,5.75),dashed);
label("1.75 Ft.",(1.75,(4+5.75)/2),E);
[/asy]

To find the perimeter of this shape, we can start by taking the sum of the lengths of the three exposed ends of the rectangular sheet of glass (because we cannot include the length of the end attached to the semicircle). The sum of the lengths of the exposed ends is: $4+4+3.5=11.5$.

We must now find the length of the edge of the exposed portion of the semicircular sheet of glass. This will be exactly one-half the circumference of the circle that we would get if we were to place two of the semicircular sheets of glass together. We can find the circumference of this circle by multiplying $\pi$ by the circle's diameter (3.5). When we do this, we get $3.5\pi$. After multiplying this value by $\dfrac{1}{2}$, we get $1.75\pi$.

To finish the problem, we add $11.5+1.75\pi$. When we round this to the nearest integer, we get $17$. We now know that approximately $17$ feet of weather-stripping will be needed to go around the glass.
[/font][/hide]

[hide=Problem 10][font=Century Schoolbook]What is the difference between the largest and the smallest values of $x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*} 2x-6 \le 6\\ 4x+3 \ge 13 \end{eqnarray*}
We start this problem by isolating $x$ in both inequalities.

We can simplify $2x-6 \le 6$ by adding $6$ to both sides. When we do this, we get:

$$2x \le 12.$$

We then divide both sides by $2$ to finish isolating $x$. This gives us:

$$x \le 6.$$

Now that we have finished simplifying the first inequality, we repeat the process with the second, $4x+3 \ge 13$. We start by subtracting $3$ from both sides. This gives us:

$$4x \ge 10$$.

We then divide both sides by $4$, giving us:

$$x \ge \dfrac{10}{4}$$.

We can simplify the fraction on the right to get:

$$x \ge \dfrac{5}{2}$$.

We now know that the smallest value of $x$ satisfying the system is $\dfrac{5}{2}=2.5$ and that the largest value of $x$ satisfying the system is $6$. The difference between these values is $6-2.5=3.5$.

(Note that, when isolating the variable, we performed inverse operations in the reverse order of operations. Although not related to this problem, multiplication (or division) of both sides of an inequality by a negative number would reverse the signs of the numbers, requiring us to do the same to the sign of the inequality for it to remain true.)

[/font][/hide]
[font=Century Schoolbook]I also have a question. In general, is it recommended that I center important equations on their own lines in order to make them look as professional as possible? Is there any general advice for improving the appearance of my solutions? This will also be my last solution for today. Thank you.[/font]

[hide=Problem 5][font=Century Schoolbook]Eric's Florist Shop sells roses for $20$ dollars per dozen, daises for $5$ dollars per dozen, and chrysanthemums for $10$ dollars per dozen. At these rates how much would an arrangement of $9$ roses, $6$ daises, and $18$ chrysanthemums cost in dollars?

We know that the price of each item (as with most items for sale in real life) is directly proportional to how much of that item is purchased. Because of this we can create equations of the form:

$$\dfrac{\text{Price}_1}{\text{Quantity}_1}=\dfrac{\text{Price}_2}{\text{Quantity}_2}$$,

in which we use our knowledge of the price for a certain amount of items and the number of items to determine the price for a different amount of items.

We will start with the $9$ roses. We currently know that they cost $20$ dollars per dozen, and we want to solve for the price of $9$. As a result, we create an equation in which $x$ represents the price we want to find:

$$\dfrac{20}{12}=\dfrac{x}{9}$$.

(Note that we have the prices in the numerators and the quantities in the denominators.)

When we have an equation with one fraction on either side of the equals sign, we can take cross-products (the product of one fraction's numerator with the other fraction's denominator) to eliminate the inconvenient denominators. When we do this, we get:

$$20\times 9=12x$$,

or

$$180=12x$$.

To isolate x, we simply divide both sides of our equation by $12$ to get:

$$15=x$$.

We now know that $9$ roses would cost $\$ 15$.

We repeat a similar process with the daises, which are sold for $5$ dollars per dozen. We create our equation with $x$ representing the price that we would like to find:

$$\dfrac{5}{12}=\dfrac{x}{6}$$.

We then take cross products to get:

$$5\times 6=12x$$,

or

$$30=12x$$.

We then divide both sides of the equation by $12$. This gives us:

$$\dfrac{30}{12}=x$$,

which can be simplified to

$$\dfrac{5}{2}=x$$.

We now know that $6$ daises would cost $\$ 2.50$.

Finally, we repeat the process with the $18$ chrysanthemums, which are sold for $10$ dollars per dozen.

We create our equation:

$$\dfrac{10}{12}=\dfrac{x}{18}$$.

We then take cross products, which gives us:

$$180=12x$$.

We already solved this equation, so we know that $18$ chrysanthemums would cost $\$ 15$.

We now add the prices:

$$15+15+2.50=32.50$$

An arrangement of $9$ roses, $6$ daises, and $18$ chrysanthemums would cost $\$ 32.50$.
[/font][/hide]
[font=Century Schoolbook]Actually, I had time to do one more.[/font]

[quote=TheSubway]
[font=Comic Sans MS]In general, is it recommended that I center important equations on their own lines in order to make them look as professional as possible? Is there any general advice for improving the appearance of my solutions?[/font][/quote]
Nice solutions!  Displaying important equations on their own lines is good.  I would have written less detail for things I consider routine such as solving linear equations, but it depends on your audience.  Writing up solutions to problems is a good exercise.

[font=Century Schoolbook]I appreciate your advice and will ensure that I apply it to my written solutions! Unfortunately, I did not have the opportunity to see this earlier. While my explanations for solving linear equations may have been appropriate for an audience being introduced to the process of solving them; here, it would be more appropriate to summarize the steps quickly and efficiently.[/font]

[font=Century Schoolbook]I had time to write three new solutions on July 1.[/font]

[hide=Problem 8][font=Century Schoolbook]
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length. If the width is 2, how many units are in the length of the diagonal?

If we make $l=\text{Length}$, $d=\text{Diagonal}$, and $w=\text{Width}$, we can create the equation below based on the provided information:

$$\dfrac{l}{w}=\dfrac{d}{l}.$$

We are told that the width is $2$, so we can put $2$ in the place of $w$ in our equation:

$$\dfrac{l}{2}=\dfrac{d}{l}.$$

If we take cross-products, we get a more attractive version of the equation:

$$l^2=2d.$$

We then use the Pythagorean Theorem to create a second equation showing the relationship between the width, length, and diagonal length: $4+l^2=d^2$. A diagram showing this relationship is below:

[asy]
unitsize(1cm);
path T,Q;
T=(0,0)--(2,0)--(0,sqrt(2sqrt(5)+2))--cycle;
Q=(0,sqrt(2sqrt(5)+2))--(2,sqrt(2sqrt(5)+2))--(2,0)--cycle;
draw(T);
draw(Q);
label("$\huge{l}$",(0,(sqrt(2sqrt(5)+2))/2),W);
label("$\huge{2}$",(1,0),S);
label("$\huge{d}$",(1,(sqrt(2sqrt(5)+2))/2),NE);
[/asy]

We now have a system of equations:

$$l^2=2d$$
$$4+l^2=d^2.$$

We see that $l^2$ is in both equations, and that it is equal to $2d$, so we can replace it with $2d$ in the bottom equation so that we can solve for $d$, the length of the diagonal. We now have:

$$4+2d=d^2.$$

We can rewrite this as:

$$0=d^2-2d-4.$$

After finding the roots of this quadratic, we find that $d=\sqrt{5}+1$. (Note that we do not include the negative root, since length cannot be negative.)
[/font][/hide]

[hide=Problem 6][font=Century Schoolbook]Problem 6: The Diophantine Equation

Given the equation $\frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ such that $m>n$ and $m$ is not a multiple of $n$.

We want to eliminate denominators from both sides, so we multiply both sides by $15mn$. This gives us:

$$\dfrac{15mn}{m}+\dfrac{15mn}{n}=\dfrac{2\cdot15mn}{15}=15n+15m=2mn.$$

For our convenience, we multiply both sides by $2$ (so that the coefficient of 2mn is a perfect square appropriate for factorization) and move the terms $15n$ and $15m$ to the right side of the equation with subtraction. We now have:

$$0=4mn-30n-30m.$$

We want to factor the right side in the form $(2m-15)(2n-15)$, although this is not possible without a constant of $225$ added to the right side of the equation. To ensure that the equation remains true, though, we must add $225$ to the left side, as well:

$$225=(2m-15)(2n-15).$$

We must now list pairs of integer factors of $225$ and set them equal to both $2m-15$ and $2n-15$. After performing this process and experimenting with different pairs of factors, we find that the factor pair $25\cdot9$, as well as $m=20$ and $n=12$ satisfy the restrictions.

By adding the values of $m$ and $n$ that we found, we find that the sum $m+n$ equals $32$.

[/font][/hide]

[hide=Problem 9][font=Century Schoolbook]Problem 9: Application of Vieta's Formulas

Suppose a number $x$ satisfies $x^2-4x-1=0$. How much greater is $x$ than its reciprocal?

Although we could solve this problem in a relatively long and disorganized way by completing the square or using the quadratic formula, the information provided by Vieta's Formulas is enough to (creatively) find the difference between $x$ and its reciprocal.

Vieta's Formulas state that, for a quadratic with a coefficient of the form $x^2+bx+c$, the sum of the roots, which we can call $r$ and $s$, is equal to $-b$ and the product of the roots is equal to $c$. Form this information, we know that:

$$r+s=4,$$

and

$$rs=-1.$$

If we use bottom equation to solve for $s$ in terms of $r$, we find that $s=-\tfrac{1}{r}$, or the negative reciprocal of $r$.

When we input $-\tfrac{1}{r}$ in the place of $s$ in the equation above, we get:

$$r-\dfrac{1}{r}=4.$$

This tells us that the difference between $r$ and its reciprocal is $4$.

We now know that $x$ is $4$ greater than its reciprocal.

Alternately, we could have done this by adding $1$ to both sides of the equation to get:

$$x^2-4x=1,$$

which can be rewritten as:

$$x(x-4)=1.$$

We could have then divided both sides by $x$ to get:

$$x-4=\dfrac{1}{x}.$$

This equation tells us that $x$ if four greater than its reciprocal.
[/font][/hide]
[font=Century Schoolbook]Note: I solved this problem by completing the square, although I soon realized that Vieta's Formulas made it much faster to solve.[/font]

[quote=TheSubway]
$$\dfrac{^1/_2}{l}=\dfrac{1}{2}\cdot\dfrac{l}{1}$$
[/quote]
This step doesn't look right.

[quote=TheSubway][hide=Problem 9][font=Century Schoolbook]Problem 9: Application of Vieta's Formulas

Suppose a number $x$ satisfies $x^2-4x-1=0$. How much greater is $x$ than its reciprocal?

Although we could solve this problem in a relatively long and disorganized way by completing the square or using the quadratic formula, the information provided by Vieta's Formulas is enough to (creatively) find the difference between $x$ and its reciprocal.

Vieta's Formulas state that, for a quadratic with a coefficient of the form $x^2+bx+c$, the sum of the roots, which we can call $r$ and $s$, is equal to $-b$ and the product of the roots is equal to $c$. Form this information, we know that:

$r+s=4,$

and

$rs=-1.$

If we use bottom equation to solve for $s$ in terms of $r$, we find that $s=-\tfrac{1}{r}$, or the negative reciprocal of $r$.

When we input $-\tfrac{1}{r}$ in the place of $s$ in the equation above, we get:

$r-\dfrac{1}{r}=4.$

This tells us that the difference between $r$ and its reciprocal is $4$.

We now know that $x$ is $4$ greater than its reciprocal.
[/font][/hide]
[font=Century Schoolbook]Note: I solved this problem by completing the square, although I soon realized that Vieta's Formulas made it much faster to solve.[/font][/quote]
Cool.  Another approach is to divide the original equation by $x$.

[font=Century Schoolbook]Thank you for your responses! I will try to review my work as soon as I have time. (I confused the $l$'s for $1$'s when typing the $\text{\LaTeX}$ on Word. I will be careful not to repeat this mistake again, although I should have noticed that it would be impossible to have a diagonal length shorter than the width!)
I also decided to add the alternate solution to Solution 9.
[/font]

[font=Century Schoolbook]I decided to make the diagrams by learning basic Asymptote and added them to Solutions 3 and 8 (since the original diagrams did not work very well on this message board). I also updated my solution to 8.
[/font]

Nice work using Asymptote.  In the answer to Problem 8, you may wish to write \sqrt, not just sqrt.

[font=Century Schoolbook]Thank you for finding that $\text{\LaTeX}$ error! I corrected it. In the future, I will look at the preview to check for errors.
[size=50]July 4, 2020 (Decided to add missing diagram to Problem 1 solution.)[/size]
[/font]

[font=Century Schoolbook]You may have noticed that I have not been very active recently. This is because I have been working on creating my own mathematics competition because of an announcement made by a school official. I hope to put it in the Art of Problem Solving by next week.[/size]
[/font]

[hide=Problem 4][font=Century Schoolbook]Problem 4:
The numbers from 1 through 8 are separated into two sets, A and B. The numbers in A are multiplied together to get $a$. The numbers in B are multiplied together to get $b$. The larger of the two numbers $a$ and $b$ is written down. What is the smallest possible number that can be written down using this procedure?

In order for the larger of the two numbers to be as small as possible, we would want it to be as close as possible to the smaller number. To get an estimate of what number we want both $a$ and $b$ to be close to, we take the product of all integers from $1$ through $8$ inclusive:

$$\prod_{n=1}^{8}n = 40320.$$

We then take the square root of $40320$:

$$\sqrt{40320}=\sqrt{1 \cdot 2^7 \cdot 3^2 \cdot 5 \cdot 7}=24\sqrt{70}.$$

We can round this to approximately $200$. Now, we check for pairs of (integer) factors $a$ and $b$ of $40320$ in which $a \approx b \approx 200$. The pair that meets this requirement is $192$ and $210$, which can be gotten with the arrangement of numbers:

$(4 \cdot 6 \cdot 8)(1 \cdot 2 \cdot 3  \cdot 5  \cdot 7).$

This indicates that $210$ is the smallest number that can be written down using this procedure.
[/font][/hide]
[font=Century Schoolbook]This is the last of the 10 solutions.[/font]


The following is similar to my solution for http://www.mathlinks.ro/phpBB/viewtopic.php?t=1622
Okay, let's start: WLOG x_1 \leq x_2 \leq ... \leq x_n.

\sum x_i 1/2 \geq  (n-1)\sum (1/x_i)1/2 <=>
n\sum (1/x_i)1/2 \leq \sum x_i 1/2 + \sum (1/x_i)1/2 = \sum (x_i+1)/x_i 1/2 ------ (X)

Now let a_i = 1/(x_i+1) , b_i = (x_i+1)/x_i1/2 = x_i1/2+1/x_i1/2.
We have x_1 \leq x_2 \leq ... => a_1 \geq a_2 \geq ...

Let i<j. I wanna prove b_i \leq b_j that is the same as x_i1/2+1/x_i1/2 \leq x_j1/2+1/x_j1/2  <=> (x_j1/2-x_i1/2)(1-1/(x_ix_j)1/2) \geq 0, that is true, because x_j1/2-x_i1/2 \geq 0 and 1-1/(x_ix_j)1/2 \geq 0 that's because x_ix_j \geq 1 <= 1/(x_i+1) + 1/(x_j+1) \leq 1.

So a_1 \geq a_2 \geq ... , b_1 \leq b_2 \leq ..., so we can apply Chebyshev's inequality:

(\sum a_i)(\sum b_i) \leq n\sum a_ib_i

\sum a_i  = 1, so \sum b_i \leq n \sum a_ib_i. a_ib_i = 1/x_i1/2, so \sum b_i \leq n\sum 1/x_i1/2, that is the same as (X)


Numbers $ 1,2,3,...,9$ are written on a circle randomly. Prove that there exists the numbers consecutively on that circle such that their sum is greater than $ 15$.

[Consecutively or adjacent? I don't know how to say it...]

"Prove that there exist consecutive numbers..." or "Prove that there exist adjacent numbers..." both work.  :)

I don't think this problem is true, just put the numbers in this order: 1,9,2,3,4,5,6,7,8.  :huh:

[quote="matt276eagles"]1,9,2,3,4,5,6,7,8.[/quote]
Actually, $ 7\plus{}8\equal{}15$.

But you're right about this being false:
1, 6, 2, 7, 3, 8, 4, 9, 5

Perhaps you meant 3 consecutive numbers? (though that's kind of easy)

[hide="Solution"]There are 9 such sums of 3 consecutive numbers, whose total is $ 3 \cdot (1 \plus{} 2 \plus{} \ldots \plus{} 9) \equal{} 135$. The average of these 9 sums is $ 15$. So it remains to check that we can't have all $ 9$ being $ 15$, but two adjacent sums can't be equal because that implies two different numbers on the circle are equal.[/hide]


real analysis
absolute value
real analysis unsolved

You've answered your question your self - it converges for all $ x$ such that $ |x|<1$. You just need to see the absolute value from a different angle - $ |a\plus{}bi|\equal{}\sqrt{a^2\plus{}b^2}$! :)

[quote="MagnusZ"]from k=1 to infinity ∑ x^k/k converges if |x|<1. For which complex x with |x|=1 does it converge?[/quote]

All $ x$ with $ |x|\equal{}1$ except $ x\equal{}1$.  This is the content of theorem 3.44 of Rudin and can probably be found in most similar analysis texts.  As alekk mentions, the proof does follow from abel summation by parts if you know it.


geometry
combinatorics unsolved
combinatorics

WLOG there is some triangle XYZ among the n points with area 1. Suppose that XYZ is the medial triangle of ABC, with X as the midpt of BC, Y the midpt of AC, Z the midpt of AB. ABC is a triangle of area 4 which contains all the points: suppose there is a point P outside of ABC; then WLOG it lies on the opposite side of BC as A, but then 1 = [YXZ] < [YXP], a contradiction.

[quote="HUY VAN"]There are n points given the plane. Any there of the points form a triangle of $area\le 1$. Show that all n points lie in a triangle of $area\le 4$[/quote]

Hmm... why does this sound familiar to me?

[quote="weiquan"][quote="HUY VAN"][quote="weiquan"]

Hmm... why does this sound familiar to me?[/quote]
What are you talking?  :D[/quote]

I did this problem before, so I was wondering where it came from...[/quote]
Oh, link?

[quote="HUY VAN"][/quote][quote="weiquan"][quote="HUY VAN"][quote="weiquan"]

Hmm... why does this sound familiar to me?[/quote]
What are you talking?  :D[/quote]

I did this problem before, so I was wondering where it came from...[/quote][quote="HUY VAN"]
Oh, link?[/quote]

Sorry, I meant that I had done it before outside of MathLinks, and I can't remember where it came from, so I was surprised to see it again.

[quote="HUY VAN"][quote="zgorkster"]I believe (though im not 100% sure) that it was an example in Problem Solving Strategies by Engel.[/quote]
Yes, this problem is an example in Problem Solving Strategies. But, can it be generalized ?[/quote]

Like how? And if it can be done, I don't think the original method can be adapted to solve it.


In a wildlife preserve in Australia, there are ten kangaroos at the beginning of Week 1. Every week, the number of kangaroos in the preserve doubles, unless there are  two hundred kangaroos or more. In that case, the park manager opens hunting season, and by the end of the week, 50 kangaroos are eliminated. The end of what week is the first time that there is the highest possible amount of kangaroos in the park?

[hide="my answer"]
I'm not sure, but I think it's 13.
[/hide]

[hide]So we're looking for the first week with 190, then double it to get 380, which I think is the maximum. This is easy to just list out.

at the end of week:

1-20
2-40
3-80
4-160
5-320
6-270
7-220
8-170
9-340
10-290
11-240
12-190
13-380

Yay, week 13.[/hide]

Hooray! Another Treething problem! :D
[hide]I just listed the weeks like Klebian did and got week #13 - 380 kangaroos[/hide]

[hide]5 weeks to get to 320
3 weeks to get to 170
1 week to get to 340
3 weeks to get to 190
1 week to get to 380, which is the highest. 13 weeks.[/hide]

[quote="jclarkemathL314159"][hide]
10
20
40
80
160
320
270
220
170
340
290
240
190
380

If you continue, it repeats, and 380 is the maximum.
Hence, it is week 13.[/hide][/quote]

That is the way I did it, and that is how I got the answer

[hide]13[/hide]


[b]The First Jifted Mathematical Competition[/b]
There are two levels for this year’s competition.  Each level has four rounds.  Level one is primarily for elementary through middle school level students, while level two is intended for high school level students.  Those who qualify to compete in level one can decide to compete in level two, but students who are designated to compete in level two cannot compete in level one.  In each round, the students score will be calculated and added to a total score.  By the end of round three, the individuals with the top ten scores will face of in a final round.  The top three individuals after this final round will advance to a fifth round.  The fifth round is a round in which the top three individuals from both levels will compete against each other to decide the final places.  A student is allowed to use any non-intelligent resource to aide them in answering questions.  Calculators are only permitted for specific rounds.  Those who break any rules will be disqualified.

The following are excepted forms of answers.  Fractions must be simplified to where the numerator and denominator have a greatest common factor of one.  Radicals must be simplified and rationalized.  Algebraic expression must be simplified.
[b]
Questions, comments, and solutions should be e-mailed to jiftedland at gmail dot com.[/b]

[b]When submitting solutions please have the subject in the form Level [X]: Round [X]: [Name]: [Grade][/b]

One point will be deducted for solultions not in that form.

Awards may be given for the top ten contenders in both levels!

Check out the JMC site for more information:
http://yashar.biz/jmc/

Edit: Nevermind this....Link is working fine.


[b][color=red]Prove that for every positive integer n, there exists a unique ordered pair (a,b) of integers such that,
          [i]n[/i] = 1/2 * (a+b-1)(a+b-2) + a[/color][/b]

[color=darkblue][size=150] :wink: Rooster.Track[/size][/color]

Sorry,but I dont think I understood your question fully. :blush:
$ n\equal{}\frac{(a\plus{}b\minus{}1)(a\plus{}b\minus{}2)}{2} \plus{} a$,
If this is the question,
take n=5
a=-5,b=11
a=-10,b=17
Both satisfy(there will be infinite pairs,I guess).There is no unique pair.
OR did you mean
there exists a unique ordered pair (a,b) of [b]positive [/b]integers?

Here's my solution, I hope its correct.

[hide]I'm assuming $ a,b$ are positive integers. Making the substitution: $ c\equal{}(a\plus{}b\minus{}2)$ We have the constraint that $ c$ is a non-negative integer and $ a<\equal{}c\plus{}1$. Having a unique pair of $ (a,b)$ is equivalent to having an unique pair of $ (a,c)$.

Denote $ T_k$ as the $ k$th triangular number. The given  expression is: $ n \equal{} T_c \plus{} a.$ Now let $ T_i < n <\equal{} T_{i\plus{}1}$ for some $ i$. If we set $ c<i$ then the maximum value the $ RHS$ can attain is $ T_{i\minus{}1} \plus{} i \equal{} T_i < n$ and if $ c>i$ the minimum value of the $ RHS$ will be $ T_{i\plus{}1} \plus{} 1 > n$ so $ c \equal{} i$ and $ a\equal{}n\minus{}T_i$. Since $ T_i < n <\equal{} T_{i\plus{}1} \equal{} T_i \plus{} i\plus{}1$, we have $ 0 < a(\equal{}n\minus{}T_i) <\equal{} i \plus{}1 \equal{} c\plus{}1$ which was our required condition. [/hide]


find all $x,y\in\mathbb R$:
\[\sin^2(x+y)=\cos^2x+\cos^2y\]

Hmm... isn't that true for any $x+y \in [ - \frac{\pi}{2} , \frac{\pi}{2} ]$ ?

no,
let $x=y=0$.
the LHS is $0$, while the RHS is $2$.

[hide="My solution"]
Playing around with the equation we get the following identity:

[tex]\sin^2(x+y)-\cos^2x-\cos^2y=-2\cos (x+y)\cos x\cos y[/tex]

[hide="How?"]First two terms:

[tex]\sin^2(x+y)-\cos^2x=\left (\sin (x+y)-\cos x\right )\cdot\left (\sin (x+y)+\cos x\right )=\\=\left (\sin (x+y)-\sin\left (x+\frac{\pi}{2}\right )\right )\cdot \left (\sin (x+y)+\sin\left (x+\frac{\pi}{2}\right )\right )=\\=2\cdot\cos\frac{2x+y+\frac{\pi}{2}}{2}\cdot\sin\frac{y-\frac{\pi}{2}}{2}\cdot 2\cdot\sin\frac{2x+y+\frac{\pi}{2}}{2}\cdot\cos\frac{y-\frac{\pi}{2}}{2}=\\=\sin\left (2x+y+\frac{\pi}{2}\right )\cdot\sin\left (y-\frac{\pi}{2}\right )=-\cos (2x+y)\cos y[/tex]

Adding the third term:

[tex]-\cos (2x+y)\cos y-\cos^2y=-\cos y\cdot\left (\cos (2x+y)+\cos y\right )=\\=-2\cos y\cos\frac{2x+y+y}{2}\cos\frac{2x+y-y}{2}=-2\cos (x+y)\cos x\cos y[/tex][/hide]

So now it's all rather simple:

[tex]-2\cos (x+y)\cos x\cos y=0\ \Longleftrightarrow\\\Longleftrightarrow \left (x=\frac{\pi}{2}+k\pi\right )\ \vee\ \left (y=\frac{\pi}{2}+k\pi\right )\ \vee\ \left (x+y=\frac{\pi}{2}+k\pi\right ),\ k\in\mathbb{Z}[/tex][/hide]

Is it correct?

The answer is $x=\frac{\pi}{2}+2l\pi,y=\frac{\pi}{2}+2m\pi,x+y=\frac{\pi}{2}+2n\pi\ (l,m,n\in\mathbb{Z})$

[quote="aappaa"]Is it correct?[/quote]
those are the right solutions. if $x$ and $y$ are angles of a triangle that would mean it has to have one right angle $A$,$B$, or $C$. anyway thanks aappaa and if there is anyone with a solution shorter tha his i would greatly apreciate its posting ;)

peeta

Why is it [tex]\boxed{2}k\pi[/tex] kunny? Isn't the period just [tex]k\pi[/tex] in this case?


number theory
greatest common divisor
number theory proposed

[color=blue]I saw this nice question long ago but i found a great answer for it and I will tell it to you tomorrow:

Let X(i) be a sequence of natural numbers such that :

X(1) = 1

X(n+1) = [ e * (X(n)!) ]      \\ _where e = 2.71828128..

prove that for any natural numers n,m :

gcd( X(m) , X(n) ) = X( gcd(m,n) ) .

it seems hard but i have an easy answer for it.[/color]

hi arman
i would try it

shayan dashmiz ;)

you can use the sequence $u_{n}=nu_{n-1}+1,u_{1}=2$ because we have $u_{n}=[e*n!]$ for all$ n \in N^{+}$ :D  :D

$U_n$ doesnt equal$ [e\times n!]$ it is equal $[e\times U_{n-1}!]$ :D


conics
parabola
geometry
geometric transformation
reflection
analytic geometry
symmetry

I wonder that a plane can covered by n parabolas? Why?

No. It's not hard to prove that for every parabola there exist an angle with less than $\frac{360^\circ}{n}$ degrees  such that parabola is completely inside it. As we can't cover the plane using $n$ angles  with  less than $\frac{360^\circ}{n}$ degrees, our proof is complete.

[quote="Goblin"]No. It's not hard to prove that for every parabola there exist an angle with less than $\frac{360^\circ}{n}$ degrees  such that parabola is completely inside it.[/quote]

Could you post a proof ?Thank you

Did you try to prove it yourself? Note that a parabola is convex, so it lies completely under or completely above any line that is tangent to it. We know that the equation of a parabola is $y^{2}= 2px$. Take any point on it, compute the tangent there and its intersection with $Ox$. Note that the angle formed between it and its reflection in $Ox$ goes to $0$ as $x$ goes to $\infty$.

[quote="1=2"]Is it possible to solve this problem using the geometric definition of the parabola?[/quote]
Take a parabola. Select the origin of coordinate system a $y = ax^2$ wherer $a>0$.
For inner region $y \geq ax^2$.

$y = kx + b$ (A line which is not parallel to $y-$axis) and the parabola meet at two points, or one point, or no point.
It can be show mathematically as:
All $x$ satisfying $kx+b \geq ax^2$ will be covered by the parabola.
$ax^2-kx+b \leq 0$. Discriminant $\triangle = D = k^2-4ab$.
If $D < 0 $, the line does not cut the parabola.
If $D > 0 $, the line cuts the parabola at two points $(x_1,y_1)$ and $(x_2, y_2)$ .
If $D = 0$, the line cuts the parabola at one point.
In all situations, only a segment of the line is inside the parabola. Its length is $d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$, or $d = 0$.

If there are finite parabolas, each will cover only a segment with $d_i$ long of a choosen line $y=kx+b$. The total is $d=d_1+d_2+\dots+d_n < \infty$.

If there is a line that is not parallel to all of the symmetry axes of the parabolas, only a finite segment of this line will be covered by finite parabolas.
So can we choose the line?
There are infinite lines that are passing through a fixed point $P$. Only $n$ of them are parallel to the symmetry axes of $n$ parabolas.  So we can choose a line that is not parallel neither of the symmetry axes of $n$ parabolas.

This problem was proposed in [url=https://vjimc.osu.cz/problems]6. Vojtěch Jarník International Mathematical Competition, Category 2 Problem 1[/url] in 1996.

[b]Official Solution:[/b]
Suppose that there exists a finite system $\mathcal{S}$ of parabolas, which cover the plane. The number of parabolas is $n$. Take two parabolas from $\mathcal{S}$ which intersect. These parabolas have at most four intersection points. We choose another parabola from $\mathcal{S}$ which covers at least one of the intersection points. Hence we have either a new intersection point or part of a parabola which is not covered. We do this iteration process for all parabolas from $\mathcal{S}$. At the end of this iteration process we get one intersection point or part of a parabola which is not covered. So we have a contradiction. $\blacksquare$

Also it is [url=https://artofproblemsolving.com/community/c6h486994p2728787]Turkish TST 1999/6[/url].


[quote="thanhnam2902"]Find root of this system: $ \begin {cases} x^3 \plus{} 2y^2 \minus{} 4y \plus{} 3 \equal{} 0 \\
x^2 \plus{} x^2y^2 \minus{} 2y \equal{} 0\end {cases}$[/quote]

We assume that the roots are real; then \[ x^3 \plus{} 2y^2 \minus{} 4y \plus{} 3 \equal{} 0 \Longleftrightarrow x^3 \equal{} \minus{} 2(y\minus{}1)^2 \minus{} 1 \le \minus{}1 \Longrightarrow x \le \minus{}1.\] The first equation yields $ y^2 \minus{} \frac{2}{x^2}y \plus{} 1 \equal{} 0$, and this is a quadratic in $ y$ and the discriminant must be positive: $ \left(\frac{2}{x^2}\right)^2 \minus{} 4 \ge 0 \Longrightarrow \minus{}1 \le x \le 1$. Thus $ (x,y) \equal{} \boxed{(\minus{}1, 1)}$.

[quote="azjps"][quote="thanhnam2902"]Find root of this system: $ \begin {cases} x^3 \plus{} 2y^2 \minus{} 4y \plus{} 3 \equal{} 0 \\
x^2 \plus{} x^2y^2 \minus{} 2y \equal{} 0\end {cases}$[/quote]

We assume that the roots are real; then
\[ x^3 \plus{} 2y^2 \minus{} 4y \plus{} 3 \equal{} 0 \Longleftrightarrow x^3 \equal{} \minus{} 2(y \minus{} 1)^2 \minus{} 1 \le \minus{} 1 \Longrightarrow x \le \minus{} 1.
\]
The first equation yields $ y^2 \minus{} \frac {2}{x^2}y \plus{} 1 \equal{} 0$, and this is a quadratic in $ y$ and the discriminant must be positive: $ \left(\frac {2}{x^2}\right)^2 \minus{} 4 \ge 0 \Longrightarrow \minus{} 1 \le x \le 1$. Thus $ (x,y) \equal{} \boxed{( \minus{} 1, 1)}$.[/quote]

Thank you very much! :P


probability
geometry
function
rhombus
probability and stats

Hi, I have a convex polygon of n vertexes. Inside the polygon, there are
k sites (or points). Does anyone know that answer to three problems listed below.

1. What is the expected average distance between two sites.
2. What is the expected shortest distance between two sites.
3. What is the expected longest distance between two sites.

I know there exists a solution to this problem if the n convex polygon is
replaced by a circle. But n vertex polygon is a more general.

Any help is highly appreciated. Thanks, ice

#1 does not really require many points: the answer will be the same as for $k=2$, where you can compute it by setting up an appropriate double integral. #2 and #3 are much more difficult. I do not believe that there exists any nice formula even for an equilateral triangle (as usual, I'll be only happy to be proved wrong :)). By the way, here is a nice question. Suppose that you have two convex sets $A\subset B$. It seems "intuitively obvious" that all three quantities you mentioned should be not smaller for $B$ than for $A$. Prove or disprove this "obvious claim". (for non-convex sets it is, clearly, false: just take the unit circumference, where the avarage distance between points is greater than $1$ and add a small disk near its center).

We can at least get some asymptotic information about #2; as $\epsilon\to 0$, the probability that the distance between two randomly chosen points is less than $\epsilon$ is comparable to $\frac{\pi\epsilon^{2}}{K}$, where $K$ is the area of the figure.
The probability that all pairs are at distance greater than $\epsilon$ is thus approximately $(1-\frac{\pi\epsilon^{2}}{K})^{\binom{n}{2}}\approx e^{-(n^{2}-n)\frac{\pi\epsilon^{2}}{2K}}$. That is the cumulative distribution function of the magnitude of a 2-d normal distribution centered at zero with variance $\frac{K}{n^{2}-n}$ in each direction; the mean is a constant times $\sqrt{\frac{K}{\pi(n^{2}-n)}}$, and thus our average shortest distance is asymptotically proportional to $\frac{\sqrt{K}}{n}$, with a proportionality constant that doesn't depend on shape.

The asymptotics of #3 are more complicated; the maximum distance tends to the diameter, but the speed of convergence depends on shape. Convergence is much faster with a square than with a slightly imbalanced rhombus, for example.

ice_man, could you give some references to the solution in the case of the circle? or some hints? Very interested.


I have a square of side $ 1m$.
An ant, starting from one corner of the square, walks along the sides to reach the diagonally opposite corner travelling $ 2m$.
Now, I create a step-like structure as shown in the figure, the ant has to walk the same distance again, $ 2m$.
I create more steps, again the same distance, $ 2m$.

I go on doing this tilll number of steps, $ n \to \infty$ and the distance to be travelled remains $ 2m$.
But the final path is clearly the diagonal of the square with length $ \sqrt2 m$.
$ \therefore 2 \equal{} \sqrt2$

What is wrong with this result?

The fault is here: "But the final path is clearly the diagonal of the square with length $ \sqrt2 m$."
The final path is not the diagonal!

There is a very good explanation here:

http://everything2.com/e2node/2%255E.5%2520%253D%25202.

Actually, the flaw lies in an unjustified passage to the limit. It is true that the stairway paths $ \gamma_n$ converge to the diagonal path $ \gamma$ in many reasonable senses (say, uniformly) but it does not follow that so do the lengths $ L(\gamma_n)$. In other words, the length of the path does not depend on the path continuously. All that can be proved is that it is semicontinuous from below (i.e., you can not [i]decrease[/i] the length much by slightly perturbing the path, or, equivalently, the length of the limit does not exceed the limit of the lengths). The presented geometric argument thus proves that $ \sqrt 2\le 2$, but not much more.