i sorta derived this from the "Turtles" topic. anyways, there are 3 ants, ant A says there are 2 ants behind me, and B says there is one ant in front and behind me, and ant C says there are 2 ants ahead of me and 1 behind me.How is this possible? no, ant C is not lying. this riddle might be kinda stupid, but oh well.

there is a fourth ant. It simply wasn't mentioned by the first two or the problem. I posted that in the turtles topic too.

That doesn't work.  All I can think of is that they are going sideways with some unmentioned ants.  Directions they're facing doesn't help either.

So there are three ants. Does that mean that there cannot be a fourth?

Or even better, there is a big number 1 behind them all...

[quote="seamusoboyle"]We could be on the surface of a sphere![/quote]
you're warm!!

we could be in hyperspherical space?  :P

or we could be on a circular line....but I prefer hyperspherical space

is it of any importance that now we are talking of ants and not turtles anymore?
 :)

yea, that's it!!... i know, it's kinda stupid.

darn....I wanted it to be something like hyperspace curved in the fifth dimension lol.

Actually that's what a circular line is...1 dimensional space curved in the second dimension. Any n dimensional space curved in the n+1 dimension would work here.

I guess the hyperdimential concept could make this problem a possiblity,without one ant over the other

[quote="GoBraves"]But my answer makes more sense and its a better answer!!![/quote]
I would agree, if only the riddle hadn't said that there were exactly three ants (well actually it just said 'three ants' but gr8tsk8r clarified that there were exactly three afterwards, I believe).

Hi guys, kinda new here...

I can't help myself but to react on this topic... How lucky the ants are, they were the center of discussion here.

I think there's somethin wrong with the problem. It should be clear you know to avoid opinions :)


algorithm videos function Duke college factorial Mafia

[b] FOR THOSE WHO ARE READING THIS THREAD FOR THE FIRST TIME
AOPSTI is usually held Friday and Saturday. 8PM EST over AOL Instant Messenger. We ask participants to show up at around 7:45PM so they can receive invites. Expect the games to last till about 9:30.
Wanna join? Just leave your AIM in this thread and come online at the appropriate time. Or talk to Lucky (AIM: hezgotzbeenz)[/b]

Based on the success of the AOPS Trivia Challenge, I have compiled a list of 30 or so questions for use for the first ever "AOPS Trivia Invitational".

Basically, you may remember whitehorseking88 planning something like this many many months ago but never getting around to actually doing it. Well... I have the questions and am ready to administer them.

This would last about an hour and would occur over AIM (Since that seems to be what most everyone has). I'm still not sure whether I would do it in teams (probably) or individual.

I have not set a date yet because I need to see if there is interest for this first.

Yes, it says the AOPS Trivia [i]Invitational[/i] but truth be told, [b]you're all invited [/b].

So is anyone interested? Reply here or contact me on AIM (hezgotzbeenz) if you are. I'd need at least 8 people or so for this to run smoothly.

(And if you can't make it this time and there's enough interest, I'd be perfectly fine with holding more of these events)

[b] Quick Stats After 30 Trivia Invitationals  [/b]

Most Games Played (by Games)
GoBraves - 22
yif man 12 - 19
hello - 17
DarkKnight - 15
mcalderbank - 15
metsfan001 - 14
hspotter2002 - 13
paladin314159 - 12
chesspro - 12
whitehorseking88 - 12

Most Games Played (by Questions)
GoBraves - 646
yif man 12 - 533
DarkKnight - 431
hello - 398
mcalderbank - 384
metsfan001 - 380
paladin314159 - 345
hspotter2002 - 342
chesspro - 339
Lucky707 - 270

Total Points
GoBraves - 1950
yif man 12 - 1949
metsfan001 - 1926
Lucky707 - 1475
DarkKnight - 1447
paladin314159 - 1417
Aznphatso - 1086
hello - 913
mcalderbank - 867
chesspro - 860

Points Per Game
Lucky707 - 163.89
metsfan001 - 137.57
paladin314159 - 118.08
Aznphatso - 108.60
yif man 12 - 102.58
Ankur87 - 97.25
DarkKnight - 96.47
miraculouspostmaster - 96.00
Chigr - 94.60
JSRosen3 - 94.00

Points Per Question
Lucky707 - 5.463
metsfan001 - 5.068
Aznphatso - 4.180
paladin314159 - 4.107
miraculouspostmaster - 3.934
ComplexZeta - 3.807
yif man 12 - 3.657
Ankur87 - 3.458
DarkKnight - 3.357
scrambled - 3.255
whitehorseking88 - 3.224
Dutchfreak81 - 3.167
Chigr - 3.153
JSRosen3 - 3.133
GoBraves - 3.019
bakwardz41 - 2.867
PoignantPianist1 - 2.733
YashaBK - 2.694
henryyangrui - 2.548
chesspro - 2.537

Best Record (by Wins)
chesspro - 10-1
DarkKnight - 8-3
yif man 12 - 8-9
sheepwarrior - 7-2
GoBraves - 7-12
Lucky707 - 6-3
metsfan001 - 6-4
mcalderbank - 6-7
Aznphatso - 5-2
paladin314159 - 5-3
hello - 5-9

Winning %
bakwardz41 - 100.0% (1-0)
MiChFrEaK15 - 100.0% (1-0)
weiyangmiami - 100.0% (1-0)
chesspro - 90.9% (10-1)
sheepwarrior - 77.8% (7-2)
DarkKnight - 72.7% (8-3)
aznphatso - 71.4% (5-2)
Lucky707 - 66.7% (6-3)
paladin314159 - 62.5% (5-3)
metsfan001 - 60.0% (6-4)

The 100 Club -
(People whose highest scores exceed 100, highest score is listed beside name)
yif man 12 - 201
metsfan001 - 201
Lucky707 - 189
Aznphatso - 183
GoBraves - 173
paladin314159 - 154
miraculouspostmaster - 138
DarkKnight - 134
Ankur87 - 133
ComplexZeta - 131
mcalderbank - 128
JSRosen3 - 120
whitehorseking88 - 118
chesspro - 113
hspotter2002 - 112
Chigr - 110
hello - 106
sheepwarrior - 104

MVP's
metsfan001 - 7
hezgotzbeenz - 6
yif man 12, aznphatso, paladin314159 - 3
GoBraves, miraculouspostmaster, Ankur87 - 2
ComplexZeta, JSRosen3 - 1

Overall Rating (using Lucky's algorithm ... based on capped experience and 50% of rating is based on points per question)
(all ratings above 70 are shown)
Lucky707 - 96.15
metsfan001 - 92.62
Aznphatso - 86.15
paladin314159 - 84.40
yif man 12 - 83.64
DarkKnight - 79.70
GoBraves - 77.86
Ankur87 - 76.91
mirauclouspostmaster - 73.20
chesspro - 71.95
Chigr - 71.85
whitehorseking88 - 70.14

Awesome. I'm in. I'm looking forward to being the first to lose or getting the lowest or whatever.

Sounds fun! :) But I'll lose of course :P

[hide]Im in, adentro, po, binnen,dentro,μέσα,внутри[/hide]

IM= AmeemShk

Hmm... lol, I count 8 people already. [b] This does not mean that sign-ups are closed [/b]

The time is this Monday at 8 PM EST.

If you want in, I'll be on at ~7:45 PM. Send me a message and I'll invite you in.

Once in, I will decide on the teams and we will go at it for an hour.

Leave your AIM here if you want an invitation at the time of the AIT. This goes for all interested people, not just for the ones that replied above.

If it's Monday night, I won't be able to make it.  Sorry.

Don't worry about it. I'll hold more of them.

SO FAR: (TOTAL: 8 with 1 maybe)
plokoon
metsfan
GoBraves
yif man 12
hello
qwertyeq
mccalderbank
whitehorseking88
sheepwarrior ???

If you'd like to join, just leave your AIM here so I can invite you on Monday at 8PM EST.

I'll sign up, whoever wants to be my partner, im fine with it

I'll play.

EDIT: actually, i'll be gone friday, so i cant play.

AOPSTI is scheduled for this Friday at 8 EST

It won't be a typical AOPSTI, rather it will be more like the old format, except with buzzing. ie. most questions are categories or regular questions.

Come if you'd like.

Er... which format are you hosting under?

[b] Final Results of Tournament 2 If Anyone Cares [/b]

ROUND ONE:
*Phatso/Penguin 49, Hello/Potter 15
Mets/Dark 37, Montoya/Leo 32
Ameem/Ankur 29, Yif/Rosen 44
Sponge/Perry 22, Chigr/Paladin 58

ROUND TWO:
**Mets/Dark 41, Paladin/Chigr 42
***Phatso/Penguin 38, Yif/Rosen 34

CHAMPIONSHIP:
Phatso/Penguin 51, Chigr/Paladin 26

* - Denotes match was split into halves. Phatso faced Hello 1v1 in first half and Penguin faced Potter 1v1 in second half.
** - Denotes match went to a 3 question overtime.
*** - Denotes match was not only split into halves but went to a 5 question overtime with all players participating.

I've been gone for a while.
I don't understand the new format, so I'm just hosting under the two teams format. Come and separate into two teams. You know, the old way. :roll:

Yay, haven't had trivia for a long time. Still at 8est?

New way's just each person buzzes if they know and each team can only buzz once on a question.

[quote="white_horse_king88"]Anyone up for another AOPS trivia invitational? I will host tomorrow and same time as always. 7:30 - 9 EST
Of course, I'd prefer it earlier. That's up to you.[/quote]

I wouldn't like it earlier, since the mock AMC B is ending at 7:15.

I maybe won't do it because I'm not seeing much interest.

i would play, but ill probably have a unexpected event when you do the game


abstract algebra
function
Ring Theory
group theory
superior algebra
superior algebra unsolved

How do I prove that the rings Z2 x Z2 AND Z4 are not isomorphic?
From this, is there a ring homomorphism Z4 to (Z2 x Z2) and is there a ring homomorphism (Z2 x Z2) to Z4?
Note, for me, a ring homomorphism takes 1 to 1.
Thanks in advance for help.

Use the classification theorem of finite abelian groups to prove that the two aren't even isomorphic as groups.
Edit: if you don't know the theorem, show one is cyclic and the other is not. Also, you can show that in $ \mathbb{Z}_4$ there is a nilpotent element where as there are no nilpotent elements in $ \mathbb{Z}_2 \times \mathbb{Z}_2$ The second part shouldn't be so hard once you understand the structure of the rings better.

I figured out the isomorphism part. I'm guessing that there does exist a ring homomorphism for both in the second part. But how do I define the functions? f(x) = x? I don't get how you set up the function for the product ring. Obviously, I'm just learning abstract algebra.

Well in this specific case, you really only have four elements so you could in principle write out the function element by element. So you said that for this problem, homomorphism means that it takes 1 to 1 (and 0 to 0 also).
This takes care of 2 of the four elements. That only leaves us with two more to mess with.
So lets list the elements of $ \mathbb{Z}_2 \times \mathbb{Z}_2$
We have: (0,0) (1,1) (1,0) (0,1)
What elements do we have in $ \mathbb{Z}_4$? Well we have 0,1,2,3

We know that f(0) =  (0,0) and f(1) = (1,1) so where can we send 2 and 3?
Well, 2 = 1+1 so the only place we can send f(2) is f(1+1) = f(1)+f(1) = (1,1)+(1,1) = 0

Now what about 3 = 1+1+1. by a similar argument f(3) = (1,1)+(1,1)+(1,1) which is (1,1)

this is automatically a group homomorphism because we defined everything based on where the generator goes. does this make a ring homomorphism though?

Lets check f(2*3) = f(2). This should equal f(2)*f(3). This is in fact true.
How about f(1*x)? Well this is f(1)*f(x) but f(1) is the unit so this of course is f(x)
How about f(0*x)? Well f(0) is the 0 in the image so this is also equal f(0).

Try the reverse process yourself and see if theres a map going the other way.

Edit: if you're curious the argument i just posted is expressed in a more general phenomenon known as a "universal property". Heuristically, it says that if you can find two projections of a ring onto two other rings, there is a unique map from that ring onto the product of the two rings (satisfying some "niceness" conditions). In my opinion, this would be the quickest way to show there is a map from $ \mathbb{Z}_4$ to $ \mathbb{Z}_2 \times \mathbb{Z}_2$


function
algebra
domain
real analysis
real analysis unsolved

Does there exists a function $f: \mathbb{R}\setminus \{ 0 \}\to \mathbb{R}\setminus \{ 0 \}$ such that
$f(f(x))=-\frac1x$

$f(x)=\frac{x-1}{x+1}$ is interesting, but it's got the wrong domain and the wrong range.

There are such functions, but they can't be continuous, or even continuous except at a finite set of points. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=113408]here[/url] or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=52068]here[/url] for a pair of closely related problems.


Find the condition of $a_{0},a_{1},a_{2}$ such that the sequence ${a_n}$ satisfies:
                      $a_{n+2}=\frac {a_{n+1}a_{n}a_{n-1}}{6a_{n}a_{n-1}-11a_{n-1}a_{n+1}+6a_{n}a_{n+1}}$  for all $n \in N \geq 1$  has the infinite integer terms. :)


inequalities
function
logarithms
number theory proposed
number theory

Let $ H$ denote the set of those natural numbers for which $ \tau(n)$ divides $ n$, where $ \tau(n)$ is the number of divisors of $ n$. Show that

a) $ n! \in H$ for all sufficiently large $ n$,

b)$ H$ has density $ 0$.

[i]P. Erdos[/i]

(a) has already been discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=586&t=290291]here[/url].

(b) Let $S_p$ denote the set of positive integers with maximal prime divisor $p$ and $T_p$ denote the set of positive integers with all prime divisors $\le p$ (so obviously $S_p,T_p$ both have zero density within $\mathbb{N}$). If $n\in H_p=S_p\cap H$, then $\tau(n)\mid n\implies v_p(n)+1\mid n$, so $v_p(n)+1>1\in T_p$. Hence $H_p$ has zero density within $S_p$ (because $T_p$ has zero density within $\mathbb{N}$), so averaging over all primes, $H=H_{p_1}\cup H_{p_2}\cup\cdots$ (should have?) zero density within $\mathbb{N}=S_{p_1}\cup S_{p_2}\cup\cdots$.

I'm clueless as to how to rigorize (b), and I think the whole argument might be wrong in the first place, so can someone check it?

[b]Edit:[/b] Actually, this is trivial, because for all $\epsilon>0$ there exists $N$ such that for all $n>N$ and $k\ge1$, $|H_{p_k}(\le n)|\le \epsilon|S_{p_k}(\le n)|$, so because both $|H_{p_k}(\le n)|,|S_{p_k}(\le n)|$ eventually become zero we can just sum these inequalities up.

But we do need to prove $|H_{p_k}(\le n)|/|S_{p_k}(\le n)| = o(1)$ more rigorously (for fixed $p$). Let $\Psi(x,y)$ denote the $y$-[url=http://en.wikipedia.org/wiki/Smooth_number#Distribution]smooth number counting function[/url]. Because $\Psi(x,p_k)$ is the number of nonnegative integer solutions to $a_1\log{p_1}+\cdots+a_k\log{p_k}\le \log{x}$, simplex volume comparisons (both lower and upper lattice point bounds are easy to estimate) yield
\[\Psi(x,p_k)=\frac{\log^k{x}}{k!\log{p_1}\cdots\log{p_k}}+O(\log^{k-1}{x}).\]Because all numbers in $S_{p_k}$ and $H_{p_k}\subset S_{p_k}$ can be written as a positive power of $p_k$ times a $p_{k-1}$-smooth number, it's easy to see that because $T_{p_k}$ has zero density within $\mathbb{N}$ and $v_p(H_{p_k})+1\subseteq T_{p_k}$,
\begin{align*}
\frac{|H_{p_k}(\le n)|}{|S_{p_k}(\le n)|}
&\le \frac{\sum_{r=1}^{o(1)\log_{p_k}{n}}\Psi(n/p_k^r,p_{k-1})}{\sum_{r=1}^{\log_{p_k}{n}}\Psi(n/p_k^r,p_{k-1})} \\
&= \frac{O(\log{n}\log^{k-2}{n})+\sum_{r=1}^{o(1)\log_{p_k}{n}}\log^{k-1}{n/p_k^r}}{O(\log{n}\log^{k-2}{n})+\sum_{r=1}^{\log_{p_k}{n}}\log^{k-1}{n/p_k^r}} \\
&= \frac{O(\log{n}\log^{k-2}{n})+\sum_{r=1}^{o(1)\log_{p_k}{n}}\log^{k-1}{n}}{O(\log{n}\log^{k-2}{n})+\sum_{r=1}^{\log_{p_k}{n}}\log^{k-1}{n}} \\
&= \frac{O(\log^{k-1}{n})+o(1)\log_{p_k}{n}\log^{k-1}{n}}{O(\log^{k-1}{n})+\log_{p_k}{n}\log^{k-1}{n}} \\
&= o(1),
\end{align*}where we have used the fact that $\log^{k-1}{n/p_k^r}=\log^{k-1}{n}+O(\log^{k-2}{n})$.


Can anyone help me with this series problem? We've never done word problems and I guess I'm setting it up incorrectly becuase I'm having it diverge to infinity.

A ball drops from a height of 15 feet. Each time it hits the ground, it bounces up 65 percents of the height it fall. Assume it goes on forever, find the total distance it travels.

Thanks =)

-----
Sorry, I always forget that part!
I know it's the summation of a series from 1 (first drop) to infinity
Each time the height is 65% or 0.65 of the height before so the series would be:
15 + 9.75 + 6.3375 + ...
For the series I was doing this is 15(.65n)
But I'm not sure this really makes sense. Or if I'm even going about this right! Sorry, we just started this stuff Friday I'm still pretty new at it =o)

Why don't you write down what you've done?  Then we can see where you've gone wrong.

15+2[(15)(.65)+(15)(.65)^2+...]
that's just 15 plus twice a geometric series- how did you get that it diverges?


[quote="GameBot"]Each bottle of perfume at The Perfume Parlor contains $ \frac3{16}$ ounces of perfume. How many of these bottles of perfume could be made from 96 ounces of perfume?[/quote]

Our answer is $ 96 \div \frac{3}{16} \equal{} \boxed{512}$.


show that for every integer $n\ge 6$ , there exists a convex hexagon wich can be dissected into exactly  $n$ congruent triangles.

[color=red][Moderator edit: See also http://www.kalva.demon.co.uk/apmo/asoln/asol905.html for this problem.][/color]

trivial. and an old problem. (I remember doing it 4-5 years ago)
The trick is that you can use right angled triangles with sides a, b and (na) with $\left(na\right)^2=a^2+b^2$
(you only require n to be integer $\geq 2$) to make a rectangle of sides b and na. you need 2n triangles for this.
then you add two of these triangles to the sides to get a convex hexagon.
so we can do it for any $\left(2n+2\right)\geq 6$ triangles.
-Ali


floor function
inequalities
Euler
integration
calculus
function
arithmetic sequence

Let $ A \equal{} \sum_{k \equal{} 1}^{10^{289}}{\frac {1}{\sqrt [17]{k}}}$

Determine the integer part of $ A$

 :)

i believe $ \lfloor A \rfloor \equal{} \frac {17}{16}\cdot 10^{17\cdot 16} \minus{} 1$.

:lol:  Yes that's the correct answer.

I wonder if your method is the same as mine???

What did you d0?

I was working with this kind of problem, but with square and cubic roots so I generalized an inequality wich works now for every real greater than 1, and of the form $ \frac {1}{p}$ where $ p$ is an integer greater than one.

 :)

A straightforward solution would be just to bash the problem with Euler-Maclaurin (Note that the term is almost the same as $ \int_1^{10^{289}} \frac{1}{\sqrt[17]{x}}\,\mathrm dx$) and estimate the remainder, which is most likely less than one.

Here's another interesting problem:

Find all $ (a,n,m)\in(\mathbb{Z}^{+})^3$ for wich the integer part of $ B=\sum_{k=1}^{n^{m(a+1)}}{\frac{1}{k^{\frac{a}{a+1}}}}$ is a multiple of $ 2007$.

 :)

I don't know much about calculus so I haven't read with much attention bertram's post, but suddenly I realized that my inequality give the same result as that integral, in fact my inequality is like an integration, so this nice result came to my mind:

Let $ f$ be a continuous, decreasing function in the interval $ [a,b]$ wich contains $ x$, $ x \plus{} \alpha$ and $ x \minus{} \alpha$.

$ \frac {f(x \plus{} \alpha) \plus{} f(x)}{2} \le f(x) \le \frac {f(x) \plus{} f(x \minus{} \alpha)}{2} \Leftrightarrow \frac {1}{\alpha}(\alpha\frac {f(x \plus{} \alpha) \plus{} f(x)}{2}) \le f(x) \le \frac {1}{\alpha}(\frac {f(x) \plus{} f(x \minus{} \alpha)}{2})$

If the function is convex:
$ \frac {1}{\alpha}\int_x^{x \plus{} \alpha}{f(x) \mathrm dx}\le\frac {1}{\alpha}(\alpha\frac {f(x \plus{} \alpha) \plus{} f(x)}{2})\le f(x)$

Similar if $ f$ is concave, increasing..., when you have an arithmetic progression, and you sum this inequalitys for consecutive values of the progression you are like telescoping so it may be very useful for certain functions and values, as the problems posted before.

From this point the only advantage that my inequality has is that it gives and upper and lower bound for the same function, and it can be easily proved by AM GM.

Here it is:

$ \frac {k}{k \minus{} 1}((n \plus{} 1)^{\frac {k \minus{} 1}{k}} \minus{} n^{\frac {k \minus{} 1}{k}}) < \frac {1}{n^{\frac {1}{k}}} < \frac {k}{k \minus{} 1}(n^{\frac {k \minus{} 1}{k}} \minus{} (n \minus{} 1)^{\frac {k \minus{} 1}{k}})$

I hope someone replies to the problems, with a complete solution.

 :)


σημερα ειναι τα γενεθλια του ελληνικου φορουμ.ΕΝΑΣ χρονος ελληνικο....καλη συνεχεια σε ολους λοιπον.... :clap:

Χρονια πολλά !  :lol:

πότε θα :gathering: για να το γιορτάσουμε?
χρόνια πολλά...

Θανο ,ρε φιλε μακαρι να ειχαμε μια ηλεκτρονικη τουρτα...ή μηπως ειμαι ακραιος?  :P

Λοιπόν φίλες κι φίλοι συμπληρώνεται σήμερα 4/9/06 ένας χρόνος Ελληνικό μαθηματικό φόρουμ χρόνια του πολλά κι πριν κλείσω αυτό το μήνυμα θα ήθελα να αναφερθώ σε αυτόν τον 1 χρόνο ο οποίος φαντάζομαι να μας ήταν εποικοδομητικός Πιστεύω ότι ήταν ένας χρόνος με καλά στοιχεία που αναβάθμισαν το ελληνικό φόρουμ όπως 1 Η προσπάθεια κι η προσήλωση των μελών του φόρουμ η οποία ήταν  αξιοθαύμαστη   κι 2 η δημιουργία του σαιτ του Αλέξανδρου η οποία σηματοδοτεί την κορύφωση μιας υπερπροσπάθειας από μέρους του να βοηθήσει στην διάδοση ωφελίμων βιβλίων για Ολυμπιάδες Τέλος θέλω να ευχηθώ καλή συνέχεια στο φόρουμ με την ίδια προσπάθεια κι προσήλωση κι θέλω επίσης να πω ότι η δική μου συνεισφορά  ήταν μικρότερη των περιστάσεων κι υπόσχομαι να βελτιωθώ
Ευχαριστώ πολύ κι χρόνια πολλά ξανά
Μετά τιμής
Αμβρόσιος Σαββίδης (ambros)
ps υπερβάλλεις κώστα

Πως σας φαίνεται  :P  :D

[quote="RAKAR"]Χρονια πολλα λοιπον και απο εμενα,για τον εναν χρονο ελληνικο forum!!!!
A,χρονια πολλα και για κατι αλλο.....  7 χρονια alter!!!! :rotfl:
Α,ωραια η τουρτα αλλα ειστε σιγουροι οτι ειναι τουρτα γενεθλιων,γιατι εμενα μου φαινεται για τουρτα αποκριας...... :rotfl:

ΡΑΛΛΗΣ :D[/quote]

 :D  :lol:

για ποια τούρτα μιλάς???

ρε παιδια ειστε υπεροχοι. ρε αποστολη μπραβο...αυτο ειναι 7 χρονια αλτερ ναιιιιιιι...χαχχαχαχαχα και, λιγη μουσικη καποιος   :lol:  :rotfl:

na zisis aopsgf kai y polla
oo na gineis me topic polla
ktl

κριμα να μην ειμαι εκεινη την ημερα εκει.δεν πειραζει  ειμαι τωρα τουλαχιστον.φανταστειτε οτι το βρικα καταλαθος το μαθλινκς.μεσω αναζητησης  :o  φανταζομαι οι περισσοτεροι ετσι θα το βρηκαν.ο Αμβροσιος εκανε κατι σημαντικο:ενα απολογισμο....σωστος ομως εκει που μιλα για τον εαυτο του ειναι αδικος.πιστεψε με ρε μεγαλε εχεις δωσει πολλα .Αμβοσιος for ever !

Χρόνια πολλά και να τα χιλιάσουμε!!!!!!!!!!!!
 :thumbup:

Πότε θα κόψουμε την τούρτα; :lol:

P.S.Ράλλη είδες που σου έλεγα "χρόνια πολλά".Κάτι ήξερα...! :wink:

Edit:Ουφ,στο τσαφ πρόλαβα! 12:00 am είναι η ώρα!  :P

Kathws i giorti krataei (opws lene) 40 meres, kathws den exw alli dikaiologia gia tin kathisterisi mou, kai telos epeidi stous fisikous vazw kai to 0 (ara simera kleinei 1 xrono to forum) thelw na euxithw sto elliniko forum xronia polla kai sinexws na anavathmizetai apo idees, atoma kai askiseis. Auto pou euxomai omws olopsixa einai na voithisei tous ellines mathites gia na anavathmistei i olimpiaki paideia sti xwra mas... as min perimenoume tous "megalous" na kanoun kiniseis gi'auto... as prospathisei o kathenas mathitis monos tou na parei ta efodia gia na sinexisei!! Se auti tin prospatheia tou, oloi emeis tha stathoume sinodiporoi!!

Na eiste kala,
Alexandros

Αγαπητοί φίλοι έχετε όλοι ανεξαιρέτως δίκιο επίσης εκπληκτική η τούρτα ευχαριστώ κώστα για τα καλά σου λόγια κι θα συνεχίσω με πιο όρεξη
χαιρετώ

Να πω και εγώ τα χρόνια πολλά στο <<ελληνικά>> .Ήταν ένας πολυτάραχος χρόνος θα μπορούσα να πω .Ξεκινήσαμε με ενθουσιασμό μετά τίποτα και τώρα έχουμε 6 μήνες που πάλι είμαστε σε ένα πολύ καλό επίπεδο .Έχουν συζητηθεί φοβερές ασκήσεις , έχουν λυθεί από σπουδαία μέλη (μην ξεχνάτε ότι έχουμε αρκετούς medalist ΙΜΟ αλλά και διαγωνισθέντες σε ΙΜΟ) αλλά και άλλους κάποιους λίγο πιο μέτριους (π.χ εγώ) που όμως προσπαθούμε μπας και κάνουμε κάτι .
Ας συνεχίσουμε έτσι με περισσότερες ασκήσεις που θα λύνονται κάποια στιγμή όμως και με μεθόδους που όλοι τις επιζητούμε .
Θα ήταν επίσης καλό να γίνουμε λίγο πιο οργανωμένοι (χωρίζοντας το φόρουμ σε υπο-φόρουμ )(αρκεί να ζητήσουμε από τον Vornicu ).
Na είστε καλά όλοι και κάθε επιτυχία

Υ.Γ. Ελπίζω ο Στέργιος και ο Νίκος (Ράπανος ) να γραφουν κάτι τις που και που .

κι ένας καινούριος

>χρήστος
>θεσσαλονίκη

sorry ξέχασα να κάνω upload... enjoy!
christinho αν θέλεις υπάρχει ένα topic για να συστηθείς και να μπεις στον κατάλογο των μελών. Απλώς στείλε ένα post. Δε σε βάζω με δική μου πρωτοβουλία γιατί την τελευταία φορά που το έκανα έμπλεξα (με M stedes)

Ωραία τούρτα!  :rotfl:
Εσύ την έφτιαξες; Κι εκείνα από πάνω τί είναι; Ελιές;  :rotfl:

ναι και όχι! είναι κάτι σαν μπισκοτάκια

Μπισκοτάκια μμμ, μου άνοιξες την όρεξη!
Αλήθεια,το Μ που έγραψες πιο πάνω τι σημαίνει;;;;  :maybe:

το F τι σημαίνει?
[color=#f8f8f8]http://home.comcast.net/~wolfand/[/color]

Δεν νομίζω να εννοούσες αυτό όταν το έγραφες.
[color=#f0f0f0]http://www.humor.gr/modules.php?name=News&file=article&sid=45[/color]  :mad:

Γιατί απλά δεν κολλάει σ'αυτά που γράφεις.

έμπλεξα με σένα... ΔΕΜ?

ΦΚΕ!
Αλλά γιατί το έγραψες;

για να μην μπλέξουν και άλλοι... :rotfl:
ΚΝ&SD τώρα, οκ? bye!


$\triangle ABC$ is a right triangle with $\angle A=90^\circ$. $K$, $L$ are the points outside of $\triangle ABC$ such that $\triangle KAB$ and $\triangle LAC$ are equilateral triangles.

Let $M$ be the intersection of $KC$ and $LB$. Find $\angle KML$.

EDIT- Since my previous figure was a little bit confusing, I attached a new picture. ($AB$ is not necessarily equal to $AC$.)

Since KA=BA, LA=CA, and angle KAC=LAB (Both are equal to 60+90), triangles KAC and LAB are congruent.
Since angle AKC=ABL, quadrilateral AKBM is concylic.
Then we get angle BMK=BAK=60. So, angle KML is 180-60=120

Is it $120$ degrees? The way I did it was that since triangle $ALC$ is equilateral, all the angles are $60$, therefore, angle $LAB$ was $150$ and angle $ALB$ and angle $ABL$ are $15$, and then since angle $ABC$ and $ACB$ are equal, they both are $45$, so then angle $MBC$ and $MCB$ are each $30$ and so angle $BMC$ is $120$ and using verticle angles, angle $KML=\boxed{120}$.

P.S. What's the latex code for angles?

[quote="krustyteklown"]You can't assume that the two angles of the right triangle are equal, cuz it might be a 30,60,90 triangle, for example. But 120 is right.[/quote]

I assumed it was a $45,45,90$ triangle because triangles $KAC$ and $BAL$ were congruent using the diagram.

[quote="Iversonfan2005"]
P.S. What's the latex code for angles?[/quote]

it's \angle $\angle$
for degrees you can use ^{\circ}, EX: $120^{\circ}$

also i do not understand how you can make such an assumption that the right triangle is 45-90-45. or anything else....

there is nothing that confirms that, and you are not allowed to do such things in math, only because the diagram looks like it.. :huuh:

The educated guessing strategy is much faster than all of your slow solutions :P It only takes $\le 3$ seconds.

First notice that it's talking about equilateral triangles. Then notice that it's an obtuse angle. No other angle measurements are given. $120^{\circ}$ comes to mind. :D

Actually, that's a good strategy before you start, so that you'll have an $\approx$ for the answer.

Sorry that my picture caused some confusions :oops:
But anyway, $120^\circ$ is correct.

EDIT- I changed my picutre now. :)

[quote="krustyteklown"]Then we get angle BMK=BAK=60.[/quote]

Where did that come from? :? :blush:

[quote="nr1337"]When I did the problem, I also thought "Well, let's just say the right triangle is 45-90-45..."[/quote]
That's actually a good strategy especially if the problem only asks the answer, not complete solution :laugh:

[quote="krustyteklown"]It's from the cyclic-ness of AKMB[/quote]

Okay I see...I'm inexperienced with cyclicness. Can you explain how you proved that it was cyclic? :blush:

[quote="krustyteklown"]Since angle AKC=ABL, quadrilateral AKBM is concylic.[/quote]

Well yeah, I still don't understand how that proves anything. ;)

[quote="shobber"]We don't need the right angle. The conclution is true for any triangles.[/quote]

Nice, I didn't see that, although I didn't use the right angle in my proof, hehe.

There are more solution paths given that $ABC$ is a right triangle.  My original method, before I read the replies to the question, did utilize the right angle.  Since I didn't see that triangles $KAC$ and $BAL$ were equilateral, I said that $K$, $A$, and the midpoint of $CL$ were collinear (beacause $60^\circ + 90^\circ + 30^\circ = 180^\circ$).  $KA$ is perpendicular to $CL$ because triangle $ACL$ is equilateral, so the altitude and median to side $CL$ will be concurrent.  Hence, triangle $KCL$ is isoceles, making $\angle KLA = \angle ACK$.  Since triangle $KBL$ is isoceles for similar reasons, $\angle KLA = \angle ALB = \angle ACK$.  The latter two of these angles being equal tells us that quadrilateral $AMCL$ is cyclic, so $\angle LAC = \angle LMC = 60^\circ$, which tells us that $\angle KML = 180^\circ - 60^\circ = 120^\circ$.


is there a formula for a question where there r 6 people standing in a line and 3 wont stand next to each other. how many ways are there for them to stand

yeah there is

so, first you do the total amount of ways that they could go in order which is

6! or 6*5*4...*1 or 720

Then, you find the amount of ways that they could stand together for this, pretend that the three are one block, so pretend there are 4 people, so 4! is 24. But, the 3 could be in different orders within their group, so it is

4!3! or 144

Then you do 720-144, or 576.

Edit.

I'm quite sure 4!3! doesn't equal 140. :lol:

Sorry if this is spam.

how i would do it is
there are 3 people that can stand anywhere in 6 ways. the other 3 people can go inbetween them in 4 ways. so 6*4=24

did i misread it or somethhing? can two be together but just not all 3?

u got the first part right, with 24, but remember, they can be ordered in different ways inside of their 3-person group. so it is 24*3! or 144.

One other thing that you missed is the 3 are NOT supposed to be together, not that they are.

if the "o"s wont go together then you could have:

ononon
ononno
onnono
nonono

and both the n's and the o's could be arranged 6 ways so 4*6*6=144


inequalities
function
inequalities unsolved

let aa,b,c be positeve real number prove that
          \sum a 2 (b+c) \geq 2 (a 2 b 2 c 2 )^1/3*(a+b+c) :?  :(

Can you tell me the source of the problem ?
I think you should tell more clearly about it .
If you don't know the source , write Unknown.

[quote="Minh Thang"]By Cauchy
a 2 b+a  2 c+abc \geq 3(a^5*b^2*c^2)^(1/3)
The similar  \sum (sym)a 2 b+3abc \geq 3( \sum a^5*b^2*c^2)^(1/3) =2( \sum a^5*b^2*c^2)^(1/3)+( \sum a^5*b^2*c^2)^(1/3) \geq 2( \sum a^5*b^2*c^2)^(1/3)+3abc[/quote]

I think this should be

The similar  \sum (sym)a 2 b+3abc \geq 3 \sum (a^5*b^2*c^2)^(1/3) =2 \sum (a^5*b^2*c^2)^(1/3)+ \sum (a^5*b^2*c^2)^(1/3) \geq 2 \sum (a^5*b^2*c^2)^(1/3)+3abc.

Nice solution indeed (and yes, it's true). Somehow I failed to note your reply two months ago and kept searching for a solution on my own. Here is what I came up with:

We have to show the inequality

$a^{2}\left( b+c\right) +b^{2}\left( c+a\right) +c^{2}\left( a+b\right) \geq 2\root{3}\of{a^{2}b^{2}c^{2}}\cdot \left( a+b+c\right) $

for all positive numbers a, b, c.

This is obviously equivalent to $f\left( a,b,c\right) \geq 0$, where the function f(a, b, c) is defined as follows:

[tex]f\left( a,b,c\right) =\frac{a^{2}\left( b+c\right) +b^{2}\left( c+a\right) +c^{2}\left( a+b\right) }{a+b+c}-2\root{3}\of{a^{2}b^{2}c^{2}}[/tex].

Now, we want to reduce the problem of proving $f\left( a,b,c\right) \geq 0$ to a two-variable inequality by "mixing variables". The best way to do so is to replace both of the variables b and c by their geometric mean $\sqrt{bc}$; in fact, if we compute the difference $f\left( a,b,c\right) -f\left( a,\sqrt{bc},\sqrt{bc}\right) $, we see that the nasty $\root{3}\of{...}$ radicals cancel each other. Also, it's wise to consider $a^2$, $b^2$, $c^2$ instead of a, b, c here, so that the $\sqrt{...}$ radicals disappear, too. Here is the solution in detail:

Instead of proving $f\left( a,b,c\right) \geq 0$, we will show $f\left( a^{2},b^{2},c^{2}\right) \geq 0$. This is, of course, the same, since a, b, c are assumed positive.

We have

[tex]f\left( a^{2},b^{2},c^{2}\right) =\frac{a^{4}\left( b^{2}+c^{2}\right) +b^{4}\left( c^{2}+a^{2}\right) +c^{4}\left( a^{2}+b^{2}\right) }{a^{2}+b^{2}+c^{2}}-2\root{3}\of{a^{4}b^{4}c^{4}}[/tex].

We will calculate the difference $f\left( a^{2},b^{2},c^{2}\right) -f\left( a^{2},bc,bc\right) $. Since $a^{2}b^{2}c^{2}=a^{2}bcbc$, the radicals cancel each other, and we get

[tex]f\left( a^{2},b^{2},c^{2}\right) -f\left( a^{2},bc,bc\right) =\frac{a^{4}\left( b^{2}+c^{2}\right) +b^{4}\left( c^{2}+a^{2}\right) +c^{4}\left( a^{2}+b^{2}\right) }{a^{2}+b^{2}+c^{2}}-\frac{a^{4}\left( bc+bc\right) +b^{2}c^{2}\left( bc+a^{2}\right) +b^{2}c^{2}\left( a^{2}+bc\right) }{a^{2}+bc+bc}[/tex].

This simplifies to

[tex]f\left( a^{2},b^{2},c^{2}\right) -f\left( a^{2},bc,bc\right) =a^{2}\left( b-c\right) ^{2}\frac{2b^{3}c+a^{2}b^{2}+3b^{2}c^{2}+2bc^{3}+2bca^{2}+a^{4}+a^{2}c^{2}}{\left( a^{2}+b^{2}+c^{2}\right) \left( a^{2}+2bc\right) }[/tex],

what is obviously non-negative. Hence, $f\left( a^{2},b^{2},c^{2}\right) \geq f\left( a^{2},bc,bc\right) $. Now, it remains only to prove that $f\left( a^{2},bc,bc\right) \geq 0$. Denoting $a^{2}=x^{3}$ and $bc=y^{3}$, we get

[tex]f\left( a^{2},bc,bc\right) =f\left( x^{3},y^{3},y^{3}\right) =2y^{3}\left( x+y\right) \left( x-y\right) ^{2}\frac{x^{3}+y^{2}x+y^{3}}{x^{3}+2y^{3}}[/tex],

what is $\geq 0$ again, so we are done.

Of course, it's by far not so nice as your proof, but when I have seen yours, I had already written up the mine, and I didn't want it to get lost, hence this posting...

  Darij

Or: [tex]a^2(b+c)+b^2(c+a)+c^2(a+b)=(a+b+c)(ab+bc+ca)-3abc[/tex], and so we have to prove that [tex](a+b+c)(ab+bc+ca-2\sqrt[3]{a^2b^2c^2})\geq 3abc[/tex], which is true because [tex]a+b+c\geq 3\sqrt[3]{abc}[/tex] and [tex]ab+bc+ca\geq3\sqrt[3]{a^2b^2c^2}[/tex]. ;)

Hmm ! Chebychev inequality should help:
    We have
   $ \sum a^2(b+c) = \sum bc(b+c) $
     WLOG, we assume $a \geq b \geq c$. Thus
                  $ab \geq ac \geq bc $ and $a+b \geq a+c \geq b+c $
    Hence
     $ RHS \geq \frac {1}{3}(bc+ca+ab)(a+b+c) \geq LHS $


A square and a triangle have equal perimeters.  The lengths of the three sides of the triangle are $ 6.2$ cm, $ 8.3$ cm, and $ 9.5$ cm.  The area of the square is

\[ \textbf{(A)}\ 24 \text{ cm}^2 \qquad
\textbf{(B)}\ 36 \text{ cm}^2 \qquad
\textbf{(C)}\ 48 \text{ cm}^2 \qquad
\textbf{(D)}\ 64 \text{ cm}^2 \qquad
\textbf{(E)}\ 144 \text{ cm}^2
\]

Then the square has perimeter $ 24$.  Hence each side has length $ 6$, meaning the area of the square is $ 36$.

Just to clarify, to get that the perimeter is 24, you need to add 6.2+8.3+9.5=24.


limit
function
integration
inequalities
calculus
real analysis
real analysis unsolved

Show that

if $ f \in L_{\infty}$ and $ f \in L_q$ for some q,

then $ f \in L_p$ for all $ p > q$ and $ \| f \|_{\infty} \equal{} \lim_{p \to \infty} \|f\|_p$

suppose that $ f$ is a bounded measurable function on $ [0,1]$
we have $ \parallel{}f\parallel{}_{p} \equal{} (\int_{[0,1]}|f|^{p})^{\frac {1}{p}}\leq (\int_{[0,1]}\parallel{}f\parallel{}_{\infty}^{p})^{\frac {1}{p}} \equal{} \parallel{}f\parallel{}_{\infty}$
Hence $ \overline{lim}_{p\rightarrow\infty}\parallel{}f\parallel{}_{p}\leq \parallel{}f\parallel{}_{\infty}$

for $ \epsilon > 0$ let $ E \equal{} \{x: |f(x)| > \parallel{}f\parallel{}_{\infty} \minus{} \epsilon\}$

$ \parallel{}f\parallel{}_{p}\equal{}(\int_{[0,1]}|f|^{p})^{\frac {1}{p}}\geq (\int_{E}|f|^{p})^{\frac {1}{p}}\geq (\parallel{}f\parallel{}_{\infty} \minus{} \epsilon\})m(E)^{1/p}$ , from here you have $ \underline{lim}_{p\rightarrow\infty}\parallel{}f\parallel{}_{p}\geq \parallel{}f\parallel{}_{\infty}\minus{}\epsilon$

No, you do not have to assume that $ m(E)<\infty$ or that $ f$ is a function on $ [0,1]$ or whatever.  Use Holder's inequality to control the $ p\minus{}$norm $ (p>q)$ of $ f$ in terms of $ \|f\|_\infty$ and $ \|f\|_q$ instead.

[quote="blahblahblah"]Huh?  You are implicitly assuming that $ f$ is a function from $ [0,1]$ to $ \mathbb{R}$ (or $ \mathbb{C}$), and then using the fact that the $ L^p$ norm can be controlled strictly by the $ L^\infty$ norm in this setting.  This is not true if $ f$ is a function on $ \mathbb{R}$ or on some other measure space $ \Omega$ in which the underlying space has infinite measure.  There you need to use the assumption that there exists $ q < \infty$ with $ f\in L^q(\Omega)$.  In this setting (as I said, you can use Holder's inequality to get the bound $ \|f\|_p \leq \|f\|_q^\alpha\|f\|_\infty^\beta$ for $ p > q$ (figure out what $ \alpha$ and $ \beta$ have to be).

e: the post I replied to was deleted while I was composing this message.[/quote]

I can't figure it out..

Sorry but.. could you explain that explicitly?

I think we don't have to use Holder's inequality:

Since $ f\in L_{\infty}, |f|\le \|f\|_{\infty}$ a.e.

So $ |f|^p\le \|f\|_{\infty}^{p\minus{}q}.|f|^q$

Hence  $ \|f\|_p \leq \|f\|_q^{q/p}\|f\|_\infty^{1\minus{}q/p}$ Done!

Sure.  I think of that as Holder's inequality with exponents $ 1$ and $ \infty$.

Notice that 1234567a didn't use the fact that $ \Omega$ has finite measure in his proof of the reversed inequality :wink: .

[quote="limsup"]blahblahblah:

 Ok from that it is clear that $ \lim_{p \to \infty} \|f\|_p\le \| f \|_{\infty}$,  but what about the reversed inequality?[/quote]

Well, pick $ t\in [0,\|f\|_\infty)$.  By Chebyshev's inequality, the integral of $ |f|^p$ over $ A_t: \equal{} \{x: f(x)\geq t\}$ is at least $ t^pm(A_t)$, and (by the definition of $ \|f\|_\infty$), $ m(A_t) > 0$.  So the $ L^p$ norm of $ f$ is at least $ tm(A_t)^{1/p}$.  For fixed $ t$, letting $ p\to\infty$ shows that $ \liminf_{p\to\infty} \|f\|_p \geq t$ (since $ m(A_t)^{1/p}\to 1$).  Since $ t\in [0,\|f\|_\infty)$ is arbitrary, $ \liminf_{p\to\infty} \|f\|_p \geq \|f\|_\infty$.


Suppose I have 6 shirts, 4 ties, and 3 pairs of pants.  If an outfit requires a shirt and pants, and can either have a tie or not have a tie, how many outfits can I make?

[hide]This gets a little confusing because of the tie/no tie part. It seems like you could have 4 ties and 4 no ties, but there is only 1 true no tie![/hide]

The answer is easily seen to be $ 6\cdot(4\plus{}1)\cdot3\equal{}\boxed{90}$.

I get confused by this also.
The questions says: "Suppose I have $ 6$ shirts, $ 4$ ties, and $ 3$ pairs of pants." then "how many outfits can I make?" I think this last statement should be replaced by "how many way?" because if you have $ 3$ pairs of pants that's mean that you can't make more than $ 3$ outfits because you have only $ 3$ pants???

[hide=My personal solution :)]The choices for having a tie are $6 \cdot 3 \cdot 4 = 72$. The choices without the ties are $6 \cdot 3 = 18$. $72+18=\boxed{90}$.[/hide]

I did the same as g1zq.
[hide= Here’s How!]With ties: $6 \cdot 3 \cdot 4 = 72$
Without ties: $6 \cdot 3 = 18$
Altogether: $72 + 18 = \boxed{90}$[/hide]

[quote=limath1014][hide][hide] :wacko: Do not understand [/hide][/hide][/quote]

[hide=Hint]
Use the Multiplication Principle in Counting. That makes things much easier.  :)
[/hide]

[hide=my sol..?]I thought of considering the ties as Tie1, Tie2, Tie3, Tie4, and NoTie (Tie5). So, there are essentially 5 ties. 6 x 3 x 5 = 90[/hide]

uh

don't post solutions that are the same thing others have posted, even abstractly

its sort of a postfarm :/

[hide=My sol]I figured since [b]not[/b] wearing a tie counts as a "clothing" I added that to the tie count of $4$ and got,

[center]$6 \cdot 5 \cdot 3 = \boxed{90}$.[/center]

And we're done. $\smiley$[/hide]

[quote=son7][hide=My sol]I figured since [b]not[/b] wearing a tie counts as a "clothing" I added that to the tie count of $4$ and got,

[center]$6 \cdot 5 \cdot 3 = \boxed{90}.$[/center]

And we're done. $\smiley$[/hide][/quote]

fixed the slashes

edit: oh oops


real analysis
geometry
rectangle
induction
analytic geometry
real analysis solved

Let $X$ is a measurable set and $f : X\times \mathbb{R} \to \{0,1\}$. It is known that for each $a \in X$ we have $f(a,y)=0$ except countably many $y\in \mathbb{R}$. Also for each $b\in R$, $f(x,b)=0$ for almost all $x\in X$. Is it true that $f(x,y)=0$ a.e. in $X\times \mathbb{R}$?

There exists a non-measurable set $E$ on the plane such that every straight line intersects $E$ at no more than 2 points. So the answer is "No".

I deleted my reply the other night because I thought there was something wrong with it, but I don't think there was :).

Here's a construction that works here, I think. It's set with Lebesgue external measure $\ge 1$ (actually, it's not measurable, but we don't need that for the problem) which every horizontal or vertical line cuts in at most one point.

Let $\alpha$ be the smallest ordinal of continuum cardinal ($c$), and let $(C_u)_{u<\alpha}$ be an indexing of all countable unions of open rectangles with sides parallel to the axes, rational vertices, and total area $\le 1$ (there are clearly $c$ such unions). We can now perform transfinite induction as follows: for every ordinal $u<\alpha$ choose one of the $c$ vertical lines which have not been chosen at any previous step ("previous step" here means having chosen the line for some $u'<u$) and which have $c$ points outside $C_u$, and pick one of the $c$ points on this line which do not belong to $C_u$ and whose $y$ coordinates have not been chosen before (it's not hard to argue that we can indeed find $c$ vertical lines which satisfy these conditions).

The set formed by all the chosen points is our set: it's intersected by every horizontal or vertical line in at most one point, and it's not contained in any countable union of open rectangles with total area $\le 1$, meaning that it must have Lebesgue outer measure $\ge 1$.

P.S.

liyi, is "mikhail" (from the source field) our friend Myth? :)

[quote="grobber"]
liyi, is "mikhail" (from the source field) our friend Myth? :)[/quote]
Yes, I forgot to use his nickname here.  :blush:


probability
analytic geometry
graphing lines
slope

Can energy of activation be negative?
My opinion
[hide]
Yes
[/hide]

Well technically, energy is a vector.  In phsyics, negative energy means it is lost/gained depending on the situation.  To me, a negative activation energy implies energy is lost, which doesn't make sense to me.

Opinions:

Based on the Arrhenius Equation: $ k \equal{} Ae^{\frac { \minus{} E_a}{RT}}$ we can show that if $ E_a < 0$ then qualitatively a reaction proceeds slower as the temperature increases. Now I am sure these reactions do exist. However I am unsure of how valid this argument is. For one thing the approximate law of Arrhenius is, supposedly, formulated for the classical notion of $ E_a > 0$.

After a bit of searching I found:

$ 2 \ce{NO} \plus{} \ce {O2}$ $ \rightarrow 2 \ce{NO2}$

is a reaction whose rate decreases with temperature.

In fact I believe, from a bit more searching, that the explanation lies in the fact that the $ E_a$ value is only

[i]apparently[/i] negative. That is to say we have to look at the reactio mechanism.

$ 2 \ce{NO}$ $ \rightleftharpoons \ce{N2O2}$

$ \ce{N2O2} \plus{} \ce{O2}$ $ \rightarrow 2 \ce{NO2}$

The second process is the rate limiting step and we can thus derive the rate equation (I will leave that up to you and denote $ K_c$ as the equilibrium constant and $ k'$ as the rate constant for the second process).

$ R \equal{} k[NO]^{2}[O_2]$ where $ k \equal{} k'K_c$. This is the rate equation. Now $ K_c$ decreases with increasing temperature (dimerization is bond formation which is exothermic) and hence explains the anomalous behaviour of the process.

My conclusion would be that an actual negative $ E_a$ does not exist. After all the notion of the activation energy is just a model to explain the thermodynamics of a chemical reaction. No model is infallible.

[quote="BanishedTraitor"]
After a bit of searching I found:

$ 2 \ce{NO} \plus{} \ce {O2}$ $ \rightarrow 2 \ce{NO2}$

is a reaction whose rate decreases with temperature.

In fact I believe, from a bit more searching, that the explanation lies in the fact that the $ E_a$ value is only

[i]apparently[/i] negative. That is to say we have to look at the reaction mechanism.

$ 2 \ce{NO}$ $ \rightleftharpoons \ce{N2O2}$

$ \ce{N2O2} \plus{} \ce{O2}$ $ \rightarrow 2 \ce{NO2}$
[/quote]
Yes I was expecting someone to post that reaction ( My Prof also told me that )
But one has to accept that Eact is negative bcos we define it as such an expr.
Can someone answer this question.
What can you say abt the energy of activation of an exothermic reaction?
a. negative  b. positive  c. zero  d. cant say

If the energy of activation of a reaction is negative, that would appear to imply that the reactants take an intermediate form more stable than the final product.  It would seem that the reactions to and from the stable form need to be studied independently (since one is exothermic and the other endothermic); I believe this is related to BanishedTraitor's point.

[quote="t0rajir0u"]If the energy of activation of a reaction is negative, that would appear to imply that the reactants take an intermediate form more stable than the final product. [/quote]

Maybe we should empahasize it is [i]thermodynamically[/i] more stable.

Anyhow, we might have to wait for Carcul or someone else to make further comments.

I do agree that your reaction has a negative Ea.

But one clarification is the reaction exothermic???

I did not say that my reaction has a negative $ E_a$. I basically said that I don't believe that Arrehnius' model involving an energy profile (also the equation $ k\equal{}Ae^{\frac{\minus{}E_a}{RT}}$) is applicable in this case and we should treat it separately using rate equations.

I am not sure about the thermodynamic properties of the overall process. All I know is that the dimerization equilibrium should be exothermic (forward) because it is a bond forming process.

My answer: no, unless you have some really weird substance.  Think about what the activation energy of a reaction actually is.  It's the amount of energy difference between a [i]reactant[/i] (or intermediate) and the [i]transition state[/i].  So a negative activation energy would mean that the reactant has a higher energy than the transition state.  But this can only happen if our reactant is at a relative maximum on the potential energy curve, meaning that it would be extremely reactive... kind of like balancing a ball on the point of a needle, any slight disturbance would be enough to cause reaction.

I searched the net this is what [b]wikipedia[/b] said :-

[b][i]In some cases rates of reaction decrease with increasing temperature. When following an approximately exponential relationship so the rate constant can still be fit to an Arrhenius expression, this results in a negative value of Ea. Reactions exhibiting these negative activation energies are typically barrierless reactions, in which the reaction proceeding relies on the capture of the molecules in a potential well. Increasing the temperature leads to a reduced probability of the colliding molecules capturing one another (with more glancing collisions not leading to reaction as the higher momentum carries the colliding particles out of the potential well), expressed as a reaction cross section that decreases with increasing temperature. Such a situation no longer leads itself to direct interpretations as the height of a potential barrier.

[/i][/b]

Though I totally agree with greg. I think that would contradict one more theory relating to kinetics
which states that all the reactant molecules in that T.S. have a choice or it is like a see-saw.
It can either go to Product side or come back to reactants. If Ea were to be negative then this
whole theory of toppling fails as the T.S. becomes very stable and the Product will be the intermediate T.S. which is very absurd.

Hence the next question might be whether there is an alternative model which can add to/replace Arrehnius' Model of a Reaction (involving an activation energy barrier and a transition state/activated complex).

I agree to your views everyone but the fact is that Eact has been defined as so , so one must accept that Eact is negative in those cases or find another alternate model for this.  :P

[quote="JRav"]Well technically, energy is a vector.[/quote]

As far as I know, energy is a scalar quantity, not a vector. Regarding the issue of this thread, here's my opinion: the activation energy $ E_{act}$ is usually defined as the minimum amount of energy required for a chemical reaction to occur, where that energy is to be provided by collision between species. With this definition at hand, then only two things can happen: $ E_{act}$ is positive, and so there is in fact an energy barrier to overcome in order for reaction to occur, or $ E_{act} \equal{} 0$, in which case there is no barrier for the reaction to occur. Most of the processes with $ E_{act} \equal{} 0$ are reactions which involve the "capturing" of two species A and B, where the "complex" AB formed corresponds in the usual potential energy diagram to a "potential well", and not to a "potential hill". Obviously, for these reactions increasing the temperature correspond to decrease the probability of the two species A and B to capture one another in a colision, because now they have increasing energy. Therefore, for these reactions increasing the temperature leads to a decreasing reaction rate. However, if we make a plot using Arrhenius equation (ln k versus 1/T) then we will obviously get a positive slope, which means that $ E_{act} < 0$, but we already know that in this case the activation energy (as we defined above) is zero, because species A and B do not have to overcome any energy barrier. Geting a negative $ E_{act}$ is just a result of modeling the reaction with Arrhenius equation.

Yes all that is ok but what answer will you write in an exam which asks the foll. question:
Can Eact be negative?

The post I wrote above is a possible answer to that question.

Okay I understand your post but in an objective question if there are only 2 options a. can be negative
b. cannot be negative   c. Cannot be determined  and you have to write an ans and there is no question of reasoning with the examiner then what will be your answer?  :maybe:

Still my answer is that it will be positive unless the equation like the
one you referred to comes up.We only need to go with the majority.
Suppose in an organic reaction the product is asked we always take the
major product as the answer even though other options do occur in the reaction.

theiritically speaking i would say yes but practically know
well it clearly depends what u want to choos reactant species as
say if my reactant species were a carbocation then if the final product say is a halide which obviously has a lower potential energy than the internediate hence this transition has a negative Activation Energy .
i remember reading Olah's super Acid experiment where proabbly he was able to 'capture' a Carbocation hence if i were to take this as a reactant then the my activation energy would be negative

No.1 first understand what we are arguing.
When you a cabocation is chosen as the reactant then this reaction is a barrierless reaction and trying to define a hypothetical Ea for it is wrong.The carbocation is very reactive and so don't talk of an
Ea for it.

@Chemrock
i was answering madness's question to whether activation energy can be negativein a question paper
well tah depends on what u want to choose as ur reactant and as far as a carbocation goes it can equally revert back to the thing fromn which it originated from so there is a choice factor here also but both lead top a path of negative activation energy
for ex let's consider dehydration were the removal of $ \ce{H2O}^{\plus{}}$ is reversible
Here the Carbocation has a choice of reverting back to the original thing from where it is formed and that can in some sense have a negative activation energy
though this thing itself is meaning less(i agree with Carcul on this)
But if i were asked this 'indian' Question Paper i would write yes and state this

No its an objective paper with " NO MARKS MADE ANYWHERE ELSE IN THE ANSWER SHEET "  :P   :D

I would go for $ \text{NO}$.

After reading the Chemistry dictionary I think the Activation Energy is [i]defined[/i] to be positive.

I apologize for reviving this thread. But something I don't quite understand struck me. It concerns the potential well. So are we saying consider the following reaction:

$ A\plus{}B \rightarrow [AB] \rightarrow P$

The "transition" state lies in the potential well?

the conversion of ethyl free radical on reaction with hydrogen bromide to give ethane and bromine free radical has negative Eact

for a reaction to happen , it should have some activation energy...

I don think that Eact is negatiuve....

but it comes out to be in one of those examples Banished Traitor has given  :huh:

that is just a hypothetical concept....

and @ madness- where is the example...

see the third post of this thread da  :)
And if the model is not infalliable, then we mustn't use the model but as we are using it, we need to use it here also n say that the activation energy comes to be negative here  :)

actually the result i stated was due toa research work by P.W.Seakings,M.J.Pilling and co


What's AM-GM???
it's too new for me.
please explain!!!

AM-GM is very famous inequality in maths, and you should learn it if you want to do competition problems. It states the following: Let $x_1,\ldots,x_n$ be non-negative real numbers. Then $\frac{x_1+\ldots +x_n}{n}\geq \sqrt[n]{x_1\cdot \ldots \cdot x_n}$, i.e. the arithmetic mean of non-negative numbers is larger or equal than the geometric mean of those numbers. Equality holds if and only if $x_1=\cdots =x_n$.


trigonometry
ratio
quadratics
geometry
angle bisector
similar triangles
algebra

Source: A Mathematical Mosiac--by Ravi Vakil

A regular decagon of side 1 is inscribed in a circle.  What is the radius of the circle?

This is pretty easy once you get the right idea.

Yes, but I forgot to mention no calculators.

Hint: [hide]Why is your answer the golden ratio???? Think about that[/hide]

Let $O$ be the radius, and $\overline{AB}$ be a side of the dodecagon, so we have a isosceles triangle $\triangle OAB$ with $AB = 1$, $\overline{OA} \cong \overline{OB}$, $\angle O = \frac{\pi}{5}$, and $\angle A \cong \angle B$.  Now draw the angle bisector from $A$ intersecting $\overline{OB}$ at $C$, henceforth $\triangle ABC \sim \triangle OAB$ and $BC = \frac{1}{OA}$.  Using the angle bisector theorem, $OA - \frac{1}{OA} = 1$.  Solving, we find that $OA = \frac{1 + \sqrt{5}}{2}$.

That is a very, very clever way to do it.  My way (or Ravi Vakil's way) was different

[hide] You have a triangle with angles 72, 72, and 36, where the side opposite the 36 has lenght 1. Now draw a pentagon and draw all the diagonals in.  Let the sides of the inner pentagon have length 1, and fill in the measures of all the angles.  Locate one of the 72, 72, 36 triangles.   Use similar triangles and the quadratic formula, you find the length of the sides opposite the 72 degree angles are ${1 + \sqrt(5)}/2$  [/hide]

Can you explain how to "Use similar triangles and the quadratic formula" to find out the answer??

Yeah, well call the outer pentagon ABCDE and the inner pentagon formed by the intersection of the diagonals FGHIJ.

Ok, now, we want to find the lenghts of the sides of triangle AHI given that $AB=1$.  Sorry, in my last post I accidentally wrote that the lengths of the inner pentagon have length 1--I meant the outer pentagon--sorry.  Now, let $x$ denote the length of  $AC$.  Okay, now triangle AIB is similar to traingle ABC, so $AI/AB = AB/AC$.  Now, we substitute $AB=1$ and $AC=x$ to get that $AI=1/x$.  Now tirangle IBC is isosceles, so $IC=BC$.  Since $BC=1$, $IC=1$.  Now, $x=AC=AI+IC=1/x + 1$, so $x^2 - x - 1 = 0$.  The roots of this quadratic equation are x = $(1+\sqrt{5})/2$ and $(1-\sqrt{5})/2$.  Since distance is positive, the only solution is $x=(1+\sqrt{5})/2$.


What percent of the natural numbers from 1 to 100, inclusive, have at least one digit which is a 5?

You have 10 numbers with a 5 in the units digit, and 10 with a 5 in the tens digit. The only number you have double counted is 55, giving you $ \boxed{19}$


1. Find all $ f: \mathbb {R}\to\mathbb {R}$ such that
\[ f(x^2 \plus{} y) \plus{} f(f(x) \minus{} y) \equal{} 2f(f(x)) \plus{} 2y^2,\]
happens for all $ x,y\in \mathbb {R}$.

2. Find all $ f: \mathbb {R}\to\mathbb {R}$ such that for all $ x,y\in \mathbb {R}$, we have
\[f(f(x) \plus{} y) \equal{} 2x \plus{} f(f(y) \minus{} x).\]

[quote="mathVNpro"][u]Problem 1:[/u] Let $ f: \mathbb {R}\longrightarrow \mathbb {R}$. Find all $ f$ such that:
$ f(x^2 \plus{} y) \plus{} f(f(x) \minus{} y) \equal{} 2f(f(x)) \plus{} 2y^2$, $ \forall$ $ x,y\in \mathbb {R}$.[/quote]

Let $ P(x,y)$ be the assertion $ f(x^2 \plus{} y) \plus{} f(f(x) \minus{} y) \equal{} 2f(f(x)) \plus{} 2y^2$

$ P(x,0)$ $ \implies$ $ f(f(x))\equal{}f(x^2)$

$ P(x,f(x)\minus{}x^2)$ $ \implies$ $ f(f(x))\plus{}f(x^2)\equal{}2f(f(x))\plus{}2(f(x)\minus{}x^2)^2$ and so, since $ f(f(x))\equal{}f(x^2)$ : $ 2(f(x)\minus{}x^2)^2\equal{}0$ and so $ \boxed{f(x)\equal{}x^2}$

And it's easy to check that this function indeed is a solution.

Probleme 1:

let $ P(x,y)$ be the assertioon : $ f(x^2\plus{}y)\plus{}f(f(x)\minus{}y)\equal{}2f(f(x))\plus{}2y^2$

denote $ a\equal{}f(0)$

$ P(0,0)\implies f(a)\equal{}a$

$ P(0,a) \implies a\plus{}a\equal{}2a\plus{}2a^2 \implies a\equal{}0$

so $ f(0)\equal{}0$

$ P(x,f(x))\implies f(x^2\plus{}f(x))\equal{}2f(f(x))\plus{}2f^2(x)$ (1)

$ P(x,\minus{}x^2)\implies f(f(x)\plus{}x^2)\equal{}2f(f(x))\plus{}2x^4$  (2)

$ P(0,x)\implies f(x)\plus{}f(\minus{}x)\equal{}2x^2$  (3)

from (1) ,(2) and (3) we get $ f(x)\equal{}x^2 ,\forall x$

Mehdi

[quote="pco"][quote="mehdi cherif"]Probleme 1:

let $ P(x,y)$ be the assertioon : $ f(x^2 \plus{} y) \plus{} f(f(x) \minus{} y) \equal{} 2f(f(x)) \plus{} 2y^2$

denote $ a \equal{} f(0)$

$ P(0,0)\implies f(a) \equal{} a$

$ P(0,a) \implies a \plus{} a \equal{} 2a \plus{} 2a^2 \implies a \equal{} 0$

so $ f(0) \equal{} 0$

$ P(x,f(x))\implies f(x^2 \plus{} f(x)) \equal{} 2f(f(x)) \plus{} 2f^2(x)$ (1)

$ P(x, \minus{} x^2)\implies f(f(x) \plus{} x^2) \equal{} 2f(f(x)) \plus{} 2x^4$  (2)

$ P(0,x)\implies f(x) \plus{} f( \minus{} x) \equal{} 2x^2$  (3)

from (1) ,(2) and (3) we get $ f(x) \equal{} x^2 ,\forall x$

Mehdi[/quote]

Hello Mehdi !

Quite nice :) (I appreciated the usage of assertion 3 for solving $ f(x)^2 \equal{} x^4$)

Congrats.
(and sorry to have posted some seconds before you) :)[/quote]

thank's mr.Patrick  :)

[quote="mathVNpro"][u][u]Problem 2:[/u] Let $ f: \mathbb {R}\longrightarrow \mathbb {R}$. Find all $ f$ such that:
$ f(f(x) \plus{} y) \equal{} 2x \plus{} f(f(y) \minus{} x)$, $ \forall$ $ x,y\in \mathbb {R}$[/quote]

Let $ P(x,y)$ be the assertion $ f(f(x)\plus{}y)\equal{}2x\plus{}f(f(y)\minus{}x)$

$ P(\frac{f(v)}2,u\minus{}f(\frac{f(v)}2))$ $ \implies$ $ f(u)\equal{}f(v)\plus{}f(\text{something})$ and so $ f(u)\minus{}f(v)\in f(\mathbb R)$ $ \forall u,v$

So $ f(f(x)\plus{}y)\minus{}f(f(y)\minus{}x)\equal{}2x\in f(\mathbb R)$ and so $ f(x)$ is surjective.

If $ f(y_1)\equal{}f(y_2)$, then comparing $ P(x,y_1)$ and $ P(x,y_2)$, we get $ f(y_1\plus{}f(x))\equal{}f(y_2\plus{}f(x))$ and, since $ f(x)$ is surjective, $ f(x\plus{}T)\equal{}f(x)$ $ \forall x$ with $ T\equal{}y_1\minus{}y_2$

Comparing then $ P(x,y)$ and $ P(x\plus{}T,y)$, we get $ T\equal{}0$ and so $ f(x)$ is injective.

Then $ P(0,x)$ $ \implies$ $ f(x\plus{}f(0))\equal{}f(f(x))$ and, since $ f(x)$ is injective, $ f(x)\equal{}x\plus{}f(0)$

And it's immediate to check back that this indeed is a solution

Hence the result : $ \boxed{f(x)\equal{}x\plus{}a}$ $ \forall x$

[quote="mathVNpro"]
[u]Problem 2:[/u] Let $ f: \mathbb {R}\longrightarrow \mathbb {R}$. Find all $ f$ such that:
$ f(f(x) \plus{} y) \equal{} 2x \plus{} f(f(y) \minus{} x)$, $ \forall$ $ x,y\in \mathbb {R}$.

Best regard,
[b]mathVNpro[/b][/quote]
$ P(x,\minus{}f(x))\rightarrow f(0)\equal{}2x\plus{}f(f(\minus{}f(x))\minus{}x)$, so $ f$ is surjective.
Let $ a\in \mathbb R$ such that $ f(a)\equal{}0$.
$ P(a,y)\rightarrow f(y)\equal{}2a\plus{}f(f(y)\minus{}a)\Rightarrow f(y)\minus{}a\equal{}a\plus{}f(f(y)\minus{}a)$
Since $ f$ is surjective, for each real $ x$, there exist a real $ y$ for $ x\equal{}f(y)\minus{}a$. Then, $ x\equal{}a\plus{}f(x)$ for every $ x\in \mathbb R$.
The result  $ f(x)\equal{}x\plus{}c$, $ c\in \mathbb R$

[quote="mr.danh"][quote="mathVNpro"]
[u]Problem 2:[/u] Let $ f: \mathbb {R}\longrightarrow \mathbb {R}$. Find all $ f$ such that:
$ f(f(x) \plus{} y) \equal{} 2x \plus{} f(f(y) \minus{} x)$, $ \forall$ $ x,y\in \mathbb {R}$.

Best regard,
[b]mathVNpro[/b][/quote]
$ P(x, \minus{} f(x))\rightarrow f(0) \equal{} 2x \plus{} f(f( \minus{} f(x)) \minus{} x)$, so $ f$ is surjective.
Let $ a\in \mathbb R$ such that $ f(a) \equal{} 0$.
$ P(a,y)\rightarrow f(y) \equal{} 2a \plus{} f(f(y) \minus{} a)\Rightarrow f(y) \minus{} a \equal{} a \plus{} f(f(y) \minus{} a)$
Since $ f$ is surjective, for each real $ x$, there exist a real $ y$ for $ x \equal{} f(y) \minus{} a$. Then, $ x \equal{} a \plus{} f(x)$ for every $ x\in \mathbb R$.
The result  $ f(x) \equal{} x \plus{} c$, $ c\in \mathbb R$[/quote]

Quite OK. Nice and quick , and better than mine !

Congrats!  :)

How can I edit my text thus he saty like yours( R -> R) hehe
Thanks


[quote="scpajmb"]Let $x,y,z \ge 0,x+y+z = 1,n \in N,n \ge 2$
Prove that
$n^{n}y+y^{n}z+z^{n}x \le \frac{{n^{n}}}{{(n+1)^{n+1}}}$[/quote]
It must is $x^{n}y+...$  :wink:


An equilateral triangular of side $ 1$ is cuted into $ 3$ pieces, which are arrangened such that they form a square. What is the maximum lenght of the square?

[quote="Fraenkel"][hide]
calculate the area of the triangle.
It is the area of the corresponding square.
Take the square root and u have the length of a side of the square.
[/hide][/quote]

Following...

The formula for area is $ \frac{s^2\sqrt{3}}{4}$

we have $ \frac{\sqrt{3}}{4}$ as the area.

Thus the side of the square is $ \sqrt{\frac{\sqrt{3}}{4}}\equal{}\boxed{\frac{3^\frac{1}{4}}{2}}$

And that can be simplified $ \implies \boxed {\frac{13}{8}}$ or $ \boxed {1\frac{5}{8}}$

[quote="isabella2296"]And that can be simplified $ \implies \boxed {\frac {13}{8}}$ or $ \boxed {1\frac {5}{8}}$[/quote]

He was saying 3 to the 1/4th power, not 3 and 1/4. 3 to the 1/4th power is equal to the 4th root of 3.

$ 3^{\frac 14} \equal{} \sqrt[4]{3}$

well it can be simplified to 3/16 by quarting the top and the bottom.

Umm... I think that would be equivalent of raising the whole fraction to the 4th power, which is definitely NOT the same as the original fraction!


So it seems my school is probably going to only offer the B test. Is there any way for me to take the A test at another school, or should I just beg the head of the math department to change it? It's really discouraging, especially since when I talked to them about HMMT they said "We'll think about it" as the first registration deadline passed silently. The same thing will probably happen here.

Either ask politely, and if they reject, say that you are going to take it at a different school, and that's final.

[quote="worthawholebean"]It's really discouraging, especially since when I talked to them about HMMT they said "We'll think about it" as the first registration deadline passed silently.[/quote]

Ask the head of the math department every other day until he/she gives you an answer.

The problem is that there really isn't anywhere good to take the A nearby. All the other schools are taking it during the school day, and I can't leave then, and the colleges also only offer the B.

The advice in AMC Frequently Asked Question #4 generally applies here.
See http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Ask a local college or university to give the A date.  Offer to pay for the registration.

--  Steve Dunbar
MAA Director, American Mathematics Competitions


A point $ (x,y)$ is randomly selected such that $ 0 \le x \le 8$ and $ 0 \le y \le 4$. What is the probability that $ x\plus{}y \le 4$? Express your answer as a common fraction.

For $ x\plus{}y\le 4$, the region within the rectangle with vertices $ (0,0)$, $ (4,0)$, $ (4,8)$, and $ (0,8)$ is the isosceles right triangle with vertices $ (0,0)$, $ (4,0)$, and $ (0,4)$.

Geometric probability of this occurring is just $ \frac{\frac{1}{2}(4)^2}{(8)(4)}\equal{}\frac{8}{32}\equal{}\boxed{\frac{1}{4}}$.


$ABC$ is a triangle with $AB=min\{AB,AC,BC\}$ and $P$ is a point in the interior of the triangle .
Prove that $PA+PB+PC<AC+BC.$

There are different ways to solve this problem. Here is one of them:

Let E and F be the intersections between the parallel trhough P to AB with AC and BC respectively.
We have that AP < AE+EP  and BP < BF+FP. By similitude EF < min{EC, FC}. Then, since PC < max{EC, FC} we are done.

Nice solution, sprmnt21!
I meant to the function $f(P)=PA+PB+PC-AC-BC$ is convex in $\Delta ABC.$
Hence, $sup f$ for $P\in AB\cup AC\cup BC.$ :)


Does anyone know where I can find some copies of the old Arml-Nysml books?  I have the newest one from 1994-2002 with the power contests and I have ordered the one from 1989-1994.  I m pretty sure the older ones are out of print but I was hoping that maybe someone knew of where I could find some.

I think this question has been asked here several times, but I haven't heard of anyone who got a satisfactory response.  Perhaps all you people interested should petition ARML to re-release the older books, since there may be a real market for them.  I believe NYSML is (or has?) independently come out with a book covering the same period as the most recent ARML book.  Although NYSML tends to be easier than ARML, it has identical format and so is  a good alternate source of problems to look at.  Other than that, though, your only bet is to find someone who already owns it and photocopy.

I tried to order the 1989-1994 ARML volume on Amazon this summer.  Though Amazon said the book would ship in a month or two, they periodically sent me a message saying that the order has been delayed.  After about 4 months, I cancelled the order.

Go to their website, E-mail the director, and if they have any left (they had some in stock), he'll send one to you.  Thats how i got mine


USNCO 2006/23

For which exothermic reaction is $ \Delta E$ more negative than $ \Delta H$?

(A) $ Br_2 (l) \rightarrow Br_2 (g)$
(B) $ 2C(s) \plus{} O_2 (g) \rightarrow 2CO (g)$
(C) $ H_2 (g) \plus{} F_2 (g) \rightarrow 2HF (g)$
(D) $ 2SO_2 (g) \plus{} O_2 (g) \rightarrow 2SO_3 (g)$

The correct answer is D,

but I keep thinking its be B since you have more moles of gas produced, which would increase pressure and hence do work on the surroundings, which would make E=q+w more negative... :S

We are looking for when $ \Delta E < \Delta H$

We know that $ \Delta E \equal{} \Delta H \plus{} W$

Where $ W\equal{}P \Delta V$

Thus, in order for the first inequality to hold, $ W<0 \rightarrow P \Delta V <O$ or simply

$ \Delta V <0 \rightarrow V_f<V_i$

Basically now we go back to the elementary concept in stoichiometry, that in a chemical reaction, the coefficients can represent moles, molecules, and volume units.

Long story short, we're looking for the reaction where the sum of the coefficients of the gas reactants in the products is less than the sum of the gas reactant coefficients.

a: 1- 0 = 1>0 ... nope
b: 2-1 = 1> 0 ... nope
c. 2-2 = 0 which isnt greater than 0... nope
d. 2-3 = -1<0... and there you have it

[quote="feynman_wannabe"]

Where $ W \equal{} P \Delta V$

[/quote]

Hmmmm, not sure but I though it's supposed to be $ W \equal{} \minus{} P \Delta V$. :maybe:

By the way the $ E$ in this case is internal energy right?

EDIT...
OK my first response is flawed twice to get a correct answer...

[quote] Thus, in order for the first inequality to hold, [/quote]

yea.. that should be $ W>0 \rightarrow P\Delta V<0$ etc...

Thanks for the catch banished traitor..

Is the change in internal energy related at all to the change in free energy?
Also the change in free energy for the first reaction (A), would be zero right?

$ \Delta G^o \equal{} \Delta H^o \minus{} T\Delta S^o$

I guess you could substitute variables in and  find a relation

and if $ \Delta G \equal{} 0$, then the reaction is in equilibrium.

Okay, I'm really sorry for reviving an old thread, but I was really confused.

[quote="feynman_wannabe"]It doesnt really matter in this case whether its PdV or -PdV in this case since we just need E<H.

So... if you wanna get technical...

Let $ W\equal{}\minus{}k$ where $ K>0$

Thus we have

$ \Delta E \plus{}k\equal{} \Delta H  \rightarrow k\equal{} \Delta H \minus{} \Delta E\equal{}\minus{}P\Delta V$ Which must be greater than 0 (E<H => H-E>0). Therefore, we need $ k>0 \rightarrow P\Delta V <0$

etc..

And yes... $ \Delta E$ would refer to the Internal energy of the system. This whole premise of this question is the first law of thermodynamics[/quote]

I thought that the equation was $\Delta H = \Delta E + P \Delta V$, so that $\Delta H - \Delta E = P \Delta V$. So why does it say $ k=\Delta H-\Delta E=-P\Delta V $?


The perimeter of a square is 44 meters. How many square
meters are in the area of the square?

hello, let $ a$ the side length of the given square then we have $ 4a\equal{}44 \Leftrightarrow a\equal{}11$ and the area is $ a^2\equal{}121m^2$.
Sonnhard.


First of all, I apologise to Harazi and Arne. From now on, I'll write full solutions. :oops:
  Here is an inequality. From what I know so far this problem has a brute force solution. I have found a nice solution for this one, but no more comments, here it is:

   Let $a,b,c>0$ with $abc=1$.Prove that: $2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ ;)

$\frac{a}{b}+\frac{b}{c}+\frac{b}{c} \ge 3\sqrt[3]{\frac{ab}{c^2}} = \frac{3}{c}$
$\frac{a}{b}+\frac{a}{b}+\frac{b}{c} \ge 3\sqrt[3]{\frac{a^2}{bc}} = 3a$
Sum up, $\frac{a}{b}+\frac{b}{c} \ge a+\frac{1}{c}$
Add up similar inequalities and done.

Very nice. This is shorter than my solution. ;)


inequalities
logarithms
function
inequalities proposed

$a,b,c$ are positive numbers.prove that :
 $(\frac{a+b+c}{3})^{a+b+c}\geq a^{b}b^{c}c^{a}$

Beautiful!!!
Taking $\ln$ of both sides we get: \[(a+b+c)\ln\frac{a+b+c}{3}\geq\sum b\ln a\iff\]  \[\iff \sum b\ln a-b\ln\frac{a+b+c}{3}\leq 0\iff\]  \[\iff \sum\frac{b}{a+b+c}\ln\frac{a}{\frac{a+b+c}{3}}\leq 0.\] Applying Jensen for concave function $\ln t$ we get \[\sum\frac{b}{a+b+c}\ln\frac{a}{\frac{a+b+c}{3}}\leq \ln \frac{\sum\frac{ab}{a+b+c}}{\frac{a+b+c}{3}}.\] We have to prove \[\ln \frac{\sum\frac{ab}{a+b+c}}{\frac{a+b+c}{3}}\leq 0\iff\]  \[\iff \frac{\sum\frac{ab}{a+b+c}}{\frac{a+b+c}{3}}\leq 1 \iff\]  \[\iff \sum\frac{ab}{a+b+c}\leq \frac{a+b+c}{3}\iff\]  \[\iff 3\sum ab\leq (a+b+c)^{2}\iff 3\sum ab\leq a^{2}+b^{2}+c^{2}+2\sum ab\iff\]  \[\iff a^{2}+b^{2}+c^{2}\geq ab+bc+ca,\] which is true by rearrangement and we are done.

We may suppose that $a+b+c=3.$ Then by weighted AM-GM, \[a^{b}b^{c}c^{a}\leq \left(\frac{ab+bc+ca}{3}\right)^{a+b+c}\leq \left(\frac{(a+b+c)^{2}}{9}\right)^{a+b+c}=\left(\frac{a+b+c}{3}\right)^{a+b+c}.\]

both of your solution are very nice.and i think this problem is so nice too.thank you.if anyone have another solution plz send it.thanks.

here is my solution for this problem:
   we have:
      $\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1$
and we have:
   $a+b+c=\frac{(a+b+c)^{2}}{a+b+c}=\frac{a^{2}+b^{2}+c^{2}+2ab+2ac+2bc}{a+b+c}\geq 3(\frac{ab}{a+b+c}+\frac{bc}{a+b+c}+\frac{ca}{a+b+c}) \geq 3(a^{\frac{b}{a+b+c}}\times b^{\frac{c}{a+b+c}}\times c^{\frac{a}{a+b+c}})$
so we get :
    $(\frac{a+b+c}{3})^{a+b+c}\geq a^{b}b^{c}c^{a}$
and we R done.

[quote="Lovasz"]More general, see http://www.mathlinks.ro/Forum/viewtopic.php?t=32451[/quote]thanks alot. :D


function
floor function
ceiling function

From AoPS I:

If $ f(x) \equal{} \lfloor{x}\rfloor$ and $ g(x) \equal{} \lceil{x}\rceil$, express $ g(x)$ in terms of $ f(x)$ and $ x$.

Basically, I got stuck on the second step: (I got the rest of the proof)

1. If $ x$ is an integer: then $ \lfloor{x}\rfloor \equal{} \lceil{x}\rceil \equal{} x$

2. If $ x$ isn't an integer than $ \lfloor{x}\rfloor \plus{} 1 \equal{} \lceil{x}\rceil$ (it's "common sense", but how do we prove this?)

I'm probably missing something really basic :)

For 2, let $ \lfloor x\rfloor\equal{}x\minus{}\{x\},$ where $ \{x\}$ is the fractional part of $ x.$ Then let $ \lceil x\rceil\equal{}x\plus{}(1\minus{}\{x\}).$ It becomes obvious that \[ (x\minus{}\{x\})\plus{}1\equal{}x\plus{}(1\minus{}\{x\}).\]

Not really sure if this is rigorous or not.

since x is not an integer.
x= y+p/q(p<q).
thenfloor of x is y
and ceiling of x is y+1.
y+1-y=1.


Let $0<a\neq 1.$ Solve the system of equations:
\[\begin{cases}a^{x}+a^{y}=\frac{1}{2}\\ x+y=1\end{cases}\]

[hide]I think $b$ is $a$, if so  then system is equivalent to $a^{x}+a^{y}=\frac{1}{2},a^{x}\cdot a^{y}=a$. This is easy![/hide]

Is that first equation supposed to be $a^{x}+b^{y}= \frac{1}{2}$?

No, it's my typo  :blush:  And N.T.TUAN's post is correct. Thanks :)


Two cards are dealt from a standard deck of 52 cards.  Find the probability both cards will have the same color.

If you find Getting Started problems very easy, please leave this challenge problem for less experienced students to solve.

Source: Alabama State Contest

[hide]The probability of picking the first card is 1.  It doesn't matter what the first card is, since it never says it has to be a paticular color.  The probability of picking the second card with the same color as the first is 25/51, since one card has already been picked, the one with the color you want for the second card.[/hide]

What does AL mean?  Is it a typo or what?

[quote="h_s_potter2002"]What does AL mean?  Is it a typo or what?[/quote]

AL is the postal abbreviation for "Alabama," the state where the problem comes from.

Is this right?

One card is dealt. It does not matter what color that is. There are now 25 more cards of that color left in the deck, so the probability of picking one is 25/51

There are originally 26 cards of each color in the deck. Whichever color the first card is, there are 25 of the same color in the deck that now has 51 cards. The answer is 25/51.


[quote="13375P34K43V312"]There are 120 $n$-digit palindromes. What is the sum of all possible values of $n$?[/quote]
[hide="huh"]Wouldn't it be zero since there has to be a factor of 9 palindromes (the first and last digits)[/hide]

Think again.
How about this problem: How many four digit palindromes are there?
Before actually doing it, ask yourself what we're actually counting.

We're counting the number of digits the first and last can be, and the number of digits the middle two can be?
Which would make it 90?

I agree with bpms - unless I'm misreading the problem...

Ok I'm probably being stupid like usual but the first digit has 9 choices...120 isn't divisible by 9?

Oh wow, oops, I screwed up. Blahhh ok yes it is $9\cdot10^{n-1}$ k just ignore this thread from now on then

actually, it's 9 * 10^((n-2)/2), and take the ceiling of the exponent (round it up)

[quote="nutz_for2.718281828"][hide="solution"]The number of n digit palindromes is floored(n/2)*9, but 120 is not a multiple of 9. There are 0 values of n, so we can't find the sum of all possible values of n.[/hide][/quote]
Um, I think we all pointed that out, no need to spam.


geometry
circumcircle
perpendicular bisector
geometry unsolved

In $ \triangle ABC$($ AB<AC$).Let $ D$ be the midpoint of $ BC$,and $ E$ be the point on the median $ AD$.
Let $ EF \perp BC(F \in BC)$.
$ P$ is a point on the segment $ EF$.
Let $ M,N$ be the foot of perpendiculars from $ P$ to sides $ AB,AC$ respectively.
Prove that $ M,N,E$ are collinear iff $ \angle BAP\equal{}\angle PAC$

[quote="KDS"]In $ \triangle ABC$($ AB < AC$).Let $ D$ be the midpoint of $ BC$,and $ E$ be the point on the median $ AD$.
Let $ EF \perp BC(F \in BC)$.
$ P$ is a point on the segment $ EF$.
Let $ M,N$ be the foot of perpendiculars from $ P$ to sides $ AB,AC$ respectively.
Prove that $ M,N,E$ are collinear iff $ \angle BAP = \angle PAC$[/quote]
[hide="Solution"]Let $ I\in AB$, $ K\in AC$ such that $ IK\parallel BC$ and $ E\in IK$. Then, $ E$ is the midpoint of $ IK$ and $ EF$ is perpendicular bisector of $ IK$.
$ \angle IMP=\angle IEP=\angle KEP=\angle KNP=90^o \Rightarrow$ two quadrilateral $ IEPM$ and $ PENK$ are cyclic.
Suppose that $ M,N,E$ are collinear. Then, $ \angle PME=\angle PIE=\angle PKE=\angle PNE$
$ \Rightarrow \angle PMN=\angle PNM\Rightarrow PM=PN\Rightarrow \angle BAP = \angle PAC$
Suppose that $ \angle BAP = \angle PAC$ . Then, $ P$ is the intersection of the bisector of angle $ A$ and perp bisector of segment $ IK$, so $ P$ is the mipoint of arc $ IK$ (which doesn't contain $ A$) of the circumcircle $ (AIK)$.
We have that $ \angle MEP+\angle NEP=(180^o-\angle AIP)+(180^o-\angle AKP)=180^o$, which implies that $ M,N,E$ are collinear.
We are done!  :D [/hide]


find the inverse matrix of A
$ A\equal{}\left[\begin{array}{ccccc}1&1&1&\cdots&1\\1&\epsilon&{\epsilon}^2&\cdots&{\epsilon}^{n\minus{}1}\\1&{\epsilon}^2&{\epsilon}^4&\cdots&{\epsilon}^{2(n\minus{}1)}\\\vdots&\vdots&\vdots&\cdots&\vdots\\1&{\epsilon}^{n\minus{}1}&{\epsilon}^{2(n\minus{}1)}&\cdots&{\epsilon}^{(n\minus{}1)^2}\end{array}\right]$
here $ \epsilon\equal{}cos\frac{2\pi}n\plus{}i sin\frac{2\pi}n$
 :lol:

This is a [url=http://en.wikipedia.org/wiki/Vandermonde_matrix]Vandermonde Matrix[/url].

That's not a very useful thing to say. This is the Fourier matrix; $ A^*A\equal{}nI$, so its inverse is $ \frac1n$ times its conjugate.


logarithms
combinatorics proposed
combinatorics

Let $ m$ be a positive integer, and let $ S = (a_{1} = 1,a_{2},..., a_{m})$ be a set of positive integers. Prove that there exists a positive integer $ n$ with $ n \leq m$ and a set $ T = ({b_{1},b_{2},..., b_{n})}$ of positive integers such that
 (a) all the subsets of $ T$ have distinct sums of elements
 (b) each of the numbers $ a_{1},a_{2},..., a_{m}$ is the sum of the elements of a subset of $ T$

I think  $ 1, 2, ..., 2^k$  works, where  $ k \equal{} \log_{2}(\max\{a_i\})$

[quote="Namdung"]I think  $ 1, 2, ..., 2^k$  works, where  $ k \equal{} log_{2}(max{a_i})$[/quote]

It doesn't work when $k>m$.

Posted verbatim at JBL's first link - the poster did not even take the trouble to change notations. The problem is indeed interesting, but this does not mean we have to repeat it forever ...
The mentioned link above sends, in turn, to JBL's second link, where I presented a proof.

Topic locked; there is no interest to reopen the discussion in a new thread - all contributions must go to former postings.


Find the minimum value of $ a_{1}a_{2}a_{3}+b_{1}b_{2}b_{3}+c_{1}c_{2}c_{3}$, where the nine numbers are permutations of $ 1,2,3,...,9$.

[hide="Suprisingly simple solution"]We have by AM-GM that the expression is greater than $ 3\cdot \sqrt[3]{9!}\approx 213.98$ for every permutation.  Thus, the theoretical minimum is 214.  But we can actually achieve this, with all the permutations which give us a final expression like $ 9\cdot8\cdot1+7\cdot5\cdot2+6\cdot4\cdot3$.[/hide]

If this problem were given to you on a contest, how would you go about calculating the value you got with no calculating instruments? Since it's so close to the lower bound, approximating might be difficult because you might go over.

Like, suppose someone taking the contest was really bad at computation (there are many at the olympiad level). Would you be able to recommend a way for them to use this method to write up a solution? It seems like there might be a better way to solve this.

The solution process without being able to compute cube roots would look like this:

Discover the arrangement $ 9\cdot8\cdot1+7\cdot5\cdot2+6\cdot4\cdot3$.  Remark to oneself that 72, 70 and 72 are very close together, suggesting that we might be able to argue by AM-GM.  Observe that in order to show that 214 is the minimum, it suffices to show that there is some lower bound on the possible values which is larger than 213.  From AM-GM, note that every possible value must be at least $ 3\sqrt[3]{9!}$.  Suppose that there is a possible value $ x \leq 213$.  Then $ x^{3}= 27\cdot 9! \leq 213^{3}$, so $ 9! \leq 71^{3}$.  Now compute $ 9! = 326880$ and $ 71^{3}= 357911$, a contradiction.  Thus there are no solutions less than or equal to 213, so 214 is indeed the minimum.


calculus
integration
probability and stats

Let $ W$ be standard $ 1$-d Brownian Motion, $ T$ a stopping time with $ E(T) < \infty$.  Prove that $ E(W_T) \equal{} 0$ and $ E(W_T^2) \equal{} E(T)$.  Thanks

use that $ W_{\min{t,T}}^2 \minus{} \min(t,T)$ is a martingale to show that $ W_{\min{t,T}}$ is a uniformly integral martingale, hence the conclusion.


calculus
integration
function
derivative
calculus computations

Hello Everyone  :)

I'm kind of stuck on this exercise because I'm not sure whether or not I can integrate. The exercise:
Let $ f$  be a differentiable function on $ \mathbb{R}$ whose derivative satisfies: $ |f'(x)|<\frac{5}{6}$ for all $ x \in \mathbb{R}$.
Define the sequence: $ a_0 \equal{} 0$ and $ a_n \equal{} f(a_{n\minus{}1})$ for any natural $ n$. Prove that the sequence $ (a_n)$ converges.

If I can integrate $ f'(x)$ then I think I can prove that $ |{a_{n\plus{}1} \minus{} a_n}| < \frac{5}{6}|a_n \minus{} a_{n\minus{}1}|$. However, not every function that has an antiderivative is Riemann Integrable. So, is it possible to integrate here?

Thanks in advance!
Perrin

The right way around this one is to avoid integrals entirely: use the Mean Value theorem to get $ |f(x)\minus{}f(y)|\equal{}|f'(c)|\cdot |x\minus{}y|$ for some $ c$, and $ |f'(c)|<\frac56$.

Of course! I was thinking about integrals so much that I completely missed it. Thank you very much, you've really helped me!   :lol:


group theory
superior algebra
superior algebra unsolved

Major problems with this proof:

Let G be a finite cyclic group of order n. If d is a positive divisor of n, prove that the equation x^d=e has exactly d distinct solutions in G, where e is the identity in G.

Assume you have an x = g^m such that x^d = 0. Then g^md = 0, so n | md, so n/d | m, so f | m, so all solutions are of the form g^(in/d), of which there are d such elements (as g^n = e, and nothing less than n does that).


calculus
integration
function
trigonometry
calculus computations

Consider the function $ f(\theta) \equal{} \int_0^1 |\sqrt {1 \minus{} x^2} \minus{} \sin \theta|dx$ in the interval of $ 0\leq \theta \leq \frac {\pi}{2}$.

(1) Find the maximum and minimum values of $ f(\theta)$.

(2) Evaluate $ \int_0^{\frac {\pi}{2}} f(\theta)\ d\theta$.

hello, we get $ f(\theta)=%Error. "signum" is a bad command.
(\sin(\theta))\sin(\theta)-\frac{1}{4}%Error. "signum" is a bad command.
(\sin(\theta))\pi$ and
$ -\frac{1}{4}\pi\le f(\theta)\le 1-\frac{1}{4}\pi$, further we have $ \int_{0}^{\frac{\pi}{2}}f(\theta)\,d\theta=1-\frac{1}{8}\pi^2$.
Sonnhard.

[quote="1605272"]Let $ x \equal{} \cos t$
$ f(\theta) \equal{} \int_0^{\frac {\pi}{2}}|\sin t \minus{} \sin\theta|dt$
$ \equal{} \int_0^{\theta}(\sin\theta \minus{} \sin t)dt \plus{} \int_{\theta}^{\frac {\pi}{2}}(\sin t \minus{} \sin\theta)dt$
$ \equal{} (2\theta \minus{} \frac {\pi}{2})\sin\theta \plus{} 2\cos\theta \minus{} 1$

(1)$ f'(\theta) \equal{} (2\theta \minus{} \frac {\pi}{2})\cos\theta$

let$ f'(\theta) \equal{} (2\theta \minus{} \frac {\pi}{2})\cos\theta \equal{} 0$,We get $ \theta \equal{} \frac {\pi}{4}$,Caculate

$ f(\frac {\pi}{4}) \equal{} \sqrt {2} \minus{} 1, f(0) \equal{} 1, f(\frac {\pi}{2}) \equal{} \frac {\pi}{2} \minus{} 1$

so

$ \max_{0 < \theta < \frac {\pi}{2}}f(x) \equal{} 1,\min_{0 < \theta < \frac {\pi}{2}}f(x) \equal{} \sqrt {2} \minus{} 1$

(2)$ \int_0^{\frac {\pi}{2}}f(\theta)d\theta$
$ \equal{} \int_0^{\frac {\pi}{2}}[(2\theta \minus{} \frac {\pi}{2})\sin\theta \plus{} 2\cos\theta \minus{} 1]d\theta$
$ \equal{} 4 \minus{} \pi$[/quote]


I have a spreadsheet that has three columns:

1. Loan Amount, 2. Fees ($\$$ amt), and 3. Fees % (fee amt/Loan amount)

and at the last row, there's an average of each column.

Here's my question: Why doesn't the (average Loan Amt/average Fee Amt) = average fee % ?

All of the averages work out (ie. The bottom rows are "correct"), but I would think that the Avg Loan amount / Avg Fee Amt = Avg Fee %. What's up? Thanks in advance!

[quote="TRAN THAI HUNG"]:oops: I'm not sue I think that
The average fee%=average fee amt/average loan amt

Math is my life  :gathering:[/quote]

I believe you're correct, but I'm having trouble firguring out why. I've ran through a few examples and most of them worked out.

Is there an exaplaination?

Okay, I've figured it out. My brain is very wonky today... :blush:

The total fee over the total loan amount will give you the average percentage.  This number will (probably) be [i]different[/i] than the number you get by averaging the percentages for each loan.  This is an example of [url=http://mathworld.wolfram.com/SimpsonsParadox.html]Simpson's Paradox[/url]


geometry
geometric transformation
rotation
LaTeX

I'm not very sure about how to find the angle for this one.

Express the angle that the second hand on a watch turns through 1s. a) in degree measure b) in radian measure in exact form.

[quote="Mathematicus"][hide] There are 60s in a minute (or in 1 rotation) and also 360 degrees in one rotation. 1 sec. is 1/60 of a rotation and 1/60 of 360 degrees (one rotation) is 6 degrees.
To solve for radians, remember that 180 deg. = pi radians. So (6/180)*pi or (1/30)*pi is the answer (I really need to learn LaTeX!))[/hide][/quote]

[hide="helping out mathematicus :D"][b]Degrees[/b]:$360^\circ(\frac{1}{60})=\boxed{6^\circ}$

[b]Radians[/b]: $(6^\circ)(\frac{\pi}{180^\circ})=\boxed{ \frac{\pi^{rad}}{30}}$.[/hide]
(sorry if the text is huge, theres somethin wrong...)


Do you think that you can determine which is morally better --- letting someone die or making someone die --- in terms of medical field?

[quote="ZetaX"]Why do people always think there is a "one-line-exact-rule" to follow¿  :wink:

This strongly depends on the circumstances. In general, decissions should ba made by wagering the different possible outcomes.[/quote]

The origin of this question stems from an article I read in [i]Journal of Applied Philosophy[/i]. Like, this was the example they gave, which wasn't exactly medical, but related:

Let's say we have two people -- Smith and Jones. Smith intentionally drowned his nephew (I think?) for money. Jones, with same intention, tried to do this, but the nephew hit his head, fell in the water, face down and unconscious, and Jones watched the child die. So, the former case was killing someone while the latter was watching someone die. The article then went to see if we can justify which one was more moral or not.

[quote="Silverfalcon"]But, to me, life is something that shouldn't be taken away unless by God.[/quote]  Because reducing questions to determining the will of a god has worked so well to generate consensus all throughout human history?

Ethics is very hard.  It seems to me that toy examples such as the one you mentioned typically just illustrate the unlikelyhood that there are objective moral principles that apply in all circumstances.

[quote="Silverfalcon"] We can't create life, so we shall not take one away.[/quote]

But what if someone has a terminal illness and begs a doctor to humanely kill him or her, in order to avoid more pain that would serve no cause?

Thanks JBL. I agree... This is a field that I just walked on recently by reading one article.

isabella2296, do you know that what you are just asking is one of the most often debated question in bioethics? I don't think you should expect an answer, and thus, I won't answer it especially when I don't have an adequate amount of information.

P.S. Can JBL lock this? I think that I'm not going to see any too much intellectual debate in this thread just because the concept is too difficult to answer.


limit
inequalities
inequalities proposed

Prove that for every $ n \in N$ :

$ \sqrt{1\plus{}\sqrt{2\plus{}\sqrt{3\plus{}...\plus{}\sqrt{n}}}}<2$

Let $ (a_n)$ be a sequence ,and $ a_n >a_{n\minus{}1}$ ,such that

$ a_n\equal{}\sqrt {1 \plus{} \sqrt {2 \plus{} \sqrt {3 \plus{} ... \plus{} \sqrt {n}}}}$

I can not prove ,but I think it is true that  $ \lim_{n \rightarrow \infty}{a_n}\leq2$

observe that your inequality is equivalent to

$ \sqrt{2\plus{}\sqrt3{....\plus{}\sqrt{n}}}<3$

$ \Leftrightarrow\sqrt{3\plus{}\sqrt{4..\plus{}\sqrt{n}}}<7$
....
$ \sqrt{n}<(\text{something})$

Let be the sequence $ (a_n)$ such that: $ a_{n\plus{}1}\equal{}a_n^2\minus{}n$ and $ a_0\equal{}2$

we need only prove that $ n<a_{n\minus{}1}^2$

wich is equivalent to $ a_{n\minus{}1}^2\minus{}(n\minus{}1)>1 \Leftrightarrow a_n>1$ which is true since $ a_n>a_0$

[quote="Hamddi"]Let be the sequence $ (a_n)$ such that: $ a_{n \plus{} 1} \equal{} a_n^2 \minus{} n$ and $ a_0 \equal{} 2$

why is that ?[/quote]

observe that :  $ \sqrt{1\plus{}\sqrt{2\plus{}\sqrt{3\plus{}...\plus{}\sqrt{n}}}}<2\equal{}a_0 \Leftrightarrow \sqrt{2\plus{}\sqrt{3\plus{}...\plus{}\sqrt{n}}}<3\equal{}a_1$

$ \Leftrightarrow \sqrt{3\plus{}...\plus{}\sqrt{n}}<7\equal{}a_2 \rightarrow......\sqrt{n}<a_{n\minus{}1}$

 :lol:

[quote="Hamddi"]Prove that for every $ n \in N$ :

$ \sqrt {1 \plus{} \sqrt {2 \plus{} \sqrt {3 \plus{} ... \plus{} \sqrt {n}}}} < 2$[/quote]

$ \sqrt{k\plus{}2\sqrt{k\plus{}1}} < 2\sqrt{k}$ , $ k\equal{}1,2\dots$

$ \Rightarrow \sqrt{n\minus{}1\plus{}\sqrt{n}} <  \sqrt{n\minus{}1\plus{}2\sqrt{n}} < 2\sqrt{n\minus{}1}$

$ \sqrt{n\minus{}2\plus{}\sqrt{n\minus{}1\plus{}\sqrt{n}}} < \sqrt{n\minus{}2\plus{}2\sqrt{n\minus{}1}} < 2\sqrt{n\minus{}2}$
$ \dots$
$ \sqrt {1 \plus{} \sqrt {2 \plus{} \sqrt {3 \plus{} ... \plus{} \sqrt {n}}}} < 2\sqrt{1}$


does anyone have good recommendations for geometry books (not problem collections, but rather for learning the subject)?  i would really like to know of some for geometry, something that starts from the beginning again but perhaps more rigorous and well-organized than the books you get in most high school classes.  Coxeter's books dont really start the beginning either.  if anyone knows of any others, please tell (it'd be nice if I could d/l them online too, as I don't really want to pay right now  :D ).

thanks.

I would have to say either of the AOPS books...I mean, they're not $just$ geometry but they really teach it well. If Volume 1 seems to easy for you, then try volume 2.

i dont think there is a book that fits your description. I'd say either take the Intro Geo here or use Coxeter and study hard!

Theres an interesting Geometry book im studying from that they use at Phillip Exeters. Its called Geometry Plane and Solid or something like that.

[quote="Palytoxin"]that's true, or perhaps read the all-time classical Euclid's Elements![/quote]
True. But it is too old. I'd say if you want to learn Olympiad geometry, don't read Elements. But if you are interested in geometry and would like to feel the beauty of geometry, i would strongly recommand it.


Triangles $\Delta ABC$ and $\Delta DEF$ are inscribed in the same circle. Show that $\sin{A}+\sin{B}+\sin{C}= \sin{D}+\sin{E}+\sin{F}$ iff the two triangles have the same perimeter.

:huh: Pre-Olympiad?

[hide="I mean, really..."]$\sin A={a\over 2R}$ etc.[/hide]

Hm you're probably right. But the moderators in the other forums are quite picky when it comes to requiring certain formulas.


Let $X$ be a set of $2k$ elements for some positive integer $k \geq 2$.Let $F$ be a family of $k$-element subsets of $X$ such that every $(k-1)$-element subset of $X$ is contained in exactly one member of $F$.Prove that $K+1$ is a prime.

Suppose it is composite and take a prime $p|k+1$. Fix some $M \subset X$, $|M|=k-p$. The number of $k-1$-element subsets of $X$ containing $M$ is, on one hand
\[ {2k-(k-p) \choose k-1-(k-p)} = {k+p \choose p-1} \]
and, on the other hand, it is also $(k-(k-p)) \left|\{R\in F: M\subset R\}\right|=p \left|\{R\in F: M\subset R\}\right|$, so
\[ p|{{k+p}\choose {p-1}} = \frac{(k+p)!}{(k+1)!(p-1)!} \Rightarrow p!|(k+2)...(k+p), \]
contradiction.


[quote="luiseduardo"]In the figure below, ABCD and EFGH are squares of side 48 cm. Knowing that A is the midpoint of EF and G is the midpoint of DC, determine the area outlined in cm².

[img]http://img43.imageshack.us/img43/8441/calsn.jpg[/img][/quote]

the area will be  1704 cm2 :wink:

[quote="luiseduardo"]In the figure below, ABCD and EFGH are squares of side 48 cm. Knowing that A is the midpoint of EF and G is the midpoint of DC, determine the area outlined in cm².

[img]http://img43.imageshack.us/img43/8441/calsn.jpg[/img][/quote]

hi all   :wink:
[url=http://www.servimg.com/image_preview.php?i=355&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk21.jpg[/img][/url]
with my best wishes
mr mohamed
almonkez

[quote="luiseduardo"]In the figure below, ABCD and EFGH are squares of side 48 cm. Knowing that A is the midpoint of EF and G is the midpoint of DC, determine the area outlined in cm².

[img]http://img43.imageshack.us/img43/8441/calsn.jpg[/img][/quote]
[size=150][color=red][b]another solution without trigonometry notice that AG^2 = 24^2 (1+4) [/b][/color][/size]
[url=http://www.servimg.com/image_preview.php?i=356&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk15.gif[/img][/url]
[size=150][color=blue][b]with my best wishes
mr mohamed
almonkez[/b][/color][/size]

Thanks mr. mohamed ;)
Very good solution :)


Prove that there are no positive integers $a$ and $b$ such that for all different primes $p$ and $q$ greater than $1000$, the number $ap+bq$ is also prime.

Assume that such integers $a,b$ exist. Let us fix $p=1009.$
Then $1009a+bq$ is prime for every prime $q>1009.$
Choose any prime $r$ not dividing $b.$ Then the congruence $1009a+bx \equiv 0\pmod{r}$ has a solution $x\equiv c\pmod{r}.$
According to Dirichlet's theorem, there are infinitely many primes in the sequence $c+r,\ c+2r,\dots.$ Let $q>1009$ be a prime from this sequence. Then $1009a+bq>r$ and it is divisible by $r$ (thus, not prime).
Contradiction proves that there are no such $a,b.$

You don't need Dirichlet. There are infinitely many primes greater than 1000. Thus there are 2 distinct primes $ p<q$ that are the same residue modulo $ a\plus{}b$. Say $ q\equal{}p\plus{}(a\plus{}b)k$. Then $ ap\plus{}bq\equal{}ap\plus{}bp\plus{}bk(a\plus{}b)\equal{}(a\plus{}b)(p\plus{}bk)$ is not prime.

Luke
See my puzzle blog at [url]http://bozzball.blogspot.com/[/url]