conics
parabola
analytic geometry
graphing lines
slope
quadratics
parameterization

$ AB$ is the [hide="Normal Chord"] with end points $ A$ and $ B$[/hide] [color=white].[/color]of parabola: $ y^2 \equal{} 4ax$ of which $ S$ is the focus.
Prove that locus of centroid of $ \Delta SAB$ is given by:
$ 36ay^2(3x \minus{} 5a) \minus{} 81y^4 \equal{} 128a^4$

This is so ugly it's almost cute. Make up your mind, if the chord is $ PQ$ or $ AB.$
[url=http://books.google.com/books?id=yMW9vRHYcIcC&pg=PA203&lpg=PA203&dq=normal+chord+parabola&source=web&ots=NVGz0U-6EG&sig=NSPjk8HCbvDfE-VLccO9XiryXjg&hl=en&sa=X&oi=book_result&resnum=5&ct=result]Sidney Luxton Loney[/url] says on page 203 that it is $ PQ.$

Coordinates of focus $ S$ of a parabola $\mathcal P$ with equation $ y^2 = 4ax$ are $ (x_S, y_S) = (a, 0).$ [hide=Why ?]Point $C \equiv (4a, 4a)$ is on $\mathcal P,$ at distance $\overline{CS} = \sqrt{(3a)^2 + (4a)^2} = 5a$ from point $S \equiv (a, 0)$
and at the same distance $\delta (C, d) = 4a + a = 5a$ from line $d$ with equation $x = -a.$
By definition, $S$ is the parabola focus and $d$ its directrix.[/hide]
Coordinates of an arbitrary point $ A$ on the parabola are $ (x_A, y_A) = \left( \tfrac {p^2}{4a}, p \right),$
 where $ p$ stands for an arbitrary parameter.

Slope of parabola tangent $ t$ at $ A$ is $ m = \left. \tfrac {\text dy}{\text dx}\right|_A = \left( \left. \tfrac {\text dx}{\text dy}\right|_A \right)^{ - 1} = \tfrac {2a}{p}.$
Perpendicular $ n$ to $ t$ at $ A$ has equation $ y - p = - \tfrac {1}{m} \left( x - \tfrac {p^2}{4a} \right) = - \tfrac {p}{2a} \left( x - \tfrac {p^2}{4a} \right).$

The get coordinates $ (x_B, y_B)$ of $ B,$ substitute the parabola equation $ x = \tfrac {y^2}{4a}.$
This yields a quadratic equation $ y^2 + \tfrac {8a^2}{p} y - ... = 0.$ for $ y_A, y_B.$
One root $y_A = p$ is known, the other root is therefore $ y_B = - \left( p + \tfrac {8a^2}{p} \right)$ and $ x_B = \tfrac {1}{4a} y_B^2 = \tfrac {1}{4a} \left( p + \tfrac {8a^2}{p} \right)^2.$

Coordinates $ (x, y)$ of the centroid $G$ of $ \triangle SAB$ are:

$ x = \tfrac {1}{3}(x_S + x_A + x_B) = \tfrac {1}{12a} \left( \tfrac {64a^4}{p^2} + 2p^2 + 20a^2 \right),$
$ y = \tfrac {1}{3}(y_S + y_A + y_B) = \tfrac {8a^2}{3p}.$

To get equation of the locus of $G$, eliminate the arbitrary parameter $ p$
by substituting $ p = \tfrac {8a^2}{3y}$ from the 2nd equation into the 1st one:

$ 12ax = 9y^2 + \tfrac {128a^4}{9y^2} + 20a^2,$

$ 108axy^2 = 81y^4 + 128a^4 + 180a^2y^2,$

$ 108axy^2 - 180a^2y^2 - 81y^4 = 128a^4,$

$ 36a (3x - 5a)y^2 - 81y^4 = 128a^4,$

$ \boxed {36ay^2(3x - 5a) - 81y^4 = 128a^4}\ .$
____________________________________________________________________________________________

[quote="yetti"]This is so ugly it's almost cute. Make up your mind, if the chord is $ PQ$ or $ AB.$[/quote]

Sorry about that!
:oops:
It is $ AB$. I have corrected it!


$ a,b,c\geq 0$. $ S=\sum{\sqrt{\frac{ab}{(b+c)(c+a)}}}$
Find $ S_{min}$ and $ S_{max}$

http://www.mathlinks.ro/viewtopic.php?t=158550
$ S\leq \frac{3}{2}$

Actually, the value is the same as CosA+CosB+CosC
first notice that x/(y+z)=a,z/(x+y)=b, y/(z+x)=c
sastify ab+bc+ca+2abc=1
now let's say root(ab)=p, root(bc)=q, root(ca)=r
then p^2+q^2+r^2+2pqr=1
therefore there exists angle A+B+C=180so that p=cosA q=cosB r=cosC

therefore the given value is equal to p+q+r which is cosA+cosB+cosC
from then it becomes trivial. :wink:


If your computer with all your old transcripts and challenge sets recently died, what would you do?

cant u go to "my classes" and find the stuff there?

Yeah, but they're from a really long time ago (like, classes I took a year ago), so they aren't even there anymore.

Anyway, I used the old hard drive and connected it by USB to another computer, so I have the problems. Turns out I didn't have one of my transcripts either way, but I guess that's my fault.

Sorry to revive.

Is there any way to get old transcripts from my old classes?  My hard drive crashed on my old computer, so I don't have them any more.

I don't think you can get your old transcripts anymore.  I took a class over the summer, but even though that was only about four months ago the transcripts are not there. :(

Have someone who was in your class pm or email you transcripts.

Hmm... that was what some people in WOOT did last year...


what is a
diameter
radius
point
centeral angle
chord
 :?:

Always remember that [url=http://www.google.com]Google[/url] is your friend.

A [url=http://www.google.com/search?q=define%3A+diameter&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]diameter[/url]:

[asy]draw(unitcircle);
draw((-1,0)--(1,0));[/asy]

A [url=http://www.google.com/search?q=define%3A+radius&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]radius[/url]:

[asy]draw(unitcircle);
draw((0,0)--(1,0));[/asy]

A [url=http://www.google.com/search?q=define%3A+point&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]point[/url]:

[asy]dot((0,0));[/asy]

A [url=http://www.google.com/search?q=define%3A+central+angle&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]central angle[/url]:

[asy]draw(unitcircle);
draw((0,0)--(1,0));
draw((0,0)--(0,1));[/asy]

A [url=http://www.google.com/search?q=define%3A+chord&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]chord[/url]:

[asy]draw(unitcircle);
draw((1,0)--(0,1));[/asy]


HMMT MIT college counting distinguishability Harvard AMC

Hey, do any people on this forum want to meet at the HMMT coming up next Saturday? I think it'd be interesting to meet some of you in person.

Ideas? Comments? Suggestions?

lots of people were saying they wanted to meet, so post here if you want to.

Let's see...how about "look for a really short Asian with a weird accent" :lol:

Where do this meet happen at? I still don't know where to go to for this competition...

did you register tare?

I've never been there before, so i don't know how this works (where it's held, what time is free, etc).  The simplest thing to do would be to go up and talk to tycoon when he gets his 1st place award.

[quote]The simplest thing to do would be to go up and talk to tycoon when he gets his 1st place award.[/quote]

..or you can find me sulking in the corner while everyone else is going up to get awards.  :mrgreen:

Well, I was there last year, although that was at MIT and this year its at Harvard. The thing wasn't run particularly smoothly and there was a long delay at the beginning because there were not enough rooms to house teams for the team round, but I don't know about this year. Maybe we could meet before the contest, like during registration or something? The only other possible time would be during lunch, but since our teams are all in different places then, that might be difficult.

[quote="cats..."]i thought for lunch you just go to random places and find some restaurant or something?[/quote]

Yeah, the "random places" part is a problem though. :D Although last time they ordered a lot of dominos pizza and just sold those on site.

I'm not sure but depending on the location I'll bet there's a restaurant or two

Yes I did register with my high school. But they said that we're going to meet onsite and that we were to individually figure out where this competition is held since no one at my school knew. They said it'll be on the website but it's not...so I'm wondering where we originally meet

Um we'll need to have a distinguishable figure if we're going to find each other...how about we keep screaming "artofproblemsolving.com" until we find another person screaming "artofproblemsolving.com" too? :lol: It'll be the ultimate advertisement...buncha kids screaming it all over the place...I wonder if Joel or Eve hear us :lol:

The worst case scenario is that I still won't know where the competition is held the day it is held...so please post any info on the whereabouts of where it begins! Thank you.

...you do that.

I found it at the site before.  http://web.mit.edu/hmmt/ww  w/faq.shtml#boston

It says "We will occupy Harvard Science Center Hall C for the entire day; students may hang out and store their goods here. It will be the location of the Guts Round, as well as the awards ceremony.

"  Look at the map, or maybe Joel could help dictate where that is.

Registration will be held at Lobby 16. See here: http://web.mit.edu/hmmt/www/faq.shtml . There is a link to a map there too.

Hmm, I was thinking that since people (including me) would be reluctant to post too much information on the forum, maybe everyone who was interested could like PM me his/her name, school/team name, and what he/she is going to wear to the contest. Then I would PM a compiled list of information to everyone who wants to meet.

http://www.map.harvard.edu/  is a map of the Harvard area.
http://www.map.harvard.edu/level2.cfm?mapname=camb_allston&tile=F7&series=N should be a map centered on the Science Center.

It's pretty easy to find -- big concrete building with a layered roof and big black steel beams sticking out of the side making it look like some kind of spider or something.

Did any of you guys get an award last year? What subject? I know that I sound stupid when I ask this, but how smart are the people there? i.e. have most of them made it to AIME, USAMO, MOP, IMO?

And yes, I would like to meet at HMMT.

I think there was 1 USAMO winner there last year who won it.

Do people like my idea for a meeting?

im not sure what your idea is, yet.  Is it to meet at registration, at lunch, ... ?

If you look at the list of winners of past years, there are many recognizable names.

Whenever we meet a bunch of people we'll try to find each other.

I've been placed on the stupid team so if anyone sees a team acting/being particularly stupid that's probably us :lol:

[quote]im not sure what your idea is, yet. Is it to meet at registration, at lunch, ... ? [/quote]

I was refering to:

[quote]Hmm, I was thinking that since people (including me) would be reluctant to post too much information on the forum, maybe everyone who was interested could like PM me his/her name, school/team name, and what he/she is going to wear to the contest. Then I would PM a compiled list of information to everyone who wants to meet.[/quote]

We could then meet at registration I guess.

[quote="cats..."]My teacher signed me up for 1 round that i don't want to be in... is it possible to change?  Do i send an email, or handle it the day of the contest?[/quote]

Same deal with me. However, I think that we can just change it the day of the contest. The coordinators are usually flexible...I think.

Aww...there must be other people who are going to this thing.

Hmm... I don't think I should go. Boston is quite far from Michigan :)

[quote="mathnovice"]Hmm... Did you guys meet? I had not idea who to look for. Anyway, did any of you get an award? My school came in 4th to last place, I think-- or at least on the guts round. One team had a 0 consistently throughout the round-- so they probably never even showed up. I guess that makes use in 3rd the last place. However, (1) we had only 6 people; (2) we are a new team; (3) this was the first math competition that any of the team went to; (4) I coached the team and I am pretty stupid. Thus, we had 4 things going against us. How I wish I went to Exeter, AAST, or TJ....[/quote]

The school system I went to started out this way -- not even placing in competitions in Alabama.  It did however improve over a period of years and eventually did well in many competitions.

What you are doing is great -- plugging yourself and other students into that culture of problem solving.  You're also teaching which is...the best way to learn how to solve problems.  Keep up the good work.

i won something for countdown, but that doesn't really count.

Allison and Nukular both did well.

I thought it was neat seeing this site mentioned and giving prizes.

We didn't meet, but Tare was running around with some sheet that had artofproblemsolving on it, and found me somehow.  And I sat next to allison when taking the individual rounds...  I'm pretty sure i scared her.

Poopies, I was one problem off from getting in the top 10 in algebra and geometry :/ Oh well. I met Tare, and saw Nukular and Alison when they won stuff at the prize ceremony.

We oughta make an artofproblemsolving team next year. We would own. If Alison and Nukular are still around and want to do it, and cats.., Tare, and I all did it along with a couple other borderline-USAMO qualifiers, we could definitely contend with Exeter and TJ.

Tied for fifth in the general. I also got a deck of round playing cards (with planets on the back) for being one of the Countdown winners. Team didn't do so well. Oh well.

cats, were you the person who finished Countdown in two questions?

My go was really quick, and yeah it was only 2 rounds.  I don't know if that was the only one with 2 rounds though.

I was wearing a shirt and tie - that should help.

I was looking forward to trying to knock you off in the winner's round. My round was three questions. Should've gotten the second one right, but made a really stupid mistake.

i wish they had a second round... i always wanted to do countdown before, but i never got a chance to in mathcounts.

The 2 questions i got were easier than most of the other questions they gave.

Yeah, mine seemed easier than the others too. That, and countdown rounds may be my area of specialty at math contests.

What school do you go to?  What were your questions?  Mine were like: 679 (or some number like that, i dont remember exactly) is the sum of 2 primes.  Find their product.  Then the other was like  f(1)=1, f(2)=2, f(3)=5, f(4)=10, find f(5).

Did you ever do countdown for mathcounts?

I go to Derryfield. One of my questions involved four square pens arranged side by side, and if the perimeter of the entire rectangular pen was 120, what was the area of one of the square pens (144). Another was if a + b is less than 5, and b  :ge: 2, then how many values of a are there. There was also one that asked for the greatest a such that 1/a was greater than 5/26. For some reason, I answered 4 instead of 5.

[quote="Tare"]I've seen Tom Bellu...(sorry, I dcan't spell your name) at the award ceremony, maybe he's Nukular.

I've seen both cats... and Fierytycoon.

Actually that blue hand-out was given to everyone...I thought it was neat :)[/quote]

You can't spell it and they can't pronounce it :). It's "Belulovich," but I really don't see why its so hard to pronounce... its completely phonetic.

Yeah, I'm kinda angry at myself for putting an extra 0 in both parts of the fraction for number 10 on the algebra... (I wrote 2001... / 2001.... instead of 201.../201...), so that kind of surprised me as to why I got a 33 instead of a 41. I don't even want to bring up the stupid things I did on combinatorics (like misreading "8x8" as "6x6")

On a side note, did anyone use the  :Sigma:n=1 to :inf:  1/n2=  :pi: 2 / 6 on the team question asking you to prove  :stddev: (1) / 1 + :stddev: (2) / 2 + :stddev: (3) / 3 + ... + :stddev: (n) / n :le: 2n ? If you do, I'm pretty sure you get that this inequality is strict, and substantially stronger than they make it.

Nukular, I wasn't at this contest, but based on reading the team question in your post, I also solved it using :pi:2/6.  Maybe they wrote "2" to avoid giving away the method.

[quote="mathnovice"]Hmm... Did you guys meet? I had not idea who to look for. Anyway, did any of you get an award? My school came in 4th to last place, I think-- or at least on the guts round. One team had a 0 consistently throughout the round-- so they probably never even showed up. I guess that makes use in 3rd the last place. However, (1) we had only 6 people; (2) we are a new team; (3) this was the first math competition that any of the team went to; (4) I coached the team and I am pretty stupid. Thus, we had 4 things going against us. How I wish I went to Exeter, AAST, or TJ....[/quote]

what school are you from?  This is the first yr that BCMT went too, and I'm quite happy with how they did.  If you need any help with anything, you might be able to ask me (email me).

On a seperate note, I tried to greet several people (Alison included, though she look freaked out).  I was the guy in the CTY Staff shirt running all over the place getting pissed at everyone and everything.

Hm... I seem to be rather talked about.

I think an AOPS.com team would be a great idea, but seeing as I'll be in college next year and hopefully will help run HMMT, no way I'd be on it.

And don't worry, neither of you scared me, or anything.

And thanks to you guys for the scholarship, although I don't know how I'm going to use it.

[quote="Alison"]Hm... I seem to be rather talked about.[/quote]

Don't look, but they are talking about your mom in the newspaper thread.  :mrgreen:

[quote]I think an AOPS.com team would be a great idea, but seeing as I'll be in college next year and hopefully will help run HMMT, no way I'd be on it.[/quote]

Poo...

[quote]And thanks to you guys for the scholarship, although I don't know how I'm going to use it.[/quote]

lol, yeah, I really liked how AoPS had tons of advertising at the HMMT, but for practical purposes, I don't see how Tiankai Liu is going to need to take any of the courses here, heh.

I'd love to be on the AoPS team since neither of our two teams even reached three-digits on the guts round and got a 15 on the team round :lol:

Even if, say, Tiankai doesn't want to take courses, at least he'll probablly look at the site and register.  That'd be a good thing.


ARML
algebra
polynomial
quadratics
pigeonhole principle
geometry
probability

[quote="lingomaniac88"][size=150][b]Problem 1:[/b][/size]

The letters $A$, $B$, and $C$ represent different digits, $A$ is a prime and $A-B=4$.  If the number $AAABBBC$ is a prime, compute the ordered triple $(A,B,C)$.[/quote]

Lessee...
[hide]
C can't be 3 or 9 or 5 or else AAABBBC isn't prime, A=5,7, so B=1,3, so we have 7773331 and 5551117. If you do some divisibility stuffs with 5551117 it's divisible by 11 so then it is (7,3,1).[/hide]

ARML 2006

Compute the four-digit positive integer $N$ whose square root is three times the sum of the digits of $N$.

[hide]First find out that the square root of x is a multiple of 3, than that it's a multiple of 9, then that it's a multiple of 27, using the divisibility rules for 3 and 9.  Thus, the number must be a multiple of 729, and a perfect square.  There are only two of those: 2916 and 6561, so just check both, and get 2916.[/hide]

Here's one from my county's math league meet:

Take a number, with digits ab (not necessarily distinct).  if you multiply that number by 23, you get the number 1ab1.  find the value of ab.

[quote="B-tone_Party"][hide]First find out that the square root of x is a multiple of 3, than that it's a multiple of 9, then that it's a multiple of 27, using the divisibility rules for 3 and 9.  Thus, the number must be a multiple of 729, that is a perfect square.  There are only two of those: 2916 and 6561, so just check both, and get 2916.[/hide]

Here's one from my county's math league meet:

Take a number, with digits ab (not necessarily distinct).  if you multiply that number by 23, you get the number 1ab1.  find the value of ab.[/quote]

[hide]$1000+100a+10b+1 = 23(10a+b)$
$1001+100a+10b = 230a+23b$
$1001 = 130a+13b$
$77 = 10a+b$
$a = 7$, $b = 7$, $ab = \boxed{49}$[/hide]

[TJ ARML Practice] Let $f(x) = x^{2}+4x$. Find all values of $x$ such that $f(f(x)) = f(x)$.

[hide]First, find both values of x where f(x)=x.  Second, expand the expression into the 4th degree polynomial $x^{4}+8x^{2}+19x+12=0.$  Factor out the two values you found earlier, and the remaining quadratic is simple to solve, giving you -4, -3, -1, and 0.[/hide]

The two solutions to $x^{2}-4x-19=0$ are a and b.  The quadratic equation with roots 1/a and 1/b can be expressed as $x^{2}+cx+d=0$.  Find c+d.

[hide]The equation with the reciprocal of roots is with reversed coefficients so $-19x^{2}-4x+1$, divide by $-19$, $x^{2}+\frac{4x}{19}-\frac{1}{19}$ so $\frac{3}{19}$[/hide]

Problem 6
ARML 2006 (yeah, those are posted on the website, otherwise I have no other resources)

Let $N$  be an integer such that $1\le N\le 2006^{2}$. Compute the number of values of $N$ such that there exists a factor of $N$, call it $a$, such that $|\sqrt{N}-a|\le 1$.

[quote="B-tone_Party"][hide]That was a bit tricky.  Would 6016 be correct? Basically, I took all of the multiples of a from $a^{2}-a$ to $a^{2}+2a$ for each number, removed the $a^{2}-a$ for each because they were equal to $(a-1)^{2}+(a-1)$, and removed $2006^{2}+2006$ and $2006^{2}+4012.$[/hide]

For how many integers n is $n^{2}+4n+109$ a perfect square?[/quote]
[hide]$(n+2)^{2}+105$
we need two perfect squares that differ by 105.
Case I: They differ by 1
$n^{2}-(n-1)^{2}=2n-1$
n=53, 52 and 53.
Case II: They differ by 3
$n^{2}-(n-3)^{2}=6k-9$
19/16
Case III: They differ by 5
$n^{2}-(n-5)^{2}=10n-25$
13, 8
Case IV: They differ by 7
$n^{2}-(n-7)^{2}=14n-49$
11/4
So 4.[/hide]

[hide]Actually, there are eight: they can also differ by 15, 21, 35, or 105 (I never said n had to be positive.) [/hide]

[size=150][color=red][b]PROBLEM 8[/b][/color][/size]
ARML 2006(=P)
Let $\{a_{i}\},\{b_{i}\},\{c_{i}\}$ be the sequence, from left to right, of elements in the 2005th, 2006th, 2007th rows of the Pascal's triangle, respectively, with the leftmost element occuring at $i=0.$ Compute $\sum_{i=0}^{2006}\frac{b_{i}}{c_{i}}-\sum_{i=0}^{2005}\frac{a_{i}}{b_{i}}$

edit: heh, I didn't include the diagram and forgot to add it's pascal's triangle  :blush:  sorry

Oh wow, I was like I'm feeling stupid...
[hide]
I expanded the general terms and stuff and got $\sum_{n=0}^{2006}\frac{2007-n}{2007}-\sum_{n=0}^{2005}\frac{2006-n}{2006}=\frac{2008}{2}-\frac{2007}{2}=\boxed{\frac{1}{2}}$[/hide]

But I did this really quickly and I have to go.

[quote="white_horse_king88"]I think it's about that time of year again. joml is probably very busy in Cambridge, so, maybe someone can fill in for him this year and kick this off?

Not sure if I have the resources to do so. But if nobody is willing, I'll try.[/quote]

Yeah, I'm pretty busy.  But if you guys run out of problems or something, just PM me and I'll post some from the ARML books.

Btw, I stickied the thread.

OHHHHHhhhhhh...so the [b]product[/b] of the $3$ numbers has to be greater than $1989$

well, it is a perfectly reasonable problem, in my opinion, you have that $19^{3}-19*k^{2}=n> 1989$ you know that the LHS is divisible by $19$, so is the RHS, but then you can strengthen your bound to say that $19^{3}-19*k^{2}=n\ge 1995$, the first multiple of $19$ after $1989$, at this point, you get $361-j^{2}\ge 105\implies 256\ge k^{2}$, immediately, $k=16$

Okay, I'll post another one:

[b]PROBLEM 37[/b]
In quadrilateral $ABCD$, $m\angle DAB=90^\circ$, $m\angle BCD=135^\circ$, $BC=3$, and $CD=2\sqrt{2}$.  Compute the largest possible area of $ABCD$.

I have to say....this is a pretty tricksy problem...

the min of x+1/x for x>0, real, is 2, and then we must have that

$tan 7x=cot 7x=sin 6x=cos 4x=\pm 1$

[if you don't see why, play around for a bit, this is a problem you should try for yourself :!: ]

so anyway, we have a whole lot of stuff now... i am going to use degrees since it looks nicer...

$7x=45+180a$, $6x=90+360b$, $4x=360c$, using that last condition, we have $x=0,90,180,270, 360$

none of those satisfy the first condition, so no solution

for negative, we have $7x=-45+180a$, $6x=-90+360b$, $4x=180+360c$, the last condition gives $x=45, 135, 225, 315$, the second condition gives $x=45, 105, 165, 225, 285, 345$, checking $\boxed{x=45}$, works in the first equation

[quote="perfectnumber628"]So, Law of Cosines says $BD=\sqrt{29}$, and that's the hypoteneuse of triangle $ABD$.  That triangle has the largest possible area when it's isosceles (think of inscribing it in a semi-circle) so the area of $ABD$ is $\frac{1}{4}(29)=\frac{29}{4}$.  For the other triangle, $\frac{1}{2}ab\sin C$ says the area is 3.  So the answer is $\frac{41}{4}$.

So here's [b]Problem 38[/b]
Find all values of $x$ for $0$ to $2\pi$ such that
$\tan 7x-\sin 6x=\cos 4x-\cot 7x$[/quote]
$\tan 7x+\frac{1}{\tan 7x}=\sin 6x+\cos 4x$
$\frac{\sin 7x}{\cos 7x}+\frac{ \cos 7x}{\sin 7x}=\sin 6x+\sin 4x$
$2=\sin 14x(\sin 6x+\sin 4x)$
Note that the maximum possible value of the RHS is 2.

[b]Problem 39[/b]

Find the largest prime that divides the sum of the 120 permutations of any five digit number.

well that is an odd problem since we are unsure what to do with 0 digits...

we can evaluate the sum directly if we ignore that, the expected value for any digit is $\frac{1}{6}(\sum a)$, then we have that there are 5 place values for that to occur so multiply  $11111$, and there are 120 combinations, so our sum is: $(\sum a)*20*11111$, note that $\sum a$ can take values of $7$ or $11$, so there is no prime factor that is surely added by that, so we have $\boxed{222220}$

[hide]If the number is abcde, where a,b,c,d,e are the digits, then each digit appears in each of the 5 place values 4!=24 times, so the sum of the 120 permutations is $24\cdot11111\cdot(a+b+c+d+e)$.
Prime factorize, $11111=41\cdot271$, so the the largest prime that always divides the sum is $\boxed{271}$.[/hide]

I see I did a stellar job of reading the problem. Also I should have $\frac{1}{5}\sum a$, that gives the same result as pianoforte.

[own i guess...this is still classic...]

Problem 40

Provided that $a+b+c=1$, find the minimum of $a^{2}+4b^{2}+9c^{2}$.

#41 (own, easy):
Let $s(x) =$ the sum of the digits of $x$ for all positive integers $x$. If $s(a)=10$ and $s(b)=13$, find the sum of all possible values of $s(a+b)$.

[hide="solution to 41"]
We have $s(n)\equiv n\pmod{9}$ so $s(a+b)\equiv a+b\equiv s(a)+s(b)\equiv 5\pmod{9}$. Also, $s(a+b)\leq s(a)+s(b)$, since if $a=d_{0}+10d_{1}+100d_{2}+\cdots$ and $b=e_{0}+10e_{1}+100e_{2}+\cdots$, then $a+b=(d_{0}+e_{0})+10(d_{1}+e_{1})+100(d_{2}+e_{2})+\cdots$, and carrying replaces 10 of one digit with 1 of the next, so it decreases sum of digits.

Therefore, all possible values are $\boxed{5,14,23}$. (These are attainable with $a=19,b=94$, $a=190,b=94$, and $a=1900,b=94$ respectively.)
[/hide]

Problem 42 (own)

Let $S$ be the set of strings of three binary digits (including those that begin with 0). Compute the number of permutations of $S$ that preserve the xor operation. (If $*$ denotes xor, then
\[0*0=1*1=0,\ 1*0=0*1=1 \]
and the xor of two binary strings is the same as xor'ing each of its digits.)

I don't really understand the problem...Compute the number of permutations [i]in[/i] S?

Preserve the xor  operation? What is being xor -ed with what?

What does preserve mean in this instance?


Let $A,B \in {\mathbb R}^{n \times n}$ with $A \neq B$. If $\exists p,q \in {\mathbb N}^*$ such that $A^{p+1}=B^{q+1}$ and $A^pB=B^qA$. Prove that $det(A^p+B^q)=0$.  :ninja:

It is enough to notice that $(A^p+B^q)(A-B)=0$, from where it follows that $det(A^p+B^q)=0$

You say that if $(A^p + B^q)(A-B)=0 => det(A^p + B^q)det(A-B)=0 =>det(A^p+B^q)=0 $???????????????? If u say like this IS FALSE.

Here is my sollution:

Supose that $det(A^p+B^q) \neq 0 => (\exists) (A^p+B^q)^{-1}$. Clearly using our ipotesis we have $(A^p+B^q)(A-B)=0$. If
we multiply in the left part with inverse of $A^p + B^q$ we obtain $A-B=0 <=> A=B$, which is a contradiction. So, $det(A^p+B^q)=0$.

(q.e.d.)

[quote="jianuovidiu"]You say that if $(A^p + B^q)(A-B)=0 => det(A^p + B^q)det(A-B)=0 =>det(A^p+B^q)=0 $???????????????? If u say like this IS FALSE.
[/quote]
Of course i didn't mean this: I meant that if $CD=0$  (*) and $D \neq 0$ then if we denote by $v_1,...,v_n$- columns of $D$ then from (*) it follows that $Cv_1=Cv_2=...=Cv_n=0$, therefore $C$ has zero eigenvalue,which means that it's determinant is zero.
BTW,it is the same you did.


Anyways, I have 4 questions from last years contest and seeing as the contest is tomorrow can you guys explain a solution to the 4.
Thanks in Advance  :D
Heres the link
http://www.math.ksu.edu/~soibel/2007-7-8.pdf

[hide="4"] The measure of each of the vertices is 180 - the two angles below them. Each of these two angles measure 180 - one of the angles in the pentagon. Also, two adjacent vertices in the star have angles below them that measure the same by verticle angles. There are $ 108\cdot3$ degrees in the sum of the angles of a pentagon. So cummulatively, the sum of all five angles would be $ 180\cdot5 \minus{} 2(180\cdot5 \minus{} 540) \equal{} 180$. [/hide]

Edit: Argh, where did I go wrong in my argument? I'll have to check that over...

[hide="1"]Well, the first thing that came into my mind was $ \frac {x(x \plus{} 1)}{2} \equal{} \frac {1}{2}x^2 \plus{} \frac {1}{2}x$...[/hide]
[hide="2"]$ 43x \plus{} 75y \equal{} 8(3x \plus{} 7y) \plus{} 19(x \plus{} y)$[/hide]
[hide="3"]Between the numbers, there are 1004 multiples of 2^1, 502 multiples of 2^2, 251 multiples of 2^3, 125 multiples of 2^4, 63 multiples of 2^5, 31 multiples of 2^6, 16 multiples of 2^7, 8 multiples of 2^8, 4 multiples of 2^9, 2 multiples of 2^10, and 1 multiple of 2^11. This leads to an answer of 1004 + 502 + 251 + 125 + 63 + 31 + 16 + 8 + 4 + 2 + 1 = 2007. Woot, it's the year! [/hide]
edit: oops.

Bruteforce, you just add up all the numbers, you don't multiply them by their respective powers since they were already counted for the lower powers also.

for number 1, I was considering some possible values for x a,b, and c. I found that if x=1 and a=0.5 b =1.5 and c =1, it would still satisfy the integer value of the expression.
so $ 0.5x^2\plus{}1.5x\plus{}1$  Now I know that this works, but my question is how to prove this algebraically?

[hide="question 1"]

n(n+1)/2= 0.5n^2+0.5n+0 is always an integer, since one of n, and n+1 is divisible by 2 ( since they can't be both odd), so the denominator is always a factor of the numerator.

So we cannot conclude a,b,c are always integers.

[/hide]
[hide="question 2"]

19 is a factor of 3x+4y, so is a factor of 8*(3x+7y)=24x+56y, and is also a factor of 19x+19y+8*(3x+7y) =(19+24)x+(19+56)y
= 43x+ 75y

[/hide]

[hide="question 3"]
The powers of 2 up to 4014 are 1,2,4,8,16,32,64,128,256,512,1024,2048

the numbers of 2's that divide 4014! is floor(4014/2048)+floor(4014/1024)+...+floor(4014/2) and the # of 2's that divide 2007! is floor(2017/1024)+floor(2014/512)+...+floor(2014/2) by a simple counting argument on the factors of m! Where m is 4014 and 2007. Each of the floors represent how many numbers have that power of 2 as a factor. Try and ill in the rest of the argument for practice.

The answer is those 2 numbers subtracted from each other, since the product is 4014!/2007!. I'm much too lazy to do this, but in a contest, you would actually have to calculate this.
[/hide]

[hide="question 4"]

Any angle of one of the 5 triangles is 180- the sum of the other 2 angles=(180- one of the angles)+(180- the other angle)-180=2 adjacent angles of the Pentagon-180 (actually try and figure this out with a diagram, it makes sense). Summing up the 5 angles cyclically will equal the 2*(sum of the angles of a pentagon)-5*180=540*2-5*180=6*180-5*180=180

[/hide]

[quote="neurosurgeon1118"]for number 1, I was considering some possible values for x a,b, and c. I found that if x=1 and a=0.5 b =1.5 and c =1, it would still satisfy the integer value of the expression.
so $ 0.5x^2 \plus{} 1.5x \plus{} 1$  Now I know that this works, but my question is how to prove this algebraically?[/quote]
You can start with the base case a=b=.5, c=0. (Denote it as f(x).) Note that from here, you can add or subtract any integer to a, b, or c, as this will only add or subtract an integer value. You can prove that f(x)=f(-.5-x) algebraically. Therefore, it suffices only to prove that f(x) is an integer for all whole numbers x. This can be done through induction, or nothing that $ f(x)\equal{}\binom{x\plus{}2}{2}$ for all whole numbers x, which implies that it is an integer value.


Express the sum $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{4}\plus{}\cdots\plus{}\frac{1}{10}$ as a decimal rounded to the nearest hundredth.

Evidently you're supposed to add them all up and get something $ \approx \boxed{2.93}$

EDIT: originally forgot the $ \frac{1}{10}$


Let $ a,b,c\geq0$ s.t. $ a\plus{}b\plus{}c\equal{}1$.Prove that:
$ \sqrt{a\plus{}b^2}\plus{}\sqrt{b\plus{}c^2}\plus{}\sqrt{c\plus{}a^2}\geq2$

[quote="Inequalities Master"]Let $ a,b,c\geq0$ s.t. $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \sqrt {a \plus{} b^2} \plus{} \sqrt {b \plus{} c^2} \plus{} \sqrt {c \plus{} a^2}\geq2$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106782


For a certain problem on a challenge set, I want to display a visual on the problem to support my solution. I need to display a cube with a 5x5 grid on all the sides. (Basically, the cube is split up into 125 small congruent cubes and I want to show how the bigger cube has been split up.) Thanks in advance!


It’s been a few years since I studied algebra and need a little help. Can someone show me the working involved to solve for x, y, z.

 0.1x + -0.2y + 0.3z = 0
-0.4x + -0.5y + 0.6z = 0
 0.7x + 0.8y + 0.9z = 5

Thanks.

[hide="First of all"]
Get rid of the decimals:
$ x \minus{} 2y \plus{} 3z \equal{} 0$
$ 6z \minus{} 4x \minus{} 5y \equal{} 0$
$ 7x \plus{} 8y \plus{} 9z \equal{} 50$
[/hide]
[hide="Solution"]
Express $ y$ from the first equation and substitute into the second. Now you should get a connection between $ x$ and $ z$, and express e.g. $ y$ and $ z$ from $ x$. Now substitute $ y$ and $ z$ into the third equation and you get $ x$ and after $ x$, you easily get $ y$ and $ z$.
$ x \equal{} \minus{} 0.625 \equal{} \frac { \minus{} 5}{8}$ , $ y \equal{} 3.75 \equal{} \frac {15}{4}$ , $ z \equal{} 2.7083333... \equal{} \frac {65}{24}$ [/hide]

why does one need to remove decimals? :?:
secondly,you changed      0.7x + 0.8y + 0.9z = 5  to  7x + 8y + 9z = 0

It's easier for me to do without decimals. Sorry for the mistake(solutions are correct).

@analytic

[b]The solutions are correct![/b]
You should think before you post!!!
Try substituting my values.
My only mistake is that I didn't write $ 50$, I wrote $ 0$, but I have done this on paper, and that's a mistake in writing. :wink:

Let's not forget how to solve using matrices.

(Um....I'm too lazy too make a bunch of matrices here, but you can use Cramer's Rule, inverses, etc.)

No problem. Just ask when you need help.


If $a^{b}=b^{a}$ for two real numbers $a$ and $b$, then find $a$ as a function of $b$.

A web site (by one of the teachers from my school) for reference:
[url=http://www.qc.edu.hk/math/Advanced%20Level/y^x_x^y.htm]click here[/url]

Thanks , I found what I have searching for.

I posted a bunch at

http://www.mathlinks.ro/Forum/viewtopic.php?p=519815

conceive

What about something like this

a=b^(a/b) = e^(a/b*lnb) <=>

1=a*e^(-(a/b*lnb)) // * (-1/b*ln2)  <=>

-1/b*lnb = -a/b*lnb*e^(-(a/b*lnb))  <=>

-a/b*lnb=W(-1/b*lnb) <=>

[b]a = -b/lnb*W(-1/b*lnb) [/b]

Where W=[url=http://functions.wolfram.com/ElementaryFunctions/ProductLog/]ProductLog[/url]


So, i want to prepare for the AIME. I have done some past AMC 10's off of AoPsWiki and i am doing pretty well at them. But, here are a few questions:
1. Is Vol 2 more helpful for AIME preparation or ACoPs (i already have all the intro's)?
2. Are there any more recommended resources?
3. If i take the AMC 10 and i have a score above 120 but i'm not in the top 1% or vice versa, do i still make the AIME?

3) yes.  The cutoffs vary a little from year to year.  The most you would need to qualify for the AIME is 120 on the AMC 10, or 100 on the AMC 12.  I believe this year the cutoff was around 118 for the AMC 10 and 97.5 for the AMC 12.

does anyone else have an answer for 1 and 2?

Acops is art and craft of problem solving.
We seem to be in a similar position, so heres what im doing.
Aops V2 thoroughly (every prolbem) Past AIMEs and then ACoPs.
Have you read all the intros? What have you been getting on the amc 10's? Have you done Volume 1?

2)All the AMC 8/10/12s are good prep(the last five AMC 8 problems, the AMC 10s, and the first 20 or so AMC 12 problems), AoPS volume 1 and 2, and ACoPS. If you can find more highschool competitions, those would probably be good preparation too.


Find $A,B$ such that :

$\sin A + \sin B + \sin (A+B) = 1 + \cos A + \cos B$

Do you want [u]all[/u] solutions, because that will be complicated.

A very easy solution to this is $(0,\frac{\pi}{2})$.

Here is a graph of the two sides of the equation.  The contacts of the blue and the red are solutions.

As you can see, there are many straight lines containing solutions.

The general equation of these lines is $x+y=(2n+\frac{1}{2})\pi$ for all integers n

Now, to detemine the general equation of those sinusoidal curves...

[hide="Here is answer"]$x+y\in \left \{2k\pi +\frac{\pi}{2}\ |\ k\in Z\ \right \}$ or $\sqrt 2\cdot \cos \frac{x-y}{2}=\sin \left(\frac{x+y}{2}-\frac{\pi}{4}\right).$

[u]Remark.[/u] The second relation is equivalently with the relation

$2\cos \frac{x-y}{2}=\sin \frac{x+y}{2}-\cos \frac{x+y}{2}\Longleftrightarrow \tan \frac x2+\tan \frac y2=3+\tan \frac x2\tan \frac y2.$[/hide]

In fact, the solution is:
$\sin A + \sin B + \sin (A+B) = 1 + \cos A + \cos B$
$\longrightarrow 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}+2\sin\frac{A+B}{2}\cos\frac{A+B}{2}=\sin^2\frac{A+B}{2}+\cos^2\frac{A+B}{2}+2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
$\longrightarrow (\sin^2\frac{A+B}{2}+\cos^2\frac{A+B}{2}-2\sin\frac{A+B}{2}\cos\frac{A+B}{2})+2(\cos\frac{A+B}{2}\cos\frac{A-B}{2}-\sin\frac{A+B}{2}\cos\frac{A-B}{2})=0$
$\longrightarrow (\sin\frac{A+B}{2}-\cos\frac{A+B}{2})^2+2\cos\frac{A-B}{2}(\cos\frac{A+B}{2}-\sin\frac{A+B}{2})=0$
$\longrightarrow (\sin\frac{A+B}{2}-\cos\frac{A+B}{2})(\sin\frac{A+B}{2}-\cos\frac{A+B}{2}-2\cos\frac{A-B}{2})=0$
From here, you yourslef can catch the result,can't you? :lol:


Can someone prove the following identity?

$ \sum_{k\equal{}0}^n 2^k {2n\minus{}k \choose n}\equal{}2^{2n}$

[quote="andrsncpr"]Can someone prove the following identity?

$ \sum_{k \equal{} 0}^n 2^k {2n \minus{} k \choose n} \equal{} 2^{2n}$[/quote]

[hide="Hint"] Let $ S_n \equal{} \sum_{k \equal{} 0}^n 2^k {2n \minus{} k \choose n}$ then show that $ S_n \equal{} 2^{2}S_{n \minus{} 1}$.[/hide]

I used quite a bit of algebra so I might have made some mistakes.

[hide="Solution"]

$ S_n \equal{} \sum_{k \equal{} 0}^{n} 2^{k} \binom{2n \minus{} k}{n} \equal{} 2^{0} \binom{2n}{n} \plus{} 2^{1} \binom{2n \minus{} 1}{n} \plus{} 2^{2} \binom{2n \minus{} 2}{2} \plus{} ......... \plus{} 2^{n \minus{} 1} \binom{n \plus{} 1}{n} \plus{} 2^{n} \binom{n}{n}$

Now use the Hockey Stick Formula:

$ S_n \equal{} (2^{1} \minus{} 1) \binom{2n \minus{} 1}{n \minus{} 1} \plus{} (2^{2} \minus{} 1) \binom{2n \minus{} 2}{n \minus{} 1} \plus{} (2^{3} \minus{} 1) \binom{2n \minus{} 3}{n \minus{} 1} \plus{} ..... \plus{} (2^{n \plus{} 1} \minus{} 1) \binom{n \minus{} 1}{n \minus{} 1} \equal{} 2 \binom{2n \minus{} 1}{n \minus{} 1} \minus{} \binom{2n}{n} \plus{} 2^{2} S_{n \minus{} 1} \equal{} 2^{2} S_{n \minus{} 1}$

Now just show that $ n \equal{} 0$ is true and go from there.[/hide]

P.S. In case you were wondering

[hide="Identity"]

$ \binom{k}{k} \plus{} \binom{k\plus{}1}{k} \plus{} ...... \plus{} \binom{n}{k}\equal{} \binom{n\plus{}1}{k\plus{}1}$[/hide]

Thanks a lot.  I'm really impressed by the way you applied the Hockey Stick Identity. BTW can there be a proof by some counting argument(e.g. committee selection)?

Here you go  :) :

[hide="Counting Argument"]Both sides count the number of subsets of $ \{1,2,\ldots ,2n \plus{} 1\}$ with at least $ n \plus{} 1$ elements.

LHS:  For some subset $ S$ of $ \{1,2,3,\ldots 2n \plus{} 1\}$ with at least $ n \plus{} 1$ elements, let $ X \equal{} 2n \plus{} 1 \minus{} k$ be the $ n \plus{} 1$th element of $ S$ when $ S$ is arranged in increasing order.  $ X$ must be at least $ n \plus{} 1$ and at most $ 2n \plus{} 1$, so we need to take $ k$ from 0 to $ n$.  We need to choose $ n$ elements out of the $ 2n \minus{} k$ numbers less than $ X$ - there are $ \binom{2n \minus{} k}{n}$ ways to do this.  Then we can choose any subset of the $ k$ numbers greater than $ X$ - there are $ 2^{k}$ ways to do this.  Summing from $ k \equal{} 0$ to $ k \equal{} n$, the number of subsets $ S$ is $ \sum_{k \equal{} 0}^{n}2^{k}\binom{2n \minus{} k}{n}$ as desired.

RHS:  The number of subsets is $ a \equal{} \binom{2n \plus{} 1}{n \plus{} 1} \plus{} \binom{2n \plus{} 1}{n \plus{} 2} \plus{} \cdots \plus{} \binom{2n \plus{} 1}{2n \plus{} 1}$.  Then
$ 2^{2n \plus{} 1} \equal{} \binom{2n \plus{} 1}{0} \plus{} \binom{2n \plus{} 1}{1} \plus{} \cdots \plus{} \binom{2n \plus{} 1}{n} \plus{} \binom{2n \plus{} 1}{n \plus{} 1} \plus{} \cdots \plus{} \binom{2n \plus{} 1}{2n \plus{} 1}$
$ 2^{2n \plus{} 1} \equal{} \binom{2n \plus{} 1}{2n \plus{} 1} \plus{} \binom{2n \plus{} 1}{2n} \plus{} \cdots \plus{} \binom{2n \plus{} 1}{n \plus{} 1} \plus{} \binom{2n \plus{} 1}{n \plus{} 1} \plus{} \cdots \plus{} \binom{2n \plus{} 1}{2n \plus{} 1}$
$ 2^{2n \plus{} 1} \equal{} 2a$

Therefore $ a \equal{} 2^{2n}$ as desired.  So the original identity is true.[/hide]


Does anyone know much about the world's population situation? Is overpopulation really a problem? If so, could you direct me to some books, websites, etc.?

And if so, what is being done to control the increase in population as of right now? I know China has a one-child policy (though I'm not entirely sure on all the details of that regulation), but are any other countries doing anything to stop the increase in population growth?

I dont know much about the world's situation, but I can give you a few details on India......we're seriously Overpopulated.....to get an idea on the indian situation, here are a couple of websites --

http://www.mts.net/~gcg/resources/sasia/index03.html

[url=http://72.14.203.104/search?q=cache:YyEOBh6ji6QJ:serendip.brynmawr.edu/local/scisoc/environment/seniorsem03/Overpopulation_in_India.pdf+Indian+overpopulation&hl=en&gl=in&ct=clnk&cd=2]Big URL[/url]

I think there is some two child norm or something, but then enforcement is really not all that prominent and still our illiteracy is rather high...

[i]Edited by Moderator: please use the [u]URL[/u] button for long URLs.[/i]

I am no expert on the subject either, but as always I find it my duty to recommend wikipedia [url=http://en.wikipedia.org/wiki/World_population#Rate_of_population_increase]link[/url]

It offers past population numbers, evolution for each continent, and predictions about the population

in 1987 we became five billion,in 1999 we became six billion, right now we just crossed 6.5 billion

right now in Belgium we have crossed the 10.5 million barrier, quite surprising as our population is aging, but the increase is mainly due to immigration probably, not births.

[url=http://www.rpi.edu/~eglash/eglash.dir/ethnic.dir/race/chimps/chimps.htm]diversity[/url]
this article claims all of us here are genetically more similar than a group of apes living together, because we almost got wiped out in the past

[url=http://www.math.hawaii.edu/~ramsey/People.html]number of people[/url]

this article claims that there have been 96 billion people on earth

[url=http://vhemt.org/]voluntary human extinction movement[/url]
these people think you should only pass on your genes when you are smart enough, and you are smart enough when you realise you shouldn't make children  :roll:

They don't have an official opinion about abortion, they oppose violence, and expect their members not to make children, while they are allowed to have created children before they became a member.

I recommend the book Collapse, by Jared Diamond.  It's primarily concerned with the effects of ecological problems on civilization survival, but overpopulation figures quite strongly in it.  (The total human impact on the world is a product of the individual impact by the number of people, of course.)  I personally suspect that the more serious problems of the near future, in terms of global significance, will be caused by an increase in individual impact in China and India then by total increases in the human population.  (A small amount of number-crunching indicates that Chinese economic growth alone (Chinese population growth is quite low at this point) will lead to an almost-doubling of world consumption in the next 30 or so years.  That is, before 2050 China seems likely to be consuming as much of everything per year as the entire world is right now.)  Of course, this can also be read as saying that the world is already badly overpopulated and we just haven't realized it yet because so many people have exceedingly low standards of living.

Actually, in the U.S., social security is a big problem. Guess why? There are not enough people entering the work force to support the aging Baby Boomer generation. Chew on that.

The formula for human population growth on earth is:

$P(t)=\frac{P(0)e^{rt}}{1+\frac{P(0)(e^{rt}-1)}{K}}$

P(t) is population after t years
P(0) is population when t=0
r is the annual growth rate
K is the Earth's carring capacity.

This type graph of this creates a S-shape similar to the graph shown below. Eventually, The growth rate will decline depending on what K equals. But K could equal anywhere from 7 billion to 1 trillion or more depending on how far science can prolong life.

If nothing is done, our population will explode, then level off. At that point, an Ebola outbreak or any other fatal and contagious disease could wipe out the human population.

There are several solutions to population overcrowding:

I: We build skyscrapers several miles high on this planet and we live on other planets. Just like they did a long time ago in a galaxy far, far away.
Problems: We haven't even got to Mars, or invented travel that takes us quickly to other planets, or built skyscrapers that tall.

II: We stop using medicines and reduce our life span.
Problems: Few people are willing to suffer or die to reduce overcrowding; all people in the medical field are unemployed.

III: Encourage stupid people to try to get a Darwin Award.
Problems: Most stupid people aren't that stupid.

IV: Every couple has exactly 2 children when their life is exactly halfway over. This will stabilize our population well.
Problems: No one knows when half their life is over. Also triplets can ruin the plan.

V: Pay Antarctica a lot of money so the penguins will allow us to live in Antarctica. This would provide temporary relief to overcrowding.
Problems: Penguins don't want humans settling in their country, Penguins have no need for money, Humans can't speak Penguin, Antarctica is too cold for most people.

Clearly all of these solutions have problems, (Some bigger than others.) Hopefully some genius out there in the future will come up with a good solution.

Unless something really bad like a viral outbreak happens, my bet is that no one will care enough to do anything, we will continue to grow an cover the planet, and then lots of people will die from malnutrition or stupidity or something, and everyone will be really sad. But we will eventually stabilize.

[quote="tokenadult"]Contraceptive methods (I am excluding abortion, as that is not contraception) seem to work well enough these days that many countries have natural DECREASE in population expected in the next fifty years. Maybe eventually lots of countries will be that way, although there may be local exceptions here and there.

I think JBL's point about aggregate ecological impact is more pertinent, although technology (and changes in a country's political and economic system) can amelioriate those too. The Soviet Union was never as prosperous as the United States, but after glasnost we discovered that it was much, much more polluted. A sound legal system and free elections can do a lot to nudge economic development in more sustainable directions.

P.S. I can remember, back in 1982, hearing a French student describe China as a place without pollution (in contrast to Taiwan, where we were having that conversation soon after separate visits to China). In 2001, I had the astounding experience of hearing a Ukrainian-American, who had just visited Hong Kong, tell me when I was again living in Taiwan that Taipei then was cleaner than Hong Kong--something I never would have imagined would be possible in the early 1980s. Political and legal systems matter. Dictatorship not only destroys human freedom and dignity, it ruins the environment.[/quote]

The reasoning method of brainwash-ee is really undiversified and ridiculous.

Buffalo, your last comment doesn't make any sense.  Even if it did make sense, it wouldn't be clear what you were responding to.  Even if it were clear what you were responding to, it doesn't appear that your response constitutes anything more than an ad hominem.

[quote="tokenadult"]I think JBL's point about aggregate ecological impact is more pertinent, although technology (and changes in a country's political and economic system) can amelioriate those too.[/quote]
Well, this is certainly true in many places.  China, for instance, currently relies on relatively outdated, inefficient and dirty coal-burning technology.  The switch to more efficient and cleaner-burning coal plants would certainly reduce pollution and reduce overall resource consumption.  But it doesn't seem likely to me that technological advances could possibly offset the consumption increase inherent in raising the world-wide standard of living -- it's not even clear if current world consumption levels are long-term sustainable, after all.

[quote="tokenadult"]The Soviet Union was never as prosperous as the United States, but after glasnost we discovered that it was much, much more polluted. A sound legal system and free elections can do a lot to nudge economic development in more sustainable directions.[/quote]
I hate to stick up for dictatorships here, but serious top-down governance can have substantial positive environmental effects, as well.  The history of the Dominican Republic under Balaguer, for instance, is that of a dictator who was serious and successful about caring for the environment.  Even China has had its successes in this regard -- I believe that after the 1998 floods, loggings was effectively banned throughout much of the country, the sort of thing we can't do in a Western Democracy.  (One can debate how important that is for the world environment, since China has in a sense just exported its logging to Southeast Asia and other areas, but from the point of view of the Chinese enviorment, it's certainly a plus.)  One shouldn't forget the green wall around Beijing, either.
Of course, this requires that the absolutist leadership actually care for the environment and act on that care.  China is currently far more interested in pursuing economic growth, and so the Chinese environment will continue to suffer.

Also, frankly, I'm skeptical that elections are the tools which drive economic protection.  Saving the environment is rarely a make-or-break political issue (just see the current American government, for instance), although I suppose that the ability of the Greens to join in parliamentary majorities in Europe might make a difference there.  I'm inclined to view the regulatory powers and individual ability to sue polluters as much more significant factors in this regard.

(Side note: most of my factual knowledge regarding the D.R. and China in this post comes from [u]Collapse[/u], the book I mentioned above.)

The simple answer to your question is that the current population of humans on the earth is not a problem.

Like most simple answers, this answer can be misleading or not applicable in some cases where it applies.  In some areas, such as the back of some pickup trucks, some bedrooms, some houses, some parties, some beaches, some airplanes, some companies the population of humans in the space is too much, by some standards.

The question as applied to food is the same.  There is currently an overproduction of food for the human population.  Distribution is the issue, as well as local government corruption, fellow people willing to allow or even work towards starving others.  The same thing occurs with violence.  Let's get "those" to leave, so, "we" can take the property they are occupying.  We see this in Africa, Iraq, neighborhoods in developed nations, including the US.

If one considers that an animal takes approximately 20 times the amount of food it produces, one can take the percentage of food that is represented by animal products and know that one can increase the food supply roughly 20 times simply by growing food for people rather than for feed animals.  However, in some cases the land may not be easily converted to people editable food.

If one examines human densities (people per square mile) to address this question and one assumes that one could re-distribute people without any cost/ wars/ or political constraints (the purple people don't want to be in the same nation as the people that worship the stars that make up Orin's belt), one can put the entire human population in a small state of the United States, Russia, China, or elsewhere, assuming a human density equal to that of Tokyo.

The world has been blessed with prophets of doom.  In the 1950s the world was going to end in wars over the lack of food.  Irrigation, chemicals to reduce the loss to insects, increase yields, changes in farming techniques have greatly increased food output, at the same time the amount of land that is farmed has decreased significantly.  I had an article in a US national newspaper raising alarm with the upcoming oil crisis.  The proven reserves would only last 15 years, then we would be out of oil and need to revert back to other sources of energy, with the resulting collapse of civilization.  The newspaper headline was in the 1930s.  Proven reserves, assume today’s technology, today’s prices and an acceptable risk that the oil field will produce sufficient profit to cover the development costs applied to the oil field.

This goes on and on.  The future is before us.  It is bright.  If you are concerned, don't just sit there, do something yourself -- go to work for a relief agency, send letters/e-mails to support those trying to stop wars and government actions that put people at risk of starving, turn the house down a degree or two in the winter, use the windows rather than the air conditioner, or turn it up a degree or two in the summer (radiation heat loss is related to the difference in absolute temperature taken to some high power, like 5), tell your Mom that you will walk home one day a week or from one store in the shopping center to another.  Don't do jackrabbit starts and stops when driving.  In short, if you care, don't just think, show you care by taking some actions.  It would be better to take little steps that you do for a long time than an one time larger step.  (Did you even turn off the light after you left the room?  It doesn't take much to make a difference, if you really want to help and not just complain or get glory for doing something.)

This may seem far from your question, but the “burden” from humans on earth is directly related to how much each consumes.  I’m tired of folks saying that they are convinced that global warming is a real problem, but they want someone else to fix it, while they do nothing different to support their conviction.

If you want to find more on this look up “commons.”  It is the title for this kind of thing.  Studies show that things like commons, where everyone benefits, with little or no costs are desirable, but no one is willing to pay themselves, they want to urge others to pay for the commons that they can enjoy.

There are generally various kinds of recourse against polluters in countries with a free civil legal system, and it is especially that kind of private action under ordered liberty that is lacking in countries with dictatorships.

[quote="mysmartmouth"]Google the Voluntary Human Extinction Movement, or something like that.  I wonder if those people are serious.[/quote]

I already posted that link. ;)

OOPS!!  :blush:  I guess great minds think alike.  Lol.


Putnam
number theory proposed
number theory

Call $ S(n)$ is sum of all positive divisor of n
Know that
$ S(n)\equal{}2^{2k\plus{}1}n\plus{}1$  k is a nonnegative  integer
Prove that n is a perfect square

Sum of divisors denote $ \sigma (n)$.
Obviosly, that $ \sigma(n)$ odd if and only if $ n\equal{}2^{l}m^{2}$, were m is odd.
I think, that equation $ \sigma(n)\equal{}2^{2k\plus{}1}n\plus{}1$ had not solutions.

[quote="TTsphn"]Can you prove that n is a perfect square.[/quote]
Can you find n, satisfyed your equation?
Let $ n\equal{}2^{l}m^{2}$, m -odd. If l odd, then $ 3|\sigma(2^{l})|\sigma(n)$, therefore from $ \sigma(n)\equal{}2^{2k\plus{}1}n\plus{}1$ we get $ n\equal{}1\mod 3$, but from $ n\equal{}2^{l}m^{2}\equal{}1\mod 3$ we get l is even - contradition.

[quote="TTsphn"]Call $ S(n)$ is sum of all positive divisor of n
Know that
$ S(n) \equal{} 2^{2k\plus{}1}n\plus{}1$  k is a nonnegative  integer
Prove that n is a perfect square[/quote]
See here http://www.mathlinks.ro/Forum/viewtopic.php?t=70618 . In this one has a similary problem.

My method work for $ \sigma(n)\equal{}an\plus{}b$, were $ a\equal{}2\mod 6,\ b\equal{}1\mod 6$.

It is the same method .
This problems take from Putnam for k=0 and when I solve it I general this problem.


inequalities
quadratics
algebra
quadratic formula
number theory

Prove that if $ a,b,c>0$ then $ a^5\plus{}b^5\plus{}c^5 \geq 5abc(b^2\minus{}ac)$.

When does equality hold?

$ a^5 \plus{} b^5 \plus{} c^5\ge b^5 \plus{} 2\sqrt {a^5c^5}$.

Let $ \sqrt {ac} \equal{} x$.

It suffices to show that $ b^5 \plus{} 2x^5 \ge 5x^2b(b^2 \minus{} x^2)$

$ b^5 \minus{} 5x^2b^3 \plus{} 5x^4b \plus{} 2x^5 \ge 0$

$ (b \plus{} 2x)(b^2 \minus{} bx \minus{} x^2)^2 \ge 0$, which is clearly true.

Equality holds when $ a \equal{} c$ and $ b \equal{} \frac {1 \plus{} \sqrt {5}}{2}a$ (the quadratic formula also yields $ b \equal{} \frac {1 \minus{} \sqrt {5}}{2}a$ but this is a negative value, so it cannot equal $ b$).

Nice problem BTW, where did you get it?

Cheers,

Rofler

Thanks for your nice solution Rofler.
I got the problem from Introductions to Number Theory and Inequalities by CJ Bradley.  Next to the problem it says (DM) which I'm guessing refers to some contest but I don't know which one.


This problem posted here http://www.mathlinks.ro/Forum/viewtopic.php?p=198167#p198167  :wink:


Given $ a, b, c, k > 0$ and $ a\plus{}b\plus{}c\equal{}3$. Prove that: $ (a\plus{}b)\sqrt[k]{\frac{ab\plus{}1}{c^2\plus{}1}}\plus{}(b\plus{}c)\sqrt[k]{\frac{bc\plus{}1}{a^2\plus{}1}}\plus{}(c\plus{}a)\sqrt[k]{\frac{ca\plus{}1}{b^2\plus{}1}} \ge 6$
 :)

[quote="nguoivn"]Given $ a, b, c, k > 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that: $ (a \plus{} b)\sqrt [k]{\frac {ab \plus{} 1}{c^2 \plus{} 1}} \plus{} (b \plus{} c)\sqrt [k]{\frac {bc \plus{} 1}{a^2 \plus{} 1}} \plus{} (c \plus{} a)\sqrt [k]{\frac {ca \plus{} 1}{b^2 \plus{} 1}} \ge 6$
 :)[/quote]
This general problem exists a nice solution by Am-Gm  :lol:


can you tell me how is called the program with witch you write math's symbols

LaTeX

When you want to use it on these forums, enclose your math in dollar signs.

[code]Like This: $a^2 + b^2 = c^2$[/code]
becomes

$a^{2}+b^{2}= c^{2}$


How many ways are there of arranging 6 A's, 6 B's, and 6 C's around a circle.  The generalization of this problem would be to count the number of ways of arranging N objects around a circle; of these objects, M1 are of type 1, M2 are of type 2, ... and Mn are of type n (so these Mn objects are indistinguishable from each other), where M1+M2+...+Mn=N.

[quote="mathwizarddude"]I think the answer is $ 17!/(6!)^3$.[/quote]

How did you get it?

[hide="How he probably got it"]There are $ 17!$ ways to seat $ 18$ different people around a round table. Here we are considering three groups of six people the same, so we are dividing by $ 6!$ three times.[/hide]

[hide="Why it is not right"]A sequence like $ ABCABCABCABCABCABC$, which is periodic, is counted multiple times in this method.[/hide]

Here is a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194927]related[/url] problem, though not the same. The generalized version I don't think is going to have any nice closed form due to the difficulty of describing all periodic sequences in general. The problem you've shown is a lot of computation using the ideas in that topic.

(Also, please only posts topics in one forum. Even if it's not quite the right one, the mods will sort it out. Thanks.)

Look up polya's counting method (or enumeration theorem)... This is a bit of a pain because you need the cycle index of 18 which is annoyingly composite. There is a more or less general solution to this but it would come out as a summation.

$ \frac{1}{n}\sum_{d|n}\phi\left(\frac{n}{d}\right)\cdot (w_{\frac{n}{d}})^{d}$

where $ w_d\equal{}x_1^d\plus{}x_2^d\plus{}...\plus{}x_m^d$ where we have $ m$ colors and $ n$ total beads ($ x_i$ are variables in the generating function).

In this case, $ n\equal{}18$, and we want the $ (x_1x_2x_3)^6$ coefficient. I don't feel like computing the sum but in principle, it is easy.

$ d\equal{}1,2,3,6,9,18$

\[ \frac{1}{18}\cdot (\phi(18)(w_{18})\plus{}\phi(9)(w_9)^2\plus{}\phi(6)(w_6)^3\plus{}\phi(3)(w_3)^6\plus{}\phi(2)(w_2)^9\plus{}\phi(1)(w_1)^{18})\]

So the answer is \[ \frac{1}{18}\cdot\left(\phi(6)\frac{3!}{1!1!1!}\plus{}\phi(3)\frac{6!}{2!2!2!}\plus{}\phi(2)\frac{9!}{3!3!3!}\plus{}\phi(1)\frac{18!}{6!6!6!}\right)\]

\[ \equal{}\frac{1}{18}\cdot (12\plus{}180\plus{} 1680\plus{} 17153136)\equal{}953056\]


I have a query. Can there be infinite perfect numbers. Actually this was based on "alchemist"'s question . Can there be a polynomial where P(n) rep sum of divisors of n. The obvious proof is Prime Numbers.

But  Perfect number can also be a good proof but I dunno if there are infinite of them. Any good proofs fellow proving or contradicting it?

Hint;
Perfect Numbers are those whose sum of divisors is equal to the number itself. Or including that number itself, it is twice that number.

eg: Factors of 6 - 1,2,3
1+2+3 = 6
+6 = 12

28 = 1 + 2 + 4 + 7 + 15

[quote="madness"]Sorry to post the obvious, but another thing is that there are no odd perfect nos.  :D
  :D[/quote]
Why is this so obvious?
to my knowledge the question on odd perfect nos: is an unsolved one .it has
been shown that all odd perfect numbers are larger than .(10^21)            but there existence has not been disproved yet.

also if (2^p) -1 is a prime(mersenne), (2^p-1)*(2^p -1) is a perfect number.(easy to show).
however its not known wether infintely many  mersenne primes exist but it is thought that it does. this will lead to  infinetly many perfect numbers

sry got confused with another thing never mind  :oops:   :)

alter ego, i don think that formula holds good cause try putting p=2...
$ {2^p}{\minus{}}{1}$  =3 which is prime... but 3*3=9 is not a perfect number

yes that's because the formula is $ 2^{p}(2^{p}\minus{}1)$ :wink:

no pardesi is right bcos its an ISI sum  :P   :rotfl:

[quote="valeriummaximum"]why are u hidden madness????[/quote]

Because everybody else is hidden and he just follows the crowd.

Dont be a stereotype, be a prototype.

CHUMMA JUJUBI  :P   :rotfl:  :rotfl:
simply da was bored so..........
and btw i lost some parts of me chem book so im also in ncert website

[quote="valeriummaximum"]why are u hidden madness????[/quote]

shreyas avanukku [b][i]STOMACH BURN![/b][/i]

Ana rohit unakku ethu knojam over.
Enna varasolli nee eppadi eruntha :mad:

rangavukka?  :rotfl:  :rotfl:
shre : its not for the purpose of copying da everything i do has a sense of purpose  :wink:

[quote="pardesi"]yes that's because the formula is $ 2^{p}(2^{p} \minus{} 1)$ :wink:[/quote]
isnt the formula $ 2^{p\minus{}1}(2^{p} \minus{} 1)$ ?

e.g.$ 2^{5\minus{}1}(2^{5} \minus{} 1)$ = $ 31*16  \equal{} 496$, a perfect number

yes may be... :lol: ...i aint sure of it just know something like that exist which i have seen before


calculus
derivative
function
limit
trigonometry
real analysis
real analysis unsolved

Let $ f: U \rightarrow R$ be differential for all $ x \in U$ and let $ f_{n}(x)\equal{}n(f(x \plus{} \frac{1}{n}) \minus{} f(x))$ then is it true that $ f_n(x)$ converges uniformly to $ f'(x)$.

The right adjective form is "differentiable".

The derivative of a function which is differentiable everywhere does not have to be uniformly continuous. A simple counterexample: $ f(x)\equal{}x^3$ on the whole line.

your right the first impression that I got was that, but the only if part is kinda not I have shown that $ f(x) \equal{} x^3$ is uniformly convergent just by showing it was Cauchy.

[quote="fedja"]If one just looks at the question for just a second, one gets the impression that, for the functions defined on the entire real line, it is true if and only if the derivative $ f'$ is uniformly continuous. The "if" part is easy: just use the mean value theorem. How about the "only if" part? Is this impression correct? ;)[/quote]

No, that's not uniformly convergent. If $ f(x)\equal{}x^3$, we have $ n\left(f(x\plus{}\frac1n)\minus{}f(x)\right)\equal{}3x^2\plus{}3\frac{x}{n}\plus{}\frac1{n^2}$. That $ 3\frac{x}{n}$ is unbounded for any $ n$, so we don't get uniform convergence. I meant that to be a counterexample for your original question, as well as for uniform continuity of the derivative.

A derivative that simply grows rapidly at $ \infty$ isn't badly behaved enough to have a chance at uniform convergence the wrong way.

Please plug in $ a_n$ into x and tell me what $ a_n \rightarrow a$ as $ n \rightarrow \infty$ so that  $ \lim_{n \to \infty} f_n(a_n) \neq f(a)$

Your right lets make $ U$ a bounded set in $ R$

That just calls for a different example. Pick a differentiable function with a discontinuous derivative, like $ f(x)\equal{}\begin{cases}x^2\sin\frac1x& x\neq 0\\ 0&x\equal{}0\end{cases}$

The functions $ f_n$ are continuous, so they can't converge uniformly to the discontinuous function $ f'$.

Or better yet lets make $ f'(x)$ bounded in in $ U$

In your example $ f(x)$ is not differentiable at $ x \equal{} 0$. If $ f'(x)$ is not continuous at $ x\equal{}a$ then $ f(x)$ is not differentiable at $ x\equal{}a$.

:what?: Fraenkel, you need to refresh your memory about differentiability. You may even have a function differentiable at a point and discontinuous everywhere else. Jmerry's function is differentiable everywhere including $ 0$, only the derivative at $ 0$ is not given by the same formula as everywhere else; it is just $ 0$.

Now, how about my question? (I should confess that, however strange it may seem, I still don't know the answer :blush:).

fedja, It sounds like your question is a better question I will work on it now

That question is a lot harder. It is quite possible to create oscillating functions that have $ f_n$ close to $ f'$ for some $ n$ for the "wrong" reasons; if $ \alpha$ is a nontrivial (complex) root of $ \alpha\equal{}n(e^{\alpha/n}\minus{}1)$, the real and imaginary parts of $ e^{\alpha x}$ have exact equality. That may be useful in attempting to construct examples; each such function falls apart for larger $ n$, but splicing together several versions might have a chance.

If it had fallen apart only for larger $ n$, that would be no problem. But it falls apart for smaller $ n$ as well. Actually, the "only if" part holds  :lol:. I do not want to spoil the fun by posting the solution immediately (if nobody finds it within a couple of weeks, I'll post it), but here is a small
[hide="Hint"]
If $ h\equal{}\frac 1{2n}$ for large enough $ n$, what can you say about $ f'(x)\minus{}f'(x\plus{}h)\equal{}2f'(x)\minus{}(f(x)\plus{}f(x\plus{}h))$ if you know that $ f'(t)\approx \frac{f(t\plus{}H)\minus{}f(t)}{H}$ for all $ t$ and $ H\equal{}h,2h$? ;)
[/hide]
If one finds this hint too revealing, he may try to show that $ f'$ is uniformly continuous if there exists [b]some[/b] sequence $ h_n\downarrow 0$ such that $ \frac{f(x\plus{}h_n)\minus{}f(x)}{h_n}$ converges to $ f'$ uniformly on $ \mathbb R$. :)

Thanks fedja I kinda see where its going please give me a little more than 2 weeks.

What do you mean by uniformly converging the wrong way.

[quote="jmerry"]No, that's not uniformly convergent. If $ f(x) \equal{} x^3$, we have $ n\left(f(x \plus{} \frac1n) \minus{} f(x)\right) \equal{} 3x^2 \plus{} 3\frac {x}{n} \plus{} \frac1{n^2}$. That $ 3\frac {x}{n}$ is unbounded for any $ n$, so we don't get uniform convergence. I meant that to be a counterexample for your original question, as well as for uniform continuity of the derivative.

A derivative that simply grows rapidly at $ \infty$ isn't badly behaved enough to have a chance at uniform convergence the wrong way.[/quote]

If $ f_n(x)$ uniformly converges to $ f(x)$ and $ a_n$ converges to $ a$ then $ f_n(a_n)$ uniformly converges to $ f(a)$ but what does this tell you about the convergence of $ f(a_n)$ to $ f(a)$ is this uniform as well???

It's not clear what you are asking. $ f(a_n)$ is a single numerical sequence, so what can be "uniform" here? :?

sorry I guess I was thinking it put not writing it $ a_n\equal{}x\plus{}\frac{1}{n}$

Then no: just take $ f_n(x)\equal{}f(x)\equal{}x^3$ for all $ n$. Then $ f_n(x\plus{}\tfrac 1n)$ do not converge to $ f(x)$ uniformly.

[quote="Fraenkel"]fedja i have given up on your problem. it is too hard for me.

I would like to see your solution.[/quote]

I am confused   :blush:  fedja did give you a counter example

@1234567a: Read the entire thread; you'll notice that we decided to change the original problem to a more meaningful one somewhere in the middle.

@ Fraenkel: Matiasp asked me to give him more than 2 weeks and to give up after just 5 days is a bit too early (it is the hard problems which you have to solve to really learn something and I gave you a fairly decent hint), so I prefer not to post a solution right now (but don't worry, I'll not forget about it and will post it eventually unless somebody else beats me to it :)).

its been over two weeks.
I anxiously await in hungry anticipation.

OK, let's start with the easiest version of the problem. Suppose that $ \frac{f(x\plus{}h)\minus{}f(x)}h$ converges to $ f'$ uniformly as $ h\to 0\plus{}$. Write
\[ 2\frac{f(x\plus{}2h)\minus{}f(x)}{2h}\equal{}\frac{f(x\plus{}2h)\minus{}f(x\plus{}h)}{h}\plus{}\frac{f(x\plus{}h)\minus{}f(x)}{h}\]
The left hand side is $ 2f'(x)\plus{}o(1)$ while the right hand side is $ f'(x)\plus{}f'(x\plus{}h)\plus{}o(1)$ as $ h\to 0\plus{}$ where $ o()$ is uniform in $ x$. Comparing, we get $ f'(x\plus{}h)\equal{}f'(x)\plus{}o(1)$ with uniform $ o()$ as $ h\to 0\plus{}$, which is just a restatement of the definition of uniform continuity.

In the original problem, this leads only to $ f'(x\plus{}\tfrac 1{2n})\equal{}f'(x)\plus{}o(1)$ as $ n\to\infty$. One more step is needed. Can you find it now? ;)

Since nobody has answered, I'm posting the solution for $ 1/n$ (but not for general $ h_n$, this will remain "open" until somebody else solves it :P).

So, we know that $ f'(x\plus{}h_n)\minus{}f'(x)\equal{}o(1)$ as $ n\to\infty$ with $ h_n\equal{}\frac 1{2n}$. Now, let us take any $ h$ and write
\[ \begin{aligned}
f'(x\plus{}h)\minus{}f'(x)&\equal{}\frac{f(x\plus{}h\plus{}h_n)\minus{}f(x\plus{}h)}{h_n}\minus{}\frac{f(x\plus{}h_n)\minus{}f(x)}{h_n}\plus{}o(1)
\\
&\equal{}\frac 1{h_n}\Bigl.[f(t\plus{}h_n)\minus{}f(t)]\Bigr|_{t\equal{}x}^{t\equal{}x\plus{}h}\plus{}o(1)
\\
&\equal{}\frac h{h_n}[f'(t\plus{}h_n)\minus{}f'(t)]\plus{}o(1)\equal{}o(1)
\end{aligned}\]
with some $ t\in\mathbb R$, provided that $ |h|\le h_n$, $ n\to\infty$. (The Mean Value Theorem was used at the last step).


Jack and Jill got into a fight. The probability that they will see each other on the $n$th day from now is $\frac{3}{3n+5}$. What is the probability that they will not see each other for the next 31 days?

[hide="solution"] OK, so the probability that they won't see each other on the $n$th day is $1-\frac{3}{3n+5}= \frac{3n+2}{3n+5}$. Note that the denominator can be expressed as: $3(n-1)+2$, so each denominator will equal the following numerator. (Here we have $\frac{5}{8}\cdot \frac{8}{11}\cdot \frac{11}{14}\cdot...$). Since to find the total probability you multiply the probabilities for each day together, everything cancels except for the first numerator and the last denominator. The first numerator is $3(1)+2 = 5$ and the last denominator is $3(31)+5 = 98$. Thus, the final probability is: ${\frac{5}{98}}$ [/hide]


linear algebra
matrix
linear algebra unsolved

how can one prove elementarily (i.e: no Jordan form etc ..) that if
$\begin{pmatrix}A&0\\0&M\end{pmatrix}$ and $\begin{pmatrix}A&0\\0&N\end{pmatrix}$ are similar ($A,M,N$ are square matrices), then $M$ and $N$ are similar.

I had to prove this elementary in trying to prove the "uniqueness" of Jordan decomposition. Anyone with a nice approach ?

$B$ is similar to $C$ if an only if $\exists$ and invertible $P$ such that $BP=PC.$ Let's block-structure $P$ as follows:

$P=\begin{bmatrix}U&V\\X&Y\end{bmatrix}.$

Then, $\begin{bmatrix}A&0\\0&M\end{bmatrix}\begin{bmatrix}U&V\\X&Y\end{bmatrix}= \begin{bmatrix}U&V\\X&Y\end{bmatrix}\begin{bmatrix}A&0\\0&N\end{bmatrix}$

$\begin{bmatrix}AU&AV\\MX&MY\end{bmatrix}= \begin{bmatrix}UA&VN\\XA&YN\end{bmatrix}$

Right away, we have that $MY=YN.$ All we need to know to finish this is that $Y$ is invertible.

Hmm.... Haven't been able to prove that yet, so I don't have it.

I thought of this approach but $Y$ doesn't need to be invertible: just take the identity matrix for each of the original block matrices and any invertible matrix for the one giving the similarity relation (in this case you can easily get $Y=0$)


Hello!

I suppose that these are not very difficult, but I get stuck with them.

Which of the following represents a valid formula for an alcohol (please explain your answer)?

1) $CH_{3}-CH = CH-CH_{2}-OH$

2) $CH_{2}= CH-OH$

3) - 4) Please see the attachment.

Thank you!

[quote="Sebi Crisan"]Which of the following represents a valid formula for an alcohol[/quote]

Assuming that "valid formula" means not only that the compound is an alcohol but also a [i]stable[/i] alcohol that may exist in solution, then (remembering that an alcohol is an organic compound that have in its structure the functional group $-OH$):

1) Yes, this is an alcohol.
2) The formula represents an alcohol, but this structure is too reactive to exist in solution, where it tautomerizes to the more stable acetaldehyde.
3) This is also an alcohol.
4) This is not an alcohol, but a ketone (because it have the functional group C=O between two carbons).


[quote="harazi"]Find the smallest value of a^2+b^2+c^2+d^2+e^2+abcde(1/a+1/b+1/c+1/d+1/e) when a,b,c,d,e>0 have sum 1. What about the general case? I don't have a solution to the last question.
    Harazi[/quote]
 f(a,b,c,d,e)=a^2+b^2+c^2+d^2+e^2+abcde(1/a+1/b+1/c+1/d+1/e)
 we will prove that f[(a+b)/2,(a+b)/2,c,d,e]<=f(a,b,c,d,e)
 this is equivalent to (a-b)^2*(2-ed-cd-ce)>=0(obvious)
 with the same technique we can prove that
 f(a,b,c,d,e)>=f[(a+b+c+d)/4,(a+b+c+d)/4,(a+b+c+d)/4,(a+b+c+d)/4,e]
 the problem becomes so simple now :)  :)  :)  :)
 I think that we can also solve the general problem similarly!!

I find this problem very interesting and I would like to see your solution, Harazi.

Thank you, very much.

I don't have it. That is why it is in the open questions forum. But try what silversunboy said.

The general problem was created by me a long time ago in http://www.diendantoanhoc.org and once posted in Mathlinks, too, but no one shows a solution.

In my Vietnamese book in 2006, it was solved. The answer is that, if $a_{1}+a_{2}+...+a_{n}=s$ then
\[\sum_{i=1}^{n}a_{i}^{2}+\prod a_{i}\sum_{i=1}^{n}\frac{1}{a_{i}}\]
attains the minimum if and only if $n$ numbers equal, $n-1$ numbers equal and the rest is $0$, $n-2$ numbers equal and the rests (two numbers) are $0$.

I solve it my mixing variable.


Is there by chance a winning strategy for the chopstick game? If not, then is there at least a way to mathematically look at the problem in terms of number theory?

Thanks in Advance

Assuming we're talking about this: http://en.wikipedia.org/wiki/Chopsticks_(hand_game)

According to that artcle, "Chopsticks is a solved game in which the second player can always force a win when playing with the original rules." Not sure what exactly that strategy is, but I'll think about it

Ah, I played this incessantly in third grade. I used to have this winning strategy that always worked as the second player, and sometimes as the first - I think this is the one grn_trtle mentioned. Since it's been a while, I don't quite remember the exact method, but I'll try to outline it the best I recall.

First, the game only works when you have splits. There are 3 basic rules to splits - 1) you can split however you like, 2) you can bring a dead hand back to life with these, and 3) the only finger counts that can be split are those that are divisible by 2. So you can't go 1,2 to 2,1.

You should go second, and try to avoid tapping your opponent's hands as much as possible. I really don't remember much else, but at the end, you want to end up with your opponent having (1,0) and you having (1,1).

[quote="isabella2296"]You should go second, and try to avoid tapping your opponent's hands as much as possible. I really don't remember much else, but at the end, you want to end up with your opponent having (1,0) and you having (1,1).[/quote]

[quote="ckck"]I think it depends on which variation you play this game in.
I play the variation where you take the higher numbers in mod 5 so you have to hit a hand exactly to 5...which gives more options xD[/quote]

What izzy is saying is that if you have (1,1) and your opponent has (1,0) [u]AND ITS YOUR OPPONENTS TURN NOW[/u] then your opponent must use his 1 to add to one of your 1's, as you cant split with a (1,0), at least how I play you cant do that. So now it looks like:
You:
(2,1)
Opponent:
(1,0)

then  its your turn. You split the (2,1) to get:
You:
(3,0)
Opponent:
(1,0)

Again, your opponent must hit your (3,0) so your at:
You:
(4,0)
Opponent:
(1,0)

Since its your turn, you win by hitting his (1,0) with your (4,0).

As for how to get to the  (1,1)/(1,0) split when its your  friends turn, Im at a loss.

If your friend had (1, 4) somehow and you hit his 4, he has  (1, 0) and you have (1, 1).

I'm not 100% sure about this, but I'm under the impression that it is a draw with the mod 5 rule, and with splitting however you want (without any knubs, quarters, etc. )

The outcome of the game really depends on what gametype you play. No one set of rules can encompass them all.

Also, is it just me, or are the people playing "knubs" only joking around when they do it?


Hi, does anyone how to solve the following problems:

[color=green]In each of the following cases, determine if the given function is bijective.  If the function is bijective, find its inverse.

(a) The function f : R -> R defined by f(x) = 2x-3
(b) The function f : Z -> Z defined by f(x) = 2x-3[/color]

I don't really know what's the difference between (a) and (b).
I've done part (a), anything wrong?

Ans: (a)
Let yER, then if xER, then
y is the image of x under f <=> f(x)=y
                                       <=> 2x-3=y
                                       <=> x=(y+3)/2
so each yER is the image under f of a unique image of xER.
Therefore, f is bijective and has an inverse (y+3)/2

Thanks very much!!!!  :lol:

Part (a) is fine.

For part (b), does there exist an integer $x$ such that $2x - 3 = 0$?


$ a\equal{}5^{1000}\text{sin}(1000b)$ where $ \text{sin}b\equal{}\frac{3}{5}$
1.Prove that $ a\in Z$
2.Find $ a (\mod5)$

Obviosly $ cos b \equal{} \frac {\pm 4}{5}$, therefore $ a \equal{} Re(3\pm 4i)^{1000}\in Z$.
Therefore $ a \equal{} (164833 \pm 354144i)^{5^3} \equal{} 3^{5^3} \equal{} 3\mod 5$.


modular arithmetic
logarithms
search
inequalities
complex analysis
complex analysis unsolved

If $ (a_n)$ is a real sequence, we denote $ S_n(a)\equal{}\frac{1}{n}\sum_{k\equal{}1}^n e^{\i a_n}$. Clearly we have $ |S_n(a)|\leqslant 1$.
If $ a_n \equal{}n\theta$ with $ \theta \not \equiv 0 \pmod {2\pi}$, we get a much better bound $ |S_n(a)| \leqslant \frac{C_{\theta}}{n}$.

I am interested in asymptotic bounds of $ |S_n(a)|$ for simple sequences like:
1. $ a_n\equal{}n^2$
2. $ a_n\equal{}\sqrt{n}$
3. $ a_n\equal{}\ln n$

More general results are also appreciated.

If you search for "exponential sums", you'll find that there are entire books written on this subject, and [url=http://www.ma.utexas.edu/users/voloch/expsums.html]courses taught.[/url] Let's not take average and work with $ S_N\equal{}\sum_1^N \exp(2\pi i a_n)$. Weyl's inequality gives $ O(\sqrt{N})$ for the first of your sequences. The other two grow slowly.. The logarithm is so slow that the last 10% of the terms in the sum will have pretty much the same argument, so the trivial estimate $ O(N)$ is actually sharp. With $ a_n\equal{}\sqrt{n}$ the bound can't be better than $ O(\sqrt{N})$ (the last $ 1/\sqrt{N}$% of the terms have about the same argument), but I don't have a proof of $ O(\sqrt{N})$.


algebra
function
domain
Ring Theory
superior algebra
superior algebra solved

Is the set of reals Unique Factorization Domains (UFD)?
Is the set of reals Principal Ideal domain (PID)?
What is irreducible elements of reals?

 :(  :?
Can someone help me like to very-very stupid one?

Best regards
Tanja

The real numbers form a field, and every field is both a PID and a UFD.
PID: the only ideals are (0) and (1), both principal.
UFD: every non-zero element of a field is a unit.  So every non-zero element is a unit times a product of irreducibles: in fact, an emtpy product, since there are no irreducibles.

Thank You very much :lol:

Best regards
Tanja :omighty:


modular arithmetic
function
combinatorics unsolved
combinatorics

At a  conference take part an even number of persons.They sit at a round table.After a break,they sit down again,occupying different positions.
Prove that there are 2 persons such that the number of persons that lie between them remains the same.
What about an odd number of persons?

[hide]Assume that the persons are vertices of a $ 2n$-gon whose vertices are numbered $ 1,2,\ldots,2n$. Consider everything  modulo $ 2n$. Let $ f(i)$ denote the position of person $ i$ after the break, for all $ i\in\{1,2,3,\ldots,2n\}$. The number of persons that lie between $ a$ and $ b$ remains the same if and only if $ a\minus{}b\equiv f(a)\minus{}f(b)\pmod{2n}$, or $ a\minus{}f(a)\equiv b\minus{}f(b)\pmod{2n}$.

By way of contradiction, assume that there do not exist $ a$ and $ b$ such that $ a\minus{}f(a)\equiv b\minus{}f(b)\pmod{2n}$. Hence the numbers $ i\minus{}f(i)$ must be $ 0,1,2,\ldots,2n\minus{}1\pmod{2n}$. The sum is $ n(2n\minus{}1)\pmod{2n}$. However $ f(1),f(2),\ldots,f(n)$ is a permutation of $ 1,2,\ldots,n$. Thus the sum $ \sum i\minus{}f(i)$ must be $ 0$. Contradiction.

The result is not true for odd number of persons. For example $ 0,1,2,3,4\rightarrow0,2,4,1,3$.[/hide]

I don't get that relation for the number of persons that lie between a,b to be the same:  a-b=f(a)-f(b) (mod 2n).How did you get that.
For example,if a=1,b=2 and they interchange positions (f(a)=2,f(b)=1),it is clear that the number of persons between them remains the same.
But then the relations becomes -1=1(mod 2n) ,0=2 mod 2n,which is false!

And by the way,there are two posibilities for the number of persons between a and b(it depends on what side we look),it could be b-a-1 or 2n-1-b+a if a<b.(i think we should take the minimum value).And these two posibilities are not congruent mod 2n...

Just read "if" instead of "if and only if". The "only if" direction is wrong, but never used.

Topic will be merged with http://www.mathlinks.ro/Forum/viewtopic.php?t=250448 & http://www.mathlinks.ro/viewtopic.php?t=60414.

  darij

Similar to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=119174&sid=94efb9c4ce506231515e07ca360a9f4c#p119174]ISL 2001 C2[/url].

Let $a_1, a_2, \cdots , a_n$ be the original positions of the persons and let $b_1 , b_2 , \cdots , b_n$ be the positions after they shift. Suppose $a_i - b_i \neq a_j - b_j$ for all $i \neq j.$ Then, the function $f(x) = a_x - b_x$ is bijective modulo $n,$ meaning $\{f(1), f(2), \cdots , f(n)\}$ is equivalent to $\{1, 2, \cdots , n\}$ modulo $n$ in some order. Then, note that
\begin{align*}
\displaystyle \sum_{i=1}^{n} f(x) & \equiv \displaystyle \sum_{i=1}^{n} i \\
& \equiv \dfrac{n(n+1)}{2} \pmod{n}.\end{align*}
However, since $(b_1, b_2, \cdots , b_{n})$ is a permutation of $(1, 2, \cdots , n)$ which is also a permutation of $(a_1, a_2, \cdots , a_{n}),$
\[\displaystyle \sum_{i=1}^{n} a_i = \displaystyle \sum_{i=1}^{n} b_i \implies \displaystyle \sum_{i=1}^{n} a_i - b_i = 0.\]
Hence, we have $n \mid \dfrac{n(n+1)}{2} \implies 2 \mid n+1,$ contradiction. $\blacksquare$

It is false for odd numbers. In general, for any odd $n$, consider the permutations $(1, 2, \cdots , n)$ and $(2, 4, \cdots , 2n).$ The latter is a permutation of $(1, 2, \cdots , n)$ because $2x \equiv 2y \pmod{n} \iff x \equiv y \pmod{n}$.


Hi everybody,

the forum for Iranian community has been created, I am happy to be the first one to post here. I congratulate you for the 4th place in IMO 2005.

Congratulation on your opening Iranian Community! :)  :)

Amirhtulsa, you had a good job.

kunny

:w00tb:  :yup:  hmm.. at last we have a community for Iranians,Congradulations. :yup:  :w00tb:

i would like to thank all the Admins for creating us this comunity...

AT LAST!!! :)  :)

i don't know if we can use the Farsi font on here, i'll try to type in Persian from now on..

haha hello to all ,Thank`s to all of the admins for thier generous respond ,Now let`s speak persian.

Salam be hamegi ,khob hala ke baramon inja ro dorost kardan , chi kar konim? :D

Salam Bacheha, mabaraket tun,  hala metunim degeh farsi harf bezanim!!!


My solution for the cowrun problem in the USACO Silver contest failed for every test case except the first two  :(
I read the analysis and the solution seems different from mine but my logic seemed correct too so I was wondering if someone could please correct my thinking.

[hide]
Problem 7: Running [Neal Wu, 2007]

The cows are trying to become better athletes, so Bessie is running
on a track for exactly N (1 <= N <= 10,000) minutes. During each
minute, she can choose to either run or rest for the whole minute.

The ultimate distance Bessie runs, though, depends on her 'exhaustion
factor', which starts at 0. When she chooses to run in minute i,
she will run exactly a distance of D_i (1 <= D_i <= 1,000) and her
exhaustion factor will increase by 1 -- but must never be allowed
to exceed M (1 <= M <= 500).  If she chooses to rest, her exhaustion
factor will decrease by 1 for each minute she rests. She cannot
commence running again until her exhaustion factor reaches 0. At
that point, she can choose to run or rest.

At the end of the N minute workout, Bessie's exaustion factor must
be exactly 0, or she will not have enough energy left for the rest
of the day.

Find the maximal distance Bessie can run.

PROBLEM NAME: cowrun

INPUT FORMAT:

* Line 1: Two space-separated integers: N and M

* Lines 2..N+1: Line i+1 contains the single integer: D_i

SAMPLE INPUT (file cowrun.in):

5 2
5
3
4
2
10

OUTPUT FORMAT:

* Line 1: A single integer representing the largest distance Bessie
        can run while satisfying the conditions.

SAMPLE OUTPUT (file cowrun.out):

9

OUTPUT DETAILS:

Bessie runs during the first minute (distance=5), rests during the
second minute, runs for the third (distance=4), and rests for the
fourth and fifth. Note that Bessie cannot run on the fifth minute
because she would not end with a rest factor of 0.

[/hide]

My idea was that best (m, n) = MAX(best (m - 1, n - 1) + val (m), best (m - 1, n + 1)) where best (m, n) is the farthest the cow could have ran at minute m with n exhaustion points
(with best (m, 0) = MAX (best (m - 1, 0), best (m - 1, 1))), because the cow could have done nothing and gone from n + 1 exhaustion points to n exhaustion points or ran and gone from n - 1 exhaustion points to n exhaustion points.

I guess the logic was incorrect though so could someone please hint me where I went wrong?

"She cannot commence running again until her exhaustion factor reaches 0."

I don't think your recurrence takes that into account.

Ohhhhhhhhh  :wallbash:
I missed a very key sentence there! Ouch! Thanks for the help!


superior algebra
superior algebra unsolved

Let $K$ be a field of characteristic $p$, with $p\equiv1(mod4)$
a).Prove that -1 is a square of an element of K;
b).Prove that any non-zero element of K can be written as the sum of squares of three non-zero elements in K;
c).Can 0 be written as a sum of the squares of three non-zero elements from K?

Cool! I was a contestant then (in $12$'th grade :)). I'll leave it to the others. Just one correction: I think that in $b)$ you mean "the sum of the squares of three non-zero elements" (you just wrote "the sum of three non-zero elements").

maybe i'm not right, but in my condition of this problem:"the sum of three non-zero elements" is written-i've just downloaded it from the Mathlinks-if you sure about "squares" i'll correct it

If it were just "the sum of three non-zero elements", the question would be silly (something you'd do in a classroom when you first teach students what a field is, not something to propose for a national olympiad) :).

i edited it now, i'm sorry for such a stupid supposion, but it needs to be detailed(it maybe happened at some other olympiad, of course not in Romanian)


so my cousin invented this game called word challenge. Here's what you do:
first, i type a letter. then the next person who posts types a letter if my letter and letter when combined produce a word, then the next poster is out. If not, then the next person posts and when the three letters get combined and they produce a word, the last person loses and soo on. But, say the first person says "a" the next person says "z" than the next person can say word challenge and ask the person before him what word he ways trying to make. if what the person says before him is a real word, then the person who says word challenge is out. if not, then the second person is ou. people need to sign up. then i'll say who goes first second third etc(this will be done like who sign's up first). [b]PLEASE DO NOT CHEAT IN ANY WAY OR IT WON'T BE FUN![/b].

Is the game you're talking about similar to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180167&search_id=283718584]this one[/url]?

I know that (and I don't want you to revive it), I just wanted some clarification because the rules were a bit confusing.

Now that we have that settled, I will sign up for the game.


http://www.acsl.org/

Has anyone here competed in this?

tried some.

awkwardly the open programming problems seem a lot easier than the short ones :?

I've seen it, but it doesn't look as prestigeous as the USACO, or IOI.

But I would still like to participate in it, but it seemed kind of easy to me.

I participated in ACSL from 2001 until 2004. The questions aren't hard, and yes, the competition definitely isn't as prestigious as USACO; but it's somewhat fun nonetheless. I was league first place (meaning, perfect score on the league rounds - there are lots of people who get them) 2001-2002 and 2003-2004. My team entered the all-star contest in 2001-2002, and while we sucked horribly in the programming portion, I got a perfect score in the written portion. I was awarded a PERL book and MS Flight Simulator Pro 2002.

[quote="Osiris"]USA Computing Olympiad?[/quote]

Thank you.  That word "Olympiad" means "way above my level", though.  The ACSL is probably a bit easier,  though I have never seen a USACO problem.


conics ellipse ratio hyperbola parabola trigonometry

What is an ellipse with an eccentricity of zero?

It is a circle. How about an ellipse with an eccentricity of 1?

I forgot how eccentricity was measured. What was the definition of eccentricity?

Don't all ellipses have eccentricities 0<e<1?, and hyperbolas have eccentricities e>1? So then parabolas have e=1.

Oh, and now I see....the lower the eccentricity of an ellipse, the more circle-like-shaped it looks, right?
i guess thats why the eccentricity of a circle is 0.

[quote="ProtestanT"][quote="BanishedTraitor"]$ \text{Eccentricity}$ is the ratio of the distance of a point from the focus to the perpendicular distance of the corresponding directrix.[/quote]
Then how can eccentricity ever be $ 0$? In a circle, the points still have a distance to the focus...[/quote]

Well, ask yourself where the directrices of a circle are (theoretically).

Rather than restrict ourselves to the term "ellipse," we should talk about the eccentricity of a [i]conic section[/i] generally.

[quote="gauss1181"]the lower the eccentricity of an ellipse, the more circle-like-shaped it looks, right?
i guess thats why the eccentricity of a circle is 0.[/quote]

Yes.  This is the motivation for the definition of eccentricity; one application of the concept is to apply it to the elliptical orbits of the planets as a way to measure how close they are to being circular.

A concrete algebraic manifestation:  the polar equation of a conic section is given by

$ r \equal{} \frac{k}{1 \plus{} e \cos \theta}$

where $ e$ is the eccentricity.


inequalities
trigonometry
function
algebra unsolved
algebra

Find all solutions in real numbers:

\[   \{\begin{array}{c} sin x + cos y \geqslant \sqrt{2} \\ sin y + cos z \geqslant \sqrt{2} \\ sin z + cos x \geqslant \sqrt{2}   \end{array} \]

You add up all the inequalities and get: $\sin{x}+\cos{x}+\sin{y}+\cos{y}+\sin{z}+\cos{z}\geq3\sqrt{2}$. We can assume $x,y,z\in[0,2\pi]$. From the inequalities it follows that sinx, cosx and so on must be all positive so $x,y,z\in[0,\frac{\pi}{2}]$. The maximum of the function $\sin{x}+\cos{x}$ in this interval is $\sqrt{2}$ for $x=\frac{\pi}{4}$. Since the equality holds it follows: $x=y=z=\frac{\pi}{4}$.

Nice problem!

Misha


function
calculus
integration
calculus computations

Let $f: [a,b]\rightarrow R$ be an integrable function. Suppose that $g: [a,b]\rightarrow R$ and there is a finite set $S\subset[a.b]$ such that $f(x)=g(x)$ if $x\notin S$

Show that $g$ is integrable and that its integral equals to integral of $f$

Partition into tiny chunks, especially around each bad point. The function's still bounded, and finitely many bad points contribute an arbitrarily small error term ($2Mn\delta$, where $M$ is a bound on the function, $n$ is the number of bad points, and $\delta$ is the width of the subintervals, which can be made arbitrarily small.

I still dont get what you meant, could u explain more in detail, thx alot.


Cast $ n\geq 2$ dices at a time, let $ L$ be the L.C.M. of the outcome. Denote the probability such that $ L\equal{}k$ by $ p_k.$
Find $ p_4,\ p_6.$

Well, okay, I guess I am not above a little elementary probability:

We have $ L \equal{} 4$ if and only if the dice we throw include at least one 4 and all other numbers are 1 or 2.  There are $ 3^n$ ways to avoid throwing any 3, 5 or 6, but of these there are $ 2^n$ that do not include any 4, so the answer is $ \frac {3^n \minus{} 2^n}{6^n} \equal{} \frac {1}{2^n} \minus{} \frac {1}{3^n}$.

We have $ L \equal{} 6$ if and only if the following occurs: we throw no 4 or 5 and either we throw at least one 6 or we throw at least one of both 2 and 3.  There are $ 4^n$ throws with no 4 or 5.  Of these, $ 2^n$ have no 2 and no 6, $ 2^n$ have no 3 and no 6, and just one has no 2, 3 or 6.  Thus the answer is $ \frac {4^n \minus{} 2 \cdot 2^n \plus{} 1}{6^n} \equal{} \frac {2^n}{3^n} \minus{} 2\cdot \frac {1}{3^n} \plus{} \frac {1}{6^n}$.

In general, the probability that $ L$ [i]divides[/i] $ m$ is $ \frac {i^n}{6^n}$, where $ i$ is the number of elements of $ \{1, 2, \ldots, 6\}$ that divide $ m$.  Then we may compute the probability that $ L$ [i]equals[/i] $ m$ by using Mobius inversion.  (This method works for dice of arbitrary sides.)

I have nothing to say more than this additionally . :).


geometry
trigonometry
perimeter
area of a triangle
trig identities
Law of Cosines
algebra

Is it possible to find the area of a triangle when all you know is the side lengths?

One problem was this: Determine the area of a triangle with side lengths of 7, 8, and 11.

Use Herons formula: $[ABC]=\sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semiperimeter and $a,b,c$ are the sides respectively.

[quote="The QuattoMaster 6000"]Yes, you use Hero's Formula:
[hide] If the triangle has side lengths a, b, and c, then the area is $\frac{\sqrt{(a+b-c)(a+c-b)(b+c-a)(a+b+c)}}{4}$.  Plugging everything in for your problem, we get $2\sqrt{15}$. [/hide][/quote]
Using your formula, I got $2\sqrt{195}\approx27.93$

Also, what is the semiperimeter?

The semiperimeter is half of the perimeter of the triangle.  You are right about that area, sorry about that!

you mean the semiprimeter is half of the PERIMETER!!
not the area!!

Sorry!  Edited, I am not focusing.   :blush:

[quote="me@home"]To solve this without knowing the formula, simply find an angle (using law of cosines) then then use half a b sin c. This in fact is exactly how to derive the formula in general, through plugging in (I challenge you to derive it, it is not hard and it is nice)[/quote]
Another method of deriving is to set up a system of equations in terms of a side length and the altitude to it. It only takes pythagorean :)

[quote="me@home"]That would be unimaginably messy I think.

Whereas, its only substitution then like 3 steps simplification in my method :wink:[/quote]
I know, I was just showing how heron's can be derived using only the simplest of methods.