calculus
integration
geometry
limit
ratio
geometric series
calculus computations

(1) For integer $n=0,\ 1,\ 2,\ \cdots$ and positive number $a_{n},$ let $f_{n}(x)=a_{n}(x-n)(n+1-x).$ Find $a_{n}$ such that the curve $y=f_{n}(x)$ touches to the curve $y=e^{-x}.$

(2) For $f_{n}(x)$ defined in (1), denote the area of the figure bounded by $y=f_{0}(x), y=e^{-x}$ and the $y$-axis by $S_{0},$ for $n\geq 1,$ the area of the figure bounded by $y=f_{n-1}(x),\ y=f_{n}(x)$ and $y=e^{-x}$ by $S_{n}.$ Find $\lim_{n\to\infty}(S_{0}+S_{1}+\cdots+S_{n}).$

1) Calling $ g(x) = e^{ - x}$, trivially $ g'(x) = - e^{ - x} = - g(x)$ for all $ x$, so $ f'_n(x_n) = - f_n(x_n)$ at point $ x = x_n$ where $ y = f_n(x)$ and $ y = e^{ - x}$ touch.  Therefore,
\[ - f'_n(x_n) = a_n(2x_n - 2n - 1) = a_n(x_n - n) - a_n(n + 1 - x_n) = a_n(x_n - n)(n + 1 - x_n) = f_n(x_n) = a_n( - x_n^2 + (2n + 1)x_n - n^2 - n),
\]
and since $ a_n > 0$,
\[ x_n^2 - (2n - 1)x_n + n^2 - n - 1 = 0,
\]

\[ x_n = \frac {2n - 1 + \sqrt {4n^2 - 4n + 1 - 4n^2 + 4n + 4}}{2} = n + \frac {\sqrt {5} - 1}{2},
\]
where we have selected the positive root since $ f_n(x_n)$ would be negative (and thus it could not touch $ e^{ - x}$ which is always positive) for the negative root.
Now, we have $ f_n(x_n) = - f'_n(x_n) = a_n(2x_n - 2n - 1) = a_n(2n + \sqrt {5} - 1 - 2n - 1) = a_n(\sqrt {5} - 2),$
or
\[ a_n = \frac {e^{ - x_n}}{\sqrt {5} - 2} = \frac {e^\frac {1 - \sqrt {5}}{2}e^{ - n}}{\sqrt {5} - 2}.
\]
2) It is trivial to show that curves $ y = f_n(x)$ and $ y = f_{n - 1}(x)$ meet at $ x = n$ and $ y = 0$, or the limit that we are looking for is
\[ \int_0^\infty e^{ - x}dx - \sum_{n = 0}^\infty\int_n^{n + 1}f_n(x)dx.
\]
The first integral is easily calculated to be equal to 1, while
\[ \int_n^{n + 1}f_n(x)dx = a_n\int_n^{n + 1}( - x^2 + (2n + 1)x - n^2 - n)dx = a_n(\frac {n^3 - (n + 1)^3}{3} + (2n + 1)\frac {(n + 1)^2 - n^2}{2} - n^2 - n) = \frac {a_n}{6}.
\]
Finally, the limit asked may then be written as
\[ 1 - \frac {1}{6}\sum_{n = 0}^\infty a_n = 1 - \frac {1}{6}\frac {e^\frac {1 - \sqrt {5}}{2}}{\sqrt {5} - 2}\sum_{n = 0}^\infty e^{ - n},
\]
and sinde the sum is a geometric series starting at 1 and having ratio $ e^{ - 1}$, the sum equals $ \frac {1}{1 - e^{ - 1}} = \frac {e}{e - 1}$, yielding the final result
\[{ 1 - \frac {e^\frac {3 - \sqrt {5}}{2}}{6(\sqrt {5} - 2})(e - 1)}.
\]

That's right, you have a good calculation. :)

[quote="daniel73"]\[ \int_n^{n \plus{} 1}f_n(x)dx \equal{} a_n\int_n^{n \plus{} 1}\left( \minus{} x^2 \plus{} (2n \plus{} 1)x \minus{} n^2 \minus{} n\right)dx \equal{} a_n\left(\frac {n^3 \minus{} (n \plus{} 1)^3}{3} \plus{} (2n \plus{} 1)\frac {(n \plus{} 1)^2 \minus{} n^2}{2} \minus{} n^2 \minus{} n\right) \equal{} \frac {a_n}{6}.
\]
[/quote]

You can calculate such as

$ \int_n^{n \plus{} 1}\left( \minus{} x^2 \plus{} (2n \plus{} 1)x \minus{} n^2 \minus{} n\right)dx\equal{}\minus{}\int_{n}^{n\plus{}1} (x\minus{}n)(x\minus{}n\minus{}1)\ dx$

$ \equal{}\frac{1}{6}(n\plus{}1\minus{}n)^3\equal{}\frac{1}{6}.$


You have 75 minutes to do this, you may use a calculator.

On problem number 20, for the answers, please change all square roots of 3 to square roots of two! On problem number 15, please disregard the first answer, it is incorrect! Problem number 10 is stupid. You get 6 free points for answering it randomly.

[quote="Pakman2012"]Comments

First ten are too hard
Second ten are too easy
Last 5 are waaay too easy.

Number 5 appears to be poorly worded at the end.
Number 9 and 10 are too calculator intensive.
18 is entirely too vague.
20 is solved by a simple formula, exactly not like the AMC.
22, likewise.
23 through 25 are also easily bruteforcable.

You can do better than that.[/quote]

I'm going to have to agree with that  :roll:

18 was [i]way[/i] to vague (though "y is not a vowel" kind of gives it away).

This is kind of ignorant of me, but we PM you the answers, right? Do you want just the letters of the answer, or the answer itself, or both?

Can you check the answer for #23, I'm not getting any of them.

I got one of the answers for #23

And I'll give you a hint:
[hide]The answer is not Q.
[/hide]

After foolishly overlooking one of the sets in 23, I think I'm going to have to make that a free question too.

@fireninja: Just pm me the letters that you got.

@Leetspeak: This is due December 8, but you should sit down and do it in one 75 minute period of time, because it is a AMC test.

Could you please make it due saturday, since I won't have time to do it until then. Sorry.

[quote="fireninja0"]Three questions:

On #17, could you clarify what "B" is? Is it a digit or a number? and what about "A" and "C" too?

I have a feeling that "A" represents 10, "B" represents 11, and "C" represents "12." Am I correct?
------
On #8, it says "What is the expected number of days that it will snow?" Is it just me, or is this problem missing something? Like the number of days in the period of time you're talking about.
------
On #21, is there a typo? I'm talking about angle CDE being equal to 90 degrees.[/quote]

17. This is in base 23, remember. What would the number 11 be in base 23?

8. One year

21. That should be CDB

@anirudh I'll still grade your test if you submit it too me on the weekend

Am I just being dumb, or is the right answer for 11 not one of the choices?

[quote="mustafa"][quote="fireninja0"]
17. This is in base 23, remember. What would the number 11 be in base 23?
[/quote][/quote]
Sorry, I think you misunderstood. My question is what do you mean when you say:
$1839B_{14}$?
Do you mean:
$1839_{14}\cdot B_{14}$
or
$1839B_{14}$ where B is the "ones" digit.
Then what's A, B, and C where it says: $CAB_{23}$?
Sorry for my ingorance. :wink:

I think I got the answer: A is a one-digit representative of the value 10, B is a one-digit rep of 11, and so on.

I haven't gotten any answers in a really long time, but just in case someone just forgot to do it (or because I'm lazy right now), I'm going to really truly officially end it at midnight Wednesday, at which time the top 5 scorers and the answers will be posted.


function
inequalities
integration
linear algebra
matrix
analytic geometry
real analysis

suppose that $u: \mathbb{R}^{n}\to \mathbb{R}$ is a smooth function that satisfies
$\sum_{i,j}\partial_{i}( A_{i,j}(x) \partial_{j}u)(x)=0$
on $\mathbb{R}^{n}$, where $A_{i,j}(x)$ is uniformly elliptic. How can one prove an inequality like
$\| \nabla u \|_{\infty, B(0,R)}\leq \frac{C}{R^{N+2}}\int_{B(0,2R)}|u(x)-\overline{u}_{0,2R}|^{2}dx$
where $\overline{u}_{0,2R}$ is the mean value of $u$ over $B(0, 2R)$, and $C$ is a constant? (I am not sure it is correct, but I suspect that such an inequality exists ?)
If someone has an idea, even for the case where the matrix $A_{i,j}$ is constant, I am interested :-)

I think it is true for the constant case:
By homogeneity, it suffices to prove it for $R=1$, and by a suitable change of coordinate, it suffices to show it for $A=I_{d}$, wich is more or less classical.

This is false unless you impose some regularity assumptions on $A_{ij}$, in addition to uniform ellipticity. Indeed, a weak solution of a uniformly elliptic equation in divergence form is in $C^{\alpha}$ for some $\alpha>0$ (De Giorgi, Nash, Moser), but no better than that. A simple example is $u(x)=x_{1}|x|^{\alpha-1}$, which solves such an equation with the uniformly elliptic matrix $A(x)=I+\frac{C_{\alpha}}{|x|^{2}}x\otimes x$ (I don't want to work out the value of $C_{\alpha}$). Of course, this $u$ is not smooth, but this does not matter: smoothing things near the origin, we can get a solution with large sup-norm of the gradient.


What is the 100th digit of the decimal representation of $\frac{1}{7}$?

[hide]The pattern for $\frac{1}{7}$ is $0.\overline{142857}$.

100/6=16 with a remainder of 4. That means the 100th digit is $\boxed{8}$.[/hide]

[hide]

0.142857142857142857...

100/6 leaves a remainder of 4. the 4th digit after the decimal point is 8, so [color=red][b]8[/b][/color]
[/hide]

[quote="nonie"][quote="mtms5467"][hide]The pattern for $\frac{1}{7}$ is $0.\overline{142857}$.

100/6=16 with a remainder of 4. That means the 100th digit is $\boxed{4}$.[/hide][/quote]

[hide="...."]I think  you meant 8 because 8 is the 4th digit in the repeating block. (The remainder isn't the answer...it tells you which number is the answer.)  [/hide][/quote]
[hide="."]whoops, that was a typo, I guess I wasn't thinking.......but I did mean 8 for the answer.[/hide]

[hide]$\frac{1}{7}=0.\overline{142857}$ After n digits, the digit will be 1 if n/6 has a remainder of 1, 4 if n/6 has a remainder of 2, and so on. 100/6=16R4, and therefore the 100th digit is 8[/hide]

[hide]$\frac{1}{7}=.\overline{142857}$.

100 leaves a remainder of 4, when divided by 6.

So it is the fourth digit in the block, which is 8.

$8$[/hide]

[hide]
$\frac{1}{7}=\overline{142857}$
So, divide 100 by 6. There is a remainder of 4, so the answer is $\boxed{8}$[/hide]

[hide]I say [size=200][b]8[/b][/size]. I don't know how you get four[/hide]

[size=75][color=darkred]Please hide your answers. Thanks.
-nebula42[/color][/size]


Women are not pieces of meat, and as such you, and everyone else, should probably refrain from posting threads of this nature.

@blahblahblah.I didn't meat that the girls are pieces of meat :!:  :!:  .

[quote="Tiks"]Ok!!!
@blahblahblah.I havn't ment that the girls are pieces of meat :!:  :!:  .[/quote]

The way this thread was going you certainly were treating them that way.  (and he does not mean 'piece of meat' in a literal sense, by the way).

I am also a piece of meat! SHOUT OUT TO ALL YOU PIECES OF MEAT OUT THERE

While the meats gather, the butcher sharpens his knife :)
[quote="Same Tiks"]Guys it's again me :blush: ,as Mathlinks do not allowed me to post reply sooner than 72 hours I will post some photos from this user :wink: .[/quote]The whole point of this restriction was for you to understand that it's not a wise idea to post consecutively like this  :!:  :|

[quote="Valentin Vornicu"]
[quote="Same Tiks"]Guys it's again me  ,as Mathlinks do not allowed me to post reply sooner than 72 hours I will post some photos from this user  .[/quote]The whole point of this restriction was for you to understand that it's not a wise idea to post consecutively like this [/quote]
OK!
I am sorry,it was a bad idea   :oops_sign:  :cleaning:  :|

I have seen the girl from Korea, she received a gold medal at IMO 2005

do you know: actually they have some durıng the traınıngs but they, but un the ımo hıstory of ıran they dont have any gırls who partıcıpated ın ımo.

davron

Davron, you are wrong, some Iranian girls participated in IMO.
For example two of them where Maryam Mirzakhani (She participated two times with scores of 41,42 an amazing job!!!) and Roya Beheshti

Maryam Mirzakhani And Roya Beheshti were in 1994. (2 girls at one IMO).
Mirzakhani also partipated at IMO 1995.
Also Sepideh Mirrahimi (another girl) particpated in IMO 2002.
We had 3 girls in IMO yet.

[quote="Aizana"][quote="Davron"]:-) Tiks I think our friend blahblahblah is right, if someone between girls see this maybe they will not be glad.

davron[/quote]

maybe...  it depends on what kind of photo it was.
     right now I'm just :)[/quote]

That were the pictures from the official IMO page :) . And the guys above just exaggerated all the matter :ninja: .


Hi, everyone,

Some people here are current college students, some college alumni, and some college applicants. Most people who post on AoPS are math-liking people and look for colleges that fit well for math-liking students. What involvement in team sports is there in the college you attend, or the college you want to attend, among math-liking students? Do you play team sports in college?

What sport do you play? (American) football? Ultimate Frisbee? Water polo?

Thanks for any perspective you can provide on the balance between sports and math.

I love playing sports. I'm very tall and decently athletic, so I can pick up new games easily, but I'm not good enough at any sport to be on a club or varsity team. In a perfect world, I'd probably spend 3-5 hours each week playing team sports; in reality, that only happens on occasion.

My favorite sports are soccer (football) and ultimate frisbee. I also sometimes play basketball and volleyball.

During the semester, I'm usually on several intramural teams at once, so I usually play two or three 1/2 - 1 hour games (total, of various sports) each week.

I play basketball at the rec center with my friends, but there's no real math connection there since they are all chemistry/engineering majors.  :oops:

If you're looking for a good connection between statistics and sports, start playing fantasy sports. You can number crunch that to your hearts content.

Thanks for the replies. Please tell me more about intramural soccer at Princeton.

There are three intramural sports "seasons" at Princeton: fall, winter, and spring. I hope to do at least two sports every season.

All of the winter ones are indoor; there's outdoor soccer in fall and spring and indoor soccer in the winter. Frisbee is only in the fall. Volleyball is every season, I think. I'm not sure about basketball.

Residential colleges and eating clubs field most of the IM teams. There are usually a few grad college teams too, and sometimes one from Spelman (the independent housing) and/or the Brown Co-op.

Intramural sports are free and require no try-outs. The level of play is not very competitive, so anyone can do it. Most of my volleyball team last year had never played before, and we still won some of our games.

Each team has about 2 games per week, and games are usually 30 min - 1 hour depending on the sport.

What else is there to know about IM sports? They're fun, I can say that for sure! ;-)

Thanks for the details on intramural sports. That definitely looks like a good way to keep balance in the life of a busy student.

I did Varsity XC last year and may do it again this year.  It was a very nice experience.  At most state schools, I wouldn't have even made the team.  Here, it's a little easier to make the team (easier...but still not 'easy' =D).  IM's are pretty thriving here as well.  Most of the people I know/know of play a sport at least once a week or so.  If you want to talk about balancing math and athletics though, here's a short aside.  Before taking the Putnam, Daniel Kane got his blood flowing by doing jumping jacks :)

Anyway, after XC in the fall I played a bunch of IM's in the spring for my fraternity.  Over the semester I did all of the following: hiking, ice climbing, softball, basketball, soccer, water polo, ultimate frisbee, football, unihoc, dodgeball, weightlifting, breakdancing, rope climbing, biking and (of course) running.  That's a huge list, but (a) I love being active and playing sports and (b) some of those I just played a game or two of such as water polo and soccer (both of which I played for the first time!).

This summer I have played very few sports...partially due to my commute and partially due to running so much for cross country.


algebra
polynomial
limit
algebra unsolved

I have two polynomials:
$P(x)=2x^{3}-x^{2}+\frac{1}{27}=2(x-\frac{1}{3})^{2}(x+\frac{1}{6})$
and
$P_{\epsilon}(x)=2x^{3}+(4\epsilon-1)x^{2}+(2\epsilon^{2}-2\epsilon)x+\frac{1}{27}-\epsilon^{2}$,
where $\epsilon$ can be arbitrarily small. Let $P_{\epsilon}(x_{0})=0, x_{0}\geq 0$.The problem is - I want to show that $\lim_{\epsilon\to0}x_{0}=\frac{1}{3}$. How to do it without such awful means as Cardano's formulas? The result is very intuitive (two polynomials can be arbitrarily close to each other, so their roots can also be arbitrarily small).


Well... I'm an 8th grader, so this was my last year in Mathcounts. Seeing as how I can no longer participate in it, does anyone know of math competitions that high schoolds would do?

AoPSWiki has a [[list of United States high school mathematics competitions]]. :)

There are tons of competitions available, as i_like_pie provided, but if you feel like you're missing MATHCOUNTS, you can always go back to your old middle school in your spare time and help coach the kids there  :) . I believe this is what chess64 did in his freshman year of high school and it helped us a lot.

Yeah... That sounds like a good idea. There's sitll alot I can learn from Mathcounts.

And thanks Ihatepie.

try ARML... You might meet some old mathcounts friends there.  And obviously the AMCs.

where can you get in an ARML team, or do you just make one?

ask ppl in your state, its difference from state to state, and try to find out who ur state coordinator is.

[quote="Ihatepie"]i like pie?

anyways, I would say that your best bet is with the AMC's as they are really nice.
Next year, I'm gonna coach my school team if we have one. My school isn't very involved in MC.[/quote]

Oops  :rotfl:

I've never seen Ilikepie before, so I guess when i was writing the response I remembered your name  :P

hint for peeps in AL (including Ryuk! )

The ARML guy talked right before cd, remember? I think his site is [url]http://www.mistacademy.com[/url]


[quote="Lovasz"]Let $x,y \in (0,1)$ such that $x+y=1$. Finf the least value of the expression:
\[ T=x^x+y^y  \][/quote]
$f(x)=x^x$ is convex. Hence, $x^x+y^y\geq2\cdot\left(\frac{x+y}{2}\right)^\frac{x+y}{2}=\sqrt2.$

[quote="jin"]Just $\sqrt{2}$.[/quote]
Thank you, jin. :)


number theory
decimal representation
IMO Longlist
Digits
Last digit
modular arithmetic
IMO Shortlist

We take $100$ consecutive natural numbers $a_{1},$ $a_{2},$ $...,$ $a_{100}.$ Determine the last two digits of the number $a_{1}^{8}+a_{2}^{8}+...+a_{100}^{8}.$

has this got a solution with inequalities or is it just here by accident? :)

anyway, since we only need the remainder mod 100, and $(100k+x)^8=x^8 \bmod 100$, we only want the remainder mod 100 of $1^8+...+100^8$. (All multiples of 10 can be removed as well.)

To know the remainder mod 100 it is sufficient to know it mod 4 and mod 25. (via Chinese Remainder Theorem)

mod 4 we see it is obviously 50=[b]2[/b]

mod 25 we have that $f(x)=x^8$ goes as follows:
0 -> 0
1 -> 1
2 -> 6
3 -> 11
4 -> 11
5 -> 0
6 -> 16
7 -> 1
8 -> 16
9 -> 21
10 -> 0
11 -> 6
12 -> 21
13 -> 21
14 -> 6
15 -> 0
16 -> 21
17 -> 16
18 -> 1
19 -> 16
20 -> 0
21 -> 11
22 -> 11
23 -> 6
24 -> 1
25 -> 0
which repeats from here since $(25k+x)^8=x^8 mod 25$, but I bet there are much more elegant ways to show that it ends in 30...

I did... but next to proving the pattern 0-12 is symmetric to 13-25 I think there must be more tricks to avoid the case analysis... seems a quite ugly solution like that :blush:

since $gcd(8,\phi(25))=4$, we can replace the sum of the 8-th powers mod 25 by the sum of the 4-th powers mod 25. it is clear that we can leave all terms divisible by 5 and that $x^4\equiv 1\; mod\; 5$ if $gcd(x,5)=1$. since $(5k+x)^4\equiv x^4-5kx\; mod\; 25$, we see that each of the numbers $1, 6, 11, 16, 21$ occurs the same number of times, thus $\frac{20}{5}=4$ times, thus the sum of all 8-th powers mod 25 is $4(1+6+11+16+21)=20$. taking this 4 times(for the numbers $1,...,25$,$26,...,50$,$51,...,75$,$76,...,100$), we get 5 as the remainder mod 25, thus the last two digits are 30.

not much nicer, but without case analysis.

Peter


Ok....this is kinda late but.......WHAT SHOULD I DO TO PRACTICE?!?!?!?  lol :blush:
Any last-minute things I can do? :)  :D

relax. actually, you're in the same state as me, so don't relax, cram like crazy, stay up till 4 in the morning friday night studying, and then fall asleep during the test.

Mr. Rusczyk once stated in a WOOT class to not do any math at all the day before USAMO.

So you probably shouldn't do that much math the day before state...

umm do math...just not extensive math...hmm maybe 1 or 2 sprints for practice...well this depends on what u define as extenseive preparation like mine was atleast 4-5 sprints, another 100 or so problems and a bunch of other crap...day before 1-2 sprin?

oh yeah ask ur parents to give u a pep talk so u dont feel bad afterwards............ that sentense was really corny...... but its what my dad did for me throughout the entire award ceremony...... :rotfl:

we were cracking up like crazy afterwards.......

hmm, oh yeah, eat ur favorite food the day before.....hmm pizza, spaghetti, chocolate, etc...a

No, just don't do any math at all. Why? Because it makes your creativity go down unless you can [i]truly absorb[/i] the material and techniques that you learn.

hehe, if u study loads, and go to sleep at 10, you might have the same thing that happened to me... I was subconsciously solving the problems...

it was pretty funny though, our school had two teams, 8 total, and everyone took a place in the top ten. the competitors going to nats were all from my school. and the coach was too.

ha ha...
nike

I think it's best to do quite a bit of extensive practice starting a few weeks out. Then, for the day before, just relax. Don't do any homework. I think it's best to actually do a little bit of review. (go over a few tricky concepts, play a little[not too much] FTW, and look over the MC bible). Be conservative throughout the day and make sure you go to sleep reasonably early, but don't break your routine too much. Hope you do well! GL

Heh, i was solving problems from Math Olympiad Challenges the day before state in the car.

tehe i have it tomorrow too! Just relax and take it easy. Dont do any test. You should have prepared earlier and by doing more the night before you might hurt your score.
List of people going to nats tomorrow im MI:

1st place: Hirsh
2nd: Akhil
3rd: Chanukya
4th: No idea (hopefully me)

who the heck is blackhole? Cuz ur top 3 list for MC is completely off... I mean sorry just saying. It's not even funny how inaccurate that is.

[quote="nikeballa96"]pewpew...mewmew
haha i bet you're from missouri.[/quote]

nope, that's this loser dork (sike..maybe) from my school (not in MO).

hay, URC, thats mean...

I WOULD have made it to nats, and knocked killer out, if i had not mixed up my numbers 011 from 100 (BINARY)
it WAS 2:2, and i couldve beaten him...I didnt know how to do those multiple ratios.

[quote="blackhole"]

1st place: Hirsh
2nd: Akhil
3rd: Chanukya
4th: No idea (hopefully me)[/quote]

Wow... Actually pretty accurate.

Ya it was lol

1st akhil
2nd hirsh
3rd michael
4th chanukya

...and 6th echen who was a beast

and our team got 2nd woot

dude take out names...
congratulations to all who made nats.


Roll a regular die, with numbers 1,2,3,4,5, and 6 on it, four times in a row, and add the four numbers.  Then repeat this many times.  What number do you expect to appear as the sum the most often?

[hide="click to reveal stuff"]
I found the average of 1, 2, 3, 4, 5, and 6. Which is 3.5
Then i did 3.5*4 and that gives me $\boxed{14}$which is the answer.[/hide]

[hide]If you roll 2 die, the most common sum will be 7. So with 4 die, the most common sum would obviously be 7*2=14[/hide]

[hide="spoiler"]
There are 2 pairs of 2 dice. Each pair sums to 7. so 7*2=[hide="have you done the problem?"]14[/hide][/hide]

[hide]
The most common sum when you roll two die is $7$. Thus, the most common sum when you roll four die is $14$.[/hide]


Each of $ p$, $ q$, $ r$, $ s$ can either be $ 10$ or $ \minus{}10$. The number of values that $ p\plus{}2q\plus{}4r\plus{}5s$ can have is...?

There are $ 16$ possible sums obtained by assigning $ 0, 1, 2, 3$ or $ 4$ minus signs to the $ 10$s. Of these $ 0$ occurs twice:$ (p, q, r, s) \equal{} (10, \minus{}10, \minus{}10, 10), (\minus{}10, 10, 10 \minus{}10)$; $ 20$ occurs twice: $ (\minus{}10, 10, \minus{}10, 10), (10, 10, 10, \minus{}10)$; $ \minus{}20$ occurs twice: $ (10, \minus{}10, 10, \minus{}10), (\minus{}10, \minus{}10, \minus{}10, 10)$. So there are $ 16 \minus{} 3 \equal{} 13$ different possibilities.


[hide="giveaway hint"]
the maximum is 0  :P
[/hide]

I'm new to this, but I'll try to post a (half) solution.
Take AM-GM in three variables on a, b, and c.
$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$
$\frac{1}{3}\geq \sqrt[3]{abc}$
$\frac{1}{27}\geq abc$

Shoot. I am hopelessly stuck. Any help from other solvers?

[quote="miyomiyo"]Consider the non-negative reals a, b, and c with a+b+c=1.  Compute {a, b, c} so that abc - b is maximum.[/quote]
[hide]
Since they're non-negative then $a,b,c\in [0,1]$ so then obviously $abc\le b$ so then we say 0 can be achieved which it can because $abc-b=b(ac-1)$ so you either have $\{1,b,1\}$ or $\{a,0,c\}$ but 1+1+b>1 so the first solution doesn't work so it's just $\boxed{\{a,0,c\}}$[/hide]

Er... I don't think {1,b,1} isn't possible

Whoops yeah, forgot about the condition -.-

[quote="miyomiyo"]Whoops, sorry - I meant to write abc - b/2[/quote]
[hide="but"]
even the maximum of $abc-\frac{b}{4}$ is 0
[/hide]

arrrg i keep mis-typing.  That's supposed to be b^2, not b/2

[hide="now it is like eww"]
\[abc-b^{2}=b(ac-b)\leq b\left( \left( \frac{1-b}{2}\right)^{2}-b\right)\]
(by AM-GM on $a,c$)
Now we must maximize a cubic in terms of $b$ on the interval $[0,1]$. When $b=0$, cubic is 0, when $b=1$, cubic is negative. The local maximum can be found by setting the derivative equal to 0, that is, $3b^{2}-12b+1=0$, which has a root of $2-\frac{\sqrt{33}}{3}$. Then $abc-b^{2}$ will be maximized when
\[b=\frac{6-\sqrt{33}}{3}\]
\[a=c=\frac{\sqrt{33}-3}{6}\]
Now just plug them in to find the maximum... i'm too lazy to do it by hand, so i throw it into Mathematica and get
\[\boxed{\frac{11\sqrt{33}-63}{18}}\]
Anyone going to verify the whole solution?  :rotfl:
[/hide]

[quote="scorpius119"]Anyone going to verify the whole solution?  :rotfl: [/quote]

Looks fine.  Only part that I didn't see at first is why you assumed that the maximum of the LHS and RHS in the first inequality would be the same.  You should mention that if $b$ is fixed the maximum is attained when $a = c$ :P


topology
function
real analysis
real analysis unsolved

Suppose that we have two different norms that induce a same topology. Prove or disprove that these two norms are equivalents.

At first I thought this to be not true, but in fact I think it is:

Let $ (X, \|\cdot\|_1)$ and $ (X, \|\cdot\|_2)$ be the two spaces. We are given that the norms induce the same topology, but then the identity function is a continuous linear map from one to the other. But then the identity map is bounded both ways, which means that the norms are equivalent :)


let $C_{1},C_{2},C_{3}$ be the boundaries of three convex polygons with $n$ sides.
if the sets of points $C_{1}\cap C_{2},C_{2}\cap C_{3},C_{3}\cap C_{1}$ are finite, find the maximum number of elements of the set $C_{1}\cap C_{2}\cap C_{3}$.


The results of a survey of the 40 families at West ES are shown in the graph. During the survey, each family was asked how many children were in their family. What was the average of the families' responses? Express your answer as a decimal to the nearest tenth.
[asy]size(200); defaultpen(linewidth(0.8));
add(scale(2,5)*grid(4,2));
yaxis("\small{Number of Families}",0,10,Ticks(Step=5));
draw(Label("\small{Children per Family}",align=4*S),(0,0)--(8,0));
for(int i = 1; i <= 4; ++i){
 label((string)i,(2*i,0),S);
}
void drawBar(real x, real height){
 filldraw((2*x-0.5,0)--(2*x-0.5,height)--(2*x+0.5,height)--(2*x+0.5,0)--cycle,gray(0.7));
}
drawBar(1,10); drawBar(2,10); drawBar(3,10); drawBar(4,10);[/asy]

By definition, it is \[ \frac{10\cdot 1\plus{}10\cdot 2\plus{}10\cdot 3\plus{}10\cdot 4}{40}\equal{}\boxed{2.5}\]
Of course, you can tell just by looking that the average is 2.5 as well

Since each was equal, the average is just the average of the four outcomes, 1, 2, 3, and 4, making the answer clearly $ \boxed{2.5}$


algebra
function
domain
Science Olympiad

I had aim for about 2 weeks.  It brought in so many viruses on my computer that i eventually uninstalled it.  Now i have msnm.

[quote="Valentin Vornicu"][quote="4everwise"]Hey, maybe somebody on AoPS used my account name for AIM. DOubt it though, as it isn't that cool.  ;) [/quote]So you decided to do it too? ...  :?[/quote]Yeah, I tried using my AoPS username.

[quote="noneoftheabove"]...AIM doesnt bring people viruses. It just has like one pop up every time you log on.[/quote]
perhaps you haven't heard of the AIM virus?  :?

[quote="inteluser"]it makes u tlk lik this!!!!!!!!!!!!!!!!!!!!!!!!!!!!!11

Awful thing, that AIM virus.[/quote]

 :rotfl:

I have AIM. Doesn't give me viruses.
And the AIM virus is stupid. Don't click on the link...

I have AIM. I think only people who click links that lead to places they don't know are the only people that get viruses. I haven't gotten any viruses but a bunch of my friends have.

Well you only get the viruses from other people when they say hey click this and you click it.

[quote="4everwise"]Any of you guys here use AIM? I just created an account today because it's convenient for Science Olympiad meetings over the web.

If anybody would like to add me to their (probably long buddy lists), my username is: Myths Apprentice

Everything else I thought of had been taken already ( :(   :noo:  :ewpu:), but it is a cool username. My AoPS account name didn't even work! I hope the AoPS moderator mithsapprentice won't file a lawsuit. :lol:

Hey, maybe somebody on AoPS used my account name for AIM, because I tried using it, but it was taken. Doubt it though, as it isn't that cool.  ;)

This post is getting really long. I should stop.  :|[/quote]

I think I figured out why 4everwise won't work.
Maybe I'm wrong, but maybe it's because you can't use numbers at the beginning of a sn. Or did you already know that?

Well whatever you do, get AimFix now.
Yeah, I have aim.

who DOESN'T have aim?

haha, well around where I live, everyone everywhere has an sn.
But that might because we're ungrateful high-school bums who sit around asking "wsup" every ten seconds to the same people, because we're bored.

i use aim, i only got one virus, from a link i clicked on (stupid myself), i fixed by uninstall and reinstall aim (didn't affect other programs).

around where i live, everyone has an aim

the only problem with aol aim is it's so ugly, so i uses trillian(on windows) and iChat( on mac)

i personally can't live without AIM- i like it soo much. i get to keep in touch with people from high school quite easily :)

Actually, i don't.

I'm the biggest geek to walk the earth. :(

Now I gotta turn gothic to find my identity...thanks a lot everyone... :mad:

Slipknotttttttttttttttttt :evilgrin:

I use AIM.  Funny stuff there Tedward.  That's great.

I use GAIM (intended for linux but available for windoze) and it's secure, but probably not as secure as Trillian. Both are free for download, check them out. I usually use GAIM to partake in team efforts for homework, and to play the max pratt game (use chatrooms).

guys get in the hips like rashad9607 :P

I use Gaim as well, quite nice

I'd put my sn up here, but I'm always away and I find it difficult to talk with people I don't know.  Also, my buddy list has more than 200 buddies.  I'll have to clean it up someday.

i use AIM all the time, but itll be tough to get my sn (hint click the AIM thing..)

I've never gotten a virus cuz i can tell the dif between a bad link and a good one.

theres that one like _www.pic1253.com/sdjlkfdsajfldfsalfkjsdf/pic1253.com

Is obviously a virus.. for one, its slashses, so the pic1253 must be a valid picture file (if its really a pic). .com is there to fool u, but a domain name cant have slashes. It would be .jpg or .png or .gif. or .bmp or any of the other various formats.

In addition, i dont think pic1253.com is a cleverly designed domain.. lol.

I've had aiM, but I've never really used it.  The only time I've used it was when I had to collaborate on a project...

I use gaim as well.

I am always on, yet always invisible...

I recently found out that with linked sns, you came login to your main account invis from your other account  :lol: if that makes sense...

i use aim now,  my username is:  mathlete92

Sn: BaseballJoe2002

It's an outdated sn (i.e. I don't play baseball anymore) but I'm too lazy to get a new one right now.

my sn is: [hide]monoversketcher
or
minus150degrees[/hide]

[quote="ToBe"]hey guys just a heads up - now i use Trillian as my AIM client (it connects to the AIM network) and it is SO much better - many more features. Google it, its pretty sweet.[/quote]
Trillian isn't bad, but the Pro version, which you have to pay for, is a lot better than basic. But it requires relatively a lot of resources (6-7 megs of memory on my computer). Gaim is pretty nice too. Free and has a lot of features Trillian doesn't have. It's just that Trillian looks better.


Let $X=\{A_{1},...,A_{n}\}$ be a set of distinct 3-element subsets of the set $\{1,2,...,36\}$ such that

(a) $A_{i},A_{j}$ have nonempty intersections for all $i,j$
(b) The intersection of all elements of $X$ is the empty set.

Show that $n\leq 100$.

Determine the number of such sets $X$ when $n=100$

http://www.mathlinks.ro/Forum/viewtopic.php?t=107181
http://www.mathlinks.ro/Forum/viewtopic.php?t=25362

  darij


Prove that there exists no more than one polynomial of degree n which commutes with the quadratic polynomial $ P(x)\equal{} x^2\plus{}px\plus{}q$

Notice that the funtions $ P(x)$ and $ Q(x)$  are commute if $ P_{o}Q\equal{}Q_{o}P$

The following article [url]http://nyjm.albany.edu/j/2007/13-5v.pdf[/url] is a real overkill, but a quite interesting and well-written one on commuting polynomials. The result above is a theorem of Rivlin (see page 6 of the above), and generalizes to existing at most $ k \minus{} 1$ polynomials of degree $ n$ commuting with a given polynomial of degree $ k > 1$.

Also see [url]http://books.google.com/books?id=oLKlk5o6WroC&pg=PA31&lpg=PA31&dq=commuting+polynomials&source=bl&ots=WyRMto3wdX&sig=QbvWEK0WAzE0JH5JsRlgevHJb9c&hl=en&ei=jwqqSszTJdOKsAafzJnxBw&sa=X&oi=book_result&ct=result&resnum=7#v=onepage&q=commuting%20polynomials&f=false[/url] for the original Russian problem.


Thirteen blue and six green hexagonal tiles were used to create the figure
below. If a new figure is created by attaching a border of green tiles with
the same size and shape as the others,
what will be the difference between the total number of green tiles and the
total number of blue tiles in the new figure?
[asy]/* AMC8 2004 #15 Problem */
path hex=rotate(30)*(dir(360)--dir(60)--dir(120)--dir(180)--dir(240)--dir(300)--cycle);
pen lightcolor=palegreen;
pen darkcolor=heavyblue;
filldraw(hex, darkcolor,black);
filldraw(shift(sqrt(3),0)*hex, darkcolor,black);
filldraw(shift(sqrt(3)*2,0)*hex, darkcolor,black);
filldraw(shift(sqrt(3)*2.5,1.5)*hex, darkcolor,black);
filldraw(shift(sqrt(3)*3,3)*hex,darkcolor,black);
filldraw(shift(sqrt(3)*2.5,4.5)*hex, darkcolor,black);
filldraw(shift(sqrt(3)*2,6)*hex, darkcolor,black);
filldraw(shift(sqrt(3),6)*hex, darkcolor,black);
filldraw(shift(0,6)*hex, darkcolor,black);
filldraw(shift(sqrt(3)*-0.5,4.5)*hex, darkcolor,black);
filldraw(shift(sqrt(3)*-1,3)*hex, darkcolor,black);
filldraw(shift(sqrt(3)*-0.5,1.5)*hex, darkcolor,black);
filldraw(shift(sqrt(3),3)*hex,darkcolor,black);
filldraw(shift(sqrt(3)*.5,1.5)*hex,lightcolor,black);
filldraw(shift(sqrt(3)*1.5,1.5)*hex,lightcolor,black);
filldraw(shift(sqrt(3)*2,3)*hex,lightcolor,black);
filldraw(shift(sqrt(3)*1.5,4.5)*hex,lightcolor,black);
filldraw(shift(sqrt(3)*.5,4.5)*hex,lightcolor,black);
filldraw(shift(0,3)*hex,lightcolor,black);[/asy]

I believe we would attach 18 more green tiles so their are 24 green tiles and 11 blue tiles so the difference is 13.


A wheel 60 ft in diameter makes one revolution in 3 minutes. Find the speed of a seat on the rim and the angular speed.

$ D \equal{} 60 \text{ ft} \implies C \equal{} 60 \pi \text{ ft} \implies v \equal{} \frac {60 \pi \text{ ft}}{3 \text{ min}} \equal{} \boxed{20 \pi \frac { \text{ ft}}{ \text{ min}}}$

$ \omega \equal{} \frac {2 \pi}{3 \text{ min}} \equal{} \frac {2 \pi}{180 \text{ sec}} \equal{} \boxed{\frac {\pi}{90} \text{ sec}^{ \minus{} 1} \equiv \frac{1}{180} \text{ Hz}}$

[i]Sharkman, if this applies, please don't post homework exercises on the AoPS forum without trying them yourself, feel free to look things up on Wikipedia or the AoPS Wiki (or your own textbook) for your initial effort.[/i]

Also, this should go into HSB...


im a little frightened by the qualifying test.

does it have to be very easy for me in order to get in?

-jorian

READ KIDDUMS READ. Nah It is supposed to be challenging, not to read though.  Xxx xxxxx xxx xxxx xxxx xxx xx xxxx xxx xxxx.  Xx xxx xxxxx xx xxx xxx xxxxxx.

I realised typos

[color=red]
[size=75]Comments concerning difficulty of problems have been removed.[/size][/color]

I disagree, well the optional is hard, so maybe that is considered part tho it is optional.

[quote="daermon"]isn't the optional the one where you make something up?  so it can be as easy or hard as you make it? :P
u think the last one is easy?  wow
are u a probability fiend or something?[/quote]

it is something along lines of with same spinner make solitaire or 2 player game and analyze....too time consuming without imagination...i find the one before this the hardest the one where you have to create a conjecture about the shape i think i'll just create an obvious one and hope to get some credit for that...

did you try your hand at all of them?

[quote="daermon"][color=red]Discussing relative difficulty of problems or parts of problems is not allowed.  -- Thomas / MathPath[/color][/quote]
[color=red]Discussing relative difficulty of problems or parts of problems is not allowed.  -- Thomas / MathPath[/color]


Sretno svim ekipama! Ocekujemo izvjestaje sa lica mjesta... kakva je hrana, smjestaj? (Vidim da prisutni u Madridu (na mornikovom poznatom topicu) izvjestavaju o karfiolu i paelli  :lol: ).

P. S. Ne morate mi svi slati razglednice (ona koja bi trebala, zna to vec-iako joj moja razglednica iz Vijetnama jos nije stigla :) ).

Još samo 35-tak sati do polijetanja!!!  :P

Krivo! Još manje od 25 sati do polijetanja!  :D

Sretan put svima, vidimo se u Madridu!

ustvari, za manje od 12 sati  :rotfl:
ove godine za sada : 2 srebra (RMM i BMO) i 1 zlato (max poena - JBMO).
znaci, fali i bronza, ali necu se ljutiti ako upadne srebro. srecno svima! vidimo se!

[quote]Još samo 35-tak sati do polijetanja!!! Razz[/quote]

Jos samo 2.5 sata   :pilot:

[quote]ustvari, za manje od 12 sati Rotfl
ove godine za sada : 2 srebra (RMM i BMO) i 1 zlato (max poena - JBMO).
znaci, fali i bronza, ali necu se ljutiti ako upadne srebro. srecno svima! vidimo se![/quote]

Ja imam bronze sa balkanijada, srebro sa mediteranske Dakle samo mi jos zlato fali da dopunim kolekciju  :gathering:  :wow:

pozdravlja vas [b]mornik[/b]. u njegovoj sobi je osvajač zlatne medalje. jest da to nije sam mornik, al jebiga bliže ne može :D

(za one koji (još) ne znaju, on je u sobi sa već drugim vijetnamcem. prvi nije znao riječi engleskog pa je došo drugi. uglavnom nije nam dosadno.)

Sjajno :) Sa mnom je svojeveremeno (u Sloveniji 2006) u sobi bio Bolivijac, kojeg sam poslije drugog dana pitao koliko je uradio- odgovor je bio "three-four". Bio sam fasciniran, sve do objave rezultata, kad sam vidio da je doticni imao jedan poen  :lol:

Nadam se da nije vruce... Ovdje kod nas je bas zahladnilo.

Po ovome sto cujem iz Madrida, bice gusto... Tezak prvi dan (ma jes' malo sutra, ali eto: cini mi se da je raspored kao po zeljama (bar tako bi meni bilo): G A NT :) ).

BTW, zanimljive price herr Grinberga i ZetaX-a na forumu o zadacima  :rotfl:  :rotfl:  :rotfl:

Zadatke sam, istina, vidio samo na spanjolskom, ali ne cine se teskim. Au contraire.

Da,ali ako danas urade solidno imamo mozda i 3-4 medalje...to bi bilo extra :)

Pa jesi ti vidio kombinatorike, hercegovac?  :lol: Bice gusto...

A sad ozbiljno: svidjaju mi se zadaci drugog dana. Po mom cejfu. (ovo se odnosi i na lazne zadatke Fedora Petrova i na ove "prave" ) :lol:

Ja ih nesto nisam puno zagledao ali me ova funkcionalna NERVIRA(vjerovatno zato sto ih slabo radim :lol: ).Ali izgleda da su nam max 2 medalje...nadam se da nije tako ali...takve su vijesti-nagađanja...

Hehe... Kad sam rekao kombinatorike, mislio sam na one Fedorove  :lol:
Ova s lampama nije losa, pozabavicu se s njom malo. Geometrija je na prvi pogled lijepa (i superteska, sigurno). A funkcionalna djeluje zeznuto, ali samo na prvi pogled.

I heared that Jelena had solved 6.problem with angle chasing :!:

De nemoj bogati tog engleskog... A ako je to istina, imate oboje pice od mene: ti za mustuluk, a Jelena za podvig.

Nije da ti ne valja engleski (daleko od toga), nego dosadi kad ga citas na svim drugim podforumuma na MLu-pa jos i ovdje. Da, Vedran mi je pomenuo da je ovo njegovo maslo... Stvarno, bolje zvuci Angle chasing od npr. g[b]o[/b]njanje uglova i sl. :lol:

[color=green]Moderator se izvinjava zbog zachatavanja teme.[/color]

kakvo ba moje maslo, nisam ja kriv sto je jelena tako izjavila  :D

Hrvati vjerojatno 2-3 medalje. Ako je vjerovati pronozama na forumu pragovi ce biti visoki.

Slobodno saljite i prognoze rezultata ostalih ex-Yu drzava  :)

[quote="hercegovac"]Nije valjda da mi ne valja engleski :blush: .Salim se.Morao sam jer nisam htio da mijenjam [color=red]chasing[/color]:Tu informaciju sam dobio od Vedrana :D[/quote]

Ko ti je kriv sto si povjerovao (u Gavrilo´s principle) hihihihihi
Zadaci su bili uradivi, realno govoreci, granice ce biti visoke.

Da li je neko rjesavao 5? Koje su jos ideje osim dvodimenzionalnih nizova?

Sve je to fino...samo vi nama javite kako je prosla koordinacija prvi dan,tj.koliko imate poena :D

ok....evo bih parcijalnih rezultata(problemi 1,2,3,6 redom):

 BIH1 7 4? 0 0
 BIH2 7 1 1 1
 BIH3 1 1 0 0
 BIH4 7 0 0 0
 BIH5 1 1 0 0
 BIH6 5 5 1 0

  we wait for 4th and 5th problem...

Jelena ima 20,Salem 18,Vedran17,Ana8,Emina 3,Marko 2.....nadam se da je ovako :lol:

Stan' malo, od Bilece Vlado... ovo je novi izvjestaj? Daj malo detalja :)

Evo konacnih rezultata
BIH1 740700 18
BIH2 711701 17
BIH3 110000 2
BIH4 700100 8
BIH5 110100 3
BIH6 551720 20

Nadam se da ce biti tri medalje...sve su prilike da ce bronza biti oko 17...

Slozicu se sa procjenom praga. Ovo su tri bronze (+jedna pohvala ;)).

Tja... sto su ti olimpijci sa iskustvom  :lol:

Moje cestitke ekipi, i to slijedecim redom: BIH6, BIH1, BIH2, BIH4, BIH5, BIH3.  :)

cestitam svim ucesnicima i ekipama!
molim vas objavite granice CHIM ih saznate!

u potpisu: teodorova mama,
predsednik pokreta I MAME ZASLUZUJU RASPUST

Kad sam prije mjesec dana rekao Vedranu da ce Jelena biti bolja od njega njegov odgovor je bio---- :rotfl: ----Sta sad kazes Veca?

Sad ce on sve svaliti na koordinatore i naseg lid. i deputija  :D

Veca je tek 3. u nasem timu (tezak kix).

Evo i pragova (za Tedijevu mamu):

Bronze-15
Silver-22
Gold-31.

Cestitke svima!!!

Ako jos niste vidjeli, sve je vec objavljeno na

http://www.imo-official.org/year_info.aspx?year=2008

Mi smo zadovoljni... Ali Nini je falio bod do bronce a meni 2 do srebra. Ali bit ce bolje sljedece godine!

Cestitke svima.Posebno za Luku!Lik je uzeo zlato!Svaka cast!

Vedrane,nakon ovoga nemas sanse da budes[color=red] deputy[/color] na nekom narednom IMO-u :!:

ooo cestitke svim koji su uzeli neku medalju....  :D ja sam u svom zadnjem nastupu uzeo ono po stom sam dosao...a moj uljarevicu iskusni ucesnice s IMO  :D  :D moja referentna linija je: http://www.imo-official.org/country_hall.aspx?code=BIH&column=awards&order=desc&gender=hide&nameform=western (samo ce uskoro dodati ovu godinu)....a ti se spremaj da prodjes sljedece godine a ne brini za to ko je deputy  :)  :) a jelena da iduce godine prodje bez revizije  :P

Pa sad meni ce biti dosta jedna IMO pa da te prestignem na ovoj GLUPOJ listi...na kojoj je jedan Uljarevic svjetlosnim miljama daleko ispred svih :wink: .A ne moras sad komentarisati Jelenu koja je ispred tebe daleko odmakla :wink:

[color=green]Moderator vas najljepse moli da smirite dozivljaj i diskusiju nastavite ili uz kaficu, ili na msnu, ili da je uopce ne nastavljate. Kako god, ovdje joj nije mjesto.

Brisanje postova i/ili zakljucavanje teme ce uslijediti ukoliko se ovakva konverzacija nastavi.[/color]

doci do IMO-a je uspeh. osvojiti bilo sta tamo (a da niste iz zemlje u kojoj vas spremaju preko cele godine) je uspeh. Zato pre svega cestitke SVIM ucesnicima IMO iz bivse YU.
A SRBIJI VELIKO  :clap: Luki sve cestitke!!! Mariji takodje uz malecko  :wallbash: (tu je Putnam).
Dusanu i Teu zelim zlato sledece godine. Aleksandru ce ovogodisnje iskustvo pomoci da se spremi sledecu godinu. BRAVO sjajnim koordinatorima - Dusanu i Djordju! Lep provod na dodeli veceras!


inequalities vector geometry percent calculus integration trigonometry

I know that at least in Ontario these concepts aren't at all covered in the high school curriculum. Are they taught anywhere else? What kind of university course would introduce these if they're not introduced in high school?

[quote="max_tm"]I know that at least in Ontario concepts such as these aren't at all covered in the high school curriculum. Are they taught anywhere else? What kind of university course would introduce these if they're not introduced in high school?[/quote]

They're not taught here (Ireland) either. Inequalities are briefly covered, but only so far as "any square is greater than or equal to 0" stuff. The proof of AM-GM in two variables came up on the paper one year, but it wasn't advertised as AM-GM.

I'm not in university myself, so can't really answer your second question. I have however came across Cauchy-Schwartz (in vector form) in a Linear Algebra course, as part of the basic vector stuff (Cauchy-Schwartz in the form "the product of the sum of the squares is greater than or equal to the square of the sum of the products" then appeared as an exercise; but it trivially follows from the vector form).

Hah, this reminds me of this cartoon I once saw that I should've downloaded:
It showed a train with the words school cirriculum on it and tracks labelled language arts, biology, chemistry, i.e.
Then it showed three tracks that the train didn't go over (missing rail) labeled geometry, inequalities, and number theory.
To the side one man was saying to another "One percent of the track missing won't derail the train, will it?"

yeah it will
inequalities, geometry, and number theory are some of the kewlest kinds of math :(

You're right, as far as I know, no standard school curriculum has specific aims(as in, listed in the syllabi) to explore these topics, among other things.
The treatment of number theory and combinatorics is very lax as well. Not sure about how they handle these at college level, try posting on the college forum.

Incidentally, part of the reason you don't see many inequalities is that many of them require an extremely large amount of work to prove.  Holder's and Minkowski's inequalities, for example, are very far from simple to prove, and so the way they would be introduced would be as a revelation from on high.  ("The teacher spoke, and then there was Minkowski's inequality.")  This, I think, is a very good reason for excluding them from curricula.

Some of the inequalities, such as Jensen's, have their most natural home in the context of calculus or analysis.  In fact, it was in my analysis course in college where I first saw a proof of either Minkowski or Holder (although both were solved in their integral, rather than summation, forms).

As for combinatorics, it doesn't fit very well into the Algebra-Geometry-Algebra 2-Trig-Calculus scheme, which is unfortunate.  In New York State, the older integrated curriculum definitely spent some time on binomial coefficients, Pascal's triangle, etc., but I don't know what things will look like now that they're adopting the AGA2TC system.

At least you guys have a course in geometry... Up here it just gets sprinkled into the curriculum in the middle of grade 9 and grade 12.


Anjali has 230 quarters, 300 dimes and 165 nickels. Niki has 210
quarters, 316 dimes and 173 nickels. How many
more cents does Anjali have than Niki?

Reduce each person's coins.
Anjali has 20 more quarters, while Niki has 16 more dimes and 8 nickels.
Anjali has 20*.25=4 more dollars in quarters,
while Niki has 16*.1=1.6 more dollars in dimes and .4 more in nickels.
Anjali has 3 more dollars or 300 more cents.


geometry
circumcircle
symmetry
geometry unsolved

Suppose M and N are distinct points in the plane of triangle ABC such
that
AM : BM : CM = AN : BN : CN.
Prove that the line MN contains the circumcenter of 4ABC.

[color=darkblue]This problem can be found in Prasolov's Plane geometry.
Here is a proof from the book.

You can use the following result

Let $ x \plus{} y \plus{} z \equal{} 0$. Then the locus of the point $ P$ satisfying $ xAP^2 \plus{} yBP^2 \plus{} zCP^2 \equal{} u$ is a line. ($ u,x,y,z$ are constant reals).

Then

Let $ AM: BM: CM \equal{} p: q: r$. Then all points $ X$ for which $ (q^2 \minus{} r^2)AX^2 \plus{} (r^2 \minus{} p^2)BX^2 \plus{} (p^2 \minus{} q^2)CX^2 \equal{} 0$ lie on a line.
But points $ M,N,O$ satisfy this condition.[/color]

Alternatively, one can show that the locus of points P such that PN/PM is constant is a circle. By symmetry, that circle has center on MN.


Problem:
Let a,b,c be the lengths of a triangle. Show that:
$ \sum_{cyc}\frac{a}{b\plus{}c} < 2$

Me being a total newb at inequalities, I've gotten it this far:

Use ravi substiution with a = x+y, b = y+z, c = z+x as usual. This gives:

$ \sum_{cyc}\frac{x\plus{}y}{x\plus{}y\plus{}2z}\equal{}
\sum_{cyc}\frac{s\minus{}x}{s\plus{}x}$ where s is the semiperimeter. Unfortunately, now I'm stuck. Any help?

[quote="kops723"]Problem:
Let a,b,c be the lengths of a triangle. Show that:
$ \sum_{cyc}\frac {a}{b \plus{} c} < 2$

Me being a total newb at inequalities, I've gotten it this far:

Use ravi substiution with a = x+y, b = y+z, c = z+x as usual. This gives:

$ \sum_{cyc}\frac {x \plus{} y}{x \plus{} y \plus{} 2z} \equal{} \sum_{cyc}\frac {s \minus{} x}{s \plus{} x}$ where s is the semiperimeter. Unfortunately, now I'm stuck. Any help?[/quote]

 Note that,If $ 0<x<y; z>0$ then $ \frac x{y} < \frac{x\plus{}z}{y\plus{}z}$
=> $ \sum \frac a{b\plus{}c} < \sum\frac {a\plus{}a}{a\plus{}b\plus{}c}\equal{}2$
done!  :blush:

[quote=Chronoz]Sorry for this: Can someone tell me why this is called ravi substitution?[/quote]

Imagine you're going to work with an inequality using sides of a triangle.
If there's no substitution to be used then you would have to use three inequalities
a+b > c
a+c > b
b+c > a
And that's ugly and do not help solving problems... This is the part where Ravi Substitution is useful...
If you use that:
a+b-c = x
b+c-a = y
c+a-b = z
You already have that x,y,z >0 and , therefore, you can use Geometric and Arithmetic inequalities without worries...
Hope that i could give an insight about the topic
Sorry for the bad english , im not a native


integration
induction
function
calculus
factorial
real analysis
real analysis unsolved

let $b_{m,n}=\int_{0}^{1}x^{m}(1-x)^{n}dx$
Show that $b_{m,n}=\frac{m!n!}{(m+n+1)!}$

I know you can integrate by parts to make a relation and then use induction.

Is there a way to show this without induction?

Not really. We can relate it to the gamma function by substitution, but that integral also requires induction. Factorials are inherently inductive objects.

That's a recursive formula- without induction, it doesn't tell you anything.

Not true, there are methods to solve one dimensional relations that do not involve induction. I think some general methods apply to two dimensional ones too. Induction is good for when the answer is known, but I would like to know more general ways to find an answer.

The following problem will help you, Ish.

[color=blue]Tokyo Denki University entrance exam 2004[/color]

Let $m,\ n$ be positive integers and $I(m,\ n)=\int_{0}^{1}x^{m}(1-x)^{n}dx$.

(1) Find $I(m,\ 1)$.
(2) For $n\geq 2$, express $I(m,\ n)$ in terms of $I(m+1,\ n-1)$.
(3) Express $I(m,n)$ in terms of $m$ and $n$.

Another approach: Probably we can use binomial theorem.

For part 2 of the above problem I get:
$ I(m,n)\equal{}\frac{n}{m\plus{}1}I(m\plus{}1,n\minus{}1)$

Is this correct? I'm not sure because I can't seem to arrive at the correct answer for part 3 (even though I think I'm using the correct method)... :(

The result follows with a direct application of the Beta function identity $ \beta (x,y) \equal{} \frac{{\Gamma (x) \cdot \Gamma (y)}}
{{\Gamma (x \plus{} y)}},\,x,y > 0.$

[quote="Black"]The result follows with a direct application of the Beta function identity $ \beta (x,y) \equal{} \frac {{\Gamma (x) \cdot \Gamma (y)}} {{\Gamma (x \plus{} y)}},\,x,y > 0.$[/quote]

Plus a little inductive proof to show that $ \Gamma(n)\equal{}(n\plus{}1)!$, yes. :)

That's a trivial proof. Why do we need to do that since we want to get a direct result? Those are things that we already know and of course, all of them require a proof, but not now, it makes no sense.

Of course it's easy- but that doesn't mean we can ignore the induction. The definition of the factorial is a recursion, and that implies that any equation with factorials on one side needs induction [b]somewhere[/b] in its proof. The simplest proof of the original statement is a straightforward induction, as Ish mentioned in the first post:
$ b_{m,n}\equal{}\int_0^1 x^m(1\minus{}x)^n\,dx$
$ \equal{}\left[\frac1{m\plus{}1}x^{m\plus{}1}(1\minus{}x)^n\right]_0^1\plus{} \int_0^1 \frac{n}{m\plus{}1}x^{m\plus{}1}(1\minus{}x)^{n\minus{}1}\,dx\equal{}\frac{n}{m\plus{}1}b_{n\minus{}1,m\plus{}1}$
$ \equal{}\cdots \equal{} \frac{n(n\minus{}1)(n\minus{}2)\cdots 1}{(m\plus{}1)(m\plus{}2)\cdots (m\plus{}n)}\int_0^1 x^{m\plus{}n}\,dx$

That is easily converted to the factorial form- and the ... is induction, even though I didn't say so explicitly.

We can't even introduce the integral without induction. For instance, we don't know that the sum of $ n$ numbers is well-defined without induction. (and for that matter, what is $ n$ without induction?). I don't understand what the objection to induction is. It's a very natural thing, and after a while even larger types of induction are carried out without explicit mention.

I think Black's answer avoids induction in the sense that Ish wants.

I didn't have an "objection" to induction, I just think that when answering a question like "How do I prove X?" we have to be careful sometimes to avoid answering "Use theorem Y, which says X".


induction
number theory unsolved
number theory

Let a sequence $ \{a_n\}_{n\equal{}1}^{\infty}$ be defined such that $ a_1\equal{}1$, and for $ k\geq 1$, $ a_{k\plus{}1}$ is the least positive integer not equal to $ a_i$ for any $ 1\leq i\leq k$ such that $ k\plus{}1$ divides $ a_1\plus{}a_2\plus{}...\plus{}a_{k\plus{}1}$.

Is there any positive integer not in the sequence $ {a_n}$?

I have not found a solution yet, just post here my brainstorming

Remark that such a sequence an is determined uniquely

Let (un) be a sequence satisfying:

u1=1;

u(n+1) = (u1+...+un)/n if   (u1+...+un)/n not belongs to {u1,...,un} and u(n+1)=(u1+...+un)/n +(n+1) otherwise

I think we could show that   un have the same properties as an. Hence an=un for all n

Then we could show that un covers the set of all positive integers

[quote="mto"]

I think we could show that   un have the same properties as an. Hence an=un for all n

Then we could show that un covers the set of all positive integers[/quote]

Please show these two things

I posted only my brainstorming, i did not say that 's my solution. I wrote out 20 first elements of this sequence without discovering generic
formulation... :(

Let $ b_n\equal{}\frac{a_1\plus{}a_2\plus{}...\plus{}a_n}{n}$. We will prove by induction that for all $ n \geq 2$:
$ b_{n\minus{}1}\plus{}1\geq b_n\geq b_{n\minus{}1}$,
$ b_n\leq n$.
We have $ b_1\equal{}1, b_2\equal{}2$. So we may assume that $ n\geq3$.
We have $ nb_n\equal{}a_1\plus{}a_2\plus{}...\plus{}a_n\geq a_1\plus{}a_2\plus{}...a_{n\minus{}1}\equal{}(n\minus{}1)b_{n\minus{}1}\equal{}nb_{n\minus{}1}\minus{}b_{n\minus{}1} >$ $ nb_{n\minus{}1}\minus{}n\equal{}n(b_{n\minus{}1}\minus{}1)$ so $ b_n>b_{n\minus{}1}\minus{}1$ and $ b_n\geq b_{n\minus{}1}$.
To prove $ b_n\leq b_{n\minus{}1}\plus{}1$ we show that $ a_n\leq n\plus{}b_{n\minus{}1}$. But $ b_i\leq b_{i\minus{}1}\plus{}1$ implies $ a_i\leq i\plus{}b_{i\minus{}1}<n\plus{}b_{n\minus{}1}$ for $ 2\leq i\leq n\minus{}1$. So $ n\plus{}b_{n\minus{}1}$ is not equal to $ a_i$ and is not equal to $ a_1\equal{}1$.
So $ a_n\leq n\plus{}b_{n\minus{}1}$. And of course $ b_n\leq b_{n\minus{}1}\plus{}1\leq n\minus{}1\plus{}1\equal{}n$.
So the sequence $ \{b_n\}$ is nondecreasing and contains every positive integer as $ b_{2k\minus{}1}\geq \frac{1\plus{}2\plus{}...\plus{}(2k\minus{}1)}{2k\minus{}1}\equal{}k$.
If the number $ m$ isn't in the sequence $ \{a_n\}$ and $ b_s\equal{}m$ then $ a_{s\plus{}1}\equal{}m$, contradiction.
So the sequence $ \{a_n\}$ contains every positive integer.


Determine the solutions ($x,y,z,w$) of the equation\[x^3+y^3+z^3=wx^2y^2z^2\]in natural numbers $x,y,z,w$

Sorry, I can't understand it. Are you sure it is a correct solution?


analytic geometry
combinatorics theorems
combinatorics

Let $\lambda=(\lambda_1,\ldots,\lambda_m)$ with $\lambda_1+\ldots+\lambda_m=d$ and $\lambda_1\geq\lambda_2\geq\ldots\geq\lambda_m$.
Define $l_i=\lambda_i+m-i$ and $\mu_i=|\{j|\lambda_j\geq i\}|$.
Now define $h_{ij}=\lambda_i+\mu_j-i-j+1$ for all $i=\overline{1,m}$ and $j=\overline{1,\lambda_i}$.

Prove: \[ \prod_{i=1}^m(l_i!)=\left(\prod_{1\geq i<j\geq m}(l_i-l_j)\right)\cdot\left(\prod_{i=1}^m\prod_{j=1}^{\lambda_i}h_{ij}\right).  \]
The equality can actually be split as an equality for all the terms with fixed $i$.

[i]To make the problem digestible:[/i]
Consider the $xy$ plane with the $x$ coordinate increasing to the right and the $y$ coordinate increasing downwards. Notice that this is not the standard drawing of the plane. Now consider the block formed by the squares that have the bottom right corner of coordinates $(i,j)$ with $1\geq i\geq m$ and $1\geq j\geq\lambda_i$ for all $i$. It should look like an upside-down half a pyramid.
For all $i$, $\lambda_i$ is the length of the $i$-th line, $\mu_j$ is the length of the $j$-th column, and $h_{ij}$ is the number of $1\times 1$ squares that lie below on the same column or to the right on the same line of the square on $(i,j)$ including this square.
I don't have a nice interpretation for $l_i$ but it has the nice property that $l_1>l_2>\ldots>l_m$ even though $\lambda_1\geq\ldots\geq\lambda_m$.

A nice interpretation for the l_i can be :
the l_i are simply the h_ij of the first column.

If anyone interested in a very nice probabilistic proof of Hook Length  Formula, may be look at the following paper:

   A probabilistic proof of a formula for the number of Young Tableaux of a given shape, Adv. in Math., 31 (1979), 104-109 (with C.
   Greene and A. Nijenhuis).

  you can download the paper from the homepage of the author, one of the best combinatorial mathematicians, Herbert S. Wilf

  as follows:

     http://www.math.upenn.edu/~wilf/

   in the section of "papers". You can also find lots of other interesting stuffs in combinatorics, including free ebooks of the author.

Just wondering... is it possible that amfulger confused the dimension of the irreducible representation of $S_n$ with the dimension of the corresponding irreducible representation of $\mathrm{GL} V$ ? Because the formula definitely does not hold, and looks like a mixup of the hook-length formula with the Weyl character formula for $\mathrm{GL} V$.


Compute how many ordered sets (m,n,p) of integers there are such that:
$ m \plus{}3y \plus{} 5p \leq 600$
PuMaC 2007

infinite; maybe you mean the number of positive integer solutions?


there are some similar triangles in the figure. look for those.

[quote="Rep123max"]Abcd is a trapezoid with the measure of base AB twice the measure of base DC. Point E is the intersection of the diagonals. The measure of diagonal AC is 11. Find the length of segment EC.
[/quote]

Drawing it out with parallel bases AB and DC, you find that triangle AEB :sim: triangle CED.  This is because <BAE=<ECD by parallel lines, <AEB=<CED by vertical angles, and <EBA=<CDE by parallel lines.  AB/CD=2x/x=2 (the ratio of similarity).  Applying that to diagonal AC, AE=y, so EC=y/2.  AE+EC=AC=y+y/2=11 ---> (3/2)y=11 --->y=22/3.  EC=y/2=11/3

EC=11/3


1. The segment AB has length 5. On a plane containing AB, how many straight lines are at a distance 3 from B?
  2. In triangle ABC, D is a point on the extension of BC, and F is a point on the extension of AB. Yhe bisector of ACD meets the extension of BA at E, and the bisector of FBC meets the extension of AC at G, as shown in the diagram below. If CE=BC=BG, what is the measure of ABC
  3. ABCD is a parallelogram and P is a point inside triangle BAD. If the area of triangle PAB is 2 and the area of triangle PCB is 5, what is the area of triangle PBD?
  4.The non-zero numbers a,b,c,d,x,y and z are such that x/a=y/b=z/c  is the value of xyz(a+b)(b+c)(c+a)/abc(x+y)(y+z)(z+x)  ?
  5. On level ground, car travels at 63 kilometres per hour. Going uphill, it slows down to 56 kilometres per hour. Going down hill, it speeds up to 72 kilometres per hour. A trip from A to B by this car takes 4 hours, when the return trip from B to A takes 4 hours and 40 minutes. What is the distance between A and B ?
  6. The square ABCD has side length 2. E and F are the respective midpoints of AB and AD, and G is a point on CF such that 3CG = 2GF Determine the area of triangle BEG.
  7. Determine x+y where x and y are real number such that (2x+1)^2 + y^2 + (y-2x)^2 = 1/3
  8.

[quote="watcharapon"]
  6. The square ABCD has side length 2. E and F are the respective midpoints of AB and AD, and G is a point on CF such that 3CG = 2GF Determine the area of triangle BEG.
  [/quote]

[hide="Solution"] Let $ K$ is point on $ FD$ such $ GK$ perpendicular to $ FD$. Determione $ h$ is the length of altitude from point $ G$ to $ BE$($ \triangle BEG$). It's clear that $ h\equal{}AK$.
We have that $ 2FK\equal{}3KD$. Therefore $ FK\equal{}0.6$
$ h\equal{}AK\equal{}AF\plus{}KF\equal{}1.6$
$ S_{\triangle BEG}\equal{}\frac{1}{2}\cdot BE \cdot AK\equal{}0.8$[/hide]

[quote]7. Determine x+y where x and y are real number such that (2x+1)^2 + y^2 + (y-2x)^2 = 1/3
[/quote]
We use Cauchy-Schwarz inequality:
$ (2x\plus{}1)^2 \plus{} (\minus{}y)^2 \plus{} (y\minus{}2x)^2 \geq 1/3.(2x\plus{}1\plus{}(\minus{}y)\plus{}y\minus{}2x)^2 \equal{} 1/3.$
So, equality occures when x=y=-1/3.
Hence x+y=-2/3.

For 3, note that for any point P, [APB]+[CPD]=[DPA]+[BPC], where[] denotes area. (If you don't see why, drop perpendiculars from P to each side).
And [BPD]=1/2(total area) -[APB]-[APD]
The answer ends up being 2  :).


Salut tout le monde,

J'aimerais savoir si le problème ci dessous peut se résoudre avec des calculs de sommes et de produits,
voir de la divisibilité style spé maths...
En fait je n'ai pas abordé le cours d'arithmétique (la partie 1 :D qui me fait trop rire tellement elle est longue,
je comprends pourquoi le doute subsiste sur une éventuelle partie 2...)

Si quelqu'un voit qu'il faut aller chercher plus loin (cad dans la fameuse Partie 1), je le remercierai de me prévenir pour que je ne cherche pas pour rien et que j'attende de la lire (meme si c'est instructif de chercher
il faut quand meme gagner du temps et je finis à penne le chapitre sur les inégalités)

Find all représantations of number 1982 as the sum of several naturel numbers a1, a2, ...., an   ( n>1 )
wich form an arithmetic progression with odd common difference, such ai is divisible by i+1 for all i .

cad(traduction incertaine et imcomplete de ma part) Trouver les représentations du nombre 1982 comme
somme des termes d'une suite arithmétique (an) ( n>1) " !!!!  with odd common difference !!!! ", tel que pour tout i, ai soit divisible par i+1.

Merci d'avance.

'with odd common difference' = dont la raison (de la suite arithmétique)est impaire.

Pour le reste... :P

Pierre.


[quote="Hokkage"]From the SDMO 2005: Find with proof all solutions $(x,y,z)$ in integers to the following Diophantine equation: $2005x^3=y^3+25z^3$
Could someone check if they got the same solution as me after they solve it?
[hide="Is my solution correct?"]Since the coefficients for x and z are multiples of 5, then we can rewrite the equation using the substitution $y=5b$ giving $2005x^3=5^3b^3+25z^3$. Now, since the coefficients of b and z are multiples of 25, then $x=5a \Rightarrow 2005\times5^3a^3=5^3b^3+25z^3$. Now both b and a have coefficients which are multiples of $5^3$ so $z=5c$ and, continuing in this way, the equation has no solution other than the trivial $(0,0,0)$ by infinite expansion. Is this valid?[/hide][/quote]

yes you can do it this way . It is call infinity descent .

Ah, right. I thought infinite expansion sounded a little weird. Infinite descent makes much more sense.
So it's right?


If $ n$ is an integer and $ n > 1$, then show that $ 2^n \minus{} 1$ cannot be divisible by $ n$.

Posted before many times
http://www.mathlinks.ro/Forum/viewtopic.php?t=150382


geometry
3D geometry
sphere
rotation
analytic geometry

A sphere is placed in a box in the shape of a cube, in which the length of any side is equal to the diameter of the sphere.  The box has four polyhedra glued to the corners of the box.  Each of these polyhedra consists of four 1x1x1 cubes that are glued together in such a way so that it looks like this:
[img]http://i15.photobucket.com/albums/a390/IronCleft/blocks.jpg[/img]
Obviously, you can't see the fourth one, because it is behind the others.
The sphere touches each of the three corners of the polyhedra that do not come in contact with the box.  Compute the volume of the sphere.

The radius is 1 so the volume is $\pi\frac{4}{3}$

So the dimensions of the box are 2x2x2.  But how can you fit four of the polyhedra (with a volume of 4) that I showed you before into a 2x2x2 box (with a volume of 8)?  You can't!

The polyhedra are inside the box? I didn't realsie that. I thought they were somehow outside the box.

This clears the problem up considerably. Now it makes sense.

[hide]
Let r be the radius of the sphere.
the equation is $(x-r)^2+(y-r)^2+(z-r)^2=r^2$
$(2-r)^2+(1-r)^2+(1-r)^2=r^2$
$3r^2-8r+6=r^2$
$r^2-4r+3=0$
$(r-3)(r-1)=0$

The radius of 1 solution doens't work so r=3 so the answer is
$36\pi$

[/hide]

the length of the big box isn't known, at leats not yet.  We know that the lengths of the polyhedra are 1 for eahc box side so in that case we are just dealing with points.  The diameter between sides of the big box and the diameters between opposing polyhedra.

I just set the center of the sphere at r,r,r where r is the radius. You can translate and rotate all the coordinates however you feel like.

Hey, 0714446459923, how did you get [hide]$144\pi$?[/hide]
[hide]If $r=3$, the volume of the box is $(3\cdot2)^3$ or 216.  And $216<144\pi$.[/hide]

If $r=3$, then the volume should be $36\pi$. :D

oops I accidently multiplied it by 4 twice. I have fixed it now.


I no longer receive mail notifications on my watched topics. Moreover, the "watched topics" section in my profile is empty, and even the newly added topics don't show up. What's the catch?

That menu is not yet operational. It's a fault of my own that you were able to see it :)

I will disable it. Your watched topics are there. You are probably not receiving notifications because they were added to the spam list (check it please) when the server changed its IP.

I've just received the notifications from the two newly added "test" topics (i.e. the topics I don't really want to watch, but used them to try the thing out) - that probably means that the server IP address is OK (no spam blockage) - but no older topics ever appear, though I can see that they're active.

Moreover, and I think that's important - the links in the mails lead to [i]AOPS[/i] and [b]NOT[/b] to ML, which forces me to manually visit the corresponding ML topics in order to make them "visited" and eligible for next notification. That kind of "bug" was also present before. What's the thing?


What are the conditions that must be satisfied in order that two source interference
fringes may be clearly observed?

don't know what you mean by two interference fringes, but an interference pattern will be observed whenever light passes through some opening.  This can be explained with Huygen's principle, where each point on the opening will act as a point source of light waves.  (Basically you can think of it as a single slit is actually an infinite number of infinitely small slits).  An interference pattern is still observed with a single slit, it's just that it's much more pronounced with two slits because of the distance between them, meaning greater differences in path lengths, meaning a greater chance for the path lengths to be different by a multiple of wavelength + wavelength/2 (which would produce complete destructive interference).

The two sources used must be coherent so that theres a  constant phase difference between them, only den ull get to see a const fringe pattern, :idea:
Also if l is the wavelength of the light used and d is the slit width, s is the size of the source and S is the distance between the plane of the slits and the source, s/S<l/d
[hide]I donot know the reason for the second one, i just know it as a fact. :maybe: [/hide]


limit
floor function
calculus
calculus computations

Give examples of (a) a sequence $ (x_n)$ of irrational numbers having a limit $ \lim x_n$ that is rational, and (b) a sequence $ (r_n)$ of rational numbers having a limit $ r_n$ that is an irrational number.

So (a) $ x_n \equal{} \frac{1}{\sqrt{n}}$ where $ n$ is prime.

And (b) $ 1.141, 1.1412, 1.1413....$ which tends to $ \sqrt {2}$?

Are these correct?

(a) is correct, but notationally awkward in the way it used the letter $ n.$ My choice would probably have been $ x_n \equal{} \frac {\sqrt {2}}n.$

The problem with (b) is that the sequence isn't well-specified. From what you wrote, I can't tell what the 4th term of the sequence is, or the 7th term. Perhaps you meant to make it the decimal approximations to $ \sqrt {2}.$ We could say this by letting $ x_n \equal{} 10^{ \minus{} n}\left\lfloor 10^n\sqrt {2}\right\rfloor.$ Then the first few terms would be $ 1.4, 1.41, 1.414, 1.4142, 1.41421, \dots.$

I like the recursively defined sequence $ x_1 \equal{} 2,$ $ x_{n \plus{} 1} \equal{} \frac12\left(x_n \plus{} \frac2{x_n}\right).$ Here, the first few terms are $ 2, \frac32,\frac {17}{12},\frac {577}{408},\dots$ This is clearly a sequence of rational numbers and converges to $ \sqrt {2}$ -- converges extremely rapidly.

Here's another one for (a): How about $ x_{n}\equal{}\pi^{1/n}$.  Has $ n\rightarrow\infty$, the exponent tends to 0, so the entire expression tends to 1.


linear algebra
matrix
vector
greatest common divisor
number theory unsolved
number theory

5. Given integers $ a,b,c,d,u,v,m,n$ satisfying $ ad\minus{}bc\equal{}\pm1, u \equal{} am \plus{} bn, v \equal{} cm \plus{} dn$ prove that $ (m,n) \equal{} (u,v)$.

How about: $ u\equal{}2\cdot 3\plus{}4\cdot 5\equal{}26$ and $ v\equal{}6\cdot 3\plus{}8\cdot 5\equal{}58$ then $ (m,n) \equal{} (3,5)\equal{}1$ but $ (u,v)\equal{}(26,58)\equal{}2$ ?

Sorry I forgot to mention $ ad - bc = \pm1$

Could someone see if there's any error/ambiguousness in my proof?

[hide="Proof"]

Let $ (m,n)=d$.
Since $ d|m$ and $ d|n$, $ d|mx+ny$ for any integers $ x$ and $ y$.
In other words $ d$ divides any linear combination of $ m$ and $ n$.
Thus, $ d|am+bn = u$ and $ d|cm+dn=v$.
Therefore, $ d = (m,n)\le(u,v)$ by definition of the greatest common divisor of $ u$ and $ v$. (If $ d > (u,v)$, then $ (u,v)$ wouldn't be true)

Now we want to show $ (u,v)\le(m,n)$ in order to prove $ (u,v)=(m,n)$.
Assume a matrix $ A=\begin{pmatrix} a' & b' \\ c' & d'\end{pmatrix}$ where $ a'=\frac{d}{ad-bc}, b'=\frac{-b}{ad-bc}, c'=\frac{-c}{ad-bc},$ and $ d'=\frac{a}{ad-bc}$.
We then multiply $ A$ by a vector $ \begin{pmatrix} u \\ v \end{pmatrix}$ to get

\begin{eqnarray*} \begin{pmatrix} a' & b' \\ c' & d'\end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix} & = &  \begin{pmatrix}\frac{du}{ad-bc}-\frac{bv}{ad-bc} \\ \frac{-cu}{ad-bc}+\frac{av}{ad-bc}\end{pmatrix} \\
& = &  \begin{pmatrix}\frac{dam+dbn-bcm-dbn}{ad-bc} \\ \frac{-cam - cbn + cam + adn}{ad-bc}\end{pmatrix} \\ & = & \begin{pmatrix}\frac{m(ad-bc)}{ad-bc} \\ \frac{n(ad-bc)}{ad-bc} \end{pmatrix} \\ & = & \begin{pmatrix} m \\ n \end{pmatrix}\end{eqnarray*}

Note that since $ ad-bc=\pm1$, the elements of $ A$ are all integers. Thus $ m$ and $ n$ can be written as a linear combination of $ u$ and $ v$ (e.g. $ m = a'u + b'v, n  = c'u + d'v$).

Thus, using the same idea as in the beginning of the proof, let $ (u,v)=g$.
Since $ g|u$ and $ g|v$, $ g$ divides any linear combination of $ u$ and $ v$.
So $ g|a'u+b'v = m$ and $ g|c'u+d'v = n$.
Therefore, $ g = (u,v)\le(m,n)$ by definition of the greatest common divisor of $ m$ and $ n$.

$ (u,v)\le(m,n)$ and $ (m,n)\le(u,v)$ concludes that $ (m,n) = (u,v)$.

[/hide]

u=am+bn
v=cm+dn
cu-av=(cb-ad)n=+/-n
du-bv=(ad-bc)m=+/-m
The first two equations imply that (m,n) divides (u,v)
The third,fourth eq imply that (u,v) divides (m,n)
So (m,n)=(u,v)


I want to take a picture of my six friends standing in two rows of three. How many different photos can I take if the tallest person must stand in the back row, the shortest person must stand in the front row, and two of my friends are identical twins who I can't tell apart? (The identical twins are neither the tallest nor the shortest.)

There are $ 3$ places for the tallest person, $ 3$ places for the shortest person, $ 4$ places for one of the other people who's not one of the twins, $ 3$ places for the other person who's not one of the twins either, and finally, the twins are fixed. This yields a total of $ 3\cdot3\cdot4\cdot3\equal{}\boxed{108}$ ways.

[hide][hide][hide]there are 3*3*4*3=108 ways[/hide][/hide][/hide]


geometry
rectangle
induction
combinatorics unsolved
combinatorics

A given rectangle $ R$ is divided into $mn$ small rectangles by straight lines parallel to its sides. (The distances between the parallel lines may not be equal.) What is the minimum number of appropriately selected rectangles’ areas that should be known in order to determine the area of $ R$?

Not sure how to go about proving this but wouldn't it be

[hide]
WLOG assume $ m \leq n$. Then, by selecting diagonally down, you can ascertain the area of the first section. Call the height the distance fallen. There will be m squares in this. Then the remaining dimensions of the rectangle can be determined from selecting $ n\minus{}m$ squares in the remaining horizontal. Hence $ m\plus{}(n\minus{}m)\equal{}n$
[/hide]

Hmmm.  I may be interpreting the problem wrong, but I think that we are only given each of the small rectangles areas, not the width and length.  For instance, if $ m\equal{}n\equal{}2$, then if I only give you two of the areas, there is no way to determine the area of $ R$.

I think the answer is $ m\plus{}n\minus{}1$.  Suppose $ R$ is cut into $ m$ rows and $ n$ columns.  Then if we are given an entire row of rectangles' areas and an entire column of rectangles' areas, then it is not hard to see that we can produce the area of $ R$.

To show this is optimal, we proceed by induction.  We induction on $ m\plus{}n$.  If $ m\equal{}n\equal{}1$, the proof is trivial.  Now given an $ m\times n$ rectangle, suppose that it could be done with at most $ m\plus{}n\minus{}2$ areas.  Then we know there is a column or row that contains a known area but which not all the areas are known; otherwise, we would have to have at least $ m\plus{}n\minus{}1$ areas.

Without loss of generality, assume that this occurs at the far left column.  Then we can swap the rows so that all the known rectangles occur at the top-leftmost part of $ R$.  As denoted below, call their union rectangle $ A$.  Suppose there are $ 0<a<m$ small rectangles that make up $ A$.  Then we can partition $ R$ into four big rectangles $ A,B,C,D$ as shown below:

$ \begin{array} [b] {cc}
A&B\\
C&D\\
\end{array}$

Consider the bottom-rightmost rectangle $ D$ of dimension $ (m\minus{}a)\times (n\minus{}1)$.  By assumption, there are at most $ m\plus{}n\minus{}2\minus{}a$ known squares in this rectangle.  Note, however, that we have no information about $ C$, and so if we do not know $ D$, there is no possible way to obtain $ R$ by the induction hyopthesis.  But by the induction hyopthesis, we need at least $ m\plus{}n\minus{}a\minus{}2$ areas in $ D$ to determine its area, and thus all remaining $ m\plus{}n\minus{}a\minus{}2$ squares must be in $ D$.  But this means that none are in $ B$, and so we do not know either $ B$ or $ C$.  Hence by the induction hypothesis, we cannot determine $ R$ again.  Thus we need at least $ m\plus{}n\minus{}1$ areas, and this completes the induction.

This proof may require that the $ 2\times 2$ case is proved first, but this is really not that difficult.


geometry
ratio
geometric transformation
homothety
trigonometry
geometry unsolved

A "Pacman fractal" is built in the following way:

1) Draw a circle with radius 10
2) Delete a circle sector of angle 1rad (the mouth of Pacman)
3) Inscribe another Pacman between the mouth of the first and the original circle of the first Pacman, with the mouth at the same direction than the first.
4) Inscribe another Pacman between the mouth of the second and the original circle of the first Pacman, with the mouth at the same direction than the first
n) etcetera...

A) Find the distance between points A and B
B) Find a formula for this distance between the points of the mouth of any Pacman
C) Find the total area of the Pacman Fractal

Welcome to the forum!

Er...not very hard, and not an olympiad geometry sort of problem...
You need to get a kind of feel for problem difficulty.  If you don't know the solution to a problem, you may have a tendency to either overestimate or underestimate its difficulty and put it in a place where it doesn't belong.  It's okay, but unless you know the problem is from an olympiad or contest of that difficulty, it's probably best to stick to the Basics, Intermediate, and Pre-Olympiad sections.  (FYI, this problem is probably somewhere around the easier problems of Intermediate)

The ratio between of homothety between any pacman and the next smaller pacman is clearly $ 1 \plus{} \tan{\frac {1}{2}}$.  So from here it's pretty simple...

sorry, you are wrong...

There is not homothety ratio between any two pacman because there is [i]not[/i] isomorphism between them :wink:

remember that they are inscribed between the mouth of the last pacman and the original circle, not between the mouth and the last pacman circle...

another solution?

PS: sorry... not very good at english :(
PS2: obviously, all pacman are equal, but there isn't an exact proportion between their sizes

This problem doesn't seem to be well defined...Is the angle of the mouth the same for each pakman? If that is true, then K8107 is probably correct (it would be easy to calculate anyway). If that is not the case, then could you clarify?

a) Yes, the angle is the same for all Pacman
b) There isn't any exact homothethy proportion that apply for all Pacman, so K81o7 is wrong
c) any solution (with proof)?

Oh right good point. The right side of the pakman is always bounded by the larger circle... Okay. I still don't think that it is that hard though. You should be able to set up a recursion for the center of the mouth and then it is pretty easy.

Oh okay right I see.  My mistake.

Yeah you can do it recursively.  Define the location of the center and/or radius of the $ n$th pacman, and then set up an equation for the same properties for the $ n\plus{}1$st pacman.

Sorry to revive, but can someone attempt this again


calculus
derivative
function
calculus computations

What are the partial derivatives of the following functions:  (g,h are real function)

f(x,y,z)= k(g(x+y), h(y, x-z)) and

f(x,y) = r(x,g(x), h(x,y))

Just use the chain rule:
$\begin{aligned}\frac \partial {\partial x} f(x,y)&=\\& \frac {\partial k} {\partial g} (g(x+y), h(y,x-z))\cdot  g'(x+y)+\frac {\partial k}{\partial h}(g(x+y), h(y,x-z))\cdot \frac {\partial h}{\partial w} (y,x-z)\end{aligned}$
(here $\frac {\partial h}{\partial w} (y,x-z))$ denotes the partial derivative of $h$ with respect to the second argument; $g'$ is the usual derivative of the function of one variable).

The general rule is that
$\frac \partial {\partial x}F(g_1,\dots, g_m)=\sum_{j=1}^m \frac {\partial F}{\partial g_j}\cdot \frac{\partial g_j}{\partial x}$ where $\frac{\partial F}{\partial g_j}$ denotes the partial derivative of $F$ with respect to the $j$-th argument. If some $g_j$ is a function of several other functions, this rule should be applied once more to expand $\frac{\partial g_j}{\partial x}$. If $g_j$ depends only on one argument, the corresponding partial derivative becomes an ordinary one.

Hope this helps. Try the second problem yourself now to see whether you got it.

BTW, it is a typical question for Calculus Computations and Tutorials, so I'm moving it there  :)


AMC USA(J)MO USAMO inequalities calculus geometry

[quote="chess64"]If you have a complete solution but it is not the shortest or easiest solution, do you still get 7 points or do you lose 1 or 2 (or 3 or 4 or 5 or 6 or 7 or ... 8 :lol:)?[/quote]

If it is rigorous and correct and the graders can understand it you will receive 7 points. I will point out though that longer solutions are less likely to be correct. Why? Suppose you have a 0.1% chance of making an error with every word.

A nice, short, 100 word solution has a 90.48% chance of being completely correct.

If you have a really long, inelegant solution (say 500 words maybe?), you have only a 60.64% chance of being completely correct. Incorrectness most likely results in a 6 (or maybe even a 5).

Also if you do weird stuff or have silly logic or something, a grader is more likely to want to try to understand what you said if your overall solution is short (of course, ideally they would care either way, but people are human). Still, don't worry too much. The biggest loss probably is that you have to spend a bunch of time writing up your solution instead of thinking about other problems.

Generally when you brute-force a problem, it's either 0 or 7 points, 7 of course if you rigorously proved it, but 0 if you made any mistakes. Brute-forcing might include doing inequalities with calculus or geometry problems with too much algebra.

[quote="chess64"]If you have a complete solution but it is not the shortest or easiest solution, do you still get 7 points or do you lose 1 or 2 (or 3 or 4 or 5 or 6 or 7 or ... 8 :lol:)?[/quote]

You'd probably get 7 points if you solution were totally correct. However, many first time USAMOers are quite surprised when they lose a lot of points because of tiny holes in their proofs.

Small holes. Reminds me of a MOP problem last year

Prove that if two squares with side length 0.9 are placed completely on a unit disc, they overlap.

Small holes got a lot of people on that one...

Yep, that must have been one of the winners for most non-rigorous proofs.

[quote="dogseatcheese"]For doing proofs, sometimes, when doing a very detailed proof, it gets very long. Are points taken off for that? Because like, to make a short proof you 1. have a really smart solution or  2. you skip a couple of explanations that are like really simple (and the judges will probably understand your step without explanation)... Number 1 rarely happens, so for number 2, how much can you expect a judge to know for you to skip an explanation. Becuase if you have a step without an explanatoin adn the judge doesn't understand.... what happens?[/quote]

As has already been stated, a completely correct solution gets full points regardless of how long it is.  As long as the grader understands the solution, and as long as the solution is rigorous, length is not a factor.  Of course, if the grader does not understand the solution because of its length...

Then again, remember that the graders aren't stupid.

Then again, remember that they aren't mind readers either.  :P


7th graders:

If I have 8 students on my mathcounts team, in how many different ways can I choose a captain, a vice captain, and a secretary?

8th graders:

If I have 8 students on my mathcounts team, in how many different ways can I choose a captain and two co-vice captains? How about 3 co-captains and two co-vice presidents?

I am an 8th grader, so I qualify to do this.

[hide="I think I have the answer"]336 ways of different ways you can choose a captain and two co-vice captains.(8*7*6)

Next one, 6720 ways of different ways you can choose 3 co-captains and two co-vice presidents (8*7*6*5*4)[/hide]

jogaman, can you please hide your answers? When you type a post, there should be a "Hide" button. Just highlight what you want to hide and click the button.

[hide]If you're doing the 7th grade problem, you're right. Can you show how you got it? (remember, hide it)[/hide]

DK, [hide]make sure you only count one way once. Distinguish between the instances where order matters and does not matter.[/hide]

[quote="Chinaboy"]
DK, [hide]make sure you only count one way once. Distinguish between the instances where order matters and does not matter.[/hide][/quote]

Didn't I?? :huh:

Nope. Try to figure out why. A problem doesn't have to be just permutations or just combinations

Chapter 6 of Intro to Counting and Probability might help if you can't figure it out.

[quote="Chinaboy"]jogaman, can you please hide your answers? When you type a post, there should be a "Hide" button. Just highlight what you want to hide and click the button.
[/quote]
woops....
sorry
I didn't know about the hide thingy.. :oops_sign:

Haha, no biggie. Funny and cute smiley face!

[quote="Chinaboy"]jogaman, can you please hide your answers? When you type a post, there should be a "Hide" button. Just highlight what you want to hide and click the button.

[hide]If you're doing the 7th grade problem, you're right. Can you show how you got it? (remember, hide it)[/hide]

[hide]I multiplied 8,7,and 6, using the permutation formula...( i think.. :w00tb: )[/hide]

[quote="Chinaboy"]Nope. Try to figure out why. A problem doesn't have to be just permutations or just combinations[/quote]

Umm.. You are confusing me.. Now I think you use combination, because 3 people out of 8 people regardless of orders. AHHHHHHH!! :cursing:

[quote="danb80"]Hint:  This is not a single step problem.

Chapter 6 of Intro to Counting and Probability might help if you can't figure it out--but work it again before you look![/quote]

So what you mean is that this problem can not be solved in one equation??

One equation, yes, but made up of more than one step.

But isn't question #1 right? Because any of 8 people can be captain.
2 co-vice captain would be 7*6 or is it 7C2??

[quote="themathmaster770"]jogaman are you a 7 or 8 grader?[/quote]
I'm a 7th grader... Is your last name Song??  :gleam:

[quote="Chinaboy"][quote="DonkeyKong"][quote="Chinaboy"]Ok, you guys haven't solved this yet.

8th Graders, split the problem in two. First find the number of ways to find a captain, then find the number of ways to find 2 co-vices from the remaining 7.

You have to be flexible. Not everything is just permutation or just combination.[/quote]
Blah... I think I kind of get it now...
But I am not sure...
[hide="First problem"]Number of ways to pick a captain: 8
Number of ways to pick 2 co-captains:42=(7*6).
So I guess you add them up...to get 50?[/hide]

[hide="Second problem"]Number of ways to pick 3 co-captains : 8*7*6
Number of ways to pick 2 co-vice presidents:5*4
So I guess you add them up to get 336?[/hide][/quote]

Why are you adding them though? They're completely dependent events.

For second question, CO-captains, so does order matter among the captains?[/quote]

Well, one first one, I added it,because I've already did multiplying and I assumed that you told me that it was wrong when you said "make sure you only count one way once. Distinguish between the instances where order matters and does not matter. " But now I think I actually got it right on the first one..

Hmmm...... so  on the second one, would the answer be $ \ _8P_3 \times 5 \times 4 ?$

No no.

Adding/Multiplying and Combo/Permu are totally different.

Adding: Indepedent Events
EX: What is the probability of rolling a 2 or a 3 on a dice?
ANS: 1/6 + 1/6 = 1/3

Multiplying: Dependent Events
EX: What is the probaility of rolling a 2 and then rolling a 3?
ANS: 1/6 * 1/6

_________________________

Permutation: Order matters
EX: How many 3 digit numbers can we make if we use 3 different digits from 1-9?
ANS: P(9,3)

Combination: Order doesn't matter, and this is where you don't want to count the same thing more than once because ABC and CBA are the same thing when order doesn't matter. That is why you divide the permutation by k!
EX: How many ways are there to choose 3 numbers from 1-9?
ANS: C(9,3)
If you had done P(9,3), you would've had 123, 132, 213, 231, 312, 321, each of which are the same thing, so you have to divide P(9,3) by 3! to make sure that 123 is counted only once.

__________________

For #1, there are 8 ways to choose the captain. From the remaining 7, there are C(7,2) = 21 ways to choose the 2 co-vices. Are these dependent? (Do you know the shirt/jeans problem that asks how many outfits you can make? Pretend that the captain is the shirt and the co-vices are the jean.)

For #2, why use permutations when they are COs?

[quote="Chinaboy"][quote="DonkeyKong"]Blah... I stink.... :(
But I think I get this little by little..

#1: Yes, I think they are independant, because you are choosing 1 captain then you have 7 people to choose 2 people to be co-captains.

#2: Oh... I don't know why I did that.. really... It should be combination instead of permutation, because (1,2) and (2,1) is same..right?[/quote]

No no, this comes little by little.

However, the two aren't independent.

1. Think of this as the shirts/pants problem. If I had 8 shirts and 21 pants, how many different costumes can I make? So, now, change the shirts into choices for captain, and the pants into choices of co-captains.

2. Yes, COs are the same. This time, the number of choices for co-captains is the number of shirts you have. The number of choices for co-vices is the number of pants you have.

Does that make more sense?[/quote]

Oops.. I meant to put dependent on #1(really)... Sorry..

Yes, it makes perfect sense now..
Can you post other problems like this, if you have time? So I can work on them?

for one captain and two co-vice captain
[hide]168
im pretty sure its wrong so im not bothering to put the solution
tell me if im right[/hide]

for 3 co-captains and 3 co-vice captains
[hide]560
i think my answer is ridiculously large.... whatever[/hide]

I have 12 people in my Mathcounts class. If there are three girls that will absolutely not bring snacks, how many different ways can I have 1 person bring snacks and 1 person bring drinks (if people who bring drinks cannot bring snacks?)?
[hide]quick guess, 72??[/hide]

What about 1 person for snacks and 2 people for drinks?
[hide]still maintaining the three stubborn girls..... uh, im guessing 28??[/hide]

What if in addition to the 3 girls that won't bring snacks, 2 boys won't bring drinks. How many different ways can I have 1 person bring snacks and 1 person bring drinks?
[hide]42, another guess[/hide]

martin, dont get mad at me because im just guessing
its 8:30 at night and my parents are not going to be happy if i stay longer
and i still wanna look at other posts

Spiders, good job at the first two (post solutions when you have time.)

For the other three, I'm sorry if I wasn't clear. The girls who can't bring snacks can bring drinks.

Try the last three again when you have time.

Keep up the good work!

[quote="Chinaboy"]Ya, definitely, here's a couple:

1. I have 12 people in my Mathcounts class. If there are three girls that will absolutely not bring snacks, how many different ways can I have 1 person bring snacks and 1 person bring drinks (if people who bring drinks cannot bring snacks?)?

2. What about 1 person for snacks and 2 people for drinks?

3. What if in addition to the 3 girls that won't bring snacks, 2 boys won't bring drinks. How many different ways can I have 1 person bring snacks and 1 person bring drinks?[/quote]
I have a weird feeling in my stomach.....

Here are my answers..

[hide="#1"]$ 9\cdot11\equal{}99$ways[/hide]

[hide="#2"]$ \ 12\cdot \binom{11}{2}\equal{}660$ways[/hide]

[hide="#3"]$ 9\cdot10\equal{}90$ways[/hide]

[quote="spiders_hate_chestnut"]i disaggree on DonkeyKong's answer for #2

[hide]
i got 55
[/hide]
can i show the solution during the meeting??[/quote]
You can post your solutions here... if you don't mind..

For #2, I think there are more than 55 ways...
Here's my explanation:
[hide]

Any 12 people can bring snacks, so $ 12$ ways.

Then we can choose 2 people out of 11 people who can bring drinks. So $ \ _{11}C_2$
Since they are dependent, you multiply them together..
----------------------------------------------------------------------------- [/hide]
That solution does not regard that 3 girls won't bring snacks.

But even though we apply that 3 girls won't bring snacks, it would be $ \ 9\cdot \binom{11}{2}\equal{}495$ ways..

ya, the 3 girls still can't bring snacks in #2.

For #3...

What if the person you chose for snacks is one of the 2 boys? What if the person isn't one of the 2 boys? Do you see how you can't just multiply 9*10? (This is still assuming that no one person can bring both snacks and drinks)

Good job one #1 and #2!

[quote="Chinaboy"]

For #3...

What if the person you chose for snacks is one of the 2 boys? What if the person isn't one of the 2 boys? Do you see how you can't just multiply 9*10? (This is still assuming that no one person can bring both snacks and drinks)
[/quote]
Huh??

9 comes from (total amount of people)-(3 girls)

10 comes from (total amount of people)-(1 boy or girl who brought snack)

What is wrong with it?

Say I chose someone for snacks that is not one of the 5 forbidden children. How many choices are left for the drinks? What if that someone I chose for snacks is one of the two boys. How many choices are left for the drinks? Now are those two numbers the same? I bet not. So, you'll have to break the cases up and do casework.

[hide="My solution"]well, if you choose a person that is willing to bring either drink or snack, and assign him/her to bring a snack first, there is 9 ways to choose from.
but if you are choosing another person to bring a drink, there are 9 people left (12-2 for the two stubborn boys and -1 more because the person bringing snack cant bring drink), so there are 9 ways.
so 9x9=81.[/hide]
now what?


1/x +1/y =2/15. what is the sum of all possible integer values of x?

[hide="sfft"]
$ 15(x\plus{}y)\equal{}2xy$ by cross multiplying
$ x(15\minus{}2y)\minus{}7.5(15\minus{}2y)\equal{}\minus{}112.5$ by rearrangement, then subtracting 112.5 from both sides
$ (x\minus{}7.5)(15\minus{}2y)\equal{}\minus{}112.5$
$ (15\minus{}2x)(15\minus{}2y)\equal{}225$
Now just solve for x and y, don't forget negatives.

[/hide]

[hide="or if you dont want the decimals"]
$ 15x\plus{}15y\equal{}2xy$
$ 2xy\minus{}15x\minus{}15y\equal{}0$
$ 4xy\minus{}30x\minus{}30y\equal{}0$
$ (2x\minus{}15)(2y\minus{}15)\equal{}225$
Solve from there
[/hide]

[hide]There are an infinite amount of integer solutions for $ x$. The sum is $ \infty$. (The problem only specifies that $ x$ is an integer.)[/hide]

eh, although since for every integer value of x, there's a corresponding y, you would have the sum of all the integers.  not sure if you can call that $ \infty$


let a,b,c be positive real numbers such that $ 2b \minus{} a;2c \minus{} b;2a \minus{} c > 0$
prove that:$ \sum {\frac{1}{{\sqrt {2b \minus{} a} }}}  \le \frac{{\sqrt {3{{(a \plus{} b \plus{} c)}^3}} }}{{ab \plus{} bc \plus{} ca}}$

[quote="Toan_VN_LC"]let a,b,c be positive real numbers such that $ 2b \minus{} a;2c \minus{} b;2a \minus{} c > 0$
prove that:$ \sum {\frac {1}{{\sqrt {2b \minus{} a} }}} \le \frac {{\sqrt {3{{(a \plus{} b \plus{} c)}^3}} }}{{ab \plus{} bc \plus{} ca}}$[/quote]
It's wrong. Try $ a\equal{}3$ and $ b\equal{}c\equal{}5.$

[quote="arqady"][quote="Toan_VN_LC"]let a,b,c be positive real numbers such that $ 2b \minus{} a;2c \minus{} b;2a \minus{} c > 0$
prove that:$ \sum {\frac {1}{{\sqrt {2b \minus{} a} }}} \le \frac {{\sqrt {3{{(a \plus{} b \plus{} c)}^3}} }}{{ab \plus{} bc \plus{} ca}}$[/quote]
It's wrong. Try $ a \equal{} 3$ and $ b \equal{} c \equal{} 5.$[/quote]
I'm sorry
$ \sum {\frac {1}{{\sqrt {2b \minus{} a} }}} \ge \frac {{\sqrt {3{{(a \plus{} b \plus{} c)}^3}} }}{{ab \plus{} bc \plus{} ca}}$


I've only played the DS and right now, all of my consoles are made by Nintendo. So, until I actually see and play a PSP, I have to go with the DS.

Both are bad (too expensive) that's why a computer is the best. :D

I would say the PSP but they both cost so much. My friend has both but I haven't got to see/play either of them.

psp graphics are like ps2 graphics... but i think the psp gets banged up ezier and gets dirtier ezily too (prob) both are cool though... expensive i'd rather have an xbox  :D

Well, a PSP is better for different reasons. PSP have better games and more capabilities than the DS. The DS has been out for several months yet the PSP already has better games. Also, you can listen to music, store photos, watch movies, and more with the PSP. And then there is that screen-it has more pixels than both the DS screens combined! The only setback about the PSP is that it's really expensive.  :(

Linux for the DS is coming out soon...that'll make one (heck) of a PDA!

My friend brought his PSP to school yesterday and he let me play it. THe graphcis are AMAZING. He brought wipeout pure (racing game) and the graphics were very very good. He also had music on it so we listened to some of it.

[quote="sheepwarrior"]but i think the psp gets banged up ezier and gets dirtier ezily too (prob)[/quote]

It comes with a carrying case, and I presume that people would take care of something that expensive.

Yeah, it is mighty expensie..$250 last time I checked. Hopefully the price will drop in a year, but I really do not feel the need to get a hand game playing console. It would be such a time waster, and I am afraid that I would begin to play it everywhere, because the graphics and the games are so good!