geometry
quadratics
trapezoid
trigonometry
similar triangles

A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices.  Find the value of $n$ if the the area of the small square is exactly 1/1985.
[asy]
size(200);
pair A=(0,1), B=(1,1), C=(1,0), D=origin;
draw(A--B--C--D--A--(1,1/6));
draw(C--(0,5/6)^^B--(1/6,0)^^D--(5/6,1));
pair point=( 0.5 , 0.5 );
//label("$A$", A, dir(point--A));
//label("$B$", B, dir(point--B));
//label("$C$", C, dir(point--C));
//label("$D$", D, dir(point--D));
label("$1/n$", (11/12,1), N, fontsize(9));[/asy]

Let $A''$ be the foot of the perpendicular from $A'$ to $AC'$. Then $\angle AA'A''=\angle DAC'$ (perpendicular sides), so $\triangle AA'A''\sim\triangle C'AD$, therefore $A'A'': AA'=AD: AC'.$

Since $AA'={1\over n}, AD=1, AC'=\sqrt{1+\left({n-1\over n}\right)^2}$, we get

$A'A''=\frac{{1\over n}}{\sqrt{1+\left({n-1\over n}\right)^2}}=\frac{1}{\sqrt{n^2+(n-1)^2}}$

Hence the area of the small square is $A'A''^2=\frac{1}{n^2+(n-1)^2}$

This gives the equation

$n^2+(n-1)^2=1985\iff n^2-n-992=0\iff n=\frac{1\pm 63}{2}$

Since $n$ is positive, we get $n=32$

[hide="Solution"][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=831&sid=3cb1846f71cebd85e6f8d75257883829[/img]
Let the top-left vertex of the inner square be $P$ and $BP=m$, so that $AP+\sqrt{1-m^2}$ by the Pythagorean Theorem. Additionally, $\triangle BAP\sim \triangle D'AP$ because $AP$ is the altitude to the hypotenuse of right triangle $D'AB$.

By similar triangles $\frac{m}{1}=\frac{\sqrt{1-m^2}}{1-1/n}$. Simplifying, $m^2=\frac{n^2}{n^2+(n-1)^2}$. However, note that $\frac mn$ is equal to the length of a side of the interior square, $\frac1{1985}$.

Thus, $\frac mn=\frac{1}{\sqrt{n^2+(n-1)^2}}=\frac{1}{\sqrt{1985}}$, and $(n-1)^2+n^2=1985$. Solving the quadratic, $n=32$.
[/hide]

Vague notation (the points with primes aren't defined). Please update.

[quote="4everwise"]However, note that $ \frac mn$ is equal to the length of a side of the interior square, $ \frac1{1985}$.
[/quote]

How do you know that? I don't see the connection ... ...
(Yes, I realize this topic is very old...tried to find out why above true, but couldn't...)

For convenience I'll redefine 4everwise's points, but the solution will remain the same.
[asy]defaultpen(fontsize(8));
pair A=(0,1), C=(31/32,1), P, D=(0,0), B=(0,31/32), Q;
P=foot(A,D,C);Q=foot(B,D,C);
draw(A--C--D--A--P);
draw(B--Q);
label("$A$",A,(-1,1));label("$D$",D,(-1,-1));label("$D'$",C,(1,1));label("$P$",P,(0,1));label("$C'$",B,(-1,-1));label("$Q$",Q,(-1,0));[/asy]
Define $ DP\equal{}m$. $ AD\equal{}1\implies AP\equal{}\sqrt{1\minus{}m^2}$. $ AD'\equal{}1\minus{}\frac{1}{n}$, and we know $ \triangle ADP\sim \triangle D'AP$. Therefore we have $ \frac{m}{1}\equal{}\frac{\sqrt{1\minus{}m^2}}{1\minus{}\frac{1}{n}}$
Also $ \triangle ADP\sim\triangle C'DQ\implies \frac{AC'}{AD}\equal{}\frac{PQ}{PD}\implies \frac{\frac{1}{n}}{1}\equal{}\frac{PQ}{m}\implies PQ\equal{}\frac{m}{n}$.

*Revive wand*

This is probably a really stupid question, but why is the area of the little square not just $(1/n)^2/2$?

That would imply the angles of the trapezoid near the edge of the square are $ 45^\circ $ and $ 135^\circ $ but they are not because that would imply $ \overline{AP} = \overline{PD'} $ which is clearly false.

So I guess the square in the center is slightly "skewed"?

I'm not sure what you mean by skewed. If you mean not at a $ 45^\circ $ angle to the unit square when extended, then yes.

Not to delve off topic, but why does your post have 21 edits? Sorta funny...

[color=darkred][b][u]PP[/u].[/b] A small square is constructed inside a square of area $1$ by dividing each side of the unit square into $n$ equal parts, and then connecting
the vertices to the division points closest to the opposite vertices.  Find the value of $n$ if the the area of the small square is exactly $\frac {1}{1985}$ .[/color][asy]
pair A=(0,1), B=(1,1), C=(1,0), D=origin;
draw(A--B--C--D--A--(1,1/6));
draw(C--(0,5/6)^^B--(1/6,0)^^D--(5/6,1));
pair point=( 0.5 , 0.5 );
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$1/n$", (11/12,1), N, fontsize(9));[/asy]
[b][u]Proof[/u].[/b] Let $M\in (AD)$ and $N\in (AB)$ so that $AM=NB=\frac 1n\ ,\ m\left(\widehat{ADN}\right)=\phi $ and the sidelength $l$ of the small square.

Observe that $\boxed{l=AM\cdot \cos\phi}\ (*)\ , \tan\phi =\frac {AN}{AD}=\frac {n-1}{n}$ . Therefore, $\cos\phi =\frac {1}{\sqrt{1+\tan^2\phi}}=$ $\frac {n}{\sqrt{n^2+(n-1)^2}}\stackrel{(*)}{\implies}$

$\frac {1}{\sqrt {1985}}=\frac 1n\cdot \frac {n}{\sqrt{n^2+(n-1)^2}}\iff$ $2n^2-2n+1=1985\iff$ $n(n-1)=992=32\cdot 31\iff$ $\boxed{n=32}$ .

[hide="Solution"]

A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices.  Find the value of $n$ if the the area of the small square is exactly 1/1985.
[asy]
pair A=(0,1), B=(1,1), C=(1,0), D=origin;
draw(A--B--C--D--A--(1,1/6));
draw(C--(0,5/6)^^B--(1/6,0)^^D--(5/6,1));
pair point=( 0.5 , 0.5 );
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$1/n$", (11/12,1), N, fontsize(9));[/asy]
[hide="Solution"]

We find after doing a lot of algebra that the two intersection points on one side of the square are $(\frac{n^2-2n+1}{2n^2-2n+1}, \frac{n^2-n}{2n^2-2n+1}), (\frac{n^2-n}{2n^2-2n+1}, \frac{n^2}{2n^2-2n+1})$.  The distance between these two points is going to be $\sqrt{\frac{(n-1)^2+n^2}{(2n^2-2n+1)^2}}$.  This is the side length of the square, so the area of the square is $\frac{(n-1)^2+n^2}{(2n^2-2n+1)^2}=\frac{1}{1985}$.  Therefore, $2n^2-2n+1=1985$.  This gives us after dividing by $2$ and factoring, $(n-32)(n+31)=0$, so $n=\boxed{32}$. [/hide]
 [/hide]

$\color{blue} \boxed{\textbf{Unique SOLUTION}}$

Consider the parallelogram $BLDK,$
Let $l=1$ be the length of the big square and $k$ be the length of small square, then
$[BLDK]=\frac{1}{n} \times l$
Again,
$[BLDK]=k \times \sqrt{l^2+(\frac{n-1}{n})^2}$
So,
$$\frac{1}{n} \times l=k \times \sqrt{l^2+\frac{n-1}{n}^2} \implies k^2=\frac{\frac{l^2}{n^2}}{l^2+(\frac{n-1}{n})^2}=\frac{\frac{1}{n^2}}{\frac{n^2+(n-1)^2}{n^2}}=\frac{1}{n^2+(n-1)^2}\implies 2n^2-2n+1=1985 \implies (n-32)(n+31)=0 \implies n=32 $$

one she joined early, also she is a classroom moderator so she answers a lot of questions from students

[quote=bbojy][quote=lian_the_noob12]$\color{red} \boxed{\textbf{Unique SOLUTION}}$

Consider the parallelogram $BLDK,$
Let $l=1$ be the length of the big square and $k$ be the length of small square, then
$[BLDK]=\frac{1}{n} \times l$
Again,
$[BLDK]=k \times \sqrt{l^2+(\frac{n-1}{n})^2}$
So,
$$\frac{1}{n} \times l=k \times \sqrt{l^2+\frac{n-1}{n}^2} \implies k^2=\frac{\frac{l^2}{n^2}}{l^2+(\frac{n-1}{n})^2}=\frac{\frac{1}{n^2}}{\frac{n^2+(n-1)^2}{n^2}}=\frac{1}{n^2+(n-1)^2}\implies 2n^2-2n+1=1985 \implies (n-32)(n+31)=0 \implies n=32 $$[/quote]

you revived this thread after 11 years :read:[/quote]

Hehe :)

I did not like this problem. Basically, we can do length chasing. And get that $n^2-n-992=0$, and hence, the positive solution is $\boxed{32}.$


http://www.youtube.com/watch?v=OirGSlHqONA

I thought it was some computer animated thing until the end.

[quote="anirudh"]that's not very hard to make. I know how to make one in Premiere Pro 2, Flash Pro, and Avid. Anyway, its good[/quote]

How can u make like a cardboard thing with Premiere Pro 2, or Flash Pro or Avid.

There's Flash Programming. Premiere has a lot of sophisticated features that are somewhat hidden. Elements can't do it though

I have one of those dragons in real, and I must say that this video is not made with some programs but just filmed normally.
You should also get one, plans can be found on the web :D

who came up with the original plans for the dragon? that takes a lot of ingenuity come up with that thing.

That is just plain incredible. :lol: Any idea what they are called or where I can get such a dragon?

No seriously, I've seen a lot of optical illusions in my life and most didn't trick me or I found logical. But this one is really awesome.

I just had to find and make one  :D . Here's the link, the pdf's in the middle of the page somwhere.
[url=http://www.grand-illusions.com/opticalillusions/dragon_illusion/]http://www.grand-illusions.com/opticalillusions/dragon_illusion/[/url]

this one rivals the dragon one in terms of how cool it is. when i finished this one i looked around, and it was the strangest thing. check it out.

http://www.youtube.com/watch?v=vzSRVgF501M&mode=related&search=


Show that a/(2a+b) + b/(2b+c) + c/(2c+a) < 1, iff a,b,c are positive real different nr.

\[2 \sum_{cyc}{\frac{a}{2a+b}}= \sum_{cyc}{(1-\frac{b}{2a+b})}\]

So we have to prove that \[\sum_{cyc}{\frac{b}{2a+b}}> 1\]

Trivial by Cauchy:

\[\sum_{cyc}{\frac{b}{2a+b}}\geq \frac{(a+b+c)^{2}}{(a+b+c)^{2}}=1\]

Equality occurs iff $\frac{1}{2a+b}=\frac{1}{2b+c}=\frac{1}{2c+a}$ iff $a=b=c$, so no equality case.
QED


Dar mosallase ABC,nimsaze zaviyeye B ,khatti ra ke az vasate AC va vasate ertefae marsoom bar AC migozarad dar noghteye M ghat mikonad ,N vasate Nimsaze zaviyeye B ast ,sabet konid ke nimsaze zaviyeye C ,nimsaze zaviyeye MCN niz hast... :!:

vasate zele $ AC$ ro $ F$ va vasate ertefae $ BH$ ro $ F'$ minamim chon $ N$ vasate  $ BD$ ast kafie ke neshoon bedim $ BM$ mozdavaje hamzaviyeye mianeye nazire $ BD$ to mosallase $ BCD$ ee baraye in kafie neshon bedim $ {DM \over BM} = {DC^2 \over  BC^2}$

be sadegi mishe did ke $ \vartriangle DMF   \sim \vartriangle NMF'$ hamchenin kheili tabloee ke $ NF' = {HD \over 2}$  banabarin

$ {DM \over BM} = {DM \over {DN + NM}} = {DF \over FH}$ amma

$ {DF = {b \over 2}  - {{bc} \over {a+c}} = {{b(a-c)} \over {2(a+c)}}}$ va hamchenin

$ FH = {{(a-c)(a+c)} \over 2b}$

pas $ {DF \over FH}  = {b^2 \over {(a+c)^2}} =  {DC^2 \over  BC^2}$


calculus
integration
real analysis
real analysis solved

Let  $ n$ be an odd positive integer and   $ T_n(z)=\displaystyle \frac{\left(z+\sqrt{z^2-1}\right)^n +  \left(z-\sqrt{z^2-1}\right)^n}{2}\; . $
[b]
If $ S_n(x):=\displaystyle \sum\limits_{k=0}^{n}T_{n}\left((2k-n)x\right){n
\choose k}^{2} $ , find  $\; \; \displaystyle \int\limits_{a}^{b} x^2S_n(x)dx \; . $ [/b]

$S_n(x)=0 \; \; ,\; \forall x \in {\mathbb R}. $ To see this , use the fact that $ T_m(-y) = (-1)^m T_m(y)$ as well as equality
\[\sum\limits_{k=0}^n A(k)=\sum\limits_{k=0}^n A(n-k) \; . \]

And what is then the purpose of the problem? Why do we have to compute the integral? I do not see probably obvious things.

[quote="flip2004"]$S_n(x)=0 $forall x in R [/quote]

So $\int_{a}^{b}x^2.S_n(x)dx=0$


last year it was a TI - 84 silver so itll be something different this year.
maybe TI 89 titanium?
:starwars:  oh man isnt this one cool

"We" implies everyone here made it to Nationals :(

[quote="biffanddoc"]Does anyone have an idea?

If it is a TI-84 plus silver I will die because I don't need a third one...
I hope its a TI-89 titanium[/quote]
We get calculators for GOING to nats?
AWESOME! I need at least another... I lost my TI-83 Plus I got for making it to state last year... -_-

I have a broken-down Ti-83 without the back.  I AM HOSTING A MATHJAM JOIN THE PARTY AT ZMASTR34!!!!!!!!!

You got a calculator for getting to state?!

My calculator is dysfunctional...it can do calculations, but at random moments (I had around 8 so far) the calculator's screen goes off as if it were turn off, then it doesn't turn on.  After you try for a while (mess with batteries or jam on repeatedly), it turns on but it reads

RAM Cleared
TI-83 plus

I stocked up on programs to help me for nationals, but I shouldn't bother anymore...

Don't count out TI84+ Silvers again just because they gave them out last year.  I've been to the last five(and had students go the 2 before that) and this is what they gave out:

2005 TI84+ Silver
2004 TI89
2003 TI83+ Silver
2002 TI83+ Silver
2001 TI89
2000 TI89
1999 TI92

Whatever they give out, we should all be grateful for it.  You can always sell it on ebay if you already have 2 of the same thing :)

Three 84 silvers... :rotfl: I'll take one if you don't want it ;)

Unless they give out antique calculators or something, I'm guaranteed a better calculator. I have a TI-82 :roll:

I don't mind trading in my 83 plus for a normal 83 even,

I have the worst calculator on my whole team, (including the individuals and those that didn't or didn't want to go)

[quote="kyyuanmathcount"]I'm assured a better calculator, having a TI-83 myself.

However, last time I tried to use a higher model, I failed in finding the "on" key :(.[/quote]

[hide="AHEM!!!"][size=200]HAA HAA HAA :rotfl: thats funny[/size][/hide]

[hide]I have done that also with a friends calculator  :| [/hide]

[quote="binonunquineist"]they gave out ti92s before!?!?!

wow. i already have 2 ti83+s, 3 ti84 silvers, and a ti89T, so i dont really need another one of those. chances are they wont give out ti92s, but maybe i'll be able to get another ti89T since i dont really need another ti83 or 84.[/quote]

Why would you want a TI-92? They're outdated and banned from most places.

They are cool and plenty of people wouldn't use a new 84 silver or something, they would continue to use their old calculator for competitions (except me who was lucky it never had problems during a competition)

BTW, do they hand out the calculators at nationals or are we supposed to get them at state?

Nationals

I think it's also after all the rounds

They usually give the coaches the calculators Saturday when the coaches pick up the test results.

I want the new one before the test.  I don't want my calculator to mess up and end up using a TI-30...those things feel weird to me now.

[quote="k8reindeer"]Is it true the calculators actually say 'mathcounts' on them in big red letters!?
I would be completely clueless about anything other than a TI83 if it weren't for my math class. everybody else seems to have 84s and everybody else seems to come to me with questions so I am actually ok in the 84.
by 'ok' I mean 'horrible', according to how some other people I know can use thier calculators.
I've only recently discovered about the scientific notation and converting fractions to decimals, things that I could do with my eyes shut on my old (non-graphing) calculator.
also, a plea for help:
when I turn on my calculator and try to graph some thing, there is a 99% chance it will tell me something I really do not want to hear, such as 'invalid dim' or something. this happens when I try such simple things as y=x. help!!!!! what is wrong with my calculator!!?!??!? most often it simply will not graph it, but there are times when it won't give me a table either!![/quote]

Hit the window button, and be sure that Xmin<Xmax and Ymin<Ymax. :) Last year the calculator was not emblazoned with MATHCOUNTS; they were stock TI-84+Silvers.

does your window look right

if it does, do you display a list that cannot be fully pictured in the graph...

I think there is black lettering in the back, but it isn't that bold, I am not sure...

[quote="samath"]Last year, we got the calculators as soon as we got to the hotel.  But I couldn't get the box open in time so I had to use my old one on the test  :D[/quote]

LOL. There's a reason why they call them blister packages... ;)

I don't understand TI-89s... If I got one I'd continue to use my old calculator.

Are you sure?

en.wikipedia.org/wiki/TI-89

It shows all the cool stuff it can do

Yeah, but after two years of using 82s, 83s, and 84s, I'm not used to the 89 layout.

I'm still using my TI-83+SE

The TI-84 has a nice layer of dust after sitting on the shelf with my 11 triangle trophies.

You have eleven?

I struck out on all countdown round trophies...

I have 10, but only because I have been in Mathcounts for 2 years officially, because in 6th Grade I didn't get selected officially, so I just took the contest for practice.

frost13 wrote

[quote]2005 TI84+ Silver
2004 TI89
2003 TI83+ Silver
2002 TI83+ Silver
2001 TI89
2000 TI89
1999 TI92 [/quote]

Correction to frost13's list--1999 was actually TI89.  (I was there with two of my students.  The TI-89 was very new at the time and I remember that they had a good time figuring out all the bells and whistles during a prolonged airport layover delay on the way home.)

According to a fellow coach and parent of a student who had gone to Nats in 1997 and 1998, it was a TI92 in both those years.  Because the 92 had a QWERTY keyboard and was not usable on SATs, APs, math contests, etc., TI came up with the 89.  (Ironically, my kids now use the Dvorak keyboard whenever they can!)

As for whether it says MATHCOUNTS on the case, I know the 89's given out in 1999 and 2000 said MATHCOUNTS in red letters on the plaster cover that slides onto the case.

I'm not sure if every year's calculators had covers with the MATHCOUNTS logo though--I seem to recall a recent year in which students reported that their calculators did not have the logo.

nice catch Sophia,

My first year, my top kid had been the year BEFORE and got the 92.  I had forgotten he hadn't received it my first year, but the year before.

Somceone else asked about the Red lettering on the calculators.  I have 5 clacs from Nats and I am pretty sure 3 of them have the lettering and the last 2 came is the hard plastic cases you would finon the shelves in stores with no special lettering.  Not sure why they stopped it?

frost

[quote="ehehheehee"]How do they decide what calculator do you get?

Does a CEO go: I had a bad day today, I will give them each TI 82s

Or: I am happy today, they will get TI-89s

Or is it like: We have a excess of TI-89s lets give them to the mathletes...[/quote]
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl: Maybe we should anonymously send the CEO a box of donuts the week before.

They usually give out a "new" model.
My guess is the TI-84+SE

Ask Mithsaprentice to guess. Mithsaprentice guessed the USAMO cutoff correctly.

I think Texas Instruments donates them as they are also a sponsor. I mean they give calculators to Texas state qualifiers and countdown round makers.

[quote="biffanddoc"][quote="kyyuanmathcount"]I would count on getting a better model, after all, isn't Lockheed Martin doing well right now?[/quote]
 :( The C-5 plane, which is a military cargo plane, and one of the biggest in the world, crashed this morning.  Supposedly it is unusable again.
[hide="sad part"]
It is owned by lockeed martin :(  :(
Were gonna get TI-30s[/hide][/quote]

It's probably the military's fault, not Lockeed Martin's. Also, they probably aren't going to get sued.

[quote="samath"][quote="Xantos C. Guin"]They usually give out a "new" model.
My guess is the TI-84+SE[/quote]

Well, depends on how you define "new."  That's what we got last year.[/quote]

That's why the word new is in quotes.

Texas Instruments Donates them, not lockheed martin...

It says here what we get:
[url]http://www.mathcounts.org/webarticles/articlefiles/536-LM_MC_Awards_06.pdf[/url]
Turns out we're getting TI 84+ Silver Edition calcs.

Last year it was TI-83 plus silver edition ones.

No, last year it was TI-84+SE's.

I know since I got one.

:stretcher:
We are getting TI-84+ Silver again (on mathcounts.org prizes section)

Darn, thats my third one  :(


superior algebra
superior algebra unsolved

Let R=Z/ (p^n)Z with p prime n integer

find all nilpotent elemts of R.

Describe the units of R

What's your idea on this¿ Try some cases, make a conjecture, (dis)prove it...


algebra
polynomial
Vieta
system of equations
algebra proposed

Here's an idea that popped to my mind when I saw this. Using Vieta, you can set up a system of equations, assuming that A,B,C,D, and E are roots.  Then you see that the system is not compatible. :) I'll think more on it.

EDIT: Nice problem by the way.
Masoud Zargar


[quote="Interval"]To get from his high school to his home, Jamal travels 5.0 miles east
and then 4.0 miles north. When Sheila goes to her home from the same
high school, she travels 8.0 miles east and 2.0 miles south. What is the |
measure of the shortest distance, to the nearest tenth of a mile, between|
Jamal’s home and Sheila’s home?[/quote]

[hide="solution"]Let the school be at the origin.  Jamal's house is at point (5,4) and Sheila's house is (8,-2).  Use the distance forumula to get $\sqrt{3^{2}+6^{2}}=\sqrt{45}$.[/hide]

[quote="Interval"]What words in the question indicated that the distance formula was needed here?[/quote]
Um, the word distance.

[quote="Interval"]Are there word problems that include the word distance without having nothing to do with the distance formula?[/quote]
Um, I can do that for anything really. Here's an example:
Bob and Joe are standing 8 miles apart. What is the distance between them?

Although you can use the distance formula for that to I suppose.


floor function
function
ceiling function
combinatorics unsolved
combinatorics

prove for every natural $n$:

$\sum_{k=0}^n \binom{n}{k}2^k \binom {n-k}{ \lfloor \frac{n-k}{2} \rfloor}=\binom {2n+1}{n}$

give me ONE natural n such that the identity is true....

Try everything U want.
I will give U 42123214798 try it .
I checked it was true

dude, there was no 2^k in there before! :mad:

A really beautiful problem!

[color=blue][b]Problem.[/b] Let $n$ be a nonnegative integer. Prove the identity

$\sum_{k=0}^n \binom{n}{k} 2^k \binom{n-k}{ \left\lfloor \frac{n-k}{2} \right\rfloor}=\binom{2n+1}{n}$.[/color]

Here is my [i]solution[/i]. I hope it is correct ;) .

We will color integers. Six colors are avaliable; due to lacking phantasy, I am denoting these colors by $0$, $1$, $2$, $3$, $0'$ and $1'$.

A [i]4-coloring[/i] of the set $X = \left\{1, 2, \ldots, n\right\}$ will mean a function which assigns to every integer from the set $X$ one of the colors $0$, $1$, $2$ and $3$. Such a 4-coloring will be called [i]nice[/i] if $x_0=x_1$ or $x_0=x_1+1$, where, for every color $i$, the number $x_i$ is the number of all integers from the set $X$ with color $i$. This condition is equivalent to $x_0=\left\lceil\frac{x_0+x_1}{2}\right\rceil$, or, equivalently, to $x_1=\left\lfloor\frac{x_0+x_1}{2}\right\rfloor$; in other words, this is equivalent to requiring that, among the $x_0+x_1$ integers from the set $X$ with color $0$ or $1$, either (if $x_0+x_1$ is even) exactly $\frac{x_0+x_1}{2}$ have color $0$ and $\frac{x_0+x_1}{2}$ have color $1$, or (if $x_0+x_1$ is odd) exactly $\frac{x_0+x_1+1}{2}$ have color $0$ and $\frac{x_0+x_1-1}{2}$ have color $1$.

On the other hand, a [i]2-coloring[/i] of the set $Y = \left\{1, 2, \ldots, 2n + 1\right\}$ will mean a function which assigns to every integer from the set $Y$ one of the colors $0'$ and $1'$. Such a 2-coloring will be called [i]nice[/i] if exactly $n$ integers get the color $1'$.

[color=blue][b]Lemma 1.[/b] There are exactly $\sum_{k=0}^n \binom{n}{k} 2^k \binom{n-k}{ \left\lfloor \frac{n-k}{2} \right\rfloor}$ nice 4-colorings of the set $X = \left\{1, 2, \ldots, n\right\}$.[/color]

[i]Proof of Lemma 1.[/i] How to construct a nice 4-coloring of the set $X = \left\{1, 2, \ldots, n\right\}$ ? Here is one way to do this: First we choose some integer $k$ with $0\leq k\leq n$, and select $k$ integers from the set $X$ which will get one of the colors $2$ and $3$. There are, of course, $\binom{n}{k}$ possibilities to do that. Then, color these $k$ integers with the colors $2$ and $3$; for each of these $k$ integers, this makes $2$ possibilities, so that, altogether, we have $2^k$ possibilities to color these $k$ integers. Now, the remaining $n - k$ integers from the set $X$ have to be colored with one of the colors $0$ and $1$, and we have to fulfill the additional condition that $x_1=\left\lfloor\frac{x_0+x_1}{2}\right\rfloor=\left\lfloor\frac{n-k}{2}\right\rfloor$ (note that $x_0+x_1=n-k$, since $x_0+x_1$ is the number of all integers from the set $X$ which will get one of the colors $0$ and $1$), i. e. that exactly $\left\lfloor\frac{n-k}{2}\right\rfloor$ of them will get the color $1$. So it remains to select $\left\lfloor\frac{n-k}{2}\right\rfloor$ out of $n - k$ given integers. Of course, there are $\binom{n-k}{ \left\lfloor \frac{n-k}{2} \right\rfloor}$ ways to do that. Once all these selections are done, a nice 4-coloring of the set $X$ is uniquely specified.

Hence, altogether, there are exactly $\sum_{k=0}^n \binom{n}{k} 2^k \binom {n-k}{ \left\lfloor \frac{n-k}{2} \right\rfloor}$ nice 4-colorings of the set $X$, and Lemma 1 is proven.

[color=blue][b]Lemma 2.[/b] There are exactly $\binom{2n+1}{n}$ nice 2-colorings of the set $Y = \left\{1, 2, \ldots, 2n + 1\right\}$.[/color]

[i]Proof of Lemma 2.[/i] A nice 2-coloring of the set $Y = \left\{1, 2, \ldots, 2n + 1\right\}$ is a 2-coloring which satisfies the condition that exactly $n$ integers from the set $Y$ get the color $1'$. Thus, in order to construct a nice 2-coloring of the set $Y$, it is enough to choose $n$ integers from the set $Y = \left\{1, 2, \ldots, 2n + 1\right\}$. Since there are $\binom{2n+1}{n}$ possibilities to do this, there must be exactly $\binom{2n+1}{n}$ nice 2-colorings of the set $Y$, and Lemma 2 is proven.

[color=blue][b]Lemma 3.[/b] There exists a bijection between the set of nice 4-colorings of the set $X = \left\{1, 2, \ldots, n\right\}$ and the set of nice 2-colorings of the set $Y = \left\{1, 2, \ldots, 2n + 1\right\}$.[/color]

[i]Proof of Lemma 3.[/i] Consider a nice 4-coloring of the set $X = \left\{1, 2, \ldots, n\right\}$. Let $x_0$, $x_1$, $x_2$, $x_3$ be the numbers of all integers from the set $X$ with colors $0$, $1$, $2$, $3$, respectively. Clearly, $x_0+x_1+x_2+x_3=n$.

Now, let's color the set $Y$ as follows:
For any integer $k$ with $1\leq k\leq n$,
- the integer $k$ will get the color $0'$ if it has the color $0$ or $2$ in the set $X$, and it will get the color $1'$ if it has the color $1$ or $3$ in the set $X$;
- the integer $n + k$ will get the color $0'$ if the integer $k$ has the color $0$ or $3$ in the set $X$, and it will get the color $1'$ if the integer $k$ has the color $1$ or $2$ in the set $X$.
Finally, the integer $2n + 1$ will get the color $0'$ if $x_0+x_1$ is even, and the color $1'$ if $x_0+x_1$ is odd.

Thus, we have obtained a 2-coloring of the set $Y$. Why is this 2-coloring nice?

Well, let's count how much integers from $Y$ have got the color $1'$. Remember that, for any integer $k$ with $1\leq k\leq n$, the integer $k$ gets the color $1'$ if it has the color $1$ or $3$ in the set $X$, and the integer $n + k$ gets the color $1'$ if the integer $k$ has the color $1$ or $2$ in the set $X$. So, altogether, there will be $\left(x_1+x_3\right)+\left(x_1+x_2\right)=2x_1+x_2+x_3$ integers from the set $\left\{1, 2, \ldots, 2n\right\}$ colored with the color $1'$. Additionally, the number $2n + 1$ gets the color $1'$ if and only if $x_0+x_1$ is odd. So, altogether, the number of integers from the set $Y = \left\{1, 2, \ldots, 2n + 1\right\}$ with the color $1'$ is $2x_1+x_2+x_3$ if $x_0+x_1$ is even, and $2x_1+x_2+x_3+1$ if $x_0+x_1$ is odd. Now, since our 4-coloring of the set $X$ was nice, $x_1=\left\lfloor\frac{x_0+x_1}{2}\right\rfloor$, so when $x_0+x_1$ is even, we have $x_1=\frac{x_0+x_1}{2}$, and thus the number of all integers from the set $Y$ with color $1'$ is $2x_1+x_2+x_3=2\frac{x_0+x_1}{2}+x_2+x_3=x_0+x_1+x_2+x_3=n$ in this case. When $x_0+x_1$ is odd, we have $x_1=\left\lfloor\frac{x_0+x_1}{2}\right\rfloor=\frac{x_0+x_1-1}{2}$, and thus the number of all integers from the set $Y$ with color $1'$ is $2x_1+x_2+x_3+1=2\frac{x_0+x_1-1}{2}+x_2+x_3+1=x_0+x_1+x_2+x_3=n$ in this case. Hence, in every case, the number of all integers from the set $Y$ with color $1'$ is $n$, and thus the 2-coloring of the set $Y$ is nice.

So we have found a way to define, for every nice 4-coloring of the set $X$, a corresponding nice 2-coloring of the set $Y$. Actually, this will be our bijection. But in order to show that it is a bijection, we have to prove that, being given any nice 2-coloring of the set $Y$, we can uniquely reconstruct the corresponding nice 4-coloring of the set $X$.

This can be done as follows: Consider a nice 2-coloring of the set $Y = \left\{1, 2, \ldots, 2n + 1\right\}$. Now let's construct a 4-coloring of the set $X = \left\{1, 2, \ldots, n\right\}$. This will be done by reverting the procedure used above to obtain a nice 2-coloring of $Y$ from a nice 4-coloring of $X$:
An integer $k$ with $1\leq k\leq n$ will get
- the color $0$ if the integers $k$ and $n + k$ both have color $0'$ in the set $Y$;
- the color $1$ if the integers $k$ and $n + k$ both have color $1'$ in the set $Y$;
- the color $2$ if the integer $k$ has color $0'$ in the set $Y$ and the integer $n + k$ has color $1'$ in the set $Y$;
- the color $3$ if the integer $k$ has color $1'$ in the set $Y$ and the integer $n + k$ has color $0'$ in the set $Y$.

The proof that this 4-coloring of the set $X$ is nice is more or less the reverse argument of the above proof that the 2-coloring of $Y$ is nice. I won't show it now, since I am tired and you, dear reader, probably too. Anyway, hope that it is clear now that we have a bijection, and Lemma 3 must hold.

So, now, by Lemma 3, there exists a bijection between the set of nice 4-colorings of the set $X = \left\{1, 2, \ldots, n\right\}$ and the set of nice 2-colorings of the set $Y = \left\{1, 2, \ldots, 2n + 1\right\}$. But, after Lemma 1, the former set has $\sum_{k=0}^n \binom{n}{k} 2^k \binom{n-k}{ \left\lfloor \frac{n-k}{2} \right\rfloor}$ elements, and by Lemma 2, the latter set has $\binom{2n+1}{n}$ elements. Thus, $\sum_{k=0}^n \binom{n}{k} 2^k \binom{n-k}{ \left\lfloor \frac{n-k}{2} \right\rfloor}=\binom{2n+1}{n}$, and the problem is solved.

  darij

Darij, nice idea  :)

Actually this problem appeared in "102 Combinatorial Problems" by Titu Andreescu with a beautiful proof. I just give a sketch. There are $2n+1$ people - $n$ girls, $n$ boys and a teacher  :D We have $n$ tickets and want to give out all - right side is just giving out randomly these tickets.

To get left side we group in pairs, each girl with exactly one boy.  Left side has the following interpretation then. We choose $k$ pairs and give one ticket to one of the person from the couple ($2^k \binom{n}{k}$ such possibilities). Then we choose $\lfloor \frac{n-k}{2}\rfloor$ pairs and give each pair two tickets. If $n-k$ is odd we give the last ticket to the teacher. Hence the result  :)


let $ ABC$ be a triangle ,prove that;
$ r_{a}\plus{}r_{b}\plus{}r_{c}\equal{}r\plus{}4R$

just represent $ r_a, r_b, r_c, R$ equation of $ r$ and the sides of the triangle.

$ \mbox{If: } p \equal{} \frac {a \plus{} b \plus{} c}{2}\Rightarrow r_a \plus{} r_b \plus{} r_c \minus{} r \equal{} \frac {S}{p \minus{} a} \plus{} \frac {S}{p \minus{} b} \plus{} \frac {S}{p \minus{} c} \minus{} \frac {S}{p} \equal{} \\
\equal{} S.\left(\frac {1}{p \minus{} a} \plus{} \frac {1}{p \minus{} b} \plus{} \frac {1}{p \minus{} c} \minus{} \frac {1}{p}\right) \equal{} S.\left(\frac {2p \minus{} a \minus{} b}{(p \minus{} a)(p \minus{} b)} \plus{} \frac {p \minus{} (p \minus{} c)}{p(p \minus{} c)}\right) \equal{}\\
\equal{}S.\left(\frac{c}{(p\minus{}a)(p\minus{}b)}\plus{}\frac{c}{p(p\minus{}c)}\right)\equal{}Sc.\frac{p(p\minus{}c)\plus{}(p\minus{}a)(p\minus{}b)}{p(p\minus{}a)(p\minus{}b)(p\minus{}c)}\equal{}\\
\equal{}Sc.\frac{p(2p\minus{}a\minus{}b\minus{}c)\plus{}ab}{S^2}\equal{}\frac{abc}{S}\equal{}4R.$


[quote="digitalboy1"]Let $A_1A_2A_3$ be a triangle and $M$ be an interior point. The straight line $MA_1,MA_2,MA_3$ intersect the opposite sides at the points $B_1,B_2,B_3$ respectively.
Prove that if the areas of triangle $A_2B_1M,A_3B_2M,A_1B_3M$ are equal, then $M$ coincides with the centroid of triangle $A_1A_2A_3$[/quote]

[hide]
Let $[A_2B_1M]=[A_3B_2M]=[A_1B_3M]=a$  and

let $[B_1A_3M]=x$, $[B_2A_1M]=y$, $[B_3A_2M]=z$. Then

$\frac{a+x}{a+y}=\frac{z}{a}$ --------- (1)

$\frac{a+y}{a+z}=\frac{x}{a}$ --------- (2)

$\frac{a+z}{a+x}=\frac{y}{a}$ --------- (3)

From (1), (2) and (3), we get $xyz=a^3$ and $xy+yz+zx=3a^2$  The equality condition of AM-GM gives us $x=y=z=a$. This implies that $B_1,B_2, B_3$ are midpoints.

[/hide]

[quote="fuzzylogic"][quote="shobber"]what is the equality condition of AM-GM?[/quote]

AM>=GM. If AM=GM, then all the elements are equal.[/quote]
o i see, thank you!


inequalities
function
inequalities proposed

Let $ x\plus{}y\equal{}s,\ xy\equal{}t$, then $ x>0$ and $ y>0\Longleftrightarrow s>0$ and $ t>0$ and $ s^2\geq 4t$.
Now L.H.S. of the given inequality can be written as .....

I have no time now, because on boarding.

Here is my solution:

Let $ {A = x^4 + y^4 + x^3y - 5x^3 - 5y^3 + 15x^2 + xy^3 + x^2y^2 - 5x^2y + 15y^2 - 5xy^2 + 15xy - x - y + 3}$.
$ \bigstar$If $ x\neq y$, we have:
\begin{align*} (x - y)\times A & = \color{blue}x^5\color{black} - \color{red}y^5\color{black} - \color{blue}5x^4\color{black} + \color{red}5y^4\color{black} + \color{blue}15x^3\color{black} - \color{red}15y^3\color{black} - \color{blue}x^2\color{black} + \color{red}y^2\color{black} + \color{blue}3x\color{black} - \color{red}3y\color{black} \\
& = \left(\color{blue}x^5 - 5x^4 + 15x^3 - x^2 + 3x\color{black}\right) - \left(\color{red}y^5 - 5y^4 + 15y^3 - y^2 + 3y\color{black}\right) \\
& = f(\color{blue}x\color{black}) - f(\color{red}y\color{black})\color{magenta}\text{ where }f(t) = t^5 - 5t^4 + 15t^3 - t^2 + 3t \end{align*}
Since $ f'(x) = 5x^4 - 20x^3 + 45x^2 - 2x + 3 = \color{green}\frac {5}{4}\color{magenta}(2x^2 - 4x + 1)^2\color{black} + \color{green}\frac {4}{5}\color{magenta}(5x + 1)^2\color{black} + \color{green}\frac {19}{20}\color{black} > 0$, $ f(x)$ is an increasing function, so $ A = \frac {f(x) - f(y)}{x - y} > 0$.
$ \bigstar$If $ x = y$, $ A = 5x^4 - 20x^3 + 45x^2 - 2x + 3$, and it's clearly positive with the transfomation above.
:)

[quote="can_hang2007"][quote="vanhoadh"]$ \bigstar$[u][b]Problem:[/b][/u] Prove that $ {x^4 \plus{} y^4 \plus{} x^3y \minus{} 5x^3 \minus{} 5y^3 \plus{} 15x^2 \plus{} xy^3 \plus{} x^2y^2 \minus{} 5x^2y \plus{} 15y^2 \minus{} 5xy^2 \plus{} 15xy \minus{} x \minus{} y \plus{} 3 > 0}$ for any $ x,y\in\mathbb{R}$.
________
At first i must use the following transformation to solve it:
[hide="Equation"]
$ x^4 \plus{} y^4 \plus{} x^3y \minus{} 5x^3 \minus{} 5y^3 \plus{} 15x^2 \plus{} xy^3 \plus{} x^2y^2 \minus{} 5x^2y \plus{} 15y^2 \minus{} 5xy^2 \plus{} 15xy \minus{} x \minus{} y \plus{} 3$
$ \equal{} \frac {\left(10x^2 \plus{} 5xy \minus{} 25x \plus{} 2\right)^2}{100} \plus{} \frac {3\left(10y^3 \minus{} 50y^2 \plus{} 15xy^2 \minus{} 50xy \plus{} 167x \plus{} 148y\right)^2}{20\left(5\left(3y \minus{} 5\right)^2 \plus{} 376\right)} \plus{}$
$ { \plus{} \frac {3\left(15\left(209y \minus{} 10\right)^2 \plus{} 209\left(20y^2 \minus{} 25y \plus{} 2\right)^2 \plus{} 276888\right)\left(5\left(2y \minus{} 5\right)^2 \plus{} 171\right)}{167200\left(5\left(3y \minus{} 5\right)^2 \plus{} 376\right)} > 0}$
[/hide]But then I found a very nice solution by polynom...  :P[/quote]
I think it is better for us to rewrite it as
\[ (x^4 \plus{} y^4 \plus{} x^3y \plus{} xy^3 \plus{} x^2y^2) \minus{} 5(x^3 \plus{} y^3 \plus{} x^2y \plus{} xy^2) \plus{} 15(x^2 \plus{} y^2 \plus{} xy) \minus{} x \minus{} y \plus{} 3 > 0
\]

\[ \Leftrightarrow (x \plus{} y)^2(x^2 \plus{} y^2 \minus{} xy) \plus{} x^2y^2 \minus{} 5(x \plus{} y)(x^2 \plus{} y^2) \plus{} 15(x \plus{} y)^2 \minus{} 15xy \minus{} (x \plus{} y) \plus{} 3 > 0
\]
Denote $ p \equal{} x \plus{} y,r \equal{} xy$, then it becomes
\[ p^2 (p^2 \minus{} 3r) \plus{} r^2 \minus{} 5p(p^2 \minus{} 2r) \plus{} 15p^2 \minus{} 15r \minus{} p \plus{} 3 > 0
\]

\[ \Leftrightarrow f(r) \equal{} r^2 \minus{} (3p^2 \minus{} 10p \plus{} 15)r \plus{} p^4 \minus{} 5p^3 \plus{} 15p^2 \minus{} p \plus{} 3 > 0
\]
Since $ \frac {3p^2 \minus{} 10p \plus{} 15}{2} \ge \frac {p^2}{4}$, then it is easy to check that $ f(r)$ is decreasing for all $ r \le \frac {p^2}{4}$, therefore
\[ f(r) \ge f\left( \frac {p^2}{4}\right)
\]
Or on the other hand, if we denote the left hand side of the original inequality as $ f(x,y)$, then
\[ f(x,y) \ge f\left( \frac {x \plus{} y}{2},\frac {x \plus{} y}{2}\right)
\]
It suffices to prove that
\[ f(t,t) > 0
\]
where $ t \equal{} \frac {x \plus{} y}{2}$.
\[ \Leftrightarrow 5t^4 \minus{} 20t^3 \plus{} 45t^2 \minus{} 2t \plus{} 3 > 0
\]
The rest is easy. :)[/quote]
I edited my post.  :oops:


The arithmetic mean of 16 and a number is equal to the geometric mean of the two numbers. What is the number?

Call that number x. Now, we have the equation:
(16+x)/2 = sqrt(16x)
Just some more manipulation:
16+x = 2sqrt(16x)
Square both sides:
256+32x+x^2 = 4*16x = 64x
More manipulation:
x^2-32x+256=0
Factoring:
(x-16)^2=0
Therefore, the number is 16.


LaTeX Harvard college AMC USA(J)MO USAMO

Congratulations to all the team members!

Taking a look at his signature, and then checking the list, one could deduce that he probably is in :) Congratulations!

dude, jason, you're so diligent about making sure mathlinks knows about us!  this year, i'm not going to be a lamer and miss the midnight ihop runs.

- kenan

Not so diligent - I couldn't bear to post the B problems from this year.  So much typing and $\LaTeX$...

Btw. Djole, how did your national competition go? :D

@[b]Thaakisfox[/b]: I don't want to get too much off topic, so I'll just state that I did pretty well (:first:). I'll post the problems in a new topic.

Btw, I hope someone will post some problems from the USAPhO  :roll: And will the US participate at the IPhO this year?

The first section of the USAPhO was posted a few weeks ago.  The second section has long problems.  The AAPT requests that we not share the problems from camp.

I believe the AAPT is deciding whether or not the US participates in the next couple of days.

[quote="Kent Merryfield"]Four Californians. The one from my part of the state (from Harvard-Westlake) is someone I don't know. As for the others: Henry Tung had the highest AMC/AME index in the state (138/12=258).[/quote]wow, henry's a beast.  i wonder how usamo is going for him.

[quote="inteluser"]I believe the AAPT is deciding whether or not the US participates in the next couple of days.[/quote]if we don't, i'm going to be _pissed_.  safety really is not an issue, and i have many philosophical objections to missing an ipho for PR/political reasons.  also, stephen hawking is a guest speaker this year (he confirmed his appearance like two days ago), which is quite compelling.


function
calculus
derivative
vector
limit
quadratics
linear algebra

Let A be an n by n matrix. Define $ f: \mathbb{R}^n \to \mathbb{R}$ by $ f(\mathbf{x}) \equal{} A\mathbf{x} \cdot \mathbf{x} \equal{} \mathbf{x}^T \cdot A\mathbf{x}$.

Show that f is differentiable and
$ Df(\mathbf{a}) \mathbf{h} \equal{} A \mathbf{a} \cdot \mathbf{h} \plus{} A \mathbf{h} \cdot \mathbf{a}$

So, do I want to prove that f has continuous partial derivatives or do I want to go back to the definition of differentiability or what?

Here, you want the definition What is that definition in a general normed vector space? The derivative $ Df(p)$ is a linear operator such that $ \lim_{x\to 0}\frac{|f(x\plus{}p)\minus{}f(p)\minus{}Df(p)(x)|}{|x|}\equal{}0$, or equivalently $ f(x\plus{}p)\equal{}f(p)\plus{}Df(p)(x)\plus{}o(|x|)$.

You have a candidate derivative. Calculate the remainder, and estimate its size.

Is there a trick to estimating

$ \frac{A\mathbf{h} \cdot \mathbf{h}}{|\mathbf{h}|}$

?

[quote="jmerry"]The numerator is a quadratic form- a linear combination of terms of the form $ x_ix_j$. That's comparable to a constant multiple of $ |h|^2$.[/quote]

OK. Let $ M$ be the maximum of the absolute values of the coefficients of the quadratic form. Then Cauchy-Schwarz says that $ \mathbf{h} \cdot \mathbf{h'} \leq |h|^2$ where h' is a permutation of h. Then its obvious that the limit is 0. Thanks.

Wait...is it Cauchy-Schwarz? I didn't think this through completely in my last post.

$ \left|\sum_{k\leq j}^n h_k h_j c_{kj} \right| \leq M \left|\sum_{k\leq j}^n h_k h_j\right| \leq ?$

You definitely need a larger $ M$- this is about matrix norms. If you're using the Euclidean norm, the corresponding matrix norm is most natural: $ \|A\|_2 \equal{} \max_{v\neq 0}\frac {|Av|_2}{|v|_2}$ is the largest singular value, or equivalently the square root of the largest eigenvalue of $ A^TA$. This is often hard to measure exactly, so we can instead use the trace of $ A^TA$, which is the sum of the squares of the entries of $ A$.

Equality here is possible-let $ A$ be the $ n\times n$ matrix of all ones, and $ h$ be the vector of all ones.


So what did you guys accomplish this summer?

Got a bit of math done... (not much)
some science

summer work  :roll:

and more things.. too tired to list and type

I was REALLY lazy and ended up only doing a little of what I had planned.

Eh I guess summer isn't really over yet, but school starts in about a week and a half for us.

LOL I typed up over 200 pages of math into latex form. That's what I've been doing all summer.

Did a LOT of math.  Also some other things...(ie continued my regular violin/karate study)

In the summer camp, there were like three IMO bronze medallists and still I was the best in logic. Yo!
Oh and also opened a band with this other guy from the camp. Sum total of our musical talents is he can sing a bit lol.

some AP bio work...a little of AMC...and ah, a lot of computer time.

You're mistaken. Even I didn't believe it at first, but some people actually find studies fun. Eugh!

[quote="Ignite168"]this topic makes me sad  :(   :noo:

what have these parents done to their children to ruin all of the fun that's supposed to be summer?[/quote]

I find everything I mentioned fun.  If I were doing what my parents want, I would be learning to drive, and then driving them around.  :D

[quote="Ignite168"]this topic makes me sad  :(   :noo:

what have these parents done to their children to ruin all of the fun that's supposed to be summer?[/quote]

You've got it all wrong. We work 24/7 on math because we are rebelling against our families who want us to have active social lives  :wink:

In all seriousness, none of us would be devoting so much of our time to something we didn't enjoy.

[quote="Prime factorization"]We work 24/7 on math because we are rebelling against our families who want us to have active social lives  :wink:[/quote]

Hehe, well actually one of us is working on math because our parents [i]won't[/i] let them have a social life... :dry:

[quote="Nerd_of_the_Ages"]Got legend of zelda twilight princess and played it for 30 hours in 10 days...

Found a new interest in number theory...

Started kong fu classes...

Staying up until 12:00 AM every night doing math...

Keeping my regular piano classes...

Not much really  :D[/quote]

I've already played Twilight Princess for over 100 hours. And don't even ask me about Resident Evil 4, I've beaten that at least 40 times  :D

well, youve had it longer than me, i only got it like 15 days ago. I also beat super paper mario in a week.

Yeah, I got mine a few months ago. Can't wait for Metroid Prime, Smash Bros., Super Mario Galaxy, but they're all coming out when I start college and I'll practically have every second of my life busy for school  :(

Guitar Hero 2 and Ninja Gaiden Black.  'Nuff said.

Well, I don't have a PS2 or PS3, so I've never played GH before, but I'm going to get GH3 for the Wii when it comes out.

[quote="azhang"]i leveld to lvl 33 hunter on WoW?[/quote]

Ew

a) you're a hunter and b) you're level 33

So what? I just started this summer. Anyways, it's on a private server with burning crusade so hah. Free WoW

BTW ALLIANCE PWNS U

[quote="azhang"]private server[/quote]

Ew

Learned Game Theory, Started calculus, and thinking of starting physics sometime soon.


analytic geometry
probability
graphing lines
slope
number theory
relatively prime

A lattice point is a point whose coordinates are both integers. Suppose we draw a line segment from (0, 2004) to (a, b) where a and b are integers each randomly selected from 1, 2, 3, . . . , 100. What is the probability that the line segment contains an even number of lattice points?

I see a way to the solution, but it requires a bit of brute force. How can we make it more elegant? Am I close to the right track?
Hint:
[hide]The segment connecting (0, 2004) and (a, b) has a (negative) slope of (b-2004)/a.
If a lattice point (x, y) is inside that segment,
0<x<a , b<y<2004, and (2004-b)/a = (2004-y)x
If (2004-b) and a are relatively prime, there will be no lattice point inside the segment.
How many lattice points could the segment contain if a is a power of 2?
How many if a is an odd prime number?
What if it is a power of an odd prime number?
What if it's an odd number with two distinct prime factors?
Could it be an odd number with three distinct prime factors?
What if it has odd factors and power of two factors?[/hide]


how would you relate the difficulties of the 2007, 2008, and 2009 and f=ma exams to each other?

and anyone know what the required scores were for each to continue to the next round?

Except for #26-38 on 2007, I'd say that each year gets a little bit more difficult.
Wiki says cutoff for 2008 was 16.

I believe 2009 was 14.75, or somewhere near there.

on the 2007 f=ma are the first 25 questions the equivalent of the 75 min. 25 question prelim exam now?
---people who actually took it weren't given only 75 min. for all 38 questions...just the first 25 right?

anyone else know or think they know previous cutoff scores or approximates of them?


[quote="mustafa"][hide="1"]5:50. At 5:50, each mink doubled, so the total minks also doubled.
[/hide]
[hide="2"]7:50 Starting with 2 minks cut out one step in the process, when 1 mink doubled, so it would take 10 less minutes[/hide][/quote]

for ur number 1 I think u had a typo, look at it.

[quote="Ihatepie"]There is a new type of mink on the market that reproduces itself in 10 minutes. Each new mink is exactly like the original.

1.If you were to put one mink in the cage at 6 p.m. and 2 hours later the cage was full, when was the cage half full?

2. If you put 2 minks in a cage at 6 p.m. when would the cage be full?[/quote]

Minks are vertebrates; they reproduce sexually, even the supposed new ones on the market. Interestingly, I have never seen documentation of the new minks... Are you sure that is scientifically accurate?

:harhar:

Other than that:

[hide]1. Ten minutes prior, a.k.a. 7:50 p.m.
2. 7:50 p.m.[/hide]

i dunno if it is scientific but ur answers r right


For the ACT reading and writing sections, which do you guys prefer?  :)
1.  first read and try to comprehend the passage, then answer the questions.  or
2.  look at the questions and then find them in the passage  (without reading)

read first, then see the questions. If you look for answers, you won't know much details or flow of the passage probably.

1) definitely. Context is so important.

You can read the questions quickly before you read, if only to help subliminally.

I'll third what the other two have said.

Yea read first and then answer for all the passages except Science. Science questions are mostly, if not all, detail questions, and it's really easy to skim around and find answers.

I usually ignore the context questions and look at the details, so if I come across them, I can make a little note.


[hide]Or treat it as a quadratic in $3^{x}$.  Let $X = 3^{x}$ and you have the inequality $X^{2}-12X+27 > 0$[/hide]

Full solution:
Let $3^{x}= a$.
Then we can rewrite the desired as
$a^{2}+27 > 12a$
$a^{2}-12a+27 > 0$
$(a-9)(a-3) > 0$
Now we do something called sign analysis.
------------0----3------9-----------
(a-9)     Neg       |   Neg  |  Neg          | Pos
(a-3)     Neg       |   Neg  |  Pos          | Pos
Product  Pos       |   Pos   |  Neg         |Pos

So this is true for all $a \in (-\infty, 3) \cup (9, \infty)$
We don't include three or nine in the interval because they make it equal to zero, not greater than zero.
So we have two cases.
$3^{x}> 9$ or $3^{x}< 3$
Taking both $\log_{3}$, we have $\boxed{x > 2 \vee x < 1}$


probability
function
advanced fields
advanced fields unsolved

consider a random walk on $\mathbb{Z}^d$, such that a particle can goes to each of its $2d$ nearest neighbours with a probability $\frac{1}{2n}$. What is the probability that a particle starting from the origin comes back after exactly $n$ jumps ?

$\frac{1}{2d}$ instead of $\frac{1}{2n}$

This exercise was my exam, I'm searching my paper ...

Sorry my exam was not on this question but on a random walk on $\mathbb{Z}^d$ (questions about harmonic functions)

http://arxiv.org/PS_cache/math/pdf/0001/0001057.pdf
You have the result for dimension 1, 2 and 3 starting page 93.
(It's the subject of the "groupe de lecture" of a friend)

ho ! thank you very much !
I'm also studying random walks :)


$V_{a}$,$V_{b}$ and $V_{c}$ are medians of an $ABC$ triangle. Prove that $\frac{3}{4}(a+b+c)<V_{a}+V_{b}+V_{c}<a+b+c$

In the triangle $ABC$, first prolong $V_{a}$, to $A_{1}$, such that $AA_{1}=2V_{a}$, then $CA_{1}=c$, in the triangle $AA_{1}C$, we have $2V_{a}<b+c$, similarly: $2V_{b}<a+c$ and $2V_{c}<a+b$, then $V_{a}+V_{b}+V_{c}<c+b+a$, other side let G the centroid and $M_{a}$ the midpoint of $BC$, let $M_{a}A_{2}$, a parallel to $GC$, and $A_{2}$ is in the line $AC$, we have for the relation 2:1 of the centroid, $M_{a}A_{2}=V_{c}$, then in the triangle $AM_{a}A_{2}$ we have the relation $V_{a}+V_{c}>\frac{3}{2}b$, similarly $V_{b}+V_{c}>\frac{3}{2}a$ $V_{a}+V_{b}>\frac{3}{2}c$, and we are done


trigonometry quadratics function calculus derivative

[quote="Chocoboom"]Determine the minimum value of $f(x)$ where
$f(x) = (3\sin x-4\cos x-10)(3\sin x+4\cos x-10)$.[/quote]
[hide]
let $A=3\sin x-10$ then,
$f(x)=(A-4\cos x)(A+4\cos x)$.
or
$f(x)=A^{2}-16\cos^{2}x$.
now sub in $3\sin x-10$ for $A$ and converting $\cos^{2}x$ into
$1-\sin^{2}x$, we get:
$25\sin^{2}x-60 \sin x+84=0$, which is a quadratice equation.
the min. of this function occurs at $\sin x=\frac{6}{5}$.
however, sin can never be greater than 1.
therefore, the min. occurs when $\sin x =1$...
answer: $49$....
is this right?
[/hide]

yes its correct.. just check it w/ a graph

i took the derivative of it but yea both ways work

[quote="jhcreinhardt"]yea good problem.  im sure many people would ignore the trig part and choose 48 as the minimum.[/quote]

What do you mean by ignore the trig part?

Edit: O never mind, you meant the thing with sin(x)>1

Ok Got ya


calculus
function
derivative
trigonometry
calculus computations

I am having difficulty understanding why general substitution doesn't work in composite function derivatives, e.g. the chain rule. for instance,

$y=\sin{x}$ => $y^{`}=\cos{x}$  (1), and by the chain rule,
$y=\sin{2x}$ => $y^{`}=2\cos{2x}$    (2)

But substituting $u=2x$ gives $y=\sin{u}$ => $y^{`}=2\cos{u}$, which makes (2) a contradiction of (1). Help please?

Both formulas $y'=2\cos 2x$ and $y'=\cos u$ are correct. However, the derivatives that appear in the formulas are different. In the first one $y'$ is the derivative of $y$ with respect to $x$, also denoted $\frac{dy}{dx}$. In the second $y'$ is the derivative of $y$ with respect to $u$, also denoted $\frac{dy}{du}$. The chain rule says: \[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\] In this case $\frac{du}{dx}=2$.


geometry
geometric transformation
homothety
ratio
inequalities
geometry unsolved

Consider triangle $ ABC$, and three squares $ BCDE$, $ CAFG$ and $ ABHI$ constructed on its sides, outside the triangle. Let $ XYZ$ the triangle en-closed by the lines $ DE$, $ HI$ and $ FG$
prove that $ S_{XYZ}\geq (13\plus{}4\sqrt{3})S_{ABC}$

Denote $X \equiv GF \cap IH,$ $Y \equiv IH \cap ED,$ $Z \equiv ED \cap GF.$ It's well known that $AX,BY,CZ$ are the symmedians of $\triangle ABC$ $\Longrightarrow$ $\triangle ABC$ and $\triangle XYZ$ are homothetic with homothetic center the symmedian point $K$ of $\triangle ABC.$ Hence the homothety coefficent $k$ equals to the ratio of the distances from $K$ to $YZ$ and $BC.$ Thus if $d_a$ denotes the distance from $K$ to $BC,$ we have

$k=\frac{d_a+a}{d_a}=\frac{a \cdot d_a+a^2}{a \cdot d_a} \Longrightarrow k=1+\frac{a^2}{2 \cdot [KBC]}.$

But $\frac{[KBC]}{[ABC]}=\frac{a^2}{a^2+b^2+c^2} \Longrightarrow k=1+\frac{a^2+b^2+c^2}{2 \cdot [ABC]} \Longrightarrow$

$\frac{[XYZ]}{[ABC]}=k^2= ( 1+\frac{a^2+b^2+c^2}{2 \cdot [ABC]} )^2.$

Combined with Weitzenbock's inequality $a^2+b^2+c^2 \ge 4 \sqrt{3} [ABC],$ we get ${\frac{[XYZ]}{[ABC]} \ge  ( \frac{4 \sqrt{3}}2}+1 )^2=13+4\sqrt{3}.$


function
irrational number
arithmetic sequence
number theory open
number theory

The odd digits of $\sqrt{2}$ in the decimal representation form a rational or an irrational number?
      What can we say in general about the digits with places forming an arithmetic progression of an irrational number?

Does anyone knows if this problem was solved? If yes can anyone post a solution? I am very interested in this result :)

Interesting question... if it was rephrased in terms of continued fractions, though, it would be much easier :wink:

As for the general case, you could have something like this: $3.0104010509020605030...$, or, on the other hand, something like this: $3.3114411559922665533...$. However, I would guess that the odd digits of $\sqrt{2}$ make an irrational number, as you could probably find the $n$th digit as a function of $n$, and prove that the odd values of that function are not eventually periodic either (I know that you can find the $n$th digit of $\pi$ as a function of $n$).

I'm curious about the function which gives the nth digit. Can zeb please write down the function? :)

In the first example - for odd digits I used 0, for even digits I used digits of pi. In the second one, I did something similar.


Hi guys, try this one.
   There are 16 camels of different height. Is it possible to arrange them in a line in such a way that if any 11 of them step out, the other five will be not arranged in the increasing or decreasing order?
  I think it is hard. At least for me! :(

Yes, it is possible. Arrange them like this:

13,14,15,16,9,10,11,12,5,6,7,8,1,2,3,4

Try to prove yourself that this arrangement is ok (I hope you see the pattern...). Now you will be able to solve it not just for $16$ and $11$ but for $n^2$ and $n^2-n-1$ ;)


When $ y^2 \plus{} my \plus{} 2$ is divided by $ y \minus{} 1$ the quotient is $ f(y)$ and the remainder is $ R_1$.  When $ y^2 \plus{} my \plus{} 2$ is divided by $ y \plus{} 1$ the quotient is $ g(y)$ and the remainder is $ R_2$.  If $ R_1 \equal{} R_2$ then $ m$ is:

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \minus{} 1 \qquad \textbf{(E)}\ \text{an undetermined constant}$

[hide]$ y^2\plus{}my\plus{}2\equal{}(y\minus{}1)(y\plus{}(m\plus{}1))\plus{}(m\plus{}3)$, and $ y^2\plus{}my\plus{}2\equal{}(y\plus{}1)(y\plus{}(m\minus{}1))\plus{}(3\minus{}m)$.  Hence $ 3\minus{}m\equal{}m\plus{}3$ and $ m\equal{}0$, so the answer is $ \boxed{A}$ [/hide]


logarithms
calculus
integration
limit
inequalities
real analysis
real analysis unsolved

Find the limit of $ x_n\equal{}1\plus{}\frac{1}{2 \cdot \ln 2}\plus{}\frac{1}{3 \cdot \ln 3}\plus{}...\plus{}\frac{1}{n \cdot \ln n}$ when $ n \rightarrow \infty$, where $ \ln x\equal{} \log _e x$.

[url=http://en.wikipedia.org/wiki/Cauchy_condensation_test]Cauchy condensation[/url] is your friend.

Use integral test of convergence. Easy to see that integral $ \int_{\alpha}^{\infty}\frac{dx}{x\ln x}$ for $ \alpha>1$ is $ \infty$. And this means that $ \lim_{n\to\infty}x_n\equal{}\infty$.

Hmmm... another solution, which is more "high-schoolish" so to speak is by using Lagrange.
Note that $ (\ln(\ln x))'\equal{} \frac{1}{x \ln x}$.
Then by Lagrange you get: $ \ln(\ln (k\plus{}1))\minus{}\ln \ln k\equal{}\frac{1}{c \ln c}$ for $ c\in (k,k\plus{}1)$
So, you get by using the ineq that $ \frac{1}{k \ln k}<\ln \ln (k\plus{}1)\minus{}\ln \ln k<\frac{1}{(k\plus{}1)\ln(k\plus{}1)}$
But actually you use only the second ineq because you get this way that $ x_n>1\plus{}\ln\ln(n\plus{}1)$ and hence the conclusion.

Thank you for the solutions. Infact I was trying to find a sequence $ a_n$ such that:

$ (1)$ $ x_n\equal{}1 \cdot a_1 \plus{} \frac{1}{2} \cdot a_2\plus{} \frac{1}{3} \cdot a_3\plus{}...\plus{}\frac{1}{n} \cdot a_n$

$ (2)$ $ x_n \rightarrow \infty$ and $ a_n \rightarrow 0$, when $ n \rightarrow \infty$

Since $ x_n\equal{}1\plus{}\frac{1}{2 \cdot \ln 2}\plus{}\frac{1}{3 \cdot \ln 3}\plus{}...\plus{}\frac{1}{n \cdot \ln n}$ requires the given conditions, it means that I was inspired when I tried this sequence.

[quote="dina013"]
So, you get by using the ineq that $ \frac {1}{k \ln k} < \ln \ln (k \plus{} 1) \minus{} \ln \ln k < \frac {1}{(k \plus{} 1)\ln(k \plus{} 1)}$
[/quote]

I think the inequality should be upside-down (typo) :D, but it's the same thing:

\[ \frac {1}{k \ln k} > \ln \ln (k \plus{} 1) \minus{} \ln \ln k > \frac {1}{(k \plus{} 1)\ln(k \plus{} 1)}\]

How can we use $ \lim_{n\to\infty}\sum_{r\equal{}1}^{n}f\left(\frac{r}{n}\right) \equal{} \int_{0}^{1}f(x)dx ?$


geometry
circumcircle
cyclic quadrilateral
geometry solved

Let $ABC$ and $DEF$ be two triangles such that $A+ D = 120^{\circ}$ and $B+E = 120^{\circ}$. Suppose they have the same circumradius. Prove that they have the same 'Fermat length'.

How would you define Fermat length?

Do you mean $AF +  BF + CF$ where $F$ is the Fermat point of the triangle?

Yup $ \min ( PA + PB + PC )$ where $P$ is a point in the triangle.

[color=blue][b]Problem.[/b] The [b]Fermat length[/b] of a triangle is defined as follows: Let ABC be a triangle, and let BA'C, CB'A, AC'B be equilateral triangles erected outwardly on the sides BC, CA, AB of triangle ABC. Then, it is well-known that the segments AA', BB', CC' have the same length. This length is called the [b]Fermat length[/b] of triangle ABC. (Note that if all angles of triangle ABC are $\leq 120^{\circ}$, then this length equals AT + BT + CT, where T is the first Fermat point of the triangle ABC.)

Now, let ABC and DEF be two triangles with respective angles A = < CAB, B = < ABC, C = < BCA, D = < FDE, E = < DEF and F = < EFD such that A + D = 120º and B + E = 120º. Also, assume that the triangles ABC and DEF have the same circumradius. Prove that the triangles ABC and DEF have the same Fermat length.[/color]

[i]Solution.[/i] Erect an equilateral triangle AC'B outwardly on the side AB of triangle ABC. Then, the Fermat length of triangle ABC is CC'. Now, if we prove that the Fermat length of triangle DEF is also CC', then the problem will be solved.

Let the lines AC' and BC' meet the circumcircle of triangle ABC at the points X and Y (apart from A and B). Then, since the triangle AC'B is equilateral, < ABC' = 60º, so that, in the cyclic quadrilateral ABYX, we have < AXY = 180º - < ABY = < ABC' = 60º. In other words, < C'XY = 60º. Similarly, < C'YX = 60º. Hence, the triangle XC'Y has two 60º angles, so it is equilateral; i. e., the triangle XC'Y is the equilateral triangle erected outwardly on the side XY of triangle YXC. Hence, the Fermat length of triangle YXC is CC'.

Since the points C, B, X and Y lie on one circle, < CXY = < CBY = 180º - < ABC' - < ABC = 180º - 60º - B = 120º - B. On the other hand, B + E = 120º yields E = 120º - B. Hence, < CXY = E. Similarly, < CYX = D. Thus, the triangles YXC and DEF are similar. These two triangles also have the same circumradius (in fact, since the points X and Y lie on the circumcircle of triangle ABC, the circumradius of triangle YXC is the circumradius of the triangle ABC, but the latter equals to the circumradius of the triangle DEF); hence, these two triangles are congruent. Consequently, these two triangles YXC and DEF have the same Fermat length. Since we know that the Fermat length of triangle YXC is CC', we conclude that the Fermat length of triangle DEF is CC', and this, as shown above, solves the problem.

  darij


email
search
linear algebra
matrix
inequalities
linear algebra unsolved

If $ A$ is weakly diagonally dominant, without
loss of generality, for some $ t$ ($ 1\leq t\leq n$),
$ |a_{tt}| > \sum\limits_{j \equal{} 1 \\
j\neq t}^n|a_{tj}|$ , and $ a_{it}\neq 0$, $ i \equal{} 1,2,\cdots, n$.
Prove or disprove $ detA\neq 0$
If you have got a solution, please email me
  miwalin@163.com

[quote="JC_math"]Take a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1488690471&t=32062[/quote]

I don't see how this should help.

@miwalin: I rename your matrix $ A$ as $ M$. WLOG assume that $ t \equal{} 1$ (in fact, transpose the $ 1$-st row with the $ t$-th row, and then transpose the $ 1$-st column with the $ t$-th column). Also, WLOG assume that all diagonal entries $ m_{ii}$ are positive (otherwise, just switch the sign of the respective rows). Then apply the inequality
$ \det M \geq \prod_{i \equal{} 1}^n\left(m_{i,i} \minus{} \sum_{j \equal{} i \plus{} 1}^n\left|m_{i,j}\right|\right)$
which I proved in http://www.mathlinks.ro/viewtopic.php?t=176225 . The right hand side is a product with all its factors positive, so you obtain $ \det M>0$. Of course, this does not mean that the determinant of your matrix is actually $ >0$, because you have switched the sign of some rows and interchanged some rows and some columns - but it is still $ \neq 0$, which is what you wanted to prove.

  darij

[quote="darij grinberg"][quote="JC_math"]Take a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1488690471&t=32062[/quote]
I don't see how this should help.[/quote]
Darij,
The link refers to Gershgorin theorem, which seems very close to the question.


integration
calculus
calculus computations

In the derivation of the conduction of heat in a one-dimensional rod, I encountered the following equation (followed from the equation $ \phi(a,t) \minus{} \phi(b,t) \equal{} \minus{}\int_a^b \frac{\partial \phi}{\partial x} dx$)

$ \int_a^b \left(\frac{\partial e(x,t)}{\partial t} \plus{} \frac{\partial \phi(x,t)} {\partial x} \minus{} Q(x,t)\right) \ dx \equal{} 0$

where $ e(x,t)$ is the thermal energy density, $ \phi(x,t)$ the heat flux, and $ Q(x,t)$ the heat energy per unit volume generated per unit time, provided $ a, b$ constant.

It is said that the integrand must itself be zero for the integral to be zero, but I don't see why.

I think what the text probably said (or should have said) is that because the integral vanishes [b] for every choice of [/b] $ a,b$, the integrand must be $ 0$.  This is a weaker variant of the so-called Fundamental Lemma of the Calculus of Variations, and is proved in much the same way.

Assuming the integrand is continuous, it is negative on at most a countable union of intervals $ S_{ \minus{} }$ and positive on at most a countable union of intervals $ S_{ \plus{} }$.  Assume the integrand is nonzero at any point, and integrate it over each of $ S_{ \minus{} }$ and $ S_{ \plus{} }$ to arrive at a contradiction.


integration
calculus
trigonometry
logarithms
conics
parabola
real analysis

I have a problem where there are two towers, each 143 meters high, and 1280 meters away from each other.  There is a cable between these two (starting from the top of each), that just barely touches the ground in the middle.

I have to do three things:  find an equation in the form of $y=kx^2$, find the arc length, then find how much more the sag would increase if the length of the cable increased by .05%.

When I solved for k, I came up with $y=.000342x^2$, then plugged this into the arc length formula:

$\int_{-640}^{640}\sqrt{1+(.000684x)^2}dx$

I had trouble solving this on my own, but plugged it into my calculator and came up with 1319.78 meters as the length of the cable.

To do the third part of the question (where the length increases), I set up a new equation y=kx^2 as follows:

$\int_{-640}^{640}\sqrt{1+(2kx)^2}dx=1320.44$

This is where I am stuck.  Can anyone tell me if I am correct for the answers I did get, and also how to evaluate this type of integral?

Thanks a lot.

To evaluate this integral, use the hyperbolic trig substitution jmerry mentioned in your post about $\int\sqrt{1+x^2} \ dx$, and use the identity $\sinh^{-1} x = \ln\left(\sqrt{1+x^2}+x\right)$

Also, the $k$ I got was slightly different; $k=\frac{143}{640^2} \approx .000349$.

Alright.  What about if the length of the cable increases?  How do I find out the increase in sag?

Thanks.

I think you're on the right track, calculate the new $k$, then determine how far below the ground the parabola would go, I guess this is what they mean by "sag", I'm not 100% sure though.


Hey i have a question (a pretty darn long one)

Professor pipkin has engineered two new species of bacteria, blue and red, that behaves in the following way: after one minute, every blue bacterium turns into 6 red ones and every red bacterium turns into 1 blue and 1 red bacterium. During one of his experiments, professor put 1 blue and 1 red bacterium into a test-tube, so after one minute there were 1 blue bacterium and 7 red ones, and after two minutes there were 7 blue bacteria and 13 red ones. prove that after n minutes there will be

(3\5) x (3^n)+(2/5)x(-2)^n

I really need a solution to this so thanks a million if you tried![/code][/u]

I think this is easiest to prove by induction. Do you know how to do that?

Are you sure,  this is not the same "home work"   asked here:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=52223]www.artofproblemsolving.com/Forum/viewtopic.php?t=52223[/url]

You may be able to see all the solutions too. :)


integration logarithms probability function calculus derivative algebra

Newton's Law of Cooling states that an object whose temperature is different from the surrounding environment will heat up (or cool down) at a rate proportional to the difference
$ (\mbox{temperature of the environment}\minus{}\mbox{temperature of the object})$
Use the proportionality constant $ k\equal{}2$.

a) Write an equation describing the temp of an object placed into a $ 160^{\circ}$ oven:
$ \frac{dT}{dt}\equal{}2(E\minus{}T)$ where $ E$ is the environment temp (some constant).
I believe I am correct for this first part but if I'm wrong, please tell me.

b) If you place a pot of $ 25^{\circ}$ water into the oven, how long does it take to reach 100?
$ \frac{dT}{dt}\equal{}2(E\minus{}T)\Rightarrow \int\frac{dT}{2(E\minus{}T)}\equal{}\int{dt}\Rightarrow T(t)\equal{}160\minus{}Ce^{2t}$.  Since $ T(0)\equal{}25$, we have solve that $ C\equal{}135$ so the equation becomes
$ T(t)\equal{}160\minus{}135e^{2t}$.  Solving for $ T\equal{}100$,
$ \minus{}60\equal{}\minus{}135e^{2t}\Rightarrow \frac{4}{9}\equal{}e^{2t}\Rightarrow 2t\equal{}\ln{\frac{4}{9}}\Rightarrow t\equal{}\ln\frac{2}{3}$.
So I believe that $ \boxed{t\equal{}\ln{\frac{2}{3}}}$ is the answer.  However, it's easy to see that this is negative leading me to believe the correct answer is the inverse of that.  Where is my error?

Your error is in solving the differential equation; that's $ e^{\minus{}2t}$, not $ e^{2t}$. Exponential growth away from equilibrium is rarely physically plausible.

When I teach first-order differential equations for the very first time, I always teach two methods: separation of variables for dealing with separable equations, and the multiplying factor method for dealing with linear equations. It is entirely possible for an equation to be both separable and linear, and that is indeed the case with this one. When that happens, I usually express a preference for treating it as a linear equation and using the multiplying factor method. I express that preference because in my experience, the probability of human error is higher with separation of variables and lower with multiplying factors - especially for constant coefficient linear equations. [Note that JRav used separation of variables.]

To treat it as a linear equation:

$ \frac{dT}{dt}\equal{}2E\minus{}2T$

$ \frac{dT}{dt}\plus{}2T\equal{}2E.$

This is in standard form, with a coefficient of $ 1$ multiplying the derivative. Now take the coefficient of the other term with the function to be solved for, integrate it, and exponentiate it to get the multiplying factor. In this case, we get that $ e^{\int 2\,dt}\equal{}e^{2t}.$ Multiply through by that.

$ e^{2t}\frac{dT}{dt}\plus{}2e^{2t}T\equal{}2e^{2t}E.$

If we've done it right (and this is very checkable), the left hand side should be the derivative of a product:

$ \frac{d}{dt}\left(e^{2t}T\right)\equal{}2e^{2t}E.$ Now integrate.

$ e^{2t}T\equal{}e^{2t}E\plus{}C.$

Is there an initial condition? Say, at time zero, we have $ T\equal{}T_0.$ Then at time zero, we have $ T_0\equal{}E\plus{}C,$ so $ C\equal{}T_0\minus{}E.$

$ e^{2t}T\equal{}e^{2t}E\plus{}(T_0\minus{}E).$ Divide by $ e^{2t}:$

$ T\equal{}E\plus{}(T_0\minus{}E)e^{\minus{}2t}.$ Putting in our specific numbers:

$ T\equal{}160\minus{}135e^{\minus{}2t}.$

It takes a little longer to write all of that out than to use the method of separation of variables, but each step is a little more transparent.

Of course, we also use common sense, as jmerry indicated. All solutions to this differential equation tend to $ E$ as $ t\to\infty.$ That's just what we expect to happen.

Since I used separation of variables, can you point to where in my solution I mixed up $ e^{2t}$ and $ e^{\minus{}2t}$?

[quote="JRav"]
$ \frac {dT}{dt} \equal{} 2(E \minus{} T)$[/quote]

which can be rewritten as $ \frac {dT}{dt} \equal{}\minus{}2(T\minus{}E)$.


Given a prime $ p$ and integers $ a,b,c \in \mathbb{Z}$ such that

$ p \cdot a^n\plus{}b\equal{}p \cdot b^n\plus{}c\equal{}p \cdot c^n\plus{}a$

Prove that $ a\equal{}b\equal{}c$

[quote="Sergey"]Given a prime $ p$ and integers $ a,b,c \in \mathbb{Z}$ such that

$ p \cdot a^n \plus{} b \equal{} p \cdot b^n \plus{} c \equal{} p \cdot c^n \plus{} a$

Prove that $ a \equal{} b \equal{} c$[/quote]

If two of $ a,b,c$ are equal, then it's easy to see that $ a\equal{}b\equal{}c$.

Assume that $ a \neq b, b \neq c , c \neq a$,

Then for $ n \ge 2$, $ pa^n\plus{}b\equal{}pb^n\plus{}c \Leftrightarrow p(a^n\minus{}b^n)\equal{}c\minus{}b$
$ \Leftrightarrow |a\minus{}b|<p(|a\minus{}b|)(|a^{n\minus{}1}\plus{}a^{n\minus{}2}b\plus{}...\plus{}b^{n\minus{}1}|)\equal{}|c\minus{}b|$

Similarly we get $ |b\minus{}c|<|a\minus{}c|$ and $ |c\minus{}a|<|a\minus{}b|$

Combine these three we get $ |a\minus{}b|<|b\minus{}c|<|c\minus{}a|<|a\minus{}b|$, contradiction.

For $ n\equal{}1$, the question becomes $ a\plus{}b\equal{}b\plus{}c\equal{}c\plus{}a$, which clearly leads to $ a\equal{}b\equal{}c$

[quote="stephencheng"]
$ |a \minus{} b| < p(|a \minus{} b|)(|a^{n \minus{} 1} \plus{} a^{n \minus{} 2}b \plus{} ... \plus{} b^{n \minus{} 1}|) \equal{} |c \minus{} b|$
[/quote]

A beautiful solution, but it seems that you missed the case $ |a|\equal{}|b|$, i.e. $ a\equal{}\minus{}b$ when $ n$ is even, though.. Yeah, this case is really simple. Thanks!  :wink:

I think that this is IMO Shortlist, since this was on Brazilian TST.

Isn't it a bit weird that condition that p is a prime isn't used at all?

As dgreenb801 pointed out, my solutions was wrong. Sorry  :blush:

$ p|0$ is not a contradiction, any number divides 0 except 0. $ p|1$ is, though.


AMC
AIME
ARML
summer program
MathPath
Mathcamp
Stanford

A Korean Sophomore

1400s on the old, 2100-2200s on the new SAT
5 on APCalc BC
4 on Physics B AP

3.85 GPA
already 200hr community services (all at hospital)

Honor society, Math Team
AMCs: 140.5 10A last yr, 116 12B last yr, 5 AIME last yr
132 10A this yr, 118 12B this yr, 9 AIME this yr

ARML: Washington A team 2004 B division champion
Mu Alpha Theta: 3rd in nationals in Ciphering, 4th overall last yr.
MathPath participant 2001
Applied for HSSCIM and MathCamp this yr...

that is about it....

will ihave chance when comes senior yr at collegessuch as Stanford, MIT, CIT, and Harvard...

and what do i need to improve on if i cannot

You definitely have a chance, but (not to put you down) so does everyone.
When you apply, definitely stress those high AMC/AIME scores. That's respectable.

All I heard during my college search was that colleges want someone "well-rounded"

I would suggest to spread out yourself to the other subjects a little. Try a little different community work.

Also, when someone is so strong in math, the essays become a little more important. Know your essay topics early and spend a lot of time writing them. To write mine, I went through at least ten drafts with my former English teachers of various topics. It can be a tremendous boost to have an essay that sticks out.

With that said, good luck and I hope you get to the college of your choice.

you got a great chance

Do you play sports? college love students who are fit and play sports

Those AIME scores look good. Harvard says explicitly that it looks for BOTH "well rounded" and "well-lopsided" students. A few of the other top schools are that way too: if you are at the very top of a field you are passionate about, you don't have to be completely balanced. You are male, right? East Asian male applicants sometimes have to overcome a prejudice that they are being pushed by their parents into certain narrow fields rather than following their own path. Maybe the most important thing you can do in choosing extracurricular activities and such--be sure to talk to your parents about this--is simply to make sure to do what you are really personally motivated to do rather than do what reputedly looks good on a college application. Choose a field in which to excel, set a personal standard of excellence, and dare to be yourself. Being yourself and NOT being obviously worried about getting into college might be the most appealing quality you could have to get you into college--that's kind of a paradox, isn't it?

So you took AP Calc and Physics last year? What math this year?

[quote="juicybooty911"]how smart are you?[/quote]
colleges don't give IQ tests....they look at grades/scores/community service, etc.

as to the original question, xxreddevilzxx would have a pretty good chance if he keeps his grades up in challenging classes in all subjects.

why did you take AP physics B
you shouldve waited... unless you were sure you were gonna get a 5

[quote="eryaman"]Hmmm...you're a sophomore and got a 9 on AIME, so you'll probably qualify for USAMO.  That might help.  Taking non math/science related AP's might help...such as english or social studies.[/quote]

well our school offers IB, so i am only planning on taking three AP's:

Calc BC(already)
Physics C(studying right now)
CS AB(about to begin)

and rest are IBs:
English, History, Spanish, Biology, Calculus (AGAIN :( ) and Psychology.

Your chances depend on how well you do in things that are not math related.  Certainly you can help your chances out a LOT by going to MOP and by making USAMO again (you're 25 points from the cutoff, so you'll have to work to make sure that you make it again next year).  Given that you're not currently one of the superstars of the math competition world, your admission will then hinge on a lot of other things - how well you do in your courses, how hard your courses are, what extracurricular activities you are engaged in, how well your essays are written, etc, etc.

Just remember that there will probably be ~100 people applying to university in the same year as you who will have made USAMO as well; there are not that many spaces for math majors at the elite schools.

[quote="JBL"]Probably because he just took the class.  Since the AP is scaled, it is impossible for everyone to get a 5, and thus your logic supports only the conclusion that no one at all should take the exam.  (Or perhaps only one single individual.)[/quote]

well what i meant was that he would have more time to study if he waited and should've probably taken the class junior or senior year. That gives him a better chance of getting a 5


Two players alternately write the numbers $ 1,2,...,7$ in the cells of the figure on the picture. After the figure is filled up, the sum of numbers along each line on the picture is calculated. If three of these sums are equal, the first player wins; otherwise the second player wins. Which of the players has a winning strategy?


Given a circle with center O,and radius R.A and B are 2 points outside the (O).OA=2R.Determine a point M on (O) such that (MA+2MB) is a minimum
Thanks!

MA+2MB is minimum when 2R+2MB is minimum, or when R+MB is minimum, or OM+MB is minimum. So O, M, B must be colinear. Then M is the point where OB intersects the circle.

MA+2MB is minimum when 2R+2MB is minimum.Why?
I don't understand.

red_dog soluted this problem
@chien than :  :spam


geometry
Support
inequalities
3D geometry
tetrahedron
ratio
analytic geometry

1) Let T be the set of ordered triples (x, y, z) of nonnegative real numbers that lie in the plane x + y + z = 1. Let us say that (x,y,z) supports (a,b,c) when exactly two of the following are true: x >= a, y >= b, z >= c. Let S consist of those triples in T that support (1/2, 1/3, 1/6). The area of S divided by the area of T is m/n, where m and n are relatively prime positive integers. Find m + n.

2) A map T takes a point (x, y) in the first quadrant to the point (Sqrt(x), Sqrt(y)). Q is the quadrilateral with vertices (900, 300), (1800, 600), (600, 1800), (300, 900). Find the greatest integer not exceeding the area of T(Q).

Here's Number 1:

[hide]First, we need to find the area of T, denoted by A_T, which in this case when we plot it is an equilateral triangle with vertices (0,0,1), (1,0,0), and (0,1,0). Now, since (x,y,z) supports if exactly 2 are true, then we consider the possible cases:

x>=1/2, y>=1/3

OR

x>=1/2, z>=1/6

OR

y>=1/3, z>=1/6

When we translate these inequalities to the plane, and then considering their intersections with T, then we see that the area of S is the sum of three equilateral triangles, with sides 1/6, 1/3, and 1/2. (Do you see why?)

Then, it's all downhill from here.

The answer is 7/18. [/hide]

Response (in spoiler):

[hide]Hmm...I know how to get the area of T, and I see that after I plot the inequalities, that I get three triangles intersecting at the (x,y,z) point (1/2, 1/3, 1/6). How did you prove that these three triangles were equilateral with sides 1/2, 1/3, 1/6 though?[/hide]

[hide]By similarity, u kno the angles are all 60 degrees. ANd you get the lengths of the sides by extending the other sides towards u from each respective axis, and then subracting the edges to get 1/3, 1/2, and 1/6. Darn, these questions are hard without actual pictures....[/hide]

[hide]Hmm...imagine this diagram. Have the positive x-axis going east, positive y-axis going north, and positive z-axis coming out towards you. Let us take one of the triangles. Let plane z = 1/6 and plane y= 1/3 intersect at segment m (it is actually a line, but let segment m be the part of the line that is within the tetrahedron formed by the boundaries of x+y+z <= 1, or the part with 0 <= x <= 1/2). Let the left part of the plane y=1/3 (the line on the plane directly above the z-axis) be line n, and let the part of this line that satisfies 1/6 <= z <= 1 be segment n. now Sqrt(m^2 + n^2) is the length of a side of oneof the equilateral triangles, right? But, after noticing that m = 1/2 and n = 1 - 1/6 = 5/6, we find that a side of the triangle is not one of 1/2, 1/3 or 1/6. Are you seeing the picture that I am seeing?[/hide]

Hmm, be careful.
n=1-1/6-1/3 (or is it 1-1/6-1/2)

Anyway, you forgot to subtract one thing. GO back over it again,and see if u understand what i mean...

Oh, right. I had subtracted the 1/6 segment from  0 <= z <= 1/6, but I still have to subtract the segment from the other end, from (some number) <= z <= 1. Now that segment is 1/3. To derive that, we can draw a segment from the tip of n down to the z-axis such that they intersect perpendicularly, forming similar triangles. But still, with m=1/2 and n=1/2, we find that a side of the triangle that we are examining is 1/Sqrt(2). Doing the same for the other two triangles, we find the area ratio to be 7/18.

By the way, to avoid confusing the wandering reader, I think you (in your first post) meant that the elements of {1/2, 1/3, 1/6} are the lengths of m, n for each of the three triangles, not their actual sides, correct?

yea, but it sorta ends up being the sides of the triangles, so, its' the same thing

For problem 2, figure out the equations of the sides of the quadrilateral. They are y = 3x, y = x/3, y = 2400 - x, y = 1200 - x, in the wrong order. After applying T a point (x,y) => (x',y') = (sqrt(x),sqrt(y)), hence x = (x')^2, y = (y')^2. Substitute these new values into the equations. Two of them (with slope -1) become circles, and two remain linear (after taking roots). The answer is therefore the area of one sector minus another.

[quote]Um, no. THe sides are of the triangles of S are 1/6, 1/2, 1/3[/quote]

What was wrong with my reasoning:

[quote]Well, not really, since the sides are equal to Sqrt(m^2 + n^2), so, letting m,n = 1/2 for example, the side would be Sqrt((1/2)^2+(1/2)^2) = Sqrt(1/4 + 1/4) = Sqrt(1/2) = Sqrt(2)/2, right?[/quote]

[color=cyan]The problem is that all of the triangles should be equilateral.  I think there is a flaw in your diagram.  Try and draw it out, if you can, and you should see lots of equilateral triangles.  No right triangles.[/color]

But the planes x = 1/2, b= 1/3, z = 1/6 must intersect at right angles..thus the right triangles. I think you may have misunderstood me. I know that the triangles on the plane x+y+z=1 are equilateral, but the triangles which I am using to derive their sides are right triangles. The sides of the equilateral triangles are not 1/2, 1/3, 1/6, but Sqrt(2) *1/2, Sqrt(2) *1/3, Sqrt(2) * 1/6. Anyways, I get the right answer...  8-)

[color=cyan]On further conference, darkquantum and I figured out what was going on.

You are in fact perfectly correct.  He and I took a step to simplify calculations: instead of considering the sides of the actual triangle, we merely considered the ratios between segments formed, which was all we needed since this was a probability question.  Your calculations and the results they lead you to are all exactly right.[/color]

[quote]Your calculations and the results they lead you to are all exactly right.[/quote]

Woohoo! Thanks for the help guys.

[quote]Two of them (with slope -1) become circles, and two remain linear (after taking roots). The answer is therefore the area of one sector minus another.[/quote]

Waait. Ok, first of all, general question: are equations of the form x + y^2 = k (as we have here) really circles? What is their radii? Secondly, how did you arrive at that the area of T(Q) is the difference of the areas of the circles?[/quote]

I did not say that x + y^2 = k is a circle. I said that x^2 + y^2 = k is a circle. The radius is sqrt(k). I also didn't say that the area of T(Q) is the difference of areas of circles. I said it was the difference of areas of sectors. A sector is the part of a circle subtended by two different rays emanating from its center. In our case the sides of the two sectors in question are the lines which were transformed into lines.

Duh.

[quote]I did not say that x + y^2 = k is a circle. I said that x^2 + y^2 = k is a circle. The radius is sqrt(k). [/quote]

My bad, I misunderstood your first post.

[quote]I also didn't say that the area of T(Q) is the difference of areas of circles. I said it was the difference of areas of sectors. A sector is the part of a circle subtended by two different rays emanating from its center. In our case the sides of the two sectors in question are the lines which were transformed into lines. [/quote]

Oh, I thought you were using the term "sector" in the informal sense--as any part or section.

Ok, I have found the answer now. Just one question though: the two lines intersecting one of the circles form two different sectors of the same area. How come we only take into account one of the sectors, and not both of them?

Also, for the second problem, why can't we simply take the square root of each x, y coordinate and then use determinants to figure out the area?

[quote="JBL"][color=cyan]What call is there for rudeness?[/color][/quote]
Ignore his rudeness. He's always like that :evil:
*sigh*

[quote]I think you can.[/quote]

No, I tried and got a negative number.

The area of the image in the second question is made up of sectors of a circle. It's not a quadrilateral. The method you suggest will give you the area of a quadrilateral whose vertices are the images of the vertices of the given quadrilateral. It will not give you the area you seek.

Hmm, the question says that the resulting figure is a "map" of T(ABCD). What exactly does this mean? Does this mean that we must take the square root of the segments too? Or can we simply take the square roots of the points and assume that the map is a rectangle (like the original figure)?

Every single point (x,y) in the plane is mapped to (sqrt(x),sqrt(y)). When you apply the transformation to the quadrilateral, you apply it to every point in the quadrilateral. That means the points on the edges too.

Well at least you're not hiding it :)
It's OK to be yourself. Just not in the public :P (kidding!)


calculus
integration
function
calculus computations

Find the following indefinite integral:
[img]http://i82.photobucket.com/albums/j274/boyhiphop160/Linh%20tinh/tichphan.jpg[/img]

Yes, this is one of the classic functions whose primitive cannot be expressed as a finite sum of elementary functions.


[quote="griffin08"]A gecko is in a room that is 12 feet long, 10 feet wide and 8 feet tall. The gecko is currently on a side wall (10' by 8'), one foot from the ceiling and one foot from the back wall (12' by 8'). The gecko spots a fly on the opposite side wall, one foot from the floor and one foot from the front wall. What is the length of the shortest path the gecko can take to reach the fly assuming that it does not jump and can only walk across the ceiling and the walls? Express your answer in simplest radical form.[/quote]

[hide]

As you can see, he is 6 feet higher than the fly, because since the height is 8 and he is 1 foot from the top and the fly is 1 foot from the bottom so 8-1-1=6 feet

Then, he is 22 feet from the side of the fly as he takes one foot to get to the 12' by 8', then 12 ft to get to the other 10' by 8' and then from there 9 ft from the fly.

so you make a triangle of

22 ft as 1 leg and 6 ft as the other leg

6^2 + 22^2=c^2   520=c^2

c=sq.root 520 or 2 root 130

I might be wrong but I think the answer is 2*sq.root of 130[/hide]

ihatepie, $ 2\sqrt{40}$ can be further simplified to $ 4\sqrt{10}$.

:blush: I mistyped the answer

Does any1 else have answers to this problem?

/me got same answer as unimpossible

I'm amazed that the last sprint national question was this easy. I expected it to be nearly impossible.  :D

[quote="unimpossible"]Well the general approach to this would be to make a net and find the most direct way on the net.

if you make a net, i think that the shortest way is the hypotenuse of a right triangle of legs 6 and 22, i get $ 2\sqrt130$[/quote]

How?

I get $ 4\sqrt{29}$ with my net. (i dont think thats right since I havent tried anyother nets)

There are many types of nets that can be formed, I think theres more than one case to consider.

[quote="abacadaea"]the fly is one foot from the front wall, not the back i did the same thing[/quote]

huh?  that still wouldnt make it 2sqrt130 would it?

I may be wrong but doesn't the triangle thing have the gecko jump to the fly? Because the problem states that the gecko can only move along the floor, ceiling, and walls.

The Answer Key says the answer is [hide]$ 2\sqrt{113}$...[/hide]

I'm still not clear why the answer is $ 2\sqrt{113}$...I keep getting the answer $ 4\sqrt{29}$...

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=483130557&t=37202]Here[/url]
Scroll down to rcv's post to see a diagram.


geometry
angle bisector
area of a triangle
Heron\u0027s formula

Source: 2000 AMC 12-Number 19

In triangle $ABC$, $AB=13, AC=15, BC=14$. Let $D$ denote the midpoint of $BC$ and let $E$ denote the intersection of $BC$ with the angle bisector of $BAC$. What is the area of triangle $ADE$?

[hide]
$\frac{AB}{AC}=\frac{BE}{EC}=\frac{13}{15}$

So assume $BE=13x$, $CE=15x$.

$13x+15x=14 \Rightarrow x=0.5$

Since $BD=7$, so $ED=0.5$

So $\triangle{AED}=\frac{1}{14} \triangle{ADB} = \frac{1}{28} \triangle{ABC}$

Use the Heron's formula: $\triangle{ABC} = \sqrt{21 \cdot 7 \cdot 8 \cdot 6}= \sqrt{7^2 \cdot 3^2 \cdot 4^2}=3\cdot 28$.
Hence $\triangle{AED} = 3$.[/hide]

[hide="Pretty much the same solution"] You know that $BE$ is $\frac{13}{28}BC$.  Thus, $DE=\frac{BC}{28}$, because $BD=\frac{14}{28}$.
Use Heron's formula to find that the area of $ABC$ is
$\sqrt{21*8*7*6}=4*3*7=84$.  Thus, the area of $ADE$ is $3$[/hide]


Each row of the Misty Moon Amphitheater has 33 seats.  Rows 12 through 22
are reserved for a youth club.  How many seats are reserved for this club?

We only want the rows and then its easy. There are 11 rows and 33 seats in each so the answer is 363.


Find the coordinate of the point on a horizontal number line that is $ 5$ units from point $ G$ whose coordinate is $ 1$, and $ 3$ units from point $ H$ whose coordinate is $ \minus{}7$.

The point is 5 units from G  (1), so it can be $ \minus{}4$ or $ 6$.

The point is 3 units from H (-7), so it can be $ \minus{}10$ or $ \minus{}4$.

The point satisfying both is $ \boxed{\minus{}4}$.


algebra
polynomial
calculus
integration
algebra unsolved

hey guys give me a neat solution for this problem
let r and s be integer s \geq 0.prove that ther exists an interval I of length 1/s and a polynomial P(x) with integral coefficients such that for all x  \varepsilon  I we have
-1/s 2  < P(x)-r/s < 1/s 2
help!

hint :
 I=(l,k) ,k-l=1/s ,l<k<1
 +s=1, let P(x)=r
 +s>1 ,chose m is natural number such that : k^m<1/s
       and   n is large enough : a^n <1/(rs), a=1-l^m
so let P(x)=r/s*[1-(1-lx^m)^n]

I think, we should correct A1lqdSchool's hint (I don't se where he get integer polynomial).

Suppose $s>1$, $I=[\frac{1}{2s},\frac{3}{2s}]$. Let $P(x)=\frac{r}{s}\cdot(1-(1-sx^{m})^{n})$, where $m$ and $n$ are large enough that $x^{m}<\frac{1}{2s}$ on $I$ and $\left(1-s\cdot \frac{1}{(2s)^{m}}\right)^{n}<\frac{1}{rs^{2}}$.


function
probability
expected value
combinatorics unsolved
combinatorics

1. Consider integers $ n \geq 0$, $ b \geq 2$, and $ k \geq 1$. As a function of $ n, b$, and $ k$, what is the average number of blocks of length $ k$ among all $ b^n$ strings of length $ n$ over the $ b$-letter alphabet $ {1, 2, . . . , b}$?

It seems to be escaping me :(

"Blocks of length k" means "k consecutive copies of the same letter," with a block of length k + 1 counting as two blocks of length k?  If so, just use linearity of expectation ;)

I... have no idea what linearity of expectation is :(
And no (unless I misinterpretted), a block is of any length. Just consecutive symbols
IE let A = a*a be the block of little a's, similarly for B,C.
then $ aabbbbbcccabbaabababbccbabcbacabaaabbbcccacccccccccccccccccccccccccccccc$ = $ ABCABABABABCBABCBACABABC$
(hopefully I got that right ^_^)

... which doesn't answer my question: does a block of length k + 1 include two blocks of length k, or zero blocks of length k?

Linearity of expectation: the expected value of a sum is equal to the sum of the expected values.  The events you should look at (assuming the answer to the above question is "two") are "the first k entries form a block," "the second through k + 1th entries form a block," and so on.

A block of length k+1 does not have two "blocks" have length k. It is just one block of length k+1.


Haha - I was just so tempted to make a counter topic and see how many people visit anyways, after I warn them not to.

And I want to see how many people even bother to click on the link to this thread which is absolutely pointless.

AIME results make me happy!

C'mon, stop it please!. The first one was funny, the second one was acceptable, but the third one is just too much.

On the other hand, you are not taking in consideration that some people review all the new topics just using the Next>>> button.

Best,


For which integers $ n$ we can write $ n$ as the sum of two squares? Do you know a general algorithm which allows us to do this?

It is necessary and sufficient that for any prime $ p \equiv 3 \bmod 4$ dividing $ n$ the greatest power of $ p$ that divides $ n$ is even.  See the proof [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]here[/url].  One general algorithm is testing values of $ a$; another involves factoring $ n$.  For example,

$ 2005 \equal{} 5 \cdot 401 \equal{} (1^2 \plus{} 2^2)(1^2 \plus{} 20^2) \equal{} (1 \cdot 1 \pm 2 \cdot 20)^2 \plus{} (1 \cdot 20 \mp 2 \cdot 1)^2 \equal{} 41^2 \plus{} 18^2 \equal{} 39^2 \plus{} 22^2$.

This has a relation to fermat's two square theorem and also to another result attributed to Lagrange if i'm not wrong..
..that general algorithm is a very nice application...

Yes; both results are discussed in the link.


geometry
geometric transformation
combinatorics proposed
combinatorics

Let $ M$ be a subset of $ \{1,2,...,2008\}$ with the following property: For any three elements $ x$, $ y$ and $ z$ $ (x < y < z)$ of $ M$, $ x \plus{} y$ does not divide $ z$. Determine the largest possible size of $ M$. Justify your claim.

The largest possible size for $ M$ is $ 1005$. The proof consists of two steps:

1) The subset $ \{1004,1005,\ldots,2008\}$ fulfills the properties, i.e. the largest possible size for $ M$ is greater or equal to $ 1005$.

2) Let $ M$ be a subset for which the largest possible size is assumed and let $ m\equal{}\min M$. 1) implies $ m\leq 1004$. Denote $ M^*\equal{}\{x\in M\setminus\{m\}|x\leq 2008\minus{}m\}$. We clearly have $ |M^*|\geq |M|\minus{}m\minus{}1$, since we have removed at most $ m\plus{}1$ elements from $ M$. Moreover, all elements of the translation set $ m\plus{}M^*$ have values between $ m\plus{}2$ and $ 2008$ and must have empty intersection with $ M$, i.e. we have $ |\{x\in M|x\geq m\plus{}2\}|\leq |\{m\plus{}2,m\plus{}3,\ldots,2008\}|\minus{}|m\plus{}M^*|\equal{}2007\minus{}m\minus{}|M^*|$, which implies

$ 2|M|\equal{}|M|\minus{}2\plus{}|M|\minus{}m\minus{}1 \plus{}3\plus{}m\leq |\{x\in M|x\geq m\plus{}2\}|\plus{}|M^*|\plus{}3\plus{}m\leq 2010$. This yields $ |M|\leq 1005$


25) Calculate the number of carbon atoms in a mole of sucrose, C12H22O11.

1)12

2)1.3 x 1025

3)6.02 x 1023

4)6.62 x 1024

5)7.22 x 1024

No answers just procedures

if by 1024 you mean $10^{24}$, then 5th one is right.!

It goes like this...

1 mole = 6.022 X1023 molecules

1mole sucrose X  (6.022X 1023 molecules sucrose)  X  12 atoms Carbon
                                       1 mole sucrose                       1 molecule sucrose

 The moles cancels out and the molecules cancels out. Your base units are atoms of Carbon.

(6.022 X 1023)  X   12 atoms Carbon  =  72.3 X 1023  ->  7.23 X1024

I'll $\LaTeX$ it up for clarity on the factor labeling.

$\left(\frac{1\text{mol Sucrose }\left(\text{C}_6\text{H}_12\text{O}_6\right)}{1}\right)\cdot\left(\frac{6.022\cdot 10^{23}\text{ molecules Sucrose }\left(\text{C}_6\text{H}_12\text{O}_6\right)}{1\text{mol Sucrose }\left(\text{C}_6\text{H}_12\text{O}_6\right)}\right)\cdot\left(\frac{12\text{ atoms C}}{1\text{ molecule Sucrose }\left(\text{C}_6\text{H}_12\text{O}_6\right)}\right)$

Hmmm.... OK, it broke the line.  Just imagine it.