An integer $ n>101$ is divisible by $ 101$. Suppose that every divisor $ d$ of $ n$ with $ 1<d<n$ equals the difference of two divisors of $ n$. Prove that $ n$ is divisible by $ 100$.

I didn't quite understand, let $ d$ be the largest divisor of $ n$, then $ d=d_{1}-d_{2}\Rightarrow d1>d$?

[quote="Fuaran"]Nope.
Suppose d is the maximum divisor of n. All divisors of n (besides n) satisfy: d1<=n/2.
Then if d<=n/2 we have d=n-d1>=n/2 (n is the largest divisor of n besides d).
And that means d=n/2.
n is divisible by 2. There's no contradiction here.[/quote]
But then $ d=\frac{n}{2}$ is $ d_{i}-d_{j}$ for some $ i,j$ by condition of the problem,which shows that $ d_{i}>\frac{n}{2}$ ?

[quote="goc"]yes, and that $ d_{i}$ is exactly $ n$.
every divisor SMALLER than $ n$ is a difference of SOME(including $ n$) 2 divisors of $ n$.  :wink:
so $ \frac{n}{2}=n-\frac{n}{2}$ and its good
edit:
i think i've found a solution
[hide]
we know $ n=101k$ for some $ k$ and we know $ k>1$ from the condition so
$ k=a-b$ where $ a,b$ are divisors of $ n$. assume $ 101\not|a$ then from $ a|n=101k$ and since $ 101$ is prime we have $ a|k$ which isn't true...so $ a=101a_{0}$ and now we get that $ 101a_{0}|n=101k$ so $ a_{0}|k$ so $ k=la_{0}$ and from that it's easy to get that $ b=ma_{0}$ so $ k=(101-m)a_{0}$
$ n=101(101-m)a_{0}$. since $ b$ is a divisor of $ n$ we have that $ m|101(101-m)$ for some $ m$ such that $ 1\leq m\leq 100$ but again since $ 101$ is prime it can only happen when $ m=1$
so this works if $ 101$ is replaced with prime $ p$ and $ 100$ with $ p-1$
i didn't find an example that such a number exists yet...[/hide][/quote]
To [u]goc[/u]: Good job
To [u]pco[/u]: Very convincing

ok,seems that i'm just a lazy bastard for not checking the trivial case... :blush:
thanks pco...


abstract algebra
superior algebra
superior algebra unsolved

Something related, Let H be any group with an automorphism f such that the only fixed point  of "f" is "e" and such that ff(x)=x for all x in H. Prove that H is abelian and f(x)x=e for all x in H.

I think $H$ should be finite, right? :? If this is indeed the case, we could do this:

If $x_1^{-1}f(x_1)=x_2^{-1}f(x_2)$, then $f(x_1x_2^{-1})=x_1x_2^{-1}$, meaning that $x_1=x_2$. This shows that $x\mapsto x^{-1}f(x)$ is injective, and thus surjective, since $H$ is finite. Now take any $t\in H$ and some $x$ s.t. $x^{-1}f(x)=t$. We have $tf(t)=x^{-1}f(x)[f(x)]^{-1}(f\circ f)(x)=x^{-1}(f\circ f)(x)=1$, which means that for all $t\in H$ we have $f(t)=t^{-1}$. The fact that $H$ is abelian follows from this.


This problem was  found in a  romanian  book, (Mircea Fianu, Olimpiadele bucurestene de Matematica, page 36) but the solution is  so  long,,,, Do you think  there is a shorther one?
Let $a,b,c>0$ such that $a|b+c-a|+b|c+a-b|+c|a+b-c|=a^2+b^2+c^2.$ Prove that $a=b=c.$
Thank you!

Nickolas ,

    I took  $a , b , c$ as sides of a triangle( so absolute values have no role) , put $a+b-c = x , b+c-a = y , c+a-b = z$ ect ,  and the proof was easy. I have not tried for the other cases. I had not time. I ' ll try it in the evening.

  [u]Babis[/u]


In the upcoming year, im going to be teaching the mathcounts team at my old middle school, but being that it wasnt offered when I was there, I dont know what Im supposed to teach.
If someone could please post the main topics and give an example of 1 or 2 of the more popular types of questions asked for each topic, I would be very appreciative. Also, a general overview of the competitions would be very helpful as well.

Use the coaching kit at http://mathcounts.org/Program/program.html

I'm still learning what jelyman is asking about myself, so I would appreciate other, more experienced coaches jumping in here with additional suggestions.

Well I can hardly be considered "an experienced coach". My MC experience just consists of 3 years of competition, but I'll be teaching a MC class (not on AoPS) this year (and ppl are actually paying to be in the class...what a surprise) so I'll make a comment

[quote="jelyman"]I think that coahcng kit is the most useless idea ever invented. Its just schedules. Anyways, if anyne can answer my original question, I would be greatly appreciative.[/quote]

jelyman, in my personal opinion the coaching kit is useless unless u've got no idea what ur doing and neither does anyone else u know...

HOWEVER, it does include an EXCELLENT toolbox that covers just about every formula that will be covered in the MC competitions...if u can make ur students understand those then they'll have everything they need. after that, just practice...and apply those formulas...doing problems is the best way. MC resources can be purchased at http://www.mathcounts.org or they're available online in some places...

Ooh, I liked the coaching kit I got from my MathCounts coach last year. So many interesting formulas :) .

Let me give a tip: hand out snacks. My MathCounts coach does that, and most people like snacks.

[quote="Treething"]
Let me give a tip: hand out snacks. My MathCounts coach does that, and most people like snacks.[/quote]

Sounds nuts to say, but this is very important when you're bringing new people in.  Don't overlook it.

Try to make it social - try some team games in addition to you just standing up there and teaching.   If you stand up there and lecture all the time, your team will get small in a hurry.

As for teaching - start from the basics.  Don't assume anything is too easy.  If you want to teach them mods, do base numbers first.  If you want to teach them combinations and permutations, do 'using multiplication to count' first.  And so on.

[quote]Let me give a tip: hand out snacks. My MathCounts coach does that, and most people like snacks.[/quote]

I coach ARML, not Mathcounts. But yes, by all means provide snacks. Once you're up and established, maybe you can get some parents to help you with that. After all, there's a pretty widely followed tradition along those lines for youth sports teams.

[quote="Kent Merryfield"] Once you're up and established, maybe you can get some parents to help you with that. After all, there's a pretty widely followed tradition along those lines for youth sports teams.[/quote]

This is how we handle it at the San Diego Math Circle.

I m in high school so we don't do mathcounts.  My teacher usually has food at the meetings like pizza and hamburgers and we still don't have all that high of turnouts.  It s probably because math is "nerdy" and some of the kids on the team are...well very nerdy to say the least.  I usually never make a practice until halfway into the second quarter of school since I have cross country until then.  So I can safely say that I am not scaring away the people  :D ( I think I am one of the least likely persons on my team to scare ppl away but its still nice to have proof).  I think we ll have to start recruiting people for our jv team (our varsity is pretty good this year).  I think I might start talking to my teacher about ways we can expand the team.

Well the point of all of this was that food alone wont necessarily make your team larger.  It never hurts though!

[quote]If someone could please post the main topics and give an example of 1 or 2 of the more popular types of questions asked for each topic, I would be very appreciative. Also, a general overview of the competitions would be very helpful as well.[/quote]

Hmm... I don't know how much I can help you with that but if you want to look some not-textlike questions, meaning some creative questions, scan through the SC post I did long time ago. There is first one which many people competed (winner was theone853) and second one which very few people did. It is good test with questions though. And if you still want more, look at the my competition which I'll post the questions for the round.
That's official competition I made and for details, read the another thread.

 :cool:  Plus, don't worry too much. Pretty sure you'll do fine.

Also, try hiring helpers (or getting volunteers) to help out with the advanced kids...

Oh, if you get food, give some to me. ;)

Food is key. I remember at my second Mathcounts practice, I found a huge stash of starbursts. As soon as that news spread around, we had 4 new members the next week.  :D

Do some practice tests to see what level each person is at. You'll ned to know which 4 are best to take on your team.

my math teacher was the coach of the california mathcounts team (won first place 3 years in a row)  ;) .

she doesnt do much. i think she is kind of lucky because our school has many intelligent students, but all she does is give warm-ups and workouts and previous mathcounts contests

My MathCounts coach writes out problems that people don't understand...

What's the point? Try not to bump meaninglessly.

[quote=falcon22]Sorry since this is off-topic - but you both have identical avatars! :wow:[/quote]

[img]https://i.kym-cdn.com/entries/icons/facebook/000/007/666/_57c8a1a431a592af806925e57258202f.jpg[/img]

@above that's creepy if i die tonight I'm suing you

[quote name="falcon22" url="/community/p17297959"]
Sorry since this is off-topic - but you both have identical avatars! :wow:
[/quote]

The avatar was copied, not a coincidence.

My gosh MSM is gonna get locked at this rate. Please don't let that happen.

[quote=HumanCalculator9][quote=falcon22]Sorry since this is off-topic - but you both have identical avatars! :wow:[/quote]

yeah they both copied a certain 3Blue1Brown thumbnail[/quote]

I see.

Anyway, stop posting in this thread - I've reported the post, it should be handled soon! :)


calculus
integration
trigonometry
calculus computations

If $ \displaystyle m$ is a positive interger, show that

$ \displaystyle \int_0^{\frac{\pi}{2}} \cos ^{m}x \sin^{m}x \,dx\: =\: 2^{-m}\int_0^{\frac{\pi}{2}} \cos^{m} x \,dx$


Emily used a standard metric ruler to measure the length of a piece of yarn she was using for a project. The yarn was as long as 24 of the smallest divisions on the ruler. How many centimeters long was the piece of yarn?

[hide]$2.4$ i counted them out on the ruler[/hide]

[hide]the smallest measure on a metric ruler is millimeters. so 24mm=2.4cm.[/hide]

Well, you could also think that if you count 24 dividers, the length is 23.  There are two dividers around 1 mm, three around 2mm, so I guess you could also say 2.3mm

The smallest mark on a metric ruler is a millimeter. On an "inch ruler" it's 1/8 of an inch.

So the yarn is 24 mm, which is 2.4 cm.  :)


algebra
polynomial
calculus
integration
absolute value
superior algebra
superior algebra unsolved

Let P(z) be a polynomial with complex coefficients and complex roots. Is it true that the coefficient in P(z) with largest absolute value always is greater (that is, has greater absolute value) than the absolute value of the roots of P(z)?

Sorry ZetaX, I messed one thing up. I meant to write integral coefficients (and complex roots).   :)


calculus
integration
function
calculus computations

If f and g are continuously differentiable functions for all real numbers, which of the following definite integrals is equal to f(g(4))-f(g(2))?

${ \int_{2}^{4}{f'(g(x))}}\ dx$
${ \int_{2}^{4}{f(g(x))f'(x)}}\ dx$
${ \int_{2}^{4}{f(g(x))g'(x)}}\ dx$
${ \int_{2}^{4}{f'(g(x))g'(x)}}\ dx$
${ \int_{2}^{4}{f(g'(x))g'(x)}}\ dx$

use the chain rule; what is $ [f(g(x)]'$?

you don't know what $ \frac {d}{dx} f(g(x))$ is?  it is very hard to do integral calculus if you don't know essentially the most important result in differential calculus.

in particular if you think hard enough you probably do know what the answer is, and i am not inclined to answer this sort of multiple choice question for you with no apparent effort on your part.

Calculus BC?  November?  Either your teacher doesn't review material at all or you haven't been paying attention in class.  In either case,

$ [f(g(x))]' \equal{} f'(g(x)) g'(x)$

Using the Fundamental Theorem of Calculus, can you solve the problem now?


Let $ V$ be a subspace of the complex inner product space $ E$, let $ T: V \rightarrow E$ be linear, and define $ Q(x) = <T(x),x>$ for $ x \in V$.  Show that

(i) If $ Q(x) = 0$ everywhere, then $ T(x) = 0$ for all $ x$.

(ii) If $ Q(x)$ is always real, then $ T$ is hermitian.

(ii) Well if $ Q(x)$ is real for all $ x$ then

$ Q(x)=<T(x),x> = <x,T(x)> = <T^{*}(x),x>$

for all $ x$ thus $ T(x)=T^{*}(x)$

(i) If $ Q(x)=0$ everywhere then choose a nonzero $ x$ such that $ Q(x)=0$. Now since $ Q(x)=0$ we have that

$ <Q(x),x> =0$ for all $ x$. Also $ <Q(x),x> = x^{*}Q(x)=\left(x^{*}\right)^{2}T(x)$

since $ Q(x)=x^{*}T(x)$. now since we assumed that $ x\neq 0$ we also have that $ x^{*}\neq 0$ thus $ T(x)=0$ for all $ x$.


In a triangle ABC, D is a point on side BC such that DB = 14, DA = 13 and DC = 4. Knowing that the circle circumscribed to the triangle ADB is equal to the radius of the circle circumscribed to the triangle ADC, calculate the area of triangle ABC.

Portuguese question:

Em um triângulo ABC, seja D um ponto sobre o lado BC tal que DB = 14, DA = 13 e DC = 4. Sabendo que o círculo circunscrito ao triângulo ADB tem raio igual ao do círculo circunscrito ao triângulo ADC, calcule a área do triângulo ABC.

Applying the formulae for radius we concluded that b=c=15 and the area is 36*15^.5

since for any $ \triangle ABC$ we have $ R\equal{}\frac{abc}{4L}\equal{}\frac{abc}{4.\frac12ab\sin X}\equal{}\frac{c}{2\sin X}$ , then

$ \frac{c}{2\sin X}\equal{}\frac{b}{2\sin (180\minus{}x)}\equal{}\frac{b}{2 \sin X}$ , so , $ b\equal{}c\equal{}k$ , ($ X\equal{}\angle{ADB}$)

then by stewart , we get $ 4k^2\plus{}14k^2\minus{}14.4.18\equal{}13^2.18\rightarrow k^2\equal{}225\rightarrow k\equal{} 15\equal{}b\equal{}c$

so by Heron we get $ L\triangle \equal{} \sqrt{24.9.9.6}\equal{}108$

Well ... the area is $ 8\sqrt{3}\minus{}6$
How to do it ?

[quote="luiseduardo"]Well ... the area is $ 8\sqrt {3} \minus{} 6$
How to do it ?[/quote]
How come ?
Or maybe there are some mistakes in my solution huh ?

No, I think your solution that's correct, but the question wrong  :o
I copy two questions differents in one ...  :blush:
I'm sorry, but thanks for your help.


javab: $2$
hanooz montazere rahe hal haye ghashange shma hastam. :D

[quote="amirhtlusa"]age as $[$ $]$ manzooret $\lfloor$ $\rfloor$ hast,
soel kheyli rahateh  ;)[/quote]
daghighan manzooram hamoone va khob man ham neveshtam ke kheili rahate vali doost daram ba rahe hal haye motafavet va ghashang hal beshe. :D

$\\\lim_{x\rightarrow 2}\lfloor x+0.6\rfloor\\ =\lfloor x+0.6\rfloor |_{x=2}\\ =\lfloor 2.6\rfloor=2$

EDIT: Mitooni $\delta -\epsilon$ estefade koni.

Masoud Zargar


Now that the test is over, I'd like to ask anyone who took it, or who didn't to show me how they proved this problem. I did it really quickly, but I think I may be wrong.
----------------------------------------------------------------
BAMO (Bay Area Math Olympiad) 2007
Question 5 of 5

Two sequences of positive integers $x_{1}, x_{2}, x_{n}...$ and $y_{1}, y_{2}, y_{n}...$ are given such that $\frac{y_{n+1}}{x_{n+1}}> \frac{y_{n}}{x_{n}}$
for each $n \geq 1$. Prove that there are infinitely many values of $n$ such that $y_{n}> \sqrt{n}$.
----------------------------------------------------------------
[hide="Possible Solution"]
Let $y_{n}= n$, and $x_{n}= n+1$

Our sequence then looks like
Y: 1, 2, 3, 4, 5, 6
X: 2, 3, 4, 5, 6, 7

The ratio of terms is like this:
$\frac{1}{2}< \frac{2}{3}< \frac{3}{4}< \frac{4}{5}< \frac{n}{n+1}$
This satisfies the conditions, because $lim_{n \to \infty}\frac{n}{n+1}$ converges to one.

Thus, we want all $y_{n}> \sqrt{n}$
Because $y_{n}= n$, we get $n > \sqrt{n}$, and because N is positive, we can multiply both sides by it without changing direction or multipying by zero.
Thus, we get $n^{2}> n$, which is true for all real numbers.
The cardinality of the set of all real numbers is infinite.
Therefore, there are infinite solutions.
QED.
[/hide]
If you read my proof, can you comment on either what I did wrong, or what I could do better? I'm looking for some constructive criticism, as this was my first ever Olympiad.

[hide]The problem with your solution is that you are assuming that each sequence is a certain type of sequence. Another set of sequences could be $x_{n}=F_{n}$ where $F_{n}$ is the nth Fibonacci number and $y_{n}=3^{n}$. I'm sorry, but I don't have a proof.
[/hide]

[b]Edit: I think the directors of BAMO would appreciate if we wait till at least tommorow to discuss the exam.[/b]

Don't you just have to show that if you can find two sequences with the first property (ratios one) that there are an infinite amount of solutions to the second?

i.e. does it matter what the series are, as long as they match the properties?

[quote="DiscreetFourierTransform"]does it matter what the series are, as long as they match the properties?[/quote]

Of course it does.  You need to show that [b]any possible sequences that satisfy the first condition[/b] also satisfy the second.

Edit:  You seem to have written the problem incorrectly.  Should it be $y_{n}> \sqrt{n}$?

yeah i just got home from taking the test; wait till at least tomorrow

[hide="it is the next day now"]
We are given $\frac{x_{n}}{y_{n}}>\frac{x_{n+1}}{y_{n+1}}$. (It is better to have the $y$'s in the denominators; you'll see why soon.) After clearing denominators,
\[x_{n}y_{n+1}-x_{n+1}y_{n}>0\Rightarrow x_{n}y_{n+1}-x_{n+1}y_{n}\geq 1\]
We now have the sharpened inequality
\[\frac{x_{n}}{y_{n}}\geq \frac{x_{n+1}}{y_{n+1}}+\frac{1}{y_{n}y_{n+1}}\]
We now prove infinitely many $n$ with $y_{n}>\sqrt{n}$ by contradiction. Suppose that for some $N$, $y_{n}\leq \sqrt{n}$ for all $n\geq N$. Then for any $n\geq N$, we get
\[\frac{x_{n}}{y_{n}}\geq \frac{x_{n+1}}{y_{n+1}}+\frac{1}{\sqrt{n(n+1)}}\]
\[>\frac{x_{n+1}}{y_{n+1}}+\frac{1}{n+1}\]
So we end up with
\[\frac{x_{N}}{y_{N}}>\sum_{k=N+1}^\infty \frac{1}{k}\]
 :o THE HARMONIC SERIES DIVERGES!!! CONTRADICTION!!! YAY!!!
[/hide]


geometry
perimeter
combinatorics unsolved
combinatorics

Nine 1*1 cells of a 10*10 squre are infected. In one time unit, the cells with at least two infected neighbors(having a common side)become infected.Can the infection spread to the whole squre?

It seems that if the 9 cells are in one row, the answer is obviously "no".

It's been discussed before here: the idea is that whenever a new cell is infected, the perimeter of the infected region decreases (or stays the same), so since in the end you want the whole board to be infected, thus having a perimeter of $4\cdot 10$ for the infected region, you cannot achieve this starting with $9$ squares, which have a perimeter $\le 4\cdot 9$.


geometry
rectangle
perimeter
LaTeX
calculus
derivative
calculus computations

I don't quite understand what I am doing wrong.. but the answer I am getting is not correct.
And this is my last problem and its due online at 11pm tonight =/...
I get this problem right and I get a 100% on the assignment!
I went to my math lab at my school and this is the answer/ work I got.
Any help is appreciated!
----------

The problem asks:
[b]A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 25 feet?[/b]

So this is how I approached it: we need the areas of the rectangle and the semi circle... so we get:

A = bh + pi(b/2)^2 = bh + (pi/4)(b^2)

for simplicity, b = x (easier on the eyes :))

A= xh + (pi/4)(x^2)

so I got that h (the height) is = (25-(1 + pi)x)/2

this means that A(x) = (25x - (1+pi)x^2)/2  + (pi/4)(x^2)

So from here I differentiated the equation and got...
[25x - (1 + pi) x^2]/2 + (pi/4)(x^2) =
(25x)/2 - (1/2)(x^2) - (pi/2)(x^2) + (pi/4)(x^2) =
i solve for x...

25/2 - x - pix + (pi/2)x

and then i attempt to solve for x and get...
x= (-25/2)/(-1 - pi + pi/2)

and i get that x is equal to 4.862306620604284

I then plug that into my area equation and that should be my answer for the problem...
which is:
[b]42.210429215162904[/b]

Any help? Suggestions?
Thanks once again.
Sorry if my work is hard to understand.. hopefully I made no typing errors.

I know it is later, but here is a way to solve the problem using first derivatives and maximums

We have two equations
$ P$ is perimeter. $ A$ is area. $ l$ and $ w$ are length and width respectively.

If the semicircle's diameter is the width of the rectangle, then we can say that the radius is half the width ($ r=\frac{w}{2}$).

The area of the semicircle is $ \frac{1}{2}\pi r^{2}$
Area of Semicircle(plugging in for $ r$) is $ \frac{\pi w^{2}}{8}$

$ P=2l+w+\frac{\pi w^{2}}{2}= 25$
Solving for $ ;$ we get
$ l={\frac{25-\frac{w \pi}{2}-w}{2}= \frac{25}{2}-\frac{w \pi}{4}-\frac{w}{2}}$

Next, the area of the window
$ A = wl+\frac{\pi w^{2}}{8}$

Plug in for $ l$

$ A = w({\frac{25}{2}-\frac{w \pi}{4}-\frac{w}{2})+\frac{\pi w^{2}}{8}}$

$ A = \frac{(-\pi-4)w^{2}+100w}{8}$

We want to maximize this, so take the first derivative and set it equal to zero

$ \frac{dA}{dw}= \frac{25}{2}-\frac{(\pi+4)w}{4}= 0$

Solve for $ w$
$ w = \frac{50}{\pi+4}$

Plug into our equation for $ l$
$ l = \frac{25}{\pi+4}$

Now to find the area, plug $ l$ and $ w$ back into our original area equation

$ A = 43.74 ft^{2}$

And to check to make sure it is right, plug $ l$ and $ w$ back into the perimeter equation to make sure it equals 25 ft, which it does.

i was fortunately able to get it before the deadline! yay =)

but i do like your method of solving it ankur... my way seems more confusing.

thanks!

[quote="JBL"]1. Learn LaTeX.  There is a very helpful FAQ in the LaTeX forum.

2. Right in your first step, you used the formula for the area of a circle, not a semicircle.[/quote]

 ill be sure to look it up.
thanks for the help.


I am running a tourney, as my graduation from newbdom. The first five contestants are lucky, as I already have the first five places down:

1. biblioman
2.
3.
4.
5.

This is for anyone, but [i]post your rating[/i]! Else I don't know who to match you against. I want suggestions on the minimum number of contestants, but it starts the day after I have the agreed number.

WARNING: I am semi-active, so IF I do not post for 7 days, I personally will pass on the duties of running the tournament to somebody else, so PM me if the deadline has passed, and I will choose the person. who will be the person who PMed me. Feel free to pass the title down as needed.

I just realized I haven't said the format. Here goes:

[b]FORMAT[/b]:

2 and only 2 people against each other, with lowest rating against 2nd lowest, 3rd against 4th, et cetera up to 2nd highest vs. highest. Winner moves on. 3 games per deciding match, which means that you play 2 games. If one person (A) has beaten the other (B) both times, then A moves on. Else, there is one more match. NO COUNTDOWNS. PERIOD. 10 questions, 45 seconds each, Responses style.

More moderators would be appreciated.

My rating is 1340-1360 depending on the day.

Thanks,
PLEASE SIGN UP!,
biblioman

Join, even though I stink at FTW.  :P

Rating: 1160-1180 (depends when you catch me.)
FTW test rating: 1231

You should totally have the tourney happen in FTW test. :P

I sign up, rating:1050-1100 Depends on the day.
1. biblioman
2. westiepaw
3. monkeygirl13
4.
5.

Hi westiepaw and monkeygirl,

Thanks! Spread the word, because I would like to have 8 or 16 people.

Thanks! Again, if you could spread the word, I would appreciate it. My current goal is 8 sign-ups.

1. biblioman
2. westiepaw
3. monkeygirl13
4. ksun48

Matchups:

Monkeygirl13 vs. westiepaw
biblioman (me!) vs. ksun48

I'll join.

1. biblioman
2. westiepaw
3. monkeygirl13
4. ksun48
5. $ \text{\LaTeX}$

Current rating is 1570, high is 1663.

I'll join if you change scoring type.
I don't like responses :wink:
My current rating is 1345
Test rating is 1291
high for normal is 1444...hmmmmmmm i seem to have lost a lot recently :(

Okay, ranking or time?

1. biblioman
2. westiepaw
3. monkeygirl13
4. ksun48
5. Chompy

Thanks!

Did you guys kick me out?  :(

If not: Time pwns rating.

Oh, sorry Latex:
Lets retry this:

1. biblioman
2. westiepaw
3. monkeygirl13
4. ksun48
5. $ LaTeX$
6. Chompy

Sorry, I could not get your name to look right, LaTeX.

All right, take votes:

ksun48 and biblioman vote Ranking
$ LaTeX$ votes Time.

As for the time we start, it will be established the day I get 8 people. It will most likely be the day after that, in the evening, about 7:00-8:00 PM.

I'll join iff [that's not a typo] you reduce the time to 25 seconds or less.  Either ranking or time is fine with me.

Tentatively... what do you guys think? I vote that we remain at 45 seconds, but that is just me.

1. biblioman
2. westiepaw
3. monkeygirl13
4. ksun48
5. LaTeX
6. Chompy
7. (depending on opinions) ernie

I beat monkeygirl today.

New rating: 1170

$ LaTeX$ 2-0 slightly beats ksun48

BFTR1M3KvL

[size=75][color=red]i'm not exactly sure what the above means, but i'm pretty sure you don't have to post it 10+ times.
please refrain from doing that in the future
~vallon[/color][/size]

$ \text{\LaTeX}$ defeats ksun48 in 2 games. 15-5, 10-7.

All right. I PMed ernie and lotsofmath. If he doesn't get a reply within 2 days, we move on to Round 2.

I will be away this weekend, so we will start Round 2 today or next Monday.

EDIT: If he doesn't hear from him within 2 hours, I will send out the PMs so that we aren't held up.

Wait until next Monday, its not entirely their fault if they can't find a suitable time. Of course, if lotsofmath is intentionally avoiding a game (with I doubt), then we start Round 2 today.

[quote="lotsofmath"]I PM'd ernie yesterday, asking for a suitable time. He hasn't responded yet.[/quote]

And I shall respond here.  How about tomorrow?

All right. Round two starts on Monday. We are patiently waiting. Yes, I stole that from you, ernie.  :P

All right. I will send the PMs out. And thank you, ernie.

EDIT: The PMs will go out Monday.

I PMed the Round 2 contestants with their matches. Also, we are NOT placing everybody.

Just wondering, is this dead? I haven't seen lotsofmath or biblioman on in a while.....

No, it is not.

NOTE TO PARTICIPANTS IN ROUND 2: PM ME YOUR RESULTS!!

-biblioman


Fie numarul natural $A$. Notam cu $s$ suma cifrelor lui $A$ cu $S$ suma cifrelor lui $5\cdot A$ si fie $T=(2\cdot S-s): 9$.
Sa se arate ca $T$ este egal cu numarul de cifre impare din scrierea lui $A$.

fie A format cu cifre din $\{0 , 1\}$.
atunci $5\cdot A$ e clar de determinat si cu siguranta
$S=5\cdot s$ => $T=(10s-s): 9=s$ adica numarul de 1
din A care este si numarul de cifre impare.
sa presupunem ca mai adaugam un 2 la multimea de apartententa
a cifrelor.(unde $k$ numarul cifrei nou adaugate la multimea
de aparatenenta a cifrelor)
acel 2 va influenta pe
$s'=s+k*2$
$S'=S+k$(deoarece 2*5=10 => 1+0=1 => adaugam cate un
1 pentru fiecare 2)
deci $T'=(2S'-s'): 9=( 2S+2k-s-2k): 9=(2S-s): 9=T$
sa presupunem acum ca adaugam un 3 la multimea de cifre.
$s'=s+k*3$
$S'=S+k*(1+5)$ (de la 15 care apare din 5*3)
atunci $T'=(2S'-s'): 9=(2S+k*12-s-k*3): 9=(2S-s+k*9): 9=T+k$
pentru 7
$s'=s+k*7$
$S'=S+k*8$(5*7=35 => 3+5=8)
avem $T'=(2S'-s'): 9=(2S+k*16-s-k*7)=T+k$
pentru 5
$s'=s+k*5$
$S'=S+k*7$(5*5=25 => 2+5=7)
avem $T'=(2S'-s'): 9=(2S+k*14-s-k*5)=T+k$
lafel si pentru 9 si pentru celelalte numere pare...
de ex pentru $8$ avem $s'=s+k*8$
si $S'=s+k*4$
cazurile care au ramas se fac lafel
(initial am vrut sa fac o inductie dar nu prea iese...deoarece
suma_cifre(5*cifra) - cifra   nu se imparte la 9,decat daca se considera
fiecare caz in parte)

Cam multe cazuri...
Exista o solutie mult mai eleganta. :)

Eu zic sa mai asteptam putin. Problema mi-a placut mult si nu cred ca merita sa-i dau solutia asa repede.

Fie $f : \mathbb N \to \mathbb N, \, f(n) = \sum_{k=0}^{m}a_{k}$, unde $n = \sum_{k=0}^{m}a_{k}10^{k}, \, a_{k}\in \overline{0,9}, \, \forall k \in \overline{0,m}$.
Folosind aceasta notatie, $T_{A}= \frac{2 \cdot f(5A)-f(A)}9$.

Fie $A = 10p+q$, unde $p,q \in \mathbb N, \, q \in \overline{0,9}$.
\begin{eqnarray*}2 \cdot f(5A)-f(A) &=& 2 \cdot f(50p+5q)-f(10p+q) \\ \ &=& 2 \cdot f(50p)+2 \cdot f(5q)-f(10p)-f(q) .\end{eqnarray*}
Avem
\begin{eqnarray*}2 \cdot f(5q)-f(q) &=& 2 \cdot f \left( 10 \cdot \left\lfloor \frac{q}2 \right\rfloor+10 \cdot \left\{ \frac{q}2 \right\}\right)-f(q) \\ \ &=& 2 \cdot \left\lfloor \frac{q}2 \right\rfloor+20 \cdot \left\{ \frac{q}2 \right\}-q \\ \ &=& \begin{cases}0, \ \ \ \ \ \ q \ \textrm{par}\\ 9, \ \ \ \ \ \ q \ \textrm{impar}\end{cases}\end{eqnarray*}
Asadar, $T_{A}= T_{p}$ daca $q$ este par si $T_{A}= T_{p}+1$ daca $q$ este impar.

Acuma problema poate fi rezolvată prin inducţie.


function
algebra
domain
analytic geometry
graphing lines
slope

This was a problem that wasn't getting much attention [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180938]here[/url].

I quickly noticed that $ \boxed{x\equal{}1}$ is a solution, but I'm still curious as to how we can manipulate or work with this more elegantly (and not brute-forcing it to such an extent of cubing it again and again until we get something workable and then check the possible solutions :wacko: ).

[hide="Completion of the above"][b]Sign of x:[/b] If $ x < 0$, then since $ x \plus{} 7 > x$, we must have that $ 2x \minus{} 1$ is positive. However, the latter cannot be true for $ x < 0$. Since, clearly, $ x \neq 0$, we must have that $ x > 0$.

[b]Term-wise rates of change:[/b] Now, note that the rate of change of $ (x \plus{} 7)^{\frac {1}{3}}$ is less than that of $ x^{\frac {1}{3}}$ because the cuberoot function is concave-down over the positives (imagine a cuberoot graph - it gets less steep).

Too, since the cuberoot function is everywhere increasing, $ (2x \minus{} 1)^{\frac {1}{3}}$ has a positive rate of increase as well.

[b]Overall rate of change:[/b] Thus, the combined rate of increase of $ \sqrt [3]{x} \plus{} \sqrt [3]{2x \minus{} 1}$ is greater than that of $ \sqrt [3]{x \plus{} 7}$, and thus $ f(x) \equal{} \sqrt [3]{x} \plus{} \sqrt [3]{2x \minus{} 1} \minus{} \sqrt [3]{x \plus{} 7}$ is strictly increasing.

[b]Conclusion:[/b] Since $ f(x)$ is strictly increasing over the valid domain, there can be at most one solution. We notice that $ \boxed{x \equal{} 1}$ is said solution, and we are done.[/hide]

There is a strategy at this kind of equations . I'm not quite sure that it works here .

First we note like this : $ \sqrt[3]{x\plus{}7}\equal{}a$ $ \sqrt[3]{2x\minus{}1} \equal{}b$ and $ \sqrt[3]{x}\equal{}c$ .
We have that $ a \minus{} b \equal{} c$ , $ a^3 \equal{} c^3 \plus{} 7$ , and $ b^3 \equal{} 2c^3 \minus{}1$ .
We have to solve this sistem of 3 equations now .

[quote="The Zuton Force"][hide]Now, note that the rate of change of $ (x \plus{} 7)^{\frac {1}{3}}$ is less than that of $ (x \plus{} 7)^{\frac {1}{3}}$ because the cuberoot function is concave-down over the positives (imagine a cuberoot graph - it gets less steep).[/hide][/quote]

 :huh: Is that a typo? Anyway, I don't get your logic...

Fixed, the rate of change of $ \sqrt[3]{x\plus{}7}$ is strictly less than the rate of change of $ \sqrt[3]{x}$.

Here's the same logic, heuristically (and possibly somewhat inaccurately worded).

The solution(s)s to that equation is/are the root(s) of $ f(x) \equal{} \sqrt[3]{x} \minus{} \sqrt[3]{x\plus{}7} \plus{} \sqrt[3]{2x\minus{}1}$

(1) Consider the graph of $ \sqrt[3]{x}$ over the positive numbers. As $ x$ increases, the steepness decreases.

(2) Thus, for any $ x$, $ \sqrt[3]{x}$ has greater slope than $ \sqrt[3]{x\plus{}7}$.

(3) Thus, the slope of the difference $ \sqrt[3]{x} \minus{} \sqrt[3]{x\plus{}7}$ wrt x is positive.

(4) Also, $ \sqrt[3]{2x\minus{}1}$ has positive slope. Thus, their sum, $ \sqrt[3]{x} \minus{} \sqrt[3]{x\plus{}7}\plus{}\sqrt[3]{2x\minus{}1}\equal{}f(x)$, has positive slope. Since $ f(x)$ is strictly increasing, it can only cross the $ x$ axis once.

(5) Thus, there is only one solution.

Nice logic.  :)
Though, is there a way to obtain $ x\equal{}1$ other then using brute force or the system of equations.

[hide="Another approach"]
We have $ \sqrt[3]{x\plus{}7}\equal{}\sqrt[3]{x}\plus{}\sqrt[3]{2x\minus{}1} \equal{} 2 \frac{\sqrt[3]{x}\plus{}\sqrt[3]{2x\minus{}1}}{2} \leq 2\sqrt[3]{\frac{3x\minus{}1}{2}}$

Cubing gives $ \frac{x\plus{}7}{8} \leq \frac{3x\minus{}1}{2} \implies x \geq 1$

Can we make the other side (ie, $ x \leq 1$) in a similar way?
[/hide]


limit
logarithms
calculus
calculus computations

$ \lim_{n\to \infty} \left(\frac{\sqrt[n]{a}\plus{}\sqrt[n]{b}}{2} \right)^n$

The full result is this:

Fix $ a,b>0$ with $ a\ne b.$ Define $ M(p)\equal{}\left(\frac{a^p\plus{}b^p}2\right)^{\frac1p}$ for $ p\ne0$ and define $ p(0)\equal{}\sqrt{ab}.$ Then $ M(p)$ is continuous on $ \mathbb{R}$ (which notably includes the statement that $ \lim_{p\to0}M(p)\equal{}M(0),$ which is what this problem asks for). Also: $ M(p)$ is strictly increasing, $ \lim_{p\to\infty}M(p)\equal{}\max(a,b),$ and $ \lim_{p\to\minus{}\infty}M(p)\equal{}\min(a,b).$

This is a pretty famous collection of results. After all, $ M(0)<M(1)$ is AM-GM and $ M(\minus{}1)<M(1)$ is AM-HM.

Just to look at the part asked for here:

$ \ln M(p)\equal{}\frac1p\ln\left(\frac{a^p\plus{}b^p}2\right)\equal{}\frac1p\ln\left(\frac{e^{p\ln a}\plus{}e^{p\ln b}}2\right)$

For $ p$ near zero, $ e^{p\ln a}\equal{}1\plus{}p\ln a\plus{}O(p^2)$ and $ e^{p\ln b}\equal{}1\plus{}p\ln b\plus{}O(p^2).$

Add those together to get that

$ \frac{e^{p\ln a}\plus{}e^{p\ln b}}2\equal{}1\plus{}p\left(\frac{\ln a\plus{}\ln b}2\right)\plus{}O(p^2)$

$ \ln\left(\frac{e^{p\ln a}\plus{}e^{p\ln b}}2\right)\equal{}p\left(\frac{\ln a\plus{}\ln b}2\right)\plus{}O(p^2).$

Divide that by $ p$ and take the limit at $ p\to0$ to get that

$ \lim_{p\to0}\ln M(p)\equal{}\frac{\ln a\plus{}\ln b}2$

Exponentiate that to get that $ \lim_{p\to 0}M(p)\equal{}\sqrt{ab},$ which is the geometric mean of $ a$ and $ b.$

Thanks Kent Merryfield!!!
I like this solution...

Letting $ M(p) \equal{} \left(\frac {a^p \plus{} b^p}2\right)^{\frac1p}$ as Kent did, we could also set $ f(p) \equal{} \ln {\frac{a^p\plus{}b^p}{2}}$ and note that $ \lim_{p\minus{}>0} \ln{M(p)} \equal{} \lim_{p\minus{}>0}\frac{f(p)\minus{}f(0)}{p} \equal{} f'(0)$ by definition of the derivative. We have $ f'(p) \equal{} \frac {2}{a^p\plus{}b^p}\frac{a^p\ln{a} \plus{}b^p\ln{b}}{2}$, and so $ f'(0)\equal{}(\ln{a} \plus{}\ln{b})/2$ as desired.


trigonometry
inequalities
geometry solved
geometry

Prove that in any triangle we have:

$\sum\left(a^2\left(b+c-a\right)\right) \leq 3abc$

thank you!!

[b]here is a solution:[/b]

$3abc-a^2b-a^2c+a^3-b^2c-ab^2+b^3-ac^2-bc^2+c^3\geq 0$ we get that
$\sum\left(a^3-ab^2-ac^2+abc\right)\geq 0$  which gives us : $\sum \left(a\left(a^2-b^2-c^2+bc\right)\right)\geq 0$ which is equivalent to : $\sum\left(a\left(bc-2bc\cdot\cos A\right)\right) \geq 0$ $\Longleftrightarrow$ $abc\left[3-2\left(\sum\left(\cos A\right)\right)\right] \geq 0$ which is true.
we have equality if a=b=c.

[b]Solution 2:[/b]

thinking more over this one I came with another solution:

$\sum\left(a\left(ab+ac-a^2-bc\right)\right)\leq 0$ gives us $\sum a\left(b-a\right)\left(a-c\right)\leq 0$ we multiply it now by 2 and arrange it like so:

$\sum\left(\left(b-a\right)\left[a\left(a-c\right)+b\left(c-b\right)\right]\right) \leq 0$ which gives:

$\sum\left(\left(b-a\right)^2\left(c-a-b\right)\right) \leq 0$ which is true cause c<a+b , a<b+c and b<c+a

cheers!!!

This works for all positive reals a, b, c. It's Schur's inequality:

$\sum a\left(a-b\right)\left(a-c\right) \geq 0$.

grobber's ideea just makes the 2nd proof a lot shorter!!
10x!

cheers! :D


geometry
3D geometry
prism
rectangle
calculus
perimeter

the surface area and volume of a prism are given find the surface area and volume of a prism with (a)double, (b)triple, and (c) half the dimensions of the given prism:

1) S.A.= 23ft.squared
V.= 12ft. cubed
1a)
1b)
1c)

2) S.A.= 61.7m. squared
V= 28.2m. cubed
2a)
2b)
2c)
 :)  :lol:  :wink:

You don't need to find the original dimensions.

Think about it. If you double the side lengths of a rectangle, the rectangle's area is multiplied by 4 (because $ 2 \cdot 2\equal{}4$). For the volume, you multiply another side by the "scale factor" or 2, so it is multiplied by 8.

In general, if you have all the sides of a figure multiplied by the scale factor, you have to multiply its lenght/area/volume/..., which is in n dimensional space, by $ \text{)scale factor)}^n \cdot \text{original length/area/volume}\equal{}\text{new length/area/volume}$

I hope I was clear enough.

As PowerOfPi said before, you actually don' t need to know the dimensions of the prisma. You just calculate the size of the new one, according to the first. But if you really need some kinda calculus, let' s say we have the surface area of prisma = 2 × area of base  +  perimeter of base × H, where H = the height of the prisma.
If you are asked for the double area of this one, you simply multiply 2 with 2 × area of base  +  perimeter of base × H, which is exactly doubling the size of the first prisma. As for the volume result, is the same procedure...


hey all... encountered a small problem with related rates. i can't really get started with this problem because i'm not sure how to solve for what i'm looking for. can someone help me out?

the volume of a rectangular box with square base remains constant at 400cm^3 as the area of the base increases at a rate of 13cm^2/sec. find the rate at which the height of the box is decreasing when each side of the base is 14cm long.

Call the area of the base $A$ and the height $h$, then we have $h=\frac{400}{A} \Rightarrow \frac{dh}{dt} = -\frac{400}{A^2}\cdot\frac{dA}{dt} = -\frac{400}{(14^2)^2}\cdot 13 \approx -.135 \ \text{cm/s}$.

hey thanks a lot haha i forgot about this thread but yeah i figured it out also with the help of a friend... but thanks for the reply anyway! appreciate it!


Let $K$ be a compact of the euclidean space $\mathbb{R}^{n}$ and let $m$ be a positive integer.
If $\text{vol}(K) \geq m,$ there exist at least $m+1$ distinct points $v_{1}, \ldots, v_{m+1}$ of $K$ such that $v_{i}-v_{j}\in \mathbb{Z}^{n}$ for all $i, j.$

Basic idea is the same as in :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=minkowski&t=2456

Pierre.


58 is the number of commutative semigroups of order 4.

can anyone explain me what does this mean

First of all, this is clearly in the wrong forum, as semi-groups are not a standard high school topic in any country I know of.  Second, I suggest you do a search of the web for the definition of "semi-group," "order of a semi-group," and "commutative semi-group."  (Mathworld is a good place to start, although you could probably go check the college topics - algebra section, as well.  Wikipedia also is sometimes helpful.)  At that point, I imagine the meaning of the statement should be clear, although perhaps not how to prove it.

ı've tried them but explanations are so complicated so ı don't want to read it.
also what is the real place of this post? if you'll say it to me, i'll send it there? thanks

Well, I mean, of course it's complicated -- it's from a fairly advanced branch of mathematics.  I can give you a brief explanation, if you like:

Order 4 means 4 elements.  So, we have a semi-group with 4 elements.  Semi-group means that we have a binary operation, usually denoted * (although you might sometimes see +, I'm not really sure what the convention for semigroups is).  So we have the set {a, b, c, d} and an operation * such that if you multiply any two elements in the set, you get a third element in the set.  The only limitations here are that the operation be commutative and associative.  So, the theorem you stated is that there are 58 distinct ways you can set up such a semi-group.  Frankly, that seems like a lot to me, but I'll take your word for it.

Either "other problem solving topics" or "college playground" "superior algebra" would have been a more appropriate place for this post.


How many bijection $ X\to Y$ satisfy $ |X|\equal{}|Y|\equal{}m$


MATHCOUNTS
geometry
circumcircle
perpendicular bisector

Hint
[hide]Find the intersection points.
Then, find the perpendicular bisectors of the triangle.  These will intersect in the center of the circle.  Then, find the radius of the circle.  Then use the equation (x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center of the circle.[/hide]

[hide]is it x^2+y^2=13, if im right ill explain a solution[/hide]

usaha,
Yes.  That's what I got.

I had posted my "hint" before I realized how incredibly convenient the problem was...

Its mathcounts, usually its convenient...

Well, I understand the method that archimedes1 described in his hint, but how would you do the problem quickly?

Like we said, the problem is very convenient.  You can even just look at the intersection points and from there you can find the answer, if you're experienced with geometry.

The circumcenter of a triangle (center of the circle that circumscribes the triangle) is located at the intersection of the perpendicular bisectors of the triangle.

You can see this if you imagine that the perpendicular bisector of AB is the locus of all points that are equildistant from A and B.  Since the circumcenter from be equildistant from A, B, and C, the intersection of the three perpendicular bisectors marks the point equildistant from A, B, and C.

[quote="kheg_k"]How do u get the answer after getting the points of intersection?[/quote]
[hide] We can find the points of intersection to be (2,-3), (2,3), and (-3,2).  Notice that $x^{2}+y^{2}=13$ for all of these points.  Thus, the circle that circumscribes these points is the circle that has an equation of $x^{2}+y^{2}=13$. [/hide]


inequalities
rearrangement inequality
inequalities unsolved

m>n>0 and a,b,c>0 .Prove that
$ \frac{a^{m}}{b^{n}}\plus{}\frac{b^{m}}{c^{n}}\plus{}\frac{c^{m}}{a^{n}}\geq a^{m\minus{}n}\plus{}b^{m\minus{}n}\plus{}c^{m\minus{}n}$

It is difficult?I don't think so :lol:

Hint:
$ (m\minus{}n)\frac{a^{m}}{b^{n}}\plus{}nb^{m\minus{}n}\ge ma^{m\minus{}n}$
by
$ xa\plus{}yb\ge a^{x}b^{y}$
for $ a,x,b,y>0,x\plus{}y\equal{}1$

[quote="zhaobin"]It is difficult?I don't think so :lol:

Hint:
$ (m\minus{}n)\frac{a^{m}}{b^{n}}\plus{}nb^{m\minus{}n}\ge ma^{m\minus{}n}$
by
$ xa\plus{}yb\ge a^{x}b^{y}$
for $ a,x,b,y > 0,x\plus{}y \equal{} 1$[/quote]
thank u 4 your solution but... can you make it more detail ? I'm a newbie

I think zhaobin mean to sum these three ineq:
$ (m\minus{}n)\frac{a^{m}}{b^{n}}\plus{}nb^{m\minus{}n}\ge ma^{m\minus{}n}$
$ (m\minus{}n)\frac{b^{m}}{c^{n}}\plus{}nc^{m\minus{}n}\ge mb^{m\minus{}n}$
$ (m\minus{}n)\frac{c^{m}}{a^{n}}\plus{}na^{m\minus{}n}\ge mc^{m\minus{}n}$

[quote="imathsvn"][quote]
$ xa\plus{}yb\ge a^{x}b^{y}$
for $ a,x,b,y > 0,x\plus{}y \equal{} 1$[/quote]
Can you prove this ?[/quote]
Weighted AM-GM inequality!
[i]Case 1:[/i] x,y arw rational numbers
Set $ x\equal{}\frac{m}{m\plus{}n},y\equal{}\frac{n}{m\plus{}n}, m,n\in N$
By AM-GM inequality, we have:
$ ma\plus{}nb\ge (m\plus{}n)a^{\frac{m}{m\plus{}n}}b^{\frac{n}{m\plus{}n}}$
It follows that: $ ax\plus{}by\ge a^{x}b^{y}$
[i]Case 2:[/i] x,y arw real numbers, there exist two sequences rational numbers $ (r_{n})_{n\ge 0}$ and $ (s_{n})_{n\ge 0}, r_{n}\rightarrow x, s_{n}\leftarrow y, r_{n}\plus{}s_{n}\equal{}1$.
So:
$ ar_{n}\plus{}bs_{n}\ge a^{r_{n}}b^{s_{n}}\Leftrightarrow ar_{n}\plus{}b(1\minus{}r_{n})\ge a^{r_{n}}b^{1\minus{}r_{n}}$
Taking the limit when $ n\rightarrow\propto$, we have $ xa\plus{}yb\ge a^{x}b^{y}$

I thinks it is very easy if you use Rearrangement Inequality for two sequences:
$ a^{m}\geq b^{m}\geq c^{m}$
$ \frac{1}{a^{n}}\leq\frac{1}{b^{n}}\leq\frac{1}{c^{n}}$

Choosing a positive rational sequence a1,a2,.. such that Lim an (n->oo) =x and letting bi=1-ai, we get limbn (n->oo) =y .Okay ,The problem 's really upset
Thank u 4 yr response !


Let us take the three integers $ 1$, $ 2$ and $ 3$.

They can be arranged in $ 6$ ways
$ 123$
$ 132$
$ 213$
$ 231$
$ 312$
$ 321$

For each permutation, if it is represented as ($ a$, $ b$, $ c$), compute :
$ S\equal{} (a\minus{}b)^2 \plus{} (b\minus{}c)^2$. So we get the following:

$ 123$ $ \Rightarrow$    $ 2$
$ 132$ $ \Rightarrow$    $ 5$
$ 213$ $ \Rightarrow$    $ 5$
$ 231$ $ \Rightarrow$    $ 5$
$ 312$ $ \Rightarrow$    $ 5$
$ 321$ $ \Rightarrow$    $ 2$
The average value of $ S$ over the $ 6$ permutations here is $ \frac{24}{6}$ = $ 4$.

Can you compute the average value of $ S$, for the general case where you take the integers $ 1$,$ 2$,$ 3$,....$ n$ for any value of $ n$?

The general formulate for $ S$ being: $ (a_1\minus{}a_2)^2 \plus{} (a_2\minus{}a_3)^2 \plus{} (a_3\minus{}a_4)^2  \plus{} .... \plus{} (a_{n\minus{}1} \minus{} a_n)^2$

Are you talking about the average (when using the formula for S) of [b]all arrangements[/b] of 1,2,3,.....n?
If so, how would you find all the arrangements because there were 6 arrangements in your original but there are infinitely many in this set of numbers (1,2,3...n)?

There are $ n!$.  The key here is linearity of expectation; show that it suffices to compute the average value of $ (a_1 \minus{} a_2)^2$ and multiply by $ n \minus{} 1$.


summer program
Mathcamp
Ross Mathematics Program
PROMYS
email
MathPath
number theory

Hi, I thought I should open a thread that directly asks all of you to compare the [url=http://www.math.ohio-state.edu/ross/introduction.html]Ross Program[/url] (and its sister program the [url=http://www.claymath.org/programs/outreach/PROMYS/]PROMYS[/url] program) to [url=http://www.mathcamp.org]MathCamp[/url].

The thing that I find puzzling, for planning purposes, is that the minimum age for MathCamp is lower than it is for PROMYS, and thus my son is going to have one summer when MathCamp is the only logical choice for a summer math program for him (unless lightning strikes and he qualifies for MOP that year, which I'm not planning on). But I seem to recall threads here suggesting that after one year of MathCamp, especially a year at MathCamp taking its number theory courses, there isn't a lot left to learn at Ross/PROMYS.

I'm not trying to stir up a fight here, as I have heard MANY good things about the Ross/PROMYS programs, and I have a high regard for the alumni of those programs who post here, but who has made the head-to-head comparison? After a person has been to one year of MathCamp, there is plenty more to go back for for at least two more years (according to what I have read on this site).

I'm the kind of guy who could readily be persuaded to send my children to a different program the following year just for the sake of variety, but after a first year of MathCamp, what would be the best different program to go to?

[quote="zscool"]I don't know about PROMYS, but the Ross program does allow age exceptions on both ends, so a 13 year old can go as long as he or she is qualified.[/quote]

Thanks for letting me know that. I particularly had PROMYS in mind, because I think I read once (on an email list where Alison's mom posts) that PROMYS has a hard younger age limit because of Boston University dorm rules. If that doesn't apply to Ross, and I can thus disregard the age rules I see on the Ross Web site, that opens up another possibility for that summer. It happens that A+MATH has a summer birthday that falls just after the program start dates for most summer programs. So when a program says, "Must be age [x] by [date of program start]," he always has to wait an "extra" year by strict application of the program rules. For example, this year he will be just a few weeks too young for MathPath, which is our top plan for next year. Next year, he will be just too young for the strict interpretation of the rules for MathCamp, which would be our plan in 2006, but if he is legal for Ross in 2006 I might rather have him try Ross. (We know a Ross alumnus from the 1970s locally, and I am pretty impressed with what he knows about math.)

Rules, rules. I'm a big guy for "rules are rules," and not trying to test the limits, but one always has to take into account exceptional situations. I guess I'll just have him fill out an application form and do an application quiz and see what he decides after he gets a response. It may be that my son is stretching the lower age limit in one of his summer programs this year, but the program didn't seem to have any hard limit on its Web site, and they let him in.

Thanks for the tip. Just one more thing to think deeply about.

Hi!  I'm one of the directors of Mathcamp.

I'm always a little surprised that there isn't more crossover between Mathcamp and Ross/PROMYS: though there are some exceptions, people who come to Mathcamp once tend to keep coming back, and the same for Ross or PROMYS.   Part of it, no doubt, is wanting to see old friends again, but I think that experiencing the variety of different programs is an admirable thing to do.  There's enough going on at Mathcamp that you can come for a summer, stay away from number theory classes, and still be challenged by Ross's offering; conversely, there's still plenty to do at Mathcamp if one has previously been to Ross/PROMYS.  I think there are few enough people who are alums of both programs that there's no consensus on a preferred order for trying the programs, but I would be very interested if there's someone who feels otherwise!


You can find the topic again with a search. If it's your topic that was moved, take the hint about where to post next time.

Try this: look up your own profile. Click on "Find all posts by (your username)". Since the one you're wondering about should be one of your more recent posts, it should be near the top.

But what if it's a topic that I haven't posted in?

[quote="Zakary"]When a topic is moved, how can we find out where it goes?[/quote]
Doesn't the topic remain in the forum it was originally posted in, but with a blue arrow thingy instead of the new post indicator?

Moderators have an option of whether or not to insert the "blue arrow thingy." In practice, it's used only small portion of the time - too many topics get moved to make it the default. And the topic itself doesn't remain - the "blue arrow thingy" is just a pointer, not the topic itself.

To answer Zachary: you can go through the profile of a user that you remember having posted in the topic and do what I suggested above, or you can use the search feature to search for some word or words you are sure were used in the topic. You can also pretend you're a moderator - if you were going to move this, what would be the most likely destination?

Alternatively: is there some reason why the topic might have been deleted altogether? One possible reason for that would be if it contains information about some contest that is still in progress.

[quote="Zakary"]When a topic is moved, how can we find out where it goes?[/quote]

Wait when a topic is moved there is an arrow and also the link...click the link and look at what forum it is in? unless im missing something?  :maybe:

D3m0n, go read the post directly above yours.

Ok.  Thanks.  It looks like I'll have to go searching for the topic.  I was just wondering that if, in case the topic was moved and I couldn't find the topic in the original forum, there was some way of locating it.  Hopefully I have a good memory then.  Thank you!!!

EDIT:  Oh, and I didn't immediately see where the search button at the top was was. >_>  :rotfl: <_<  But I see it now.

Can't you just click the topic title, and you will be redirected to the forum that it was moved to? It worked for me.

Why don't people read posts in the threads they respond to?  1=2, go read Kent Merryfields post in this thread.


Let $ ABC$ be an acute trianle.Prove that:
$ \frac{h_a}{l_a}\plus{}\frac{h_b}{l_b}\plus{}\frac{h_c}{l_c} \ge \frac{21}{10}\plus{}\frac{9r}{5R}$

[quote=Mateescu Constantin]This one is not hard if we take into account the following [url=http://artofproblemsolving.com/community/c1756h1019905_geometrical_inequality_with_sum_of_h_al_a]super sharp inequality[/url]:

\[\boxed{\frac {h_a}{l_a} + \frac {h_b}{l_b} + \frac {h_c}{l_c} \ge 1 + \sqrt{\frac {s^2 + r^2 + 2Rr}{2R^2}}}\, .\]

After a couple of computations it remains to show that in any [b]acute-angled[/b] triangle the following inequality holds:

\[50\Big(s^2 + r^2 + 2Rr\Big) \ge \Big(11R + 18r\Big)^2.\]

But the latter can be proven without too much effort via [url=http://artofproblemsolving.com/community/c6h342626][b]Walker[/b]'s inequality[/url], i.e. $\boxed{s^2\ge 2R^2 + 8Rr + 3r^2}$ as well as the known inequality $\boxed{s\ge 2R + r}$ which holds in acute or right-angled triangles. Specifically, we can show that:

\[
\begin{array}{rcr}
\displaystyle 50\Big[\Big(2R^2 + 8Rr + 3r^2\Big) + r^2 + 2Rr\Big] \ge \Big(11R + 18r\Big)^2 & \text{holds for} & 2 \le \displaystyle \frac Rr \lessapprox 2.95238 \phantom{.} \\\\
\displaystyle 50\Big[\Big(2R + r\Big)^2 + r^2 + 2Rr\Big] \ge \Big(11R + 18r\Big)^2 & \text{holds for} & \displaystyle \frac Rr \gtrapprox 2.39773.
\end{array}
\][/quote]

But

\[\boxed{\frac {h_a}{l_a} + \frac {h_b}{l_b} + \frac {h_c}{l_c} \ge 1 + \sqrt{\frac {s^2 + r^2 + 2Rr}{2R^2}}}\, .\]

is yang xue zhi and yinghua yan inequality(In 1999.11)

see "Inequality Research" P594

Mateescu Constantin:

say you e_mainl,and me:
wshr987@163.com,

and I post you "inequality  Research" .pdf


calculus
derivative
real analysis
integration
linear algebra

What is a good calculus book for self study for a junior entering senior year? I understand limits and derivatives pretty well, they're kinda easy, and want to continue so that next year i can learn calc 2.

It depends how advanced you want to go.  Stewart's book is considered one of the standards, but I've never used it.  If you want a really hard-core, rigorous treatment of calculus, you have to go with Spivak's book.  I've heard Apostol is good but I don't know anything about it.

Oh, right. It's Principles of Mathematical Analysis by Walter Rudin.

[quote="Cyrazeno"]Also, there is Intro to Calculus and Analysis by Richard Courant and Fritz John.
Perhaps even ... baby rudin.[/quote]

That's really bad advice. Rudin's book is by no means meant for an "introduction to Calculus".

My advice is to try George Thomas' [u]Calculus[/u] (my favorite intro to calculus book) or James Stewart's [u]Calculus[/u]. If you know everything in these books already, then perhaps try Michael Spivak's [u]Calculus[/u] or Tom Apostol's [u]Calculus, Vol. 1: One-Variable Calculus with an Introduction to Linear Algebra[/u]. Rudin should be saved until you've mastered all of the topics of introductory calculus, and perhaps even (computational) multivariable calculus.

Whooaa I just looked at Spivak and it is WAY more interesting then all the multivariable computations I've been at recently.

But yeah don't do it if you have no idea what integration is...

[quote]Whooaa I just looked at Spivak and it is WAY more interesting then all the multivariable computations I've been at recently.
[/quote]
Spivak has a really rigorous book but if you sit down with it, you can get a lot out of it.  Some of the problems are really, really interesting.


hmm, just a random thought, can't we just treat every country on a map, through some topological transformation, as a square (or any fixed shape) of a different size? This way, the map (or plane) is composed of whole bunch of squares of different side lengths. :?

Just curious.. How does that help in simplifying the problem?  :?

so then we can start to coordinatize the plane and play with equations? (i don't know, otherwise i'd have published my results already :P ) i'm just wondering if the people who actually have an idea what's going on can tell me whether or not this is helpful. :lol:

I see.. I agree with your square argument..

Intuitively, it seems right, and probably is, like many math problems. I just don't see where to go from it, like you said, I would be working on my paper right now if I knew that.  :D

okay, what's wrong with this:

Take an arbitrary square ABCD in the plane. Let O be its centre. Coordinatize the plane with O as the origin, x and y-axes perpendicular to BC and AB, respectively. Label the direction of positive y-axis N, nega y-axis S, posi x-axis E and nega x-axis W. Now we only consider the squares that touch square ABCD, and are to the N or to the E of square ABCD. Let D{N} and D{E} denote these two sets of squares. Temporarily suppose that there is no other square in the plane except square ABCD, and D{N}. Then we can color D{N} alternatively with only two colors s.t. any two neighbouring squares are of a different color. Similarly, if only square ABCD, and D{E} were in the plane, we could color D{E} with only two colors as well. Now consider the existence of all of square ABCD, D{E} and D{N}. Let D{N_E} denote the most eastern square in D{N}, similarly, D{E_N} denote the most northern square in D{E}. Then we have two cases:

Case 1: D{N_E} is the same as D{E_N}. In this case, it's just a square that shares exactly point B with square ABCD. Hence it requires a different color. And D{N} and D{E} each requires another different color, so we need 4 colors in total.

Case 2: D{N_E} touches D{E_N}. This way we only need to give each of them a distinct color from each other and from ABCD. then D{N} and D{E} can all be colored alternatively using the same colors. So we need only 3 colors in total.

This shows that for any arbitrary square, we can use at most 4 colors to color all the squares touching it in two neighbouring directions. But now we add in the other squares inductively, and using the same argument, we can ensure using still at most 4 colors in total. Right??? :P  :?  :D

I don't find the square statement intuitively right. How will you for example do the situation of croatia-bosniaherzegovina-jugoslavia? one country surrounded by only 2 other countries that touch eachother at both sides?

hmm, i see, but by "squaring" the map, we can only end up with MORE countries, with the extra countries assigned arbitrary colors. So in that regard, suppose that there exists a map could not be colored with 4 colors, then "squaring" it in the plane must also not be able to be colored with 4 colors...isn't it? So it's like a generalized case....:?

p.s. So my argument above is correct IF we can indeed convert a map of countries into squares in a plane???!!! :D :P  :D

I don't get it... can you give me a practical explanation of how you would do do for the thing I listed? I don't think adding countries is a generalisation! Because if 2 adjacent countries cannot have the same color, you are avoiding this by putting in another country? :?

yeah i see the problem, an even bigger problem is what if a country is COMPLETELY within another? Then we can't very well take it out.... :( damn... :(

[quote="al.M.V."]yeah i see the problem, an even bigger problem is what if a country is COMPLETELY within another? Then we can't very well take it out.... :( -... :([/quote] That is not a problem since that is completely trivial, it doesn't influe the amount of colors ;) but indeed, I'm afraid your lemma is false.


calculus
derivative
algebra
system of equations

Solve in $ \mathbb{R}$ the following system of equations:
$ \begin{array}{l}x^{3}\plus{}y^{3}\plus{}z^{3}\equal{} 0\\ x^{5}\plus{}y^{5}\plus{}z^{5}\equal{} 0\\ x^{2007}\plus{}y^{2007}\plus{}z^{2007}\equal{} 0.\\ \end{array}$

I don't know how to find the solutions you actually have to think on to figure out, but couldn't x, y, and z just equal 0?

If x,y,z is solution, then -x,-y,-z is solution. So wlog assumume that two of them (WLOG x and y) are non-negative and z is negative. Let z=-a and now x,y and a are non-negative,
$ x^{3}\plus{}y^{3}\equal{}a^{3}$
$ x^{5}\plus{}y^{5}\equal{}a^{5}$
$ t\equal{}x^{3}$,$ r\equal{}y^{3}$
$ t^{5/3}\plus{}r^{5/3}\equal{}(r\plus{}t)^{5/3}$
One possibility is that t=0, the other is
r/t=k
$ 1\plus{}k^{5/3}\equal{}(k\plus{}1)^{5/3}$
$ 1\plus{}k^{5/3}\minus{}(k\plus{}1)^{5/3}\equal{}0$
LHS is 0 for k=0 and it's derivative is
$ 5/3.k^{2/3}\minus{}5/3(k\plus{}1)^{2/3}<0$, so its decreasing and cannot be 0 for k>0.
In both cases one of r, t is 0, so we know either x,y,or z is 0. The solutions are
[0,t,-t], [t,0,-t], [t,-t,0], $ t\in\mathbb{R}$


I know this is a stupid question...

I'm thinking of taking a class soon, but I don't know how to get on the classes.

I mean, how do you go to class?

Thanks.

It's ok. Everybody has to learn sometime.I didn't now for sometime too.
Click 'classroom' on the left hand side. If you are logging in during your class time it should say that the classroom is loading. This takes some time so click on embedded classroom. It has no pop-ups, that is the only difference.
Hope this was helpful.

You can also go to My Classes and there's a link to the class, especially if there is a math jam running

When you enroll, you'll receive an email with some instructions about how to get to the class homepage.  The homepage has a Course Information document that explains many details about the courses, including how to reach the classroom.

And if some reason the both the embedded classroom and the regular classroom does not work, you can always view a transcript of the class while it is going on. :)


Prove if a,b,c>0  then
$ 2(\sum\frac {a \plus{} b}{c}) > \equal{} \sum\frac {a^2}{bc}\plus{}9$
This inequality is true sure and it is very easy :P

[quote="tranquoc"]Prove if a,b,c>0  then
$ 2(\sum\frac {a \plus{} b}{c}) > \equal{} \sum\frac {a^2}{bc} \plus{} 9$
This inequality is true sure and it is very easy :P[/quote]
Are you sure it is true? Try $ a\equal{}b\equal{}1,c\equal{}3$. :)


geometry
calculus
search
superior algebra
superior algebra theorems

Hello,  I need to  develop  a project on the  close link  b/w algebra  and  geometry
If you can provide some simple algebraic identities proved with geometry(preferably Euclidean)
Please avoid use of calculus beyond the elementary level

you can try to get your hand on this book: [url]http://www.amazon.com/exec/obidos/tg/detail/-/0883857006/002-6477257-6168037?v=glance[/url]
or you can search for some proofs without words via google


Prove that there exist infinitely many positive integers $ k$ for which equation $ n^m\minus{}m^n\equal{}k$ has unique solution in positive numbers.

[quote="rogue"]Prove that there exist infinitely many positive integers $ k$ for which equation $ n^m \minus{} m^n \equal{} k$ has unique solution in positive numbers.[/quote]
$ \forall k>0$: $ (n,m)\equal{}(k\plus{}1,1)$ is solution
then then equation has unique solution
$ \iff$ $ \forall m>1$ we have $ k\neq n^{m}\minus{}m^n$
$ \iff$ $ k\in E\equal{}\mathbb{N}^*\minus{}\{n^m\minus{}m^n: \ m>1\}$

so we must pove that $ E$ can't be finite
i continue after  :wink:

We will prove that if $ k \equal{} 8m \plus{} 3$ or $ k \equal{} 8m \plus{} 5$ then the equation has unique solution in positive integers.
we have $ n^{m} \minus{} m^{n} \equal{} k$ . now we have 2 case :
1)$ m$ is odd and $ n$ is even . now if $ m\geq 3$ then $ n^{m}\equiv 0(mod8)$ and $ m^{n}\equiv 1(mod8)$ . so $ k\equiv 7 (mod8)$ . which is contradiction .
if $ m \equal{} 1$ then $ n \equal{} k \plus{} 1$
2)$ m$ is even and $ n$ is odd . now $ n\geq 3$ because $ n^{m} > 1$
$ m^{n}\equiv 0(mod8)$ , $ n^{m}\equiv 1 (mod8)$ so $ k\equiv 1(mod8).$ which is contradiction :)
so the equation has unique solution $ (m,n) \equal{} (1,k \plus{} 1)$ in positive integers .


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hello,

Let G be a finite group and a,b two elements of order p prime, with a
not in <b>. Show that G contains at least p^2-1 elements of order p.

I am stuck in the case where all the p-Sylow subgroups of G are cyclic
(there are at least p+1 such p-Sylows, but they may overlap, which
prevents me from concluding).

Thanks for your help,

--
Julien Santini

A very useful theorem, which I proved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22490]here[/url], says that if $m$ divides the order of $G$, the number of elements of $G$ with order dividing $m$ is divisible by $m$.

For this problem, we note that the number of elements of order $p$ must be $\equiv-1\mod p$ and must also be divisible by $p-1$. This means that the number of such elements is of the form $(p-1)(kp+1)$ for some nonnegative integer $k$, and the smallest such value greater than $p-1$ is $p^2-1$.

Actually, the special case of prime $m$ is quite easy:
Consider the set $S$ of all ordered $m$-tuples $(a_1,a_2,\dots,a_m)$ of elements of $G$ with product $a_1a_2\cdots a_m=e$. Let $n$ be the number of elements of $G$. Clearly, $S$ has $n^{m-1}$ members, which is a multiple of $m$ since $m|n$. The shift operator $T$ acts by $T(a_1,a_2,\dots,a_{m-1},a_m)=(a_2,a_3,\dots,a_m,a_1)$.
Decompose $S$ into orbits under $T$. All have size 1 or $m$, and the orbits of size 1 are $m$-tuples $(a,a,\dots,a)$ with $a^m=e$. Simple counting verifies that the number of such elements is divisible by $p$, and we are done.

The Sylow theorems are a waste of time on this problem.

"The Sylow theorems are a waste of time on this problem."

The problem is directly connected to a generalizaton of one of the Sylow theorems: "whenever p^r divides the cardinal of a group G, where p is of course prime, then the number of sungroups of G with cardinal p^r is 1 mod p" -which makes the problem trivial-.


geometry
parallelogram
similar triangles

ABCD is a parallelogram. Points Q, R, S, and T are the midpoints of AB, BC, CD, and DA, respectively. If ST=8 and BD=14, then AC+QT=?

[hide]Use similar triangles (i know, i hate them too)
$\frac{ST}{AC}= \frac{1}{2}= \frac{QT}{BD}$

$AC =16, QT =7$
$AC+QT=23$[/hide]

[size=75][color=darkred]Don't forget to hide your answers/solutions.
-nebula42[/color][/size]


Let  a.b >0. Show  that
1/2 (a+b) 2 +1/4(a+b) \geq a \sqrt b+b \sqrt a

The left-hand side equals a2/2 + b2/2 + ab + a/4 + b/4. Now note that
a2/2 + ab/2 + a/8 + b/8 \geq 4(a2/2 ab/2 a/8 b/8)1/4 = a\sqrtb by AM-GM.
Similarly
b2/2 + ab/2 + a/8 + b/8 \geq 4(b2/2 ab/2 a/8 b/8)1/4 = b\sqrta by AM-GM.
Adding these inequalities gives the result.

Here my solution :
        1/2 (a+b) 2  + 1/4(a+b)  \geq  a \sqrt b + b \sqrt a
<=> 2(a-b) 2  + b(2 \sqrt a - 1) 2  + a(2 \sqrt b - 1) 2   \geq  0 (obvious)


linear algebra
matrix
logarithms
inequalities
function
inequalities proposed

Prove that:$\frac{a^{3}}{b}+\frac{b^{3}}{c}+\frac{c^{3}}{a}\geq a\sqrt{ac}+b\sqrt{ba}+c\sqrt{bc}$
where $a;b;c$ are positive real numbers
Can you prove it by many solution?

Let $\vec{x}= \begin{bmatrix}\ln a\\ \ln b\\ \ln c\end{bmatrix}$, $P= \begin{bmatrix}3 &-1 & 0\\ 0 & 3 &-1\\-1 & 0 & 3\end{bmatrix}$ and $Q= \begin{bmatrix}\frac{3}{2}& 0 & \frac{1}{2}\\ \frac{1}{2}& \frac{3}{2}& 0\\ 0 & \frac{1}{2}& \frac{3}{2}\end{bmatrix}$. Then the inequality rewrites as
$\sum_{i}\exp\left(\left[P\cdot\vec{x}\right]_{i}\right)\geqslant \sum_{i}\exp\left(\left[Q\cdot\vec{x}\right]_{i}\right)$.
Since the matrix $P$ is nonsingular, we can make a substitution $\vec{y}= P\cdot\vec{x}$. Then $\vec{x}= P^{-1}\cdot\vec{y}$ and the inequality becomes
$\sum_{i}\exp\left(\left[\vec{y}\right]_{i}\right)\geqslant \sum_{i}\exp\left(\left[Q\cdot P^{-1}\cdot\vec{y}\right]_{i}\right)$.
A direct calculation shows that the matrix $Q\cdot P^{-1}$ is doubly - stochastic (the entries of the first row are $\frac{15}{26}, \frac{5}{26}, \frac{3}{13}$; the second and third row are its cyclical shifts), so $\vec{y}\succ Q\cdot P^{-1}\cdot\vec{y}$ and by the Karamata inequality for the convex function $f(x)= \exp\left(x\right)$ we are done.

since$(3.-1.0)$ majorize$(1.\frac{1}{2}.\frac{1}{2})$ and . ur inequlitie is equivalent to $\sum_{sym}a^{3}b^{-1}\geq \sum_{sym}ab^{\frac{1}{2}}c^{\frac{1}{2}}$ so ur inequaliti holds .

[quote="arkhammedos"]since$\(3.-1.0)$ majorize$(1.\frac{1}{2}.\frac{1}{2}$ and . ur inequlitie is equivalent to $\sum_{sym}a^{3}b^{-1}\geq ab^{\frac{1}{2}}c^{\frac{1}{2}}$ so ur inequaliti holds .[/quote]
But the sum at the left hand side is not symmetric.

help me!
I need about five or six solutions!

Easy I think.

By Cauchy,

\[LHS.(ab+bc+ca) \ge (a^{2}+b^{2}+c^{2})^{2}.\]

Moreover, we have certainly,

\[2RHS \le a^{2}+b^{2}+c^{2}+ab+bc+ca.\]

Sorry but I don't think this one is a very nice and good problem.

I know this solution on "Math for children magazine"
Do you have other solution?


LaTeX
integration
calculus
trigonometry
logarithms
limit
modular arithmetic

[quote="fanzha"]Hey, I just want to create a thread for testing tex.

[tex]\large i=i_{0}[e^{\frac{\alpha_{A}n\Im}{RT}\eta} - e^{\frac{\alpha_{C}n\Im}{RT}\eta}][/tex][/tex][/quote]

use \displaystyle{.....} around your tex, it will make it look much better:

[tex]\displaystyle{i=i_{0}[e^{\frac{\alpha_{A}n\Im}{RT}\eta} - e^{\frac{\alpha_{C}n\Im}{RT}\eta}]}[/tex]

hmm... maybe that particular sequnce of characters, it won't :o.

Fractions in exponents tend not to work very well.

I also noticed that regular latex formatting tags like  \large or \small don't work.  Is there a reason or is the same effect reproducible some other way?

You can use \large and \small in an mbox.
Thus normal size is [tex]\displaystyle \int_{0}^{1}\dfrac{x^{4}\left( 1-x\right) ^{4}}{1+x^{2}}dx=\dfrac{22}{7}-\pi[/tex]
Tiny size is
[tex]\mbox {\tiny \displaystyle \int_{0}^{1}\dfrac{x^{4}\left( 1-x\right) ^{4}}{1+x^{2}}dx=\dfrac{22}{7}-\pi}[/tex]
and Large size is
[tex]\mbox {\Large \displaystyle \int_{0}^{1}\dfrac{x^{4}\left( 1-x\right) ^{4}}{1+x^{2}}dx=\dfrac{22}{7}-\pi}[/tex]

[quote="Spoon"]use \displaystyle{.....} around your tex, it will make it look much better:

[tex]\displaystyle{i=i_{0}[e^{\frac{\alpha_{A}n\Im}{RT}\eta} - e^{\frac{\alpha_{C}n\Im}{RT}\eta}]}[/tex][/quote]

Here's one way to make the exponents clear - using \mbox, as mentioned in the previous post. Try the various sizes \large, \Large, \LARGE, \huge, \Huge
This is what \LARGE looks like
[tex]\mbox{\LARGE \displaystyle{i=i_{0}[e^{\frac{\alpha_{A}n\Im}{RT}\eta} - e^{\frac{\alpha_{C}n\Im}{RT}\eta}]}}[/tex]

The full range of sizes is:
\tiny, \scriptsize, \footnotesize, \small, \normalsize, \large, \Large, \LARGE, \huge, \Huge

This is my first time using Latex...heh

[tex](3-25)^2(x+3)=5[/tex]

BTW...how do you use integrals and other symbols?

[tex]\mbox {\Large \displaystyle \int_{1}^{\sqrt[3]3}}z^{2}dz\ast\cos{\dfrac{3\pi}{9}}=\ln{\sqrt[3]e}[/tex]

Did that work?

Here are a bunch of the more common symbols:

[tex]\displaystyle\sum_{n=0}^\infty a_n \prod_{k=1}^n b_k[/tex]

[tex]\displaystyle \lim_{t\to \infty} \int_{-t}^t e^{-x^2}\; dx=\sqrt{\pi}[/tex]

[tex]\displaystyle 19\equiv 3 \pmod 4[/tex]

[tex]\displaystyle n\in\mathbb{Z}\subset\mathbb{Q}[/tex]

testing \times vs. \cdot

[tex]\mbox{\Large a\cdot b ... a\times b}[/tex]

*EDIT* - DUH! lol

\ldots instead of three periods is the preferred way to enter an ellipsis.

[tex]\ldots \mathrm{~v.s.~} ...[/tex]

Hm. I usually use \cdots (at least for multiplication), as in [tex]n!=n\cdot(n-1)\cdots 2\cdot 1[/tex].

how does \ work, which parts does the \ modify?

\ introduces a macro, but some things don't take \s. The best examples of things that don't take \s are ^ (superscript) and _ (subscript). More or less everything else that you don't want to see in plain text will take a \, such as \sum for [tex]\sum[/tex], \log for [tex]\log[/tex] (note that this is not the same as log without the \, which would look like [tex]log[/tex]), \sqrt{\pi} for [tex]\sqrt{\pi}[/tex], etc.

Also, you may have noticed that a { or a } won't appear in your text normally.  If you want them, you have to do \{ and \} (and I think also \[ and \], although I don't remember for sure).

You don't need to do \[ and \] to get [ and ]. I think those do something funny with spacing, but I have never used them.

The characteristic polynomial $ f(\lambda)$ of the
$ 3\times 3$ matrix
\[ \begin{pmatrix}a & b & c\\ d & e & f\\ g & h & i\end{pmatrix}\]
is given by the equation
\[ f(\lambda) \equal{}\begin{vmatrix}\lambda\minus{}a &\minus{}b &\minus{}c\\ \minus{}d &\lambda\minus{}e &\minus{}f\\ \minus{}g &\minus{}h &\lambda\minus{}i\end{vmatrix}.\]

[hide="Unparseable or potentially dangerous latex formula.Error: 3"]$ \frac{1}{4\pi}\oint_{\Sigma}\frac{1}{r}\frac{\partial U}{\partial n}ds$[/hide]

There is now a [url=http://www.artofproblemsolving.com/Forum/index.php?f=224]Test Forum[/url] for this sort of thing, rather than reviving an ancient thread.


Well since Vista has been launched, and it seems that it includes a nicely done nav. panel with embedded RSS feed reader (as IE 7 and other modern browsers do) I thought we might use this opportunity to launch the AoPS - MathLinks RSS 2.0 Feeds.

For AoPS we have [code]http://www.artofproblemsolving. com/Forum/feed.php[/code] (which feeds topics from the HighSchool Forums) and for MathLinks we have [code]http://www.mathlinks. ro/Forum/feed.php[/code] (which feeds topics from the Olympiad forums).

More feeds will be available soon aswell.

That sounds good!
Ummm...what is an RSS feed?

Google for RSS (I'm sure you'll find tons of info about it). To use it the simplest way would be to input those links above in the code tags (without the space between them) in IE 7 and subscribe to the feed.

Here's how it looks on my computer
[img]7080[/img]

[quote="solafidefarms"]How about blog RSS feeds???[/quote]Will come soon.

Is it something like google reader?
Like I don't have to go to the site to read it?
(In fact, when I used to have google desktop, and had visited a friends blog, I used to get the blog entries on it.)

You can use Google to read the feed (although if you do that you might as well go on the forum as you do now :D). It's recommended for Windows Vista (and/or) Internet Explorer 7 users.

Isn't the feed just the same thing as viewing the regular web page?
how do i add feeds on IE7?

[b]Instructions for Firefox 2.0+
[/b]

Go to this link:
[url]http://www.artofproblemsolving.com/Forum/feed.php[/url]
or
[url]http://www.mathlinks.ro/Forum/feed.php[/url]

Select "Live Bookmarks" in the drop down box.

Hit "Subscribe Now".

You should now see an orange square which has a dot and 2 quarter circles on the top part of your browser, under the address bar. Click it, and a drop down box of recent post titles will appear. You can click on one of those titles, and it will take you to the topic.

[b]Instructions for IE 7
[/b]

Go to this link:
[url]http://www.artofproblemsolving.com/Forum/feed.php[/url]
or
[url]http://www.mathlinks.ro/Forum/feed.php[/url]

Hit "Subscribe to this feed".

Now, press your favorites icon and tab over to Feeds. Then click the Art of Problem solving one. It takes you to a page where you can preview all of the posts and if you click on their titles you can go to those topics.

You might want to add them to your Gmail web clips :D

I didn't want to post the links because both links will point to the same site (the one that the user is using to access the forum) :)

[quote="#H34N1"]It says somethng about not proper authorization[/quote]

It's giving us Sorry_auth_view_album when we click on the picture link you posted.

When I click on the firefox link, it takes me to a bunch of code with the following error message at the top: "This XML file does not appear to have any style information associated with it. The document tree is shown below."

[quote="#H34N1"]That's what its saying.[/quote]

I know.  I was trying to get Valentin's attention...

Indeed it is a problem. I may be a gallery mod, but I can't even see Valentin's pictures.

I dont' understand, are you having trouble with the image posted or with the RSS feed?

Edit: the picture was indeed problematic, as I put it my personal album pictures :D Fixed now, you should be able to see it.

RSS Feeds work properly for me at this point on both Vista and XP and on both IE 7 and Firefox 2.0

I am having no problems with the RSS feeds, I only had problems with the picture.  :lol:

[quote="#H34N1"]What about RSS feeds for comments?

Why need RSS feeds when http://www.artofproblemsolving.com/Forum/marquee.php?ls=4&width=800&height=600[/quote]

Umm...RSS is like a newspaper.  Multiple news stories from multiple sources are delivered to one place, so you can read them all at once in one place.  You can also use RSS to pop up a message on your computer desktop every time a new thread is posted, if you are so inclined.

There's a bunch of downloadable RSS readers that do that.  I know the old version of Pluck used to do it, but it got annoying, so I uninstalled it.

[quote="#H34N1"]What about RSS feeds for comments?

Why need RSS feeds when http://www.artofproblemsolving.com/Forum/marquee.php?ls=4&width=800&height=600[/quote]Feeds are a totally different thing than that marquee. That marquee is a list of the latest topic links.

The comments you will see by visiting the site :D

[quote="Valentin Vornicu"]
For AoPS we have [code]http://www.artofproblemsolving. com/Forum/feed.php[/code] (which feeds topics from the HighSchool Forums) and for MathLinks we have [code]http://www.mathlinks. ro/Forum/feed.php[/code] (which feeds topics from the Olympiad forums).

More feeds will be available soon aswell.[/quote]

Are there more feeds yet?  It would be awesome to be able to choose feeds on a forum-by-forum basis.

Thanks for all your hard work!

Are there different feeds now?

I recently tried to access the links in the first post and they no longer work :(.

What? They do work. http://www.artofproblemsolving.com/Forum/feed.php and http://www.mathlinks.ro/Forum/feed.php

Is it possible to get the feed of a particular forum?

eg: Combinatorics forum?

Sorry I was wondering, but is it possible to get a feed of the middle school and particular forums?

Thanks in advance


can someone please explain breifly what are the scaling principles? I was looking an old uil calc test.

thanks

it's a proportion...
if you are scaling a 3-d object, then you will have to cube it
2-d, then square it

ex:
a turkey (a 3-d object) that is 11 feet long feeds 10 people, how many people can a 15 feet turkey feed??

it would be:
let x be the answer
11^3/10 = 15^3/x
as you can see, the proportion of the turkey must be cubed because it is in a 1-d format.

next for 2d:
a 7 feet tall pyramid has a 15 sqr feet base, then what would be the base of a 9 feet tall pyramid.

ans:
7/(sqrt of 15) = 9/(sqrt of x)
this is because the base is already cubed, so you're looking for the length of the base in proportion with the height....


Se da un triunghi $\ ABC$ in care notam $\ G$ centrul de greutate si $\ L$ punctul lui Lemoine, $\ AM,BN$ mediane si $\ R,S$ doua puncte in interiorul triunghiului $\ ABC$(nu pe laturi).Daca exista relatia \[(a^{2}-b^{2})(S(CAR)-S(ABR))(S(ABC)-S(BSC))+(b^{2}-c^{2})S(BCR)(S(BSC)-S(ABS))+(c^{2}-a^{2})S(CAS)(S(CAR)-S(ABR))=0\] demonstrati ca dreptele $\ LG,MR,SN$ sunt concurente.

Prin $\ S(XYZ)$ intelegem aria triunghiului $\ XYZ$.

Impartim tot monstrul ala la $S_{ABC}$, dupa care folosim [b]coordonate baricentrice[/b] (Daca in punctele $A,B,C$ se pun greutatile $S_{BTC}, S_{CTA}, S_{ATB}$, atunci centrul de greutate al sistemului e chiar $T$.). Urmeaza niste calcule plictisitoare, dupa care (cel putin teoretic) obtinem concluzia.

Acuma n-am timp sa fac calculele. Poate diseara sau maine.

Da, asta era si ideea, Tiberiu.Asa am compus-o, folosind coordonatele baricentrice.Poate ceva mai greut era sa stii ca $\ L\left(\frac{a^{2}}{a^{2}+b^{2}+c^{2}},\frac{b^{2}}{a^{2}+b^{2}+c^{2}},\frac{c^{2}}{a^{2}+b^{2}+c^{2}}\right)$

Se poate si mai general: Daca $P$ are coordonate $\left( x, y, z \right)$ (nu neaparat normalizate), atunci izogonalul sau (nu sunt sigur cum ii zice exact in romana) are coordonate $\left( \frac{a^{2}}{x}, \frac{b^{2}}{y}, \frac{c^{2}}{z}\right)$.

Da, punctul ala se numeste izogonal.Mai exista si puncte izotomice, dar oricum..


calculus
integration
derivative
calculus computations

$ \int\frac {a}{cos^2ax}$

answer:tanx+C

I got the answer but how do I apply the double angle formula to get the same result.
If it cannot be integrated then it what cases can I apply the double angle formula to integrate

$ cos2ax \equal{} 2cos^2ax \minus{} 1$

$ acos^2ax \equal{} \frac {cos2ax \plus{} 1}{2}$

$ \int\frac {a}{cos^2ax} \equal{} 2a\int\frac {1}{cos2ax \plus{} 1}$

how to proceed from here to get $ tanax \plus{} c$

--------------------------------------------------------------------------

for $ \int{cos^2x}$

can I say =$ \minus{} \frac {cos^3x}{sinx}$ and proceed

But if I use double angle formula

$ \int{cos^2x}$=$ \int\frac {cos2x \plus{} 1}{2}$

does not equal the above expression! wacko wacko!

[quote="Bos1234"]how to proceed from here to get tanax + c[/quote]

Make the substitution tan(2ax/2) = tan(ax) = t.

[quote="Bos1234"]can I say = $ \frac {\cos^3x}{\sin x}$ and proceed[/quote]

Where did that cos^3x/sinx came from?

I don't think you need to use the double angle formula for the first one. It  will only make things more complicated. Set $ u\equal{}ax\equal{}>du\equal{}adx$ so you have to integrate $ 1/cos^2 u$ with respect to u. But $ 1/cos^2 u\equal{}sec^2u\equal{}d/du(tanu)$
So the Integral that you want is $ tanu\plus{}c\equal{}tan(ax)\plus{}c$ where c is a constant.
For the second one you cannot do what you said. If you take the derivative of $ \minus{}cos^3x/sinx$ it is not equal to $ cos^2x$
For the second one you have to use the double angle formula. You can do it from here right?

[quote="Carcul"][quote="Bos1234"]how to proceed from here to get tanax + c[/quote]

[quote="Bos1234"]can I say = $ \frac {\cos^3x}{\sin x}$ and proceed[/quote]

Where did that cos^3x/sinx came from?[/quote]

sry I misunderstood the theory :blush:

ok thanks I can do it from here


Welcome to the Mock AMC 12 9/09! [hide="85 participants?"]

bargello
rd5493
isabella2296
Poincare
bokagadha
PowerOfPi
professordad
Ihatepie
narik
modularmarc101
themorninghlighttt
AIME15USAMO
TachyonPulse
zapi2007
basketball9
AwesomeToad (assuming you're gonna take it)
Complex_Ninja
pascal12: preference Friday
athunder
camath08, time assuming
Thunder365
maybach
CyanRedYellow
veezbo: preference Friday
Triple-AAA
limac
rts2007
AlphaBetaTheta
NuncChaos
modx07
lightning
AIME15
mathemonster
Jongao
bbgun34
CountdownKing
SnowEverywhere
MrBob
varunrocks
wormABC
mathrat, time assuming
crazypianist1116
Paiev
yaofan
SorcerorofDM (and his 9 friends)
Stokes93
Mewto55555
shentang
FantasyLover
mathemagician1729
bulutcocuk
Brut3Forc3's math club
alkjash
zhou_qiang
pumpkinpi
blackbelt14253
Sephiroth
winternights1
andersonw
dnkywin
cognos599
zserf
westiepaw
Just_Beginner
dgreenb801
BOGTRO
iYOA
n0tn0t
David2143
pytheagle
infinite_sigma
ProtestanT
15!!!
nohomo332
gogators
EstaticPotter
Bacteria
[/hide]
[hide="Official Friday participants"]
David2143
pascal12
veezbo
countdownking
themorninglighttt
Snoweverywhere
Bacteria
professordad
[/hide]

You may still take it even if you have not signed up. Please take the Saturday time if possible. However, we will be lenient and let you submit before the beginning of Sunday. It would be nice if you followed the time limit though.  :wink:  :wink:

[b]OFFICIAL TEST DATES:[/b]
(Alternate) Friday: 6:00 P.M.-7:15 P.M. / 7:30 P.M. EDT (3:00-4:15/4:30 PDT)
Saturday 9/19 4:00 P.M.-5:15 P.M. / 5:30 P.M. EDT (4:00-5:15/5:30 PDT)

[b]You will get 75 minutes to officially take this exam. [/b] However, you may turn it in late (Saturday night) and we'll still grade it, but please put how much time it took you to complete the mock AMC 12 for statistics purposes.

[b]How you should take the test:[/b]
Download the file. Print it out (hopefully double-sided); close your computer (come back on if you have questions, there may be updates); shut yourself from everyone (take some food/drinks if you need to), and time yourself! Once you're done, get back online and get ready to submit.

[b]How to submit:[/b]

Put your answers in blocks of five. Blank answers get a -. For instance, if the answer to every question except for #25 (you left that one blank) was J, you'd submit your answers like this:

JJJJJ JJJJJ JJJJJ JJJJJ JJJJ-

Afterwards, complete the survey. There are five questions. We will give you extra time to complete it. (You should take only five extra minutes at most for the survey). Remember, your input will benefit future mock AMCs!

Finally, PM Caelestor and gfour84 (faster grading), or email us both at caelestor@gmail.com (alternate: caelestor@aol.com) and king-of-cards@cox.net.

[b]Remember, failure to follow any of these instructions may delay everyone's results![/b]

Hello. I looked at some of your blog questions and some of them are really a lot harder than any AMC 12 I have ever seen! Will this be 100% a killer AMC 12?

shoot i'm a friday participant?

err well... i know for sure i won't be available tonight.. maybe saturday, but i doubt it. :/ stupid father.
sorry~

To all ppl signed up today: please check either your PM or email (the one you used to sign up on aops). Have fun! :wink: If you have issues, please contact either me @ caelestor@gmail.com or king-of-cards@cox.net.

Grr, I hate typos. Note in #15, $ \frac {\sqrt {b}}2$ should be $ \sqrt {b}$. That will be edited in Form B.

For the rest of you looking, please do not discuss the test material until Sunday 5:00 A.M. EDT. Feel free to prattle about its difficulty.  :rotfl:

[quote="limac"]I may have missed it...but where do we download the test from?[/quote]

I don't recall you specifying the Friday time. Nevertheless, your email does not exist apparently.
Take it tomorrow.

[quote="veezbo"]I am a retarded buffoon  :mad:  :wallbash_red:

Sorry, it completely slipped my mind. If however you are changing the turn-in date (and hopefully keeping the test up?) until Sunday, then I'll still be able to take it tomorrow.[/quote]
#
I emailed you the test. Just take it and give yourself a reasonable amount of time (preferably the official 75 minutes).

Caelestor, oh, I meant like where can I download the test from, but I want to take it on Saturday (today).

Hello. Sorry, I didn't take the test yesterday even though I was one of the Friday participants. Is it alright if I do it today and submit it by tonight?

Are you sending it to us via pm? Or posting it here?

Email or PM, I think.

EDIT: [b]Is it going to start in half an hour?[/b] Email or PM?

READDDDY?

There really is nothing more to say . . . Everyone ENJOY, and good luck (the typos from yesterday have been corrected)

would a number like 121 be modular for number 18?

Well, why would it not be (thats a yes).

and also, in the first question, is there meant to be a minus sign in front of the 2009?

Does the polygon in 8 have to be convex?

If it does not say, then why would it have to be?

Hello everybody,

The test was apparently very difficult. Thus, we will try to make it easier in the revision.

Note that we will be less rigid and take submissions until tomorrow afternoon.

so I can solve the problems I didn't have time to do today?

[quote="Just_Beginner"]The test is really hard...  :lol: . When we submit our answers, how do we type blanks? Should we just use Xs or something like that?[/quote]

Blanks get a -.

Oh oops! I just read the directions again.  :blush:

If you get over 90 on this test, be proud of yourself (unless you cheated  :P ).

We may offer another "form" of this test later, probably an easier version of the current one.

Will there be a solutions thread up soon for discussing answers?

In passing through the instructions (I am not trying to take this
mock exam or the actual AMC test), you have inconsistent grading.

"A correct answer is worth 6 points, an unanswered question is
worth 1.5 points, and an incorrect answer is worth no points."

1) Any unanswered question should be marked as "incorrectly
answered" because it did not get the correct mark by it.

2) With that said, all unanswered questions should get no points
as well because they are incorrect answers.

3) a) Look at this disparity in an example:  Student A gets 4 problems
correct, 8 problems incorrect, and leaves 13 problems unanswered.
6(4) + 8(1.5) + 13(0) = 36 . . . (Score)

b) Student B turns in an exam that has no answered questions at all.
25(1.5) = 37.5 . . . (Score)

Student B's score is higher than Student A's, but Student B shows
no apparent knowledge.

Is it possible for you guys to post the answer key since the submitting deadline is over already? I just realized today that there was going to be a Mock AMC 12 -_-.

Arrange your tan: Err, this is the exact same way the AMC is graded. Your example is wrong, student A receives 43.5 points. If person A were to receive 36 points with 4 correct answers, he would need 13 wrong and 8 blank, and this doesn't really show any knowledge either. This grading isn't inconsistent, the same concept is used on tons of tests. It discourages guessing (since guessing has a lower expected value).

Regarding the actual test, I totally failed. Probably shouldn't have taken it at 1 am Friday night. :maybe:

gfour84 is doing something in excel. He likes to do statistics. Expect the full results tomorrow, maybe? (You should've received a grade).

And for statistics reasons, try Form B of this test. Hopefully, this is more like a regular mock AMC 12 of yore (note this is still harder than a regular AMC 12). Hey, we all could use some practice.

Meanwhile, we will open up a tentative results page.

When do we have to get the Form B answers turned in? I know that I probably can't get it in until the weekend unless I get really lucky and my teachers decide to go easy on the homework.   :P

No need to turn in. Just compare your answers once the answer key is completed, and post a response saying how much you improved. Since about 80-90% of the questions are the same, there's no need to time yourself.

Don't worry, the Mock AMC 10-09 will resemble the actual test easier. Hey, we've got about 10-15 questions unused from the previous one.


geometry
trapezoid
parallelogram
geometry unsolved

[color=darkblue]Let $ABCD$ be a trapezoid $(AB\parallel CD)$ for which exists an interior point $P$ so that

$m(\widehat{ABP})=2\cdot m(\widehat{ADP})$ and $m(\widehat{DCP})=2\cdot m(\widehat{DAP})\ .$

Prove that $\boxed{\ m(\widehat{PAD})-m(\widehat{PDA})=\frac{A-B}{2}\ }$ and in this case

$m(\widehat{BAP})+m(\widehat{ADP})=\frac{A+B}{2}$, $m(\widehat{CDP})+m(\widehat{DAP})=180^{\circ}-\frac{A+B}{2}\ .$

[b]Particular case.[/b] If the quadrilateral $ABCD$ is a parallelogram, then $\frac{A+B}{2}=90^{\circ}$ and

$m(\widehat{BAP})+m(\widehat{ADP})=m(\widehat{CDP})+m(\widehat{DAP})=90^{\circ}$, i.e. $AB=BP=PC=CD\ .$

[b]Remark.[/b] See http://www.mathlinks.ro/Forum/viewtopic.php?t=106618 [/color]

I've got a solution but I don't have time to put it


geometry
similar triangles
Pythagorean Theorem

In the adjoining figure, points $B$ and $C$ lie on line segment $AD$, and $AB$, $BC$, and $CD$ are diameters of circle $O$, $N$, and $P$, respectively. Circles $O$, $N$, and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$. If $AG$ intersects circle $N$ at points $E$ and $F$, then chord $EF$ has length

[asy]
size(250);
defaultpen(fontsize(10));
pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1];
draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7));
dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P);
label("$A$", A, W);
label("$B$", B, SE);
label("$C$", C, NE);
label("$D$", D, dir(0));
label("$P$", P, S);
label("$N$", N, S);
label("$O$", O, S);
label("$E$", E, dir(120));
label("$F$", F, NE);
label("$G$", G, dir(100));[/asy]

$\textbf {(A) } 20 \qquad \textbf {(B) } 15\sqrt{2} \qquad \textbf {(C) } 24 \qquad \textbf{(D) } 25 \qquad \textbf {(E) } \text{none of these}$

[hide="Solution"]Draw a perpendicular from point N to EF, and let them meet at $M$. Thus, $MF=\frac 12 EF$. Also notice that $\triangle NMF$ is right and $NF=15$, as a radius of circle $N$.

Now, we apply similar triangles to see that $\frac{MN}{GP}=\frac{AN}{AP}$ and $\frac{MN}{15}=\frac{45}{75}$. Solving this, $MN=9$.

Now, we apply the Pythagorean theorem of triangle $NMF$ to get that $MF=12$, and $EF=24$. [/hide]

[hide]Yeah dude, just use similar triangles you know!

$GP=15$
$AP=75$
$AG=15\sqrt{26}$ (not really important, unless you wanna brute force)

Drop a perpendicular from point $S$ on $EF$ to point $N$. Similar triangles.

$\frac{AN}{AP}=\frac{NS}{PG}$

$\frac{45}{75}=\frac{NS}{15}$

$NS=9$

Here was the tricky part of the problem, realizing that $NSF$ was right.  At least to me

$9^2+x^2=15^2$
$x=12$

Since $N$ was the midpoint of the diameter, they're both $ES$ and $SF$ are equal
$12+12=\boxed{24}$
$\boxed{C}$[/hide]

Is the picture available, because the problem means nothing without the diagram in the resource section?  :(

[geogebra]b5bdb0e94ffadadf1db9e9a8b821ba77327ac27f[/geogebra]

Here's a rough drawing of it.


function
linear algebra
matrix
advanced fields
advanced fields unsolved

Show that if f is a diffeomorphism M -> N and g is C^1 close enough to f (and is
smooth), then g is a diffeomorphism.

I am not even sure what "C^1 close enough" means.

Thanks.

It means close enough in the $ C^1$ norm, that is both $ |f\minus{}g|$ and $ |f'\minus{}g'|$ are bounded by a constant that you can take as small as you wish.
The tools to use: (1) inverse function theorem; (2) the set of invertible matrices is open (under any kind of matrix norm).

I wasn't quite able to prove it...Could you please give me more details?
thanks


inequalities
number theory
relatively prime
algebra unsolved
algebra

let s dinote the set of rational numbers in the interval(0,1),dertermine with proof if there exists a subset T of S such that erery element in s can be uniquely witten a the sum of finitely many distinct elements in T

The answer, I think, is no.  Suppose that $ T$ exists.

[b]Lemma.[/b] If $ a$ and $ b$ are elements of $ T$ for which $ a>b$, then $ a\ge 2b$.

[i]Proof.[/i]  Suppose otherwise.  Then $ a\minus{}b < a < b$.   There exists a set $ R \subset T$ such that $ \sum_{r\in R}r\equal{}b\minus{}a$; clearly, $ a,b\notin R$.  Then
\[ a \equal{} b\plus{}(a\minus{}b) \equal{} b \plus{} \sum_{r\in R}r, \]
contradicting the hypothesis that $ a$ has a [i]unique[/i] representation as a finite sum of elements of $ T$.  $ \blacksquare$

It follows from the lemma that every nonempty subset of $ T$ has a maximal element.  In particular, if $ \alpha$ is the supremum of a nonempty subset $ R$ of $ T$, then we can pick $ a\in R$ larger than $ \alpha/2$, and for any $ b\in T$ for which $ b>a$, we have the inequality $ b\ge 2a > \alpha$, so $ b\notin T$; hence $ a$ is the maximum of $ T$.

We can form a sequence $ a_0, a_1, \dotsc$ of elements of $ T$ recursively, so that
\[ a_k \equal{} \max T\setminus \{a_j\}_{0\le j <k} . \]
By the lemma, the $ a_k$ tend to zero, so they must include every element of $ T$.  In fact, by the lemma, $ a_i \le 2^{\minus{}i}a_0$, for all indices $ i$.

Now, we have two cases.

[b]Case I :[/b]   For all indices $ i$, $ a_i \equal{} 2^{\minus{}i}a_0$.
Let $ p$ be an odd prime relatively prime relatively prime to the denominator of $ a_0$.  Then $ 1/p$ cannot be expressed as a sum of finitely many elements of $ T$, a contradiction.

[b]Case II :[/b]  For some index $ k$, $ a_k < 2^{i}a_0$.
Then $ \sum_{i\ge 1}a_i < a_0$.  Pick a rational $ a$ between $ \sum_{i\ge 1}a_i$ and $ a_0$.  Then $ a$ cannot be expressed as a sum of finitely many elements of $ T$, the same contradiction.

We have a contradiction in each of the only two possible cases.  Therefore $ T$ does not exist.  $ \blacksquare$

Does anybody know where this problem comes from originally?  I know it's from MOP, but I think they borrow a lot of problems.


You enter a raffle to win a cruise from Miami to St.Croix valued at $3000. It costs $5 to buy one ticket . You are eager to take the cruise and buy twenty tickets. Suppose 2000 tickets are sold.
a)  What is the probability that you will win the cruise?
b)  If X=amount of the money made or lost by entering the raffle. What is the expected value of X? Explain the meaning of this result in terms of the $100 ticket purchase.

a) probability of winning is $\frac{20}{2000} \implies \frac{1}{100}$

b) not sure, but expected money to be earnt may be $\frac{1}{100} \cdot 3000 = 30$

However you spent $100$ so you would expect to lose $70$


If f(x)= 0^x

What is the limit of f (x) for x => 0(+)? (Right limit)

What is the limit of f (x) for x => 0(-)? (Left limit)

What is the limit of f (x) for x => 0?    (Limit at x=0)

Thanks
leonsotelo

$ f(x)\equal{}0^{x}$?  That's just $ f(x)\equal{}0$, or the $ x$-axis.

[quote="JRav"]$ f(x) \equal{} 0^{x}$?  That's just $ f(x) \equal{} 0$, or the $ x$-axis.[/quote]

No, 0^x doesn't exist for nonpositive numbers, so the limit from the right side is 0, from the left side is undefined, so the limit at x=0 is also undefined.

The limit of x^x as x tends to zero (from the right) is 1. In other words, if we want the x^x function to be right continuous at 0, we should define it to be 1.

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