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I emailed mr jaye about being a TA, but didnt get an answer yet.

I'll be there :)
Geehoon, it'll be fine... you know you're VERY well qualified!!!

[quote="Geehoon"]Who's going to CCNY?[/quote]

is it too late to apply? :?

officially, it is late, but you might be able to cut edges if you talk to mr jaye  :)
also, there are many cool programs outside of NYC, are not considering any?

I remember when I was there, Lana joined after like first two weeks.
having fun trying.

I am trying also haha

I MIGHT visit mathcamp during puzzle hunt weekend, if i can afford money for the jet blue

[quote="Geehoon"]I remember when I was there, Lana joined after like first two weeks.[/quote]
did she go as student or TA? that might be different from your situation

I think Geehoon is talking about CCNY 2 summers ago, because I rememeber Lana only showed up after about 2 weeks into the program...

Oh...and I believe she went as a student...

yes i was talking about the summer of 2003, and she was a student.

I will be going to HCSSIM!!!
                                             Stan

there's nothing wrong with doing programs again.  there's always more to learn.  i hope to be attending the ccny summer scholars thing "again" (that is, if mr. jaye chooses to accept my one week-late app), but not really because i foolishly took the science program last year.

i know haha..
 i was just making fun of HCSSIM.. :lol:

yay!!!
problem solved!!

[hide="mr.jaye said"]
From: Daniel Jaye <dannyjaye@gmail.com>
To: Geehoon Hong <geehoonhong@gmail.com>
Date: May 18, 2005 10:17 PM
Subject: Re:

We have secured some small funding to support your employment this summer. You will not be doing this for a lot of money, but rather for your love of mathematics and the math team. I hope this makes you happy.

See you soon.

~danny jaye
[/hide]

I'm so happy.
Thanks Joel, Maria, Oana, and many people who cared about and helped me =)
I'll see you [b]at CCNY[/b].

Excellent!  Now I'm going to have a [i]really[/i] hard choice for picking my TAs :)

Actually, I should figure out how the TAs get chosen/assigned.  For that matter, I should figure out what the heck I'm teaching.  Hmmm...

yay HCSSiM for this summer

my summer plan works perfectly this year too.
9th grade- CCNY summer scholar program :!:
10th grade- HCSSiM :!:
11th grade- MathCamp 2006 or RSI(if possible) :?:
12th grade- trip to KOREA.. :?:

i hope to see some nyc math team pp in HCSSIM this year.

I will most likely take a trip to visit both HCSSiM and CCNY.  Yellow Pig day seems to be on the weekend this year, so I don't have to use vacation days! Wohoo!

And yes.. I have traded my soul in and joined corporate America.  (though some might argue that I had no soul to begin with, but they are plainly wrong!!!, I think)

I'll be giving a talk at HCSSiM on July 14th.  To any students going there this summer, please stop by to say hello!

We had a orientation today at CCNY.
It was little hot in the room, but it was fun.

It was nice to see everybody again.
I'm glad that I'm going to be working with many many nice people :D
I am very excited.

so, summer is here... everyone is away somewhere. it's amazing how quickly this board lost posters!

i just got back from korea.

i'll (try to) post more here  :D

hey bubala how are you doing there? :D
i hope youre enjoying the camp as much as i did last summer.

HCSSiM is awesome...Just had a prime time from Ravi Boppana yesterday, a.k.a. Ravi B... Showeda whole new perspective on probability...I'm here with Steven Yu, Andrew Choi, and many other NYC people...Having so much fun...I hope CCNY is being good to you other NY people...But stop using the names Prime Time and Interesting Test...

It's being fantastic.
I'm thankful that I'm working with the best people in the world.

by the way..
"PRIME TIME" does not belong to hcssim.
if you google prime time, I dont think hcssim appears in first 10 pages.

true....

14th page 5th website

anyways...good luck, geehoon hyung with ccny!!

Stan, I enjoyed meeting you and the other students at HCSSiM.  Glad you enjoyed the probability talk.

For the CCNY folks, I'll see you on August 8 and 9 when I give a "Prime Time" talk.  The tentative topic is algorithms for testing primality.  Please introduce yourself to me then if we haven't met before (or even if we have...).

[quote="Ravi B"]For the CCNY folks, I'll see you on August 8 and 9 when I give a "Prime Time" talk.[/quote]I'm looking forward to it, very much :D

Although Prime Time may not be "owned" by Hampshire, if you guys use the term Interesting Test, which happens to be on the second page of the google search, then you should also refer to 3:00 in the afternoon the approximate time everyone leaves, Quiet Time...Prime Time has been used by HCSSiM for I believe the majority of it's 33 years, and for CCNY, a program for only 4 - 5 years, to use it, I find a bit disrespectful... I believe that if you ask Mr. Jaye how he "came up" with that name, he'll have to say that it was from the Summer Studies...

I'm TA-ing at CCNY as well, but with the science program.
I get to go outside in the morning, so I can fully appreciate the air conditioning in the afternoons.
I look forward to your talk, Ravi!

I hope everyone at HCSSiM is having a wonderful summer.

Stan, don't be so pushy. Kelly has trademarked neither "Interesting Test" nor "Prime Time." Until he does, quiet down. They're both perfectly normal English phrases, albeit capitalized.

Well...

I spent most of July in Beijing, as a part of a culture program.  It was enjoyable, if not a tad hot at times.


Is f(x) = xlnx uniformly continuos for x >= 1?

I have looked at
Abs(f(x+h)-f(x)) (h>0)
= Abs((x+h)ln(x+h)-xlnx)
= Abs(xln(x+h/x)+hln(x+h))
= xln(x+h/x) + hln(x+h) (since x >= 1, h > 0)
...

I want to show that lim (x -  :inf:) (xln(x+h/x) + hln(x+h)) approach :inf: for any h > 0 no matter how small.
I know that lim (xln(x+h/x) + hln(x+h)) > lim xln(x+h/x).
I substitute x with 1/t which lead to lim (t - 0+) ln(1+ht)/t which approach  :inf:.

The reason I didnt find it before, was that I forgot lim (t - 0+) and thought of it as lim (t -  :inf: ) :blush: :blush:  :blush:


[i]This was given to me from a friend who studies economics.Of course not Analysis , but where should I pose it ?Any way , any help is welcome![/i]

$300$ people must elect one of three persons $A , B, C$. The first $200$ of them gave $90$ votes to $A$ , 60 to the second , $40$ votes to the third and  $5+5 = 10$ votes were invalid or white. To be elected from the first round, one person must have at least $127$ votes.Ih how many ways can  vote the other $100$ people , in order to elect person $A$ in the first round ?

 [u]Babis[/u]

[i]Is also proposed in the combinatorics section , but a solution is not sent so far.[/i]

It doesn't belong here. Be patient and wait for an answer in the first location.


algebra
polynomial
symmetry
induction
calculus
derivative
algebra solved

Let $ n$ be a positive integer. Fill all the polynimials with degree $ 2n+1$ satisfying the conditions:
       For all $ k \in n$ ,the gragh of $ y=p^{(2k)}(x)$ has symmetry point.

I can't be sure right now, but I think the answers is: $P(x)=\displaystyle \sum_{i=0}^n a_i(x-k)^{2i+1}+t$. I also think that we can prove this by induction: take a polynomial of the above form, integrate it twice and we get something with the same form, except that it has a term of the form $A(x-k)^2$. We then show that $A=0$ in this case.

I'm not sure, so forgive me if I'm talking nonsense here.. :)

I don't know if we can prove that, but if $P$ has a center of symmetry, it means that there is some $k$ for which $P(x)+P(2k-x)=2P(k),\ \forall x\in \mathbb R$. Let $Q(x)=P(x+k)-P(k)$. Then $Q(x)+Q(-x)=0$, so all the exponents of $x$ in $Q$ are odd, so $\displaystyle P(x)=\sum_{i=0}^n a_i(x-k)^{2i+1}+P(k)$, so all the derivatives of even order of $P$ have a symmetry point, so the question is equivalent to finding those polynomials which have a symmetry point.

Is this Ok?

opp! I forgot to type the condition of k : $ k >0$.
I got the solution:
   \[ P(x)= \sum^{n}_{i=1}(x-a)^{2i+1}+\frac{(P''(a))^2}{2}x^2+Cx+D\]


If $a,b,c \ge 0$, then prove that:
\[\sqrt{a^{2}+bc}+\sqrt{b^{2}+ca}+\sqrt{c^{2}+ab}\geq \sqrt{a^{2}+ab}+\sqrt{b^{2}+bc}+\sqrt{c^{2}+ca}. \]
And if $a \geq b \geq c \geq 0$, show that:
\[\sqrt{a^{2}+ab}+\sqrt{b^{2}+bc}+\sqrt{c^{2}+ca}\leq \sqrt{a^{2}+ac}+\sqrt{b^{2}+ba}+\sqrt{c^{2}+cb}. \]

[quote="Lovasz"]If $a,b,c \ge 0$, then prove that:
\[\sqrt{a^{2}+bc}+\sqrt{b^{2}+ca}+\sqrt{c^{2}+ab}\geq \sqrt{a^{2}+ab}+\sqrt{b^{2}+bc}+\sqrt{c^{2}+ca}. \]
[/quote]
Let $c=\min\{a,b,c\}$.
Hence, $\sqrt{c^2+ab}+\sqrt{a^2+ac}\geq\sqrt{a^2+ab}+\sqrt{c^2+ac}\Leftrightarrow$
$\Leftrightarrow(c^2+ab)(a^2+ac)\geq(a^2+ab)(c^2+ac)\Leftrightarrow$
$\Leftrightarrow(b-c)(a^2-c^2)a\geq0$, which is true.
Hence, $\sum_{cyc}\sqrt{a^2+ab}\leq\sqrt{c^2+ab}+\sqrt{a^2+ac}+\sqrt{b^2+bc}$.
Id est, it remains to prove that $\sqrt{c^2+ab}+\sqrt{a^2+ac}+\sqrt{b^2+bc}\leq\sum_{cyc}\sqrt{a^2+bc}$
or $\sqrt{a^2+ac}+\sqrt{b^2+bc}\leq\sqrt{a^2+bc}+\sqrt{b^2+ac}$
or $(a^2+ac)(b^2+bc)\leq(a^2+bc)(b^2+ac)$ or $c(a+b)(a-b)^2\geq0$. Done!


inequalities
calculus
calculus computations

hi i really dont understand how to do this problem

using the (lower case delta), E definition of limit (rigorous definition)
show that limit as x approaches 2  x^3=8

hint: showthat for 1<x<3 it followsthat (x^3-8) <= 19(x-2)

please please!!!! help me out, and show me how to do this
thank you so much

Basically, what you have to do is show that for any $\epsilon > 0$, there exists a positive $\delta$ such that $|x-2|<\delta$ implies $|x^{3}-8| <\epsilon$, so following the hint, we have

$|x^{3}-8| \le |19(x-2)| = 19|x-2|$, so if we chose $\delta = \frac{\epsilon}{19}$, then $|x-2|<\delta$ implies $|x^{3}-8|<\epsilon$.

As for proving that hint, note that for $1<x<3$, the following inequality holds $(x-3)(x-2)(x+5)\le 0$, from which the desired inequality follows.


The problem was posted again by me but I could not find the link so I will post the solution.

By Cauchy we have

$\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1} =\frac{a^3}{a^2b^2+a^2}+\frac{b^3}{b^2c^2+b^2}+\frac{c^3}{c^2a^2+c^2} \geq \frac{(\sum a\sqrt{a})^2}{\sum a^2b^2+ \sum a^2}$

By using the condition $a^2+b^2+c^2=1$, we see that it is sufficient to show

$4(\sum a^2)^2\geq3(\sum a^4+ 3\sum a^2b^2)$

But this reduces to

$\sum a^4\geq\sum a^2b^2$ which is true

...........  :?:

[quote="silouan"]
By using the condition $a^2+b^2+c^2=1$, we see that it is sufficient to show

$4(\sum a^2)^2\geq3(\sum a^4+ 3\sum a^2b^2)$

[/quote]

Sorry...       could you explain this part of your solution??

Carlos Bravo :)
Lima-PERU

Carlos, just square $(a^{2}+b^{2}+c^{2})^{2}$ and that's all. ;)

Similar problem:
Let $ a;b;c$ be positive real numbers satisfying $ a\plus{}b\plus{}c\equal{}1$
Prove that:
$ \sum\frac{a}{4b^{2}\plus{}1}\geq (a\sqrt{a}\plus{}b\sqrt{b}\plus{}c\sqrt{c})^{2}$


summer program
PROMYS
Ross Mathematics Program

I have solved all of the Ross Problems pretty much, but for PROMYS only 1 completely,2 completely,3a, 3b(have the answer but can't prove it, 4(not sure how much rigor is needed, I just said why the proof was wrong), 7 completely, 8 completely, 9 completely, 10 completely.

I have a few questions:
1. Should I send in my Promys app now, or give it a few weeks for #5 and 6 to come. That is does the cost of letting more time elapse outweigh the benefit 2 more solutions will give me? I have excellent solutions for 1,2,7,8,9,10.
*all solutions are LaTexed

Thankyou, I would appreciate comment from PROMYS vets and such (torajirou etc.)

Strive for having more problems done.  The problems are pretty important to the application process, as I understand it.

OK. I just got 4 so now I have 1,2,4,7,8,9,10, missing 5-6, 3b.


for example, I want to read the file (good.txt) that contains:
3
3
9
6
3
1
7
8
2

any help is appreciated, thank you!

I think I may need to use 2 dimensional arrays too if I use loops.

Any Hints or comments?

how do I fix my code to print all the numbers in a 3x3 box? thanks!

How about this?

int i,t;

for(i=1; i<=9; i++)
{
  infile >> t;
  cout << t;
  if (i%3) cout << " ";
    else cout << endl;
}

thank you, kubus. But Can you explain your code a little more.
what is i%3 mean?

Is there any possible way you  can do it with the nested loop and array?

i%3 means the remainder of x after division by 3: if we went through a number of elements divisible by three, give a newline, otherwise just give a space.

Using nested loops, you could do also this:

for(i=1; i<=3; i++)
{
for(j=1; j<=3; j++)
{
infile >> t;
cout << t;
cout << " ";
}
cout << endl;
}

(This would give you a space at the end of each line, however.) Of course there are many ways to do this, but using an array is not necessary - you have to remember only one number a time.
But your code was almost OK, except that you didn't print any newlines. Your second pair of loops could go like this, for example:

for (r = 0; r < 3; r++) {
for (c = 0; c < 3; c ++){
cout << a[r][c] << " ";
}
coud << endl;
}

your explaination is very  helpful.  Thanks again.

One last question.

333
222
222
333
555
333

Can anyone show me What I did wrong in my code and how to fix it?

Thanks in advance.

Can anyone tell me what went wrong?

Any inputs are really appreciated.

Why are you using a 3x6 array? You should be using a 3x2 array as there are only 2 numbers in each row.

jmadsen wrote
"You should be using a 3x2 array as there are only 2 numbers in each row."

Can you explain this part a little more, I am not quite understand what you mean.

It is too late now I guess but anyway.... your program assigns numbers to the right cells of your array (from [0,3] to [2,5]) but you don't print them in the right order! You do this:

[code]
for (r = 0; r < 3; r++) {
for (c = 0; c < 3; c ++){
cout << a[r][c] << " ";
}
cout << endl;
}

for (r = 0; r < 3; r++) {
for (c = 3; c < 6; c ++){
cout << a[r][c] << " ";
}
cout << endl;
}
[/code]
but you really want do this:

[code]
for (r = 0; r < 3; r++) {
for (c = 0; c < 6; c ++){
cout << a[r][c] << " ";
}
cout << endl;
}
[/code]
(Think about the difference between these two programs...)

I finally understand it!  Thanks so much.


Let $ ABC$ be an acute triangle.Find all the points $ M$ on the plane such that
$ (\frac{BC}{MA})\overrightarrow{MA}\plus{}(\frac{AC}{MB})\overrightarrow{MB}\plus{}(\frac{AB}{MC})\overrightarrow{MC}\equal{}0$


Prove that, for real $ a, b$,
\[ \frac{\mid{a\plus{}b}\mid}{1\plus{}\mid{a\plus{}b}\mid}\leq{\frac{\mid{a}\mid}{1\plus{}\mid{a}\mid}\plus{}\frac{\mid{b}\mid}{1\plus{}\mid{b}\mid}}\]

$ RHS \ge \frac{ 1}{\frac{1}{|a|\plus{}|b|}\plus{}1} \ge \frac{ |a\plus{}b| }{|a\plus{}b|\plus{}1}\equal{}LHS$

I have seen a problem
$ \frac{c}{1\plus{}c}\leq \frac{a}{1\plus{}a}\plus{}\frac{b}{1\plus{}b}$
There $ a,b,c\geq 0$,and $ c<b\plus{}a$
Is your way that you created it
The new problem is harder than that one.Any proof?

$ \frac{a}{1\plus{}a}\plus{}\frac{b}{1\plus{}b}\ge \frac{a\plus{}b}{1\plus{}a\plus{}b} \equal{}\frac{1}{1\plus{}\frac{1}{a\plus{}b}} \ge \frac{1}{1\plus{}\frac{1}{c}}$


I believe my team took a TI-89 and mutliplied the numbers, looked at the decimal at the end and was like tchah this problem is dead. We thought to just consider the mod 16 residues and multiply them, but we didn't really want to, because it was more mistake-prone. I think that was the intended method, it comes out to be around 10 digits long if you only multiply the numbers $ \pm1,2,\pm3,\pm5,\pm7$.

But it was a pretty stupid problem, in my opinion.

Those are the residues if I'm not mistaken.

[quote="Math Geek"]Why put negatives for the numbers and why put 1, when it's not prime?  Also, why only multiply those numbers? :huh:[/quote]

Because if you type in 11 instead of -6, there are more digits in the number, and more chance you'll get overflow.

i think you'll get overflow anyway though. you probably need to find the residue mod 16 every once in a while.

[hide="Outline"]Write all the primes out.

Write a new list, having them all mod 16.

Notice that 13*11 and 3*5 are both -1 in mod 16, so we can replace each such pair with -1.

Cross out every pair of -1's.

Then, the rest is just brute force, really not too bad.  I did this one in just a few minutes.  [/hide]

With cabamap, you can do this is <90 seconds on a TI-84

an application on the calculator that calculates large numbers

The result number is too large to deal with on a calculator.

However, one should realize that if a=16*p + q, then
a*b = a*q (mod 16)

So in fact, with calculator or a computer program, you can
multiply the next prime, mod 16 to get result, multiply again, etc.
to eventually get the correct result.

Still better way would be to get residual of each prime mod 16, then
multiply each one and mod 16 of the result along the way.

Certainly I would like to see a smarter way without using computer.

Jeff
[quote="Math Geek"]Does anyone who went to nationals remember that team question that asked:

"What is the remainder when the product of all prime numbers less than 100 is divided by 16?"

Is there a smart way to do this without a calculator?  I know there should be one, but my team and I were way to desperate and had to type out every prime number onto our TI-84 Silver Editions and multiply them.  Afterwards, we used my amazing modulo program on my calculator, but I guess the variables that I stored did not come up when I used the program. :oops_sign: :bomb:  :cursing:  :furious:  :stretcher:[/quote]

Is there a formal name for cabacamp, because I can't find it on the internet?  It sounds pretty useful to me. :lol:

[quote="youdonotknowme"]is the answer 2?[/quote]
this is definitley going to be marked spam

was that a guess? because our school has something where if we dont know the answer we guess 2


calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

find all points on the given curve y=f(x) where the tangent line is horizontal.

y= 1/(3sin^2(x)+2cos^2(x))

[hide="Hint"]Horizontal Lines have a slope of...[/hide]


Which year's problem #15 of AIME do you guys think is the easiest?

ONE of them I thought of was 1984's...but I highly doubt it...

2006 II wasn't terrible... at least, you could brute-force it relatively quickly. It was one of the three problems I solved that year. :D

AIME 1999 has extremely easy #15 but extremely hard other problems. Also, AIME 2008 A #15 is easy.

last years AIME II wasnt that bad...i almost got it :)

I find 1988 #15 and 1995 #15 to be pretty easy, but combinatorics is my strongest area so I may be biased.

2003 II #15 is a classic example of a problem that looks much harder than it actually is.

AIME $ 1983$ had a cool geometry problem that wasn't too hard at $ \# 15$.


Where can I get fancyhdr.sty? What steps do I follow after downloading?

Thanks.    :)

It depends on your TeX distribution.

MiKTeX 2.4 or earlier: Package Manager, right-click on fancyhdr and choose Install

MiKTeX 2.5 (with auto-install): Just include \usepackage{fancyhdr} in your document and MiKTeX will install it the first time you compile

Other tex distributions: Download from [url=http://texcatalogue.sarovar.org/entries/fancyhdr.html]CTAN[/url] and install fancyhdr.sty in /texmf/texmf/tex/latex/fancyhdr and refresh the database.


trigonometry
inequalities
algebra
High school olympiad

Find all real numbers $ a$ and $ b$ such that
\[ 2(a^2 \plus{} 1)(b^2 \plus{} 1) \equal{} (a \plus{} 1)(b \plus{} 1)(ab \plus{} 1).
\]
[i]Valentin Vornicu[/i]

$(a^2+1)(1+1)\ge (a+1)^2$
$(b^2+1)(1+1)\ge (b+1)^2$
$(a^2+1)(b^2+1)\ge (ab+1)^2$
Multiplying these we get that $4(a^2+1)^2(b^2+1)^2\ge \left((a+1)(b+1)(ab+1)\right)^2$ but the relation fromt he hypotese tells us that these are equal. Hence we must have equality in all those 3 inegalities. We get $a=b=1$.

Another solution : we set $a= \tan x, \, b= \tan y$  and we have

$\\ 2 = ( \sin x + \cos x ) (\sin y + \cos y ) ( \sin x \sin y + \cos x \cos y ) \Rightarrow \\ \\ 2= 2 \sin \left ( x+ \frac{ \pi}{4} \right ) \sin \left ( y+ \frac{ \pi}{4} \right ) \cos (y-x)$

From this equation if  $y,x \in [0,2 \pi )$ we get $y=x= \frac{ \pi }{4}$ or $x =\frac{ \pi }{4} ,y= \pi + \frac{ \pi }{4}$ .

Both of them give $a=b=1$ .

[quote="kunny"][quote="shyong"]another method:

$2(a^2+1)(b^2+1)=(a+1)(b+1)(ab+1)$ $\Longleftrightarrow (b^2-b+2)a^2-a(b+1)^2+(2b^2-b+1)=0$

Since $b^2-b+2=(b-\frac{1}2)^2+\frac{7}4>0$ So the discriminant

$(b+1)^4-4(b^2-b+2)(2b^2-b+1)=-(b-1)^2(7b^2-2b+7)\leq 0$ . :?

Hence there are only real roots when $b=1$ and this gives us $a=1$[/quote][/quote]

anything wrong with that factorization ?? It seems ok to me :)

[quote="delta"]Shyong, I think Kunny was puzzled (amased) how u factored this. So am I :?[/quote]

actually , you can either expand everything out and do the rational roots theorem things . But somehow , from above post , i already know that $b=1$ is the answer . So i guess the discriminant should looks something like $(b-1)^2(pb^2+qb+r)$ . Then you use substitute for some value for left and right and solve simultaneous equation to get $p,q,r$ .  :P

Let $a,b,c$ be reals. [url=https://artofproblemsolving.com/community/c6h1707389p11009343]Prove that[/url]$$(1+a^2)(1+b^2)(1+c^2)\ge(1+ ab)(1+bc )(1+ca )$$
$$\implies$$
Let $a,b$ be real numbers. Prove that$$2(a^2+1)(b^2+1)\ge (a+1)(b+1)(ab+1)$$


geometry
3D geometry
quadratics
LaTeX
algebra
quadratic formula
AMC 8

a.)How do you find the volume of a cone?

b.)Simplify (a+b)(a-b).

c.)Simplify (a+b)^2

d.)What is the quadratic formula? (i encourage you use latex even tho i dont kno how to use it)

e.)Define vertical angles.

f.)Prove why a triangle has 180 degrees

IF I CAN THINK OF MORE CHALLENGING 'QUESTIONS' THEN I WILL POST THEM!

Generally, challenge problems asking for proofs belong at least one level up among the five problem-solving forums. The Getting Started Forum is for real beginners at problem-solving, e.g., United States middle school students or Chinese elementary school students, who usually work with problems involving finding a numerical answer rather than with problems involving showing a proof. ([url=http://www.unl.edu/amc/e-exams/e4-amc08/amc8.html]AMC 8[/url] is an excellent example of the appropriate level for this forum.)

I'll leave this thread where it is, because the other problems are easier, but I'll start "promoting" more problems to higher-level forums if they don't look like Getting Started problems.

[hide]a.

b. (a+b)(a-b)      a(a-b) + b(a-b)

c. (a+b)(a+b)      a(a+b)+b(a+b)

d.    (-b :pm:  :sqrt: (b^2 - 4ac))/2a

e. when two lines cross vertical angles are a pair of congruent oppisit angles formed by the two lines.

f. Draw a line through one vertex of the triangle make it parallel to the base of the triangle. Using alternate interior angle we see all the angles of the triangle lie on that straight line. Sum of the angles on straight lines is 180, so the sum of the angle is 180.

note: I didn't to it all in Latex so people can still work it on their own.[/hide]

[hide]a) 1/3bh

b)a :^2: -b :^2:

c)a :^2: +2ab+b :^2:

d)

-b :pm:  :sqrt: b :^2: -4ac

                    2a

e) angles constructed by the same two lines which intersect at exactly one point which are not tangent

f)angles = 180(n-2) where n is the number of sides

Have fun![/u][/tex][/hide]

[quote="tokenadult"]The Getting Started Forum is for real beginners at problem-solving, e.g., United States middle school students or Chinese elementary school students, who usually work with problems involving finding a numerical answer rather than with problems involving showing a proof. [/quote]

Chinese students begin to prove geometry problems at Junior 1, which is the 1st year of middle school.

[quote="paganinio"]Chinese students begin to prove geometry problems at Junior 1, which is the 1st year of middle school.[/quote]

Yes, which is what puts Chinese seventh graders at the Intermediate level in terms of our levels of challenge problems here.

This is a bit of off topic here, but there's a lot of chinese people who just came form china and they are all in Algebra 2...

I'm a year younger (9th) and is in Pre-Calc...

I wonder why?

I'm ashamed that I'm not in calc.


The area of triangle $ XYZ$ is 8 square inches.  Points $ A$ and $ B$ are midpoints of
congruent segments $ \overline{XY}$ and $ \overline{XZ}$.  Altitude
$ \overline{XC}$ bisects $ \overline{YZ}$.  What is the area
(in square inches) of the shaded region?
[asy]/* AMC8 2002 #20 Problem */
draw((0,0)--(10,0)--(5,4)--cycle);
draw((2.5,2)--(7.5,2));
draw((5,4)--(5,0));
fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey);
label(scale(0.8)*"$X$", (5,4), N);
label(scale(0.8)*"$Y$", (0,0), W);
label(scale(0.8)*"$Z$", (10,0), E);
label(scale(0.8)*"$A$", (2.5,2.2), W);
label(scale(0.8)*"$B$", (7.5,2.2), E);
label(scale(0.8)*"$C$", (5,0), S);
fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white);[/asy]

Area of $ XAB$ is one quarter from area of $ XYZ$ because of the formula: $ area\equal{}\frac{absin\gamma}{2}$, so twice as the area of the shaded region is $ 8\minus{}2\equal{}6$
Answer: $ \boxed{3}$

Or you can not that XCY is half of XYZ, and the small triangle side length to XCY is 1/2, so area ratio is 1/4. So shaded region is 3/8 of entire thing, or area 3.

An even easier, visual way to do this quickly is to note with the diagram that the top triangles fit perfectly 3 times into the bottom trapezoids. Logically knowing that the area is an integer (via the lack of a specification of whether to "Write the answer as a decimal to the nearest tenth" etc.), we can tell  that the area of a top triangle is one, and thus area of a bottom trapezoid is 3.


1. A, B, C, D are positive integers such that each is a factor of the sum of the other three. Prove that at least two of the integers are equal.

2. The points A, B, C, D are not in the same plane. $ \angle{ABC}\equal{}\angle{BCD}\equal{}\angle{CDA}\equal{}90^\circ$. Prove that $ \angle{DAB}$ is acute.

3. Find all solutions to $ x^{7}\plus{}7^{7}\equal{}(x\plus{}7)^{7}$.

4. ABC is equilaterial with height $ h$. A circle with radius $ h$ is tangent to BC at Y and intersects AB and AC at X and Z, respectively. Prove that arc XYZ is $ 60^\circ$.

[hide="2 maybe"]WLOG $ B \equal{} (1,0,0)$, $ C \equal{} (0,0,0)$, $ D \equal{} (0,a,0)$ for some positive real $ a$. Then we can prove that $ A \equal{} (1,a,b)$ for a nonzero real $ b$, and use some vectors to get the answer.[/hide]

[quote="boo"]1. A, B, C, D are positive integers such that each is a factor of the sum of the other three. Prove that at least two of the integers are equal.

2. The points A, B, C, D are not in the same plane. $ \angle{ABC} \equal{} \angle{BCD} \equal{} \angle{CDA} \equal{} 90^\circ$. Prove that $ \angle{DAB}$ is acute.

3. Find all solutions to $ x^{7} \plus{} 7^{7} \equal{} (x \plus{} 7)^{7}$.

4. ABC is equilaterial with height $ h$. A circle with radius $ h$ is tangent to BC at Y and intersects AB and AC at X and Z, respectively. Prove that arc XYZ is $ 60^\circ$.[/quote]
[hide="3 reduced to a cubic"]
Let x=y-3.5
$ 42y^{6}\plus{}3001.25y^{4}\plus{}22059.1875y^{2}\plus{}1050.4375\equal{}7^{7}$
Let $ y^{2}\equal{}k$
$ 672k^{3}\plus{}48020k^{2}\plus{}352947k\equal{}7^{7}*16\minus{}7^{5}$
$ 96k^{3}\plus{}6860k^{2}\plus{}50421k\equal{}7^{6}*16\minus{}7^{4}$
$ k\equal{}7a$
$ 96a^{3}\plus{}980a^{2}\plus{}129a\equal{}7^{3}*16\minus{}7$[/hide]

[hide="3"] If $ x > 0$ then clearly the RHS is strictly greater than the LHS.

If $ \boxed{x \equal{} 0}$ we have a solution.

If $ \minus{}7 < x < 0$ then $ 7^7 \equal{} (x \plus{} 7 \minus{} x)^7 \equal{} (x \plus{} 7)^7 \plus{} (\minus{}x)^7$ and clearly the LHS is strictly greater than the RHS.

If $ \boxed{x \equal{} \minus{}7}$ we have another solution.

If $ x < \minus{}7$ then let $ y \equal{} \minus{}x$; then $ 7^7 \plus{} (\minus{}x\minus{}7)^7 \equal{} (\minus{}x)^7 \equal{} (7\minus{}x\minus{}7)^7$ and clearly the RHS is strictly greater than the LHS.

Those are our only two real solutions.  If you're wondering why this seems to imply that there are $ 5$ complex solutions, one of the above roots has an even multiplicity.

If you want to find the complex solutions, divide out these roots from diophantient's cubic. [/hide]

[quote="diophantient"][quote="boo"]1. A, B, C, D are positive integers such that each is a factor of the sum of the other three. Prove that at least two of the integers are equal.

2. The points A, B, C, D are not in the same plane. $ \angle{ABC} \equal{} \angle{BCD} \equal{} \angle{CDA} \equal{} 90^\circ$. Prove that $ \angle{DAB}$ is acute.

3. Find all solutions to $ x^{7} \plus{} 7^{7} \equal{} (x \plus{} 7)^{7}$.

4. ABC is equilaterial with height $ h$. A circle with radius $ h$ is tangent to BC at Y and intersects AB and AC at X and Z, respectively. Prove that arc XYZ is $ 60^\circ$.[/quote]
[hide="3 reduced to a cubic"]
Let x=y-3.5
$ 42y^{6} \plus{} 3001.25y^{4} \plus{} 22059.1875y^{2} \plus{} 1050.4375 \equal{} 7^{7}$
Let $ y^{2} \equal{} k$
$ 672k^{3} \plus{} 48020k^{2} \plus{} 352947k \equal{} 7^{7}*16 \minus{} 7^{5}$
$ 96k^{3} \plus{} 6860k^{2} \plus{} 50421k \equal{} 7^{6}*16 \minus{} 7^{4}$
$ k \equal{} 7a$
$ 96a^{3} \plus{} 980a^{2} \plus{} 129a \equal{} 7^{3}*16 \minus{} 7$[/hide][/quote]
Okay, I just realized that it's trivially obvious to note that x=0 is a solution, meaning we can simplify this...
[hide]
x=0 means a=1.75
Dividing by 4a-7, we get
$ 24x^{2}\plus{}287x\plus{}534.5\equal{}0$, but the discriminant is clearly negative, meaning only  x=0 [/hide]

Not just $ x \equal{} 0$. If you go back through your substitutions from $ a$ to $ x$, you get a second solution since reversing $ y^2 \equal{} k$ gives two possible values for $ y$. So your solution does get both $ x \equal{} 0$ and $ x \equal{} \minus{}7$, which is correct.

[quote="MellowMelon"]Not just $ x \equal{} 0$. If you go back through your substitutions from $ a$ to $ x$, you get a second solution since reversing $ y^2 \equal{} k$ gives two possible values for $ y$. So your solution does get both $ x \equal{} 0$ and $ x \equal{} \minus{} 7$, which is correct.[/quote]
Oops, I wasn't paying attention.


Functional Analysis
advanced fields
advanced fields unsolved

$1<p<\infty$, $K$ a convex subset of $L^p$, $f \in L^p$. show that there is a unique $g \in K$ such that $|f-g|_p=inf_{h \in K} |f-h|_p$

Both the existence and the uniqueness of $g$ follow from the fact that $L^p$ is uniformly convex: for any $\varepsilon>0$, there is a $\delta>0$ such that whenever $\|f\|=\|h\|=1$ and $\|f-g\|\ge\varepsilon$, we have $\left\|\frac{f+g}2\right\|\le 1-\delta$.

By a theorem of Milman and Pettis, all uniformly convex Banach spaces are reflexive, and for every reflexive Banach space the existence holds. On the other hand, the uniqueness follows from the fact that $L^p$ is strictly convex (because it's uniformly convex), so if $\|f-g_1\|=\|f-g_2\|=d(f,K),\ g_1\ne g_2\in K$, then $\left\|f-\frac{g_1+g_2}2\right\|<d(f,K),\ \frac{g_1+g_2}2\in K$, which is absurd.

yes, that is. And do you have a nice prove of the uniform convexity of $L_p$ for $1<p<\infty$ ? I have only seen an horrible one and I can't find a natural proof by myself.


calculus integration limit logarithms parameterization Euler trigonometry

is it true that:
$\int_0^{+ \infty} \frac{ln(t)sin(t)dt}{t} = - \frac{\pi \gamma}{2}$  ?

$\gamma$ is euler's  constant approximately 0.5772156649....
or $\gamma = \lim_{n \rightarrow \infty} \sum_{k=1}^n \left ( \frac{1}{k} \right ) - \ln \, n$ .

I claim that integral is true. Improper integral by using parameter :) .

yes beginner, it is Euler constant  :)  Could you show us your inetegral parameter solution ?

No, alekk, I don't have any solution  :( . I'm just guessing the numerical value of that integral. What I say about parameter is just my approach. I was in hurry in my last post , cause i was in exam (I'm offline for about three days :) ). sorry for the cause of confusion. i'm stuck in limit of the sum, after using the expansion and integrated.

you mean this $\lim_{n \rightarrow \infty} \sum_{k=0}^{\infty} (-1)^k n^{2k+1} \left( \frac{(2k+1) \ln n - 1}{(2k+1)^2 (2k+1)!} \right )$ I was assuming that the series are uniformly convergent. So term by term integration is allowed.
Anyway, you're from Indonesia. Apa Kabar. Lokasimu di mana?

I think you mean $\lim_{n \rightarrow \infty} \sum_{k=1}^{\infty} (-1)^k n^{2k} \left( \frac{(2k) \ln n - 1}{(2k) (2k)!} \right )$, But maybe this is not useful.
Some calculation using Laplace Transformation will make the euler constant occurs.
I'm in Jakarta :lol: .

alekk do you a " taupinal MP*" solution for this  integral ?

Use contour integral to prove that $\int_{0}^{\infty}t^{x-1}\sin tdt=\Im \int_{0}^{\infty}t^{x-1}e^{it}dt=\Im \int_{0}^{\infty}i^{x}s^{x-1}e^{-s}ds=\Gamma\left(x\right)\sin\frac{\pi x}{2}$,
$\Gamma\left(x\right)\sim\frac{1}{x}-\gamma+o\left(x\right)$ as $x\to 0$, thus $\int_{0}^{\infty}\frac{\ln t\sin t}{t}dt=\lim_{x \to 0}\frac{d}{dx}\int_{0}^{\infty}t^{x-1}\sin tdt=-\frac{\pi\gamma}{2}$.


Consider the set $ \{1,\dots,n\}$. Find a integer $ k$ so that for every $ k$-subset of $ \{1,\dots,n\}$ we have two distinct elements $ a,b\in subset$ with $ a|b$.

(on similar lines of http://www.mathlinks.ro/viewtopic.php?p=1104753 )

very very old well-known problem by erdos.
$ [n\plus{}1/2]$
think about chains of elements $ 2^i k$

You might mean "find the least integer such that ...". Otherwise the problem maxes no sense ($ k\equal{}n\minus{}1$!).


Well ,What about this idea:

Lets choos a team from the mathlinks members which are going to IMO ,and they will be the mathlinks TEAM .

Some thing like ,Considering mathlinks a country:

We can take some exams as the TST`s of mathlinks (perhaps the Mathlinks contest can be the TST )

and the First 6 ,ones will be the team of mathlinks .

You know ,I bet mathlinks team will be able to beat china in IMO ,SO

What do you think? :D  :D

Hehe, nice idea but that won't happen. I think that the best 6 students in IMO from AoPS - MathLinks this year all got Perfect scores, so they obviously beat any team :P

And you've probably got members of the Chinese national team here. If not, then at least people who're doing the Chinese TST.


calculus
integration
number theory unsolved
number theory

It is given that:
\[25qk^3=q^4+k^4\quad k,q\in\mathbb Z\quad\gcd(k,q)=1\]
Notice that $25qk^3=q^4+k^4\Rightarrow k^3(25q-k)=q^4\Rightarrow k^3|q^4$. Since $\gcd(k,q)=1$, we know that $k=1$ or $k=0$. It is easily shown that both cases have no integral solutions.

Please excuse me if I am wrong.

That looks right... it seems fairly obvious there are no solutions if $(q, k)=1$. But in fact there are no solutions even without that condition except $0,0$.

Suppose $q, k$ are numbers satisfying the equation. Let $a, b$ be non-negative integers such that $2^a \| q, 2^b \| k$. Then $q=2^aq_2, k=2^bk_2$, with $q_2, k_2$ odd. Then the equation is

\[25qk^3=q^4+k^4 \\
2^{a+3b}5^2q_2k_2^3=2^{4a}q_2^4+2^{4b}k_2^4\]

If $a \neq b$, then suppose that $a < b$. Then $2^{a+3b} \| LHS, 2^{4a} \| RHS$, which of course doesn't give equality. You can similarly prove it if $a > b$.

If $a=b$, then

\[5^2q_2k_2^3=q_2^4+k_2^4\]

Impossible because LHS is odd and RHS is even. The only case where you can't apply this argument is when infinite powers of 2 divide one of the numbers, which means it is $0$.

[quote="Laucha"]Find $q,k \in \mathbb{Z}$, $(q:k)=1$, which satisfy: $5^2 q k^3 = q ^4 + k^4 $

I can't find a solution  :([/quote]
If $t$ is a prime which divides $q$ then $t$ doesn't divides $k$ but by ur equation must divide $k^4$ so, divides $k$, which is a contradiction, so $q=1$, the same is posible to do with $k$, then $q=\pm 1$ and $k=\pm 1$ or $k=q=0$, so clearly the unique solution is $q=k=0$.

B.R.

RX TCM


function
trigonometry
induction
calculus
calculus computations

1) Expand in powers of $ x$  the function $ \sin^{2}x \cos^{2}x$ and find the region of convergence.

2) Prove that $ \sum_{n\equal{}0}^{N\minus{}1}\cos nx\equal{}\frac{\sin\frac{Nx}{2}}{\sin \frac{x}{2}}\cos \frac{(N\minus{}1)x}{2}$.

Hi. 2. can be done by mathematical induction, using the fact that the difference of two sines is twice the cos of half the sum by the sine of half the difference.

2)

Clearly true for $ n\equal{}0$.

Assume true for $ n\equal{}N$, then

$ 1\plus{}\cos x \plus{} ... \plus{} \cos (N\minus{}1)x \plus{} \cos Nx \equal{} \frac{\sin \frac{Nx}{2}}{\sin \frac{x}{2}}\cos\frac{(N\minus{}1)x}{2} \plus{} \cos Nx$

$ \equal{}\frac{\sin\frac{Nx}{2}\cos\frac{(N\minus{}1)x}{2} \plus{} \sin\frac{x}{2}\cos Nx}{\sin\frac{x}{2}}$

$ \equal{}\frac{\sin(2N\minus{}1)x \plus{} \sin x \plus{} \sin(2N\plus{}1)x \minus{} \sin (2N\minus{}1)}{2\sin \frac{x}{2}}$

$ \equal{}\frac{\sin(2N\plus{}1)x \minus{} \sin x}{2\sin \frac{x}{2}}$

$ \equal{}\frac{\sin\frac{N\plus{}1}{2}\cos\frac{Nx}{2}}{\sin\frac{x}{2}}$

1)  $ \sin^2{x}\cos^2{x} = \frac {1}{4}\sin^2{2x} = \frac {1}{4}\frac {1 - \cos{4x}}{2}.$ And just use  $ \cos{4x} = 1 - \frac {(4x)^2}{2!} + \frac {(4x)^4}{4!} - \frac {(4x)^6}{6!} + \cdots$. The convergence radious is $ \infty$.

2)Without using mathematical induction(two methods);
 1.
  $ \sum_{n = 0}^{N - 1}\cos{nx} = 1 + \cosh + \cos2x + \cdots + \cos(N - 1)x = Re(1 + e^{ix} + e^{2ix} + \cdots + e^{(N - 1)x}) = Re(\frac {e^{Nix} - 1}{e^{ix} - 1}) = Re(\frac {e^{\frac {Nix}{2}}}{e^{\frac {ix}{2}}}\frac {\sinh{\frac {Nix}{2}}}{\sinh{\frac {ix}{2}}}) = Re(\frac {\sin{\frac {Nx}{2}}}{\sin{\frac {x}{2}}}e^{\frac {(N - 1)ix}{2}}) = \frac {\sin{\frac {Nx}{2}}}{\sin{\frac {x}{2}}}\cos{\frac {(N - 1)x}2}.$
  2. Let's think about an identity; $ \sin{\frac {k + 1}{2}x} - \sin{\frac {k - 1}{2}x} = 2\sin{\frac {x}{2}}\cos{\frac {k}{2}x}$.
     Substitute $ k = 0,2,4,6,\cdots ,2(N - 1)$ and sum all of them at each side. The left hand side is the telescoping addition, and the right hand side includes what we want. After simple arrangement, we get the goal.


calculus
integration
trigonometry
calculus computations

Show

(i) $\int_{0}^{2}375 x^{5}(x^{2}+1)^{-4}\, dx = 2^{n}$ for some $n \in \mathbb N$.
(okay i admit this is not something from an integral table...)

(ii) $\int \frac{dx}{x^{n}\sqrt{ax+b}}=-\frac{\sqrt{ax+b}}{(n-1)bx^{n-1}}-\frac{(2n-3)a}{(2n-2)b}\int \frac{dx}{x^{n-1}\sqrt{ax+b}}\, dx$, $n \neq 1$

(iii) $\int \frac{\cos^{m}x}{\sin^{n}x}=-\frac{\cos^{m+1}x}{(n-1)\sin^{n-1}x}-\frac{m-n+2}{n-1}\int \frac{\cos^{m}x}{\sin^{n-2}x}\, dx$, ($n \neq 1$)

torajirou, you forgot to multiply by a factor of $g'(\theta) = \sec^{2}\theta$ when you substituted $x=g(\theta)$; i.e. your integral should have been

$I = 375 \int_{0}^{\alpha}\sin^{5}\theta \cos \theta \, d \theta$, $\alpha = \arctan 2$,
which will give you the correct answer.

Alternatively, you might want to try the substitution $u = x^{2}+1$ which, I think, is a little nicer for computation.


integration
real analysis
real analysis unsolved

Let $ A: L^2(0,1)\rightarrow L^2(0,1)$ be difined: $ Ax(s)\equal{}\int_{0}^{1}st(1\minus{}st)x(t)dt$. Find spectrum of $ A$

an easy one-standard task-note that you have finite-dimensional operator: please write what you've done and waht problems you have with this task and I'll help you to solve them.


Let there be a triangle $ ABC$ and $ P$ on it`s circumcircle. Prove that Simson line for $ P$ passes through the middle of $ (HP)$, where $ H$ is the ortocenter of the triangle.

not hard at all.
posted many times in ML ,Use search function.

Let $P$ be in the arc $BC$.Let $P_{1},P_{2},P_{3}$ be the altitudes from $P$ to the sides.$AB,BC,CA$, respectively.And let $A_{1}$ be the foot of altitude from $A$. Assume that $BA_{1}\geq BP_{2}$.Let Simson line intersect the lines $HA_{1},HP,HB$ at $K,M,L$.
Then from parallel lines $\angle KHL =\angle KHM+\angle MHL= \angle MPP_{2}+\angle MPP_{3}=\angle P_{2}PP_{3}$ and $\angle P_{2}P_{3}P= \angle KLH$.Then $\bigtriangleup PP_{2}P_{3}\sim \bigtriangleup HKL$ and $\bigtriangleup PP_{2}M \sim \bigtriangleup HKM$.So
$\frac{HK}{PP_{2}}=\frac{KM}{P_{2}M}=\frac{LK}{P_{2}P_{3}}=\frac{LM}{P_{3}M}$
$\frac{LM}{P_{3}M}=\frac{HK}{PP_{2}}=\frac{HM}{PM}$
So $HP_{3}\parallel PL$ and $HL \parallel PP_{3}\Longrightarrow HLPP_{3}$ is a paralelogram.Hence Simson line goes trough midpoint of $HP$


This forum is awefully quiet for having recently finished an AMY packet.  Is anyone interested in discussing part of it?

Or is everyone too busy working on the Law of Sines/Law of Cosines packet?  :P

Sure, I'll discuss. So far I've gotten all the easy problems and all the medium problems except 9.4, 9.7, 9.9, which have me stumped :huh: .
Is there a mistake for 3.3? AB^2*CD^2=AB^4, but either AB<AC and/or AB<BD, so AB^4<AC^4+BD^4, contradiction.
I've been busy doing last year's packets as well (since I didn't enroll at the beginning) so I haven't gotten really far on this one yet.
For the first segment I was able to do most of the easy/medium problems (except E26, E31, M8, M14, M24,31,33,35) and a few hard problems (I did H1,2,3,9,12,16,18,27,31). Thought there were a bit too many typos/errors.
My AIM is oldheneel.

I meant discussion of the sums and products packet. We should probably hold off on discussion of the geometry packet until after the e-mail deadline. I will say, however, that I think there were a few errors scattered throughout the packet, mostly writing $ AB\cdot CD$ rather than $ AB\cdot BC$.

I got all of the E problems on the Sums/Products thing and 27 of the H problems. The M problems were too hard for me  :D


analytic geometry
factorial
geometry
geometric transformation
reflection
graphing lines
slope

On a coordinate plane, point P is located at (1,2) and point Q is located at (2,3).  If point R is located at some point on the line x=-1, what is the minimum value of PR+QR?

Thanks anyone for the help  :)

ch1n353ch3s54a1l, that's a good start! But what is the minimum value?

[hide] $PR = \sqrt{(x+x)^2 + (y+y)^2}$ so if x1 = -1, x2 = 1 $\sqrt{\frac{17}{4}}$ is distance; you know it's the midpoint so you get the same thing for the other side, then you get 2\sqrt{{17}{4}} as the min value [/hide]

[quote="57_factorial"][quote="ch1n353ch3s54a1l"]The shortest length of PR+QR would be where R's y coordinate is the midpoint between P and Q.[/quote]

Are you sure that would be the case? Since the point must be on the line x=-1, wouldn't this mean that the midpoint (2-root 26) could not be directly taken as the y-coordinate?[/quote]The midpoint claimed by ch1n353ch3s54a1l is at $(-1,2.5)$. But, ch1n353ch3s54a1l, is this y-coordinate midpoint REALLY going to give the minimum value of $PR+QR$? Because $PR\ne QR$... and $x=-1$ is not the midpoint of $PR+QR$ if you calculate the lengths...

I can't prove it...but the midpoint of the y coords is 2.5 and the coordinate is (-1,2.5) and as the point decreases or increases, the lengths get larger. I tried the base points (-1,2) and (-1, 3) and those were greater. Anything exceeding those y values are obviously larger.

would you have to draw a circle from that midpoint in the y position and see where it would become tangent to the line or is there an easier way (if that even works)?

But that doesn't make any sense I still agree with what ch1n353ch3s54a1l said...

Actually, 10000th User and 57_factorial are correct; $(-1,2.5)$ does not give the shortest distance.

[hide="Hint"]
Reflect one of the points (either $P$ or $Q$) over the line $x=-1$.
[/hide]

Ok.
I'm not sure if this is the right approach, but...
If i follow frt's hint and reflect point (1,2) across x=-1,  i drew two right triangles, one of length 2,x, and $\sqrt{x^2+4}$ and another with lengths 3,y, and $\sqrt{y^2+9}$. THen $\sqrt{x^2+4}+\sqrt{y^2+9}=\sqrt{26}$ and $x+y=1$ and then I got to the equation $42y^2+62x^2+2x^2y^2=x^4+y^4+25$ and if $a=x^2, b=y^2$ then $42b+62a+2ab=a^2+b^2+25$ and then i'm stuck.

wow that's way too complicated...  maybe you can just find the line which goes across the two points and intersects the -1 line?  ie: y = mx+b, we have (-3, 2) - the reflected point - and (2, 3) - now we get: 1/5 for slope using slope formula and plugging in the second point:
3 and 3/5 for b, or the y intercept.  So the point of R is:
y= 1/5x + 2+3/5
x = -1, plugging in we get:
2 and 2/5
So that is R
Isn't my picture beautiful  :lol:

That's right, but all we need to find is $PR+QR$ :roll:

[quote="frt"][hide="Hint"]
Reflect one of the points (either $P$ or $Q$) over the line $x=-1$.
[/hide][/quote]

Hey, that's what I did  :D  (probably this came after many problems like this).

[hide="La respuesta incorrecta"]$\sqrt{7^2+1^2}=5\sqrt2$. [/hide]

[quote="236factorial"]Hey, that's what I did  :D  (probably this came after many problems like this).

[hide="I get"]$\sqrt{7^2+1^2}=5\sqrt2$. [/hide][/quote]Let's make this short, I guess a lot of people haven't met reflections yet. First I want you guys to draw pictures to see. If you don't draw because you are lazy or no paper or no pen, you can't learn well.

Reflect point $P$ over the line $x=-1$ and let this new point be $P'$, so $P'$ will have coordinates $(-3,2)$ because $P$ is 2 units away from the line and we keep $y$-coordinates the same because $PP'$ is perpendicular to it.

Now pick a couple of points (anything, random, that's fine, just somewhere) on the line $x=-1$ as candidates for $R$, you can label them $R_1,R_2,\ldots$, etc.

Draw lines from $P'$ to those points R's and draw lines from $Q$ to those R's too.

Draw the line $P'Q$. Now, do you see why we initially reflected $P$?

From here you guys do the rest and GOOD LUCK, otherwise ...............

BTW, 236factorial, don't think I forgot about the quote above, your answer is wrong :D

The shortest distance between two points is a straight line.  If we reflect one of the points over the line $x = -1$, then the distance will be the same as $PR + QR$, since we're merely reflecting one part of the line.
Let's reflect one to get the points $(-3,2)$ and $(2,3)$, which has distance $\sqrt{26}$.  Therefore this is the smallest possible distance.

[quote="10000th User"]BTW, 236factorial, don't think I forgot about the quote above, your answer is wrong :D[/quote]

Hmm.... it would have been nice for me to know that x=-1 was vertical, not horizontal, and that is why I spend, on average, 1.5-1500 seconds more per problem involving x=k and y=k.


quadratics
number theory unsolved
number theory

The sequence $ \{a_n\}$ is defined by $ a_0\equal{}2,a_1\equal{}1$ and $ a_{n\plus{}2}\equal{}a_{n\plus{}1}\plus{}a_n$ ,for all $ n\geq 0$.Prove that if $ p$ is a prime factor of $ a_{2k}\minus{}2$ for some $ k\in N$ then $ p$ is a factor of $ a_{2k\plus{}1}\minus{}1$

[quote="forevercontinue"]$ a_n \equal{} (\frac {1 \plus{} \sqrt {5}}{2})^n \plus{} (\frac {1 \minus{} \sqrt {5}}{2})^n$
Let $ u \equal{} \frac {1 \plus{} \sqrt {5}}{2}, v \equal{} \frac {\sqrt {5} \minus{} 1}{2}$. Hence $ uv \equal{} 1,u \minus{} v \equal{} 1$
$ \Rightarrow a_{2k} \minus{} 2 \equal{} u^{2k} \plus{} v^{2k} \minus{} 2(uv)^k \equal{} (u^k \minus{} v^k)^2 \Rightarrow u^k \equiv v^k(mod p)$
Moreover
$ a^{2k \plus{} 1} \minus{} 1 \equal{} u^{2k \plus{} 1} \minus{} v^{2k \plus{} 1} \minus{} 1 \equiv u^{2k \plus{} 1} \minus{} u^{2k}v \minus{} 1 \equal{} u^{2k}(u \minus{} v) \minus{} 1 \equal{} u^{2k} \minus{} 1$
$ u^{2k} \minus{} 1 \equal{} u^{2k} \minus{} (uv)^k \equal{} u^k(u^k \minus{} v^k) \equiv 0(mod p)$
Q.E.D[/quote]
Why:$ u^k\equiv v^k(mod p)$?Where $ u^k$ be not integer.So not using modulo.
I think we should write:$ a_{2k} \minus{} 2 \equal{} a_k^2$ and prove that:$ a_k| (a_{2k \plus{} 1}\minus{}1)$

[quote="math10"]Why:$ u^k\equiv v^k(mod p)$?Where $ u^k$ be not integer.So not using modulo.
I think we should write:$ a_{2k} \minus{} 2 \equal{} a_k^2$ and prove that:$ a_k| (a_{2k \plus{} 1} \minus{} 1)$[/quote]

The proof fails for $ p\equal{}2,5$. Other then that, it works.
If 5 is a quadratic residue mod p, then $ u$ and $ v$ are elements of the field $ \mathbb{Z}/p\mathbb{Z}$ and the argument works unmodified. Otherwise, we need to be working in $ \mathbb{Z}/p\mathbb{Z}[\sqrt5]$ which is isomorphic to $ \mathbb{F}_{p^2}$, the field with $ p^2$ elements. The same argument works in this field.

See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=229442]here[/url] for more information (scroll down to the problem 3)


Ross Mathematics Program
summer program
Mathcamp

Hello guys,
I'm a sophomore and I would like to go to a summer program during next year or something interesting and prestigious.
I am an international student from Europe, and I would like to go to something like this in universities from Ivy League, but not only.
Do you any suggestions or ideas?
Thanx in advance,
ExceLLence

I forgot to mention that I also need financial aid :/

ISSYP is free.  You only need to pay for the travel to and from Canada.  :)
http://www.perimeterinstitute.ca/en/Outreach/Students/ISSYP/

I am an alumnus of it and I know a few other alumni from the program.  All of us agree, that is was the best program we have ever done like that and would love to go back.

The Ross Program

http://www.math.ohio-state.edu/ross/

has international students and it has financial aid (which I think the international students may obtain). It is not expensive even at list price.

How much does Ross/SSP/ISSYP etc. weigh in your college application?

I also highly recommend Canada/USA Mathcamp (mathcamp.org). It accepts international students, and it provides financial assistance.


search
quadratics
function
number theory

Just go to google, and type free math ebooks or free ebooks.

I've put something in [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=99001[/url]

There used to be a sticky in the Round Table section which listed a lot of online resources. But now I can't find it. You guys can also try to do some search in case you really find it!

[quote="Inspired By Nature"][quote="guile"]http://homepage.mac.com/kaotech/Free_Books.html[/quote]
That's an awesome site guile! :D
Really detailed.[/quote]

I agree. Even though there's only 5 Math ones, the others could come in handy.

[quote="guile"]http://homepage.mac.com/kaotech/Free_Books.html[/quote]

nice and useful and good  :)  :)  :)

[url]http://www-cse.ucsd.edu/~maxal/e-books.html[/url]
[url]http://www.gallup.unm.edu/~smarandache/math.htm[/url]

shobber, I've tried but I can't find it.

Some More Sites:

http://www.e-dsp.com/free-ebooks/
http://www.generallibrary.com/Math.html

And check this out, books in full text.  :D

http://www.xpmath.com/ebooks/math_ebooks.php

great links. do you guys know anywhere that i can get problem books for Algebra I level? Like, not much explanation, just straight forward algebra practice. i teach my brother algebra (like help him with homework, study for tests, etc) and a workbook like that would be great.

 just so u have an idea of what i'm talking about, workbooks with 1) solving/factoring quadratic equations 2) radicals and frational/whole number exponents 3) graphing and functions

stuff like that.

thanks in advance :D

[quote="SplashD"][quote="akimatsu"]well the local library or any book store is not going to be of much use if i cant speak the local language here is it? i'm japanese but i live in china[/quote]
This would a great time to learn since Japanese is very silimar to Chinese.[/quote]
Indeed. I once met a Japanese. He could read all the Chinese letter and could understand everything written on paper, though he couldn't speak Chinese.

Great links guys, try this one,
http://dmoz.org/Science/Math/Publications/Online_Texts/

The best book on Number Theory is that timless classic, Hardy's Number Theory. If you can find it, you're in luck.

[quote="Astronum"]Would you know some online free ebooks about number theory that you could suggest?
I know, I know, you have Introduction to Number Theory, but I had a bad experience ordering something from USA since I never received what I asked, but, fortunetely, got the money refunded.

* I am afraid I did not post in the appropriate category, but could not find it. Sorry.[/quote]
lib.org.by


Did I do this problem correctly?

I'm having a hard time getting it to show correctly. The √ sign extends over x+3 only. Could you show me how I should write it so it shows in this forum correctly please?

√(x+3) + 3 = 9

subtract 3 from each side

√(x+3) = 6

√(33 + 3) = 36

√36 = 6

x = 33

That is correct. To make the sqrt, do

[code]$\sqrt{x}$[/code]

To make

$ \sqrt{x}$.

what BOGTRO was showing you is an example of LaTeX, there is a link to more LaTeX examples on the left panel

Yeah... X=33 is correct.

But i can show you a faster way to do it...

[hide]First you would want to subtract 3 from each side. Then square each side and you would come up with

X+3=36
 -3    -3
---------
X=33

Well thats a quick way not like what you said you did. Lolz :rotfl: [/hide]

[quote="rt_08"]hey  u got x+3=36
so x=36-3
x=33
i hope thats

the answer

![/quote]
Yep it is!!!

Also if you wish to do cube roots you do $ \sqrt[3]{x}$ (rollover the example to see its code)

YOu can do similar things for nth roots, $ \sqrt[n]{1}\equal{}1$

Please don't revive old threads - this is from May.


Let ABCD  be  a  convex  quadrilateral  and  M,N,K,L  be  mid  point of  sides  AB,BC,CD,  and  DA  respectively.Let  R  and   T   points  be  mid  point of  AC  and  BD  diagonals.If  area(MRKT)=area(NTRL)  show  that  one  of  diagonal  of  ABCD  quadrilateral divide the  area of   ABCD  quadrilateral equal  two areas.

Let $ Q \equiv AC \cap BD.$ Then it is clear that $ MRKT$ and $ NRLT$ are both parallelograms.  If $ [MRKT] \equal{} [NRLT],$ then $[\triangle MRT] \equal{} [\triangle NRT],$ which implies that $ MN \parallel RT,$ but $ MN \parallel AC.$ Hence, $ R \equiv Q.$ If $ Q$ bisects the segment $ DB,$ then the distances from $ D,B$ to $ AC$ are equal $ \Longrightarrow$ $ [\triangle ABC] \equal{} [\triangle CDA].$


Let M be any point in the plane of \[ \Delta ABC\] then find the minimum of \[ MA^2\plus{}MB^2\plus{}MC^2\]

is the ans by any chance [hide]\[ \frac{9R^2}{4}\][/hide]

[quote="anorithdialga"]\[ MA^2 \plus{} MB^2 \plus{} MC^2
\]
[/quote]
What does the above expression mean?  Aren't M, A, B, C points?

[quote="mathwizarddude"][quote="anorithdialga"]
\[ MA^2 \plus{} MB^2 \plus{} MC^2
\]
[/quote]
What does the above expression mean?  Aren't M, A, B, C points?[/quote]

MA is the line joining the points M and A .

now r u clear.   stop :spam:
if u have a solution just post it

[quote="anorithdialga"][quote="HTA"][hide="hint"]using lebnitz's formula $ \sum MA^2 \equal{} \sum GA^2 \plus{} MG^2$[/hide][/quote]

can u explain in details . and state the formula and its proof?

:wow:[/quote]
i'm sorry it should be $ \sum MA^2 \equal{} \sum GA^2 \plus{} 3MG^2$ anyway ,here is the proof for lebnitz's formula
$ \sum MA^2 \equal{} \sum GA^2 \plus{} 3MG^2$
$ \sum (vecMG\plus{}vecGA)^2 \equal{} \sum GA^2 \plus{} 3MG^2$
$ 3MG^2\plus{} \sum GA^2 \plus{} 2vecMG(\sum vec GA) \equal{} \sum GA^2 \plus{} 3MG^2$
it is true since \sum vec GA = 0
so $ \sum MA ^2 \geq \sum GA^2$
sorry i dont know how to type vector so vec GA means vector GA

[quote="apratimdefermat"]how does it help the original problem[/quote]
since $ \sum MA^2 \equal{} \sum GA^2 \plus{} 3MG^2$
and $ MG^2 \geq 0$ so ................. $ \sum MA^2 \geq \sum GA^2$

"[
\overrightarrow {AB}
\]"

this is the code for a vector .again thanks for the solution


vector
geometry
geometric transformation
rotation
linear algebra
matrix
rotation matrix

I feel like this problem should be simple, however I'm a bit stuck. I need to express a vector with respect to a new basis (that includes a new origin). Here is what I'm given:

Original basis: standard, orthogonal Euclidean basis (e1 = [1,0], e2 = [0,1])
Vector: v = [2,2]

The new basis (which is a right-handed basis) has an origin at (5,2) with e_new_1=[1/sqrt(2), 1/sqrt(2)] and e_new_2=[-1/sqrt(2), 1/sqrt(2)] (aka, it represents a counter-clockwise rotation of 45 degrees)

I need to express 'v' in terms of the new basis. I can do it if the new basis were just the 45 degree rotation, which gives a rotation matrix of [[1/sqrt(2), -1/sqrt(2)][1/sqrt(2), 1/sqrt(2)]. However, I do not know how to add in the change of origin. I feel like just adding it to the result does not give the right answer.

You need to [i]subtract[/i] the new origin (why?).


algebra
polynomial
linear algebra
matrix
function

[i]
    Let $ P(x)$ be a polynomial with $ deq P \ \equal{} \ n \minus{} 1$  ( $ n$ is a fixed positive integer ), $ P(x) \ \in \ \mathbb{R} [x]$

   The matrix $ A$ is defined as follow :  $ A \ \equal{} \ (P(i \plus{} j \minus{} 1))_{1 \ \leq i;j \leq n}$

             Evaluate $ |A|$ :) [/i]

OK, this is pretty neat. Solution hidden due to large matrices and so others can solve it if they want to.

[hide="Solution"]As a point of notation, let $ Df(k)\equal{}f(k\plus{}1)\minus{}f(k)$; $ D$ is the forward difference operator.

We apply row operations to $ A$, leaving the determinant unchanged. First, subtract the $ (n\minus{}1)$th row from the $ n$th, the $ (n\minus{}2)$nd from the $ (n\minus{}1)$th, and so on until we subtract the second from the first. The matrix becomes $ \begin{bmatrix}P(1)&P(2)&P(3)&\cdots&P(n)\\DP(1)&DP(2)&DP(3)&\cdots&DP(n)\\ DP(2)&DP(3)&DP(4)&\cdots&DP(n\plus{}1)\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ DP(n\minus{}1)&DP(n)&DP(n\plus{}1)&\cdots&DP(2n\minus{}2)\end{bmatrix}$
Repeat this for the bottom $ n\minus{}1$ rows, then the bottom $ n\minus{}2$ rows, and so on. Eventually, we reach $ A'\equal{}\begin{bmatrix}P(1)&P(2)&P(3)&\cdots&P(n)\\DP(1)&DP(2)&DP(3)&\cdots&DP(n)\\ D^2P(1)&D^2P(2)&D^2P(3)&\cdots&D^2P(n)\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ D^{n\minus{}1}P(1)&D^{n\minus{}1}P(2)&D^{n\minus{}1}P(3)&\cdots&D^{n\minus{}1}P(n)\end{bmatrix}$, which has the same determinant as $ A$.

Now, each of these new $ D^iP$ functions are polynomials, and their degrees decrease. We can do the same thing horizontally using column operations. We pass through $ \begin{bmatrix}P(1)&DP(1)&DP(2)&\cdots&DP(n\minus{}1)\\DP(1)&D^2P(1)&D^2P(2)&\cdots&D^2P(n\minus{}1)\\ D^2P(1)&D^3P(1)&D^3P(2)&\cdots&D^3P(n\minus{}1)\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ D^{n\minus{}1}P(1)&D^{n}P(1)&D^{n}P(2)&\cdots&D^{n}P(n\minus{}1)\end{bmatrix}$ on our way to $ A''\equal{}\begin{bmatrix}P(1)&DP(1)&D^2P(1)&\cdots&D^{n\minus{}1}P(1)\\DP(1)&D^2P(1)&D^3P(1)&\cdots&D^nP(1)\\ D^2P(1)&D^3P(1)&D^4P(1)&\cdots&D^{n\plus{}1}P(1)\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ D^{n\minus{}1}P(1)&D^{n}P(1)&D^{n\plus{}1}P(1)&\cdots&D^{2n\minus{}2}P(1)\end{bmatrix}$, which has the same determinant as $ A$.

But since $ D^kP$ has degree $ n\minus{}k$, everything below the antidiagonal is zero. The determinant is $ (\minus{}1)^{n\minus{}1}$ times the product of the antidiagonal elements, each of which is $ D^{n\minus{}1}P(1)$. We can get that in more familiar terms for this degree $ n\minus{}1$ polynomial; if $ P(x)\equal{}a_{n\minus{}1}x^{n\minus{}1}$ plus lower-degree terms, the $ (n\minus{}1)$th difference is the constant $ n!a_{n\minus{}1}$, just like the $ (n\minus{}1)$th derivative. Thus the overall determinant is $ (\minus{}1)^{n\minus{}1} (n!)^n (a_{n\minus{}1})^n$.

This is valid anywhere the determinant makes sense; $ \mathbb{R}$ is completely irrelevant, and we don't even have to divide by anything.[/hide]


Prove that if $ a$ and $ b$ are both add integers, then $ 16| a^4 \plus{} b^4 \minus{} 2.$

Hint
[hide]
Write $ a^4 \plus{} b^4 \minus{} 2$ as $ a^4 \minus{} 1 \plus{} b^4 \minus{} 1 \equal{} (a \plus{} 1)\cdot (a \minus{} 1)\cdot (a^2 \plus{} 1) \plus{} (b \plus{} 1)\cdot (b \minus{} 1)\cdot (b^2 \plus{} 1)$
[/hide]


Solve the equaltion
$sin(x-\frac{\pi}{4})-cos(x+\frac{3\pi}{4})=1$
satisfy the inequaltion
$\frac{2cos7x}{cos3+sin3}>2^{cos2}x$
Note : $2$ ,$3$ in $cos2$ $sin3$ and $cos3$ is radian.

hello, do you mean
$\frac{2\cos(7x)}{\cos(3)+\sin(3)}>2^{\cos(3x)}$?
Sonnhard.

[hide="1."] Using the identity $\cos\theta=\sin(\frac{\pi}{2}-\theta)$, we obtain

$\sin(x-\frac{\pi}{4})-\sin(-\frac{\pi}{4}-x)=1$.

Using the identity $\sin\alpha-\sin\beta=2\sin(\frac{\alpha-\beta}{2})\cos(\frac{\alpha+\beta}{2})$, we obtain

$2\sin x\cos(-\frac{\pi}{4})= 1 \implies \sin x=\frac{1}{\sqrt{2}}$.

Therefore, $x=\frac{\pi}{4}+2\pi n, \frac{3\pi}{4}+2\pi k , n, k\in \mathbb{Z}$.[/hide]


Hi! Read on an other forum and unsolved: Determine $\displaystyle{\sum \frac{x^n}{(n!)^p}}$ for $p$ integer.

$p$ -is fixed and $\sum_{n=0}^{\infty}$ or $\sum_{n,p>0}$?


I finished quadratic residues and i'm inquring what type of questions will i use it if you have a question post it

How about this?

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1229308&search_id=1704383035#1229308[/url]

[quote="21 7 15"]The problem was nice but i did not from an aops book i learnt from wikipedia so no i don't understand their solution is not good anyone with a problem and nice solution or a blog where i can see problems tell me[/quote]You can hardly learn anything properly from wikipedia (online encyclopedia written by other users). And without a proper grasp of the material, do you think you can solve any of the harder problems? So you learn to run without even learning to walk first?

@Mewto55555:  :rotfl:

[hide="Hint for mine"]
What are the quadratic residues mod 4?
[/hide]

Solution for mewto55555's problem.

[hide] it ends in 99, so it's 3 mod 4. Square numbers can only be 0, or 1 mod 4.

You can't add any two squares to get 3 mod 4[/hide]


conics
trigonometry
analytic geometry
hyperbola

The polar equation $ r\equal{}\frac{8}{2\minus{}\sin\theta}$ represents one of the four conic sections.
What conic section is represented?
What is/are the coordinates of the focus/foci?
Give the coordinates in rectangular form

In converting to rectangular coordinates, keep in mind the following: $ r\equal{}\sqrt{x^2\plus{}y^2},x\equal{}r\cos\theta,y\equal{}r\sin\theta$ (why?). Now try utilizing these substitutions in the given problem.

[quote="Quevvy"]The polar equation $ r \equal{} \frac {8}{2 \minus{} \sin\theta}$ represents one of the four conic sections.
What conic section is represented?
What is/are the coordinates of the focus/foci?
Give the coordinates in rectangular form[/quote]
[hide]Don't like fractions, so multiply both sides by the denominator to get $ 2r \minus{} rsin(\theta) \equal{} 8$
This is equivalent to $ 2\sqrt {x^{2} \plus{} y^{2}} \minus{} y \equal{} 8$
$ 2\sqrt {x^{2} \plus{} y^{2}} \equal{} y \plus{} 8$
$ 4x^{2} \plus{} 4y^{2} \equal{} y^{2} \plus{} 16y \plus{} 64$
$ 4x^{2} \plus{}3y^{2} \minus{} 16y \equal{} 64$
And one can recognize this as a general form for an ellipse.[/hide]

Hyperbola? if you just graph the polar equation, it is obviously not a hyperbola...

Sorry, edited.
Also note that any polar form of a conic section is in the form $ \frac{l}{1\plus{}ecos(\theta)}$, where e is the eccentricity of the conic, and l is the semi-latus rectum. So knowing this we could transform that equation into
$ r \equal{} \frac {4}{1 \minus{} \frac {1}{2}cos(\theta)}$. Since the eccentricity is $ \frac {1}{2}$, you can tell from this that the conic section is an ellipse.

I found an equation of $ \frac{x^2}{\frac{64}{3}}\plus{}\frac{(y\minus{}\frac{8}{3})^2}{\frac{256}{9}}\equal{}1$ for an ellipse. Is that right?


Simplify $ \displaystyle\frac{24t^3}{15t^4}\cdot \frac{5t^8}{3t^6}$.


hello, calculating the solutions of $ x^2\plus{}\frac{1}{x^2}\equal{}23$ we get
$ x_1\equal{}\minus{}\frac{1}{2}\plus{}\frac{1}{2}\sqrt{21}$
$ x_2\equal{}\minus{}\frac{1}{2}\minus{}\frac{1}{2}\sqrt{21}$
$ x_3\equal{}\minus{}\frac{5}{2}\plus{}\frac{1}{2}\sqrt{21}$
$ x_4\equal{}\minus{}\frac{5}{2}\minus{}\frac{1}{2}\sqrt{21}$
now you can calculate $ x^7\plus{}\frac{1}{x^7}$
Sonnhard.

[hide="A lot easier on calculation"]$ x^2 + 2 + {1\over x^2} = 25\implies x + {1\over x} = \pm 5$

Instead of $ \pm 5$, we'll write $ k$ for shortness.

$ (x^2 + {1\over x^2})(x + {1\over x}) = x^3 + x + {1\over x} + {1\over x^3}\implies x^3 + {1\over x^3} = 23k - k = 22k$

$ (x^3 + {1\over x^3})(x^2 + {1\over x^2}) = x^5 + x + {1\over x} + {1\over x^5}\implies x^5 + {1\over x^5} = 22k\cdot 23 - k = 505k$

\begin{eqnarray*}(x^5 + {1\over x^5})(x^2 + {1\over x^2}) = x^7 + x^3 + {1\over x^3} + {1\over x^7}\implies x^7 + {1\over x^7} = 505k\cdot 23 - 22k = 11593k = \pm 57965\end{eqnarray*}[/hide]


Given an infinite sequence $ \{a_n\}$, the sum of which is irrational, prove that there exist no infinite subsequences such that their sum is rational.

Part 2: If there exist subsequences that are rational, what can be said about them?
Part 3: Replace "rational" with "algebraic" and "irrational" with "transcendental" above.

[quote="t0rajir0u"]I don't understand.  Let $ a_n \equal{} \frac {1}{2^n}$; then the set of sums of (finite or) infinite subsequences of $ \{ a_n \}$ is the set of reals in $ (0, 1)$, but $ \sum a_n \equal{} 1$.[/quote]

Sorry, I worded it wrong. See above.

Still wrong.  Let $ a_1 \equal{} \pi, a_{n \plus{} 1} \equal{} \frac {1}{2^n}, n \in \mathbb{N}$.


probability
algebra
polynomial
geometry
3D geometry
trigonometry

1. Suppose that $ sin\theta\plus{}cos\theta\equal{}0.8$. What is the value of $ sin(2\theta)$

2. Three digits $ a, b,$ and $ c$ are selected from the set $ \{0,1,2,3,4,5,6,7,8,9\}$, one at a time, with repetition allowed. What is the probability that $ a>b>c$

3. The polynomial $ P(x)\equal{}2x^4\minus{}x^3\minus{}7x^2\plus{}ax\plus{}b$ is divisible by $ x^2\minus{}2x\minus{}3$ for certain values of $ a$ and $ b$. What is the sum of $ a$ and $ b$

4. Find the ordered 4-tuple of positive integers (a, b, c, d) so that:

$ \frac{37}{13}\equal{}a\plus{}\frac{1}{b\plus{}\frac{1}{c\plus{}\frac{1}{d}}}$

5. The faces of a cube are colored red and blue, one at a time, with equal probability. What is the probability that the resulting cube has at least one vertex $ P$ such that all three faces containing $ P$ are colored red?

[hide="1"]
$ \sin \theta \plus{} \cos \theta \equal{} .8\\
(\sin \theta \plus{} \cos \theta)^2 \equal{} .64\\
\sin^2 \theta \plus{} \cos^2 \theta \plus{} 2 \sin \theta \cos \theta \equal{} .64\\
1 \plus{} 2 \sin \theta \cos \theta \equal{} .64\\
2 \sin \theta \cos \theta \equal{} \minus{}.36\\
\sin 2\theta \equal{} \minus{}.36$
[/hide]
[hide="2"]
First of all, the three digits must be distinct. There is a $ \frac{9}{10} \cdot \frac{8}{9} \equal{} \frac{4}{5}$ probability of this occuring. Then, the digits must be in decreasing order. This only occurs in $ \frac{1}{6}$ of the cases, so the total probability is $ \frac{2}{15}$.
[/hide]
[hide="4"]
Since $ \frac{1}{b\plus{}\frac{1}{c\plus{}\frac{1}{d}}}$ is a value between 0 and 1, $ a\equal{}2$. Then:
$ \frac{11}{13}\equal{}\frac{1}{b\plus{}\frac{1}{c\plus{}\frac{1}{d}}}\\
\frac{13}{11}\equal{}b\plus{}\frac{1}{c\plus{}\frac{1}{d}}$
Similar logic makes $ b\equal{}1$. Then:
$ \frac{2}{11}\equal{}\frac{1}{c\plus{}\frac{1}{d}}\\
\frac{11}{2}\equal{}c\plus{}\frac{1}{d}\\
c\equal{}5, d\equal{}1$.
Therefore, the answer is $ (2, 1, 5, 2)$.
[/hide]

[hide]

3.

If you use long division, you will eventually get to x^2-2x-3 divides 5x2 + (a+9)x +b. We know that x^2-2x-3 must be multiplied by 5 to satisfy our division. If we multiply and change the signs to subtract, we get:

5x^2 + (a+9)x + b

-5x^2 + 10x + 15

So a+9 = -10, b = -15

a= -19, b = -15, and their sum is -34.

[/hide]

[hide="3"]
Since $ P(x) \equal{} (x \plus{} 1)(x \minus{} 3)Q(x) \equal{} 2x^4 \minus{} x^3 \minus{} 7x^2 \plus{} ax \plus{} b$, then $ P( \minus{} 1) \equal{} \minus{} 4 \minus{} a \plus{} b \equal{} P(3) \equal{} 72 \plus{} 3a \plus{} b \equal{} 0$. So we have two systems + two equations: subtraction gives $ 76 \plus{} 4a \equal{} 0 \Longrightarrow a \equal{} \minus{} 19, b \equal{} \minus{} 15$. Thus $ a \plus{} b \equal{} \minus{} 34$.
[/hide]
[hide="5"]
[list]
[*]$ 6$ faces red, all work, $ 1$ way
[*]$ 5$ faces red, all work, configurations determined by which face is missing, so $ 6$ ways
[*]$ 4$ faces red, only those with two adjacent blue faces work, so configurations are determined by number of edges, or $ 12$ ways
[*]$ 3$ faces red, all three red faces must be at the same vertex, and there are $ 8$ vertices.[/list]
The probability is $ \frac {1 \plus{} 6 \plus{} 12 \plus{} 8}{2^6} \equal{} \frac {27}{64}$, which suggests theres an easier way .. ?
[/hide]

[quote="the future"]Uhh number 2 should be 120/1000=12/100=6/50=3/25^^^

Since the total cases involves some where all digits are not distinct.[/quote]
Actually, the problem asks for $ a>b>c$, so all the digits must be distinct. :D

[quote="happyme"]1. Suppose that $ sin\theta \plus{} cos\theta \equal{} 0.8$. What is the value of $ sin(2\theta)$

2. Three digits $ a, b,$ and $ c$ are selected from the set $ \{0,1,2,3,4,5,6,7,8,9\}$, one at a time, with repetition allowed. What is the probability that $ a > b > c$

3. The polynomial $ P(x) \equal{} 2x^4 \minus{} x^3 \minus{} 7x^2 \plus{} ax \plus{} b$ is divisible by $ x^2 \minus{} 2x \minus{} 3$ for certain values of $ a$ and $ b$. What is the sum of $ a$ and $ b$

4. Find the ordered 4-tuple of positive integers (a, b, c, d) so that:

$ \frac {37}{13} \equal{} a \plus{} \frac {1}{b \plus{} \frac {1}{c \plus{} \frac {1}{d}}}$

5. The faces of a cube are colored red and blue, one at a time, with equal probability. What is the probability that the resulting cube has at least one vertex $ P$ such that all three faces containing $ P$ are colored red?[/quote]

these look familiar, any specific source?

[quote="Brut3Forc3"][quote="the future"]Uhh number 2 should be 120/1000=12/100=6/50=3/25^^^

Since the total cases involves some where all digits are not distinct.[/quote]
Actually, the problem asks for $ a > b > c$, so all the digits must be distinct. :D[/quote]

Yes but not the [u]entire set[/u] (denominator)

The numerator is 120, but the denominator [u]does not[/u] have to be all distinct
You, in your solution, assumed the denominator was all distinct number too.

You cant calculate that since the question isnt conditional probability.

I think problem 4 is too easy.
a = 2, b = 1, c = 5 and d = 2


AB and AC are tangent to circle O at B and C, respectively, and D is a point on the minor arc BC. Lines BD and AC meet at E, and lines CD and AB meet at F. If BF = a, CE = b, and EF = x, prove that $ x^{2}\equal{}a^{2}\plus{}b^{2}\minus{}ab$

Tangent of $(O)$ through $D$ cuts $BC$ at $P.$ Thus, $E,F,P$ are collinear on the Lemoine axis of $\triangle BDC.$ Let $AB=AC=L.$ By Menelaus' theorem for $\triangle ABC$ cut by $\overline{PFE},$ we have

$\frac{a}{L-a} \cdot \frac{L-b}{b}=\frac{PB}{PC}=\frac{DB^2}{DC^2}=\frac{a^2}{b^2} \cdot \frac{EB^2}{FC^2} \Longrightarrow \ \frac{EB^2}{FC^2}=\frac{b(L-b)}{a(L-a)} \ \ (1)$

On the other hand, by Stewart theorem for the cevians $EB,FC$ in $\triangle AFE,$ we have

$EB^2=\frac{x^2L+aL(L-a)-a(L-b)^2}{L-a} \ , \ FC^2=\frac{x^2L+bL(L-b)-b(L-a)^2}{L-b}$

$\frac{EB^2}{FC^2}=\frac{L-b}{L-a} \cdot \frac{x^2L+aL(L-a)-a(L-b)^2}{x^2L+bL(L-b)-b(L-a)^2} \ \ (2)$

From  $(1)$ and $(2)$ we have then

$\frac{x^2L+aL(L-a)-a(L-b)^2}{x^2L+bL(L-b)-b(L-a)^2}=\frac{b}{a} \Longrightarrow \ x^2=a^2+b^2-ab$


trigonometry
calculus
calculus computations

On $x-y$ plane, find the numbers of intersection points denote by $N(r),$between the curve $y=\cos \left(\sqrt{\frac{\pi}{2}}x\right)$ and the circle with radius $r$ centered the origin.


Prove that the equation of the circle of which the points $ (x_1, y_1)$ and $ (x_2, y_2)$ are the ends of a chord of segment containing an angle $ \theta$ is
$ (x \minus{} x_1)(x \minus{} x_2) \plus{} (y \minus{} y_1)(y \minus{} y_2) \pm \cot \theta [(x \minus{} x_1)(y \minus{} y_2) \minus{} (x \minus{} x_2)(y \minus{} y_1)] \equal{} 0$

Also tell me in which section in the forum should i post this sum to get more replies.......  :maybe:

[url=http://www.artofproblemsolving.com/Forum/index.php?f=150]this[/url] place would do for easy ones (im quoting you man :D doesnt mean its easy for me :D ) :)

madness give me some hints to solve this da  :|

edit: yes your figure is right( There is no figure given in the problem)

consider the two points as 2 complex numbers and let the center be the origin...then the problem becomes simple....

I'll give the algo, I think it should give the answer. Let any point on the circle be $ (x,y)$

Obviously the angle is $ \frac{\theta }{2}$.
$ m_1 \equal{} \frac {y \minus{} y_1}{x \minus{} x_1}$
$ m_2 \equal{} \frac {y \minus{} y_2}{x \minus{} x_2}$

But $ \frac {m_1 \minus{} m_2}{1 \plus{} m_1m_2} \equal{} |\tan \frac{\theta }{2}|$

Solve. :D

shreyas i didnt get you   :huh:

which angle are you talking about?

Oops. Sorry, make it $ \theta/2$. I thought they subtend the angle on the circumference. ;)

But I think the angle should be theta at the circumference. I'm not sure the above will give me the answer.

why arent consecutive posts allowed before 10 minutes interval?   :mad: :bomb:

The equation of the circle with $ (x_1,y_1)$ and $ (x_2 , y_2)$ as diameter is:
$ S: (x\minus{}x_1)(x\minus{}x_2) \plus{} (y\minus{}y_1)(y\minus{}y_2)\equal{}0$

And the line passing thro them is:
$ L: \frac{y\minus{}y_1}{y_2 \minus{}y_1} \minus{} \frac{x\minus{}x_1}{x_2 \minus{}x_1}\equal{}0$

So the equation of the circle passing thro them is,
$ S \plus{} \lambda \cdot L\equal{}0$

I posted the problem in the geometry forum ....but could not quite understand the solutions given :(
See here
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=197146

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=197146

see my soln here


In the figure, AB is an [b]extensible massless[/b] string at the ends of which two equal masses m are attached.Another body of mass m is imparted a velocity $ v_0$ making an angle of $ 30^o$ with AB. AB= $ l$.Find the velocity of A,B &C & the cm of A&B after the impact of B and C.Also find the angular velocity of the system about the cm after the impact of B and C.
(the whole arrangement is placed in a horizontal plane)

Is it possible to solve the problem with no information about the force due to the [b]extensible [/b] string?
 Is it an elastic force?
Or there is no force at all?

The stretching of the extensible string will give rise to an elastic force ...............and this was obvious from the string being elastic. No further info was given.

Who said the string is elastic? It could be in a plastic or rupture phase.
In case of elasticity isn't there the need to know the elastic constant?


function
integration
inequalities
calculus
calculus computations

Show that exists at the most a single continuous function $ f: \left[{0,1}\right] \times \left[{0,1}\right] \to \mathbb{R}$ that satisfies

$ f(x,y)=666+\int_{0}^{x}\int_{0}^{y}{f(u,v)}~dvdu$

The difference of two such functions will satisfy a similar equation without 666. Try to come up with a version of [url=http://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality]Gronwall's inequality[/url] that applies here.


Let $ a,b,c \ge 0$ . Prove that :

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{b \plus{} a} \ge \frac {3}{2} \plus{} \sqrt {2}.\frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{(a \plus{} b)^2 \plus{} (b \plus{} c)^2 \plus{} (c \plus{} a)^2}$

@: It's very easy ..... Stronger than Nesbit

Nice to see you hophinhan!!!
$ Ineq \Leftrightarrow \sum (\frac{a}{b\plus{}c}\minus{}\frac{1}{2}) \geq \sqrt{2}\frac{\sum (a\minus{} b)^2}{\sum (a\plus{}b)^2}$
$ \Leftrightarrow \sum \frac{a\minus{}b\minus{}(c\minus{}a)}{2(b\plus{}c)} \geq \sqrt{2}\frac{\sum (a\minus{} b)^2}{\sum (a\plus{}b)^2}$
$ \Leftrightarrow \sum (a\minus{}b)^2(\frac{1}{(a\plus{}c)(b\plus{}c)}\minus{}\frac{2\sqrt{2}}{\sum (a\plus{}b)^2}) \geq 0$
WLOG we assume $ a \geq b \geq c$. It's easy to show that $ S_b,S_c \geq 0$
$ S_a\plus{}S_b\equal{}2(2\minus{}\sqrt{2})(a^2\plus{}b^2\plus{}(2\plus{}\sqrt{2})c^2\minus{}\sqrt{2}(ab\plus{}bc\plus{}ca)$
$ \Rightarrow S_a\plus{}S_b\equal{}2(2\minus{}\sqrt{2})(\frac{\sqrt{2\minus{}\sqrt{2}}}{2}(b\plus{}a)\minus{}\sqrt{2\plus{}\sqrt{2}}c)^2\plus{}\frac{2\plus{}\sqrt{2}}{4}(a\minus{}b)^2 \geq 0$
The inequality is proved!!!!


inequalities
function
inequalities solved

Come on, this inequality is well known and also very simple. All you have to do is to observe that ai verifies the inequality 2>=ai+(1-1/1980^2)/ai. The rest is simple: sum these inequalities and aplly the AM-GM inequality for two numbers.

It is the same to Russian 1978 ineq :)))

We will prove that if $a_1,a_2,\dots,a_n\in [a,b]$ $(0<a<b)$ then
\[(a_1+a_2+\dots+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n})\leq
\frac{(a+b)^2}{4ab}n^2\]

P=(a_1+a_2+\dots+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n})=
(\frac{a_1}{c}+\frac{a_2}{c}+\dots+\frac{a_n}{c})(\frac{c}{a_1}+\frac{c}{a_2}+\dots+\frac{c}{a_n})\leq
\frac{1}{4}(\frac{a_1}{c}+\frac{c}{a_1}+\frac{a_2}{c}+\frac{c}{a_2}+\dots+\frac{a_n}{c}+\frac{c}{a_n})^2

Function f(t)=c/t+t/c have its maximum on [a,b] in a or b. We will choose
c such that f(a)=f(b). c=\sqrt{ab}. Then f(t)\leq \sqrt{a/b}+\sqrt{b/a}. Then

P\leq n^2(\sqrt{a/b}+\sqrt{b/a})^2 /4= n^2(a+b)^2/4ab.


This is mainly a question on terminology.

Is it appropriate to say that fermentation produces 2 ATP as it is inclusive of glycolysis or is fermentation strictly the process of NADH being oxidized to NAD+ and thus it is not appropriate to say fermentation produces 2 ATP.

Oooh, this is a very good question and I'm glad you asked it. It's important to discuss these types of things.

I know that in most Bio 1 classes, the professors teach that glycolysis comes first, produces 4 ATP (2 net ATP), and results in NADH and 2 pyruvate molecules. If oxygen is present, the pyruvate enters the mitochondria and transition reaction and is converted to acetyl-CoA which will enter the Kreb's Cycle and create some ATP. Most of the ATP, however, is produced by oxidative phosphorylation, using the energy from the electron transfer from NADH to the $e^{-}$ system to create a proton gradient and, thus, ATP is produced by chemiosmosis.

Sorry for that tangent into general respiration...I just love discussing it. Now to your question:

Fermentation produces energy. If producing energy was not part of the fermentation process, it would totally not work...at all. Of course, the ATP of fermentation comes from [i]glycolysis[/i] but, in general, the ATP produced can be attributed to fermentation. The main function of what we normally refer to as fermentation by itself is just to regernate $\text{NAD}^{+}$ for glycolytic pathway by reducing pyruvate eventually into a product such as ethanol or, in our case, lactic acid, although there are other known products of fermentation pathways.

Hope I've been of some help.

So does the fermentation pathway include the glycolysis pathway? If so, I think it is appropriate to say fermentation also produces the net 2 ATP.

Glycolysis involves one oxidation step (for the trioses), so the pathway from glucose to (two molecules of) pyruvate does convert two NAD+ to NADH. A fermentation, by defintion, proceeds without any net oxidation, so the pyruvate must be reduced to close the loop and make it a fermentation:

$\text{pyruvate}+\text{NADH}+\text{H}^+\longrightarrow \text{lactate}+\text{NAD}^+$

This is a typical pathway in vertebrate skeletal muscle. As indicated, the whole glucose to lactate pathway produces a net 2 ATP per glucose.

And it's awfully inefficient. Further oxidizing pyruvate to acetyl Co-A and then burning the acetyl Co-A to $\text{CO}_2$ and water in the Krebs TCA cycle results in a net of something in the neighborhood of 34 or 36 ATP per glucose, mostly by oxidative phosphorilization in the mitochondria. The lactate can be exported from the muscle to the liver, where at the cost of a little more ATP, it can be sent back up the glycolysis pathway in reverse (as gluconeogenesis) to produce glucose again.

There are many different fermentation pathways. In yeast, pyruvate is not reduced to lactate but rather to ethanol and carbon dioxide. Of course, yeast only do that if they are deprived of oxygen; given oxygen, they will employ oxidative respiration, and use far less glucose in the process, because of the greater energy yield.

[I may be a mathematician, but I'm married to a biochemist.]

[quote][I may be a mathematician, but I'm married to a biochemist.][/quote]

Is your wife as strong at math as you are at biology?  :D

[quote="LordoftheMorons"]No. Glycolosis is not part of fermentation. Fermentation's only purpose is to regenerate NAD+.[/quote]

that's not fully correct!,
the pathway of glycolosis that happens in aerobic respiration is similar to those in fermentation, with the exception of production of lactate in fermentaiton and pyruvate in aerobic respiration.

So would you say fermentation generates 2 ATP?

Or would you say it is not fermentation, but Glycolysis that generates 2 ATP?

fermentation.
(glycolysis and fermentation are not separate!)

Ah okay I see. Fermentation (both alcohol and lactic acid) pathways are inclusive of the glycolysis pathway, thus it is appropriate to say that fermentation produces 2 ATP.

However, through my experiences with biology texts, I have noticed there is a trend to generally say "Fermentation" is only the process of regenerating NAD+ while Lactic Acid Fermentation or Alcohol Fermentation refers to the both fermentation and glycolysis pathways.


YAHOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!!

IT S STARTING!

You can sign up, then wait for the next game.

Then again, he might be leaniant since it is a small game.

I hope I m mafia again :ninja:

or another power role

EDIT: finally i posted without having the view the newest post

You are in, and I close signups here. I'm typing out role pm's. There will be a fifth citizen then.

Wow, a werewolf mother, that's cool. So this is like a game of Mafias vs. Werewolfs?

cool

thnx for including me in the game Peter VDD

It is. :) Start the game, if there are any questions ask me. There are 16 players and a healer role is also added in.

Identities of killed persons will be revealed to the killers only. (so day lynch for everyone, mafia kill to mafia only and werewolf kill to werewolf only, unles WW mother targets mafia). This does not apply to WW slayer, since he can only kill people of one type. (death of werewolf mother is announced always, of course)

Yo people, this is my very first game online of mafia (I did they play it in 2002 at IMO, but that was only with a detective)

This is pretty wicked!  :ninja:   I hope I will not do anything dumb..nut I have a hidden trumph card:

[hide]I am the only one who can physically reach out to our moderator and force him to give me more power  :rotfl:  :rotfl: ..... :| .. :o [/hide]

[quote="life=tennis+math"]I assume that it is nighttime now. Assuming that it is: :help: How long is the night? Can you please put it at 72 hours (assuming that it starts right now, otherwise 48 hours if it starts tomorrow) because this weekend I will be in Canada and I doubt that I will be able to access the Internet during this time. Thank you![/quote]

Please disregard this message. My brain was not functioning due to too much chlorine in my body system (I had been swimming for about two hours).

Klebian maybe the citizens are permitted to get drunk and then go on killing sprees as the nighttime activity.

Okay, a few questions, aside from the citizen's power role (already asked):

What is the policy on votes?  Are we allowed to switch our votes at any time, or not at all, or only once, or what?

Should the Network Masons also suicide when caught by the Werewolf Mother, in line with the Mafia suiciding so the other Mafia don't have their identities blown?

Can the Werewolves make an attack on the same night they're bitten?

Are the Citizens made aware who is killed by what (i.e. you say, for instance, 142857 was killed by Mafia, Adunakhor was killed by Werewolves, bubka was killed by Werewolf Slayer)?

In the same vein, will we be told if a Mafia suicides to avoid the clutches of the Werewolves?

If the Werewolf Mother goes outside to bite someone and is killed by either the Werewolf Slayer or the Mafia, will the bite be successful (i.e., does the person the Werewolf Mother targets get killed)?

Finally, if the Werewolf Mother stays inside, can it still be killed by a joint Angel/Healer overdose, or does that fall under "cannot be killed during the night by any means"?

OK, that's enough questions, I guess.  This looks like it'll be an interesting game, can't wait for it to start!

EDIT: Okay, thought of one more...

If someone is targeted by, say, the Mafia, and saved by the Angel/Healer, will you say something like "wobster109 was attacked by the Mafia, but he was rescued"?

[quote="Klebian"]er...
you say you have 14 people, but the list has 15 people, and alkara1 isn't one of them...

[quote="Peter VDD"]Citizen will not be a boring role in next game. Don't worry. Even the citizens will have a night role. :) (this as a way to prevent inactivity -- don't worry, it will work... and it will be fun!)[/quote]

???[/quote]fredbel6 is added, so we're 16, and the citizen night role will not be necessary, that was for in a much larger game (25+)

[quote="dts"]What is the policy on votes?  Are we allowed to switch our votes at any time, or not at all, or only once, or what?[/quote]switch without abuse. No hop-around without reason.

[quote="dts"]Should the Network Masons also suicide when caught by the Werewolf Mother, in line with the Mafia suiciding so the other Mafia don't have their identities blown?[/quote]no. if that means the werewolves know all three masons then too bad. (not that they'll first kill the masons anyway...)

[quote="dts"]Can the Werewolves make an attack on the same night they're bitten?[/quote]no. until end of night, they keep their own role. At end of night, they die and re-emerge as werewolf. Their normal ability for that night still goes on.

[quote="dts"]Are the Citizens made aware who is killed by what (i.e. you say, for instance, 142857 was killed by Mafia, Adunakhor was killed by Werewolves, bubka was killed by Werewolf Slayer)?[/quote]I'm still thinking on that. I'll go for the one with most tactical options probably.

[quote="dts"]In the same vein, will we be told if a Mafia suicides to avoid the clutches of the Werewolves?[/quote]The answer will probably be the same as above.

[quote="dts"]If the Werewolf Mother goes outside to bite someone and is killed by either the Werewolf Slayer or the Mafia, will the bite be successful (i.e., does the person the Werewolf Mother targets get killed)?[/quote]Yes.

[quote="dts"]Finally, if the Werewolf Mother stays inside, can it still be killed by a joint Angel/Healer overdose, or does that fall under "cannot be killed during the night by any means"?[/quote] Indeed, she cannot be killed in that way either. She is far away in her hole outside the city and cannot die from things happening in the city. :)

[quote="dts"]OK, that's enough questions, I guess.  This looks like it'll be an interesting game, can't wait for it to start![/quote]it's started, go ahead! ;)

[quote="dts"]Is it day?[/quote]
Yes, it is.

w00t this will be interesting :lol:

one question: doesn't a game usually start with night?

[quote="142857"]I've a question. If the mafia and ww kill the same person who was saved by both the angle and the healer at night, will that person die?[/quote]
It really doesn't matter, that person will die of overdose.

[quote="PeterVDD"]

[quote="dts"]If the Werewolf Mother goes outside to bite someone and is killed by either the Werewolf Slayer or the Mafia, will the bite be successful (i.e., does the person the Werewolf Mother targets get killed)?[/quote]
Yes. [/quote]

Wait, so the person bit will be killed?

OK, now I have my own questions.

If the werewolf mother goes outside and is saved by both angel and healer, [i]then[/i] will she die?

So if werewolf slayer targets a werewolf, we're announced that a werewolf dies or not?

If werewolf slayer targets werewolf mother and there are already 3 other werewolves, are we told that these 4 people died and that they were werewolves?

Are we told when someone (without naming who) has been bit by werewolf mother?

Can angel or healer save from being bit by werewolf mother?

Can angel or healer save from being killed by werewolf slayer?

If a network mason targets a player and finds out that he is also a mason, the next night, they'll be able to talk and try 2 different people to see if either is mason right?

Can werewolf slayer die by mafia?

Is werewolf slayer innocent?

Can werewolf and werewolf mother communicate at night?

Important question: if werewolf mother bites inspector/healer/angel, then we won't know that this power role was bitten, right?

So essentially all our power roles could be werewolves? Then we wouldn't know how close we were to winning/losing?

another question:
If the ww mother bite and the mafia kill the same person, will the person die or become a ww?

w00t!

this is my first time playing, so yea

it will be a bit rocky, but hey?

Ok, ready to rock... How long will the night period last (when it finally begins :))?

[quote="fredbel6"]Peter could you just add everything in the initial post about who can know what and is notified in what manner of what?

I have so many questions!

Am I correct there is only 4 bad guys?

Is time of great essence here?  I mean when I played mafia with just a detective nobody really had to hurry, there was a plan for everything and one thing came after the other.
For instance, the werewolf mother, does she immediately  have to inform the moderator when night starts about her decision to go out or not?  Are all other reports ignored by the moderator until he gets such a pm?  And how many times can the slayer hit someone at night?  And what if
- night starts
- slayers pm's , says he wants to hit B
- B, who is the mother, pm's moderator to say she is going out
Is this a hit or not?

Does a protect by angel or healer imply also a protect from mother werewolf bites?

Are we gonna get the full description of events after the game?

Where does the allegiance lie of a civilian, inspector, angel...
Does he care for himself or ALL the nonbaddies?
I mean, is sacrifice tactically interesting?
IF his own survival is top priority, it might actually be interesting for him to get bitten, if he thinks the werewolves are taking over, to become one
And what if one does still act in the interest of his civilian group and sabotages the werewolf attempt?[/quote]

I can answer most of your questions.

1) As of now, there are 4 bad guys.
2) At night the order of 'pm's do not matter. They can bearranged by Peter in the order that makes sense.
3) (Continuation of 2) She does die.
4) No. Angel or healer protectsfrom kills only.
5) If Peter cares to give it, yes.
6) Any player is supposed to show complete allegiance to the group he plays for: Innocent, Mafia or Werewolves. That is part of the game ethics.
7) After becoming weresolf, he wins only if the werewolves win. So...

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

[quote="klebian"]Wait, so the person bit will be killed?

OK, now I have my own questions.

If the werewolf mother goes outside and is saved by both angel and healer, then will she die?

So if werewolf slayer targets a werewolf, we're announced that a werewolf dies or not?

If werewolf slayer targets werewolf mother and there are already 3 other werewolves, are we told that these 4 people died and that they were werewolves?

Are we told when someone (without naming who) has been bit by werewolf mother?

Can angel or healer save from being bit by werewolf mother?

Can angel or healer save from being killed by werewolf slayer?

If a network mason targets a player and finds out that he is also a mason, the next night, they'll be able to talk and try 2 different people to see if either is mason right?

Can werewolf slayer die by mafia?

Is werewolf slayer innocent?

Can werewolf and werewolf mother communicate at night?

Important question: if werewolf mother bites inspector/healer/angel, then we won't know that this power role was bitten, right?

So essentially all our power roles could be werewolves? Then we wouldn't know how close we were to winning/losing?
[/quote]

I may answer some of your questons.

1) Angel or healer can save from slayer kills.
2) W.S. can die by mafia.
3) W.S. is innocent.
4) W. and W.M. can of course communicate at night!

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

Now for my own questions:

[color=red]1) [b]Least serious[/b]: Do werewolves continue their original roles?
2) [b]Somewhat more serious[/b]: A mafia vs. werewolves game ends when one party is eliminated. Does that mean slaying the werewolf mother ends the game?
3) [b]MOST SERIOUS[/b]: [i]Is the werewolf mother female[/i]?[/color]

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

Oh by the way, if you still haven't got it, [color=blue]I am here[/color].

All questions are answered in the first post. The game starts below this line. Further rules questions should be asked to me.

================================================

Are dead people allowed to speak in these splits?


parameterization
calculus
integration
real analysis
function
complex analysis
real analysis solved

[quote="liyi"]$f$ is absolutely continuous on $\mathbb{R}$
Prove that there exist two non-negative constants $A$ and $B$ such that
$|f(x)|\leq A|x|+B$ for all $x\in \mathbb{R}$[/quote]
take $A$ such that $|y-x| \leq A => |f(y)-f(x)| \leq 1$ and $B$ such that $|f(x)| \leq B$ on $[-1;1]$

Maybe I dind't understand your question.

Otherwise the statement is wrong, since e.g. $f(x)=e^x$ or $f(x)=x^2$ are absolutely continuous on $\mathbb R$ but not bounded by $A|x|+B$ for any $A,B\in\mathbb R$?

Absolute continuity - without any reference to a specific measure - means, that $f(x)$ can be written as an (parameter-)integral w.r.t  Lebesgue-measure, e.g. $x^2=81+\int_9^{x} 2u\;du$...

It seems that your question originates from complex analysis, and 'absolutely continuous' has to be changed into 'entire + ...' or something else (I can't remember)?

Hm,
Leon

I'm pretty sure liyi was referring to something like [url=http://planetmath.org/encyclopedia/AbsolutelyContinuousFunction.html]this[/url], in this case. $f(x)=x^2$ is not absolutely continuous, and neither is $e^x$. First of all, the function should be uniformly continuous, and your examples are not.

Dear grabber

The definitions given on the linked page coincide with my (read theorem 2): an absolutely continuous function is a function with (Lebesgue- or Radon-Nikodym-) derivative.

Absolutely continuity on $\mathbb R$ does not imply uniform continuity. If instead liyi means** a.c. on a closed interval $[a,b]$, the statement is trivial, since we may chose arbitrary large constants $A$ and $B$.

Sincerely,
Leon

*(do you [like me] hate people that throw notions like an octopus squirts ink?)
**(please excuse my english)

[quote="Leon Parsek"]Absolutely continuity on $\mathbb R$ does not imply uniform continuity.[/quote]
Why?

[quote="Leon Parsek"]Absolute continuity - without any reference to a specific measure - means, that $f(x)$ can be written as an (parameter-)integral w.r.t  Lebesgue-measure, e.g. $x^2=81+\int_9^{x} 2u\;du$...
[/quote]
Dear Parsek,

In this way, the integrand (which is $2u$ in your example) should be *integrable on $\mathbb{R}$*.

Dear Liyi, dear grabber

I have now discovered  our/my/your problem.

1. Absolute continuity (w.r.t. Lebesgue measure $\lambda$) is a notion of measure theory, where it is used for a measure $\mu$ if $\lambda(A)=0$ implies $\mu(A)=0$ for measurable sets $A$. Using this definition even unbounded measures $\mu$ are allowed - I first supposed, that this definition was meant. The Radon-Nikodym theorem (theorem 2 of the link) says that  $\mu(A)=\int_A f\;d\lambda$ for some function $f$ that is locally $L_1(\mu)$ (here we have our problem). Thus my examples are a.c. in this sense (see e.g. Mr.Rudin's book).

2. But you are both right, if you restrict your attention to finite (real-valued) measures $\mu$. For these measures the Radon-Nikodym theorem says, that $\mu(A)=\int_A f\;d\lambda$ for some proper $L_1(\mu)$-function $f$. Moreover in this case the alternative definition, grabber gave in his link, applies: $\mu$ is a.c. iff for every $\epsilon>0$ there is a $\delta>0$, so that  $|\lambda(A)|<\delta$ implies $|\mu(A)|<\epsilon$.  These functions are clearly uniformly continuous.

And after sleeping $n$ nights, with $n>0$, I  now  think that Liyi actually meant this definition, as grabber suspected. So let's forget the more general definition and focus on absolutely continuous functions, 2nd definition, and try to solve problem.

Regards,
Leon

I just recognized that alekk gave a solution to the problem 250 years ago (sorry for the confusion).

Here is my version. According to the uniform continuity of $f$ we can find an $A>0$, so that $|f(x)-f(y)|<1 $ whenever $|x-y|<1/A$. Then letting $A|x|\leq n\leq A|x|+1$ we get $|f(x)-f(0)|=|\sum_{k=1}^n f(xk/n)-f(x(k-1)/n)|$
$\leq\sum_{k=1}^n |f(xk/n)-f(x(k-1)/n)|\leq n\leq A|x|+1$,
 so that $|f(x)|\leq A|x|+B$ with $B=|f(0)|+1$.

Leon

PS: Can we sharpen the result, if we use absolute continuity instead of the weaker condition of uniform continuity?

hmm. I was wondering that why it needs "absolutely continuous" which is stricter than "uniformly continuous"...
So I posted it...


1
Find all positive integer solutions x and n of the equation $x^2 + 615 =2^n$ .

2
Find all pairs of positive integers $x,y$ which satisfy the equation
$2x^2+5y^2=11(xy-11)$

3
Find all pairs of integers $x,y$ which satisfy the equation
$x^2(y-1)+y^2(x-1)=1$

Seems too easy and I'm tired, maybe I made a mistake...
[EDIT: In fact I was tired, and made a mistake :)]

2)rewrite as:
$(5y-x)(2x-y)=121$
now it is easy