trigonometry
analytic geometry
graphing lines
slope

[quote="buzzer11"]Convert the equation $ y=x$ into polar form.

I have a question about this, but I'll present it after an answer is supplied.[/quote]

[hide="Solution"]The graph is a line that makes a $ 45^{\circ}$ degree angle with the x-axis. So our equation is $ \theta=45^{\circ}$. Note that $ r$ doesn't matter.[/hide]

[hide]
$ y = x$
$ r\sin{\theta}= r\cos{\theta}$
$ \sin{\theta}= \cos{\theta}$

Therefore $ \theta = 45^{\circ}?$[/hide]

Thank you. my question was whether $ \theta = 225$ would be included in the graph, since on the Cartesian plane, it is...

$ \theta = 45^{\circ}$ encompasses all values of $ r$ - even negative ones.  Part of the issue is that polar points don't have a unique representation (by which I mean $ (r, 45^{\circ}) = (-r, 225^{\circ})$), which is part of why polar graphs often cross themselves.

If a line passes through the origin and point $ (\cos(\theta), \sin(\theta) )$, the slope of this line is $ \tan(\theta)$. Therefore:

$ y = \tan(\theta)x$

If you know your $ 45-45-90$ triangle well, the $ \tan(45) = 1$.

Thus, your answer is $ 45^{\circ}$.

You can also generate an infinite amount of answers with the tangent trigonometric reduction formulae. For instance, using $ \tan(180p+\theta) = \tan(\theta)$ with $ p=1$ and $ \theta = 45$, we get $ 225^{\circ}$.


You have to get more badges, and to get badges you have to get more problems correct.  :ninja:

[quote="Joe10112"]When do we new videos?

What % do we pass stuff

i only have 6 videos, and i want more!!!!

can n e one help me????[/quote]

Well, you have to do at least 4 Q's and 50% to pass stuff.

if you want more videos, keep on doing the problems and the videos will appear (ooh! magic!  :) )

[quote="Joe10112"]@ Davidd. No, i have over 50%, and still have not passed a subject

and usually how many problems until you get a video?[/quote]

Maybe not, well I did a lot of problems yesterday, maybe over 100, and that was my first day using alcumus, but currently I have 19 vids, maybe you get videos for doing problems on that subject.

you have 19 videos???/??//

i only have 6, and have done 300+ problems

wats your alcumus score

mine is around 33.

fail.

to get more movies, you have to answer the questions that you are getting now correctly
once alcumus thinks you are ready for the next lessons, it will provide you with additional videos


logarithms
trigonometry
function
calculus
calculus computations

Find the Fourier series of $f(x)=\ln \left( 2\cos\frac{x}{2} \right)$.

I'm sure liyi means the function $f(x)$ which is $\ln\left(2\cos\frac x2\right)$ on the interval $(-\pi,\pi),$ extended from there to be periodic of period $2\pi.$ One way to write this so it would be valid for all $x$ would be to write it as $\ln\left|2\cos\frac x2\right|.$

The fact that I recognize this function means I should leave the problem for someone else to solve. A couple of things to note:

1. $f$ is an even function, so the Fourier series should involve only cosines. Alternatively, if you use the complex form of the Fourier series, $\hat{f}$ should be real and even.

2. The Fourier series shouldn't converge absolutely. We could say this as $\hat{f}\not\in\ell^1.$ A reasonable guess might be that $\hat{f}(k)=O\left(\frac1k\right).$

A byproduct: $\frac{x}{2}=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}\sin nx}{n}\quad \forall x \in \left(-\pi,\pi\right)$


calculus
integration
function
analytic geometry
graphing lines
slope
geometry

$ F(x)$ is a differentiable function such that

$ F'(a\minus{}x)\equal{}F'(x)$

for all $ x$ satisfying $ 0\leq x\leq a$.

Evaluate $ \int_{0}^{a}F(x)dx$ and give an example of such a function.

Let $ F(x)\equal{}c$, a constant function.  Then $ F'(x)\equal{}0$ everywhere where it's defined, so $ F'(a\minus{}x)\equal{}0$ as well.  In that case,
$ \int_{0}^{a}F(x)\,dx\equal{}\int_{0}^{a}c\,dx\equal{}cx\bigg|_{0}^{a}\equal{}ca$.

Hi :) !!

than we such that: $ \int_a^b g(t)dt \equal{} \int_a^b g(a \plus{} b \minus{} t)dt$ with $ g$ continue in $ [a,b]$

than:

$ \int_0^a F(x)dx \equal{} [xF(x)]_0^a \minus{} \int_0^a xF'(x)dx \equal{} aF(a) \minus{} \int_0^a (a \minus{} x)F'(a \minus{} x)dx$

we such that$ F'(x) \equal{} F'(a \minus{} x)$

than:

$ \int_0^axF'(x)dx \equal{} a\frac {F(a) \minus{} F(0)}{2}$

than $ \int_0^a F(x)dx \equal{} a\frac {F(a) \plus{} F(0)}{2}$

thanks :)
PS: i calcul that quickly than may be exist error

Since the slope is constant on all $ 0 \le x \le a$, we have a linear function $ F(x)\equal{}mx\plus{}c$, for constants $ m$ and $ c$.

Then $ \int_0^a mx\plus{}c\,dx \equal{} \left[\frac12mx^2\plus{}cx\right]_0^a \equal{} \boxed{\frac12ma^2\plus{}ca}$.

[quote="undefined117"]Since the slope is constant on all $ 0 \le x \le a$, we have a linear function $ F(x) \equal{} mx \plus{} c$, for constants $ m$ and $ c$.

Then $ \int_0^a mx \plus{} c\,dx \equal{} \left[\frac12mx^2 \plus{} cx\right]_0^a \equal{} \boxed{\frac12ma^2 \plus{} ca}$.[/quote]

Hi i think that i do a generisation of this proprety

for all $ f$ differentiable function with $ f'(a\minus{}x)\equal{}f'(x)$ we have:

$ \int_0^a f(t)dt \equal{} \frac{1}{2}af(a) \plus{} \frac{1}{2}af(0)$

than if we take your function we see that is correct because

if we applicate directly we find the same result

and thinks

Well, isn't that because the area under a linear function is a trapezoid? That is, $ \int_{0}^{a}f(t)dt \equal{} \frac {1}{2}af(a) \plus{} \frac {1}{2}af(0) \equal{} \frac {f(0) \plus{} f(a)}{2} \cdot a$.


Find all functions  such that $(f(x)+f(y))(f(z)+1)=f(xz-y)+f(x+yz)$ for all real $x, y, z$.

Find inspiration on the following way  :)

[url=http://www.kalva.demon.co.uk/imo/isoln/isoln025.html]IMO 2002 n°5[/url]


Does anyone know the meaning of this error?
[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]
Cannot execute the command.
I get this error when I try to convert to PDF. I get similar errors when I try to convert to DVI and PS.
I use TeXnicCenter and MiKTeX.

ave you told TexnicCenter where to find the pdf, dvi and ps viewers? You do this by clicking on Build, Define Output Profiles and clicking on the Wizard button in the bottom left corner. [url=http://www.youtube.com/watch?v=-h8Oi1oCTQg]This video[/url] will show you what to do.

Check also that in Build, Define Output Profiles, Viewer tab that the path is correct.

Hmm, no, that's not it. I already tried that and it said the same thing.

Which PDF viewer (and version) are you using? Have you checked the path is OK? Is the pdf created and can you view it without using TexnicCenter? Are there any errors in your document, shown in the output window at the bottom of TexnicCenter? Does this minimal document work:
[code]\documentclass{article}
\begin{document}
Hello world
\end{document}[/code]

[quote][b](1)[/b]Which PDF viewer (and version) are you using? [b](2)[/b]Have you checked the path is OK? [b](3)[/b]Is the pdf created and can you view it without using TexnicCenter? [b](4)[/b]Are there any errors in your document, shown in the output window at the bottom of TexnicCenter? [b](5)[/b]Does this minimal document work:
[code]
\documentclass{article}
\begin{document}
Hello world
\end{document}
[/code]
[/quote]
(1): I am using Adobe Acrobat Reader 8.1.2.
(2): I hadn't before, but after I checked I noticed that it was all caps. I fixed it back to not all caps, even though it might not do anything. Weird...
(3): No, the PDF is not created.
(4): Nothing at all is in the bottom frame.
(5): Nope, doesn't work.
Forgot to mention one thing: Adobe Reader comes up and says, "Cannot open file. The file cannot be found."

[quote="WrathOfMobius"](4): Nothing at all is in the bottom frame[/quote]Nothing at all? No text whatsoever?
[quote="WrathOfMobius"]Forgot to mention one thing: Adobe Reader comes up and says, "Cannot open file. The file cannot be found."[/quote]That's the important bit - rather than the stuff about [DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]. It means that the document hasn't been created.

It looks like there's a problem with your latex installation. What files, if any, are generated? If your tex file is called example.tex then there should be a whole load of files called example.log, example.aux etc. The most important one is example.log which tells you what latex is doing. That is the text that appears in the bottom frame.

Do you know how to use the command line? Can you use pdflatex.exe from there and it will tell you what is happening?

You need to look for "Command Prompt". Run it then type
pdflatex -version
and you should get some response. If not then use the full path of pdflatex.exe as shown in TexnicCenter,with quotes if the path includes spaces.
If you get version information then type
(apth if required)pdflatex c:\put_path_here_of_this_file\example.tex
and see if it compiles correctly.

I got the version, and when I tried to compile it through command prompt, it looked like it was working until it hit the last 5 lines. They then say, "0 words of memory in BibTeX. 0 pages created." No .aux or .log files are created. And the PDF isn't created, either. I tried moving it to another computer. Fail. I tried moving it off my flash drive onto the actual hard drive. Fail. I even tried making a new folder and copying it into there in case for some weird reason it decided to not take it from that folder. FAIL. It's worrysome that it's coming up with this, as on the second computer it had been installed for many months and had worked until a couple of weeks ago.
Pretty soon, I'm just going to handwrite and scan it and see if someone can decode my handwriting.

Can you compile the minimal document from the command line and then copy and paste the full output to here? It shouldn't ba a problem with the installation of latex as it is happening on two computers, but we will see from the output.

To copy from the command line, right-click on the window, choose Mark, highlight the area you want and press Enter.

[quote="WrathOfMobius"]Hmm... I copied it to my dad's laptop and it worked! Don't know why.[/quote]What is [b][i]it[/i][/b]?  What did you copy? Does the other computer use the same version of MiKTeX or even another LaTeX distribution? What was the output from the command line for the minimal document on your computer?

IT is the file I was trying to compile.
Everything is the same version- we downloaded it just last night (well, technically, this morning since it was past midnight)

OK, then how about letting us know what was the output from the command line for the minimal document on your computer? This should have information on what is going wrong.
It's important to use the minimal document so as to eliminate any package problems.
[code]\documentclass{article}
\begin{document}
Hello world
\end{document} [/code]


How many different three-member committees can be chosen from a group of ten people?

This is simply $ \binom{10}{3} \equal{} \boxed{120}$. To learn more about combinations, click [url=https://www.artofproblemsolving.com/Forum/weblog_entry.php?t=218699]here.[/url]

This answer is wrong.  It should be 720, which is $10\times9\times8$


My friends, teachers and I have all been having an argument over this problem:

I'm in a car driving along at 70 mph.  I throw a baseball out the back window at 70 mph.  To an onlooker on the side of the road, what would the ball appear to do?  Also, there is no wind other than that of the moving car and baseballs air resistance.

Some of my friends say the ball would fall straight down, others say it would curve.  Help me please! :|  :?:  :!:  :?:

I think that it would fall in a straight line because there is no force driving the baseball to go curved.

Just my thought. :)

Does the ball go perpendicular to the car or does it go behind the car?

I think it would drop straight down.  Cause if you're going 70 mph in one direction and you throw it 70 mph in the other direction the forces would cancel each other out and it would drop straight down.  I dunno though, maybe you should go try it lol :)

This is complicated by the fact that the car will carry a certain amount of air along with it. If at the instant the ball left your hand, it were to enter still air unaffected by the car, then it would have zero velocity in still air and would fall straight down (as seen from the side of the road). But if it starts out in air affected by the car at the moment it leaves your hand, then that wind will carry it forward. From your point of view, it left your hand traveling backwards at 70 mph (in your frame of reference), but encountered air resistance and slowed down slightly, so that a fraction of a second later, it was traveling (to you) backwards at maybe 67 mph. To the observer at the side of the road, the ball started out at zero velocity, but within a fraction of a second, the air traveling with the car blew it forward so that it has a velocity of, say, 3 mph in the same direction as the car. Its path would then curve downwards toward the ground.

If you compensated for this effect by throwing it at whatever speed (say, 73 mph) needed to compensate for the effect of the air traveling with the car, the observer at the side of the road would see it start out traveling at 3 mph in the direction opposite the car, but that the wind blowing with the car would bring it quickly to a near standstill, from which it would fall (in now lighter wind further away from the car) nearly straight down. In this case, the observer at the side of the road would see the path curve in the opposite sense to the previous case.

Try drawing pictures of all of this.

It actually all depends. Say there was a giant 500 square foot mirror on each side of the car blocking the ball from wind. That would definitely affect the ball's flight :P .

[quote="white_horse_king88"]The ball will indeed travel beyond the onlooker and curve downward along its path.

Same - why don't you fly back when you jump on an airplane or a bus?

This will answer the previous question.[/quote]

But who does jump on an airplane, or a moving bus for that matter?

the air resistance slows it down so no matter how little it will affect the overall motion and therefore the ball will fall in a slightly curved manner


Let $k$ and $m$ be positive integers. Show that \[S(m, k)=\sum_{n=1}^{\infty}\frac{1}{n(mn+k)}\] is rational if and only if $m$ divides $k$.

This problem is considered in [url=http://lib.mexmat.ru/forum/viewtopic.php?t=6490&start=0]russian forum[/url]

Juna (or someone else), would you mind translating the solution for us? :)

Actually, I don't think that this problem was solved in that topic,at least I didn't feel that it was after reading that topic :huh: . They are manly concerned with calculating the sum \[S(m;1)\] and I don't think that they are done even with that work :razz: .

[quote="Peter"]Juna (or someone else), would you mind translating the solution for us?[/quote][url=http://babelfish.altavista.com/babelfish/trurl_pagecontent?lp=ru_en&trurl=http%3a%2f%2flib.mexmat.ru%2fforum%2fviewtopic.php%3ft%3d6490%26postdays%3d0%26postorder%3dasc%26start%3d0%26sid%3df50c905e7c75e82bcb199c28cd410c23]Try so[/url]
[quote="Tiks"]Actually, I don't think that this problem was solved in that topic,at least I didn't feel that it was after reading that topic[/quote]
I did not speak that problem is solved completely. But is there proved that amount transcendental if $k|m$

The question is equivalent to proving that the expression $\gamma +\psi(x)$ is irrational for $x\in\mathbb{Q}-\mathbb{N}$ where $\gamma$ is the [i]euler-mascheroni constant[/i] and $\psi$ is the [i]digamma function[/i]. Just refer a paper by M.Ram Murty and N. Saradha that proves that in this case they must in fact be transcendental. $\square$


Prove that:
$ O_1O_5,O_2O_6,O_3O_4$ are concurrent.
See:


Hi -- I don't know if Silverfalcon has any plans, but if we have a "math chat" this week can it be on friday?

Okay sign up here.  Please leave your S/N!  Also I propose a new rule.  People should be allowed to ask harder questions as long as they are willing to fully explain it to people who don't understand. Silver?

I think I already answered that people are able to take long time as they want to understand the question.

But let me point one thing:  It doesn't do any good to say to other person like lecturing, okay?  It is because it doesn't feel so good to be lectured...

K but
explaining  :!=: lecturing

See you friday.

I'll try to be there but I'm not sure if I'll be home.  i might know more as the time nears.

I'll also try to be there, but I might be busy... I think that raising the level to anything somewhat easily explainable is fine.  I agree that explaining and lecturing are two very different things.  Explaining problems that people have tried themselves is helpful, while lecturing just gets boring ;)

Yep.  Two different things.  I think I got confused...


greatest common divisor
number theory
relatively prime

Find every positive integer n such that $n^3+n^2+n+1$ is square

Hint:

[hide]Notice that the given expression is equal to $(n + 1)\left(n^2 + 1\right)$.
What do you know about the greatest common divisor of these factors?
What does that imply?
When is the second factor a square?[/hide]

[hide]Also  before any  one simply says n=0,1   watch out ...
[hide]Check out when n=7 [/hide][/hide]

Let me try to complete the solution:

$\gcd (n+1,n^2+1) = \gcd (n+1, 1-n) = \gcd (2, 1-n)$
Therefore, $\gcd (n+1,n^2+1) = 2$ when n is odd and $\gcd (n+1,n^2+1) = 1$ when n is even.

[b]Case 1 when n is odd[/b]
Let n = 2m -1, where m is a positive integer.
$(n+1)(n^2+1) = 2m(4m^2-4m+2)=(2)^2 \cdot m (2m^2-2m+1)$
Note that $\gcd (m,2m^2-2m+1) = \gcd (m,1) = 1$.
Therefore, both $m$ and $2m^2-2m+1$ have to be perfect squares.
Let $m=k^2$ for some positive integer k.
Then $2m^2-2m+1 =2k^4-2k^2+1 =(k^2)^2+(k^2-1)^2$
...... and I don't know how to proceed here.
In my knowlege, I only know that $k^2=1$ and $k^2=4$ are possible solutions.
That is, k = 1 or k = 2, showing n = 1 or n = 7.[b]
Someone please help to finish this part.[/b] :lol:

[b]Edit:Further explanation for this part[/b]
The general solution for $x^2+y^2=z^2$ is of the form $x=2tuv,y=2t(u^2-v^2),z=2t(u^2+v^2)$ where t, u and v are integers.
Note that $\gcd (k^2,k^2-1)=\gcd(1,k^2-1)=1$.
Either $(k^2,k^2-1)=(2uv,u^2-v^2)$ or $(k^2,k^2-1)=(u^2-v^2,2uv)$.

[b]Case 2 when n is even[/b]
Let n = 2m, where m is a positive integer
$(n+1)(n^2+1)$ with $n+1$ and $n^2+1$ being relatively prime.
Then both $n+1$ and $n^2+1$ should be perfect squares.( in view of the unique factorization theorem)
However, $(2m)^2<4m^2+1=n^2+1<4m^2+4m+1<(2m+1)^2$, which proves that $n^2+1$ cannot be a perfect square.
Therefore, $n^3+n^2+n+1$ cannont be a perfect square in this case.

ya!
\[ \gcd(n+1,n^2+1)=\gcd(n+1,n^2+1-(n-1)(n+1))=\gcd(n+1,2) \]
so if $n$ is even then $\gcd$ is 1
imply $n+1$ and $n^2+1$ must be square
but $n^2+1$ is not square is $n\geq2$

hence $n$ must be odd
and $\gcd$ of them is 2
$n: = 2k-1$
\[ (n+1)(n^2+1)=(2k)(4k^2-4k+2) \]
and $\gcd(k,2k^2-2k+1)=1 \qquad (\because \gcd\ of\ them\ is\ 2)$
then $k$ and $2k^2-2k+1$ must be square again

 \[ 2k^2-2k+1=(k-1)^2+(k^2) \]

Oh..no..
luimichael  post it already!!

55+
now I found one topic that solving in same problem (may be not complete)
http://www.mathlinks.ro/Forum/viewtopic.php?t=85454

So, it is left  to solve the equation: $(k^2)^2+(k^2-1)^2=z^2$ where k and z are non-negative integers.

In view of the general solutions of $x^2+y^2=z^2$(that is, the Pythagorean triples), and the fact that $k^2$ and $k^2-1$ are relatively prime, we may assume:

($k^2=2uv,k^2-1=u^2-v^2$) or ($k^2-1=u^2-v^2,k^2-1=2uv$) for some integers $u$ and $v$.

For the first case, we will arrive at
$u^2-v^2+1=2uv \Longleftrightarrow (u-v)^2-2v^2=-1$, while the second case leads to $(u-v)^2-2v^2=1$.

Both equations are classified as the standard form called [i]Pell's Equation[/i].

Finally, all solutions are resorted to the solutions of the [i]Pell's Equation[/i].


For any positive integer n, prove that there exists a multiple of 2^n whose each digit (base 10)
is not zero.

(i really want to know how to solve problems like these)

[hide="hint"]How many powers of 2 are there such that none of their digits are 0?[/hide]

[hide]
We can find a sequence $ a_n$ such that $ 2^n\mid a_n$ and each $ a_n$ has only the digits 1 and 2. We will define our sequence recursively. Take $ a_1\equal{}2,a_2\equal{}12,a_3\equal{}112$. The way we get $ a_{n\plus{}1}$ from $ a_n$ is if $ a_n/2^n$ is odd, put the digit 1 in front of $ a_n$, or if it is even put a 2.

To prove this works, suppose $ a_n\equal{}k2^n$.
Case 1: k odd
$ a_{n\plus{}1}\equal{}10^n\plus{}k2^n\equal{}2^n(k\plus{}5^n)$ and $ 5^n\plus{}k$ is even.
Case 2: k even
$ a_{n\plus{}1}\equal{}2*10^n \plus{}k2^n\equal{}2^{n\plus{}1}(5^n\plus{}k/2)$

1 and 2 can by an odd and an even digit, and the argument still works.
I'd bet there's a better way, but this works.
[/hide]

Well if someone could prove my hint then the problem would be essentially done...

I don't understand caffineboy's solution.
Why put a 1 before the term when its odd and 2 when it's even?
Also, when putting the digit 1 before the a term 2^n you canot simply add 10^n because 2^n and 10^n don't always have the same number of digits
i.e. let n = 5
2^n + 10^n = 10032      (not 132)
This defeats the purpose of the question as you are looking for numbers without zeros
[/hide][/list][/code][/quote]

caffeineboy constructed the solution to the problem "Prove that there exists a $ n$-digit number with only the digits $ 1$ and $ 2$ such that the number is divisible by $ 2^n$". That sequence then satisfies the question as it clearly they will not have any $ 0$s.

In fact, the very sequence caffeineboy described is unique; in other words, there is only one $ a_n$ with $ n$ digits, each digit being 1 or 2, that is divisible by $ 2^n$.

The sequence is also sequence [url=http://www.research.att.com/~njas/sequences/A053312]A055312[/url] in OEIS.

A related sequence:  there is only one $ b_n$ with $ n$ digits, each digit being odd, that is divisible by $ 5^n$.  I believe this was an AIME problem.

And surprisingly, this sequence does not appear in OEIS. I used the search terms "5,75,375,9375" which are the initial terms of the sequence.

perhaps it could be added?
there is a place to do so

You do it. "Stevenmeow" looks like a more interesting author than "Yongyi". :D


inequalities
function
calculus
integration
topology
real analysis
real analysis solved

Let $f,g$ be piecewise continuous non-negative (I do not remember exactly if they must be non-negative) functions such that all integrals that follow exist. Prove that
   $(\int_{0}^{\infty}{\sup_{a+b=t, 0\leq a,b\leq t}{f(a)g(b)}})^2\geq 4\cdot\int_{0}^{\infty}{f^2(t)dt}\cdot\int_{0}^{\infty}{g^2(t)dt}$

Have a look here page 2 to 10 you will find the proof and applications

http://picard.ups-tlse.fr/~annales/sommaires/textes/barthe.pdf

For function  one variable

$f,g,h$:$R\rightarrow ]0,+\infty[$ continous, $0\leq k \leq 1$ such that
$\int_{R}f,\int_{R}g$ are finite
Suppose $\forall x,y \in R$, $h(kx+(1-k)y)\geq f^{k}(x)g^{1-k}(y)$

Then $\int_{R}h \geq (\int_{R}f)^{k}(\int_{R}g)^{1-k}$

Proof (Source Barthe)

Defined $x(t),y(t)$ on $[0,1]$ by

$\int_{-\infty}^{x(t)}f(u)du= t \int_{R}f$
$\int_{-\infty}^{y(t)}g(u)du= t \int_{R}g$

$x'(t)f(x(t))=\int_{R}f$
$y'(t)g(y(t))=\int_{R}g$

$x,y$ are strictly increasing and injectives

Let $z(t)=kx(t)+(1-k)y(t)$ is strictly increasing

$z'\geq x'^{k}y'^{1-k}$

Then $\int_{R}h \geq \int_{0}^{1}h(z(t))z'(t)dt\geq
\int_{0}^{1}h(kx(t)+(1-k)y(t))x'^{k}(t)y'^{1-k}(t)dt \geq $
${\int_{0}^{1}f^{k}(x(t))g^{1-k}(y(t))(\frac{\int f}{f(x(t))})^{k}(\frac{\int g}{g(y(t))})^{1-k}=(\int f)^{k}(\int g})^{1-k}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $a,b,c$  be elements of a group $G$ of odd order. If $a^{2}b^{2}=c^{2}$, prove that $ab$ and $c$ are in the same coset of the commutator subgroup $K$.

Are you sure it's from AMM? It's pretty obvious :).

Let $H\lhd G$ be a normal subgroup of $G$ s.t. $G/H$ is abelian (in particular, we can take $H=G'$, the derived or commutator subgroup of $G$). For $x\in G$, let $\bar x$ be its class modulo $H$. We then have $(\bar a\bar b)^2=\bar a^2\bar b^2=\bar c^2$. Since $G/H$ has odd order, the map $x\mapsto x^2$ is injective, so $(\bar a\bar b)^2=\bar c^2$ implies that $\overline{ab}=\bar a\bar b=\bar c$, as desired.


calculus
derivative
function
calculus computations

A man starts walking north at 4ft/s from a point P.Five minutes later a woman starts walking south at 5ft/s from a point 500 ft due east of P.At what rate are the peolpe moving apart 15 minutes after the woman starts walking?

ANyone?

Hello there. This related-rates problem can be solved in a straightforward manner. Let (0, 240(t+5)) be the position of the man and (500, -300t) be the position of the woman at time t, t being the number of minutes after the woman begins walking. Calculate the distance between the two people using the distance formula, and calculate the derivative of that distance function at t=15.


inequalities
function
inequalities proposed

Prove that with all$a,b,c>0$we have
   $\frac{a+b+c}{\sqrt[3]{abc}} \ge 2+\sqrt[4]{\frac{27(a^4+b^4+c^4)}{(a+b+c)^4}}$

That´s an homogeneous inequalitie then we can make a=px, b=py e c=pz, where p=a+b+c and obtain the following:

  $\frac{1}{\sqrt[3]{xyz}} \ge 2+3\sqrt[4]{\frac{x^4+y^4+z^4}{3}}$

   Notice now that x+y+z=1 and I considered a function of two variables making z=1-x-y and I had prove that the critical points occur when x=y. Itsn´t difficult to evaluate this hessian, althoug it´s very long...

   Another idea? Thank´s!

Can you post your solution? my solution isn't long but It isn't simple.


Hey guys:) How many of you will participate at IMO 2007 and will enroll at MIT? I am asking this because I look forward to meeting you in Vietnam :) ...Until now I know at least 2 other mathlinkers :)

So great!!! Could you tell me who will attend MIT? I have a friend who attended MIT last year but the takes the IOI, not IMO.

[quote="ciprian"]Until now I know at least 2 other mathlinkers :)[/quote]

I think one of them is Tiks,right? :lol:

Who is the second one?

You're right Tiks :)  The other one is Sasha :)
Anybody else? :maybe:

Teddy from the U.S. team will be at MIT.

[quote="ciprian"]Hey guys:) How many of you will participate at IMO 2007 and will enroll at MIT? I am asking this because I look forward to meeting you in Vietnam...Until now I know at least 2 other mathlinkers[/quote]

Oh Ciprian, I just figured out that this means you are coming to IMO 2007 too,right?(I know I am not attentive :oops:)  Congratulations,very well done!!!
Looking forward to meeting you in Hanoi!!

[quote="Tiks"]
Oh Ciprian, I just figured out that this means you are coming to IMO 2007 too,right?(I know I am not attentive :oops:)  Congratulations,very well done!!!
Looking forward to meeting you in Hanoi!![/quote]

Again you're right. I'm going to IMO 2007 (strange though...haven't you read the e-mail reply?). Thx and congrats to you too Tiks :)

[quote="Zubr"]Sorry, if I participate in IMO and take silver can i entered to MIT without exams?[/quote]

No.

[quote="Zubr"]Sorry, if I participate in IMO and take silver can i entered to MIT without exams?[/quote]

Actually I am a silver medalist too,but I took a lot of exams(and not only exams) to get into MIT :D ;nevertheless they say an IMO medal helps a lot in entering to MIT. :wink:

@ciprian: No ciprian,I heven't received any e-mail from you for a long while. I guess something is wrong either with my e-mail or yours. Could you try to send me an e-mail again?


calculus
integration
induction
algebra proposed
algebra

[color=blue] [b]Find all $x \; ,\; x>1$ such that  $\; \; 1= \left[\; x\cdot n!\; \right] - n \left[\; x\cdot (n-1)!\; \right] \; \; \; ,\; \; \; \forall n\in \{2,3,...\}\; .$
[/b][/color]

[size=75] The symbol $[\; \cdot \; ]$ denotes integral part.  [/size]

[quote="Rust"] ......It  is  obviosly $x=e$ .....           [/quote]
[b] Prove  this ! (that is that  $e =$ solution !) [/b] But...this is the unique solution ? Try if , e.g.  $x_1: = e+2341$ is solution (!?)

$It \ is \ obviosly, \ if \ x \ is \ solution, \ then \ x+m \ (m \ an \ integer) \ give \ solution. \ All \ solution \ may \ be \ written \ x=e-2+m, \ m=1,2,3. \dots$.

"obviously", eh? :roll:
Mind proving these "obvious" things?

[quote="Rust"]$It \ is \ obviosly \ x=e=1+\frac{1}{1!}+ \frac{1}{2!}+\dots.$[/quote]
It is also obvious that you don't want to post a (correct) solution, so stop posting as long as you are not willing to!
At the moment you remember me a bit to RobertuX... (read his postings when you don't know what I mean...)

we have $[xn!]=[nx(n-1)!] \geq n[x(n-1)!]$ and egality if only if:"if $x(n-1)!=[x(n-1)!]+r$ $\Rightarrow$ $0\leq r<\frac1n$"
so the natural positive x solutions are:
$x=\frac{m}{(n-1)!}+r$ $m\in \mathbb{N}$ and $0\leq r<\frac{1}{n!}$

we have $[xn!]= n[x(n-1)!]+1$ $\Rightarrow$ if $x(n-1)!=[x(n-1)!]+r$ so $\frac1n\leq r<\frac2n$
so the natural positive x solutions are:
$x=\frac{m}{(n-1)!}+r$ $m\in \mathbb{N}$ and $\frac{1}{n!}\leq r<\frac{2}{n!}$

I hope you will correct me if i have also mistakes


1. A standard deck of cards contains 52 cards, half of which are red, half which are black. You have a well- shuffeled deck of cards and draw 2 at a time. If both cards are red, put them onto a "red pile". If both are black, put them on a "black" pile. If you have two cards of different colors, tear them up and throw them in the trash can. What is the probability that you will have exactly the same number of cards in each of the two piles on the table when you come to the end of the deck?

2. Ashok's mother usually picks him up from school at 3 PM. But today, she was not there, so he began walking home at the rate of 3 miles per hour. After 24 minutes of walking, he met his mother who drove him the rest of the way home. They arrive 22 minutes later than usual. What is his mother's average driving speed?

3. Determine the constant a >0 such that the graphs of the two curves y = 2/x and x 2 +y 2  = a 2  intersect in only 2 points.

4. A circle entirely in the first quadrant is tangent to the y-axis and the line y =x/2, and the center of the circle is on the line y=2. Find the radius of the circle.

4.
[hide]see diagram
draw in perpendiculars to $y$ axis and line $y=\frac{x}{2}$
draw in line parallel to $x$ axis to form right triangle with legs $a, 2a, a\sqrt{5}$
radius of circle is $\frac{a+2a-a\sqrt{5}}{2}$
$\frac{a+2a-a\sqrt{5}}{2}+2=a$
$3a-a\sqrt{5}+4=2a$
$a(\sqrt{5}-1)=4$
$a=\frac{4(\sqrt{5}+1)}{4}=\sqrt{5}+1$
radius is $a-2=\boxed{\sqrt{5}-1}$[/hide]

Hi, I have an easier solution to #5 (In my opinion)
[hide]
The lines from the radius of the circle that are perpendicular to the tangents will extend from the back of the cirlce and will meet at the same distance.

We have the tangent lines as the y coordinate and y=1/2x.
The lines perpendicular to the tangent lines's equations are y=2 and y=-2x+b, and b is unknown.

But since we know that they are same distance. we form a right triangle using the other set of lines, meeting at 4,2. sqr(4^2+2^2)=sqr(5). therefore the y intercept is sqr(5).

We find the center by systems of equations y=-2x+sqr(5) and equation y=2, so therefore the radius is sqr(5)-1.[/hide]

Kinda hard to explain without a picture..

[quote="1234567890"]2. Ashok's mother usually picks him up from school at 3 PM. But today, she was not there, so he began walking home at the rate of 3 miles per hour. After 24 minutes of walking, he met his mother who drove him the rest of the way home. They arrive 22 minutes later than usual. What is his mother's average driving speed?[/quote]

[hide="Solution"]Let the average driving speed be x mph , and the distance from home to school in miles be y.

3mph*24minutes=6/5miles

$\frac{5y-6}{5x}\cdot 60+24=\frac{y}{x}\cdot 60+22$

$30y-36+x=30y$

$x=36$

Another way is to notice that the distance from home to school is irrelevant, so we can just set it to something.[/hide]

Another way to do 2..
[hide]Another way to do it is just notice that if they were 22 minutes later than normal, the mother would have driven the distance that the boy on foot traveled in 2 minutes. Then we just make the equation $\frac 1{30}*r = \frac 65$ and solve for $r$ which is $36$.[/hide]

Hi, I have a solution to #2,
[hide]
The time the boy walked is 24 but they arrived 22 minutes later. Think of it this way: when the boy doesn't walk at all or at 0 mph, they would have arrived 24 minutes late. Therefore the child was walking at 2/24=1/12 the rate of the driving, so the driving speed is 3*12=36 mphh.[/hide]

1
I got something ugly.
[hide]The probability that you get two reds is $\frac 14$. The probability that you get two blues is $\frac 14$. The probability that you get different cards is $\frac 12$. Suppose that you never tear any of the cards up. Then the probability that that happened is $\left( \frac 14 \right)^{26}$. But since that's 26 events in total, the events can be rearranged in $\frac {26!}{13!13!}$ different ways. Suppose you tear two of the cards up. In order to have an equal number of red and blue cards, then, you must tear two more cards up because then there would be an odd number of blue cards and an odd number of red cards. The probability that you tear four up and get the same number of blue and red is $\left(\frac 14\right)^{24} * \left(\frac 12\right)^2 * \frac {26!}{12!12!2!}$. Then we can go on and on tearing up more and more cards until we have only zero red cards and zero blue cards. So then the answer would be the sum of $\left(\frac 12\right)^{52} * \frac {26!}{13!13!} + \left(\frac 12\right)^{50} * \frac {26!}{12!12!2!} * \left(\frac 12\right)^{48} * \frac {26!}{11!11!4!} + \left(\frac 12\right)^{46} + \frac {26!}{10!10!6!} + \left(\frac 12\right)^{44} * \frac {26!}{9!9!8!} + \left(\frac 12\right)^{42} * \frac {26!}{8!8!10!} + \left(\frac 12\right)^{40} * \frac {26!}{7!7!12!} + \left(\frac 12\right)^{38} * \frac {26!}{6!6!14!} + \left(\frac 12\right)^{36} * \frac {26!}{5!5!16!} + \left(\frac 12\right)^{34} * \frac {26!}{4!4!18!} + \left(\frac 12\right)^{32} * \frac {26!}{3!3!20!} + \left(\frac 12\right)^{30} * \frac {26!}{2!2!22!} + \left(\frac 12\right)^{28} * \frac {26!}{24!} + \left(\frac 12\right)^{26}$.[/hide]

Let it suffice to say that [hide]there is an equal number of reds and blacks in a deck[/hide] :)

Um, tarquin.. umm, isn't the probability of getting 2 reds 26/52 * 25/51?

[quote="jimhu"]Hi, I have an easier solution to #5 (In my opinion)
[/quote]

you meant number 4....

btw i think your solution is a bit easier and more interesting too...

Why, thank you very much  :) . Although my solution was a little more trick/geometry oriented more than algebra, it's not necessary to work with variables at times.

Number 1 was clever :D

[hide]Suppose we only get n pairs of red cards.  Then we will throw away 52-2n red cars and 52-2n black cards, leaving us with n pairs of black cards.  So our probabilty is 1.[/hide]

That's my favorite problem I've solved all day.... I needed a break from pre-olympiadness...[/hide]

[quote="djimenez"][quote="1234567890"]3. Determine the constant a >0 such that the graphs of the two curves y = x/2 and x 2 +y 2  = a 2  intersect in only 2 points.
[/quote]

Those two curves always intersect in exactly two points for any value of $a>0$.[/quote]

Could it be a typo?  (Maybe something other than y = x/2, which I usually don't consider a curve, although I guess it is...but it doesn't curve...hehe.)

Geez, I get his solution.  How did I not see that?  It's really simple, jimhu.
Explaining it is another story...

Say before putting everything into the piles, you divide the deck into 26 pairs.  Now you take every pair that has one red and one black and you remove them.  Since you removed one red and one black each, the amount of reds and blacks are still equal.  So all the pairs have to be both red and or both black, and since the amount of reds and blacks are equal, the number of red sets and number of black sets are equal, thus the probability of them being equal is 1.

[hide="Number 3"]
2/x is obviously closest to the origin where the x and y values are equal. Graph it yourself if this doesn't make sense.

y = 2/x
y = x

x = 2/x
x :^2: = 2
x = sqrt(2)

So the circle's radius is the distance from the origin to the point (sqrt(2), sqrt(2)). The radius is [b]2[/b].
[/hide]

[hide="Cheap way of doing number 2"]
Well, since it doesn't specify anything about the length of the trip home, you can just set it to six-fifths of a mile, which happens to be how far Ashok walks in 24 minutes. That means Ashok's mom drives six-fifths of a mile in 2 minutes, meaning she drives at 36 miles per hour.
[/hide]

[quote="jimhu"]What? I don't get what you are quite saying or your way of solving that problem. How can both black and red be just 52-2n?[/quote]

Heh, I made a mistake.
[hide]
SInce we get a total of n red pairs, we have 2n red cards in the red pile.  That means we have 52/2-2n = 26-2n red cards left that get thrown away.  Whenever we throw away a red card, we also throw away a black card, so we have to throw away 26-2n black cards, leaving us with 2n black cards for the black pile.  The answer is still the same.[/hide]

Hopefully that explains it better.

Yes, I believe the answer to #1 is 1. Each card will either be torn or placed into a pile. Getting a black and red deducts an equal amount from both red and black. Thus there can never be a discrepancy in the total in each deck.


Can anyone explain to me how you explain the sphere formula of a sphere with radius r?

I'm not sure which one you need, but here's both (radius=r):

Volume: 4/3(pi)r^3 (four thirds pi r cubed)
so, say you have a radius r of 6, then plug in 6 for r.
4/3(pi)216
288(pi)= volume of a sphere with a radius of 6.

Surface area: 4(pi)r^2 (four pi r squared)
so, say again you have a radius of 6.
4(pi)36
144(pi)= the surface area of a sphere with a radius of 6.

does that answer your question?

[quote="alanchou"]Getting the formula, on the other hand, probably shouldn't be discussed here[/quote]
It should have been in Classroom Math.

The formula for the volume of a sphere was known to Archimedes; here is [url=http://mathcentral.uregina.ca/QQ/database/QQ.09.99/partridge1.html]his method[/url], which is an interesting read.


hey

Suppose that a and b are positive integers and that when b|a, a quotient of q and remainder of are obtained, with r > 0. Determine the quotient and remainder when b|-a and prove that your answer is correct using the statement of Division Algorithm.

Please someone help me with this, i really need it.

Think about it.  What is -a in terms of a?  Okay, so then if a is qb+r, what is -a?

And no, this is not an olympiad problem...

Ok so
Since $b|a$ then $a = bq+r$ where $0<r<b$
If $b|-a$ then $-a = (-q)^{'}b+r^{'}$ where $0<-r^{'}<b$

I know some how i need to convert the 2nd equation into first which will prove that 2nd part is correct. I've tried so many times but i can't figure out, so please give me some hints, at least do couple of stpes please. My other question is do i need to show algebrically how i got $-a = (-q)^{'}b+r^{'}$? Please help me, tons of thanks in advance!

[quote="Farenhajt"]$a=qb+r\implies-a=-qb-r=-qb-b+b-r=-(q+1)b+(b-r)$

Hence $q'=-(q+1),r'=b-r$ for $r\neq 0$ and $q'=-q,r'=0$ for $r=0$[/quote]
[quote="salubadshah"]
but $r \neq 0$
[/quote]

and therefore, $q'=-(q+1),r'=b-r$.

We work together to write a logical sentance![/quote]

one last final question, i understood everything but doesn't the solution only gives what remainder and quotient. It doesn't prove using divison alogrithm that what we found it correct? Or does it because we drove from the original statment which is driven by division alogrithm?

Please just calrify this, thanks!


Indicate whether the statement is always true or sometimes false.Justify your answer with a logical argument or a counterexample.
(a)If $ A$ is a singular $ n$x$ n$ matrix,then $ Ax \equal{} 0$ has infinitely many solutions.
(b)If $ A$ is a singular $ n$x$ n$ matrix,then the reducedrow-echelon form of $ A$ has at least one row of zeros.

I think these two are true.But I can't understand what the sentence,"Justify your answer with a logical argument", means.
Can someone give me some examples?
Thanks~!

[hide="Argument for Part A"]Since $ A$ is singular, it has a reduced echelon form $ B$ with a row of zeroes (see part B for justification). Here, you have a matrix with more columns than rows, so necessarily, there are infinitely many solutions to the equation $ Bx \equal{} 0$, so therefore there are infinitely many solutions to the equation $ Ax \equal{} 0$.

[/hide]

[hide="Argument for Part B"]Assume that $ A$ is singular but some echelon form of $ A$ has no row of zeroes. The determinant of that echelon form of $ A$ is nonzero, which implies that the determinant of $ A$ is nonzero, a contradiction. Therefore, $ A$ being singular implies that all echelon forms of $ A$ have a row of zeroes.

[/hide]

Excuse me,how do you get the determinant of the echelon form of $ A$ is nonzero :?:

Due to some echelon forms of $ A$ has no row of zeroes?

[quote="skywalkerJ.L."]Excuse me,how do you get the determinant of the echelon form of  is nonzero[/quote]

Since there are no zero-rows, there can't also exist zero-columns, from the definition of row echelon form of a matrix. Also, the non zero rows of a row echelon form are linearly independent. Therefore, its determinant is non-zero.


function
geometry
3D geometry
tetrahedron
parameterization
symmetry
real analysis

In R^3, let A, B, C, and D be the vertices of a (regular, for definiteness) tetrahedron, and let T be the subset of R^3 made up of these four vertices and the six line segments which connect them. Suppose that f:T->R is a continuous function.

Prove that there exists a b in R whose preimage f^-1 ({b}) contains at least four points of T.

If at all possible, please provide explanations along with your proof. Thank you.

Let's start by assuming that $f(A),f(B),f(C),$ and $f(D)$ are four different real numbers. Because of the symmetries of the tetrahedron, we can make the same argument no matter what order these numbers are in. So, WLOG, assume $f(A)<f(B)<f(C)<f(D).$

Now choose $y$ such that $f(B)<y<f(C).$

Think of a continuous function $\gamma$ which maps $[0,1]$ onto the segment $BC$ in such a way that $\gamma(0)=B$ and $\gamma(1)=C.$ (In other words, $\gamma$ is a parameterization of the segment $BC.$) Then $f\circ\gamma$ is a continous function on $[0,1]$ such that $f\circ\gamma(0)<y<f\circ\gamma(1).$ By the intermediate value theorem, there exists a point $x\in(0,1)$ and hence a point $\gamma(x)$ on the segment $BC$ such that $f(\gamma(x))=y.$

Now repeat the same argument for the segment $AC,$ the segment $BD$, and the segment $AD.$ We have four points, one on each segment, in $f^{-1}(\{y\}).$

The only things we now have to cope with are cases in which two or more of the vertices get mapped to the same real number.

Suppose $f(A)<f(B)=f(C)<f(D).$

Now, how does $f$ behave on the segment $BC?$ If it's constant then we have infinitely many points that $f$ maps to the same value. So suppose $f$ is not constant on $BC.$ Suppose there exists a point $E$ in the segment $BC$ such that $f(E)>f(C).$ Now choose $y$ such that $f(C)<y<\min(f(D),f(E)).$ We can find, by an intermediate value argument similar to the last one, four points in $f^{-1}(\{y\}):$ One in each of the segments $BD,CD,CE$ and $DE.$ If there's no point $E$ in which $f(E)>f(C),$ then there must be an $E$ such that $f(E)<f(C).$ Now we use $y$ such that $\max(f(A),f(E))<y<f(C)$ and find one point in $f^{-1}(y)$ in each of the four segments $AB,AC,BE,$ and $DE.$

If $f(A)=f(B)<f(C)\le f(D)$ then we can use $f(A)<y<f(C)$ and find points on $AC,BC,AD,$ and $BD.$ $f(C)=f(D)$ is similar.

Now suppose $f(A)=f(B)=f(C)<f(D).$ Again, we must assume that $f$ is nonconstant on the segments $AB,AC,$ and $BC,$ else we get infinitely many equal values.

If there's some $E$ on some one of these segments - WLOG, on $AB$ - such that $f(A)<f(E),$ then use $f(A)<y<\min(f(E),f(D))$ to find points in the inverse image of $y$ on $AE,EB,AD,$ and $BD.$ Else we have points $E,F,$ and $G$ on $AB,AC,$ and $BC$ such that each of $f(E),f(F),$ and $(G)<f(A).$ Now use $\max(f(E),f(F))<y<f(A)$ to find points in the inverse image of $y$ on each of $AE,EB,BF,$ and $FC.$

The case $f(A)<f(B)=f(C)=f(D)$ is similar.

Finally, if $f(A)=f(B)=f(C)=f(D),$ we're already done.

--

Surely, there's a less piecemeal argument to be found. This is just brute force.

How can you make the conclusion that f(A) < f(B) = f(C) = f(D) is similar, is it because you use the same process or because of symmetry?

Also, how can you go from the work you have shown to saying the if f(A) = f(B) = f(C) = f(D) we're done, what allows you to make that conclusion?

The case $f(A)<f(B)=f(C)=f(D)$ can be reduced to $f(A)=f(B)=f(C)<f(D)$ by replacing $f$ with $-f$ and renaming the vertices.
If $f(A)=f(B)=f(C)=f(D)$, then $f^{-1}(f(A))$ contains at least four points, namely $A$, $B$, $C$, and $D$.

-----
I keep thinking "[url=http://en.wikipedia.org/wiki/Morse_theory]Morse theory[/url]" ... and not getting anywhere.  :(


I thought I might bring this up. How do you normally mark corrections of typos, etc. on Messenger?

For example, how I do it:

<user1> I'm really hapy today
<user1> *happy

Basically, I put a star before the corrected word. I won't normally do this when I'm not correcting an error, so it's become a habit for me.

Now, if I make two or more errors, then I do this:

<user1> I'm really hapy today
<user1> *hppy
<user1> **hapy
<user1> ***happy

Because then it's a correction of a correction, etc.

How do you people do it? I'd like to know if there's some sort of standard among people.

[quote="Kalle"]/s/hapy/happy[/quote]

Really, you do all that?

But I'm confused. Do you mean this:

<user1> I'm really hapy today
<user1> /s/hapy/happy

: or something else?

I do stuff like this...

<guy>: So... how's life?
<me>: Oky
<me>: *oay
<me>: *okay
<me>: I hate typos

or...

<guy>: So... how's life?
<me>: Oky
<me>: I mean oay
<me>: Wait... I mean okay
<me>: I hate typos

Yeah.. stuff like that.  :)

I don't know, but I also got myself used to *... It's probably the most common

I also use the star. Always one though. If I make another mistake I stick with one.

I put the star after the corrected word.

[quote="sdkudrgn88"]<user1> I'm really hapy today
<user1> *hppy
<user1> **hapy
<user1> ***happy

Because then it's a correction of a correction, etc.

How do you people do it? I'd like to know if there's some sort of standard among people.[/quote]

This.

I put the star after the word....

ex:

helo
heello*
hello*


For positive reals $ x,y,z$ and reals $ 0<a\leq b\leq c$,
\[ b\left( \left( x\plus{}y\plus{}z\right) ^{2}\minus{}4zx\right) \left( c\minus{}a\right) ^{2}\minus{}4xyc\left( b\minus{}a\right) ^{2}\minus{}4yza\left( c\minus{}b\right) ^{2}\geq 0\]

No reply?  :(
I'm learning how to use SOS but none of those 5 criteria is fulfilled. Am I missing something? Or maybe there's a counterexample?

[quote="dyd"]For positive reals $ x,y,z$ and reals $ 0 < a\leq b\leq c$,
\[ b[\left( x \plus{} y \plus{} z\right) ^{2} \minus{} 4zx] \left( c \minus{} a\right) ^{2} \minus{} 4xyc\left( b \minus{} a\right) ^{2} \minus{} 4yza\left( c \minus{} b\right) ^{2}\geq 0\]
[/quote]
I will help you. Since $ c \ge b \ge a>0,$ we have
$ b(c\minus{}a)^2 \minus{}c(a\minus{}b)^2\minus{}a(b\minus{}c)^2 \equal{}(c\plus{}a)(c\minus{}b)(b\minus{}a) \ge 0.$
Also, it is clear that $ (x\plus{}y\plus{}z)^2 \minus{}4zx >0,$
so we get
$ b(c\minus{}a)^2[(x\plus{}y\plus{}z)^2\minus{}4zx] \ge [a(b\minus{}c)^2\plus{}c(a\minus{}b)^2][(x\plus{}y\plus{}z)^2\minus{}4zx],$
and hence, it suffices to show that
$ a(b\minus{}c)^2[(x\plus{}y\plus{}z)^2\minus{}4zx\minus{}4yz]\plus{}c(a\minus{}b)^2[(x\plus{}y\plus{}z)^2\minus{}4zx\minus{}4xy] \ge 0,$
or
$ a(b\minus{}c)^2(x\plus{}y\minus{}z)^2 \plus{}c(a\minus{}b)^2(y\plus{}z\minus{}x)^2 \ge 0.$
Of course, this is true.

Thanks!! Your proof makes me think twice before SOS-ing blindly without inspecting the condition of variables  :blush:


function
real analysis
topology
real analysis unsolved

Q1: Prove the following statements are equivalent. $ E$ is a set, $ \forall{\epsilon}\in{\mathbb{R}}$, $ \exists{O}$ s.t. $ m_*(O\minus{}E)<\epsilon$, where $ O$ is open and $ E\subset{O}$; $ \exists{F}$ s.t. $ m_*(F\minus{}E)<\epsilon$, where $ F$ is closed and $ F\subset{E}$. ($ m_*$ means exterior measure)
It seems simple, but I could not get it...

Q2: Prove that a measurable function is the limit of a sequence of continuous functions almost everywhere.
My idea is to use continuous function revised from a sequence of step functions which converge to a measurable function, but I found my construction not very rigorous...

Q1: Both statements equivalent to Lebesgue measurability of $ E$.

Q2: Why isn't your construction very rigorous?

Q1: I need to clarify on this. The question asks about proving each statement directly from the other, although it's clear they're the definitions of Lebesgue measurability.
Q2: Approximating step function by a continuous function works fine on $ \mathbb{R}^2$, but I skipped the proof on higher dimension space, on which I don't know how to construct continuous functions approximating step functions... I think there's a short proof from Lusin theorem. However, the two proofs I found are both on graduate text books. Does anyone know an accessible proof to undergraduate based on Lusin theorem?

I found the solution to Q1, while I still have no clue about Q2.

Sps we have $ E\subset{O^*}$open and $ \forall{\epsilon}>0$, there is $ m_*(O^{*}-E)<\epsilon$, we want to derive the other statement from this.

Since this is the definition of Lebesgue measurability $ \Longrightarrow$ $ E$ is measurable $ \Longrightarrow$ $ E^c$ is measurable $ \Longrightarrow$ $ \exists$open set $ O$ with $ E^{c}\subset{O}$ s.t. $ m_*(O-E^c)<\epsilon$.

$ O^c$ is closed, and we define $ F=O^c$. $ x\in{F}=O^{c}\Longrightarrow$$ x\notin{O}\Longrightarrow{x\in{E}}\Longrightarrow{F\subset{E}}$

$ m_*(E-F)=m_*(E\cap{F^c})=m_*(E\cap{O})=m_*(E\cap{(E^{c}\cup{(O-E^c)}))=m_*(\emptyset\cup{(O-E^c)})=m_*(O-E^{c})<\epsilon}$
The other direction is very similar to this.

[quote="emchong16"]However, the two proofs I found are both on graduate text books. Does anyone know an accessible proof to undergraduate based on Lusin theorem?[/quote] One doesn't need Lusin theorem.

Approximate your function by continuous ones [i]in measure[/i] (which is easy; you can use Lusin theorem if you really want), then take a subsequence which converges almost surely.


Here is a sequence of numbers:
a_1 = 1, a_2 = 2 + 3, a_3 = 4 + 5 + 6, a_4 = 7 + 8 + 9 + 10 ....
Find a formula for a_n

alright, can you show the intermediary work?

[hide="My Solution"]
Where notice that every $ a_k$ denotes the sum of the integers from $ k$ to $ k \plus{} (k \minus{} 1) \equal{} 2k \minus{} 1$. We can use the formula for the sum of an arithmetic sequence to find $ a_k$.

We have $ a_k \equal{} \frac{(2k \minus{} 1\minus{} k \plus{} 1)(k \plus{} 2k \minus{} 1)}{2} \equal{} \boxed{\frac{k(3k \minus{} 1)}{2}}.$
[/hide]

well i do not know if i am doing something wrong but the first term is always n(n+1)/2 +1. so we can call that x.
so we have x+1,x+2......x+n.
so we apply the formula
so
n(n(n+1)/2+1+n(n+1)/2+n)/2=
n(n(n+1)+n+1)/2=
n(n+1)^2/2.
so I do not know if what the first person did was incorrect?

[quote="varunrocks"]but the first term is always n(n+1)/2 +1[/quote]
The first term is $ \frac{n(n\color{red}\minus{}\color{black}1)}2\plus{}1.$ :)

You can always check and see if your formula is correct. For example, we know that $ a_4\equal{}7\plus{}8\plus{}9\plus{}10\equal{}34.$ If you substitute $ n\equal{}4$ into your formula and don’t get 34, you’ll know that your formula is wrong.

Yes, elizabeth is right. I actually based my formula off of the pattern shown in the first 2 terms, which turned out to be different from the actual pattern  :blush:.

Ohhhhh sorry for giving an incorrect solution.

I noticed $ \frac{n(3n\minus{}1)}{2}$ is the $ n$th pentagonal number. Is this a coincidence, or is there a way to prove it (that the sequence of numbers is the sequence of pentagonal numbers)?

That formula doesn't represent this sequence, so it would be impossible to prove that it does. I wrote an incorrect solution.

i see what elizabeth is saying, but how does she get the final answer?

[hide="Corrected Solution"]
Note that $ a_n$ consists of the sum of integers from $ \frac {n(n \minus{} 1)}{2} \plus{} 1$ to $ \frac {n(n \minus{} 1)}{2} \plus{} 1 \plus{} n \minus{} 1 \equal{} \frac {n(n \minus{} 1)}{2} \plus{} n$. Using the formula for an arithmetic series, we have

$ a_n \equal{} \frac {n\left(\frac {n(n \minus{} 1)}{2} \plus{} 1 \plus{} \frac {n(n \minus{} 1)}{2} \plus{} n\right)}{2} \equal{} \frac {n(n^2 \minus{} n \plus{} 1 \plus{} n)}{2} \equal{} \boxed{\frac {n(n^2 \plus{} 1)}{2}}.$

If you don't see how I made my first conclusion, think about how we know what the first number in the sum is for any $ a_n$. For each $ 1 < k < n$, $ a_k$ has $ k$ positive integers in its sum and note that every $ a_k$ continues the sum from where $ a_{k \minus{} 1}$ left off. Therefore, there are $ \frac {n(n \minus{} 1)}{2}$ integers that have been summed already, and you start from the next integer which is $ \frac {n(n \minus{} 1)}{2} \plus{} 1$.
[/hide]

Hello. So the first term is going to be $ 1\plus{}2\plus{}\dots \plus{}(n\minus{}1)\plus{}1\equal{}\frac{n(n\minus{}1)}{2}\plus{}1$.

Therefore, $ a_n\equal{}n \left(\frac{n(n\minus{}1)}{2} \plus{}\frac{n\plus{}1}{2}\right)\equal{}\frac{n}{2} \cdot (n^{2}\plus{}1)$.


[quote="diophantient"][quote="ckck"]A certain amount of money saved for 1 year at a certain annual interest rate yielded
$\$$7.50. If the principal had been $\$$25 more and the interest rate 1% less, the interest would have been the same. Find the interest rate.[/quote]
[hide]
$Pr=750$
$(P+25)(r-1)=750$[/hide][/quote]
[hide="now that I've gotten over my lazyness"]
$Pr-P+25r=775$
$25=25r-P$
$25(r-1)=p$
$25(r-1)r=750$
$r(r-1)=30$
$r=6$
6[/hide]

[quote="ckck"][quote="Art of Owna"][quote="mets501"]Is the interest compounded quarterly, monthly, or continuously?[/quote]

Well the question says annula interest rate..I can do this but I have to use a caculator for the quadratic formula as my answer is [hide]solution to 0.25x^2-0.0025x-0.075=0, so I get x to be about .5527 so around 55.27%...[/hide][/quote]
The answer is a whole number.  :)[/quote]

Oops..it said $25 more, not$0.25 more, I gotta read better..


percent
search
logarithms
number theory proposed
number theory

What percent of the powers of 2 have 1 as their leftmost digit?
Does 100, 1000 or 10000 appear as the leftmost digits in the decimal representation of any power of 2?
Any given sequence of digits may appears as the leftmost digits in the decimal representation of any power of 2?
Many thanks grobber for sharing this link:
http://www.princeton.edu/~mathclub/

You're welcome :).

Your second and third questions have been answered on the forum before, I believe. It's not that hard to show that unless $a$ is a power of $10$, every combination of digits can appear to the left of $a^n$ for some $n$.

Edited: replaced "unless $a$ is a power of $2$" with "unless $a$ is a power of $10$".

[quote="grobber"]You're welcome :).

Your second and third questions have been answered on the forum before, I believe. It's not that hard to show that unless $a$ is a power of $2$, every combination of digits can appear to the left of $a^n$ for some $n$.[/quote]

I can'y find that topic :(.
If $a$ is not a power of $2$ or it is $2$ this is possible all the time?
If $a$ is a power of $2$, what can we say about this combination of digits?

[quote="grobber"]I apologize. That's a stupid typo I made. I meant to say "if $a$ is not a power of $10$". I suppose it makes more sense now :). I can't remember any actual topics, but I'm sure it's been discussed.[/quote]

I will search more carefully. :)

[quote="Myth"]Clearly answer for the first question is $100\log_{10} 2$. :?[/quote]

Honestly, I doubt.  :?

Myth is right.
In fact, the percent of the powers of $2$ wich have $d\in[1,9]$ as their leftmost digit is $100\log_{10}(1+1/d)$.

[quote="blang"]Myth is right.[/quote]
As always?  :lol:  :lol:  :lol:  :D  :D  :D

Maybe the $100$ which is there to turn the fraction into a percentage was confusing? :)

It follows immediately from the equidistribution of $\{na\}$ in $(0,1)$ when $a$ is irrational. The first digit of $2^t$ is $d$ iff there is a natural $k$ s.t. $d\cdot 10^k\le 2^t<(d+1)10^k$. Taking $\log_{10}$ out of this, we get $k+\log_{10}d\le t\log_{10}2<k+\log_{10}(d+1)$. Since $\log_{10}2$ is irrational, we get the desired result.

Indeed, that is why I said "Clearly answer for the first question is $100\log_{10}2$"  :lol:  :lol:  :lol:


hehe. i'm a rookie in arml (just starting this year). for indivual rankings, do they only use the indivual round's contest or do they use any others (power relay,...?) also, wut r the best resources to pwn in arml? where would i find old contests (i found a few, but is there some website that's collected a whole bunch of the old ones?) thx for your time and help!  :)

Individual rankings are based on individual score only. Team rankings are based on all scores.

ARML questions are not collected online anywhere (copyright), though the last few years can be found, there's a thread around here that has the links (search it or something). ARML publishes a few books of old tests, and old questions can be found in the various ARML marathons and practice series, and the websites of quite a few ARML teams.

Individual rankings are computed by individual score only, with a special tiebreaker round to determine the top 10.  The tiebreaker consists of one question, and you get one chance to answer it, but you can answer whenever you like.  First correct answer ranks highest.  Everyone who answers incorrectly stays tied, so there are sometimes (usually) more tiebreaker questions to finish breaking the ties.  You only compete in the tiebreaker if you're potentially up for top 10 in the country or a site award (top individual at your site outside of the top 10), and rankings don't get computed for anyone else.  ARML is mainly a team contest; the individual rankings are generally considered to be "on the side".

For practice, I suggest getting an ARML book or two.  You might also try old HMMT/SMT contests; I used to use those a lot for ARML training.

how much do u need to get correct to be in the tiebreaker round? do u have to write our ur solution and ur answer or just a specific numerical answer? thx again!  :)

The last few years a 7 out of 8 was enough to make tiebreakers. I'm not sure how it'll turn out this year since they're changing it to 10 questions on the individual round.

As for solutions, all answers in ARML are strictly numerical.  The sole exception is the Power Question (one of the team parts), where you have to write out proofs for all answers.

do u think the problems will be easier this year, or so that ppl (most n00bs including me) can get a few right but not a lot of them like it is now?

Prior to 2006 (or maybe 2005), they only gave awards to the top 3 individuals, not top 10.  In 2002 and 2004, a perfect score of 8/8 was required to make tiebreaks.  In 2003 and 2006, one person in the contest got a perfect score, so people with 7's got to tiebreak for second place.  In 2005 and 2007, no one got a perfect score, so 7's made the tiebreakers and could finish first.  Also in 2005, there were very few 7's at the Iowa and Las Vegas sites, so people with 6's were invited to tiebreakers because they were potentially in the running for the site award; this, however, is a very rare anomaly.  Not sure about 2008, since I wasn't there.

I believe ARML has stated that the goals behind increasing the individual to 10 problems were both to give everyone a shot at getting at least one problem right, and to provide more differentiation between the top scorers.  Based on that, I'm guessing they'll be adding one easy problem and one hard problem.  So it will be slightly easier to get a couple of problems right, but much harder to get a perfect score.

One measure of how easy the easier problems are: average score among all teams (Most teams have the full 15 members)
In 2008, the two easiest problems were at 10.9 and 8.2
In 2007, the two easiest problems were at 9.9 and 8.9
In 2006, the two easiest problems were at 9.5 and 8.5
In 2005, the two easiest problems were at 11.4 and 10.1

Expect the two or three easiest problems to be in that range; anybody can get it, but some people will make mistakes. It's already true that individuals have to be actively unlucky to be shut out, and another easy problem would make that less common.

How hard it is to get a perfect score varies greatly year-to year based on exactly how hard the hardest problems. In 2006, only three people got #8, which makes that the hardest ARML individual question ever. Last year's hardest question (#8) was at a much more reasonable 43 right answers, with all of the top ten teams having somebody get it right. This led to 10 perfect scores (plus one Taiwanese), perfect for the tiebreaker. That's not really normal- it's the easiest hardest question since 2002, and also an unusually high number of perfects.

For an expanded version of what jmerry said, see post #63 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=207240&start=60]here[/url], and also go back a page on that topic to post #47. (Any slight differences between my numbers and jmerry's have to do with the completeness of the data; they're not important.)

on the relay round, what's with the T=TNYWR (something like that)? wut does that mean? :)

TNYWR="the number you will receive"; it's the number you get handed from the person in front of/next to you.

Of course, you start working long before you know what TNYWR is, and solve as much of the problem as you can before you get passed a number. Ideal would be to derive a formula such as your answer being, say, $ \frac{T^2\minus{}1}{4}$ where $ T\equal{}$TNYWR. You can't always have that, of course, but there's usually something you can be doing.

To add to that, even if you get a number and work it out, it's still helpful to find a general formula to check your answer.

what would u say avg is at arml?i guess it depends on the yr...but think like avg year?if that helps....

If you mean individual scores, on our teams (we have 4) the average is usually around a 3. That's out of the 8 that's used to be the contest, though, and I'm not sure how representative our teams are of the rest of the participants.

Perhaps someone with more data could answer more precisely

Follow the link I gave in post #9 and then divide the numbers you see by 15, since there are 15 members on a team (although not all teams have the full 15). From that it looks like the average individual score out of 8 possible over the last four years has been about 3. It was closer to 2 in 2004 and even lower in 2003.

The 2009 ARML individual exam will have 10 questions, not 8. We have no experience with 10-question ARML exams, so we'll learn something this year.

Edit: yes, cyberspace, you're reasonably representative. (What organization are you from - Lehigh?)

Greetings to one and only kaido kid,
I able to give good arml help,
Tank u vary much for listen, saar,


hi i just find out about amc tests and i wanna take tests
so wat shouold i study w/ ? btw im looking forward taking amc10 next yr  recommand me someding puh lease!

AoPS Vol. 1 is a good start. Vol. 2 is useful if you're concentrating on the last few problems.

aops vol 1 for what youre working on. But get aops2 also because youll be using it soon.

Also, study past AMC tests. Some are on AoPS Wiki.

[quote="kimyejin2000"]thanz for the help,  ill get v.1[/quote]

get v2 cuz of the discount. You'll purchase it eventually anyways.


inequalities
real analysis
real analysis unsolved

It is a useful theorem.

$ B$ is a Banach space. $ U\in B(X)$$ \exists \lambda_1,\lambda_2\in[0,1)\forall x\in X \parallel{}Ux \minus{} x\parallel{} \leq \lambda_1\parallel{}x\parallel{} \plus{} \lambda_2\parallel{}Ux\parallel{} \Rightarrow U^{ \minus{} 1}\in B(X)$, i.e. $ U$is invertable.

WLOG we can assume that $ \lambda: \equal{} \lambda_1 \equal{} \lambda_2\in [0,1)$. Obviously $ U$ is injective. Then plug the given relation into the inequality $ \parallel x\parallel\le\parallel Ux \minus{} x\parallel \plus{} \parallel Ux\parallel$ which yields that $ \parallel Ux\parallel\ge m\parallel x\parallel$ for all $ x\in X$ where $ m \equal{} \frac {1 \minus{} \lambda}{1 \plus{} \lambda}>0$. This condition is equivalent to that the inverse of $ U$ is continuous (defined on the range $ R(U)$ of $ U$) with $ \parallel U^{ \minus{} 1}\parallel\le m^{ \minus{} 1}$.

but, how to porve $ R(U)\equal{}H$. That's the difficult point of the problem. Only we prove it, we have $ U\in B(X)$.

Here a partial result: Choose a minimal $ \lambda$ such that the key inequality holds. If $ \lambda<\frac{1}{3}$, then $ U$ is surjective.

[hide="Proof"]If $ R(U)$ is a proper subspace of $ X$, then it is a closed proper subspace and for every $ \varepsilon\in(0,1)$ there is some $ x_\varepsilon\in X$ of norm 1 such that  $ \varepsilon\le\parallel Ux\minus{}x_\varepsilon\parallel$ for all $ x\in X$, by Riesz lemma. Hence \[ \varepsilon\le\parallel Ux_\varepsilon\minus{}x_\varepsilon\parallel\le\lambda(1\plus{}\parallel U\parallel)\le \frac{2\lambda}{1\minus{}\lambda}\]Taking $ \varepsilon\to 1$ yields $ \lambda\ge \frac{1}{3}$.[/hide]

Thus it is open what happens to $ R(U)$ if the minimal $ \lambda\ge \frac{1}{3}$.

[quote="lambda"]but, how to porve $ R(U) \equal{} H$. That's the difficult point of the problem. Only we prove it, we have $ U\in B(X)$.[/quote]

here $ R(U)\equal{}X$ . not $ R(U)\equal{}H$


function
integration
limit
logarithms
Euler
calculus
real analysis

$f(x)=\int_0^x \frac{e^{-t}-1}{t} dt$
Determine $\lim_{x->\infty}f(x)$.

[quote="Moubinool"]$\int_{x}^{+\infty}\frac{e^{-t}}{t}dt=\int_{0}^{x}\frac{1-e^{-t}}{t}dt-\ln x -\gamma$ for x>0

$\gamma$ Euler constant[/quote]

$\int_{0}^{+\infty}\frac{1-e^{-t}}{t}dt=\gamma$

[quote="heartwork"]
$\int_{0}^{+\infty}\frac{1-e^{-t}}{t}dt=\gamma$[/quote]
No, this integral diverges with logarithmic speed at infinity (see Moubinool's post).


LaTeX
inequalities
inequalities unsolved

let $x,y,z$ positive reals

prove or disprove that $\left(\frac{a}{a+b}\right)^{3}+\left(\frac{b}{b+c}\right)^{3}+\left(\frac{c}{c+a}\right)^{3}\leq \frac{9}{8}$

${\left(\frac{c}{a+c}\right)^{3}+\left(\frac{a}{a+b}\right)^{3}+\left(\frac{b}{b+c}\right)^{3}\leq \frac{9}{8}}$

${\left(\frac{1}{\frac{b}{a}+1}\right)^{3}+\left(\frac{1}{\frac{c}{b}+1}\right)^{3}+\left(\frac{1}{\frac{a}{c}+1}\right)^{3}\leq \frac{9}{8}}$

${\left(x=\frac{b}{a}\right)\land \left(y=\frac{c}{b}\right)\land \left(z=\frac{a}{c}\right)}$
${x y z=1}$

${\frac{1}{(y+1)^{3}}+\frac{1}{(z+1)^{3}}+\frac{1}{(x+1)^{3}}\leq \frac{9}{8}}$

I Will make some Latex calculation here and post the continuation here

Suppose

$a = 2048 c$
$b = 1024 c$

the equation fail

have a lot of values like this

it's just look in the my last inequality

:)

You can prove that :
 $\sum(\frac{a}{a+b})^{n})\geq \frac{3}{2^{n}}$

see my last equation

if n=3

so can be greater than 9/8  or less

so, it dont hold

we can find a counter example easily

its false


Prove that $ a^{2}\plus{}b^{2}\equal{}7abn$ has no positive integer solutions for a, b, n.

[hide="Hint"]
Use $ a^2\plus{}b^2\equal{}0\mod 7\Leftrightarrow a\equal{}b\equal{}0\mod 7$ and infinite descent.
[/hide]

[quote="bertram"][hide="Hint"]
Use $ a^2 \plus{} b^2 \equal{} 0\mod 7\Leftrightarrow a \equal{} b \equal{} 0\mod 7$ and infinite descent.
[/hide][/quote]
That's not a hint, that's the whole solution :D

My hint is in the other topic, so here comes the solution :P

[hide]
From number theory we know that $ x^2\plus{}1\neq 0\mod 7$ for all integers x. But from $ a^2\plus{}b^2\equal{}7abn$ it's clear that $ a^2\plus{}b^2\equal{}0\mod 7$, and if one of the two is not zero (WLOG b), then there is a $ b^*$ with $ bb^*\equal{}1\mod 7$ and $ (ab^*)^2\plus{}1\equal{}0\mod 7$ which is a contradiction, so $ 7|a,7|b$ and setting $ x\equal{}\frac a7$ and $ y\equal{}\frac b7$ it's obvious that $ x^2\plus{}y^2\equal{}7xyn.$
 Now choose $ (a,b)$ satisfying the equation with $ a\plus{}b$ minimal; as both are positive $ \frac{a\plus{}b}7<a\plus{}b,$ and as desired we get a contradiction.

Alternatively, one could have used infinite descent and shown that we will always be able to construct smaller pairs which is impossible in the positive integers.[/hide]


ratio
real analysis
calculus
calculus computations

when finding what values of x a power series converges for, it is common to use the ratio test. however, suppose that we use the ratio test on a power series and find an interval of convergence, $I$. We know that the power series converges absolutely (from ratio test thm.) on $I$. At this pt. the problem is usually finished.

however, doesn't this ignore intervals on which the power series is conditionally convergent? how do we know we have found all intervals of convergence (the original power series converging)?

Theorem on power series: If $\sum_{n=0}^{\infty}a_n(x-x_0)^n$ converges for $x=x_1,$ then it converges [b]absolutely[/b] for all $x$ such that $|x-x_0|<|x_0|.$

This theorem assures us that there can be no "intervals of conditional convergence." If you're considering $x$ to be a real variable, there are only the two endpoints of the interval of convegence. If you're working in the complex plane, the set of convergence is a disk, and the points that remain to be investigated are on the boundary circle of that disk.

Here's another way to put it: suppose that the ratio test gives a clear radius-of-convergence result for a power series. (It doesn't always, which is why the limsup version of the root test is theoretically a superior test. But for many practical problems, there's no difference.) To summarize the meaning of a ratio test or root test result:

If the limiting ratio (or limiting root) is less than 1, then the series converges absolutely (by comparison to a convergent geometric series).

If the limiting ratio (or limiting root) is equal to 1, then the test is inconclusive becuase the series cannot be usefully compared to a geometric series.

If the limiting ratio (or limiting root) is greater than 1, then the series [i]is necessarily divergent becuase the terms do not tend to zero.[/i] In fact (the absolute values of) the terms tend to infinity at at least a geometric rate.

There's a little bit more to investigate, but it only involves the boundary of the interval (or disk) of convergence - nothing more.

Gallery of examples - same radius of convergence in each case. I've assumed we're talking about real variables - complex variables would allow a few other interesting examples and would change the discussion of some of these.

$\sum_{n=0}^{\infty}x^n$ converges (absolutely) only on $(-1,1).$

$\sum_{n=1}^{\infty}\frac{x^n}{n^2}$ converges absolutely (and uniformly) on $[-1,1].$

$\sum_{n=1}^{\infty}\frac{x^n}n$ converges absolutely on $(-1,1)$ and conditionally at $x=-1.$

$\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+1}}{2k+1}$ converges uniformly on $[-1,1].$ The convergence is absolute on $(-1,1)$ and conditional at $\pm1.$

ok, thanks for the clarification. is that thm. from analysis?

[quote="maokid7"]what exactly is conditional convergence?
i do not see how it is any different from absolute convergence.[/quote]

Conditional convergence means $\sum a_i$ converges but $\sum |a_i|$ diverges. Absolute convergence means they both converge. These terms are usually used with alternating series, as far as I know.


What's the smallest integer multiple of 49 whose digits are all the same?

[/hide]
is it 777777
if it is then
[hide]

We will derive a divisibility by 49 rule.

1=1(mod49)
10=10(mod 49)
100=2(mod49)
1000=20(mod49)
10000=4(mod49)
100000=40(mod 49)
...

It will repeat eventually, but all we need is the first 6

In order for it to be divisible by 49 since all the digits are the same is for (1+10+2+20...)n to be divisible by 49.

N<10, so n=7 is the best.

1=1
1+10=11
1+10+2=13
1+10+2+20=33
1+10+2+20+4=27
1+10+2+20+4+40=77 which is divisible by 7 while the rest aren't

so if you multiplt 111111 by 7 you get 777777[/hide][hide][/hide]

yes, thats the right answer. Thanks ehehheehee. :)

Or:
[hide]
the powers of tens mod 7 forms the sequence 1, 3, 2, -1, -3, -2, ... Since all the digits are the same, its number of digit must be multiples of 6.  Since we want it to be able to be divided by 2 7's, just take 777777.
[/hide]

[quote="lotrgreengrapes7926"][quote="Inspired By Nature"][quote="t0rajir0u"][hide="More generally"] The minimal case is obviously achieved if we can factor out the $7$ and consider the smallest repunit divisible by $7$.
Fermat's Little Theorem tells us this is $\frac{10^{6}-1}{9}= 111111$ (it's not hard to verify that $11, 111$ don't work).  Hence we want $777777$. [/hide][/quote]
 i get what you are saying, but whats repunit?[/quote]
"A repunit is a number consisting of copies of the single digit 1."
http://mathworld.wolfram.com/Repunit.html[/quote]
thank you lotrgreengrapes7926.


parameterization
linear algebra
matrix
combinatorics unsolved
combinatorics

There are some people in a meeting; each doesn't know at least 56 others, and for any pair, there exist a third one who knows both of them. Can the number of people be 65?

as posed, the question begs the answer no.
suppose the number of people is $65$.consider the number of triples $(x,y,z)$ where $x\leftrightarrow z, y\leftrightarrow z,x\neq y$. on the one hand the number of such triples is $\sum_{(x,y),x\neq y} |N(x)\cap N(y)|$ where $N(x)=\{z|x\leftrightarrow z\}$. on the other hand,for each vertex $z$ we can pick any two of its neighbors and that gives such a configuration so the number of such triples is also $\sum_{z\in V} d(z)(d(z)-1)$ where $d(z)=|N(z)|$.
 but also by assumption, $d(z)\leq 8,\forall z,|N(x)\cap N(y)|\geq 1\forall x,y$ so we have
   $65\cdot 64\leq \sum_{z\in V} d(z)(d(z)-1)\leq 65\cdot 8\cdot 7$
$\Rightarrow 65\cdot 64\leq 65\cdot 56$ and that is a contradiction.

Hmm... this problem wants to mean "for any pair [i]who doesn't know each other[/i], there exist a third one who knows both of them", doesn't it? Otherwise it would be trivial.

oh yes, then the problem definitely becomes much more non-trivial(although the wording of the problem as posed does not suggest that in any way).
what one can see easily is that in such a graph(if it exists) every vertex has degree $8$ and also, for any $2$ adjacent vertices, there are no common neighbours and for any $2$ nonadjacent vertices they share a unique common neighbor.(in other words this has to be a strongly regular graph of degree $8$ with parameters $\lambda=0,\mu=1$).one way to see this is to note that such a graph actually has diameter $2$.then just count for any vertex,its neighbors and non-neighbors.
and such a graph does not exist by the so-called integrality condition.this integrality condition is used for instance in the proof of the friendship theorem-if in a group of people for every pair of persons there is a unique common friend, then all the people have a common friend.
also graphs with degree $k$ and $k^2+1$ vertices and diameter $2$ exist only if $k=2,3,7$(and possibly $57$-though this case is still an unsolved problem).these graphs are called Moore graphs.for $k=2$ it is simply $C_5$;for $k=3$ it is the Petersen graph and  for $k=7$ it is the Hoffman Singleton graph.all these are also unique such graphs.

Could you explain a bit more about the "integrality condition"?

ok the integrality condition is the following:
  if we have what is called a strongly regular graph with parameters $(v,k,\lambda,\mu)$(i.e a graph on $v$ vertices,regular with degree $k$, with the follwing two conditions:
   (a) for every two $\it{adjacent}$ vertices,$v,w$ the set $N(v)\cap N(w)$ has the same size $\lambda$.(here $N(x)$ denotes the set of neighboring vertices to $x$.)
   (b) for every two $\it{nonadjacent}$ vertices, $v\neq w$ the set $N(v)\cap N(w)$ has size $\mu$.

then the following holds(the integrality condition):
 \[\frac{1}{2}\left(v-1\pm\frac{(v-1)(\mu-\lambda)-2k}{\sqrt{(\mu-\lambda)^2+4(k-\mu)}}\right)\] are integers.
this holds because if you try to compute the dimensions of the eigenvalues of the adjacency matrix of such a graph these will the dimensions(it is difficult to explaim more-you can check the book 'An Introduction to graph theory' by D West for more details-check under strongly regular graphs.)
  here,$v=65,k=8,\lambda=0,\mu=1$ and so the denominator in the integrality expression is not an integer.

in general, i don't know of any simpler proof about the non-feasibility of such highly symmetric structures withou t using some such algebraic relations.

I dont know if this solution is valid so if anyone can check it ,it would be appreciated.Degree of every vertex is 8 ,there are no triangles and for two vertices that are not connected there is unique vertex connected to both ( mentioned in post above ).Now we count  number of closed walks (of lenght $7$) $v_0,v_1,v_2,\dots,v_7$ where $v_i$ and $v_{i+1}$ are connected and $ v_0=v_7$ . Let $f(7)$ be that number  . We first show that this number is divisible by 7.If $v_0,v_1,v_2,\dots,v_0$ is closed walk of lenght 7 so are  $v_1,v_2,v_3,\dots,v_1$ ; $v_2,v_3,v_4,\dots,v_2$  ; $\dots$ ; $v_6,v_1,v_2,\dots,v_6$ , so there are groups of $7$ and $f(7)$ is divisible by $7$.Fix vertex $v_0$ ,and we count number of closed walks with $v_0,v_1,v_2,\dots,v_0$. If $v_5=v_0$ then the number is $8f(5)$ and if not number is $8^5-f(5)-X$ where $X$ is the number of time when $v_5$ and $v_0$ are connected .Lets find $X$.Note that if$v_5$ is connected to $v_0$ then  $v_4$  is not neighbour of $v_0$(or else there would be triangle) We now look at $v_4$ . Number of walks in $v_4$ is connected to  $v_0$  is $Y$ and number of walks it is not is $Z$.Next $Z=8^4-Y$ ,and we need to find $Y$. $Y= 8*7^2$ because from $v_0$ t we have $8$ possibilities for $v_1$,from $v_1$ to $v_2$ there are $7$ possibilities  (if we  go back  then it is impossible for $v_0$ and $v_4$ to be connected),from $v_2$ to $v_3$ there are $7$ possibilities(again we cant go back) and from $v_3$ to $v_4$ there is unique vertex that is connected to $v_0$ and $v_3$.From that we have $Z=8^4-8*7^2$.Next observation is that we can go to $v_5$ , so that $v_5$ is  connected to $v_0$  if $v_4=v_0$.Call  number of walks like this $T$.Then $T=8*15$.The key is that $X=Z+T=8^4-8*7^2+8*15$ because there are two ways to get to $v_5$ , $v_4$ to be $v_0$ or not neigbour of $v_0$ and we covered them both  .So $f(7)=8f(5) + 8^5-f(5)-X=7f(5)+8^5-(8^4-8*7^2+8*15)\equiv   -1 \pmod{7}$ .Summing over all $v_i$ we get $-65\equiv 0\pmod {7}$ contradiction.(this took 3 hrs)


Hi all,

Is there any way to increase the font size any larger than with the \Huge command?

For example, if I hypothetically wanted

[code]^h_k\bar{V}^{(m)}_{{x}:\vbox{\hrule height .1pt
\kern 1.5pt
\hbox{$\scriptstyle{n}\kern 1pt$}}
\vrule width .1pt}
[/code]
to take up a full page (literally).

I also tried using the scalefnt package, which can get it a bit bigger, but its still a far cry from a full page.

In the preamble add
[code]\usepackage{type1cm}[/code]Then change the font size to however large you want it by for example:
[code]\fontsize{72}{86}\selectfont[/code]which makes the font size 72pt and the skip (distance from baseline to baseline) 86pt.
You can make these values as large as you like.


Hi,
I'm in 12th grade, and I'm wondering what math contests are available once I'm out of high school...this is my last year taking the AMC 12..

Thanks!


[quote="Temperal"][hide="hint"]
Let $ D$ be the point where $ k$ intersects $ AB$ and $ O$ be the centroid. $ \triangle ADO\cong \triangle BDK$ by AAS. Now we can use algebra bash and practically get the length of every segment with the properties of the centroid.[/hide]

That seems too easy... Maye I'm assuming something that I can't.[/quote]

Um, where is point $ K$? Also, we can't have ADO and BDK congruent unless AD=BD, i.e. D is the midpoint of AB. Clearly this isn't necessarily true.

Wait - darn it. I confused $ k$ and one of the medians, which is why I got messed up. Sorry. (point $ K$ was also suppoed to be point $ O$ in my post)

So does anyone have any ideas?  Or is there not enough information

[quote="firecricket91"]Line k is drawn through the centroid of triangle ABC.  Points B and C are on the opposite side of line k from point A.  The perpendicular distances from A to B are k are 10 and 6 respectively. Find the perpendicular distance frmo C to k.[/quote]
[hide="Solution"]
Let the feet of the perpindiculars from $ A$, $ B$, and $ C$ to $ k$ be $ X$, $ Y$, and $ Z$ respectively.  Also, let $ G$ be the centroid at let $ M$ be the midpoint of $ BC$.  Also, let $ W$ be the foot of the perpindicular from $ M$ to $ k$.  Hence, $ \triangle AGX\sim \triangle AGW\implies \frac {AG}{MG} \equal{} \frac {AX}{WM}$.  Since $ G$ is the centroid of $ \triangle ABC$, we have that $ \frac {AG}{MG} \equal{} 2$.  Thus, $ \frac {AX}{WM} \equal{} \frac {AG}{MG} \equal{} 2\implies AX \equal{} 2WM$.  Also, $ BYZC$ is a trapezoid with $ BY\parallel MW\parallel CZ$ and $ MB \equal{} MC$.  Hence, $ MW$ is a median of trapezoid $ BYZC$, so $ \frac {BY \plus{} CZ}{2} \equal{} MW\implies 10 \equal{} AX \equal{} 2MW \equal{} BY \plus{} CZ \equal{} 6 \plus{} CZ\implies CZ \equal{} \boxed{4}$. [/hide]


Let $ P$ be an internal point of triangle $ ABC$. The line through $ P$ parallel to $ AB$ meets $ BC$ at $ L$, the line through $ P$ parallel to $ BC$ meets $ CA$ at $ M$, and the line through $ P$ parallel to $ CA$ meets $ AB$ at $ N$.

Prove that $ \frac{BL}{LC}\times \frac{CM}{MA}\times \frac{AN}{NB}\leq \frac{1}{8}$

Denote $ AP\cap BC\equal{}D,PN\cap BC\equal{}L'$ then
$ \frac{LC}{BL}\cdot \frac{MA}{CM}\cdot \frac{NB}{AN}\equal{}\frac{LC}{BL}\cdot \frac{PA}{PD}\cdot \frac{L'B}{CL'}\equal{}\frac{LC}{BL}\cdot \frac{L'C}{L'D}\cdot \frac{L'B}{CL'}\equal{}\frac{LC}{BL}\cdot \frac{L'B}{L'D}$
$ \equal{}\frac{(b\plus{}c\plus{}d)(a\plus{}b\plus{}c)}{ac}\Big (BL\equal{}a,LD\equal{}b,DL'\equal{}c,L'C\equal{}d: ac\equal{}bd$ because $ \frac{a}{b}\equal{}\frac{AP}{PD}\equal{}\frac{d}{c}\Big)$
$ \geq \frac{(d\plus{}2\sqrt{bc})(a\plus{}2\sqrt{bc})}{\sqrt{abcd}}\geq\frac{2\sqrt{2d\sqrt{bc}}\times 2\sqrt{2a\sqrt{bc}}}{\sqrt{abcd}}\equal{}8$
Equality when $ a\equal{}2b\equal{}2c\equal{}d$, it mean $ P\equiv G$!

Again for the record, this is Question 5 BrMO 1 2008


integration
analytic geometry
vector
function
calculus
geometry

I prefer working problems from conservation of energy:

In this case, there is not friction, so there is not only conservation of energy, but also conservation of mechanical energy.

Mechanical energy of the system is a constant $ C$, that you can set to 0 by astutely defining U= 0 at a certain height of the block.

I'll consider a spring horizontally attached to a block.

Then, mechanical energy =
\[ (1) \frac {1}{2} mv^2 \plus{} \frac {1}{2}kx^2 \equal{} constant
\]
So as the block goes from x=0, to $ x \equal{} x_0$, the spring does $ \minus{} \frac {1}{2} k{x_0}^2$ work on the block, so the blocks KE becomes 0.  However, the kinetic energy lost by the block is used to strectch the spring.  In fact, the block "pushes" on the spring in the direction of motion, thus doing positive work on the spring.  The spring elastic energy thus increases by $ \frac {1}{2} k{x_0}^2$

You can solve it similarly using equation (1)

Your statement is incorrect.  I think it should be:

-Work done by an external body (whose energies are not included in the equation) on an internal body = PE gained by the [i]internal[/i] body + KE gained by the [i]internal[/i] body

By your last equation(if spring is attached to block horizontally)

        -(-0.5kx^2) = 0(PE gained by internal body)
                                      + 0-0.5mv^2(KE gained by internal body)

                Now the velocity comes out to be [b]imaginary.[/b]

       And please explain it to me when the spring is attached to the block

vertically(which was my initial case).

First of all you have to be aware of your reference system. The actual formula, in horizontal motion of a spring comes this way:

Let's consider a block of mas $ m$ at his rest lenght in some position that we will consider, from our reference system, to be $ x \equal{} 0$. Hooke's law says that the force on the body by the spring is proportional to the distance to the natural lenght, that is:
\[ m\frac {d^2x}{dt^2} \equal{} \minus{} kx
\]
In other words, if you move the spring in a positive direction the force is negative: It will counter-act, trying to recoil to his original position (in the negative direction). The work done in a little displacement $ dx$ is:
\[ dW \equal{} m\frac {d^2x}{dt^2} dx \equal{} \minus{} kx dx
\]
So now we integrate to obtain the total work:
\[ \int^{b}_{a}dW \equal{} W_{b} \minus{} W_{a} \equal{} \int^{b}_{a}m\frac {d^2x}{dt^2} dx \equal{} \int^{b}_{a} \minus{} kx dx
\]
We can write $ \frac {d^2x}{dt^2} dx \equal{} \frac {dv}{dt}dx \equal{} v dv$ so:
\[ \Delta W \equal{} m\int^{b}_{a}v dv \equal{} \int^{b}_{a} \minus{} kx dx
\]
And this way we get:
\[ \Delta W \equal{} \frac {1}{2}mv_{b}^2 \minus{} \frac {1}{2}mv_{a}^2 \equal{} \minus{} \frac {1}{2}kx_{b}^2 \plus{} \frac {1}{2}kx_{a}^2 \equal{} \minus{} \left(\frac {1}{2}kx_{b}^2 \minus{} \frac {1}{2}kx_{a}^2 \right) \equal{} \minus{} \Delta U
\]
Where we considered the potential energy of a spring to be defined as:
\[ U(x) \equal{} \frac {1}{2}kx^2
\]
So now you see that, if you get your coordinate system right, you get your solution in the right way (in other words if you know what's ACTUALLY happening, as you defined your coordinate system, you can know what your solution [b]means[/b]). Either if you consider the potential energy to be positive or negative, that's all up to you, depending on your coordinate system. The soultion also applies to vertical motion (just consider $ x \equal{} z$, or $ x \equal{} y$ depending on how you call your vertical axis).

How can we take final PE of the spring to be negative, when after being compressed or elongated, energy is being stored in it?

It's all up to the coordinate system you use. You can make the final PE negative if you like, but as long as you KNOW it's energy stored on it you know what your solution means.

Our KNOWING or NOT KNOWING will not prove a mathematical equation correct.
You have to show it mathematically.
The + or - sign matters a lot, because of that one side of a whole equation can change and what is actuallly right we may be proving wrong and vice versa.

[b]Please give a more satisfactory mathematical explanation of what you are trying to say[/b]

Yugos, I think what you're asking is related to this:

As long as you state conservation of energy correctly, it should go like this:

$ PE_i \plus{} KE_i \plus{} Work\equal{} PE_f \plus{} KE_f  (1)$

Let's consider a 1-dimensional motion (it can easily be extended to 3-D).  A block of mass m is moving in a force field (I assume you know what I mean).  By force field I mean that for every  x coordinate, I can assign a force vector from the field, F that acts in either directions (F has positive-negative directions or signs, however you want to put it).

THen you need to construct your PE formula for that force field.  How to do it?

Well pick a reference point A, at the coordinate x = a.

From the formula
\[ F \equal{} \minus{} \frac {dU}{dx}
\]
Where $ U$ is the potential energy, it follows
\[ U(x) \equal{} \minus{} \int_{a}^{x} F \cdot dx (2)
\]
What Yugos said was that the point a you chose is arbitrary.  Obviously, at the point a, the potential energy is 0.  But all that matters to you are changes in potential energy.  It will be seen when we plug in the potential energy function you have in (2), where PE (x) = U(x).

Once you have carefully followed the guidelines outlined, you can plug in your newly made nice function PE (x) into the formula in (1), and that formula will be correct if you compute your work done by an external object correctly.

How to compute it?

1. Know which forces are called "external"  When you defined PE, you took in account forces that from the spring, from a gravitational field, whatever...  But maybe, you forgot that your hand also acts upon the block. Let's say your hand, acts like another field, with a force F'(x), which you didn't compute into your PE equation.  Then, sth is missing in your force field, and thus you have to account for that loophole in the work external.

2. The work external is an integral of the force external with respect to x.  It is positive when F and x are in the same direction, negative when F/x are opposite.

Now, let's transform our nasty energy conservation into YOUR equation. Suppose that KE at the times you want to study are both 0. Then the equation becomes:
\[ PE_f \minus{} PE_i \equal{} Work external
\]
or
\[ Work external \equal{} \Delta PE of that body
\]
Work through the steps I outlined with the force field F = -kx, for a block at x = 0, and see what happens (basically what Qboy did... I don't see anything missing from his explanation)

PS: As far as I'm conserved, the construction of a potential energy function PE (x) is meaningful only when the force field that causes those changes in potential energy at different x coordinates are conservative fields meaning, if you start at x=a, when a PE (a) = A and 0 KE, and you end at point x = a, your PE (a) = A, you don't have extra energy to play with, and move around (meaning KE = 0 again).

[quote="yugos"]We say that
Work done by an external force on a body = -(P.E. of that body).

Consider this case :-

When a block is suspended from a spring (spring constant 'K'),
the spring does a work -(1/2)K(x^2) on the block(Natural to Compressed/Elongated form), therefore the block(external force) does a work (1/2)K(x^2) on the spring, but the spring has a P.E. of (1/2)K(x^2) finally.

I cannot understand how this case explains this above statement.

[b]Please rectify me where I am wrong.[/b][/quote]

The basic definition you stated is wrong . Work done by external force is equal to change in PE. The work done by internal forces = -PE of that body

[quote="integration"]The thing about internal/external is a question of semantics.  Of course, energy is conserved in the universe, but since you are not omnipotent, you select what is considered your system, which may be not isolated, and thus have to consider what is outside "external"  Thus, external/internal labels is based on what is convenient, and you have to keep consistently with that label throughout the problem[/quote]

That's exactly what I meant.

What I gave to you, yugos, was exactly a [b]satisfactory mathematical explanation
[/b] of what i'm trying to say: As long as you know what you are doing mathematically, signs will come out alone. It's you who will have to interpret them.

Recall that in the re-derivation that I did on my first post, I defined my function $ U(x)$ to be always positive. This way, the signs only come out depending on my reference system: How i plugged the forces in my coordinate system.

yugos, please read the comment that "integration" and I discussed just a post before yours.

Conjurer was wrong. yugos, your original statement about work was correct.

Conventionally, we like to define potential as higher potential = higher energy. Thus, objects will, of themselves, go towards an area of low potential. This means a force must be pushing in the opposite direction of the gradient of the potential, implying $ F \equal{} \minus{} \frac {dU}{dx}$. Integrating dx to get work, we find that $ \int F dx \equal{} \minus{} U$.

[quote="yugos"]Conjurer, if my statement is wrong, then explain to me this :-

Consider ground level as 0 PE level and y-axis upwards.
When you run up a flight of stairs to a height 'h', change in PE is mgh-0=mgh.
Here we are doing work against gravity(external force), so work done by gravity=-mg(h-0)=-mgh.

So here it comes out that work done by external forces (neglecting friction) =
-mgh[/quote]

Thats because gravity is internal. Your system is man + earth  and the gravitational force between them is hence internal. ( Remember , we need atleast 2 bodies to actually define potential energy).

I have checked my book too.

Thank you Conjurer, I have now understood that I had been making mistake in considering the correct system.


If $ x \equal{} \sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} \cdots}}}}$, then:

$ \textbf{(A)}\ x \equal{} 1 \qquad\textbf{(B)}\ 0 < x < 1 \qquad\textbf{(C)}\ 1 < x < 2 \qquad\textbf{(D)}\ x\text{ is infinite}$
$ \textbf{(E)}\ x > 2 \text{ but finite}$

[quote="Brut3Forc3"]If $ x \equal{} \sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} \cdots}}}}$, then:
$ \text{(A)}\, x \equal{} 1 \qquad\text{(B)}\, 0 < x < 1 \qquad\text{(C)}\, 1 < x < 2 \qquad\text{(D)}\, x\text{ is infinite} \\
\text{(E)}\, x > 2 \text{ but finite}$[/quote]
x is the positive root of $ x^2\minus{}x\minus{}1\equal{}0$
So That's C

[quote="Brut3Forc3"]If $ x = \sqrt {1 + \sqrt {1 + \sqrt {1 + \sqrt {1 + \cdots}}}}$, then:
$ \text{(A)}\, x = 1 \qquad\text{(B)}\, 0 < x < 1 \qquad\text{(C)}\, 1 < x < 2 \qquad\text{(D)}\, x\text{ is infinite} \\
\text{(E)}\, x > 2 \text{ but finite}$[/quote]
To clarify what the above poster said, rewrite the radical as $ \sqrt{1+x}$. Then:

$ x=\sqrt{1+x}$
$ x^2=1+x$
$ x^2-x-1=0$
${ x=\frac{1 \pm \sqrt{1+4}}{2}=\frac{1}{2} \pm \frac{\sqrt{5}}{2}}$
$ x$ has to be positive (it is greater than $ \sqrt{1}$) and so ${ x=\frac{1}{2} + \frac{\sqrt{5}}{2}}$. Since $ 2 < \sqrt{5} < 3$, we have ${ 1 < \frac{\sqrt{5}}{2}} < 1.5$ and thus ${ 1.5 < \frac{1}{2} + \frac{\sqrt{5}}{2}} < 2$, which corresponds to C.


geometry
trapezoid
parallelogram
Pythagorean Theorem

In trapezoid $ABCD$ with bases $\overline {AB}$ and $\overline {CD}$, we have $AB=52$, $BC=12$, $CD=39$, and $DA=5$. The area of $ABCD$ is \[ \text {(A) } 182\qquad \text {(B) } 195\qquad \text {(C) } 210\qquad \text {(D) } 234\qquad \text {(E) } 260. \]

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=505[/img]

[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=506[/img]

Draw the line through $D$ parallel to $\overline {BC}$. It divides the base $\overline{AB}$ into two segments with length 13 and 39, and it divides trapezoid $ABCD$ into a right triangle with sides 5, 12, and 13 and a parallelogram. The area of the trapezoid is the area of the right triangle, $(1/2)(5)(12)=30$, plus the area of the parallelogram. Let the altitude from $D$ to $AB$ divide the hypotenuse of the right triangle into two segments of length $x$ and $13-x$, so that $5^2-x^2=12^2-(13-x)^2$ by the Pythagorean Theorem, so $x=25/13$, so the altitude from $D$ to $AB$ has length $\sqrt{5^2-(25/13)^2}=60/13$. Finally, the area of the parallelogram is $39(60/13)=180$, so the area of the trapezoid is $30+180=\boxed {210}$, which is answer choice C.
[/hide]

chess64, you can also find the altitude simply by expressing the area of the triangle in two ways: $\frac{1}{2}\cdot 5\cdot 12=\frac{1}{2}\cdot 13\cdot h$, so $h=\frac{60}{13}$ ;)

[hide="Yet another solution"]
Extend $AD$ and $BC$ to intersect at $E$.
$\triangle AEB\sim\triangle DEC$ and $\AB\parallel DC$, so $DE: AE=CE: BE=DC: AB$.
So $AE=20$ and $BE=48$.
Since $\triangle AEB$ is a 5-12-13 triangle, $\angle AEB=90^\circ$.
Therefore, the area of $\triangle AEB$ is $\frac{1}{2}\cdot 20\cdot 48=480$.
The area of $\triangle DEC$ is $\frac{DC^2}{AB^2}(\triangle AEB)=\frac{9}{16}\cdot 480$.
Hence the are of the trapezoid is $\left(1-\frac{9}{16}\right)480=\boxed{210}$
[/hide]

[quote="frt"]chess64, you can also find the altitude simply by expressing the area of the triangle in two ways: $\frac{1}{2}\cdot 5\cdot 12=\frac{1}{2}\cdot 13\cdot h$, so $h=\frac{60}{13}$ ;) [/quote]

Yeah, you're right. I can't believe I didn't see that. :wallbash: Also, once I found the height, I could have just multiplied it by 59. :roll:


In triangle $ ABC$ you are given $ AB \equal{} 2 \times AC$. $ D$ lies on $ AB$ and $ E$ lies on $ BC$ so that angle $ BAE$ equals angle $ ACD$. $ AE$ and $ CD$ intersect at $ F$. Find the angles of triangle $ ABC$ if triangle $ CFE$ is equilateral.

[hide]CFE is equilateral, so <CFE=60 degrees. <AFC is a supplement of it, so <AFC=180-60=120. Let <ACF=<DAF=x and <CAF=y. We have x+y+120=180, since they are all in triangle ACF, and it follows that x+y=60. However, we also know that <CAB=x+y; thus <CAB=60. Now, we draw a segment from C to the midpoint of AB, call it X. AX=BX, since we are given that AB=2AC. Extending this, AX=BX=AC. We know that <CAB=<CAX=60, already. This means that triangle CAX must be equilateral. From this follows <BXC=180-60=120, and <ACX=60. We know that <ACB=x+60 (because CFE is equilateral), so <BCX=x. Since CAX is equilateral, CX=BX, and triangle CXB is isosceles. Thus, 120+x+x=180->x=30, and <ACB=x+60=90. Thus, the angles of the triangle are $ \boxed{30,60,90}$.[/hide]