MATHCOUNTS
number theory
least common multiple
algebra
polynomial
AoPSwiki
partial fractions

Was this from last years chapter countdown round?

Not sure... as I saw this in a local math competition (in Singapore).

[quote="tiddlewink0"]Find the value of : 1 - (5/6) + (7/12) - (9/20) + (11/30) -(13/42) + (15/56) - (17/72) + (19/90)

I assume there is a simple way of doing this which I am missing (rather than the brute force computation).[/quote]

[hide="Actually very simple (click here)"]Note that we can re-write this as:

$ 1-\frac{2+3}{2\cdot 3}+\frac{3+4}{3\cdot 4}-\frac{4+5}{4\cdot 5}+...+\frac{9+10}{9\cdot 10}$

Notice that:
$ \frac{x+(x+1)}{x(x+1)}= \frac{1}{x}+\frac{1}{x+1}$, so we may again re-write this as:

$ 1-1/2-1/3+1/3+1/4-1/4-1/5+1/5+1/6+...-1/8-1/9+1/9+1/10$

Canceling like crazy,
$ 1-1/2+1/10 = \boxed{3/5}$[/hide]

[quote="kimmystar94"]wow..
u ppl r amazing..
seriously..
i wouldn't ever find that type of pattern...
how will I ever make it to nationals or even state???  :([/quote]

Well, to notice the pattern, the numerators were odd numbers, so that gives it away...

Every odd number can be expressed as the sum of two consecutive integers.

P.S. I never made nationals.

Partial Fractions is a [u]very[/u] powerful method when it comes to evaluating sums of seemingly complicated fractional expressions. It is even applicable in calculus. As a problem solver, it would be helpful to keep it under your problem solving belt (or your 1337 bag)  :D

Aha, so this is the famous partial fraction decomposition.  That's pretty cool.  I think I remember seeing it in AoPS II
kimmystar94, I don't think any MATHCOUNTS question would require partial fraction decompsition.

I guess I did mine the longer way... I think I got everything...
[hide]
$ So,$ $ the$ $ question$ $ is$ $ :$
$ 1-(\frac{5}{6})+(\frac{7}{12})-(\frac{9}{20})+(\frac{11}{30})-(\frac{13}{42})+(\frac{15}{56})-(\frac{17}{72})+(\frac{19}{90})$

I took the LCM for the denominators which was 5040 (all them went into it). Then multiplied the numerators by the number which was multiplied to the original denominators to get 5040.

So:   $ (\frac{5040-4200+2940-2268+1848-1560+1350-1190+1064}{5040})$

And last I simplified it down to $ \frac{3}{5}$

* Lots of simplifying!  :)

[/hide]

ok..
i'll keep this in mind..
(i'll probably forget it eventually... i'm sorry!!)
but thanks..
another useful tip in the mathcounts competition... ^-^

Is there more to the partial fraction thingy or is this basically a complete demonstration of it? (btw, it's not in Aops Wiki) What's the difference between this and the telescoping series thing?

erdokangerem, there's a nice bit on it in AoPS vol. 2.  You can expand it to variables and polynomials and cool stuff like that.

uh, I have no idea how that relates to any of this, and you shouldn't be reviving threads with something that useless, but since its back up anyways...

[quote="archimedes1"]there's a nice bit in AoPS vol. 2.[/quote] I don't have AoPS vol.2 (though I should probably get it), so can you demonstrate some of these properties?


real analysis
function
integration
real analysis unsolved

If we are given a function $F(x)$ on $\mathbb{R}$, and we assume that $f(x)$ is continuous on $\mathbb{R}$ with

$F(x) = \int_\mathbb{R}f(y)f(x+y) dy$

then what can we say about $f(x)$?

I am aware this is a pretty vague question -- I suppose one could start by assuming some conditions on $F(x)$ and see what can be deduced about $f$. I'd be interested in whatever people have to say about this.

Assuming that $f\in L^{2}$, we have $F(x)=\int_{\mathbb R}\widehat f(\xi)\widehat f(-\xi)e^{2\pi i\xi x}\,d\xi$, so, basically, we can recover the values of $\widehat f(\xi)\widehat f(-\xi)$ from this information and not much more unless we make some extra assumptions about $f$.


calculus derivative function algorithm algebra polynomial trigonometry

So I have a set of points, we'll just use some from x^3 as an example:

0,0
1,1
3,27
6,216

The actual points I am using may not necessarily be perfectly fitting onto any given function.

Anyways, I need a simple algorithm to determine/estimate the second derivative at both of the end points, which would, of course, be 6x = 0 and 36. Any ideas on how to do this?

Do a polynomial interpolation (or polynomial fit, whichever is appropriate), and then calculate the second derivative.

In this case, a polynomial interpolation is appropriate, and you can use finite differences find that $ y\equal{}x^3$ is a good model for the numbers.

Well see, this is actually for a polynomial interpolation program I am building. For this, you must set the second derivatives at the end point and I was looking for a good way to estimate this.

I'll say this: polynomial interpolation is a terrible idea. It introduces way too much oscillation, and is especially unstable at the outer edges. Derivatives make all of the flaws worse.

[hide="A miscalculated example"]: Consider $ f(x) \equal{} \sin x$ on $ [ \minus{} \pi,\pi]$, estimated with an interpolating polynomial $ p$ that matches it at each multiple of $ \frac {\pi}{6}$. We have $ f'(\pi) \equal{} \minus{} 1$ and $ f''(\pi) \equal{} 0$, with both derivatives never exceeding $ 1$. Calculating in a spreadsheet, I get that $ p'(\pi)\approx 1.08$ and $ p''(\pi)\approx 7.68$.[/hide]

The bottom line: Find a method better suited to the form of your data, which doesn't require estimating derivatives as an intermediate step.

While I completely agree with the first and the third paragraph jmerry wrote, the claims he made in the second paragraph are unbelievable. Since $ f'\minus{}p'$ has at least one root $ x_k$ on each interval $ [\pi\minus{}\tfrac{\pi k}6,\pi\minus{}\tfrac{\pi(k\minus{}1)}6]$ ($ k\equal{}1,\dots,12$), we have $ |f'(\pi)\minus{}p'(\pi)|\equal{}\frac {|f^{(13)}(\xi)|}{12!}\prod_{k\equal{}1}^{12}|\pi\minus{}x_k|\le \left(\frac \pi 6\right)^{12}\approx 0.000425$ and $ |f''(\pi)\minus{}p''(\pi)|$ can be just about 20 times bigger. So the values $ 1.08$ for $ p'(\pi)$ and $ 7.68$ for $ p''(\pi)$ reflect nothing but presence of a programming error :(.

I found the bugs- I was interpolating the wrong values, and then dividing by $ \frac {\pi}{6}$ instead of dividing when I rescaled. The corrected values are $ \minus{} 1 \minus{} 2.8\cdot 10^{ \minus{} 5}$ for $ p'$ and $ \minus{} 3.3\cdot 10^{ \minus{} 4}$ for $ p''$.

That doesn't really illustrate the point, so I would have chosen a different example if I'd spotted this first.

So, what were those original values from? I forgot a division by $ k$ at the $ \pm\frac{k\pi}{6}$ points, making my function $ f(x)\equal{}\frac{\pi}{6 |x|}\sin x$ (and zero at the origin). That gives $ f'(\pi)\equal{}\minus{}\frac16$ and $ f''(\pi)\equal{}\frac1{3\pi}$, while $ p(\pi)\approx 3.94$ and $ p'(\pi)\approx 102$.

Discontinuities [i]really[/i] break polynomial interpolation.

The classical one is $ f(x)\equal{}\frac 1{x^2\plus{}1}$ on a sufficiently long interval (if I remember it right $ [\minus{}5,5]$ is more than enough). Then the $ L^\infty$ norms of the interpolating polynomials with the equidistant nodes grow exponentially, which makes me think that their derivatives at the endpoints behave no better.

My sheet is set up for odd functions and for 13 points, so I'll tweak that. New $ f$: $ \frac{15x}{x^2\plus{}9}$ on $ [\minus{}6,6]$, interpolated at integers. We have $ f'(6)\equal{}\minus{}\frac13$ and $ f''(6)\equal{}\frac{4}{225}$.
The interpolating polynomial $ p$ has $ p(6)\equal{}2$ (as it should be), $ p'(6)\approx \minus{}1.98$, and $ p''(6)\approx \minus{}10.1$.

Roundoff error could theoretically be a problem, but the terms that mostly cancel down to the above numbers are merely in the hundreds.


What's goin on with the array environments?

\[\begin{bmatrix}
a & b & c \\
d & e & f
\end{bmatrix}\]

That was mentioned in one of the other forums.  Valentin said he would get around to it.  Here's the link to that post: [url]http://artofproblemsolving.com/Forum/viewtopic.php?t=20450&postorder=asc[/url]


quadratics MATHCOUNTS geometry HMMT algebra polynomial function

[i]You can skip this paragraph...[/i]
Recently my math team went to a regional competition and we saw people solving mundane problems quickly through the graphing calculators. I think that "mundane problems" are self-explanatory.  :lol:

We are preparing for the state final and we needed help.

[b]If you have any kind of calculator tips and tricks, please share![/b]

Please include a short description of what it is and what it does.  For example,
it's a program that solves quadratic equations so I just type in a, b, c and presto,
it tells me the roots.  Or, maybe it's a lesser known functionality you discovered
that is really useful.

I personally am against calculators but I guess it's better to know... :roll:

Isn't it against the rules to use pre-made programs in mathcounts?

[quote="b-flat"]You are not supposed to need a calculator program to solve any problem in mathcounts.  Here is one that gives the prime factorization of a number.  I will post one that simplifies radicals later.

[hide="Prime Factorization"]:1$\rightarrow$ Y
: “Input Number? “,A
:A$\rightarrow$ B
:1 $\rightarrow$dim(LAB
:End
:2 $\rightarrow$X
:Goto 3
:Lbl 5
:1+X$\rightarrow$ X
:If A=1:Goto 2
:If B/A>A:Goto 2
:Lbl 3
:Lbl 6
:If X=1: Goto 2
:If fPart(A/X)=0
:Then
:If dim(LAB)>1:1+dim(LAB)$\rightarrow$dim(LAB)
:X $\rightarrow$LAB(Y)
:Y+1$\rightarrow$ Y
:A/X$\rightarrow$ A
:If fPart(A/X)≠ 0:Goto 5
:Goto 6
:End
:Goto 5
:Lbl 2
:Output(1,1, LAB
:Delvar LAB
[/hide][/quote]

You don't need any fancy programs to do prime factorization with a TI-89  :P

you can use any calculator in mathcounts, as long as it doesn't have a qwerty keyboard or isn't a cellphone, palm pilot, etc. It's not like people check each calculator to see what programs you have.

Like Frank said, the TI-89 has almost all the features anyone in undergrad or less could possibly want from a calculator, pre-built. It can solve equations, differentiate, take limits, handle matrices, factor, expand, approximate, find zeroes, do summations, find local max/min... anything. And it's not that much more expensive than a regular TI-83. If the competition allows you to use it, I definitely reccomend the investment for a TI-89 Titanium calculator.

[quote="k8reindeer"]something that I find useful is the prime factorization of the current year. this is easy to program or store some other way and comes up every so often, enough that it pays to know. 2006=2*17*59[/quote]
Here are the prime factorizations of the next few years (for the future use)... :D
$2007=3^2\cdot 223$
$2008=2^3\cdot 251$
$2009=7^2\cdot 41$
$2010=2\cdot 3\cdot 5\cdot 67$
$2011$ is a prime!

Anyway, since TI-89 is extremely powerful, some competitions don't allow the use of TI-89. (I know some contests that allow TI-83s but not 89s.)
So, it is still good to have TI-83. I personally like TI-83 because it's easy to handle :)

[quote="frt"]
Anyway, since TI-89 is extremely powerful, some competitions don't allow the use of TI-89. (I know some contests that allow TI-83s but not 89s.)
So, it is still good to have TI-83. I personally like TI-83 because it's easy to handle :)[/quote]

Yeah. If you're considering buying a graphing calculator and are serious about math, then definately spend the extra 20 on getting the TI 89. It might be barred in some Algebra II classes (it is in mine and rightfully so) but some don't. Not that geometry was hard, but I used it all the time in geometry since it put all numbers into simplest radical form, while everyone else had to do it by hand.

It's also good to keep a TI 83 or 84 or 86 for those competitions that don't allow the 89. You can get one pretty cheap off Ebay.

Back to the 89T. Basically, it can save tons of time when doing homework, and in math contests, because it can solve equations for you and stuff like that (factor, expand, etc.). In the HMMT gutz round, we had 2 89s among 5 people and basically when a calculator was needed, this was the only one we used to save time, because it could solve those systems that no one wants to do by hand ("mundane problems", as you call it). People BS some hard questions on the AMC 12s with 89s, and solve them in like a minute.

One thing though. Don't go into a competition for the first time having never used the 89. It's utter suicide, because it's a bit hard to use at first (I know a kid who had an 89 but couldn't get it to give rounded answers).

Anyone have a site with all the useful tricks of the 89? I know the basics, like differentiate, integrate, solve for variables, factor a polynomial, but I don't know more of the "hidden" ones like prime factorization. How do you prime factorize on an 89 anyway?

Did your calculator not come with a user's manual?

I have a ti-89 and the user manual didn't explain anything to me!!!!!
It was just a bunch of previews.

Quadratic Formula Program

:Prompt A,B,C
:DISP:(-B+sqrt(B^2-4AC))/(2A)
:DISP:(-B-sqrt(B^2-4AC))/(2A)

That is very simple. I have rarely used programs in my 3 years in MATHCOUNTS Competitons. But I usually create some during the competiton when needed.

Come on!!! Be honest!!!! the best mathematicians do not need that tricks!!!!!!!!

and since we aren't the best mathmaticians in the world, we do need tricks

[quote="boxedexe"]Mathematics is nto abtou your current mathematical knowledge.  If you truly like mathematics, you woudl not say that. Math is all about the beauty of the human mind, not a calculator.

Masoud Zargar[/quote]

I think the same...

[quote="dts"][quote="Watson"]$factor(n)$ factors a number.
$isPrime(n)$ returns true or false if the number is prime or not.
$rref(matrix)$ returns a reduced row echolon form of the matrix. It is used for solving systems
of liner equations.

These three functions are very powerful, and will save vast amounts of time.[/quote]

You can actually use the "solve" function to solve systems of linear equations...[/quote]

But which is easier/faster to type?

solve(3a+4b+c-19d=12 and b/2+4c+d-a=0 and 8b-9/2c+3a-d=-5 and a-b/15+c+10d=12,{a,b,c,d})

vs.

[3,4,1,-19,12;-1,1/2,4,1,0;3,8,-9/2,-1,-5;1,-1/15,1,10,12]     <enter>
rref(ans(1))

On an 89, numbers are much easier to type than letters (except x,y,z,t), so the second will probably be quicker and easier than the first.  Also, mistakes are easier to find when you don't have to scroll through the history entry.


This is a great read for a mathematician, about the math jokes in the Simpsons.  I suggest you read it, it's pretty...informative:

http://100cia.com/noticias/index.php?subaction=showfull&id=1149928198&archive=&start_from=&ucat=12

BTW:  THe title is in Spanish, but the article is in perfect English.

[quote="Xantos C. Guin"]I saw that episode and tried it on my calculator. It worked, atleast I thought it did.[/quote]

Hmm...what do you mean that it worked?  You actually thought of Fermat's Last?  Or it worked as in it added up correctly?

im never gonna watch the simpsons the same way again. :P

I think it's the one where he falls through the back of a book case into a 3-d world.

Yea, it's just a calculator roundoff error.

Did anyone read to the end of the article, where they mentioned those three numbers? I know the first is the sum of the first 13 non-negative powers of 2, and the second is a perfect number. Dunno about the third.

is there a site that gives special properties of many numbers?

Hmm, I typed $1922^{12}-1841^{12}-1782^{12}$ into my calculator and got $7 \text{E} 29$.  :D  But, yeah, if you evaluate each side separately, you get that they are equal.

Btw, don't look [url=http://www.mathsci.appstate.edu/~sjg/simpsonsmath/baseball.html]here[/url] unless you want to see what's interesting about those numbers :P

[quote="Elemennop"]I think it's the one where he falls through the back of a book case into a 3-d world.[/quote]
Unless I recall wrongly, the episode is called $\text{Homer}^3$.

Picture:
[img]http://www.mathsci.appstate.edu/~sjg/simpsonsmath/3dhomerandfermat.jpg[/img]

Hey look at Danbert's picture: P=NP :lol:
Anyone figure what's interesting among these numbers: 8,191, 8,128, or 8,208?

8128 is perfect :lol: , 8191 is a mersenne prime and a factor of the perfect number 33550336..... don't know about 8208.

Huh.  I already posted a link that tells you what is special about those three numbers.  (see my previous post)

[hide="edit"]I LOSE!!![/hide]

A series of good Futurama/Simpsons math related interviews can be found [url=http://www.cs.appstate.edu/~sjg/futurama/]here[/url] (it's mentioned in the article as well). The [url=http://www.cs.appstate.edu/~sjg/futurama/dxcinterview.html]interview with David Cohen[/url] is particularly good. It includes a link to the [url=http://www.cs.appstate.edu/~sjg/futurama/nearmiss.html]source[/url] of his Fermat near-miss finding program.

[quote="10000th User"][quote="paladin8"]Haha. My calculator is stupid. It thinks Fermat's Last Theorem is false. I win.[/quote]:rotfl: Look at my signature :D[/quote]

Sweet! :D


Hi,
I need some help.  :( Can't seem to get the correct answer.

Solve the formula for the variable mentioned.

l.  H=3V / llr^2  (for V)

2. E=1/2 mv^2 (for m)

I believe what is being asked here will not produce a result but will rearrange the formula so that the formula can be used to solve for the variable mentioned.

If you could please show the steps that would be great.   :)

Thanks.

Trish

Is that an 11 in #1? If so,

$H=\frac{3V}{11r^2}$. We want to remove everything on the right side except for the V. Multiply by 11 to remove the 11 in the denominator.
$11H=\frac{3V}{r^2}$. Multiply by $r^2$ for the same reason.
$11Hr^2=3V$. Divide by 3
$\frac{11}{3}Hr^2=V$. And there you go.

Maybe you want to try the second one on your own first?

are they physics formulas??
The second one looks like the theorem of kinetic energy with initial velocity $v_0=0$.

[quote="Trisha"]Hi again,
Thanks for your help.  There isn't a (1) in the formula it should be (Pi) 11r^2. Will this change to outcome?  I don't know how to make those special characters on the computer.  :(     Answer book has this:

Answer:  H IIr^2 / 3 =V

Thanks

Trish[/quote]

What TheBB put for number one is equivalent to what your textbook has. For number two:
[hide]m=2E/(v^2)[/hide]

For the second equation $E=\frac{1}{2}mv^2$ we have
$2E=mv^2$
$m=\frac{2E}{v^2}$.

Thanks Jimmy,
I know the answer it's in back of the book.  What I really need are the steps but I figured it out.  I have another problem, please give the step taken to solve it.  Thanks again.

Problem:  a=180-360/n (solve for n)

answer:  -360/a-180

Step 1: Subtract 180 from both sides  (a-180 =-360/n)

Step 2: Multiply to get rid of "n" this is where I have a  problem[color=darkblue][/color]

ok
your problem is:$a=180-\frac{360}{n}$
You already got: $a-180=-\frac{360}{n}$
Then multiply both side with $n$, we get:
$n(a-180)=-360$
Dividing both side with $(a-180)$, we shall obtain:
$n=\frac{-360}{a-180}$.


geometry
circumcircle
Euler
search
cyclic quadrilateral
projective geometry
geometry proposed

Given triangle $ ABC$. Let $ A_1,B_1,C_1$ be the midpoints of $ BC, CA, AB, A_2B_2C_2$ be the pedal triangle of triangle $ ABC. A_1B_1\cap A_2B_2\equal{}\{Z\}, B_1C_1\cap B_2C_2\equal{}\{X\}, A_1C_1\cap A_2C_2\equal{}\{Y\}$, let $ O$ be the center of $ (A_2B_2C_2)$. Prove that $ O$ is the orthocenter of triangle $ XYZ$.

the collinearity of $ Y,Z,A$ is well-known .similarly we have this kind of collinearity for $ B,C$.
also we have the collinearity :
$ (C_1A_2,C_2A_1),B,X,Z$ which can be proven by many ways(e.g. menelaus)...
it implies that the polar of $ Y$ wrt $ (O)$ is the line $ BZ$ which implies $ OY \perp BZ$
which implies $ OY \perp XZ$ because of the collinearity mentioned.
do the same thing for the other sides and the result follows.

Hi,shoki,the more general case of this problem is: replacing "the pedal triangle of $ \Delta ABC$" with "the pedal triangle of any inner point $ P$ with respect to $ \Delta ABC$"

P.S.livetolove212 may mean the same thing,and I apologize if repeating the post

Dear [b]Hxy[/b],
You are same as me. Note that Fontene's theorem is expressed in general case.

HI,my friend,could you please post your nice solution for this tough masterpiece?Thank you so much!!! :lol:
The solution I knew was quite strange and unbelievable :P

P.S.I found this problem in the homework of 2007 China IMO Team~

Dear [b]Hxy09[/b],
The [b]Fontene[/b]'s theorem said that $ A_2X, B_2Y, C_2Z$ concur at the intersection of 9-point circle of triangle $ ABC$ and Pedal circle of triangle $ A_2B_2C_2$. Then my problem is only a corollary if we use [b]Brocard[/b]'s theorem for cyclic quadrilateral:
Given a cyclic quadrilateral  $ ABCD$ and its circumcircle $ (O). AC\cap BD\equal{}\{X\}$, $ AB\cap CD\equal{}\{Y\}, AD\cap BC\equal{}\{Z\}$ then $ O$ is the orthocenter of triangle $ XYZ$.

It can also be proved that $ AX,BY,CZ$ are perpendicular to the Euler Line of triangle $ ABC$.
And the center of $ A_2B_2C_2$ is the nine point center of $ ABC$ and lie on the EUler Line of triangle $ ABC$. By the collinearity that Shoki mentioned, we can get that $ N$ is the orthocenter of $ XYZ$.

The first part that $ CZ$, etc are perpendicular to the Euler Line can be proved using Pascal's Theorem and the result on Poles and Polars previously mentioned in of the posts. Pascal's for proving that $ A_2B_1$ and $ A_1B_2$ are concurrent with the Euler Line and  the Poles Polars result for the circle considered (nine point circle).

[quote="livetolove212"]Dear [b]Hxy09[/b],
The [b]Fontene[/b]'s theorem said that $ A_2X, B_2Y, C_2Z$ concur at the intersection of 9-point circle of triangle $ ABC$ and Pedal circle of triangle $ A_2B_2C_2$. Then my problem is only a corollary if we use [b]Brocard[/b]'s theorem for cyclic quadrilateral:
Given a cyclic quadrilateral  $ ABCD$ and its circumcircle $ (O). AC\cap BD \equal{} \{X\}$, $ AB\cap CD \equal{} \{Y\}, AD\cap BC \equal{} \{Z\}$ then $ O$ is the orthocenter of triangle $ XYZ$.[/quote]

Wow,the solution I knew also applied Fontene's theorem,but I just didn't know its name~
Could you please post the proof for Fontene's theorem,thank you so much :)

P.S the proof for Fontene's theorem I knew is quite complicated and unthinkable :(

[quote="livetolove212"]As I have known, the solution of Fontene's theorem was posted by Yetti on Mathlinks:
http://www.mathlinks.ro/viewtopic.php?search_id=683193611&t=173756
It's also the solution in the book "Advanced Euclidean geometry" by A.Johnson. I think it's very nice and not long :).[/quote]

Thank you so much for your nice link :lol:
The solution I mentioned in previous posts was almost the same :D
[url=http://fmn.xnimg.cn/fmn044/20100311/2150/p_large_2NQn_5d75000118882d0e.jpg]Page 1[/url]
[url=http://fmn.xnimg.cn/fmn036/20100311/2150/p_large_InFs_0f90000030c02d0d.jpg]Page 2[/url]


AMC USA(J)MO USAMO search Harvard college LaTeX

Any Intel STS Finalists, Intel ISEF winners, Siemens winners, USAMO winners, IMO medalists, Davidson Fellows and other prestigious competitions winners here?

Yes, of course there are.  Do you have anything more to say than just that question?

Erm, can you guys help us (the beginners)? Like Q&A or something. Anyone else, if you wanna ask anything, ask away!

hmmm, this reminds me of the many ppl who are open to questions on cogito, the forum for ppl who qualified through CTY (SET) and other talent search programs.  Maybe we could set something up like that...

can you give more information about these programs and competitions? are they available for international students?  :maybe: can someone describe them more? :)
thanks
JgFi

If you have questions, find appropriate fora or threads and [i]ask the questions[/i].  Or, even better, first do a thorough search to see if the questions have been asked and answered here before.  Asking whether you can ask questions serves little purpose.  (So far, it has resulted in 6 posts that have accomplished exactly nothing.)  On a similar note, [url=http://www.google.com]here's[/url] someone who is very adept at answering questions like JgFi's.

Ok JBL...Although I understand the logic behind your argument, you are just too DAMN RUDE!!!! This is not a way to approach people you know; particularly when we have never meet each other (Good first impression I might add!). One of the things in life is to know how to approach people and you have really blew any good image that anyone might have. I know that you are from Harvard, but that does not entitle you to boss people around! We (Not everyone here, ok!) are just beginners in this thing, so of course we are VERY CURIOUS and EXCITED on all these opportunities and GREAT EXPERIENCES that they will bring. So, this is a first step question that is logical for anyone to ask. By replying impolitely, you have SUCCESSFULLY bring to an end any hope/spirit that a beginner could have (Not to me, though! I am strong!) in continuing any endeavor related to this beautiful/elegant subject matter. Is that what they teach you at Harvard? Then, I am sure I don't wanna go to Harvard. What I want is a free-thinking (but with reasoned) society which have limitless opportunities for me to explore my mind and the world around me; to reach my fullest potential; a society where I can truly be myself (in a lawful way) without any fear whatsoever...A society of people who loves learning and helps each other to reach their fullest potential and certainly not a society that impedes the learning process of its member.

So, there you go. I guess I have added something of value (howsoever small) in this worthless thread. Thank you for even replying. I am honored!

My two cents about forum usage:
--------------------------------------------------
I'd just like to mention a few things.  Not trying to pass judgment or act as some superior figure, but just some things that I've noticed over the years as being a part of this message board...

First, the internet is a horrible medium for conversing.  In particular, on message boards you have a bunch of people that don't know each other.  You don't know the other person's personality and thus the tone of posts is very rarely communicated well.  I don't believe JBL's posts were intended to be as rude as you perceived them as.  Yes, they were a little rough and could have been worded a little more kindly, but he was mainly trying to give you advice that could ultimately help you out moreso in the long run.  My advice to everyone is to really think about what you're reading and posting.  Imagine how it would sound if you said it out loud.  And when you do come across someone that you find rude, the best thing to do is to ignore them (rather, their rude comments) as much as possible.  If they're intentionally being rude, they'll derive pleasure from seeing you react abrasively to their comments.

Also, there's also the issue of younger members lacking the experience that the older members have.  Older members, for the most part, wish that younger  members would just RTFM and use the forum as efficiently as possible without creating new posts.  Newer members usually just ask if they don't know something.  Personally, I think it's fine for new members to ask about things that they could figure out from doing a search.  I don't believe that a forum should be used as a reference.  I feel that it should be more of community where people come and go.  New members come in, old members leave.  Old members help out the new members.  People communicate, learn, etc.  There are certain things that should be compiled into easy to navigate resources such as the LaTeX guide, the contests section, and so forth.
--------------------------------------------------

Now, back to the original topic...

I don't fall under any of those categories, but I'll still let you know a little bit of what I've learned about.  First, AoPS has a really good list of contests with links compiled [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]here[/url].

Next, what specifically are you interested in?  Do you want to get into research and possibly enter one of those competitions?  Do you want to do really well in math competitions?

Usually, if you're wanting to get into research, try asking professors at a nearby college to see if any of them would be willing to take you in.  It's probably a long shot, but worth a try nonetheless.  There are usually programs in individual states which match up high school students with professors where they usually have you working with a grad student for a summer.

I know quite a few people that excelled at science competitions and math competitions.  They're all very hard working and have been for pretty much their whole lives.  Though, they're also usually pretty "normal".  They like to have fun, play games, etc.  They tend to have a few things they're really passionate about and have incredible dedication toward those things.

That's a pretty quick overview...  Anything more specific you'd like to know?

Hi. I'm a DF 10k winner and I've gotten a 2nd at ISEF (and a major special award). I know I didn't win tops in either, but maybe I could answer your questions... if I know what they are ;).

[quote="IReallyHateMath"]So, there you go. I guess I have added something of value (howsoever small) in this worthless thread.[/quote]  Presumably, your goal in starting this thread was to learn something about the competitions you named, and it is in this sense (relative to what I guess to be your motives) that the thread is valueless.  If I wanted to learn something about X from this forum, I would first
1) search for threads about X.
If step 1 fails to yield the information I sought, I would proceed to
2) post a new comment or thread asking questions about X.

I have no way of knowing if you've completed step 1.  At least in this thread, you have utterly failed to complete step 2.  It seems to me that this is a much more likely explanation for the lack of useful or interesting information on this thread than anything I have done, especially since both of my previous posts contained language that was intended to get you to perform step 2.  Even a Harvard education doesn't prepare me to read your mind and answer the questions you haven't asked.  :wink:  (Perhaps I should put this emoticon after every comment I make, so as not to offend anyone else?   :wink: )

Seriously, IReallyHateMath, are you new to the Internet? I mean, if a simple "rude" post is enough to send you into a frenzy, then lurk moar.

Also, posting about that frenzy isn't going to make things better...

STS Finalist.  What do you want to know?  There's a fair bit of information available on the [url=http://www.societyforscience.org/sts/]official site[/url].


[quote="Alumen"]Andrew (serialk11r) made me join and post this :P

Um... if you were in mathcounts 2006 there's a slim chance you know me, I placed 6th in state that year. Otherwise you probably have no idea who I am... I'm a freshman at Aragon HS (where participation in math-related stuff's been like nonexistant for years) so aside from Andrew, and the names of some of you "crazy geniuses", I don't think I know anyone here.

So err hello?[/quote]
Hey, you're gonna help revive this forum okay? :D

Anthony Lu right?

(this is non-spam) WELCOME!

Quatto is right next to me :'( .

[quote="serialk11r"][quote="CircleSquared"]Welcome to ..... how should I describe it.[/quote]
Spamworld?[/quote]

You're getting soft!

Hello, and welcome to the SFBA forum :)

We're pretty awesome here. Don't forget to come to our ARML meetings when they happen.. Most of us go to that, and it's a great place to meet in person :)

[quote="happyme"][quote="CircleSquared"][quote="happyme"]When are they?

and Where?

and would you mind if a stupid person like me tagged along?[/quote]

It's kind of odd how I was about to ask those same three questions.[/quote]

Great minds think alike,

Or in our case...[/quote]
heh don't worry, they wouldn't mind if you were "dumb" or smart.
Speaking of ARML, I really hope the meeting times aren't the same as last year...

they prolly are, they've been for all the years i went


quadratics
algebra
quadratic formula
geometry
power of a point
AMC

In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If $AG = 2,GF = 13, FC = 1$ and $HJ = 7$, then $DE$ equals

$\text{(A)} \ 2 \sqrt{22} \qquad \text{(B)} \ 7 \sqrt{3} \qquad \text{(C)} \ 9 \qquad \text{(D)} \ 10 \qquad \text{(E)} \ 13$

[hide](here $B$ is the "C" on the left)

Let $h$ be the length of $AH$. By Power of a Point, $h(h+7)=2*15\Rightarrow h=3$ by inspection.

Since the side length of the triangle is 16, $JB=6$.

Let the lengths of $BD,\ DE,\ EC$ be $w,\ x,\ y$ respectively. $w+x+y=16$ and by Power of a Point, $w(w+x)=78$ and $y(y+x)=14$.

Rewriting the last two equations in terms of $w$ and $y$, $w(16-y)=78$ and $y(16-w)=14$. Subtracting the equations, $16(w-y)=64\Rightarrow w=y+4$.

Substituting into the first equation, $(4+y)(16-y)=78\Rightarrow y^2-12y+14=0$. From quadratic formula, the roots are $6\pm \sqrt{22}$ so $w=10\pm \sqrt{22}$ and $x=\mp 2\sqrt{22}$.

Since $x$ is positive, $x=2\sqrt{22}\Rightarrow \boxed{A}$.[/hide]


geometry
trapezoid
incenter
geometric transformation
homothety
ratio
geometry unsolved

$ABCD$ is a trapezoid with $AB || CD$. There are two circles $\omega_1$ and $\omega_2$ is the trapezoid such that $\omega_1$ is tangent to $DA$, $AB$, $BC$ and $\omega_2$ is tangent to $BC$, $CD$, $DA$. Let $l_1$ be a line passing through $A$ and tangent to $\omega_2$(other than $AD$), Let $l_2$ be a line passing through $C$ and tangent to $\omega_1$ (other than $CB$).

Prove that $l_1 || l_2$.

Let \[ \ell_3 \parallel \ell_2, A\in \ell_3, AD\cap BC=K, \ell_3\cap BC=Y, \ell_2\cap AD=X, \ell_3\cap CD=L, \ell_2\cap AB=M \] The quadrilateral $CXAB$ are circumscribed, then $MC+AK=CK+AM$. Then $AL+AK=CK+CL$ and the quadrilateral $CDAY$ are circumscribed, hence $\ell_3=\ell_1$.

Fagot you wrote that
[quote="fagot"] The quadrilateral $CXAB$ are circumscribed, then $MC+AK=CK+AM$. .[/quote]

but in my figure this doesnot hold since if I take equivalently this equality I found that a positive number is equal with a negative.Could you please explain or post the figure in which you solve the problem ?Thanks

Another solution:

Denote by $K$ the point of intersection of $AD$ and $BC$. The center of $\omega_2$ is the incenter of $DCK$. Center $E'$ of $\omega_1$  is the image of the center $E$ of the excircle of $DCK$ ,corresponding to the edge $DC$, trough o homothety of center $K$, and ratio $KA/KD$. It's easy to check that triangles $KDI$ and $KCE$ are similar. Since $KA/KD=KE'/KE$ the two figures $K,I,A,D$ and $K,C,E,E'$ are similar, hence $\angle KIA\equiv \angle E'CK$. We have that $\angle DAI=\frac{\angle DKI} 2+\angle AIK=\frac{\angle DKI} 2+\angle E'CK \Rightarrow \\ 2\angle DAI=\angle DKI+\angle 2\angle KCE'$. (1)
The angle between tangents from a pooint to a circle is twice the angle between a tangent and the line joining that point to the ccenter of that crcle. Hence, the angle between $BK$ and $\ell_2$ is $2\angle KCE'$ and the angle between $KD$ and $\ell_1$ is $2\angle DAI$. From (1) we can get that $\ell_1$ and $\ell_2$ intersect $KC$ at same angles, hence are paralell.

[quote]but in my figure this doesnot hold since if I take equivalently this equality I found that a positive number is equal with a negative.Could you please explain or post the figure in which you solve the problem ?Thanks[/quote]
The segments with signs: if $D\in [AX]$ and $A\in [DK]$ then $AK-CM=CK-AM$, hence $AK-AL=CK-CL$ and $CDAY$ inscribed, e. t. c. [quote]the two figures $K,I,A,D$ and $K,C,E,E'$ are similar[/quote]
the two figures $K,I,A,D$ and $K,C,E',E$ are similar
[quote]$\angle DAI={\angle DKI\over 2}+\angle AIK$[/quote]
$\angle DAI=\angle DKI+\angle AIK={\angle DKC\over 2}+\angle AIK$...... $2\angle DAI=\angle DKC+2\angle KCE'$.

[quote="xirti"]Another solution:

Denote by $K$ the point of intersection of $AD$ and $BC$. The center of $\omega_{2}$ is the incenter of $DCK$. Center $E'$ of $\omega_{1}$  is the image of the center $E$ of the excircle of $DCK$ ,corresponding to the edge $DC$, trough o homothety of center $K$, and ratio $KA/KD$. It's easy to check that triangles $KDI$ and $KCE$ are similar. Since $KA/KD=KE'/KE$ the two figures $K,I,A,D$ and $K,C,E,E'$ are similar, hence $\angle KIA\equiv \angle E'CK$. We have that $\angle DAI=\frac{\angle DKI}2+\angle AIK=\frac{\angle DKI}2+\angle E'CK \Rightarrow \\ 2\angle DAI=\angle DKI+\angle 2\angle KCE'$. (1)
The angle between tangents from a pooint to a circle is twice the angle between a tangent and the line joining that point to the ccenter of that crcle. Hence, the angle between $BK$ and $\ell_{2}$ is $2\angle KCE'$ and the angle between $KD$ and $\ell_{1}$ is $2\angle DAI$. From (1) we can get that $\ell_{1}$ and $\ell_{2}$ intersect $KC$ at same angles, hence are paralell.[/quote]

Sorry, I'm newbie but could you(or someone) explain me why triangle $KDI$ and $KCE$ are similar?

[quote="fagot"]Let \[\ell_{3}\parallel \ell_{2}, A\in \ell_{3}, AD\cap BC=K, \ell_{3}\cap BC=Y, \ell_{2}\cap AD=X, \ell_{3}\cap CD=L, \ell_{2}\cap AB=M\] The quadrilateral $CXAB$ are circumscribed, then $MC+AK=CK+AM$. Then $AL+AK=CK+CL$ and the quadrilateral $CDAY$ are circumscribed, hence $\ell_{3}=\ell_{1}$.[/quote]
If $\ AD\parallel BC$ then...?

[quote="M4RI0"]Could someone post a picture :?:[/quote]
Here is a picture.
[img]http://img522.imageshack.us/img522/3210/chinatst2006p1qa8.png[/img]

[size=150]To fagot:[/size]
Could you or somebody else explain your solution in details on my image?


does anyone know if there's a 2007 georgia arml website? it's not that important, i'm just wondering.

Not yet. Steve does this as time allows. You could always offer to put it together and send it to him!  :D

We do have a 2006 one, I'm pretty sure. It's somewhere on the AoPSWiki...

Well, exactly, it needs updating for '07.

kinda off topic, but what kind of tournaments do you guys think would be places where ARML coaches see good possible team members?

Probably the ones you are already thinking about. I no particular order (well, a bad calendar order) and certainly not an inclusive list:

Kennesaw State (at your school)
Lassiter
Rockdale
Ga. Tech
Furman
State

for the younger students:
JV State

You should definitely plan to take the AMC (10 or 12) and try to qualify for the AIME.

Fulton

Also, you should probably do varsity tournaments as opposed to JV so that you will be competing with former ARMLers. I would think that it is better to get 5th place and beat some good ARMLers at a tournament like Lassiter than it is to get 1st place at a random tournament with no one there.

I don't think I agree completely with Eric. (Hmmm, I think I've said something like that before.) In some tournaments, Arvind will have to compete varsity. Lassiter is probably one of those and certainly Cobb was. But when he can compete JV, then maybe that's OK. Once you get to a tournament that defines JV as 10th and below, he's still competing against ARMLer's and other good folks.

Okay, you should listen to Mr. Fulton because he's an ARML coach, so he knows what they're looking for, but one of these days, I'm going to say something that you will actually agree with, Mr. Fulton.  :lol:

Well, I've disagreed with both you and Miles, so perhaps I'm pushing my luck! :-)

[quote="Arvind_sn"]Somebody disagreed with Miles?  :o  :o  :o

That's a first.

Anyway, thanks guys for your help.[/quote]

what?

i've strongly disagreed with his fashion sense for years

I think you hold some sort of record for reviving old topics. :wink:

[quote="Benjamin Hu"]Same here. Apparently someone told someone else on our team that he has a powdered wig, which is ridiculous...[/quote]
It is not ridiculous.  In fact, I have two.

[quote="Karth"]Okay you should seriously stop making fun of people who have done nothing whatsoever to you.  :|[/quote]

i'm too cowardly to make fun of epople who confront me

Oh. Sorry, Miles. For some reason, I question almost every thing I hear (unless it has a logical explanation to back it up, or at least a very persuasive one).


geometry
angle bisector
geometry unsolved

How do we divide a circle in 9 parts?(geometrical construction )

If you meant equal parts, ... here comes a quote from wikipedia:

"...Although a regular nonagon is not constructible with compass and straightedge there are methods of constructing that are very close approximations..."

http://en.wikipedia.org/wiki/Enneagon

A very good ( I think ) approximately construction of a regular [b][size=100]9-gon[/size][/b], has already been presented, in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=154939]angle bisector[/url], posted by [b][size=100]yuyang[/size][/b].

Unfortunately, I have not in mind any proof and the approximation has been checked by computer.

Kostas Vittas.


calculus
real analysis
real analysis unsolved

The following is regarding to one of my first posts on this website.  No one could solve it in a proper way, without gussing and checking.  It was previously posted in the Getting Started Forums.  Rushil wrote that there is a calculus method to solve this.  Would someone please solve this using calculus?
\[ 3^x+2^x=35 \]
Thanks in advance. :)

Masoud Zargar

What calculus can tell you:

By the intermediate value theorem, there is exactly one real solution. This solution can be numerically approximated to any desired accuracy by many different methods.

What calculus can't tell you: The answer is 3.

I've been seeking for a purely mathematical (without guessing and checking) for such equations for 2 years now.  I believe that there must be a way to solve such equations.


The hypotenuse of a right triangle is $ 10$ inches and the radius of the inscribed circle is $ 1$ inch.  The perimeter of the triangle in inches is:

$ \textbf{(A)}\ 15 \qquad
\textbf{(B)}\ 22 \qquad
\textbf{(C)}\ 24 \qquad
\textbf{(D)}\ 26 \qquad
\textbf{(E)}\ 30$

[hide]
The area of the triangle can be found as $ rs$ where $ r$ is the inradius and $ s$ is the semi-perimeter.
Since the inradius is $ 1$, the area will be equal to the perimeter. You can easily see that a the triangle is $ 6\minus{}8\minus{}10$ right triangle.
So the perimeter is $ 24 \implies \boxed C$. [/hide]

[hide="Umm..."]
$ c\equal{}a\minus{}r\plus{}b\minus{}r$, so $ 10\equal{}a\plus{}b\minus{}2 \implies P\equal{}a\plus{}b\plus{}c\equal{}10\plus{}2\plus{}10\equal{}22$ or $ \boxed{\textbf{(B)}}$.
[/hide]

hello, after the fromula $ r\equal{}\frac{a\plus{}b\minus{}c}{2}$ we get $ a\plus{}b\equal{}12$ and from here $ a\plus{}b\plus{}c\equal{}22$, so the correct answer is $ \text{B}$.
Sonnhard.


[quote="Phelpedo"]This is a method for probability problems that has been bothering me lately. An AIME problem (2000 B i think?) goes like this:

3 numbers: $x_{1}, x_{2}, x_{3}$ are chosen from the set {1,2...2000} and removed without replacement. Then 3 numbers $y_{1}, y_{2}, y_{2}$ are chosen. What is the probability that $y_{i}> x_{i}$? The solution claimed that the probability would be the same if the set was {1,2,3,4,5,6} Could someone provide some insight into this? It doesn't seem very natural.

Recently this problem came up: You are given 10 cards. 8 of them are numbered 1-8, one is one of the numbers 1-8, and the other is marked with an X. If you reveal the pair of numbers, you win; otherwise, you lose. What is the probability of winning?

Again, the solution claimed that having 10 cards was equivalent to having 3 cards.

So my question is: in general, when and why are you allowed to make these sort of assumptions?[/quote]
For the first problem, the numbers will always be in some order $a<b<c<d<e<f$ (where $a,b,\dots,f$ are a permutation of $x_{i}$ and $y_{i}$), so you can just assume that $\{a,b,\dots,f\}$ is $\{1,2,\dots,6\}$.

However, I'm not completely sure about the second problem. Could you explain it in more detail?

In general, you are allowed to make these sort of assumptions whenever they do not change the substance or the significant conditions of the problem.

[quote="Phelpedo"]For the second problem: 8 cards numbered 1-8. The 9th is a repeat. The 10th is marked with an X. You flip over cards; if you reveal the repeat, you win; if you reveal the X, you lose. What's the probability of winning.

Apparently, it's the same for any number of cards.

For the first problem, I'm not really convinced by that justification. The specific value of a,b,c,d,e,f doesn't really matter, I suppose, but one might think that the extra values wedged in between would affect the probability.[/quote]

For the first one, think about it this way. There are $\binom{2000}{3}$ ways to choose the six $x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}$. For any of these six chosen, there are certain arrangements such that our desired condition holds (obviously it doesn't matter how far apart each is from the others, just the order they are in). Furthermore, the probability of these certain arrangements occurring is equal no matter which six we choose. That's why we can say it is the same as choosing from $\{1,2,3,4,5,6\}$.

I assume you mean $\binom{2000}{3}$. I am also not quite sure what you mean by

[quote] Furthermore, the probability of these certain arrangements occurring is equal no matter which six we choose. [/quote]

For the second one- the general idea seems to be that if we are trying to find the probability of event A and event B, where one of them has to occur, then we can throw out all elements that contribute neither to A nor B. However, this needs to be proved.

[quote="Phelpedo"]I assume you mean $\binom{2000}{3}$. I am also not quite sure what you mean by

[quote] Furthermore, the probability of these certain arrangements occurring is equal no matter which six we choose. [/quote]
[/quote]

Say we chose $1, 2, 900, 1500, 1999, 2000$. The probability that we can choose the $x$'s and $y$'s such that $y_{i}> x_{i}$ is the same as if the numbers were $5, 10, 15, 500, 1200, 1700$. It's not important which set of six we choose, so the probability for any one set of six is the same as the probability for all of them.

Oooh. Okay, that does make sense. But what about the second problem, and a generalization of this thingy?

In the second problem, suppose the duplicate card is 8. You need to turn over both 8s to win; you need to turn over the X to lose. Suppose you turn over the 7. That didn't change anything. You are no closer to winning or losing. The ratio of probabilities of turning over an 8 or an X are the same no matter how many extra cards there are; turning over one of the extra cards doesn't do anything. So you may as well assume there are no extra cards.

There's no generalization of "how to be clever when solving problems." I guess the way to simplify these particular problems is to consider what actually matters in the problem and what doesn't. In the first one, it's similar -- the other 1994 numbers don't do anything at all, so why are they even there?

What I meant by generalization was this (mentioned earlier): when finding the probability of event A vs. event B, can you ALWAYS disregard events that contribute to neither A nor B?

In the context of the above problems, it would be like ignoring numbers besides 1-6 or ignoring cards that aren't duplicates or X's.

Yes. If you're looking for $\frac{P(A)}{P(B)}$ and event $C$ happening affects $P(A)$ and $P(B)$ but not their ratio, then you can disregard the possibility of $C$.

[quote="Phelpedo"]3 numbers: $x_{1}, x_{2}, x_{3}$ are chosen from the set {1,2...2000} and removed without replacement. Then 3 numbers $y_{1}, y_{2}, y_{2}$ are chosen. What is the probability that $y_{i}> x_{i}$? The solution claimed that the probability would be the same if the set was {1,2,3,4,5,6} Could someone provide some insight into this? It doesn't seem very natural.[/quote]

We are allowed to assume the set is $\{1,2,3,4,5,6\}$ because all we care about is the relative values of the numbers.  For any six numbers, the probability is the same.  You could solve this generally using variables, but it's easier to just assume the set is something easy to work with, like $\{1,2,3,4,5,6\}$.

Here are two other examples where this technique is useful:

AIME 2004 #6:  An integer is called snakelike if its decimal representation $a_{1}a_{2}a_{3}\cdots a_{k}$ satisfies $a_{i}<a_{i+1}$ if $i$ is odd and $a_{i}>a_{i+1}$ if $i$ is even. How many snakelike integers between 1000 and 9999 have four distinct digits?

AHSME 1996 #12:  Four distinct points, $A$, $B$, $C$, and $D$, are to be selected from $1996$ points
evenly spaced around a circle. All quadruples are equally likely to be chosen.
What is the probability that the chord $AB$ intersects the chord $CD$?


log base(2x) of(4x) + log base(4x) of(16x) = 4

Solve for x:

Sorry, but that solution is wrong.

You factored 2y^2 + 3y = 0 incorrectly.

The zeros of the function are 0 and -3/2 not -2/3.  That will lead you to an answer of 1 or radical 2 over 4.

Hurray for elementary arithmetic mistakes.  :D


quadratics function conics parabola calculus

The function $ x^2 \plus{} px \plus{} q$ with $ p$ and $ q$ greater than zero has its minimum value when:

$ \textbf{(A)}\ x \equal{} \minus{} p \qquad\textbf{(B)}\ x \equal{} \frac {p}{2} \qquad\textbf{(C)}\ x \equal{} \minus{} 2p \qquad\textbf{(D)}\ x \equal{} \frac {p^2}{4q} \qquad\textbf{(E)}\ x \equal{} \frac { \minus{} p}{2}$

[hide="Click for solution"]
The minimum value of the function is at the turning point of the graph where $ x\equal{}\minus{}\frac{p}{2}$ or $ \boxed{\textbf{(E)}}$.
[/hide]

[hide="Solution"]
It is a well-known fact that the vertex of a parabola $ ax^2\plus{}bx\plus{}c$ is found at $ \frac{\minus{}b}{2a}$, so the answer is obviously $ \boxed{\textbf{(E)}\ x\equal{}\frac{\minus{}p}{2}}$.

If you are familiar with differential calculus, you could also go with:
$ 2x\plus{}p\equal{}0\implies \boxed{\textbf{(E)}\ x\equal{}\frac{\minus{}p}{2}}$.
[/hide]

Notice, however, that it is $ E$ only because the coefficient of $ x^2$ is positive.

Otherwise, the parabola opens up instead of down and instead of having a minimum and no maximum would have a maximum and no minimum.


function
real analysis
real analysis unsolved

Equation $y$=1+$\sum_{i=1}^{\infty}g_{i}\left(  \frac{x}{y}\right)  ^{i}$ defines a function $y$=1+$\sum_{n=1}^{\infty}a_{n}x^{n}$ near $(x,y) $=$(0,1)$.
Prove that:
a). $a_{n}$ is the linear combination of all the monomials like $% %TCIMACRO{\dprod \limits_{i=1}^{j}}%
%BeginExpansion
{\displaystyle\prod\limits_{i=1}^{j}}
%EndExpansion
g_{k_{i}}^{e_{i}}$ with $\sum_{i=1}^{j}e_{i}k_{i}$=$n$ and $k_{\alpha}\neq
k_{\beta}$ for $\alpha\neq\beta$.
b). Call array $\left(  n;e_{1},...,e_{j}\right)  $ the index of the monomial $%
%TCIMACRO{\dprod \limits_{i=1}^{j}}%
%BeginExpansion
{\displaystyle\prod\limits_{i=1}^{j}}
%EndExpansion
g_{k_{i}}^{e_{i}}$. The coefficient $C\left(
%TCIMACRO{\dprod \limits_{i=1}^{j}}%
%BeginExpansion
{\displaystyle\prod\limits_{i=1}^{j}}
%EndExpansion
g_{k_{i}}^{e_{i}}\right)  $ of $%
%TCIMACRO{\dprod \limits_{i=1}^{j}}%
%BeginExpansion
{\displaystyle\prod\limits_{i=1}^{j}}
%EndExpansion
g_{k_{i}}^{e_{i}}$ in $a_{n}$ is completely determined by its index,
\[
C\left(
%TCIMACRO{\dprod \limits_{i=1}^{j}}%
%BeginExpansion
{\displaystyle\prod\limits_{i=1}^{j}}
%EndExpansion
g_{k_{i}}^{e_{i}}\right)  \text{=}C\left(  n;e_{1},...,e_{j}\right)
\text{=}\frac{\left(  -1\right)  ^{\left(  S-1\right)  }\left(  n+S-2\right)
!}{(n-1)!%
%TCIMACRO{\dprod \limits_{i=1}^{j}}%
%BeginExpansion
{\displaystyle\prod\limits_{i=1}^{j}}
%EndExpansion
e_{i}!}.
\]
where $S$=$\sum_{i=1}^{j}e_{i}.$

Lagrange's inversion formula for power series should be relevant here.

The result looks similar, but I don't think the series reversion could help much, at least not in an obvious way.

[quote="Myth"]Buffalo!
Why have you put it Combinatorics Section? :what?:[/quote]
This is a typical combinatorics problem, analysis is nothing but its expression.

Anyway I have alredy moved it.
We can say that it belongs to both fields  :P


MATHCOUNTS
calculus
limit
inequalities
trigonometry
analytic geometry
symmetry

A 10-foot wide hallway and a 5-foot wide hallway intersect in the corner of a building.  What is the maximum length pole that could be carried past the corner without tilting it?

This is assuming that the pole's width is negligible, right?

you keep the pole horizontal so it is a 2-dimensional problem, not 3D.  Do the problem as if the pole is too heavy and you can't lift it off the floor :)

I believe this is a wild guess, but is it  15:sqrt:2? sorry, i cant really explain how I got it since its kinda graphical...

[hide]
The longest pole possible will be the one that scrapes the corner and both walls exactly as it is halfway done rotating (45 degrees). See the picture.

By 45-45-90 triangles, we can get the length of the pole as $10\sqrt{2}+5\sqrt{2}=15\sqrt{2}$.[/hide]

[quote="EFuzzy"][hide]15\sqrt{3}$.[/hide][/quote]
Are you sure? In he AoPS book, the hypotenuse is  :sqrt: 2 * the side length

the hypotenuse of an isosceles right triangle is always the length of a leg times root2.  Use the Pythagorean Theorem with x as the legs to prove it to yourself.

This is no Mathcounts problem; this is a calculus problem. The critical angle for the pole isn't 45° or 60° or indeed any special angle. The actual maximum length for the pole is (answer deleted). The techniques for finding that are not available at this level, so it's not surprising that none of you have that answer.

You could try posting it over in the Calculus Computations and Tutorials forum.

When the pipe makes a 45 degree angle with the wall, that's the time when the pipe will be the tightest in the room.

Moved to Calculus Computation and Tutorial. After checking the problem, there seems to be no method to do this problem that Mathcounts will ever put on their problems (hopefully).

Treething and EFuzzy: a pole that touches both walls and the inside corner at a 45° angle is too long. It cannot be pushed from there any further into the narrower hallway.

It's time for the middle schoolers to step out of the way; let those just learning calculus have a chance.

Make it the general version. Suppose the two hallways (which are at right angles to each other) have respective widths $a$ and $b$. Let the length of the longest pole that can be carried around the corner be $\ell(a,b).$ What is the general formula for $\ell(a,b)$? Here are some properties that formula should have:

1. If $\lambda>0,\,\ell(\lambda a,\lambda b)=\lambda \ell(a,b).$ (Also known as "positively homogeneous of degree  1.") More simply, if $a$ and $b$ are given in a particular unit of measurement - say, feet -  then $\ell(a,b)$ will be in the same unit of measurement.

2. $\ell(a,b)=\ell(b,a).$ If you can go forwards around the corner, you can also go backwards.

3. $\ell(a,a)=2\sqrt{2}a.$ If the hallways are the same width, then we have the 45° angle that the middle schoolers wanted it to have.

4. $\lim_{b\to0^+}\ell(a,b)=a.$ If one hallway is of microscopic width, you have to carry the pole sideways in the other hall to fit it in.

By the way, it's neat to note that Holder's inequality gives an immediate solution to this problem without the need to use calculus.

[quote="Kent Merryfield"]Here's the dual problem: let a tall, vertical-sided building be located on flat ground. A fence of height $a$ is located a distance $b$ away from the building. What is the shortest ladder that you can lay over the top of the fence and still touch the ground and the side of the building?[/quote]

Expressed in terms of a and b?

kunny and blahblahblah: would you both please show your methods?

I believe gauss202 posted this problem before and called it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=12563]Moving In[/url].

I supplied a solution there using Holder's inequality, as blahblahblah suggested.

Yeah, I thought that ravi's solution using Holder in that thread was really neat.

The calculus solution is pretty straightforward, but it's a bit of a pain in the butt to derive, insofar as we can find the numerical value of maximum-length ladder pretty easy using some basic, basic calculus, but it is much harder to find it in terms of $a$ and $b$ exactly (especially if you work with trig)

Or at least it was the last time I did max/min problems like this.

All right,Kent Merryfield.

Let $P,Q$ be the points of inter section between the line passing through the fixed point $R(a,b)$,which lie in the first quadrant, with a gradient $ m<0\ $ and $x$ axis, $y$ axis respectively.We will find the minimum value of the segment $PQ$.Since the equation of the line passing through the point $R(a,b)$ is

$y=m(x-a)+b\  (m<0)\ $, we get two coordinates of $P\left(a-\frac{b}{m},0\right),Q\left(0,-am+b\right)$.

Therefore $PQ^2= \left(a-\frac{b}{m}\right)^2+(am-b)^2$,setting this equation $f(m)$,

$f'(m)=2\left(a-\frac{b}{m}\right)\cdot\frac{b}{m^2}+2a(am-b)=\frac{2a}{m^3}(am-b)\left(m^3+\frac{b}{a}\right)$.as $a>0,b>0,m<0$,we have $\frac{2a}{m^3}(am-b)>0$.By this the sign of $f'(m)$ is coincides  with that of $m^3+\frac{b}{a}$.Setting $f'(m)=0$,we obtain the value of a negative number $m=-\sqrt[3]{\frac{b}{a}}$.
Thus we have $f'(m)<0$ for $m<-\sqrt[3]{\frac{b}{a}}$, $f'(m)>0$ for $-\sqrt[3]{\frac{b}{a}}<m<0$,we can obtain the minimum value of $f(m)$ when $m=-\sqrt[3]{\frac{b}{a}}$.

Then $f\left(-\sqrt[3]{\frac{b}{a}}\right)=f\left(-a^{-\frac{1}{3}}b^{\frac{1}{3}}\right)=\left(a+a^{\frac{1}{3}}b^{\frac{2}{3}}\right)^2+\left(a^{\frac{2}{3}}b^{\frac{1}{3}}+b\right)^2$

$=\left[a^{\frac{1}{3}}\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)\right]^2+\left[b^{\frac{1}{3}}\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)\right]^2=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^3$
Consequently the desired minimum value of segment  $PQ$  is $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}$.

kunny

[quote="blahblahblah"]Yeah, I thought that ravi's solution using Holder in that thread was really neat.[/quote]Thanks.  When I took the AP Calculus BC exam way back when, I decided as a joke to solve all the free-response questions I could without calculus.  It mostly worked, although on one question my non-calculus solution got bogged down, and I had to resort to calculus.  I ended up getting a 5 (out of 5): I guess the examiners didn't care whether calculus was used or not.

doesnt holder's inequality require p and q to be both greater than 1?? ravi used -2 and 2/3..?

[quote="Ravi B"]When 0 < p < 1 and q is negative, Holder's inequality holds with the sign reversed.[/quote]

why is that though? doesnt the final inequality end up having a sqrt of a negative number??

Most of the classic single-variable calculus max-min problems can be usefully recast as constrained several variable optimaztion problems for which the technique of Lagrange multipliers can be used. This is often the way to the most intuitively pleasing algebraic forms of the answers.

For the pole-corner problem, the pole must be no longer than the shortest length of a segment between the two outer walls that touches the inside of the corner; the ladder over the fence version calls for the same computation.

We are trying to find $ x$ and $ y$ such that the segment between $ (x,0)$ and $ (0,y)$ passes through the point $ (a,b)$ and that minimizes the length of the segment.

By similar triangles, $ \frac {x \minus{} a}b \equal{} \frac {a}{y \minus{} b}$. This can be rearranged to the constraint $ (x \minus{} a)(y \minus{} b) \equal{} ab.$ We now seek to minimize the length, or more conveniently, the square of the length. So the problem has become:

Minimize $ f(x,y) \equal{} x^2 \plus{} y^2$ subject to $ g(x,y) \equal{} (x \minus{} a)(y \minus{} b) \equal{} ab.$

We set $ \nabla f \equal{} \lambda \nabla g.$

$ 2x \equal{} \lambda(y \minus{} b)$
$ 2y \equal{} \lambda(x \minus{} a)$

Eliminating $ \frac {\lambda}2$ between these two equations, we get

$ x(x \minus{} a) \equal{} y(y \minus{} b).$

But from the constraint equation, $ y \minus{} b \equal{} \frac {ab}{x \minus{} a}$ and $ y \equal{} \frac {bx}{x \minus{} a}.$

$ x(x \minus{} a) \equal{} \frac {bx}{x \minus{} a}\cdot\frac {ab}{x \minus{} a}.$ Multiply through by $ (x \minus{} a)^2:$

$ x(x \minus{} a)^3 \equal{} ab^2x,$ or $ (x \minus{} a)^3 \equal{} ab^2$

So, $ x \equal{} a \plus{} a^{\frac13}b^{\frac23} \equal{} a^{\frac13}\left(a^{\frac23} \plus{} b^{\frac23}\right).$

By symmetry, $ y \equal{} b \plus{} a^{\frac23}b^{\frac13} \equal{} b^{\frac13}\left(a^{\frac23} \plus{} b^{\frac23}\right).$

Then $ x^2 \plus{} y^2 \equal{} \left(a^{\frac23} \plus{} b^{\frac23}\right)^3.$

Take the square root and we are done.

The usual recipe of differential geometry works as well. With the coordinate origin O at the building corner and the x-, y-axes along the building walls, the pole of length L touching the walls at distances p, q from the corner is a line with an equation

$\frac x p + \frac y q = 1$

or, since $q = \sqrt{L^2 - p^2}$,

$F(x, y, p) = \frac x p + \frac{y}{\sqrt{L^2 - p^2}} - 1 = 0$

To find the envelope of this family of lines (or any curves) depending on the parameter p, eliminate p from the following two equations

$F(x, y, p) = 0$ and $\frac{\partial F}{\partial p} = 0$

$\frac{\partial F}{\partial p} = -\frac{x}{p^2} - \frac y 2 \frac{(-2p)}{(\sqrt{L^2 - p^2})^3} = -\frac{x}{p^2} + \frac {yp}{(\sqrt{L^2 - p^2})^3} = 0$

$\frac{x}{p^3} = \frac {y}{(\sqrt{L^2 - p^2})^3}$

Rearranging the line equation and taking its cube,

$\frac{(p - x)^3}{p^3} = \frac{y^3}{(\sqrt{L^2 - p^2})^3}$

Dividing the two equations, in order to calculate p,

$\frac{(p - x)^3}{x} = y^2$

$p = x + (xy^2)^{\frac 1 3} = x + x^{\frac 1 3} y^{\frac 2 3}$

$p^2 = x^2 + 2 x^{\frac 4 3} y^{\frac 2 3} + x^{\frac 2 3} y^{\frac 4 3}$

Squaring the equation of the line and substituting for $p - x$ and $p^2$,

$\frac{(p - x)^2}{p^2} = \frac{y^2}{L^2 - p^2}$

$(p - x)^2(L^2 - p^2) = y^2 p^2$

$x^{\frac 2 3} y^{\frac 4 3} (L^2 - x^2 - 2 x^{\frac 4 3} y^{\frac 2 3} - x^{\frac 2 3} y^{\frac 4 3}) = y^2(x^2 + 2 x^{\frac 4 3} y^{\frac 2 3} + x^{\frac 2 3} y^{\frac 4 3}) =$
$= x^{\frac 2 3} y^{\frac 4 3} (x^{\frac 4 3} y^{\frac 2 3} + 2 x^{\frac 2 3} y^{\frac 4 3} + y^2)$

Reducing the equation by $x^{\frac 2 3} y^{\frac 4 3}$, rearranging, and taking its cube root,

$L^2 = x^2 + 3 x^{\frac 4 3} y^{\frac 2 3} + 3 x^{\frac 2 3} y^{\frac 4 3} + y^2 = (x^{\frac 2 3} + y^{\frac 2 3})^3$

$L^{\frac 2 3} = x^{\frac 2 3} + y^{\frac 2 3}$

Hence, the envelope is an astroid with the main axis L. To find the maximum possible length L of the pole, simply substitute the coordinates of the hallway corner $[a, b]$ into the equation of this astroid:

$L^{\frac 2 3} = a^{\frac 2 3} + b^{\frac 2 3}$

$L = (a^{\frac 2 3} + b^{\frac 2 3})^{\frac 3 2}$

And...what if we allow tilting the pole? (3D version of the problem)? what would be the maximum length, say that that the height of the hallway is h?
I think the answer is simple: if we allow tilt, then the optimal strategy is ti keep one end of the pole on the floor and the other end on the ceiling, both of them on the margins.
And then the wanted length would just be a diagonal of a rectangle which has h as one side and exactly the above computed minimum length as the other side,
Am i right?


modular arithmetic
number theory
relatively prime
number theory proposed

[hide="possible excellent solution :lol: "]look at the numbers base 2 :D
so then notice that $2^{341}-2=11111....1110b$ and 341=101010101b.  So then we see that 3*341=11b*101010101b=1111111111b.  We then easily see that $3*341|2^{341}-2$ and hence, $341|2^{341}-2$
Q.E.D.
[/hide]

[quote="shyong"]this is just Fermat Little theorem since $341$ is a prime and $(341,2)=1$  ;)[/quote]
yeah shyong is right ,I forgot to say : $(341,2)=1$
sorry

[quote="TomciO"]$341$ is prime? That's interesting :).[/quote]

haha silly mistake

$341=31\cdot 11$ :blush:

[quote="TomciO"][quote="parinaz.sa"][quote="TomciO"]$341$ is prime? That's interesting :).[/quote]
we didn't say that $341$ was prime number, we said if $341$ and $2$ were comprime numbers :$(341,2)=1$ we could use Fermat Little Theorem!!!!!!![/quote]

Are you sure :>?[/quote]
yeah I'm sure that it's right and $341$ isn't prime number!!and $341$ and $2$ are comprime numbers :$(341,2)=1$

Actually I had Fermat's Little Theorem on mind. $341$ has to be prime to apply FLT.

[quote="mathmanman"]Indeed, 341 is a 2-[url=http://en.wikipedia.org/wiki/Fermat_primality_test]Fermat liar[/url]. ;)[/quote]
 :)

Does $n$ divide $b^n-b$ for $b$ relatively prime to $n?$

If $n$ is prime, then this is true by Fermat's Theorem. But sometimes this is true if $n$ is not prime. In that case, we say that $n$ is a [b]base-[i]b[/i] pseudoprime[/b].

For any particular base $b,$ pseudoprimes are much less common than primes, but there are infinitely many of them. As it happens, 341 is the smallest base-2 pseudoprime and the smallest pseudoprime to any base.

If $n$ is a base-$b$ pseudoprime for all $b$ relatively prime to $n,$ then $n$ is called a [b]Carmichael number.[/b] This time, 341 is not a Carmichael number; the smallest Carmichael number is 561.

Could Fermat's Theorem be used as a means to test whether a number was prime? In other words, for a few possible values of $b,$ we compute $b^{n-1}\pmod{n}$ and ask whether it is 1. If $n$ fails this test, it will not be prime, and that does eliminate most non-primes. It's and attractive idea, becuase computing $b^{n-1}\pmod{n}$ would be fairly cheap in terms of computational complexity. The problem is that a number that passes one such test could be a pseudoprime, and a number that passes several such tests could be a Carmichael number. We just don't get a conclusive test that way.

At our institution, the course number for the elementary number theory course is MATH 341.


Prove that for the 2nd order recursive sequence $n = \{0,1,6,35,204 \dots \}$ that the product of any two adjacent terms is a triangular number.  (A second order recursive series has the relation $n_{a} = An_{\left(a-1\right)} + Bn_{\left(a-2\right)}$).

Only a step towards a solution.
Well, we have the series $a_0=0, \ a_1=1$ and $a_n=6a_{n-1}-a_{n-2}$. Any triangular number is in the form $\displaystyle \frac{n(n+1)}{2}$. Multiplying two adjacent terms of the sequence, we get $(6a_{n-1}-a_{n-2})(6a_{n-2}-a_{n-3})=36a_{n-1}a_{n-2}-6a_{n-1}a_{n-3}-6a_{n-2}^2+a_{n-2}a_{n-3}$. Don't know where to go from here.

[quote="frt"][hide="Idea"]From $a_n=6a_{n-1}-a_{n-2}$, use characteristic equation to find the explicit formula for $a_n$. Then express $a_n\cdot a_{n-1}$ in terms of $n$ and show that the expression is in the form of a triangular number.[/hide][/quote]
Another way, longer but using more basic math, is to  write $2a_{m}a_{\left(m+1\right)} = n_{m}(n_{m} + 1)$ and using symultaneous equations is to find a formula for the argument $n_{m}$as a function of $a_{m} \text{and} a_{\left(m+1\right)}$ and prove that it holds for all $m$.  It worked for me and allowed me to generalized the problem.


calculus
integration
logarithms
real analysis
real analysis theorems

While working an integral problem I arrived at the following expression:
$ \int{\frac{\ln |t-a|}{(t^2+a^2)}dt=\int{(\frac{1}{t(t^2+a^2)} + \underbrace{\frac{a \ln|t-a|}{t^2(t^2+a^2)}}} - \frac{a \ln|a|}{t^2(t^2+a^2)})dt\;\;}$ (1)

Carrying the underbrace term of the expression to the left side results in this expression:

$ \underbrace{(1-\frac{a}{t^2})}\int{\frac{\ln (|t-a|}{(t^2+a^2)}= \int{\frac{1}{t(t^2+a^2)}- \frac{a \ln|a|}{t^2(t^2+a^2)}dt \;\;}}$ (2)

What do I do with the underbrace factor in expression (2) ?  Is or does the underbrace factor actually need to be operated on by the integral  -->  "brought to the right side"?

Maybe expression (1) should written like this instead?
$ \int{\frac{\ln (|t-a|}{(t^2+a^2)}dt=\int{(\frac{1}{t(t^2+a^2)} + \underbrace{\frac{a \ln\frac{t-a}{a}}{t^2(t^2+a^2)}}} )dt\;\;}$ (3)

Comments?

Sorry, I seem to have severe difficulties making sense of what you wrote in your post (maybe because it is the end of a long day, maybe for some other reason) :( What exactly are you aiming at? The equation (1) doesn't look correct at all, so I do not see any point in trying to derive something else from it. Also I don't quite understand how you draw terms in the integrand outside the integral sign... Can you step back a bit and show where you started? :)

[quote="fedja"]Sorry, I seem to have severe difficulties making sense of what you wrote in your post (maybe because it is the end of a long day, maybe for some other reason) :( [/quote]
No need to apollogize.  I think your severe difficulties are the result of my not understanding the proper way to factor the expression.
[quote="fedja"]What exactly are you aiming at? The equation (1) doesn't look correct at all, so I do not see any point in trying to derive something else from it. [/quote]
Getting to equation (1) is not the problem.  Moving to equation (2) or (3) is where the question lies

[quote="fedja"]
Also I don't quite understand how you draw terms in the integrand outside the integral sign... Can you step back a bit and show where you started? :)[/quote]
I think this answers the underlying question.  --> It is improper to factor integral expressions in a manner that takes the variables outside of the integral operator.

Thanks!!

[quote="fedja"][quote="vector2"]
Getting to equation (1) is not the problem.  Moving to equation (2) or (3) is where the question lies
[/quote]
Well, what I wanted to say is that there is something fishy about (1) already: if two pure integrals are equal, the integrands should coincide as well and they obviously don't. So either some terms are missing in the equation or it is wrong altogether. In both cases, there is no point to try to get anything from (1).[/quote]

I think you are correct.  I noticed something wrong as I was creating / posting my steps for your review... I'll make that correction and see if it becomes "not fishy".


inequalities
LaTeX
number theory unsolved
number theory

let n in N and m is a number such that: m>=2 and a^m -1 divisble n for every (a,n)=1
prove that: n<=4m(2^m-1)

[quote="nguoikhonkho"]let $n \in \mathbb N$ and $m$ is a number such that: $m \geq 2$ and $a^{m-1}$ divisible by $n$ for every $(a,n)=1$
prove that: $n \leq 4m(2^{m}-1)$[/quote]

Use the [b]LATEX[/b], please...   :wink:


superior algebra
superior algebra unsolved

If a1, a2, ......a(n) belong to a group, what is the inverse of a1a2....an?

[quote="julien_santini"]$a_{n}^{-1}\dots a_{1}^{-1}$  :D[/quote]

thanks, can u explain more!!

You've already asked this trivial  question here.

http://www.mathlinks.ro/Forum/viewtopic.php?t=112017

Verify julien_santini's solution by checking the definition (of course, of an inverse element); there is nothing else to do.


Suppose that $ m$ and $ n$ are positive integers such that $ 75m \equal{} n^{3}$. What is the minimum possible value of $ m \plus{} n$?

$ \textbf{(A)}\ 15 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 5700$

[quote="Thales418"]$m=0$ and $n=0$ is a solution....

humpf...

or

$75 = 5.5.3$

So  $n = 5.3$

and $m+n=60$[/quote]

actually, the question asks for positive integers, 0 is NOT a solution...

[hide="Solution"]  $75$ factorizes as $5^{2}\times 3$.  If $75m=n^{3}$, so the exponents of each of the prime factors of $75m$ must be multiples of 3.  Then $75m=3^{3}\times 5^{3}\Longrightarrow m=3^{2}\times 5$ and $n=\sqrt[3]{75m}=3^{\frac{3}{3}}\times 5^{\frac{3}{3}}=3\times 5=15$ and so $m+n=45+15=60$.[/hide]

Just factor the 75
we get 5^2 x 3
so we need another 5 and 3^2
so 9 x 5 = 45 and n = 15
15 + 45 =[b]60[/b]

[hide = Solution]
So we know $75 = 25$ x $3$ so we have $5^2$ x $3$ x $m$ $=$ $n^3$. So it follows that the minimum value of $m$ would be $5$ x $3^2$. Where $n$ $=$ $5$ x $3$ or $15$. So we have $45 + 15 = \boxed{60}$
[/hide]


Prove that $ a^4b\plus{}b^4c\plus{}c^4d\plus{}d^4a \geq abcd(a\plus{}b\plus{}c\plus{}d)$ with $ a;b;c;d>0$

[quote="boy own me"]Prove that $ a^4b \plus{} b^4c \plus{} c^4d \plus{} d^4a \geq abcd(a \plus{} b \plus{} c \plus{} d)$ with $ a;b;c;d > 0$[/quote]

[hide]

Consider , by AM_GM

$ {23a^4b \plus{}7 b^4c \plus{} 11c^4d \plus{} 10d^4a \over 51} \ge \sqrt [51] {a^{102}b^{51}c^{51}d^{51}}\equal{}a^2bcd$

from which the desired follows easily.

[/hide]


if the sum m+n+p of three natural nos is divisible by 6, then prove $m^3+n^3+p^3$ is also divisible by 6.

[quote="mathmanman"]Congruences :D
We have :
$0^3 \equiv 0 [6]$
$1^3 \equiv 1 [6]$
$2^3 \equiv 2 [6]$
$3^3 \equiv 3 [6]$
$4^3 \equiv 4 [6]$
$5^3 \equiv 5 [6]$

Hence the result ;)[/quote]

I couldn't get your solution until I did the problem myself. So can you tell me how your solution explains the answer.

[hide]If $6|m+n+p$ then for $6|m^3+n^3+p^3$ this should also be true: \[ 6|(m^3+n^3+p^3)-(m+n+p) \]
Which is true since $6|n^3-n$[/hide]

mathmanman , i didnt really get your soln , can you xplain it again???

We have the six congruencies that he gave:

$0^3 \equiv 0[6]$
$1^3 \equiv 1[6]$
$2^3 \equiv 2[6]$
$3^3 \equiv 3[6]$
$4^3 \equiv 4[6]$
$5^3 \equiv 5[6]$

and we know that $a+b+c \equiv 0[6]$

Because $a \equiv a^3[6]$. $b \equiv b^3[6]$, and $c \equiv c^3[6]$, we know that $a^3 + b^3 + c^3 \equiv 0[6]$.

I thought that his solution was pretty clear, but I guess not.

Idea is to see that $n^3 = n\ mod\ 6$
Then $m^3 + n^3 + p^3 = m+n+p \ mod\ 6$

Alternative algebraic solution is to see that $m,n,p$ can't be all odd and to use

$m^3 + n^3 + p^3 = 3 \cdot m \cdot n \cdot p + (m+n+p)(m^2+n^2+p^2 - m \cdot n - n \cdot p - p \cdot m)$

Thanks to Hamster1800 and tµtµ for having explained my solution ;)

I also thought it was clear enough.. :oops:


function
quadratics
algebra
polynomial
algebra unsolved

Find all functions $f$ defined on natural numbers (yes, zero is a natural number) such that, for all $n \in N$, we have:

$f(f(n)+n)+f(f(n)-n) = 8n$

[quote="ashwinrk_jain"]$f(n)$ must be the linear function function because R.H.S don't have any quadratic term then the coe. of quadratic and more powers must be zero...
now we assume that $f(n) = an+b$ and substitue in the equation and solve it. we get $a=+2,-2$ $b=0$ hence the function is $f(n) = 2n$ and $f(n) =-2n$.[/quote]

wrong way, my friend!

try to prove that f is linear! isn't easy!  :wink:

That proof isn't valid, not all functions can be written as polynomials (for instance the busy beaver function etc.).

Let $f(f(n)+n)+f(f(n)-n)$ be called $g(n)$. First, simplifying $g(0)$ yields $f(f(0))=0$. Let $f(0)=a$ then simplifying $g(a)=f(f(f(0))+f(0))+f(f(f(0))-f(0))=f(f(0))+f(-f(0))=f(-f(0))=f(-a)=8a$.
I'm just realizing something is missing - are we told that $f(n)\geq n$ for all $n$? Otherwise the problem is meaningless since $f(n)-n$ can be zero then it would be undefined. Is this assumed?

Well I'll assume it. Then since $a$ is defined we have $f(-a)$ is defined so $-a\geq 0\implies a=0$ so $f(0)=0$.
Also $f(n)\geq n\implies f(f(n)+n)+f(f(n)-n) = 8n\geq 2f(n)\implies 4n\geq f(n)\geq n$. Also assume for some $n$ $f(n)=n$ then $f(f(n)+n)=f(2n)=8n$. But from $f(f(2n)+2n)+f(f(2n)-2n)=16n\geq 2f(2n)$ we see equality must hold, i.e. $f(10n)=f(f(2n)+2n)=f(2n)+2n=10n, f(6n)=f(f(2n)-2n)=f(2n)-2n=6n$. So far $f(n)=n\implies f(2n)=8n, f(6n)=6n, f(10n)=10n$
This also implies $f(3n)>3n$ and $f(5n)>5n$ and ...

If we assume there is an $f(n)=4n$ then $8n=f(3n)+f(5n)\implies f(3n)=3n, f(5n)=5n$.
I just saw the elopes posted, any ideas to continue my method?

edit:continuing

instead of $n$ business just suppose $f(1)=1$. then all the above shows $f(2)=8,f(6)=6,f(10)=10$ and $f(3)\geq 4$, $f(5)\geq 6$. Now $f(5)=6\implies f(11)=39$ and $f(3)=4\implies f(7)=23$

i did exactly the same things of you, my friend me@home, but not concrete results!  :(

the function g(x) = f(x) - x have nice properties, but nothing to kill the problem!


Is there a way to send a private message to multiple people?
It would make the tournament messages a little bit easier.

We had a bit of trouble dealing this with the AoPS Team Chess Tournament.

The only way to mass message people would be to copy and paste a message and pm to all the users.

You could also get everyone's email and mass message them there, which is easier.

Yeah, thats what I've been doing.
Once I get everyone's email from my site, I can just make a list and send it to everyone.
Thanks

There's no multiple PMing right now because that makes it way too easy to spam.

One way to allow multiple PMing would be to make it an option but to require image verification in that case (and limit the total number of recipients to 10 at a time or something).

I would be in favor of something like that in general, but there are certain other things that I think are more important, so I'm not going to be terribly surprised or upset if this never happens or doesn't happen for a long time. ;)

Lots of PM spam is neither commercial nor automated (think contest cheaters).  What would deal with the problem without a spam hazart is the creation of opt-in PM lists/groups with only the list founder (or some predetermined finite group of users) empowered to PM all group members.

Ooh -- I like JBL's idea. That addresses the specific needs of the people in this forum much better than my suggestion does.

[quote="JBL"]What would deal with the problem without a spam hazart is the creation of opt-in PM lists/groups with only the list founder (or some predetermined finite group of users) empowered to PM all group members.[/quote]

This is a good idea, but what if a certain event requires a group member to mass reply back?

I think a good solution, going along with JBL's idea, would be to having an operative code word that can be distributed to members of a list. If you type in the code word correctly, you'll effectively join a "chat" where everyone can mass pm each other.


function
inequalities
algebra proposed
algebra

Find all monotonically increasing or monotonically decreasing functions $f: \mathbb{R}_+\to\mathbb{R}_+$ which satisfy the equation $f\left(xy\right)\cdot f\left(\frac{f\left(y\right)}{x}\right)=1$ for any two numbers $x$ and $y$ from $\mathbb{R}_+$.

Hereby, $\mathbb{R}_+$ is the set of all positive real numbers.

[i]Note.[/i] A function $f: \mathbb{R}_+\to\mathbb{R}_+$ is called [i]monotonically increasing[/i] if for any two positive numbers $x$ and $y$ such that $x\geq y$, we have $f\left(x\right)\geq f\left(y\right)$.
A function $f: \mathbb{R}_+\to\mathbb{R}_+$ is called [i]monotonically decreasing[/i] if for any two positive numbers $x$ and $y$ such that $x\geq y$, we have $f\left(x\right)\leq f\left(y\right)$.

Is this a joke? :)

f(f(y)/x) = f(f(x)/y) so f(y)= f(x)/x y so f(x)=k/x where k>0.
so f(x)=1/x
that's it...
-ali

[quote="ali"]Is this a joke? :)
[/quote]

No, it's a trap  :P

[quote="ali"]Is this a joke? :)

f(f(y)/x) = f(f(x)/y) so f(y)= f(x)/x y so f(x)=k/x where k>0.
[/quote]

a little question: what does monotonically increasing mean? i found that it doesn't mean strictly increasing (in a romanian book, where it said: $x_{1}<x_{2} \Rightarrow f(x_{1}) \leq f(x_{2})$), so the function isn't injective as you supposed. i think this may be the trap.

well i think the main mystry is hidden behind these four words:

increasing, non-decreasing AND decreasing, non-decreasing.

Am i right?

if so then the problem statement says strictly increasing

:rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

Have I said "strictly monotonically"? No, f is just monotonically increasing or decreasing, what means that either $x\geq y\Rightarrow f\left(x\right)\geq f\left(y\right)$ for all x and y or $x\geq y\Rightarrow f\left(x\right)\leq f\left(y\right)$ for all x and y.

So the trap perfectly worked, as it did on the TST (I believe I was not the only one who fell into this trap)...

  darij

So, is it the same as "non-increasing" and "non-decreasing"?

Now the problem is difficult.  :(

Wow, seems like i missed quite a lot here.

The fuction is constant between f(x)/y and f(y)/x for all x and y.
CAse1. f(x) is non-decreasing.
Then the above implies that f(x) is a constant between 0 and c. where c< inf only if f be bounded above by some finite number.
So either f(x)=1 or f is bounded above by S.
Now we can easily show that f(x) is constant between 0 and f(x)/x for all x. Since f(X) is bounded above, f(x)/x is not. and therefore
f(x)=1 again solving the problem.

Case2. f(x) is non-increasing.
Note that for any c,if y= f(c/x)/x then f(y)=1/f(c)=f(f(c))
This implies that f(z)=1/z for all z iin R(f).
Let f(x) be bounded by L above. Then f(x) is bounded by 1/L below.
Clearly then y.f(y) is not bounded and this implies that
since f( f(y).y/c) = f(f(c)). The value of function is constant between f(c) and inf and also between f(c) and 0,
thus f(x)=1.
Otherwise f(x) is not bounded and we may also assume that f(x).x is bounded by S and L.
So S/x>f(x) > L/x.

 Remains.... COMING SOON :lol:

i hope i didn't miss something here! i did this quickly
assume that $f$ is a monotically increasing!
if we take $x=1$ we see that $f(y)f(f(y))=1$ for each $y$
then for each $y$ $f(y)\neq0$
consider $x>y$
so $f(x)>=f(y)$ and $f(f(x))>=f(f(y))$ and by multiplying these two inequalities with elaboration with $f(x)f(f(y))$
then  $f(x)(f(f(x))-f(f(y)))+f(f(y))(f(x)-f(y))=0$

so we have $f(x)=f(y)$
then $f:\mathbb{R}_+\to\mathbb{R}_+$
         $   x--->1$

I don't understand this solution? Can you elaborate?

Here's what I did for increasing function. The decreasing function I havent finished. This seems hard for problem 1. And also sorry for the bump :D

Let $P(x,y)$ denote the assertion $f(xy)f(\frac{f(y)}x)=1.$ $P(\frac{f(x)}x,x)$ and $P(\frac 1{f(x)},f(x))$ gives $f(1)=1.$ Then $P(f(x),x)$ implies $f(xf(x))=1.$ Since $f$ is increasing, $xf(x)$ approaches $\infty$ as $x\to \infty$, implies $f\equiv 1;$ clearly works.

We get $f(1)=1$ as above. Let $\mathcal K$ be the set of all $x\in \mathbb R^+$ such that $f(x)=1.$ If $k\in \mathcal K$, then $1/k \in \mathcal K$, as well. Let $\mathcal K$ be bounded by some $M,$ then decreasing implies $f(x)=1$ for all $x$ such that $M\geqslant x \geqslant 1/M$ or for all $x$ such that $M>x>1/M.$ For any $x\in \mathbb R^+$ pick $y$ such that $M>xy>1/M,$ so $M \geqslant f(x)/y \geqslant 1/M$, which implies $f(x)=1/x,$ a working solution. If $\mathcal K$ is unbounded then $f\equiv 1,$ clearly works.


just use $(a+b+c)^{2}\geq 3(ab+bc+ca)$ and Schur $\sum a^{3}+6abc\geq (\sum ab)(\sum a)$

It is problem 10 from Polish Mathematical Olympiad from this year. I'll write the official solution.
At first we will prove that \[\text{(I)}\; a^{3}+b^{3}+c^{3}+6abc\geq(a+b+c)(ab+bc+ca).\] After expanding we have \[\sum a^{3}-\sum a^{2}b+3abc\geq 0\] which is equivalent to \[\sum a(a-b)(a-c)\geq 0\] and it is true by Schur. Now we will prove that \[\text{(II)}\; (a+b+c)^{2}\geq 3(ab+bc+ca).\] It is equivalent to \[\sum a^{2}\geq \sum ab \iff \sum (a-b)^{2}\geq 0\] which is obviously true and we have \[\text{(II')}\; a+b+c\geq \sqrt{3(ab+bc+ca)}.\] Using inequalities $\text{(I)}$ and $\text{(II')}$ and condition $ab+bc+ca=3$ we get \[a^{3}+b^{3}+c^{3}+6abc\geq(a+b+c)(ab+bc+ca)\geq \sqrt{3(ab+bc+ca)}(ab+bc+ca)=9\] wich ends the proof.

Edit: I see I was late.


calculus
derivative
function
limit
calculus computations

$f(x) = |ax-1| + |ax+1|; \quad a >0$

Is $f(x)$ differentiable at $x=0$? If so. Find $f'(0)$

We check left and rigth derivative at $x=0$ .

The function can be written as  $f(x)=a|x-\frac{1}{a}|+a|x+\frac{1}{a}|$ .

Notice that since  $h \to 0$ eventually we have $|h|<\frac{1}{a}$ and $f(0)=2$ .

For the left derivative we have:

$\\ f'_{\ell}(0) = \lim_{h \to 0-} \frac{f(h)-f(0)}{h} = \lim_{h \to 0-} \frac{a|h-\frac{1}{a}|+a|h+\frac{1}{a}|-2}{h} =\\ \\ \\ \lim_{h \to 0-} \frac{-a(h-\frac{1}{a})+a(h+\frac{1}{a})-2}{h} =\lim_{h \to 0-} \frac{2-2}{h} = 0$

and for the right derivative we have

$\\ f'_{r}(0) = \lim_{h \to 0+} \frac{f(h)-f(0)}{h} = \lim_{h \to 0+} \frac{a|h-\frac{1}{a}|+a|h+\frac{1}{a}|-2}{h} =\\ \\ \\ \lim_{h \to 0+} \frac{-a(h-\frac{1}{a})+a(h+\frac{1}{a})-2}{h} =\lim_{h \to 0+} \frac{2-2}{h} = 0$

Since $f'_{\ell}(0) =f'_{r}(0), \,\,\, f'(0)$ exists and $f'(0)=0$ .

I hope it's correct...check again the signs.  :D  :)

Showing differentiability can be done easily: the functions $|ax - 1|$ and $|ax + 1|$ are differentiable everywhere except at $\frac 1a$ and $- \frac1a$, respectively, and so also at zero.  Thus their sum is differentiable at zero, as well.

Even simpler, the question of differentiability at zero reduces to the question of differentiability on the open interval $(-\frac1a, \frac1a)$ and on this interval $f \equiv 2$.

[quote="NaZAf"]That's correct!

but there was no need to find $f'_{\ell}$ and $f'_{r}$ since $a \left( h+\frac{1}{a} \right )>0$ for some $\mathsf{ V_{\epsilon}0}$ & $a \left( h-\frac{1}{a} \right )<0$ for some $\mathsf{ V_{\epsilon}0}$[/quote]

Yes i saw it after i posted all that thing above...  :D


Right now, something came up to my mind.  We know that $\frac{0}{0}=\text{undefined}$.  When we have $\frac{0}{x}=0$, where $x\in\mathbb{R}$, $x\neq 0$.  When we solve for $x$, we get $x=\frac{0}{0}=\text{undefined}$, but in our case, $x\in\mathbb{R}$, $x\neq 0$.  Does this mean that from now on, we'll be saying that $\frac{0}{x}=\text{undefined}\neq 0$, where $x$ holds the conditions above?

Masoud Zargar

when you are at $\frac{0}{x}=0$, you cannot go to $x=\frac{0}{0}$, that is dividing by 0, the correct step would be to multiply by x, and you would get $0=0$, in other words, any nonzero value for x would work

[quote="boxedexe"]Sorry, I'm out of total confusion now.  I see that if $x=\text{undefined}$, then $\frac{0}{\text{undefined}}=0$, correct?[/quote]

nope, that is still undifined.

When it comes to division by zero etc. common logic on known operations breaks down! So even if it seems obvious that you can cross multiply, the operations is wrong and undefined.... dunno why!!!

When you multiply both sides of a relation by 0, you're automatically making the thing true...

When we say x=0/0 it means that x may be any

real number

you can see it in "lim(x)" when it becomes 0/0

In other words in an equation when we find

x=0/0 it means that we have not found x yet!

I think what h_d_hadi is saying is that we assign a "form" 0/0 in limits, for example, the limit of $\frac{sin(x)}{x}$ as x approaches 0.

However, we are not really defining a value at 0, just a very close value which seems to be approached.


What topics you thing that are more important to study in geometry(to do an olympiad)?

thanks

...........
Gabriel

To me, I am going to try focusing on practicing math tournaments and learning about how many things come about, not just memorize certain things. And in preparing, triangles circles solids and basic rules and recognizing the style of word problems are probably what you should study more if you aren't comfortable with a lot of problems. This relates to me since I am also taking geometry. I started 2 weeks studying this kind of stuff and practicing old tests, and I have already improved quite a bit.

How do you get better at geometry in general?


What are SAT's in AMERICA???? I know in here in England they are Exams held every 2 years in school.

The [url=http://www.collegeboard.com/student/testing/sat/about/SATI.html]SAT Reasoning Test[/url] is one of two competing college entrance tests used by most colleges in the United States. The other test is the [url=http://www.actstudent.org/]ACT[/url]. Both tests are developed and distributed by private, nonprofit organizations (with HUGE budgets), and individual colleges decide which test, if either, they prefer as a factor in admission.


function
vector
calculus
integration
derivative
calculus computations

G:C[0,1] into R
G(f) = sumfrom o to 1 squarerootf(x)
Study the differentiability of G at 0.

What, exactly, do you mean by a function from an infinite-dimensional vector space being differentiable?
What do you mean by that sum?

This question doesn't make enough sense to answer.

the sum is the integral from 0 to 1 of the function squarerootabs(f)
the space C[0,1] IS ENDOWED with the sup norm
differentiable means the existence of a continuous linear fuction that approxmimate correcly the difference G(0+f)-G(0)
It means that difference is a linear funtion plus 0(norm f )

I think i have found something:
1. if we calaculate de derivative at o according to any direction we'll find that it deosnt exist and so is the derivative because they must be equal

2.if we cnsider the sequence Fn(x)=x/n then it cnverges uniformilly to o but nor is it for G(Fn(x))/norm(Fn) when n coverge to infinity.

[quote="jmerry"]What, exactly, do you mean by a function from an infinite-dimensional vector space being differentiable?[/quote]
http://en.wikipedia.org/wiki/Frechet_derivative

I see it now it's easy, G doesnt have directional derivative at o function, so it's not differentiable at o.


What is the sum of the first eleven positive odd integers?

The sum of the first n positive odd integers is n^2. Therefore the answer is 11^2=[b]121[/b].

The first 11 positive integers are: 1,3,5,7,9,11,13,15,17,19,21. The sum of that is 121

You could also use the formula:   S=n/2+(2aX(d-1)). Where A is the first number in the sequence, N is the number of things in the sequence (11), and D is the difference. 2, -2.

the sum of the first $n$ odd numbers is $n^{2}$. so $11^{2}$ is $121$


How long does it take until they recieve and grade answers on average?  Just wondering, Im new.  Thanks

So when do you think our scores/grader feedback will be online?

[quote="rrusczyk"]Yep 3-4 weeks is what we're aiming for.  It will still be at least a week before we can be confident we have all the papers.[/quote]

So they are instantaneously uploaded when they are finished being graded...?

Is the 3-4 weeks the time from when the round ends, or from when you recieve everyone's papers? I mean, does grading not even begin until all papers have been recieved?

They will be uploaded within 2 days of being graded, so long as our software is nice to us.

That 3-4 weeks is from the deadline date.  We have grading dates set on Oct 18 and Oct 23.  We expect we'll have them all finished then, and will start uploading the results on the 25th.  As we've never done it before, we don't know how long that will take, but we hope it will go smoothly.  Our guess is that most will have their results on the 25th, some on the 26th, and a handful much later because it will take us while to figure out whose papers they are (some of you didn't put any identification on papers you faxed - if you don't see a little green checkmark next to Round 1 in the scores column by next Tuesday or so, I'm probably talking to you).

i'm just wondering if the 0 in my profile means i really have only 0 points becasue after checking the answers i see i have most of them right. Might it be becacuse that you guys aren't finished with all the gradings yet? ;)  :maybe:

The papers are not graded yet.  Notice all the - marks; those are for ungraded problems.


geometry
symmetry
summer program
MathPath
ratio

From Dr. Zeitz's Symmetry and Recasting Point of View Worksheet at Mathpath

1.) A square is inscribed in a circle which is inscribed ina square. Find the ratio of the area of the large square to the small square.

2:1 as a well known fact:

[hide]The side of the smaller square is $\frac{d\sqrt2}{2}$, where d is the diameter of the circle. The small square's area is $\frac{2d^2}{4}=\frac{d^2}{2}$. The large square has side d. Its area is $d^2$. The ratio would be

$\frac{d^2}{\frac{d^2}{2}}=\frac{2}{1}$. [/hide]

that's correct also, although it is more conventional to see it in the form 2:1. Same thing though


I got 112.5 on the AMC10 A by doing various stupid mistakes (#5,19,10) and leaving 22-23-25 blank. Do you guys have advice on preparing for the ten B?

The easiest points are mistakes, work on eliminating those, get more organized (read rrusczky's article if you haven't), don't trick yourself into being content with an answer unless you're reasonably sure.

The article's here: [url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php[/url]

Also, while the time factor on the AMC 12 is really shaky, time is actually pretty manageable on the AMC 10. I would suggest spending 10-15 minutes at the end checking your work, even if you still have a few blank.

A little anecdote: Last year, I was aiming for a perfect 150 on the AMC 10. I finished everything except #21 in just 45 minutes. I spent the last 30 minutes on #21 and still ended up missing it (I wasted so much time trying to brute-force it), along with #23 and 2, so I got a 132. Moral of story: Don't overextend yourself. As the saying goes, don't bite off more than you can chew.


Find $n$ if $n^3$ and $n^4$ together contain all the digits (0,1, $\ldots$ ,9) exactly once.

[hide]
If $n>21$, then $n^3$ and $n^4$ will have too many digits for it to work. So guess and check, going down from 21, until you find that $n=18$ works.
[/hide]


abstract algebra
group theory
number theory
greatest common divisor
superior algebra
superior algebra unsolved

1Describe $Hom_Z (Z_m,Z_n)$
2Proove that $Hom_Z (Z_m,Z_n) \simeq Z_{(m,n)}$

i will point out only important details in the proof so
that i can ask the questions linked to those details

proof :
$f: Z_m \to Z_n$ with $f(x)=xf(\hat{1})$
we find that $\overline{0}=f(\hat{0})=f(\hat{m})=m\overline{b}$
now we want to find all $\overline{a} \in Z_n$ so that $m\overline{a}=\overline{0}$
if $a<n$ $ma \equiv 0 (mod n) \Rightarrow$ $\exists b \in Z$ so that
$ma=nb$(1)
Consider now $d=(m,n)$

1) $d=1$
 $(1)\Rightarrow n|a \Rightarrow \overline{a}=\overline{0}$
2) $d>1 \Rightarrow \frac{n}{d} | a$

      $\frac{n}{d} |a,a<n$ with $\overline{a}$ has the property that$m\overline{a}=\overline{0}$
$a\in \{ \overline{0},\frac{\overline{n}}{d},\frac{\overline{2n}}{d},...,\frac{\overline{(d-1)n}}{d} \}$
now it says that $Hom(Z_m,Z_n)$ is cyclic generated by the homomorphism
wich coresponds to class $\overline{\frac{n}{d}}$ .

now the questions
is the homomorphism wich coresponds to the class n/d $f(x)=x\cdot \frac{n}{d}$ ?
how is $Hom(Z_m,Z_n)$ cyclic and how are we sure that it is closed ?
is $Hom(Z_m,Z_n)=\{h|h: Z_m\to Z_n$ with $h(x)=x\frac{in}{d}$  and $i=\overline{1,d-1} \}$ and has $Hom(Z_m,Z_n)$ d elements ?

[quote="spx2"]
is the homomorphism wich coresponds to the class n/d $f(x)=x\cdot \frac{n}{d}$ ?
[/quote]
Yes.
[quote="spx2"]
how is $Hom(Z_m,Z_n)$ cyclic and how are we sure that it is closed ?
[/quote]
Well, you've got a proof that the class of n/d is a generator. And it is obvious that you can add homomorphisms between abelian groups (even so that Ab becomes an abelian category).
[quote="spx2"]
is $Hom(Z_m,Z_n)=\{h|h: Z_m\to Z_n$ with $h(x)=x\frac{in}{d}$  and $i=\overline{1,d-1} \}$ and has $Hom(Z_m,Z_n)$ d elements ?[/quote]
Almost. $i$ lies between the natural numbers $0$ and $d-1$.

I wan't to describe the whole thing in another way. I suggest the following preparation: There's a natural bijection
\[ \hom_Z(Z_m,G) \cong \{g \in G : mg = 0\} \]
for every Group G (written additive). This is clear: Map $f \in \hom_Z(Z_m,G)$ to $g : = f(\overline{1})$ and map $g \in G$ with $mg = 0$ to $f : Z_m \to G , \overline{z} \mapsto zg$. The point is, that a homomorphism defined on a cyclic group is entirely determined by the image of the generator. The single rescriction to this image is that is m-torsion.

For abelian $G$, the set on the right side becomes a subgroup of G, the $m$-torsion-subgroup of $G$, and the bijection becomes an isomorphism (simple calucation).

Now let $G=Z_n$. We have to determine the $\overline{z} \in G$ with $m\overline{z}=0$, i.e. $n | mz$. Consider the gcd $d$ of $m$ and $n$. Then the condition $n | mz$ is equivalent to $n/d | z$. So the $m$-torsion-subgroup of $G$ is cyclic and generated by $\overline{n/d}$. This element has order $d$, hence also the group. This shows
\[ \hom_Z(Z_m,Z_n) \cong Z_d \]
By the way, these arguments show also
\[ \hom_Z(Z_m,Z_n) \cong Tor_{1}^{Z}(Z_m,Z_n) \]
and this is also isomorphic to the tensor product $Z_m \otimes_{Z} Z_n$


Consider these three equations

(1) $ y\equal{}x^2\minus{}x\plus{}1$

(2) $ y\equal{}\frac{x^3\plus{}1}{x\plus{}1}$

(3) $ (x\plus{}1)y\equal{}x^3\plus{}1$

and their respective graphs, $ G_1$, $ G_2$, and $ G_3$.  List $ G_1$, $ G_2$, and $ G_3$ in an appropriate order and at the two places between them insert symbols chosen from $ \equal{}$, and $ \subset$ to indicate their relationships ("equals" and "is a proper subset of") between the sets of points on the graphs $ G_1$, $ G_2$, and $ G_3$.

If $ x\plus{}1 \neq 0$, $ G_1\equal{}G_2\equal{}G_3$.

For $ x\plus{}1\equal{}0$, the values of $ y$ :
undefined for equation (2) ;
$ \equal{}3$ for equation (1) ;
can be any value for equation (3)

So $ G_2 \subset G_1 \subset G_3$ is the answer.

How does

[quote="stephencheng"]For $ x \plus{} 1 \equal{} 0$, the values of $ y$ :
undefined for equation (2) ;
$ \equal{} 3$ for equation (1) ;
can be any value for equation (3)[/quote]

help you draw the conclusion that

[quote]So $ G_2 \subset G_1 \subset G_3$ is the answer.[/quote]

Thanks!

Equation (1) includes the point(-1,3) whereas eqn. (2) does not.
equation (3) includes all points on the line $ x\equal{}\minus{}1$ i.e. of the form (-1,k)


License plates in the newly established state of Atlantis consist of three letters followed by three digits.  The first letter must be a vowel, and no letter can be repeated.  The first digit must be one of $ \{1,2,4,6\}$, which corresponds to the location the license was issued.  No digit may be repeated.  How many possible license plates exist?

first one - aeiou- 5
second- 25 other letters
third- 24
fourth 1246- 4
fifth- 9 other digits
sixth- 8

$ 5*25*24*4*9*8\equal{}864000$

You selected one of the ten digits when choosing the fourth digit. Since digits can't be repeated, there are 9 more digits to choose from for the fifth digit.


Just wanted to see if i was right or wrong
11253 base 7 is 2880 base 10


Fie [b]ABC[/b] un triunghi inscris intr-un cerc . Fie[b] M,N,P[/b] punctele de intersectie dintre cerc si bisectoarele din [b]A,C,[/b]respectiv[b] B[/b]. Demonstrati urmatoarele inegalitai:
a) \[3R^{2}\le MI^{2}+NI^{2}+PI^{2}\le 12R^{2}-36r^{2}\]
b) \[MI \cdot NI \cdot PI \le R^{3}\]
c) \[MI \cdot NI+MI \cdot PI+NI \cdot PI \le 3R^{2}\]
d) \[\sum{\frac{1}{{MI \cdot NI}}\ge \frac{3}{{R^{2}}}}\]
e) \[\frac{{MI^{2}}}{a}+\frac{{NI^{2}}}{b}+\frac{{PI^{2}}}{c}\le \frac{{9R^{2}}}{{4S}}\]

[hide="Hint"]$MI=MB=MC=2R\sin\frac{A}{2}\Longleftrightarrow$ $MI^{2}=\frac{4R^{2}(p-b)(p-c)}{bc}=$

$\frac{4aR^{2}(p-a)(p-b)(p-c)}{abc(p-a)}=$ $\frac{4aR^{2}pr^{2}}{4Rpr(p-a)}=$ $Rr\cdot \frac{a}{p-a}$ a.s.o.[/hide]

Asa este. Practic am ascuns in inegalitatile cu lungimile laturilor  inegalitai trigonometrice. Deci pentru a demonstra inegalitatile date de mine trebuiesc demonstrate inegalitatile trigonometrice (unele cunoscute, altele nu). Propun in continuare sa fie demonstrate inegalitatile ce reies din observatia facuta de domnul Virgil Nicula.