function
group theory
abstract algebra
algebra unsolved
algebra

[quote="haus"]Find all non-decreasing functions $ f : \mathbb{R} \rightarrow \mathbb{R}$ such that $ f(x \plus{} f(y)) \equal{} f(f(x)) \plus{} f(y)$ for all reals $ x$ and $ y$.

I apologize for double-posting. I proposed it also in the Preolympiad forum, but no answer was given.[/quote]

[color=green]Let $ P(x,y)$ the property $ f(x \plus{} f(y)) \equal{} f(f(x)) \plus{} f(y).$

- $ P(0,0$ implies $ f(f(0))\equal{}f(f(0))\plus{}f(0)$ then $ f(0)\equal{}0$.

- $ P(x,0)$ implies $ f(f(x))\equal{}f(x)$.

- $ P(x,f(y))$ impies $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$.

Because $ f$ is a non-decreasing function so $ f(x)\equal{}kx,k\ge 0$. Then $ k\equal{}0$ or $ k\equal{}1$.[/color]

I think that $ P(x,f(y))$ gives us $ f(x\plus{}f(y))\equal{}f(x)\plus{}f(y)$?

$ f\equiv 0$
or
$ f(x) \equal{} x$ for all $ x\in\mathbb{R}$
or
$ f(x) \equal{} {1\over a}[ax \plus{} b]$ for all $ x\in\mathbb{R}$
for some $ a\not \equal{} 0,b\in[0,1)$.

Here $ [x] \equal{} \sup\{ n\in\mathbb{Z}: n\le x\}$ for all $ x\in\mathbb{R}$.

The proof though is quite tedious.

[quote="Erken"]I think that $ P(x,f(y))$ gives us $ f(x \plus{} f(y)) \equal{} f(x) \plus{} f(y)$?[/quote]

[color=green]Oh, sorry. A big mistake  :blush: .[/color]

OK, let me outline my proof. (though I usually don't like to do that)

First prove $ A: \equal{} \{x\in\mathbb{R}: f(x) \equal{} x\}$ is a subgroup of $ (\mathbb{R}, \plus{} )$.

Next prove, that all subgroups of $ (\mathbb{R}, \plus{} )$ are either dense or of the from $ t\mathbb{Z}$ for some $ t\in\mathbb{R}$.

Then for $ A$ dense prove $ f(x) \equal{} x$ and for $ A \equal{} 0\mathbb{Z} \equal{} \{0\}\Rightarrow f\equiv 0$.

For $ A \equal{} t\mathbb{Z},t\not \equal{} 0$ prove that $ I: \equal{} \{x\in\mathbb{R}: f(x) \equal{} 0\}$ is
a half-open intervall of length $ |t|$ containing $ 0$.
($ I \equal{} [r,s)$ or $ (r,s]$ with $ s \minus{} r \equal{} |t|$)

Finally prove if $ f$ is a solution for the problem, then so is $ g(x): \equal{} {1\over r}f(rx)$ for any $ r\not \equal{} 0$
and also $ h(x): \equal{} f(x \plus{} s)$ for any $ s\in\mathbb{R}$ with $ f(s) \equal{} 0$.

Use this to assume $ A \equal{} \mathbb{Z}$ and $ I \equal{} [0,1)$, and follow $ f(x) \equal{} [x]$.


Ring Theory
vector
superior algebra
superior algebra unsolved

Let $ (A, \plus{} ,*)$ be a ring with the property that there exist n ,positive integer, so that $ 1 \plus{} 1 \plus{} 1 \plus{} ... \plus{} 1 \equal{} 0$, where $ 1$ repeats n times,and let $ p$ be the smallest prime divisor of $ n$.Prove that $ A$ can't be written as a union of $ p$ subrings(excluding the trivial subrings) of its.

Reunion?  Do you mean union?

Let's see if we can mimic the proof of the simpler version, which involves vector spaces.  My proof uses the primality of $ p$ but not the fact that it is the smallest prime factor.  So perhaps someone will find a mistake.

Suppose our subrings are $ R_1,\ldots R_k$, $ k\leq p$.  Without loss of generality, we can assume $ n\equal{}p$, otherwise consider $ \frac{n}{p} R$ and $ \frac{n}{p}R_i$.

Also without loss of generality suppose that no subset covers $ R$.  Then there is some $ a\in R_1$ such that $ a$ lies in no other $ R_i$.  Pick $ b\in R_2$ with the same property.

Now consider the set $ \{a\plus{}\lambda b | \lambda \in \mathbb{Z}/p\}$.  This set contains $ p$ elements, so we have 2 cases:

1) Each $ R_i$ contains exactly one element.  Then we have some $ a\plus{}\lambda b \in R_2$, but $ b\in R_2$, so $ a\in R_2$, contradiction.

2) Some $ R_i$ contains more than one element.  Then we have $ a\plus{}\lambda b, a\plus{}\lambda' b \in R_j$, so $ \lambda'' b \in R_j$ for some $ j$ and $ \lambda'' \not \equal{} 0$.  But this implies $ b\in R_j$, so $ a\in R_j$, so $ j\equal{}1$,  But $ b\in R_j$, so $ j\equal{}2$ contradiction.


ARML mathleague.org AMC USA(J)MO USAMO geometry email

Hi everyone,

Just a quick note to let you know a little about how ARML team selection will work this year.  Instead of wasting time in our April and May practices and having only four chances to register three scores for team selection, students who are interested in joining the ARML team this year will have six testing opportunities throughout the year, from which each student’s top three scores will be combined with his or her AMC scores to determine who is on this year’s A and B teams.  This will free us to spend more time in our spring practices focusing on content, and I have already begun talking to the other coaches and some of the students about what we can do to maximize the effectiveness of those practices.

More information about how team selection will work is available on the team website at http://mathleague.org/armlsfba.php.  Please note that the first test is on November 8.  I apologize for the short notice, but I have been trying to pin down locations for all six tests at once.  A couple of locations are still not finalized, but I figured I couldn’t wait any longer to get this information out.  Because the testing will be done at actual math contests, each test will have a registration fee; but all fees you pay at contests hosted by mathleague.org this year will be credited toward your registration fees for ARML if you end up joining the team.  We will also be using these contests as fundraisers to help offset the costs of ARML this year, so please encourage your friends to participate in the contests!

I want to win ARML this year and have TWO A teams in the top ten next year, and I am really counting on all of you to help us achieve these goals.  I am sending a direct mailing to all schools in Northern California that participated in last year’s AMC, as well as to all USAMO qualifiers from these schools who were not on last year’s ARML team.  I could use your help to recruit more students and get them excited about starting a new SFBA dynasty.  And if any of you have connections to corporations in the Bay Area that might be interested in sponsoring the team financially I would welcome any assistance in getting me in contact with them.

This letter is being sent to the SFBA-ARML mailing list and posted on the Bay Area forum at http://www.artofproblemsolving.com.  Anyone at AoPS who is interested in keeping up with information about ARML should sign up for the mailing list on the team website.  If you would like to discontinue receiving emails about ARML, you can find information on unsubscribing from the mailing list at the same place.

Tim Sanders
SFBA-ARML Head Coach

This sounds great!
Are any of the locations and times for the first tryout competition public?  I can't find any of them...

all the information about the tests themselves can be found at http://mathleague.org/california.  that page is also linked from the team website i cited above.  again, sorry for the short notice on the first one but i hope to see as many of you there as possible!

I'm a little worried about this--it sounds like our presence at these contests will have to be school-organized. Our school has never directly lent assistance to the ARML team and I don't want to try to make it start now. More specifically, I don't want the (very few) people here who want to do ARML to have their opportunities tied up with whether we can find a parent/teacher chaperone for so many contests (and this first one on such short notice). We've had heaps of trouble getting a chaperone in the past. Is there any way we could dodge the chaperone requirement? Failing that...Gunn people? Can we send a combined team to this thing?

I believe it also says you can make the Chaperone a parent.. so just pick a parent from your team :)

Our parents are worse than our teachers. You don't know what I went through trying and failing to send a chaperoned team to Julia Robinson last year.

I agree with bookaholic. It's a pain to get to a math competition through our school. If we go the chaperone method, we have to organize within ourselves, which is even more troublesome. I liked it better when we just went to ARML practices by ourselves, and screwed everyone else.

And please people, don't spam. This is a RELEVANT thread.

The last part of that analysis is...interesting, bowei

I think most of that new system is awesome. I agree that the school organized section will be difficult for people to do; i'm not sure if our school even know that ARML exists.

I'm slightly confused about the tryout format too... so that page about the "high school championship" == sfba arml tryouts? Does that mean we have to go to all of those and participate in them, which will determine if we qualify or not?

no you don't have to go to all of them.  i'm giving you six opportunities to register three scores - this way everyone has more flexibility in which ones they can go to.  i'm definitely with you on the issue of a chaperone; i know how difficult it can be to find adults who are willing to help.  if you sign up on your own you can show up on your own - i just don't want a teacher signing up two dozen students and organizing things on that side but failing to show up at the contest to help with the extra grading that will entail.  :)  that said, i would love to see you all organize teams from your schools.  the overall idea is i'm definitely trying to make things easier for you all this year, so please feel free to continue sending constructive suggestions..

Okay, cool, thanks! :) I think we'll end up with a group of about 5 people--is that too many to not also bring an adult along? If not, do we leave the "sponsor" box blank?

You sent something to all Northern California schools that participated in AMC - does that include AMC 8?

If not, could you please send something to Harvest Park Middle School?  I think my MC coach would appreciate it.

Another question--the Nov. 8 testing will include all formats listed under "test descriptions" except the Power Question, right? What order will they be in? I have people who might have to leave early. Thanks!

Wait so uh... what is the contest next Saturday? How do we register? Do we need to fill out any forms or something for ARML?

go ahead and register even if you don't have an adult or have to leave early or whatever; we'll do everything we can to make this as easy as possible for everyone.  of course all of you who are not bringing adults to help have to be patient if the grading takes longer than anticipated.  :)  we've sent letters to a few middle schools that participate in amc 10 and 12, and we'll also be sending out a mailing to registered mathcounts schools later this month.  on that note, if any of you middle schoolers are willing to put me in touch with your teachers about possibly hosting a mathcounts-style contest at your school, please email me privately (tim@mathleague.org) so i can get the ball rolling..


trigonometry
factorial
trig identities
Law of Cosines

A regular polygon of $2n$ sides is inscribed in a circle of radius $5.$ The sum of the squares of the distances from any point on the circle to the vertices of the polygon is $600.$ Compute the value of the integer $n.$

Edit: Sorry guys, it's inscribed in a circle of radius $5.$

Here's what I have so far. Too tired to continue. Been doing math all day. (and some chess :P)

The concept might help, but I probably made a careless error in the math:

Without loss of generality, let the chosen point be a vertex of the polygon.
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=885&sid=6f457b2dbecd1fac0931754cedba9856[/img]
Note that if the polygon has $2n$ sides, then the square of the distance from one vertex to the nearest vertex is $2\cdot4^2\left(1-\cos \frac{180^\circ}{n}\right).$ (Law of Cosines) Similarly, the distance from one vertex to the second closest vertex is $2\cdot4^2\left(1-\cos \frac{2\cdot 180^\circ}{n}\right).$ (Red) Performing the same calculation around the circle, the sum of the squares of [b]all[/b] the segments is $8^2+2\sum^{n-1}_{k=1}2\cdot4^2 \left(1-\cos \frac{180^\circ \cdot k}{n}\right)=400.$

.... There must be a better way. I'll come back to this one. ;)

Edit: Regarding a circle of radius 4.  :blush:

Roots of unity maybe? (Remember interm trig class :D)

[quote="4everwise"]Performing the same calculation around the circle, the sum of the squares of [b]all[/b] the segments is $8^2+2\sum^{n-1}_{k=1}2\cdot4^2 \left(1-\cos \frac{180^\circ \cdot k}{n}\right)=400.$

.... There must be a better way. I'll come back to this one. ;)[/quote]

I think that should be $600$.  ;)

After some manipulation, the equation becomes

$64+64\sum_{k=1}^{n-1}(1-\cos\frac{180k}{n})=600$
$8+8\sum_{k=1}^{n-1}(1-\cos\frac{180k}{n})=75$
$8+8(n-1)-8\sum_{k=1}^{n-1}\cos\frac{180k}{n}=75$
$8n-8(\cos\frac{180}{n}+\cos\frac{360}{n}+\cdots +\cos\frac{180n-360}{n}+\cos\frac{180n-180}{n})=75$

The first term cancels with the last term, and so on.  If $n$ was odd, all the terms would cancel out, leaving us with $8n=75$, which is impossible for integer $n$.  Then $n$ is even, and there is one term remaining.  We have

$8n=75+8(\text{some real number between -1 and 1})$

Since $n$ is even, the only possibility is $n=10$.

Not sure about that one...  :(   There should be an easier way, especially since it's a #4.  Maybe cyclic quads and Ptolemy?

I like the diagram, by the way.  :lol:

Cool, so my method did work.

I'll try to find something simpler. :)

AHA! There [i]is[/i] a nice solution to this problem!  :D

$4)$ Without loss of generality, let the chosen point be a vertex of the polygon. Note that we can create $n-1$ inscribed right triangles so that the sum of the squares their legs, in addition to the square of the diameter, is equal to $600!$ (not factorial)
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=886&sid=adda2dfd34605f6dcb6f8d512f4a22c8[/img]

Furthermore, the sum of the squares of two legs in a right triangle is equal to the square of the hypotenuse. (In this case the diameter) Hence, $(n-1)\cdot 10^2+10^2=600\implies n=\boxed{6.}$

Very nice solution!  :)  I guess we should have thought of Pythagorean when the problem asked for the sum of squares...  :lol:

yea thats wat i did

its also nice to note that this only works because there are exactly an even number of sides, cuz then each vertice has a vertice directly opposite of it and hence gives us those right triangles


analytic geometry
graphing lines
slope
geometry
power of a point
radical axis
geometry unsolved

supose that  $ABCD$
is arectangle .M is a point on the side BC.let the txo circles of diameters AM and BC intesect
 at N that is distinct from B ; let L be the point of intersection  of NB and CD .
 prove that ML is perpendcular to AC.

Let $LA$ meet the circle with diameter $AM$ at $A_1$.  Then $\angle{LA_1M}=90^\cdot$, hence $LA_1MC$ is cyclic.
Consider the inversion $I$ of pole $L$ and power $k=LC^2$.  Then $I(C)=C$ and since $LC^2=LN\cdot{LB}={LA_1}\cdot{LA}$ we conclude that $I({A_1})=A$. Consequently $I(M)\in{AC}$.
Let $LM\cap{AC}=\{K\}$.  Then $LN\cdot{LB}=LK\cdot{LM}$, hence $K$ belongs to the circle circumscribed to $\triangle{BMN}$ $\Longleftrightarrow$ $\angle{AKM}=90^\cdot$ $\Longleftrightarrow$ $LM\perp{AC}$.

Obviously, $ BN$ is radical axis of the two circles, hence $ LC^2\equal{}LN\cdot LB \equal{} LP \cdot LM$, $ P$ being the intersection of $ LM$ and the circle $ (ABM)$, consequently $ PM \perp AP \ ( \ 1 \ )$. $ P$ being on $ LM$, from $ LC^2\equal{}LM\cdot LP$ we get $ CP \perp LM  \ ( \ 2 \ )$ ([b]the slope theorem?[/b]), from (1) and (2) $ A, P$ and $ C$ are collinear and $ LM \perp AC$, done.

Best regards,
sunken rock


trigonometry
Euler
function
algebra
polynomial
geometry
complex analysis

[color=darkblue]Let $ u_n \equal{} \displaystyle \left ( 1 \plus{} \frac {1}{2^2} \plus{} \frac {1}{3^2} \plus{} \frac {1}{4^2} \plus{} ... \plus{} \frac {1}{n^2} \right)$.
[size=150]Find[/size] $ lim u_n$ ?[/color]

OR

$ x \minus{} \frac{x^3}{3!} \plus{} ... \equal{}  \sin x \equal{} x \prod_{n \equal{} 1}^\infty\left(1 \minus{} \frac{x^2}{\pi^2 n^2}\right)$

Compare coefficient of $ x^3$ on both sides:

Coefficient of $ x^3$ in $ x \prod_{n \equal{} 1}^\infty\left(1 \minus{} \frac{x^2}{\pi^2 n^2}\right)$ is the coefficient of $ x^2$ in $ \prod_{n \equal{} 1}^\infty\left(1 \minus{} \frac{x^2}{\pi^2 n^2}\right)$. Clearly the coefficient of $ x^2$ is $ \minus{}\sum_{n\equal{}1}^{\infty} \frac1{\pi^2 n^2}$.

Coefficient of $ x^3$ in power series expansion of sin is $ \minus{}\frac1{3!}$.

So:

$ \minus{}\sum_{n\equal{}1}^{\infty} \frac1{\pi^2 n^2} \equal{} \minus{}\frac1{3!}$

$ \sum_{n\equal{}1}^{\infty} \frac1{\pi^2 n^2} \equal{} \frac1{6}$

$ \sum_{n\equal{}1}^{\infty} \frac1{n^2} \equal{} \frac{\pi^2}{6}$

[b]P.S:[/b] Actually this is a proof by Euler for which he won some award. I consider this a very beautiful proof :P

In general an entire function may not be written as such a product of its zero factors though, one requires an exponential factor in front of the product (as well as convergence factors inside the product to make it converge, but here it clearly converges). I mean, in terms of writing such a product, what is the difference between $ \sin x$ and $ e^{x}\sin x$? I find it quite weird that Euler did not notice this; I know that the theory of analytic functions was not developed, but it's so easy to give a counterexample to his construction. I know that a lot of people would argue that Euler had the right idea, just not the tools to express it as rigorously as we require today, but in this case I really think there is a huge omission in logic. I think the easiest theorem that allows one to immediately conclude that $ \sin x$ has no exponential factor is Hadamard's factorization theorem (apply it to $ \frac {\sin \sqrt {x}}{\sqrt {x}})$, and that's fairly deep (well not that deep, but fairly nontirival).

In terms of these things, one explanation why it works for polynomials is that when you divide by the linear factors you get a bounded function, and an entire bounded function is constant by Liouvilles theorem, so no exponential factor is needed (just a constant). In fact, it's a large area of study how different types of growth bounds affect the exponential factor in front of the product (in a more generalized sense), and one may consider Liouville's theorem the first and easiest result in such matters.

I was finishing off the semester in the course "Early History of Mathematics" by spending a lecture talking about Leonhard Euler. And there's so much to say! The seven bridges of Königsberg, $ 641\mid 2^{2^5}\minus{}1,$ $ V\minus{}E\plus{}F\equal{}2,$ and so on. I presented the argument that isomormphism wrote above for $ \sum_{n\equal{}1}^{\infty}\frac{1}{n^2}$ and then I decided on the spur of the moment to go after $ \sum_{n\equal{}1}^{\infty}\frac{1}{n^4}.$ The lesson: don't do that without preparing your notes beforehand - I had a factor of 2 mistake somewhere that I didn't fix until I got back to my office. But you can make that work, and the sixth power, and beyond. Euler went quite a ways beyond.

Kalle I think you're underestimating Euler a bit.  :P

Though named after Lagrange, it was Euler that developed much of the early theory of polynomial interpolation in the Lagrange form.  Taylor's theorem was also known and a popular tool at the time, so Euler no doubt had an explicit bound on the error of Lagrange interpolants.  He would surely have used it to note that the Taylor series of $ \sin$ decays fast enough to ensure that global error decreases as more points are added.  It may not have amounted to proof, but Euler's spectacular intuition certainly wasn't based on faith, either.

Alright, maybe I am not talking about Euler, but rather about the Wikipedia-author-type's understanding of what's essential to the argument. Anyway, I don't think that proving the product expansion of $ \sin$ is just "rigorizing the idea", but actually the most interesting and important step. I'm sure a lot of people would disagree with me, seeing as the argument is often outlined without even mentioning the importance of the growth order of $ \sin x$. Maybe my viewpoint is what it is because I don't find the value of $ \sum \frac {1}{n^2}$ interesting in any way, really.

[quote="Kent Merryfield"]then I decided on the spur of the moment to go after $ \sum_{n \equal{} 1}^{\infty}\frac {1}{n^4}.$[...]But you can make that work, and the sixth power, and beyond. Euler went quite a ways beyond.[/quote]
Please, can you give me some hints to evaluate these sums? I understand that I have to use Euler's method, but I don't know how  :huh:

Thanks

Euler, continued ...

(I was going to do this anyway, whether GioMott asked for it or not. And I may have botched the lecture, but I do know how to do it.)

We start with the identity (yes, Kalle, the truth of it is not trivial):

$ \frac{\sin x}{x}\equal{}\left(1\minus{}\frac{x^2}{\pi^2}\right) \left(1\minus{}\frac{x^2}{4\pi^2}\right) \left(1\minus{}\frac{x^2}{9\pi^2}\right)\cdots\equal{} 1\minus{}\frac{x^2}6\plus{}\frac{x^4}{120}\plus{}\cdots$

If we multiply out the infinite product to make it a sum, we get the following for the $ x^2$ terms:

$ \minus{}\sum_{k\equal{}1}^{\infty}\frac{x^2}{k^2\pi^2}\equal{}\minus{}\frac{x^2}6$

from which we conclude that $ \sum_{k\equal{}1}^{\infty}\frac1{k^2}\equal{}\frac{\pi^2}6.$

(So far, this is just what isomorphism has already said, above).

Now look at the $ x^4$ terms. To get $ x^4,$ we must use the term other than $ 1$ of each factor precisely twice. So we pick each (unordered) pair of integers exactly once. The result is

$ \sum_{j<k}\frac{x^4}{j^2k^2\pi^4}\equal{}\frac{x^4}{120}.$

Of course we will multiply both sides of this by $ pi^4$ and divide both sides by $ x^4.$ Let's do one more thing: multiply both sides by $ 2,$ thus changing from the "upper triangle" $ j<k$ to the set $ j\ne k,$ which is the whole array of ordered pairs $ (j,k)$ except for the main diagonal:

$ \sum_{j\ne k}\frac{1}{j^2k^2}\equal{}\frac{\pi^4}{60}$

The next step is to add the "main diagonal" terms to both sides. This completes the array on the left.

$ \sum_{j\equal{}1}^{\infty}\sum_{k\equal{}1}^{\infty}\frac1{j^2k^2}\equal{}\frac{\pi^4}{60} \plus{}\sum_{k\equal{}1}^{\infty}\frac1{k^4}.$

But the left hand side is now just $ \left(\sum_{k\equal{}1}^{\infty}\frac{1}{k^2}\right)^2.$ Hence

$ \left(\frac{\pi^2}6\right)^2\equal{}\frac{\pi^4}{60}\plus{} \sum_{k\equal{}1}^{\infty}\frac1{k^4}$

$ \sum_{k\equal{}1}^{\infty}\frac1{k^4}\equal{}\frac{\pi^4}{36}\minus{}\frac{\pi^4}{120}\equal{} \frac{\pi^4}{90}.$

[color=green][To be continued in the next post.][/color]

Now let's look at the $ x^6$ term, which requires three nontrivial factors of the infinite product.

$ \minus{}\sum_{i<j<k}\frac{x^6}{i^2j^2k^2\pi^6}\equal{}\minus{}\frac{x^6}{5040}$

$ \sum_{i<j<k}\frac1{i^2j^2k^2}\equal{}\frac{\pi^6}{5040}.$

Now multiply by $ 3!\equal{}6$ to free up the order in that sum:

$ \sum_{i,j,k\text{ distinct}}\frac1{i^2j^2k^2}\equal{}\frac{\pi^6}{840}.$

We want to add in the missing terms on the left, in which the $ i,j,k$ are not all distinct. If we add in $ \sum_{i\equal{}1}^{\infty}\frac1{i^4}\sum_{k\equal{}1}^{\infty}\frac1{k^2}$ that will take care of the case $ i\equal{}j.$ Add in two more terms just like it, and we take care of $ i\equal{}k$ and $ j\equal{}k$. But in the process, we have added the terms for which $ i\equal{}j\equal{}k$ three times, so we must subtract off twice that sum in order to have those terms only once. The result of all this is:

$ \sum_{i\equal{}1}^{\infty}\sum_{j\equal{}1}^{\infty}\sum_{k\equal{}1}^{\infty} \frac1{i^2j^2k^2}\equal{}\frac{\pi^6}{840}\plus{} 3\sum_{j\equal{}1}^{\infty}\frac1{j^4}\sum_{k\equal{}1}^{\infty}\frac1{k^2}\minus{} 2\sum_{k\equal{}1}^{\infty}\frac1{k^6}$

$ \left(\frac{\pi^2}6\right)^3\equal{}\frac{\pi^6}{840}\plus{}3\cdot \frac{\pi^4}{90}\cdot\frac{\pi^2}6\minus{} 2\sum_{k\equal{}1}^{\infty}\frac1{k^6}$

$ 2\sum_{k\equal{}1}^{\infty}\frac1{k^6}\equal{}\frac{\pi^6}{840}\plus{}\frac{3\pi^6}{540} \minus{}\frac{\pi^6}{216}$

Setting aside a generous amount of scratch paper for the fraction arithmetic, we arrive at
\[ \boxed{\sum_{k\equal{}1}^{\infty}\frac1{k^6}\equal{}\frac{\pi^6}{945}}\]
It is theoretically possible to obtain $ \sum_{k\equal{}1}^{\infty}\frac1{k^n}$ for any positive even $ n$ this way, although you cannot skip ahead and the arithmetic and counting arguments get more daunting at each step. I believe Euler pushed it all the way to $ n\equal{}26.$

I will add that methods based on Fourier series and methods based on the residue theorem of complex analysis can also reach all of these sums. While there is a considerable arithmetic burden to each of those methods as well, I think each of them ultimately takes less work than Euler's method for $ n$ larger than about $ 6.$

[quote="jmerry"]There are more refined methods that relate the values of $ \zeta(2n)$ to the [url=http://en.wikipedia.org/wiki/Bernoulli_number]Bernoulli numbers[/url]. Those have a recursive formula (easily determined from the definition) which depends linearly on the preceding values- a substantial improvement over the polynomial complications of the Euler argument.[/quote]

Thank for help. :lol:


geometry
perimeter
AMC
AIME
trigonometry
Pythagorean Theorem
AIME I

In quadrilateral $ABCD, \angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD},$ $AB=18, BC=21,$ and $CD=14.$  Find the perimeter of $ABCD$.

[hide]Drawing the figure in one can see that the missing side is merely $\sqrt{18^2+21^2+14^2}=\sqrt{961}$ which luckily is a tricky square root in which the digits reverse in both the squareroot and the integer so $CD=31$ leaving us with $31+18+21+14=84$[/hide]

[hide="Click to reveal"]Drawing this leaves 2 right triangles ABC and ACD. Find AC via Pythagorean theorem and use the answer to find AD. Add all the sides.[/hide]

Edit: I think the problem states AB is equal to 18 not AD.

[hide]
If you draw the figure correctly, you should get a right triangle $ABC$ with hypotenuse $\overline{AC}$ and right triangle $ACD$ with hypotenuse $\overline{AD}$, and legs $\overline{AC}$ and $\overline{DC}$. Therefore $\overline{AD}=\sqrt{18^{2}+21^{2}+14^{2}}$, which comes out to $\sqrt{961}=31$. Since the question asks for the perimeter, $18+21+14+31=\fbox{84}$[/hide]

[quote="chess64"]Err, from what I heard, a lot of people didn't know that $\sqrt{961}=31$. Just found that really weird... :?[/quote]

Even if someone didn't, they know that $30^{2}=900$, and just have to add 30 and 31 to get $31^{2}=961$.

[quote="ShoeFactory"]They want to lift the confidence of test takers that are not so adept at math. They did the same thing with the AMC 12 this year.[/quote]
I actually took about 10 seconds to think about #1 on the AMC, to make sure I hadn't misread and oversimplified that problem  :blush:

[quote="JJ89"]This one was way too easy.  They shouldn't have wasted one of their 15 slots on it.[/quote]

I dunno. If I took the AIME I, I wouldn't have minded seeing this question as number 1...  :D it's like a warmup/freebie question. Or an SAT math question.

Seriously that was just meant to be a problem for the thousands of people who qualified for the AIME (given that it was pretty easy to qualify but hard to score very high this year on the AMCs).

[hide="solution"]
Using the Pythagorean Theorem twice gives us

\begin{eqnarray*} AD &=& \sqrt{14^2+AC^2}\\ &=& \sqrt{14^2+18^2+21^2}\\ &=& \sqrt{961}\\ &=& 31 \end{eqnarray*}

Thus, the perimeter of $ABCD$ is $18+21+14+31=\boxed{084}$.[/hide]

Haha my friend did #1 (it took him about 10 minutes too) and then he went to sleep. It was pretty funny actually. :rotfl: o_0 and then my other friend who didn't take it, and he's not very good at math either did it in 30 seconds, lol.

[quote="amcavoy"]I agree, this was an easy problem for the AIME.  There were a couple others that seemed to be quite easy.  Was this year's AIME I generally easier than those in the past?  This is the first year I have done the AMC/AIME, but it seems that (from this site and Kalva) past problems were more difficult for the most part.[/quote]

Although it was easy, it wasn't as easy as last year's. That was quite easy... the first 12 were pretty much gimmies...

[quote="zanttrang"][quote="krustyteklown"][quote="zanttrang"]Is it just me, or was this an incredibly easy problem, even for an AIME #1?  I was kind of disappointed that they wasted a whole problem with this, since it didn't even require thinking (and the other early problems, while relatively straightforward, at least required you to apply something more complicated than the Pythagorean theorem).[/quote]

#14 was pythagorean theorem : |[/quote]
I'd be extremely impressed if you could show me how to solve #14 exclusively using the Pythagorean Theorem  :P I meant using the Theorem as the exclusive solution method, not a tool to help find the solution.[/quote]

You could get pretty darn close to solving #14 using only the Pythagorean Theorem. Just by projecting everything onto the plane containing the broken leg and the height, you might need a little trig to find one length, but all the other steps were Pythagorean Theorem. Of course you quite a bit of algebra and arithmetic to work through, but those are the basics.

lol...i got this problem wrong...i did $\sqrt{361}$...

[quote=sugar_rush][quote=RedFireTruck][hide=Solution]
$AC=\sqrt{AB^2+BC^2}=\sqrt{18^2+21^2}=3\sqrt{85}$.

$DA=\sqrt{AC^2+DC^2}=\sqrt{(3\sqrt{85})^2+14^2}=31$.

$P_{ABCD}=AB+BC+CD+DA=18+21+14+31=\boxed{084}$
[/hide][/quote]

lol you should call it "solution without words"[/quote]

[quote=sugar_rush][quote=RedFireTruck][hide=Solution]
$AC=\sqrt{AB^2+BC^2}=\sqrt{18^2+21^2}=3\sqrt{85}$.

$DA=\sqrt{AC^2+DC^2}=\sqrt{(3\sqrt{85})^2+14^2}=31$.

$P_{ABCD}=AB+BC+CD+DA=18+21+14+31=\boxed{084}$
[/hide][/quote]

lol you should call it "solution without words"[/quote]

i didnt notice my solution didnt have words earlier lol. i just typed what was in my head.

Solved with [b]amuthup[/b], [b]js104[/b], [b]crazyeyemoody907[/b], [b]ETS1331[/b], [b]dottedcaculator[/b], [b]v4913[/b], [b]CT17[/b], [b]CyclicISLsclesTrapezoid[/b], [b]squareman[/b], and [b]bluelinfish[/b].

Only one side is unknown. The Pythagorean Theorem gives:
$$
AD^2 = AC^2 + CD^2 = AB^2 + BC^2 + CD^2 = 961 \implies AD = 31,
$$
resulting in an answer of $18 + 21 + 14 + 31 = \fbox{084}$.


function counting distinguishability geometry parallelogram

How do you show that a function or a map is a bijection. Can anyone give me an example?

Im not very familiar with bijections. I know 1-1 correspondences, which i think are virtually the same thing. I think a bijection is if:
Given two sets A and B, for every a in A, there exists exactly one b in B such that f(a)=b for some function f.
Ex.
A={1,2,3,4} B={1,4,9,16}. $ f(x)\equal{}x^2$, since 1^1=1, 2^2=4 ect.

I may be wrong, since i dont know that much.

[b]Note:[/b] That's one of them. There are 24 different bijections between your set A and your set B. We can pick any of 4 elements in B to correspond uniquely with the "1" in A, any of the remaining 3 in B to correspond with the "2" in A, etc.

A bijection, as he said, is just a one-to-one correspondence. They do not necessarily have to be related by some simple function; often, the elements of the sets are not numbers, as is the case in many combinatorial arguments.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204790]This is an example[/url] of a bijection that is not what you would normally think of as a function -- read my first post in that thread after understanding the problem.

i'm pretty sure this doesn't belong around here...

Why not, Poincare?

Bijections are a powerful tool to prove that two things are the same in number. The word is just a name to the common intuition involved in combinatorial problems.

You could look at almost any combinatorial thingy as a bijection. The ways to order ten symbols in sequence, with 6 of one type and 4 of another, the same types of symbols being indistinguishable, is in bijection with the ways to choose 4 objects from a group of 10 distinct objects, which is also in bijection with the 4-element subsets of a 10-element set. Thus there are the same number of each of these.

Of course, there are less trivial examples. The parallelograms formed by the edges in a triangular lattice of side length $ n$ is in bijection with the ways to choose four points from the same side of a triangular lattice of side length $ n\plus{}1$. Thus the number of parallelograms is $ 3\binom{n\plus{}2}{4}$.


If the two internal angle bisector of a triangle are equal, then prove that the triangle is an isosceles triangle.
(prove by Euclidean Geometry)

Thank you........

Well if you have two bisected angles with an equal value, then you have two equal angles - which leads to having two equal and oppsite sides to each corresponding angles.

[hide]Suppose $ \widehat{B} > \widehat{C}$ and $ BD, CE$ are two bisected angles with an equal value. We will prove that $ \widehat{C} > \widehat{B}$
Two triangles $ DBC$ and $ ECB$ have $ BD = BE, \widehat{DBC} > \widehat{ECB}, BC$ is the common side $ \Rightarrow CD > BE$
Let $ BDCD'$ be a parallelogram.We have $ BD' = CD > BE$ and thus $ \widehat{BED'} > \widehat{BDE'}$
Because $ CE = BD = CD', ED'C$ is an isosceles triangle and $ \widehat{CED} = \widehat{CD'E}$
$ \Rightarrow \widehat{BEC} = \widehat{BED'} + \widehat{D'EC} > \widehat{BD'C} = \widehat{BD'E} + \widehat{ED'C}$.
In the other hand, we have
$ \widehat{CED} = \widehat{BAC} + \widehat{ECA} = \widehat{BAC} + \frac {\widehat{BCA}}{2}$
And $ \widehat{BD'C} = \widehat{BDC} = \widehat{BAC} + \frac {\widehat{CBA}}{2}$
$ \Rightarrow \frac {\widehat{BCA}}{2} > \frac {\widehat{CBA}}{2}$
Or $ \widehat{C} > \widehat{B}$
Thus $ ABC$ is an isosceles triangle[/hide]


superior algebra
superior algebra unsolved

Knowing

$ x\plus{}y\plus{}z\equal{}3$
$ x^2\plus{}y^2\plus{}z^2\equal{}5$
$ x^3\plus{}y^3\plus{}z^3\equal{}7,$

determine the value of $ x^4\plus{}y^4\plus{}z^4$

Set $ s_1\equal{}x\plus{}y\plus{}z, s_2\equal{}xy\plus{}yz\plus{}zx, s_3\equal{}xyz$. Then you have:

$ 3\equal{}s_1$
$ 5\equal{}s_1^2\minus{}2s_2 \implies s_2\equal{}2$
$ 7\equal{}s_1^3\minus{}3s_1 s_2\plus{}3s_3 \implies s_3\equal{}\minus{}\frac{2}{3}$

Now $ x^4\plus{}y^4\plus{}z^4 \equal{} s_1^4 \minus{} 4s_1^2 s_2 \plus{} 4s_1 s_3 \plus{} 2s_2^2 \equal{} 9$.


Let $a , b , c > 0$ and $2abc+ab+bc+ca = 1$ . Prove that
\[\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le \frac{3}{2}\]
.

Edit : Yes , i've noticed that is $\le$ .

You can make the substitution $cos(x)=\sqrt{ab}$,$cos(y)=\sqrt{bc}$,$cos(z)=\sqrt{ca}$ with $x,y,z$ angles of an acute triangle.And maybe the inequality is $\sum \sqrt{ab}\leq \frac{3}{2}$


[hide="well i guess"]1958[/hide]
well now that i look at it it seems to be
[hide]1952[/hide]

I dunno. It was from a chapter countdown that i watched and then i also saw it on saab today

No. that's not right

hmm...
[hide]
well it is
$\frac{2^{2009}*5^{1957}}{2^{58}}$
or
$2^{1951}*5^{1957}$
or
$100^{1951}*5^{6}$
so then we now we have the power of 10
with 1952 digits times 125^2
right? i have to get to the stuff i am supposed to be doing but then you have to go further.
more importantly: how will multiplying by 5^6 affect this and
well are these the right steps
[/hide]

Is it [hide]1956?[/hide] because if you get that you will have 1952 digits time 125^2, which has 5 digits, so you add them, but you have to subtract one, because lets look at an easier case: 25*100, thats 2500, which is 2+3-1 so now, its 1952+5-1, which is 1956, so anytime you multiply a number with a  number in the form 10^n the number of digits in the resulting number is the number of digits in that number+n

strange how no one got the same answer...

[hide]Doing some subtraction of exponents, the value of this expressions is

$2^{1951}\cdot 5^{1957}=10^{1951}\cdot 5^{6}$.

You can quickly recognize that $5^{6}=125^{2}$ which will have 5 digits.  Thus, adding 1951 zeroes to that, we get an answer of $\boxed{1956}$.

unless made an "Allen Error"[/hide]

[quote="#H34N1"]How many digits are in the value of the following expression $2^{2009}\times5^{1957}\div 4^{29}$?[/quote]

[hide]
it's $1956$
$2009-48=1951$
$5^{6}$ has 5 digits
[/hide]

$1950$ is incorrect. ProtestanT's 10 year old solution is correct.

ok... I guess if it's to correct an error it's okay


geometry
rectangle
rhombus
geometry proposed

This is a problem I made up myself a few months ago. Once you realize what it’s all about, it should be a piece of cake. :P

Let O be a fixed point, and let $ a$ be a fixed positive real number. Let Q be a variable point such that the length of OQ is positive and less than $ 2a$. Then let P and R be points satisfying the following conditions:

$ \color{white}.\quad\color{black}\begin{array}{rl}
\mbox{(i)} & \mbox{OQPRO is a rectangle.}\\
\mbox{(ii)} & \mbox{If PQ is shifted parallel to itself to P}'\mbox{Q}'\mbox{such that P}'\mbox{Q}'\\
{} & \mbox{and PR mutually bisect each other, then PP}'\mbox{RQ}'\mbox{P is a}\\
{} & \mbox{rhombus with sides of length }a.
\end{array}$

Prove that the locus of P is a circle of radius $ 2a$ with centre O.

[quote="Jane_Bennet"]This is a problem I made up myself a few months ago. Once you realize what it’s all about, it should be a piece of cake. :P

Let O be a fixed point, and let $ a$ be a fixed positive real number. Let Q be a variable point such that the length of OQ is positive and less than $ 2a$. Then let P and R be points satisfying the following conditions:

$ \color{white}.\quad\color{black}\begin{array}{rl} \mbox{(i)} & \mbox{OQPRO is a rectangle.} \\
\mbox{(ii)} & \mbox{If PQ is shifted parallel to itself to P}'\mbox{Q}'\mbox{such that P}'\mbox{Q}' \\
{} & \mbox{and PR mutually bisect each other, then PP}'\mbox{RQ}'\mbox{P is a} \\
{} & \mbox{rhombus with sides of length }a. \end{array}$
Prove that the locus of P is a circle of radius $ 2a$ with centre O.[/quote]
[hide="Solution"]
Notice that $ PR\perp PQ$ and thus $ PR\perp P'Q'$.  Hence, $ PP'RQ'$ is a quadrilateral with perpindicular diagonals that bisect each other.  The second condition implies that that $ PP'RQ'$ is a parallelogram.  This implies that $ PP'RQ'$ is a rhombus.  Now, allow $ PR$ and $ P'Q'$ to intersect at $ O$.  Hence, by the second given condition, we see that $ a \equal{} PP' \equal{} \sqrt {PO^2 \plus{} P'O^2}$.  Notice that $ PO \equal{} \frac {PR}{2}$ and $ P'O \equal{} \frac {PQ}{2}$.  Hence, $ a \equal{} \sqrt {\frac {PR^2 \plus{} PQ^2}{4}} \equal{} \frac {OP}{2}\implies OP \equal{} 2a$.  Hence, $ P$ is on the circle with radius $ 2a$ and center $ O$. [/hide]

Happy New Year to all!

The solution I have in mind is [hide="this:"]It is a well-known fact (then again, maybe not so well known, because I had to discover it myself :P) that the sum of the squares of the diagonals of a rhombus with sides of fixed length is constant. More precisely, if the rhombus has sides of length $ a$, then the sum of the squares of its diagonals is $ 4a^2$. In this case

[img]http://i188.photobucket.com/albums/z198/fairjane/GeometryProblem.png[/img]

the diagonals the rhombus PP′RQ′ are congruent to the sides OQ and QP of the right-angled triangle OQP. Hence $ |\text{OP}|^2\ \equal{}\ 4a^2$. :cool:[/hide]


calculus
integration
abstract algebra
inequalities
real analysis
real analysis unsolved

Ok, stupid question.  I figured this out last week, but I can't retrace my steps.  Suppose $ T: L^2 \mapsto L^2$ is a compact, symmetric integral operator (with real, symmetric kernel) and eigenvectors $ \phi_k$.  Why is the sum of the squares of the norms of the eigenvalues finite?  It has something to do with the finite-rank  approximations of the kernel with the eigenvectors, but I just don't see it

Not every compact self-adjoint operator is Hilbert-Schmidt. You definitely skipped some condition in your question. I assume that it is the square integrability of the kernel. If so, just use orthogonality of eigenvectors $ \varphi_k$ and Bessel's inequality applied to the orthogonal system $ \Phi(x,y)\equal{}\varphi(x)\varphi(y)$.


Find all pairs of positive integers $ (p,q)$, where $ p,q$ are primes and such that

$ p^q \equal{} q^p\plus{}2$

$ p\equal{}1$ is not solution, $ q\equal{}1\to p\equal{}3$ -solution.
$ p\equal{}2\to q^2\equal{}2^q\minus{}2$ - no solution. $ q\equal{}2\to 2^p\plus{}2\equal{}p^2$ - no solution.
If $ p,q\ge 3$, then $ 3\le p\le q\minus{}2$. But $ p^q\minus{}q^p>2$ in these case. Therefore $ p\equal{}3,q\equal{}1$ - unique solution.

[quote="Ahwingsecretagent"]Sorry this is such a trivial problem. I couldn't think of Mihailescu's theorem when I tried to do this so I thought it might be relatively difficult.
Thanks anyway.[/quote]
I don't use these theorem.
Obviosly $ p,q$ must be odd (else $ p^q\minus{}q^p$ odd or divisible 4).
And $ p^x\minus{}x^p$ increase for $ x>p$.


geometry
circumcircle
geometric transformation
reflection
geometry proposed

Let $A'B'C'$ be orthic triangle of $ABC$ prove that Brocard axes of three triangle $AB'C',BC'A',CA'B'$ are concurrent at a point.
[hide="Hint"]Do you recognize this special point ?[/hide]

Let $O,K,H$ be the circumcenter, symmedian point and orthocenter of $\triangle ABC.$ $O_A,O_B,O_C$ are the midpoints of $HA,HB,HC$ (circumcenters of AB'C', BC'A', CA'B') and $D,E,F$ are the midpoints of $BC,CA,AB.$ $X,Y,Z$ are the projections of $K$ on $BC,CA,AB.$ $K_A,K_B,K_C$ are the symmedian points of $\triangle AB'C',$ $\triangle BC'A',$ $\triangle CA'B',$ respectively $\Longrightarrow$ $O_AK_A,$ $O_BK_B,$ $O_CK_C$ are Brocard axes of $\triangle AB'C',$ $\triangle BC'A',$ $\triangle CA'B',$ respectively.

Let the tangents of $(O)$ at $B,C$ meet at $A_0.$ Then $\triangle ABC \sim \triangle AB'C'$ are similar with corresponding symmedians $AKA_0$ and $AK_AD$ $\Longrightarrow$ $AK_A:DK_A=AK:A_0K$ $\Longrightarrow$ $KK_A \parallel A_0D \perp BC$ $\Longrightarrow$ $K_A \equiv AD \cap KX.$ Since $XK$ passes through the midpoint of $YZ$ (well-known), then the intersection of the line passing through $K$ and the midpoint of $YZ$ with the A-altitude $AD$ of $\triangle AYZ$ is nothing but the orthocenter $K_A$ of $\triangle AYZ.$

Since $O_A$ is the orthocenter of $\triangle AEF,$ then it follows that $O_AK_A$ is the Steiner line of the four-line $AB,AC,EF,YZ.$ Hence, Miquel point $A_1$ of this four-line, which is the 2nd intersection $A_1$ of the circles $\odot(AEF)$ and $\odot(AYZ)$ (projection of A on OK) is then the pole of $O_AK_A$ WRT $\triangle AEF$ and $\triangle AYZ$ $\Longrightarrow$ $O_AK_A$ goes through the reflection of $A_1$ on $EF,$ which is nothing but the anti-Steiner point of $OK$ WRT $\triangle DEF,$ in other words the orthopole of $OK$ WRT $\triangle ABC.$ So we conclude that $O_AK_A,$ $O_BK_B,$ $O_CK_C$ concur at the orthopole of $OK$ WRT $\triangle ABC,$ which clearly falls on the 9-point circle $\odot(DEF).$


Find all functions $ f: N\to N$ satisfying $ f_n(n) \equal{} n$ for every non negative integer $ n$.

[quote="mavropnevma"]Shall we assume $ f_n$ is the $ n^{\textrm{th}}$ iterate of $ f$ ? (usually $ f \circ f \circ \cdots \circ f$ is denoted by $ f^n$).[/quote]
$ f_n$ is the $ n^{\textrm{th}}$ iterate of $ f$

The conditions imply that $ f(0)$ may be taken arbitrarily, $ f(1) \equal{} 1$, and for any $ n>1$, $ n$ must belong to a cycle $ n, f(n), f_2(n), \ldots, f_d(n) \equal{} n$ of length $ d \mid n$.
In particular, when $ n$ is prime, its cycle must be of length $ 1$ (i.e. $ f(n) \equal{} n$), or length $ n$.

Therefore we may build $ f$ by taking $ f(0)$ arbitrary, $ f(1)\equal{}1$, then for the first $ n$ for which $ f$ has not yet be defined, a cycle of length $ d$ a divisor of $ n$, containing other elements divisible by $ d$.


A metric calendar has 1 metric year equivalent to our calendar year of 365 days.  A metric year is divided into 10 equal metric months; a metric month is divided into 10 equal metric weeks; a metric week is divided into 10 equal metric days.  To the nearest day of our calendar, how many days are there in 4 metric months, 5 metric weeks and 8 metric days?

[hide]

in one metric year is 1000 days, and 4 metric months, 5 metric weeks and 8 metric days is 458 days = .458 year

.458 * 365 = 167.17

167 days.

[/hide]


If on the website, it is written that my USAMO forms were not received, could/should I fax them in again (and have my teacher re-fax the T-II forms on the 28th? Because he is on vacation)

Sorry if this post is redundant, I sent an email several days back but haven't received a reply.

Yes, I was wondering this as well. My teacher says he's faxed the T-II form in, but it says he hasn't on the updated pdf.

My teacher faxed the T-II form and a S-III form of my friend on the due day, Apr. 23rd, but neither of these are received according to the pdf. Should I ask her to send it again?

I tried faxing my S-III form but got no dialtone on either line, so I mailed it on the 21st.  It should have arrived on the 23rd or 24th, but I have no way of knowing whether it was processed because they haven't updated their list...

Well I guess I'll just try faxing it from my teacher's fax tomorrow...

Yeah, my teacher mailed on like, the 16th or so, and apparently the S-III and T-II forms haven't been received. I got an email back from the AMC people saying that they noted that the forms are on the say, but I find it hard to believe that it would take more than a week to arrive.

I would actually give them a call first to see what they have to say, instead of just faxing in again. However, the AMC offices are probably extremely pressed right now, so you may end up not getting anyone to talk to. If that happens, then just try faxing again.

I faxed my S-III form once on the 22nd and again on the 23rd when the list didn't show that mine had been received. Should I just keep sending it?

If you are unsure, [b]call the AMC office[/b]!  Don't post here the day before the contest and hope that your question gets answered.

I know that we live in an electronic age, but there are these quaint devices called "telephones" that can often get you an instant response.

The "received forms" list on the AMC web site was updated again this morning, so check there first, but at this stage, roughly 24 hours before the start of the contest, a phone call is in order if anything is not right yet.

Yeah, I called them early this morning, they said to check the website around noon their time (10 AM Cali), and voila, my forms were in.


In equilateral triangle ABC, point P is chosen so that the perpendiculars to the sides have lengths 4,5,6.
Find the inradius.

[color=cyan]well, what have you thought about so far?  what good do distances from a point to a segment do?[/color]

[hide]We have a triangle ABC and an interior point P.  Let the length of each side be x.  Thus AB = BC = CA = x.

AP, BP, and CP divide ABC into three smaller triangles, APB, BPC, and CPA, so that APB + BPC + CPA = ABC.  We can now calculate the area of ABC in two ways:

- as the area of an equilateral triangle with side x.  Thus A(ABC) = x^2*sqrt(3)/4.

- as the sum of the areas of three triangles, each with base x and height 4, 5, and 6, respectively.  Thus their areas are 4x/2, 5x/2. and 6x/2, so APB + BPC + CPA = 15x/2.  Thus A(ABC) = 15x/2

Equating these two expressions together, we have

x^2*sqrt(3)/4 = 15x/2

x*sqrt(3)/4 = 15/2 (we can do this since x can't equal 0)

x = 60/(2*sqrt(3)) = 30/sqrt(3) = 10*sqrt(3)

Now, we can use a big clunky formula to find the inradius, or we can do this:

The area of any polygon equals half the product of its apothem (in the case of the equilateral triangle, the apothem is the inradius) and its perimeter.  Thus

A = ap/2

(10*sqrt(3))^2*sqrt(3)/4 = a(30*sqrt(3))/2

300*sqrt(3)/4 = a*15*sqrt(3)

75*sqrt(3) = 15a*sqrt(3)

a = 5.

Thus the inradius has length 5.  Answer.[/hide]

I hope that explains enough.

[color=cyan]That was quite a bit more complicated then necessary.  The important issue is just that the triangle's area is the same no matter what.  So that the sum of the lengths 4, 5, and 6 multiplied by their respective sides must equal the sum of the inradii multiplied by their respective sides.  Since all the sides are the same, this means that the sum of the three lengths 4, 5, and 6 must equal the sum of the three inradii, and thus the length of the inradius is 5.

Two questions:
1) If the point P is not necessarily inside the triangle, what are all possible values for the inradius?
2) Not positive I know the answer to this one, but give it a shot -- in triangle ABC, a <= b <= c.  What is the point P inside the triangle for which the sums of the three distances to the sides is maximized?[/color]

well
i believe the easiest answer is just apply the formula:

A=rs

[quote="JBL"]
2) Not positive I know the answer to this one, but give it a shot -- in triangle ABC, a <= b <= c.  What is the point P inside the triangle for which the sums of the three distances to the sides is maximized?[/quote]

If you haven't seen this problem before, try its simplest case (for an equilateral triangle a = b = c) - the result will probably surprise you!

Well, for an equilateral triangle, it's the same everywhere in the triangle since we can break the triangle into smaller pieces and use 1/2bh for the area, and we can easily see it's the same everywhere. But for a non-equilateral triangle, it's not so easy. When you try to Cauchy it, the inequality goes the wrong way, although the minimum isn't trivial either, as far as I know.


Find all polynomials $f,g$ which are both monic and have the same degree and
\[f(x)^2-f(x^2)=g(x).\]

Let $n$ be the degree of $f, g$. If $n=1$, we have $f(x)=x+c$, so $f(x)^2-f(x^2)=2xc+c^2-c$, which must be a monic polynomial of degree $1$, so $c=\frac{1}{2}$.

Now suppose $n>1, f(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$. The expansion of $f(x)^2$ has a term $2a_{n-1}x^{2n-1}$, which must be $0$, since $f(x^2)$ has terms of even exponent only and $2n-1>n$ for $n>1$. Let $M$ be the largest $k \leq n-1$ for which $a_k$ is nonzero. Then the expansion of $f(x)^2$ has a term $2a_kx^{n+k}$, but $f(x^2)$ only has terms with exponent $2n, 2k$ and lower. Suppose $k>0$. Since $a_k \neq 0$, we have $n+k>n$ so $n+k=2n$ or $n+k=2k$. This gives $n=k$, absurd. Thus $k=0$, so $f(x)=x^n+c$. Then we can check that $c$ must be $\frac{1}{2}$.

yes! its also my solution.

As you noticed there is no need to say that f is monic.(it follows directly after the expansion.)


linear algebra
matrix
superior algebra
superior algebra unsolved

Suppose $ A$ is fixed matrix $ nxn$ on $ C$ if matrixs Satisfying  next relations :
1) $ T^{t}A+AT=0$ form the Algebra Li. (Concerning a usual matrix Li bracket)
2) if $ T$ is nilpotent Matrix Satisfying conditions Li from 1). then $ \exp(T)^{t}A%Error. "expT" is a bad command.
=A$

Please stop calling your problems "problem #[number]", use meaningful titles that describe the problem instead.


A composite number $n$ is [i]Duffinian[/i] if none of its positive factors (except 1) is a factor of the sum $s$ of its proper factors (all factors except itself; includes 1).
For example, 15 is not Duffinian because the sum of its proper factors is $s=9$ and $3$ divides $9$.
Let $p$ be a prime and $k$ a positive integer greater than or equal to 2. Show that $p^k$ is Duffinian.

[hide]
the sum $s$ of the proper factors of $p^k$ is $s=1+p+p^2+p^3+...+p^{k-1}$. Clearly, $s \equiv 1 \mod p^i$, $\forall$ $1<i<k$. For this reason $\not \exists i$ s.t. $p^i|s$, and so $p^k$ is duffinian[/hide]


geometry
3D geometry
tetrahedron
Pythagorean Theorem

A cube with sides of length 4 cm has three face diagonals drawn from a vertex. These three diagonals, along with three other face diagonals, form the edges of a new (6-edged) solid. What is the number of cubic cm in the volume of the new solid? Express your answer as a mixed number.

[hide]I am not certain about all of the information in this reply.  When you draw the figure, you will see that the resulting solid is a tetrahedron with edges length 4sqrt(2).  This has a base of an equilateral triangle, whose area is 8sqrt(3).  The altitude of the tetrahedron is drawn from the centroid of the triangle, which is 2sqrt(6)/3 from a slant height.  The slant height is length 2sqrt(6).  Using Pythagorean theorem, you get that the altitude is 8sqrt(1/3).  Finding the volume now is simple.  Solve for the volume and you get 64/3, or 21+1/3

[/hide]


algebra
polynomial
number theory proposed
number theory

Find all positive integers $p$, such that the polynomial $4x^2+p$ admits a prime value for each of the numbers $0,1, ...,(p-1)$.
Happy new year  :)  :)  :)

I cannot bring it to the end though I guess considering the following estimate:

$\prod _{i=0} ^{p-1} (p+4i^2) \leq \binom{2(p-1)}{p-1}\binom{4(p-1)}{2(p-1)}\binom{8(p-1)}{4(p-1)}...\binom{2^a(p-1)}{2^{a-1}(p-1)}$ (where $a$ is chosen to be the smallest integer satisfying $2^a(p-1) \geq 4p^2-7p+4$)

one is able to get just few cases to check. Any ideas?

Fix $p \in \mathbb{Z}^+$ and let $T_p(x) : = 4x^2 + p$. You want $T_p(x)$ to be a prime for any $x = 0, 1, \ldots, p-1$. Then obviously $p \in \mathfrak{P}$ and $p > 2$. Suppose $p = 4k+1$, for any $k \in \mathbb{N}$. Then $T_p(k) = 4k^2 + 4k+1 = (2k+1)^2$, and subsequently $T_p(k)$ is not a prime. So you need $p \equiv 3 \bmod 4$, viz $p = 8h + 3$ or $p = 8k+7$, where $h$ and $k$ are non negative integers. By means of some computations, you quickly check that making $h = k = 0$ (i.e., $p = 3$ or $p = 7$) works. So in the following admit $h, k > 0$, to argue $T_{8h+3}(h+1) = (2h+1)(2h+7)$ is composite.

Moreover, observe $T_{8k+7}(x) = (2x-1)(2x+1) + 8(k+1) = (2x-3)(2x+3) + 8(k+2)$; consider that $k+1$ and $k+2$ cannot be powers of $2$ at the same time (while $k > 0$) and let $q$ be any odd prime dividing $k+1$ or $k+2$. Hence assume $x_1 = \frac{q-1}{2}$ and $x_2 = \frac{q+1}{2}$, if $q \mid (k+1)$; $x_1 = \frac{q-3}{2}$ and $x_2 = \frac{q+3}{2}$, otherwise. By design, it's apparent $\gcd(T_{8k+7}(x_1), T_{8k+7}(x_2)) \ge q$ and $0 \le x_1 < x_2 < p$ in both cases. So $T_{8k+7}(x)$ can't be a prime for each $x = 0, 1, \ldots, p-1$. Well, guys, I think that's enough...

P.S.: merry Christmas, mathlinkers. It's not too late, is it? :? You know, Christmas is a scent in the air and a feeling in the heart, not just a date on a calendar. :oops:


Let $A,P,B$ lie on a line in that order. Find the locus of point $Q$ such that $\frac{AP}{PB}=\frac{AQ}{QB}$.

[quote="rnwang2"]It follows from the Angle Bisector Theorem that $\overline{QP}$ is the angle bisector of $\angle{AQB}$.[/quote]

Yes.

But, can you identify the locus more clearly? (Is it a circle, a line, or what? And where?)

Put the point $A$ on the origin, the point $P$ at $(p, 0)$, and the point $B$ at $(p,0)$.  Let $Q$ be at $(x,y)$.  The equation now says $\frac{p}{b-p}=\frac{\sqrt{x^{2}+y^{2}}}{\sqrt{(b-x)^{2}+y^{2}}}$

$\Rightarrow \frac{p^{2}}{b^{2}-2bp+p^{2}}=\frac{x^{2}+y^{2}}{b^{2}-2bx+x^{2}+y^{2}}$.  Let $k=x^{2}+y^{2}$, and cross multiply.

$\Rightarrow kp^{2}+b^{2}p^{2}-2bp^{2}x=kb^{2}-2bpk+kp^{2}\\ \Rightarrow bp^{2}-2p^{2}x=kb-2pk\\ \Rightarrow k=\frac{bp^{2}-2p^{2}x}{b-2p}$

Substitute back in for $k$ and solve for $y$.

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

Or, in interms of the original line segments, $y=\sqrt{|AP|^{2}\frac{|AB|-2x}{|PB|-|PA|}-x^{2}}$

[quote="maggot"][quote="Altheman"]we have that $\frac{AQ}{QB}$ is constant, so the locus is a circle (an appolonius circle), evidently, this circle passes through P[/quote]

How to prove it?[/quote]

this is a common problem, the easiest proof is using a cooridnate system, also you can consider the tangent to the circumcircle of ABQ at Q, this hits line AB at a constant point, P, and PQ is constant, thus it is a circle, these facts can be verified using triangle similarity


Hello.  I'm trying to make some posts in several topics, including one for a class I'm teaching.  But I keep getting a message that says

Consecutive posts not allowed for less than 72 hours

I didn't used to get this message.  Do you know how I can fix this?  I really need to finish posting solutions in the topic for the class.

Thanks,
Catherine

24 hours it's best: See, at the world cup topics, 3 days is too much,  lot of things can pass in a WC in only 3 days

[quote="Valentin Vornicu"]It's an anti-spam measure. Users are not allowed to post consecutively in the same topic (but they can post on different topics of course) unless 72 hours have passes.

We might reduce the time from 72 to 24 hours though, I'm not sure yet which is best.[/quote]
But what about posting several sets of problems in the National Olympiad Section?

It will not let me edit, either.  I get the same message.

I post regularly in both the Summer class topic I am teaching in the Advanced Physics Forum, and also in the homework topics for the online class that I am taking.  It will no longer let me do either.  I cannot edit or reply to comments about posts for the AoPS homework, and much more seriously, I cannot post the rest of the solutions my students have been waiting for in physics.

This is for me a serious problem.  I chose the Art of Problem Solving for our physics forum because of the excellent atmosphere here that seemed so conducive to running a mini-forum for my students, where I can give information, post results, and post solutions.  I also chose it because it of the ability to do posts with equations and to edit those posts if the author finds a typo or mistake.  These are also among the reasons I have been recommending the site and its various resources to the parents of my students over other math sites.

I should also note my students and I pay for the Art of Problem Solving classes we take here, a fee I had assumed included the ability to discuss problems with other students online.

How can we have a discussion when we can't respond to posts, questions, or homework problems for three days, or even one day, or if we have to start a new topic for every comment we want to make on the subject in the previous topic?  It won't even let me edit.

If we can't arrange a way to deal with this situation, I will have to go elsewhere than the Art of problem Solving to do these things.

Sincerely,
Catherine Asaro, Coach
Howard Area Math Program

[quote="Valentin Vornicu"]You cannot post consecutively [b]in the same topic[/b]. It's best one edits the post if it is the last one rather than to post again.

I will remove the restriction on the National Olympiad forums anyway, good call.[/quote]

Well, it looks like I may be able to move our physics forum to yahoo groups.  It doesn't have the features for doing equations in LaTex that are available here, so I will have to see what I can figure out.  We will be sorry to have to go elsewhere; Art of Problem Solving had begun to feel like a community. :(

Creating and using additional user names would just be adding more busy work to my already overloaded schedule.

[quote="Ravi B"]Could you create an AoPS blog for your physics class?

(A cheesy way to get around the consecutive posts limit is to create a second username.  But Val might not like that.)[/quote]

[quote="Ravi B"]Could you create an AoPS blog for your physics class?

(A cheesy way to get around the consecutive posts limit is to create a second username.  But Val might not like that.)[/quote]That's the whole point of the limit.

Not being able to edit your posts is a bug that will be fixed (I thought I fixed it , but you are saying it's not fixed, i'm going to check that).

The restriction is logical. Why would one need to post twice [b]consecutively[/b] in a topic, unless that person is either a) a moderator (posting one post of his own, and then a post dealing with some moderating issues) b) a person that likes to spam a lot.

Again you can post as many times as you want in a topic, it's just that you cannot post consecutively (so without anyone else posting) for 72 hours.

Creating another user to avoid this is not a wise idea  :!:

Editing works just fine, the bug is fixed indeed as I specified above.

Personally I feel it's the moderator's job to take care of doubleposts, and the software limits are just an annoyance. Sure, it will prevent some unknowing people from many-posting, but it will annoy experienced users who have a good reason to many-post, and the real spammers will just make another username.

What is a good reason to many-post consecutively [b]immediately[/b] (you can post consecutively after a number of hours has passed, for example I've lowered the timeframe to 36 hours right now)?

I cannot find one really, except very [b]rare[/b] cases, in which the user can ask me to disable the restriction for him in that specific forum.

From http://www.artofproblemsolving.com/Wiki/index.php/Netiquette, entry 3:
[quote]For the most part, double posting is highly discouraged. ...are some exceptions to this rule such as when one is posting a problem ... and then posts the solution right after that... [/quote]

I think it's the moderator's duty to make sure double-posting doesn't occur. Perhaps instead of preventing double posts have a special icon for the moderators for topics in which someone double-posts, flagging it for review?

Immediate double posts are likely to be the result of technical problems (If my connection drops the moment I hit "submit", I'll reconnect and hit it again). There's nothing wrong with blocking posts for a few minutes to avoid this, and it also cuts down on certain kinds of spam.
On the other hand, 36 or 72 hours feels excessive. If I come back a day later to fix a mistake or complete a solution, I want to do that in a new post, so any other readers notice the change.

I agree with jmerry. In general, I think that the restrictions should be user-oriented: rather than to disallow [b]everyone[/b] to make two consecuitive posts for 36/72/whatever hours, the system should somehow identify the users who double post a lot and let the admins (or even mods) review the corresponding topics to see whether the double posting is justified. If it is not not, then you may just send a (pre-composed, if you want) PM to them and, only if it doesn't work and the double posts are clear spam, impose that 36 hour restriction on [b]that particular user[/b]. The same applies to other anti-spam measures. The spammers are relatively few and well-known to everybody, so taking a directed personal action seems more advisable than to create new universal laws. It's all my humble opinion, of course...

Okay, due to peer pressure the time has been minimised to 1 hour. And that's as low as I'll go :)

Valentin, many thanks for your flexibility on this and for your response to my situation with the class.  I much appreciate it.  :-)

I think this has been mentioned elsewhere, but when adding a problem for the resources section, one needs to sometimes add a note that should not be included in the problem.  For this, an immediate double post is needed.  However, I've never noticed this time restriction...maybe there is a bypass somewhere.

Moderators bypass it by default. In plus you should use the tag remark [hide="Remark"]Put here comment[/hide] for the problems in the resources section.


What are your opinions towards the legalization of euthanasia?

i'm happy that in my country it hasn't been legalized yet. I think that everyone should be able to choose himself what to do of his life.

Euthanasia and abortion are two completely different topics. Care to change the subject line of this post? Thanks.

[quote="EUCLA"]I think that everyone should be able to choose himself what to do of his life.[/quote]I think you missed the point of the topic there :)
Anyway the topic is a bit on the edge, so to avoid any problems it is going to be locked ;)


Which one is better?? or Are there differences...(besides the book cover, author, and the title :lol: )

Both are excellent books, they teach you very special techniques to specific types of Olympiad problems. The main difference is that [i]Treasures[/i] is slightly easier than [i]Challenges[/i]

I would definitely get "Treasures" first.

I went trhough the whole challenges book first, and after that I got the treasures. Treasrues has more variety of subjects, in the other hand, challenges is very good at algebraic stuff, which helped me a lot  :)

I can get a preview of the book somewhere?

Go to http://www.amazon.com, they usually have previews for all of the books.


hai
   can u please explain about \index in latex.
   give some example

   how we use that tag the document
   explain deeply

See [url=http://research.silmaril.ie/latex/chapter7.html#indexes]Indexes and glossaries[/url] and [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=lshort-english]The Not So Short Introduction to LaTeX2e[/url] section 4.3. You can choose which language to read it in.


induction
geometry
graph theory
combinatorics solved
combinatorics

I really liked this one when I read the article dedicated to it:
We are given n lines in the plane, in general postion, no 2 paralel, no 3 concurent.
Prove that we can find n-2 triangles with sides on this lines, with no intersection points of lines in their interior.
Good Luck ! :) :D

Yes, it is a very nice problem and we used it for our correspondence program last year.

Let $ \Delta_1, \Delta_2,..., \Delta_n$ be the lines.
Since no two of the lines are parallel, each of the $ \Delta_i$'s for $i=1,2,...,n-1$ meets $ \Delta_n$ in a point, say $M_i$. And since no three are concurrent, the $M_i$'s are pairwise distincts.
Wlog, we may assume that the ordering has been chosen so that $M_1,M_2,...,M_{n-1}$ are in that order on $ \Delta_n$.

For $i=1,...,n-2$ the lines $\Delta_i, \Delta_{i+1}, \Delta_n$ form a triangle $T_i$, which is not necessary one of the faces of the planar graph representation given by the lines (the triangles we are looking for), but these triangles have pairwise no common interior point.
Now, to get to the conclusion, it suffices to remark that if we have a triangle $T$ with sides belonging to the lines, then $T$ contains a face which is a triangle : Let's consider the lines $ \Delta_i$ in the increasing order of their subscript.
If $\Delta_1$ does not meet the interior of $T$ then consider $\Delta_2$.
If $\Delta_1$ meets the interior of $T$ then $ \Delta_1$ meets two different sides of $T$ since no three lines are concurrent. It follows that $ \Delta_1$ is the border of a smaller triangle included in $T$. Then consider $ \Delta_2$.
And so on....

Pierre.

It is not difficult to produces a configuration for which there are exactly $n-2$ triangles. Thus $n-2$ is best possible for a lower bound.

A related problem for which I do not have a solution. I think it may be a difficult one :
What is the maximum number of triangular faces?

Pierre.

Hi!
Unfortunately, I have to disappoint you, but your solution is wrong. At least, in the article from where I read this, this solution, with the idead that triangles don't dissapper, when a line cuts them, was given as a wrong one. The thing is that you count some triangles twice... Really, this problem is very hard...
Is this solution someway official, or is it somebody's that seems right?

You are right, pff... :(
I've seen the article in Kvant.

Pierre.

If you read the article, then I think you also read the solution....
Did you like it ? :)
If you didn't read it, I will post it after a while :)

I think the solution may interest some people here, so post it when you will have some time.

Pierre.

The basic idea is the following, try to move lines, line $i$ with speed $v_i$. Lines are moved parralel to their initial postions. The idea is to move lines, but to keep the sizes of the triangles. So, to keep the triangle created by lines $i,j,k$, we will have a linear relation between $v_i,v_j,{}_v{}_k$. If triangles are less than n-2, we will have a nonzero solution for $v_i$ 's, so we can move some lines. But a collision will take place, I mean some line will pass through the intersection of another 2, showing a triangle that was not taken in account.
That's it :)

I dont understand, could you please explain it a little better? sorry, maybe I am stupid :maybe:  or sleepy :)

[quote="DusT"]
The basic idea is the following, try to move lines, line $i$ with speed $v_i$. Lines are moved parralel to their initial postions. The idea is to move lines, but to keep the sizes of the triangles. So, to keep the triangle created by lines $i,j,k$, we will have a linear relation between $v_i,v_j,{}_v{}_k$. If triangles are less than n-2, we will have a nonzero solution for $v_i$ 's, so we can move some lines. But a collision will take place, I mean some line will pass through the intersection of another 2, showing a triangle that was not taken in account.[/quote]
I'm not sure I understand this. Let triangle formed by lines $i,\ j,\ k$ be one of the triangles with no intersection points in its interior. Now, we move the lines and keep the size of this triangle. But how do we know that some intersection point would not "step" into this triangle (which would still have the same size, but couldn't be counted anymore)? Could you explain this proof?

I solved this in another way:

Let $AB$ be an arbitraty segment with $A$ and $B$ being some intersection points. $AB$ is on the border of exactly two sections (one of which could be infinite). Let in one of these sections be $\angle A+\angle B<180^o$. In that case, we color the "side" of the segment $AB$ inside the section red, in the other case blue (to explain this better, imagine that segments are walls and sections are rooms, and for one room we color the side of the wall which is in that room). We do the same for the other side of $AB$ and repeat this for all segments. It is obvious that every segment has one red and one blue side (that's because if for one section of some segment $PQ$ holds $\angle P+\angle Q<180^o$ then for the other section of the same segment is $\angle P_1+\angle Q_1=(180^o-\angle P)+(180^o-\angle Q)=360^o-(\angle P+\angle Q)>180^o$). This means that the number of segments equals to the number of red sides. One could easily check that for any triangle all of its edge's sides inside it are red. Now, we shall prove that if for some polygon with $a\geq 4$ edges holds that more than one sides of its egdes are red, then they must be next to each other. Suppose that inside $A{}_1A{}_2...A_n$ polygon sides of $A{}_1A{}_2$ and $A{}_iA{}_{i+1}$ are red. Then follows $\angle A_1+\angle A_2<180^o$ and $\angle A_i+\angle A_{i+1}<180^o$, so

$(n-2)180^o=(\angle A_1+\angle A_2)+\angle A_3+...+(\angle A_i+\angle A_{i+1})+...+\angle A_{n-1}+\angle A_n<180^o+180^o+...+180^o+...+180^o+180^o=(n-2)180^o$

what is clearly a contradiction. From this immediately follows that in any polygon with $a\geq4$ edges at most two of its edge's sides are red.

Let $D$ be total number of segments and $F$ total number of finite sections. Denote by $m$ the number of triangles. Counting the number of red sides (remeber that this number equals to the number of segments) we get:
$D\leq 2F+m$
Knowing that $D=n(n-2)$ and $F=\frac {(n-1)(n-2)} 2$ (both identities are well-known from combinatoric, the first one is obvious and the second one could be easily proven by induction) one gets:
$n(n-2)\leq 2\frac{(n-1)(n-2)} 2+m$,
what means
$m\geq (n-2)(n-(n-1))=n-2$. QED.

Very nice!
I know that this problem initially was stated for $n/3$ and $2n/5$ triangles, and only after some time it was proven that the number of triangles is at least $n-2$ !

What about Pierre's question: What is the maximum number of triangular faces?

I have found a quite trivial proof of an upper bound equals to ${ \frac {(n - 1)(n - 2)} 2}$ (an even ${ \frac {n(n - 3)} 2}$ for $ n>3$...) :wink:

Anyone for a better bound?

Pierre.

:wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:
What a nut am I.... as far as I know the best known upper bound is $ [\frac {n(n\minus{}2)} 3]$ where $ [\cdots ]$ denotes the integer part.
But this bounds follows trivially from Bojan Basic's solution above, since each triangle has three red sides so that $ 3m \leq n(n\minus{}2)$.

Pierre.

See
Triangles in Euclidean Arrangements, S. Felsner and K. Kriegel
for a proof basically identical to the one Bojan gave.

Here is my proof:

We need a little lemma:
In each convex polygon that is not a triangle there exist adjacent edges $ AB$ and $ BC$,
such that $ \angle A \plus{} \angle B > 180^o$ and $ \angle B \plus{} \angle C > 180^o$ (all angles are interior angles
of the polygon).

The lemma can be proven using Bojan's idea: First prove that there is one edge with this
property and then show that for one of its adjacent edges the same property must hold.

Now comes the main proof:
Let $ t$ denote the number of triangular faces and $ f \equal{} \frac {(n \minus{} 1)(n \minus{} 2)}{2}$ the total number
of faces. We construct a bipartite graph $ G$ with one vertex partition $ \mathfrak{T}$ consisting of all
triangular faces and one vertex partition $ \mathfrak{P}$ consisting of all non-triangular faces
$ (t \equal{} |\mathfrak{T}|,f \minus{} t \equal{} |\mathfrak{P}|)$.
The edges of this graph are defined as follows: In every non-triangular face $ P\in\mathfrak{P}$
(a convex polygon) pick two adjacent edges $ AB$ and $ BC$ such that $ \angle A \plus{} \angle B > 180^o$ and
$ \angle B \plus{} \angle C > 180^o$.
Let $ X$ denote the intersection point of the two lines running through $ A$ and $ B$
($ X$ lies outside of $ P$), and $ Y$ the intersection point of the two lines running through $ B$
and $ C$ ($ Y$ lies outside of $ P$). The two triangles $ ABX$ and $ BCY$ are non-overlapping.
They are not necessarily triangular faces, but certainly each of them contains a triangular
face. In the bipartite graph $ G$ connect $ P$ to these two triangular faces.
Hence the degree of every vertex from $ \mathfrak{P}$ in $ G$ is 2.
What is the degree of every vertex from $ \mathfrak{T}$? Consider such a vertex, corresponding to
a triangular face $ ABC$ and let $ l_1,l_2,l_3$ denote the three lines bounding the triangle $ ABC$.
Let $ g$ be one of the remaining $ n \minus{} 3$ lines. Each of the triples $ (l_1,l_2,g)$, $ (l_1,l_3,g)$ and $ (l_2,l_3,g)$
creates exactly one triangle (not necessarily a face), but only exactly one of these three triangles
contains the triangular face $ ABC$. Hence the degree of every vertex from $ \mathfrak{T}$ is at most $ n \minus{} 3$.

Doublecounting the edges of the bipartite graph $ G$ we get $ t(n \minus{} 3)\geq 2(f \minus{} t)$. Solving for
$ t$ gives $ t\geq \frac {2f}{n \minus{} 1} \equal{} n \minus{} 2$. q.e.d.

Torsten

[quote="Toasti"]I think the inequality in Bojan's proof should read $ D\leq 2F \plus{} 3m$[/quote]

No.
A triangle is a bounded face, so two red segments for each triangle have been counted in $ 2F$. It remains to add one red segment per triangle, so the term $ m$.

Pierre.

Yes, you are right. I just realized my mistake when reading the paper I mentioned above.

Thanks,
Torsten

and the general problem where the lines can also concur is an unsolved problem (i.e determining the lowest upper bound)
http://www.research.att.com/~njas/sequences/A006066

[quote="Toasti"]See
Triangles in Euclidean Arrangements, S. Felsner and K. Kriegel[/quote]

Do you have the exact reference (paper or link), please?

Pierre.

Have a look here:
http://citeseer.ist.psu.edu/felsner98triangles.html
There you find a linked PS and PDF file.

Torsten

Hey friend i want some advice, do you know where i can find just the Kvant journal? U know i'm very interested in seeing it. Even i saw Die WURZEL, Elm ve Texnika and Matematik Dunyasi but yet i haven't seen Kvant.

and the solution.
Proof is by induction. Let's take $ n \equal{} 3$. then so we have to have one triangle which is trivial to see.
Now let us $ n \equal{} k$ and assume there are $ k \minus{} 2$ triangles with no intersection point interior of them. Let us draw the $ k \plus{} 1$th line. After the new line is drawn, there must be at least one triangle without any intersection point its interior. The worst state is to have the least area triangle. So we have then at least $ k \minus{} 1$ such triangles and we are done with this inductive step.

This is exactly what the paper in Kvant is about : destroying all such attempts of proof :wink:

Pierre.


There was a mile time poll in FF, so it gave me an idea of this.

Ummm... I didn't want to reply at first cuz I'm like a horrible sprinter... lol. I'm a 12 year old and can only do a 17 or 18 second 100 meters. That was last summer... maybe I'm faster now but probably not.

I voted "Never timed myself" or yourself or whatever. :P

I seriously would have gone so much quicker in teh Olympics, but I was winning, so I made the stupid mistake of raising my arms. I guess I just like teasing other sprinters  :P

very honest, mewto. XP

I had an 18 sec 100m last year
and i bumped it down to about 15 flat this year.

I'm really happy.

I suck at short distance. Only 15-16s. :blush:  :blush:

I ran around my unusually large house in 11 seconds. Does that count?

[quote]I can't find a 100m track.
[/quote]

I just measured it out on a road. You don't need a track. haha lol and no, running around your house doesn't count.

Sub-12s.  2 more seconds and I'm in teh olimpix.  No way it's worth the effort.

Ugh, I only had ~14 or so a few months ago, but I think I've improved.

11.6 second 100 meter dash is my personal best.

No.

It's 100 meter sprint and 200 meter dash. :P

i know somone with a 100m time of 48 seconds. seriously. shes lazy and and walked the whole thing. my 100m time is 18.23 :blush:

[quote="lifesapain"]Sub-7. I'm Usain Bolt[/quote]

even Usain Bolt can't run 7 seconds :roll:  :P

[quote="Poincare"][quote="hunter34"]No legs =/[/quote]

really?[/quote]

i cut them off :evilgrin:

[size=59]jk[/size]


trigonometry
geometry
AMC
AIME
complex numbers
AoPS Books
\/closed

[quote="SplashD"][quote="bpms"]My trig needs HUGE improvement. Are you guys planning on making a book on trig soon? I am mad that I just missed the class deadline.
Thanks.[/quote]

I would think that's included in geometry or algebra[/quote]I mean complicated trig. Like for example on one aime problem (#3) I had to draw out a right triangle and set up complex equations because I didn't know one identity.

well if you talk to Mr. Rusczyk he might still let you in the class if it isn't too far along already.

We plan to have an Intermediate Algebra / Trig book at some point, but not until late 2007 at the earliest, and more likely in 2008.

AIME-level trig problems are the meat of our Intermediate Trig / Complex Numbers online course.  As was alluded to in this thread, the enrollment deadline for the class has already passed -- the class is already in its 3rd week.  We sometimes accept late enrollments, but only over the phone.

Have you read AoPS book 2 trig chapters?  It's a great foundation, and if you do all the problems, you should be ok for most trig problems, though those that are 11-15 on AIME might be hard.

I don't think it will be ok with the admins.

In addition to the AoPS books there are some nice notes here http://www.tjhsst.edu/~tmildorf/math/Trigonometry.pdf

[quote="13375P34K43V312"]I've typed up some notes from the Trig and Complex Numbers class. If it's ok with the admins, I'll give you the notes to the full trig lectures (not complex numbers) that I wrote.[/quote]It depends.  If these are indeed your own notes, and not just transcripts or other such rewriting of the lectures, then that's fine.  The classes are copyrighted, however, so please  do not distribute the lectures themselves or anything that is essentially a reworded version of the lectures.

They are indeed my own notes. I follow the given problems and concepts and stuff and transfer them into pdf, but leave some of the concepts that I am already comfortable with, or are really easy, out.

Um, I now feel comfortable with trig since chapter 2 of AoPS v.2 has finally sunken in. So, thanks for the notes they look interesting.


geometry
circumcircle
geometric transformation
reflection
geometry unsolved

$ ABCD$ is a cyclic quadrilateral. The projections of the intersection of its diagonals to sides $ AB$ and $ CD$ are $ E, F$, respectively. Show that the line $ EF$ is perpendicular to the line containing the midpoints of the sides of $ BC$ and $ DA$.

[hide="I highly doubt this is a good solution, but..."]

We define the circumcircle of $ ABCD$ to be the unit circle. Then, we assign each vertex a complex value: the points $ A, B, C, D$ have complex values $ a, b, c, d$, respectively. Then the value of the intersection of diagonals $ AC$ and $ BD$ is equal to:

$ \dfrac{ac(b \plus{} d) \minus{} bd(a \plus{} c)}{ac \minus{} bd}$

We can find the complement of this by noting that it is constructed with respect to points $ A, B, C, D$, and thus, its complementary point has the value given by the construction with respect to the complementary points of $ A, B, C, D$. Thus, if we denote the value of the intersection of diagonals as $ x$, then:

$ \bar{x} \equal{} \dfrac{\bar{a}\bar{c}(\bar{b} \plus{} \bar{d}) \minus{} \bar{b}\bar{d}(\bar{a} \plus{} \bar{c})}{\bar{a}\bar{c} \minus{} \bar{b}\bar{d}}$

Notice that for any number $ p$ on the complex unit circle, $ \bar{p} \equal{} 1/p$. Thus,

$ \bar{x} \equal{} \dfrac{\frac {1}{a} \frac {1}{c} (\frac {1}{b} \plus{} \frac {1}{d}) \minus{} \frac {1}{b} \frac {1}{d} (\frac {1}{a} \plus{} \frac {1}{c})}{\frac {1}{a}\frac {1}{c} \minus{} \frac {1}{b}\frac {1}{d}}$

$ \equal{} \dfrac{(b \plus{} d) \minus{} (a \plus{} c)}{bd \minus{} ac}$

Now we use the fact that the value of a projection of a point $ P$ onto the chord $ MN$ of the unit circle has the value $ p^\prime \equal{} \frac {1}{2} (m \plus{} n \plus{} p \minus{} mn\bar{p})$.

$ e \equal{} \frac {1}{2} (a \plus{} b \plus{} x \minus{} ab\bar{x})$

$ f \equal{} \frac {1}{2} (c \plus{} d \plus{} x \minus{} cd\bar{x})$

Also, the midpoints of $ BC$ and $ DA$ obviously have values $ \frac {b \plus{} c}{2}$ and $ \frac {a \plus{} d}{2}$, respectively.

If two lines $ P_1P_2$ and $ P_3P_4$ are perpendicular then $ \frac {p_1 \minus{} p_2}{p_3 \minus{} p_4}$ must be pure imaginary.

Suppose we try to find the value of this expression for the line $ EF$ and the line connecting the midpoints of $ BC$ and $ DA$. We denote this expression by $ z$.

$ z \equal{} \dfrac{\frac {b \plus{} c}{2} \minus{} \frac {a \plus{} d}{2}}{\frac {1}{2} (a \plus{} b \plus{} x \minus{} ab\bar{x}) \minus{} \frac {1}{2} (c \plus{} d \plus{} x \minus{} cd\bar{x})}$

$ \equal{} \dfrac{(b \plus{} c) \minus{} (a \plus{} d)}{(a \plus{} b) \minus{} (c \plus{} d) \plus{} \bar{x} (cd \minus{} ab)}$

$ \equal{} \dfrac{b \plus{} c \minus{} a \minus{} d}{a \plus{} b \minus{} c \minus{} d \plus{} (cd \minus{} ab)\dfrac{(b \plus{} d) \minus{} (a \plus{} c)}{bd \minus{} ac}}$

$ \equal{} \dfrac{(b \plus{} c \minus{} a \minus{} d)(bd \minus{} ac)}{(a \plus{} b \minus{} c \minus{} d)(bd \minus{} ac) \plus{} (cd \minus{} ab)(b \plus{} d \minus{} a \minus{} c)}$

Notice that $ z$ is pure imaginary if and only if $ z \equal{} \minus{} \bar{z}$. Then we compute $ \minus{} \bar{z}$.

$ \minus{} \bar{z} \equal{} \minus{} \dfrac{(\frac {1}{b} \plus{} \frac {1}{c} \minus{} \frac {1}{a} \minus{} \frac {1}{d})(\frac {1}{b} \frac {1}{d} \minus{} \frac {1}{a} \frac {1}{c})}{(\frac {1}{a} \plus{} \frac {1}{b} \minus{} \frac {1}{c} \minus{} \frac {d})(\frac {1}{b}\frac {1}{d} \minus{} \frac {1}{a} \frac {1}{c}) \plus{} (\frac {1}{c} \frac {1}{d} \minus{} \frac {1}{a} \frac {1}{b})(\frac {1}{b} \plus{} \frac {1}{d} \minus{} \frac {1}{a} \minus{} \frac {1}{c})}$

$ \equal{} \dfrac{(acd \plus{} abd \minus{} bcd \minus{} abc)(bd \minus{} ac)}{(bcd \plus{} acd \minus{} abd \minus{} abc)(ac \minus{} bd) \plus{} (acd \plus{} abc \minus{} bcd \minus{} abd)(ab \minus{} cd)}$

We wish to show that $ z \equal{} \minus{} \bar{z}$. We do this as follows:

$ \minus{} 2a^2b^2cd \plus{} 2a^2bc^2d \plus{} 2ab^2cd^2 \minus{} 2abc^2d^2 \plus{} acd(ac^2 \minus{} a^2c \plus{} cd^2 \minus{} c^2d) \plus{} abd(a^2b \minus{} ab^2 \plus{} b^2d \minus{} bd^2) \minus{} bcd(b^2d \minus{} bd^2 \plus{} cd^2 \minus{} c^2d) \minus{} abc(a^2b \minus{} ab^2 \plus{} ac^2 \minus{} a^2c) \equal{} \minus{} 2a^2b^2cd \plus{} 2a^2bc^2d \plus{} 2ab^2cd^2 \minus{} 2abc^2d^2 \plus{} acd(ac^2 \minus{} a^2c \plus{} cd^2 \minus{} c^2d) \plus{} abd(a^2b \minus{} ab^2 \plus{} b^2d \minus{} bd^2) \minus{} bcd(b^2d \minus{} bd^2 \plus{} cd^2 \minus{} c^2d) \minus{} abc(a^2b \minus{} ab^2 \plus{} ac^2 \minus{} a^2c)$

$ \Rightarrow \minus{} 2a^2b^2cd \plus{} 2a^2bc^2d \plus{} 2ab^2cd^2 \minus{} 2abc^2d^2 \plus{} acd(ac^2 \minus{} a^2c \plus{} cd^2 \minus{} c^2d) \plus{} abd(a^2b \minus{} ab^2 \plus{} b^2d \minus{} bd^2) \minus{} bcd(b^2d \minus{} bd^2 \plus{} cd^2 \minus{} c^2d) \minus{} abc(a^2b \minus{} ab^2 \plus{} ac^2 \minus{} a^2c) \plus{} a^3bcd \minus{} a^3bcd \plus{} ab^3cd \minus{} ab^3cd \plus{} abc^3d \minus{} abc^3d \plus{} abcd^3 \minus{} abcd^3 \equal{} \minus{} 2a^2b^2cd \plus{} 2a^2bc^2d \plus{} 2ab^2cd^2 \minus{} 2abc^2d^2 \plus{} acd(ac^2 \minus{} a^2c \plus{} cd^2 \minus{} c^2d) \plus{} abd(a^2b \minus{} ab^2 \plus{} b^2d \minus{} bd^2) \minus{} bcd(b^2d \minus{} bd^2 \plus{} cd^2 \minus{} c^2d) \minus{} abc(a^2b \minus{} ab^2 \plus{} ac^2 \minus{} a^2c) \plus{} a^2b^2c^2 \minus{} a^2b^2c^2 \plus{} a^2b^2d^2 \minus{} a^2b^2d^2 \plus{} a^2c^2d^2 \minus{} a^2c^2d^2 \plus{} b^2c^2d^2 \minus{} b^2c^2d^2$

$ \Rightarrow a^3bcd \minus{} a^2b^2cd \plus{} a^2c^3d \minus{} a^3c^2d \plus{} ab^2cd^2 \minus{} abcd^3 \plus{} ac^2d^3 \minus{} ac^3d^2 \plus{} a^3b^2d \minus{} a^2b^3d \plus{} a^2bc^2d \minus{} a^3bcd \plus{} ab^3d^2 \minus{} ab^2d^3 \plus{} abcd^3 \minus{} abc^2d^2 \minus{} a^2b^2cd \plus{} ab^3cd \minus{} abc^3d \plus{} a^2bc^2d \minus{} b^3cd^2 \plus{} b^2cd^3 \minus{} bc^2d^3 \plus{} bc^3d^2 \minus{} a^3b^2c \plus{} a^2b^3c \minus{} a^2bc^3 \plus{} a^3bc^2 \minus{} ab^3cd \plus{} ab^2cd^2 \minus{} abc^2d^2 \plus{} abc^3d \equal{} a^2b^3c \minus{} a^2b^2c^2 \plus{} ab^3d^2 \minus{} a^2b^3d \plus{} a^2bc^2d \minus{} abc^2d^2 \plus{} b^2c^2d^2 \minus{} b^3cd^2 \plus{} a^2b^2c^2 \minus{} a^2bc^3 \plus{} ab^2cd^2 \minus{} a^2b^2cd \plus{} a^2c^3d \minus{} ac^3d^2 \plus{} bc^3d^2 \minus{} b^2c^2d^2 \minus{} a^3b^2c \plus{} a^3bc^2 \minus{} a^2b^2d^2 \plus{} a^3b^2d \minus{} a^3c^2d \plus{} a^2c^2d^2 \minus{} abc^2d^2 \plus{} ab^2cd^2 \minus{} a^2b^2cd \plus{} a^2bc^2d \minus{} ab^2d^3 \plus{} a^2b^2d^2 \minus{} a^2c^2d^2 \plus{} ac^2d^3 \minus{} bc^2d^3 \plus{} b^2cd^3$

$ \Rightarrow (acd \plus{} abd \minus{} bcd \minus{} abc)(a^2b \minus{} ab^2 \plus{} ac^2 \minus{} a^2c \plus{} b^2d \minus{} bd^2 \plus{} cd^2 \minus{} c^2d) \equal{} (a^2b^2c \minus{} a^2bc^2 \plus{} ab^2d^2 \minus{} a^2b^2d \plus{} a^2c^2d \minus{} ac^2d^2 \plus{} bc^2d^2 \minus{} b^2cd^2)(b \plus{} c \minus{} a \minus{} d)$

$ \Rightarrow (acd \plus{} abd \minus{} bcd \minus{} abc)(bd \minus{} ac)(a \plus{} b \minus{} c \minus{} d) \plus{} (acd \plus{} abd \minus{} bcd \minus{} abc)(cd \minus{} ab)(b \plus{} d \minus{} a \minus{} c) \equal{} (bcd \plus{} acd \minus{} abd \minus{} abc)(ac \minus{} bd)(b \plus{} c \minus{} a \minus{} d) \plus{} (acd \plus{} abc \minus{} bcd \minus{} abd)(ab \minus{} cd)(b \plus{} c \minus{} a \minus{} d)$

$ \Rightarrow (acd \plus{} abd \minus{} bcd \minus{} abc)[(a \plus{} b \minus{} c \minus{} d)(bd \minus{} ac) \plus{} (cd \minus{} ab)(b \plus{} d \minus{} a \minus{} c)] \equal{} (b \plus{} c \minus{} a \minus{} d)[(bcd \plus{} acd \minus{} abd \minus{} abc)(ac \minus{} bd) \plus{} (acd \plus{} abc \minus{} bcd \minus{} abd)(ab \minus{} cd)]$

$ \Rightarrow (acd \plus{} abd \minus{} bcd \minus{} abc)[(a \plus{} b \minus{} c \minus{} d)(bd \minus{} ac) \plus{} (cd \minus{} ab)(b \plus{} d \minus{} a \minus{} c)](bd \minus{} ac) \equal{} (b \plus{} c \minus{} a \minus{} d)[(bcd \plus{} acd \minus{} abd \minus{} abc)(ac \minus{} bd) \plus{} (acd \plus{} abc \minus{} bcd \minus{} abd)(ab \minus{} cd)](bd \minus{} ac)$

$ \Rightarrow \dfrac{(acd \plus{} abd \minus{} bcd \minus{} abc)(bd \minus{} ac)}{(bcd \plus{} acd \minus{} abd \minus{} abc)(ac \minus{} bd) \plus{} (acd \plus{} abc \minus{} bcd \minus{} abd)(ab \minus{} cd)}$

$ \equal{} \dfrac{(b \plus{} c \minus{} a \minus{} d)(bd \minus{} ac)}{(a \plus{} b \minus{} c \minus{} d)(bd \minus{} ac) \plus{} (cd \minus{} ab)(b \plus{} d \minus{} a \minus{} c)}$

$ \Rightarrow \minus{} \bar{z} \equal{} z$

so we are done.

[/hide]

Let I, J be the midpoints of $ AD, BC. X=AC\cap BD, EX\cap DC=\{N\}, FX\cap AB=\{M\}$
We need to show that $ \vec{IJ}.\vec{EF}=0$
$ \Leftrightarrow \frac{1}{2}(\vec{AB}+\vec{DC})(\vec{EX}+\vec{XF})=0$
$ \Leftrightarrow \vec{AB}.\vec{XF}+\vec{DC}.\vec{EX}=0$
$ \Leftrightarrow AB.XF.cos \angle BMX+DC.EX.cos \angle XND=0 (*)$
But $ \Delta AXB\sim \Delta DXC$ then $ \frac{EX}{AB}=\frac{FX}{DC} \Rightarrow EX.DC=FX.AB$
${ \angle BMX=180^o-\angle ABX-\angle MXB=180^o-\angle ABX-(90^o-\angle XDC}=180^o-\angle XCD-(90^o-\angle XAB)=\angle XNC$
Then cos $ \angle BMX=-cos \angle XND$
therefore (*) is true. We are done!

JUST LIKE THE IMAGE :$ {T,R,J,K}$ are midpoints of $ {CE,DE,BE,AE}$; by $ {\angle CTE =\angle BGE=90}$ ;we get :$ {FT =\frac {1}{2} CE = NR}$; $ {GR=\frac{1}{2}BE=NT}$;
and easy to get :$ {\angle FTN=\angle FTE+\angle ETN=2\angle DCA+ETN=2\angle ABD+\angle ERN =\angle ERG+\angle ERN=\angle NRG}$  $ {\Rightarrow}$
$ {\triangle NFT\cong \triangle GNR}$ so $ {FN=GN}$; thus,  $ {FN=GN ,GM=MF}$ $ {\Longrightarrow}$ $ {MN}$ is perpendicular to $ {FG}$ and $ {MN}$ bisect $ {FG}$ .
[hide][color=red]This prblem is quite like CMO 2009/DAY1 /P1(see:http://www.mathlinks.ro/viewtopic.php?p=1371287#1371287)[JUST THE FIRST PART][/color][/hide]

[hide]
Let $ M, N$ be the midpoints of $ AD, BC$, respectively. Let $ P$ be the intersection of diagonals. Let $ Q, R$ be the reflections of $ P$ over $ AB, CD$ respectively. We know that the quadrilaterals $ AQBP$ and $ DPCR$ are directly similar and are kites. Then the midpoints of the lines connecting corresponding points of the two similar figures forms a figure similar to the original, i.e. $ MENF$ is similar to both $ AQBP$ and $ DPCR$. Therefore $ MENF$ is a kite and $ EF \perp MN$, as desired.
[/hide]


Can someone help me in proving the following:
Suppose that $ {X_{i}}$ is sequence of iid $ Ber(\alpha)$ r.v's and $ Y$ is a nonnegative integer valued r.v. independent of $ X_{i}$'s then $ E\left(\left(\sum_{n \equal{} 1}^{Y}X_{n}\right)_{[k]}|\,\,Y\right) \equal{} \alpha^kY_{[k]}$ where $ X_{[k]} \equal{} X(X \minus{} 1)(X \minus{} 2)\,.\,.\,.(X \minus{} (k \minus{} 1))$

\[ E\left(\left(\sum_{n \equal{} 1}^{Y}X_{n}\right)_{[k]}|\,\,Y\right)\equal{}\sum_{m\equal{}k}^\infty E\left(\left(\sum_{n \equal{} 1}^{m}X_{n}\right)_{[k]}\right)I_{[Y\equal{}m]},\]
 and for each $ m\geq k$,
\[ E\left(\left(\sum_{n \equal{} 1}^{m}X_{n}\right)_{[k]}\right)\equal{}\sum_{j\equal{}k}^m \frac{j!}{(j\minus{}k)!}\frac{m!}{(m\minus{}j)!j!}\alpha^j(1\minus{}\alpha)^{m\minus{}j}\equal{}\frac{m!}{(m\minus{}k)!}\alpha^k.\]
We are done.


algebra
polynomial
vector
linear algebra
matrix

How would you find the dimension of the subspace of $\mathbb{P}_{3}$ spanned by the subset $\{t,t-1,t^{2}+1\}$?

How would you find a basis and dimension of the given subspace ${\mathbb{R}^{3}}$:

$\{[a,a-b,2a+3b]|a,b, \in \mathbb{R}\}$ as a subspace of $\mathbb{R}^{3}$?

Thanks

For the first question, are the three polynomials independent? That is, if $at+b(t-1)+c(t^{2}+1)$ is identically zero as a polynomial, must we have $a=b=c=0?$ If they are independent, then they span a set of dimension 3. If not, then one of them is a linear combination of the others; delete it as redundant and try again with what is left.

In the second case, we can write a typical member of that set as $au+bv$ for certain vectors $u$ and $v$ (which you should write down specifially). What does that tell you?

yes they are independent because we get the RREF.

For the second one the dimension is 2, because we have two dimensional vectors in three dimensional space.

The basis would be $a\begin{bmatrix}1\\1\\2 \end{bmatrix}+b\begin{bmatrix}0\\-1\\3 \end{bmatrix}$


vector
superior algebra
superior algebra unsolved

Here is the second (and fortunately the last) problem I'm struggling with, this time from algebra and functional analysis.

Let $ L$ be $ n$-dimensional space., $ M \subset L$ - $ m$-dimensional subspace. Prove that there exist linear functional $ f_1 , ..., f_{n\minus{}m} \in L$ such that $ M\equal{}\{l \in L : f_1 (l)\equal{}...\equal{} f_{n\minus{}m} (l) \equal{} 0 \}$.

Thank you once again for help.

Nobody1111

this does not belong to superior algebra, it's just trivial linear algebra. you may restrict to the case $ K^m \subseteq K^n$, and then everything is clear.

[quote="Nobody1111"]Here is the second (and fortunately the last) problem I'm struggling with, this time from algebra and functional analysis.

Let $ L$ be $ n$-dimensional space., $ M \subset L$ - $ m$-dimensional subspace. Prove that there exist linear functional $ f_1 , ..., f_{n \minus{} m} \in L$ such that $ M \equal{} \{l \in L : f_1 (l) \equal{} ... \equal{} f_{n \minus{} m} (l) \equal{} 0 \}$.

Thank you once again for help.

Nobody1111[/quote]

Hello Nobody1111.

Choose a basis $ a_{1}, ..., a_{m}$ of $ M$. Complete it to a basis $ B$ of $ L$ with vectors $ b_{1}, ..., b_{n \minus{} m}$.
For $ i \equal{} 1,\ 2,\ ...,\ n\minus{}m$, let $ f_{i}$ the linear functional defined by $ f_{i}(x) \equal{}$ the scalar coefficient of $ b_{i}$ in the decomposition of $ x$ along the basis $ B$.

Friendly,
P.


algebra
polynomial
inequalities
algebra proposed

Let $a_1,a_2,\dots,a_{n-1}$ denote arbitrary integers.then the polynomial $\frac{x^n}{n!}+a_{n-1}\frac{x^{n-1}}{(n-1)!}+\dots +a_2\frac{x^2}{2!}+a_1x\pm 1$ is irreducible over rationals.

I sent to Myth an article of Filaseta where he proves some more general assertion. Myth, could you please put the article?

WOW! I see you posted in the proposed section algebra? Sam-n, do you have short or easy solution for this monster?

yes this problem is due to Schur, in my material this problem is solved with Ideal theory approach (Strange solution! :( ) so if you want i post it but i want another approach if it is possible :D


How do you think the last few problem on the AMC 12 compare to the AImE?

This was discussed before.

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=AIME+AMC+difficulty&t=70117

As a result, this thread is locked.


The topic title really should be self-explanatory.

No, but I do a lot of Flash work and store it on my Photobucket account.

http://www.photobucket.com/albums/v259/drspyder

[quote="GoBraves"][hide]No but I did save a ton of money on my car insurance by switching to Geico[/hide][/quote]

That's a good one! :cool: