Find the value of k so that 2x^2 -3x -3k = 0 has only one root for x

For any quadratic polynomial of the form ax^2+bx+c, where a is not equal to 0, the equation has only one root if that very root has multiplicity of two. For this to happen the discriminant (b^2-4ac) must exactly equal zero.

In this equation we have: $ a\equal{}2; b\equal{}\minus{}3; c\equal{}\minus{}3k$

$ b^2\minus{}4ac\equal{}0\Longrightarrow(\minus{}3)^2\minus{}4(2)(\minus{}3k)\equal{}0\Longrightarrow9\plus{}24k\equal{}0$

$ k\equal{}\boxed{\dfrac{\minus{}9}{24}}$


\sum (k=1..n) a_k=0
 \sum (k=1..n) (a_k)^2=1

find a function f(n) such that for any sequence a_k satifying the above conditions, |a_k|<=f(n) and there exists at least one sequence a_k with at least one term a_i such that equality occurs ie a_i=f(n)

I think f(n)=((n-1)/n)1/2. Is that obvious?

i agree for even k.  but what if k is odd?  can we make the bound tighter?

My answer is correct in all cases.
a1,...,an-1=-(1/n(n-1))1/2, an=((n-1)/n)1/2.


I recently read this short story called "The Lottery" by Shirley Jackson (sp?). Anyway, it's about winning lottery means getting hit by stones and maybe killed I think. You can find the full version of the story in Google. My question is, why is it called "lottery" when it's bad? I mean, isn't lottery something that you buy and gives you money (big money!)?

I'm asking this because I wonder if there are other meanings of lottery or maybe there were different types of lotteries in small towns in past or something.

Hmm.. Thanks.  :D

After you win the big money, you will begin to worry that the robbers will come your house at night,  you will begin to worry that your relatives and friends get to know it and come to share your money, you will begin to worry that your wife or husband will take the money away secretly, you will begin to worry................

At last, the worrying kills you.

Erm, shobber, no.  That has nothing whatsoever to do with the story nor is it a correct answer to his question.

Go look in a dictionary!  The word has two major meanings: one is a contest like those for which you can win various sums of money (where the winners are determined either by choosing tokens or playing pieces of some sort, some of which are pre-determined winners, or by a random process such as the drawing of numbers).  The alternate major meaning is [i]any[/i] activity or situation whose outcome depends on chance (or, possibly, fate) alone.

[quote="JBL"]The alternate major meaning is [i]any[/i] activity or situation whose outcome depends on chance (or, possibly, fate) alone.[/quote]

Yeah, I just read that in dictionary.com. But still, my question is this: Why there were such events? I know this is a fiction but still, why would anybody in any town do this "random event" to throw stones at "selected person"? Are they doing this for fun? Or maybe punish them for their bad acts (which doesn't make sense since you don't know who will get selected)? The purpose of whole "lottery" is simply vague and retarded. Maybe someone can tell me what it is.

[quote="juicybooty911"]I read that story in 7th grade. I didn't like it too much, it's really creepy. The meaning of lottery in that book is opposite of what we thinks today.[/quote]

It is horror so it is bit creepy.. especially when you had to see it in 7th grade. I don't think the whole story is bad story though. I just don't understand why would author create such event? What is the purpose?

I think it was a witch hunt, or something like a human sacrifice to ensure a good harvest. I guess the story was set in a time of superstition.

Or, perhaps, the author is trying to make some sort of political or social point- that's generally the case in literature. Why don't you look up some analyses on the story? That would be a good place to start.

[quote="catcurio"]The lottery is done as an annual sacrifice for a good harvest. The oldest man (I forgot his name) says something that indicates that. It's set in a rather primitive society, so they have those superstitions. I think the name of the lottery is purposely ironic, since it's the opposite of what we associate with lotteries now. The story is also an allegory, and the author is criticizing several aspects of human society.[/quote]

That's what I was guessing. Either religious or whole town sacrifice like ancient people would do. Thanks for all your help, I think I finally got it.


[tex]x^4+y^4+z^4 \ge x^2yz+xy^2z+xyz^2[/tex] by Rearrangement, with equality holding when x=y=z. To get the desired inequality, divide by xyz

[quote="beta"]Show that [tex]\displaystyle\frac{x^3}{yz} + \frac{y^3}{zx} + \frac{z^3}{xy} \geq\ x + y + z[/tex] for any positive reals x, y, z. When do we have equality?[/quote]

This is from Canadian MO 2002.

The problem is basically a corollary of the Muirhead's inequality: Since [tex]
(3, - 1, - 1) \succ (1,0,0) [/tex], so [tex]\sum_{\rm{sym}} x^3 y^{-1} z^{-1} \geq \sum_{\rm{sym}} x [/tex].

Note that [tex]a^2+b^2\geq\ 2ab [/tex]

What I did was [tex]x^4+y^4\geq\ 2x^2y^2[/tex]
                         [tex]y^4+z^4\geq\ 2y^2z^2[/tex]
                         [tex]x^4+z^4\geq\ 2x^2z^2[/tex]
add all them up
[tex]2x^4+2y^4+2z^4\geq\ 2x^2y^2+2y^2z^2+2x^2z^2[/tex]
divide by 2
[tex]x^4+y^4+z^4\geq\ x^2y^2+y^2z^2+x^2z^2[/tex]

[tex]x^2y^2+y^2z^2\geq\ 2xy^2z[/tex]
[tex]x^2z^2+y^2z^2\geq\ 2xyz^2[/tex]
[tex]x^2z^2+x^2y^2\geq\ 2x^2yz[/tex]
add all them up and divide by 2
we have
[tex]x^2y^2+y^2z^2+x^2z^2\geq\ xy^2z+xyz^2+x^2yz[/tex]
transitive prop.
[tex]x^4+y^4+z^4\geq\ xy^2z+xyz^2+x^2yz[/tex]
divide xyz on both side give up the desired inequality

You can use Rearrangement, you can use Power Mean, or you can use (certainly the prettiest way):

(x^4+x^4+y^4+z^4)/4 >=x^2yz

(x^4+y^4+y^4+z^4)/4 >=xy^2z

(x^4+y^4+z^4+z^4)/4 >=xyz^2

And sum to obtain the desired result.


I simply started this thread  :P  I think it is necessary to clear the cutoff  :)

Great with single digits in chemistry, even if I end up getting a top 100 rank in IIT, I wont be ranked. :D

And please merge this with the JEE thread. Only solns need a sep thread.

same here with a slight modification for me it is physics and not chemistry though my Open test doesn't suggest it(bad solutions :mad: ). :stink:  :stink:

Is anybody sure about the existence of such a cutoff, I don't remember seeing such a thing in the form. And POST ONLY IF YOU KNOW, NOT YOUR OPINION.

And is the cutoff there for each paper or for a subject overall???

I think the term 'was' might be correct. :D  :D

Cut offs exist for each subject in IIT JEE - Mathematics, Physics and Chemistry. There is an overall cut off as well. And one has to necessarily clear the cut off for individual subjects and only then will he/she be considered for ranking. I know of people who have comfortably cleared the overall cut off, enough to get a rank in the top 1000 and fall short in say one subject, and hence don't make it to the list at all.
Sathej

No way i might clear the phy cut off!!!  :(  And how much will the cut off be in genl for a paper like fiitjee  :?:

I think overall a 35 in each subject will be the min.cutoff for the FIITJEE(though some marks should be reserved for their bad solutions :rotfl: ) I think u will cross the cutoff just make sure that u and ofc me study and practise physics more to ensure a good rank in IITJEE

dei u also use this oxford university often  :P

dei pardesi...........
Oxford university oru periya university nu oru paechuku kooda sollakudatha da...........

rombathan ovara pannare

enathan periye universitya erundhakoodai....adhu enodu copyright da...engu ethukon oru 5 dollar thali vidu :D

[quote="valeriummaximum"]don forget me.....
i am as bad in phy and maths as the vice chancellor of oxford university[/quote]
[size=200]ENGLISH NV[/size]

I think this year around 2.5 lakh students are appearing for the IITJEE.We should atleast get the top 1 percentile to go by stats to get through(EVEN IF CIVIL ENGINEERING IS OK).Coming to the FIITJEE point of view,percentiles are not going to be affected by the wrong solutins.i have also heard that most of the students of Andhra and Kota donot write FIITJEE. keeping in mind these facts we should be getting atleast 98 or 99 percentiles in all our tests
(I SHOULD MENTION THAT MOST OF GOOD sTUDENMTS OF THE PREVIOUS YEARS HAVE BEEN CONSISTENTLY GETTING THE 99 PERCENTILES)

The 98 and 99 percentiles of the fiitjee exam have got close to an average of 240-250.
Cutoffs on average would be
Physics-65
Chemistry-95
MAths-75
(approx.)

I should say these marks are based on the facts,so SHREYAS shouldnt scold me for genuineity

I beg to differ with Nitish. This is because FIITJEE has got nothing to do with your JEE performance. And if the people you are talking about got high marks in FIITJEE,you must have noticed their JEE ranks. Satish got a standard 99 plus percentile and ended up 199!!, which is pathetic for his standards. Vinod got 96+ and got 599, again not that great given his standards. Arjun got 123, quite his standards but only after something like 94+ in FIITJEE.. Morever in FIITJEE, the equivalent cutoff will be around 35 per subject as Rak mentionaed. Cos in IIT, atleast 20 percent of the people writing are taken in and hence the cut off per subject is not kept too high. I think neither for Rak or Baby in Phy, or Shre n I in chem, the subject cut off clearance will be a big deal.. This is because the IIT standards are definitely easier than FIITJEE and your cut off even if is 60, is easily achievable per subject. And coming to the point of Sathej, yeah there might have been people who have cleared overall but missed individual cut off, but that is in incredible cases. I guess if one has prepared for JEE sincerely for 2 years, done the best stuff, been consistent through out, have worked out stuff from the best place available, and have been dedicated to your guru,  just don't worry. Rak, Baby, Shre, Shruthi, Physici Olympiaci,Valerium, Nitish, I know all of us safely satisfy these necessities.. And I guess so do Pardesi, Anit Sahu n many others all over India..  FIITJEE is just a practice test series and not a yardstick measure. And if you clear 35 genuine marks in FIITJEE per subject, in spite of his wrong solutions for the other questions, I think you are well on track to clear atleast the subject cutoffs for JEE.. Jus keep cool, do wat you have to, and the result will come searching for you..
Kalakkaron :read:  Jaikkaron...  :10:

i spent most of my past two years with physics doing completing almost all the possible classical eletcrodynamics(took most part of last year),Optics,Mechanics(a bit of lagrangian),and a hell lot of thermodynamics :D
and u acl this consistency

[quote="Chapplikka"]Kalakkaron Jaikkaron... [/quote]

Same old dialogue da to no avail . Adhu A-M-M-A-N  T-R-Y  :P  :rotfl:   :rotfl:

Well written. I dunno about Anitsahu, but rest of us yes.
[hide]Obviously because I don't know him that well. You couldn't have thought otherwise[/hide]

What I'd like to know is that is the cutoff for the entire subject (Paper 1 +2) or for individual paper. :)

Well, getting a rank of 199 or thereabouts is definitely commendable. As for FIITJEE, yes, it is extremely rare for both Andhra and Kota students to take it. They seriously don't care about these exams. Some of them really don't know the existence of FIITJEE (some who have even made it to the IITs). So, a 99 percentile of course does not guarantee a top 50 rank or whatever in JEE. And as for subject cut offs, it is not the 'incredible cases' who fail to clear them. I know of atleast 4 people personally who failed to clear the subject cut off 2 years back, 3 of them in Chemistry and one in Physics and hence failed to make it. And one of those students had got a total which would put him somewhere in the top 300, if only he had cleared the cut off in Chemistry, which happened to be a bit high in 2006. Well, having said that, clearing subject cut offs is not too high a matter of concern and people giving a fair emphasis on all three subjects could do it with reasonable ease.
Sathej

@Sathej: What I meant by incredible was those rare cases. And I personally feel, one who can't clear the IIT JEE cutoffs, doesnt deserve to be in the IIT.. And yeah, 199 is a good enough rank, but when you are born for greater things, its just obvious that you can't help but feel disappointed with that rank.. There was this guy, Swarun Kumar from PS, who was one of the most brilliant people of his batch, in Chennai, but he could manage only a rank in the 700s.. A disappointment for his calibre.. And he was batch topper in IIT-M, right from Sem1.. Why did he flunk JEE?? Many reasons, including school performance being insisted, not proper training in the city at that time.. Wat else could you expect?? There was this guy Rohit Singh (IPhO gold medalist last year and a good friend of mine from IMOTC) who was sad that he got a rank of 47
(!!) in JEE, bcoz he expected a rank of 1.. I mean it is natural to blame one with a rank below your dignity, be it 47, or 199, or whatever..
And coming back to FIITJEE, I've got this friend Dvijotham(AIR 35 in 2006) from IIT-B, who was from Ramiah in Andhra, and he told me taht people there not writing FIITJEE were all rumours and they do write FIITJEE.. I know people in Kota don't write FIITJEE, as I've got a cousin studyin there, and above that last year AIR1 Achin Bansal, asked an interviewer what FIITJEE was.. :rotfl:
Yeah, but I am firm in the stand that you actually can't read much in to your FIITJEE percentiles..

Well, its correct that people who can't clear the cut offs don't deserve to be in the IITs. But one can argue saying that almost nothing of the Chemistry you learn at high school has any bearing on what you do at IIT, if you take a course like Computer Science. So, the system of examining a person who wants to do Computers in Chemistry is itself a rather questionable idea.

Coming to ranks, well, getting an AIR 1 does not mean 'you are born for great things' :) And getting a back rank in JEE, wll, brilliant people do not necessarily clear JEE with a good rank and not tall those who clear JEE iwth a top rank are brilliant. There are people who get a rank of 20 and struggle to even complete the BTech course. The reasons for brilliant people getting an ordinary rank are not merely lack of training and school pressure. A rank in IIT JEE is not an end-all in itself. One has to aspire for something beyond. It is just a competitive exam and no amount of training or lack of pressure can 'guarantee' one a good rank. More so, with the 'terrible' pattern of JEE these days. Further, in life there are lots of things beyond the IITs and a rank is simply a short-term achievement.

And well, Andhra people, a majority do not write FIITJEE - atleast 75% and this is not a myth. They get FIITJEE papers and papers from almost all coaching classes and solve them in their institutes and do not appear for the tests as such. And btw people from Ramiah are rarely seen nowadays :) its largely people from elsewhere.


real analysis
LaTeX
algebra unsolved
algebra

[quote="al.M.V."]p!q!(n+p+q-1 C p+q+1).  :D[/quote]
Can you give a proper soln. and not just the answer?!!!!

[hide]I believe you might have commited a mistake I had done! That's why where's the soln?[/hide]

desired sum = p!q!*{Sum (r=1 to n-1) (r+q-1 C q)*(n-r+p-1 C p)}. But I'll show that Sum (r=1 to n-1) (r+q-1 C q)*(n-r+p-1 C p) = (n+p+q-1 C p+q+1). This is essentially two ways of selecting p+q+1 people from a total of n+p+q-1 people, each labelled 1, 2 ,..., n+p+q-1 to form a committee. Trivially, we have the RHS as one way. On the other hand, suppose we have chosen 1 <= a_1 < a_2 < ... < a_p+q < a_p+q+1 <= n+p+q-1. We consider a_q+1. Once a_q+1 is assigned a number, there are ((a_q+1)-1 C q) ways to assign a value to each of a_1, ... , a_q. And there are (n+p+q-1-(a_q+1) C p) ways to assign a value to each of a_q+2, ... , a_p+q+1. However, for the sum to be non-zero, a_q+1 can only be from q+1 to n+q-1. Hence summing over all possible values of a_q+1 gives the desired result. (Note: what actually happens is we sum over variable (r+q) instead of r, but this is okay since q is a constant.)  :)

[quote="al.M.V."]desired sum = p!q!*{Sum (r=1 to n-1) (r+q-1 C q)*(n-r+p-1 C p)}. But I'll show that Sum (r=1 to n-1) (r+q-1 C q)*(n-r+p-1 C p) = (n+p+q-1 C p+q+1). [\quote]

This was exactly what I had reached and written the result in the next step later thinking it was wrong since I did not have a proof for it.

Anyway,thanks for clarifying things out!!!

[quote]This was exactly what I had reached and written the result in the next step later thinking it was wrong since I did not have a proof for it.

Anyway,thanks for clarifying things out!!![/quote]

You are welcome! :D

$\sum_{r=1}^{N-1}{{{r+q-1}\choose{q}}\cdot{{N-r+p-1}\choose{p}}} = {{N+p+q+1}\choose{p+q+1}}\ (1)$
I write another combinatorial argument to prove this identity.
We can put b bals in u urns in ${b+u-1}\choose{u-1}$ ways. After transforming $(1)$we get:
$\sum_{r=0}^{N-2}{{{r+q}\choose{q}}\cdot{{N-2-r+p}\choose{p}}} = {{N+p+q-1}\choose{p+q+1}}$
Let's consider $N - 2$ balls which should be put in $p + q + 2$ urns. There are ${N - 2 + p + q + 2 - 1}\choose{p+q+2-1}$ ways to do that.
We can count the number of ways in a different way. We can put $r$ balls in the first $q+1$ urns in ${r+q+1-1}\choose{q+1-1}$ ways. Remaining $N-2-r$ balls need to be placed in the last $p+q+2-(q+1)=p+1$ urns. We can do this in ${N-2-r+p+1-1}\choose{p+1-1}$. Thus the number of placements with r balls in the first q+1 urns is ${{r+q}\choose{q}}{{N-2-r+p}\choose{p}}$. When we sum numbers of placements for all $r$ we get left side of $(1)$.
And we are done.
Another similar identity is the [b]Vandermonde convolution formula[/b].
$\sum_{i=0}^k{n\choose{i}}{m\choose{k-i}}={{n+m}\choose{k}}$
al.M.V.: Learn $\LaTeX$!


geometry
geometric transformation
circumcircle
search
geometry proposed

Let $ ABC$ be an acute triangle, and $ P$ be its Fermat point (i.e. point such that $ \angle APB\equal{}\angle BPC\equal{}\angle CPA\equal{}120^{\circ}$) Let $ X,Y,Z$ be the midpoints of arcs $ BPC,CPA,APB$ respectively. Prove that

(a) $ XYZ$ is equilateral.

(Part (b) to follow soon  :) )

[quote="28121941"]These points X,Y,Z are the centers of the equilateral triangles constructed internally to the triangle over the sides.
I affraid the result is more old than you expected (but this is not a bad thing!); I have located the result in an old russian book by Efremov (New geometry of triangle) Odessa 1908, with a romanian translation by M.Miculitza.
So, you had rediscovered an old theorem. Congratulations![/quote]

Wow!

But I'd like to know your solution... my one is quite ugly.

Or the solution of Efremov

I am enclosing a pdf file with the picture of the Efremov to clerify the exposition.
Let P' the second point of Fermat of ABC (that is, the intersection of the circles circumscribing the equilateral triangles constructed internally over the sides odf ABC), of respective centers X,Y,Z (I am using your notation; that of Efremov is different, but with the picture is easy of follow).
Then AP',BP',CP' passes through the midpoints of the sides YZ,ZX,XY.
The sides of the triangle XYZ are perpendicular to AP',BP',CP' and therefore form angles of 60º each other; therefore XYZ is equilateral.
This is the solution of Efremov.

This is a direct result of  Napoleon Theorem
see here:http://mathworld.wolfram.com/NapoleonsTheorem.html
 :lol:

[quote="discredit"]The problem seems nice enough  :wink:
But it's too old and well-known  :(

[quote="28121941"]
Then AP',BP',CP' passes through the midpoints of the sides YZ,ZX,XY.
The sides of the triangle XYZ are perpendicular to AP',BP',CP' [/quote]

Please explain both lines here  :roll:
And your $ P'$ is $ U'$ in the picture?

Moreover, plane geometry, I don't see how this follows directly from Napoleon's theorem.[/quote]
Well,if you have read the content of the link,you will find the statement "Analogous theorems hold when equilateral triangles erected internally on the sides ……………………" ,read it carefully , you will find it

[quote="discredit"]The problem seems nice enough  :wink:
But it's too old and well-known  :(

[quote="28121941"]
Then AP',BP',CP' passes through the midpoints of the sides YZ,ZX,XY.
The sides of the triangle XYZ are perpendicular to AP',BP',CP' [/quote]

Please explain both lines here  :roll:
And your $ P'$ is $ U'$ in the picture?

Moreover, plane geometry, I don't see how this follows directly from Napoleon's theorem.[/quote]
U' in the figure is the second Fermat point

OK, I think I see, but there isn't a proof in that link anyway  :)

[quote="Carlez Tevos"]Let $ ABC$ be an acute triangle, and $ P$ be its Fermat point (i.e. point such that $ \angle APB \equal{} \angle BPC \equal{} \angle CPA \equal{} 120^{\circ}$) Let $ X,Y,Z$ be the midpoints of arcs $ BPC,CPA,APB$ respectively. Prove that

(a) $ XYZ$ is equilateral.

(Part (b) to follow soon  :) )[/quote]

Three lines da,db,dc pass through A,B,C and perp to PA,PB,PC repectively. Denote A'=db∩dc, B'=da∩dc, C'=da∩db. Then, triangle A'B'C' is equilateral, and PA',PB',PC' are the diameters of the circles (BPC),(PCA),(PAB).
Thus, the bisectors A'X,B'Y,C'Z of angles A',B',C' of triangle A'B'C' are concurrent at the center O of triangle A'B'C'.
So X,Y,Z,O all lie on the circle whose diameter is OP.
Hence, triangle XYZ is equilateral  (because ∠A'OB'=∠B'OC'=∠C'OA'=120)

Hmm... actually I think it holds true too this:

If we construct points $ A',B',C'$ such that $ A'B'C'$ is equilateral, and $ B$ lies on segment $ A'C'$, $ C$ on $ A'B'$, and $ A$ on $ B'C'$. Then the center of triangle $ A'B'C'$ lies on the circumcircle of $ XYZ$.

[quote="plane geometry"][quote="discredit"]It's a challenge. Can anybody prove it?[/quote]
in mr.danh's reply you can find the solution directly[/quote]

Please give a bit more detail. mr. danh constructed a particular equilateral triangle.

[quote="plane geometry"]PROOF：
P is the similar center of △RST and △A’B’C’
∠C’PT=∠O’PO    C’P/O’P=TP/OP
△OO’P ∽ △TC’P
DONE![/quote]

What are your points? And why the similarity?

P is the Miquel point of △RST and △A'B'C' so P is the similar center of the two triangles
Because we have △OO’P ∽ △TC’P
angle OO'P = angle PTC' =90 this means O' lies on the circle which contains D,E,F,O,P and with OP its diameter

R,S,T stands for A',B',C' of mr.danh solution above , the particular equilateral triangle mentioned by discredit.
You can click the figure to enlarge it to see it clearly and in my figure R is missing,sorry for that

[quote="plane geometry"]Moreover, P is the Miquel point of △RST and △A'B'C'[/quote]

Hmm... What is Miquel point, and why is P so? And the similarity center.

It would be nice if you give more details, as I'm too lazy to think about it now  :P

[quote="28121941"]To Carlez Tevoz: If at any triangle ABC you pick arbitrary points U,V,W in the sides BC,CA,AB, respectively, then the three circumcircles (AVW), (BUW) and (CUV) meet at a point called Miquel point. This is the so called theorem of Miquel (but if you search it in the net, probably appears other results too)[/quote]
Yes, thank dear 28121941 to answer Carlez Tevoz.
Actually i am lazy typing,too  :lol:

[quote="28121941"]The number of july 2008 of the British journal "The Mathematical Gazette", vol.92, number 524 , includes a very interesting paper by C.J. Bradley on "The Fermat point configuration". The proofs included are synthetic and with areal coordinates.[/quote]
wow,can you share the pdf version?thank you in advance!

Dear plane geometry, I affraid I will be violating the copyright rules if I do a such thing....you can contact the editor of The mathematical Gazette, Dr. Gerry Leversha, gazette@m-a.org.uk and ask him for this.

And how come $ P$ the similarity center?


is it possible to write the natrual numbers in a string such that after $n$ numbers this string is a palendrone?

i.e.   $123456\dots{n}$ is a palendrone

o yes...sorry about the spelling error.  :oops:

What`s palindrome??? Is what we call in spanish "capicúa"??

E.G.: 101, 20902, 1145411 (they`re the same number written from right to left and from left to right)???.

Jose:  Yes, that is a palindrome.

[hide="Solution"]
First, definitions.  For two strings of digits $s_1, s_2$ let us define $app(s_1, s_2)$ to be the result when $s_2$ is appended to the end of $s_1$ (ex.: $app(123, 456) = 123456$).  For a string of digits $s$ define $inv(s)$ to be the inversion of the string (ex.: $inv(123) = 321$).  Finally, for a string of digits $s$ define $first(s, l)$ to be the string that contains the first $l$ digits of $s$.

Claim:  $n = 1$.

Proof:  Define the sequence $a_1 = 1, a_k = app(a_{k-1}, k)$, which is precisely the string in the problem.  If $n > 1$ then $n$ must be of the form $inv(a_k)$ for some integer $k > 1$ in order that $a_n = inv(a_n)$.

Now, consider the previous term to be appended to the string, $n-1$.  Because $n = inv(a_k) = k...321$ we know $n-1 = k...320$.  The string ends with $app(n-1, n) = k...320k...321$ and because $a_n = inv(a_n)$ we require that $inv(app(n-1, n)) = first(a_n, 2k)$.  This is impossible - we can write $inv(app(n-1, n)) = 123...k023...k$, and because we append $k+1$ immediately after $k$ when constructing $a_n$ a $0$ cannot be present there.

Hence, for $n > 1$ no such $n = a_k$ exists.  Q.E.D.
[/hide]


Let $ a_1,a_2,...,a_k$  be positive real numbers and $ m_1 ,m_2 ,...,m_k$ natural numbers ,then prove that

$ a_{1}^{\frac{m_1}{n}} \plus{} a_{2} ^{\frac{m_2}{n}} \plus{}...\plus{}a_k ^{\frac{m_k}{n}}\leq \frac {m_1}{n}a_1  \plus{}\frac{m_2}{n}a_2 \plus{}...\plus{}\frac{m_k}{n}a_k$

where $ n\equal{}m_1 \plus{} m_2 \plus{} ...\plus{}m_k$

http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Generalizations
See the part "Generalizations"

It is just the weighed AGM inequality, which can be proved by weighed Jensen ....


A small stone with mass 400g is projected vertically upwards from the bottom of a pond full of water with 10ms^-1. As the stone moves through the water it experiences a constant resistance of magnitude 3N. Assuming the stone does not reach the surface of pond. Find :

A) The greatest height above the bottom of the pond that the stone reaches,
B) Speed of stone as it hits bottom of pond on its return,
C) Total time taken for the stone to return to its initial position on the bottom of pond

(oops.. forgot to add this, assume g= 9.8ms^-2)

The main problem is that if you are going to use the equations of motion, you need to take into account the fact that acceleration is not simply $ g$.  In particular, the downward force acting on the stone is its own weight AND the the resistance.  That is, $ mg\plus{}3$ (in Newtons).


AMC USA(J)MO USAMO MATHCOUNTS trigonometry videos

What are your predictions for the four IL competitors competing for Nats on Friday, May 9?

Any thoughts? Who do you think would be in top 57? Countdown? Prizes? Anything you predict.

You guys better all tie for first with perfect scores.
And somehow tie during countdown.
And get a perfect team score.  :wink:

I predict(teams):
1st-Massachusetts
2nd-Texas
3rd-Indiana
4th-California
5th-Illinois
6th-New Jersey
7th-Connecticut
8th-Michigan

Why? What made you think IL would be that high?

you believe in Josh Dement, right?

Or the other three: Adam Kalinich, John Park, or Ravi Jagadeesan?

Actually, these are guesses that are made only from my extent of knowledge. I think those states are like the most "intelligent", Illinois basically by the city of Chicago, you always hear about these people from Chicago doing these things. Oh, and Mass. because thats where I live and Con. because that's where I used to live. :lol:

Michigan's going to be higher than that.
Hopefully.

Well, I don't think my predictions will be even close to right since [size=75]I am bad predictor :oops: [/size]

My prediction.

1. Indiana.
2. Texas
3. California (seems like an incredibly strong team this year)
4. Massachusetts
5. North Carolina (ALL 4 members made USAMO)
etc.

Except i'm biased for indiana. lol

Don't be biased, mathcrazed. Let's be honest. :wink:

I think Adam Kalinich would be the best Illinoisian in Nats.

Hey guys, any Illinoisian AoPSers out there??? I don't see why you guys are here if you're not from Illinois like me. No offense, though. :wink:

1. Indiana
2. Texas
3. Massachusetts
4. North Carolina
5. Michigan
.
.
.
.
.
57. Illinois (just kidding)

1.Texas (sorry Indiana people :P )
2.Indiana
3.Massachusetts
4.California
5.Michigan
6.North California

Let's consider the mathematical maturity of each team. Oh. We win. =P.

Anyone predict Illinois individuals? We're not here to talk about OTHER teams, just the members of Illinois.

[quote="the future"]Im not in mathcounts anymore (good thing cause i sucked)
but ill just say from prior history and knowledge of some of the competitors

1. indiana
2. texas
3. michigan
4. massachusetts
5. north carolina
6. virginia
7. pennsylvania
8. new york
9. illinois
10. minnesota[/quote]

Whoa, why did you put Michigan so high?
I think...
Texas, Indiana, California, Florida, Minnesota, and then Michigan.

[quote="mathcrazed"]I need to go practice mathcounts instead of usamo. muffins. You're going down mr. david lu. oh. burn. lol[/quote]
Hey mathcrazed, I'm NOT A MUFFIN! HHOW MANY TIMES DO I HAVE TO SAY THAT TO PEOPLE????

Anyone have predictions for Illinois INDIVIDUAL PEOPLE???

I predict TX,IN,NJ,MA,NC,CA,and MN in some order 1 thru 7

I think these will comine for 10 of the top 12

[quote="uldivad9"][quote="mathcrazed"]I need to go practice mathcounts instead of usamo. muffins. You're going down mr. david lu. oh. burn. lol[/quote]

All right. My personal goal has been changed from "Do better than last year" to "beat mr. lyndon ji". :D oh. retaliatory burn in self-defense.
Anyways, I need to practice usamo instead of practicing video games.[/quote]
No, I asked what people thought about Illinois. I'M NOT CONSIDERING OTHER TEAMS, PEEPS! :wink:


geometry
similar triangles
geometry proposed

[quote="ThAzN1"]Let $ABCDE$ be a cyclic pentagon with $AC=CD=DA$.  Let $AC$ and $BD$ meet at $P$, and let $AD$ and $CE$ meet at $Q$.  Prove that

\[\frac{AQ\cdot AB}{AP\cdot AE} = \frac{AE+ED-QE}{AB+BC-AP}.\]

[actually, I'm not even sure if this is right haha, so I guess I'll have to check it later.][/quote]

i think the $AP$ should be $BP$...

by Ptolemy's theorem, $AE+ED-QE = CQ$ and $AB+BC-BP = DP$, so we are to prove

\[\frac{AQ\cdot AB}{AP\cdot AE} = \frac{CQ}{DP}.\]

indeed, by similar triangles, we get $\frac{AQ}{AE} = \frac{CQ}{CD}$ and $\frac{AB}{AP} = \frac{CD}{PD}$, and multiplying these two equations, we get the desired result.


function
limit
real analysis
real analysis unsolved

Although I don't understand why can't you find the solution of this problem in a book it's a nice problem. Here is the solution I made up now:

Let's suppose that $f$ has no fixed point. Hence we must have that $\forall x\in[0,1]$ $f(x)>x$ or $\forall x\in[0,1]$ $f(x)<x$ or $\exists x_0\in(0,1],y_0\in[0,1)$ s.t. $f(x_0)-x_0<0<f(y_0)-y_0$.
In the first case we will then have that $\displaystyle 1\geq f(1)=\lim_{x\to 1}f(x)\geq 1$ so $1$ is the fixed point. In the second $\displaystyle 0\leq f(0)=\lim_{x\to 0}f(x)\leq 0$ so $0$ is the fixed point.
If we find ourselves in the last case by taking $x_{n+1}=x_n$,$y_{n+1}=\frac{x_n+y_n}{2}$ if $f(\frac{x_n+y_n}{2})-\frac{x_n+y_n}{2}\geq 0$ or $x_{n+1}=\frac{x_n+y_n}{2}$,$y_{n+1}=y_n$ if $f(\frac{x_n+y_n}{2})-\frac{x_n+y_n}{2}< 0$ $\forall n\in N\cup\{0\}$ we will have that $(*)$$f(x_n)-x_n\leq 0\leq f(y_n)-y_n$ with

$x_0\leq x_1\leq ...\leq x_n\leq ...\leq y_n\leq ...\leq y_1\leq y_0$(if $x_0<y_0$) and $\displaystyle\alpha=\lim_{n\rightarrow\infty}{x_n}=\lim_{n\rightarrow\infty}{y_n}$ with $\alpha\in(0,1)$ so in $(*)$ $\displaystyle\lim_{x_n\nearrow\alpha}{(f(x_n)-x_n)}= f(\alpha)-\alpha\leq 0\leq f(\alpha)-\alpha=\lim_{y_n\searrow\alpha}{(f(y_n)-y_n)}$ $\Rightarrow$ $f(\alpha)=\alpha$
or
$y_0\leq y_1\leq ...\leq y_n\leq ...\leq x_n\leq ...\leq x_1\leq x_0$(if $x_0>y_0$) and $\displaystyle\alpha=\lim_{n\rightarrow\infty}{x_n}=\lim_{n\rightarrow\infty}{y_n}$ with $\alpha\in(0,1)$ so in $(*)$ $\displaystyle f(\alpha)-\alpha=\lim_{x_n\searrow\alpha}{(f(x_n)-x_n)}\leq 0\leq \lim_{y_n\nearrow\alpha}{(f(y_n)-y_n)}=f(\alpha)-\alpha$ $\Rightarrow$ $f(\alpha)=\alpha$

Therefore the proof is complete.
If you have found another solution I would be interested in seeing it.

[quote="perfect_radio"]Let $f : \left[ 0,1 \right] \to \left[ 0,1 \right]$ a function which is continuous in $0$ and $1$ and which has lateral limits in each point $x_0 \in \left( 0,1 \right)$. Also, \[{ \lim_{x \nearrow x_0} f(x) \leq f(x_0) \leq \lim_{x \searrow x_0} f(x)} , \, \forall x_0 \in \left( 0,1 \right) .  \]

Prove that $f$ has a fixed point.[/quote]

WLOG $f(0) > 0$ and $f(1) < 1$, since otherwise we have found a fixed point of $f$.

Set $x^* = \sup \{t \in [0,1] | f(t)>t\}$. Then of course $x^* < 1$, and also  $0 < x^*$, since otherwise $0 < f(0) \le \lim_{x \searrow 0}f(x) \le \lim_{x \searrow 0}x = 0$.

By the first assumption on $f$, $x^* \le \lim_{x \nearrow x^*}f(x)\le f(x^*)$.

Now by construction $f(x) \le x$ for all $x > x^*$ and therefore also $f(x^*) \le \lim_{x \searrow x^*}f(x) \le x^*$.

Combining these estimates, we obtain
\[ x^* \le \lim_{x \nearrow x^*}f(x)\le f(x^*)\le \lim_{x \searrow x^*}f(x) \le x^* \]
and therefore $f(x^*) = x^*$.

Maybe u should take a second look over your solution but overall it is an elegant solution.


hi again rachid

I see that you didn't understand me ,any way see her:

[url]http://www.men.gov.ma/dajc/notes/notesannee/%20notes%202007/note%20127.2007.doc[/url]

Fixed. Spaces will be removed at submission.


function
floor function
Asymptote
calculus
integration

How can you graph a function with a floor value?
(like $ \lfloor x \rfloor \minus{}x$)

Hello

You can generate the path with the operator ^^

O.G.

int(7.5)=7

does that help?

P.S. $ \lfloor x\rfloor \minus{}x\equal{}\minus{}\{ x\}$ where $ \{ x\}$ is the fractional part of $ x$. Now it is much easier to graph on paper.

No I actually meant graphing on Asymptote...
like function floor x floor or something like that...

Since the functions Floor, floor, ceil, Ceil can be used in Asymptote
(see the Mathematical functions part of the documentation)
the only problem is to consider non continuous functions ?
For my students (Riemann integral) I gave some graphs of
such functions which are piecewise constant.
http://math.mad.free.fr/wordpress/?page_id=89

O.G.

Yes the code is/was not on my page. My goal was to show such a graph
to my student so my code is not clean (I am not Philippe Ivaldi). One year later I am not able
to read it and perhaps there are bugs !
You can find the file here : http://math.mad.free.fr/wordpress/wp-content/uploads/escalier2.asy

Roughly speaking I define a structure for this functions
struct fctesc {
  real [] sub ;
  real [] vsub;
  real[] vpt;
}

sub is the subdivision
vsub is the value in ]x_i,x_{i+1}[
vpt is the valeur at x_i

If you are interested I can add some comments to my file.

Good luck
O.G.

Thanks Olivier for the compliment :-)
Here a simpler method but less general than yours:
[asy]size(10cm,0);
import graph;

xlimits( -5, 5);
ylimits( -5, 5);
yaxis( "$y$" , Ticks());
xaxis( "$x$", Ticks());

path[] fl=sequence(new path(int n){
    return (n,n)--(n+1,n);},
  5);

fl.append(shift(-5,-5)*fl);
draw(fl);
for (int i=0; i < fl.length; ++i) {
  dot(point(fl[i],0));
  dot(relpoint(fl[i],1),UnFill);
}[/asy]


If real numbers $a,b,c$ satisfy the two conditions

1) $a+b+c=2$

2) $ab+bc+ca \geq abc+1$

then the inequality

$(ax+by+cz)^2 \geq 4((1-a)yz+(1-b)zx+(1-c)xy)$

holds for each reals $x,y,z$

Thank you very much Zxingtan for your solution.

It is a great pleasure for me to study it.


In how many ways can 24 students be seated in a classroom of 30 desks?

[hide="...."]Um, isn't this just $ 30C24$ or is there something else to it?[/hide]

[quote="DiscreetFourierTransform"]Yes, I believe that seems correct.

Another (virtually identical) way to approach it is to say "How many ways can I pick the desks that are not being sat at"? That's $ \binom{30}{6}$, which is identically equivalent to what you said.[/quote]
Waaah, but the answer key said the answer was $ \frac{30!}{6!}$. Is the answer key wrong, or are we doing something wrong? D:

I believe their answer is right, think of the question this way: How many permutations can we have of a set of 30 elements, where 6 of the elements are identical (the empty seats)?

[quote="DiscreetFourierTransform"]I'm not sure, but I don't think this question cares about the ORDERING of the desks, therefore we should not be concerned with permutations..[/quote]
It doesn't care about the ordering of the desks, but it does care about the ordering of the students.

For example, consider the following arrangements (where - denotes an empty desk and A,B,C, etc. are individual students)

ABCDEFGHIJKLMNOPQRSTUVWX------
XWVUTSRQPONMLKJIHGFEDCBA------

These should be considered as two different arrangements as far as this question is concerned.

Think of it like this in how many ways can six empty deks remain if twenty four are occupied where order is not considered.
I.e 30!/6!.


[b]Let $ \Delta ABC$ be a right triangle of area 1. Let $ DEF$ be the points obtained by reflecting A, B, and C respectively in their opposite sides. Find the area of $ DEF$.[/b]

It's the same.  Dilations/augmentations are the only transformations that alter size.

...

if I understood the problem corectly then  my answer is 3...

here's why let D be the reflected C (where C=90°)...

we notice CEF=ABC (1)

now let DH be the height od DEF...

DH=3*CH...

from here ...

DEF=3...

have i understood the problem corectly?...

that is indeed the correct answer wichking.

Unless I misunderstand, 3 doesn't make sense:

Let $ A\equal{}(0,1)$, $ B\equal{}(1,0)$, and $ C\equal{}(2,1)$

To find the area set up the matrix as follows:

[ 0 1 1 ]
[ 1 0 1 ]
[ 2 1 1 ]

Then find the determinant:

| 0 1 1 |
| 1 0 1 | = 2
| 2 1 1 |

The area equals $ 1/2$ of the determinant:

$ 1/2*2\equal{}1$

Now rotate the points 90 degrees:

[ 0 1 ]     [ 0 -1 ]
[ 1 0 ] * [ 1  0 ]
[ 2 1 ]

      [ 1  0 ]
 =  [ 0 -1 ]
      [ 1 -2 ]

Now find the area of this triangle:

| 1  0  1 |
| 0 -1  1 | = 2: $ 1/2*2\equal{}1$
| 1 -2  1 |

The area doesn't change.
If I am wrong please explain. Thankyou.


[color=darkblue]Let $AB_{1}C_{1}$, $AB_{2}C_{2}$ be two triangle so that $B_{1}=B_{2}=90^{\circ}$ and $AB_{1}=AB_{2}$.
Define the point $P$ for which $PB_{1}\perp AC_{1}$ and $PB_{2}\perp AC_{2}$.
Prove that $AP\perp C_{1}C_{2}$.

[b]See[/b][/color]
http://www.mathlinks.ro/Forum/viewtopic.php?t=91654
http://www.mathlinks.ro/Forum/viewtopic.php?t=100004
http://www.mathlinks.ro/Forum/viewtopic.php?t=100021

First: $(AB_{1})^{2}=(AH_{1})(AC_{1})$ and $(AB_{2})^{2}=(AH_{2})(AC_{2})$

From the statement we have $AB_{1}=AB_{2}$, then: $(AH_{1})(AC_{1})=(AH_{2})(AC_{2})$, from these we have that $C_{1}H_{1}H_{2}C_{2}$ is cyclic.

:arrow: $\measuredangle AH_{1}H_{2}=\measuredangle H_{2}C_{2}C_{1}=\alpha$ and $\measuredangle AH_{2}H_{1}= \measuredangle H_{1}C_{1}C_{2}=\theta$

$AH_{1}PH_{2}$ si also cyclic  :arrow: $\measuredangle AH_{1}H_{2}=\measuredangle AH_{3}H_{2}=\alpha$ and $\measuredangle AH_{2}H_{1}=\measuredangle AH_{3}H_{1}=\theta$

Then it follows that $C_{1}H_{1}PH_{3}$ and $C_{2}H_{2}PH_{3}$ are cyclics.

So: $AP \perp C_{1}C_{2}$  $Q.E.D.$


Determine the best constant $ k$ for $ a\geq k,\ b\geq k,\ c\geq k$ such that:

$ ab\plus{}bc\plus{}ca\plus{}6k(k\minus{}1)\geq (3k\minus{}2)(a\plus{}b\plus{}c)\plus{}\frac{1}{a}(k\minus{}c)\plus{}\frac{1}{b}(k\minus{}a)\plus{}\frac{1}{c}(k\minus{}b).$

The best possible constant, $ k_{min}\equal{}0.$

what is the definition of best constant?is it the greatest or the least? if it is the greatest , then it´s not $ 0$.

[quote="kunny"]What I want you to determine is the value of $ k$.[/quote]
$ k\equal{}1$ clearly satisfies the given inequality.

The problem becomes more restrictive as $ k$ becomes smaller, so he wants the smallest possible value of $ k$.