OK Guys. Listen up. I am a 4th year high school student (graduating in 2006). Till Class 10( I'm in class 12 now) , I was an avid computer enthusiast - learning HTML, XML, Javascript, a bit of C and Networking... but all this ended in Class 11. If you know, in India, after class 10, almost everyone interested in engineering attends tuitions and prepare for IIT-JEE-- and I was no exception. So, for the past year, I was totally cut off from computing. But now, it seems my interest is returning. As I was seeing the different competitons like Informatics Olympiad and APs etc. whcih might be useful in US admissions( just keeping all otions open!) I got the first volume of Knuth's "Art of Computer Programming" recently and am searching for knowledege! So I need your help on a no. of things(the previous para just established the context!)

1) Is it realistic for me to consider taking part in these programming competitions an AP exams etc. now? I'm willing to spend at most 2 hours $\pm$ 1 hour per day!

2) Which are the different languages that I can learn now. Which are the best? Please recommend online pages where I can get the best compilers for the different languages asa well as documentation for learning these langugages(besides mathlinks.ro of course!!)

3) I believe the competitons focus more on finding algorithms etc. and analysing them rather than just purely writing long , tedious programs. Am I right? If yes, please recommend the best books/online pages on ALgorithms etc. and general computing that will be helpful. Does Knuth's Volume I cover the requisite material for IOI?? Which other books are there? I f somebody who lives in india can suggest locally available books, that ill be the best!

4) What else do I need to do if the answer to (1) is yes??

[size=150][b]Does nobody have any views on my pitiful situation??????[/b][/size]

i dont know too much since im not much of a programmer myself, but here are my answers:
1) The AP tests have a reputation of being relatively straightforward, so that is definitely doable in a year. Try the USACO contests at ace.delos.com/usacogate. I've learned a lot from doing the excercises and problems on that site.
2)This is a matter of taste. For AP you need to know java. For USACO you need java, c, or pascal.  I use c++. the best free compiler as far as i can tell is the bloodshed devc++ (or something like that...google it).
3) USACO is a great introduction as far as I can tell.


invariant
quadratics
algebra
polynomial
linear algebra
matrix
vector

Suppose $ A\in M_{n}(\mathbb{R}).$ Suppose that the quadratic polynomial $ x^{2}\plus{}\alpha x\plus{}\beta$ for real $ \alpha,\beta$ is a factor of the characteristic polynomial of $ A.$ (For the sake of nontriviality, you might as well assume this quadratic is irreducible over the reals.)

Prove that the subspace $ E\equal{}\{v\in\mathbb{R}^{n}: (A^{2}\plus{}\alpha A\plus{}\beta I)v\equal{}0\}$ is at least two-dimensional.

Now suppose $ Q$ is an orthogonal real matrix whose first two columns lie in $ E.$ (Given two independent vectors in $ E,$ we can always construct such a $ Q$ by the Gram-Schmidt process.) What can you say about the structure of $ Q^{T}AQ?$

Let's replace $ \mathbb R$ with an arbitrary field $ \mathbb F$. Suppose that $ p_{A}$ factors as $ gh$, where $ g$ is an irreducible polynomial of degree $ m$.
The conclusion should be that $ \text{dim}\ker\left( g(A)\right)\geq m$.

Now assume that we can find $ v\in\ker\left( g(A)\right),\, v\neq 0$.
The idea is to prove that $ v, Av, A^{2}v,\ldots, A^{m\minus{}1}v$ are linearly independent.
If not, then there is $ q\in\mathbb F[X]$ such that $ q(A) v \equal{} 0$ and $ \deg q < m$.
But $ g(A) v \equal{} 0$ and $ g,q$ are coprime, so $ I_{n}v \equal{} 0$, contradiction.
However, $ v, Av,\ldots, A^{m\minus{}1}v$ are all in $ \ker\left( g(A)\right)$, so this is enough to prove the claim.

My question is: How do we know that $ v$ exists? Of course, we could consider the algebraic closure of $ \mathbb F$, but I want to avoid such complications. I just can't move on to the second part 'til I figure this out.

By Cayley-Hamilton, $ g(A)h(A)\equal{}0.$ If $ g(A)$ were invertible, then $ h(A)\equal{}0.$ Can you do anything with that?

Let's try this again. Forget Cayley-Hamilton for now, and stop talking about the characteristic polynomial.

$ A$ has a minimal polynomial, $ M_{A}$ such that $ M_{A}(A)\equal{}0.$ We get this just from noting that $ \{I,A,A^{2},\dots,A^{r}\}$ must be linearly dependent for some $ r$ and finding the smallest $ r$ for which that is true. [Cayley-Hamilton tells us that $ m(A)$ divides the characteristic polynomial of $ A,$ but I said we're staying away from that for now.]

Now let's change the second sentence of perfect_radio's post to "Suppose that $ M_{A}$ factors as $ gh,$ where $ g$ is an irreducible polynomial of degree $ m.$" Change the second sentence of my original post to say "minimal polynomial" instead of "characteristic polynomial."

We know that $ g(A)h(A)\equal{}0.$ If $ g(A)$ were invertible, then $ h(A)\equal{}0.$ But that contradicts that $ m_{A}$ was the minimal polynomial for $ A.$ Therefore we conclude that $ g(A)$ is singular and there exists nontrivial $ v\in\ker(g(A)).$

OK, perfect_radio - can you take it from here?

Actually, I can't say too much about $ Q^{T}A Q$. If $ \dim E$ is [i]precisely[/i] $ m$, then it's pretty obvious that the resulting matrix has a $ (n\minus{}m)\times m$ zero sub-matrix in the left-down corner. Other than that, I didn't find nothing :dry:

Can we prove the existence of v this way?:

suppose that only v=0 belongs to Ker(g(A)), so dim(Ker(g(A))) = 0. Because n = dim(Ker(g(A)))  + dim(Im(g(A))) we conclude that dim(Im(g(A))) = n => rank(g(A)) = n => det(g(A)) \= 0. This contradicts the hipothesis that g(x) is a factor of P_A


function
inequalities
induction
Cauchy Inequality
algebra unsolved
algebra

Let there be $ f: (0; \infty) \to \mathbb{R}$ with the propriety that :
$ \frac{f(x)\plus{}f(y)}{2} \equal{} f(\sqrt{xy})$ , $ \forall x , y \in (0, \infty)$ .

Prove that :

$ \frac{f(x_1) \plus{} f(x_2) \plus{} \dots \plus{} f(x_n)}{n}\equal{} f(\sqrt[n]{x_1x_2 \dots x_n})$ , $ \forall x_i \in (0; \infty) \text{ with } i \equal{} \overline{1,n}$ .

[hide="Hint"] Similarly with ways of prove Cauchy inequality: by Cauchy induction
Prove equality is true with $ k$ then true with $ 2k$, and true with $ k$ then true with $ k\minus{}1$! Then OK![/hide]

[quote="tdl"], and true with $ k$ then true with $ k \minus{} 1$! Then OK![/hide][/quote]

Would you please post your solution for this step ? Thanks

$ f(\sqrt[n\minus{}1]{x_1\cdots x_{n\minus{}1}})\equal{}f(\sqrt[n]{x_1\cdots x_{n\minus{}1}\sqrt[n\minus{}1]{x_1\cdots x_{n\minus{}1}}})\\
\hspace*{1.1in}\equal{}{f(x_1)\plus{}\cdots \plus{}f(x_{n\minus{}1})\plus{}f(\sqrt[n\minus{}1]{x_1\cdots x_{n\minus{}1}})\over n}$

$ \Rightarrow f(\sqrt[n\minus{}1]{x_1\cdots x_{n\minus{}1}})\equal{}{f(x_1)\plus{}\cdots \plus{}f(x_{n\minus{}1})\over n\minus{}1}$

Let $ x=e^u,y=e^v$ then
$ \frac{f(e^u)+f(e^v)}{2}=f(e^{\frac{u+v}{2})}$
Consider $ g(x)=f(e^x)$ then
$ g(x)+g(y)=2g(\frac{x+y}{2})$
It is the Jensen's equation!


I took both the SAT and ACT in the fall of my sophomore year, and I am satisfied with the scores I got on both, so I don't want to retake either exam. I am looking to go to college after my junior year, btw. So i've been looking at a lot of admission applications, and I'm starting to get the feeling that I'll have to retake the tests anyway because they're going to be "out of date" or something by the time colleges will want to look at them, since everyone says I should take the tests spring of my junior year or fall of my senior year. Should I retake the tests? I'd appreciate any help/advice anyone could give me on this. Thanks  :)

Don't value my advice that much, because I have none, zippo, zilch nonya, zero, etc. experience on that. I just got into high school, but if it were me, I'd just do whatever I felt like doing. If I was going to think that if I didn't take it, I'd be screwed for college, then I'd take it. But if I was pretty sure it wouldn't make a difference, I wouldn't retake it. If I was in the middle, I might take it just to be sure but I dunno.

The important thing to check on the SAT I is which version (the "old," used through January 2005, or the "new," used beginning March 2005) is required by your favorite college(s) for people enrolling in the year you want to enroll. This shouldn't be an issue for the ACT in quite the same way.

One of the books I mention in the "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16376]Books on College Admission[/url]" thread suggests never taking the SAT I more than three times--taking it four times or more looks obsessive. But taking it twice is very normal, and it will be especially normal for people who have one "old" SAT I score to take the "new" SAT I at least once too.

Good luck in your applications.

You should really check with the school that you want to go to.  I just went to an MIT info session and they said that they are flexible and will take pretty much anything (i.e. old sat, new sat, or act).  The day after the MIT session I went to one that had Harvard, Penn, Georgetown, and Duke and they were pretty much the same.  So I get the impression that for most schools it wouldn't really matter.


We know about the numbers $ a_0, a_1\ldots, a_{n\minus{}1},a_n$ that $ a_0\equal{}a_n\equal{}0$ and that $ a_{k\minus{}1}\minus{}2a_k\plus{}a_{k\plus{}1}\ge 0: \,\forall 1\le k\le n\minus{}1$. Prove that $ \forall a_k\le 0$.

assume that $ a_k>0$ for some $ k$. then $ (a_n\minus{}a_{n\minus{}1})\plus{}\cdots \plus{}(a_{k\plus{}1}\minus{}a_k)\equal{}\minus{}a_k<0$, while $ (a_k\minus{}a_{k\minus{}1})\plus{}\cdots \plus{}(a_1\minus{}a_0)\equal{}a_k>0$. so there exists $ i>j$ for which $ a_i\minus{}a_{i\minus{}1}<a_j\minus{}a_{j\minus{}1}$, contradicting the hypothesis.

The condition can be rewritten as $a_n-a_{n-1} \ge a_{n-1}-a_{n-2} \ge \cdots \ge a_1-a_0$.Now let $j$ be the minimal index such that $a_j>0$.Then $a_{n}-a_{n-1} \ge a_{n-1}-a_{n-2} \ge \cdots \ge a_{j}-a_{j-1}>0 \implies a_n>a_{n-1} \cdots>a_j>0$,a contradiction.


Solve this ecuation:   \[ \sqrt{1+a*x}=x+\sqrt{1-a*x} , a \in R  \]

The answer is as follows.

For $|a|=1$ or $a=0,\ x=0.$
For $|a|<1$ and $a\neq 0,\ x=\pm 2\sqrt{1-a^2}.$
For $|a|>1,$ no solutions.

[hide]Rewrite equation as $\sqrt{1+ax}-\sqrt{1-ax}=x$ $(1)$, and multiply both sides by  $\sqrt{1+ax}+\sqrt{1-ax}$ to obtain $2ax=x(\sqrt{1+ax}+\sqrt{1-ax})$. Clearly $x=0$ is a solution gor all $a$. Dividing by $x$ we have $2a=\sqrt{1+ax}+\sqrt{1-ax}$ $(2)$ so if $a \leq 0$ there are no more solutions.
By adding $(1)$ and $(2)$ we get $x+2a=2\sqrt{1+ax}$ so $x+2a \geq 0$. From this we obtain $x^2=4(1-a^2)$ so when $a \geq 1$ no more solutions.
Else $x_1=-2\sqrt{1-a^2}$  and   $x_2=2\sqrt{1-a^2}$.
Obviously $x_2+2a \geq 0$ and checking when $x_1+2a \geq 0$ leads to $a \geq \frac{1}{\sqrt2}$.
Finally $x=0$ is a solution for all $a$
if $a \leq 0$ or ${a \geq 1}$ no more solutions
for $0<a<\frac{1}{\sqrt 2}$  solution is $x_2$
and for $\frac{1}{\sqrt 2} \leq a<1$ solutions are $x_1$ and $x_2$.[/hide]

4 delta:
I don't understand why for \[ 0<a<1/\sqrt{2}  \] just x2 is the solution..


At the beginning of a trip, the mileage odometer read $ 56200$ miles.  The driver filled the gas tank with $ 6$ gallons of gasoline.  During the trip, the driver filled his tank again with $ 12$ gallons of gasoline when the odometer read $ 56560$.  At the end of the trip, the driver filled the tank again with $ 20$ gallons of gasoline.  The odometer read $ 57060$.  To the nearest tenth, what was the car's average miles-per-gallon for the entire trip?

\[ \textbf{(A)}\ 22.5 \qquad
\textbf{(B)}\ 22.6 \qquad
\textbf{(C)}\ 24.0 \qquad
\textbf{(D)}\ 26.9 \qquad
\textbf{(E)}\ 27.5
\]

[hide="Click for solution"]
The trip was $ 57060\minus{}56200\equal{}860$ miles long and $ 12\plus{}20\equal{}32$ gallons of gasoline were used during the trip.  Thus the average number of miles per gallon was $ \frac{860}{32}\equal{}26.9$, or $ \boxed{\textbf{(D)}}$.
[/hide]

I don't see anywhere in the question that the driver always fills the tank when the tank is empty. Without that assumption, it is virtually impossible to solve this question.

Yeah its kinda weird, but its standard convention to assume that he waited until it was empty to fill

Sorry to revive!

I don't see how the first 6 gallons are irrelevant (as seen [url=https://artofproblemsolving.com/wiki/index.php?title=1986_AJHSME_Problems/Problem_19]here[/url]), but the last 20 gallons are relevant. At the beginning of the trip, the driver uses the first 6 gallons. At the end of the trip, it makes sense that he fills up his tank, but he doesn't use those 20 gallons at all!

Can somebody help me figure out why the given solution at the link says so?

The 6 gallons at the beginning were used earlier, he replaced them before the trip. He used the 32 gallons during the trip and refilled them afterwards.

But he didn't use the 20 gallons at the end, did he?

He didn't use those 20 gallons that he loaded into the tank, but he used 20 gallons that needed to be refilled afterwards.

Oh. Hmmm.

That makes more sense now (thanks Archeon :)), but I think the question was worded confusingly. Just like gr8song and Math1331Math were discussing around 2 years ago, you have to make that assumption.

I don't see any assumption that has to be made, and the question isn't really worded that badly, but it did take me reading it twice to actually get what was going on, so I guess I can see where the confusion comes from.

I think there were 3 words "up" missing after each word "filled".

This is 1986 AJHSME Problem 19 just saying

[url]https://artofproblemsolving.com/wiki/index.php/1986_AJHSME_Problems/Problem_19[/url]


When a problem says taht a bunch of variables are permutations of a set of integers, what does that mean? That all of x, y, z.... are different integers of the set?

No, it probably means that each variable is a list of integers in some order.  I'd have to see the context, though.

Suppose you want to say that the solutions of a system of equations in three variables are (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1). Then we could say that the solutions are permutations of (1,2,3) so that we have less writing to do.


geometry
incenter
geometric transformation
reflection
perpendicular bisector
geometry unsolved

Perpendicular bisector of $ FX_a$ meets the line $ FN$ at the incenter $ I.$ Draw the incircle $ (I)$ with center $ I$ and radius $ IF$ and the nine-point circle $(N)$ with center $ N$ and radius $ NF.$ Tangent to $ (I)$ through $ X_a$ meets $ (N)$ at the midpoint $ M_a$ and the foot of the A-altitude $ H_a$ (decide which are $ M_a$ and $ H_a$ according to your sketch). Construct the reflection $ Y_a$  of $ X_a$ about $ M_a$  and the reflection $ J_a$ of $ X_a$ about $ I.$ Vertex $ A$ is the intersection of $ J_aY_a$ with the perpendicular to $ H_aX_a$ through $ H_a.$ Draw the two tangents from $ A$ to $ (I)$ to complete $ \triangle ABC.$


If squares are erected externally on the sides of a parallelogram, prove their centers are the vertices of a square.

First we show that they have the same length (the sides of the "square") and then we show that the angles are all right.

Now, let the vertices be A,B,C,D and the centers of the squares X,Y,Z,W with X on A,B, Y on B,C, etc.

Then it is trivial to show that the triangles XBY and YCZ are congruent by (SAS), so XY=YZ. Then cyclically, we have all of the sides equal.

Now, we know the angle BYC is 90 degrees, so it suffices to show that angle CYZ is equal to XYB, since one of the angles is taken away, and one of the angles is given to the 90 degrees (draw a diagram, you will see). But the two triangles CYZ and XYB are congruent by (SSS) so we are done. Then cyclically, we have our square.


Last night at Princeton, there was a screening of a preliminary version of a documentary that's being done about the IMO.  It focuses mostly on the US IMO team from 2006, but the entire Canadian team from that year also appears in it several times, including, to my surprise and delight, an entire clip on Canmoo! (Basically Richard Peng talking about how Arnav stole Canmoo, and the American team confirming in disbelief that the Canadians had brought a stuffed moose to IMO.)  The documentary also includes Peng Shi answering some questions about cryptography, Peng-style: "Try every possible combination...It will take a very long time."  So keep your eyes out for "Hard Problems," being produced for the next Sundance Film Festival (January 2008), starring your favourite local math olympians  :D

I don't know when, but we plan to sell DVD copies of the documentary on our website.  There was a screening recently at a joint AMS-MAA meeting here in San Diego, but I was unable to make it.

Hahahahaha, great video.  Richard would like to point out that the footage is used without his consent and says "LAME".

Looks like they cut out Peng's speech though, sad, I would have liked to see it.

[quote="Sunny"]Hahahahaha, great video.  Richard would like to point out that the footage is used without his consent and says "LAME".  [/quote]

He was speaking to a camera ... what did he expect? ...

Just because I speak to a camera, or am in a picture, doesn't mean I want it published!!  Hehehe.

It's actually a pretty good and unintentionally funny video......


How would one go about rearanging the following equation into its rectangular form for a circle:

r=(hcos(x)+ksin(x))

to...

r^2=(x-h)^2 + (y-k)^2

are there certain trig identities i should be using?

There are four major identities for use in converting an equation between rectangular and polar.  They are:

$x^{2}+y^{2}= r^{2}$
$tan(\theta) = y/x$
$y = rsin(\theta)$
$x = rcos(\theta)$

Therefore,
$r = 2hcos(\theta)+2ksin(\theta)\\ r^{2}= 2rhcos(\theta)+2rksin(\theta)\\ x^{2}+y^{2}= 2hx+2ky\\ x^{2}-2hx+y^{2}-2ky = 0\\ (x^{2}-2hx+h^{2})+(y^{2}-2ky+k^{2}) = h^{2}+k^{2}\\ (x-h)^{2}+(y-k)^{2}= h^{2}+k^{2}$
This is the equation for a circle with center $(h, k)$ and radius $\sqrt(h^{2}+k^{2})$.


I might also add that Runpeng was sick today, so his performance was undoubtedly affected. So at the beginning of 2009, I was like, to Yang, "whatever you do, remember to not get 5th." This is, of course, in reference to last year: 5th place by 1 point, and then 1st unofficial countdown. :P

Hey, Congrats to your team that's really impressive...but it isn't the first time.  I looked quickly at the 2006 and 2007 results and it has actually happened in:

Arkansas, Utah, South Dakota, Guam and West Virginia

I also know in Maryland a few years ago all 4 were from the same school and they placed 3rd at NATS@=!!!

That doesn't make it less impressive for you though.  That's a great job!!!!

[quote="maxschindler"]CONGRATULATIONS ALL!

1st place-Max Schindler
2nd-Runpeng Liu
3rd- Victor Wang
4th- Stella Schindler
5th- Yang Liu
7th-Haohang Xu

Note: all of the above are from the same school. First time EVER a nats team has consisted of 4 people from the same school.

CD: Max (1st) Victor (2nd)[/quote]

Justin Wang=mathswimmer

[quote="jjx1"]I made nats :)[/quote]

Congrats. :)

@jjx1: Many people look forward to cussing you out at Nationals and not getting muted.  :)

BTW, I saw the John Burroughs Mathcounts coach at the chinese restaurant that I ate at yesterday evening...  :rotfl:  :huh:

These two twins made Nationals in 2005 (?) and both made it into Countdown.

oh, so you didn't make Nationals?  :rotfl:

[quote="jjx1"]meh, none of you will see me and recognize me as jjx1.[/quote]

[b]FACEBOOK[/b]

There were siblings in 06 as well, Choi or something.

Meh. I made nats. Mewto and tinytim, you know who I am. I am Can Delving [color=#F0F0F0]or some anagram of it. [/color]

"Can Delving"?? Is that when you delve around the interior of a can? What are you looking for?

[quote="27r"][quote="jjx1"]I made nats :)[/quote]

You said you didn't even make states.[/quote]

yeah, jjx, you told me you didn't make states.
i sense a liar?


superior algebra
superior algebra unsolved

Let $K = \mathbb{Q}( a^{\frac{1}{n}})$, where $a \in \mathbb{Q}$ and suppose $[K: \mathbb{Q}] = n$.Let $E$ be any subfield of $K$ and let $[E: \mathbb{Q}] = d$.Prove that $E = \mathbb{Q}( a^{\frac{1}{d}})$

Let $\alpha = a^{\frac{1}{n}}$
So ${1,\alpha,\alpha^{2},....,\alpha^{(n-1)}}$ forms a $\mathbb{Q}$ basis of $K$.
Now $\alpha^{k}$ has $\frac{n}{g.c.d(n,k)}$ distinct conjugates,so if an element $\beta \in K$ has exactly $d$ distinct conjugates then if one expresses $\beta$ in the above basis all co-efficients except those of $\alpha^{i}$ where $g.c.d(n,i) = \frac{n}{d}$ are zero.
Hence $\beta \in \mathbb{Q}(a^{\frac{1}{d})}$ and hence the result follows.......


function
trigonometry
algebra
polynomial
search
algebra proposed

Find the least positive integer $n$ such that $\cos\frac{\pi}{n}$ cannot be written in the form $p+\sqrt{q}+\sqrt[3]{r}$ with $p,q,r\in\mathbb{Q}$.

[i]O. Mushkarov, N. Nikolov[/i]

[hide]No-one in the competition scored more than 2 points[/hide]

[quote="Rust"]posted before $n=7$.[/quote]
Can you give a link?

[url=https://artofproblemsolving.com/community/c6h148539]A topic with a bad title.[/url]

The degree of the minimal polynomial of $\cos{\pi/7}$ is $3$. To those with more experience, it is obvious that $p+\sqrt{q}+\sqrt[3]{r}$ cannot satisfy an integer cubic polynomial unless $\sqrt{q}\in \mathbb{Q}$, i.e. $q$ is a square. For the others, once you have observed and showed this rather standard fact, everything follows smoothly. For then remains to show that $\cos{\pi/7}$ does not take the form $p+\sqrt[3]{r}$. The latter number has algebraic conjugates $p+\omega^{j}\sqrt[3]{r}$ for $j=0,1,2$, $\omega = e^{2\pi i/3}$, which gives you the minimal polynomial for $p+\sqrt[3]{r}$; and writing down the minimal polynomial of $\cos{\pi/7}$, you'll be done by comparing the coefficients.

To show the well-known fact that $p+\sqrt{q}+\sqrt[3]{r}$ can't satisfy an integer cubic unless $q$ is a square, one way is to look at this thread https://artofproblemsolving.com/community/c6h27089. The "theorem" I had stated there implies this a very, very special case :) .

Does there exist a more elementary solution?  :roll:


If we solve a functional equation using substitution, how do we know if we have found all solutions? For example:

If, for all $x$, $f(x) = f(2a)^{x}$ and $f(x+2)=27f(x)$, then find $a$. (AoPS Vol. 2, #116).

[hide="Answer Key"] The answer key says to set $x=2a$ and solve to get $a=\frac{1}{2}$.[/hide]

Are there other ways to solve such functional equations?

Ah...so one-to-one is key in this case. Thanks!

[quote="t0rajir0u"]In this particular example, we know that $f(x)$ is exponential, $f(0) = 1$, and $f(2) = 27$, so the base is not zero or one (in fact, $f(x) = 3^{\frac{3x}{2}}$!).  This means that $f(x)$ is one-to-one.

We also know that $f(1) = f(2a)$, so $1 = 2a$.  No other solutions are possible :)[/quote]

How do we know that? What if the function was:

$f(x)=1$ if $x$ is rational $0\le x <2$
$f(x)=2$ if $x$ is irrational $0\le x <2$

and the other values are interpolated from this

say it were continuous, the function is still not necessarily exponential

[quote="Kalle"]If $b = f(2a)$ then $f(x) = b^{x}$. $f$ is exponential.[/quote]

Yes.  The condition "for all $x$" is rather strong in this case, and you won't find many other functional equations this easy :)


Show that n! can't be square of integer for n>1 and without use Bertrand's theorem...... :huh:

Let p prime and $\frac{n}{2}<p\le n$. Suth prime exist, when n>1. Obviosly $ord_{p}(n!)=1$ odd, therefore $n1\not =m^{2}$.

$ord_{p}(n!)$ is an order so if you say that $ord_{p}(n!)=1$ this means that $n!^{1}\equiv_{p}1$, you probably wanted $v_{p}(n!)$ as multiplicity of $p$ in factorization of $n!$ :D

bye

[quote="Rust"]Let p prime and $\frac{n}{2}<p\le n$. Suth prime exist, when n>1. [...][/quote]
In fact, I'm almost sure that Αρχιμήδης 6 has been quite clear: he wrote "without appealing to Bertrand's postulate"... :maybe:

Υes without use this theorem.If sb can prove that the product of n successive integers isn't a square of integer(accept of 1!)tthen it's obviously that the problem have been solved.But hou you prove this without Bertrand's theorem? :wink:

[quote="SpongeBob"]He didn't use any known theorem, he said that there is some prime number between $\frac{n}2$ and $n$, nad that's OK, now if you look at $1\cdot 2 \cdot 3 \cdots (n-1) \cdot n$ then $p|n!$ but $2p>n$ so $p^{2}\not | n!$, and that's all, none theorem is used in this...[/quote]But we cannot deduce there is a prime between $\frac{n}{2},n$ without using any theorem.

It's obviously that we can't suppose that betwen n/2 and n there is a prime without Bertrand's theorem! :maybe:

Yah, I'm having a discussion about it, too, with my friends... (although with different questions, but using this theorem)
Is this Bertrand's thing a postulate or a theorem ?? 'Cause I've read in two books. They say Conjecture and Postulate....  :huh:
Thx for any helpful replies  :)

[quote="RDeepMath91"]
Is this Bertrand's thing a postulate or a theorem ?? [/quote]
It's a theorem, it was demonstrated first by Pafnuti Chebyshev. There's an absolutely simple proof for this involving binomials. As far as I know, it's due to Erdos.

Ι don't believe that without Bertrand's the problem is unsolved. :D

An old theorem of Erdos states that the product of consecutive integers is never a perfect square.  So this would yield an easy solution to our problem  except Erdos uses Sylvester's theorem:

The product of $k$ consecutive integers is divisible by a prime greater than $k$, which is a generalization of Chebychev's theorem.

However Erdos' proof of Chebychev's theorem simplifies drastically since if $(2n)!$ is a perfect square, then so is $\binom{2n}{n}.$
As a consequence all prime factors of $\binom{2n}{n}$ must be less than or equal to $\sqrt{2n}.$

Can anyone show:
The product of $k$ consecutive integers with $n, n+1, \dots n+k-1$ is never a perfect square given $n \le k.$

The following estimate might be useful (or not sharp enough)

The product of all primes less or equal than $n$ is less that $4^{n}.$

Reference:

MR0000022 (1,4e)
Erdös, P.
Note on products of consecutive integers.
J. London Math. Soc. 14, (1939). 194--198.


Find all real solutions to the equation $4x^2 - 40 \lfloor x \rfloor + 51 = 0$.

Denote $[x]=n\in\mathbb{Z},\{x\}=\alpha\in[0,1)$. Then the equation becomes

$4(n+\alpha)^2-40n+51=0\iff \alpha=-n\pm\sqrt{10n-{51\over 4}}$

We must have $10n-{51\over 4}\geqslant 0\iff n\geqslant{51\over 40}\implies n\geqslant 2$ and also $0\leqslant\alpha <1$, therefore we take only the plus sign and have two inequalities:

1. $-n+\sqrt{10n-{51\over 4}}\geqslant 0\iff 10n-{51\over 4}\geqslant n^2\iff n^2-10n+{51\over 4}\leqslant 0$

The roots are $1.5$ and $8.5$, hence the inequality is satisfied for $2\leqslant n\leqslant 8$

2. $-n+\sqrt{10n-{51\over 4}}<1\iff 10n-{51\over 4}<(n+1)^2\iff n^2-8n+{55\over 4}>0$

The roots are $2.5$ and $5.5$, hence the inequality is satisfied for $n\leqslant 2$ or $n\geqslant 6$

Now $(n\geqslant 2)\land(2\leqslant n\leqslant 8)\land (n\leqslant 2\lor n\geqslant 6)$ gives $n\in\{2,6,7,8\}$

From $x=n+\alpha=\sqrt{10n-{51\over 4}}=\frac{\sqrt{40n-51}}{2}$ we find the corresponding values for $x$:

$x\in\{\frac{\sqrt{29}}{2},\frac{3\sqrt{21}}{2},\frac{\sqrt{229}}{2},\frac{\sqrt{269}}{2}\}$

[hide="Alternate Method"]Let $a$ be an integer such that $a=\lfloor x \rfloor \le x.$ Substituting, our equation becomes \[ 4a^2-40a+51=(2a-3)(2a-17)\le 0.  \]$(2a-3)(2a-17)$ is only less than $0$ when $2a-3$ is positive and when $2a-17$ is negative. This means $\frac32 <a< \frac{17}{2},$ and we must test $a=2,3,4,\ldots,8$ to see if they satisfy the inequality. Because $x-a$ cannot be greateer than or equal to $1,$ $a=3,4,5$ don't work.

Now, we solve for $x.$ From $4x^2-40a+51=0,$ $x=\frac{\sqrt{40a-51}}{2}$ for $a=2,6,7,8.$ Therefore, $x=\frac{\sqrt{29}}{2},\frac{3\sqrt{21}}{2},\frac{\sqrt{229}}{2},\frac{\sqrt{269}}{2}.$ [/hide]

Note that the equation is impossible for $x\le0$

For $x>0$,I will use the trivial inequalities $x-1<[x] \le x$
$0=4x^2-40[x]+51 \ge 4x^2-40x+51=(2x-3)(2x-17)$,so we obtain $1.5 \le x \le 8.5$

$0=4x^2-40[x]+51 < 4x^2-40(x-1)+51=4x^2-40x+91=(2x-7)(2x-13) \Rightarrow x>6.5$ or$x<3.5$.

Combining these bounds for $x$ we obtain the following possible values for $[x]:3,2,1,6,7,8$

Thus the required solutions are:$\frac{\sqrt{189}}{2},\frac{\sqrt{229}}{2},\frac{\sqrt{269}}{2},\frac{\sqrt{29}}{2}$(Note that for $[x]=3$ we obtain $x=\frac{\sqrt{69}}{2}>\frac{\sqrt{49}}{2}=3.5$ which is not a solution)

Note that we have $4x^2+51 = 40 \lfloor x \rfloor.$ Since $40 \lfloor x \rfloor \leq 40x$, we know that $4x^2+52 \leq 40x$. After subtracting $40x$ and factoring, this inequality becomes $$(2x-3)(2x-17) \leq 0 \implies \frac{3}{2} \leq x \leq \frac{17}{2}.$$ Since $\lfloor x \rfloor$ must be an integer, it must equal one of the integers from $1$ to $8$, inclusive. We take cases on what $\lfloor x \rfloor$ equals. We can plug the value of $\lfloor x \rfloor$ into the original equation and find the resulting value of $x$.

[list]
[*]$\lfloor x \rfloor=1:$ We get $4x^2=-11$; not possible
[*]$\lfloor x \rfloor =2: $ We get $4x^2=29 \implies x=\frac{\sqrt{29}}{2}$; works
[*]$\lfloor x \rfloor =3:$ We get $4x^2=69 \implies x=\frac{\sqrt{69}}{2}$; $x > 4$ so not possible
[*]$\lfloor x \rfloor =4:$ We get $4x^2=109 \implies x=\frac{\sqrt{109}}{2}$; $x> 5$ so not possible
[*]$\lfloor x \rfloor =5:$ We get $4x^2=149 \implies x=\frac{\sqrt{149}}{2}$; $x> 6$ so not possible
[*]$\lfloor x \rfloor =6:$ We get $4x^2=189 \implies x=\frac{3\sqrt{21}}{2}$; works
[*]$\lfloor x \rfloor =7:$ We get $4x^2=229 \implies x=\frac{\sqrt{229}}{2}$; works
[*]$\lfloor x \rfloor =8:$ We get $4x^2=269 \implies x=\frac{\sqrt{269}}{2}$; works
[/list]
Hence, the solutions are $x=\frac{\sqrt{29}}{2}, \frac{3\sqrt{21}}{2}, \frac{\sqrt{229}}{2}, \frac{\sqrt{269}}{2}$.

Alternative, by solving for $x=\frac{\pm \sqrt{40[x]-51}-4[x]}{2}$ and bounding
$0 \leq x-[x] <1$ and use increasing/decreasing functions.

If anyone could help me to line up two inequalities in an align* environment that would be helpful.

[hide=Solution]The idea is to use bounding to find the possible values of $\lfloor x\rfloor$ and then just plug them in.

To do this, we bound:

\begin{align*}
\lfloor x\rfloor^2 &\leq x^2 < (\lfloor x\rfloor +1)^2\\
4\lfloor x\rfloor^2 &\leq 4x^2 < 4\lfloor x\rfloor^2 + 8\lfloor x\rfloor +4\\
4\lfloor x\rfloor^2 &\leq 40\lfloor x\rfloor-51 < 4\lfloor x\rfloor^2 + 8\lfloor x\rfloor +4\\
4\lfloor x\rfloor^2 -40\lfloor x\rfloor +51 &\leq 0 < 4\lfloor x\rfloor^2 -32\lfloor x\rfloor+55\\
(2\lfloor x\rfloor -3)(2\lfloor x\rfloor -17) &\leq 0 < (2\lfloor x\rfloor -5)(2\lfloor x\rfloor -11)\\
\end{align*}

This implies that $\lfloor x\rfloor\in\{2,6,7,8\}$.

Plugging each of these in gives: \[\boxed{\left\{\frac{\sqrt{29}}{2}, \frac{3\sqrt{21}}{2}, \frac{\sqrt{229}}{2}, \frac{\sqrt{269}}{2}\right\}}\]

respectively.
[/hide]

This problem is a familiar one since lhs is greater than $4x^2-40x+51$ now factoring we get a bound which is as follows 3/2<=x<=17/2 now we do some casework and get the answer

[quote=4everwise][hide="Alternate Method"]Let $a$ be an integer such that $a=\lfloor x \rfloor \le x.$ Substituting, our equation becomes \[ 4a^2-40a+51=(2a-3)(2a-17)\le 0.  \]$(2a-3)(2a-17)$ is only less than $0$ when $2a-3$ is positive and when $2a-17$ is negative. This means $\frac32 <a< \frac{17}{2},$ and we must test $a=2,3,4,\ldots,8$ to see if they satisfy the inequality. Because $x-a$ cannot be greater than or equal to $1,$ $a=3,4,5$ don't work.

Now, we solve for $x.$ From $4x^2-40a+51=0,$ $x=\frac{\sqrt{40a-51}}{2}$ for $a=2,6,7,8.$ Therefore, $x=\frac{\sqrt{29}}{2},\frac{3\sqrt{21}}{2},\frac{\sqrt{229}}{2},\frac{\sqrt{269}}{2}.$ [/hide][/quote]

[quote = The part that I don't understand]Because $x-a$ cannot be greater than or equal to $1,$ $a=3,4,5$ don't work.[/quote] Why is this true?

bump.....

Could someone please give me some insight as to the reasoning behind the above solution, any help would be appreciated!


It depends on what account you have on XP. For example, my dad's the admin so only he can creat passwords. Now if you are talking about a person not in your house creating a password then that could be a serious problem.

[quote="jeez123"]Hello guys, I am currently using Windows XP and I really want to password-protect my computer from intruders. Well, I tried to create a password via the control panel, but "someone" recently found a way to get in nevertheless... are there any other ways for me to lock my comp up?

Thanks![/quote]

make another account, and MAKE IT so that it's a regular account. Also, disable all the other good stuff so the regular person can't get in with skillful ways


algebra unsolved
algebra
system of equations

Find all solutions in real numbers of:

\[   \{\begin{array}{ccc} \frac{4x^2}{1+4x^2} = y \\ \frac{4y^2}{1+4y^2} = z \\ \frac{4z^2}{1+4z^2} = x  \end{array} \]

Clearly, any solution is formed with non-negative numbers.

If $x = 0$, we easily deduce that $y=z=0$.
Conversely, $x=y=z=0$ is a solution.

Now, assume that $x,y,z > 0$.
Then, just use that $1+4x^2 \geq 4x$ so that $y \leq x$.
Similarly, we have $z \leq y$ and $x \leq z$.
It follows that $x=y=z$.
Returning to the initial equations we deduce that $x=y=z = \frac 1 2$, which is indeed a solution.

Pierre.


function
induction
floor function
Putnam
modular arithmetic
search

I have a solution by induction, but I'm curious to see if there's a non-inductive solution.

Prove that
\[\left\lfloor \frac{n+2^{0}}{2}\right\rfloor+\left\lfloor \frac{n+2^{1}}{2^{2}}\right\rfloor+\left\lfloor \frac{n+2^{2}}{2^{3}}\right\rfloor+...+\left\lfloor \frac{n+2^{n-1}}{2^{n}}\right\rfloor = n \]
for any positive integer $n$.

It can also be shown using induction by noting that each term increases by 1 iff
$n \equiv 1 \pmod{2}$
$n \equiv 2 \pmod{4}$
...
$n \equiv 2^{k}\pmod{2^{k+1}}$
...
and it's clear that exactly one of these statements will be true.

[quote="amirhtlusa"][quote="seamusoboyle"]I got a nice solution using binary, but im in too much of a rush right now...[/quote]

i think theres exists even a nicer solution than that!

[hide="hint"]
recall the identity:
$\lfloor x\rfloor+\lfloor x+\frac{1}{n}\rfloor+\lfloor x+\frac{2}{n}\rfloor+...+\lfloor x+\frac{n-1}{n}\rfloor=\lfloor nx \rfloor$
[/hide]

[hide="solution"]
by the identity mentioned in the hint we have:
$\left\lfloor x+\frac12\right\rfloor= \left\lfloor 2x\right\rfloor-\left\lfloor x\right\rfloor$.

$\sum_{i=0}^{\infty}\lfloor \frac{n+2^{i}}{2^{i+1}}\rfloor= \sum_{i=0}^{\infty}\lfloor \frac{n}{2^{i+1}}+\frac{1}{2}\rfloor$
by identity above:
$= \sum_{i=0}^{\infty}\lfloor \frac{2n}{2^{i+1}}\rfloor-\lfloor \frac{n}{2^{i+1}}\rfloor=$ $\sum_{i=0}^{\infty}\lfloor \frac{n}{2^{i}}\rfloor-\sum_{i=0}^{\infty}\lfloor \frac{n}{2^{i+1}}\rfloor= \lfloor \frac{n}{2^{0}}\rfloor= n$[/hide]

it can also be shown via induction, show that if it holds for n, then it will hold for 2n and 2n+1.
edit: this problem appeared in University of Washington Putnam  team selection test.[/quote]

I dont assume any of those inequalities. mine looks at how you write the number n out in binary. ie divide out the highest power of two continousally...

The solution using Hermites identity is quite nice. Id be interested to see yours seamusboyle. (I apologize for the lack of apostraphes; every time I hit the apostraphe button it brings up the Mozilla search, no idea why).

seamusoboyle, i know exactly what you are talking about.
in fact the binary expansion solution, was my solution, when i first wrote the solution for our test. :)

Woah woah woah! Thats two names! Seamus O' Boyle! And does anyone know how to delete posts? My double post is annoying me, and i cant find the option :(

there's an icon on the bottom right of ur post .. i believe..umm do u mind posting ur binary solution

This is IMO 1968/6.  This problem is pretty straight forward when considering binary numbers.  Also, the Hermite identity is a nice exercise in periodicity. :) Seamus O' Boyle, you cannot delete posts when someone has already posted after you.


function
limit
real analysis
real analysis unsolved

Let $f$ be a bounded continuous function on $(a,\infty)$. For any $T>0$ show that there exists $x_{n}\to\infty$ such that $\lim_{n\to\infty}(f(x_{n}+T)-f(x_{n})) = 0$

$g(x)=f(x+T)-f(x)$ is bounded and continuous. If it changes sign infinitely often near $\infty$, then by the Mean Value Theorem we're done. If not, WLOG suppose that it's positive. Then either $\liminf_{x \to \infty}g(x) > 0$, which implies $f$ unbounded, contradiction, or $\liminf_{x \to \infty}g(x) = 0$, which gives a possible sequence.


$ b_1\equal{}1,b_n\equal{} n^{f(n)}b_{n\minus{}1},(n\geq2),f(n)\equal{}2^{1\minus{}n},\Longrightarrow b_n<3.$


Everytime I try to remove a contributor, I get this error message:

SQL requests not achieved

DEBUG MODE

SQL Error: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 3

SQL Request: DELETE FROM phpbb_weblog_contributors WHERE weblog_id = 717 AND user_id IN ( )
Line : 261
File : class_blog_options.php

What do I do?

[quote="xxazurewrathxx"]What do I do?[/quote]
Obviously not using the search function.

[quote="xxazurewrathxx"]I have searched, I don't think anyone has given a conclusive answer to the question yet, besides, "we're working on it". I was hoping that there's been an update to the situation since then.[/quote]
Remember me that I should always start new threads everytime I've something to say or want soemthing older than five minutes. Oh sorry for not starting a thread concerning this post.  :roll:

Yes it's probably better to reply to one of the existing threads and say "This problem is still occurring, here's what I was doing when it happened, any updates?" instead of posting it again (and this really is like the 4th or 5th time, it happened to me the other day and I searched for it and saw so many so I decided not to post anything....).

Problem is now fixed. Small glitch, so small that it was hard to find :)


Are you allowed to do USAMTS if you are in 6th grade?

Yes, there is no lower grade limit.  As long as you haven't graduated high school, you may participate.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let p be a prime and n is a positive integer.
Find the number of $p$-Sylow groups in a symmetric group $S_n$ .

It was posted before - but not solved:
[url=http://www.mathlinks.ro/Forum/topic-56995]www.mathlinks.ro/Forum/topic-56995[/url]

Sorry,can anyone show me a formular in a general case of this problem?I think it is a very nice problem.I mean the problem not to describe all the $p$ Sylow subgroups ,COUNT them !
For most simple cases , i have some answers:
1/$n$<$2p$ then the answer is $n!$/$(n-p)!p(p-1)$
2/$n=2p$ then the answer is $(2p)!$/$(2(p(p-1))^2)$


suppose you have the generally functional A Who can you find B in the case that f is independent of x?
[img]http://img413.imageshack.us/img413/930/functionaalbisoi4.jpg[/img]

Thanks.

This isn't the place for calculus. I'm sure your post will be moved to the calculus forum where someone can help.


Vom nota cu $f(n)$ numarul de 1-uri din reprezentarea binara a lui $n$.Demonstrati ca exista un sir $(a_{i})_{i\geq 1}$ astfel incat $\lim_{i\rightarrow \infty}\frac{f(a_{i}^{2})}{f(a_{i})}$=0

Nu s-a gandit nimeni??? Mie mi se pare interesanta...da nu prea am idei...cum ar treb sa arate sirul ala :(  Sper sa posteze pana la urma cineva o solutie:) Intre timp sper sa ma mai gandesc si eu (intre 2 teze si doua aplicatii :( )

Fie $a_{2}= (1101111)_{2}= (111)_{10}$ si $a_{n+1}= 2^{2^{n+1}}a_{n}-1, \, \forall n \geq 2$. $a_{n}$ arata asa in baza $2$: $11$, apoi $0$, apoi $111$ ($2^{2}-1$), apoi $0$, apoi $1111111$ ($2^{3}-1$), apoi $0$, $\ldots$, apoi $0$, apoi $11 \ldots 1$ ($2^{n-1}-1$), apoi $0$, apoi $11 \ldots 1$ ($2^{n}$).

$1^\circ$ $f \left( a_{n}\right) = 2^{n+1}-n$

$2^\circ$ $f \left( a_{n}^{2}\right) = \frac{n^{2}-n+6}2$

Se poate demonstra destul de usor $1^\circ$, dar $2^\circ$ are deocamdata statutul de conjectura (merge pentru [b]câteva[/b] [i]n[/i]-uri). Chiar daca $2^\circ$ nu se va dovedi a fi adevarata, exista sanse bune ca $\lim_{n \to \infty}\frac{f \left( a_{n}^{2}\right)}{f \left( a_{n}\right)}= 0$.

Cum m-am gandit monstrul de mai sus? Trisand un pic, bineinteles. Am facut un program care imi calcula raportul $\frac{f(x^{2})}{f(x)}$ pentru $x$-urile intre $2$ si $5 \cdot 10^{4}$.

Pentru numarul $x_{3}= (28415)_{10}= (110111011111111)_{2}$ am obtinut cea mai mica valoare, dupa care mi-a venit ideea de mai sus.

Ok :) Imi pare bine ca s-a gandit cineva. In lipsa totala a unor cunostinte de informatica, nu prea am avut rabdare sa ma joc cu calculatorul de buzunar (ajunsesem pe la 65 :blush: ). Ideea data de perfect_radio e frumoasa si poate in curand apare cineva cu o solutie completa :)


How many scalene triangles with perimeters less than $ 16$ have sides whose measures are integers?

[hide="Answer"]
9
[/hide]

I suppose we could use casework, but that would take a REALLY long time, anyone have a fast solution?

EDIT: Thanks!  By the way, this was #10 on Target :D

Brute force is the best way. For example, none of the sides can be $ 1$. After that, note that if $ 2$ is one of the sides, then the other sides must be $ n$ and $ n\plus{}1$, where $ 3\le n\le6$ (the second half comes from the problem restrictions). If $ 3$ is the smallest side, then quick casework gives $ \{(3,4,5),(3,4,6),(3,5,6),(3,5,7)\}$, and if $ 4$ is the smallest side, we similarly get $ \{(4,5,6)\}$. $ 5\plus{}6\plus{}7\equal{}18\ge16$, so $ 5$ can't be the smallest side. Adding up, this is $ (6\minus{}3\plus{}1)\plus{}4\plus{}1\equal{}\boxed9$ triangles.


trigonometry integration calculus logarithms function geometry

A ball is thrown from the ground with initial velocity $ v_o$ making an angle $ \alpha$ with the horizontal. Find the total distance traveled by the ball (neglect air resistance).

Here's my opinion

[hide]$ dx \equal{} v_0\cos\alpha dt$

$ dy \equal{} v_0\sin\alpha dt \plus{} gtdt$

$ ds \equal{} \sqrt {dx^2 \plus{} dy^2}$

The time that the ball flies is $ \Delta t \equal{} \frac {2v_0\sin\alpha}{g}$

The distance that the ball travels

$ s \equal{} \int ds \equal{} \int_0^{\Delta t} \sqrt {dx^2 \plus{} dy^2}$[/hide]

There is something wrong there. Also, I want to see the final answer.

[hide="yet another opinion"]
clearly the distance travelled is
$ \int \sqrt {1 + f'(x)^{2}}dx$ where $ y = f(x)$ is the equation of the motion of the body
clearly $ f'(x) = \tan \alpha - \frac {gt}{2v_{o}\cos \alpha}$
hence the problem reduces nothing  but integration an integral of the form
$ \int x\sqrt {1 + {b(x - c)}^{2}dx \bigstar}$
after substituting $ \frac {dx}{dt} = v_{0}\cos \alpha$
the integral of $ \bigstar$ can be solved by substituting
$ b(x - c) = u$
thus $ \bigstar \equiv \int u \sqrt {1 + u^{2}}du + \int \sqrt {1 + u^{2}} du$
leaving out the constants obviuosly.
the first one can be done by substituting $ u^{2} = x$ and second one is in by itself a standard form
P.S. i  would like to see a tricky proof of this  :)
[/hide]

yes of course that was the method :wink:

Assuming the same level for shooting point and target

[hide="we get:"]
\[ {v_0^2 \over g}(sin\alpha \plus{} cos^2\alpha.arc \,sinh \,tg\alpha)
\]
[/hide]

That is very close to [hide="my answer"]$ L \equal{} \frac{v_o^2}{2g}\left[2 \sin \alpha \plus{} \cos^2 \alpha \ln\left(\frac{1\plus{} \sin \alpha}{1 \minus{} \sin \alpha}\right)\right]$.[/hide]

Sorry, I forgot the factor $ \,cos^2\alpha\,$.

Now,  $ \,arc\,sinh \,tg \alpha\,$ is "short' to $ {1\over 2}ln\,{{1 \plus{} sin\alpha}\over{1 \minus{} sin\alpha}}\,$

Yes, that's right. But its "$ argsinh(\tan \alpha)$", not "$ \arcsin(\tan \alpha)$".

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

http://mathworld.wolfram.com/InverseHyperbolicSine.html   :The inverse hyperbolic sine sinh^(-1)z (Beyer 1987, p. 181; Zwillinger 1995, p. 481), sometimes called the area hyperbolic sine (Harris and Stocker 1998, p. 264) and sometimes denoted arcsinhz (Jeffrey 2000, p. 124), is the multivalued function that is the inverse function of the hyperbolic sine. The variants Arcsinhz or Arsinhz (Harris and Stocker 1998, p. 263) are sometimes used to refer to explicit principal values

Thanks for the links. However, in my country we write "argsinh" which stands for "argument of the hyperbolic sine", and that is the notation I use. "Arcsin" means (in my country) "(the) arc (whose) sine (is)".


inequalities
trigonometry
incenter
geometric inequality
inequalities proposed

I think it can be solved by using trigonometry!Put [tex]cosA=\frac{\sqrt{yz}}{2},cosB=\frac{\sqrt{xz}}{2},cosC=\frac{\sqrt{xy}}{2}[/tex].Since hypothesis we see:[tex]A,B,C[/tex] are 3 angles of a triangle...But the rest is not easy!

Replacing $x,y,z$ by $1/x,1/y,1/z$, the statement becomes as follows:
If $x,y,z>0$ such that $x+y+z+1=4xyz$, then
$1+x+y+z \ge xyz+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.
For $x+y+z=constant$ and $xyz=constant$, by n-1EV Principle it follows that the expression $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is maximal when $x\le y=z$. Thus, it suffices to prove the inequality for $y=z$, when the hypothesis condition $x+y+z+1=4xyz$ becomes $y=\frac{x+1}{2x}$.
Have you a nice solution to this nice problem, Levi ?

This algebraic inequality is obtained from the following geometric inequality in a triangle $ABC\ ,$
$\ 1+\frac{b+c-a}{a}+\frac{c+a-b}{b}+\frac{a+b-c}{c}\le \frac{2r}{R}+\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}$
by the system of substitusions $b+c=a(x+1), c+a=b(y+1), a+b=c(z+1)\ \ (*)$. The condition of the compatibility of the system $(*)$ is the relation $xy+yz+zx+xyz=4$ and his solution $(a,\ b,\ c)$ is
$\frac{a}{(y+2)(z+2)}=\frac{b}{(z+2)(x+2)}=$ $\frac{c}{(x+2)(y+2)}=\frac{a+b+c}{(x+y)(y+z)(z+x)}t\ge 0$
and we can to suppose that $t=1$.
We show easy that $a,\ b,\ c$ are the lenghts of the sides a triangle iff $x,\ y,\ z>0$ a.s.o.
Remark: From the distance between the orthocentre $H$ and the centre $I$ of the incircle,
$HI=4r(R+r)+3r^2-s^2$ results the inequality $s^2 \leq  4R(R+r)+3r^2$ which is equivalently with the up geometric inequality.

I think I got the most simple proof of this problem,still using Trigonometry:After some conclusions and the initial relationship ,it becomes:
$(2-xy)(2-yz)(2-zx)\leq x^2y^2z^2(*)$
Substituting $\sqrt{yz}=2sin\frac{A}{2}$,... then we have $A+B+C=\pi$.And (*) becomes:
$cosAcosBcosC\leq (1-cosA)(1-cosB)(1-cosC)$
which is well-known.

[quote="Virgil Nicula"][color=darkred]Prove that in any triangle $ABC$ there is the following inequality :[/color]

$\boxed{\ \ 1+\frac{b+c-a}{a}+\frac{c+a-b}{b}+\frac{a+b-c}{c}\le \frac{2r}{R}+\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\ }\ \ (*)$[/quote][color=darkblue][b][u]Proof[/u].[/b] Prove easily that  $\left\|\begin{array}{c}\sum\frac{b+c-a}{a}=\frac{2}{abc}\cdot\sum bc(p-a)=\frac{p^{2}+r^{2}-8Rr}{2Rr}\\\\ \sum \frac{a}{b+c-a}=\frac{1}{2(p-a)(p-b)(p-c)}\cdot \sum a(p-b)(p-c)=\frac{2R-r}{r}\end{array}\right\|\ .$ Thus ,

$(*)$ $\Longleftrightarrow$ $\frac{p^{2}+r^{2}-8Rr}{2Rr}+\frac{R-2r}{R}-\frac{2R-r}{r}\le 0$ $\Longleftrightarrow$ $p^{2}\le 4R(R+r)+3r^{2}\ \ (**)$

what is truly because the distance between the incenter $I$ and the orthocenter $H$ is given by

$\boxed{\ HI^{2}=4R(R+r)+3r^{2}-p^{2}\ }\ge 0\ .$ In conclusion , $(*)$ $\Longleftrightarrow$ $(**)\ .$[/color]

[quote="younesmath2012maroc"]Using transformations  inequality is shown to:
$ a, b, c >0,a^2+b^2+c^2+abc=4\Rightarrow a^2+b^2+c^2+a^2b^2c^2 \geq abc+b^2c^2+c^2a^2+a^2b^2 $[/quote]

Further, considering that $ abc=4-a^2-b^2-c^2 $ obtain:
$ a^2+b^2+c^2+a^2b^2c^2 \geq abc+b^2c^2+c^2a^2+a^2b^2\Leftrightarrow a^2+b^2+c^2+a^2b^2c^2 \geq 4-a^2-b^2-c^2+b^2c^2+c^2a^2+a^2b^2 $
$ \Leftrightarrow a^2b^2c^2+2(a^2+b^2+c^2)\geq 4+b^2c^2+c^2a^2+a^2b^2 $ Q.E.D

[quote="Virgil Nicula"]Let $ x,y,z>0$ such that $xy+yz+zx+xyz=4$ . Prove that $ 1+x+y+z\le xyz+\frac 1x+\frac 1y+\frac 1z\ .$[/quote]

Because
 $ 1+x+y+z\le xyz+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ . $ $\Leftrightarrow (2-xy)(2-yz)(2-zx)\leq x^2y^2z^2 $
demonstrated inequality can be written as:
$ x,y,z>0,xy+yz+zx+xyz=4\Rightarrow (2-xy)(2-yz)(2-zx)\leq x^2y^2z^2 $

Very nice problem! Congratulations to Virgil Nicula!
My current solution:
Problem is equivalent to:
$ x,y,z>0, x+y+z+1=4xyz\Rightarrow 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le x+y+z+\frac{1}{xyz} $
With $ p+x+y+z,q+xy+yz+zx,r=xyz $  it return to:
$ q\ge p\ge 3r\ge 3,p+1=4r\Rightarrow 1+\frac{q}{r} \le\frac{1}{r}+p\Leftrightarrow $
$ 2+\frac{q}{r} \le\frac{1}{r}+p+1\Leftrightarrow 2+\frac{q}{r} \le\frac{1}{r}+4r $
$ \Leftrightarrow 2r+q\le 1+4r^2\Leftrightarrow q\le 4r^2-2r+1\Leftrightarrow 4q\le p^2+3 $
(1)    $ \Leftrightarrow q\le\frac{p^2+3}{4} $
By Schur of three degree we have:
$ (x+y+z)^3+9xyz\ge 4(x+y+z)(xy+yz+zx)\Leftrightarrow p^3+9r\ge 4pg\Leftrightarrow $
(2)  $ q\le\frac{4p^3+9p+9}{16p} $
It is sufficient that:
$ \frac{4p^3+9p+9}{16p}\le\frac{p^2+3}{4}\Leftrightarrow p\ge 3 $
Which is true!!

[quote=sandumarin]Very nice problem! Congratulations to Virgil Nicula!
My current solution:
Problem is equivalent to:
$ x,y,z>0, x+y+z+1=4xyz\Rightarrow 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le x+y+z+\frac{1}{xyz} $
With $ p=x+y+z,q=xy+yz+zx,r=xyz $  it return to:
$ q\ge p\ge 3r\ge 3,p+1=4r\Rightarrow 1+\frac{q}{r} \le\frac{1}{r}+p\Leftrightarrow $
$ 2+\frac{q}{r} \le\frac{1}{r}+p+1\Leftrightarrow 2+\frac{q}{r} \le\frac{1}{r}+4r $
$ \Leftrightarrow 2r+q\le 1+4r^2\Leftrightarrow q\le 4r^2-2r+1\Leftrightarrow 4q\le p^2+3 $
(1)    $ \Leftrightarrow q\le\frac{p^2+3}{4} $
By Schur of three degree we have:
$ (x+y+z)^3+9xyz\ge 4(x+y+z)(xy+yz+zx)\Leftrightarrow p^3+9r\ge 4pg\Leftrightarrow $
(2)  $ q\le\frac{4p^3+9p+9}{16p} $
It is sufficient that:
$ \frac{4p^3+9p+9}{16p}\le\frac{p^2+3}{4}\Leftrightarrow p\ge 3 $
Which is true!![/quote]


LaTeX
inequalities
inequalities proposed

prove that  if abc=1 then we have   a^3 +b^3 +c^3 +2ab\ a^2 +b^2       2ac\a^2+c^2  +2bc\b^2+c^2  >= 6                                                                                it can be solved by sos but i want another solution

 :ninja:

[quote="fermat3"]prove that if abc=1 then we have a^3 +b^3 +c^3 +2ab\a^2+b^2+2ac\a^2+c^2+2bc\b^2+c^2  >= 6
I can be solved by sos but i want another solution.[/quote]
I think you meant:

[quote]prove that if $abc=1$ then we have:
\[ a^3+b^3+c^3+\frac{2ab}{a^2+b^2}+\frac{2ac}{a^2+c^2}+\frac{2bc}{b^2+c^2} \ge 6  \][/quote]

You should learn $\LaTeX$, it's easy!

Here's my solution.

We knew the well-known inequality:
\[ \frac{a^3+b^3}{2} \ge \left(\frac{a^2+b^2}{2}\right)^{\frac{3}{2}}  \]
Then:
$LHS \ge \sum_{cyc} \left( \frac{2ab}{a^2+b^2}+\left(\frac{a^2+b^2}{2}\right)^{\frac{3}{2}} \right) \ge \sum_{cyc} \left( 2\sqrt{(x^2+y^2)^{\frac12} \cdot \frac{2ab}{\sqrt{2^3}}} \right) \ge \sum_{cyc} \left( 2 \sqrt{\sqrt{2xy}\cdot \frac{2}{2\sqrt{2}} ab}\right) = 2\sum_{cyc} (ab)^{\frac34}$
$\implies LHS \ge 6$ by AM-GM and $abc=1$
Q.E.D.


Hello,
Does anybody know how class rank is determined?
thanks
:)

Class rank is determined by two factors: your grades and the difficulty of the course your are taking.

For AP/GT courses, an A earns you 6 points, a B earns you 5 points, and so on.

For honors courses, an A earns you 5 points, a B earns you 4 points, and so on.

And for standard courses, an A earns you 4 points.

Your class rank is found by averaging your final(?) grades of each year and comparing that with your peers.

-----------

That's how my high school does class rankings, but yours might be different, as the difficulty of the courses might have different points.

Some schools do GPAs with 100 being the most possible in the lower leveled classes and higher leveled classes able to earn above that.  Aside from scale differences, it is based on grades earned in classes taken.

Each high school sets its own rules. Some high schools do not rank students.


Are there any two 8-digit numbers (which have the same digits but in different order), whose sum is equal to 20062007?


geometry
algebra
polynomial
induction
analytic geometry
invariant
advanced fields

Show that $ d\plus{}1$ distinct points $ p_1,...,p_{d\plus{}1} \in \mathbb{A}^n(\mathbb{Q})$ always impose independent conditions on polynomials in $ P_{n,d}$

[quote]Definition:
Given $ S$ $ \subset \mathbb{A}^n(k)$, the number of conditions imposed by $ S$ on polynomials of degree $ \le d$ is defined $ C_d (S): \equal{} dim P_{n,d} \minus{} dim I_d (S)$, where $ I_d (S)$ is the vector space of polynomials of degree $ \le d$ vanishing at each of the points.
$ S$ is said to imposed independent conditions on $ P_{n,d}$ if $ C_d (S) \equal{} |S|$[/quote]

Try induction on $ n$: For $ n \equal{} 1$ you get $ I_1(S) \equal{} 0$, because every polynomial of degree $ \le d$ vanishing at $ d \plus{} 1$ points vanishes identically. $ \dim P_{1,d} \equal{} d \plus{} 1$ is easy.

Assume that the claim is true for $ n \minus{} 1$. Given $ q_1,\ldots,q_{d \plus{} 1}\in\mathbb Q^n$ we need to show that the linear map $ \Phi_{n,d}^{(q_i)}: P_{n,d}\rightarrow \mathbb Q^{d \plus{} 1}, f\mapsto (f(q_i))_{1\le i\le d \plus{} 1}$ is surjective, because $ I_{n,d} \equal{} \ker(\Phi_{n,d}^{(q_i)})$ and $ C_d((q_i)) \equal{} \dim P_{n,d} \minus{} \dim I_{n,d} \equal{} rk(\Phi_{n,d}^{(q_i)})$.

Wlog we can assume that the $ q_i$'s are in sufficiently general position, i.e. that the projection $ \pi_n: \mathbb Q^{n}\to\mathbb Q^{n \minus{} 1}$ dropping the last coordinate produces $ d \plus{} 1$ pairwise different points $ q'_i \equal{} \pi_n(q_i)$. If this is not the case, then decompose $ \mathbb Q^n$ in a hyperplane and a one-dimensional subspace such that the induced projection on the hyperplane produces pairwise different points. After a linear base change we can assume that the $ q_i$'s are in sufficiently general position. Note that the assertion to be proven is invariant under linear base changes such that everything is fine after our little manipulation.

Now the induction step is easy: Consider the restriction of $ \Phi_{n,d}^{(q_i)}$ to $ P_{n \minus{} 1,d}$ which coincides with $ \Phi_{n \minus{} 1,d}^{(q_i')}$, but this map is surjective by the induction hypothesis.


Hi,

If I check the option log in automatically, it loges in, but next time I enter the site:

[u]First:[/u] It doesnt log in automatically,
[u]Also[/u]: It even doesn't want to log in(when I enter username/password I see the first page http://www.mathlinks.ro/ without the "Welcome FOURRIER!") , so I get obliged to use Internet Explorer for some time...

Thank you for helping me.

 :)

Do you have cookies enabled? What version of Opera are you using?

Yes, cookies are enabled,
My opera version is 9.23.8808.0 .

I can't recreate the bug. It must be something in your browsers' settings I can only assume.
As a suggestion: try downloading Opera 9.24 (which is the latest stable version) and reinstalling it. That might fix the problem.


function algebra domain trigonometry symmetry

For each function state the:
-domain
-range
-is it one to one
-what symettry it has (y-axis, x-axis, origin, or none)
-is it even, odd or neither?

1.  $ y \equal{} x^3\minus{}5x$
2. $ y \equal{} \sqrt{x^2\minus{}4}$
3. $ y \equal{} \frac{2}{x\plus{}3}$
4.  $ y \equal{} 2 \cos{x}$

[hide="#1"]
Domain: All real numbers.
Range: (-Infinity, Infinity)
Not one-to-one
Origin symmetry
Odd
[/hide]

[hide="#2"]
Domain: All real numbers >= 2
Range: [0, Infinity)
One-to-one
Y-axis symmetry
Even
[/hide]

[hide="#3"]
Domain: All real numbers except for -3
Range: (-Infinity, 0) or (0, Infinity)
One-to-one
Origin symmetry
Odd
[/hide]

[hide="#4"]
Domain: All real numbers.
Range: [-2, 2]
Not one-to-one
Y-axis symmetry
Even
[/hide]


inequalities
algebra
polynomial
trigonometry
induction
inequalities proposed

this is again a wonderful problem.

let $-1\leq p_i\leq 1$, $i=1,...,n$ be real numbers. define $T_k=\prod_{i=1,i\neq k}^{n} |p_i-p_k|$. prove that

\[\sum_{k=1}^{n} \frac 1{T_k}\geq 2^{n-2}\]

when do we have equality(i ask this question since once you've answered this question you know where to go in order to prove it)?

I have already posted a solution for this problem, using polynomials. I agree, great problem!

Would you mind posting it one more time?
I don't find it, but I would love to see the solution to this beautiful problem

i guess harazi's solution is the same as mine, so here it goes:

first assume that $1\geq p_1>...>p_n\geq -1$. define

$P(x)=\sum_{k=1}^{n} \prod_{i=1,i\neq k}^{n} \frac{x-p_i}{|p_i-p_k|}$.

the polynomial $P(x)$ has degree $n-1$, leading coefficient $\sum_{k=1}^{n} \frac 1{T_k}$ and $P(p_i)=(-1)^i$. this last condition is the key to proving that the leading coefficient has to be quite large. now, we need some trick to show that it is actually at least $2^{n-2}$. define

$Q(x)=\cos( (n-1)\arccos x )$ for $-1\leq x\leq 1$ (the n-1'th chebyshev polynomial).

it is quite easy to see by induction that this is actually a polynomial of degree $n-1$ in $x$ with leading coefficient $2^{n-2}$. it obviously has the wonderful property that $|Q(x)|\leq 1$ for $-1\leq x\leq 1$. now it's quite clear what we have to do: define

$R(x)=P(x)-Q(x)$.

we want to show that it has a nonnegative leading coefficient. we know that $R(x)$ has degree $n-1$ and since $P(p_i)=(-1)^i$ and $|Q(p_i)|<1$ after slightly changing the variables we know that there are zeroes between $p_1$ and $p_2$, $p_2$ and $p_3$, ..., $p_{n-1}$ and $p_n$, in total $n-1$ real zeroes in $[-1; 1]$. therefore there are no other zeroes and since $P(p_1)=1>Q(p_1)$ this inequality holds for all $x\geq p_1$. this gives the estimate of the leading coefficients, QED.


analytic geometry
probability
expected value
probability and stats

Assume that in the beginning we are at the origin of $\mathbb{Z}^{n}$. At every step we move randomly one unit in the coordinate system, i.e. we move from the point $(z_{1},z_{2},\cdots,z_{i},\cdots,z_{n})$ to the point $(z_{1},z_{2},\cdots,z_{i}\pm 1,\cdots,z_{n})$ for some $1 \leq i\leq n$ and the probability that we choose a specific index $i$ is $1/n$ for every $1\leq i\leq n$. Also with probability $1/2$ we decrease the index and with the same probability we increase it. Calculate the probability $P(n,m,q)$ such that after $m$ steps in $\mathbb{Z}^{n}$ we are $q$ steps away from the origin.

(I don't know even a solution for the case $q=0$)

The exact formula is a mess, but the central limit theorem can be used to do approximations for large $m.$

If you know $P(1, m, q)$, you can get $P(n, m, q)$ by summing $P(1, a_{1}, b_{1})\cdot P(1, a_{2}, b_{2}) \cdot\cdots\cdot P(1, a_{n}, b_{n})$ over all solutions in non-negative integers to $a_{1}+a_{2}+\ldots+a_{n}= m$ and $b_{1}+b_{2}+\ldots+b_{n}= q$.  For fixed small values of any of the three variables, this is something that maybe you could evaluate and not find horrific.

There are some other very interesting questions to ask. One such question: what is the probability of the walk eventually returning to the origin? Or (closely related), what is the expected number of returns to the origin over the lifetime of the walk? For those questions, it very much matters what $n,$ the dimension, is.


I was born in washington and I'm in there right now!!! i'm doing aops during math class...heehee...my teacher doesn't care... :lol:  :P

This forum... is like, dead. I'm from Washington... lived here... for like my whole life.
:P


Prove that numbers of the form \[\frac{a_{1}}{1!}+\frac{a_{2}}{2!}+\frac{a_{3}}{3!}+\cdots,\] where  $0 \le a_{i}\le i-1 \;(i=2, 3, 4, \cdots)$ are rational if and only if starting from some $i$ on all the $a_{i}$'s are either equal to $0$ ( in which case the sum is finite) or all are equal to $i-1$.

I just wonder how this can be correct... if all are equal to $ i\minus{}1$ gives rational number $ q\in\mathbb{Q}$, then putting all equal to $ \frac{i\minus{}1}{2}$ instead simply gives you $ \frac{1}2 q\in\mathbb{Q}$...

[quote="Peter"]I just wonder how this can be correct... if all are equal to $ i\minus{}1$ gives rational number $ q\in\mathbb{Q}$, then putting all equal to $ \frac{i\minus{}1}{2}$ instead simply gives you $ \frac{1}{2} q\in\mathbb{Q}$...[/quote]

I don't have the book right now, I'm gonna check it later!

[quote="ideahitme"][quote="Peter"]I just wonder how this can be correct... if all are equal to $ i \minus{} 1$ gives rational number $ q\in\mathbb{Q}$, then putting all equal to $ \frac {i \minus{} 1}{2}$ instead simply gives you $ \frac {1}{2} q\in\mathbb{Q}$...[/quote]

I don't have the book right now, I'm gonna check it later![/quote]Have you had time to check already? :)

The $ a_i$ have to be integers (everything else cannot make sense!). And then this can be proved by mimicing the proof that $ e$ is irrational.

[quote="Peter"]I just wonder how this can be correct... if all are equal to $ i \minus{} 1$ gives rational number $ q\in\mathbb{Q}$, then putting all equal to $ \frac {i \minus{} 1}{2}$ instead simply gives you $ \frac {1}2 q\in\mathbb{Q}$...[/quote]
$ \frac {i \minus{} 1}{2}$ is not integer for even $ i$ therefore we can not put all $ a_{i} \equal{} \frac {i \minus{} 1}{2}$
sorry :oops: when I begin writing this post,zetaX dosen't post...

[quote="ZetaX"]The $ a_i$ have to be integers (everything else cannot make sense!). [/quote]That makes sense, thanks!

Actually, what you probably want to do first is to prove that every rational number has a unique "factorial base" expansion.  The rest of the proof should then proceed analogously to the proof for rational numbers in base $ b$.


How many distinct ordered pairs of positive integers $ (m,n)$ are there so that the sum of the reciprocals of $ m$ and $ n$ is $ \frac14$?

In other words, $ \frac {1}{m} \plus{} \frac {1}{n} \equal{} \frac {1}{4}$
$ \frac {1}{m} \plus{} \frac {1}{n} \equal{} \frac {1}{4} \equal{} \frac {m \plus{} n}{mn}$
Thus $ mn \minus{} 4m \minus{} 4n \equal{} 0$.
$ mn \minus{} 4m \minus{} 4n \plus{} 16 \equal{} (m \minus{} 4)(n \minus{} 4) \equal{} 16$

There are 3 unique ways of expressing 16 as a product of two positive integers: 1*16, 2*8, and 4*4.

Therefore, [b]3[/b].

(in case you were wondering they correspond to (5,20),(6,12),(8,8))

I believe the answer is $ 5$--minor technicality.

He is referring to the fact that (5,20) and (20,5) both work.

i was doing this problem on FTW and it said that the answer is 5

The answer is supposed to be $5$ because we can have $m - 4 = 1,2,4,8,16$; there are $5$ factors of $16$ and $m - 4$ has a one-to-one correspondence with $m$ so $\boxed{5}$ should be the answer.


I'm trying to get my school to register for the AMC 12 B, and I have a few questions about it.

1. Who can proctor the test? Can it be any adult, like a parent, or does it have to be someone associated with the school?

2. Is there a time period that the test has to be taken within? Or can it be taken any time on the official date?

Thanks. Any suggestions for convincing reluctant schools (we tried to take the B test last year, and failed...) would also be really appreciated. (Taking the AMC at another school won't be possible.)

[quote="RenaudianOntology"]1. Who can proctor the test? Can it be any adult, like a parent, or does it have to be someone associated with the school?[/quote]

Any responsible adult can proctor the test, except someone who is related to a test-taker.  IT MUST be given in a public building; if you can't do it in your school, you must do it in a library or other public building.

[quote="RenaudianOntology"]2. Is there a time period that the test has to be taken within? Or can it be taken any time on the official date?[/quote]

There is no set time; any time on Feb 16 is fine (but it must be on that day).

Keep in mind that the latest possible registration date is Feb 7.  For more information, see the [url=http://www.unl.edu/amc/index.html]AMC website[/url].

I suggest you go to the AMC Website
http://www.unl.edu/amc
then on the right side, about mid-way under What's New, choose the link for the 2005 AMC 10 A/AMC 12 A Teachers' Manual.  The rules, instructions and policy statements are all there, starting on about page 5.

The 2005 AMC 10 B/ AMC 12 B Teachers' Manual will be identical except for date changes and contest name changes.

The direct link to jump there now is:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2004-5pub/05TM1012/TM05hs.html

Steve Dunbar
AMC Director

Can someone explain instruction #1? "Take the unopened contest package(s) and Certification Form to your Principal (or equivalent) to certify that the package was opened within an hour before the contest. Have the Principal sign the Certification Form at that time."

Does it really have to be the principal, or could it be a math teacher or math department head? The school principal isn't here very often, and I'm not sure what to do if we didn't take the exam during the school day.

Thanks.

The person who kept custody of the test forms might not in all cases be the school's principal, but it should be SOME identifiable person who signs the form to indicate that no one else had access to the test materials until the test was administered. The person to sign the form is the person who would take responsibility for making that statement truthfully if there were any doubts about the security of the test.

I haven't yet found a proctor for the B exam, but went to the online registration form anyways just to see what was on it, and all it shows me is that my school has already registered for the A exam. It doesn't show any form for a B exam. Does this mean I can't register online, or is there some way to get the B form?
Also, if I can somehow manage to register online, is the latest time to register 11:59 pm on February 7th? (The deadline is February 7th.)

Thanks

If you try registering online for the B test, after already having registered for the A test, it won't let you.  You have to fax in your registration. And either put credit card number (I think) or if you're paying by check, that the check is in the mail (That's what the AMC people told me to do.) I'd fax it in today...The latest it would probably be okay would be 5:00PM Central time on Monday, if you really want to do it at the latest possible moment...


In AoPS City, there is only one area of trees, so if there is a forest fire, all the trees will catch on fire. The probability of a tree in AoPS City surviving in the year 2009 is 13/14. The probability of there being a forest fire is 1/52. The probability of a tree dying if there is a forest fire is 1/1. The probability of a tree in AoPS City dying by some cause other than a forest fire is 3/52. Trees A and B are both in AoPS City. What is the probability that tree A will die, but tree B will not die in 2009?

All of the rest of the stuff is filler information. We don't care how or why a tree dies, just whether or not it does. We quickly pick up on the key sentence:

The probability of a tree in AoPS City surviving in the year 2009 is 13/14.

Thus the probability of A dying is $ 1\minus{}\frac{13}{14}\equal{}\frac{1}{14}$

The probability of B living is $ \frac{13}{14}$

Therefore the probability of A dying and B living is $ \frac{1}{14}\cdot\frac{13}{14}\equal{}\frac{1\cdot13}{14\cdot14}\equal{}\boxed{\frac{13}{196}}$

That is correct. I tried to include alot of irrevelant information to throw you guys off.

I believe it was an exercise for other people to solve.

Yeah, that's why it is called warm-up. Otherwise I would have titled it Help!!!.

1.  To make a new thread, click New Topic, in any one of the forums. (there are many!)

2.  Some basic places you should go to are FTW (a multiplayer game), Alcumus (single player), and maybe some AoPS classes.

3.  All the main stuff is on the left side above the graphic.

[quote="myyellowducky82"][quote="Cheese12345"]Quick Q. Im new here and i have a couple questions.
1. How do you make a post?
2. What are some of the basic controls on this site?
3. Which links should i click on in the beginning?

TY 2 all the ppl that answer. And sorry if this thing bothers u on ur post  :maybe:[/quote]
It won't bother anyone =]
Next time, (goes for everyone too) if you have a question about the Forum in general, please post it in the Questions, Suggestions, and Announcements part of the Forum. (This will avoid lots of spam in the future.)
Thanks, and enjoy AoPS =][/quote]

He didn't know how to make a new topic.

[quote="12markkram34"]Your problem has a problem. Like you said, the probability that a tree dies is 1/14. However, in the rest of the problem you say that the probability that it dies by fire is 1/52 and by any other cause is 3/52. However, 1/52 + 3/52 = 4/52, 4/52 = 1/13, and 1/13 is not equal to 1/14. Its fine to put filler info, but it should be correct.[/quote]

Oh, actually I was concentrating on it.

:fool: I keep forgetting that 4(14) is 56. Sorry. Now please stop posting on this thread.


[quote="i_like_pie"]A cylinder is inside a cube, which is inside a sphere, which is inside a cube, which is inside a sphere. If the smaller sphere has radius $x,$ what is the ratio of the volumes of the bigger sphere and the cylinder?[/quote]

[hide]Becuase the radius of the sphere is x, the distance from the center of the smaller cube to one of its corners is also x.  Therefore, the distance from the center of the cube to the center one of its faces is x/(3^(1/2)).  This is also the radius of the cylinder.  Its hight is twice this or 2x/(3^(1/2)).  This means that its volume is 2pi*x^3*3^(1/2)/9.

The radius of the sphere is also the distance from the center of the larger cube to one of its faces.  Therefore, the distance from the center of this cube to one of its corners is x*3^(1/2). This is also the radius of the larger sphere.  Thus, the volume of the larger sphere is 4pi*x^3*3^(1/2).

Therefore the ratio of the volume of the larger sphere to that of the cylinder is 2:1/9 or 18:1.[/hide]

[hide]We write all the important lengths and volumes in terms of x:
Side length of larger cube: 2x
Space diagonal of larger cube:$2\sqrt{3}x$
Radius of larger sphere: $\sqrt{3}x$
Volume of the larger sphere: $\frac{4}{3}\pi(x\sqrt{3})^{3}=4\pi\sqrt{3}x^{3}$

Space Diagonal of smaller cube: 2x
Side length of smaller cube: $\frac{2x}{\sqrt{3}}$
Height of cylinder: $\frac{2x}{\sqrt{3}}$
Radius of cylinder: $\frac{X}{\sqrt{3}}$
Volume of cylinder: $(\frac{x}{\sqrt{3}})^{2}\pi\times (\frac{2x}{\sqrt{3}})=\frac{2x^{3}\pi}{3\sqrt{3}}$

Then the ratio of the volumes is $4\pi\sqrt{3}x^{3}: \frac{2x^{3}\pi}{3\sqrt{3}}$, or $18: 1$.[/hide]

You're both correct. :)

@ samath - You forgot the last } of \frac for volume of cylinder.


search
function
combinatorics unsolved
combinatorics

How many ways are there to write either $ 0$ or $ 1$ in each cell of a $ 4$x$ 4$ board so that the product of numbers in any two cells sharing an edge is always $ 0$?

Posted recently (and with a source listed, even!).  Search functions are wonderful things.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=444709738&t=169651

I solved it but I can't write it because I used very many diagrams
the answer is 1234


If $ a,b,c$ are positive real numbers such that $ abc\equal{}1$, then

$ \frac {12a\plus{}7}{2a^2\plus{}1}\plus{}\frac {12b\plus{}7}{2b^2\plus{}1}\plus{}\frac {12c\plus{}7}{2c^2\plus{}1} \le 19$.

[quote="Vasc"]If $ a,b,c$ are positive real numbers such that $ abc \equal{} 1$, then

$ \frac {12a \plus{} 7}{2a^2 \plus{} 1} \plus{} \frac {12b \plus{} 7}{2b^2 \plus{} 1} \plus{} \frac {12c \plus{} 7}{2c^2 \plus{} 1} \le 19$.[/quote]
It is just equivalent to your previous inequality, VasC:
If $ x,y,z>0,xyz\equal{}1$, then
\[ \sum \frac{7\minus{}6x}{x^2\plus{}2} \ge 1\]
In here $ x\equal{}\frac{1}{a},y\equal{}\frac{1}{b},z\equal{}\frac{1}{c}$ ;)

[quote="can_hang2007"][quote="Vasc"]If $ a,b,c$ are positive real numbers such that $ abc \equal{} 1$, then

$ \frac {12a \plus{} 7}{2a^2 \plus{} 1} \plus{} \frac {12b \plus{} 7}{2b^2 \plus{} 1} \plus{} \frac {12c \plus{} 7}{2c^2 \plus{} 1} \le 19$.[/quote]
It is just [b]equivalent to your previous inequality[/b], VasC:
If $ x,y,z > 0,xyz \equal{} 1$, then
\[ \sum \frac {7 \minus{} 6x}{x^2 \plus{} 2} \ge 1
\]
In here $ x \equal{} \frac {1}{a},y \equal{} \frac {1}{b},z \equal{} \frac {1}{c}$ ;)[/quote]
Can you show me the previous inequality of VasC.

Is my proof right? :lol:
$ \frac {12a\plus{}7}{2a^2\plus{}1}\plus{}\frac {12b\plus{}7}{2b^2\plus{}1}\plus{}\frac {12c\plus{}7}{2c^2\plus{}1} \le 19$
$ \sum\limits_{cyc} {9 \minus{} \frac{{12a \plus{} 7}}{{2a^2  \plus{} 1}}}  \ge 8$
$ \sum\limits_{cyc} {\frac{{2(3a \minus{} 1)^2 }}{{2a^2  \plus{} 1}}}  \ge 8$
$ \sum\limits_{cyc} {\frac{{(3a \minus{} 1)^2 }}{{2a^2  \plus{} 1}}}  \ge 4$
By Holder
$ LHS \ge \frac{{(3a \plus{} 3b \plus{} 3c \minus{} 3)^2 }}{{2a^2  \plus{} 2b^2  \plus{} 2c^2  \plus{} 3}}$
So we need
$ a^2  \plus{} b^2  \plus{} c^2  \minus{} 18(a \plus{} b \plus{} c) \plus{} 18(ab \plus{} bc \plus{} ac) \minus{} 3 \ge 0$ which can be shown by Mixing-variable :lol:

Very nice and correct your solution, T.T. Hung. :lol:

[quote="TRAN THAI HUNG"]Is my proof right? :lol:
$ \frac {12a \plus{} 7}{2a^2 \plus{} 1} \plus{} \frac {12b \plus{} 7}{2b^2 \plus{} 1} \plus{} \frac {12c \plus{} 7}{2c^2 \plus{} 1} \le 19$
$ \sum\limits_{cyc} {9 \minus{} \frac {{12a \plus{} 7}}{{2a^2 \plus{} 1}}} \ge 8$
$ \sum\limits_{cyc} {\frac {{2(3a \minus{} 1)^2 }}{{2a^2 \plus{} 1}}} \ge 8$
$ \sum\limits_{cyc} {\frac {{(3a \minus{} 1)^2 }}{{2a^2 \plus{} 1}}} \ge 4$
By Holder
$ LHS \ge \frac {{(3a \plus{} 3b \plus{} 3c \minus{} 3)^2 }}{{2a^2 \plus{} 2b^2 \plus{} 2c^2 \plus{} 3}}$
So we need
$ a^2 \plus{} b^2 \plus{} c^2 \minus{} 18(a \plus{} b \plus{} c) \plus{} 18(ab \plus{} bc \plus{} ac) \minus{} 3 \ge 0$ which can be shown by Mixing-variable :lol:[/quote]
Very nice your solution.  :lol:


Prove that if $ p$ is a prime number,then $ 7p \plus{}3^{p} \minus{}4$  is not a perfect square.(JBMO 2007)

When posting a problem from a national, regional or international competition, please check whether that problem has already been posted [see the Resources/Contests section!]

http://www.mathlinks.ro/viewtopic.php?p=874754

Hi,Thank you hsiljak,I've searched but I couldn't find it .I think the name of the topic should presents information about the problem inside.
How could I find the problem from JBMO ,when the name of the topic you've posted was "prime number implies non-square number"

As a matter of fact, the name of that topic I linked to could be used as a typical example of [b]good titling[/b].

[quote="pohoatza"]prime number implies non-square number
JBMO 2007, Bulgaria, problem 4
[/quote]

The title of the topic tells about the type of the problem inside, and the subtitle informs us about the source. You could have simply used the '[url=http://www.mathlinks.ro/search.php]Search[/url]' button to find it (keyword could have been [i]jbmo 2007[/i]), or even better:

[url=http://www.mathlinks.ro/resources.php]Contests/Resources[/url] -> [url=http://www.mathlinks.ro/resources.php?c=1&cid=21]Junior Balkan MO[/url] -> [url=http://www.mathlinks.ro/resources.php?c=1&cid=21&year=2007]2007[/url]->[url=http://www.mathlinks.ro/viewtopic.php?p=874754]Problem 4[/url]

Good day,
sorry for asking this question here, bt i could not find an appropriate place.
I need the contact information for 13th junior balkan mathematics olympiad. I mean, the contact information of the organisators of the competition.
i will be very happy if you can help me.
Thank you.

Bosnia and Herzegovina is the host, maybe hsiljak knows more.
I have also one question: will there be a web site for JBMO 09? Last year there wasn't one...

Samanyolu was PMed with something that looks like contact info (people who know how organized Bosnian Mathematical Society is, know why it's hard to give any kind of information about them  :lol:  ). And as far as I can expect, there could and should be some sort of website- i don't expect it to be something extremely special, but it should list the problems and results.


geometry
modular arithmetic
geometric transformation
reflection
homothety
circumcircle
geometry unsolved

[hide="Hint"]The currency point is also the [b]Feurebach point[/b] wrt $ \triangle ABC$ due to the fact that $ \triangle K_aK_bK_c\sim \triangle M_aM_bM_c$, where $ M_a,M_b,M_c$ respectively are the midpoints of $ BC,CA,AB$. I will be back with more detailed solution.[/hide]

Back with a full solution.

Let $ A_0,B_0,C_0$ be the tangency points of $ (I)$ with the triangle sides $ BC,CA,AB$ respectively, where $ (I)$ is the incircle of $ \triangle ABC$. Denote $ M_a,M_b,M_c$ respectively by the midpoints of $ BC,CA,AB$. We have $ (IB_0,IK_a)\equiv (IB_0,IA_0) \plus{} (IA_0,IK_a)$ $ \equiv (CB_0,CA_0) \plus{} (DA_0,DK_a)$ $ \equiv (CB_0,CA_0) \plus{} 2(DA_0,DA)$ $ \equiv (CA,CB) \minus{} 2(DA,AC) \minus{} 2(CA,CD)$ $ \equiv (CA,CB) \minus{} (AB,AC) \minus{} 2(CA,CB)\equiv (BC,BA) \pmod \pi$. Hence $ (IB_0,IK_a)\equiv (BC,BA)$. With the same argument, we also have $ (IB_0,IK_c)\equiv (BA,BC)\Longrightarrow (IB_0,IK_c)\equiv \minus{} (IB_0,IK_a) \pmod \pi$, which implies that $ K_c\mapsto K_a$ through the reflection by $ IB_0$. Therefore $ K_aK_c\bot IB_0\Longrightarrow K_aK_c\parallel{}M_aM_c$. Argue the same we also have $ M_aM_b\parallel{}K_aK_b$, $ M_bM_c\parallel{}K_bK_c$. Hence $ \triangle M_aM_bM_c \sim \triangle K_aK_bK_c$. Moreover, $ K_aM_a,K_bM_b,K_cM_c$ are concurrent at the point $ F$ such that $ \triangle M_aM_bM_c\mapsto \triangle K_aK_bK_c$ through a homothety with center $ F$, let us denote this homothety by $ \mathcal {H}_{F}$. Hence $ \mathcal {H}_{F}: (M_aM_bM_c)\mapsto (K_aK_bK_c)$, or in other word, $ (\mathcal {E})\mapsto (I)$, where $ (\mathcal {E})$ is the $ 9$- point circle wrt $ \triangle ABC$. But we have already know that $ (I)$ externally tangents to $ (\mathcal {E})$, which implies that $ F\equiv (I)\cap (\mathcal {E})$. Therefore,  $ K_cM_c,K_aM_a,K_bM_b$ are concurrent at a point $ F\in (I)$. This $ F$ is also known as the [b]Feurebach point[/b] wrt $ \triangle ABC$.
Our proof is completed then $ \square$

[quote="mathVNpro"] $ \triangle K_aK_bK_c\cong \triangle M_aM_bM_c$, where $ M_a,M_b,M_c$ respectively are the midpoints of $ BC,CA,AB$. [/quote] I believe that $ \triangle K_aK_bK_c, \triangle M_aM_bM_c$ are centrally similar, not congruent. :|

Denote KcF∩BC=L , KbE∩BC=K
L is the reflection of A with respect to CF   K is the reflection of A with respect to BE
Thus JL=JK  =>  KcKb//LK  => KcKb//BC
Analogously, KcKa//CA  KaKb//AB
⊿KcKaKb∽⊿CAB
KcC,KaA,KbB,OI are concurrent at their inner similitude center

Back to our problem:
It is obvious,that ⊿MaMbMc∽⊿KcKaKb∽⊿CAB  with corresponding sides parallel
Thus MaKa MbKb McKc are concurrent at Fe
Therefore, IO pass through Fe
Notice the circumcenter of ⊿MaMbMb is the nine-point circle of ⊿ABC whose circumcenter is O’  =>  Fe is the Feurebach point

[quote]Let $ D. E, F$ be the feet of the angle bisectors of angles $ A, B, C$, respectively, of triangle $ ABC$, and let $ K_a, K_b, K_c$ be the points of contact of the tangents to the incircle of $ ABC$ through $ D, E, F$ (that is, the tangent lines not containing sides of the triangle).
Prove that the lines joining $ K_a, K_b, Kc_$ to the midpoints of $ BC, CA. AB$ respectively, pass through a single point on the incircle of $ ABC$.[/quote]
Here goes my solution.
It is enough to prove that $ \triangle K_aK_bK_c$ and $ \triangle PQR$ are centrally similar. ($ PQR$ is the medial triangle of $ ABC$). Because the center of homothety $ T$ of incircle and nine point circle is the Feuerbach point of $ \triangle ABC$. So, If $ \triangle K_aK_bK_c$ and $ \triangle PQR$ are centrally similar, $ T$ will be their center of homothety and the problem will follow.

$ X,Y,Z$ be the points where the incircle touch the sides of the triangle $ ABC$.
Now we have
\begin{align*} \angle K_cK_bK_a & = & \angle K_cK_bX + \angle XK_bY + \angle YK_bK_a \\
& = & \left (\frac \pi2 - \frac C2 - A\right ) + \left ( \frac \pi2 - \frac B2\right ) + \left ( \frac \pi2 - \frac A2 - C \right ) & = & B \end{align*}
Analogously chasing other angles we can say that $ \triangle K_aK_bK_c \sim \triangle PQR$.

Now we shall prove that, $ K_cK_a \parallel RP$
We have $ \measuredangle(IY,K_cK_a) = \left (\pi - A - 2C\right ) + \left (\frac \pi2 - B\right ) = \frac \pi2 - B$ also $ \measuredangle (IY,RP) = \frac \pi2 - B$.
So, $ K_cK_a \parallel RP$ and the proof is complete. :)

Dear Mathlinkers,
see also for example :
http://perso.orange.fr/jl.ayme  vol. 4    Symétriques de (OI) par rapport aux côtés des triangles de contact et médian p. 10
Sincerely
Jean-Louis

[b]REMARK-[/b] As [b]plane geometry[/b] have said that $ CK_c,AK_a,BK_b$ are concurrent is a special case of this problem:

[b][color=darkblue]"Let $ \triangle ABC$ with $ (I)$ is its incircle. $ A_0,B_0,C_0$ are the tangency point of $ (I)$ with $ BC,CA,AB$ respsectively. Let $ A_1,B_1,C_1$ be the points on $ BC,CA,AB$ such that $ AA_1,BB_1,CC_1$ are concurrent. From $ A_1,B_1,C_1$, let $ A_2,B_2,C_2$ be the points on $ (I)$ such that $ AA_2,BB_2,CC_2$ are the second tangents from $ A_1,B_1,C_1$ to $ (I)$. Prove that: $ AA_2,BB_2,CC_2$ are concurrent."[/color][/b].

Let $ X,Y,Z$ be the tangency points of $(I)$ with $ BC,CA,AB.$ It is clear that $ YK_b \parallel ZX$ and $ ZK_c \parallel YX.$  Therefore, $ \angle K_cZX \equal{} \angle ZXY \equal{} \angle K_bYX$ $ \Longrightarrow$ $ K_cK_b$ is parallel to the tangent $ BC$ to $ (I)$ at $ X.$ By similar reasoning, this means that $ \triangle K_aK_bK_c$ and $ \triangle ABC$ are homothetic $\Longrightarrow$ $ \triangle K_aK_bK_c$ and the medial triangle $ \triangle M_aM_bM_c$ are homothetic. Therefore, $ M_aK_a,$ $M_bK_b,$ $M_cK_c$ concur at their homothetic center, which is the exsimilicenter of their circumcircles $ (I)$ and the nine-point circle $(N),$ i.e. the Feuerbach point of $\triangle ABC.$

[hide="Another approach"]Let $ X (0: p \minus{} c: p \minus{} b).$ Parallel through $ X$ to $ YZ$ meets the incircle $ (I)$ at

$ K_a  (( \minus{} a \plus{} b \plus{} c)(b \minus{} c)^2: (a \minus{} b \plus{} c)b^2: (a \plus{} b \minus{} c)c^2)$

Then the line joining $ M_a(0: 1: 1)$ and $ K_a$ has equation

$ (b^2 \plus{} c^2 \minus{} a(b \plus{} c))x \plus{} (b \minus{} c)( \minus{} a \plus{} b \minus{} c)y \plus{} (a \minus{} b \minus{} c)(b \minus{} c)z \equal{} 0$

Which clearly contains the Feuerbach point

$ F_e (( \minus{} a \plus{} b \plus{} c)(b \minus{} c)^2: (a \plus{} c \minus{} b)(a \minus{} c)^2: (a \plus{} b \minus{} c)(a \minus{} b)^2)$

By similar reasoning, we conclude that $ M_aK_a,M_bK_b,M_cK_c$ meet at $ F_e.$[/hide]

Let (O) be the circumcircle of triangle ABC. Ay is the tangency of (O). Mx is the tangency of the nine-point circle (E) of triangle ABC. G is the centroid of triangle ABC.
By angle charsing, it's easy to prove that $ DK_a//Ay$
The dilatation $ D^{\frac { \minus{} 1}{2}}_G$: $ A\rightarrow M, B\rightarrow N, C\rightarrow P \Rightarrow \Delta ABC\rightarrow \Delta MNP$
Then $ Ay//Mx$
So $ DK_a//Mx$
Similarly for $ EK_b, FK_c$ and we get $ K_aK_b//MN, K_bK_c//NP, K_aK_c//MP$
And it's easy to prove that $ DK_a, EK_b, FK_c$ are concurrent at J.
The Dilatation $ D^{\frac{r}{r_E}}_J$: $ M\rightarrow K_a, N\rightarrow K_b, P\rightarrow K_c$ then J lies on IE (J is the tangent of (I) and (E))
$ \Rightarrow$ QED


Are we allowed to send in applications on May 1 or must it be before May 1?
(online application)

I think that you can send it in on May 1st as well, but I'm not sure.

Quote from an email from Administrative Coordinator John Woodman:

[quote]You can proof it on May 1, but try to get it in asap.[/quote]

So they'll take it on May 1st. Hope this helps!

Ah.  This is what I get for not reading the forum much during the end of application season.

Yes, May 1 is fine. :)


calculus
integration
function
logarithms
calculus computations

I need some help with this type of exercise. There are several of them, and I can't really solve a single one. Here's one that looks simple:

For what values of $a$ does the generalized integral $\int_0^{\infty}\frac{\ln(1+x^2)}{x^a}dx$ converge?

You need to find a good function $g$ to compare with, but no matter what I choose I get a limit which tends to 0, and isn't all that useful. How should I think?

Other similar ones are: $\int^{\infty}_0\frac{\arctan x}{(1+x)^a(1+x^2)x^a}$, $\int_0^{\infty}\frac{\ln(1+x)}{x^a + x^{a+2}}$. I'm looking for the general idea of what to do. I am lost.

Look at your first example. That integral is improper for two reasons - it's potentially improper at zero, and it's improper at infinity. Break that into two pieces and consider each piece separately.

As $x\to0,\,\ln(1+x^2)$ looks like $x^2.$ Hence at that end, we're comparing to

$\int_0^bx^{2-a}\,dx.$ This converges when $2-a>-1,$ or $a<3.$

As $x\to\infty,\,\ln(1+x^2)$ looks like $2\ln x.$ Hence at that end, we're comparing to

$\int_c^{\infty}x^{-a}\ln x\,dx.$

The logarithm is of very little significance to that; it converges when $a>1.$ Thus the whole integral converges when $1<a<3.$

Some hints for your other integrals:

$\arctan x$ looks like $x$ as $x\to0$ and like a constant as $x\to\infty.$

$\ln(1+x)$ looks like $x$ as $x\to0$ and like $\ln x$ as $x\to\infty.$

$(1+x)^a$ looks like $1$ as $x\to0$ and like $x^a$ as $x\to\infty.$

$x^a+x^{a+2}$ looks like $x^a$ as $x\to0$ and like $x^{a+2}$ as $x\to\infty.$

Thank you!

I like formalizing the arguments a bit more, but it's not very difficult to do once you've identified what to do. They are all pretty easy to solve using your train of thought, and then formalizing it only takes a bit more effort. Thanks again.

edit: are they not called "generalized integrals" in english, but rather "improper"?

Yes, they're called "improper" in English.


geometry
geometric transformation
analytic geometry
graph theory
combinatorics solved
combinatorics

Hello

my question concerns planar grpahs (graphs that can be drawn in 2D withouth the edges crossing each other)
if F is the number of faces (not counting the exterior along), prove that F is really the minimal number of cycles in the graph you need to at least cross each edge once (i think that it suffices is obvious)
can anyone help? (i need it for kirchoff...)

I'm not sure I understood the question, but look for a planar graph with 4 faces (not counting the exterior one) which are triangles just like a big triangle divided into 4 smaller triangles with the medians. With the 3 smaller triangles containing the vertices of the big one, you cover each of the edges...

Pierre.

thanks pierre,

u made me realise that I made a bad translation of my physics problem to graph theory

of course, in an electrical system, i can't allow vertices with only two edges as they are only connection points and aren't of any importance

so i rephrase my problem :
show that in a graph with F faces (not counting the exterior)  and with all vertices having at least three edges, F is really the number of minimal cycles you need

now your example isn't a counterexample anymore

Ok, so I will try to give you another one.

Consider the planar graph whose vertices are the 9 points O(0,0), A(2,0), B(4,0), C(4,2), D(2,2), E(0,2), F(1,1), G(2, 1) and H(3,1) (in an orthogonal system of coordinates), and with 16 edges :
OA, AB, BC, CD, DE, EO, OF, FD, FE, AG, GD, GH, HA, HD, HC, HB.
This graph has 8 faces (without the exterior one). Each vertex has degree at least 3.

But you may cover all the edges with only 7 cycles (note that none of them contains a subcycle) :
OFEO
EFDE
OAGDFO
GDHG
DCHD
CBHC
ABHA

Pierre.

:?
i see

that seems to be a good counterexample

but where does that leave me with my kirchoff trouble?
suppose this is an electric circuit, and i have a battery somewhere and resistances on every edge, and i want to calculate the current in every edge

then my book says : first express that the sum of currents in all vertices except one have to be zero (the last one is not independent of the other conditions)
then express that the difference in electric potential in every closed circuit is zero, and do it for as many cycles you need to have each edge at least once

i know this is no place for physics but it's pure about algebra and equations

so what does this mean, or am i missing something here?


let $ a,b,c$ is nonegative.not two of which by zero.find best $ k \ge 0$ such that:
$ \frac{a\plus{}kb}{c\plus{}kb}\plus{}\frac{b\plus{}kc}{a\plus{}kc}\plus{}\frac{c\plus{}ka}{c\plus{}ka} \ge 3$
 i found $ k\sim 2,876121406$

[quote]$ \frac {c \plus{} ka}{c \plus{} ka}$[/quote]

is it ok?

[quote="tuantam1lan"]let $ a,b,c$ is nonegative.not two of which by zero.find best $ k \ge 0$ such that:
$ \frac {a \plus{} kb}{c \plus{} kb} \plus{} \frac {b \plus{} kc}{a \plus{} kc} \plus{} \frac {c \plus{} ka}{c \plus{} ka} \ge 3$
 i found $ k\sim 2,876121406$[/quote]
it is :
let $ a,b,c$ is nonegative.not two of which by zero.find best $ k \ge 0$ such that:
$ \frac {a \plus{} kb}{c \plus{} kb} \plus{} \frac {b \plus{} kc}{a \plus{} kc} \plus{} \frac {c \plus{} ka}{b \plus{} ka} \ge 3$

this is one problem nice and hard ! you can post your solution  :wink:

[quote="Guest"][quote="tuantam1lan"]let $ a,b,c$ is nonegative.not two of which by zero.find best $ k \ge 0$ such that:
$ \frac {a \plus{} kb}{c \plus{} kb} \plus{} \frac {b \plus{} kc}{a \plus{} kc} \plus{} \frac {c \plus{} ka}{c \plus{} ka} \ge 3$
 i found $ k\sim 2,876121406$[/quote]
Exactly speaking, the best $ k$ is the unique real root of the following equation:

$ k^7 \minus{} 4k^6 \plus{} 8k^5 \minus{} 16k^4 \plus{} 16k^3 \minus{} 28k^2 \plus{} 4k \minus{} 4 \equal{} 0.$[/quote]

How did you get it? :rotfl:

I use a computer program called BOTTEMA which is designed for inequality-proving.

[quote="Guest"][quote="tuantam1lan"]let $ a,b,c$ is nonegative.not two of which by zero.find best $ k \ge 0$ such that:
$ \frac {a \plus{} kb}{c \plus{} kb} \plus{} \frac {b \plus{} kc}{a \plus{} kc} \plus{} \frac {c \plus{} ka}{c \plus{} ka} \ge 3$
 i found $ k\sim 2,876121406$[/quote]
Exactly speaking, the best $ k$ is the unique real root of the following equation:

$ k^7 \minus{} 4k^6 \plus{} 8k^5 \minus{} 16k^4 \plus{} 16k^3 \minus{} 28k^2 \plus{} 4k \minus{} 4 \equal{} 0.$[/quote]

I think you are wrong, counter-example: a=c>b. :P

[quote="iqtimo"]it is :
let $ a,b,c$ is nonegative.not two of which by zero.find best $ k \ge 0$ such that:
$ \frac {a \plus{} kb}{c \plus{} kb} \plus{} \frac {b \plus{} kc}{a \plus{} kc} \plus{} \frac {c \plus{} ka}{b \plus{} ka} \ge 3$

this is one problem nice and hard ! you can post your solution  :wink:[/quote]

It is not hard,you can try. :P

[quote="shfdfzhjj"][quote="Guest"][quote="tuantam1lan"]let $ a,b,c$ is nonegative.not two of which by zero.find best $ k \ge 0$ such that:
$ \frac {a \plus{} kb}{c \plus{} kb} \plus{} \frac {b \plus{} kc}{a \plus{} kc} \plus{} \frac {c \plus{} ka}{b \plus{} ka} \ge 3$
 i found $ k\sim 2,876121406$[/quote]
Exactly speaking, the best $ k$ is the unique real root of the following equation:

$ k^7 \minus{} 4k^6 \plus{} 8k^5 \minus{} 16k^4 \plus{} 16k^3 \minus{} 28k^2 \plus{} 4k \minus{} 4 \equal{} 0.$[/quote]

I think you are wrong, counter-example: a=c>b. :P[/quote]
What do you mean? What does your so-called "counter-example" mean?

[quote="Guest"][/quote][quote="shfdfzhjj"][quote="Guest"][quote="tuantam1lan"]let $ a,b,c$ is nonegative.not two of which by zero.find best $ k \ge 0$ such that:
$ \frac {a + kb}{c + kb} + \frac {b + kc}{a + kc} + \frac {c + ka}{b + ka} \ge 3$
 i found $ k\sim 2,876121406$[/quote]
Exactly speaking, the best $ k$ is the unique real root of the following equation:

$ k^7 - 4k^6 + 8k^5 - 16k^4 + 16k^3 - 28k^2 + 4k - 4 = 0.$[/quote]

I think you are wrong, counter-example: a=c>b. :P[/quote][quote="Guest"]
What do you mean? What does your so-called "counter-example" mean?[/quote]

But you have edited your post,professor Guest.  :roll:


Can anyone help me with the below problems?

1. Show that for any list of ten distinct positive integers, there is a consecutive sequence of some of them such that the sum is divisible by 10.

2. Find the number of positive integers p=<80 that are divisible by at least one of the numbers 2,3,5 and 7.

[hide="1."]

Let the integers be $a_1, a_2, \ldots, a_{10}$. Consider the sums

$S_1 = a_1$,
$S_2 = a_1+a_2$
...
$S_{10} = a_1+a_2+\cdots+a_{10}$.

If any of them is divisible by $10$, we are done. So suppose none of them are. Then they are all $1,2,\ldots,9 \pmod{10}$. But since there are ten of them by Pigeonhole we have some two congruent modulo $10$. Let these be $S_i$ and $S_j$ with $i < j$.

We have $S_j-S_i$ divisible by $10$ but that is just $a_{i+1}+a_{i+2}+\cdots+a_j$, which is a consecutive sequence, so we are done.[/hide]

And, just a little remark, one can easily generalize the first one (i.e. : n integers instead of 10), and it is solved exactly as [b]paladin8[/b] did for the case n=10. :)


Engineers place a sidewalk along the perimeter of a circular park so that the radius of the park is increased by $ 20\%$. By what percent is the park's area increased?

The area increases with the square of the radius, so the park increases by $ (\frac {6}{5})^2 \equal{} \frac {36}{25} \equal{} \fbox{44}\%$. We use $ \frac{6}{5}$ since that fraction represents an increase of $ 20\%$.


[quote="MithsApprentice"]I'm not sure if this is a bug or just something that went wrong on that post. Anyways, take a look at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=205272#p205272[/quote]

Everything I get is a window telling me

Sorry, but only users granted special access can read topics in this forum.

What forum is that on? I mean, many of the national communities fora are closed to the general public (At least Georgia's Forum is :D ).

Quite so. This is our state forum. I was aiming the link at the admin (mostly Valentin since he apparently does most of the tech support around here...). :D

It's the old emoticons in the $\LaTeX$ code bug :) I've fixed Mathew's post.


Hey all,

I have a sorting algorithm that works only on integers, and I need to expand it to give the same order for double precision numbers. To do this, I need to map my doubles, which are on (-inf,inf) to a range where the "exponent bits" of the double are 0. Which means the range from (I think) [2^-52,1]. However, while the doubles could be anywhere on the real number line, most of them will be clustered very near 0, at r < 2^-52. So I need a non-linear map that will take miniscule numbers to larger ones, and large numbers to small ones, while still preserving their order.

If it makes things any easier, you can consider only the domain +/- (10^-19,10^4), which is where 99.99% of the numbers will be.

Once I have that I can easily map them to long integers with a simple multiplication and apply the sorting algorithm, but I am having trouble working out an appropriate map. Any suggestions are appreciated.

Once caveat is that this map can't be computationally intensive. For example, a shift of arctan(x) or some variation of log(x) would achieve what I want, but it takes too long to compute to make this sorting algorithm worthwhile.

I have no idea if there is such a map in existence, but if anyone has ideas it would be greatly appreciated.

By the way, if this is in the wrong place feel free to move it.

Why not do something like take an 8th or 16th root?  Should be relatively computationally gentle while still having the same sort of shape as arctan near 0, no?  (I guess really I mean $ \text{sign}(x) \cdot \sqrt[8]{|x|}$ to account for negative numbers.)

It's a possibility, but roots are still pretty intensive computationally when you do a lot of them. So not really as fast as I would like if there are other alternatives. Roots do have the nice property of mapping numbers < 1 to larger numbers < 1, though. Hmm, wonder if I can find an algorithm that approximates roots quickly while preserving order.

exactly what is this "sorting function"?
you should try to make it applicable to doubles, because if you shift the numbers to longs, numbers that are too close together will become equal, which would result in loss of precision

The sorting function is z-order sorting

[url]http://en.wikipedia.org/wiki/Z-order_(curve)[/url]

And yeah, that'w why I was hoping to map my numbers to somewhere between 2^-52 and 1, which I would then multiply by a constant factor before shifting to long integers in order to keep enough precision.

Sorry, the URL didn't work for some reason

try [url]http://en.wikipedia.org/wiki/Z-order_(curve)[/url]

Arg - just copy paste into the browser and be sure to include the closing bracket, I can't wrestle with this forum software ^_^

Also, could someone explain to me why simple bit interleaving works for integers but fails for doubles? Since the exponent is stored before the mantissa in a double (ignoring the sign bit for the moment) should you not be able to treat the exponent bits and mantissa bits as an 11 bit and 53 bit integer and order them with simple bit interleaving? This seems logical to me, but it doesn't work.

Another question:

If I used Newton's method with only one or two iterations in order to approxiate roots, will that preserve the order? I suspect the answer would be dependent on the initial guess somehow though.

this is probably because the exponent commonly does not have a "sign" bit.
-1 is commonly (at least on my comp) stored as 111...11
-2 would be stored as 111...10
-3 : 111...101
etc.
while positive numbers would be stored as
000...01, 000...10 etc.
however, any implementation that utilizes this would be unportable

[quote="KBriggs"]I can't wrestle with this forum software[/quote]  It was a bug that's now been fixed -- you can also always go back and edit your posts (for up to a day) using the "edit" button in the lower-right corner of your post.

Thanks JBL

@lyn:

exponents in double precision numbers are actually stored as unsigned 11 bit integers and a 'bias' of -1023(or maybe -1024?) is applied to them to give negative exponents. So I am not quite sure how it having a sign bit or not would change anything. Perhaps you could clarify what you meant?

yea that's basically what i meant (in terms of bits).
when you take -1, it underflows and is equivalent to -1 + 1024
but if you compare the bits, 1 would be less than -1..., which i don't think you want

To my understanding, the exponent is stored as an unisgned 11 bit integer, and only biased when the number is interpreted as a double. Since the number is never interpreted during the bit-interleaving, the bias is irrelevent and the comparison will be between positive numbers between 1 and 2046... so negatives should never come up at all.

Am I misunderstanding how the bias is applied?