Putnam
geometry
geometric transformation
reflection
AMC
USA(J)MO
USAMO

I know the Putnam is a very stringently graded test.  I got 20 points last year and that was more than I was expecting.  This year, I had at least 2 perfect solutions (A2 and A4), and some more solutions some of which I'm not sure count as holey or not (A1, A3, B1, B2, B3).  What score did I get?  1.  My coach says this has happened before and thinks that the graders might get a little sloppy when they reach the midwest and start handing out 0's with understandable carelessness.  What do you people think?  Should we start photocopying our team's solutions before we submit them?  Do you think the graders frequently find minor quibbles, such as begging the question by using the axiom of choice?

Yes, I think they do often take off many point for minor quibbles, so students often overestimate their scores by a large amount. As far as being stricter on the midwest schools, I find that unlikely. I don't imagine they grade it by region. (Why would they do that?) As for photocopying solutions, what would you accomplish by doing that? There isn't an appeals procedure, so there's nothing you can do once you get your score back. And you won't even know where you lost your points.

[quote="ComplexZeta"]I don't imagine they grade it by region. (Why would they do that?) As for photocopying solutions, what would you accomplish by doing that?[/quote]

They wouldn't do it on purpose.  It would be since no one actually shuffles the tests (coach's idea).  Though, that [i]would[/i] implicate that the packages arrive in order...

[quote="ComplexZeta"]As for photocopying solutions, what would you accomplish by doing that? There isn't an appeals procedure, so there's nothing you can do once you get your score back. And you won't even know where you lost your points.[/quote]

We would appeal to the director (Klazinski?) and when we find out the reason for taking off points, I guess we could argue with him, though I doubt they would change anything even if they graded it differently.

Okay, here are my solutions.  Maybe you could tell me where I went wrong.  I think Kedlaya mentions my same solution for A2 in the "official" ones.  It is just a little downing to discover the clever trick and get no credit.

[b]B2:[/b] Assume $B=\{b_{i}|\rm{Bob\ wins\ the\ heap\ with\ b_{i}\ stones}\}$ has finite cardinality $k$.  Now, we will show that Bob wins the heap where $n=(b_{1}\times b_{2}\times ... \times b_{k})-1$.  Assume Alice could turn it over to a cold position (p is prime).

$(b_{1}\times b_{2}\times ... \times b_{k})-1-(p-1)=b_{i}$

$\Rightarrow b_{i}|((b_{1}\times b_{2}\times ... \times b_{k})-p)\Rightarrow b_{i}|p\Rightarrow b=1,p$

But, $1$ is not in $B$, so $b=p$.  This means

$b_{1}\times b_{2}\times ... \times b_{k}=2p\Rightarrow b_{m}=2,p$ for all m, but $|B|\ge3$ since $\{3,8,11\}\subseteq B$.  Contradiction.

[b]B4:[/b] Simple inclusion-exclusion.  The steps generally have 3 terms.

$n$ is a maximum $n!$ times.
$n-1$ is $n!-2\times 1(n-1)!$
$n-2$ is $n!-2\times 2(n-1)!+1\times 2(n-2)!$
....
$2$ is $n!-2\times (n-2)(n-1)!+(n-3)(n-2)(n-2)!$
$1$ is $n!-2\times (n-1)(n-1)!+(n-2)(n-1)(n-2)!=0$

Sum and divide by $n!$ for the average.

$n!(avg)= n\times n!-2\left(\frac{(n-1)n}{2}\right)+(1\times 2+2\times 3+...+(n-2)(n-1))(n-2)!$

omitting a few calculations,

$=\frac{n+1}{3}$

It's always a little depressing to think that you wrote perfect solutions for $n$ problems and then get less than $10n$ points. But it happens to a lot of people (including myself); you're nowhere near alone.

(Those are actually A2 and A4)

On A2: Did you include an argument that every winning position for Alice is one move away from a winning position for Bob? If you didn't, that's not a proof.

On A4: That notation requires some explanation; the problem refers to maxima at $k$, while you're counting by the value of each maximum. Was that explanation in your solution? Also, the counts are not immediately obvious; they require some argument.

[quote="jmerry"]On A2: Did you include an argument that every winning position for Alice is one move away from a winning position for Bob? If you didn't, that's not a proof. [/quote]

I might have thought that this argument was included in calling a position "hot" or "cold" for an impartial game.

[quote="jmerry"]On A4: That notation requires some explanation; the problem refers to maxima at k, while you're counting by the value of each maximum. Was that explanation in your solution? Also, the counts are not immediately obvious; they require some argument.[/quote]

This is largely why it would be advantageous to photocopy my solutions before submission.  I said I was counting the total number of times a maximum occurs.  Then, the enumeration is clear.  I also remember explaining the counting more than I did here.

Why do they make it a one day exam if they expect some kind of perfect solutions, and easily filled in omissions are penalized so heavily? It doesn't seem like any book offers the kind of detailed explanations of proofs as the Putnam graders require. It's not a mathematical journal, it's a math competition. The way Putnam is graded doesn't seem to reflect the way math is done.

Of course not -- it's also not intended to.  If they  were trying to test you on the way math is done, they would give you a set of questions, 5 years of graduate school, 2 years of post-doc, at least 6 years of tenure track, and whoever publishes the solutions first, wins.

More seriously, your memory of what you wrote is probably just wrong, and you probably just made mistakes.  (Another possibility, of course, is that you mis-attached your identification stickers, or they lost some of your solutions, etc.)

Can your coach ask the Putnam director to check whether your test papers were looked at?

[quote="juggle7"]Would not stating that A2 was an impartial game dock 9 points?  9 points because the "counts are not immediately obvious"?  I remember fairly well what I wrote, so I expect that they're just being picky.[/quote]
As far as I can tell, I lost 8-10 points on A1 by losing a factor of 2 somewhere. They will crush your soul even for little mistakes.

Hmmm... I've been writing 1-2 page solutions for questions that really only need a few lines to explain to the average math person, and have been getting most of the points I want. But it leaves me very little time to do more than 3 solutions per half, which means it's going to be really hard to improve unless I take some chances. Does anyone here regularly write short proofs and yet still get most of the points?

Yea... last year, I wrote full page solutions, one per half, and got a 20.  But, I can solve more problems!  That's depressing.  My writing skills have [i]never[/i] been quick.  It takes 10 hours per page in English.  Math goes more quickly.  Same way with reading - I spent 3 or 4 years just reading, yet I read at the same speed I did in 3rd grade.

I've written solutions of various lengths, from half a page to two or three pages of that booklet.  (Of course, it depends on how large one writes & use of diagrams, etc.)  Writing more (assuming you write nothing wrong) means you are less likely to leave out important things and therefore less likely to lose points by omission.

sometimes is being concise, being elegant, which will give you more points, not?

[quote="ComplexZeta"]I frequently wrote solutions well under a page and got full credit or close.[/quote]

That presumes that you know what they will count off on.

Does anyone know how the strictness of Putnam grading compares to that of the USAMO?

Neither set of standards is public knowledge, but I've always felt that Putnam grading is much tougher- I overestimated my Putnam scores far more than my USAMO scores. The first year I took the Putnam, I thought I had 7 problems and one partial; that was 40 points.


Are the strings of letters supposed to be digits?

Its imposssible  :rotfl:  :rotfl:  :rotfl:

[quote="viterick"]Find a + b + c + d if  (abcde) x 4 =  (edcbe)
(a number with 5 digits)x4 =(a number with 5 digits)[/quote]

The smallest number abcde is 10000 and the largest is 24999.
Note: Since e multiplied by 4 is a number which also ends in e, [b]e=0 (0 x 4 =0) [/b]

Now going back to
(abcde) x 4 =  (edcbe), replace e with 0 gives:
(abcd0) x 4 =  (dcb0) ([b]I have taken out the first e since it's a zero.[/b])

[b]BUT NO FIVE DIGIT WHOLE NUMBER CAN BE MULTIPLIED BY 4 TO GIVE A FOUR DIGIT NUMBER, THEREFORE WE HAVE A CONTRADICTION.[/b]

[u][b]So unless abcde is a decimal, this is NOT POSSIBLE.[/b][/u] :)

It's probably a safe bet that the original poster meant $ 4\cdot (abcde)_{10} \equal{} (edcba)_{10}$.

[hide]We know that $ abcde$ must be between 10000 and 25000.  We also know that $ e$ must be at least 4, because $ edcba$ is at least 40000.

If $ e$ is 4, then $ a$ must be 1, which is a contradiction because any number ending in 4 that is multiplied by 4 ends in 6.
If $ e$ is 5, then $ a$ must be 1, which is a contradiction because any number ending in 5 that is multiplied by 4 ends in 0.
If $ e$ is 6, then $ a$ must be 1, which is a contradiction because any number ending in 6 that is multiplied by 4 ends in 4.
If $ e$ is 7, then $ a$ must be 1, which is a contradiction because any number ending in 7 that is multiplied by 4 ends in 8.
If $ e$ is 9, then $ a$ must be 2, which is a contradiction because any number ending in 9 that is multiplied by 4 ends in 6.

Therefore, the only possible value for $ e$ is 8, and the only possible values for $ abcde$ are integers between 20000 and 22500 that end in 8.[/hide]

Given that it's supposed to be $ 4(abcde)_{10}\equal{}(edcba)_{10}$,
[hide]we have
\[ 0\equal{}39999a\plus{}3990b\plus{}300c\minus{}960d\minus{}9996e\]
\[ \Rightarrow 0\equal{}13333a\plus{}1330b\plus{}100c\minus{}320d\minus{}3332e\]
We see $ a$ is even. But $ a$ can't be at least 4, that would make $ 4(abcde)_{10}$ way too large. So $ a\equal{}2$, and reducing mod 5 says $ e\equiv 3\pmod{5}$. As $ (ebcda)_{10}\geq 40000$, $ e$ can't be 3, so $ e\equal{}8$.

We now have (after dividing by 10) $ 32d\equal{}133b\plus{}10c\plus{}1$. Now $ b$ is odd, but $ b$ can't be at least 3 (that would make right hand side too large) so $ b\equal{}1$, giving $ 16d\equal{}5c\plus{}67$. This means $ d\equiv 2\pmod{5}$ and $ d>2$, so $ d\equal{}7$ and $ c\equal{}9$.

So the unique answer is $ \boxed{21978}$.[/hide]


inequalities
search
inequalities unsolved

Anh Cuong proved(topic: "Inspired by Nesbitt and solved by brute force" in proposed ineq forum) that for $a,b,c$ reals all different we have

$A= (\frac{a+b}{a-b})^2+(\frac{b+c}{b-c})^2+(\frac{c+a}{c-a})^2 >2$

My question is :

Does it exists a positive real $k$ such that $A>k>2$ ?

Well, we certainly cannot go higher than 3. Consider the case $a=1, c=b^2, b \rightarrow \infty$.

And indeed, 2 is maximal. Consider $a=-b, c=0$.

Dear manlio, that is not my inequality, we can decude it easily from Nesbit inequality. Just take [tex] a+b=z,b+c=x,a+c=y [/tex]. :)

Dear Anh Cuog, I really don't understand how to deduce it from

Nesbitt

$\sum \frac{a^k}{(b+c)^k} \geq 3(1/2)^k $  for $k\geq 2/3$

as after your substitution it becomes

$\sum \frac{z^2}{(x-y)^2}$

which is not Nesbitt.

Thank you very much.

[quote="manlio"]Anh Cuong proved(topic: "Inspired by Nesbitt and solved by brute force" in proposed ineq forum) that for $a,b,c$ reals all different we have

$A= (\frac{a+b}{a-b})^2+(\frac{b+c}{b-c})^2+(\frac{c+a}{c-a})^2 >2$[/quote]

Actually, this should be

${(\frac{a+b}{a-b})^2+(\frac{b+c}{b-c}})^2+(\frac{c+a}{c-a})^2 \geq 2$.

Only for positive a, b, c the $\geq$ sign can be always replaced by a > sign.

The proof is really cool:

${(\frac{a+b}{a-b})^2+(\frac{b+c}{b-c}})^2+(\frac{c+a}{c-a})^2 - 2$
$= (\frac{a^3+b^3+c^3+3abc-b^2c-c^2b-c^2a-a^2c-a^2b-b^2a}{(b-c)(c-a)(a-b)})^2$,

what is clearly $\geq 0$, and thus

${(\frac{a+b}{a-b})^2+(\frac{b+c}{b-c}})^2+(\frac{c+a}{c-a})^2 \geq 2$.

If additionally, the numbers a, b, c are positive, then

$a^3+b^3+c^3+3abc-b^2c-c^2b-c^2a-a^2c-a^2b-b^2a>0$

(after Schur; in fact, equality holds in Schur only if two of the variables are equal, what is prohibited for our numbers a, b, c), and thus

${(\frac{a+b}{a-b})^2+(\frac{b+c}{b-c}})^2+(\frac{c+a}{c-a})^2 - 2$
$= (\frac{a^3+b^3+c^3+3abc-b^2c-c^2b-c^2a-a^2c-a^2b-b^2a}{(b-c)(c-a)(a-b)})^2$

is strictly > 0, and we have

${(\frac{a+b}{a-b})^2+(\frac{b+c}{b-c}})^2+(\frac{c+a}{c-a})^2 > 2$.

[quote="manlio"]Nesbitt

$\sum \frac{a^k}{(b+c)^k} \geq 3(1/2)^k $  for $k\geq 2/3$[/quote]

This is very interesting! I knew Nesbitt only for $k\geq 1$. What is the proof for $k=\frac23$ ?

  Darij

It is at

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=nesbit&t=5520

This is just incredible! Dont you guys read before posted problems? This inequality was posted several times till now!

Actually I had no chance to find the thread since I searched for Nesbitt while the thread used Nesbit. Thanks Manlio!

  Darij

I think that like subject , we should write the same inequality instead of useless words like "hard,nice,easy,..." ;)

[quote="Myth"]@harazi. It is highly difficult to search for problems, especially because of titles like "Nice", "Easy", and so on ;)
Another thing that thre are to many topics. It is impossible to know all discussed problems.[/quote]

There are some people giving a reference for the problems but still making up nice thread titles as X1.001,..., X151.  :D

Dear manlio, I think you are right. Firstly,I mean the inequality:
[tex] \left|\frac{a}{b-c}\right|+\left|\frac{b}{a-c}\right|+\left|\frac{c}{a-b}\right| \geq 2 [/tex] but this is weaker than
[tex] \left(\frac{a}{b-c}\right)^2+\left(\frac{b}{a-c}\right)^2+ \left(\frac{c}{a-b}\right)^2\geq 2 [/tex] :D
Thank you very much for telling me about that. :)

Here is my approach :
Put $\frac{a+b}{a-b}=x$,so on...Then it's easy to check that $(x-1)(y-1)(z-1)=(x+1)(y+1)(z+1)$<=>$xy+yz+zx=-1$.
From it we see it's very easy:$x^2+y^2+z^2\geq -2(xy+yz+zx)=2$ as desired.
The equality doesn't occur in this case!


geometry
3D geometry
function
real analysis
real analysis unsolved

Prove that for each normal space with countable base,$ X$ , there exists function $ f: X\rightarrow I^{\infty}$ such that $ f: X\rightarrow f(X)\subseteq I^{\infty}$ is a homeomorphism.

Just to make sure our notation lines up: this says in other words that every normal second countable space can be embedded in the Hilbert cube?

Thanks.

[quote="HilbertThm90"] every normal [b]second[/b] countable space [/quote]

Basically yes, although I don't understand what you mean by second (bold word). I believe that other statement ( $ \Leftarrow$) holds as well but the proof includes some serious tools like transfinite nduction, and so on.

Some people use [b]first countable[/b] to mean the space has a countable local basis, and [b]second countable[/b] for countable basis.


logarithms
algorithm
function
calculus
calculus computations

Does this look right to you and is the intersection with the x-axis a rational number? I properly isn't right as i realised $e^{ln(x)^2}$ isn't the same as $x^2$. So could you plz show me how to get rid of those two $ln$s, thx.

$(ln(x))^2-3ln(x)=0$

$-x^3+x^2-1=0$

..... $\ln x ( \ln x -3)=0$ so the solutions are $x_1=1$ and $x_2=e^3$ and $e^3$ isn`t rational. Why would you want to get rid of them?

Of course, when will i ever learn. The reason i wanted to remove $ln(x)$ was that i thought it would make the task easier for me. Can you take a look at these two tasks for me.

$1)$

$ln(x)+x=0$

This time i believe it isn't possible to pull out $x$, is it? If so, i should be able to do the rest of the task once $x$ has been moved outside a pair of brackets.

$2)$

$f(x)=\sqrt{3^x+(ln(x))^2}$

$\frac{d}{dx}f(x)=(\sqrt{3^x(ln(x))^2}$

$\frac{d}{dx}f(x)=\frac{1}{2\sqrt{3^x+(ln(x))^2}}*\frac{d}{dx}(3^x+(ln(x))^2)$

$\frac{d}{dx}f(x)=\frac{1,0986*3^x+(\frac{1}{x})^2}{2\sqrt{3^x+(ln(x))^2}}$

$\frac{d}{dx}f(2)=1,6462$

I have a feeling this is wrong as it doesn't seem to be the right $\frac{d}{dx}f(2)$.

Do everything piece-by-piece when applying chain rule, you may lose track of your work.

$f(x)=\sqrt{3^x+(\ln x)^2}$

$\frac{df}{dx}=(\sqrt{3^x+(\ln x)^2})'$

$\frac{df}{dx}=\frac1{2\sqrt{3^x+(\ln x)^2}}\cdot(3^x+(\ln x)^2)'$

$\frac{df}{dx}=\frac1{2\sqrt{3^x+(\ln x)^2}}\cdot(3^x\ln 3+2(\ln x)(\ln x)')$

$\frac{df}{dx}=\frac1{2\sqrt{3^x+(\ln x)^2}}\cdot\left(3^x\ln 3+\frac{2(\ln x)}{x}\right)$

Easier said than done :D Thx for the help though. And i have another task i can't do :blush: I checked if i could make a common factor outside the brackets, didn't work here. I soon ran empty of ideas - maybe you could give me a hint. Hopefully i can take it from there.

$ln(x)+x=0$; $x>0$

There's exactly one root, which is between 0 and 1. You can't "solve the equation" algebraically, so instead look to find an approximate solution numerically.

Newton-Raphsons algoritm, i got it, thx... :)

Can you confirm if this is right? I get a result that could be wrong, not sure though. I have checked and rechecked it but can't find what is wrong. However, the problem is found on the 7 digit. So i wanted to know if the calculator had a minor calculation error or if i did it wrong.

$f(x)=ln(1+\sqrt{1-x^2})$

$f'(x)=\frac{1}{1+\sqrt{1-x^2}}*(1+\sqrt{1-x^2})'$

$f'(x)=\frac{1}{1+\sqrt{1-x^2}}*\frac{1}{2\sqrt{1-x^2}}*(1-x^2)'$

$f'(x)=\frac{1}{1+\sqrt{1-x^2}}*(-\frac{2x}{2\sqrt{1-x^2}})$

$f'(x)=\frac{1}{1+\sqrt{1-x^2}}*(-\frac{x}{\sqrt{1-x^2}})$

When i use the calculator to calculate $\frac{df}{dx}$ with the function that finds $\frac{df}{dx}$ automatically i get $-0.3094015$

When i calculate it myself $-0.3094010768$

How can this be? I have a TI-84 plus...


ratio
analytic geometry
geometry
rectangle
conics
ellipse

so there is a problem what is wrong with my code? Please Help.
Don't mind the strange variables.. The System.out..is to check if it works and it doesn't :(

import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.Rectangle;
import java.util.Scanner;
import javax.swing.JFrame;
import javax.swing.JPanel;

public class CoolArt {

	public static void main(String[] args) {
		JFrame frame = new JFrame();
		frame.setSize(500,500);
		frame.setTitle("Golden Ratio");
		frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
		Golden lion = new Golden();
		frame.add(lion);
		frame.setVisible(true);

	}

}

class Golden extends JPanel{

	private int x, y, w,l;
	private int counter;
	private boolean first=true;
	public Golden(){
		Scanner woah = new Scanner(System.in);
		System.out.println("Enter the X coordinate of the top right corner");
		x=woah.nextInt();
		System.out.println("Enter the Y coordinate of the top right corner");
		y=woah.nextInt();
		System.out.println("Enter the width");
		w=woah.nextInt();
		counter=-1;
	}
	public void paintComponent(Graphics g){
		Graphics2D g2=(Graphics2D) g;
		while(counter<=10){
		if(first=true){
			Rectangle body = new Rectangle(x, y,w,w);
			g2.draw(body);
			first=false;
			counter++;}
		else{
			System.out.println("Is this working");
				if(counter%3==0){
					System.out.println("Is this working");
					x=x+w;
					l=(int)Math.sqrt(Math.pow(w/2,2)+w*w);
					w=l-(w/2);
					Rectangle square = new Rectangle(x,y,w,w);
					g2.draw(square);
					counter++;
				}
				else if(counter%3==1){
					System.out.println("hmmmm");
					y=y+w;
					l=(int)Math.sqrt(Math.pow(w/2,2)+w*w);
					w=l-w/2;
					x=x+w;
					Rectangle square = new Rectangle(x, y,w,w);
					g2.draw(square);
					counter++;
				}
				else {
					System.out.println("eheheheh");
					l=(int)Math.sqrt(Math.pow(w/2,2)+w*w);
					w=l-w/2;
					Rectangle square = new Rectangle(x+w, y,w,w);
					g2.draw(square);
					counter++;
				}}
		}
	}
}

Im trying to draw a golden Rectangle btw. Please help

I would check it out but unfortunately I don't even have a Java compiler on my machine and I haven't coded in Java for a couple of years. (And your code looks ugly without any formatting :D.) But perhaps if you told us what exactly is happening and why you think it shouldn't happen then maybe I could help you out.

I am using eclipse ganymede. Nothing is drawn when the frame pops up. I suspect that it has something to do with the for loop in the paintComponent or my complier. Using a diffferent version of eclipse, a different program would run on that, but not on mines.

[quote="turtlecloud"]
	public void paintComponent(Graphics g){
		Graphics2D g2=(Graphics2D) g;
		while(counter<=10){
		if(first=true){
			Rectangle body = new Rectangle(x, y,w,w);
			g2.draw(body);
			first=false;
			counter++;}
[/quote]
I have neither compiler installed on my computer but I think it should be
[code]if(first==true) {[/code]

I don't see any for loop in your code.

(There's a while loop though ;).)

Well looks like puuhikki solved it. (first = true) always evaluates to true so the first if-branch is taken every time.

EDIT: oh yes and sorry for my laziness :oops:.


induction
number theory proposed
number theory

In the decimal expression of a $ 2009$-digit natural number there are only the digits $ 5$ and $ 8$. Prove that we can get a $ 2008$-digit number divisible by $ 11$ if we remove just one digit from the number.

We shall prove using induction that it is possible not only for $ 2009$, but for any odd natural number $ 2n\minus{}1$. We will also use the fact that a number is divisible by $ 11$ if the difference between the sum of the digits placed on the odd places and the sum of the digits placed on the even places is divisible by $ 11$.
If $ n\equal{}2$, then there are $ 8$ such $ 3$-digit numbers: $ 555$, $ 558$, $ 585$, $ 588$, $ 855$, $ 858$, $ 885$, $ 888$. It is obvious that we can remove one digit so that remaining number is divisible by $ 11$.
Now consider the statement is true wtih $ 2k\minus{}1$ $ (1)$. Now we will prove it is also true with $ 2(k\plus{}1)\minus{}1$.

[b]case 1.[/b] In this $ 2(k\plus{}1)\minus{}1$-digit number we can find two consequent digits which are equal. Now we remove them and get some $ 2k\minus{}1$-digit number - it is obvious that this number gives the same rest as the first one by $ (\bmod{11})$. According to $ (1)$, now we can take one digit away so the rest number will be divisible by $ 11$. Inserting these two consequent digits back to its place we obtain the number, which is divisible by $ 11$.

[b]case 2.[/b] If there are no such two digits, then the $ 2(k\plus{}1)\minus{}1$ number is either $ 5858\ldots 585$, or $ 8585\ldots 858$. Now we take away the $ k\plus{}1$ digit away and obtain the number which is divisible by $ 11$ (because it will be symmetric).

Now that we have $ 2009$ - an odd number, the statement is also true for it, $ q.e.d.$

It's easy to show that the residue of a number mod 11 is the alternating sum of its digits from right to left. This means that if we have two consecutive digits that are equal, we can completely ignore them and basically take them out without changing its value mod 11. So do that to whatever our 2009 digit number is until we can't do it any more. Then we must be left with a number that has an odd number of digits and is of the form
$ 858\ldots 858$ or $ 585\ldots 585$
Take the middle digit out of this number, whichever one it is. The fact its digits alternate means that the consecutive digits step can cancel everything out, so we're left with 0 and the new number is divisible by 11. This generalizes pretty readily.


1) A line four inches from lines AB and CD
2) A circle of radius 3 inches centered at point P

b) Two points.

[quote="Interval"]Lines AB and CD are parallel to each other and 8 inches apart.
Point P is located BETWEEN the two parallel lines and 2 inches from line AB.

a) Describe the locus of points that are:

1) equidistant from lines AB and CD

2) 3 inches from point P

b) How many points satisfy BOTH conditions in part a above?[/quote]
[hide]The points equidistant from both lines form a line between them.

The points 3 inches from $P$ form a circle with radius 3 centered at $P.$

There will always be $2$ points that satisfy both conditions, because the line will always go through the circle at two points.[/hide]


[i]p[/i] and [i]q[/i] are mutually prime and  $ \frac pq \equal{} 1\minus{}\frac12 \plus{} \frac13 \minus{} \frac14 \plus{}... \minus{} \frac 1{1318} \plus{} \frac 1{1319}$

Prove that [i]p[/i] can be evenly divided by [i]1979[/i].


Unknown author!
(My note: the AB vector is noted with "vAB")

   There are in a plane 4 fixed points A,B,C,D and an mobile point M.Which is the geometrical place of the point M if:
         vMA * vMB = vMC * vMD?

Somebody hepl me!
With an idea or somethin...

first of all what does vMA*vMB stand for? the scalar product? the vectorial product?

and secondly, are you sure that this is the whole statement of the problem? because if you take A, B, C, D collinear you get something, taking them the vertices of a square would give you something else, etc.

Oh , yes i forget to specify that vMA * vMB means the scalar product of vectors MA and MB.
That's the whole statament.I take the problem from o book with all mathematical olympiads from Romania in 2000, and i think that it's a right one...

well in that case it is very likely that they have forgotten to mention some details about the problem. I know the book, I have also studied it when I was younger, and I know that is has my flaws in it :?

Are you sure that you can't solve it without any details.Probably there are some cases, you know...
Well , thanks a lot for your messages, anyway!


algebra
polynomial
superior algebra
superior algebra unsolved

Show that the ideal (x^3-x-1) in the polynomial ring Z/3Z[x] over the field Z/3Z is a maximal ideal.

[hide="Hint"]I you can show that the polynomial is irreducible, you are done.[/hide]


correct me if i'm wrong. but i think we just need the condition

$a+b+c \ge 1$

The condition $a+b+c=1$ covers the condition $a+b+c\geq 1$ so I don't think it is necessary.

[quote="NguyenDungTN"]Let $a,b,c >0, a+b+c=1$. Prove that:
$\frac{a^{2}+b^{2}c}{b+c}+\frac{b^{2}+c^{2}a}{c+a}+\frac{c^{2}+a^{2}b}{a+b}\ge \frac{2}{3}$[/quote]

Let $a,b,c >0, a+b+c=1$. Prove that:
$\frac{a^{2}+b^{2}c}{b+c}+\frac{b^{2}+c^{2}a}{c+a}+\frac{c^{2}+a^{2}b}{a+b}\ge \frac{2}{3}$ (*)

We can assume that $c=max(a,b,c)$
$(*) \leftrightarrow \sum_{cyc}\frac{a^{3}+b^{2}c}{b+c}+(a^{2}+b^{2}+c^{2}) \ge \frac{2}{3}(a+b+c)^{2}$
$\leftrightarrow f(a,b,c)=\left(\sum_{sym}\frac{a^{3}}{b+c}-\frac{ab+bc+ca}{2}\right)+\left(\sum_{cyc}\frac{b^{2}c}{b+c}-\frac{ab+bc+ca}{2}\right)+\frac{1}{3}(\sum_{sym}a^{2}-\sum_{sym}bc) \ge 0$
$\leftrightarrow f(a,b,c)=M(a-b)^{2}+N(a-c)(b-c) \ge 0$
With:
$M=\frac{(a+b+c)^{2}}{(a+c)(b+c)}-\frac{1}{2}+\frac{1}{3}$
$N=\frac{(a+b+c)^{2}(a+b+2c)}{2(a+b)(b+c)(c+a)}-\frac{1}{2}+\frac{a^{2}}{2(a+b)(a+c)}-\frac{b^{2}}{2(a+b)(b+c)}+\frac{1}{3}$
We can prove that $M,N \ge 0$, so we have done.

I think the following inequality is still true:
$\sum_{cyc}\frac{a^{3}+kb^{2}c}{b+c}+\frac{1}{3}(a^{2}+b^{2}+c^{2}) \ge \frac{3k+5}{6}(ab+bc+ca)$ with $k=\frac{31}{3}$
If this's wrong, my solution is wrong, too.

[quote="Shishkin"]$\frac{a^{2}}{b+c}=\frac{a^{3}}{b+c}+a^{2}$
Also we know that $\sum_{cyc}\frac{a^{3}}{b+c}\ge \sum_{cyc}\frac{b^{3}}{b+c}$
with Couchi it gives  $\sum_{cyc}\frac{a^{3}}{b+c}\ge \sum_{cyc}\frac{c^{2}b}{b+c}$
So $\sum_{cyc}\frac{a^{2}+b^{2}c}{b+c}\ge \sum_{cyc}a^{2}+bc \ge \frac{2}{3}(a+b+c)^{2}=\frac{2}{3}$[/quote]

Could you clarify 2nd and 3rd row of your solution please? :maybe:

$\sum_{cyc}\frac{a^{3}}{b+c}\geq \sum_{cyc}\frac{b^{3}}{b+c}$ by rearangment inequality so
$\sum_{cyc}\frac{a^{3}}{b+c}\geq \sum_{cyc}\frac{c^{3}}{b+c}$ and
$\sum_{cyc}\frac{a^{3}}{b+c}\geq \frac{1}{3}\sum_{cyc}\frac{b^{3}}{b+c}+\frac{2}{3}\sum_{cyc}\frac{c^{3}}{b+c}\geq \sum_{cyc}\frac{c^{2}b}{b+c}$ by $AM-GM$.

But the initial inequality isn't symetric so we can't assume $a\geq b\geq c$. Or I'm being stupid? :maybe:

[quote="goc"]the term $\sum_{cyc}\frac{a^{3}}{b+c}$ is symetric so with it you can do what ever you want...
let it be a lemma on the side ;)[/quote]

oops... :oops:...as harazi would say, i'm the greratest idiot of all times...(no offense harazi)

anyway, thx for the explanation goc  :wink:

[quote="manlio"][quote="arqady"][quote="NguyenDungTN"]Let $a,b,c >0, a+b+c=1$. Prove that:
$\frac{a^{2}+b^{2}c}{b+c}+\frac{b^{2}+c^{2}a}{c+a}+\frac{c^{2}+a^{2}b}{a+b}\ge \frac{2}{3}$[/quote]
$\sum_{cyc}\frac{b^{2}c}{b+c}=\sum_{cyc}\frac{b^{2}c^{2}}{bc+c^{2}}\geq\frac{(ab+ac+bc)^{2}}{a^{2}+b^{2}+c^{2}+ab+ac+bc}\geq\frac{3abc}{a^{2}+b^{2}+c^{2}+ab+ac+bc}.$
Id est, it remains to prove that
$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}+\frac{3abc}{a^{2}+b^{2}+c^{2}+ab+ac+bc}\ge \frac{2(a+b+c)}{3},$
which still true. :wink:[/quote]

You are right, Arqady.  :)[/quote]

how do you prove it? :maybe:

Put $a+b+c = p$ , $ab+bc+ca = q$ and $abc= r$.
Then expand the expression and use SHUR
$p^{3}+9r \geq 4pq$
and
$p^{2}\geq 3q$

It is a little long, but not hard  :)


Prove that $k=\sqrt{3}-1$ is the best constant in the following ineq:
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{k(ab+bc+ca)}{a^2+b^2+c^2}\ge \frac{3}{2}+k$
 ;)  ;)

[quote="silouan"]http://www.mathlinks.ro/Forum/viewtopic.php?p=524210#p524210

See above. It was posted by Vasc.[/quote]
Oh,I don't know it. ;)
And [b][u]note that[/u][/b]:the equality occur if and only if
$a=b=c$or
$2a=2b=c(\sqrt{3}+1)$or
$2a=b(\sqrt{3}+1)=2c$ or
$a(\sqrt{3}+1)=2b=2c$ ;)  ;)


inequalities calculus function algebra domain logarithms

Can someone please clearly explain Jensen's Inequality (please also include the proof  :) ) and also present a few practice problems? I'd really appreciate the help, thanks.

[quote="t0rajir0u"]
In fact, Jensen's is just a special case of an even [b]more[/b] general inequality known as the Karamata Majorization Inequality.
[/quote]

Thank you very much for explaining :-)

Hmm... could you explain the Karamata inequailty and provide examples as well? I tried looking this up, but I don't understand the examples very well  :noo:  :help:

You may find [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14975]this topic[/url] useful.

[quote="t0rajir0u"]Jensen's Inequality is actually just the definition of convexity, in a non-calculus context.

We say that a function $f$ is [b]convex[/b] if for every $a, b$ in its domain and for an arbitrary real $\lambda$ with $0 \le \lambda \le 1$ we have the inequality

$\lambda f(a)+(1-\lambda) f(b) \ge f \left( \lambda a+(1-\lambda) b \right)$

This is actually just weighted Jensen's in two variables.  Graphically speaking, the secant line between two points on the graph always lies above the graph.

For $\lambda = \frac{1}{2}$ we have the familiar

$\frac{ f(a)+f(b) }{2}\ge f \left( \frac{a+b}{2}\right)$

If a function satisfies this inequality it is not hard to show it satisfies the familiar form of Jensen's inequality:

Let $a_{1}, a_{2}, ... a_{n}$ be reals in the domain of some convex function $f$.  Then

$\frac{ f(a_{1})+...+f(a_{n}) }{n}\ge f \left( \frac{a_{1}+...+a_{n}}{n}\right)$

If $f(x)$ is concave (in which case the inequality sign in the definition is reversed) then the inequality is also reversed.  The more general (weighted) Jensen's inequality:

Let $a_{i}$ be as before, and let $\lambda_{1}, \lambda_{2}, ... \lambda_{n}$ be non-negative reals such that $\lambda_{1}+...+\lambda_{n}= 1$.  Then

$\lambda_{1}f(a_{1})+...+\lambda_{n}f(a_{n}) \ge f \left( \lambda_{1}a_{1}+...+\lambda_{n}a_{n}\right)$

In fact, Jensen's is just a special case of an even [b]more[/b] general inequality known as the Karamata Majorization Inequality.

Edit:  Many common inequalities are easily solved by Jensen's.  Here are just a few.

[b]Exercise:[/b]  Prove AM-GM.

[hide="Hint"] $f(x) = \ln x$ is concave. ;) [/hide]

[b]Exercise:[/b]  Let $a, b, c$ be positive reals.  Prove that

$\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}\ge 1$

[hide="Hint"] $f(x) = \frac{1}{x}$ is convex on the relevant domain. ;) [/hide]

[b]Exercise:[/b]  Prove the Power Mean inequality.

[hide="Hint"] This one isn't as straightforward as the others.  The function you use Jensen's on will be a power function. [/hide][/quote]
You, sir, need to help me write articles for a new inequality lessons thread I am planning. PM me if interested.

[quote="Karth"]*Bump*

Can somebody show answers for the second and third? I know how to do the first... Also, is this another version of Jensen's?

$\lambda_{1}f(x_{1})+\lambda_{2}f(x_{2})+\ldots+\lambda_{n}f(x_{n}) \ge (\lambda_{1}+\lambda_{2}+\ldots+\lambda_{n}) f(\frac{\lambda_{1}f(x_{1})+\lambda_{2}f(x_{2})+\ldots+\lambda_{n}f(x_{n})}{\lambda_{1}+\lambda_{2}+\ldots+\lambda_{n}})$

EDIT: Can somebody please post more problems regarding Jensen's Inequality and Karamata's Inequality? Thanks  :)[/quote]

Yes. In your "version" just set $\lambda'_{k}= \frac{\lambda_{k}}{\lambda_{1}+\lambda_{2}+\cdots+\lambda_{n}}$ and it's the same thing.

Here's a nice IMO problem that you can solve with Jensen's:

Given positive reals $a,b,c$, show that

$\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1$.

And from an IMO shortlist:

Let $r_{1}, r_{2}, \ldots, r_{n}$ be reals greater than or equal to $1$. Prove that

$\frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}+\cdots+\frac{1}{r_{n}+1}\ge \frac{n}{\sqrt[n]{r_{1}r_{2}\cdots r_{n}}+1}$

[quote="Karth"]Can somebody show answers for the second and third?[/quote]

[hide="2"] Let $f(x) = \frac{1}{x}$, which is convex when $x$ is positive.  Our LHS $L(a, b, c)$ can be rewritten

$L(a, b, c) = a f(a+2b)+b f(b+2c)+c f(c+2a)$

Let's consider $L'(a, b, c) = \frac{L(a, b, c)}{a+b+c}$.  The coefficients of the $f$'s clearly sum to $1$, so

$L'(a, b, c) \ge f( \frac{a(a+2b)+b(b+2c)+c(c+2a)}{a+b+c}) = f( \frac{(a+b+c)^{2}}{a+b+c}) = \frac{1}{(a+b+c)}$

From which we derive

$L(a, b, c) \ge 1$

As desired.  QED. [/hide] [hide="3"] Let $f(x) = x^{ \frac{p}{q}}$, where $p \ge q$.  Then the function is convex.

Consider some numbers $a_{1}^{q}, a_{2}^{q}, ... a_{n}^{q}$.  By Jensen's we have

$\frac{ f(a_{1}^{q})+f(a_{2}^{q})+...+f(a_{n})^{q}}{n}\ge f \left( \frac{ a_{1}^{q}+a_{2}^{q}+...+a_{n}^{q}}{n}\right)$

Or, in other words,

$\sqrt[p]{\frac{a_{1}^{p}+a_{2}^{p}+...+a_{n}^{p}}{n}}\ge \sqrt[q]{ \frac{a_{1}^{q}+a_{2}^{q}+...+a_{n}^{q}}{n}}$

As desired.  QED. [/hide]


A clock loses 10 minutes each day.  How many days will it take to reach a point where the clock will indicate the correct time?

(A) 36     (B)  72      (C)  120    (D) 132     (E) 6

Is the clock analogue or digital? In other words, does it show twenty-four hour time (like 16:45 instead of 4:45) or does it go by 12 hours (4:45 AM being displayed the same way as 4:45 PM?)

[hide="Solution"]It must be a 12-hour clock, because 144 isnt in the answers. 12 hours divided by 10 minutes is 72, thus $ \boxed{B}$. Anyone is welcome to clear up my solution.[/hide]

Suppose that the clock was stopped. It would display the right time every 12 hours, or 720 minutes. Since it loses 10 minutes per day, it needs to lose 720 minutes, so the answer is 72 days, because that is how long it will take to lose 720 minutes.


inequalities
analytic geometry
graphing lines
slope
algebra
linear equation

Let $ n$ be the number of number-pairs $ (x,y)$ which satisfy $ 5y \minus{} 3x \equal{} 15$ and $ x^2 \plus{} y^2 \le 16$.  Then $ n$ is:

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \text{more than two, but finite} \qquad \textbf{(E)}\ \text{greater than any finite number}$

[hide="Solution"]
First we put the linear equation in slope-intercept form to make it easier to interpret.

$ 5y \minus{} 3x \equal{} 15 \implies 5y \equal{} 3x \plus{} 15 \implies y \equal{} \frac {3}{5}x \plus{} 3$

Now looking at the other inequality, we notice that it refers to all the points inside a circle with radius $ 4$ that is centered at the origin. From the linear equation, we see that it lies just within the circle and definitely intersects the shaded portion more than twice. Thus, the answer is $ (E) \text{ greater than any finite number}$.
[/hide]

Question to those who did the problem: If the line passes through an interval of points, doesn't that mean there are an infinite number of points that satisfy the equation (because there are an infinite number of number between any two real numbers)?


geometry
combinatorics unsolved
combinatorics

A convex hexagon is called [i]pretty[/i] if it has four diagonals of length 1, such that their endpoints are all the vertex of the hexagon.
($a$) Given any real number $k$ with $0<k<1$ find a [i]pretty[/i] hexagon with area equal to $k$
($b$) Show that the area of any [i]pretty[/i] hexagon is less than 1.

Consider the hexagon of vertices $ A \equal{} (L,0)$, $ B \equal{} (\ell,h)$, $ C \equal{} ( \minus{} \ell,h)$, $ D \equal{} ( \minus{} L,0)$, $ E \equal{} ( \minus{} \ell, \minus{} h)$ and $ F \equal{} (\ell, \minus{} h)$, where $ 1 > L > \ell > 0$, $ 1 > h > 0$ and $ (L \plus{} \ell)^2 \plus{} h^2 \equal{} 1$.  The hexagon is convex, and its diagonals $ AC \equal{} AE \equal{} BD \equal{} DF \equal{} 1$.  The hexagon may be divided into two trapezii, with parallel sides of lengths $ 2L$ and $ 2\ell$ and altitude of length $ h$, or the area of the hexagon is $ 2(L \plus{} \ell)h$.  Clearly, when $ L \plus{} \ell \equal{} h \equal{} \frac {1}{\sqrt {2}}$ (for example for $ L \equal{} \frac {2}{3\sqrt {2}}$ and $ \ell \equal{} \frac {1}{3\sqrt {2}}$), the area of the hexagon is exactly $ 1$, and this cannot be improved in this configuration because of the AM-GM inequality.  Let now $ L \equal{} 2\ell$, $ \ell \equal{} \frac {\sqrt {1 \minus{} h^2}}{3}$.  The area of the hexagon is then $ \sqrt {h^2(1 \minus{} h^2)}$, which for
$ 0 < h < 1$ has a maximum at $ h \equal{} \frac {1}{\sqrt {2}}$, and varies continuously, with limits $ 0$ when $ h\to0$ and $ h\to1$.  Therefore, all values of the area between $ 0$ (exclusive) and $ 1$ (inclusive) are obtained in this configuration.  This solves part a), and seems to indicate that part b) should be "Show that the area of any $ \it{pretty}$ hexagon is at most $ 1$" instead of "less than $ 1$"...

In Part (b), we must prove that the area of any pretty hexagon is less than 3/2 instead of 1.


geometry
trigonometry
incenter
circumcircle
geometric transformation
power of a point
radical axis

[color=darkred]Let be given a triangle $ ABC$ and its altitudes $ AA',\, BB',\, CC'$ and orthocentre $ H.$ The incircle of triangle $ A'B'C'$ meets its side $ B'C',\,C'A',\,A'B'$ at $ M,\,N,$ and $ P$ respectively. Denote $ X=BC\cap B'C',\,Y=CA\cap C'A'$ and $ Z=AB\cap A'B'.$ Prove that:

$ 1.$ $ AM,\,BN,\,CP$ are concurrent at a point $ K$

$ 2.$ $ X,\,Y,\,Z$ are collinear.

$ 3.$ $ HK\perp \overline{XYZ}$[/color]

[hide=" some hints"]1) Follows easily from the Cevian Nests Theorem

2) a. Apply Menealos on the transversal XB'C', ZB'A', YC'A', XYZ and then apply Ceva.
    b. Follows from Desargues.[/hide]

Actually for part 1, one can seek generalisations.

It might be the cevian Nests theorem but never heard of it so sorry if it is it again.

Here, for proving AM, BN, CP are concurrent, it is enough to prove the sine version of the Ceva's theorem.
But then we would have

C'M/MB' = AC' / AB' * sin (C'AM)/sin(MAB')  which implies
 sin (C'AM)/sin(MAB') = CM'/M'B *AB'/AC'

Similar expressions are obtained for other sine of the cevian angles.

As for MC'/MB', we get the concurrency as A'M, B'N, C'P concur at the Nagel point of A'B'C'.  As for AB'/AC', we get the concurrency by concurrence of the altitudes AA'BB', CC' at orthocenter H.

And here, we have not used any orthic triangle relations . (Not necessary that the point of concurrence of the initial 3 cevians AA', BB', CC' has to be incenter of the orthic triangle A'B'C' as observed as a property in orthic triangles).

As for the collinearity of X,Y,Z, it holds for any 3 concurrent cevians AA', BB', CC' as well due to perpectivity of ABC and A'B'C' by Desargue's Theorem.

As for the third HK perpendicular to XYZ, there you do require the property of orthocenters.

$(H)$ is the incircle of $\triangle A'B'C'$ and $O$ is the circumcenter of $\triangle ABC.$ Since $HM \parallel OA,$ $HN \parallel OB$ and $HP \parallel OC,$ it follows that $AM,BN,CP$ and $OH$ concur at the exsimilicenter $K$ of $(O) \sim (H).$

$M$ is the midpoint of $\overline{BC}.$ Since $(B,C,A',X)=-1,$ we have $XB \cdot XC=XM \cdot XA'$ $\Longrightarrow$ $X$ has equal power with respect to $(O)$ and the 9-point circle $(E)$ of $\triangle ABC.$ Likewise, $Y$ and $Z$ have equal power WRT $(O)$ and $(E)$ $\Longrightarrow$ $X,Y,Z$ lie on the radical axis of $(O)$ and $(E)$ $\Longrightarrow$ $\overline{XYZ} \perp EHK.$

P.S. Alternatively, the lines AM,BN,CP concur at the cevian quotient J/H with respect to A'B'C', where J is the Gergonne point of A'B'C'.


There seems to be an error with the labeling of the x and y axises.

When no attributes are added to the graph, the labels come out fine:

[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 8, 0.5));
yaxis(-8,8,Ticks(f, 8, 0.5));
real f(real x)
{
return x^3;
}
draw(graph(f,-2,2));[/asy]

But when we add color to the graph, we get:

[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 8, 0.5));
yaxis(-8,8,Ticks(f, 8, 0.5));
real f(real x)
{
return x^3;
}
draw(graph(f,-2,2),green+linewidth(1));[/asy]

When no graph is drawn on to the labels however, the error happens again.

Is this supposed to be like this, or is it an error?

I do not remember now what the int parameter in Ticks is supposed to do but the normal usage is with 2 reals [i]Step[/i] and [i]step[/i] where [i]Step[/i] is responsible for the distance between large ticks and [i]step[/i] for the distance between small ticks. If [i]step[/i] is omitted, no small ticks appear:
[asy]import graph;
size(200);
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 2.0));
yaxis(-8,8,Ticks(f, 2.0, 0.5));
real f(real x)
{
return x^3;
}
draw(graph(f,-2,2),green+linewidth(1));[/asy]
Note that I wrote [b]2.0[/b], not [b]2[/b] :!:  The latter produces
[asy]import graph;
size(200);
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 2));
yaxis(-8,8,Ticks(f, 2, 0.5));
real f(real x)
{
return x^3;
}
draw(graph(f,-2,2),green+linewidth(1));[/asy]
because 2 is not interpreted as Step anymore but as something else...

I finally found time to look at the documentation in the graph package. The relevant arguments of Ticks in the order they appear in the routine call are described below:

[quote]
int N when automatic scaling is enabled (the default; see [automatic scaling],
page 108), divide a linear axis evenly into this many intervals, separated
by major ticks; for a logarithmic axis, this is the number of decades between
labelled ticks;

int n divide each interval into this many subintervals, separated by minor ticks;

real Step the tick value spacing between major ticks (if N=0);

real step the tick value spacing between minor ticks (if n=0);
[/quote]

The description of the "automatic scaling" on page 108 is vague and rather confusing. Besides, like many other "automatic" things, this feature has a character and the tendency to have its own ideas that may contradict yours. So, I prefer to avoid it.
The little n is OK though, so you can write something like Ticks(0,7,3.0) to put 7 small marked subintervals inside each big interval of length 3.

Anyway, writing Ticks(8,0.5) as you initially did hardly makes any sense at all because the parameters will be perceived as N and Step and, since they contradict each other, only the value of N will be used.


[hide="Simply..."] We want to find the sum of every fourth coefficient in $P(x) = (x+1)^{4n}$.  This is just

$\frac{P(i)+P(-1)+P(-i)+P(1)}{4}$

Now, $(1+i)^{4}=-4, (-1+1)^{4}= 0, (-i+1)^{4}=-4, (1+1)^{4}= 16$.  Thus our desired sum is

$\frac{2 \cdot (-4)^{n}+16^{n}}{4}$

Except when $n = 0$, in which case that $P(-1)$ term is just $1$, so our sum is $1$.  [/hide]

Oh, right, that is true by DeMoivre's Theorem right?

It's not really about De Moivre's.

[hide="The relevant lemma"] Let $P(x) = x^{k}$ WLOG.  Let $\omega$ be a primitive $n^{th}$ root of unity.  Then

$P(\omega)+P(\omega^{2})+...+P(\omega^{n}) = 0$

Unless $n | k$, in which case it equals $n$ (because every term is equal to $1$).

(Can you figure out why?)  [/hide]


Let {(1 + x^2)^2}.(1 + x)^4 = ∑(r = 0 to 8)ar.x^r, then show that a1, a2, a3 are in A.P.

NOTE: In ar, r is a subscript.

So in other words, the question is asking whether you know how to multiply polynomials?

See [[Latex]] or http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 for information on using LaTeX on the forum.


hello
who can tell me what is the answer of 0^-1??????

[quote="Hexaditidom"]What about 0^0?[/quote]
It is quite meaningless untill you define what you mean.

[quote="Myth"]0 is non-invertible element of field $\mathbb{R}$.  :lol:  :lol:  :lol:[/quote]
ok but why when x>0 we say so 1/x>0  ?????   ;)

Well if you define $0^0$ to be the limit as x gets large of x^(1/x) your limit will approach 1.  Thus, $0^0$ is 1. Well not really of course but you get the idea.


search
number theory unsolved
number theory

Prove that sequence $ a_{n}\equal{} 2^{2^{n}}\plus{}1$ and $ b_{n}\equal{}6^{2^{n}}\plus{}1$ contains infinitely many composite numbers
where $ n$ is natural number.

At least the question for $ a_{n}$ being composite infinitely often is open.

[quote="ZetaX"]The question was posted before and is about showing that at least one of the sequences contains an infinite number of composite numbers.[/quote]
This problem has been discussed in http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1595647203&t=143722.


calculus
derivative
integration
real analysis
limit
algebra
function

I'm studying a partial regularity result of a parabolic system. The first encounter is the Steklov average. Does anyone know some text about the subject? I searched the internet but got nothing. Indeed, there are some properties that I can not prove myself. For example, how to prove that $ u_{ht}\equal{}(u_t)_h$, where $ u_{ht}$ is the derivative in t of $ u_h$ and $ u_h$ is the Steklov average of u defined below
$ u(x,t)\equal{}\frac{1}{h}\int_{t}^{t\plus{}h}u(x,s)ds$

By the fundamental theorem of calculus, $ (u_t)_h \equal{} \frac {u(x,t \plus{} h) \minus{} u(x,t)}{h} \equal{} (u_h)_t$.  Or am I misunderstanding your notation?  Using subscripts for both differentiation and averaging operators seems a bit awkward, but if I understand the meaning of the statement correctly there is nothing very mysterious about it.

This 'Steklov average' is a [i]convolution[/i] operator, so some of its properties -- including the particular question you asked about differentiation -- can be deduced by looking at convolutions in general.  Another observation about it is that it if $ f\in L^p$ then $ f_h\in L^p$ and $ \lim_{h\to 0}f_h \equal{} f$ in $ L^p$.  Moreover, each $ f_h$ is weakly differentiable in $ L^p$, which I suppose is what could be the usefulness of this operator in proving some regularity results.  Could you give a reference where you saw this operator used? I'm kind of curious to see it. :)

Thank you, JoeBlow
I was hurry in my first post. In fact, in the text, u is of class $ L^2(Q)$, where Q is a cylinder $ \Omega\times (0,T)$ ($ \Omega$ is a bounded domain in $ R^n$). When I tried to prove that property, I thought about the fundamental theorem of calculus. However, the trouble is that I have no idea if the theorem remains true for functions of class $ W^{1,2}$.
In my opinion, this kind of Steklov average is not really a convolution since it takes convolution for only one variable. Therefore, I think it's necessary to check each property carefully.
Here is an article that I found on the internet, you may want to take a look.
http://www.emis.de/journals/CMUC/pdf/cmuc9802/naumann.pdf


geometry
conics
hyperbola
geometry unsolved

Let $ M,M'$ denote the tangency points of the incircles of $ \triangle PAB$ and $ \triangle PAC$ with $ PA.$ Both circles are tangent $\Longleftrightarrow$ $ M$ and $M'$ are identical $\Longleftrightarrow$ $ AM \equal{} AM' .$

$\Longleftrightarrow AP \minus{} PB \plus{} c \equal{} AP \minus{} PC \plus{} b  \Longleftrightarrow PB \minus{} PC \equal{} c \minus{} b$

$\Longleftrightarrow$ $ P$ lies on the hyberbola $ \mathcal{H}_a$ with foci $ B,C$ passing through $ A,$ precisely on the branch passing through $ A.$ Therefore, the wanted point $ P$ is the first Soddy's point $ S_0$ of  $\triangle ABC.$ The construction of $ S_0$ was discussed [url=http://www.artofproblemsolving.com/viewtopic.php?t=307134]here[/url].


A box contains five balls numbered 1, 2, 3, 4 and 5. Three balls are drawn at random without replacement. What is the probability that the median of the values on the three chosen balls is greater than 2? Express your answer as a common fraction.

[hide="Solution"] Answer: $\boxed{\frac{7}{10}}$
I just made a list of the 10 distinct sets of chosen balls, and 7 of the sets had a median greater than 2. :D[/hide]

Though, you can list all the numbers and then figure it out, if there were more than just 5 balls, that would be very difficult. So, to figure out how many combinations there are, you would just use N! / r!*(N-r)!. Then, you have to see that if the first ball is 1, you have 2 options for second ball. If its 2, you have 3 options. if 3, you have 2 options. If it's 4 , 1 options, and 5, no options. 2+3+2=7, so 7/10

[hide]it is impossible for the median to be less than two, and the only way it can be 2 is if the first 2 numbers are 1,2 and you have 3 choices for the last. There are 5 pick 3 = 10 possible arrangements and 3 don't work so 7 do. 7/10. [/hide]

[hide]Count the complement, count how many ways to get a median of 2 or less.
If 1 is the median, there is no way to get 1 as the median.
If 2 is the median, we must of choosen 1 as one of the numbers, and we have 3 ways to pick the last number
There are 5C3=10 total ways to pick three balls.
$1-\frac{3}{10}=\frac{7}{10}$.[/hide]


If $ a_1 \equal{} 1$ and $ a_{n\plus{}1} \equal{} 3a_n \plus{} 2$,

 What is $ P(x)$ such that $ P(n) \equal{} a_n$?

hello, the homogenius part of the equation $ a_{n+1}=3a_n+2$ we can solve with the ansatz $ a_n=q^n$, from here we get $ q=3$,now we need a special
solution of the inhomogenius part of our equation this is $ a{}_{n}{}_p=-1$, so
the general form of our solution is $ a_n=C\cdot3^n-1$,while we get $ C$ from the initial condition.
Sonnhard.


A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token into a discard pile. The game ends when some player runs out of tokens. Players $ A$, $ B$, and $ C$ start with $ 15$, $ 14$, and $ 13$ tokens, respectively. How many rounds will there be in the game?

$ \textbf{(A)}\ 36 \qquad
\textbf{(B)}\ 37 \qquad
\textbf{(C)}\ 38 \qquad
\textbf{(D)}\ 39 \qquad
\textbf{(E)}\ 40$

[hide]
i brute forced the first 6 rounds to find a pattern.
after 3 rounds it becomes 14,13,12 then after 3 more 12,11,10
however at the second round, a 12 already appears, which means the game ends 2 round before a multiple of three (this is really confusing explaning). thus the answer is 37.
[/hide]

[quote="davidlizeng"][hide]
i brute forced the first 6 rounds to find a pattern.
after 3 rounds it becomes 14,13,12 then after 3 more 12,11,10
however at the second round, a 12 already appears, which means the game ends 1 round before a multiple of three (this is really confusing explaning). thus the answer is 38.
however, if BOTH need to be out, then 39, the question never specifies.[/hide][/quote]

[hide]The question says only one person needs to get out.  I did it the same way but i got 37[/hide]

you're right... found my mistake.....it's two away from a multiple of three, i did it the other way around, two form a multiple of three...edited


See what I found today!
Souce: Mathematical Olympiad Treasures
1.3.13 Suppose the sequence $a_1, a_2,...,a_n$ satisfies the following conditions: $a_1=0, |a_2|=|a_1+1|,...,|a_n|=|a_{n-1}+1|$.
Prove that $\frac{a_1+a_2+...+a_n}{n}\geq -\frac{1}{2}$.
This was very similar to #15 on AIME 2006 A!

I think paladin8 posted a solution on his blog some time ago.

http://wangsblog.com/jeffrey


[quote="Lagrangia"]Let $a,b,c \in (0,1)$ so that $a^2+b^2+c^2=1$. Prove that $12 \sqrt[4]{abc(1-a)(1-b)(1-c)} \leq 4 +(a+b+c-1)^2$.

cheers! :D :D[/quote]

[quote="nickolas"]What  do you want to  say  Socrates??[/quote]

I just wanted to revive this old unsolved problem. :)
Nothing special.

Just to give it to all problem solvers.

Where is your objection??

Well, in fact, at first, when I saw you only quoted Lagrangia, I thought the same thing as Nickolas, but then, checking the dates, I got your point! ;)

Are you sure that $a,b,c \in (0,1)$ is important?!  :huh:

Probably yes because some of $a, b, c$ could lay on $(-1; 0)$ for example.


calculus
integration
logarithms
calculus computations

bonjour and merci.  :D

show

$ \int_0^1\int_0^1\;\dfrac{x^2\minus{}1}{(1\plus{}x^2y^2)\cdot\ln^2(x\,y)}\;\textbf dx\;\textbf dy\;\equal{}\;\boxed{\dfrac{1}{2}\minus{}\dfrac{2\,\textbf C}{\pi}\plus{}\ln\left(\dfrac{\sqrt{\; 8\pi^2 \;}}{\Gamma^2\left(\dfrac{1}{4}\right)}\right)}$

where $ \textbf C$ is catalan's constant

It must be simple transformation of  Vardi's integral see here  http://www.mathlinks.ro/viewtopic.php?t=164417
it's same as how can be solved vardi's integral by using double integral?
 you already have posted hint , but not full solution.


calculus
limit
floor function
function
Asymptote
calculus computations

I have a calculus exam tomorrow and I was wondering if my "sample" problems for the test are right. I made them all and that's why I want to make sure they're correct.  :D

Determine a) Limit if it exists and b) if it's continuous.

1. $\lim_{x \rightarrow 1} x^5+x^4+x^3+x^2+x = 1+1+1+1+1 = 5$. Continuous

2. $\lim_{x \rightarrow -2} \dfrac{x^2+5x+6}{x+2} = \lim_{x \rightarrow -2} \dfrac{(x+2)(x+3)}{x+2} = \lim_{x \rightarrow -2} x+3 = -2+3 = 1$. Discontinuous and removable discontinuity.

3. $\lim_{x \rightarrow 5} \dfrac{x-5}{\sqrt{x} - \sqrt{5}} = \lim_{x \rightarrow 5} \dfrac{(\sqrt x - \sqrt 5)(\sqrt x + \sqrt 5)}{\sqrt x - \sqrt 5} = \sqrt 5 + \sqrt 5 = 2 \sqrt 5$.

4. $\lim_{x \rightarrow 100} \lfloor {x} \rfloor$ is DNE since step-function itself has JUMP discontinuity.

5. $\lim_{x \rightarrow 5} \dfrac{15}{x^3+6x^2+8x}$. Factoring out $x$, we see that the fraction becomes 15/0 so the limit is DNE (Does not exist) and this function has infinite discontinuity.

[quote="Silverfalcon"]$\dfrac{(\sqrt x - \sqrt 5)(\sqrt x + \sqrt 5)}{\sqrt x - \sqrt 5} = \sqrt 5 + \sqrt 5 = \sqrt 10$.[/quote]

How did you get $\sqrt 5 + \sqrt 5 = \sqrt 10$?

I think it should be $2\sqrt5$ instead....

For the last one, I did not get the same answer as you.

I solved, plugging in x=5

$\lim_{x \rightarrow 5} \dfrac{15}{x^3+6x^2+8x}$

$= \dfrac{15}{5^3+6(5)^2+8(5)}$

$= \dfrac{1}{21}$

That should be the limit, I believe. The function itself is not continuous on all Real numbers, but at x=5, it is....

I don't understand #5. What is infinite discontinuity?

And #3 ... :D

Eh..

Several typos.. The #3, it should've been $\sqrt{5} + \sqrt{5} = 2 \sqrt{5}$. I think I need to go back to Algebra I now..  :P

For #5, the $a$ value is wrong. It should've been $\lim_{x \rightarrow 0}$.

The infinite discontinuity is when you have a vertical asymptote. Hence, the left side may go down infinitely while the right side of the asymptote will rise up infinitely.