About how long would it take you to play every song you have memorized (that you actually USE often -- like "real" songs, not ones that you learned for a one-time thing or whatever . . . i.e. "solos") on the instrument you are most skilled on? Please name the instrument, too.

i voted for 10 to 20 minutes...
i think the last time i touched my accordian was the summer of 2004....

I could probably play for a good two hours with just piano stuff.  I'm required to do a twenty piece program for Guild auditions, which would probably constitute about an hour and a half.  Adding the tunes I do for fun, jazz stuff, and my own compositions would bring it to about two hours.

I [i]think[/i] it'd be around an hour--I'm not sure though.

Oh right, I play piano.

I could probably do between an hour and an hour and a half on vocals.

Probably 0-10 minutes...I don't have anything memorized on piano right now. :?

30 or so minutes on piano. I'm not a big memorizing person. I should memorize more stuff, but there's a lot of pieces I can recall if I play again for a couple days.

I'd say around 40 minutes on violin. Even though I haven't played for a year, I remember a lot of that stuff really well. Mainly because it isn't too advanced.

forgot to mention, I also sing.  :)  Actually, I only sang for one year, and that was more like three months. I just decided to try out for Fiddler on the Roof in 6th grade and ended up getting the role of Tevye. It was pretty unexpected, but pretty fun. I still remember all my songs from that, so that's probably another half an hour or so.

Hi~

I'd say around 2 hours on piano. I'm taking DipABRSM this year, which requires around 30 minutes of performance there already,,

Another 2 hours on clarinet. I find it easier to memorize on this instrument because it uses a single stave and flows easier than piano.

I can play French horn, but not that well... maybe another 30 minutes? All I know is Saint-Saens' Morceau de Concert and the Mozart Horn Concertos -_-;;

I could go for one or two hours on the bagpipes, but I HAVE to memorize everything because I [obviously] can't use sheet music while marching :lol: The thing is, I don't remember the names of the tunes, just the actual tune, so if somebody wants me to play a tune, they'll have to hum it  :blush: So many of the tunes are similar with just a slight nuance that it gets quite confusing sometimes.


Find n max such that the equation:
   $2x$ $+$ $4y$ $+$ $5z$ $=$ $n$
has no solution in $N*$


ratio
geometry
3D geometry
calculus
analytic geometry
integration

We have uniformly volumetrically charged cube. What's the ratio of the electrostatic potential in the middle of the cube and in one of the heads of the cube? (the potential is zero in infinity)

By dimensional analysis, we can conclude that the potential in any point of the cube is proportional to $a^{2}\rho$, where $a$ is the length of a side of the cube and $\rho$ is the charge density of the cube. Let $\varphi_{C}(a, \rho)$ be the potential in the center of the cube and $\varphi_{V}(a, \rho)$ the potential in one of the vortices. If we consider a cube of side $2a$, but charged with the same density $\rho$, from superposition we can conclude that $\varphi_{C}(2a, \rho) = 8\varphi_{V}(a, \rho)$. Since $\varphi_{C}(2a, \rho) = 4\varphi_{C}(a, \rho)$, the required ratio is
\[\frac{\varphi_{C}(a, \rho)}{\varphi_{V}(a, \rho)}= 2. \]

Here is a solution using calculus:
lets choose the sides of the cube as 1 unit each, and fit a coordinate system $(x,y,z)$ to one of the heads of the cube.
around the point  $\vec{r}=(x,y,z)$ we choose a small $dV=dxdydz$ volume. the potential by this small volume at the 0 point of the coordinate system is :
$k\frac{dV}{|\vec{r}|}$
the potential of the whole charge at the head:
$\phi_{head}=k\int\frac{dV}{|\vec{r}|}=k\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{dxdydz}{\sqrt{x^{2}+y^{2}+z^{2}}}$
similarly the potential in the middle of the cube:
$\phi_{middle}=k\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{dxdydz}{\sqrt{(x-\frac12)^{2}+(y-\frac12)^{2}+(z-\frac12)^{2}}}$
so the ratio:
$\frac{\phi_{head}}{\phi_{middle}}=\frac{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{dxdydz}{\sqrt{x^{2}+y^{2}+z^{2}}}}{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{dxdydz}{\sqrt{(x-\frac12)^{2}+(y-\frac12)^{2}+(z-\frac12)^{2}}}}=\frac12$
just the same as you got, though your solution is much more elegant.


Prove that the only sets that are both open and closed are $ R$ and the empty set $ \emptyset$.

More generally: a set $ A\ne\emptyset$ is connected if and only if the only open and closed of its subsets are $ A$ and $ \emptyset$ .
[hide="Proof"]
-- If $ A$ is connected.
Suppose for the sake of contradiction that there exist $ B\subset A$ different from $ A$ and $ \emptyset$ which is both open and closed.
Because $ B$ is closed we have that $ C \equal{} A \minus{} B$ is open. Then there exits a partition $ \{B, C\}$ of $ A$ into two open sets. Then $ A$ is disconnected, contradiction. Hence, the only subsets of $ A$ which are open and closed are $ A$ and $ \emptyset$.

--If the only subsets of $ A$ which are open and closed are $ A$ and $ \emptyset$.
Then for every $ B\subset A$ different from $ A$ and $ \emptyset$, $ B$ is only open or closed. Then its $ C \equal{} A \minus{} B$ is also only open or closed. This means that all partitions $ \{B, C\}$ of $ A$ are compound by two sets which are not simultaneoulsy open. Then $ A$ is connected.
[/hide]

Since we know that $ \mathbb{R}$ is connected by the previous result it follows that the only open and closed of its subsets are $ \mathbb{R}$ and $ \emptyset$.

:)

Thank u.
But if I only know the concepts of open and closed, then how to prove it?
A set $ O \subset \mathbb{R}$ is open if for all points $ a \in O$ there exists an e-neighborhood $ V_{e}{(a)} \subset O$.
A set $ F \subset \mathbb{R}$ is closed if it contains its limit points.

Well then this helps (for future reference, a classical proof that a segment is connected).
[hide]
Suppose for the sake of contradiction that there exists $ A\subset\mathbb{R}$ different from $ \mathbb{R}$ and $ \emptyset$ which is both, open and closed. Since $ A$ is closed, its complement, say $ B$, is open. Clearly we can find a closed interval $ J \equal{} [c,d]$ which contains points from both, $ A$ and $ B$.  Suppose with out lose of generality that $ d\in A$.
Let $ r \equal{} \sup\{B\cap J\}$.
Suppose that $ r\in A$. Since $ A$ is open it contains a whole open interval with center in $ r$, say $ (b, a)$, with $ b > a$.  Then, the fact $ a\in A$ implies $ a > x,\;\forall\; x\in B\cap J$, which is a contradiction because $ r > a$  and $ r$ was supposed to be the least real with that property.
Then $ r\in B$, so $ r<d$. Because $ B$ is also open it contains a whole open interval with center in $ r$, say $ (b,a)$, with $ b > a$, so $ a<d$ also. Then $ a\in B\cap J$ and $ a > r$, which is again a contradiction because we should have $ r > a$.

Then $ r$ is real but, $ r\not\in A\cup B$, which is another contradiction because $ A\cup B \equal{} \mathbb{R}$ by hypothesis.
Hence, there does not exists such set $ A$. This means that the only subsets of $ \mathbb{R}$ which are both open and closed are $ \mathbb{R}$ and $ \emptyset$.
[/hide]

:)

[quote="ElChapin"]Well then this helps (for future reference, a classical proof that a segment is connected).
[hide]
Suppose for the sake of contradiction that there exists $ A\subset\mathbb{R}$ different from $ \mathbb{R}$ and $ \emptyset$ which is both, open and closed. Since $ A$ is closed, its complement, say $ B$, is open. Clearly we can find a closed interval $ J \equal{} [c,d]$ which contains points from both, $ A$ and $ B$.  Suppose with out lose of generality that $ d\in A$.
Let $ r \equal{} \sup\{B\cap J\}$.
Suppose that $ r\in A$. Since $ A$ is open it contains a whole open interval with center in $ r$, say $ (b, a)$, with $ b > a$.  Then, the fact $ a\in A$ implies $ a > x,\;\forall\; x\in B\cap J$, which is a contradiction because $ r > a$  and $ r$ was supposed to be the least real with that property.
Then $ r\in B$, so $ r < d$. Because $ B$ is also open it contains a whole open interval with center in $ r$, say $ (b,a)$, with $ b > a$, so $ a < d$ also. Then $ a\in B\cap J$ and $ a > r$, which is again a contradiction because we should have $ r > a$.

Then $ r$ is real but, $ r\not\in A\cup B$, which is another contradiction because $ A\cup B \equal{} \mathbb{R}$ by hypothesis.
Hence, there does not exists such set $ A$. This means that the only subsets of $ \mathbb{R}$ which are both open and closed are $ \mathbb{R}$ and $ \emptyset$.
[/hide]

:)[/quote]
What does "$ J \equal{} [c,d]$ which contains points from both, $ A$ and $ B$" mean?
and "say $ (b, a)$, with $ b > a$"?  Why  $ a\in A$ implies $ a > x,\;\forall\; x\in B\cap J$?

[quote="IORI"]
What does "$ J \equal{} [c,d]$ which contains points from both, $ A$ and $ B$" mean?
[/quote]

There exists $ J \equal{} [c, d]$ with $ c < d$ such that $ A\cap J\ne\emptyset\ne B\cap J$.

[quote="IORI"]
and "say $ (b, a)$, with $ b > a$"?  Why  $ a\in A$ implies $ a > x,\;\forall\; x\in B\cap J$?
[/quote]

You are right about asking this, because I forgot to add something about $ a$, that it is taken such that $ a\in A$. After this clarification:
$ \forall\; x\in B\cap J$ we have $ x\le r$
Suppose that there exists $ y\in B\cap J$ such that $ y\ge a$. Then $ y\in [a,r]$ but $ [a,r]\subset A$ then $ y\in A$, contradiction. Therefore $ \forall\; y\in B\cap J, y < a$.
:)

[quote="IORI"]e...I still can't get your proof.
U wrote:  $ J \equal{} [c,d]$ with $ c < d$ s.t. $ A \cap J \ne \emptyset$, $ B \cap J \ne \emptyset$. Let $ r \equal{} sup \{B \cap J \}$
Then why "$ \forall x \in B$ we have $ x \le r$"?
And why "If there exists $ y \in B \cap J$ s.t. $ y \ge a \equal{} > y \in A$ is a contradiction"?[/quote]

I just had a mistake in writing it. Clearly, the correct statement is $ \forall\; x\in B\cap J$ we $ x\le r$.

[quote="IORI"]e...I still can't get your proof.
U wrote:  $ J \equal{} [c,d]$ with $ c < d$ s.t. $ A \cap J \ne \emptyset$, $ B \cap J \ne \emptyset$. Let $ r \equal{} sup \{B \cap J \}$
Then why "$ \forall x \in B$ we have $ x \le r$"?
And why "If there exists $ y \in B \cap J$ s.t. $ y \ge a \equal{} > y \in A$ is a contradiction"?[/quote]

I defined $ B$ as $ \mathbb{R} \minus{} A$, so $ A\cap B \equal{} \emptyset$.

:)


parameterization
real analysis
real analysis theorems

Suppose $ x \in R^n$,
I encountered the sentence:
$ |x| \rightarrow \infty \Leftrightarrow |x_1| \rightarrow \infty \vee |x_2| \rightarrow \infty \vee ... \vee |x_n| \rightarrow \infty$
Could anyone explain why this sentence makes sense? How to write it in a clearer way?
Thanks.

[quote="jmerry"]I definitely wouldn't use so many symbols.

Translated to use words:
$ |x|$ goes to $ \infty$ if and only if ($ |x_1|$ goes to $ \infty$ or $ |x_2|$ goes to $ \infty$ or ... or $ |x_n|$ goes to $ \infty$).
A step farther, we can read that chain of "or" as "at least one":
$ |x|$ goes to $ \infty$ if and only if at least one of the $ |x_i|$ ($ i\in\{1,2,\cdots,n\}$) goes to $ \infty$.[/quote]

But what does it mean by
$ |x|$ goes to $ \infty$

I mean how to define
"$ |x|$ goes to $ \infty$ "
Or how to express it in terms of a logic expression

[quote="JBL"]For all $ R > 0$, there exists $ N > 0$ such that for all $ [ [$whatever your time parameter is$ ] ]$ larger than $ N$, $ |x| > R$.  "For any arbitrarily large distance from the origin, eventually $ x$ is always that far away."
[color=green]Moderator edit: accidental Wiki link removed[/color][/quote]

If this is not nonsenses ,what is.

But since you are good in refering people to books,refer me to a book that i can find that definition.

What is the meaning of  [b]|x| if $ x\in R^n$[/b]

Usually, although the precise meaning is not for us to deduce, rather for you to provide given the context of the statement, the meaning is the following:

For $ x \in \mathbb{R}^n$ we have that $ |x|$ also denoted $ \|x\| \equal{} \sqrt{\sum\limits_{i\equal{}1}^n x_i^2}$, the Euclidean norm.

|x| on its own in $ R^n$ it has no meaning ,because it could be any norm ,unless you state clearly what norm is

It could be the maxnorm.

It doesn't actually matter what norm is meant by $ |x|.$ Why not? Because all norms on a finite-dimensional space are equivalent, so if you prove this for one norm, it holds for all norms.

In fact, the original statement is closer to the "maxnorm", $ \|x\|_{\infty}\equal{}\max_{1\le j\le n}(|x_j|)$ than it is to the Euclidean norm.

yes, but the proving process is completely different for each norm,on a certain problem.

So we have to specify which norm we are using.

You specified the maxnorm as more appropriate for the original statement


integration
logarithms
search
real analysis
real analysis unsolved

Hint
[hide]$\frac{1}{x^x}=\sum_{n\geq 0}\frac{(-1)^n}{n!}x^n{\ln x}^n$[/hide]

It has been posted before.
And I gave a proof there...
You can search it on this forum


topology
geometry
inequalities
number theory

I am not sure which forum to post this in, so i aplogize if this isn't where it should be. Anyways, i was just wondering. This is a very broad question but, what is/do you think the hardest/most difficult/most complicated etc field of mathematics that is currently in existence? I guess what i am asking is, what is the highest math such that there is no form of mathematics higher than it? Please grace me with your insight  :)

Every advanced field is hard. Different people say different things; its only a matter of opinions. For me, I find number theory hard.

I was going to say the same thing, Number Theory. It's so theoretical and expansive at times. But specifically, I've heard people say that K-Theory is very difficult. It blends topology, number theory, tensors, etc. Used in theoretical physics, like string theory. Physicists are waiting for more mathematics (like K-Theory) to develop so that they can continue to build upon string theory.

@last three people: sorry, but do you really know what you are speaking of¿
This topic seems not to concern olympiad math but "real" math.
When you meant geometry as "algebraic geometry" or something like that, I could understand it. But for the way geomatry is normally used on that forum: I disagree (not saying that olympiad geometry is easy).


[quote="Vasc"]

The general ineq is true for $a,b,c$ positive reals and $p>-2$ and $q\ge 2p+1$.

$\displaystyle \displaystyle \frac{2a^2+qbc}{b^2+pbc+c^2}+\frac{2b^2+qca}{c^2+pca+a^2}+\frac{2c^2+qab}{a^2+pab+b^2} \ge \frac{3(2+q)}{2+p}$

[/quote]

This is so nice and Vasc promised to post his solution.

As it is again solved I hope Vasc will post his solution.

Thank you very much , Vasc.

manlio, do you mean $q \le 2p+1$?
I wrote this in my exercise book.


In parallelogram $ABCD$, a line from $C$ cuts diagonal $\overline{BD}$ in $E$ and $\overline{AB}$ in $F$. If $F$ is the midpoint of $\overline{AB}$, and the area of $\bigtriangleup BEC$ is $100$, find the area of quadrilateral $AFED$.


Today we celebrate the victory of the democracy in Romania. The democratic-liberal lider Traian Basescu won with 51,23% of the votes the second round of the Presidential Elections in Romania, making him the first political figure in Romanian post-communist era which wins 3 elections in a row (2000 is elected mayor of Bucharest, in june 2004 is again elected mayor  of Bucharest in first round - his counter-candidate obtaining half of his procentage, and in december 2004 is elected president after an on-the-edge presidential run).

The exit polls showed a 50-50% never enountered situation yesterday when the voting rooms closed. Most of the voters of Basescu are young, open-minded people, with a high-level of education, living especially in urban areas. The counter-cadidate Adrian Nastase, lider of the neo-communist party in Romania, was in general voted by old people, living in rural areas, and generally with below average education.

For the first time, the President is elected for 5 years (instead of 4).

[quote="tetrahedr0n"][quote]Today we celebrate the victory of the democracy in Romania. The democratic-liberal lider Traian Basescu won with 51,23% of the votes the second round of the Presidential Elections in Romania,[/quote]

I heard the good news from my parents a little earlier today. My parents voted Basescu in the first round, but inclement weather in Detroit made it difficult to vote in the second round. Did you vote Valentin?[/quote]Of course I voted. You can also figure out with who I voted, looking at the pool of voters of the two candidates. :)

[quote="Valentin Vornicu"]You can also figure out with who I voted, looking at the pool of voters of the two candidates. :)[/quote]

I dunno. I'm kind of puzzled here.  :D  :P
Just kidding, please don't destroy me with your super-admin powers...  :D  ;)


P is a point out of ABC triangle. perpendiculars from P to BC,CA,AB meet them at D,E,F respectively.PCE,PBD and PAF have the same area.prove that the areas of these triangles are equal to area of ABC trianle.


function
algebra
polynomial
conics
algebra unsolved

find all continous fuction f:R...R   such that                    f(xf(y)+yf(x) )=2f(x)f(y)     $x,y\in\ R $

I'm new, so I'll plead ignorance for not including a formal proof.  It appears that it works for f(x) = mx where m is any real number.

f [ xf(y) + yf(x) ] =
f [ xmy + ymx ] =
f [ 2mxy ] = 2(m^2)xy

 and

2f(x)f(y) = 2mxmy = 2(m^2)xy.

Couldn't find it to be true for any other continuous functions (any other polynomials, the conic sections, trigonometric, exponential, or logarithmic functions) for all x, y in R.

Hope this gets some discussion going.


prove that (a/b+b/c+c/a)^1/2 + ((ab+bc+ca)/(a^2+b^2+c^2))^1/2>=3^1/2 + 1 holds for all nonnegative real number a,b,c.

[quote="Lee Sang Hoon"]prove that $ (a/b\plus{}b/c\plus{}c/a)^1/2 \plus{} ((ab\plus{}bc\plus{}ca)/(a^2\plus{}b^2\plus{}c^2))^1/2>\equal{}3^1/2 \plus{} 1$ holds for all nonnegative real number a,b,c.[/quote] ....
now lets solve

$ \sqrt{\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}}\plus{}\sqrt{\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}} \geq \sqrt{3}\plus{}1$
We can use ABC theorem :wink:

[quote="chien than"]$ \sqrt {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}} \plus{} \sqrt {\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \geq \sqrt {3} \plus{} 1$
We can use ABC theorem :wink:[/quote]

Can you give exact solution???

I want to know how can we apply ABC theorem!

This inequality is cyclic not symmetric how can i use ABC theroem? do I need to apply some transformation?

i have also tried to solve this problem by using some inequalities involving a^2b+b^2c+c^2a. but all faild. can some body give me some hints?


vector
probability
function
analytic geometry
probability and stats

Hi all.  I was wondering if my work for the following problem was correct:

Suppose (X,Y) is a random vector with a probability mass function p_(X,Y) of (i, j) = 1/10 for 1 ≤ i ≤ j ≤ 4.

Compute the conditional expectation of Y given X.

So for this, I realized that the possibilities of (i,j) were (1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), and (4,4).

P(1,1) = .1
P(1,2) = .1
P(1,3) = .1
P(1,4) = .1
P(2,1) = 0
P(2,2) = .1
P(2,3) = .1
P(2,4) = .1
P(3,1) = 0
P(3,2) = 0
P(3,3) = .1
P(3,4) = .1
P(4,1) = 0
P(4,2) = 0
P(4,3) = 0
P(4,4) = .1

P(X=1) = .1 + .1 + .1 + .1 = .4
P(X=2) = .1 + .1 + .1 = .3
P(X=3) = .1 + .1 = .2
P(X=4) = .1

Below I did p(x,y)/P(X=x) for each pair of coordinates listed above that had a probability not equal to 0.
For (1,1), I got 1/4
For (1,2), I got 1/4
For (1,3), I got 1/4
For (1,4), I got 1/4
For (2,2), I got 1/3
For (2,3), I got 1/3
For (2,4), I got 1/3
For (3,3), I got 1/2
For (3,4), I got 1/2
For (4,4), I got 1.

Multiplying each of these probabilities by the value of the x coordinate, I got (1/4 * 1) + (1/4 * 1) + (1/4 * 1) + (1/4 * 1) + (1/3 * 2) + (1/3 * 2) + (1/3 * 2) + (1/2 * 3) + (1/2 * 3) + (1 * 4) = 10

So I got 10 for a conditional expectation of Y given X.

Which is obviously wrong because $ Y$ never exceeds $ 4$.

"Given X" means you are told that $ X\equal{}n$ and your answer will depend on $ n$. For example, if $ X\equal{}3$, then you have either $ (3,3)$ or $ (3,4)$, hence
\[ E[Y\, |\, X\equal{}3]\equal{}\frac{3P[(3,3)]\plus{}4P[(3,4)]}{P[(3,3)]\plus{}P[(3,4)]}\equal{}3.5\]

[quote="mlok"]Which is obviously wrong because $ Y$ never exceeds $ 4$.

"Given X" means you are told that $ X \equal{} n$ and your answer will depend on $ n$. For example, if $ X \equal{} 3$, then you have either $ (3,3)$ or $ (3,4)$, hence
\[ E[Y\, |\, X \equal{} 3] \equal{} \frac {3P[(3,3)] \plus{} 4P[(3,4)]}{P[(3,3)] \plus{} P[(3,4)]} \equal{} 3.5
\]
[/quote]

Thanks!

Question - then for the conditional expectation, would the solution just be the individual expectations for each of x=1,2,3,4 done out as what you demonstrated?

i.e.:

E(Y|X=1) = 2.5
E(Y|X=2) = 3
E(Y|X=3) = 3.5
E(Y|X=4) = 4

OR is the conditional expectation each of the E(Y|X=x) multiplied by the value of x and then added up?

It's the first. Conditional expectation is not a number - it's a random variable.


Find the function $f$ that satisfies \[ xf(x)+2xf(-x)=-1. \]

[hide]
We can write another function using the cyclic property of this one:
$-xf(-x) -2xf(x) = -1$

Then, using system of linear equations we find that $f(x) = \frac{1}{x}$[/hide]


At a dance party a group of boys and girls exchange dances as follows: one boy dances with $ 5$ girls, a second boy dances with $ 6$ girls, and so on, the last boy dancing with all the girls. If $ b$ represents the number of boys and $ g$ the number of girls, then:

$ \textbf{(A)}\ b \equal{} g\qquad
\textbf{(B)}\ b \equal{} \frac{g}{5}\qquad
\textbf{(C)}\ b \equal{} g \minus{} 4\qquad
\textbf{(D)}\ b \equal{} g \minus{} 5\qquad \\
\textbf{(E)}\ \text{It is impossible to determine a relation between }{b}\text{ and }{g}\text{ without knowing }{b \plus{} g.}$

I think it is $ b\equal{}g\plus{}4$, but that is not an answer choice. Are you sure $ \mathbf{(C)}$ isn't $ b\equal{}g\plus{}4$.

To solve the problem, you can make a table.

B G
1 5
2 6
3 7
4 8
5 9
6 10
...

So we find the pattern to be $ b\equal{}g\plus{}4$.

Actually, your table implies that $ b \equal{} g \minus{} 4 \implies \fbox{(C)}$.

Yeah, stupid mistake. I was thinking the other way around, $ g\equal{}b\plus{}4$. My answer is $ \mathbf{(C)}$.


probability algorithm calculus inequalities vector logarithms function

Suppose there are 4 ships with probability $ P_i$ for the $ i^{th}$ ship of holding treasure  (Each $ P_i$ is a real number between 0 and 1).  Furthermore, there are 6 pirates who can each select at most 1 of the 4 ships and plunder its treasure with probability of success $ S_j$ for the $ j^{th}$ pirate  (Each $ S_j$ is a real number between 0 and 1).  Which ships should each pirate target to plunder in order to maximize the likelihood of the pirates stealing the treasure from the ships?  Can you generalize with $ n$ ships and $ m$ pirates?

Assume that the probability of success for each pirate is independent of its targeted ship.  Also assume that all of the pirates will select their target ships independently and will plunder the ships at the same time.  There are no restrictions to the number of pirates that can plunder a particular ship.

This is somewhat ambiguous.  If all you want is to maximize the expected quantity of treasure, the answer is given by the greedy algorithm.  The best pirate plunders the richest ship, and so forth.

[b]Edit:[/b]  As stated, shouldn't every pirate plunder the richest ship?

Sorry for the ambiguity.  Let me restate the problem with more numbers.

Suppose there are 4 ships with probabilities 50%, 30%, 15%, and 5% for the first, second, third, and fourth ship respectively of holding treasure.  It is known that exactly 1 of the ships will have the treasure and the other 3 will have nothing valuable.  Furthermore, there are 5 pirates who can each independently select at most 1 of the 4 ships and plunder its treasure with respective probabilities of success 90%, 80%, 60%, 50%, and 30%.  Which ships should each pirate target to plunder in order to minimize the likelihood that the 5 pirates walk away with no treasure?  (ie, the pirates will have no treasure if they decide to plunder the wrong ships or if the ones that plunder the ship with treasure are unsuccessful in plundering it.)

Is it possible generalize with n ships and m pirates?

So would applying the greedy algorithm mean that you send all of the pirates to the ship with the largest chance of having the treasure (the 50% one)?

Correct me if I'm wrong, but the probability of the treasure not being plundered then would be:

$ P_{plunder} \equal{}$ P(ship 1 has treasure)*P(combined ability to plunder ship 1) + P(ship 2 has treasure)*P(combined ability to plunder ship 2) + ...
$ P_{plunder} \equal{} (50\%)*(1\minus{}(1\minus{}90\%)(1\minus{}80\%)(1\minus{}60\%)(1\minus{}50\%)(1\minus{}30\%)) \plus{} (30\%)*(0\%) \plus{} (15\%)*(0\%) \plus{} (5\%)*(0\%) \equal{} (50\%)*(99.72\%) \equal{} 49.86\%$
$ P_{not plunder} \equal{} 1 \minus{} P_{plunder} \equal{} 1 \minus{} 49.86\% \equal{} \boxed{51.14\%}$

But lets say that all pirates go for the 50% ship except the pirate who has an 80% success rate.  The one with 80% success rate decides to plunder the ship that is 30% certain to contain the treasure.

The probability of the treasure still not being plundered from this pirate-ship assignment is:

$ P_{plunder} \equal{}$ P(ship 1 has treasure)*P(combined ability to plunder ship 1) + P(ship 2 has treasure)*P(combined ability to plunder ship 2) + ...
$ P_{plunder} \equal{} (50\%)*(1\minus{}(1\minus{}90\%)(1\minus{}60\%)(1\minus{}50\%)(1\minus{}30\%)) \plus{} (30\%)*(80\%) \plus{} (15\%)*(0\%) \plus{} (5\%)*(0\%) \equal{} (50\%)*(98.60\%) \plus{} (30\%)*(80\%) \equal{} 73.3\%$
$ P_{not plunder} \equal{} 1 \minus{} P_{plunder} \equal{} 1 \minus{} 73.3\% \equal{} \boxed{26.7\%}$

[quote="Tojoyk"]So would applying the greedy algorithm mean that you send all of the pirates to the ship with the largest chance of having the treasure (the 50% one)?

Correct me if I'm wrong, but the probability of the treasure not being plundered then would be:

$ P_{plunder} \equal{}$ P(ship 1 has treasure)*P(combined ability to plunder ship 1) + P(ship 2 has treasure)*P(combined ability to plunder ship 2) + ...
$ P_{plunder} \equal{} (50\%)*(1 \minus{} (1 \minus{} 90\%)(1 \minus{} 80\%)(1 \minus{} 60\%)(1 \minus{} 50\%)(1 \minus{} 30\%)) \plus{} (30\%)*(0\%) \plus{} (15\%)*(0\%) \plus{} (5\%)*(0\%) \equal{} (50\%)*(99.72\%) \equal{} 49.86\%$
$ P_{not plunder} \equal{} 1 \minus{} P_{plunder} \equal{} 1 \minus{} 49.86\% \equal{} \boxed{51.14\%}$

But lets say that all pirates go for the 50% ship except the pirate who has an 80% success rate.  The one with 80% success rate decides to plunder the ship that is 30% certain to contain the treasure.

The probability of the treasure still not being plundered from this pirate-ship assignment is:

$ P_{plunder} \equal{}$ P(ship 1 has treasure)*P(combined ability to plunder ship 1) + P(ship 2 has treasure)*P(combined ability to plunder ship 2) + ...
$ P_{plunder} \equal{} (50\%)*(1 \minus{} (1 \minus{} 90\%)(1 \minus{} 60\%)(1 \minus{} 50\%)(1 \minus{} 30\%)) \plus{} (30\%)*(80\%) \plus{} (15\%)*(0\%) \plus{} (5\%)*(0\%) \equal{} (50\%)*(98.60\%) \plus{} (30\%)*(80\%) \equal{} 73.3\%$
$ P_{not plunder} \equal{} 1 \minus{} P_{plunder} \equal{} 1 \minus{} 73.3\% \equal{} \boxed{26.7\%}$[/quote]

No, each pirate would go to whichever ship had the highest probability of containing the treasure. So if the pirate with 90% success plunders the 50%, then the 50% becomes a 5%.

Although in this case using greedy algorithm needs an order to pick the pirates in. So maybe I should really spend more time trying to solve the problem instead of randomly guessing answers.

[quote="Tojoyk"]That seems like a logical method.  But it seems like a big recursion loop where you evalute, compare, determine next ship, evalute, compare, etc.  Is there a more explicit way to determine the answer?

What if the probabilties of success for every pirate were the same?

ps, did I calculate the probabilties correctly for my scenarios?[/quote]

My method is wrong. We're not trying to find a better answer, we're trying to find an answer at all.

If the probabilities of success for every pirate were the same, then my method would work. Same if all the ships had equal chances to contain treasure.

Your probabilities look correct to me.

Why is your method wrong?  It seems recursive as in a computer program would be able to follow the alogirhm to reach the optimal answer.

As stated, the method would need to run through every permutation of the pirates.  This has the potential to actually be [b]less[/b] efficient than running through every possible assignment of pirates to ships (depending on how many more ships than pirates there are), although both methods are woefully slow.

The problem is a little more sensible when stated in terms of failure rates, since failure rates are multiplicative, so I'll restate it as follows:  there are probabilities $ P_i$ for each ship to have the treasure with $ \sum P_i \equal{} 1$, and there are probabilities $ F_i$ that each pirate fails to plunder the treasure.  Say that ship $ i$ is targeted by pirates to be labeled $ (i, 1), (i, 2), ...$.  Then we want to minimize something like

$ \sum_{i \equal{} 1}^{n} P_i F_{(i, 1)} F_{(i, 2)} ...$

over all assignments of pirates to ships.  The greedy algorithm works fine if at most one pirate is allowed per ship; this is just the rearrangement inequality.  Allowing multiple pirates to plunder a ship introduces a nonlinearity which is difficult to deal with; two pirates with failure probabilities $ F_i, F_j$ plundering a ship is equivalent to a single pirate with failure probability $ F_i F_j$.

It's not even clear to me how to solve this problem if every ship has equal probability of having the treasure.  Then the problem is something like the following: we are given some positive real numbers less than $ 1$ and we want to find the minimum value of an expression made using those numbers, $ \plus{}$, and $ \times$.

How about an easier case when there are 5 pirates who have the each have the same 50% chance of success?  Lets say that there are 3 ships with respective probabiltiies 60%, 30%, and 10% respectively of bearing the treasure.  What should be the pirates' plan to make sure that they mimimize their chances of not getting the treasure?

Correct me if I'm wrong, but following from what I understand with the greedyness, we have 5 steps:

1) We should send the first pirate to the first ship (first ship now has a 30% of holding treasure after this step)
2) We send the second pirate to the first ship again (first ship now has a 15% of holding treasure after this step)
3) third pirate to the second ship (second ship has now a 15% of holding treasure after this step)
4) fouth pirate to the first ship (first ship has now a 7.5% of holding treasure after this step)
5) fifth pirate to the second ship (second ship now as a 7.5% of holding treasure after this step).

After the 5 pirates have tried to plunder the ships, the ships have a 7.5%, 7.5%, and 10% chance respectively of still having the treasure.  Continuing this logic, a 6th pirate would attack the third ship.

I still think that there should be a more constructive way to plan out the attacks.  Like solving some equations where you get (3, 2, 0) as the answer or something?

What about a more general scenario: The pirates will have equal proababilities of success/failure, but each of the ships still had their individual probabilities $ P_i$ to bear the treasure with $ \sum P_i \equal{} 1$.  What would be the attack plan?

[quote="Differ"]If the probabilities of success for every pirate were the same, then my method would work. Same if all the ships had equal chances to contain treasure.[/quote]

I take this statement back. Looking at just two ships with 50% probability leads to trouble.

I was hoping to get some sort of algebraic solution to this problem, but I guess maybe using calculus might be necerssary to optimize the probability?

Alas, it is not even calculus. As stated, it includes the infamous bin packing problem as a partial case. Suppose that there are just 2 ships with probabilities $ P_1$ and $ P_2$ of having the treasure. Suppose we have many pirates and $ f_j$ are the [b]failure[/b] probabilities for them. Then you need to minimize $ P_1\prod_{j\in J_1}f_j\plus{}P_2\prod_{j\in J_2}f_j$ over all partitions of the set of pirates into two subsets $ J_1$ and $ J_2$. The AM-GM inequality gives the lower bound $ 2\sqrt {P_1P_2\prod_j f_j}$ for this quantity. It is attained if and only if you can find a partition satisfying $ P_1\prod_{j\in J_1}f_j\equal{}P_2\prod_{j\in J_2}f_j$. If you think of this equality for a few minutes, you'll realize that every bin packing problem can be restated this way. Pirate's life is not easy (actually, NP-hard), you know... :D

From the general case:

Each of the $ m$ pirates have individual probabilties of [b]failure[/b]: $ F_j$.  (Each pirate can only choose 1 ship to plunder)
Each of the $ n$ ships have individual probabilties of bearing treasure: $ P_i$.
Exactly one of the $ n$ ships has the treasure: $ \sum_{i\equal{} 1}^{n} P_i \equal{} 1$.

Let the group of pirates that will target the first ship be denoted as $ J_1$.  Likewise, the group of pirates that will target the second ship be denoted as $ J_2$, and so on for all of the ships.  I came up with this as my generalized probability equation representing the probability that the treasure will be successfully plundered by the pirates:

$ P \equal{} \sum_{i \equal{} 1}^{n}(P_i*(1 \minus{} \prod_{j\in J_i}F_j))$

However, I have no idea how to minimize this with the $ P_i$ and $ F_j$ as parameters.

While trying out different methods, I did realize that we can realize this equation as the dot product of two $ n$-component vectors, one of them is the vector with respective components $ a \equal{} P_i$ and the other vector is one with respective components $ b \equal{} 1 \minus{} \prod_{j\in J_i}F_j$.  And the question becomes trying to minmize the dot product of $ a$ and $ b$.  But I didn't get too far with this approach either ...

Any help would be appreciated.  Maybe Im approaching this the wrong way?

Have you read my last remark? [b]Nobody[/b] currently knows how to do it in a much better way than considering all possible assignments and just computing the numbers. If you ever come up with one, it'll be quite a revolution in computer science. :roll:

Gotcha, but what about when all of the pirates have equal probabilities?  Does there exist a more structured approach to the problem?

Then it is easy. You need to minimize the sum $ \sum P_i F^{j_i}$ in your notation. WLOG $ P_1$ is the largest of $ P_i$. Note that there should be at least one factor $ F$ on $ P_1$ (otherwise, moving it there from anywhere else will decrease the sum). Thus, the assignment for the first pirate is clear. Once it has been made, you have exactly the same problem as before but with the numbers $ FP_1,P_2,\dots,P_n$. So, again, the next pirate should be assigned to the largest of these numbers and so on. This runs in $ (n\plus{}m)\log n$ time (you need to sort the numbers in the beginning and after each step; the initial sorting costs $ n\log n$ and placing the changed number correctly takes $ \log n$ operations). Probably, you can improve the time to linear, but I do not see how to do this immediately.

Excellent, sounds like a simple program can be written to solve this semi-general case.

Here was my attempt at trying to calculate the pirate-ship distribution:

If each of the $ m$ pirates had an equal probability of failure $ F$ and the $ n$ ships had individual probabilties $ P_j$ of bearing the 1 treasure, then the probability of succesfully collecting the treasure would be:

$ (*) P_{\text{success}} \equal{} P_1*(1 \minus{} F^{r_1}) \plus{} P_2*(1 \minus{} F^{r_2}) \plus{} ... \plus{} P_n*(1 \minus{} F^{r_n}) \equal{} \sum_{i \equal{} 1}^{n}(P_i*(1 \minus{} F^{r_i}))$

where the $ r_j$ is the number of pirates who plan to attack the jth ship.

Since there are $ m$ total ships, we have a constraint that:

$ (**) \sum_{i \equal{} 1}^{n}(r_i) \equal{} m$

Since we're trying to maximize (*), which can be thought of as a function of $ r_n$ ($ n$ total variables), would it be possible to treat the $ P_i$ and $ F$ as constants (they will be constants in less general cases) and apply Lagrange Multipliers [(*) is minimized and (**) is constaint equation]?  I haven't actually tried this out yet, but if it works, we'll probably get non-integer answers for the $ r_n$, so we can just search integer solutions nearby to get the answer that we want?

Please let me know if my idea will be an utter failure before I actually try this out.

I do not see how it is simpler to implement than the algorithm I suggested (especially the last step: search for the integer solutions near the real solution). Also, if you've read my answer to your post in the Calculus forum, you might notice that the real-valued problem has not really much to do with Lagrange multipliers and requires sorting of $ P_i$ anyway. ;) So, unless $ m\gg n$, I wouldn't bother with trying to improve the running time (not in your way, at least). If $ m\gg n$, then we can try to think of something better. What are the actual values you want to work with?

In the case when the success probabilities are different, we cannot hope to get the truly optimal solution, as I explained, but we can try to get reasonably close to it. Again, it would help to have some idea of the actual numbers involved.

Ok thanks for the info.  I was just wondering if there were a more explicit way to calculate the solutions.

So how would your algorithm work if we had 10 pirates each with 40% chance of failure and 3 ships (70%, 20%, 10%) respectively?  How would we distribute the 10 pirates among the 3 ships to minimize chance of treasure leakage?

As I said, we can do better if $ m$ is much larger than $ n$ but for $ m$ smaller than or comparable to $ n$ this runs reasonably fast.

In your particular example, we'll have the following steps (at each step you replace the largest of the 3 numbers you have by 0.4 times it and add 1 pirate to the corresponding ship):

0) P: 70,20,10 assignments: 0,0,0
1) P: 28,20,10 assignments: 1,0,0
2) P: 11.2,20,10 assignments: 2,0,0
3) P: 11.2,8,10 assignments: 2,1,0
4) P: 4.48,8,10 assignments: 3,1,0
5) P: 4.48,8,4 assignments: 3,1,1
6) P: 4.48,3.2,4 assignments: 3,2,1
7) P: 1.792,3.2,4 assignments: 4,2,1
8) P: 1.792,3.2,1.6 assignments: 4,2,2
9) P: 1.792,1.28,1.6 assignments: 4,3,2
10) P: 0.7168,1.28,1.6 assignments: 5,3,2

with final probability of failure 0.7168+1.28+1.6=3.5968%

Excellent!  I think i can make a little program out of this.  Although I do agree that if we had say 500 pirates and the same three ship (70, 20, 10) configuration, a more explicit optimization approach might be quicker.


If you know how to play Nim, you don't need to click [hide="this."]  9 coins are set up in 3 rows the top one has 1 coin, the next has 3, and the last has 5.  On your turn, you may remove any number of coins from any [b]one[/b] row.  The player who removes the last coin loses. [/hide]  There have been many game theory problems based on Nim and variations of it.

[b]Question 1:[/b] It is possible to always win Nim.  To always win, should you go first or second?  And what algorithm do you use to choose your moves?

[hide="Variation 1: Extreme Nim"]  The gameplay is the same as original Nim, but the first row starts with 3 coins, the second with 5, and the last with 7.[/hide]

[b]Question 2:[/b] Is it possible to always win Extreme Nim?

More problems coming later...

[hide="Variation 2: Nim-25"]
Nim-25 is played with a 5x5 board of coins.  On your turn, you may remove up to 4 coins from any row or column.  The player that is forced to remove the last coin loses.
[/hide]
[b]Question 3:[/b] Is it possible to always win Nim-25?[/b]

Is the answer to #1,

[hide]remove 3 from the 5 coin pile so that the coins are 1,3,2 after this step, no matter what your opponent does, you can reduce the coins to 1,1,1 or 2,2 or just 1, where you win[/hide]

I have to write a paper on game theory, which i know little about (at the moment) and i stumbled upon this

[quote="Yagmoth0000"][hide]remove 3 from the 5 coin pile so that the coins are 1,3,2 after this step, no matter what your opponent does, you can reduce the coins to 1,1,1 or 2,2 or just 1, where you win[/hide][/quote]

Yeah, that's right.

There's a variation of NIM: Pearls before swine.

link:  http://www.transience.com.au/pearl3.html

Levels 2 and beyond are all NIM with more pearls. Level 1 is pointless once you figure out an algorithm.

[quote="AstroPhys"]I have a number theory book that explain how to win this with any amount of coins in any amount of columns, though it may not work with Nim-25.
The explaination they provide invovle looking at each piles' number of coins as binary.
I can post it if you want.[/quote]I have seen the solution involving binary before, but I forget it.  could you post it?

This works best if you go first, but it should still work if you go second.
First you put the number of coins of each pile into sums of powers of two.
Then you attempt to "balance" the piles so that each power of two is in a pair.

For example
You have three piles with $17 \ , \ 28 \ , \ 30$
In sums $17 = 16+1$ , $30 = 16+8+4+2$ , $28 = 16+8+4$
Then to balance, you would take you take $15$ from the pile with $17$.
This leaves you with pairs for each of the remaining powers of two.
$2 \ , \ 16+8+4+2 \ , \ 16+8+4$.
The pattern continues until there are no more coins.

Now to generalize this method, you would just need to make more than one pair of each for multiple piles.
If modified it may work for Nim-25.

@ Astrophys...

What do you mean by "Now to generalize this method, you would just need to make more than one pair of each for multiple piles. "

and also, isn't it not possible to do the powers of 2 pairs in some cases?[/quote]

By generalizing, I meant to extend the method to games of Nim with more than three piles, then having more than one pair per power of two.

According to my book, the method is applicable to all case regardless of the number of coins per pile.

Indeed, it is. The method is actually write all the numbers in binary and bitwise xor them together (so 0+0=0, 0+1=1, 1+0=1, and 1+1=0. This is the same as adding without carrying). You want to move into a position where this "nim-sum" is 0.

Example: Say the piles are
45, 32, 16, and 55

101101 = 45
100000 = 32
010000 = 16
110111 = 55
-------------------
101010

So, we need to toggle 101010 on some of them. The winning moves are:
101101 -> 000111 = 45 -> 7
100000 -> 001010 = 32 -> 10
110111 -> 011101 = 55 -> 29

Also, technically, this game should be called misere nim. In actual nim, you want to take the last stone, as the person who can't move loses.


function trigonometry algebra polynomial inequalities

write cos5 theda as a function of cos theda

[quote="Ubemaya"]Do you mean $\cos^{5}\theta$ or $\cos 5\theta$?[/quote]
probably(wow took me 5 tries to type this right) the latter.
[hide]Consider $(\cos x+i\sin x)^{5}.$ Now expand that thing using binomial thrm.That thing is equal to $\cos5x+i\sin5x$ by DeMoivre's thrm. Set the real  parts equal to each other, and boom, we're done.

oops i used $x$ instead of $\theta$[/hide]

[hide="partial solution"]From DeMoivre, $\cos5\theta+i\sin5\theta=(\cos\theta+i\sin\theta)^{5}$
$=\cos^{5}\theta+i5\cos^{4}\theta\sin\theta-10\cos^{3}\theta\sin^{2}\theta-i10\cos^{2}\theta\sin^{3}\theta+5\cos\theta\sin^{4}\theta+i\sin^{5}\theta$.
Equating real parts, $\cos5\theta=\cos^{5}\theta-10cos^{3}\theta\sin^{2}\theta+5\cos\theta\sin^{4}\theta$.
$=\cos^{5}\theta-10\cos^{3}(1-\cos^{2}\theta)+5\cos\theta(1-\cos^{2}\theta)^{2}$
I don't fell like doing the rest...[/hide]

[quote="bigboy82892"][hide="Hint."]Use $\cos{4\theta+\theta}$, $\cos{2(2\theta)}$, and $\cos2\theta$. :D [/hide][/quote]

Well, it's easier to used DeMoivre's theorem, since what if I asked you to find $\cos{783\theta}$?

DeMoivre's isn't that easy, since you have annoying $(1-\cos^{2}x)^{k}$ expressions to expand.  It would become quite tedious if I asked you to find, say, $\cos 10x$ as a polynomial in $\cos x$.

[hide="I would try this instead..."] Let $T_{n}(\cos \theta) = \cos n \theta$ be the Chebyshev Polynomials of the First Kind.  then

$T_{0}(x) = 1$
$T_{1}(x) = x$
$T_{n+1}(x) = 2x T_{n}(x)-T_{n-1}(x)$  (good lemma!  Can you prove it? ;) )

From here you can quickly calculate that

$T_{5}(x) = 16x^{5}-20x^{3}+5x$ [/hide]


If $n$ is a sum of squares, prove that $2n$ is also a sum of squares.

What a well stated problem  :roll:

If $n$ is sum of squares, then $2n= \underbrace{1^{2}+1^{2}+...+1^{2}}_{2n \text{ times}}$ is so, too ;)

You meant sum of two squares, of course.

[hide="Generalization"] $(a^{2}+b^{2})(c^{2}+d^{2}) = (ac \pm bd)^{2}+(ad \mp bc)^{2}$

This problem is the special case $c = d = 1$. [/hide]


Find all functions from reals to reals such that:
$ f(x\plus{}f(y)) \equal{} x \plus{} f(f(y))$ and $ f(2004) \equal{} 2005$
[hide="my approach"]
The way I solved this was that I showed that if $ y\equal{}2004$ then you see that $ f(x\plus{}2005) \equal{} x \plus{} f(2005)$ so I thought that the answer might be all functions of the form $ f(x) \equal{} x \plus{} k$ where k is some real, but how do I know that I have found all possible functions? Is there a general way to solve functional equations??? Like, what methods should I know?
[/hide]
Thanks.

Let $ g(x)\equal{}f(x)\minus{}x$ then from $ f(x\plus{}2005)\equal{}x\plus{}f(2005)$ we have $ g(x\plus{}2005)\equal{}g(2005)$, it mean $ g(x)\equal{}c$, in which $ c\equal{}g(2005)$: const!
Then OK!

[quote="Nawahd Werdna"]Find all functions from reals to reals such that:
$ f(x \plus{} f(y)) \equal{} x \plus{} f(f(y))$ and $ f(2004) \equal{} 2005$
[hide="my approach"]
The way I solved this was that I showed that if $ y \equal{} 2004$ then you see that $ f(x \plus{} 2005) \equal{} x \plus{} f(2005)$ so I thought that the answer might be all functions of the form $ f(x) \equal{} x \plus{} k$ where k is some real, but how do I know that I have found all possible functions? Is there a general way to solve functional equations??? Like, what methods should I know?
[/hide]
Thanks.[/quote]
Take $ y\equal{}0$ easy to check :$ f(x)\equal{}x\plus{}a$
From $ f(2004)\equal{}2005$ then $ a\equal{}1$
$ f(x)\equal{}x\plus{}1$

let $ y\equal{}2004$,we get that:

$ f(x\plus{}2005)\equal{}x\plus{}f(2005)$ (*)

now in (*) set $ x\equal{}\minus{}1$,we get that:

$ f(2004)\equal{}\minus{}1\plus{}f(2005)\Rightarrow f(2005)\equal{}2006$

$ (*)\Rightarrow f(\underbrace{x\plus{}2005}_{t})\equal{}x\plus{}2006$

$ \Rightarrow\boxed{f(t)\equal{}t\plus{}1}$

------------------------------------------------
another solution is the following one:

let $ x\equal{}\minus{}f(f(y))$ we get that:

$ f\left(\underbrace{f(y)\minus{}f(f(y))}_{t}\right)\equal{}0$

$ \Rightarrow f(t)\equal{}0$

now let $ y\equal{}t$ we get that:

$ f(x)\equal{}x\plus{}f(0)$

now let $ x\equal{}2004$ we get that:

$ f(2004)\equal{}2004\plus{}f(0)\Rightarrow f(0)\equal{}1$

$ \Rightarrow \boxed{f(x)\equal{}x\plus{}1}$


Me too! But I think this problem also sloved by using
Lagrange Interpolating Polynomial
http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html
Because $\deg{f}=2003$ and we know $f(1),f(2),...,f(2004)$.


Show that there doesn't exist natural numbers $ k$ and $ m$ that satisfy
\[ k! \plus{} 48 \equal{} 48(k \plus{} 1)^m
\]

The question $ \Leftrightarrow 15 \times \frac {k!}{6!} \plus{} 1 \equal{} (k \plus{} 1)^m$ , so $ k \plus{} 1$ coprime with any of the $ n!$, where $ 7 \le k \plus{} 1 \le k$.
Case 1:$ k \plus{} 1$ is not prime. It's easy to see that for $ k \ge 10$ it is impossible, and it's impossible for $ 6 \le k \le 9$ too after checking.
Case 2:$ k \plus{} 1$ is a prime. Then we have $ k! \plus{} 48 \equal{} \minus{} 1 \plus{} 48 \equal{} 47 \equal{} 0(mod \;\; k \plus{} 1)$ by Wilson, so $ k \plus{} 1 \equal{} 47$ , $ 46! \plus{} 48 \equal{} 48(47)^m$, which is impossible too.

Done.

Please word your solution more clearly; I think you confused some variables.
[hide="solution"]
\[ (k\minus{}1)!\equal{}48((k\plus{}1)^{m\minus{}1}\plus{}...\plus{}1)\]
If $ k\plus{}1$ is prime then by Wilson's, $ 1\equiv 48 \mod{k\plus{}1}$ so $ k\equal{}46$; now, 46!+48=48(47)^m. But 47 is not a wilson prime.
If $ k\plus{}1$ is not prime then $ 0\equiv 48\mod{k}$, a contradiction.
[/hide]


The number of diagonals of a regular polygon of 2n sides, not through the center of the circle to the polygon is given by:

a) 2n(n-3)
b)2n(n-2)
c) 2n(n-1)
d) [n(n-5)]/2

[hide="Solution"]The total number of diagonals a regular 2n sided polygon has is $ \frac {2n(2n \minus{} 3)}{2} \equal{} n(2n \minus{} 3) \equal{} 2n^2 \minus{} 3n$.  The diagonals that cross through the center of the circle to the polygon (I assume circumcenter?) are the ones that have endpoints of verticies that are diametrically opposite each other.  Each vertex (in a regular polygon with even side lengths) has another vertex that is diametrically opposite it.  So it seems like we have $ 2n$ diagonals that cross the circumcenter.  However, we over counted by a factor of $ 2$.  Thus, we want to subtract out (from $ 2n^2\minus{}3n$) $ \frac {2n}{2} \equal{} n$.  Thus, our answer is $ 2n^2 \minus{} 3n \minus{} n \equal{} 2n^2 \minus{} 4n \equal{} 2n(n \minus{} 2)\implies \boxed{\textbf{b)}}$[/hide]

$ \underbrace{\left(\binom{2n}{2}\minus{}2n\right)}_{\text{diagonals of a regular polygon of} \ 2n \ \text{sides}}\minus{}\underbrace{n}_{\text{not through the center of the circle to the polygon}}$
$ \equal{} \boxed{2n(n\minus{}2)}$


Let   A = matrix{[1, 0, 1];[0, 1, 2]; [0, 0, -1])

Find all matirx B of size 3x3 such that AB+BA = 0.

<i>Source: VUMC 2001</i>

Namdung

well if u put
B=
a b c
d e f
g h i

and solve the sistem you get B.
a lot of computation, but I think it works!

Here the  system like Lagrangia suggest us

2a+g=0
2b+h=0
i+a+2b=0
2d+2g=0
2e+2h=0
2i+d+2e=0
-2i+g+2h=0

Now your turn solve it

There is a non-computational sln:

Note that  A^2 = 1. If  AB+BA=0 then B = - ABA, so B = X - AXA, where X = (1/2)B. Conversely, if B = X - AXA then AB + BA = AX - XA + XA - AX = 0.

Thus, we can conclude all B we have to find have form X - AXA where X - arbitrary matrix order 3x3.

Namdung

[quote="Namdung"]
.....
Note that  A^2 = 1. If  AB+BA=0 then B = - ABA, so B = X - AXA, where X = (1/2)B.....

Namdung[/quote]

With your idea B is not only of the form X-AXA

From B=-ABA =>
B = B/q - (q-1)ABA/q  q any integer > 1
 => B=X-(q-1)AXA with X=B/q

Conversely if B=X-(q-1)AXA
AB+BA = AX-(q-1)XA+XA-(q-1)AX=2XA+2AX-qXA-qAX=
2/qBA+2/qAB-AB-BA=(2/q-1)(AB+BA)

we  get (2/q-2)(AB+BA)=0 => AB+BA=0

I solved the system here the solution

(a  b  c)
((2a) (2b) f)
((-2a) (-2b) (-a-2b)) where a,b,c,f are reals


[quote="Beat"]Find all natural numbers $n$, for which the inequality $a(a+b)^{n}+b(b+c)^{n}+c(c+a)^{n}\ge 0$ holds for all real numbers $a,b,c$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55723

how to prove it for n=3, and that there are no other n different from 1,3,5 that this inequality is true
?

[quote="Beat"]how to prove it for n=3...
?[/quote]
The same idea helps:
Let's $a+b=z,$ $a+c=y$ and $b+c=x.$
Hence, $\sum_{cyc}a(a+b)^{3}\geq0\Leftrightarrow\sum_{cyc}(x+z-y)x^{3}\geq0\Leftrightarrow\sum_{cyc}(x^{4}-x^{3}y+xy^{3})\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left((2x+y)^{2}(4x^{2}-12xy+11y^{2})+5x^{2}\right)\geq0.$
[quote="Beat"]... and that there are no other n different from 1,3,5 that this inequality is true?[/quote]
For $n\geq7,$ $n\in2\mathbb N+1$ try $a=1,$ $b=-2n$ and $c=n.$ :wink:


Let $PB$, $PC$ are tangents from $P$ to circle $(O)$  ($B,C$ in $(O)$).
Suppose $PO$ meets $(O)$ at $A$ and $J$ ($J$ lies on the segment $PA$). $CO$ meets $(O)$ at $E$, the bisector of angle $ACO$ meets $AC$ and $(O)$ at $Q$ and $D$ respectively. $AE$ meets $PQ$ at $X$.
Prove that $XA=XD$.

[i]Edited by Myth[/i]

[quote="Herodatviet"]  bisector of angle $ACO$ meets $AC$ and $(O)$ at $Q$ and $D$
[/quote]
 :?:

[quote="Herodatviet"]Let $PB$, $PC$ are tangents from $P$ to circle $(O)$  ($B,C$ in $(O)$). Suppose $PO$ meets $(O)$ at $A$ and $J$ ($J$ lies on the segment $PA$). $CO$ meets $(O)$ at $E$, the bisector of angle $ACO$ meets $AC$ and $(O)$ at $Q$ and $D$ respectively, $AE$ meets $PQ$ at $X$.
Prove that $XA=XD$.[/quote]

Are you sure you don't have some typos here? Anyway, I could imagine that you wanted to say "the bisector of angle $ACO$ meets $OP$ and $(O)$ at $D$ and $Q$ respectively" instead of "the bisector of angle $ACO$ meets $AC$ and $(O)$ at $Q$ and $D$ respectively"; this would make the problem, at least, look right on an imprecise sketch, but zooming in shows that the problem is still wrong in this case.

  darij

[quote="Herodatviet"]Im sorry it  is :'bisector of $ABO$ meets $AC$,$(O)$ $Q,D$ '[/quote]

Still wrong (the segments XA and XD are [i]almost[/i] equal).  :(

  darij


For what value of the constant $ a$ does the system of equations below have infinitely many solutions?   \begin{align*} 3x + 2y &= 8,\\ 6x &= 2a - 7 - 4y \end{align*}

[quote="GameBot"]For what value of the constant $ a$ does the system of equations below have infinitely many solutions?
\begin{align*} 3x + 2y & = 8, \\
6x & = 2a - 7 - 4y \end{align*}
[/quote]

Infinitely many solutions simplifies to $ 0 = 0$.
Bring all variables to the left:
$ 3x+2y=8$
$ 6x-2a+4y=-7$

Multiply first eq by 2 as you would to solve for a
$ 6x+4y=16$
Subtract the second eq from first, you get
$ 2a=23$
$ a=\boxed{11.5}$


modular arithmetic
number theory
prime numbers

Hello. I recently looked at CMO #4 and tried to solve it. I was wondering if you guys could give me tips for writing (also I am kind of happy because even though it is sad that this is one of the hardest problems that I have solved, I solved it :) ). I hope that my solution is correct, however I have more faith in the AOPS community than anything else to help me check it over. I will conceal my solution for those who want to try the problem.

[b]Problem[/b]

Prove that for every prime number $ p$, there are infinitely many positive integers $ n$ such that $ p$ divides $ 2^{n}\minus{}n$.

[hide] Firstly, I looked for a solution in terms of $ p$ with the knowledge that $ p>1$. Using Euler's Totient Theorem, we get the following.

$ 2^{\varphi(p)} \equiv 1 \pmod{p}$

Since there exist $ p\minus{}1$ numbers $ k$ such that $ 0<k<p$ such that $ gcd(p,k)\equal{}1$ when $ p$ is prime, $ \varphi(p)\equal{}p\minus{}1$. This gives the following.

$ 2^{p\minus{}1} \equiv 1 \pmod{p}$

Raising both sides to the power of $ (p\minus{}1)^{2k\minus{}1}$, where $ k$ is an integer which satisfies that $ k\ge 1$ gives the following. Note that since $ p>1$, $ (p\minus{}1)^{2k\minus{}1}>0$.

$ 2^{(p\minus{}1)^{2k}} \equiv 1^{(p\minus{}1)^{2k\minus{}1}} \pmod{p}$

$ 2^{(p\minus{}1)^{2k}} \equiv 0\plus{}1 \pmod{p}$

Now lets examine $ (p\minus{}1)^{2k}\minus{}1$ in $ \pmod{p}$.

$ (p\minus{}1)^{2k}\minus{}1 \equiv (\minus{}1)^{2k}\minus{}1 \equiv 1\minus{}1 \equiv 0 \pmod{p}$

Now we can substitute, yielding that,

$ 2^{(p\minus{}1)^{2k}} \equiv (p\minus{}1)^{2k}\minus{}1\plus{}1 \equiv (p\minus{}1)^{2k} \pmod{p}$

Rearranging yields the following, that

$ 2^{(p\minus{}1)^{2k}}\minus{}(p\minus{}1)^{2k} \equiv 0 \pmod{p}$

Therefore for all prime numbers $ p$ and positive integers $ k$, there will exist infinite solutions for $ n$ such that $ p|(2^{n}\minus{}n)$ that take the form $ n\equal{}(p\minus{}1)^{2k}$. [/hide]

Would anyone mind giving feedback on my solution, or providing a more legitimate solution.

I'd be glad to check it for you, but I don't know some of the concepts involved in your solution (like Euler's totient).  I think posting it in Pre-Olympiad might help.


[quote="GameBot"]A jar has 8 red, 9 blue and 10 green marbles. What is the least
number of marbles you can remove from the jar to guarantee
that you have three of the same color?[/quote]

The worst case scenario is 2 red,blue,green. We need to pick 1 more, so our answer is 7.


Sequence $ \{ f_n(a) \}$ satisfies $ \displaystyle f_{n\plus{}1}(a) \equal{} 2 \minus{} \frac{a}{f_n(a)}$, $ f_1(a) \equal{} 2$, $ n\equal{}1,2, \cdots$. If there exists a natural number $ n$, such that $ f_{n\plus{}k}(a) \equal{} f_{k}(a), k\equal{}1,2, \cdots$, then we call the non-zero real $ a$ a $ \textbf{periodic point}$ of $ f_n(a)$.

Prove that the sufficient and necessary condition for $ a$ being a $ \textbf{periodic point}$ of $ f_n(a)$ is $ p_n(a\minus{}1)\equal{}0$, where $ \displaystyle p_n(x)\equal{}\sum_{k\equal{}0}^{\left[ \frac{n\minus{}1}{2} \right]} (\minus{}1)^k C_n^{2k\plus{}1}x^k$, here we define $ \displaystyle \frac{a}{0}\equal{} \infty$ and $ \displaystyle \frac{a}{\infty} \equal{} 0$.

I think the narration must be wrong! You mean that Fn+1=2,then Fn=$\infty$. What's that?

I don't think that def is needed... some other one!


conics
parabola
analytic geometry
function

The equation y = (x - c)^2 defines a family of parabolas, one parabola for each value of c.  On one set of coordinate axes, graph the memebers of the family for c = 0, c = 3, c = -2.

[hide="Hint"] $f(x-c)$ is $f(x)$ shifted $c$ units right. $f(x+c)$ is $f(x)$ shifted $c$ units left. [/hide]

Thanks for the hint but can you show me how to do this type of question?

Your hint does not lead me on the right track to the answer.

to shift the parabola $y=x^{2}$ to the right, say 5 units, we use the function $y=(x-5)^{2}$. This is because if by shifting the graph to the right 5 units, putting in the value of x+5 in the shifted graph would be the same as putting in the value of x in the original. We see this works because $y=(x+5-5)^{2}=x^{2}$.

Thus shifting the parabola to the right $c$ units, we get the function $y=(x-c)^{2}$

tjhance...thanks for the tips and extra reply.


A man realizes that the product of his two sons' ages is his age and that the product of his daughters' ages is his wife's age.  He also realizes that the sum of his kids' ages will surpass the sum of him and his wife's ages in $25$ years.  He also realizes that $n$ years ago, the product of his kids' ages was less than his own age.  Compute the smallest possible $n$.

how old does the mom need to be to have children?...if its any age I get..[b]2[/b]..sounds weird.. :maybe:

When the children are ages (right now) 28,2,6,6 and the dad is 56 and mom is 36?

Im probalby wrong but I got this equaition... (a-1)(b-1)+(c-1)(d-1)=52

[hide="My original setup was"]
Man: 40
Woman: 35
Sons: 4, 10
Daughters: 5, 7
[/hide]

Note that I said the sum will SURPASS, not EQUAL.  They are very different.

Actually, it doesnt make that big of a difference

My equation would just be changed to (a-1)(b-1)+(c-1)(d-1)<52..in your case you set (a-1)(b-1)+(c-1)(d-1)=51..

which would your answer faulty because I can easily find another case:

for example-
dad-48
mom-38
boys-4,12
girls-2,19

that contradicts your answer..and still get [b]2[/b]

A better question would be to ask the largest possible value of n

Okay, then find the largeset possible $n$.


function
calculus
integration
real analysis
real analysis unsolved

f(x) is measurable on (0,1). And if we know that
f(x)-f(y) is integrable on [0,]x[0,1].

Prove: f(x) is integrable (that is f in L[0,1]).

(I am supposed to use Fubini?)

In essence, yes. What does it mean for a function $g(x,y)$ to be integrable on $[0,1]^{2}$? By Fubini, the iterated integral $\int_{o}^{1}\int_{0}^{1}g(x,y)\,dx\,dy$ exists (and is equal to the integral of $g$). In particular, $\int_{0}^{1}g(x,y)\,dx$ exists and is finite for almost all $y$, since it is an integrable function of $y$.

Applying the above to $g(x,y)=f(x)-f(y)$, $\int_{0}^{1}f(x)-f(y)\,dx=\int_{0}^{1}f(x)\,dx-f(y)$ exists and is finite for almost all $y$. For this to be true, $f$ must be integrable.

[quote="jmerry"]$\int_{0}^{1}f(x)-f(y)\,dx=\int_{0}^{1}f(x)\,dx-f(y)$ exists and is finite for almost all $y$. [/quote]
I finally work out a not straitforward pf:
Actually we don't know:
$\int_{0}^{1}f(x)-f(y)\,dx=\int_{0}^{1}f(x)\,dx-f(y)$
linearity true for integrable functions, but we haven't prove f is integrable yet.
Couter example: f=1/x, g=1/x,
$\int_{0}^{1}f(x)-g(x)\,dx=\int_{0}^{1}f(x)\,dx-\int_{0}^{1}g(x)\,dx$
is wrong.

But when C is a finite contant, and $\int_{0}^{1}f(x)\,dx$ exists,
$\int_{0}^{1}f(x)-C\,dx=\int_{0}^{1}f(x)\,dx-C$ is true.
So need some extra work...

You've given the argument already: $f(y)$ is a constant.


[quote="arqady"]1)Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$ Prove that:
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [11]{\frac {a^3b \plus{} a^3c \plus{} b^3a \plus{} b^3c \plus{} c^3a \plus{} c^3b}{6}}\]
[/quote]
It's very easy with Am-Gm, arqady. I'll post my proof later  :)

The second is obviously trues because: $ \sum\ ab(a^2\plus{}b^2) \plus{} 8abc \leq\ 14$
With $ a\plus{}b\plus{}c\equal{}3$  :maybe:

[quote="nguoivn"]The second is obviously trues because: $ \sum\ ab(a^2 \plus{} b^2) \plus{} 8abc \leq\ 14$
With $ a \plus{} b \plus{} c \equal{} 3$  :maybe:[/quote]
Very nice proof! :lol:

[quote="arqady"]1)Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$ Prove that:
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [11]{\frac {a^3b \plus{} a^3c \plus{} b^3a \plus{} b^3c \plus{} c^3a \plus{} c^3b}{6}}\]
2)Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that:
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [8]{\frac {a^3b \plus{} a^3c \plus{} b^3a \plus{} b^3c \plus{} c^3a \plus{} c^3b}{6}}\]
[/quote]
For the first,we can use this way.
$ a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b) \equal{} 3p^2 \minus{} 18 \minus{} pr \ge 3p^2 \minus{} 18 \minus{} \frac {p^2(12 \minus{} p^2)}{9}$
Finally,we need to prove that:
$ f(x) \equal{} \frac {x^{11}}{3^{11}} \minus{} 3p^2 \minus{} 18 \minus{} \frac {p^2(12 \minus{} p^2)}{9} \ge 0$
We have f’’’(x) >0
And f’(x) $ \ge f(3) \equal{} 0$
And$ f(x) \ge f(3) \equal{} 0$
Our proof are completed.Equality occur if and only a=b=c.

[quote="arqady"]1)Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$ Prove that:
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [11]{\frac {a^3b \plus{} a^3c \plus{} b^3a \plus{} b^3c \plus{} c^3a \plus{} c^3b}{6}}\]
[/quote]
Proof of weaker Inequality

Notice the identity, then use the Am-Gm Inequality and $ 2\sum a^4 \geq \sum a^3b \plus{} \sum a^3c$:

$ (a \plus{} b \plus{} c)^4 \equal{} 4(a^3b \plus{} a^3c \plus{} b^3a \plus{} b^3c \plus{} c^3a \plus{} c^3b) \plus{} a^4 \plus{} b^4 \plus{} c^4 \plus{} 6(ab \plus{} bc \plus{} ca)^2$

$ \equal{} 24\left(\frac {a^3b \plus{} a^3c \plus{} b^3a \plus{} b^3c \plus{} c^3a \plus{} c^3b}{ 6}\right) \plus{} 3\left(\frac {a^4 \plus{} b^4 \plus{} c^4}{3}\right) \plus{} 54$

$ \geq 81 \sqrt [81]{\frac {(a^3b \plus{} a^3c \plus{} b^3a \plus{} b^3c \plus{} c^3a \plus{} c^3b)^{24}(a^4 \plus{} b^4 \plus{} c^4)^3}{6^{24} \cdot 3^3}}$

$ \geq 81 \sqrt [81]{\frac {(a^3b \plus{} a^3c \plus{} b^3a \plus{} b^3c \plus{} c^3a \plus{} c^3b)^{27}}{6^{27}}}$

$ \equal{} 81\sqrt [3]{\frac {a^3b \plus{} a^3c \plus{} b^3a \plus{} b^3c \plus{} c^3a \plus{} c^3b}{6}}$

So, we have proved that:
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [12]{\frac {a^3b \plus{} a^3c \plus{} b^3a \plus{} b^3c \plus{} c^3a \plus{} c^3b}{6}}\]
:)

By the way, we can prove the following stronger inequality too: (with the same condition: $ ab \plus{} bc \plus{} ca \equal{} 3$)
\[ \frac {a \plus{} b \plus{} c}{3}\geq\sqrt [10.8]{\frac {a^{3}b \plus{} a^{3}c \plus{} b^{3}a \plus{} b^{3}c \plus{} c^{3}a \plus{} c^{3}b}{6}}\]
by using the same method above, but instead of using $ 2\sum a^4 \geq \sum a^3b \plus{} \sum a^3c$, we will use
\[ a^4 \plus{} b^4 \plus{} c^4 \plus{} \frac{(ab \plus{} bc \plus{} ca)^2}{3} \geq a^{3}b \plus{} a^{3}c \plus{} b^{3}a \plus{} b^{3}c \plus{} c^{3}a \plus{} c^{3}\]
which directly follows from schur's inequality for degree 4.  :)


projective geometry
geometry unsolved
geometry

Let $S_1$ and $S_2$ be two circumferences, with centers $O_1$ and $O_2$ respectively, and secants on $M$ and $N$. The line
$t$ is the common tangent to  $S_1$ and $S_2$ closer to $M$. The points $A$ and $B$ are the intersection points of $t$ with $S_1$ and $S_2$, $C$ is the point such that $BC$ is a diameter of $S_2$, and $D$ the intersection point of the line $O_1O_2$ with the perpendicular line to $AM$ through $B$. Show that $M$, $D$ and $C$ are collinear.

The problem is not clear to me. Should not the secant MN pass through the intersection H of the tangent t with the center line $O_1O_2$ and should not the points N, D, C be collinear? :?

Assuming this is so, let the secant MN meet the circle $S_2$ again at P and let the normal to AM through B meet the circle $S_2$ again at Q. Because of similarity of the circles $S_1, S_2$ with the similarity center H, $BP \parallel AM$, hence, $BP \perp BQ$, so that PQ is another diameter of the circle $S_2$. The diameters BC, PQ of $S_2$ meet at its center $O_2$. Consider the cyclic pentagon BPCQN as a degenerate cyclic hexagon B'BPCQN with $B' \equiv B$, so that $BB' \equiv t$. By Pascal's theorem, the intersections $H = BB' \cap NP \equiv t \cap MN,$ $O_2 \equiv BC \cap PQ,$ $D' \equiv NC \cap B'Q \equiv NC \cap BQ$ are collinear, i.e., $D' \in HO_2$. But since $HO_2 \equiv O_1O_2,$ we have $D' \equiv O_1O_2 \cap BQ \equiv D,$ the points $D \equiv D'$ are identical and N, D, C are collinear.

I read the original problem and it states that $M$ and $N$ are the intersection points of $S_{1}$ and $S_{2}$.

Assuming M, N are intersections of the circles $\mathcal S_{1},\ \mathcal S_{2},$ let $H \equiv AB \cap O_{1}O_{2}$ be their external similarity center. Let a parallel to AM through B meet $\mathcal S_{2}$ again at P, PM passes through H. Let the perpendicular from B to AM meet $\mathcal S_{2}$ again at Q. $\angle PBQ$ is right, PQ is a diameter of $\mathcal S_{2}.$ By Pascal's theorem for the degenerate hexagon BBPCQM, the intersections $H \equiv BB \cap PM,$ $O_{2}\equiv BC \cap PQ,$ $D \equiv BQ \cap CM$ are collinear, $D \in HO_{2}\equiv O_{1}O_{2}.$


geometry
trigonometry
trig identities
Law of Cosines
complex numbers

Im terrible at geometry in math contests.  Does anyone have any recommendations for some good geometry books that will help for math contests?

Thanks.

[u]Challenging Problems in Geometry[/u], by Alfred S. Posamentier and Charles T. Salkind.  Published by Dover.  Not all useful for competitions, but much of it is.

I had that book for 4 years now but i've never got thru the first 10 questions in chapter 1.  in fact, i distinctly remember i stopped at #11 which i didnt know how to do

[quote="cats..."]does that mean you didn't like it?  I read the reviews on amazon and it seemed that they thought it was good.

I bought geometry revisited today, too.  Did any of you go through it?  What did you think?[/quote]

i got that a few days ago and so far it seems pretty good, havent been able to read past section 1.2.

[quote="cats..."]I bought geometry revisited today, too.  Did any of you go through it?  What did you think?[/quote]

yeah, i went through the whole thing. i don't know how much i'll remember, or how much it'll help on competitions, but i liked it.

well, thats fine too, so long as it's interesting.

Yes, Geometry Revisited has an excellent reputation.

[i]Geometry Revisited[/i] is a terrific book, but it's rather hard. I don't know of a book that's good but less challenging unfortunately.

well revisited is not really "hard" persay, but it is somewhat interesting... the volume of questions is hardly enough to be used in a competition.  it mainly has proofs for classic theorems... and a section on projective/inversive geometry.  challenging problems in geo is better for contest preparation...  and i like the book, just didnt get around in working on the problems ;)
if you want to choose between the two, i'd definately go with challenging

If someone can learn how to win math contests (or at least the geometry parts) by reading [i]Elements[/i], I will be extremely impressed.

the "problems" book, does it contain contest problems? or is it mostly problems by the authors?

By the authors.  Maybe half of it would be competition-type stuff -- for example, it covers law of cosines, stewarts theorem, etc.  It also has other cool stuff in it too, like Simpson lines and things of interest.  It's all by the authors.  It's largely proof-based, but there are plenty of "compute" questions.

[quote="cats..."]Im terrible at geometry in math contests.  Does anyone have any recommendations for some good geometry books that will help for math contests?[/quote]

I agree with the other books already mentioned (and have them at home). The other book I would recommend is [url=http://www.amazon.com/exec/obidos/tg/detail/-/0883855100/learninfreed]Complex Numbers and Geometry[/url] by Liang-shin Hahn. That's in one of the useful MAA book series, at a very approachable reading level, full of interesting problems.

Hope this helps!


[quote="indybar"]Consider the equation $ x^2 \minus{} 2kxy^2 \plus{} k(y^3 \minus{} 1) \equal{} 0$,
where $ k$ is some integer.
Prove that the equation has integer solutions $ (x, y)$ such that $ x > 0, y > 0$ if and only if $ k$ is a perfect square.[/quote]
Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=262#p262


Prove that the numbers $\binom{2^n-1}{ i},i=0,1,...,2^{n-1}-1$,have pairwise different residues modulo $2^n$.

I don't really have time to get into the problem, but maybe this hint would help (it's a common idea for this sort of problems)

Let's preassume that the hypothesis of the problem is wrong: that there exist two values of i (i.e. k, l) for which the residues are the same. Then, you will probably get a contradiction (k=l, or something like that)... I'm not sure this will help, but try it!

This problem is from the mathlinks contests,but I haven't seen the solution yet. :(

And how does that solve the problem  :huh:


limit
trigonometry
calculus
calculus computations

hello , can you prove:

$ \displaystyle\lim_{x\to 0\plus{}}\frac{\sin x \minus{} x}{x \sin x}\equal{}0$

Does not belong in Pre-Olympiad.

[hide="Solution"]Note that both the numerator and the denominator of this limit tends to 0, so we can use l'Hospital's Rule.

$ \lim_{x\to 0^\plus{}}\frac{\sin x\minus{}x}{x\sin x}\equal{}\lim_{x\to 0^\plus{}}\frac{\cos x\minus{}1}{\sin x\plus{}x\cos x}$.

Both the numerator and the denominator of the second limit tends to 0; let us use l'Hospital's Rule again.

$ \lim_{x\to 0^\plus{}}\frac{\cos x\minus{}1}{\sin x\plus{}x\cos x}\equal{}\lim_{x\to 0^\plus{}}\frac{\minus{}\sin x}{\cos x\plus{}\cos x\minus{}x\sin x}\equal{}0$, as desired. $ \blacksquare$[/hide]

[hide="Alternate"]
As $ x \to 0$ , $ \sin{x}\equal{}x$

Thus, $ \frac{1}{\sin{x}}\equal{}\frac{1}{x}$

$ \frac{1}{x}\minus{}\frac{1}{\sin{x}}\equal{}0$

$ \frac{\sin{x}\minus{}x}{x\sin{x}}\equal{}0$ as desired.[/hide]


if there were a $ z $ common to the two sets then

$ 13 = |5-12i| \leq |5-z|+|z-12i| = 5+5 = 10 $,

a contradiction

sorry it was 8 and not 5 for the second set
try to solve it now... they say the answerar would be 4

please tell me how did you managed to solve it because that's what I want
thank you

[quote="spx2"]how many common elements do the follwing sets have ?

$ A= ${ ${z\in C , |z-5|=5} $}
and
$B= ${${ z\in C , |z-12i|=8} $}[/quote]

 A:cup:B
it would be more interesting if u do it:)))

8+5 = 13, the hypotenuse of the 5-12-13 right triangle that's formed...


[hide="un fel de solutie"]scriu asa
$\sum \frac{a^{2}+b(a+b+c)}{(b+c)(a+b+c)}\geq 2$
se mai grupeaza umpic, se mai reduce umpic
si apoi aduc la acelasi numitor si iese cu metoda cu care v`am obisnuit deja :) insa nu mai am rabdarea sa scriu toate calculele
sper sa ma credeti pe cuvant :)[/hide]

eu am inceput altfel rezolvarea cu cauchy , daka e poti sa vezi aici ca am postat-o si la preolimpiad[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=114446[/url]. se mai pot vedea si alte 2 solutii frumoase la aceasta ineg

imi retrag cuvintele, inegalitatea nu este own, in momentu de fatza nu imi gaseam culegerea cu new and old inequalities, deci cer scuze fatza de omul caruia i-am furat-o :)

eu vazusem o generalizare urata tot aici pe ml si am particularizat pt 3 variabile, sry iar.

[quote=pohoatza]Fie $a,b,c \geq  0$ cu $a+b+c=1$. Sa se demonstreze inegalitatea:

$\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b}\geq 2$[/quote]
[url]https://artofproblemsolving.com/community/c6h1965107p13604675[/url]
[url]https://artofproblemsolving.com/community/c6h1581044p9756632[/url]
[url]https://artofproblemsolving.com/community/c6h1177493p5688614[/url]


Suppose I have a list with $2007$ positive integers, including the number $2006$. When I take out the $2006$, the average of the numbers in the list becomes $2006$. What would be the average of the numbers in the list if I added a $2006$ to the [i]original[/i] list?

[hide]$\boxed{2006}$ :D The average of the new list is 2006, when you add back the 2006 you took out, it remains 2006, and if you add another 2006, it's still 2006.[/hide]

[hide]the average would still be $\boxed{2006}$[/hide]

[hide]When there are 2006 integers and the average of them 2006, that means all the integers add up to $2006^2$,  or 4024036. Add 2*2006 and divide by 2008 to get [b]2006[/b][/hide]

[hide]Is this supposed to trick us?

I get 2006,  by setting some equations.

$2006$[/hide]

i think it is supposed to be a trick.

i got [hide]2006. but i didn't do anything fancy. it kind of popped into my head as the obvious answer. so i don't know how to put how i got it, except that when i put it in the problem, it works. :huh: [/hide]

[quote="SplashD"][quote="lotrgreengrapes7926"][quote="mathgeniuse^ln(x)"][hide]So, basically if you add n to an set of numbers with an average of n, the average is still n.  Wow.[/hide][/quote] Isn't it amazing? :D[/quote]

I just love your sarcasm.
[color=white][size=59]It is sarcasm, isn't it?[/size][/color][/quote]
Yes, it is.
[color=white][size=59]That wasn't sarcastic.[/size][/color]

[quote="SplashD"]
[color=white][size=59]It is sarcasm, isn't it?[/size][/color][/quote]

I thought you told me to stop writing in white text SplashD.

[color=yellow][size=200]HYPOCRITE.[/size][/color]

Besides, I still think that drunner2007 had a good question, and if it was Countdown, people better get the answer quickly.

[quote="mathmeister22"][hide]

Doesn't adding the average into a set keep the same average?

[/hide][/quote]

[quote="mathgeniuse^ln(x)"][hide]So, basically if you add n to an set of numbers with an average of n, the average is still n.  Wow.[/hide][/quote]

...and I don't even have to say a single word.

[quote="lotrgreengrapes7926"][quote="mathgeniuse^ln(x)"][quote="mathmeister22"][hide]

Doesn't adding the average into a set keep the same average?

[/hide][/quote]

Yes!

It should in the real world.[/quote]
And imaginary and complex, too. :D[/quote]

Actually, complex would just sum everything up. It includes real and imaginary components.  So, you really just needed to say imaginary world.  As in me doing well in a Math Competition.

[hide]The sum is then 2006*2007 with 2007 numbers, so the average is still 2006.[/hide]

[quote="mathgeniuse^ln(x)"][quote="b-flat"]Here is proof for a set of $N|N\geq2$ numbers, with average of the $N-1$ numbers $n$.

[hide]Call what is subtracted $x$.  We know that
$n-x\equiv0\pmod{x}$
$n\equiv x\pmod{x}$
$n\equiv 0\pmod{x}$
$n+x\equiv x\pmod{x}$
$\therefore n+x\equiv0\pmod{x}$
QED[/hide][/quote]

Nice proof.

But, why did we need to use modular arithmetic for this one though?  We could use algebra without mods right?[/quote]

Thanks  :lol:  :lol:  :)  :)

As for proving it using algebra, it is going to be messier, and using mods is incredibly simple.  So, yes, I am pretty sure you could prove it using Algebra.


If d divisor n, then n/d divisor. It give $\sum_{d|n} d=\sum_{d|n} \frac nd .$

please remember you are posting in the advanced section.


[url=http://img138.imageshack.us/my.php?image=8104ei9.png][img]http://img138.imageshack.us/img138/8026/8104ei9.th.png[/img][/url]

Is there any relationships between theta1 and theta2?
If there is any, could anyone tell me the ratio or the equation to get one another please?

Thanks

ok, I will re write the problem like this: We have a right angled triangle $ABC$ ($\angle B = 90^{\circ}$) and $M$ is the midpoint of side $BC$. $\angle AMB = \theta_{1}, \angle ACM = \theta_{2}$, the relation between them:
$tan \theta_{1}= \frac{AB}{BM}, tan \theta_{2}= \frac{AB}{BC}$ As a consequence :
$\frac{tan \theta_{1}}{tan \theta_{2}}= 2$. Is this what you wanted?

No. What he wants is the ratio of the angles, not of their tan's

is that possible because i think you need more information?

Hey Dude, I don't think there's any ratio for the angles then...


with a,b,c is three sidelenghts of triangle

and $p=\frac{a+b+c}{2}$

prove that:

$\sum\frac{a(b+c)}{p-a}\geq8p$
goodluck :D

[hide]$a=x+y$,$b=x+z$,$c=z+y$
$\sum {(x+y)(2z+x+y) \over z} \geq 8(x+y+z)$
$\sum {(x+y)^2 \over z}\geq 4(x+y+z)$
but ${x^2 \over y}+{y^2\over x}\geq x+y$
and ${xy \over z}+{xz \over y}=x({y \over z}+{z \over y}) \geq 2x$
[/hide]

direct computation reduces this too
$ \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c} \ge 3$
By cauchy we hv to prove:
$\frac{(a+b+c)^2}{16(2ab+2bc+2ca-a^2-b^2-c^2)} \ge 3$
This is true because $\sqrt{a}, \sqrt{b}, \sqrt{c}$ are sidelengths of a triangle and the famous ineq (can anybody tell me the name of it?) $a^2+b^2+c^2 \ge 4\sqrt{3} S$

Sorry, siuhochung, but I don't understand how to deduce

[quote="siuhochung"]direct computation reduces this too
$ \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c} \ge 3$
[/quote]

and why this is Cauchy

[quote="siuhochung"]
By cauchy we hv to prove:
$\frac{(a+b+c)^2}{16(2ab+2bc+2ca-a^2-b^2-c^2)} \ge 3$
[/quote]

Your used inequality is named Weitzenbock's inequality and you can find it with many proofs in Theorem and Formulas section.

Thank you.

manlio,
\[
\begin{array}{l}
 \sum {\frac{{a(2p - a)}}{{p - a}}}  \ge 8p \\
  \Leftrightarrow \sum {a + } \frac{{ap}}{{p - a}} \ge 8p \\
  \Leftrightarrow \sum {a + } \sum {\frac{{ap}}{{p - a}}}  \ge 8p \\
  \Leftrightarrow \sum {\frac{{ap}}{{p - a}}}  \ge 6p \\
  \Leftrightarrow \sum {\frac{a}{{p - a}}}  \ge 6 \\
  \Leftrightarrow \sum {\frac{a}{{b + c - a}}}  \ge 3 \\
 \end{array}
\]
for the second question,
\[
\left( {\sum {\frac{a}{{b + c - a}}} } \right)\left( {\sum {a(b + c - a)} } \right) \ge \left( {\sum a } \right)^2
\]

$  \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c} \geq 3 $ can be proved by putting $x=a+b-c,y=a-b+c,z=-a+b+c$ which are positive and writing as:
$\frac{x+y}{2z}+\frac{y+z}{2x}+\frac{z+x}{2y}\geq\frac{\sqrt{xy}}{z}+\frac{\sqrt{yz}}{x}+\frac{\sqrt{zx}}{y}\geq 3$
Using AM-GM two times.


:D find p is positive integer so that
                x 2 +y 2 -3xy-p=0

In other words,
\[ (2x\minus{}3y)^2 \minus{} 5y^2 \equal{} 4p.\]
So $ p$ must be a product of $ 4^k$ and primes of the form $ 5t\pm 1$.


AMC AIME USA(J)MO USAMO email Support algebra

What about the problems in the Resources section?

[quote="DPatrick"]Folks, the AIME is a [b]copyrighted[/b] contest.  We cannot have PDF versions of the AIME circulating around.  Sorry.[/quote]

I'm very sorry to hear that.

Well what if you copyright the PDFs as property of AIME.

4everwise spent all that time TO GET HIS TIME=MONEY work DELETED. TIME AWASTED.  :(

He could use it himself to practice with it :wink:

Is not the format copyrighted? Its in a PDF with just problems, not scanned versions.

Also, what about the old PDF generator? When will that work? Thanks.

[quote="1/(ln x)"]True the AIME is copyrighted, but what about the AMCs, AIMEs, USAMOs, and IMOs in the Contests section?.

4everwise's AIME PDF can just been seen as if he had just copied all the AIMEs from the Contests section and put them all in one easy file instead of 20+.[/quote]
uh, 4everwise didn't make one AIME pdf, he made one per year.

[quote="anirudh"]Unfortunately  :([/quote]
When I saw that every single AIME had been compiled in one source, I almost cried. And now I regret not having saved that PDF to my computer.

Maybe 4everwise could PM or email the comprehensive PDF to various people who wanted it (cough, cough).  :police:

That would be not right either :wink:

[quote="DPatrick"]Folks, the AIME is a [b]copyrighted contest[/b]. We cannot have PDF versions of the AIME circulating around. Sorry.[/quote]

So, he can't do that.

Yeah I don't know why they waited so long to take these off. It made people waste their time. Well, not really. I know A Lot of people already downloaded them.  :lol:

Okay. People. To repeat what DPatrick said, the AIMES ARE COPYRIGHTED!!!

AoPS cannot have PDF's floating around, in PM's or public posts.

If you want to be scared, maybe read Title 17.

Please, please, do not violate what DPatrick has said.

Thanks.[/hide]

Anirudh I think we get the idea.  :lol:

Who wants to be scared? I bet you've downloaded them already and then got scared.  :rotfl:

It still seems rather silly to copyright the AIME, preventing interested students from practicing a test that was designed to inspire highschool students in mathematics.

[quote="purple plume"][quote="anirudh"]You have to buy rights? :maybe:

Then couldn't we have each individual contest have a link on top of its specific year in the Contests section?[/quote]

nothing's free in the world today, little grasshopper[/quote]

uh huh...You can get rights from a certain publisher for free :wink: I got rights from History Channel once for a documentary to use a small clip that they had. :wink:

[quote="anirudh"][quote="purple plume"][quote="anirudh"]You have to buy rights? :maybe:

Then couldn't we have each individual contest have a link on top of its specific year in the Contests section?[/quote]

nothing's free in the world today, little grasshopper[/quote]

uh huh...You can get rights from a certain publisher for free :o I got rights from History Channel once for a documentary to use a small clip that they had. :wink:[/quote]

with connections maybe. unless they are ridiculously nice...

[quote="now a ranger"][/quote][quote="anirudh"][quote="purple plume"][quote="anirudh"]You have to buy rights? :maybe:

Then couldn't we have each individual contest have a link on top of its specific year in the Contests section?[/quote]

nothing's free in the world today, little grasshopper[/quote]

uh huh...You can get rights from a certain publisher for free :o I got rights from History Channel once for a documentary to use a small clip that they had. :wink:[/quote][quote="now a ranger"]

with connections maybe. unless they are ridiculously nice...[/quote]
Are you kidding me? HC gave me permission to use a clip and I didn't have a connection

little grasshopper you have lots to learn in life. there's more to it than you think. you haven't been around long enough

[quote="PenguinIntegral"][quote="DPatrick"]
2. I didn't say that the problems were copyrighted, I said that the [b]contests[/b] were copyrighted.  Using an individual problem is "fair use" under copyright law, similar to excerpting a passage from a book.
[/quote]

Oh, my bad. So what does it mean for a contest to be copyrighted, then? What are the details of what I can and cannot do?[/quote]

Yes, could an admin elaborate on what things are allowed and what things aren't allowed for copyrighted contests?  Does the AoPS community have permission to discuss all copyrighted contests, just as long as those contests aren't reproduced for easy mass distribution?  Are pdf versions of marathons legal, like the MATHCOUNTS marathons in the MC forum?

Discussing individual problems is fine, on the forum or in the wiki.  Discussing contests in general is fine (like the post-contest discussion of the problems that occurs here).  We also have permission from the AMC to use their problems in our classes and textbooks.

Producing reproducible/transferable version of entire contests is where it crosses the line, as this interferes in a significant way with the AMC selling their product.

There is a good Wikipedia article on what is or is not "fair use" under copyright law: [url]http://en.wikipedia.org/wiki/Fair_use[/url]

hmmm On this year's(2007) AIME booklet it says, you can make copies(for non profit aid), So can we have the PDF back ? :maybe:

Again, it says you can make copies of INDIVIDUAL PROBLEMS.  The best you can get is here:

http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2007

[quote="4everwise"][quote="anirudh"]Um...aren't they still downloadable?

http://www.artofproblemsolving.com/Forum/files/aime_1983.pdf

still works. :maybe:[/quote]The old ones are still downloadable. (only 4 or so) The new versions are not. (1983-2006, and the PDF with all the tests in a single file).[/quote]

Really? I got all of them a few weeks ago and it seems like they are  still up.
(The 1983+ ones, when I was preparing for the test.)

Well, the 2007 works now so no comment from me.

Nah, don't think so. Not all of the questions even have solutions posted; some just have a discussion thread.


function
induction
algebra unsolved
algebra

Find all strictly increasing functions $f: \mathbb N \to \mathbb N$ such that
\[ f(n\plus{}f(n))\equal{}2f(n), \quad \forall n\in \mathbb {N}.\]

[hide]$ f(n\plus{}f(n))\equal{}2f(n)$

$ n\plus{}f(n)\plus{}f(n\plus{}f(n))\equal{}n\plus{}3f(n)$

$ f(n\plus{}f(n)\plus{}f(n\plus{}f(n))\equal{}f(n\plus{}3f(n))$

$ 2f(n\plus{}f(n))\equal{}f(n\plus{}3f(n))$

$ 4f(n)\equal{}f(n\plus{}3f(n))\ge f(n\plus{}3f(n)\minus{}1)\plus{}1\ge f(n\plus{}3f(n)\minus{}2)\plus{}2\ge\ldots\ge f(n\plus{}1)\plus{}3f(n)\minus{}1\ge f(n)\plus{}3f(n)\equal{}4f(n)$

Equality holds, so $ f(n\plus{}1)\equal{}f(n)\plus{}1$. Thus $ f(n)\equal{}n\plus{}a$ for all $ n$. Easy to check the this function satisfies the given conditions for any $ a\ge0$.[/hide]

we have $ f(n\plus{}1) \ge f(n)\plus{}1$ by induction we attain $ f(m\plus{}n) \ge f(n)\plus{}m$
we choose m=f(n) we have $ f(n\plus{}f(n)) \ge 2f(n)$ so f(n+m)=f(m)+f(n) with every natural m,n....

[quote="mathVNpro"]Find all strictly increase function $ f$ from $ \mathbb {N}$ to $ \mathbb {N}$ such that:
\[ f(n \plus{} f(n)) \equal{} 2f(n), \forall n\in \mathbb {N}\]
[/quote]

If $ \exists u$ such that $ f(u\plus{}1)>f(u)\plus{}1$ then $ f(u\plus{}f(u))>f(u)\plus{}f(u)$, impossible. So $ f(u\plus{}1)\equal{}f(u)\plus{}1$ $ \forall u$. hence the unique solution $ f(x)\equal{}x\plus{}a$


Let $0\leq a \leq b \leq c$ be real numbers.
Prove that $(a+3b)(b+4c)(c+2a) \geq 60abc$

EDIT: Oops, never mind  :oops:

EDIT #2: I have an extremely messy, but correct, solution:
[hide]
Let b = a+x and c = a + x + y with x,y >= 0.  Expand everything and bring everything to one side to get
5a^2 x + 20ax^2 + 15x^3 + 8a^2 y + 27axy + 27x^2 y + 16ay^2 + 12xy^2, where very term is non-negative, so it is >= 0
[/hide]

it was posted in the inequalities forum recently.

hmm can anyone do this w/ AM-GM?

i got $(a+3b)(b+4c)(c+2a) \ge 16\sqrt 6 (abc)$ by AM-GM and then multiplying them but i'm not sure on another more clever/clean way of doing this

Try and prove that

$(b+4c)(c+2a)\geq \frac {15}{16}(b+3c)(c+3a)$

err after expanding everything out i got

$bc + 19c^2 \ge 7ac + 12ab$

but i dunno wat to do from there

haha sorry im still trying to learn this proof stuff  :blush:

[quote="gotztahbeazn"]err after expanding everything out i got

$bc + 19c^2 \ge 7ac + 12ab$

[/quote]

should be $bc + 19c^2 \ge 7ac + 13ab$

Prove it using $a\leq b\leq c$

[quote="Singular"]Looks easy

$(a+3b) \geq 4a^{1/4}b^{3/4} $ by weighted amgm

similarly with other two product terms on lhs, now divide by 60 and take away the denominators in the exponents:

we have now:

$a^{55}b^{57}c^{68} \geq a^{60}b^{60}c^{60}$

which is obviously true[/quote]

Yep, that's the nicer sol :D

[quote="DPopov"]How is that weighted AM-GM?[/quote]

How is it not? (a,b,b,b)

[quote="DPopov"]Isn't that just plain AM-GM

\[\frac{a+b+b+b}{4} \geq \sqrt[4]{ab^3}\][/quote]

Indeed. But it's still called "weighted".

http://www.artofproblemsolving.com/Forum/topic-32451.html


function
calculus
derivative
calculus computations

How can I show in general that u=f(x-vt) and u=f(x+vt) satisfies the wave equation, where f is any function with second derivatives?.

I figured this out easy enough for sin(x-vt) but in general it seems elusive. Any ideas?.

Thank you,
Cody

This is simply the chain rule. Here's one way of looking at it: Consider $f : \mathbb{R}\mapsto \mathbb{R}$ and $g : \mathbb{R}^2\mapsto \mathbb{R}$ where $w=g(x,t)=x-vt.$ We are looking at $u(x,t)=f(w)=f(g(x,t)).$ Then

$\frac{\partial f}{\partial x}=\frac{df}{dw}\frac{\partial w}{\partial x}=f'(w)\cdot 1=f'(x-vt).$

Then, $\frac{\partial^2 f}{\partial x^2}=\frac{d(f')}{dw}\frac{\partial w}{\partial x}=f''(w)\cdot 1=f''(x-vt).$

Also, $\frac{\partial f}{\partial t}=\frac{df}{dw}\frac{\partial w}{\partial t}=f'(w)\cdot (-v)=f'(x-vt)(-v).$

Followed by $\frac{\partial^2 f}{\partial t^2}=(-v)\frac{d(f')}{dw}\frac{\partial w}{\partial t}=(-v)f''(w)\cdot (-v)=v^2f''(x-vt).$

Hence if $u=f(x-vt),\,\frac{\partial^2u}{\partial t^2}=v^2\frac{\partial^2u}{\partial x^2}.$

In your question, you assumed that $f$ was twice differentiable. Dealing with less than that as a hypothesis and the notion of "weak derivative" would be for another topic.


number theory
prime numbers
number theory proposed

Let $ a,n$ be positive integers such that $ a\ge(n\minus{}1)!$. Prove that there exist $ n$ [i]distinct[/i] prime numbers $ p_1,\ldots,p_n$ so that $ p_i|a\plus{}i$, for all $ i\equal{}\overline{1,\ldots,n}$.

Maybe using Marriage Theorem??? NO?? :huh:

Yes, a solution using Hall's theorem is possible. (we only need to consider the numbers which have as prime divisors only primes <n, and apply the marriage theorem to the sets of those primes for each number.)


counting
distinguishability
combinatorics unsolved
combinatorics

[quote="eleanorwang"]A bakery sells chocolate, cinnamon, and plain doughnuts and at a particular time has 6 chocolates, 6 cinnamon, and 3 plain. If a box contains 12 doughnuts, how many different options are there for a box of doughnuts?

Hint: inclusion-exclusion principle[/quote]
[hide="if donuts of the same type are indistinguishable"]
If we take 0 plain, then there's 1 way.
If we take 1 plain, then there are 2 ways.
If we take 2 plain, then there are 3 ways.
If we take 3 plain, then there are 4 ways.
So 10.[/hide]


I have a very good solution of this but I want to see others.

Let the midpoint$ M$ of the side$ AB$ of an inscribed quardiletar, $ABCD$.Let$ P $the point of intersection of $MC$ with $BD$. Let the parallel from the point $C$ to the$ AP$ which intersects the $BD$ at$ S$. If $CAD$ angle=$PAB$ angle= $\frac{BMC}{2}$ angle, prove that $BP=SD$.

My solution for Mr Stergiou
It is a very good problem!
At first, at the triangle $MAP$ the angle $PMB$ is exterior, so angle $PMB$=angle $MAP$+angle $MPA$. If angle $MAP=x$ , the angle $PMB=2x$ from the hypothesis. So $2x=x+angle MPA$. So angle $MPA=x$. Triangle $MPA$ isosceles with $MA=MP$.But $MA=MB$ because $M$ is midpoint. Now at the triangle $PAB$ the $MP$ which is medium $=MA=MP$. So from the theory the triangle $PAB$ is a right-anlged triangle with angle $BPA=90$ . But from hypothesis we have that AP// CF(F is the point where the CS intersects the AD). So angle $PSF=90$.
Let $MPB=m$.So $x+m=90$.The triangle $MBP$ is isosceles because $MB=MP$ so angle $MBP=m$. Angle $DBC=x$ because it is going to the same arc of the circle with angle DAC. So angle $B=x+m=90$. The $ABCD$ is insribed angle$B+angle D=180$ so angle $D=90$. Because angle B=angle D=90 the diagonal $AC$ pass through the $O$ which is the center of the circle.So $AC=2R$, $R$ is the ray of the circle.Now we make a trick. We extend the $CS$ until CS intersects the circle in $N$. We also form the lines $AN$ and $ND$. The angle $SNA =90$ because it is going to the middle of the circle. The $APSN$ has all of his angles=90 so it is an right-angled parallelogram.So $NS=AP$. The triangles $NSD$ and $APB$ are equals because there are right-angled and they have: $AP=NS$ and angle DNS=angle PAB=x(angle DNS=x decause it is going to hte same arc of the circle with angle DAC). Finally because $NSD$ and $APB$ are equals we  have $SD=BP$.    :)

Silouan ,  Thank you very much. I appreciate it!

Babis

I think it is a great geometry problem which is from JBMO TST 2005 from Greece.

Very beautiful solution, I just couldn’t imagine myself extending the line to F and N without motivation.. Are there any other solutions?


A human heart pumps 5 liters of blood per minute using an excess pressure of 13
kPa. Assume you have an artificial heart that is powered by a standard 12 V car battery with a
capacity of 48 Ah. Assume that the artificial heart has an efficiency of 50%.
Estimate how long the battery lasts?

Thank you!

The[hide="answers"]11 days[/hide]

perhaps I come with solutions later, but they are easy.

A more central problem, basically the same:
If heart pressure= $120\, mm\, Hg$ and the flow is $5\,dm^{3}$
Calculate heart power!

But this is not advanced phys. problem. please post these kind of problems in introductory phys. forum. :D

Seeing this problem another problem came into my mind.

You have a pipe-cylinder with base $S$ and a length $H$ underwater.How much work you must do in order to make air come out from the inferior base.$P_{0}$ is known.

Please create new thread on this. Also explain a bit more please, since I did not get the
scenario. Any figure is appreciated.


[color=darkred]A circle with the chord $ [BC]$ meets again the sides $ [AB]$, $ [AC]$ of the acute triangle $ ABC$ in the points $ F$, $ E$ respectively.

Denote the points $ M\in (BE)$, $ N\in (CF)$ for which $ \frac{MB}{ME}\equal{}\frac{NC}{NF}$ and the intersections $ P\in AN\cap BE$, $ Q\in AM\cap CF$. Prove that :

$ 1\blacktriangleright$ The quadrilateral $ MNPQ$ is cyclically.

$ 2\blacktriangleright$ The quadrilateral $ BCPQ$ is cyclically $ \Longleftrightarrow$ the triangle $ ABC$ is $ A$- is isosceles.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=165539 [/color]

$ 1\blacktriangleright$ $ \frac{MB}{ME}\equal{}\frac{NC}{NF}\equal{} k$; $ \triangle ABE$ and $ \triangle AFC$ are similar and $ \frac{BM}{NC}\equal{}\frac{BE (1\plus{}k)}{CF (1\plus{}k)}\equal{}\frac{BE}{CF}\equal{}\frac{c}{b}$, so
$ \triangle ABM$ and $ \triangle ANC$ are similar and $ \angle BAM \equal{}\angle NAM\ \ \Longleftrightarrow\ \ \angle AME \equal{}\angle FNA$ then the quad MNPQ is cyclically.

$ 2\blacktriangleright$ BCPQ cyclic $ \ \Longleftrightarrow\ \angle BCQ \equal{}\angle QPB \equal{}\angle QNM\ \Longleftrightarrow\ MN\parallel AB\ \Longleftrightarrow\ FE\parallel BC\ \Longleftrightarrow$
$ \Longleftrightarrow\angle EBC \equal{}\angle FCB\ \Longleftrightarrow\angle ABC \equal{}\angle BCA\ \Longleftrightarrow AB \equal{} AC$


Given that A $ \Diamond$ B = $ A^2 \plus{} 2A \minus{} B$, find the coefficient of the X term in the polynomial representation of the expression (X $ \Diamond$ 2) $ \Diamond$ X.

We just plug it in!

$ (X \diamond 2) \Diamond X \equal{} (X^2 \plus{} 2X \minus{} 2) \Diamond {X}$

Now when I see $ A^2 \plus{} 2A$ I naturally think of the expansion of $ (A \plus{} 1)^2$

Let's rewrite our expression: $ A \Diamond {B} \equal{} A^2 \plus{} 2A \minus{} B \equal{} A^2 \plus{} 2A \plus{} 1 \minus{} 1 \minus{} B \equal{} (A \plus{} 1)^2 \minus{} (B \plus{} 1)$

Applying this:

$ (X^2 \plus{} 2X \minus{} 2) \Diamond {X} \equal{} (X^2 \plus{} 2X \minus{} 1)^2 \minus{} (X \plus{} 1) \equal{} (X^2 \plus{} 2X \minus{} 1)(X^2 \plus{} 2X \minus{} 1) \minus{} X \minus{} 1 \equal{} X^4 \plus{} 4X^3 \plus{} 2X^2 \minus{} 4X \plus{} 1 \minus{} X \minus{} 1 \equal{} X^4 \plus{} 3X^3 \plus{} 2X^2 \minus{} 5X$

We want the X term, so our answer is $ \boxed{ \minus{} 5}$


This is my first post, and I am new to the forum. I am going out for the Guinness Book as the Worlds Fastest Human Calculator. I am wondering if there are many other people who are capable of doing this sort of thing, and if so what can it possibly be used for. I've been told that things that I have come up with can be a course in itself but i want to here what other people of math feel about this. THanks

Wow awesome! What did you have to do to get the record?
There are only a few other people in the world who can mental calculate very fast, and they have a whole library of career options open to them.

Well I havent acuqired it yet. I am still in college and waiting till I get through in order to take it. I've been on radio stations and held interviews acknowledging the attempt. I've also been looking for sponsorship with different companies in this venture, but have yet to find one. What kind of jobs would be ideal for someone with these types of mental capabilities?

There's probably not a job based on fast mental calculation but jobs in accounting could use a fast mind. To have a career just based on mental math, you could hold a teaching position to educate others how to aquire such a skill.

Yeah, there are very few jobs which require solely mental arithmetic, computers are generally faster, cheaper and more reliable

Hmm, or you could pursue a career in entertaining?

Yea thats what i was thinking. The initial first step is g etting my foot in the door in order to pursue entertainment. I guess i can go about doing this by acquiring the record. Until then I'll keep trying to gain sponsorhsip locally and see what can come of that. Thanks, oh and any other suggestions are still welcomed.

It's difficult to tell how long the entertainment business will last. That's why I would recommend a teaching proffession or a guest speaker at prominant math competitions.