How do we divide the number

8598 7  3  2 (it's exponent 732)

by nine and what's the remainder?

First, a very brief lesson in $\LaTeX$ to avoid future possiblity of confusion with exponents (replace the &s with dollar signs so it will work).  To render $a^b$ in $\LaTeX$, just type &a^b&.  If your number in the exponent has more than one digit, enclose the exponent in brackets {}, so to do 8598^732, we do &8598^{732}& (in dollar signs, rather than ampersands) to get $8598^{732}$.

As for the problem itself, since we only want the remainder, consider the following:

What is the remainder when $8598^2$ is divided by 9?
What is the remainder when $8598^3$ is divided by 9?
What is the remainder when $8598^4$ is divided by 9?
What is the remainder when $8598^5$ is divided by 9?

Can you figure what the remainder must be when $8598^k$ (k>1) is divided by 9?

Thanks for the help, but that was only an example to the real problem. But, it was still helpful.

[hide][/hide]

by the way the  base is divisible by 3 and power is more than 1 so the number is [b]divisible by 9. [/b]

(Edited later)

I think what you would do it find the remainder when you divide 8642 by 9.  Then you multiple that number by the exponent, 6420, and find the remainder when you divide that number by 9.

Quote:Fine. What about 8642^6420?

1 obviously.

Any number raised to power 6420 divided by 9 leaves a remainder of 0 or 1.

(If the base   is divisible by 3, the answer is zero otherwise 1)

(See simple things like Fermat's trick makes the whole thing very easy)

Rest in spoiler ...

[hide]And before you ask another question, the answer  to  any such problem (where  mod 9 is involved) could be gotten right away (say within  a second)

find out base modulo 9 (easy, add all digits) and power modulo 6 (again easy , you know the number is even or odd and addidng the digits give  value r modulo 3) .. and you have two very small number to deal with.[/hide]

[quote="Gyan"]by the way the  base is divisible by 3 and power is even so the number is [b]divisible by 9. [/b][/quote]

No.  The power just has to be greater than one.

[quote="wereallgonnadie"]Do it the way Dts suggested.  Find a pattern.[/quote]

In all problems like this, it is possible to find a pattern.  It's probably easier to use something to h_s_potter2002's method; first take out the largest possible number that is divisible by 9 away from the base, and then look for a pattern with powers, but it's basically the same thing.

[quote]The power just has to be greater than one. [/quote]
Yes.  you were quite quick.. I did edit that though.. :)


algorithm
function
algebra
rational function
advanced fields
advanced fields unsolved

Can you suggest me an effective rational interpolation algorithm. The
algorithm must interpolate rational function R(x)=P(x)/Q(x) (where
dgP(x)=m and dgQ(x)=n, m and n are known fixed numbers) through m+n+2
points (x_i, r_i). Also, approximative algorithms (such as Pade
approximation) are not solving the problem.

Thanks forward,

DeXter_of_Nis

[quote="dexter_of_nis"]Can you suggest me an effective rational interpolation algorithm. The
algorithm must interpolate rational function R(x)=P(x)/Q(x) (where
dgP(x)=m and dgQ(x)=n, m and n are known fixed numbers) through m+n+2
points (x_i, r_i). Also, approximative algorithms (such as Pade
approximation) are not solving the problem.[/quote]
Why not just build a system of linear equations

P(x_i) - r_i * Q(x_i) = 0

and solve it with respect to the coefficients of P(x) and Q(x)?


we consider the polynomial $ f(x) \equal{} (x \minus{} x_{1})(x \minus{} x_{2}).....(x \minus{} x_{n})$ ,where x_i are complexes numbers,
show that there exist a complexe $ z_{0}$ with l$ z_{0}$l=1 and l$ f(z_{0})$l$ > \equal{} 1/3^n$(1+l$ x_1$l)...(1+l$ x_n$l)


calculus
integration
trigonometry
symmetry
function
complex analysis

Integrate (cos x -1)/x^2 from 0 to infinity.

Since $ \cos x \minus{} 1 \equal{} \minus{} 2\sin^2\frac {x}{2}$, and the integrand has then even symmetry, we may write
\[ \int_0^{ \plus{} \infty}\frac {\cos x \minus{} 1}{x^2}dx \equal{} \minus{} \int_{ \minus{} \infty}^{ \plus{} \infty}\frac {\sin^2\frac {x}{2}}{x^2}dx \equal{} \minus{} \frac {\pi}{2}\int_{ \minus{} \infty}^{ \plus{} \infty}\frac {\sin^2(\pi y)}{(\pi y)^2}dy,
\]
where we have made the variable change $ x \equal{} 2\pi y$.  Note that the last integrand is then $ sinc^2y$, a function that is well known to integrate to 1 (*).  So the answer is $ \minus{} \frac {\pi}{2}$.

(*) For a proof that $ sinc^2y$ integrates to 1, consider function $ f(x)$ such that $ f(x) \equal{} 1$ when $ |x|\leq\frac {1}{2}$, $ f(x) \equal{} 0$ otherwise.  Its Fourier transform is $ sinc y$.  By Rayleigh's theorem, the integrals of the modulus squared of the function, and of the modulus squared of its Fourier transform, are equal, done!  Note that in this case we have defined the Fourier transform in "standard frequency", instead of in angular frequency, ie, the Fourier transform of $ f(x)$ has been defined as
\[ \int_{ \minus{} \infty}^{ \plus{} \infty}f(x)e^{ \minus{} i2\pi xy}dx
\]


Write the equation in standard form: $ y$ = $ 4x^2$ -$ 16x$ + $ 3$

and the answer is: $ 4(x\minus{}2)^2$ $ \minus{} 13$
I keep getting

[hide="this!"]$ y$ = $ 4(x\minus{}2)^2 \plus{}19$[/hide] arggh! im so frustrated! what am i doing wrong? When you factor out the 4, you'll get : $ 4(x^2\minus{}4x\plus{}4) \plus{}3$ ( i completed the square), so then, you'll get: $ 4(x\minus{}2)^2 \plus{}19$.

[quote="tridecalogism_lover"]When you factor out the 4, you'll get : $ 4(x^2 \minus{} 4x \plus{} 4) \plus{} 3$[/quote]

Well, no.  What you get is $ 4(x^2 \minus{} 4x) \plus{} 3$.  If you want to complete the square, you need to subtract appropriately to compensate for the $ 4$ in the parentheses; you'll get $ 4(x^2 \minus{} 4x \plus{} 4) \minus{} 13$ like you want.

I might be wrong, but isn't $ y\equal{}4x^2\minus{}16x\plus{}3$ already in standard form which is $ ax^2\plus{}bx\plus{}c$. I thought $ y\equal{}a(x\minus{}h)^2\plus{}k$ was considered vertex form.

Your terminology might be different, but I'm pretty sure that what they are saying is correct. You might want to read up on conic sections.


geometry
3D geometry
sphere
number theory
least common multiple

1. The equation x^3=9+46i has roots in the form a+bi where a and b are integers, find a+b.

2. Find the area enclosed by the graph of |x+y|+|x|+|y|=1.

3. The sequence of letters rolls off if these conditions are met :
   a. The sequence doesnt begin or end with 2 consecutive consonants.
   b. No 3 consecutive letters are all consonants.

   How many permutations of the word "mathematics" roll off the top?

4. Two points A and B lie on a sphere of radius 12. The length of the straight line segment joining A and B is 12 sqrt 3. Find the shortest path from A to B if every point of the path is on the sphere of radius 12.

5. How mant positive integers less than 221 have no common factor with 221 other than 1?

6. What is the greatest number of points of intersection that can occur when two circles and two straight lines are drawn on the same piece of paper?

7. Find the second smallest positive integer such that when divided by 2 leaves a remainder of 1, when divided by 3 leaves a remainder of 2, when divided by 5 leaves a remainder of 4, when divided by 7 leaves a remainder of 1.

8. A string is just long enough to be tightly tied around a spherical planet of radius 2008200820082008 meters. If the string were 2pi meters longer, how much larger is the radius of the sphere it could be tied around?

These questions are easy but may be difficult for you. I am answering two of them but the rest i encourage you to do yourself. Just think deeply if y're finding it difficult, but do it yourself.
[u]ans 7[/u]we see that the number is =-1(mod2,3,5,7)
for smallest number take the lcm of 2,3,5,7 and subtract 1 from this.this is equal to $ 210\minus{}1\equal{}209$ it will be smallest requied number. Don't just read this but properly understand this so that u can apply elsewhere.

[u]ans6[/u]
this is quite easy. let us suppose lines dont intersect. they will individually  intersect circle in 2 points. if they intersect we get a total of 5 point and there can't be more intersections.

yup  its 11. i didnt take the second circle in account earlier.

$ \boxed{\#1}$ : $ (a + bi)^3 = 9 + 46i$, Solve by equating reals to reals, imaginary to imaginary. You get to finish it on your own  :) .

[b]Another approach to (1):[/b]

Note that the modulus of $ 9 \plus{} 46i$ is $ \sqrt {9^2 \plus{} 46^2} \equal{} 13 \sqrt {13}$

Thus, the modulus of its cuberoot is $ \sqrt {13}$.

Therefore, $ a^2 \plus{} b^2 \equal{} 13$, and we have some permutation of $ (\pm 2, \pm 3)$.

Now if you look at the angle, most of the 8 possibilities are obviously false, so we only have to test out a couple of these to find that $ (a,b) \equal{} ( \minus{} 3,2)$

[quote="orangefronted"]7. Find the second smallest positive integer such that when divided by 2 leaves a remainder of 1, when divided by 3 leaves a remainder of 2, when divided by 5 leaves a remainder of 4, when divided by 7 leaves a remainder of 1. [/quote]

As the problem is written, analytic does not have a correct solution for #7.  (note the remainder is 1, not 6, when divided by 7)

And a small hint for #8:
[hide]The radius of the planet has no impact on the final answer.[/hide]

thanks for spotting it out cf249

ok, now  number is =-1(mod2,3,5)
lcm of 2,3,5 = 30
so, number according to mod(2,3,5) = 29 and it is +1(mod 7). so, it is smallest integer with the given property.
to, find second smallest,
$ 1\times1\equal{}1 \implies1\times8\equal{}1$ mod(7)
so, $ 29\times8\equal{}232$ is the answer.


Hey for all of you that celebrate it or sorta do (that's most of the people here...) :)


I was wondering if Economics is considered as a Mathematics related  career? I know maybe this should be in the Economics forum but people don't visit it a lot. Based on what I saw in the department of labor website, it says there are only 16,000 economists in America. I was wondering how hard it is to get to work in the Federal Reserve and the IMF and other such organizations.

I'm no expert on economics at all but I definately think that it is mathematics related.  i have spoken to some friends who are studying economics at a high level and they use some pretty serious maths - one of the guys studied maths to honours level just so that he had the mathematical background to deal with the economics that he wanted to do.  Game theory is one field that falls under both maths and economics - John Nash won the Nobel prize in economics for his work in game theory.  If you are interested in economics withs a maths feel try looking at some of the work of Daniel Kanheman and A Tversky ( I have no idea if I spelt their names right) at some stage.

Yes, the works of Daniel Kahneman, Amos Tversky, Gerd Gigerenzer, Cass R. Sunstein, and other authors mentioned in an [url=http://www.amazon.com/exec/obidos/tg/listmania/list-browse/-/2PA3IVBYU8SC9/]Amazon Listmania list[/url] I just searced up will give you a lot to think about in regard to how math relates to economics. Also take a look at Nassim Nicholas Taleb's book Fooled by Randomness and other books on investing recommended in another [url=http://www.amazon.com/exec/obidos/tg/listmania/list-browse/-/1CC572I1XMBSO/]Amazon Listmania list[/url].

At this point it's very difficult to make any academic contributions to Economics without knowing enough math to get into a graduate math program.

All three of the Econ grad students I know were either accepted to or attended graduate math programs.

Tversky and Kahneman's work is usually considered social science rather than what counts as "mathematical" within economics. As I recall, they were interested in the psychology of decision-making, and its deviation from rationality (in the sense of economics, game theory or probability).  Certainly they were well aware of statistics and used it both as theory and in their empirical studies.  But their work was more addressing the validity of the assumptions in various economic models, than in using heavy mathematics to draw out conclusions from those assumptions.   One of them ([b]edit:[/b] Tversky), died not long ago and there was some discussion of their joint works in the memorial articles.    Also there was some publicity when Kahneman got the Nobel prize in Economics for their joint work.

Perhaps a philosopher like Nozick could be added to that group.

I think the use of game theory and logic to assess the validity of assumptions falls as deeply within mathematics as number crunching statistical analysis.

You should read Conned Again Watson!, lots of economic logic in the form of Sherlock Holmes tales.

[quote]Also take a look at Nassim Nicholas Taleb's book Fooled by Randomness[/quote]

Taleb was the subject of a New Yorker article a couple of years ago, not long before the book came out. The piece
was not only about him and his pessimistic investment vehicle, but also had interesting comments about the relation between economic theory and the markets.  I think they quoted somebody from an investment bank as saying that it takes about 5 years to deprogram an economics PhD once they go to work in the industry.

I have seen a book math - economy by Debreu,
there was a lot of math topology /metric space /fixed point theorems...

I have read some Sherlock but dont know what it has to do with econ?

[quote="MCrawford"]the use of game theory and logic to assess the validity of assumptions falls as deeply within mathematics as number crunching statistical analysis.[/quote]

Sure, but there are also visible distinctions, however arbitrary or sociological, between disciplines and subdisciplines.  At least one of Kahneman and Tversky worked in a department of psychology, not economics or mathematics, and (whatever the value of these classifications) their work was never seen primarily as an example of mathematical economics, unlike that of Arrow, Debreu, Nash, Black-Scholes-Merton, and most of the people who have received Nobel prizes for theoretical work in economics.

Whether or not there is any objective distinction between pure mathematics, applied mathematics, computer science, operations research, theoretical physics, logic, (the list could be continued),  it certainly makes a difference for funding and other purposes. Paul Erdos used to say that computer science was just mathematics, but done in a place where it was easier to get work.

[quote="fleeting_guest"]
Paul Erdos used to say that computer science was just mathematics, but done in a place where it was easier to get work.[/quote]

Great quote.  I met Vernor Vinge the other day who moved over to the comp sci department from the math department (discrete math) at San Diego State University many years ago before he was the famous Hugo award winner author.  He said that back then the transition was easy.  He just had a different department and office.

Follow this link to see the math embedded in economic research papers:

http://www.ces.census.gov/ces.php/papers


factorial
pigeonhole principle
algebra
system of equations

[url]http://yuethomas.vze.com/files/cmo36.pdf[/url]

Written earlier today...

solved #1

Find all ordered triples (x, y, z) of real numbers which satisfy the following system of equations:

xy = z − x − y----e(1)

xz = y − x − z----e(2)

yz = x − y − z----e(3)

[hide]Two cases:

Case I, suppose x=0, then 0=z-y, so z=y

looking at e(3), we have y :^2: =-2y

                                          y=-2, or 0=z

when y or z=0, it's the same thing

so (0,-2,-2), (-2,0,-2), (-2,-2,0), (0,0,0)

Case II,

suppose x,y,z are non-zero integers

add e(1) and e(2):

we have x(y+z)=-2x or y+z=-2

add e(2) and e(3):

          x+y=-2

add e(1) and e(3):

           x+z=-2

the only set of solution to this is (-1,-1,-1)

all  solutions:

(0,-2,-2), (-2,0,-2), (-2,-2,0), (0,0,0), (-1,-1,-1)[/hide]

An alternative method to solve this problem would be as follows:

[hide]Add x + y + 1 to the first equation, this makes it xy + x + y + 1 = z + 1.  This can be expressed as (x + 1)(y + 1) = z + 1.

Perform the same transormation to the second and third equation (which are equivalent -- the only difference is the names of the variables)

Then substitute a for x + 1, b for y + 1, and c for z + 1.

Our system of equations is now simplified to

ab = c

bc = a

ac = b

I dont know of a rigerous way to prove this but I think you could show that the absolute values of a b and c are all equal to each other and either 0 or 1.[/hide]

[hide]Then just take one (say a) to be 0, and show that b and c both have to be 0. Thus in the remainder of the proof we can then not care about dividing by 0.

Rewrite a = c/b, sub into the other two equations to obtain b2 = c2 = 1. From then on it's a simple matter of casework.[/hide]

solved 1 through 3 without much difficulty... havent looked at the others yet... need sleep!... solutions later

how did you do Tom??

[hide]

#4.  for 1<=k<=p-1

k^(2p-1)+(p-k)^(2p-1)=k^(2p-1)+(-k)^(2p-1)+(2p-1)(-k)^(2p-2)*p+p^2(...)

so mod p^2, it is

k^(2p-1)+(-k)^(2p-1)+(2p-1)(-k)^(2p-2)*p

=(2p-1)*p*k^(2p-2)

=2p62*k^(2p-2)-p*(k^(p-1))^2

from FLT, (k^(p-1))^2=1^2 (mod p) so (k^(p-1))^2=u*p+1, -p*(k^(p-1))^2=-up^2-p=-p (mod p^2)

hence LHS=(p-1)(-p)/2 (mod p^2)=(p-1)(-p)/2+p^2 (mod p^2)=p(p+1)/2 (mod p^2)

#5.

Consider all factors are in the form 2^x * 3^y * 167^z, 0<=x<=200

0=z<=100

let Fi has 3 components {x,y,z}.  Fi is a multiple of Fj iff Fi.x>=Fj.x, Fi.y>=Fj.y, Fi.z>=Fj.z  [/hide] (you can tell i am a compsci guy )

[hide]

so we sort elements in S according to:

for all i<j,

Fi.x<Fj.x

OR (Fi.x=Fj.x and Fi.y<Fj.y)

OR (Fi.x=Fj.x and Fi.y=Fj.y and Fi.z<Fj.z)

Consider a set S (not necessary maximum)

suppose for consecutve index a<=i<=b, Fi.x=Fa.x, then for index b+1<=j<=d, Fj.x=Fd.x=Fa.x+c, c>0

Let the max (Fi.y, Fi.z)=m

suppose for some j, (Fj.y,Fj.z)=m also

and of course, Fj.x=Fi.x+c

this implies two things, F={Fi.x, Fj.y, Fj.z} doesnt exist (if it does it would have been a factor of Fj)

then we can reduce Fj.x=Fi.x without violating the rules of S (one isnt multiple of other)  -- hey we are just making a factor smaller

hence we can maximize S by imposing the rule Fi.x<Fj.x =>max(Fi.y, Fi.z)>max(Fj.y,Fj.z)

now for all i such that Fi.x=Fj.x, and max(Fi.y,Fi.z)=c  then we can make all i such at one of Fi.y, Fi.z=c, hence we have the following for Fi.y, Fi.z:

c, 0

c, 1

...

c, c-1

0, c

...

c-1, c

c,c

finally we can choose an x such that for all Fi.x=x, max(Fi.y, Fi.z)=100, then 99, all the way to 0.  so max of S is (201+199+...+1)=202*101/2=101*101=10201

[/hide]

for #3, an outline since i am too lazy to go thru in detail[hide]

chase some angles, you'll get BX=XD, BY=YD

hence XD YB are the longer length so the other four length are equal.  then you do more angle chasing and show that the sum of the two angles that are bisected is 180.  which imples XY is the diameter.  then using the fact that the four length are equal, their arc subtends the same angle in the center, we can prove BD is the diameter.

[/hide]

#5:

[hide]Any divisor of 2004^100 (call this big number N) has the form 3^a*167^b*2^c with 0<=a,b<=100 and 0<=c<=200. If we have 2 factors of the form 3^a*167^b*2^y and 3^a*167^b*2^z with y!=z, then the smaller divides the larger. So, for any ordered pair (a,b) with each less than 100, there is at most one divisor in S of the form 3^a*167^b*2^c. The number of ordered pairs (a,b), and therefore an upper bound for our answer, is 101*101 = 10201.

On the other hand, consider the set of numbers {3^a*167^b*2^(200-a-b) | 0<=a,b<=100}. Call this set P. Notice that since 0<=a,b<=100, we also have 0<=200-a-b<=200, so that each member of P is a divisor of N. Furthermore, I will prove that no element of P divides another element of P.

Assume otherwise, and let 3^a*167^b*2^(200-a-b) divide into another unequal divisor 3^c*167^d*2^(200-c-d). This implies

(1) a<=c

(2) b<=d

(3) 200-a-b<=200-c-d

From adding (1) and (2), a+b<=c+d. However, from (3), we get a+b>=c+d. Thus, c+d<=a+b<=c+d, so equality is obviously necessary, so we must have equality is all three of the above inequalities. Thus, a=c and b=d and the two factors are equal. Since we assumed they were unequal, we have a contradiction. Thus, P is a set of divisors of N with 101*101=10201 elements.

Since we showed that 10201 is an upper bound for our set S, and since a set has beed demonstrated that has 10201 elements, we conclude that the answer is 10201.[/hide]

I did #s 1, 2 and 4. What's your answer for #2?

14 * 5! * 4! = 40320 = 8!. Such a regular answer from such an irregular expression. I'm suspicious that there's a better way...

(That's what I got by the way.)

I just finished Number 1 and I'm looking at number 2 right now.

How would you approach this problem?

(im not tat great in problem solving so lol...i need help)

[hide]You might want to consider the two cases separately. One case would be when all the rooks are placed on squares of the corner colour (we'll say it's black); the other case would be when all the rooks are placed on white squares.[/hide]

yeah i thought about tat but are we required to use factorial in this situation?

lol nothing

im just looking at all possible situations that the rooks can be placed

for example...diagonally, zig zags, and others like tat

im slow...:(

[quote="bugzpodder"]bleh thats the one that i thought i got but completely messed up.  but other ppl from my school seem to favour the answer 14*5!*4![/quote]

Almost 100% sure, it's 14*5!*4!.

Unfortunately, I put the equivalent of 12*5!*4! - did it right, forgot to include two stupid cases.

I also missed the (0,-2,-2) triples in 1.

I almost solved 4.

I made marginal progress on 3.

I can post a full solution to 2, if anyone wants.

Consider the white case first. I've marked 4 columns and 4 rows. What can you say about the intersections of the columns and rows?

How many pieces must you place on each column/row?

How many possible locations can each piece go on a column/row?

How many possible arrangements are there in total for the white case?

The black case is analogous, but there are a few sub-cases to consider.

k thanx

I'm going to try out the problem more with the questions you gave me.

Okay, here's my solution to 2:

We chop off the last column, and consider the 8x9 truncated chessboard.  This chessboard is identical with respect to black, white rooks.  We proceed to place ONE rook in each column, so that all of the columns are now filled.

We clearly have 5 choices for the first rook, four for the second, four for the third .... down to 1 for the eighth.  This gives us 5!*4!*2 (one for black and one for white).

Now, we consider the column that has been chopped off.  This column has 5 black squares and 4 white squares.  Clearly, it can be easily 'swapped' with any of the other identical columns.  There are four identical columns, and we can use white and black squares, so now we have 5!*4!*2+5!*4!*8

Now, we consider whether we can swap this column (5 black, 4 white) with a column having (4 white, 5 black).

Case 1: Black rooks.

There are now 5 columns having 5 black, 4 white and 3 columns having 5 white, 4 black.

With the black rooks, we can permute all of the 5 rooks in the 5 identical columns such that there will be no two in the same row.

This case is thus acceptable, and we can swap the (5 black, 4 white) column with any of the (5 white, 4 black) columns.

Case 2: White rooks.

There are now 5 columns having 5 black, 4 white.  However, we are using white rooks.  We now need to distribute 5 rooks across only 4 white columns.  By Pigeonhole Principle, this is impossible, and so the white rooks case does not work.

We are thus left with 2*5!*4!+8*5!*4!+4*5!*4!=14*5!*4!

That'll work.

I was SO crunched for time for #4. I still don't understand why my initial approach did not work:

[hide]As with Bugz's solution, consider

m2p - 1 + (p - m)2p - 1, where m :le: (p - 1) / 2.

= (m + p - m) (m2p - 2 - mp - 1(p - m)p - 1 + (p - m)2p - 2)

= p (m2p - 2 - mp - 1(p - m)p - 1 + (p - m)2p - 2)

If you expand out the bracket, you get a whole lotta terms. However, since we're considering them (mod p2), any term with p can be discarded (that term, multiplied by the p out front, will be divisible by p2). We're left with:

 :equiv: p (m2p - 2 - m2p - 2 + m2p - 2) = m2p - 2p (mod p2).

Now, this isn't the right answer, as one's supposed to get m2p - 2 p (2p - 1). So why doesn't it work?[/hide]

[quote="Osiris"]Isn't this expression always divisible by p2 though?

pp - 1 * (p - 1)p / 2

(p - 1)p / 2 is obviously a whole number

expression = p2 * p p - 3 * some whole number

Am I not getting something here?[/quote]

Clearly, I made some mistake in writing that out.  Let me look at it again.

Can someone please post the problems here? I cant get the pdf to work for some reason, and I cant figure out why... It just simply wont connect.

1. Find all ordered triples (x,y,z) of real numbers which satisfy the following system of equations:
xy = z - x - y
xz = y - x - z
yz = x - y - z

2. How many ways can 8 mutually non-attacking rooks be placed on a 9x9 chessboard so that all 8 rooks are on squares of the same colour?

3. Let A,B,C,D be four points on a circle (in clockwise order), with AB<AD and BC>CD. Let the bisector of angle BAD meet the circle at X, and the bisector of angle BCD meet the circle at Y. Consider the hexagon formed by these six points on the circle. If four of the six sides of the hexagon have equal length, prove that BD must be a diameter of the circle.

4. Let p be an odd prime. Prove that:
sum(k=1, p-1) k^(2p-1) :equiv: p(p+1)/2 (mod p^2).

5. Let T be the set of all positive integer divisors of 2004^100. What is the largest possible number of elements that a subset S of T can have if no element of S is an integer multiple of any other element of S?

hope I didn't typo anything there..

hehe... i think that these problems were put directly onto a team contest at mop...
too bad i didnt see this thread earlier..


Is it possible to delete an old account (not that I am a multi (I have another account that I haven't even logged into)) if it is not activated yet? Do I have to email an admin personally about this? BTW, the username is "theVman".

Why did you create that account in the first place?

I am right between over and under 13 years old, so first I created one that said that I was under 13. But then I saw you need parents permission, you had to mail a form, ... so I made one that said I was over 13 (of course, my parents were right there and told me to make myself over 13). Also, I realized that "under 13" wouldn't be accurate for very long. I haven't even logged into it because it isn't activated.

[quote="akhil0422"]Ask an admin? levans deleted my other account for me.[/quote]

Evidently yes.  :)


Find the maximum value of $N$ such that if $p(x)=x^3+ax^2+bx+c$ is a cubic polynome with positive racines then :
$p(x)>= N * (x-a)^3$ for all $x>=0$. Find the equality conditions.

Come on guys!  For hint, the answers is $N=-1/27$


probability
search
probability and stats

Each of the 'n' passengers sitting in a bus may get down from it at next stop with probability P.Moreover, at the next stop either no passenger or exactly one passenger boards the bus. The probability of no passenger boarding the bus at the next stop being Po . Find the probability that when the bus continues on its way after the stop, there will again be 'n' passengers in the bus.

You and UTTARA are cheating on the same assignment?

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1025873771&t=306701
(Searched for the words "passenger bus".)

its same question both did not get..

my bad for having not searched the question... if its already present


[quote="Interval"]Triangle DEF is congruent to triangle D'E'F'.
If EF is represented by 6x + 4
and E'F' is represented by 2x + 20,
find the value of x.[/quote]
[hide]Since the two sides are congruent that means that $6x+4=2x+20$. Solving for x which is $\boxed{4}$.[/hide]

[quote="jli"][quote="Interval"]Triangle DEF is congruent to triangle D'E'F'.
If EF is represented by 6x + 4
and E'F' is represented by 2x + 20,
find the value of x.[/quote]

[hide="solution"]Since they are congruent, $EF=E'F'$ so $6x+4=2x+20$.  Therefore $x=8$.[/hide][/quote]

Your going to kick yourself.

CYWA

[quote="Interval"]Triangle DEF is congruent to triangle D'E'F'.
If EF is represented by 6x + 4
and E'F' is represented by 2x + 20,
find the value of x.[/quote]
[hide]Since $\triangle DEF \cong \triangle D'E'F'$, then $\overline{EF}\cong\overline{E'F'}$.
Then make an equation, $6x+4=2x+20$ combine the terms, $6x-2x=20-4$ then solve $4x=16$ so $\boxed{x=4}$[/hide]

[quote="Interval"]I get it:

x = 4 and 8, right?[/quote]
No, x=4.

jli made a mistake and said it was 8.


Okay, I get that, but then like HOW do you use it? I mean if I were given a problem of some sort and then I were to solve it, HOW would I use?
I know, I'm bugging everyone...

If a dog can run 14 miles an hour, find the number of feet it can run in a second.
The teacher said Its possible to do factor labeling.
So its like this right?
14 miles/1 hour
so...
73920 feet/1 hour
Thats All i can get to...
did i do something wrong on the multiplying?

Factor labeling (or dimensional analysis, as it's called) is commonly used in chemistry.

However, in the problem you posted, you are correct so far. With 73920 feet per hour, you can convert that into seconds. Because 3600 seconds is one hour, multiple that by 3600.

The trick is to know what units you need as your answer, and the conversion labels.

Okay, and then thats it? Or do I divide top and bottom by 3600?

$ \frac{14\text{ mi}}{1 \text{ hr}}\times\frac{5280\text{ ft}}{1\text{ mi}}\times\frac{1\text{ hr}}{3600\text{ s}}\equal{}\frac{73920\text{ ft}}{3600\text{ s}}\equal{}\frac{308}{15}\text{ fps}$

The important thing is to figure out what you want to be canceled. Then you can figure out what conversion to use, and also which one should be the numerator/denominator.

$ \bigtriangleup$

So why did you post that?

Because they were trying to get more posts, by pointing to mine.


geometry
analytic geometry
linear algebra
matrix

$ 100$ people are standing in a line and they are required to count off in fives as: '1,2,3,4,5, 1,2,3,4,5,...' and so on from the 1st person in the line. Anyone who counts '5' must leave the line. What was the original position in the line of the last person to leave?

Also prove that area of $ \triangle ABC$ with coordinates of the vertices $ (x_1,y_1), (x_2,y_2) \&$ and $ (x_3,y_3)$ is given by

$ \det$ $ \begin{pmatrix} 1 \& 1 \& 1 \\ x_1\&x_2\&x_3\\y_1\&y_2\&y_3\end{pmatrix}$

Um, wouldn't the answer to your first question be person $ 100$? or are we supposed too start over from the beginning once we get to the end?


Let $V$ be a point in the exterior of a circle of center $O$, and let $T_1,T_2$ be the points where the tangents from $V$ touch the circle. Let $T$ be an arbitrary point on the small arc $T_1T_2$. The tangent in $T$ at the circle intersects the line $VT_1$ in $A$, and the lines $TT_1$ and $VT_2$ intersect in $B$. We denote by $M$ the intersection of the lines $TT_1$ and $AT_2$.

Prove that the lines $OM$ and $AB$ are perpendicular.

$A$ is the pole of $TT_1$, so $OA\perp TT_1=BM$. $A$ is the pole of $TT_1$ and $T_2$ is the pole of $VT_2$, so $AT_2$ is the polar of $B=TT_1\cap VT_2$, so $OB\perp AT_2=AM$. From here we find $M$ to be the orthocenter of $OAB$ and we're done.

I know it's not the solution you were talking about, because (a) it's >1 line long and (b) it uses poles, so it's not suitable for 7'th or 8'th graders :), but it's a solution, right?

Anyway, it's obvious that $OA\perp TT_1$, so all we need to do is to find a short argument for the fact that $OB\perp AT_2$. I didn't think about this after seeing that it works out fine with poles :)..

Is there really a one-line solution to this?

Consider the inversion $I(O,R^2)$ and let $AO\cap{TT_1}=\{N\}$. Then $I(N)=A$, $I(T_2)=T_2$ $\Longleftrightarrow$ $AT_2\perp{OB}$.

Whoops...  :)

Anything one-line and suitable for juniors?  :D

I'd really like to know.


number theory
greatest common divisor
modular arithmetic
Divisibility Theory

What is the greatest common divisor of the set of numbers \[\{{16}^{n}+10n-1 \; \vert \; n=1,2,\cdots \}?\]

[quote="Peter"]Since for $n=1$ there stands $25$, it can be at most $25$, we will show it is $25$. Since $16^{n}$ ends in $16, 56, 96, 36, 76$, taking the remainder mod $25$ gives $16,6,21,11,1$, substract $10n\bmod 25$ from that and you have $1,1,1,1,1,\ldots$, so every element is a multiple of $25$.[/quote]

Another way: $ 16^{n}\plus{}10 n\minus{}1\equiv (1\plus{}15)^{n}\minus{}15n\minus{}1\equiv\sum_{k \equal{} 2}^{n}{n\choose k}15^{k}\equiv 0\bmod 5^{2}$, for any $ n \equal{} 1, 2,\ldots$ Since $ 16^{1}\plus{}15\cdot 1\minus{}1 \equal{} 5^{2}$, the greatest common divisor we're looking for is exactly $ 5^{2}$.

[hide="A114"]Note that when $n=1$, $16^n+10n-1=25$, so the GCD is a factor of 25. I claim the GCD is in fact 25. Note that $16^n \pmod{25}$ repeats in a cycle of length 5, namely 1, 16, 6, 21, 11. Therefore, it suffices to check that $10n-1 \pmod{25}$ repeats in a cycle 24, 9, 19, 4, 14. In fact, $10(n+5)-1 \equiv 10n-1 \pmod{25}$. Note that the corresponding terms of the cycle all to $25$, so the GCD is in fact 25.[/hide]


Given a point $ P$ in the circumcircle $ w$ of triangle $ ABC$. Prove that the segments $ PA,PB,PC$ form a triangle $ T$. Let $ R$ be the radius of the circumcircle $ w$ and let $ d$ be the distance between $ P$ and the circumcenter. Find the area of $ T$ in terms of $ R,d$.

Is this written correctly?

It seems any point $ P$ "in" the circle with sides $ \overline{PA}, \overline{PB}, \overline{PC}$ will not result in a closed figure. No closed figures implies no triangles.
If there were, by chance, a closed figure, it would not be a triangle because the point $ P$ would represent a 4th vertex.

It also seems that the same is true if the point $ P$ is "on" the circle.

If the point $ P$ is allowed to be coincident with one of the verticies of triangle $ \Delta ABC$ then there cannot be a triangle with sides $ \overline{PA}, \overline{PB}, \overline{PC}$ because the 3 sides would be orginating from a common point.

If $ P$ is allowed to be coincident with one of the vertices then distance $ d$ will be the same as the radius $ R$, noting that one of the legs does not meet the original proposition.

I didn't consider the possibilities of using congruent or similiar triangles...

I didn't see an immediate post so I thought I would post some thoughts.  Hoping to see if there is an actual real solution?

Well, actually it means you can take $ PA,PB,PC$ (you can move it  :D ) and it can form a triangle.

So you have to prove that $ PA\plus{}PB>PC, PB\plus{}PC>PA, PC\plus{}PA>PB$.  :)

I don't think $ PA,PB,PC$ always form a triangle. For example consider the following diagram:

[asy]size(150);
pair A,B,C,P;
A=(-25,0);
B=(25,0);
C=(-20,14);
P=(-17,7);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
dot(A);
dot(B);
dot(C);
dot(P);
label("A",A,W);
label("B",B,E);
label("C",C,NW);
label("P",P,W);
draw(A--P);
draw(B--P);
draw(C--P);[/asy]
Here $ PA\plus{}PC\not>PB$.

[quote="nayel"]I don't think $ PA,PB,PC$ always form a triangle. For example consider the following diagram:
[/quote]

I'm sure this is the wrong Forum to ask, but, where can details be obtained for making graphics such as the yours?

[quote="radio"]Well, actually it means you can take $ PA,PB,PC$ (you can move it  ) and it can form a triangle. [/quote]

I think I see...

[quote="vector2"]I'm sure this is the wrong Forum to ask, but, where can details be obtained for making graphics such as the yours?
[/quote]

You'll need to use the program called Asymptote.  You can either google it or look in the latex/asymptote forum.


geometry
conics
quadratics
algebra
quadratic formula

[hide]
first rearrange the equations

(1)$12x+24y = 11xy$
(2)$6x-9y = 2xy$

mutliply the second by two to obtain $12x-18y = 4xy$ and then subtract the second equation from the first so that our 12x value cancels out and we are left with

$42y = 7xy$ divide y from both side
$42 = 7x \Rightarrow \boxed{x = 6}$

plug $x=6$ into any one of the other equations to find y

$12(6)+24y = 11(6)y$
$72+24y = 66y$
$72 = 42y$
$y = \frac{72}{42}= \boxed{\frac{12}{7}}$

so our ordered pair is

$\boxed{(6, \frac{12}{7})}$
[/hide]

OMG those signs next to the problems make me think that the problem is complicated and try to do too much! in aops vol 1. I think it's called the threaded needle, but it's usually next to the problems that aren't hard but look like they are hard because of the huge needle.  :ninja:

cool. well. at least now I know I didn't miss anything that was very hard in the chapter.  :lol:

which chapter are you on? I'm hoping to get the aops texts sometime soon, im strapped for cash  :blush:

Do you have the second?

[quote="now a ranger"]I have AoPS 1 and 2 with me. I'm currently on Chapter 5 in my 2nd time around this book.  I don't have the solutions, although I wish I did. But there really is no need, as AoPS 1 problems aren't that hard. I'm just learning a lot of Algebra 2 material for a tournament coming up.

I'm strapped too.[/quote]
I looked over book 1 this spring break.  It was pretty easy but i learned a few nifty tricks.  AoPS book 2 is a bit hard, i think i'm gonna do it over the summer.  Good luck with your tournament!!!  :D

[quote="now a ranger"]Find the ordered pair $(x,y)$ that is the solution of the system

$\frac{x+2y}{xy}=\frac{11}{12}$

$\frac{2x-3y}{xy}= \frac{2}{3}$[/quote]
[hide] You can instead say that $\frac{1}{y}+\frac{2}{x}=\frac{11}{12}$ and $\frac{2}{y}-\frac{3}{x}=\frac{2}{3}$.  Now, say that $a=\frac{1}{x}$ and $b=\frac{1}{y}$.  Thus, $2a+b=\frac{11}{12}\Rightarrow 4a+2b=\frac{11}{6}$ and $2b-3a=\frac{2}{3}$.  Subtracting our equations, we obtain that \[7a=\frac{11}{6}-\frac{2}{3}=\frac{11-4}{6}=\frac{7}{6}\Rightarrow a=\frac{1}{6}\]  From here, we can figure our that since $2a+b=\frac{11}{12}\Rightarrow b+\frac{1}{3}=\frac{11}{12}\Rightarrow b=\frac{7}{12}$.  Thus, $x=6, y=\frac{12}{7}$, so our pair is $(6, \frac{12}{7})$ [/hide]

[quote="now a ranger"]Thanks! I'm in geometry though, so scoring high on an Algebra 2 is a long shot. I hope AoPS 1 might help a lot, as I've learned a lot this spring break so far, and I'm only half way done.  :lol:

I'm skipping the geometry chapters as I've already looked over them thoroughly, so I'm pretty sure I can finish the non-geometry chapters by the end of spring break.

Do you ( everyone in general ) think that AoPS vol. 1 covers mostly what you need in algebra 2 to solve Alg. 2 type problems?[/quote]
well not really. i mean maybe some things here and there *pulls out aops 1* for example most people don't learn variations in algebra 1 but i am probably wrong
AND NO NOW A RANGER
it isn't a long shot unless they have conics on there.
algebra 2=algebra 1+some stuff

[quote="now a ranger"]Thanks! I'm in geometry though, so scoring high on an Algebra 2 is a long shot. I hope AoPS 1 might help a lot, as I've learned a lot this spring break so far, and I'm only half way done.  :lol:

I'm skipping the geometry chapters as I've already looked over them thoroughly, so I'm pretty sure I can finish the non-geometry chapters by the end of spring break.

Do you ( everyone in general ) think that AoPS vol. 1 covers mostly what you need in algebra 2 to solve Alg. 2 type problems?[/quote]

I'm 2/3 of the way through with the school year, so i'm about 2/3 done with Alg 2 and we havent learned ANYTHING new except matricies which are really easy.  We did a few things with $i$ but not much.  Right now we are learning (if thats what ya wanna call it) the quadratic forumla.  I did the quadratic formula in 7th grade!  We also might cover logs, and thats about it.  Today, i got recommended for Pre-Calc Honors!  :D

[quote="now a ranger"]yeah, the only thing in algebra 2 that I know of not in AoPS volume 1 is logs+matrices.

So maybe that's why it's new.  :lol:[/quote]

I think alg I and alg II should be shortened into 1 course.

[quote="davidyko"]Mm, I missed matrices; our Algebra 2 didn't have them; actually, I wouldn't know, seeing as I skipped it and went straight to Precalc Honors, which people say is the hardest class at my school, but in fact is a breeze.
Maybe chapter 1 of AoPS Vol. II? Logs, I believe are Algebra 2, aren't they?[/quote]
I beleive logs are taught in Alg II.


calculus
integration
trigonometry
function
calculus computations

Find $ \int {\frac {e^{\sqrt {x}\sin x}}{\tan x}} dx$

How could you expect such a thing to have an elementary antiderivative?

I am solving some problems in physics. One of them leads to this integral. And I'm stuck here  :( . I hope to get some useful results.  :|

You're stuck here because this integral cannot be expressed in terms of elementary functions (although it's easy enough to, say, write down a few terms of a Taylor series).  Either you've gone down a wrong path or your question is lacking a nice answer.

hello, try this
http://www.wolframalpha.com/
Sonnhard.


search
combinatorics unsolved
combinatorics

Let $ X$ be a set with $ n$ elements, and let $ A_{1}$, $ A_{2}$, ..., $ A_{m}$ be subsets of $ X$ such that:

1) $ |A_{i}| \equal{} 3$ for every $ i\in\left\{1,2,...,m\right\}$;
2) $ |A_{i}\cap A_{j}|\leq 1$ for all $ i,j\in\left\{1,2,...,m\right\}$ such that $ i \neq j$.

Prove that there exists a subset $ A$ of $ X$ such that $ A$ has at least $ \sqrt {2n}$ elements, and for every $ i\in\left\{1,2,...,m\right\}$, the set $ A$ does not contain $ A_{i}$.

This one has a beard... http://www.mathlinks.ro/Forum/viewtopic.php?t=48298 for instance.
Please do a search for "iran* [year]" before you repost these problems.

  darij

I only posted this problem as a placeholder. See moderator forum.

Why need a placeholder? Because of the alternative formulation in the original topic? Okay, but I'm replacing your problem statement since it has buggy LaTeX.

  darij


function
calculus
derivative
algebra
functional equation
real analysis
real analysis unsolved

Find all differentiable functions $ f: \mathbb{R} \to \mathbb{R}$ such that:
\[ y \cdot f(x)\minus{}x \cdot f(y)\equal{}k \left( f(x)\minus{}f(y) \right), \forall x,y \in \mathbb{R},\]
where $ k$ is fixed real number.

[hide]Taking a derivative with respect to $ x$, we have $ y f'(x) \minus{} f(y) \equal{} kf'(x)$, so $ f(y) \equal{} f'(x)(y \minus{} k)$.  So the only possible functions are linear functions with $ x$-intercept $ k$.  It's easy to check that $ y\cdot m(x \minus{} k) \minus{} x\cdot m(y \minus{} k) \equal{} k(m\cdot (x \minus{} k) \minus{} m\cdot (y \minus{} k))$ for all $ x, y, m$, so we're done.[/hide]

[hide="Alternate method"]Just substitute $ y\equal{}0$ for $ \minus{}xf(0)\equal{}k(f(x)\minus{}f(0))$, which can be rewritten $ kf(x)\equal{}(k\minus{}x)f(0)$. If $ k\neq 0$, $ f(x)\equal{}a\minus{}\frac{a}{k}x$, where $ a\equal{}0$. In the generic functional equation, this becomes $ ay\minus{}\frac{axy}{k}\minus{}ax\plus{}\frac{axy}{k}\equal{}ak\minus{}ax\minus{}ak\plus{}ay$, which is always true.
If $ k\equal{}0$, we have instead that $ yf(x)\equal{}xf(y)$ for all $ x$ and $ y$, so $ f(x)\equal{}xf(1)\equal{}ax$ for some $ a$. Again, this satisfies the functional equation.[/hide]
No assumption of differentiability needed.


Prove that the exits no rational numbers x and y such that:

\[ (x \plus{} y sqrt(3))^n  \equal{} sqrt( 1 \plus{} sqrt (3))\] where n is a natural number.

[hide="Solution"] We make the stronger claim that $ \sqrt{1 \plus{} \sqrt{3}} \not \in \mathbb{Q}[\sqrt{3}]$.  Suppose $ \sqrt{1 \plus{} \sqrt{3}} \equal{} a$.  Then $ 1 \plus{} \sqrt{3} \equal{} a^2 \implies N(1 \plus{} \sqrt{3}) \equal{} \minus{}2 \equal{} N(a^2)\equal{} N(a)^2$, the square of a rational; but this is absurd. [/hide]


Consider the number which consists of the first 44 positive integers, so

[b]N=123456789101112.......424344[/b]

What is the remainder when [b]N[/b] is divided by 45?

The remainder when $N$ is divided by 5 is 4.  Similarly $N$ is a multiple of 9.  Note that 45 is divisible by 5 and 9.  Therefore the number must be a multiple of 9 (use rule for divisibilty by 9 i.e. add the digits up) that leaves a remainder of 4 when divided by 9.   It must also be less than 45 (remainder has to be less than what you are dividing by).  So the possible multiples of 9 are 0, 9, 18, 36.  Only 9 leaves a remainder of 4 when divided by 5.

Gah, I missed that problem cause I was retarded and said that $N$ left a remainder of 5 or something when divided by 9...I need to learn to add digits up...

Also, you can find all the answer to the USC contest on their site.  They just updated it: http://www.math.sc.edu/contest/problems.html  I don't know what their solution looks like (my adobe isn't working :( )..it might use mods which is just a shorthand of what I said above.

Wait, are you from South Carolina too, since you knew it was from the USC test yesterday?

What did you get on the test yesterday? ;)

We should probably talk via Private Messages.  I just sent you one.


Can you allow orders like 2,3,4,5,1?

[hide="If so..."]Any card can be picked first. The probability of the next card being correct is $\frac{1}{x-1}$, then $\frac{1}{x-2}$, etc. so $\frac{1}{x!}=\frac{1}{120}$. $x=5$, so Jason has $6$ cards[/hide]

But isn't there 1 correct order?

[hide]So it would be 5?[/hide]

[hide="solution"]The total possible number of orders for $n$ things is $n!$, so we need to find a number such that it's factorial equals 120. $n=5$ works.[/hide]

[hide]
theres $n!$ ways to arrange $n$ cards.
So we need to find $n$ for $n!=120$
You should know $5!=120$
Therefore theres $\boxed{5}$ cards[/hide]

[hide][i]1/n![/i]would equal 120; 1/n!=120
Solving the equation, n = 5.
Jason has [b]5[/b] cards.[/hide]

[quote="sferics"]Jason has prepared a speech for his local Student Council club, and he has created cards to help him remember it. As Jason walks up to the podium to give his speech, he trips, drops all of his cards, and begins to randomly pick them up. If the probability that he has picked them up in the correct order is 1/120, how many cards does Jason have?[/quote]

[hide]5 x!=120 then x=5[/hide]


If $ a,m,n$ are positive integers with $ a>1$ then $ a^m\plus{}1|a^n\plus{}1\Rightarrow m|n$

Let $ n \equal{} km \plus{} l,l < m$, then
$ a^n \plus{} 1 \equal{} a^l(a^m \plus{} 1)(a^{(k \minus{} 1)m} \minus{} a^{(k \minus{} 2)m} \plus{} ...( \minus{} 1)^{k \minus{} 1}) \plus{} ( \minus{} 1)^ka^l \plus{} 1 \equal{} ( \minus{} 1)^ka^l \plus{} 1\mod a^m \plus{} 1$.


AMC
USA(J)MO
USAMO
MATHCOUNTS
AMC 10
AIME

Which reason do you find yourself pursuing Mathematics over?

a)  You are fascinated by Mathematics.  You marvel at the glorius patterns and concealed secrets deeply woven within the world of numbers.  The beauitiful properties and abstract concept.  You find youself trying to understand why things work, how they work, and most of all, can they be proven.

b)  Beyond contest-level Mathematics, you have no real desire for the subject.  You simply crave the feeling of mastery and superiority.  You do math because you're good at it, and compete for the joy if winning.  You think math is intriguing, but you pursue the contests, opposed to the subject itself.

I am the a-type because I really don't care much about contests, even though they are an extra-edge for me to challenge myself rather than competing with others.

I am an "a" all the way!

I think you should repost this topic as a poll before too many people reply.

[quote="h_s_potter2002"]I don't think anyone would admit to being b.[/quote]

Just what I'm thinking.  Or most likely, if a person was B, the person might not be part of this forum.

Oh btw, you forgot an option.

C) A jar of almonds.

I go with that. :bye:

A little a - math is pretty - but mostly b. It's what makes math fun and rewarding, and the contests reward me studying math so much of the time.

Similar to solafidearms, many of my freinds admit their passion for Math, is simply their interest in contest mathematics.  Theres nothing wrong with it, but a course of choice.

mainly A, but due to a bad teacher in elementary, $\frac{\pi}{100^{100}}$th B  :P

definitely A - i like the concept of understanding math - though i cant say i dont enjoy the feeling of mastery and superiority. Yet, i dont even participate in contests, so..

70% B, 30% A. I have taken a few college classes in math, but the only reason I ever became somewhat decent at math is because of contest math. Advanced math is really boring, to be honest. Why do I like contest math? Well, it gives me an opportunity to compete and its fun.  That is probably the reason why I often stay up until 1-2 AM, doing math before homework. But I must say that I really marvel at the workings of math sometimes. I find it amazing that you can use math to solve so many problems, and the beauty of theorems.

And yes, I feel good when I "master" or conquer a problem. Especially a tough one.

for me, it's pretty much 100% "A". science and math is very interesting to me. but i do have to find a balance between personal interest and security. i hope i can get enough of both.

i don't remember winning any contests so it can't be "B". but i did earn a scholarship because of science and math.

cheers,
mike.

Some of both.  Competition math is what mostly keeps me going, rewarded, and motivated, but even being interested in competition math is the result of A.

[quote="GoBraves"]100% B. Competitions( Like USAMO) will help me in getting into a good college. Beyond that, I have no real interest for it.[/quote]

That's seriously the stupidest reason for doing anything.

[quote="MathFiend"]i would say that i'm a little bit of both, however i voted for a because that is more fitting. i love mathematics, but i also love the challenge and competition that it brings. i would never have come to love maths if it weren't for the competition and would never have come to love the subject. so i think i started out as a 'b' and now have transformed into a competitive 'a'.  :lol:[/quote]

I'd have to say this describes a lot of people, myself included.  I think this is because the younger you are, the less you care about learning, but you do like to compete.  This is how you initially get interested in math.  This, I think, is the main merit to competitions.  Then, as people strive to succeed in these competitions, they eventually begin to study math that may not be helpful in competitions.  Then they begin to realize how fun math is.

I like math because I like numbers. I can generate numbers quickly, and they are not always in one kind of fashion (some people, when they are asked to make random numbers, do 33, 44, 55, 77, etc.). I think some numbers sound better than others (3575 versus 2436) and some look better than others (236 over 267). That's mostly A. Competitions are how I got good at math and [i]started[/i] to like it, but this is all supported by choice A, which let me carry on my enjoyment of mathmatics. I count calories not because I'm overweight (*cough*), but because I like to add the numbers up (and I try to memorize the nutrition facts of everything i see for no reason - bizarre).

Well, i probably worked the opposite way of most...
I always loved math as a child...i'm told that when I was about 3, riding in the car, i'd suddenly as "How far away is that mountain?" and would get the reply "100 miles" and would do something with that.  Or just randomly blurt out "What is 3 x 3?"
I remember using up notebook after notebook after notebook computing things as a child at about the age of 6.  For example, calculating 2^100.  I kinda discovered a lot of things at that time, perhaps.  Then I found out about mathematica and was fascinated.
I never relaly liked competitions: they just made me nervous, and the problems usually weren't all that interesting (I was in 6th grade, doing this math thing which basically comes in several rounds with a certain number of problems each, totaling to 40 problems.).  Even in Mathcounts, I didn't really enjoy the math itself that much.
Then in 8th grade, I went to Mathcounts Nationals: that was quite fun, I started to like the competitions more then.  The end of the AMC10 and the AIME were what got me really interested in competition problems: they were challenging, even to think about afterwards.  However, they could all be solved with simple tools.  Hence, I spent this summer which just passed (after my freshman year) just doing math in much of my spare time, thinking, and that leads me to my current state.  I guess I take competitions more as a secondary thing.

Competitons are there to give you interesting problems to pursue and new math ideas to look at, being too competitive is unhealthy!

A little bit of both! I like math, I like math contests, but math contests aren't the only reason I do math.  I'm competitive, but not crazily competitive like some people.

To be honest, strictly B. The only thing I like Math is that it is something that I want to master, in order to win contests, make excellent scores on tests, and to look good for colleges and it's addicting. But other than that, I have no intention of pursuing mathematics as a career option, in fact, once I get to college, I will be pursuing medical school, and mathematics is not necessary for that field. If it wasn't for contests, and me needing to excell in math to do well in tests for college (SAT), I wouldn't care about math at all. In other words, I just do maths because I have to excell at it.


Strip your personality bare for Obama government job. See the questionnaire in the attachment.

Hell to the yes. What if it turned out that Rahm Emanuel's favorite hobby is to pal around with the terrorists he keeps around in his guest bathroom? That would be SOOO embarrassing. Also, then Sarah Palin would never stop talking, and I can't imagine a scenario much worse than [i]that[/i].

No, but seriously, though, the administration really can't afford to have revelations popping up here and there about tax evasions or rape charges involving high-ranking officials. I wish this level of disclosure weren't necessary, as I'm sure it does sometimes disqualify otherwise very qualified people, but the public and the media that serve them apparently require it... though it's not really stripping one's personality bare, either.

Which questions asks about all the people you cheat on your wife with? I couldn't find it


topology
algebra
function
domain
calculus
integration
analytic geometry

Construct a subset of the square I in $ R^2$ such that on the one hand its intersection with any vertical line and any horizontal line consists of at most one point, while on the other hand its closure equals I.

Maybe if you could submit a picture too..  :ninja:

Permute a Hamel basis of $ \mathbb{R}$ over $ \mathbb{Q}$, and extend to a linear isomorphism. The graph of this map is dense in the plane. Now, take a homeomorphism between the line and the open interval, and apply to the domain and range.
There are not, and will never be, pictures.

I don't know what a Hamel basis is (I can guess and I think I'm right!) or how to extend it to a linear isomorphism.

This is in the fourth section of the integration chapter of my book and unless the author is wicked, I think there must be a more elementary solution based on real analysis/calculus concepts. The book is at a level of an introductory analysis course.

As for the pictures, well, how do you do mathematics in 4 space then?  :roll:

A Hamel basis is simply a basis in the usual linear algebra sense- with no topological information, so infinite linear combinations don't exist. For an infinite-dimensional space, a Hamel basis is a very ugly object.

The reason there aren't any pictures is that the "construction" strongly requires the Axiom of Choice. I could specify infinitely many values independently, and still not know how to fill in the rest. I was just doing that to generate a bijection from the set to itself, and there are other ways.

Or just take the set of points with rational coordinates and rotate it a bit. Or. construct such countable set inductively (at each step you can add a point as close to any point you wish, which is enough to get density). Or... As long as it is just topological density and not positive outer measure, say, you can essentially do whatever you want :)

A picture of a dense set? Hmmm... That's quite a challenge indeed! :lol:

jmerry how does one extend that basis to a linear isomorphism? I don't understand what that means.

fedja I don't  understand what measure and density is yet. And I was just kidding about the picture part  :P

A Hamel basis for $ \bf R$ over $ \bf Q$ is a vector space basis (finite linear combinations; exists by axiom of choice).  Choose a permutation of such a basis.  Then there exists a unique invertible linear map that acts on the basis as that permutation.

This is an idea you should be familiar with from standard finite-dimensional linear algebra: if you choose a destination for every element of a basis, you can extend it uniquely to a linear map by simply imposing linearity.  (You get at least one map because bases are linearly independent; you get at most one because bases span.)

My professor has also suggested a most elementary approach which is a stroke of genius. After I understand it I may post it

Inductively select points $ p_{k,j} ,j \equal{} 1,...,2^k$, as follows: For k=0 pick any point $ p_{0,1}  \in I$ , and for $ k \equal{} 1,2,...$ divide I into $ 2^{2k}$equal subsquares $ I_{k,1} ,...,I_{k,2^{2k} }$of edge length $ 2^{ \minus{} k}$, and, for $ j \equal{} 1,...,2^{2k}$, pick a point $ p_{k,j}$ from $ I_{k,j}$in such a way that the point $ p_{k,j}$has x coordinate distinct from the x-coordinates of all previously chosen points $ p_{l,i}$and y-coordinate distinct from the y-coordinates of all previously chosen points $ p_{l,i}$. The (infinite) collection of points $ A \equal{} \{ p_{k,j} |j \equal{} 1,...,2^k ,k \equal{} 0,1,2,...\}$so chosen has the required property that A intersects each vertical line in at most one point and each horizontal line in at most one point, and of course A has closure equal to I because for any $ k \equal{} 0,1,2,...$ and any $ p \in I$ there is a point $ p_{k,j}$in the collection A which which has distance $ \le \sqrt 2 /2^k$from p.

So elegant and simple!

Yeap, see the second approach in my post above. :P How about rotated (by an angle with irrational tangent) $ \mathbb Q^2$? I find it even simpler. :D

Even I could have come up with the very insightful suggestion "at each step you can add a point as close to any point you wish" on my own :p

I don't see what you mean by rotate though. Maybe you should say a little more :)

[quote="§outh§tar"]Even I could have come up with the very insightful suggestion "at each step you can add a point as close to any point you wish" on my own :P[/quote]
Yes. But this is all you need to construct a countable dense set in a separable space! Take any countable dense set $ \{y_k\}$ in $ I$ (I hope, you can come up with one) and add points $ x_k$ in such a way that $ |x_k\minus{}y_k|<\frac 1k$. Then $ \{x_k\}$ will be also dense.

[quote]I don't see what you mean by rotate though. Maybe you should say a little more :)[/quote]
By rotation $ R_\theta$, I mean the transformation of the plane that maps the origin to itself and every other poits $ A$ to the point $ A'$ such that $ |OA|\equal{}|OA'|$ and $ \measuredangle AOA'\equal{}\theta$. If $ E$ is a set of points, by rotated (by angle $ \theta$) set $ E$, I mean the set of all images of the points of $ E$ under rotation $ R_\theta$.

This reminds me of Sierpinski's partition of $ I$ into two subsets $ A$ and $ B$ such that the intersection of $ A$ with any vertical line has measure zero, and the intersection of $ B$ with any horizontal line has measure zero.  :o


I have a friend who recently took a class on the internet and got approval from the high school that he could skip pre-calc.

Do most counselors allow students to skip math grades by taking online courses from sites such as this one and the John Hopkins website?

i dunno about AoPS.... at least there wouldnt be any course having a similar curriculum basis as the schools... but most schools accept transcripts from JHU or EPGY. I did alot of course in EPGY....

The way they did it at my MS was I just took the final for the class and passed it, so I got to skip. Last year, I also skipped Calculus AB to take BC because my school requires AB before BC... I just took an old AP test and (obviously) got a 5.

Well each one is different; there aren't any guarantees that your school will do it the same way, or even at all. You should go talk to the counselors.

Actually more like someone who takes care of your transcripts, e.g. for us, it is the registrar/Registration Director. She handle all my transcripts, so when i asked my IB coordinator about my EPGY Ap transcripts, she told me to see the registrar, and she added in for me. I dont think Principal or Counselor confirmation are needed.... although because i am different from you, you should go talk to your counselor first, like paladin8 said.

[quote="blahblahblah"]One of the great things about Canadian high schools is the (relative) lack of bureaucracy.  When I felt like skipping school (in grade 12) I skipped, and no one cared.  When I refused to take Math 11 because it would have bored me to death, I took Math 12 instead and a grade of 100% for Math 12 (a graduation requirement :? ) magically showed up on my transcript.

Yes, high school was great fun.[/quote]

Ok, I officially hate Canada. Ok, I had to argue my way into skipping precalculus, and then I had to fight to take the final, because guidance really does not care at all.........


If A+B+C=180, how can we prove that tan(A)+tan(B)+tan(C) = tan(A)tan(B)tan(C)?

The most straightforward way is to notice that C = 180 :^o: - (A+B). So tan C = - tan (A+B). Then use the law of tangent of sums to verify the identity.

tan (A) + tan (B) - tan (A+B) = - tan (A)*tan (B)*tan (A+B).
tan (A) + tan (B) = (1 - tan (A)*tan(B))*tan (A+B)
(tan (A) + tan (B))/(1 - tan (A)*tan (B)) = tan (A+B)

Voila!

.[hide]

tan (a+b+c) = 0 = [tan(a)+tan(b+c)]/(1-tan(a)*tan(b+c)).  So tan(a) + tan(b+c) = 0.

tan(b+c) = [tan(b) + tan(c)]/(1-tan(b)*tan(c)).

So tan(a) + [tan(b) + tan(c)]/(1-tan(b)*tan(c)) = 0.

[tan(a) - tanAtanBtanC + tan(b) + tan(c)]/(1-tanBtanC) = 0.  So tan(a)-tan(a)tan(b)tan(c)+tan(b)+tan(c)=0.[hide][/hide][/hide]


trigonometry
algebra
polynomial
binomial theorem
complex numbers

Hi, can anyone help me with this? Express cos 5x in terms of cos x using the relationship
e^ (ix) = cosx + isin x.

Thus, show that 16x^4  - 20x^2  + 5=0 has a root of x = cos (pi/10).

help me please, workings would do. I sorta know how to express cos 5x in terms of cos x, i think its using the binomial theorem? or anything shorter? after this, how do i answer the second problem? thanks again!!!:D

[hide="hint 1"]$ e^{ix}\equal{}\cos x\plus{}i \sin x$
$ e^{\minus{}ix}\equal{}\cos x\minus{}i \sin x$
what is then $ e^{ix}\plus{}e^{\minus{}ix}$??[/hide]

[hide="hint 2"]Use the reciprocal polynomial $ z^5\plus{}\frac {1}{z^5}\equal{}(z\plus{}\frac 1z)^5\minus{}5(z\plus{}\frac 1z)^3\plus{}5(z\plus{}\frac 1z)$[/hide]

[u]Expressing cos(5x) in terms of cos(x)[/u]

$ \cos x$ is the real part of $ e^{ix}$ and $ \sin x$ is the imaginary part of $ e^{ix}$

Similarly, $ \cos 5x$ is the real part of $ e^{i5x} \equal{} \left( e^{ix} \right) ^5 \equal{} (\cos x \plus{} i \sin x)^5$

So now, you expand that (with the binomial theorem) to get:

$ e^{i5x} \equal{} \cos ^5 x\plus{} 5 i \cos ^4 x \sin x \minus{} 10 \cos ^3 x \sin ^2 x \minus{} 10 i \cos ^2 x \sin ^3 x \plus{}  5 \cos x \sin ^4 x \plus{} i \sin ^5 x$

$ \cos 5x$ is the real part of that, so dropping the imaginary terms, we get:
$ \cos 5x \equal{} \cos ^5 x \minus{} 10 \cos ^3 x \sin ^2 x \plus{}  5 \cos x \sin ^4 x$

Now use $ \cos ^2 x \equal{} 1 \minus{} \sin ^2 x$ to get:
$ \cos 5x \equal{} \cos ^5 x \minus{} 10 \cos ^3 x \plus{} 10 \cos ^5 x \plus{}5 \cos x \minus{} 10 \cos ^3 x \plus{} 5 \cos ^5 x$
$ \cos 5x \equal{} 16 \cos ^5 x \minus{} 20 \cos ^3 x \plus{} 5 \cos x$

[u]Solving the polynomial:[/u]

Note that $ 0$ does not solve this polynomial. Hence, multiply the polynomial by $ x$ to introduce an extraneous solution of $ x\equal{}0$, which we will discount.

$ 16 x^5 \minus{} 20 x^3 \plus{} 5x \equal{} 0$

Note that $ x \equal{} \cos \theta$ satisfies the above equation whenever $ \cos 5 \theta \equal{} 0$ (see result of previous problem).

So now, $ x \equal{} \cos \theta$ solves $ 16 x^4 \minus{} 20 x^2 \plus{} 5 \equal{} 0$ whenever $ \cos 5 \theta \equal{} 0$ and $ \cos \theta \neq 0$

$ \theta \equal{} \frac{k\pi}{10}$ satisfies this condition whenever $ k$ is odd and not a multiple of $ 5$. Thus, $ x \equal{} \cos \frac{\pi}{10}$ is one solution to the equation. Can you find the other three?

hmm... with respect to 0<=x<=2pi, the other 3 are:

5x = 3pi/2, 5pi/2, 7pi/2. so,

x= 3pi/10, pi/2 and 7pi/10.

thanks alot for the great help!:D i was right in doing the first all along lol:))

Oops!

You forgot $ 5 \theta \equal{} \frac{9 \pi}{2} \iff x \equal{} \cos \frac{9 \pi}{10}$

You also included $ 5 \theta \equal{} 5 \frac{5 \pi}{2} \implies x \equal{} \cos \frac{\pi}{2} \equal{} 0$, which was the extraneous solution we added in by multiplying by $ x$  :P This was the one we didn't want.

Otherwise, right idea.


Prove that among 6 persons, there are always 3 who know each other or three that are complete strangers.

Are you sure this is true for all cases? What do you mean exactly by know the person?

Even Stronger: In any bicolored K6, prove that there exist two monochromatic triangles.

For this problem, you can assume that knowing is mutual.  So if A knows B then B knows A.

[hide="hint 1"]
Pigeonhole Principle
[/hide]

[hide="hint 2"]
Focus on one person (or one vertex) first
[/hide]

[quote="t0rajir0u"]"Knowing" is a binary relation that is symmetric but not transitive.

[b]Alternate statement:[/b]  Prove that in any complete graph with $ 6$ vertices colored with two colors, a monochromatic triangle exists.[/quote]

Suppose that the edges of a G_6 are red and green. Take any of the six points and call it P. At least 3 of the 5 lines which start at P are of the same color, say green. These green lines end at three points. A,B,C. if any side of the triangle A,B,C is green, then we have a green triangle. If not, we have a red triangle. In both cases we have a monochromatic triangle.

similar: A three colored G_80 contains a monochromatic quadrilateral.


geometry
ratio
analytic geometry
incenter
circumcircle
vector
trigonometry

Here is a very helpful fact from elementary geometry which is known to Romanians as [b]transversal theorem[/b]:

[b]Theorem 1, the transversal theorem.[/b] Let A, B, C be three points on one line, and let P be a point not on this line. Let A', B', C' be three arbitrary points on the lines AP, BP, CP, respectively. Then, the points A', B', C' are collinear if and only if we have

$BC\cdot\frac{AP}{A^{\prime}P} + CA\cdot\frac{BP}{B^{\prime}P} + AB\cdot\frac{CP}{C^{\prime}P} = 0$,

or, equivalently,

$BC\cdot\frac{AA^{\prime}}{A^{\prime}P} + CA\cdot\frac{BB^{\prime}}{B^{\prime}P} + AB\cdot\frac{CC^{\prime}}{C^{\prime}P} = 0$.

Hereby, all segments are directed segments.

I know three different proofs of this criterion:
- a simple proof using ratios of areas of different triangles;
- a proof using the Menelaos theorem (not very beautiful, but easy);
- an ugly proof using the linear independence of vectors.

You are encouraged to find new proofs. Also, I am sure that there are many applications of the transversal theorem. Here is one such application:

Prove, using Theorem 1, the following fact:

[b]Theorem 2.[/b] Let ABC be an arbitrary triangle, and let P and P' be two points in its plane. Let the lines AP and BP meet the lines BC and CA at the points A' and B', respectively, and let the lines AP' and BP' meet the lines BC and CA at the points A" and B", respectively. Furthermore, let Q and Q' be the isogonal conjugates of the points P and P' with respect to the triangle ABC. Then, the point Q lies on the line A"B" if and only if the point Q' lies on the line A'B'.

This fact is actually very easy to prove using trilinear coordinates, but not so easy to prove synthetically. The transversal theorem offers a very straightforward proof.

If we apply Theorem 2 to the case when the point P is the incenter of triangle ABC and the point P' is the orthocenter of triangle ABC, we get:

[b]Theorem 3.[/b] Let ABC be an arbitrary triangle. Let the angle bisectors of the angles CAB and ABC meet the lines BC and CA at the points A' and B', respectively. Let A" and B" be the feet of the altitudes of triangle ABC issuing from the vertices A and B. Then, the incenter of triangle ABC lies on the line A"B" if and only if the circumcenter of triangle ABC lies on the line A'B'.

This theorem is an extension of the problem 4 in the 2nd round of the BWM (German National Mathematical Contest) 2002. (Actually, the BWM problem 4 assumed the triangle ABC to be acute and asked to show that the incenter of triangle ABC lies on the [i]segment[/i] A"B" if and only if the circumcenter of triangle ABC lies on the [i]segment[/i] A'B'; but this immediately follows from Theorem 3.)

Another application of Theorem 1 is that one can easily derive the Menelaos theorem from Theorem 1. However, such a derivation makes sense only if one has not already used the Menelaos theorem in the proof of Theorem 1.

  Darij

[quote="darij grinberg"]

$BC\cdot\frac{AP}{A^{\prime}P} + CA\cdot\frac{BP}{B^{\prime}P} + AB\cdot\frac{CP}{C^{\prime}P} = 0$,

or, equivalently,

$BC\cdot\frac{AA^{\prime}}{A^{\prime}P} + CA\cdot\frac{BB^{\prime}}{B^{\prime}P} + AB\cdot\frac{CC^{\prime}}{C^{\prime}P} = 0$.
[/quote]

How do you manage to get 0 as everything is positive, isn't it ? Shouldn't it be 1...

The segments are directed, so negative values of some of them are possible.

I see this theorem is becoming quite popular. And it all started in 2002 :) ... ah great times :P

Can somebody show proof by Menelaus of this beatiful theorem - I'll be really thankful

I'll show it just one way; the other way I think we can get by the traditional trick of defining, for example, A'' where PA and B'C' intersect and showing that A'' and A' divide PA in the same ratio (by the given and by the first half of the theorem), so they must be the same point and thus A' lies on B'C' on desired. So anyway, assume A'B'C' are collinear and that the line they lie on intersects line ABC at X. Let F_a be AA'/A'P, etc. Then, by Menelaos, we can find AX/XB * BB'/B'P * PA'/A'A = -1, i.e. AX / XB = - F_a / F_b, i.e. AX / F_a = BX / F_b. Similarly, we can find that both of them are equal to CX / F_c. So now BC (F_a) + CA (F_b) + AB (F_c) = (BX + XC) (F_a) + (CX + XA) (F_b) + (AX + XB) (F_c) = AX * F_b + XC * F_a + BX * F_c + XA * F_b + CX * F_a + XB * F_c = 0.

I hope it's right.

$\rightarrow$  first, how does the proof using ratios of areas go?

$\rightarrow$  second, and more important, what does this mean: " linear independence of vectors " and how can it be used for solving problems in general ?

Sorry, I completely forgot about the questions in this thread! Actually, I will show two of my three proofs mentioned in post #1: the Menelaos proof (it is different from, and less nice than the one given by MysticTerminator in post #5) and the areas proof. The one using vectors is really too boring and I am not the right person to explain the notion of linear independence (here in Germany, we learn it in school, anyway it should be contained in any tutorial on linear algebra).

So here it goes.

[i]Proof of Theorem 1.[/i]

In order to prove Theorem 1, we have to show three assertions:

[i]Assertion 1:[/i] If the points A', B', C' are collinear, then $BC\cdot \frac{AP}{A^{\prime }P}+CA\cdot \frac{BP}{B^{\prime }P}+AB\cdot \frac{CP}{C^{\prime }P}=0$.

[i]Assertion 2:[/i] If $BC\cdot \frac{AP}{A^{\prime }P}+CA\cdot \frac{BP}{B^{\prime }P}+AB\cdot \frac{CP}{C^{\prime }P}=0$, then the points A', B', C' are collinear.

[i]Assertion 3:[/i] The equations $BC\cdot \frac{AP}{A^{\prime }P}+CA\cdot \frac{BP}{B^{\prime }P}+AB\cdot \frac{CP}{C^{\prime }P}=0$ and $BC\cdot \frac{AA^{\prime }}{A^{\prime }P}+CA\cdot \frac{BB^{\prime }}{B^{\prime }P}+AB\cdot \frac{CC^{\prime }}{C^{\prime }P}=0$ are equivalent.

Actually, Assertion 3 is trivial, since

$BC\cdot \frac{AP}{A^{\prime }P}+CA\cdot \frac{BP}{B^{\prime }P}+AB\cdot \frac{CP}{C^{\prime }P}$
$=BC\cdot \frac{AA^{\prime }+A^{\prime }P}{A^{\prime }P}+CA\cdot \frac{BB^{\prime }+B^{\prime }P}{B^{\prime }P}+AB\cdot \frac{CC^{\prime }+C^{\prime }P}{C^{\prime }P}$
$=BC\cdot \left( \frac{AA^{\prime }}{A^{\prime }P}+1\right) +CA\cdot \left( \frac{BB^{\prime }}{B^{\prime }P}+1\right) +AB\cdot \left( \frac{CC^{\prime }}{C^{\prime }P}+1\right) $
$=\left( BC\cdot \frac{AA^{\prime }}{A^{\prime }P}+CA\cdot \frac{BB^{\prime }}{B^{\prime }P}+AB\cdot \frac{CC^{\prime }}{C^{\prime }P}\right) +\left( BC+CA+AB\right) $
$=BC\cdot \frac{AA^{\prime }}{A^{\prime }P}+CA\cdot \frac{BB^{\prime }}{B^{\prime }P}+AB\cdot \frac{CC^{\prime }}{C^{\prime }P}$.

So it remains to prove Assertions 1 and 2.

[i]First proof of Assertion 1.[/i] We have assumed that the points A', B', C' are collinear, i. e. they lie on one line. Let the parallel to this line through the point A meet the lines BP and CP at the points M and N, respectively. Then, Thales yields $\frac{MA}{NM}=\frac{B^{\prime }A^{\prime }}{C^{\prime }B^{\prime }}$. On the other hand, applying the Menelaos theorem to the triangle ANC with the collinear points B, M, P, we get $\frac{AB}{BC}\cdot \frac{CP}{PN}\cdot \frac{NM}{MA}=-1$. Hence,

$\frac{AB}{BC}=-\frac{MA}{NM}\cdot \frac{PN}{CP}=\frac{MA}{NM}\cdot \frac{-PN}{CP}=\frac{MA}{NM}\cdot \frac{NP}{CP}$.

Since $\frac{MA}{NM}=\frac{B^{\prime }A^{\prime }}{C^{\prime }B^{\prime }}$, this becomes $\frac{AB}{BC}=\frac{B^{\prime }A^{\prime }}{C^{\prime }B^{\prime }}\cdot \frac{NP}{CP}$.

Hence, $\frac{AB}{BC}\cdot \frac{CP}{C^{\prime }P}=\left( \frac{B^{\prime }A^{\prime }}{C^{\prime }B^{\prime }}\cdot \frac{NP}{CP}\right) \cdot \frac{CP}{C^{\prime }P}=\frac{B^{\prime }A^{\prime }}{C^{\prime }B^{\prime }}\cdot \frac{NP}{C^{\prime }P}$, so that $AB\cdot \frac{CP}{C^{\prime }P}=BC\cdot \frac{B^{\prime }A^{\prime }}{C^{\prime }B^{\prime }}\cdot \frac{NP}{C^{\prime }P}$.

Since AN || A'C', Thales yields $\frac{NP}{C^{\prime }P}=\frac{AP}{A^{\prime }P}$; thus, this equation rewrites as

$AB\cdot \frac{CP}{C^{\prime }P}=BC\cdot \frac{B^{\prime }A^{\prime }}{C^{\prime }B^{\prime }}\cdot \frac{AP}{A^{\prime }P}$.

Similarly, $AC\cdot \frac{BP}{B^{\prime }P}=CB\cdot \frac{C^{\prime }A^{\prime }}{B^{\prime }C^{\prime }}\cdot \frac{AP}{A^{\prime }P}$, so that

$CA\cdot \frac{BP}{B^{\prime }P}=-AC\cdot \frac{BP}{B^{\prime }P}=-CB\cdot \frac{C^{\prime }A^{\prime }}{B^{\prime }C^{\prime }}\cdot \frac{AP}{A^{\prime }P}=BC\cdot \frac{C^{\prime }A^{\prime }}{B^{\prime }C^{\prime }}\cdot \frac{AP}{A^{\prime }P}$.

Adding these two equalities, we get

$CA\cdot \frac{BP}{B^{\prime }P}+AB\cdot \frac{CP}{C^{\prime }P}=BC\cdot \frac{C^{\prime }A^{\prime }}{B^{\prime }C^{\prime }}\cdot \frac{AP}{A^{\prime }P}+BC\cdot \frac{B^{\prime }A^{\prime }}{C^{\prime }B^{\prime }}\cdot \frac{AP}{A^{\prime }P}$
$=BC\cdot \frac{AP}{A^{\prime }P}\cdot \left( \frac{C^{\prime }A^{\prime }}{B^{\prime }C^{\prime }}+\frac{B^{\prime }A^{\prime }}{C^{\prime }B^{\prime }}\right) $.

Since

$\frac{C^{\prime }A^{\prime }}{B^{\prime }C^{\prime }}+\frac{B^{\prime }A^{\prime }}{C^{\prime }B^{\prime }}=\frac{C^{\prime }A^{\prime }}{B^{\prime }C^{\prime }}+\frac{A^{\prime }B^{\prime }}{B^{\prime }C^{\prime }}=\frac{C^{\prime }A^{\prime }+A^{\prime }B^{\prime }}{B^{\prime }C^{\prime }}=\frac{C^{\prime }B^{\prime }}{B^{\prime }C^{\prime }}=-1$,

this becomes

$CA\cdot \frac{BP}{B^{\prime }P}+AB\cdot \frac{CP}{C^{\prime }P}=BC\cdot \frac{AP}{A^{\prime }P}\cdot \left( -1\right) =-BC\cdot \frac{AP}{A^{\prime }P}$,

so that $BC\cdot \frac{AP}{A^{\prime }P}+CA\cdot \frac{BP}{B^{\prime }P}+AB\cdot \frac{CP}{C^{\prime }P}=0$, and Assertion 1 is proven.

[i]Second proof of Assertion 1.[/i] We work with non-directed segments and assume that the point B lies between the points C and A, and that the points A', B', C' lie inside the segments AP, BP, CP. Then, the point B' lies between the points C' and A'. We have to show that

$\left| CA\right| \cdot \frac{\left| BP\right| }{\left| B^{\prime }P\right| }=\left| AB\right| \cdot \frac{\left| CP\right| }{\left| C^{\prime }P\right| }+\left| BC\right| \cdot \frac{\left| AP\right| }{\left| A^{\prime }P\right| }$.

Now, let $\left|P_1P_2P_3\right|$ denote the (non-directed) area of an arbitrary triangle $P_1P_2P_3$.

Since the points A', B', C' are collinear, and the point B' lies between the points C' and A', we have $\left| C^{\prime }PA^{\prime }\right| =\left| A^{\prime }PB^{\prime }\right| +\left| B^{\prime }PC^{\prime }\right| $. Now, the area of a triangle equals $\frac12$ of the product of two sides with the sine of the angle between them; thus,

$\left| C^{\prime }PA^{\prime }\right| =\frac{1}{2}\cdot \left| C^{\prime }P\right| \cdot \left| A^{\prime }P\right| \cdot \sin \measuredangle C^{\prime }PA^{\prime }$;
$\left| A^{\prime }PB^{\prime }\right| =\frac{1}{2}\cdot \left| A^{\prime }P\right| \cdot \left| B^{\prime }P\right| \cdot \sin \measuredangle A^{\prime }PB^{\prime }$;
$\left| B^{\prime }PC^{\prime }\right| =\frac{1}{2}\cdot \left| B^{\prime }P\right| \cdot \left| C^{\prime }P\right| \cdot \sin \measuredangle B^{\prime }PC^{\prime }$,

so the equation $\left| C^{\prime }PA^{\prime }\right| =\left| A^{\prime }PB^{\prime }\right| +\left| B^{\prime }PC^{\prime }\right| $ becomes

$\frac{1}{2}\cdot \left| C^{\prime }P\right| \cdot \left| A^{\prime }P\right| \cdot \sin \measuredangle C^{\prime }PA^{\prime }$
$=\frac{1}{2}\cdot \left| A^{\prime }P\right| \cdot \left| B^{\prime }P\right| \cdot \sin \measuredangle A^{\prime }PB^{\prime }+\frac{1}{2}\cdot \left| B^{\prime }P\right| \cdot \left| C^{\prime }P\right| \cdot \sin \measuredangle B^{\prime }PC^{\prime }$.

Upon division by $\frac{1}{2}\cdot \left| A^{\prime }P\right| \cdot \left| B^{\prime }P\right| \cdot \left| C^{\prime }P\right| $, this simplifies to

$\frac{\sin \measuredangle C^{\prime }PA^{\prime }}{\left| B^{\prime }P\right| }=\frac{\sin \measuredangle A^{\prime }PB^{\prime }}{\left| C^{\prime }P\right| }+\frac{\sin \measuredangle B^{\prime }PC^{\prime }}{\left| A^{\prime }P\right| }$.

Now, by the sine law in triangles APB and CPA, we have

$\frac{\left| AB\right| }{\left| BP\right| }=\frac{\sin \measuredangle APB}{\sin \measuredangle BAP}=\frac{\sin \measuredangle A^{\prime }PB^{\prime }}{\sin \measuredangle CAP}$;
$\frac{\left| CP\right| }{\left| CA\right| }=\frac{\sin \measuredangle CAP}{\sin \measuredangle CPA}=\frac{\sin \measuredangle CAP}{\sin \measuredangle C^{\prime }PA^{\prime }}$,

and thus

$\frac{\left| AB\right| \cdot \left| CP\right| }{\left| CA\right| \cdot \left| BP\right| }=\frac{\left| AB\right| }{\left| BP\right| }\cdot \frac{\left| CP\right| }{\left| CA\right| }=\frac{\sin \measuredangle A^{\prime }PB^{\prime }}{\sin \measuredangle CAP}\cdot \frac{\sin \measuredangle CAP}{\sin \measuredangle C^{\prime }PA^{\prime }}=\frac{\sin \measuredangle A^{\prime }PB^{\prime }}{\sin \measuredangle C^{\prime }PA^{\prime }}$.

Hence,

$\frac{\left| AB\right| \cdot \left| CP\right| }{\sin \measuredangle A^{\prime }PB^{\prime }}=\frac{\left| CA\right| \cdot \left| BP\right| }{\sin \measuredangle C^{\prime }PA^{\prime }}$.

Similarly,

$\frac{\left| BC\right| \cdot \left| AP\right| }{\sin \measuredangle B^{\prime }PC^{\prime }}=\frac{\left| CA\right| \cdot \left| BP\right| }{\sin \measuredangle C^{\prime }PA^{\prime }}$.

Thus,

$\frac{\left| BC\right| \cdot \left| AP\right| }{\sin \measuredangle B^{\prime }PC^{\prime }}=\frac{\left| CA\right| \cdot \left| BP\right| }{\sin \measuredangle C^{\prime }PA^{\prime }}=\frac{\left| AB\right| \cdot \left| CP\right| }{\sin \measuredangle A^{\prime }PB^{\prime }}$.

Denoting

$k=\frac{\left| BC\right| \cdot \left| AP\right| }{\sin \measuredangle B^{\prime }PC^{\prime }}=\frac{\left| CA\right| \cdot \left| BP\right| }{\sin \measuredangle C^{\prime }PA^{\prime }}=\frac{\left| AB\right| \cdot \left| CP\right| }{\sin \measuredangle A^{\prime }PB^{\prime }}$,

we thus have

$\left| BC\right| \cdot \left| AP\right| =k\cdot \sin \measuredangle B^{\prime }PC^{\prime }$;
$\left| CA\right| \cdot \left| BP\right| =k\cdot \sin \measuredangle C^{\prime }PA^{\prime }$;
$\left| AB\right| \cdot \left| CP\right| =k\cdot \sin \measuredangle A^{\prime }PB^{\prime }$,

and we see that the equation that we have to prove,

$\left| CA\right| \cdot \frac{\left| BP\right| }{\left| B^{\prime }P\right| }=\left| AB\right| \cdot \frac{\left| CP\right| }{\left| C^{\prime }P\right| }+\left| BC\right| \cdot \frac{\left| AP\right| }{\left| A^{\prime }P\right| }$,

is equivalent to

$\frac{k\cdot \sin \measuredangle C^{\prime }PA^{\prime }}{\left| B^{\prime }P\right| }=\frac{k\cdot \sin \measuredangle A^{\prime }PB^{\prime }}{\left| C^{\prime }P\right| }+\frac{k\cdot \sin \measuredangle B^{\prime }PC^{\prime }}{\left| A^{\prime }P\right| }$,

what simplifies to

$\frac{\sin \measuredangle C^{\prime }PA^{\prime }}{\left| B^{\prime }P\right| }=\frac{\sin \measuredangle A^{\prime }PB^{\prime }}{\left| C^{\prime }P\right| }+\frac{\sin \measuredangle B^{\prime }PC^{\prime }}{\left| A^{\prime }P\right| }$;

but this is something we have already proved. So Assertion 1 is established again.

[i]Proof of Assertion 2.[/i] This is done in the very usual way one proves the converses of the Ceva and Menelaos theorems: Assume that we have $BC\cdot \frac{AP}{A^{\prime }P}+CA\cdot \frac{BP}{B^{\prime }P}+AB\cdot \frac{CP}{C^{\prime }P}=0$. Let the line B'C' meet the line AP at a point $A_{1}^{\prime }$. Then, the points $A_{1}^{\prime }$, B', C' are collinear, so Assertion 1 yields

$BC\cdot \frac{AP}{A_{1}^{\prime }P}+CA\cdot \frac{BP}{B^{\prime }P}+AB\cdot \frac{CP}{C^{\prime }P}=0$.

Compared with $BC\cdot \frac{AP}{A^{\prime }P}+CA\cdot \frac{BP}{B^{\prime }P}+AB\cdot \frac{CP}{C^{\prime }P}=0$, this leads to $A_{1}^{\prime }P=A^{\prime }P$. Since the segments are directed and since our points $A_{1}^{\prime }$ and A' both lie on the line AP, this yields $A_{1}^{\prime }=A^{\prime}$. Hence, the collinearity of the points $A_{1}^{\prime }$, B', C' becomes a collinearity of the points A', B', C', and Assertion 2 is proven.

This completes the proof of Theorem 1.

  Darij


inequalities
induction
Putnam
inequalities solved

Take a_i=(A_i) n and b_i=(B_i) n.
The inequality comes to(A_1^n+B_2^n)*...*(A_n^n+B_n^n) \geq (A_1*...*A_n+B_1*...*B_n) n.
Use Cauchy induction for this.
I think the problem was solved before on the site. It is the same or similar to one of the Putnam problems this year. That too, wasn't an original problem. Your problem is older.

let cn=bn/an.
cancel out (a1*a2*.....*an) 1/n on both side we'll only need to prove 1+(c1*c2*....*cn) 1/n  \leq [(1+c1)(1+c2)....(1+cn)] 1/n .
take n-th power to each side and expand, then use AM-GM on terms of each degree to get the result.

What  do  you  mean  ''use  AM-GM  in  each  terms? ''

You  mean  use  Newton  formula!!
Thanks


A surface is bombarded by particles, each of mass $ m$, which have velocity $ v$ normal to the surface. On average, $ N$ particles strike a unit area of the surface each second and rebound elastically. What is the pressure on the surface?

(A) $ \frac{1}{3}Nmv^{2}$
(B) $ \frac{1}{2}Nmv^{2}$
(C) $ Nmv$
(D) $ 2Nmv$

$ N\cdot A\cdot t$ equals number of particles in time t on area A.
$ p$(pressure)$ =\frac{F}{A}$
$ F=\frac{\Delta p}{\Delta t}$ this p is impulse
$ p=N\cdot A\cdot t\cdot m\cdot v$
$ F=N\cdot A\cdot m\cdot v$
$ p=N\cdot m\cdot v$

But the official answer is $ 2nmv$  :maybe:


Find a maximal value of $ k$ for which the following inequality
\[ \frac{(a\minus{}b)^2\plus{}(a\minus{}c)^2\plus{}(b\minus{}c)^2}{3}\geq k\sqrt[3]{(a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2}\]
holds for all reals $ a,$ $ b$ and $ c.$

[quote="arqady"]Find a maximal value of $ k$ for which the following inequality
\[ \frac {(a \minus{} b)^2 \plus{} (a \minus{} c)^2 \plus{} (b \minus{} c)^2}{3}\geq k\sqrt [3]{(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2}\]
holds for all reals $ a,$ $ b$ and $ c.$[/quote]
Let $ 3u \equal{} a \plus{} b \plus{} c,3v^2 \equal{} ab \plus{} bc \plus{} ca,w^3 \equal{} abc$
Then $ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} 27( \minus{} (w^3 \minus{} 3uv^2 \plus{} 2u^3)^2 \plus{} 4(u^2 \minus{} v^2)^3 )$, and $ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \equal{} 18u^2 \minus{} 18v^2$. So wlog $ w^3 \equal{} 3uv^2 \minus{} 2u^3$ and:
$ 6u^2 \minus{} 6v^2 \ge k\sqrt [3]{27 \cdot 4(u^2 \minus{} v^2)} \equal{} 3\sqrt [3]{4}k(u^2 \minus{} v^2)$. And since $ u^2 \ge v^2$ we see that $ k \equal{} \frac {6}{3\sqrt [3]{4}} \equal{} \sqrt [3]{2}$

(I know: It was a bit sketchy)

[quote="arqady"]Find a maximal value of $ k$ for which the following inequality
\[ \frac {(a \minus{} b)^2 \plus{} (a \minus{} c)^2 \plus{} (b \minus{} c)^2}{3}\geq k\sqrt [3]{(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2}\]
holds for all reals $ a,$ $ b$ and $ c.$[/quote]

Let : $ c \le b \le a$

We can assume : $ a\minus{}c\equal{}1$ , $ a\minus{}b\equal{}x$ , $ b\minus{}c\equal{}y$ , ($ x\plus{}y\equal{}1$)

$ 1^2\plus{}x^2\plus{}y^2 \ge \frac{3}{2}$ and $ 1^2x^2y^2 \le \frac{1}{16} \Rightarrow k \le \sqrt [3] {2}$


trigonometry
inequalities proposed
inequalities

[quote="Mikel2008"]Let a triangle ABC. Prove that $ \sum_{cyc}\frac {\sin A\sin B}{(\sin \frac {C}{2})^2} \ge 9$[/quote]
It's obvious after substitution: $ \sin A=\frac{2S}{bc},$ $ \sin B=\frac{2S}{ac}$ and $ \sin^2{\frac{C}{2}=\frac{1-\cos C}{2}.}$


yek namosavie hendesy.agar $a,b,c$ azlae yek mosalas bashand angah sabet conid:
$\frac{a+b+c}{2c^{2}}=\frac{cos(\frac{A}{2}) \times cos(\frac{B}{2})}{sin(\frac{C}{2})}$

[quote="Virgil Nicula"]Is correctly this identity : $\frac{a+b+c}{2c}=\frac{\cos\frac{A}{2}\cos\frac{B}{2}}{\sin \frac{C}{2}}$[/quote]
are you sure?
because i get this problem while i was solving another problem.

$\frac{\cos \frac{A}{2}\cos\frac{B}{2}}{\sin\frac{C}{2}}=\sqrt{\frac{\frac{p(p-a)}{bc}\cdot \frac{p(p-b)}{ac}}{\frac{(p-a)(p-b)}{ab}}}=\frac{p}{c}=\frac{a+b+c}{2c}\Longrightarrow$
$\sum \frac{2\cos\frac{A}{2}\cos\frac{B}{2}}{\sin \frac{C}{2}}=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge 9\Longrightarrow$
$\sum \frac{\sin \frac{C}{2}+\cos\frac{A-B}{2}}{\sin \frac{C}{2}}\ge 9\Longrightarrow \boxed{\sum\frac{\cos\frac{A-B}{2}}{\sin \frac{C}{2}}\ge 6}\ .$


trigonometry
function
algebra unsolved
algebra

Find number of solutions in non-negative reals to the following equations:
\begin{eqnarray*}x_1 + x_n ^2 = 4x_n \\ x_2 + x_1 ^2 = 4x_1 \\ ... \\ x_n + x_{n-1}^2 = 4x_{n-1} \end{eqnarray*}

Denote $x_{n+1}=x_1$, so we can write
$x_{k+1}=4x_k-x_k^2, \forall k \in \{1,2,...,n\}$
$x_{k+1}=4(1-(\frac{x_k-2}{2})^2), \forall k \in \{1,2,...,n\}$
We can easily prove that $x_k \in [0,4], \forall k \in \{1,2,...,n\}$, hence $\frac{x_k-2}{2} \in [-1,1]$, so we may put $\frac{x_k-2}{2}=\sin t_k, \forall k \in \{1,2,...,n\}$.
We can write each equation like this:
(1) $\sin t_{k+1}=\cos 2t_k$
Using (1) [b]n[/b] times we get
$\sin t_1=-\cos 2^nt_1$
The equation
(2) $\sin t=-\cos 2^nt, t\in [-\frac{\pi}{2},\frac{\pi}{2}]$
can be solved by drawing the graphs of the functions from both sides. The function in the right side has the period $T=\frac{\pi}{2^{n-1}}$ and for each interval which has precisely this length, the graphs gives two solutions for (2).
Deviding the length of the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ by $T=\frac{\pi}{2^{n-1}}$ we obtain $2^{n-1}$.
Thus the equation (2) has $2\cdot 2^{n-1}=2^n$ solutions.
The whole system has the same number of solutions, namely $2^n$.


While every scene is a movie should have importance, it is generally known that for most classes/films/etc... people remember most  of the beginning and the end.

Therefore a good movie has a good opening and ending.

Which movies have impressive openings and endings?

Openings : I'm not sure about this one, I'll just give some that I found good :

Lion King : amazing animated nature images combined with good music show you what you are in for

Team America : a bit silly, but the first few seconds is an excellent way of scaring the audience into low expectations, and a great parody on the creator's own legacy of cheapness

Fellowship of the ring : the battle with Sauron is really impressive

Ending :

my absolute favorite movie ending is that of the Japanese movie of Rurouni Kenshin
the whole movie has a much more serious undertone than the series, and the ending is a magic combination of history, action, beautiful music, history, sadness and optimism

besides that :

Sixth sense : explanation would imply spoiler

Wicker man : really really impressive ending that you will think about for the rest of your day

Spiderman (first movie) : the very last minute has very impressive shots of New York (though slightly incorrect) combined with appropriate music

Planet of the apes : explanation would imply spoiler

Memento : makes you look at things differently

Uninvited : unknown Korean horror movie with quite a last few seconds

What are your favorite movie openings/endings?

[quote="fredbel6"]What are your favorite movie openings/endings?[/quote]

My favourite opening is the one in "El juego del miedo"... just beautiful!!!!

Gandalf vs. Balrog in [i]The Two Towers[/i]  is a great beginning.

I rather like both the beginnings and the endings in Ocean's Eleven (The newer version) and Ocean's Twelve.

The ending of Unbreakable is pretty good as well.  Though it is not as good as Shyamalan's better-known movie, The Sixth Sense, it does follow quite a few of the same patterns, and it is not that bad itself.

This is a very interesting thread, and I thank the person who opened it, but oddly I can hardly think of any movie scenes that I cherish that come from either the beginning or the end of the movie. I think my all-time favorite movie scene is the scene filmed on a high-speed railroad train in Kurosawa's High and Low, which comes in about the middle of the movie.

I don't know much about film, but I like the opening of Gladiator.

George Taylor: Oh my God. I'm back. I'm home. All the time, it was... We finally really did it.
[screaming]
George Taylor: You Maniacs! You blew it up! Ah, damn you! God damn you all to hell!

Planet of the Apes : nice one, but I saw that one coming.  Not too hard, it has been parodied so many times in Simpsons, Futurama...

[quote="oooooh ya"]my favorite opening is the beginning of pulp fiction.  when the couple is sitting there at the coffee shop talking about robbing the place.  they say i love you honey bunny then stand and rob the place, then the music kicks in.[/quote]

Also, the surf guitar version of Miserlou is AMAZING. And you are right, the transition from "I love you pumpkin; I love you honey bunny" to "I'll execute every ----- ------ing one of you!!!!!!" is.... hilarious.

You know, I always liked the ending to Pirates of the Caribean (sp, I know) for a similar reason: the music! At the end! Wonderful!

The next-to-last scene of Dr. Strangelove is priceless. Definitely explantory of the second title ("How I Learned to Stop Worrying and Love the Atomic Bomb").

I also liked the beginning to Eternal Sunshine of the Spotless Mind. I like backwards movies.

The beginning and ending of Lost in Translation were both really good.

Once Upon a Time in the West has the eeriest opening ever in film history and must be my favourite. No others come close.

I really don't like films which end either with the story unfinished or which leave you more confused than at the start (2001: A Space Odyssey for example).

I have a mixed view over plot twist endings. Some of them are just poor to point of ludicrosity (Fight Club, Sixth Sense) but sometimes they work really well (The Usual Suspects). So anyway here are my favourite endings, some of them obvious, so not so:

Star Wars Episode V (How has no one mentioned this one!)
The Godfather I, II and III (Every time you know what's going to happen but it still sends chills down your spine)
The Usual Suspects
Das Boot (Saddest ending ever to a film)
Dr. Strangelove
Once Upon a Time in the West (best shootout Sergio Leone ever did)
The Pianist

I forgot to mention that the Village has a good twist at the end too.
Why don't you like the twist in Sixth sense? Or Fight Club? (If you answer here, do write in hidden content for the others)

Space Odyssey does make some sense eventually!

The pianist, I saw that ending but what was so impressive about the ending itself?
Star Wars V is really good, but I had deduced the twist there long before from endless series of parodies, just like with planet of the Apes.

As a movie lover i should definitely try  "once upon a time in the west" (no western fan though) and d"octor strangelove", and "potemkin"!

The opening of Saving Private Ryan is something special, IMHO :)

Isn't it extremely gory?  I heard a German stabs a soldier really slowly.
I am not really against violence in movies but I can't stand torture.  And I do not agree with the policy of "it's too much violence, but here we should condone it as this movie has something to say" (a policy some advocates of showing Schindler's List and Braveheart to children have)

[quote="tokenadult"]Best movie opening, and best movie ending, are both in United 93.[/quote]

You mean that controversial movie about the fourth airplane?  I definitely want to see it, but I am afraid it's one of those movies you either love or hate.

The Phantom of the Opera has the best beginning I have ever seen!!!!!!!!! Reeeeeeeaaaly impresive!!!

The Fellowship of the Ring has the best begining. The Last Samurai has the best ending.

[b]Openings.[/b] (that makes me think of chess...)
[i]Trainspotting.[/i] "Choose life. Choose a job. Choose a starter home. Choose dental insurance, leisure wear and matching luggage. Choose your future. But why would anyone want to do a thing like that?"
[i]Spun.[/i] I just like the cover of "Number of the Beast", played by Zwan Djali.
[i]Reservoir Dogs.[/i] I never knew what was "Like A Virgin" about... I do know.

I'm not sure if I'll write anything about them on the "What movies have you seen recently?" topic. I'm guessing it's a bad thing to recommend young people (as in $<<18$) movies about drug addiction.

[b]Endings.[/b]
[i]Pulp Fiction.[/i] It's really the best part of the movie.
[i]Gegen Die Wand.[/i] It's a big surprise after getting used to the typical Hollywood ending.

How surprising...bubka has not hit this thread yet...

Well I don't think I can say much that hasn't been said already, but Borat is bound to have a great start when it comes out.

you mean koda?
Evita has a really nice opening (and do not forget the PotO)
also the opening of the Die another day is quite nice

I thought koda=bubka...at least I remember bubka created another username after becoming really obsessed with brother bear...

thats right but he is going to post in this thread as koda and not as bubka... the reason is obvius...

My absolute favorite beginning would be the fellowship of the ring, the battle with all the orcs and men and elves and sauron was really awesome!  The best endings I have seen were in books  :D , but the best movie ending was in Star Wars Episode 6.

Also, on your comment that people remember the beginnings and endings best, I remember the battles :D  Sauron's orcs vrs. the world, Gandalf vs. Belrog, Helms Deep, Minas Tirith, Black Gate, Sam vs. Cirith Ungol :D  those were really cool examples from LOTR.  I actually don't remember any of their endings or beginnings, except for the opening with the battle.

I really like the endings to Oceans Eleven and Inside Man.

the falling leaves...
don't you think that there is a huge difference between hp3 and hpi, i=1,2?
first of all.... draco...!

yes and dumble mumble and lupin is so hott and the dark lord

i think though 3 is the best ending and 4 is the best opening(if you count quidcup) and if the de ring is in the ending it also has a good ending

yeah he is in my avvie do you see him he is so hott they didn't do him justice in the movies coz he is so hott and anyway hott is spelled hott(the reasoning: since theo's last name is nott nott is spelled nott and hott is hott)

TDL is kinda hott but he is kinda creepy too but as a DE i have to say something about him.

Oh but albus isn't hott i didn't mean that.

@oooooh ya

That song you are talking about is Misirlou by Dick Dale

Well I liked the beginning of Fullmetal Alchemist the Conqueror of Shambala.

the ending to independence day was nice too.

Opening: The Last of the Mohicans, no contest.

Closings: Hard to decide, but probably The Patriot.

The opening of the Phantom of the Opera movie was beautiful, as the chandelier slowly arose to unveil the grand Paris Opera House... Yes, I'm a geek!  :blush:


trigonometry geometry parallelogram rhombus ratio

Verify the following identity (a half angle formula)

tan (a/2) = sina / (1+cos a)

[hide="Hint"]Consider a parallelogram with vertices $ A(0,0)$, $ B(\cos \alpha, \sin \alpha)$, $ C(1 \plus{} \cos \alpha, \sin \alpha)$, $ D(1,0)$.

Notice that it is a rhombus![/hide]

[hide="solution"]
Consider the unit circle
 [geogebra]28cecb3c06c9f0340ec3334aa8103140c177fbc9[/geogebra]

Since I'm such a novice with geogebra i couldn't figure out how to label to angles $ \alpha$ and $ \beta$.
But because of the circle theorem we that $ \angle \alpha \equal{} 2 \angle \beta$
Since $ tan(\theta) \equal{} \dfrac{\sin{\theta}}{\cos{\theta}}$ and the diagram is a unit circle, we can see that
$ \tan \beta \equal{} \dfrac{\sin{\alpha}}{1\plus{}\cos{\alpha}}$
And we're done :lol: [/hide]

simply

sin a= 2sin(a/2)cos(a/2)
1 - cos a =2cos^(a/2)
the ratio is tan(a/2).
hi sorry just forget this post.
really sorry.


In $\triangle ABC$, $a^3+b^3=c^3$. Prove that $\angle{C}$ is acute.

well, by the pythagorean inequality theorem (I'm not sure about the name, but it is taught in high school geometry class), if $a^2 + b^2 > c^2$, then $\angle C$ is less than $90$ degrees.  since $a^3 + b^3 = c^3$, divide both side by $c$ and we get $\frac{a^3}{c} + \frac{b^3}{c} = c^2$. since $c > a$ and $c > b$ (since $c^3$ is greater than $a^3$ and $b^3$ and none of $a, b,$ or $c$ is negative), $a^2 > \frac{a^3}{c}$ and $b^2 > \frac{b^3}{c}$. adding them we get $a^2 + b^2 > \frac{a^3}{c} + \frac{b^3}{c} = c^2$, so $\angle C$ is less than $90$ degrees and it is acute.

You can also prove that $\angle C > 60^{\circ}$ (really straightforward).