vector
analytic geometry
induction
linear algebra
linear algebra unsolved

Suppose $e_1,e_2,...,e_n$ be $n+1$ vectors of an Euclidean vector space $E$ satisfied:
\[ (e_i,e_j)<0\ \ \ \forall i\not =j \]
where $(.,.)$ is the inner product (definite positive scalar product)
Prove that any $n$ vector of them formed a basis of $E$

A different solution than the one harazi posted in that other thread (but also using a problem posted by alekk some time ago :)):

Suppose some $n$ of those vectors are linearly dependent. Then there is a hyperplane in which they all lie, which determines two half-spaces of $E$, one of which contains all our $n+1$ vectors. It thus suffices to prove the following stronger assertion:

Given $n+1$ vectors $v_1,\ldots,v_{n+1}$ in an $n$-dimensional uclidean space $E$ satisfying $(v_i,v_j)<0,\ \forall i\ne j$, for every $v\in E,\ v\ne 0$, we can find $i\in\overline{1,n+1}$ with $(v,v_i)<0$.

Suppose the contrary, and let $v$ be such that $(v,v_i)\ge 0,\ \forall i$. If we can find an $i$ for which $v_i$ and $v$ are collinear, then for all $i$ we have $(v,v_i)>0$, so we can move $v$ slightly until it's no longer collinear with any of the $v_i$'s. If we express the $v_i$'s in an orthogonal basis which $v$ is part of as the $n$'th vector, all the $v_i$ will have a non-negative $n$'th coordinate, and not all the first $n-1$ coordinates equal to zero. For each $i$, let $\bar v_i$ be the projection of $v_i$ onto the orthogonal complement of $v$. The vectors $\bar v_i$ are non-null and in the orthogonal complement of $v$ viewed as an $n-1$-dimensional Euclidean space with the scalar product induced from $E$ they satisfy $(\bar v_i,\bar v_j)<0,\ \forall i\ne j$.

The above means that in an $n-1$-dimensional Euclidean space we can find $n+1=(n-1)+2$ vectors with negative mutual scalar products, but this contradicts the problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=n%2B2&t=31714]posted here[/url].

This is another proof:
It is sufficient to prove that " In an arbitrary Euclidean vector space E with finite dimention, if system of non-zero vectors $(e_1,...,e_{n+1})$ satisfied the scalar product of any two distinct vectors is negative, then every $n$ elements of them are linearly independent."
By induction on $n, n=1, n=2$ are trivially
Suppose that the statement true for $n-1\geq2$
With $n$, Consider the set consists of $n+1$ vector $(e_1,e_2,...,e_{n+1})$
First, there are no couple of vectors are linear dependent. (Why?)
Thus, we fix $e_1$
Denote $e'_i=e_i-\frac{(e_i,e_1)}{(e_1,e_1)}e_1$
Then $e'_2,...,e'_{n+1}$ are on the orthogonal complement  of $L(e_1)$ (the vector space span by $e_1$)
for every $1<i\not=j>1$,
\[ (e'_i,e'_j)=(e_i,e_j)-\frac{(e_i,e_1)}{(e_1,e_1)}(e_j,e_1)-\frac{(e_j,e_1)}{(e_1,e_1)}(e_i,e_1)+(\frac{(e_i,e_1)}{(e_1,e_1)}e_1,\frac{(e_j,e_1)}{(e_1,e_1)}e_1)  \]
\[ (e'_i,e'_j)=(e_i,e_j)-\frac{(e_i,e_1)(e_j,e_1)}{(e_1,e_1)}<0  \]
by induction hypothesis, every $n-1$ vector in $(e'_2,...,e'_{n+1})$ are linearly independent. In particular, $e'_2,...,e'_n$ are linearly independent, lie on the orthorgonal complement of $L(e_1)$. Thus, $e_1,e'_2,...,e'_n$ are linearly independent, implies, $e_1,e_2,...,e_n$ are linearly independent (why?), other subset is analogous.

Nice one!!


Let n be a positive integer and $X_{1}$ ,..........$X_{2n}$  a sequence of 2n integers where $X_{i}$ =$\pm1$,      $i$ =1,..........2n.

Define $S_{n}$ =$\sum_{1}^{n}X_{2r-1}X_{2r}$

Find the number of sequences with $S_{n}$ odd and the number with $S_{n}$ even.


A bag contains 3 tan, 2 pink, and 4 violet chips. If the 9 chips are randomly drawn from the bag, one at a time and without replacement, what the probability that the chips are drawn in such a way that the 3 tan chips are drawn consecutively, the 2 pink chips are drawn consecutively and the 4 violet chips are drawn consecutively, but not necessarily in the tan-pink-violet order? Express your answer as a common fraction.

There are 9! ways of picking any order of chips and:
3! ways of arranging the tan "bundle"
2!: pink "bundle"
4!: violet "bundle"
3!: rearranging the bundles.
So:

$ \frac{3*2*2*4*3*2*3*2}{9*8*7*6*5*4*3*2}\equal{}\frac{1}{210}$


let ABC be a triangle :

draw aline ,wich intersects AB and AC in points Pand Q such that: BP=PQ=QC

See [url=http://groups.google.de/group/geometry.college/browse_thread/thread/948b29d3346dd801/de28fab3d89a8f04]Hyacinthos messages #7199, #7201, #7203, #7210[/url].

  darij


rotation
geometry
geometric transformation
reflection
conics
parabola
function

Let's consider  the cartezian system XOY and O a point light source.Fiind the ecuation of the curve y=y(x) so that if we rotate it arround Ox we obtain a mirror so that any ray going from O that reflects goes parallel with Ox to infinite.
If you know the diameter of the mirror D and the depth h fiind the focal distance of the mirror.

I got the following answers:
$y^{2}=4fx$

$f=\frac{D^{2}}{16h}$

O is the focal point situated at a f-distance from the tip of the mirror so you should got an extra term in the parabola function

I think it must be minus(-).What is the value of focal length?

the focal lenght is sqrD/16h
if it's minus than you get x=f>0 if y=0 and f is the distance from O(focal point) to the tip than the mirror is situated with the tip in the pozitive way of Ox but i said that the rays goes to infinite not to -infinite

Yes,you're right at that point.I didn't notice that the ray goes to +infinity :)


An icosidodecahedron is a polyhedron with twenty triangular faces and twelve pentagonal faces.  Each vertex of an icosidodecahedron is identical with two triangles and two pentagons meeting at each vertex.  What is the sum of the number of vertices and the number of edges in an icosidodecahedron?

[hide]i think its eulers formula , not pascal's where if v = vertices, e = edges, f = faces, V+F=E+2[/hide]

icosadodecahedron,icosadodecahedron,icosadodecahedron

There, I said it three times fast

[quote="philB"][hide]Well.... there's a trick:

There's a formula (maybe Pascal's theorem???)out there that says" Verticies + Faces - edges = 2"
Obviously, there are 20 + 12  = 32 faces.
Over counting the edges, we see that there is 20*3 + 12*5 = 120 edges (that's a bit too much). Since every edge is shared by two figures, we should divide 120 by 2 (60). plugging this in:

Verticies + 32 - 60 = 2
 There must be 30 verticies

30 + 60 = [b]90[/b][/hide][/quote]
Euler's Formula, not Pascal's (I think)

[hide]first there are 20*3+12*5=120 edges without restrictions, but we overcounted in the polyhedron. so divide by 4 since each edge was counted twice to get 30. there are also 120 edges throughout the polygons, each edge is counted twice, so divide by 2 to get 60. 60+30=90[/hide]

[hide]Similiarly, each vertex is counted four times.... there are 120 verticies UNRESTICED, divided by four equals 30....[/hide]

are space diags where you connect vertices?

how do you figure out how many of the vertices lie on pentagons?

jorian[/hide]

[quote="13375P34K43V312"]Extension: how many space diagonals does it have?[/quote]
Isn't that just $\binom{n}{2}-5k-e$ where n is the number of vertices, k is the number of pentagons, e is the number of edges?

Yeah, that would be right.

As for the solution, we would first select 2 points, which is where the ${n\choose2}$ comes from, then subtract all of the edges, because those are generated from selecting 2 points, but don't count as space diagonals, which is where the $-e$ comes from. Finally, we have surface diagonals, the diagonals on the pentagons. There are five diagonals per pentagon, so $-5k$.

wow, the solution is pretty easy once you know how to calculate it

jorian


geometry
3D geometry
Duke
college
geometric transformation
rotation
probability

Welcome to the NC Cube Forum
Post your cubing results.
Pretty Simple. :lol:

My best time is 50.45 Seconds at the First Meeting of the Raleigh Cubing Association (RCA).

-Sanjay

141.7 seconds (I know, I'm terrible)  :P  :D

[quote="jb05"]Whatever happened to North Carolina?[/quote]

Good point. [hide=":rotfl: "]They lost to Duke by 11 points  :lol: [/hide]
I have no idea how they managed to perform so poorly.

0 seconds. Started from this position
GGG
GGG
GGG

WWWRRRYYY
WWWRRRYYY
WWWRRRYYY

BBB
BBB
BBB

OOO
OOO
OOO

PS. Who goes to the RCA?? I bet the population is 1 (you.)

My best scrambled time is some 40 or 50 seconds

My worst is... 462 hours (I left it in my drawer)

It takes me like 4 minutes... layer-by-layer :P

Now that I'm on break, maybe I should go learn some better algorithms.

[quote="PiGuy"]You know finger tricks? (most likely)[/quote]

Yes  .

[quote="PiGuy"][quote="PI-Dimension"]My worst is... 462 hours (I left it in my drawer)[/quote]

I'm sure everyones done that before. As a matter of fact. I think the record would be a few years from the time a cube is made, to the time when some kid gets mad and throws the cube away, to when finally some hobo digs the cube up from the trash and solves it before throwing it into an incinerator. What a waste.[/quote]

That's wrong Mr.Cao.

the cube is MADE finished, so the timer would start WHEN the kid messes it up

Thank you for the off-topic posts... please stop.

Are we allowed to talk about other variations of the rubik's cube because I still can't solve a rubik's revenge. The way I start is first the corners, then the center, then the layers. This proves to be very inefficient for me.

Btw, does anyone know any good websites that actually teach you how to solve a professor cube without being unobtrusively confusing.

Wait people were going off-topic? Seeing as this is essentially a spam thread, I wasn't aware it was possible to go off-topic.

Umm 5x5x5 cubes are not actually that difficult. It is fairly easy to get all the center blocks right, and once you get the interior edges, it becomes a 3x3x3. The hard part seems to be the interior edges. I'm sure there are plenty of good websites for that. I know there was one that helped me successfully solve mine... But by the time I solved it once, no less than three faces had broken off and resisted gluing back on, so I haven't used it since. Those things are designed horribly. Also, square-1 is kind of cool for showing off, even though it is easier than Rubik's cubes.

So yeah... you guys want to do some math now? Maybe? Hmm I have a related problem (well-known but whatever). If A is the group of cube transformations and B is the group of positions you can get by taking the cube apart and putting it back together, what is |B|/|A|? Better yet, what group is B/A isomorphic to?

Do you know? Do you even know how many combinations there are?

Well, I believe that once you have assembled the cube, you can enact any even permutation on the 24 visible faces of the edge cubies (that doesn't actually destroy the cubies) without disturbing the corner cubies, and similarly for the corner cubies.  So you can look at corner cubies and edge cubies independently, and in each case, you have a group isomorphic to $ \mathbb{Z}/2\mathbb{Z}$.  So B/A is the Klein four-group?

[quote="jb05"]No. You mentioned permutation of pieces, but what about orientation of the pieces? Also, your reasoning for permutations is off.[/quote]
I was actually referring to the permutations of the different visible faces of the cubies which do not destroy the cubies.  For instance, switching the orientation of one edge cubie permutes two visible faces, and so is an odd permutation of the visible faces.  I thought this took care of orientation of the cubies (I assume that we're talking about the same things when you say "pieces" and I say "cubies"), unless you are counting the orientation of the center cubies, which I thought people usually didn't care about?

It seemed to me that the edge cubies are either solvable or not, and that any edge-solvable position can be reached from any edge-solvable position without disassembling the cube and without disturbing the corner cubies, and similarly for non-edge-solvable positions; and similarly for the corner cubies.  If these conditions assumption is correct, do we not have the Klein four-group?  Or are those conditions wrong?

I believe you when you say I'm wrong, though; I don't have enough experience with Rubik's cubes to have confidence here.

I am fairly certain you can't talk about even-odd permutations of actual faces without missing stuff. The numerical answer is definitely 12 and I think you are missing a factor of Z/3Z.

hm I was doing some random contest problems earlier and I stumbled upon this one, hope you cubers have fun  :lol:

A solved Rubik's cube is a cube whose face are each painted a different color. Six fixed colors are used in constructing a Rubik's cube, though you may have noticed that they are not always painted on in the same order. How many different ways can the cube be colored? (Two "solved" cubes are colored the same if a rotation of one whole cube makes it look exactly like the other.)

Yay!!! I got a Rubik's Revenge over the weekend!!!

Here is an interesting problem.

Jack gets a scambled, solvable, Rubik's cube (the 3x3x3). He randomly turns it until it is solved. Assuming that he has unlimited time, on average, when will he solve the cube.

(for the 1x1x1, its of course, 0)

Who's Jack, and if you have unlimited time, you could take as long as wanted to. What are you asking when you say when will he finish it? Also, Jack will eventually die :| ...

He probably means that he has an equal chance of turning each layer.

That's how I interpreted it. It seems like it should be an extremely difficult problem.

I think it will probably need a computer... calculating expected value with like billions of possible configurations? ouch.

Err, if you've been reading the above posts, it would become apparent that it is highly unlikely that any of us know the answer.

Oh well. Here's another similar question: If Billy randomly turns a finished cube 10 times, what is the probability that after the ten moves, he is back to where he started?

I suspect that the moves need to go in (adjacent) pairs, because turning the cube isn't associative. Also, 10 is not divisible by 4, so it's gonna be like

$ X_1, X_1', X_2, X_2',\ldots, X_{10},X_{10}'$

where $ X_1,\ldots,X_{10}$ are moves and $ X_i'$ is the inverse of $ X_i$.

Wait... can a "turn" be a 180 degree turn? anyway, I will assume that

then there's 6 faces, each can be turned 3 ways; so 18 different moves in every situation

and I am too lazy to finish but you should get the idea

(however I may have left out ultra-tricky cases. hm.)

Hmm, this actually doesn't seem too difficult (the second question), merely a lot of boring computation...

Assuming a turn is $ 90^\circ$, there are 9 layers, each of which can be turned, and the total number of turns must be 0 (mod 4) or add to 0, so it shouldn't be too difficult to figure it out from there...

[quote="Teki-Teki"]PI-Dimension is saying that, given that a hypothetical immortal person named Jack has a Rubik's Cube, which he turns randomly*.  Given that he turns, say, once per second, how long is he expected to take in solving the cube?

@PI-Dimension: Please give a better definition of "randomly".  Depending on the interpretation, several possible answers could be correct.

Answer: Based on my definition of randomly (every possible state is equally likely at every point) I say 43,252,003,274,489,856,000 seconds.[/quote]

actually i was asking for how many turns, but it works too.

There seem to be error(s) in here (the final formula is wrong?), but I don't feel like looking for them. The point is that this problem requires some very powerful information that we don't have.

Let $ f(k)$ be the number of ways to get from 0 back to itself in $ k$ moves (the number of ordered zero-sum sequences of the eight basic operations of length $ k$). The total number of ways to do $ k$ moves from 0 is $ 8^k$, but we need to remove possibilities that pass through 0 again, hence $ g(k)\equal{}g^k\minus{}\sum_{i\equal{}1}^{k}{f(i)g(k\minus{}i)}$ or $ g^k\equal{}\sum_{i\equal{}0}^{k}{f(i)g(k\minus{}i)}$ with $ f(0)\equal{}g(0)\equal{}1$, and $ g\equal{}g(1)$ is the number of possible "moves" (in this case, 8). The expected number of turns will then be $ \frac{1}{P}\sum_{i\equal{}1}^{\infty}{\frac{g(i)i}{g^i}}\equal{}P\sum_{i\equal{}0}^{\infty}{\frac{f(i)}{g^i}}\equal{}43252003274489856000\sum_{i\equal{}0}^{\infty}{\frac{f(i)}{8^i}}$ (assuming that you eventually reach 0 with probability 1), where $ P$ is the number of positions of the Rubik's cube.

Note that I got the last equality (between the two sums) from the generating functions $ F(gx)\equal{}\sum_{i\equal{}0}^{\infty}{f(i)x^i}$ and $ G(gx)\equal{}\sum_{i\equal{}0}^{\infty}{g(i)x^i}$. We know that $ F(gx)G(gx)\equal{}\frac{1}{1\minus{}x}$, but $ G(1)$ is evidently $ P$, since we should eventually reach 0 with probability 1. Hence we can write $ G(gx)\equal{}\frac{1}{H(gx)}$ and $ F(gx)\equal{}\frac{H(gx)}{1\minus{}x}$, with $ H(1)\equal{}\frac{1}{P}$, and our desired quantity $ \frac{G'(1)g}{P}\equal{}\minus{}H'(1)gP$. However, $ H(gx)\equal{}(1\minus{}x)F(gx)$ has derivative at 1 equal to $ \frac{1}{g}\sum_{i\equal{}1}^{\infty}{i\frac{f(i)}{g^i}\minus{}i\frac{f(i\minus{}1)}{g^{i\minus{}1}}}\equal{}\frac{\minus{}1}{g}\sum_{i\equal{}0}^{\infty}{\frac{f(i)}{g^i}}$. Multiplying by $ \minus{}Pg$ yields the equality above.

Unfortunately, we can't easily finish this - it should be VERY difficult to compute $ f(i)$ in general.

[quote="not_trig"]Oh yeah, 9 layers. Ugh. I fail to think... anyway, I need a better definition of "turn"[/quote]

9 o_O I'm sorry... you guys lost me...

LETS SEE

Front, Left, Right, Back, Top, Bottom

Since you can rotate the both back and forth (CW and CCW), that means 12...

and I missing something?

[hide="hint"]

There are 43,252,003,274,489,856,000 possible ways of scrambling a cube

[/hide]

[hide="bigger hint"]

divide the former number by 2 to get the average amount of time to obtain a state...

divide that by the number of ways to rearrange the centers...

[/hide]

EDIT: WARNING!!! THIS PROBLEM HAS A 'TRICK' TO IT!! THERE ARE MORE THAN 1 SOLVED STATE OF A RUBIK'S CUBE!!!!!!!!!!!!!!!! (read second hint)

[quote="jb05"][quote="PI-Dimension"]divide the former number by 2 to get the average amount of time to obtain a state...[/quote]

Average amount of time to obtain what state?[/quote]

Any state............................................................

=( I know... I'm a terrible teacher.

The average amount of time to reach an average state in the klein four group from 0 is 1 move, but there are 4 elements. Explanation?

ok we failed on that question...

OK

What is the minimum number of permutations of a rubik's cube, with proof (answer is already in topic)


I decided to make a topic edifying people about P-adics because P-adics are cool.

In computer science terms, working with p-adics is like working with unisgned numbers. But instead of the maximum value being 65535, it is ...999999999999999999999.

In arithmetic terms, try subtracting 5 from 3. You know it's -2, but go along with me and write the numbers in a column:
[code]
  3
- 5
___

[/code]
While we're at it, write a bunch of 0s in front of the 3:
[code]
...00000000000000000000000000000000003
-                                    5
______________________________________

[/code]
Now subtract like you were taught in school. 5 is bigger than 3, so you change the 3 to a 13 by borrowing a 1 from the next decimal place. Subtract and you get an 8. Now you can't borrow 1 from the next decimal place because it's a 0 so you change it into a 9 and borrow a 1 from the next decimal place. But the next decimal place is a 0 so you change it into a 9 and borrow a 1 from the next decimal place. But the next decimal place...[code]
...00000000000000000000000000000000003
-                                    5
______________________________________
...99999999999999999999999999999999998
[/code]

Basically the point is that in p-adics positive infinite numbers are equivalent to negative finite numbers and positive finite numbers are equivalent to negative infinite numbers. Writing numbers this way is really useful because it allows you to look at infinity more closely. Here's an example.

Find the sum: [tex]\displaystyle\sum_{k=0}^{\infty}3^k=1+3+3^3+3^3+3^4+...[/tex].

Normally you would say that this diverges, but not when you're talking about p-adics. Here are a few ways to look at this problem:

Let S be the sum you're looking for. It is clear that [tex]\frac{S-1}3=S[/tex]. Solving, you get [tex]S=-\frac 12[/tex].

Another way: Convert everything into base 3. First off: in base 3, -1=...222222222222222, not ...99999999999999999. Writing the sum in base 3 is trivial if you know how to use other bases. It's ...1111111111111111111. You can see that [tex]...111111111111111111=\frac{...22222222222222222222}2=-\frac12[/tex]

If any of you has any questions or knows any cool stuff about p-adics please post. Also, does anyone know why p-adics are called p-adics?

Infinite unsigned ints... much easier to understand than the mathworld definition!  What would these be important for?

I just looked at the Mathworld definition. It is indeed a lot more complicated, but it seems to be more accurate technically.

P-adic numbers have been used to prove a lot of things, including the fact that it is possible to dissect a square into n triangles of equal area if and only if n is even. I took a class at Mathcamp that went over this proof, so I should know it, but unfortunately I didn't understand it and so I don't.

It would be nice if someone more knowledgeable on the subject would post something, because all I know about p-adics is the basics and that they are cool.

Did Gregory Galperin teach p-adics at MOP?

The Red group (the one I was in) wasn't taught p-adics and I don't think the other groups were either. I don't think Gregory Galperin was there at all, since I don't remember him. The class I took was at Mathcamp (USA/Canada) in 2003.

Galperin taught p-adics at mop 2002.... and we never used them on anything! how disappointing.


[quote="riddler"]I got your $1st$ problem... I think.
Prime no.s end with $1,3,7,9$ so
$1st$ possibility $10x+3=p$ and $p+4=10x+7$ so now there can be no prime no. in between so $sum=p+p+1+p+2+p+3$ on adding you get $3p+6$ which is multiple of $9$.
$2nd$ possiblitiy $p=10x+7$ and $p+4=10x+11$ so you get $3p+6$ as the sum of the no.  which is a multiple of $9$.
and the $3rd$ possiblility the same way
I think that is acceptable. :([/quote]
3p+6 isnt always divisible by 9. for example, what if p was 2?

[quote="kryptos"]Hi,

I just started number theory. Hopefully someone will help me cause i
just can't seem to see relationships between problems, they all look
so different! Heres 3 of the many im stuck on.

1. Prove that if p and p+4 are consecutive primes, then 9 divides the
sum of the composite numbers between them.

2. Show that there are infinitely many primes of the form 6k+5.

3. Suppose that n > 1 is an integer. If n is not the form 6k+3 then
prove that n^2 + 2^n  is a composite.

Thanks[/quote]
number 2 definitely isnt a getting started problem. It might even be
pre-olympiad level
[hide="hint for 2"]The way Euclid did his argument (assuming otherwise
and then multiplying "all" primes together and adding 1) can be
applied similarly here, but it has to be implimented more
cleverly[/hide]

1.the numbers between p and p+4 are
p+1+p+2+p+3.
which gives us
3p+6. Now it remains to prove that p=1(mod 9) or 4(mod 9) or 7(mod
9)since all of these make that divisible by 9. If p was 2(mod 9) then
the other prime would be 6(mod 9) which is divisible by 3.
Contradiction. p couldnt be 3(mod 9). if p was 4(mod 9), then the
other would be 8(mod 9). Which could be possible, since there is no
number guaranteed to divide it. Casework tells us nothing other than
these three work

problem 3 is a load of casework in mod 6.

[quote="riddler"]scrambled, how can $p=2$ eh $p+4$ is not a prime so it is not possible. they are consecutive primes.[/quote]
but you havent proved it

1. The numbers are $p, p+1, p+2, p+3, p+4$. Unless $p=3$, p isn't divisible by 3. Therefore, $p+3$ isn't divisible by 3. $p+4$ is never divisible by 3 (it's a prime larger than 3). Therefore, $p+1$ isn't divisible by 3. Since either $p+1$, $p+2$, or $p+3$ must be divisible by 3, $p+2$ is divisible by 3. The sum is equal to $3p+6=3(p+2)$, so it must be divisible by 9. If $p=3$, then $p$ and $p+4$ aren't consecutive primes. This completes the proof!

2. We'll prove by contradiction. First we need the following lemma, which is very easy to prove: every prime is of the form $6k+5$ or $6k+1$. Now suppose that for a certain $m$, 6m+5 is the largest prime of that form. Consider the number $2*3*5*.......*(6m+5)-1$, which is one less than the product of all prime up to and including $6m+5$. Notice that this number is of the form $6q+5$ for some $q$. Obviously it can't be prime (or else there would be a contradiction since it would be larger than the largest prime of its form) If this number is composite, than its prime factorization consists of primes larger than those of our product. Therefore, it has to be of the form (6a+1)(6b+1)......... But a number of this form gives a remainder of 1 upon division by 6! Contradiction!

Hope that's correct and understandable :)

[quote="Holier Than Thou"]Would the fact that $2*3*5*...*(6m+5)-1$ is also of the form $6q+5$, when $6m+5$ was assumed to be the largest of its form, be enough to prove it?[/quote]

If it's prime, then yes. If it's composite, we have to go a bit further. That's why the proof is broken up into two cases.

[quote="kryptos"]Thanks for all the replies! Least now i've got a few ideas :)
Could someone please post a solution for Q3.[/quote]
i remember a recent problem discussing it but i dont think they have solution yet
here it is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41960]2^b+b^2[/url]

[quote="Bob"]If it's prime, then yes. If it's composite, we have to go a bit further. That's why the proof is broken up into two cases.[/quote]

Ah, of course. For some reason I was thinking being in the form $6m+5$ automatically made it prime.


vector
linear algebra
linear algebra unsolved

I am so rusty in linear algebra now, so I need your help guys to clarify some point.

Is it possible for an infinite dimensional vector space $X$ to have a proper subspace $V$ such that $V$ is isomorphic to $X$?

Thanks

Sure. Every vector space has a (Hamel) basis. For an infinite-dimensional vector spact, that basis is an infinite set. Every infinite set can be placed in a one-to-one correspondence with a proper subset of itself. Use that correspondence to define your isomorphism.

Here's a concrete example of this. Let $V$ be the vector space of all polynomials. Let $V'=\{P\in V: P(0)=0\}.$

Define $T: V\mapsto V'$ by $T(P)(x)=xP(x).$ That is a linear isomorphism from $V$ onto $V'.$

It is ... remember a linear map is given entirely by the values it takes on the elements of a base ...


For what prime $n$ does there exist solutions in integers to:

$a^n+b^n=c^n+d^n$

With $a \neq c,d$
$b \neq c,d$

Do you also exclude trivial cases like $a=-b$ etc.¿


When I use "", both of them are backward, $ ""$
How can I fix this problem?

Use `` as the first symbol.

$ ``\mathfrak{Quote.}"$


AMC
USA(J)MO
USAMO
AIME
probability
AMC 10
geometry

AMC 10 Scores
How well did YOU do? (in comparison to everyone else).
EDIT: got 135 pts

[color=green]Added poll. It's bugged somehow, so just vote on the first option of the two offered. -mod[/color]

BLARGH, fail, i think i got a 132 or 138. idk at this point. most likely a 132 though.

121.5 on the 10A. Though there were some easy problems I should have gotten, overall, for me, it wasn't too bad. About a 30+ improvement from last year, and 60+ from 2 years ago. I'm just happy I made AIME.

I got a 121.5 on the 10A. Okay, big improvement, but missed some easy ones. I dont really know what index score is needed to make usamo. Maybe 6-8 for a score in the 120's? I dont really know.

I'm pretty sure that this year will have a higher usamo index than last year, unless the aime is very hard.

The amc 10 was slightly easier... got a 144 since I didn't see the (r<100) and assumed it was $ r \le 100.$

Oh well.

Amc 12 last few problems were pretty easy... Brut3Forc3 showed the me the last five amc 12 problems....

So I'm pretty sure that you should need at least an 8 on the aime to go to usamo...

150. Fail.

And the first year I made AIME too... irony, anyone?

I would've gotten a perfect if I took it...

o my gosh i think the answer for #22 is incorrect, but apart from that, i got a 132.

Answer for 22 isn't wrong.  After you roll the first die, there are 11 tiles left, but only 2 match, so answer is 2/11.

I put 1/6...

Gaaaaah 144 by missing number *17* of all questions...I don't know anybody who missed it.

144. I happened to blank out on number 21, and write B, because it looked right.

I thought the whole thing was easier than last year. I also thought that 24 and 25 were way too easy, for their numbers.

Anyway, should I take the 12B or the 10B? (Btw, I'm in eighth grade, if that changes your opinion.)

I'm taking the 12B, although that's mostly because I was the school winner last year :P . Let's pretend I wasn't. If I had gotten less than a 132 on the 10A, I probably would've taken the 10B since I hopefully would be able to score higher and that could make a big difference. However, now, the only way I can score better is getting a perfect (since there's no way I'm going to leave one blank), but there's no assurance of that. So I would still take 12B, partly because it'll be better practice for the AIME and it'll be good to get used to taking it.

I agree that 24 and 25 were far too easy for their numbers. I really liked #22 though, especially since it was one of those "logical thinking" probability problems.

I got 118.5 because I was being stupid. If I'd had another minute, I would've gotten 127.5. Oh well, there's always next year. I guess this means I disagree with Edison - not making stupid mistakes is a huge factor. But hard work still counts more than talent.

114 =(

I messed up the last one after thinking abotu the 7 case.  But I was too lazy to divide it!!

[quote="cognos599"][quote="Math Geek"]I also disagree with Edison on his statement when it applies to math competitions.  Success is more like 25% talent, 25% hard work, and 50% the act of not making careless mistakes.  I'm serious.[/quote]

depends on the math contest... i guess for mathcounts, AMC 8, AMC 10/12, and MAYBE (?), AIME, this is the case, but for competitions like USAMO, its more like 40% talent, 59% hard work and 1% the act of not making careless mistakes.[/quote]

what is "careless" then? I mean I thought I solved a problem on USAMO last year but oops, it was wrong because my steps could've made a number negative (#5). This was an oversight, but was it "careless"? :/

carelessness in the form of laziness (i.e. not justifying everything) loses points though

[quote="not_trig"]
what is "careless" then? I mean I thought I solved a problem on USAMO last year but oops, it was wrong because my steps could've made a number negative (#5). This was an oversight, but was it "careless"? :/

carelessness in the form of laziness (i.e. not justifying everything) loses points though[/quote]

In my opinion, carelessness and stupid mistakes are when you know how to do something but you are stupid and skip a step or add wrong. Like if you have a system, you solve for x, and then when you go to solve for y, you substitute x into the y variable, and.... well thats what I think being careless is

[quote="georgiamathdude"]if i got a 141, would a 7 on aime be mean a good chance of usamo, and an 8 like a gurantee?[/quote]
wow 141 nicee, 7 is a guarantee

The act of not making careless mistakes is not an independent skill. It is 40% talent and 60% hard work.

I achieved my goal!!

Now time to get ready for AIME :)

I got a 135.  I made a stupid mistake and found the wrong ratio on 17

i never got to take it :(

(my school fails)

I phailed. 105/150. This is my first AMC10, and I'm in 5th grade, but I still phailed.

I got a 120 on the dot. I had 123, and decided to put down two answers that I wasn't 100% sure on, but was pretty sure of. I got one of those right and it turns out one of the ones I thought was right, turned out being wrong. So 123-6-3+6=120

I guess it's ok for my first try.


In the game show [i]1 vs. 100[/i], one contestant must try to eliminate a mob of 100 people in 12 questions.  If successful, the contestant wins a million dollars.

Assuming the following facts:
[list]The contestant answers every question correctly.
Each mob member has a constant probability $P$ (it's the same for every mob member and for every question) of answering any single question correctly.[/list]

Compute $P$ so that the probability of the contestant winning a million dollars is $0.01$.

ur not including the probability of lifelines and such.........right?
-jorian

yeah some rules might need to be applied, but this is basic i guess

I know someone who likes the show, since Bob Saget is the host.

Forget about asking for help.  Assume that the contestant will never ask for help.  I told you, the contestant will get every question right.

Maybe I should also specify that if a mob member gets a question wrong (remember, that has a constant probability $P$ for any of the 12 questions), he/she is eliminated.  That should be enough information.

haha i know someone who is already obsessed with the show  :D


modular arithmetic
geometry
3D geometry
Diophantine Equations
pen

Find all pairs $(x,y)$ of positive integers that satisfy the equation \[y^{2}=x^{3}+16.\]

We have $x^3 = (y-4)(y+4).$ From $\gcd(y-4,y+4)$ divides $(y+4)-(y-4)=8$ and $y+4\equiv y-4\pmod{8}$ it follows that either $\gcd(y-4,y+4)=1$ or $\gcd(y-4,y+4)=8.$
In either case both $y-4$ and $y+4$ must be cubes. The only cubes with the difference 8 are (-8,0) and (0,8), implying that $y=\pm 4$ and $x=0.$

$ (y\minus{}4)(y\plus{}4)\equal{}x^3$
$ y\plus{}4 >y\minus{}4$
So we take the perpuses:
$ 1)$ $ y\plus{}4\equal{}x^3,y\minus{}4\equal{}1$(no integers solutions)
$ 2)$ $ y\plus{}4\equal{}x^2,y\minus{}4\equal{}x$(no integers solutions)
So there are no [b]positives[/b] integers solutions...

No, dimitris, your reasoning only works for prime x. For example $ 6^3$ can also be factored as $ (2*3^2)*(2^2*3)$ or similar options.

[quote=maxal]We have $x^3 = (y-4)(y+4).$ From $\gcd(y-4,y+4)$ divides $(y+4)-(y-4)=8$ and $y+4\equiv y-4\pmod{8}$ it follows that either $\gcd(y-4,y+4)=1$ or $\gcd(y-4,y+4)=8.$[/quote]

Hello, maxal!

If $y=8$, $\gcd(y-4,y+4)=4$

So we have to consider following cases.

[b]Case 1)[/b]$\gcd(y-4,y+4)=2$
$y-4=2k$, $y+4=2(k+4) (k:odd)$.
Since $4k(k+4)=x^3$, $x=2m$.Then $k(k+4)=2m^3$  which is absurd.

[b]Case 2)[/b]$\gcd(y-4,y+4)=4$
$y-4=4k$,  $y+4=4(k+2) (k:odd)$.
Since $16k(k+2)=x^3$, $x=4m$.Then $k(k+2)=4m^3$  which is absurd.

Thanks, Takeya.O.

[hide=Solution]Rewrite the equation as $x^3=(y-4)(y+4)$. We do casework based off the parity of $x,y$. It is easy to see they will both be the same parity.
-------------------------------------
Case 1: Both $x,y$ are odd.

This means both $y-4$ and $y+4$ are perfect cubes since $\gcd(y-4,y+4)=1$ by Euclidean algorithm. We can't have two positive odd cubes that differ by $8$ so there are no solutions in this case.
-------------------------------
Case 2: Both $x,y$ are even.

It is easy to see that if $2|x$ then $4|y$ because otherwise $v_2(LHS)>v_2(RHS)$. Let $y=4y'$ in the original equation and we have $16y'^2=x^3+16$. If $4\nmid x$ then we have $v_2(LHS) > v_2(RHS)$ again. Thus, $4|x$. Letting $x=4x'$ in the new equation gives $y'^2=4x'^3+1$, implying that $y'$ is odd. Letting $y'=2k+1$, $4k^2+4k+1=4x'^3+1$ which becomes $k^2+k=x^3$. This implies $k, k+1$ are both cubes by Euclidean algorithm. It follows that $k=-1,0$ are the only solutions. This results in $x=0$ which implies $y=\pm4$.
-----------------------------------
Thus, our solutions are $(x,y)=(0,4)$ and $(0,-4)$.[/hide]


Ten people are trying to guess a real number from 0 to 10 inclusive.  You can see the other players' guesses.  The eight people who are closest to the number win.

Prove that it is possible to guarantee a win if you go fifth or later.

$n (n\geq 5)$ people have picked some arbitrary real numbers in [0,10]. You pick the median of those numbers. Either you choose the correct real number (and you win), or it's greater than/less than your choice. WLOG assume it was less than your choice, the worst case is that the next $9-n$ people have closer choices than you, $9-n+\lceil\frac{n}{2}\rceil \leq  7$ people have choices as good or better than yours. So in the worst case you'll be 8th and win :)

[hide="My Solution"]Given the first four guesses $a_{1},a_{2},a_{3},a_{4}$, you can choose a number $x$ that is in the interval $(a_{2},a_{3})$.  So there are at least 2 numbers more than yours and at least 2 numbers less than yours.  Yours can be as low as the third-lowest, or as high as the third-highest.
Now, to prove that it works when you have those conditions: testing the extremes.
Assume that the third-lowest will lose.  Then the second-lowest and lowest numbers are too far away and are losers too.  But that makes at least 3 losers, and there can only be two.  Contradiction.  A similar proof exists for assuming you are the third-highest.
Therefore, we can conclude that any number in the interval $(a_{2},a_{3})$ will always work.
$\boxed{\text{Q.E.D.}}$[/hide]


algebra
polynomial
number theory unsolved
number theory

We call a positive integer an almost perfect square if it is of the form $ mn$ where the integers $ m$ and $ n$ satisfy $ m \leq n$ and $ |\dfrac mn \minus{} 1 | < \dfrac 1{2007}$ . Show that there exist infinitely many $ 6$-tuples of consecutive almost perfect squares.

$ \left| \frac{m}{n} \minus{} 1 \right| < \frac{1}{2007} \not \Leftrightarrow \left| \frac{n}{m} \minus{} 1 \right| < \frac{1}{2007}$.  The problem needs to stipulate $ m > n$ or $ m < n$.

My bad, I forgot to add that in. Edited.

why do we *need* to stipulate an order on $ m$ and $ n$??

One condition is stronger than the other.  The problem may intend the weak condition and be false with the strong.

I'll post some trivial remarks.  First off, when $ a$ is large, then $ a^2 \minus{} 1$ and $ a^2$ are almost perfect squares.  So we get two consecutive such numbers.

Here's a way to get three consecutive almost perfect squares.  Set $ a \equal{} b^2 \plus{} b \plus{} 1$.  Then we have the factorization
\[ a^2 \plus{} 1 \equal{} (b^2 \plus{} 1)(b^2 \plus{} 2b \plus{} 2).
\]
So $ a^2 \minus{} 1$, $ a^2$, and $ a^2 \plus{} 1$ are almost perfect squares when $ b$ is large.

I don't know whether this approach leads anywhere, or is a deadend.

but "...it is of the form $ mn$, where $ m,n$ satisfy $ |\frac{m}{n}\minus{}1|<\frac{1}{2007}$", without the roles of $ m,n$ distinguished, is the weaker statement. in your statement, you insist that the smaller of $ m,n$ play a particular role.

Hmm.  Point taken.

Ravi:  According to your approach, what we should be looking for is a family of six monic polynomials of degree $ 2n$, all of which factor into two monic polynomials of degree $ n$, and which differ by $ 1$.  It's an interesting idea, but I can't find a good place to start.

nobody's managed to solve this problem yet?


Have you ever heard of the Ku Klux Klan? Give your opinions on it.

yea, the KKK

they are very racist guys that are "purely" white

who would hate a group of ppl to do stuff like that? :mad: do they still exist?

There some KKK organizations in USA today, and they're legal. The members inherite everything except the actions of killing from the old KKK.
This is a good example of how free you could speak in USA, but remember NEVER talk with your friends about something from the Bible like "burning bush" in bars.

The "Neo Nazi" thread of yours has been deleted, bubka, because the nature of the post was inherently spammy.

Yes please, I think we're hear to LEARN, no to increase our post account.

Here, there's a long but interesting and complete link of KKK, with general information of the "group": http://en.wikipedia.org/wiki/Ku_Klux_Klan

[quote="José"]Yes please, I think we're hear to LEARN, no to increase our post account.[/quote]

Oh.
Just Kidding.

I hate the ku klux klan because they're racist.
This topic is a bit like the one about the racist white website.

Yes I have heard of the Ku Klux Klan.

The Ku Klux Klan is infamous for its hate towards African Americans, but it should be pointed out that they also don't accept :

- Jews
- homosexuals
- catholics (they are protestant)

The KKK is are rather "southern" matter.  The civil war (1860-1865) between the North and South of the USA basically involved the South not willing to forbid enslavement of African Americans.  Many KKK still idolize the leaders of the South like Jefferson Davis, Robert Lee, and Stonewall Jackson, and don't mind offering up a dollar or two to show that :

[url]http://users.telenet.be/frederic_bel6/Atlanta/LUC3040.jpg[/url]

For that matter, they don't accept Communists as well. Or Gypsys. They are like the Neo-Nazis escept for the hair.

Look up the Aryan Nations sometime, dude. Far, far worse. Not that I'm endorsing the KKK, definitely nothing like that, it's just that I decided to look up the Aryan Nations once, because of some local news involving the group, and it scared the holy *bleep* out of me.

Also, while the history of the KKK lies in the South, please don't get it in your minds that only the South has white supremacists. The worst of them are actually in Oregon, thanks.

nevertheless, i think the KKK is a horrible group

You mean why US is a multicultural country?
Well in the constitution it says "All men are created equal..."
It means everybody are created equal. Not just whites.

[quote="fredbel6"]Well they do have a point : it where white men who did all the hard work of occupying the land and fighting indians, all those Africans had to do is being brought there as slaves.  So why wouldn't the USA belong to whites?[/quote]

Are you kidding? Or do you seriously mean that?

[quote="bubka"]
Are you kidding? Or do you seriously mean that?[/quote]

 :huh:

Of course I am being sarcastic.

I have always wondered: Does exist any groups that defend asian supremacy? Black supremacy? Latinoamerican supremacy?
And -supposing there are- why we don't hear of them as we hear of neo-nazis or KKK?

[quote="fredbel6"]I am no expert but Black Panthers seems to be one of those.[/quote]
Actually, the Black Panthers were more of an extremist civil rights group than a black supremacy group- their mission was not to assert the superiority of African-Americans, it was to assert their equality, through whatever force they deemed necessary.  The group fell apart in the early 1970's.


I'm intending to learn C, with a focus on USACO and kernel hacking. Do ya'll have any recommendations on good books to learn C, with a pretty good reference section too?

Thanks!

Billy

I can't recommend any good books because I never read any but if you wish to focus on USACO, you should also consider learning C++ and the standard template library, they are priceless. However, I don't think these are available to kernel programmers. (May be wrong though.)

I must say that I think that books are absolutely useless in Computer Science nowadays. What i would do if I were you is that I would solve all of USACO training problems and the move on to other sites and eventually topcoder .
here are some links: http://infoarena.ro/links

Well, there are definitely a lot of online resources but that doesn't mean that books are useless. The good ones, that is.

What I mean is that everything is evolving so rapidly, new algoritms appear every day and books can't keep up with this. They are good for you to make a strong foundation of knowledge, but if he wants to take up USACO, online  resources are more appropriate in my opinion.

The C language hardly changes every day. The C99 standard is just as valid now as it was seven years ago and the new algorithms that appear every day are most likely very specialized and of no use to ordinary sane people :) - just show me an algorithm that can be used in USACO and isn't older than 25 years.

Nevertheless, spix is right about one thing - there's a lot of resources online, both C / C++ language tutorials and algorithm tutorials (as well as many more advanced articles). Google for C, C++, tutorial, STL, standard template library and then try to solve the USACO training problems. That's the way I learned C++ anyway :).

Here's an update: I ended up getting K&R, and it's been great. I've kinda got the C builtins down, but I'm still working on the syntax. C really is fun though :).

You can also [url=http://tigcc.ticalc.org/]write calculator games and stuff in C[/url]. That's how I first got into it back in middle school, when I was obsessed with my TI-89. ;)


let $a,b,c,x,y,z$ are any real numbers then let
\[A=2(a^{2}+b^{2}+c^{2}+ab+bc+ca)\]
\[B=43(x^{2}+y^{2}+z^{2})-42(xy+yz+zx)\]
\[C=(a+b-c)(x+y-z)+(b+c-a)(y+z-x)+(c+a-b)(x+z-y)\]
then prove that $AB\ge4.C^{2}$
hungktun accept this small problem gift from me.

I am not sure if I am the person Anuj Kumar sent this problem, because a bit difference in the name (oy me) he write. If that, thank you very much, Anuj Kumar. I will try it,

Today is a nice sunlight day in my hometown. I've just waken up some minute ago and think about this problem. Luckily, I found a solution instantly. I am sorry if I am not the person Anuj Kumar dedicating, here is my solution

Let $b+c=m,c+a=n,a+b=p$. By some simple transformation, we get
\[2C=\sum_{cyc}m(3y+3z-5x).\]
Applying Cauchy, we have
\[4C^{2}\le (m^{2}+n^{2}+p^{2})\left(\sum (3y+z-5x)^{2}\right).\]

Just notice that
\[m^{2}+n^{2}+p^{2}=2(a^{2}+b^{2}+c^{2}+ab+bc+ca).\]
\[\left(\sum (3y+z-5x)^{2}\right)=43(x^{2}+y^{2}+z^{2})-42(xy+yz+zx).\]

So we have done.

Good exercise for applying Cauchy  :lol: .

you are the right person. sorry for spelling mistake.


1-Is there any team who's gonna have the same stopovers "frankfort-mexico city-merida" and travels monday at early morning?
2-few days before the competition, what do you think  the most useful thing to do for a non-trained contestant could be?

[quote="wisemaniac"]2-few days before the competition, what do you think  the most useful thing to do for a non-trained contestant could be?[/quote]Rest :)

Albanian Team. Takes of from Tirana-Viena-Frankfort-Mexico-Merida, on Monday. Our Plane in Frankfort takes off at 1:00 PM.

i just checked ,our team too (and 7hours waiting in the airport)!! i think most of the teams are going to take one the 2 planes going from frankfurt to mexico monday at 13 :30 !! or am i wrong? :?  :)

We are going Amsterdam - Houston - Merida, so there are other ways to fly to Merida.

As for what to do before the test I dare say rest is a very good advice. You might try some (simple) problems for a while to keep your mind flexible. But I strongly advice against trying to cram some new (or even old) theorems in your head, since you will not have time to learn how to use them properly and they will just confuse you.

Belgian team:
11/07 PARIS CDG - MEXICO    flight AM 006    departure 10.30    arrival 16.00
11/07 MEXICO - MERIDA    flight AM 647    departure 21.35    arrival 23.00

There's long wait in Mexico City, if any team wants to join us in our waiting, you'll probably recognize us by our big Belgian flag (black-yellow-red) and the little connect 4-game we're playing. :-)

The Finnish team goes on the route Helsinki - Frankfurt - Mexico City - Merida.

We will take the plane that leaves from Frankfurt at 13:30. Even if we don't get in the same plane, I guess we will still see each other in the airport. See you there!

We (Serbia&Montenegro team) are going by Belgrade-Frankfort-Mexico City-Merida line.

[quote="Valentin Vornicu"][quote="Dave Vather"]I´m in Aereomexico 531 México City - Mérida (12:30 AM) Someone on that plane, too?[/quote]I´m already here :D Guys be aware it´s [b]really but really hot[/b] ...[/quote]
It isn´t "The withe City" just for its buildings, but also because many people wears withe clothes because of the hot


Form the result $ E[Y] \equal{} \int_{0}^{\infty} P\{Y > t\} dt$, show for nonnegative random variable $ X$, that $ E[X^{n}] \equal{} \int_{0}^{\infty} nx^{n\minus{}1} P\{X >x\} dx$.

So I start with $ E[X^n] \equal{} \int_{0}^{\infty} P\{X^n > t\} dt$

Taking $ x^{n} \equal{} t$ and deriving,gives $ dt \equal{} nx^{n\minus{}1}dx$ , thus:

$ \equal{} \int_{0}^{\infty} P\{X^{n} > x^{n} \} nx^{n\minus{}1} dx$

? $ \equal{} \int_{0}^{\infty} P\{(X^{n})^{\frac{1}{n}} > (x^{n})^{\frac{1}{n}}\} nx^{n\minus{}1} dx \equal{} \int_{0}^{\infty} P\{X>x\} nx^{n\minus{}1} dx$

I wasn't sure if my penultimate step was correct. Anyone have any ideas?

Since $ X$ and $ x$ are nonnegative, $ P[X^n>x^n]\equal{}P[X>x]$, so your penultimate step is justified.


1. ABC is an isosceles triangle with AB=BC=12. P and Q are points on AB and AC such that AP=AQ=8. PC and QB intersect at X. The area of APXQ is 8. What is the area of ABC?

2. ABC is an equilateral triangle. A point S is chosen in the plane of the equilateral triangle, and AS=3, BS=5, CS=8. Find BC.

3. What is the side length of the smallest square that can enclose 3 non-overlapping circles of radius 1 (with proof)?

4. What is the radius of the smallest sphere that can enclose 4 non-intersecting (can be tangent however) spheres of radius 1 (with proof)?

5. Three circles of equal area "cover" a unit square (side = 1). "Cover" means that each point on or in the square is under at least one of the circles. What is the minimum possible radius for the circles (with proof)?

For #3 I've got the minimum when the centers of the circles form an equilateral triangle with one of the medians lies on the diagonal of the sphere, and for #4 I've got the minimum when the centers of the spheres form the vertices of a tetrahedron, but I have trouble proving they are indeed the minimum, though...


function
algebra
IMO Shortlist
functional equation
IMO 2008
hojoo lee
IMO

Find all functions $ f: (0, \infty) \mapsto (0, \infty)$ (so $ f$ is a function from the positive real numbers) such that
\[ \frac {\left( f(w) \right)^2 \plus{} \left( f(x) \right)^2}{f(y^2) \plus{} f(z^2) } \equal{} \frac {w^2 \plus{} x^2}{y^2 \plus{} z^2}
\]
for all positive real numbers $ w,x,y,z,$ satisfying $ wx \equal{} yz.$

[i]Author: Hojoo Lee, South Korea[/i]

Put $ w \equal{} x \equal{} y \equal{} z \equal{} t$ and observe that $ f(t^{2}) \equal{} f(t)^{2}$.
Set $ k(t) \equal{} \frac {f(t)}{t}$ and consider $ pq \equal{} s ^{2}$.
You get $ \frac {f(s^{2})}{s^{2}} \equal{} k(s^{2}) \equal{} \frac {f(p^{2}) \plus{} f(q^{2})}{p^{2} \plus{} q^{2}}$
or
$ \frac {f(s)}{s} \equal{} \frac { f(p) \plus{} f(\frac {s^{2}}{p}) }{p \plus{} \frac {s^{2}}{p}}$ independent of p.
Now you can let $ p \equal{} 1$ and solve the quadratic equation resulting.
The solutions are of form
$ \frac {s \plus{} s^{ \minus{} 1} \plus{} / \minus{} \sqrt {s^{2} \plus{} s^{ \minus{} 2} \plus{} 2 \minus{} 4f(1)}}{2}$, and putting $ s \equal{} 1$ gives
$ f(1)(f(1) \minus{} 1) \equal{} 0$ so $ f(1) \equal{} 1$ and you can get $ f(s)$ taking one of the values $ s$ or $ s^{ \minus{} 1}$ at each value of $ s$. Now partition the positive reals (other than 1) into those for which $ f$ gives the identity (typical element p), and those for which $ f$ gives the inverse (typical element $ q$). The original equation contradicts the statement $ f(p^{2}q^{2}) \in ({{p^{2}q^{2},p^{ \minus{} 2}q^{ \minus{} 2}}})$, so one of those partitions must be empty. Now you can go ahead and verify that either identity (everywhere) or inverse (everywhere) are satisfactory.

Here is my solution .
Take $ x \equal{} y \equal{} z \equal{} w$ then we have
\[ f(x^2) \equal{} f(x)^2 \Rightarrow f(1) \equal{} 1
\]
Take $ x\to \sqrt {x},y\to \sqrt {y},z\to sqrt{z},w\to \sqrt {w}$ we have :
\[ (y \plus{} z)(f(x) \plus{} f(w)) \equal{} (x \plus{} w)(f(z) \plus{} f(y))
\]
Take $ x \equal{} w\sqrt {yz}$ then we have :
\[ (y \plus{} z)f(\sqrt {yz}) \equal{} \sqrt {yz}(f(x) \plus{} f(y))
\]
Take $ z\to 1$ then :
\[ \frac {f(x)}{x} \equal{} \frac {f(x)^2)}{x^2 \plus{} 1} \Leftrightarrow (f(x) \minus{} x)(x.f(x) \minus{} 1) \equal{} 0 \Leftrightarrow f(x) \equal{} x,f(x) \equal{} \frac {1}{x}
\]
Suppose exist $ a,b$ is different from 1 such that $ f(a) \equal{} a,f(b) \equal{} \frac {1}{b}$
Take $ w \equal{} 1,x \equal{} yz$ then
$ (y \plus{} z)(f(yz) \plus{} 1) \equal{} (yz \plus{} 1)(f(y) \plus{} f(z))$
Take $ y\to a,z\to b$ then
\[ (a \plus{} b)(f(ab) \plus{} 1) \equal{} (ab \plus{} 1)(a \plus{} \frac {1}{b})
\]
If $ f(ab) \equal{} ab$ then
$ b \equal{} \frac {1}{b} \Leftrightarrow b \equal{} 1$
If $ ab \equal{} \frac {1}{ab}$ then :
\[ (a \plus{} b) \equal{} a(ab \plus{} 1)  \equal{} b(ab \plus{} 1)
\]
So $ a \equal{} b$
Therefore $ a \equal{} b \equal{} 1$ (contradiction )
There are only two function satisfy condition :
1) $ f(x)\equiv x$
2)$ f(x)\equiv \frac {1}{x}$

My solution (at home) is very similar to TTsphn's one...but quite longer and messier

By the way nice solution!

Daniel

[quote="orl"]Find all functions $ f: (0, \infty) \mapsto (0, \infty)$ (so $ f$ is a function from the positive real numbers) such that
\[ \frac {\left( f(w) \right)^2 \plus{} \left( f(x) \right)^2}{f(y^2) \plus{} f(z^2) } \equal{} \frac {w^2 \plus{} x^2}{y^2 \plus{} z^2}
\]
for all positive real numbes $ w,x,y,z,$ satisfying $ wx \equal{} yz.$

[i]Author: Hojoo Lee, South Korea[/i][/quote]

hi for $ x\equal{}y\equal{}z\equal{}w f(x^2)\equal{}f(x)^2$

so  $ f(1)\equal{}1 (f(x)>0)$

supposing that $ x\equal{}w$so $ x^2\equal{}yz$ that mean:

$ f(x^2)(y^2\plus{}z^2) \equal{}x^2 (f(y^2)\plus{}f(z^2))$

$ f(yz) (y^2\plus{}z^2 )\equal{} yz (f(y^2) \plus{}f(z^2))$

for  $ y\equal{}1$ we have  : $ f(z)(z^2\plus{}1)\equal{}z(f(z^2)\plus{}1)$  wich include

 $ f(z)(z^2\plus{}1)\equal{}z(f(z)^2\plus{}1)$

==> $ zf(z) (z\minus{}f(z))\plus{}(f(z)\minus{}z) \equal{}0$

$ (z\minus{}f(z) )(zf(z)\minus{}1)\equal{}0$==>$ f(z)\equal{}z$ or $ f(z)\equal{}(1/z)$

[quote="Lepuslapis"]Note that you are not done there: there could be a horribly convoluted and nasty function which maps $ x$ to $ x$ for some real $ x$ and $ x$ to $ \frac {1}{x}$ for others. There remains to prove that no such function satisfying the conditions of the problem exists (as TTsphn did, for example).[/quote]

yes you are right we have to find contradiction for the case of  f determinated us

$ f$  :

$ f(x)\equal{}x$ for $ x \in [0.a[$

$ f(x)\equal{}\frac{1} {x}$ for $ x \in [a.\plus{}00[$

[quote="not_trig"][quote="the.sceth"]
You get $ \frac {f(s^{2})}{s^{2}} \equal{} k(s^{2}) \equal{} \frac {f(p^{2}) \plus{} f(q^{2})}{p^{2} \plus{} q^{2}}$
or
\[ \frac {f(s)}{s} \equal{} \frac { f(p) \plus{} f(\frac {s^{2}}{p}) }{p \plus{} \frac {s^{2}}{p}}
\]
independent of p.
[/quote]

Sorry, how do you get from the first equation to the second? Did you replace $ p^2,q^2,s^2$ with $ p,q,s$?[/quote]

Exactly.

Also, bear in mind that the idea of the partition was equivalent to TTsphn´s method of eliminating the posibility of $ f(t)$ being somewhere $ f(t) \equal{} t^{ \minus{} 1} \neq 1$ and somewhere $ f(t)\equal{}t \neq 1$. (I.e. as Lepuslapis pointed out.) It´s just with computation omitted.

as posted before $ f(x^2)\equal{}f^2(x)$.. we can substitute this at the hypothesis and get $ \dfrac{f(a)\plus{}f(b)}{f(c)\plus{}f(d)}\equal{}\dfrac{a\plus{}b}{c\plus{}d}$ with $ ab\equal{}cd$...

so, $ \dfrac{f(x^3)\plus{}f(x)}{2f(x^2)}\equal{}\dfrac{x^2\plus{}1}{2x}$, and it follows that $ f(x^3)\equal{}f(x)\left(\dfrac{f(x)(x^2\plus{}1)}{x}\minus{}1\right)$.

we also have that
$ \dfrac{f(x^4)\plus{}f(x)}{f(x^2)\plus{}f(x^3)}\equal{}\dfrac{x^2\minus{}x\plus{}1}{x}$

if we let $ f(x)\equal{}a$ we have that
$ f(x^4)\equal{}(\dfrac{x^2\minus{}x\plus{}1}{x})(a^2\plus{}\dfrac{a^2(x^2\plus{}1)}{x}\minus{}a)\minus{}a\equal{}a\cdot \left(\dfrac{x^2\minus{}x\plus{}1}{x}\cdot\left(\dfrac{a(x^2\plus{}x\plus{}1)}{x}\minus{}1\right)\minus{}1\right)$

but $ f(x^4)\equal{}f^2(x^2)\equal{}f^4(x)\equal{}a^4$, so it follows that

$ a^3\equal{}\dfrac{x^2\minus{}x\plus{}1}{x}\cdot\left(\dfrac{a(x^2\plus{}x\plus{}1)}{x}\minus{}1\right)\minus{}1$

this cubic equation has a negative solution and $ a\equal{}x$ and $ a\equal{}\dfrac{1}{x}$... and this can be finished as any of the solutions above...  :D

By putting $ w\equal{}x\equal{}y\equal{}z$ we have $ f(x^2)\equal{}f(x)^2$ for all $ x$.
For $ x\equal{}1$ we have $ f(1)\equal{}1$.

So $ f(w)^2\plus{}f(z)^2 \equal{} f(w^2)\plus{}f(z^2)$, and then $ \frac{f(x^2)\plus{}f(y^2)}{x^2\plus{}y^2}$ depends only on $ xy$, and so does $ \frac{f(x)\plus{}f(y)}{x\plus{}y}$.

So we can say that $ \frac{f(x)\plus{}f(y)}{x\plus{}y} \equal{} \frac{f(xy)\plus{}f(1)}{xy\plus{}1} \equal{} \frac{f(xy)\plus{}1}{xy\plus{}1}$.
By putting $ x\equal{}y$ and replacing $ f(x^2)$ by $ f(x)^2$ we get the quadratic equation in $ f(x)$
$ x\cdot f(x)^2 \minus{} (x^2\plus{}1)\cdot f(x) \plus{} x\equal{}0$
that has solutions $ f(x)\equal{}x$, $ f(x)\equal{}\frac{1}{x}$.

It remains to control that there can't be $ a,b\neq 1$ so that $ f(a)\equal{}a$ and $ f(b)\equal{}\frac1b$, as TTsphn did.

Well i did the same for $ f(x^2) = (f(x))^2$. Then $ f(1) = 1$ and so $ \displaystyle f(x) + f\left(\frac {1}{x}\right) = x + \frac 1 x$ (*). Now plug $ w = a^2, x = 1, y = a,z = a$ and we get $ \displaystyle \frac 1{f(x)} + f(x) = x + \frac 1x$(**) and since $ g(x) = x + \frac 1x$ is a strictly increasing function in $ (1,\infty)$ we obtain that $ f(x) = x$ or $ \displaystyle f(x) = \frac 1x$. From (*) and (**) we get that $ \displaystyle \frac 1{f(x)} = f(\frac 1x)$(***). Now to conclude use (***) suposing $ f(a) = a$, $ f(b) = \frac 1b$ and using the fact that $ f(x^{2^k}) = (f(x))^{2^k}$ we can assume that $ a$ and $ b$ are small enough or large enough to find that $ f(ab) \not = ab$ and $ \displaystyle f(ab) \not = \frac 1{ab}$ (pluging $ w = 1, x = ab, y = a, z = b$) which is a contradiction (in fact taking a large $ b$ and and choosing $ a$ such that $ ab<1$  one gets that $ \displaystyle f(ab)= \frac {ab + a +\frac 1b - b}{a+b}<0$).

[quote="orl"]Find all functions $ f: (0, \infty) \mapsto (0, \infty)$ (so $ f$ is a function from the positive real numbers) such that
\[ \frac {( f(w) )^2 + ( f(x) )^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}
\]
for all positive real numbes $ w,x,y,z,$ satisfying $ wx = yz.$

[i]Author: Hojoo Lee, South Korea[/i][/quote]
[hide="1)"]
Put in $ (1,1,1,1)$ to find that $ f(1) = 1$
[/hide]
[hide="2)"]
Put in $ (1,x,\sqrt {x},\sqrt {x})$. Then
\[ \frac {1 + f(x)^2}{2f(x)} = \frac {1 + x^2}{2x}
\]
Let $ f(x) = y$. This is clearly equivalent to
\[ y + \frac {1}{y} = x + \frac {1}{x}\iff x - y = \frac {x - y}{xy}\implies x - y = 0\text{ or }xy = 1
\]
Thus for all positive $ x$,
\[ f(x)\in{x,\frac {1}{x}}
\]
[/hide]
[hide="3)"]
Now suppose there were $ a,b\not = 1$ with
\[ f(a) = a,f(b) = \frac {1}{b}
\]
Substituting into the original equation shows that
\[ \frac {f(\sqrt {ab})^2 + 1}{f(a) + f(b)} = \frac {ab + 1}{a + b}
\]
From $ (2)$, we know that $ f(\sqrt{ab})$ can be one of two values:
If $ f(\sqrt {ab})^2 = \frac {1}{ab}$, then
\[ \frac {f(\sqrt {ab})^2 + 1}{f(a) + f(b)} = \frac {ab + 1}{a^2b + a} = \frac {ab + 1}{a + b}
\]
Thus $ a + b = a^2b + a\implies a = 1$, which is a contradiction.
If $ f(\sqrt {ab})^2 = ab$, then
\[ \frac {f(\sqrt {ab})^2 + 1}{f(a) + f(b)} = \frac {ab + 1}{a + \frac {1}{b}} = \frac {ab + 1}{a + b}
\]
so $ a + \frac {1}{b} = a + b\implies b = 1$, contradiction.

Thus no such $ a,b$ exist, that is
\[ f(x)\equiv x\text{ or }f(x)\equiv\frac {1}{x}
\]
[/hide]

As before, show that $ f(x)^2 \equal{} f(x^2)$.  Also similarly, substitute $ a^2, b^2, 1, ab$ to show that $ P(a, b) : f(a^2) \plus{} f(b^2) \equal{} f(ab) \left( \frac{a}{b} \plus{} \frac{b}{a} \right)$.  Now

$ P(a, b) \minus{} P(\sqrt{a}, \sqrt{b})^2 : f(a^2) f(b^2) \equal{} f(ab)^2$.

The same quadratic-solving step (letting $ ab \equal{} 1$) shows that $ f(a) \equal{} a \text{ or } \frac{1}{a} \forall a$, but now note that if $ f(a) \equal{} a, f(b) \equal{} \frac{1}{b}$ then $ \frac{a}{b} \equal{} ab \text{ or } \frac{1}{ab}$ which is clearly only possible if $ b \equal{} 1$ or $ a \equal{} 1$, respectively.

We claim that the only solutions are $f(x) \equiv x$ and $f(x) \equiv \frac{1}{x}$.

Let $P(w, x, y, z)$ denote the assertion. Then clearly $P(w, w, w, w)$ gives us that $f(w^2) = f(w)^2$. For $w = 1$ this gives $f(1) = 1$. Now we can rewrite the given as,
\begin{align*}
    \frac{f(a) + f(b)}{f(c) + f(d)} = \frac{a+b}{c+d},
\end{align*}
where we let $a = w^2$, $b = x^2$, $c = y^2$, $d = z^2$ such that $ab = cd$. Now from $P(a^2, 1, a, a)$ we see we find after some work,
\begin{align*}
    (f(a) - a)(af(a) - 1) = 0.
\end{align*}
This in turn gives $f(a) = a$ or $f(a) = \frac{1}{a}$ for all $a$ in the domain. Now to avoid the point-wise trap assume for the sake of contradiction that $f(a) = \frac{1}{a}$ and $f(b) = b$, with $a, b \neq 1$. Then from $P(a, b, 1, ab)$ we see that,

\begin{align*}
    ab^2 - a + b + \frac{1}{a} = (a+b) \cdot f(ab).
\end{align*}

Taking cases on whether $f(ab) = \frac{1}{ab}$ or $f(ab) = ab$ we can reduce them respectively to,
\begin{align*}
    (a^2 - 1)(1+ab) &= 0\\
    (b^2-1)(1+ab) &= 0
\end{align*}
The first case implies $a = 1$, the second case implies $b = 1$ so we are done.

The answer is $f(x)=x$ for all $x \in \mathbb{R^+}$ and $f(x)=\frac{1}{x}$ for all $x \in \mathbb{R^+}$. It’s easy to see that these functions satisfy the given equation. \\
Now, we shall show that these are the only solutions. First notice that letting $w=x=y=z=a$ we have that
$$\frac{2f(a)^2}{2f(a^2)}=1$$
Thus, we most clearly have that $f(x)^2=f(x^2)$. Now, proceed as follows. Notice that $f(1)=f(1)^2$ gives us that $f(1)=1$. Let $w=1,x=a^2$ and $y=z=a$. Then, notice the following
\begin{align*}
    \frac{f(1)+f(a^2)^2}{f(a^2)+f(a^2)} &= \frac{1+a^4}{2a^2}\\
    2a^2f(a^2)^2 - 2(a^4+1)f(a^2)+2a^2 &= 0\\
    a^2f(a^2)^2-(a^4+1)f(a^2)+a^2&=0\\
    (a^2f(a^2)-1)(f(a^2)-a^2)&=0\\
\end{align*}
Which means that $$f(a^2)=\frac{1}{a^2} \implies f(a)=\frac{1}{a}$$
or $$f(a^2)=a^2\implies f(a)=a$$
which gives us that indeed the solutions to the given equation are of the claimed type.

Call the assertion $P(w,x,y,z)$. Notice $P(k,k,k,k)$ gives us

\[\frac{2f(k)^2}{2f(k^2)}=1\]
\[\implies f(k^2)=f(k)^2\]

In particular, $f(1)=1$.

Now, $P(1,k,\sqrt{k}, \sqrt{k})$ gives us

\[\frac{f(1)^2+f(k)^2}{f(k)+f(k)}=\frac{1+k^2}{k+k}\]
\[\implies f(k) \in \{k, \frac{1}{k}\}\]

The only other case we must check is that for some $a,b \neq 1$, we have $f(a)=a$ and $f(b)=\frac{1}{b}$. Plugging in $P(a,b,ab,1)$ gives

\[\frac{a^2+\frac{1}{b^2}}{f(a^2b^2)+1}=\frac{a^2+b^2}{a^2b^2+1}\]

Both cases $f(a^2b^2)=a^2b^2, \frac{1}{a^2b^2}$ result in either $a=1$ or $b=1$, a contradiction.

Thus, $\boxed{f(x)=x, \frac{1}{x}}$.

We claim that $f(x) = x$ and $\frac{1}{x}$ are the only solutions. It is easy to check that these satisfy the equation.
Now, set $w = z = 1$, we get $\frac{f(x)^2 + 1}{f(x^2)+1} = 1 \implies f(x^2) = f(x)^2$
Therefore the original equation is equivalent to $\frac{f(w^2) + f(x^2)}{f(y^2)+f(z^2)} = \frac{w^2 + x^2}{y^2 + z^2}$ or $\frac{f(a)+f(b)}{f(c)+f(d)} = \frac{a+b}{c+d} \forall a, b, c, d \in \mathbb{R}^{+}$ satisfying $ab = cd$.
Substituting $(a, b, c, d) = (x, x, 1, x^2)$, we get $\frac{2f(x)}{1+f(x)^2} = \frac{2x}{1+x^2}$.
Solving this, we get a quadratic in $f(x)$ with solutions $f(x) = x, \frac{1}{x}$.
QED

We know $f(1)=1$ from substituting $(1,1,1,1)$ and $f(x^2)=f(x)^2$ from substituting $(x,x,x,x)$ (so in reality this equation is indeed symmetric). Then substitute $(1,t,\sqrt t, \sqrt t)$ to get
\[\frac{f(1)^2+f(t)^2}{2f(t)} = \frac{1+t^2}{2t} \implies (tf(t)-1)(f(t)-t)=0 \implies f(t) \in \left\{t, \frac 1t\right\}.\]

Now we handle this pointwise trap. Suppose FTSOC there exist positive constants $a \ne b$ and $a,b \ne 1$ such that $f(a) = a$ and $f(b) = \frac 1b$. Then substitute $(\sqrt a, \sqrt b, 1, \sqrt{ab})$ to get
\[\frac{a+\frac 1b}{1+f(ab)} = \frac{a+b}{1+ab}.\]

Our only possible choices for $f(ab)$ are $ab$ and $\frac{1}{ab}$, but if we plug them in and try to solve the resulting equation, we find that either $a=1$ or $b=1$, contradiction.

Hence our solution set is $\boxed{f(x)=\frac 1x, \quad f(x)=x}$, which clearly work. $\blacksquare$

Let $P$ be the given assertion. Then $P(1,1,1,1)$ means that $f(1)^2 = f(1)$ or $f(1) = 1.$ Note that $P(w, 1, \sqrt{w}, \sqrt{w})$ yields,
\begin{align*}
\frac{f(w)^2 + f(1)^2}{2f(w)} &= \frac{w^2 + 1}{2w} \\
(f(w) - \frac{1}{w})(f(w) - w) &= 0.
\end{align*}
So for any $a \in \mathbb{R}^+$, $f(a) \in \{a, \frac{1}{a}\}.$ To avoid the point-wise trap, assume FTSOC there exists $a,b$ such that $f(a) = a, f(b) = \frac{1}{b}$ for $a,b \ne 1.$ If $f(ab) = ab$ or $\frac{1}{ab}$ we get the respective contradictions of $P(a, b, ab, 1)$ as,
\begin{align*}
	\frac{f(a)^2 + f(b)^2}{f((ab)^2) + f(1)} &= \frac{a^2 + b^2}{(ab)^2 + 1} \\
	\frac{a^2 + \frac{1}{b}^2}{(ab)^2 + 1} = \frac{a^2 + b^2}{(ab)^2 + 1}  &\implies b^4 = 1 \\
	\frac{a^2 + \frac{1}{b}^2}{\frac{1}{(ab)^2} + 1} = \frac{a^2 + b^2}{(ab)^2 + 1}  &\implies a^4 = 1.\\
\end{align*}
Therefore $f(x) = x$ or $f(x) = \frac{1}{x}.$

The answers are $f(x)=x$ and $f(x)=\frac 1x$ which both.
Let $P(w,x,y,z)$ denote the assertion. Note that $P(1,1,1,1)$ gives $f(1)=1$ and $P(x,1,x,1)$ gives $f(x^2)=f(x)^2$.

[b][color=red]Claim:[/color][/b] For each $x$, we have $f(x)=x,\frac 1x$
[i]Proof.[/i] $P(x,1,\sqrt x,\sqrt x)$ gives that \[\frac{f(x)^2+1}{2f(x)}=\frac{x^2+1}{2x}.\] This is a quadratic in $f(x)$ so it has at most $2$ solutions, but $x$ and $\frac 1x$ work. $\blacksquare$
Now, if $f(a)=a$ and $f(b)=\frac 1b$ with $a,b\neq 1$, then $P(a,b,1,ab)$ gives \[\frac{a^2+\frac{1}{b^2}}{1+f(ab)^2}=\frac{a^2+b^2}{1+(ab)^2}.\] Then, $f(ab)=ab$ gives $b=1$ and $f(ab)=\frac {1}{ab}$ gives $b=1$. This finishes.

Pretty easy for p4.
First of all, letting all terms equal, find that $f(x)^2$=$f(x^2)$. Then, observe that $P(x,x,1,x^2)$ yields that $f(x)(x^2+1)=x(f(x)^2+1)$ for all positive real numbers $x$.(We get $f(1)=1$) Note that,if the function $g(x)=(x^2+1)/x$ has 2 equal values for 2 inputs $x$ and $y$, then either $x=y$ or $y=1/x$. So, $f(x)$ is either $x$ or $1/x$ for all $x$. Taking $a,b$ such that neither of them is $1$, let $P(a,b,1,ab)$. Messing with those 2 nasty cases gives us a contradiction.
Therefore, $f(x)=x$ for all $x$ or $f(x)=1/x$ for all $x$. :-D

Taking all the variables to be $1$, $f(1)=1$. Afterwards, set the variables to be equal, so $f(t)^2=f(t^2)$. Letting $(w, x, y, z)=(t, 1, \sqrt t, \sqrt t)$ gives
\[\frac{f(t)^2+1}{2f(t)}=\frac{t^2+1}{2t}.\]
It's easy to see that $f(t)=t$ and $f(t)=\tfrac1t$ are solutions -- since this is a quadratic in $f(t)$, there are at most $2$ solutions.

Now suppose that we have $f(a)=a$ and $f(b)=\tfrac1b$, for $a$, $b\neq 1$. Take $(w, x, y, z)$ as $(\sqrt a, \sqrt b, 1, \sqrt{ab})$. So
\[\frac{f(\sqrt{a})^2+f(\sqrt b)^2}{f(\sqrt{ab})^2+1}=\frac{a+b}{ab+1}.\]
The LHS is also $\tfrac{f(a)+f(b)}{f(ab)+1}$, hence
\[\frac{a+\tfrac1b}{f(ab)+1}=\frac{a+b}{ab+1}.\] There are two cases for the value of $f(ab)$ -- it's either $ab$ or $\tfrac1{ab}$.
[list=disc]
[*] $f(ab)=ab$: we get $b^2=1$ implying $b=1$. Contradiction.
[*] $f(ab)=\tfrac{1}{ab}$: we get $a^2=1$ so $a=1$. Contradiction.
[/list]
It's easy to verify that $f(t)=t$ and $f(t)=\tfrac1t$ both work. So we are done. $\square$

[hide="For storage"]

Obviously, $f(x)=x$ and $f(x)=\frac1X$ are solutions to the FE and we want to prove their uniqueness.
$P(1,1,1,1) \Rightarrow f(1)=1$
$P(x,1,x,1) \Rightarrow f(x^2)=f(x)^2$
so now we can rewrite the FE as following
 \[ \frac {\left( f(w) \right) + \left( f(x) \right)}{f(y) + f(z) } = \frac {w + x}{y + z}\] for $wx=yz$

Now consider $P(a^2,1,a,a)$ and solving the quadratic eqn. we have $f(a)=a$ or $1/a$
suppose there are reals $a,b$ none of them is $1$ satisfying $f(a)=a$  & $f(b)=1/b$ . By considering $P(a,b,ab,1)$ we have a contradiction.

$$\mathbb{Q.E.D.}$$

 [/hide]

The answers are $f(x) = x$ and $f(x) = \tfrac{1}{x}$, both of which aren't too hard to verify. We now show these are the only answers.

Let $P(w, x, y, z)$ denote the given assertion. From $P(1, 1, 1, 1)$, we have $f(1) = 1$. Then, from $P(1, x, \sqrt{x}, \sqrt{x})$, we have $\tfrac{1 + f(x)^2}{2f(x)} = \tfrac{1 + x^2}{2x}.$ This rearranges to
$$xf(x)^2 - f(x) - x^2f(x) + x = 0 \iff (xf(x) - 1)(f(x) - x) = 0.$$
Thus, for all $x$, either $f(x) = \tfrac{1}{x}$ or $f(x) = x$. Now, suppose for the sake of contradiction there exist $a, b > 0$ and not equal to $1$ such that $f(a) = \tfrac{1}{a}$ and $f(b) = b$. Then from $P(a, b, 1, ab)$, we see that
$$\frac{\tfrac{1}{a^2} + b^2}{1 + f(a^2b^2)} = \frac{a^2 + b^2}{1 + a^2b^2}.$$
Taking cases on whether $f(a^2b^2) = \tfrac{1}{a^2b^2}$ or $f(a^2b^2) = a^2b^2,$ we see that:
[list]
[*] If $f(a^2b^2) = \tfrac{1}{a^2b^2}$, then $\tfrac{1}{a^2} + b^2 + b^2 + a^2b^2 = a^2 + b^2 + \tfrac{1}{a^2} + \tfrac{1}{b^2}$, or $(a^2b^2 + 1)(b^4 - 1) = 0$, which implies that $b = 1$, contradiction
[*] If $f(a^2b^2) = a^2b^2$, then $\tfrac{1}{a^2} + b^2 + b^2 + a^2b^4 = a^2+ b^2 + a^2b^4 + a^4b^2$, or $(a^2b^2 + 1)(a^4 - 1) = 0$, so $a = 1$, contradiction
[/list]
So, either $f(x) =x$ for all $x$, or $f(x) = \tfrac{1}{x}$ for all $x$, as desired.


Duke
college
inequalities
complex numbers

dude it's a different poll

DUDE GET WITH THE PROGRAM

Do all Americans use the word 'dude' that often or is that just a stereotype? :rotfl:

[quote="not_trig"]sure dude but no one wants a poll run by Arnav regarding basketball, dude[/quote]

dude
I want a poll run by him hehe  :P

[quote="MysticTerminator"]dude at least they don't carry mooses around with them going all like eh? we're canadian what's going on eh? do you want to sample some of our moose, eh? let's go disco in this disco bar, eh?[/quote]
I don't know anyone who says eh.... :huh:
And we don't carry mooses around, they carry us. :rotfl:

huh you guys definitely carry mooses around, pip pip hail to the queen and all that rot, what

that . . . doesn't even make sense

it's like comparing an element of the real hamiltonian quarternions cross itself to the group ring formed by the Viergruppe and the complex numbers? $(4+2i, j+4k-1) > V_{4}\mathbb{C}$

IT DOESN'T MAKE SENSE, DUDE

good point

YOU CAN"T EVEN COMPARE UNC WITH CANADA

it's just TOO GOOD

edit: i mean canada is too good

wtf?

by transitivity

UNC > DUKE

and

CANADA > UNC

=> CANADA > UNC > DUKE

plus you get the same results if you use the power mean inequality

dude will you stop involving canada in this

I am pretty sure the UNC team would own canada in basketball

this is mainly because canada would not be able to shoot or play any defense since countries usually just sit there

without shooting and playing defense

It just goes to show we are peaceful. :rotfl:

this is true

but

CANADA > UNC

when it comes to ice hockey

[quote="not_trig"]how bout [size=200]NO[/size]

pwned![/quote]

amusing

guys i just happened upon this thread this is like ranting about people who say "eh","dude", and carry moose around BTW what is the plural of buffalo?

snap cracklpe pop

Popsicles

icey and popsilicious

tangy and tongue-twisting

 bite it hard

it gives fat lard,

 pleasing to the senses

now taste one and try!

it's actually quite tasty

Pop....Sickle

PopSnowSickle, stop the carnocentrism!

no.......... reallly????... i didn't know that...

Ragnarok is the guy!

the person with the tonka-tai! ( ask me what this is  :rotfl: )

he is surreptitiously special, but not in special ed

he works hopelessly hard, but he's not Farmer Ted

he's ragnarok

the man with the plan

the gold coin can.

Walk around the river is the kid with the shiver,

he eats a lot of food such as stir fried liver,

whenever he says I rock!

everyone responds with a shock,

hey? I Love You! You are my hero! My White Idol!

Then he runs wherever his legs take him,

responding with enthusiasm and sees dark slim

and TahDah! He's Michael Jackson!

[quote="ragnarok23"][quote="now a ranger"]DUDE

use correct grammar

it's buffaloes

buffalae is terminator language  :lol:[/quote]
I'm pretty sure it's buffalo, not buffaloes...[/quote]
i'll go with you seem sane

[color=orange] by putting a blank post you mean "i don't understand" or you are watching the person but i think this post is the first 1 i have put in this forum not on testing one srry :oops: [/color] oh and sorry if i hurt ur eyes and srry mystic terminator IF DUDE YOU MEANT SOMETHING THAT I DON'T KNOW.
all better, eh?

[quote="Inspired By Nature"]:spam: Try not to spam so much, eh?[/quote]

Definitely.

From now on, I will be deleting a lot more of the spam in this forum.  If people continue to spam, you may be banned from this forum.


a) For $ n \equal{} 1$, $ 2^{4} \minus{} 1 \equal{} 15$, which is clearly divisible by 15.

Assume that for n=k, $ 16^k \minus{} 1$ is divisible by 15. Let $ 16^k \minus{} 1 \equal{} 15d$

For n=k+1, $ 16^{k \plus{} 1} \minus{} 1 \equal{} 16(16^k) \minus{} 1 \equal{} 16(15d \plus{} 1) \minus{} 1 \equal{} 240 d \plus{} 15$, which is divisible by 15. Hence proved.

b) It is obvious for n=1.

For n=k, $ 3^{3k\plus{}1}\plus{}2^{k\plus{}1}\equal{}3(27^k)\plus{}2(2^k)$, let $ 3(27^k)\equal{}a$ and $ 2(2^k)\equal{}b$

For n=k+1, $ 3(27^{k\plus{}1})\plus{}2(2^{k\plus{}1})\equal{}27a\plus{}2b\equal{}25a\plus{}2(a\plus{}b)$, which is divisible by 5 since a+b is divisible by 5. And hence proved.

[hide="c"]Let $ n\equal{}1$. Then we have $ 2(7)\plus{}3(5)\minus{}5\equal{}24$, which is divisible by 24. Now, assume it is true for $ n\equal{}k$. Letting $ n\equal{}k\plus{}1$, we have (letting $ a\equal{}2(7^k)$ and $ b\equal{}3(5^k)$)
\[ 7\cdot2(7^k)\plus{}5\cdot3(5^k)\minus{}5\equal{}7a\plus{}5b\minus{}5\equal{}2a\plus{}5(a\plus{}b\minus{}5)\plus{}20.\]
Since $ (a\plus{}b\minus{}5)$ is divisible by 24 from our assumption, we only have to consider $ 2a\plus{}20$, or $ 4(7^k)\plus{}20$. Factoring, we have $ 4(7^k\plus{}5)$. We now note that since $ 7^k$ is odd and 5 is odd, $ (7^k\plus{}5)$ is even. Taking $ 7^k\plus{}5\bmod3$, we have $ 7^k\plus{}5\equiv1^k\plus{}2\bmod3$, which is clearly divisible by three. Thus, $ 4(7^k\plus{}5)$can be factored into $ 24x$ for some positive integer $ x$, making it divisible by 24. Our induction is now complete.[/hide]


Graphing $\dfrac {x-1}{x-2}>\dfrac1a$ will make it just a matter of moving x's and a's around.

Just remember to invert signs if for negative "a"  (clearly x cannot equal 1 or 2)

[quote="chess64"]Is $a \in \mathbb{N}$? :D[/quote]
No. $a \in R$.

[quote="Altheman"]multiply by $(x-2)^2$ (do you see why this is helpful?), then it just becomes
$a(x-1)>x^2-4x+4$
$0>x^2-(4+a)x+4a$
$0>(x-a)(x-4)$
from here, it is easy, note that $x\ne2$[/quote]

Your work is correct... but you mean $x \ne 4$.  Also, both terms must have the same parity... so, $x > \max(a, 4) \text{ or } x < \min(a, 4)$.  Is there a better way to write taht?

i believe that $x\ne 2$ because in the original expression, that would make it undefined

if you multiply by $(x-2)^2$ you have to do it to both sides.

[quote="Altheman"]multiply by $(x-2)^2$ (do you see why this is helpful?), then it just becomes
$a(x-1)>x^2-4x+4$
$0>x^2-(4+a)x+4a$
$0>(x-a)(x-4)$
from here, it is easy, note that $x\ne2$[/quote]
You forgot something on the left hand side... :(
$\dfrac{a(x-1)}{x-2}>1$
Multiplying $(x-2)^2$ on both sides,
$a(x-1)(x-2)>(x-2)^2$
$a(x^2-3x+2)>x^2-4x+4$
$(a-1)x^2+(4-3a)x+2a-4>0$
$(x-2)[(a-1)x-a+2]>0$
$\begin{cases}x-2>0\\(a-1)x-a+2>0\end{cases}$ or $\begin{cases}x-2<0\\(a-1)x-a+2<0\end{cases}$
$\begin{cases}x>2\\x>\dfrac{a-2}{a-1}\end{cases}$ or $\begin{cases}x<2\\x<\dfrac{a-2}{a-1}\end{cases}$

I think we have to divide cases for $a\neq 0$ to get

(i)When $a=1$ , $\{x: x>2\}$

(ii)When $a>1$ , $\{x: x<\dfrac{a-2}{a-1},x>2\}$

(iii)When $a<1$ , $\{x: \dfrac{a-2}{a-1}<x<2\}$


A cylinder has a radius of 3 cm and a height of 8 cm. What is the longest segment, in centimeters, that would fit inside the cylinder?

The longest segment is its diagonal, which is hypothenuse of the right triangle with other sides H and 2r. $ D\equal{}\sqrt{H^2\plus{}4r^2}\equal{}10cm$


Let $ a,b,c,d$ be distinct real numbers such that $ a\plus{}b\plus{}c\plus{}d\equal{}3$ and $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}5$. Compute
\[ \frac{a^5}{(a\minus{}b)(a\minus{}c)(a\minus{}d)}\plus{}\frac{b^5}{(b\minus{}c)(b\minus{}d)(b\minus{}a)}\plus{}\frac{c^5}{(c\minus{}d)(c\minus{}a)(c\minus{}d)}\plus{}\frac{d^5}{(d\minus{}a)(d\minus{}b)(d\minus{}c)}.\]

Could you help this? I could not solve it.

Denoting $ {S_k} \equal{} \sum\limits_{a,b,c,d} {\frac{{{a^k}}}{{\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)\left( {a \minus{} d} \right)}}}$
Let $ P\left( x \right) \equal{} \frac{{{x^k}}}{{abcd}},k \le 3$. Using Lagrange interpolation formula, we have \[ P\left( x \right) \equal{} \sum\limits_{a,b,c,d} {P\left( a \right)\frac{{\left( {x \minus{} b} \right)\left( {x \minus{} c} \right)\left( {x \minus{} d} \right)}}{{\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)\left( {a \minus{} d} \right)}}}\]
So \[ 0 \equal{} P\left( 0 \right) \equal{} \sum\limits_{a,b,c,d} {\frac{{{a^{k \minus{} 1}}}}{{\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)\left( {a \minus{} d} \right)}}}\]which means $ {S_0} \equal{} {S_1} \equal{} {S_2} \equal{} 0$
Let $ Q\left( x \right) \equal{} {x^3}$. We have \[ Q\left( x \right) \equal{} \sum\limits_{a,b,c,d} {Q\left( a \right)\frac{{\left( {x \minus{} b} \right)\left( {x \minus{} c} \right)\left( {x \minus{} d} \right)}}{{\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)\left( {a \minus{} d} \right)}}}\]Compare the highest coefficient, we deduce $ S_3\equal{}1$
We consider the polynomial \[ f\left( x \right) \equal{} {x^4} \plus{} {p_1}{x^3} \plus{} {p_2}{x^2} \plus{} {p_3}x \plus{} {p_4} \equal{} \left( {x \minus{} a} \right)\left( {x \minus{} b} \right)\left( {x \minus{} c} \right)\left( {x \minus{} d} \right)\]
with $ {p_1} \equal{}  \minus{} 3$ and $ {p_2} \equal{} \frac{{{{\left( {a \plus{} b \plus{} c \plus{} d} \right)}^2} \minus{} \left( {{a^2} \plus{} {b^2} \plus{} {c^2} \plus{} {d^2}} \right)}}{2} \equal{} 2$
We have
\[ \sum\limits_{a,b,c,d} {\frac{{{a^4} \plus{} {p_1}{a^3} \plus{} {p_2}{a^2} \plus{} {p_3}a \plus{} {p_4}}}{{\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)\left( {a \minus{} d} \right)}}}  \equal{} 0 \Rightarrow {S_4} \equal{}  \minus{} {p_1}{S_3} \equal{} 3\]
\[ \sum\limits_{a,b,c,d} {\frac{{{a^5} \plus{} {p_1}{a^4} \plus{} {p_2}{a^3} \plus{} {p_3}{a^2} \plus{} {p_4}a}}{{\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)\left( {a \minus{} d} \right)}} \equal{} 0 \Rightarrow {S_5} \equal{}  \minus{} {p_1}{S_4}}  \minus{} {p_2}{S_3} \equal{} 7\]

[b]EDIT: Oops... By the time I'm done with typing it, a similar solution is posted already. :wink: [/b]

Denote $ f(x) \equal{} x^5$, $ g(x) \equal{} (x \minus{} a)(x \minus{} b)(x \minus{} c)(x \minus{} d)$ and $ f(x) \equal{} q(x)g(x) \plus{} r(x)$, where $ \deg r\le \deg g$.
Note that $ r(x) \equal{} f(x) \equal{} x^5$, for $ x \equal{} a,b,c,d,e$.
So by Lagrange Interpolation Formula, we also have
\[ \sum\frac {a^{5}(x \minus{} b)(x \minus{} c)(x \minus{} d)}{(a \minus{} b)(a \minus{} c)(a \minus{} d)}\]
We can see that $ \sum\frac {a^{5}}{(a \minus{} b)(a \minus{} c)(a \minus{} d)}$ is the leading coefficient of $ r(x)$.
Now the computation part:
When $ x^5$ is divided by $ (x \minus{} a)(x \minus{} b)(x \minus{} c)(x \minus{} d)$, the leading coefficient of the remainder is:
\[ \minus{} \sum ab \plus{} (\sum a)^2 \equal{} \frac {1}{2}\sum a^2 \plus{} \frac {1}{2}(\sum a)^2\]
So the answer is $ 7$.


The spyware infecting Treething's computer creates a car commercial pop-up every 5 minutes. It also creates an Ephedra commercial pop-up every 19 minutes. An Ephedra pop-up and a car pop-up occur simultanouesly at 2:00 PM. When is the next time they will simultaneously occur?

[hide]The two commercials pop-up at the same time every 19x5=95 minutes.  Thus, 95 minutes after 2:00 is $3: 35$.[/hide]

[hide]The LCM multiple of 5 and 19 is 95, so both pop-ups will simultaneously occur every 95 minutes. If both pop-ups occur simultaneously at 2 PM, the next time they will both occur is [b]3:35 PM[/b].[/hide]

[hide]multiply the minutes together, and add it to 2:00
u get 3:35 PM[/hide]

[hide]First you need to find the least common multiple(LCM). The LCM of 5 and 19 is 95. Then you change it to hours which is 1 hour and 35 minutes(95/60). Finally, you add 1 hour and 35 minutes to 2. [size=150]3:35.[/size][/hide]

[quote="José"][quote="José"][quote="tz"][hide]The LCM of 5 and 19 is 95,
95/60=1+(35/60)=1:35
2+1:35=3:35

3:35[/hide][/quote]

[hide]3:35 hs.[/hide]

[color=blue]mod edit: please use hide tags[/color][/quote]

Don´t know how.[/quote]

[hide]Now I know[/hide]


symmetry
geometry
geometric transformation
reflection

A common light bulb (http://www.samis.it/images/lampada-big.jpg) of $W$ watts and $V$ Volts is in a circular trajectory uniform of radius $r$.

$L$ meters of the center of the trajectory of the light bulb have a torsional scale with two identical mirrors.

The distance from the mirrors to the center of the trajectory of the light bulb is equal.

If the light bulb starts to turn in an initial position randomly, give the variation of the angle of the torsional scale in the unit of time.

thales418 it would be very nice if u drew a picture of the above situation or atleast rephrase the problem specifying

1.[quote] $L$ meters of the center of the trajectory of the light bulb have a torsional scale with two identical mirrors.
[/quote]

what is $L$ metres away the mirrors or the resting point of the torsional scale

2.[quote]If the light bulb starts to turn in an initial position randomly, give the variation of the angle of the torsional scale in the unit of time.
[/quote]
how can the bulb turn randomly it has to move in a circle  of specified radius $r$ ??

No!!!!

LEt's Start Again

The Trajectory of the bulb is

$x = r. \cos(t) , y = r. \sin(t)$

The initial position of the bult is $t=a$ for some $a$

We don't know the value of $a$, so i said "initial position randomly"

For the others question i Will draw a picture.

The Purple Bars is the mirrors. The black Bar is just a think to make the mirrors together.

Suppose the mirrors and the haste does not have weigth

Does not have air too and we only have 2 dimensions

The angular velocity is constant

[quote="Thales418"]No!!!!

LEt's Start Again

The Trajectory of the bulb is

$x = r. \cos(t) , y = r. \sin(t)$

The initial position of the bult is $t=a$ for some $a$

We don't know the value of $a$, so i said "initial position randomly"

For the others question i Will draw a picture.

In 10 min i put this here[/quote]

thank u  very much :lol:  :lol:   :D

also in ur diagram specify from point are distances like initial position $a$ measured

[quote="Thales418"]In my question i forgot to put some things. sorry

and i didnt understand what are you saying

$a$ is the initial angle

can you be more clearly

thanks[/quote]

actually taht was a confusion .but that's ok well are the mirrors of ur picture violet and facing the bulb

but one thing  i liked ur  physical solution to n.t.tuan's problem very much  :D  :P  :blush:  :blush:  :lol:

[quote="Thales418"]sorry

all mirrors in the word are transparent. but JPG is a bullshift filetype

And the problem. Let's make this Semi 3d  putting the z axis from 0 to 1

the mirrors will have heigth 1

because the pressure for $m^{2}$

If you want a tip for this problem i can make one. give you one formula

NOTE: I dont have solved this problem. i creat this now. hauhua

thanks

suppose $r$ is big. the angular velocity is big and the linear velocity of the bulb is near of the ligth

To the problem become more interesting

:)[/quote]

the problem with this problem is it is created by a person who hasn't solved it(i assume) this is because i feel the mirrors should have infinite(large) lengths so that we get use symmetry otherwise the problem tends to become more complex further it if utmost necessary to have the mirrors connected by a "light rod" and other necessary "physical formalities" without these it is stupid to even attempt this problem .if u do have a sol^n [i]without[/i] these conditions  do give it a go here...

hm....

heres my partial solution, just a text

The pressure of radiation is equal to double of ligth intesi intensity  divided by ligth speed

Since the mirros have area equal to $e$

we can find the force

but the angle that the ligth with the mirros is variable for some instant $t$

but we can approximate this using the center of the mirror. this give a good approximation

So. we can calculate the force in both mirros

we can use torque to calculate the last things

just it

Does not make sense we dont approximate the angle of incidence of the ligth

we can use this assuming that $e$ is small

in physics we use approximations to become more easy

[quote]The pressure of radiation is equal to double of ligth intesi intensity divided by ligth speed

Since the mirros have area equal to

we can find the force [/quote]
i don't know how u will be able to find the force using this and without the angle of reflection (assuming elastic collisions of photons) and without symmetry arguments...

[quote]but we can approximate this using the center of the mirror. this give a good approximation

in physics we use approximations to become more easy
[/quote]

in physics u just can't approximate to make things easier it should be a logical reasoning that allows u to approximate .here u can do that iff
$r<<<n$

[quote="Thales418"]or if   e is quasi-zero

in you case if will be parallel

radiation pressure is not because the collision of fotons!!

it's because the ligth is a elletromagnet wave

fotons does not have mass. so does not have momentum!!!!

you can see it on volume 4 of halliday resnick walker

and my approximation it's because i'ts non-sense the length of the mirror be too big

we can make $e$ small and $n>>r$

and the formula i gave is rigth

the momentul is equal to the double of variation of energy of the ligth absorved divided by ligth speed

you can find on google this formula, or in the book i said

halliday resnick walker vomume 4

i dont know the title in english

http://www.physicsforums.com/showthread.php?t=87459[/quote]

well i feel u need to go through ur modern physics lessons again .i suppose u know there exist something called as de-broglie wavelength .and momentum of particles whose inertial mass is $0$ is given by

$P$ $=$ $\frac{h}{\lambda}$ $=$ $\frac{E}{c}$

where $\lambda$ is the de broglie wavelength of the particle .further you can't explain collisions on the basis of wave theory of light and i really doubt  a book like that of halliday tells so.

[quote]the momentul is equal to the double of variation of energy of the ligth absorved divided by ligth speed

[/quote]

this comes by assuming elastic collisions of photons . further if the mirror is not infinitely large then you have to consider components of the force exerted by photons

[quote="Thales418"]you are using classic physics in quantum physics

in your classic model for the collisions of the photons, the momentum is zero

the question is the ligth is tranfering energy to the mirrors and the mirrors is not absorving this energy, the reason for multiplying by 2, assuming everything is perfect

the distance of the bulb to the 2 mirrors is differente. so we have a torque

i use the wave theory to explain the energy

and the formula i gave is on the book

we are not able to say it is wrong[/quote]

who  said ur formula in "the book " is wrong i gave that clearly in my last post

 [quote]well i feel u need to go through ur modern physics lessons again .i suppose u know there exist something called as de-broglie wavelength .and momentum of particles whose inertial mass is  is given by $P$ $=$ $\frac{h}{\lambda}$ $=$ $\frac{E}{c}$

where  is the de broglie wavelength of the particle .further you can't explain collisions on the basis of wave theory of light and i really doubt a book like that of halliday tells so. [/quote]

further i suppose u have an idea what classical and what quantum m,echanics is .[i]quantum mechanics[/i] beleive on [i]"particle nature"[/i]
of light to explain the unexplained and this is where it is applieed

i suppose u have some idea about what QUANTUM MECHANICS is all about .
yes energy is transferred which i already posted in previous post saying the collision was elastic.
well if think wave theory explains it then prove it here (which i am sure u won't be able to because the entire thing is derived from quantum mechanics)
well the change in momentum of the photons  times the time taken is the force applied on the mirror which produces the torque

now please[i] don't tell [/i]me that[i] ur book tells energy produces torque [/i]and not force
   :furious:

weelll

let's start again

in m physics book dont say what you are saying. i really dont understand what are you saying.

the formula is not wrong!!!!!!!!!!!

with the ligth we need to use the particular and the wave theories together

the photon have energy because it is a eletromagnetic wave etc.....

the photon collides with the mirror crashing it. justa little bit and ping pong back etc...

so. all approximations i made is true. because we have everything Big and small etc. etc etc...

SO. i dont understand your point saying my formula is wrong

can you explain it?

[quote="Thales418"]
with the ligth we need to use the particular and the wave theories together

the photon have energy because it is a eletromagnetic wave etc.....

SO. i dont understand your point saying my formula is wrong

can you explain it?[/quote]

well u need to go through the book all once again light is supposed to be considered of quantas of energy having $0$ rest mass called $photon$
they have nothing to do with [i]electromagnetic wave[/i].

energy of a photon depend on the frequency of the constituting light given by the famous $planck's$ $law$

        $E$ $=$ $h\nu$
unlike an EM wave whos energy depends on it's amplitude .

well i never said that ur formula is wrong . i also somewhat derived it.

what i say is ur  understanding is wrong

[quote="Thales418"]u are understanding me wrong man

i dont know a derivation of this formula

but it is true. we just need it to solve. or not?[/quote]

yse u need the formulae to solve provided the mirrors are large etc.....

further what i said is ur reasoning is completely wrontg

[quote="Thales418"]i assume the mirrors are not large

so

no problem[/quote]

if the mirror sare not large assuming the bulb as a point source of light it emanates light equally in all directions which strike the mirror at varying angles hence this becomes a case of oblique collision .hence we will have to find the angle between the applied force  and the mirror or rather the incident angle.this leads to complicaton of problem.

but if the mirror is infinetly large two point on which photons collide and impart same force which are equidistant from a symmetrical line along the $y$  axis on the mirror and also equidistant from the rotational axis of the torsional scale will have their  components cancelled

pardesi

forget this problem

i cant formulate this better

so. just forget

later i will try to make the problem better etc...

but now let's close the problem

this is a hard problem. if we make assumptions we will make a lot of assumptions and approximations and the result is no torque

hehe

but thanks for your post


If I want to make a formal complaint about someone else in this forum posting stuff, what do I do?

PM or email me or MCrawford if ever you have a complaint about the posts on this forum.


Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \leq x\leq 8$. The sum of the largest and smallest values of $f(x)$ is

$\textbf {(A) } 1 \qquad \textbf {(B) } 2 \qquad \textbf {(C) } 4 \qquad \textbf {(D) } 6 \qquad \textbf {(E) }\text{none of these}$

[quote="4everwise"]Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \leq x\leq 8$. The sum of the largest and smallest values of $f(x)$ is

$\text {(A)} 1 \qquad \text {(B)} 2 \qquad \text {(C)} 4 \qquad \text {(D)} 6 \qquad \text {(E)}\text{none of these}$[/quote]

[hide]$[2,3]: f(x)=x-2+4-x-(6-2x)=2x-4$ min 0, max 2
$[3,4]: f(x)=x-2+4-x-(2x-6)=-2x+8$ min 0, max 2
$[4,8]: f(x)=x-2+x-4-(2x-6)=0$ max 0 min 0,
overall, 0 min, 2 max, $\boxed{B}$[/hide]


A one-digit number is increased by 20 when multiplied by 6. What is the one-digit number?

$ 6n \minus{} n \equal{} 5n \equal{} 20$. If $ 5 \times n \equal{} 20$, then $ n \equal{} \boxed{4}$.