conics
analytic geometry
parameterization
algebra
polynomial
hyperbola
geometry

We have a point $A$, a line $\ell$ and a circle $\left(P\right)$. Draw a line $p$ through the point $A$ such that $AB=CD$, where $B$ is the point of intersection of the line $p$ with the line $\ell$, and $C$ and $D$ are the points of intersection of the line $p$ with the circle $\left(P\right)$.

Dear Amid2
the reason why nobody solved your problem is that the solutions are generally not ruler and compass constructible.
Theycan be constructed using the common points of a circle and a conic, but this is another strory.
Kind regards. Jean-Pierre

Let the given line be the y-axis and let the given circle with radius $r$ be centered at a point $P = (p, 0)$ on the positive x-axis. Let the coordinates of the given point $A$ be $A = (x_A, y_A)$. The desired line cutting the circle $(P)$ intersects the y-axis at a point $B = (0, b)$. The equations of the circle $(P)$ and the line $AB$ are

$(P):\ \ (x - p)^2 + y^2 = r^2$

$AB:\ \ y = \frac{y_A - b}{x_A}\ x + b$

where the y-axis intercept $b$ is the unknown. The intersections of the line $AB$ with the circle $(P)$ are found by solving the above set of 2 equations:

$(x - p)^2 + \left(\frac{y_A - b}{x_A}\ x + b\right)^2 = r^2$

$x^2 - 2px + p^2 + \frac{(y_A - b)^2}{x_A^2}\ x^2 + \frac{2b(y_A - b)}{x_A}\ x + b^2 - r^2 = 0$

$\left(1 + \frac{(y_A - b)^2}{x_A^2}\right) x^2 - 2 \left(p - \frac{b(y_A - b)}{x_A}\right) x + p^2 + b^2 - r^2 = 0$

$(x_A^2 + y_A^2 + b^2 - 2by_A) x^2 - 2x_A(px_A - by_A + b^2) x + x_A^2(p^2 + b^2 - r^2) = 0$

$x_{C,D} = \frac{x_A(px_A - by_A + b^2) \pm x_A \sqrt{(px_A - by_A + b^2)^2 - (x_A^2 + y_A^2 + b^2 - 2by_A)(p^2 + b^2 - r^2)}}{x_A^2 + y_A^2 + b^2 - 2by_A}$

$y_{C,D} = \frac{y_A - b}{x_A}\ x_{C,D} + b$

We need only the differences $x_D - x_C,\ y_D - y_C$:

$x_D - x_C = \frac{2x_A \sqrt{(px_A - by_A + b^2)^2 - (x_A^2 + y_A^2 + b^2 - 2by_A)(p^2 + b^2 - r^2)}}{x_A^2 + y_A^2 + b^2 - 2by_A}$

$y_D - y_C = \frac{y_A - b}{x_A}\ (x_D - x_C)$

Hence, the distance of the points $C, D$ is

$CD = \sqrt{(x_D - x_C)^2 + (y_D - y_C)^2} = (x_D - x_C) \sqrt{1 + \frac{(y_A - b)^2}{x_A^2}} =  (x_D - x_C) \frac{\sqrt{x_A^2 + y_A^2 + b^2 - 2by_A}}{x_A}$

On the other hand, the distance of the points $A, B$ is

$AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{x_A^2 + y_A^2 + b^2 - 2by_A}$

From now on, let's denote the given coordinates of the point $A = (x_A, y_A) = (a, c)$, in order to avoid the indices. Since $AB = CD, AB^2 = CD^2$, we obtain the following equation for the y-axis intercept of the line $AB$ (the y-coordinate of the point $B$):

$a^2 + c^2 + b^2 - 2bc =$

$= \frac{4a^2[(pa - bc + b^2)^2 - (a^2 + c^2 + b^2 - 2bc)(p^2 + b^2 - r^2)]}{(a^2 + c^2 + b^2 - 2bc)^2} \cdot \frac{a^2 + c^2 + b^2 - 2bc}{a^2}$

$(a^2 + c^2 + b^2 - 2bc)^2 = 4(ap - bc + b^2)^2 - 4(a^2 + c^2 + b^2 - 2bc)(p^2 + b^2 - r^2)$

$a^4 + c^4 + b^4 + 4b^2c^2 + 2a^2c^2 + 2a^2b^2 - 4a^2bc + 2c^2b^2 - 4c^3b - 4cb^3 =$

$= 4(a^2p^2 + b^2c^2 + b^4 - 2apbc + 2apb^2 - 2cb^3)\ -$

$- 4(a^2p^2 + a^2b^2 - a^2r^2 + c^2p^2 + c^2b^2 - c^2r^2 + b^2p^2 + b^4 - b^2r^2 - 2bcp^2 - 2b^3c + 2bcr^2)$

$b^4 - 4cb^3 + \left(6(a^2 + c^2) + 4(p^2 - r^2 - 2ap)\right)b^2 - \left(4c(a^2 + c^2) + 8c(p^2 - r^2 - ap)\right)b\ +$

$+ (a^2 + c^2)^2 + 4\left(c^2p^2 - (a^2 + c^2)r^2\right) = 0$

We got a general quartic equation. Sorry, I do not know how to test a quartic equation for reducibility, but I doubt that the equation is reducible for general rational  parameters $a, c, p, r$. Putting the equation to a test, we choose $p = 5, r = 3.5, a = -4, c = 3$ (I tested it with several different sets of parameters):

$b^4 - 12b^3 + 361b^2 - 1086b + 300 = 0$

Solving numerically, we get 2 real roots $b \doteq 2.9433,\ b' \doteq 0.30733$, which lead to

$AB = CD = \sqrt{a^2 + c^2 + b^2 - 2bc} = \sqrt{b^2 - 6b + 25} \doteq \sqrt{16.0032} \doteq 4.0004$

$AB' = C'D' = \sqrt{a^2 + c^2 + b'^2 - 2b'c} = \sqrt{b'^2 - 6b' + 25} \doteq \sqrt{23.2505} \doteq 4.8219$

This can be compared with an approximate Sketchpad construction.

Dear Amid2 and Yetti
suppose A=(0,0) and the circle is centered at (d,0) with radius R; the locus of B such as the line AB intersects the circle at C,D such as AB =CD is a bicircular quartic.
The problem is to find the commo points of the line p with the quartic.
This can be done with a software allowing to draw conics (as Cabri or Geometer Sketchpad) :
If I didn't mistake, the inverse of the quartic in the circle (A,R) is the conic
4x^2/R^2+4(R^2-d^2)y^2/R^4 = 1
Draw this conic and the circle inverse of p, find their common points and inverse again these common points; you get the required points B.
Kind regards. Jean-Pierre

At first, thanks to JPE and Yetti for closing this problem, and for ZetaX for providing a link to a program which automatically computes Galois groups of integer polynomials:

http://www.math.tu-berlin.de/~kant

The problem occured before on this forum in the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=27030 .

  Darij

Dear Darij
may be, it is useful to add that if, in order to find a point, we need to solve an algebraic equation with rationnal coefficients then
1) our point is ruler and compass constructible if and only if the order of the Galois group of the equation is a power of 2
2) our point is conic constructible (common point of two constructible conics) if and only if the order of the Galois group of the equation is the product of a power of 2 and of a power of 3
Friendly. Jean-Pierre

[quote="jpe"]...Draw this conic and the circle inverse of p, find their common points and inverse again these common points; you get the required points B...[/quote]

In an inversion in the circle $(A)$, the line $m$ is carried into the circle $(M)$. The green locus of points $B$, such that $AB = CD$ and $C, D \in (P)$, is carried into the blue hyperbola. More than 5 points of the green locus have been inverted in the circle $(A)$ and then a conic constructed through 5 of the inverted points. The remaining inverted points all lied on this hyperbola. The circle $(M)$ intersects the blue hyperbola at a point $B'$, which is inverted back to the point $B$.

Thank you very much, jpe. Please, if it is not too much trouble, could you tell me how would you solve the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31034]Maximum area[/url] (disregard the quarrel) with a conic intersecting a circle ?

Regards, Yetti.

Dear Yetti
could you please precise the statement of the problem you want me to try to solve.
Kind regards JPE

The precise wording of the problem I refer to would be:

Given a circle $(O)$, a line $m$ intersecting this circle at points $B, P$ and another point $A$ on the line $m$ different from the points $B, P$. Find the lines $k, k'$ through the point $A$ intersecting the circle $(O)$ at points $C, D$ and $C', D'$, respectively, such that the directed area of the triangle $\triangle BCD$ is maximum and the directed area of the triangle $\triangle BC'D'$ is minimum.

The case when the point $A$ lies outside of the circle $(O)$ leads to a cubic equation

$qa(a - p) s^3 - a^2\left[2q^2 + p^2 - 2(a - p)^2\right]s^2 - 3qa^3(a - p) s - p^2a^4 = 0$

where the given line $m$ coincides with the x-axis, the point $B$ is the coordinate origin, the coordinates of the given points $A, O$ are $A = (a, 0),\ O = (p, q)$ (so that the radius of the circle $(O)$ passing through the coordinate origin $B$ is $r = \sqrt{p^2 + q^2}$) and $s$ is the y-axis intercept of the desired line $k \equiv CD$, i.e., the y-coordinate of the intersection $S = (0, s)$ of the desired line $k$ with the y-axis (with a normal to the given line $m$ erected at the point $B$). The cubic equation is generally irreducible and in the case of the point $A$ outside of the circle $(O)$, it has 3 real roots, one for the maximum directed area, one for the minimum directed area and one with no apparent sense.

Thank you for consideration of the problem.

Yetti


[quote="Fierytycoon"]What do you all think about the increasingly popular "Operations Research" major? Do a lot of financial firms hire Operations Research majors?[/quote]

 :ewpu: After a few of those classes you'll be  :sleeping:

More seriously, it depends on what you want to do.  If you want an interesting trading job with a hedge fund, excelling in math/physics/cs is much better than something like OR.  If you're looking for a stable, high-paying, but not as interesting job with, say, an investment bank, OR might be the way to go.

Mainly, I suggest studying something you enjoy.  Otherwise, it will be torture.  If you like OR, disregrard my teasing (my wife studied OR).  But if you don't like it, don't study it just to get into investment banking.

I cannot see any relation between Physics and the financial field. I studeid OR but had never found a good job for 6 years. the condition may be different for you guys in the U.S. :(

it's not that physics is applicable to the financial field, but in physics you learn to model natural phenomena.  that's really what physics is about.  this aspect of being able to model phenomena is very useful when it comes to trading.  (as far as i know)

Just curious, why is mathematics not enough?

I am reading a book called "How Wall Street Works" to understand more about Wall Street. I dont really understand what is the difference b/w a quant and a trader if any? Also, who do they create models for?

As far as I understood from my reading, dealers (market managers, traders are as far as I understand just other names for the same thing) do bids and ask for prices that they are willing to buy and sell for? Are they the ones who create models or who? I dont quite follow.  Thanks for ythe explanation.

quant stands for quantitative researcher.  quants are the ones who research the techniques that the traders use to trade.

as far as i know, you dont need more than a bachelor's degree to be either, it's just that getting for instance a PhD can't hurt.  in terms of the major for the bachelor's degree, i don't think there is any specified one you must have.  they do like math, CS, and physics though, but it's the interview the counts the most, and you can really get in from any other field as long as you're good.

[quote="hello"]I see. But Mr. Ruczyk was a trader, so he wasnt a quant. But both of them make approximaltely the same amount of money? Also, in that case, to be a trader only a bachelor degree is required (does it matter what subject?)? Is it only for QA that u need to have a phd in physics, not for a trader?[/quote]

This largely depends on where you are - we had people at Shaw who filled both roles, and we had plenty of 'quants' with only a bachelor's degree.  I served both roles, but didn't do nearly as much traditional quant work as others.

As for the physics link, it's partially the modeling item mentioned earlier, and partially selection bias.  Your top physics graduates have battled with and conquered ahrder problems than your top OR grads have.

[quote="hello"]But isnt a phd and bachelor's a really big difference? From what others have said it appears they mainly hire from elite colleges? So if you get a bachelor's in business field, there is also a chance of getting such a job? The pay is similar? Sorry for too many questions.[/quote]

The difference in pay between various slots highly depends on the role of the individuals.  Some strategies are more model driven, others more 'trader driven' (i.e. interpersonal contacts and soft information are very important).


inequalities
logarithms
real analysis
real analysis solved

1)Compute the limit:

lim(n-->oo) log_(n+2^m) (n+2^(m+1)); where m is natural and it is fixed! :D

2)If (an)n>=1 is a sequence of real pozitive numbers so that lim(n-->oo) (a_(n+1)-an)=r>0, compute the limit of lim(n-->oo) (Q_(n+1)-Q_n) where Q_n=sqrt((a1^2+..+an^2)/n).

3)Let (an)n>=0 be a sequence of real numbers with the property that an<=(a_(n-1)+a_(n+1))/2 for any n natural. Show that the sequence bn=(n+1)an-na_(n+1) is supperior bounded!

For 3):

We divide the given inequality by n to get: a_n/n<=(a_(n-1)+a_(n+1))/(n-1+n+1) or, in other 'words' ( :D ),
na_(n-1)+na_(n+1) >= (n-1)a_n+(n+1)a_n iff na_(n-1)-(n-1)a_n >=
(n+1)a_n-na_(n+1) iff b_(n-1) >= b_n, so (b_n)n>=1 is decreasing, and a decreasing sequence obviously has an upper bound (b_1, for example).

2):

First of all, from Cesaro, we get a_n/n->r. It's pretty easy to prove that if we find lim Q_n/a_n = l then (Q_(n+1)-Q_n)/(a_(n+1)-a_n)->l, so our limit will be r*l, so it's enough to find l=lim {n->00} Q_n/a_n. l should be equal to lim of [((a1^2+..+an^2)/n^3)*(n^2/an^2)]^1/2 = [(lim (a1^2+..+an^2)/n^3)*1/r^2]^1/2 = (by Cesaro for that ugly fraction) 1/3^(1/2). This is l, so the limit we were initially looking for is r/3^(1/2).

I'd like to add that I don't understand 1). Is there the '*' sign between (n+2^m) and (n+2^(m+1))? because if it is, the limit is obviously 00, so I think there's a mistake.

I know this problem from a gazzette and it's like so:

log_(n+2^m) (n+2^(m+1)) where (n+2^m) is the base of the logarithm and the rest is in the context! I hope it's clear enough grobber!

cheers! :D :D

Oh... It's clear now and it should have been clear from the beginning, but I missed the '_' :D Thanks!

Okay mitzah I guess i'll help you,  the problem is fairly easy,rather trivial, I don't know why there isn't a solution by now :?

 log_(n+2^m) (n+2^(m+1))=ln(n+2^(m+1))/ln(n+2^m)=(ln(n)+ln(1+2^(m+1)/n)) / (ln(n)+ln(1+2^m/n))--->1

cheers! :D :D


There is a 4 by 4 grid and you need to fill in each square with 1 of 2 colors such that the each color appears twice in each row and column. What is the least number of defined colors (already filled in) must you have so that you can fill in the rest of the squares in only 1 way. Prove your result.

I think it's $4$, with an arrangement as pictured. Can't prove it yet.

I agree, but I'm not sure how to prove it either.

I don't think that's it.

You could have

AB
BA

in the middle and you can fill in the rest in more than 1 way.

It's at random what squares you choose and what colors those squares are, so you would have to prove it for all configurations.

With 12, you might have a missing 2x2 square that can be filled in two ways:

AB
BA

or

BA
AB

Therefore, the answer is 13. At least two blanks are forced because they are the only blanks in their row or column. Then the final square is also forced.

what about

A B A B
B A    B
A      A
B B A A

[code]A B A B
B A   B
A     A
B B A A[/code]

If you want the spacing to work out, you need [code] tags.  Inside those tags, every character (including spaces) has the same width.

That configuration is simply invalid -- there are no completed squares with those 13 sub-squares colored in that way.  But that happens for much smaller values than 13.  For instance, with 3 squares you might end up with
[code]A A A[/code] which certainly ruins everything right away.

[quote="AdorableAddict"][code]X X X X
X A A X
X A A X
X X X X[/code]
X indicates unkown colors.
Why would this not work?[/quote]

because it doesn't have to be the middle ones, if you say four, any four of the squares could be colored any color.

There are two separate questions:

1) What arrangement(s), with the minimal number of squares, force the entire square to have a unique arrangement?
This is the question AdorableAddict has answered, but it's not the question SplashD wanted to ask.  The other question is

2) How many squares do you need in order to guarantee, regardless of the arrangement or coloring of those squares, that a unique completion exists?
This is the question SplashD wanted to ask, and Bictor has answered it.


i can't find couterexamples of positive integers greater than 2 which cannot be written as the arithmetic mean of two primes,is this property true?any proof?

[quote="K.W.M.A.N"]It would rather be a surprise if you did find them. I think computers checked this for very very large numbers and didn't find them...

http://www.google.be/search?hl=nl&q=goldbach+conjecture&meta=[/quote]
In other words, this is the famous Goldbach conjucture.  :)

On a related note, it has been proved that any odd number is the sum of 3 primes for $n \geq 3.33*10^{43000}$

It has also been proved that sufficiently large even number is the sum of a prime and the product of at most 2 primes.

And while im at it, in terms of primes, the closest anyone has got is that every even number is the sum of at most 300,000 primes! Slowly but surely, we'll get there!

I think it's also proven that every even number is sum of at most four primes ;)


x = a^x for some integer x and some real a.
find x explicitly in terms of a.
sorry if this is very easy and i cant c it ...

There is no way to solve it in terms of elementary functions. But there is a function called the Lambert W-function http://mathworld.wolfram.com/LambertW-Function.html that I believe it can be expressed in terms of.

note that the solution of x = a^x is the value of the power tower $a^{a^{a^{a^{\ldots}}}}$.

congratulation to silver at IChO, lukas(i hope all my information is correct...my sister told me lukas reck got gold at IChO and darij told me you are lukas reck...so if this is not correct, forget about that message)!

Peter


This is a nice little problem which arose from a research problem I was working on.

Begin with the sequence 1, 2, 3, ...
Now construct a new sequence of partial sums of the last sequence starting from the kth term.  In other words, this new sequence looks like this:

$ k, k\plus{}(k\plus{}1), k\plus{}(k\plus{}1)\plus{}(k\plus{}2), ...$

Now we create another sequence in the same manner from this one.  This new sequence then looks like this:

$ \left[k\plus{}(k\plus{}1)\plus{}(k\plus{}2)\plus{}...\plus{}(2k\minus{}1)\right],\left[k\plus{}(k\plus{}1)\plus{}(k\plus{}2)\plus{}...\plus{}(2k\minus{}1)\right]\plus{}\left[k\plus{}(k\plus{}1)\plus{}(k\plus{}2)\plus{}...\plus{}(2k\minus{}1)\plus{}2k\right], ...$

We continue to do this, each time constructing a new sequence by taking partial sums of the previous one, starting at the $ k$th term.  The question: find an explicit formula for the $ m$th term of the $ n$th sequence.  (First try, say, $ k\equal{}2$)

I liked the idea behind the solution I found, and the generalizations are very nice.

[quote="Virgil Nicula"]You can suppose that the first sequence is $ 1,1,1,\ldots$.[/quote]

That is also possible (this was what I had in the problem, and is what I used).  However, there is something about the other sequences which is not there in the 1, 1, 1, ... sequence: that's why I left it out.  But sure: you can keep the 1, 1, 1, ... sequence in as well.


Hey when do you guys think is the best time of day to take like AMC/AIME and stuff?...

For whatever, I work best between midnight and two in the morning, unless I'm dead tired.  It's really weird.

My early-morning person instincts say to start no later than 8AM.

I am not an early morning person at all, so I'd rather take it at any time between 10:00 AM and 4:00 PM.

[quote="tokenadult"]Whenever it's officially proctored so that the results count for something.[/quote]

 :)

But it is possible that the students will have some say in when the exam is proctored. When I was in high school, I got to choose when the exam was to be proctored.

According to studies by Harvard, apparently the best time is around 10:00 A.M.


algebra
polynomial
quadratics
Galois Theory
number theory
algebra unsolved

Write the roots of $x^2+x+2=0$ in the form of polynomial of 7th root of unity

Let $\zeta$ be a seventh root of unity.  We want to group the six powers of $\zeta$ into two groups so that we can combine them to give the two roots of a quadratic equation.  At this stage we won't bother about whether it''s the right one, just aim to get any quadratic.  The two groups we choose are $A = \{\zeta^1, \zeta^2, \zeta^4\}$ and $B = \{\zeta^3, \zeta^5, \zeta^6\}$.  Let $a$ be the sum of the elements in $A$ and $b$ the sum of the elements in $B$.  Then $a+b = -1$ and $ab$, after a little computation, is 2.  So $a, b$ are the roots of $X^2 + X + 2$.  This is the equation we want!

Galois theory told us how to find a quadratic field in the Galois extension $Q(\zeta)/Q$: a little algebraic number theory (elementary class field theory in fact) assured us that when we'd found the quadratic field it would be the right one.


Hello all, I am having trouble understanding this problem.

Is $ f$ defined on $ D$ continuous or discontinuous? Use the definition, not your intuition.
$ a) D\equal{}\left\{ 1,2,3,4 \right\}, f(1)\equal{}f(4)\equal{}0,f(2)\equal{}f(3)\equal{}1.$
$ b)$ Let $ F$ be defined and continuous on $ D_F\equal{}(0,1)$ and $ G$ be defined and continuous on $ D_G\equal{}(1,2)$. Define $ f(x)$ on $ D\equal{}D_F \cup D_G$ as follows
$ f(x)\equal{}F(x)$ if $ x\in (0,1)$ and $ f(x)\equal{}G(x)$ if $ x\in (1,2)$.

Thanks for any help =)

For the second one, see the [url=http://en.wikipedia.org/wiki/Pasting_lemma]pasting lemma[/url].

How would one go about with the first one?

And for the second I am not quite sure how it is possible still. Like, what about taking $ f(x)\equal{}1$ for $ x \in (0,1)$ and $ g(x) \equal{} 0$ for $ x \in (1,2)$.

Shouldn't there still be a discontinuity at $ x\equal{}1$?


A company consists of 2n+1 people..For every group of n members, there exists a member of the company who does not belong to that group,but who knows every member of the group.Acquantances are considered to be mutual..Prove that there is a member of the group who knows everybody.

Check out post number $3$ in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49197]this thread[/url] for a proof.

[quote="ali_ms2004"]again $2n+1$  :rotfl:[/quote]
What are you trying to say?


Hi,

Let $x,y,z$ be positive real numbers such that $xyz=1$. Prove that :
\[x+y+z+{1\over x}+{1\over y}+{1\over z}\leq 3+{x\over y}+{y\over z}+{z\over x}.\]

[quote="yuyang"]Hi,

Let $x,y,z$ be positive real numbers such that $xyz=1$. Prove that :
\[x+y+z+{1\over x}+{1\over y}+{1\over z}\leq 3+{x\over y}+{y\over z}+{z\over x}. \]
[/quote]

Just make the substitution $x=\frac{b}{a},y=\frac{c}{b}$ and $z=\frac{a}{c}$, and you will find that the inequality is equivalent to the well known Shure's inequality! :wink:


Does anyone else feel that the 2000 AIME is easier than others? I scored my normal score, but I had three very stupid mistakes(usually I only have one). I noticed that the statistics were missing on the site too.

I have no distinct memory of the 2000 AIME being especially easy.  I usually score within 1 of my average (except for last year's AIME I  :? ), which is why I think the AIMEs are fairly consistent.  I account for most fluctuations by:

--Test-taking environment and other factors unrelated to math
--Mood/energy (and sometimes test day adrenaline helps :))
--Slight variation in subject distribution on the test (e.g., more geometry than usual)
--I'm improving! :)

Hmmmmm....I'm improving! Hehehe...not by a long shot...


Once you have voted in the first poll, vote here.

I don't know what happen. If the first two votes are male you can just assume that the alien is male  :oops:

Who voted Alien?

Anyways, it's obvious the number of males on the forum is way way way larger than the number of females. It would be interesting to see by how much though.
I think females just don't consider math as a field for females: if they do not have a very strong desire or pressure from someone else they will not go there.

Hooray! I voted and I am a male! lol

I think lots of people on these forums are Male. People that are interested in math at my school are mostly boys.

[quote="mustafa"]The fact that this forum is predominately male kind of upholds the "males are better than females at math" theory.[/quote]

Are you retarded?

Ignite what do you mean about surviving in the real world? its not like i am a 12 year old lol

cmon, we must have more aliens on aops than that!

[hide="|V|"]=male[/hide]

I always wondered why there is no F/M indicator unter the avatars...

[quote="thatin"]I always wondered why there is no F/M indicator unter the avatars...[/quote]

There used to be, that was before the new server and before you joined

I think that girls are not pushed to do math as much as their peers.  Sophie Germain's father at first thought that she was insane and took away her math equipment, and I think a lot of that is still going on (maybe at a smaller scale, ex. discouraged from joining a club)

[quote="neelnal"][quote="k8reindeer"]yeah I miss that indicator... it helped me get my pronouns right.

i'm a girl, and i think that there is nothing about your gender that makes you naturally better or worse at math... but that doesn't explain the large majority of boys on this forum, in math competitions at higher levels, even on my school's math team. I think it has to do less with mental capacity and more with motivation: girls are less likely to be interested in math enough to do well.
there are organizations out there trying to change this, such as girlsgotech: [url]http://www.girlsgotech.org/[/url][/quote]

And yet, when they say that females are better at "essays" and language, no one makes up a whole website to counter it...

Feminists of these days think that if females are not as good at males at something, then they have to defend themselves. The truth is that each gender has its strengths and that is why there are two genders: to supplement one another.

Here is an analogy to show what I am trying to say:

A motorbike rides faster than a huge truck. Is the motorbikes then superior to the truck and does the truck have to prove that it is as fast as the motorbike? No. The truck is better than the motorbike in that its inhabitant is provided more space and is better protected if in an accident.[/quote]

I think you might've misunderstood me. I was referencing that website as an example of efforts to motivate girls to take an interest in math and science, not prove themselves equal to boys. It is okay for the truck to try to go faster.

I'd like to remind everyone that this IS the Round Table and spam will NOT be tolerated.

[quote="bookaholic"]Is it just me, or is it kind of silly to talk about an entire gender "going faster" or "being safer", to borrow from the motorbike/truck analogy? All trucks of one brand are made exactly the same. Humans? Not so much. How about we determine whether an individual is good at math by problem-solving ability, instead of gender? How about we worry about our own abilities instead of those of some group we belong to or don't belong to? Why does the particular proportion of skill x in group y matter at all? It's not going to affect the amount of skill x that any particular individual in group y has, unless their training in skill x is affected by their being in group y, which only happens when we pretend that proportions make a difference.[/quote]

Even though humans are all different, there's a very natural dichotomy that occurs: male and female.  Even on a genetic level, their DNA's have a very noticeable difference (i.e. women have an X-chromosome where men have a Y...and the Y is much smaller making it very easy to distinguish the two).  If you compare the chromosomes of two males (or two females), saying what is different between them is impossible, unless you study the sequence of of nitrogenous bases and have a genetic map of the human next to you...you'd probably want a computer as well.

When you look at the results of a math competition and you see 10 times as many boys as girls there's reason to wonder what's going on.  But the question should focus more on why there aren't as many girls interested in math as there are boys instead of why are boys better at math.  There are a lot less girls doing math and math competitions, but given the percentage that are involved, I'd have to say it seems that girls [i]can[/i] hold their own.

So why are girls less inclined to do math?  At an early stage, I believe the discrepancy is not clear.  In elementary school girls do a lot better job of keeping up with boys (they might even be ahead of them...I don't have any data, to be honest).  But then they start to lose interest around middle school/high school.  Perhaps their loss of interest is due to peer pressure or something of that form.  Or maybe it's more genetically based.

There's a bunch of literature on this.  Also, please back up any points you make in this discussion.  If you're going to make a substantial claim, you need to give the logic behind it.

These days it's much more favorable for women to go into math or science.  I think we'll see the gender difference start to level out...not overnight, but eventually.

My point isn't that consistent differences between the average members of each gender are impossible. My point is that any such differences only matter if we assume they exist and act accordingly. I don't see any need to argue over whether, all other things being exactly equal, females are less skilled at math on average. We only argue over it because some people assume that females are less skilled and consciously or unconsciously act to make sure it stays that way. The message we should be spreading isn't "there are absolutely no mental differences between the sexes" or "the sexes are very different and complementary and you better not take the role of the one you aren't". The message we should be spreading is "forget this, what you're good at has nothing to do with what your gender is good at, so just try everything, do your best, and anyone who tries to make you more like the 'average' boy/girl is silly".

I don't have anything wrong with what you're saying but talking about why women are underrepresented in mathematics is still something that should be done.  For example, if I fail a test, I'll naturally want to know why I failed it so I can go and fix those things and do better on the next test.

Talking about this issue shouldn't be taken as an attack on women but rather an inquiry into a natural situation...just as long as everyone keeps the discussion fair and dignified

Oh no, I agree with that. As I said,
[quote]I don't see any need to argue over whether, all other things being exactly equal, females are less skilled at math on average.[/quote]
Obviously all other things are not exactly equal at the moment, so this doesn't apply. I'm only arguing against saying things like "Even without social influences, girls would necessarily do worse/do equally well". The truth of such statements are completely irrelevant. But we definitely have to discuss what's going on in the current situation, which is different. I don't know enough to comment seriously on that, but I recently read something interesting that's probably related:
http://scienceblogs.com/ethicsandscience/2006/10/stereotype_threat_women_and_ma.php

well i think alien would be third gender which can be further classified to gir or boy so...why did they put alien?

How do you know the aliens aren't asexual? Or maybe they have more than two genders?

On topic, in what way are girls discouraged from going into math/science? Is it because of cultural gender roles? Or do their parents, teachers, or peers actually try to sway them? The way I got into math was that I started showing interest when I was young, so my parents encouraged me to do more advanced math, until I reached the point where I realized I liked it. Are you saying that parents are less likely to encourage mathematical development in girls? That doesn't really make sense, but then again, most people don't make sense.

they fall in love and decide they want to have a family or they just care about culture too much

[quote="funcia"]they fall in love and decide they want to have a family[/quote]
Why should this stop anyone from doing math? I want to get married and have a family and DO MATH ALL DAY EVERY DAY TOO!!! :D :D :D

More seriously, at my high school there was the typical disproportionate number of guys in AP physics and calculus and on the math team. But as far as I can tell, there isn't any discouragement on the part of the school (in fact, the gender ratio of teachers in the analytical sciences and math was at least even if not more women). So it must be social (family or peer pressures) that causes the phenomenon at my school.

I think the "math team is nerdy" deal is a pretty strong discouragement, unfortunately. It's just a little harder for us to see it because we would never look down on someone just for being on math team (and hopefully most of us are mature enough not to look down on people just for being on some other thing that we don't like, such as the [American] football team).

But really, that kind of perception doesn't change even after high school. I'm one of the officers of the Princeton Math Club, and the main problem we're always struggling with is that we don't get many members because people always assume that Math Club is only for hardcore nerdy math geeks. So we're always trying to come up with events that are legitimate for Math Club but don't reek of hardcore math or nerdiness... But still our membership is pretty small. (And a high proportion guys, but as someone mentioned earlier, this is partly because there is already a high proportion of guys in math at the university, so it isn't entirely our fault.)

[quote="Waxing Rhapsodic"]An individual girl could be as good as or better than an individual boy, for certain. But as a whole, it seems that boys are generally better at math/science and girls are generally better at writing/language/etc. For example, in my 9th grade honors English class, there are twenty girls and six boys. I wonder if there's a site to get men into writing.[/quote]

Yes, I noticed that boys tend to not fair as well in writing, but at the top levels, such as AP, in my school, there is a 5:4 girls to boys ration approximately.

But when it comes to learning a new language, such as spanish, there are 20 girls and 5 boys in the highest level class for my grade. I think girls are generally better in the writing.


Let $W(x)=(x-a_1)^2(x-a_2)^2\dots(x-a_n)^2+1$
$a_1,a_2\dots a_n\in\mathbb{Z}$
Prove that $W(x)$ is a prime polynomial. It can't be presented like: $W(x)=F(x)G(x)$, where $F$ and $G$ are polynomials with positive degree and with integer coefficients.

If you're looking for the solution[hide] it was solved in the Algebra section recently ;) [/hide]

[hide] :huh: what about 3^2 + 1 = 2*5 ?[/hide]

note that it is a prime polynomial, and not a polynomial in which all plugged in numbers will be prime (there are no such polynomials of the latter)

[quote="tµtµ"] :huh: what about 3^2 + 1 = 2*5 ?[/quote]
I said you can't factorize [u]polynomial[/u] $m^2 + 1$ in integers.
Obviously in complex you can but in integers not.

Your argument does not work. We also have $x^2+4$ irreducible over $\mathbb{Z}$, but $(x^2)^2+4=x^4+4=(x^2-2x+2)(x^2+2x+2)$ is not irreducible.

Peter

You are right.  :blush:
Any other idea ?


Find the sum of all $y$ such that $6\log_{y}3-\log_{3}y=1$.

I suck at logs so this solution is probably overly complicated
[hide]Let $y^{x}=3$ and $3^{z}=y$. Rewriting these into the bases, we have $6\log_{3^{z}}3-\log_{y^{x}}y=1\rightarrow \frac{6}{z}-\frac{1}{x}=1\rightarrow 6x-z=xz$ Rewriting these values into the upper thingys, we have $6\log_{y}y^{x}+\log_{3}3^{z}=1\rightarrow 6z-1=1$
We see that $xy=1\rightarrow x=\frac{1}{z}$ We subsitute this into the equation....
$6\log_{3^{z}}3-\log_{y^{\frac{1}{z}}}y=1 \rightarrow \frac{6}{z}-z=1\rightarrow z^{2}+z-6=0\rightarrow z=\{2,-3\}$
Since $3^{z}=y$, we have $y=\{9, \frac{1}{27}\}$. Their sum is $\frac{244}{27}$.

Which I think is wrong but oh well[/hide]

Heh I don't know...that was one of the answer choices and your solution seems to be correct...hey but what do i know

[hide="Simple one"]Just put $x=\log_{3}y$ to get a simple quadratic ${6\over x}-x=1$, and after solving for $x$, use $y_{1}=3^{x_{1}}, y_{2}=3^{x_{2}}$. Answer-wise, ragnarok23 is correct.[/hide]

[hide="Solution"]Let $x=\log_{y}3$. Using a log property, we have $\log_{3}y=\frac{1}{\log_{y}3}$. Substituting we have
$6x-\frac{1}{x}=1$ or $6x^{2}-x-1$. Solving this quadratic, we get $x=-\frac{1}{3}$ or $x=\frac{1}{2}$. Setting these equal to $\log_{y}3$, we get $y\in\{9, \frac{1}{27}\}$ so the sum is $\frac{244}{27}$.  :D[/hide]

[hide]Substitute $\log_{3}y=x$, so that $6/x-x=1$. Solving the quadratic, we get $x=2,-3$ so $y=3^{2}=9$ and $y=3^{-3}=1/27$. Their sum is $244/27$.[/hide]

This sounds alot like a problem from AIME II 2005


The rectangle with vertices $ (\minus{}1, y), (7, y), (\minus{}1, 3)$, and $ (7, 3)$ has an
area of 72 square units, and $ y$ is positive. What is the value of $ y$?

the area of the rectangle is $ (y\minus{}3)*(7\plus{}1)\equal{}72 \implies y\minus{}3\equal{}9 \implies y\equal{}12$

answer : 12


If a=bq + r
then prove that
gcd of a and b  = gcd of b and r.

Let, $G=\gcd (a,b), G'=\gcd (b, r)$

$\boxed{\text{Proof of}\ G|G'}$
Let, $a=a'G, b=b'G$
$r=a-bq=(a'-b'q)G \ \therefore G|b, G|r$
$\therefore G|G'$

$\boxed{\text{Proof of}\ G'|G}$
Let, $b=b''G', r=r'G'$
$a=bq+r=(b''q+r')G' \ \therefore G'|a, G'|b$
$\therefore G'|G$

Thus, $G|G' \ \text{and}\ G'|G \Rightarrow G=G' \iff \gcd (a, b) = \gcd (b, r)$


I was wondering if you guys wanted to start a thread where one person would give a laboratory procedure and we would have to try and figure out how to accomplish it. Give a list of materials as well please.

I guess ill begin with something fairly easy:

You are given a test tube that contains an unknown gas that is either Oxygen, Hydrogen or Nitrogen. You are also given a bunsen burner, an eudiometer, a splint, a syringe, a balance and a graduated cylinder. How could you determine the unknown gas?

Well you can light the splint, if you hear a pop, when you bring it near the gas, the gas is hydrogen.  I'm not sure how you would distinguish between nitrogen and oxygen though... anyone wanna finish this one for me?

Oxygen gas should relight a glowing (blown out) splint.

True, I wonder what the eudiometer and the other equipment is for then? Hm, well if anyone has anything else to add to that lab, they can go right ahead and I'll post another lab too.

You have seven solutions, silver nitrate, barium chloride, copper (II) nitrate, lead (II) nitrate, potassium iodide and sodium thiosulfate.  The problem is to figure out which one is which.

lol, all the other equipment were just to throw people off.

[quote]You have seven solutions, silver nitrate, barium chloride, copper (II) nitrate, lead (II) nitrate, potassium iodide and sodium thiosulfate. The problem is to figure out which one is which.[/quote]

Oh god, this lab looks way too familiar.... usnco 2008 Lab II?

Lets see, heres what i remember:

Basically, make a 7x7 grid. Then mix all the chemicals together, note that you can X out half of the boxes since adding chemical I to chemical II yields the same product as adding chemical II to chemical I.

Sodium thiosulfate forms the only black precipitate i believe
Copper II is usually blue
KI will form yellow precipitate with the Pb(NO3)2
AgNO3 and BaCl2 will form white precipitate

Now, once you figure those out, the trick is to figure out which one is which for each reaction:

KI + AgNO3 yield a precipitate, PbNO3 and AgNO3 do not

So therefore, the one that AgNO3 forms a precipitate with is KI and the other is PbNO3
the remaining solution is BaCl2.

And they are all solved for...

This is probably a really crappy explanation.. feel free to ask any questions.

Heres a handy lab procedure:

How do you determine the amount of ionizable protons present in an acid. You can use any lab materials present in a normal lab and standard reagents.

Get a strong base like sodium hydroxide.  Titrate it against the acid, recording the pH with a pH meter every mL or so.  When you see drastic changes in pH starting to occur, make the interval you record the pH of the acid closer together.  So like record them every .1 mL or something.  Then when your done, graph all your data in graph of pH vs. volume of NaOH used.  Every inflection point will be an equivalence point.  The number of equivalence points is the number of ionizable protons the acid has.  One can modify this procedure with a pH indicator such as phenolphtalein if you want to get really accurate results and determine things such as kA and what not.

New problem:  Find the enthalpy chage associated with some reaction using standard lab materials.

Perform the reaction in a test tube submerged in a known mass of water inside a coffee-cup calorimeter.  Measure the change in temperature of the water, and use its mass and specific heat to find the heat it gained/lost, which is the same as the heat the reaction lost/gained.  We're at constant pressure, so this is the change in enthalpy.

Hmm . . . you have a 1 M solution of sodium carbonate and some hydrochloric acid of unknown molarity.  Determine the molarity of the acid.

This is a pretty similar procedure to the one I outlined before.  You have 1 M Sodium carbonate solution in a beaker.  Measure out a certain volume of that solution.  When you react it with the acid, the reaction is
H+ + CO3- --> HCO3-.  Use methyl red as an indicator, which changes pH around 5.5 I think.  Titrate the weak base with the strong acid and measure how much volume of acid it takes to get to a color change.  You know the moles of carbonate in the solution, which gives you moles of acid used.  Using the moles and volume, you can find the molarity.

Okay, this isn't really a lab, but it could definitely come up on the multiple choice testing lab techniques.  Outline the differences and key points in infrared spectroscopy, visible light spectroscopy and ultraviolet spectroscopy.

[quote="feynman_wannabe"][quote]You have seven solutions, silver nitrate, barium chloride, copper (II) nitrate, lead (II) nitrate, potassium iodide and sodium thiosulfate. The problem is to figure out which one is which.[/quote]

Oh god, this lab looks way too familiar.... usnco 2008 Lab II?

Lets see, heres what i remember:

Basically, make a 7x7 grid. Then mix all the chemicals together, note that you can X out half of the boxes since adding chemical I to chemical II yields the same product as adding chemical II to chemical I.

Sodium thiosulfate forms the only black precipitate i believe
Copper II is usually blue
KI will form yellow precipitate with the Pb(NO3)2
AgNO3 and BaCl2 will form white precipitate

Now, once you figure those out, the trick is to figure out which one is which for each reaction:

KI + AgNO3 yield a precipitate, PbNO3 and AgNO3 do not

So therefore, the one that AgNO3 forms a precipitate with is KI and the other is PbNO3
the remaining solution is BaCl2.

And they are all solved for...[/quote]

I tried doing this last year and it did not work. At all. Part of it was that it was hard to tell what was a precipitate and what was colored ions in solution, and part of it was that I did not know my solubility rules. At all. Also, I noticed that after about ten minutes standing in air, a bunch of the colors started to change :-/

I'd offer an answer, but I still do not know my solubility rules. At all. To simplify a bit, though, you can certainly start with the two solutions with Cu(II) cations (there were two of them, I remember, you're missing one in the list that you posted), which you can tell because they are blue. Identify them and then use them to identify the others, rather than bashing it out with a 7 by 7 grid, is the idea...

Personally I believe making a $ 7 \times 7$ grid is the best way to go. Identifying the copper solution based on colour is the first step. I've done a similar experiment before and with a more chemicals and it was the best way to go. If the precipitate changes colour upon after a while then it is important to note that. For instance $ \ce{AgNO3}$ usually turns to a brownish colour if you leave it for long enough (or accidentally get some on yourself :P ). In fact a lot of silver precipitates will change colour because they are photo-sensitive.

Another technique I found useful though is that once you have identified what is what then test it out a bit! Mix two chemicals and see if you get the predicted outcome (only do this if you have time though). Furthermore if you know complex ions can form e.g. $ \ce{PbNO3}$ and $ \ce{KI}$ can potentially give $ \ce{PbI4 \minus{} }$ (I think it's this one) then you should be careful about the quantities you add.


I have searched over this site, but no useful thing was found.

[quote="julien_santini"]If ${P(x)=1+x+\frac{x^{2}}{2!}+\dots+\frac{x^{n}}{n!}}$ is reducible over $\mathbb{Q}$ then $n!P(x)$ is reducible as well (over $\mathbb{Q}$), meaning $n!P(x)$ is reducible over $\mathbb{Z}$ by Gauss's lemma. Now you can show this is impossible using Eisenstein's criterion and Bertrand's postulate.[/quote]

Thank you very much for your reply,but would you please say it more clearly?
I have read it from page 51 from the book <<Polynomials>> written by Prasolov(English edition), but I think that Eisenstein's criterion uses the condition that $p|a_i,i=0,\dots,n-1,$, and this can't be derived from Bertrand's postulate.
Thanks.

You're right, I didn't pay enough attention. My proof works only in the case where $n$ is prime, and Bertrand's postulate is irrelevant. I'll try to give it a second thought.

I didn't try any more, but I'm still curious about a solution for this problem. Anyone ? (must be classic I guess ?)

http://www.mathlinks.ro/Forum/viewtopic.php?t=86095
http://www.mathlinks.ro/Forum/viewtopic.php?t=64794
( http://www.mathlinks.ro/Forum/viewtopic.php?t=36701 )

Searching for irreducible in the Algebra section works pretty well...

Yeah, but as the initial poster mentionned, I'm afraid it is not of much help ...

1. where will this end if we have five threads for every problem¿
2. one of them contains what you wrote above, with explanaitions. Why is that not at least a bit helpful and why do you dislike this proof¿

[quote="ZetaX"]2. one of them contains what you wrote above, with explanaitions. Why is that not at least a bit helpful and why do you dislike this proof¿[/quote]

Well, I found this proof independently ( I say that because I also read it in another related post yesterday ). The problem is I don't think it is correct, as wuming (and another person from the other post) said.


$y=f(x)$ give $f(2y)=f(0)+(2y)^{2}.$
$y=-f(x)$ give $f(-2y)=f(0)+(2y)^{2}.$
It is easy to chek, that $f(x)=x^{2}+c$ solution.
For proof, that equation had not other solution suufficiently to show, for any y exist x, suth that $|y|=|f(x)|$.

i don`t think that the surjectivity of $f$ is very simple. in fact, i think it`s harder than the "offical" solution.

Hello,

IM(f) is a serious problem.

For example, f(x) = 0 is a solution (different from x^2 + c)

--
Patrick

The solution that I like is as follows.

Let $g(x)=f(x)-x^2$, then the functional relation becomes $g(g(x)+x^2+y)=g(g(x)+x^2-y)$ for all real $x,y$. Then $g(g(x)+x^2-g(y)-y^2+z)=g(g(x)+x^2+g(y)+y^2-z)=g(g(y)+y^2-g(x)-x^2+z)$, thus
$g(z)=g(2*(g(x)+x^2-g(y)-y^2) +z)$ for all real $x,y,z$. If $g(x)+x^2$ is constant, then we get the solution $f(x)=0$ for all real $x$. If $g(x)+x^2$ is not constant, then $g$ is periodic, say with period $T$. Then from the last equation we put $x=y+T$ to get $g(z)=g(2*(2*y*T+T^2)+z)$ for all real $y$ and $z$. But we can choose $y$ at our will to get $g(z)=g(0)$ for all real $z$ and hence $f(x)=x^2+g(0)$ for all real $x$.

If g(x) + x^2 is not constant than there exists a different from b such that a=g(x)+x^2 and b=g(z)+z^2 for some x and z.

 g(a+y)=g(a-y) (*) and g(b+y)=g(b-y) for all y.
Substitute y by a-y-b in the second relation and you obtain g(b+(a-y-b))=g(b-(a-y-b)) <=> g(a-y)=g(y+2*b-a), but from (*) g(a-y)=g(y+a) => g(y+a)=g(y+2*b-a).
 If we take z = y+a we wil get  g(z) = g(z+2*(b-a)) and because z undergoes R => g is periodic of period 2*(b-a).  :)

Hello Maky!

[quote="maky"]why is $g$ periodic ?[/quote]

Because he has $g(z)=g(2(g(x)+x^{2}-g(y)-y^{2})+z)$ for all real x,y,z
So, if $g(x)+x^{2}$ is not a constant, it exists x and y such that $g(x)+x^{2}\neq g(y)+y^{2}$
Then, if I call $T=2(g(x)+x^{2}-g(y)-y^{2})$, $T\neq 0$ and for any real z, $g(z)=g(2(g(x)+x^{2}-g(y)-y^{2})+z)$ ==> $g(z)=g(z+T)$

And g is periodic.

--
Patrick

[quote="Anto"]The solution that i like is as follows :

Let g(x)=f(x)-x^2 , then the functional relation becomes g(g(x)+x^2+y)=g(g(x)+x^2-y) for all real x,y. Then g(g(x)+x^2-g(y)-y^2+z)=g(g(x)+x^2+g(y)+y^2-z)=g(g(y)+y^2-g(x)-x^2+z) thus
g(z)=g(2*(g(x)+x^2-g(y)-y^2) +z) for all real x,y,z. If g(x)+x^2 is constant then we get the solution f(x)=0 for all real x. If g(x)+x^2 is not constant then g is periodic say with period T. Then from the last equation we put x=y+T to get g(z)=g(2*(2*y*T+T^2)+z) for all real y and z. But we can choose y at our will to get g(z)=g(0) for all real z and hence f(x)=x^2+g(0) for all real x.[/quote]

can someone turn this into $LaTeX$ and explain a bit?

[quote="maky"]all the solutions i know (including mine, in the contest) use the idea that $\mbox{Im}\, f-\mbox{Im}\, f =\mathbb{R}$. after that, it`s pure algebraic manipulations.[/quote]

can you show that?

[quote="Anto"]The solution that i like is as follows :

Let $g(x)=f(x)-x^{2}$ , then the functional relation becomes \[g\left(g(x)+x^{2}+y\right)=g\left(g(x)+x^{2}-y\right)\qquad \forall \text{ real }x,y\]Then \[g\left(g(x)+x^{2}-g(y)-y^{2}+z\right)=g\left(g(x)+x^{2}+g(y)+y^{2}-z\right)=g\left(g(y)+y^{2}-g(x)-x^{2}+z\right)\] thus
\[g(z)=g\left(2(g(x)+x^{2}-g(y)-y^{2}\right)+z)\qquad \forall\text{ real }x,y,z\]If $g(x)+x^{2}$ is constant then we get the solution $f(x)=0$ for all real $x$. If $g(x)+x^{2}$ is not constant then $g$ is periodic say with period $T$. Then from the last equation we put $x=y+T$ to get \[g(z)=g\left(2\cdot(2yT+T^{2})+z\right)\qquad \forall \text{ real }y,z\]But we can choose $y$ at our will to get \[g(z)=g(0)\qquad\forall \text{ real }z\] and hence \[\boxed{f(x)=x^{2}+g(0) \qquad\forall \text{ real }x}\][/quote]

IR is pure imaginaries.

In problem IR is $\mathbb{R}$ = set of all real numbers.

i will post here my solution. i gave it in the contest as well

the solutions are $f\equiv 0$ and $f(x)=x^{2}+a$, with $a$ real.
obviously $f\equiv 0$ satisfies the equation, so i will choose an $x_{0}$ now such that $f(x_{0})\neq 0$. i first claim that any real number can be written as $f(u)-f(v)$, with $u,v$ reals. denote $f(x_{0})$ with $t\neq 0$. then, putting in the equation, it follows that
$f(t+y)-f(t-y)=4ty$.
since $t\neq 0$ and $y$ is an arbitrary real number, it follows that any real $d$ can be written as $f(u)-f(v)$, with $u,v$ reals (take $y=d/4t$ above).
this proves my claim above.
now, let`s see that
$f\big(f(x)+f(y)+z\big)=f\big(f(x)-f(y)-z\big)+4f(x)\big(f(y)+z\big)$
and
$f\big(f(y)+f(x)+z\big)=f\big(f(y)-f(x)-z\big)+4f(y)\big(f(x)+z\big)$
(the relations are deduced from the hypothesis)
for all reals $x,y,z$. i will denote by $d$ the difference $f(x)-f(y)$.
substracting the above relations will give me
$f(d-z)-f(-d-z)+4zd=0$.
now, let`s see that this is true for all reals $d,z$, because $z$ was chosen arbitrary and $d=f(x)-f(y)$ can be choosen arbitrary because of the first claim.
now it`s easy. just take $d=z=-x/2$ and it gets that
$f(0)-f(x)+x^{2}=0$, or
$f(x)=x^{2}+f(0)$.
this obviously satisfies the relation, and this ends the proof.

First $ f(x)=0$ is solution. Now suppose $f(x) \not=0$ for some $x$. We can see that any $y\in R$ can be represented by $f(a)-f(b)$ and also $2(f(a)-f(b))$ for some reals $a, b$ (1). Now placing $f(x)=y$ gives $f(2f(x))=f(0)+4f(x)^2$. Put $y=f(x)-2f(y)$ and we get

$f(2(f(x)-f(y)))=f(2f(y))+4f(x)^2-8f(x)f(y)$
$=f(0)+4f(y)^2+4f(x)^2-8f(x)f(y)$
$=f(0)+4(f(x)-f(y))^2$.

Because of (1) we see that all solutions other than $ f(x)=0$ are $f(x)=x^2+f(0)$

[quote="ShahinBJK"]From which source did you find this question. It is International Zhautykov olympiad 2011.But you wrote it at 2007 How?[/quote]
It is Balkan MO 2007 second problem.

See here: http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=19&year=2007&sid=e95c823acb9d986b9856ea76c176582b

[quote="mavropnevma"]Not quite. Swapping $x$ and $y$, the Zhautykov writes as $f(f(x) + y) = f(y - f(x)) + 4f(x)y$ (the difference is, at the Balkan MO, the start of RHS was $f(f(x) - y)$). But, of course, the method is identical, so the problem is a spoof. See also my comment posted at the Zhautykov link [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2148157&sid=3a2481060d2b70418b0728a796017e01#p2148157[/url].[/quote]
yes i didn`t see the difference but $f:even$ can be easily found.

Does this work?
$f \equiv 0$ is trivial. Assume there exists $x_0$ such that $f(x_0) \neq 0$.
The equation is equivalent to $f(x+y)=f(x-y)+4xy$ for all $x \in \mbox{Im}\,f$, $y \in \mathbb{R}$.              (1)
Since $y$ can range over all real numbers, sub $x=x_0$ gives $\mbox{Im}\,f - \mbox{Im}\,f = \mathbb{R}$.
Let $f(x)-x^2=g(x)$. Then, (1) is equivalent to $g(x+y)=g(x-y)$ or $g((x-z)+y+z)=g((x-z)-y+z)$ for all $x,z \in \mbox{Im}\,f$, $y \in \mathbb{R}$.
Since $\mbox{Im}\,f - \mbox{Im}\,f = \mathbb{R}$, we get $g(w+y+z)=g(w-y+z)$ for all $z \in \mbox{Im}\,f$, $y,w \in \mathbb{R}$. Let $y= \frac{u-v}{2}$ and $w=\frac{u+v-2z}{2}$ for arbitrary real numbers $u, v$. Then, $g(u)=g(v)$ for all $u,v \in \mathbb{R}$ i.e. $g$ is a constant i.e. $f(x)=x^2+c$ for some constant $c$, as required.

Hi MathPanda1,

I am not sure why if you fix a real z in Im(f), then x-z can represent all reals? If not, then w is only the set Im(f)-z, which is not R. (since Im(f) is not necessarily R)

Im(f) here i mean image of f, not the imaginary part of f. I am sorry for the confusion

[quote=Anto]
Let $g(x)=f(x)-x^2$, then the functional relation becomes $g(g(x)+x^2+y)=g(g(x)+x^2-y)$ for all real $x,y$. Then $g(g(x)+x^2-g(y)-y^2+z)=g(g(x)+x^2+g(y)+y^2-z)=g(g(y)+y^2-g(x)-x^2+z)$, thus
$g(z)=g(2*(g(x)+x^2-g(y)-y^2) +z)$ for all real $x,y,z$.[/quote]

Please help me, I cannot understand. How $z$ came here. There were only $x$ and $y$

[quote=stergiu]Find all real functions $f$ defined on $ \mathbb R$, such that  \[f(f(x)+y) = f(f(x)-y)+4f(x)y ,\] for all real numbers $x,y$.[/quote]

Note that $f \equiv 0$ is a solution. Let $P(x,y)$ denote the given assertion. Suppose there exists a $t  \in \mathbb{R}$, such that $f(t) \neq 0$, then note that $g(x,y) = 2 \cdot f(x) - 2 \cdot f(y)$ is surjective as $P(t, \frac{x}{8 \cdot f(t)})$ yields $x = 2 \cdot f(f(t) + \frac{x}{8 \cdot f(t)}) + 2 \cdot f(f(t) - \frac{x}{8 \cdot f(t)})$.

Now note that $P(x, f(x))$ yields that $f(2 \cdot f(x)) = f(0) + 4 \cdot f(x)^2$.         Also, $P(x, 2 \cdot f(y) - f(x))$ yields  $f(2 \cdot f(x) - 2 \cdot f(y))$ = $f(2 \cdot f(y) - f(x)) - 4 \cdot f(x) \cdot (2f(y) - f(x)) $ =                                  $f(0) + (2 \cdot f(y) - 2 \cdot f(x))^2$. Since $g(x,y)$ is surjective, $f(x) = 2x^2 + c$. It is trivial to verify that all such functions satisfy the given condition.

[hide=Storage]We claim that $f(x)=0\forall x \in \mathbb{R}$ and $f(x) = x^2 \forall x \in \mathbb{R}$ are the only solutions. It's easy to check that they satisfy the condition. Let us now show these are the only ones.

Let $P(x,y)$ denote the assertion. Note that $P(x,0)$ together with $P(x,x^2-f(x))$ implies that $f(x)\cdot (x^2-f(x))=0$. Now either $f(x)=0 \forall x \in \mathbb{R}, f(x) = x^2 \forall x \in \mathbb{R}$, which are indeed our solutions. Otherwise, assume that $\exists a,b \in \mathbb{R}$, $a,b \neq 0$, such that $f(a) = 0, f(b)=b^2$. Note that from $P(a,b)$, we have $f(a^2-b)=b^2$. Since for any $x \in \mathbb{R}$, either $f(x) =0$ or $f(x)=x^2$, it follows that either $b^2=0$, contradiction, or $(a^2-b)^2=b^2$. The latter implies that $a^4=2a^2b$. Thus, either $a=0$, contradiction, or $a^2=2b$. Also, $P(b,-b^2)$ yields that $f(2b^2) = 4b^4$. Using this in $P(a,2b^2)$, we have $f(a^2-2b^2) = 4b^4$, so either $4b^4=0$, contradiction, or $(a^2-2b^2)^2 = 4b^4$, which implies that $a^4= 4a^2b^2$, so either $a=0$, contradiction, or $a^2 = 4b^2$. But, we already have $a^2 = 2b$, so $2b=4b^2$, it follows that $b=1/2$. To finish, note that $P(\tfrac{1}{2}, \tfrac{3}{4})$ (or $P(b,a^2-b^2)$) implies $f(-\tfrac{1}{2}) = -\tfrac{3}{4}$ which is clearly not possible. This finishes the problem. $\blacksquare$

@above $P(x,0)$ does not give any new information. Can you explain a bit more how do you obtain $f(x)(x^2 - f(x)) = 0$?

Case 1: $f(x)=0.$
Then it's obvious that its a satisfying solution.

Case 2: $f(x)\neq 0.$
$P(x,f(x))\implies f(2f(x))=f(0)+4f(x)^2.$
$P(x,2f(fy)-f(x))\implies f(2f(x)-2f(y))=f(2f(y))-4f(x)(2f(y)-f(x))$
$=f(0)+4f(y)^2-4f(x)(2f(y)-f(x))=f(0)+(2f(x)-2f(y))^2.$
$\exists r,z\in \mathbb{R} : f(z)\neq 0$
$\text{and } 2f(f(z)+x/(8f(z)))-2f(f(z)-x/(8f(z)))=r \implies f(r)=r^2+f(0),$
easy to check that it works.

This problem is from Mongolia TST 2020  :P, hilarious isn't it?

Answer: $f(x) = 0$ or $f(x) = x^2 + c$.

If for all $x$, $f(x) = 0$ then it's a solution. Thus we can assume there exists $b$ such that $f(b) \neq 0$. Let $P(x, y)$ be the given assertion. Now consider the following claim:

[b][color=red]Claim:[/color][/b] For every $a \in \mathbb{R}$, there exist $u, v \in \mathbb{R}$ such that $f(u) - f(v) = a$.

[i]Proof.[/i] $P(b, \frac{a}{4f(b)})$ gives $a = f(f(b) + \frac{a}{4f(b)}) - f(f(b) - \frac{a}{4f(b)})$, so we're done. $\blacksquare$

$P(x, f(x))$ yields $f(2f(x)) = f(0) + (2f(x))^2$. Now we'll prove that $f(a) = a^2 + c$ for all $a \in \mathbb{R}$. By claim, there exist $u, v$ such that $\frac{a}{2} = f(u) - f(v)$. Then $P(u, 2f(v) - f(u))$ yields $f(0) + (2f(v))^2 = f(2f(v)) = f(f(u) + 2f(v) - f(u)) = f(2f(v) - 2f(u)) + 4f(u)(2f(v) - f(u))$, or equivalently $f(2f(u) - 2f(v)) = f(0) + (2f(u) - 2f(v))^2$. Hence $f(a) = a^2 + f(0)$ for all $a \in \mathbb{R}$, so we're done. $\blacksquare$


A mass m is hung at the end of a string of length L.  While the other end of the string is fixed, the mass swings in a vertical circular path.  What initial tangential velocity is needed so that the tension in the string will become 0 at the top of the circle?  At 45 degrees above the horizontal?  Describe the trajectory of the mass.

For the first question, the initial velocity needed to give zero string tension at the top: The velocity needed at the top, call it $v_{T},$ is such that the acceleration of gravity, $g,$ exactly matches the acceleration needed to maintain the circular path.

Hence, $\frac{v_{T}^{2}}{L}=g$ or $v_{T}^{2}=gL.$

The kinetic energy at the bottom is $\frac12mv_{0}^{2}.$ At the top, the kinetic energy has decreased to $\frac12mv_{0}^{2}-2mgL$ because of the increase of gravitational potential energy. This gives a velocity at the top of $v_{T}^{2}=v_{0}^{2}-4gL.$ Equating our two expressions for $v_{T}^{2}$ under the assumption of zero string tension at the top, we get that $v_{0}^{2}=5gL$ or $v_{0}=\sqrt{5gL}.$

If the string tension goes to zero 45 degrees above horizontal, then the radial component of the net force must equal the force necessary to maintain that curvature to the path. If $v_{p}$ is the velocity at that point and the only force acting is gravity, then $\frac{v_{p}^{2}}2=\frac{\sqrt{2}}2g$ The $\frac{\sqrt{2}}2$ is there becuase the net force is exactly downward but we are measuring the radial component, which is $45^{\circ}$ away from downward. The same kinetic-potential energy argument as before gives us the equation

$v_{0}^{2}-2\left(1+\frac{\sqrt{2}}2\right)gL=\frac{\sqrt{2}}2gL$

or $v_{0}=\sqrt{\left(2+\frac{3\sqrt{2}}2\right)gL}.$

From that point on, the mass will fly along a parabolic free-flight path through the interior of the circle, with the string not extended to full length. There is enough information here to determine how high the highest point on that parabolic path will  be - it's higher than the point of release but lower than the top of the circle. Just track kinetic and potential energy, remembering that it will have a nonzero constant horizontal velocity component.


algebra
polynomial
function
algebra unsolved

Find all polynomials P(x) with real coefficients such that for all reals a, b c such that ab+bc+ca =0 we have the following relations :
                    $ [p(a)]^{2}+[p(b)]^{2}+[p(c)]^{2}=[p(a+b+c)]^{2}$

lol, do you want all the FUNCTIONS f or POLYNOMIALS p, btw, isn't there an IMO (2004?) problem that was exactly or extremely similar to this? :?

Really easy problem,much easier than 2004 IMO 2.
To solve the problem,let $a=b=x,c=-\frac{1}{2}x$,thus we have $2[p(x)]^2+[p(-\frac{1}{2}x)]^2=[p(\frac{3}{2}x)]^2$.If $p$ is constant,clearly $p\equiv 0$ since $p(0)=0$.Now suppose $\deg(p)=n>0$,and the coefficient of $x^n$ in $p$ is $a_n \neq 0$.it is easy to see that the coefficient of $x^{2n}$in $2[p(x)]^2+[p(-\frac{1}{2}x)]^2$ is $a^2_n(2+\frac{1}{4^n})$,and the coefficient of $x^{2n}$  in  $[p(\frac{3}{2}x)]^2 $ is $(\frac{9}{4})^n a^2_n$.therefore $2+\frac{1}{4^n}=(\frac{9}{4})^n$,let's define $f(x)=2+\frac{1}{4^n}-(\frac{9}{4})^n$,clearly $f$  decreases,and $f(1)=0$,thus $n=1$.so $p(x)=ax$,where $a$ is an arbitrary real number.obviously $p$ satisfies the property.


linear algebra
matrix
geometry
geometric transformation
rotation
vector
algebra unsolved

Hi. I need to find matrix of linear operator of rotation in 45 degree.Axle of rotation is vector(1,2,1).Please help

you mean axis not axel ?
You mean an Euclidean space $R^3$ with a standard scalar multiplication?

Then for instance an euclidean linear transformation given for linearly independent, pairwise perpendicular vectors  $(1,2,1),(-1,0,1),(1,-1,1)$ defined by the matrix:
$\left[
\begin{array}{ccc}
1 & 0 & 0\\
0 &  \frac{\sqrt{2}}{2}& -\frac{\sqrt{2}}{2}\\
0 & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}
\end{array}
\right]$ is an example.

[Edit] well if they are pairwise perpendicular then they are linearly independent in an euclidean space - so the independence didn't have to be mentioned.


Well...we can say the first disc has a thickness of 2.  So the stack has to be more than 50 units tall:

Height of Stack with x discs:

1: 2
2: 5
3: 9.5
4: 16.25
5: 26.375
6: 41.5625
7: 64.34375

So you need at least 7 discs to get the thickness of the stack 25 times the first disc.

[quote="mihail911"][b]A set of discs is being placed in a stack. The thickness of each successive disc added to teh stack is 1.5 times the thickness of the disc immediately beneath it. What is the minimum number of discs that must be placed in the stack in order for the thickness of the stack to be 25 times the thickness of the original disc?[/b][/quote]

We call the thickness of the first disc $ x$. Then, the $ n$th disc will have thickness $ 1.5^n \cdot x$. Since this will be true for any x, we can assume that $ x\equal{}1$, (We could also assume that x=2, or any number) to make the math easier. Listing the first couple discs, we get

$ 1$, $ 1.5$, $ 2.25$, $ \cdots$

or in fractions,

$ 1$, $ \frac{3}{2}$, $ \frac{9}{4}$, $ \frac{27}{8}$, $ \cdots$

If we add the first $ n$ terms, we get the following list

$ 1$,$ 2.5$,$ 4.75$,$ 8.125$,$ 13.1875$,$ 20.78125$, $ 32.171875$.

So, the first $ 7$ terms will sum to more then $ 25x$. If we didn't want to let $ x\equal{}1$, we could just make the list

$ x$, $ 2.5x$, $ 4.75x$, $ \cdots$

to get $ n\equal{}7$.

Therefore, the answer is $ \boxed{7}$.

In a CD, this way might not be so practical, so you could notice that the sequence actually looks somewhat like the Fibbonacci Sequence, for the first few terms. Then, you could easily find $ 7$.

[quote="sbanerjee"]This may sound stupid, but I got a different answer:
Let's assume that the first disc had thickness 1.
Then a stack of one coin is obviously one.
A stack of two coins is 1+1.5^2=3.25
Add 1.5^3 to get a stack of three coins: 6.625
4 coins: 11.6875
5 coins: 19.28125
6 coins: 30.671875

25 times more than the original disc is 25, and it only takes 6 coins to surpass that. so isn't the answer 6?[/quote]

You forgot to add $ 1.5^{1}$...


geometry
parallelogram
inequalities
vector
trapezoid
geometry unsolved

Let $ABCDEF$ be a convex hexagon such that triangle $ABF$ is an isosceles right triangle at $A$ (in other words, $\measuredangle BAF=90^{\circ}$ and $AB=AF$) and the quadrilateral $BCEF$ is a parallelogram, and such that $BC=1$, $AD=3$ and $DE+DC=2\sqrt{2}$.

Find the value of the area of the hexagon $ABCDEF$.

No..it's a rather nice..and not as hard problem.

Let $P$ be a point such that $CPE$ is an isosceles right triangle at P. Then from the Ptolemy's inequality for $DEPC$ we obtain
\[DP*EC \leq ED*PC+EP*CD \]
and by pluging $PE=PC=a$:
\[(DE+DC)a \geq DP* a\sqrt{2}\]
Thus
\[DP\leq 2. \]
Now consider triangle $DPA$. Here of course $PA=1$, as $PA$ is the same vector as $BC$. THis means that
\[DA \leq PD+DA \leq 2+1=3 \]
where equality hold if and only if $A,P,D$ are colinear and $DEPC$ is cyclic ( the only case to obtain equality in Ptolemy's inequality ) So $\angle CDB=90$ and $BC \parallel AB \parallel BC$. Now it's easy to compute the area. Our haxagon consists of two trapezoids: $ADCB$ and $ADEF$.Let $h_{1}$ be the distance between $BC$ and $AD$ and $h_{2}$ be the distance between $EF$ and $AD$ Now as $DEPC$ is cyclic, we have also
\[\angle PDE= \angle PDC =\frac{90}{2}=45 \]
Thus $h_{1}=\frac{CD\sqrt{2}}{2}$ and $h_{2}=\frac{DE\sqrt{2}}{2}$. So the total area is equal to
\[\frac{1}{2}(AD+BC)(h_{1}+h_{2})=\frac{1}{2}*4*2=4 \]
Maybe calculations are  wrong, but the idea is good :)


[quote="chess64"]The figure shows a (convex) polygon with nine vertices. The six diagonals which have been drawn dissect the polygon into the seven triangles: $P_{0}P_{1}P_{3}$, $P_{0}P_{3}P_{6}$, $P_{0}P_{6}P_{7}$, $P_{0}P_{7}P_{8}$, $P_{1}P_{2}P_{3}$, $P_{3}P_{4}P_{6}$, $P_{4}P_{5}P_{6}$. In how many ways can these triangles be labeled with the names $\triangle_{1}$, $\triangle_{2}$, $\triangle_{3}$, $\triangle_{4}$, $\triangle_{5}$, $\triangle_{6}$, $\triangle_{7}$ so that $P_{i}$ is a vertex of triangle $\triangle_{i}$ for $i = 1, 2, 3, 4, 5, 6, 7$? Justify your answer.

[img]6740[/img][/quote]
[hide="...solution?"]
$\triangle_{2}$ and $\triangle_{5}$ must be placed on $P_{1}P_{2}P_{3}$ and $P_{4}P_{5}P_{6}$ respectively. Then you have to put $\triangle_{1}$ on $P_{0}P_{1}P_{3}$ and $\triangle_{4}$ on $P_{3}P_{4}P_{6}$ , then $\triangle_{3}$ on $P_{0}P_{3}P_{6}$, then $\triangle_{6}$ on $P_{0}P_{6}P_{7}$ and $\triangle_{7}$ on $P_{0}P_{7}P_{8}$...$\boxed{1}$ way...which doesn't seem right [/hide]

I got [hide]12[/hide] can anyone confirm or deny this?

[hide]The locations of $\triangle_{2}$ and $\triangle_{5}$ are fixed because $P_{2}$ and $P_{5}$ are only vertices of one triangle. Then, $\triangle1$ and $\triangle_{4}$ are fixed because $P_{1}$ is only a vertex of one other triangle than $\triangle_{2}$, and similarly for $P_{4}$. Then, $\triangle_{3}$ must be the middle one, because $P_{3}$ is only a vertex for one other triangle than $\triangle_{1},\triangle_{2}, \triangle_{3}$, and $\triangle_{5}$. Then, we find only one way to place $\triangle_{6}$, and $\triangle_{7}$, so there's only 1 way.[/hide]


algebra
polynomial
function
domain
Ring Theory
superior algebra
superior algebra unsolved

How do I prove that (x,y) is not a principal ideal in C[x,y], the polynomial in the complex field?

Suppose that (x,y) was a principal ideal. Then a polynomial, say Z(x,y) generates it. This means that (x,y) = {P(x,y)Z(x,y)|P(x,y) in C[x,y]}

We will now construct a polynomial in (x,y) which is not in (Z(x,y)).
Polynomials in (x,y) are just of the form A(x,y)x + B(x,y)y  Let A(x,y) = Z(x,y) and B(x,y) = 1.

Then we get a polynomial of the form xZ(x,y) + y. This polynomial is not of the form P(x,y)Z(x,y).

More simply, just let $ A\equal{}1$ and $ B\equal{}0$.  The core of the problem is that a polynomial that divides both $ x$ and $ y$ must be a constant.

A clearer way to prove non-principal the ideal $ (x,y) < F[x,y]$ is to reason [i]structurally[/i] instead of element-wise, as below.
Doing so reveals the essence of the proof: namely, the result rests on a dimensionality argument (max length of prime chains).

Note that the ideals $ (x) < (x,y) < 1$ are distinct primes (or $ 1$) since their residue rings $ F[y] >   F   > 0$ are distinct domains (or $ 0$).

It follows that the ideal  $ (x,y)$ is not principal because nonzero principal primes are maximal among principal ideals.

This maximality is familiar from PIDs, e.g. in $ \mathbb{Z}$ or in $ F[x]$. It simply says that  $ gcd(x,p) \equal{}  p$  or  $ 1$  if $ p$ is a prime,
or, in ideal language,  $ (x,p) \equal{} (p)$ or $ (1)$, so $ (p)$ is maximal.


Every integer is colored red, blue, green or yellow. m and n are distinct odd integers such that m + n is not zero. Show that we can find two integers a and b with the same color such that a - b = m, n m + n or m - n.


The area of one face of a cube is 9 square centimeters. If the length of each edge is increased by 1 cm, what is the number of cubic centimeters in the volume of the resulting cube?

The side length of the cube is $ \sqrt{9}\equal{}3$, so the side length of the new cube is $ 4$.  The volume is thus $ \boxed{64}$.


[quote="php"]Im curently a freshman in high school but I want to know if most colleges rewuire another year of geometry. I dont know if I hate it or my teacher sucks, but im not enjoying it.
Any help?[/quote]

Actually, high school geometry is pretty much useless in that sense... BUT that doesn't mean you shouldn't learn it, because it 1) can help you gain a better understanding of math overall and 2) can help you visualize things. Of course, the only problem is it all depends on the quality of your class. Also, most high school geometry classes are utterly trivial, as in everything you learn is pretty much "obvious" anyway.

High school geometry is usually lousily taught. Many major sequences for undergraduate math majors don't require a geometry course, which means many math graduates are remarkably unfamiliar with geometry. But a good college course in geometry (almost any college course in geometry) will be a lot more interesting than almost any high school course in geometry.

[quote="krustyteklown"]yeah, i got a chinese textbook from some older cousins and learned "high school" geometry in 3 weeks

i mean, i do problem solving geo problems extra curricular, but the high school curriculum IS that small...[/quote]

Actually, it's not. That is, if you bother to learn their rigid proof structures and that kind of thing. But the real way to teach people how to write proofs is to give them non-trivial problems where it's clear what constitutes an obvious step and what doesn't.


Prove that $ \forall a,b,c \geq 0$ we have the inequality:
$a^3+b^3+c^3+3abc \geq ab(a+b)+bc(b+c)+ca(c+a)$.

this is Schur for f(x)=x increasing function. the inequality comes from expading a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b) \geq 0.


How to solve it? :D I think that LCF-RCF work here  :roll:


[url]http://www.ryanheise.com/typing-test[/url], courtesy of Ubemaya.

I got 101 WPM.

45 WPM.  :lol:

Who is this Ubemaya? Everyone seems to know about him.

96 first try
104 second try
I keep having to backspace so much.. -.-
109 third try
111 but it won't display... trying again.

Looks like I'm maxed at 54 WPM. I'm gonna keep on trying though.

Edit: 61! :lol:

You think you guy's suck? I got 48 wpm.  :(

80, 85, 82, 94.

Not horrible. I think 80 is around what most people try to obtain.

[size=200][color=red][b]69 FTW[/b][/color][/size]

i have never heard of this ubemaya person, who is she?

Whos ubemaya and why are so many people talking about him/her (whatever gender it is)?

132 WPM.  This test is pretty harsh about mistakes; http://typera.tk is a little more lenient, and has several more modes besides.  :)

[quote="t0rajir0u"]This test is pretty harsh about mistakes[/quote]
WAY too harsh
since my backspace is broken, i cant fix an error so everything afterwords is counted as wrong

i like this...

at school we have to do Turbo Typing. what's really annoying is that it marks it wrong if i put a period at the end of a sentence because  it hasn't taught me how to type a period yet.

Hi everyone, I am a new member here and decided to join after noticing some of my puzzles/games being discussion here at various times in the past. Thank you for enjoying them ;-)

As t0rajir0u noticed, the rules have some differences with TyperA, and so it is a different kind of test. With appropriate settings for the "delete" auto-key-repeat rate, I think one can develop the skill of efficient deletion. Two features also help to make deletion easier: (1) the ability delete through word boundaries and add missing commas, and (2) the displaying of what you type directly below the given text so that you can see what you're typing and more quickly react to mistakes. I think this might be more realistic since, at least in creative writing, we do normally look at what we type while we are typing it.

I would also like to add another feature that reports some statistics such as your most commonly mistyped words so that, when you type, you can be aware of which words you should slow down on.

Actually, since this is a relatively new game, and you guys are among its *very first users*, I would be very thankful to receive your feedback. If you have any ideas for how it could be improved, please let me know!

[quote="gauss1181"][quote="Nerd_of_the_Ages"]Well, don't really know a lot because I haven't been on AoPS for very long, but he's a user on AoPS (first name is albert). yeah. actually, he's mentioned quite often in the forums. especially if you look around in the blogs pages.[/quote]
Why do so many people mention things about Ubemaya? Is it because Ubemaya spams/floods or does special stuff that he's famous for?[/quote]

Two words: Meggy-Meg (or was that one)?

61 WPM.

Hi again, just a quick note about the BETA testing page for the newest version:

http://www.ryanheise.com/typing-test/beta.html

Just a couple of ideas I'm playing with.

[quote="bubka"]isn't pianoforte ubemaya?[/quote]

....no.

pianoforte=pianoforte=named Stephen.

ubemaya=ubemaya=named Albert, I think.

[quote="bubka"]isn't pianoforte ubemaya?[/quote]

 :mad:

Lol. Well, just browse around the blogs a lot and you'll find out about him


trigonometry
inequalities unsolved
inequalities

a,b are known positive numbers. $\theta \in (0,\frac{\pi}{2})$
Find the maxium $a\sqrt{sin\theta}+b\sqrt{cos\theta}$

[quote="pardesi"]by $cauchy-schwartz$ we have

${(a\sqrt\sin\theta+b\sqrt\cos\theta)}^{2}$ $\leq$ $(a^{2}+b^{2})(\sin\theta+\cos\theta)$ $\leq$ $(a^{2}+b^{2})\sqrt 2$
where the last step follows from AM GM and that $|\sin\theta|\leq 1$ and that $\theta \in (0,\frac{\pi}{2})$[/quote]
I'm sorry... that's not the right answer

Put $f(\theta)=a\sqrt{\sin{\theta}}+b\sqrt{\cos{\theta}}(0\leq \theta\leq\frac{\pi}{2})$. We have $f'(\theta)=0$ iff $\theta=\theta_{0}$, where $\theta_{0}\in (0,\frac{\pi}{2})$ such that $\tan{\theta_{0}}=\sqrt[3]{\frac{a^{2}}{b^{2}}}$. Now, by Fermat's theorem (stationary points), you need only check $f(0),f(\frac{\pi}{2}),f(\theta_{0})$.


what should i prepare if i wish to attend the mathcamp 2010?

What do you mean? I wouldn't say there's anything in specific you need to prepare with, or that you necessarily need any preparation. Just do what you normally do.
If you're asking about the application, you can find info [url=http://mathcamp.org/apply.php]here[/url].
Good luck!  :lol:


Point $P$ is outside circle $C$ on the plane. At most how many points on $C$ are $3 \text{cm}$ from $P$?

$\text{(A)} \ 1 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 4 \qquad \text{(E)} \ 8$

[hide]
B ?

Pick a point on C and draw a circle of radius 3 . Pick another point on C and draw another circle of radius 3 . Both the circles intersect at a two point . One of them is P .
[/hide]

I think it's B too. Power of a point tells us that two equal tangents are possible from circle C to a point P. Let these tangents both be 3cm, and we have our answer.


Let $ T(n)$ be the sum of the decimal digits of a natural number $ n$.

(a) Find a natural number $ N$ such that $ T(kN)$ is even for all $ k \equal{} 1, 2, . . . , 1992,$ but not for $ k \equal{} 1993$.
(b) Show that there is no $ N$ such that $ T(kN)$ is even for all $ k \in N$.

-Edit: Changed title of topic.

OK, I can solve the first part now:

a) First consider the integer $ N\equal{}10^n\plus{}1$. If $ k<10^n$, then $ T(kN)\equal{}T(10^nk\plus{}k)\equal{}2T(k)$, which is even. On the other hand, if $ k\equal{}10^n\plus{}19,$ then $ kN\equal{}10^{2n}\plus{}2\cdot 10^{n\plus{}1}\plus{}19$, so $ T(kN)$ is odd for this $ k$.

Now $ 10^{98} \equiv 0$ mod $ 1993$, so the integer $ N\equal{}\frac{(10^{98}\plus{}1)(10^{98}\plus{}19)}{1993}$ satisfies $ T(1993N)$ odd. And since $ k\frac{(10^{98}\plus{}19)}{1993}<10^{98}$ for $ k \leq 1992$, by the above $ T(kN)$ is even.

Hmm...I sort of hope that this wasn't the original solution...

And of course, the second part of the problem remains.

OK, I've got it now:

b) Let $ N$ be a positive integer with $ n$ digits and let $ K\equal{}(10^n\minus{}1)N$ and $ L\equal{}(10^{n\plus{}1}\minus{}1)N$. I then claim that if we write $ K\equal{}a10^n\plus{}b$ with $ a,b<10^n$, then $ L\equal{}a10^{n\plus{}1}\plus{}9\cdot10^n\plus{}b$. If this claim is true, then $ T(K)$ and $ T(L)$ have different parity, and thus $ T(kN)$ cannot be even for all $ k$.

To prove the claim, we write $ L\equal{}c10^{n\plus{}1}\plus{}d\cdot10^n\plus{}e$ and make three observations:

1: Since $ L\equal{}K\plus{}9\cdot 10^nN$, we have $ b\equal{}e$.
2: Since $ L\equal{}10^{n\plus{}1}N\minus{}N$, and $ N<10^n$, we must have $ d\equal{}9$.
3: Since $ 9N\equal{}L\minus{}10K\equal{}(c\minus{}a)10^{n\plus{}1}\plus{}9(10^n\minus{}b)$, and $ 0<9N<10^{n\plus{}1}$ and $ 10^n>b$, we must have $ c\equal{}a$.

And the claim is proved.


calculus
integration
limit
real analysis
real analysis solved

I have a solution to this problem, but I don't like it very much (and of course would like to see a nicer solution)

The problem is:

Suppose $g$ and $f_n$ are defined on $(0,\infty)$ and are Riemann integrable on $[t,T]$ whenever $0<t<T<\infty$.  For all $n$, $\vert f_n\vert \leq g$, and $f_n\to f$ uniformly on every compact subset of $(0,\infty)$.  We also have $\int^{\infty}_0 g<\infty$

Prove that $\lim_{n\to\infty} \int^{\infty}_0 f_n=\int^{\infty}_0 f$

Doesn't it follow directly from Lebesgue's dominated convergence theorem? :?

It is a direct application of Lebesgue theorem, isn't it?

I haven't learned anything about measure theory yet.  Shame on all of you  :D

(yes, the notes with the question state that it is a weak form of the DCT)

1.
$|f_n|\leq g$ on $[t,T]$ and $f_n\to f$, hence $|f|\leq g$ on any interval $[t,T]$. Hence $\int_0^\infty |f| dx$ exists.
So does $\int_0^\infty |f_n-f| dx$ (because it is smaller than $\int_0^\infty gdx + \int_0^\infty |f|dx$).

2.
Since $f_n\to f$ uniformly on $[t,T]$, we can prove that $\int_t^T |f_n-f| dx \to 0$.

3.
Because the improper integral is defined as limit of proper integrals, for any $\epsilon$, there exists $t,T$ such that $\left|\int_0^t f(x) dx\right| < \epsilon$ and $\left|\int_T^\infty f(x)dx\right| < \epsilon$.

4.
Write
$\int_0^\infty |f_n-f| dx = \int_0^t |f_n-f| dx + \int_t^T |f_n-f| dx +  \int_T^\infty |f_n-f| dx$
For a given $\epsilon$, by 3, there exists $t,T$ such that $\int_T^\infty |f_n-f| dx < \frac{\epsilon}{3}$ and $\int_0^t |f_n-f| dx < \frac{\epsilon}{3}$
Now $t$ and $T$ are fixed, apply 2, $ \int_t^T |f_n-f| dx  < \frac{\epsilon}{3}$ for $n$ large enough.

Hence $\left|\int_0^\infty f_n - \int_0^\infty f\right| \leq \int_0^\infty |f_n-f| dx < \epsilon$ for $n$ large enough.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hi,

How would I calculate the number of sylow_2, sylow_3, and sylow_5 subgroups of the group GL(2,Z_5), which has order 480= (2^5)*(3)*(5)? Also to describe these subgroups.

I calculated the number theoretic restrictions that each can have, but still there are alot of possibilities for each one. How would I narrow it down?

thanks

Sylow is the name of a polish mathematican.
A sylow(ian) $p$-subgroup $H$ of $G$ ($p$ a prime) is one where $|H|=p^k$ and $p^k$ is the biggest power of $p$ dividing $|G|$.

[quote]Sylow is the name of a polish mathematican. [/quote]
Are you sure ? ;)
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Sylow.html[/quote]


While reading through Richard Rusczyk's and Sandor Lehoczky's Volume II, I came across a description of a famous problem, as noted, which was to evaluate $ \frac {1}{1^{2}} + \frac {1}{2^{2}} + ...$.  In order to find its value, they used the fact that one can factor the expression $ \sinh = x - \frac {x^{3}}{3!} + \frac {x^{5}}{5!} + ...$ as $ x(1 - \frac {x}{\pi})(1 + \frac {x}{\pi})(1 - \frac {x}{2\pi})(1 + \frac {x}{2\pi})...$.  This is easily seen by noting the solutions of $ \sin{x} = 0$.  But how can one confirm, through rigorous proof, that this is indeed true, in other words, that this factorization is correct? (I'm asking for a rigorous proof of this factorization, in short).


OS - Windows XP
Browser - Opera

Whenever I press "Logout from Server" the screen turns white. It just stay's like that. And then when I use task manager/alt+f4/click the x, I get a not responding message and then it turns off.

I noticed this too, and it just started happening recently. I'm not sure why, I haven't upgraded my Opera and AoPS hasn't changed the chat software. Only thing I can think of is that there was a Java update and it has a bug in it which is causing this behavior.

i dont think there is a recent java update...but idk

Interesting. Today I logged out from a classroom, and it worked fine.

Yeah.
Usually, when I log out from the classroom, I find that it freezes the computer all together, with a white screen. However on monday, the classroom worked fine. I know I had the latest Java update and use the i think newest opera.

Aha a way to cheat the glitch
To close out of the class, just go back to the opera page, and use the method of right click closing on the bar to close the class.
Or you can close the class and as fast you can go back to the browser.

I thought you could do something like that on IE


probability
geometry
geometric transformation
reflection
function
analytic geometry
limit

Here's a pretty nice-yet easy-problem.

[u]Problem.[/u] [i]You randomly select three vertices of a regular $ n$-gon ($ n\geq 3$) to form a triangle.  What is the probabily that the triangle is obtuse as $ n\rightarrow\infty$?[/i]

[hide="Solution"][i]We are essentially choosing 3 random points on the circumference of a circle, and want the probability that the center of the circle is not contained inside the triangle.[/i]

Let the points be A, B, C, with A oriented "up". Let B be immediately ccw to A, and C to B. Now, divide the circle into left and right.

[b]Case 1:[/b] [i]B and C are on the same half[/i]

There is a $ 2 \cdot \frac {1}{2} \cdot \frac {1}{2} \equal{} \frac {1}{2}$ chance of this, which is clearly favorable.

[b]Case 2:[/b] [i]B is on the left half, C is on the right[/i]

ABC contains the center iff C lies within the arc $ A'B'$, the reflections of A and B across the center. As $ \theta \equal{} m \angle AOB \equal{} m \angle A'OB'$ ranges linearly from $ 0$ to $ \pi$, the fraction of $ ACA'$ that is favorable is $ \frac {\pi \minus{} \theta}{\pi} \equal{} 1 \minus{} \frac {\theta}{\pi}$. The average value of this function is $ \frac {1}{2}$ (the graph is a right triangle), so this is the favorable probability [u]given[/u] this case.

[b]Answer:[/b] $ \frac {1}{2} \plus{} \left( 1 \minus{} \frac {1}{2} \right) \frac {1}{2} \equal{} \frac {3}{4}$.
[/hide]

[hide="Different method, different result"]Let's choose 3 central angles, 2 of them being independent.

In order for the triangle to be scalene, we must have

$ 0<x<\pi$

$ 0<y<\pi$

$ 0<2\pi\minus{}x\minus{}y<\pi$

Sketching all of those conditions in the coordinate system gives us $ \overline P\equal{}{1\over 2}\implies P\equal{}{1\over 2}$ [/hide]

Unfortunately, your answer is, I believe, incorrect; however, I have not studied your solution.  My solution follows from the probability $ P_n$ in terms of $ n$, which is $ \frac {3(n \minus{} 4)}{4(n \minus{} 1)}$ and $ \frac {3(n \minus{} 3)}{4(n \minus{} 2)}$ for $ n$ even and odd, respectively.  Hence, the answer is $ \lim_{n\rightarrow\infty}P_n \equal{} \frac {3}{4}$.

Miscalculated the areas :roll: by my method we get $ \overline{P} \equal{} {1\over 4}\implies P \equal{} {3\over 4}$


function
real analysis
real analysis theorems

[quote="silouan"][quote="Dr. No"][quote="silouan"]What is the <<famous>> Lagrange multiplers? Is there a link about this?[/quote]

[url=http://mathworld.wolfram.com/LagrangeMultiplier.html]Yes, there is. Look here.[/url][/quote]
I went at this site but it said
<<access denied to IP adresse>>. What can I do?[/quote]Short of using a different computer on a different network -- one that allows you unrestricted access to the web -- I don't know.  In the mean time to answer your question, Lagrange multipliers is a method that allows one to calculate maxima and minima of multivariate functions. If $f$ is some function one wishes to find the extreme points of, and $g = 0$ is a constraint function that $f$ must obey, then the extreme points of $f$ occur where $\nabla f = \lambda \nabla g$.  $\lambda$ is the Lagrange multiplier.


algebra
polynomial
modular arithmetic
trigonometry
quadratics
complex numbers

Find the sum of squares of all real numbers $ r$, for which the expression
$ (1\plus{}ri)^3$ is of the form $ s(1\plus{}i)$ for some real number s, where $ i^2 \equal{} \minus{}1$

Is it $ 15$? If it is, I'll post my solution.

Ok. First, expand $ (1 \plus{} ri)^3$.

$ (1 \plus{} ri)^3 \equal{} 1 \plus{} 3ri \minus{} 3r^2 \minus{} r^3i \equal{} s \plus{} si$

Matching the real and imaginary parts:

$ 1 \minus{} 3r^2 \equal{} s$

$ 3r \minus{} r^3 \equal{} s$

We wish to get rid of $ s$, so just substitute:

$ 1\minus{} 3r^2 \equal{} 3r \minus{} r^3$

$ r^3 \minus{} 3r^2 \minus{} 3r \plus{} 1 \equal{} 0$

By Newtons Sums, we have that

$ S_2 \equal{} s_1^2 \minus{} 2s_2 \equal{} (3)^2 \minus{} (2)(\minus{}3) \equal{} \boxed{15}$.

Ohhhhh! So newton sums are things derived from symmetric polynomials

[hide="Another interpretation - not necessarily easier to calculate"]
Cubing a complex number $ z$, with angle $ \theta$ with respect to the axis, has the effect of tripling the angle.

All complex numbers in the form $ s(1 \plus{} i)$ have angle $ \frac {\pi}4$. Solving the equation $ 3\theta\equiv \frac {\pi}4\pmod{2\pi}$ we have $ \theta\in\left\{\frac {\pi}{12},\frac {3\pi}4,\frac {17\pi}{12}\right\}$.

Finally, $ \tan^2\frac {\pi}{12} \plus{} \tan^2\frac {3\pi}4 \plus{} \tan^2\frac {17\pi}{12} \equal{} (2 \minus{} \sqrt 3)^2 \plus{} 1 \plus{} (2 \plus{} \sqrt 3)^2 \equal{} 1 \plus{} 14 \equal{} 15$.

[hide="Side note"]
The values of $ \tan^2\frac {\pi}{12}$, $ \tan^2\frac {3\pi}4$, $ \tan^2\frac {17\pi}{12}$ are precisely the solutions to $ r^3 \minus{} 3r^2 \minus{} 3r \plus{} 1 \equal{} 0$!
[/hide]
[/hide]

So did you calculate the values of $ \tan(\pi/12),\tan(17\pi/12)$ from the equation $ r^3 \minus{} 3r^2 \minus{} 3r \plus{} 1$ (just solve the quadratic after getting - 1 as a root), or from something else? By the way is there another way to show the three tangent values satisfy that equation?


$ F$ and $ F'$ are the foci of an ellipse and $ P$ is any point on the ellipse. Prove that the circles described on $ FP$, $ F'P$ as diameters touch the circle which has the major axis of the ellipse as diameter.

Let $ M$ and $ N$ be the midpoints of $ PF$ and $ PF'$ respectively, and let $ O$ be the centre of the ellipse. Let the length of the major axis of the ellipse be $ 2a$.

Two circles with radius $ R_1$ and $ R_2$ touch each other if and only if $ |R_1\minus{}R_2|\equal{}\text{Distance between the two circle centres}$

Now, $ a\minus{}PN\equal{}\frac{2a\minus{}PF'}{2}\equal{}\frac{PF}{2}\equal{}MF\equal{}NO$.

So $ \odot N$ touches $ \odot O$ internally.


geometry
circumcircle
Euler
conics
hyperbola
cyclic quadrilateral
geometry proposed

Given triangle $ABC$ and $P$ is an arbitrary point. Circumcircle of triangle $PBC$ cut lines $AB,AC$ at $C_{a},B_{a}$ resp prove that three Euler's line of tringles $ABB_{a},ACC_{a}$ and $AB_{a}C_{a}$ are concurrent at point $P_{a}$ similary we have $P_{b},P_{c}$ prove that $P_{a},P_{b},P_{c}$ are colinear.

For the first part, quadrilateral $BCB_{a}C_{a}$ is cyclic and A is its diagonal intersection. (When A is inside of the circumcircle of the $\triangle PBC$, the quadrilateral $BCB_{a}C_{a}$ is considered convex, otherwise it is considered reflex). By the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107997]Euler lines in cyclic quadrilateral[/url], the Euler lines of the $\triangle ABC,\ \triangle ACB_{a}, \triangle AB_{a}C_{a}, \triangle AC_{a}B$ are concurrent at $P_{a}$, the concurrency point lying on a normal hyperbola through A and circumcenters of these 4 triangles. For the second part, the 3 concurrency points $P_{a},\ P_{b},\ P_{c}$ obviously lie on the Euler line of the $\triangle ABC.$ It appears that concurrency of the 3 circles at P is not necessary.

That's right, I inspired this problem from that problem in your link, I only remark that problem is still true when we consider intersection of opposite edge instead of consider diagonals only.
By the way we can see that problem by other way as following
Given triangle $ABC$ and $(O)$ is an arbitrary circle through $B,C,(O)$ cut lines $AB,AC$ at $C',B'$ resp then Euler lines of triangle $ABC',AB'C, AB'C'$ are concurrent at point $A'$ and locus of $A'$ (when $(O)$ move) is Euler lines of triangle $ABC$.


vector
algebra
polynomial
Vieta
complex numbers

Let  $ z_1,z_2,z_3\in\mathbb{C}$ such as  $ |z_1| \equal{} |z_2| \equal{} |z_3| \equal{} 1$ and $ z_1 \plus{} z_2 \plus{} z_3 \equal{} 1$ proove that  $ z_1 \equal{} 1$ or $ z_2 \equal{} 1$ or $ z_3 \equal{} 1$

Draw a circle $ \omega_1$ of radius $ 1$ around the origin in the complex plane. If we treat $ z_1$ as a vector, then $ z_1$ has to point from the origin to some point on the circle. If $ z_1 \equal{} 1$, we're done, so suppose $ z_1$ points to some other point $ P$ on $ \omega_1$. Now draw circles $ \omega_2$ centered at $ P$ and $ \omega_3$ centered at $ 1$ on the complex plane, both radius $ 1$. Treating $ z_2$ and $ z_3$ as vectors also, the only way for $ z_1 \plus{} z_2 \plus{} z_3 \equal{} 1$ to hold is if we can have two vectors of unit length pointing from $ P$ to $ 1$. But $ z_2$ must point to a point on $ \omega_2$, and $ z_3$ must point from a point on $ \omega_3$ to $ 1$. This means that $ z_2$ has to point to a point on $ \omega_3$ also, so $ z_1 \plus{} z_2$ lies on an intersection of $ \omega_2$ and $ \omega_3$. Since we assumed $ z_1 \neq 1$, these two circles do not coincide, so they have at most two intersection points. Notice that $ \omega_2$ and $ \omega_3$ intersect at
1. The origin. If $ z_2$ points here, then $ z_1 \plus{} z_2 \equal{} 0$ and $ z_3 \equal{} 1$.
2. At $ z_1 \plus{} 1$. If $ z_2$ points here, then $ z_2 \equal{} 1$.
If these two points are the same, then $ z_1 \equal{} \minus{}1$ and $ \omega_2$ and $ \omega_3$ only intersect once.

So in all cases, one of the complex numbers is $ 1$.

Another approach:

Note that $ |w| \equal{} 1\implies \bar w \equal{} {1\over w}$, hence

$ 1 \equal{} \overline{z_1 \plus{} z_2 \plus{} z_3} \equal{} \bar z_1 \plus{} \bar z_2 \plus{} \bar z_3 \equal{} {1\over z_1} \plus{} {1\over z_2} \plus{} {1\over z_3}$

and that implies $ z_1z_2 \plus{} z_2z_3 \plus{} z_3z_1 \equal{} z_1z_2z_3$

That in turn, by Vieta, means that there exists $ a\in\mathbb{C}$ such that $ z_1,z_2,z_3$ are the solutions of the equation $ z^3 \minus{} z^2 \plus{} az \minus{} a \equal{} 0\iff (z \minus{} 1)(z^2 \plus{} a) \equal{} 0$. The result follows.


find all function from R+ to R+ wish satisfay
$f(f(x))+af(x)=b(b+a)x$;such that $a$ and $b$
are real positif strict numbers  :rotfl:

well i tried $x=f^{-1}(x)$ and now have

$f(x)=b(a+b)f^{-1}(x)-ax$ but am not sure where to go from here.

hy arkammedos vive maroc. by the way the solution is $f\left(x\right)=bx$ you can use sequense $f^{n}\left(x\right)$. :wink:


Let $ ABCDE$ be a pentagon inscribed in a circle. Let $ A'$, $ B'$, $ C'$, $ D'$ and $ E'$ be the midpoints of the sides $ CD$, $ DE$, $ EA$, $ AB$ and $ BC$, respectively. Show that the midpoints of $ A'C'$, $ A'D'$, $ B'D'$, $ B'E'$ and $ C'E'$ all lie on a circle.

really ugly solution. i'm rusty at this whole math thing

k assume they lie on a circle centered at the origin

$ A: (x_1,y_1)$
$ B: (x_2,y_2)$
$ C: (x_3,y_3)$
$ D: (x_4,y_4)$
$ E: (x_5,y_5)$

and $ x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5 \equal{} y_1 \plus{} y_2 \plus{} y_3 \plus{} y_4 \plus{} y_5 \equal{} 0$ cuz it's centered at the origin

for the prime point things:

$ A': (\frac {x_3 \plus{} x_4}{2},\frac {y_3 \plus{} y_4}{2})$

$ B': (\frac {x_4 \plus{} x_5}{2},\frac {y_4 \plus{} y_5}{2})$

$ C': (\frac {x_5 \plus{} x_1}{2},\frac {y_5 \plus{} y_1}{2})$

$ D': (\frac {x_1 \plus{} x_2}{2},\frac {y_1 \plus{} y_2}{2})$

$ E': (\frac {x_2 \plus{} x_3}{2},\frac {y_2 \plus{} y_3}{2})$

Call the midpoints of $ A'C'$, $ A'D'$, $ B'D'$, $ B'E'$, and $ C'E'$, respectively, $ P_{1\cdots5}$

Notice that each of $ P_{1\cdots5}$ is of the form $ (\frac {x_1 \plus{} x_2 \plus{} x_3 \plus{} x_4 \plus{} x_5}{4} \minus{} \frac {x_k}{4},\frac {y_1 \plus{} y_2 \plus{} y_3 \plus{} y_4 \plus{} y_5}{4} \minus{} \frac {y_k}{4})$, where $ k$ is an integer between 1 and 5.

which equals $ ( \minus{} \frac {x_k}{4}, \minus{} \frac {y_k}{4})$

so the resultant pentagon is simply a reflection/shrinkage'd version of $ ABCDE$. it's cyclic.

I don't actually think you can assume that since the points are on a circle, that the pairwise sum of their coordinates is 0.  Try $ (0, 1), (1, 0), (0, \minus{}1), (\minus{}1, 0), \left(\frac{3}{5}, \frac{4}{5}\right)$.  These are all on a circle centered on the origin with radius 1, but the coordinates' sums are $ \left(\frac35, \frac45\right)$.

He probably meant the centroid of the pentagon was centered at the origin.


The following messages are given when I click [url=http://www.mathlinks.ro/usercp.php?mode=subs&sub=favs]Favorites[/url].

"[b]Fatal error[/b]: Allowed memory size of 33554432 bytes exhausted (tried to allocate 35 bytes) in [b]/var/www/Forum/includes/class_template.php[/b] on line [b]333[/b]"

Nothing was displayed before. (Because the error message did not appear, I did not contact it.)
And this was displayed for the first time today.

[b]Where of the site breaks down?[/b]

Seems whatever you did required a lot of memory ...  :huh:  :ninja:

Is it the problem of the server?
My friend doesn't seem to have a problem.

I don't understand a cause.


I have been stuck on this problem for quite a while, so please help

Let $ F$ be a Lebesgue-measurable set in $ \mathbb{R}^{n}$, then there exists a Borel set $ E$ such that $ E \subset F$ and $ \mu(E \minus{} F) \equal{} 0$

I already managed to prove this in the case where $ \mu(F) < \infty$ so it's the case when $ \mu(F) \equal{} \infty$ that I'm in interested in.

The key is that $ \mathbb{R}^n$ is a $ \sigma$-finite measure space. Find a sequence of sets $ A_k,$ each of finite measure, such that $ \bigcup_kA_k\equal{}\mathbb{R}^n.$ To make the rest of this work, we do need $ A_k$ to be Borel, but that's no problem - we can use balls or cubes. Now look at $ F_k\equal{}F\cap A_k.$ What can you do from there?

At some point you're going to use that a countable union of sets of measure zero has measure zero.