Find all the roots of $x^5+7=0$.

Hopefull its in the right section.

I might be doing this wrong but:
[hide]
$x^5+7=0$
$x^5=-7$
$x=\boxed{-\sqrt[5]{7}}$
[/hide]

VenDiddy: That's one of the roots. :)

$x=\sqrt[5]{-7}=7^{1/5}(\text{cis}0+2k\pi)^{1/5}$
By DeMoivre's theorem, $(\text{cis}2k\pi)^{1/5}=36^\circ, 108^\circ, 180^\circ, 252^\circ, 324^\circ$.

The answer is as follows.

Hmm, I wonder how my cis's look. :P


geometry
rhombus
circumcircle
angle bisector
geometry proposed

[color=darkred][size=134][b][u]An old nice problem.[/u][/b][/size] Let $ ABCD$ be a rhombus with $ A = 60^{\circ}$.
For a mobile point $ M\in (AC)$ define $ \{\begin{array}{c}X\in (AB)\ ,\ MX\parallel AD\\ \ Y\in (BC)\ ,\ MY\parallel AB\end{array}$. Prove that :
$ 1\blacktriangleright$ The triangle $ DXY$ is equilateral and the area of the quadrilateral $ DXBY$ is constant.
$ 2\blacktriangleright$ The centroid of the triangle $ DXY$ belongs to $ (AC)$ and $ MG = 2\cdot GO$, where $ O\in AC\cap BD$.
$ 3\blacktriangleright$ The perpendicular lines lowered from the points $ A$, $ M$, $ C$ on the lines $ XD$, $ XY$, $ YD$ respectively are concurrently.[/color]

$ 1\blacktriangleright$ the triangles $ \triangle XYB$ and $ \triangle AXD$ are congruent (L.A.L.) so $ XY \equal{} XD$ and $ \angle DXY \equal{}\angle XAD \equal{} 60$ so $ \triangle XYD$ is equilateral
$ A\equal{} A[ABCD]\minus{}A[DYC]\minus{}A[AXD] \equal{} A[ABCD]\minus{}A[ABD] \equal{} A[ABD]$

$ 2\blacktriangleright$ denote $ R: MX\cap DB$ and $ S: MY\cap DB$; $ MX \equal{} BY \equal{} BS \equal{} DR$ and $ MY \equal{} SD \equal{} RX$ so the triangles $ \triangle XMY$, $ \triangle SYD$ and $ \triangle XRD$ are congruent (L.L.L.) so the triangle $ \triangle MRS$ is equilateral and rotohomotetic with $ \triangle XYD$ then, cause MO is the angle bisector of $ \angle SMR$, $ OS \equal{} OR$ and $ MG \equal{} 2GO$

$ 3\blacktriangleright$ the perpendicolar line from A to XD meet the perpendicolar line from C to YC in P and $ U: DX\cap AP$, $ V: DY\cap CP$.
$ \angle BAP \equal{} 90\minus{}\angle ADX\minus{}60 \equal{} 30\minus{}\angle ADX$; $ \angle YCP \equal{} 90\minus{}\angle CYD \equal{} 90\minus{}(120\minus{}(60\minus{}\angle ADX)) \equal{} 30\minus{}\angle ADX \equal{}\angle BAP$
so the quad APBC is cyclyc with center D and then $ AU \equal{} UP$ and then $ \angle XPA \equal{}\angle YPC \equal{} 30\minus{}\angle ADX$;
so $ \angle XPY \equal{}\angle AYC \equal{} 120$ and so P stay on the circumcircle of $ \triangle DXY$ but $ MX \equal{} AX \equal{} XP$ and $ PY \equal{} YC \equal{} MY$ and $ \angle XMY \equal{} 120$
so the triangles $ \triangle XMY$ and $ \triangle XPY$ are congruent and symmetric with XY, so P stay on the perpendicolar line from M to XY


function
inequalities solved
inequalities

The generalized form may be (sum xi + 1)^2 >= 4(sum xi^2) when 0<= xi <= 1 for i=1,2,...,n (n is a positive integer).

This can be proved by AM>=GM.

(sum xi +1)^2 >= 4 sum xi >= 4 sum xi^2 (cuz 0<= xi <=1  <=>  xi<=xi^2.)

(when xi=1, x1=x2=...=x(i-1)=x(i+1)=...=xn=0 the equality holds.)

Thanks.

[quote]0<= xi <=1 <=> xi<=xi^2[/quote]

Not necessarily true, counterexample: take xi = 1/2, xi 2  = 1/4 but xi>xi 2

Maybe you can refine your solution a little bit?

(sum xi +1)^2 >= 4 sum xi >= 4 sum xi^2 (cuz 0<= xi <=1 <=> xi<=xi^2.)

=>

(sum xi +1)^2 >= 4 sum xi >= 4 sum xi^2 (cuz 0<= xi <=1 <=> xi>=xi^2.)

xi>=xi^2   <=>   xi(1-xi)<=0

Sorry for typing wrong, but the rest of solution is ok, I think.

Thanks

Sorry, I'm not specifically picking on you  :D  but I see that [quote]xi>=xi^2 <=> xi(1-xi)<=0 [/quote]
should be xi \geq xi^2 <==> xi(1-xi) \geq 0

The rest is neat itself

BTW, beautiful problem though.

You're right :)

(sum xi +1)^2 >= 4 sum xi >= 4 sum xi^2 (cuz xi>=xi^2  <==>  xi (1 - xi)>=0 <==> 0<=xi<=1)

This will be the last solution of the problem. Thanks!!

Another way to prove the generalisation above :
Let f(x_1,...,x_n) = 4* \sum {1 \leq i \leq n} (x_i) 2  - (1 +  \sum {1 \leq i \leq n} x_i) 2 .
This function is a convex function in each of the variables so it reaches its maximum when each of the variables is 0 or 1.
Let k be the number of x_i being equal to 1. In that case :
f(x_1,...,x_n) = 4k - (k+1) 2  = -(k-1) 2   \leq  0, as desired.

Pierre.


inequalities
LaTeX
inequalities unsolved

a_1,_a2,...,a_n,m are reel numbers. a_i\leq\fracm

Prove that or say your thinks please.

for n=2k
a_1(m-a_2)+a_2(m-a_3)+...+a_n(m-a_1)\leq \frac{nm^2}{2}

for n=2k+1
a_1(m-a_2)+a_2(m-a_3)+...+a_n(m-a_1)\leq \frac{(n-1)m^2}{2}

I'm so sorry for this. If you want to see inequalities in latex please look:   http://www.mathlinks.ro/Forum/viewtopic.php?t=69321
                                                                                Regards


function
calculus
integration
calculus computations

show that

a) fourier transform of an even function is even

b) fourier transform of an odd function is odd

c) $f$ is real function. Then $\hat{f}(-x)=\bar{\hat{f}}(x)$

[color=red]Are the converses valid??[/color]

Please help.

c) can be combined with a) and b) to produce these statements:

The Fourier transform of a real, even function is real and even.

The Fourier transform of a real, odd function is odd and pure imaginary.

The proofs are mechanical: write out the integral, make the substitution $u=-x,$ and use the definitions of even and odd.

Yes, the converses (properly interpreted) are true. A function whose Fourier transform is even is itself even, and so on.


logarithms
inequalities
number theory unsolved
number theory

Let  Theta (x) = SUM log p     (sum is primes less than or equal to x)

Prove Theta(x) <= (log 4)x

Sorry it looks messy, my code skills aren't great.

What do you mean by (log 4)x? Is it $\log_4 (x)$?
In that case, the desired inequality is 'just' $\prod_{p \leq x} p \leq 4^x$, which is a well-known result from Erdös, and which appeared sometimes here before I think.

Pierre.


In tests, if you get right, you get points, if you got wrong, you don't get points.
Some tests if you get wrong, you actually lose points.

I was thinking one think, in a actual test, how many right anwers a student actully get?
Because he can guess right, is problable (in multi choice tests).

So using a parameter like "Exactness" from 0% to 100% is a better choice?
If you are sure 76% of your answers and is rigth, you get 76% of the points, if not, you lose 76% of the points.

So, there are some kinds of avaliation you can make of the student,
If he is more sure of a answers and get wrong, there is some problem here.
If he is not sure, and get rigth, is most likely to be a moderated guess, and this matter need to be fixed.
If the is not sure and get wrong, it's a huge matter, and it need to a review.

I thoght this idea studing by myself, some questions (is not math, is brazilian law).
Some questions I was sure, and got wrong, and some I don't even understand the question.
So using this avaliation, will make my time of study better than just focusing in the wrong answers, because with this, the guesses and educated guesses can be fixed to sure things.

What do you guys think of this system of evaluation of tests?
Some improvements?
Thanks in advance

Suppose I estimate, somehow, that there is a probability $ p$ that an answer I submit is correct.  Then my expected point gain from saying I am $ q$ percent certain is $ q\left(p \minus{} (1 \minus{} p)\right)$.  Therefore, if $ p > \frac {1}{2}$, under the proposed scheme I am always best off just saying that I am 100% certain.  Likewise if $ p \leq  \frac {1}{2}$, I am best off simply not answering the question.  So the scheme is equivalent to taking points off for wrong answers, except with an added complication that an astute student won't wind up using anyway.

I think even if this scheme were adjusted in some clever way, it would boil down to essentially the same problem, especially in the context of multiple choice exams.

I have heard of systems for multiple-choice questions where you are allowed to bubble in multiple answers.  For example, let's say you have narrowed the answer to B or C.  Then instead of picking one, you can pick both.  You get less credit for a correct answer in that case, but at least you don't have to make a random guess.

JoeBlow: The objetive is not optimization! In the case is for self-study, test simulations, etc. So optimization in the result in this case is like cheating on IQ test
You can use some statistical analisys on the IQ test to try to make your score higher, but it don't mean you are more "smart".
The objetive of this test I meant to make a more real evaluation of the student.
I know that in REAL tests it's meaningless, since the student only want high scores, but in a controled situation where he wants improvement, I think is very useful

This is why multiple choice problems are a load of BS. Every teacher and grader knows this, but we keep using them because they are easy to grade. (SAT comes to mind.) If there was a cheap machine that could properly and validly grade an essay or a free response or an open answer problem as fast as it could a multiple choice problem, they would've been gone years ago. Sadly though, this will never happen, so these problems will stay.


Hmm....

1. What is the sum of the perfect squares and perfect cubes between 1 and 1000?

2. What is the sum of the perfect cubes from 15 to 18000?

Is there like a formula?

The sum of the first $ n$ squares is $ \frac{n(n\plus{}1)(2n\plus{}1)}{6}$, and the sum of the first $ n$ cubes is $ \left(\frac{n(n\plus{}1)}{2}\right)^2$.

So, for the first one:

There are 31 squares, so the sum of the first 31 squares is 10416. Similarly, there are 10 cubes, so the sum of the first 10 cubes is 3025. Added together, this is 13441. However, we overcounted! Some numbers are both squares and cubes, namely 3: 1, 64, and 729. We subtract these out to get a final answer of 12647.

Now, for the second one:

Suppose it was just 1 to 18000. There are 26 cubes there, so using the formula would give us 123201. However, from 1 to 15, we have 2 cubes, 1 and 8, so as before, we subtract these out to get a final answer of 123192.

And no there is no formula for the second but sorta for the 1st.

Also:
$ 1^3\plus{}2^3\plus{}3^3\plus{}4^3....\plus{}n^3\equal{}(1\plus{}2\plus{}3\plus{}4\plus{}....\plus{}n)^2$

@Izzy: Did you just know those two formulas off the top of your head or did u have to google them?

[quote="Thunder365"]Also:
$ 1^3 \plus{} 2^3 \plus{} 3^3 \plus{} 4^3.... \plus{} n^3 \equal{} (1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} .... \plus{} n)^2$
[/quote]

Um, that's what I put, because $ \frac{n(n\plus{}1)}{2} \equal{} 1 \plus{} 2 \plus{} 3 \plus{} \dots \plus{} n$.

Yes, I knew these. It's much more handy to memorize them  then to have to rederive them all the time. :P


Hi I am kind of stuck on this problem, i got a little bit through it but couldn't finish it. here it is.

Sin(theta)Cot(theta)Tan^2(theta)=

the possible answers are

sec(theta)

sin(theta)

tan(theta)sin(theta)

tan(theta)sin^2(theta)

tan(theta)cos(theta)

thanks for the help

[hide="hint"]$ \tan \theta \cdot \cot \theta \equal{} 1$[/hide]

yes but this is tan^2 i really don't understand this problem, could you explain?

Well isn't $ \tan ^2 \theta \equal{} \tan \cdot \tan$? How can you rewrite the original problem?

would the answer be sin(theta)tan(theta)? because cot X tan = 1 x tan = tan and then all is left is the sin? right?

Generally, it is a good idea to post as much of your solution as possible (the original poster, I mean). That way, we can check your solution, and help you along.

[hide="Solution:"]

$ \sin{\theta}\cos{\theta}\tan^2{\theta}$

$ \equal{}\sin\theta\cos\theta\tan\theta\left(\frac{sin\theta}{\cos\theta}\right)$

$ \equal{}\sin^2\theta\tan\theta$

which would be your 4th answer[/hide]

[quote="fishythefish"]Generally, it is a good idea to post as much of your solution as possible (the original poster, I mean). That way, we can check your solution, and help you along.

[hide="Solution:"]

$ \sin{\theta}\cos{\theta}\tan^2{\theta}$

$ \equal{} \sin\theta\cos\theta\tan\theta\left(\frac {sin\theta}{\cos\theta}\right)$

$ \equal{} \sin^2\theta\tan\theta$

which would be your 4th answer[/hide][/quote]

The problem is $ \sin \theta \cot \theta \tan^{2} \theta$

So, yes Kevin you are right, the answer is $ \tan \theta \sin \theta$

Whoops!  :blush:

Yes, in that case, the tan, and cot cancel. leaving 1. Multiply by sin(theta) tan(theta), and the answer is sin(theta) tan(theta).

Sorry, guys.


George is indiferrent between 100 dollars and and a bet with a .6 chance of 50 dollars and a .4 chance of 200 dollars. If U(50)=a and U(200)=b, what is U(100)?

U is just a utility function. I was thinking .6a+.4b?

Right -- indifference means E[U] is the same in both cases.


we have length $ab,bc,cd,da$ of the quadrilateral $abcd$ and we have sum of the opposite angles of it.($\angle Aangle C$)
draw the quadrilateral. :D

i think its in natan altshieler court...

this is a book with name of:[i]College Geometry:An introduction to Modern Geometry of triangle and circle , Nathan Altshiler Court[/i]

Does anybody know if we can find this book in Amazon or somewhere ? I want to buy it!

Babis

I could only find this : http://www.biblio.com/books/10206239.html

which is not the desired book :(

i think it is better that you solve the problem. ;)

stergiu ,we our selves use the farsi font of it,and dont have its original. :(

[b]To jensen[/b]

Dear Jensen !  I do not need the book for the solution. I do not care for it, because , if necessary , I can ask it from you !
.But this book is a reference book and I have seen it in the bibliografy in many other  books.I'd like to see if it deserves it's name.

  [b]To mathmanman[/b]

 Many thanks !I apreciate it.

Babis


Find all polynomial P(x) satisfying
1) $P(x^{2}+1)=(P(x))^{2}+1,\ \forall x\in\mathbb{R}$
2) $P(x^{2}-2)=(P(x))^{2}-2,\ \forall x\in\mathbb{R}$

1) Clearly$x \neq x^{2}+1$, $P(x)= \pm P(-x)$,
$P(x)$ when $x \in [1,2)$ can be determined by th values of $P(x)$ when $x \in [0,1)$
$P(x)$ when $x \in [2,5)$ can be determined by th values of $P(x)$ when $x \in [1,2)$
$P(x)$ when $x \in [5,26)$ can be determined by th values of $P(x)$ when $x \in [2,5)$
and so on. So P(x) depends only on its value in [1,2), there are infinitely many solutions.

[quote="modeler"]So P(x) depends only on its value in [1,2), there are infinitely many solutions.[/quote]
Just because it depends on those values only doesn't give $\infty$ solutions.


:rotfl:  :rotfl:
[hide="data seems insufficient"]
at the equlibrium position let the extenison of the dpring be $ x_{0}$ and the distance between the two plates of the capcitor be $ L$ then $ L \equal{} x_{0} \plus{} d$
also at equilibrium
$ qE \plus{} kx_{0} \equal{} mg$ is the plate be disturbed up by say $ dx$ then the net force on it is $ F \equal{} k(x_{0} \minus{} dx) \plus{} qE \minus{} mg \equal{} \minus{} kdx$
so this is an SHM whose $ \omega \equal{} \sqrt {k}{m}$ is independent of the speed given or any other non fixed factors
now lets say the capacitor plate is given an upward speed of $ v$ then by COE
$ \frac {kA^{2}}{2} \equal{} \frac {1}{2mv^{2}}$ thus $ A \equal{} \sqrt {\frac {k}{m}}v$
which is dependent on $ v$
thus if the $ x(t) \equal{} A\sin \omega t$ is the equation of motion of the plate then then the capcitance
$ C(t)$ of the capacitor is $ C(t) \equal{} \frac {A \epsilon_{0}}{L \minus{} x(t)}$ thus
$ \frac {C(\omega^{ \minus{} 1})}{C((2\omega)^{ \minus{} 1})} \equal{} \frac {L \minus{} x((2\omega)^{ \minus{} 1})}{L \minus{} x(\omega^{ \minus{} 1})} \equal{} \frac {L \minus{} A \sin 0.5}{L \minus{} A \sin 1}$
which is a function only  of initial speed of the capacitor plate
[/hide]

Mobile la balance illa da.. Aathuku phone pannu. :rotfl:  :rotfl:

Obviously. Not good to post personal stuff like Phone Numbers and Addresses in Public forums. Someone I knew accidentally gave his mom's mobile no in Orkut (scrapbook) and his mother got loads of prank calls and some even insulted her. :(

Doubt: Mobilae balance ellana incoming varalamla?

i asked par to remove off dat post.
Oh balance first of all got over. + Validity of the talktime also got over.
So i ll recharge it today.

[quote="madness"]how come raghav's phone no. post disappeared? :huh: mod your work? :D[/quote]

And specially you can't have nuisance cases like you calling at 12:00 in the night and asking.

Nee Verum Raghav a ella Lord Labak Raghav a.

that wasnt my point. I knew that the mod had dunnit. But actually, when i was watching the indian thread, at one point, this was the last topic posted in and the next moment, i saw that some other topic became the last posted one. But when i saw that, it didnt seem to have any new post and this topic was where it was before it was posted in. :wacko: . That seemed strange, thats why i asked :)


It might have been a grid that generalized multiplication:

Say we want to multiply 1234 times 5678
We can write it like this:

       |1000 | 200 | 30 | 4
5000|
600 |
70   |
8    |

and use that to expand.

[quote="georgeforeman666"]I remember an extra credit opportunity my teacher provided us with. He never stated what it was but he took it off the website we use so I assumed that he no longer was offering the extra credit. Now, I remember that it was a method of multiplication which invloved a grid. I think this might be lattice multiplication but are there any other methods of multiplication which involve using grids but are not lattice multiplication. Much thanks.

"The Lord is gentle"
-Franz Zelda von Kafkavich[/quote]

Hmm I remember lattice multiplication...I was in 7th grade reading my sister's 3rd grade math book... :blush: . It was quite useful but now I forgot about it  :lol: .

[quote]http://video.yahoo.com/video/play?vid=cc2559cf688df2a77342796fbfd34466.1661607&cache=1

could that be it? it involves a grid.[/quote]

Yes! That's it. Thanks.  :D   :D   :D

[quote="georgeforeman666"]LOL! Why does this work you giys?  :blush:[/quote]

it's partial products in disguise, try to figure out why.

I have no idea what partial products is but when I actually did a problem using this, I figured it out. You're just splitting up the factors into pieces based on degree and then multiplying and then adding. Might as well just use conventional method instead of drawing out the grid. This is harder and takes longer.


trigonometry
function
algebra proposed
algebra

(1) Find all real solutions of the system
$x^{2}=y+2$
$y^{2}=z+2$
$z^{2}=x+2$

(2) Find all real solutions of the system
$2x^{2}=y+1$
$2y^{2}=z+1$
$2z^{2}=x+1$
 :)  :wink:

[hide="1"]
$x^{2}=y+2$
$y^{2}=z+2$
$z^{2}=x+2$

I will rewrite a solution that I liked very much. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=73090]here[/url] (the problem is almost the same)

Let $f(t)=t^{2}-2$

The problem asks to find the triplets $(x,y,z)$ such that $f(x)=y, f(y)=z, f(z)=x.$

In other words we have to solve the equation $f(f(f(x)))=x$.

From the given equations we can see that $x,y,z\geq 2$.

Also, $\forall t>2$ we get $f(t)>t$. This means that, if we start with an $x>2$ we will get $x<y<z<x$, condradiction.

So $x,y,z\in[-2,2]$

Set $x=2\cos\theta$.
Then $f(x) = (2\cos\theta)^{2}-2 = 2(2\cos^{2}\theta-1) = 2 \cos 2\theta$

So the function $f$ maps the point $2\cos\theta$ onto $2\cos 2\theta$

$z=f(2\cos 2\theta) = 2\cos 4\theta$
$x=f(2\cos 4\theta) = 2\cos 8\theta$

and we have the equation $2\cos8\theta = 2\cos \theta \Rightarrow$

$8\theta = 2k \pi+\theta \Rightarrow \theta = \frac{2k}{7}\pi$
$OR$
$8\theta = 2m \pi-\theta \Rightarrow \theta = \frac{2m}{9}\pi$

where $k,m$ are integer numbers

$*$ Notice that we have two sets of solutions, which are on the complex plane the vertices of a regular 7-gon and 9-gon respectively.
Accurately, the solutions are the real parts of the roots of the equations $z^{7}=1$ and $z^{9}=1$
Or simply the roots of the equation $(z^{7}-1)(z^{9}-1)=0$[/hide]

Similarly we solve the system (2)

[quote="pontios"]
Similarly we solve the system (2)[/quote] :wink: yeah, why bother going into details :lol:
and of course the same applies for the system ($a\in\mathbb R$)
$2x^{2}=ay+a^{2}$
$2y^{2}=az+a^{2}$
$2z^{2}=ax+a^{2}$
:)


number theory
relatively prime
number theory unsolved

Show that every subset of ten consecutive integers, exist one that is relatively prime to the others (each one)  :P

maybe you can help

It has post on PEN.
Solution is simple but very interest.


If $x,y,z$ are reals and $x+y+z=5$ and $xy+yz+zx=3$ prove that $-1\leq z\leq\frac{13}{3}$


invariant
induction
inequalities unsolved
inequalities

M' >= M.

A proof by induction.  Essentially, M and M' are invariant when you delete rows or columns that don't contain the "answer" (the specific square a_ij that contains the value M or M').  So to prove a larger board,  we find a row/column not being used and appeal to i,j.

(note: i think the induction works <the reasoning of it>, but I did not actually test that M' >= M for small cases because I was lazy.)


If A,B,C,D are arbitrary points, BC=a,CA=b,AB=c,AD=p,BD=q,CD=r

prove that

  (-ap+bq+cr)(ap-bq+cr)(ap+bq-cr) \geq 0

By the Generalized Ptolemy Theorem we have bq+cr>ap, ap+cr>bq, and ap+bq>cr.
Thus we have the result.


linear algebra
matrix
algebra
polynomial
geometry
geometric transformation
rotation

If $ A ,B \in M_{n}(\mathbb{R})$

and $ A^{T}A\equal{}I_{n}$

$ B^{T}B\equal{}I_{n}$

[color=red]show that:[/color]

if $ det(A)\plus{}det(B)\equal{}0$

then $ det(A\plus{}B)\equal{}0$

$ \det(A \plus{} B) \equal{} \det[A(I \plus{} A^TB)] \equal{} \det(A) \det(I \plus{} A^TB)$

$ \equal{} \minus{} \det(B) \det(I \plus{} B^TA) \equal{} \minus{} \det(B \plus{} BB^TA) \equal{} \minus{} \det(B \plus{} A)$

$ < \equal{} > \det(A \plus{} B) \equal{} \minus{} \det(A \plus{} B)$

$ < \equal{} > \det(A \plus{} B) \equal{} 0$,

since the determinant of the product is the product of the determinants, and the caracteristic polynomial of a matrix and its transpose coincide.

Less abstractly:

We want to show that the sum of a rotation and a reflection is singular. Multiplying by an arbitrary rotation, the rotation is WLOG equal to $ I$. $ I\plus{}B$ is singular if and only if $ B$ has $ \minus{}1$ as an eigenvalue. The eigenvalues of a real orthogonal matrix come in conjugate pairs with product $ 1$, except for some which are $ \pm 1$. For the product of all eigenvalues to be $ \minus{}1$, at least one of those eigenvalues must be $ \minus{}1$. The reflection $ B$ thus has an eigenvalue of $ \minus{}1$, and $ I\plus{}B$ is singular.


[quote="indianamath"]i have no clue who these people are.[/quote]
[url=http://www.Wikipedia.com]Search for them Here.[/url]

[quote="calc rulz"]What about Emil Artin?[/quote]
:wallbash:  I knew I forgot somebody! And I didnt mention Iwasawa.

Ludvig Sylow is one of the only two I known on there.  :oops:

everiste galois is the founder of group theory.  He is known for his work on quintic equations.

[quote="darij grinberg"]And where is Grothendieck...  [/quote]
 :wallbash_red:

I didn't spend so much time thinking about the result options when I posted this. But I still wouldn't have picked Grothendieck. (Galois is the master of them all).

By the way, have you ever learned Grothendieck's generalized approach to Galois theory?

[quote="sylow_theory"][quote="darij grinberg"]
By the way, have you ever learned Grothendieck's generalized approach to Galois theory?[/quote][/quote]

No

where could I learn it?


geometry
circumcircle
conics
geometry unsolved

For [tex]ABC[/tex] are fixed triangle and  one any [tex]XYZ[/tex] triangle. Show that: There exists a Ceva triangle [tex]A'B'C'[/tex] of [tex]ABC[/tex] which is similar to [tex]XYZ[/tex] . (Denote [tex]A'B'C'[/tex] is Ceva triangle of [tex]ABC[/tex] when [tex]AA',BB',CC'[/tex] are concurrent and [tex]A',B',C'[/tex] are in [tex]BC,CA,AB[/tex] respectively)

actually hard to understand
Would U explain it better

This is easy to understand if we use the Miquel's pivot theorem :
There exists an unique point P such as a triangle A'B'C' inscribed in ABC is directly similar to XYZ if and only if P is the common point of the circles AB'C', BC'A', CA'B'. (If we take the inverse of P in the circumcircle instead of P, we get the inscribed triangles inversely similar to XYZ)
Now, if A'B'C' is a variable inscribed triangle directly similar to XYZ, the locus of the common points of the lines AA' and BB'' is a conic through A and B and the locus of the common points of the lines AA' and CC' is a conic through A and C; these two  conics have at least one (and at most three) other real common point U (apart A); clearly, the cevian triangle of U is directly similar to XYZ.
Same way, we get generally at least one and at most three cevian triangles indirectly similar to XYZ.
Kind regards. Jean-Pierre


[quote="Phelpedo"]A 10kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15kg package on the ground.
(a) What is the magnitude of the least acceleration the monkey must have if it is to lift the package off the ground?
(b) If, after the package has been lifted, the monkey stops its climb and holds onto the rope, what are the magnitude and direction of the monkey's acceleration and (c) the tension in the rope?[/quote]
[hide]

For part a) you have to solve the system:

$T-10g = 10a$
$T-15g > 0$

For parts b) and c) you have to solve:

$T-10g = 10a$
$T-15g =-15a$

[/hide]


geometry
cyclic quadrilateral
geometry unsolved

ABCD is a cyclic quadrilateral inscribed in circle O. Line tangent to O at A, line tangent to O at C, and line BD meets at one point. Given that $ AB\equal{}24, BC\equal{}20, CD\equal{}15$, find the length of $ BD$.

ABCD is harmonical, hence AB.CD = BC.AD

Best regards,
sunken rock


linear algebra
matrix
vector
function
linear algebra unsolved

Can someone with Maple or Mathematica or something like that loaded on their computers calculate the determinant

$\begin{vmatrix}-1&b&c&d\\a&-1&c&d\\ a&b&-1&d\\ a&b&c&-1\end{vmatrix},$ please?

I keep getting different things when I do this determinant by hand. The latest is $(c+1)(d+1)(1-ab),$ but I don't trust that.

Mathematica tells me that $\left| \ldots \right| = 1-\sum ab-2 \sum abc-3abcd$, where the first sum is over all $\binom42$ combinations and the second sum is over all $\binom43$ combinations.

When I write "Together" of $3-\sum \frac1{a+1}$ I get $-\frac{\left| \ldots \right|}{\prod (1+a)}$, so the problem is solved (although using a computer).

EDIT: Of course, one can do these computations by hand, but I was too lazy. And Kent made a suggestion. Since I started that way, I might as well finish it in the same way.

Here's a proof without a computer.  Let $\alpha = \frac{a}{a+1}$, $\beta = \frac{b}{b+1}$, $\gamma = \frac{c}{c+1}$, and $\delta = \frac{d}{d+1}$.  Our goal is equivalent to proving that $\alpha+\beta+\gamma+\delta = 1$.

Let $B$ be the matrix \[\begin{pmatrix}\alpha & \beta & \gamma & \delta \\ \alpha & \beta & \gamma & \delta \\ \alpha & \beta & \gamma & \delta \\ \alpha & \beta & \gamma & \delta \end{pmatrix}\, .\] We have the following identity connecting the matrices $A$ and $B$: \[I-A = (I-B) \begin{pmatrix}a+1 & 0 & 0 & 0 \\ 0 & b+1 & 0 & 0 \\ 0 & 0 & c+1 & 0 \\ 0 & 0 & 0 & d+1 \end{pmatrix}\, .\] We are given that $I-A$ is singular.  Thus, by the identity above, $I-B$ is singular.  In other words, $B$ has an eigenvalue of 1.  But $B$ is a rank-1 matrix, so three of its eigenvalues must be 0; the remaining eigenvalue must be its trace, namely $\alpha+\beta+\gamma+\delta$.  Thus $\alpha+\beta+\gamma+\delta = 1$, which proves our claim.

Thanks.  Also, I noticed a more pedestrian argument that works.  Namely, we can use the given vector $X$.  If we write out the matrix equation $AX = X$ as a system of scalar equations, we get equations such as \[b x_{2}+c x_{3}+d x_{4}= x_{1}\, .\] Adding $ax_{1}$ to both sides gives \[a x_{1}+b x_{2}+c x_{3}+d x_{4}= (a+1) x_{1}\, .\] We get similar equations for the other variables, so that \[(a+1) x_{1}= (b+1) x_{2}= (c+1) x_{3}= (d+1) x_{4}\, .\] In other words, for some nonzero constant $t$, we have
$x_{1}= \frac{t}{a+1}$, $x_{2}= \frac{t}{b+1}$, $x_{3}= \frac{t}{c+1}$, and $x_{4}= \frac{t}{d+1}$.  If we plug those values back into the second equation above, we get \[\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}= 1 \, .\]

[quote="Ravi B"]...In other words...ee get \[\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}= 1 \, .\] [/quote]
Very nice/proposer

Thanks.  Here's yet another, purely determinantal, approach.

Let $X$ and $Y$ be two square matrices of order $n$.  Let $I$ and $J$ be a partition of $\{1, 2, \ldots, n \}$.  Let $M(X, Y, I, J)$ be the square matrix of order $n$ whose rows indexed by $I$ match those of $X$ and whose rows indexed by $J$ match those of $Y$.  The following lemma shows how to evaluate the determinant of a sum.

[b]Lemma:[/b]  The determinant of $X+Y$ is equal to the sum of the determinants of $M(X, Y, I, J)$ over all $2^{n}$ partitions $(I, J)$.

(Proof: Determinant is a linear function of each row.)

Now let $X$ be the matrix \[\begin{pmatrix}1+a & 0 & 0 & 0 \\ 0 & 1+b & 0 & 0 \\ 0 & 0 & 1+c & 0 \\ 0 & 0 & 0 & 1+d \end{pmatrix}\, .\] Let $Y$ be the matrix \[\begin{pmatrix}-a &-b &-c &-d \\-a &-b &-c &-d \\-a &-b &-c &-d \\-a &-b &-c &-d \end{pmatrix}\, .\] Then $I-A = X+Y$.  Let's apply the lemma.  If $|J| \ge 2$, then the determinant of $M(X, Y, I, J)$ will be $0$, because $M(X, Y, I, J)$ will have at least two equal rows.  On the other hand, the cases $|J| = 0$ and $|J| = 1$ are easy to analyze.

[quote="Ravi B"]..... Here's yet another, purely determinantal, approach.
[/quote]
Thanks for these very nice solutions !  I ask you very much to send me a private message with your name. Thanking,in advance,Alex


Points $ R, S$ and $ T$ are vertices of an equilateral triangle, and
points $ X, Y$ and $ Z$ are midpoints of its sides.  How many noncongruent
triangles can be drawn using any three of these six points as vertices?
[asy]size(1.5inch,1.5inch);
pen p = white;
pair r=(0,0), x=(1,0), t=(2,0), y=(0.5,1), z=(1.5,1), s=(1,2);
dot(r); dot(x); dot(t); dot(y); dot(z); dot(s);
label(scale(0.8)*"$R$", r, N);
label(scale(0.8)*"$X$", x, NW);
label(scale(0.8)*"$T$", t, NW);
label(scale(0.8)*"$Y$", y, NW);
label(scale(0.8)*"$Z$", z, NW);
label(scale(0.8)*"$S$", s, NW);
dot((0,-0.1), p);[/asy]


$ x,y,z$ are three real positive numbers such that the sum of two of them is greater than the third one . prove:

$ \frac{x^{\minus{}1}\plus{}y^{\minus{}1}\plus{}z^{\minus{}1} }{\sqrt{(x\plus{}y\plus{}z)(x\plus{}y\minus{}z)(x\plus{}z\minus{}y)(y\plus{}z\minus{}x)}} \ge \frac{\sqrt{3}}{xyz}$

[quote="Kenny O"]$ x,y,z$ are three real positive numbers such that the sum of two of them is greater than the third one . prove:

$ \frac {x^{ \minus{} 1} \plus{} y^{ \minus{} 1} \plus{} z^{ \minus{} 1} }{\sqrt {(x \plus{} y \plus{} z)(x \plus{} y \minus{} z)(x \plus{} z \minus{} y)(y \plus{} z \minus{} x)}} \ge \frac {\sqrt {3}}{xyz}$[/quote]

Rewritre the inequality as

$ (abc)^2\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \ge 3(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$

Form schur inequality :

$ abc \ge (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$

and From the lemma : $ (a \plus{} b \plus{} c)^2 \ge 3(ab \plus{} bc \plus{} ca)$

we have

$ abc\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \ge 3abc\left(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}\right) \equal{} 3(a \plus{} b \plus{} c)$

And the conclusion follows  :)


inequalities
trigonometry
inequalities unsolved

Let be given $n+1$ real numbers $a_0,a_1,a_2,...,a_n$ satisfying:$a_1 , a_2 ,...,a_n>0,\sum_{i=1}^n a_i=1$
Prove that:
$\sum_{i=1}^n\frac{a_i}{{\sqrt{1++a_0+a_1+a_2+...+a_{i-1}}}{\sqrt{a_i+a_{i+1}+...+a_n}}}\le \frac{\pi}{2}$
(we have $a_0$ so that $%Error. "exis" is a bad command.
a_{i-1}$ when $i=1$)

There are two difficult steps in the solution:
[hide="1"]Let $\sin\alpha_i=a_0+a_1+...+a_i\forall i=0,1,2,...n$[/hide]
[hide="2"]For you[/hide]


function
number theory theorems
number theory

Define the arithmetic function $ps$ as follows: $ps(n)=1$ iff $n$ is a perfect square; $ps(n)=0$ otherwise. Prove (without using Mobius inversion) that $\mu*ps=\lambda$ (the Liouville function).

Any natural number can be written as a product of a square-free natural number and a square in one and only one way. Thus if $n=mk^{2}$, then

$(\mu*ps)(n)=0+...+0+\mu(m)ps(k^{2})+0+...+0=\mu(m)$

It is also easy to show that the Liouville function $\lambda$ is completely multiplicative; and if $m$ is square-free, then we have $\mu(m)=\lambda(m)$.

Thus we have

$(\mu*ps)(n)=\mu(m)=\lambda(m).1=\lambda(m).\lambda(k^{2})=\lambda(mk^{2})=\lambda(n)$ for any natural number $n$.

Q.E.D.


induction
number theory unsolved
number theory

I hve a new problem :
      Let $ P(x) = 2x^4 + 2x^3 + 3x^2 - 2x + 1 $
    Find $ n\in N $ such that $ \exists a : P(a)\equiv 0 (mod n)  $
   If somebody knows the solution of this above problem , pls post it for everyone to see .

The first problem:Proof by induction on $n\geq 5$
For n=5 ,it's easy to verify that: $f(m)\equiv 0(mod243)$ if and only if $m\equiv 19(mod27)$.We are done in this case.
 Assume that it holds for some $n\geq 5$.Suppose $P(d)\equiv P(b)(mod3^{n+3})$.
We have:
$d\equiv b(mod 3^n)\rightarrow b=d+h3^n\rightarrow 0\equiv P(d+h3^n)\equiv P(d)+h3^nP'(d)(mod3^{n+3})$(1)
Because:
 $d\equiv 1(mod9)\rightarrow P'(d)$ is divisible by 9 but is not divisible 27 $\rightarrow 3^nP'(d)$ is divisible by $3^{n+2}$ but is not divisible $3^{n+3}$(2)
Using (1) and (2) we dedu ce that : $h=3t\rightarrow b=d+3t3^n=d+3^{n+1}t\rightarrow b\equiv d(mod3^{n+1})$
Thus, we are done $\forall n\geq 3$.(n=3;n=4 is easy)


I saw this in one of the investigations of high school kids here in our place.  This is the rule in a certain sequence:

1.) Take any two digit number.
2.) To get the next term, add the ten's digit to twice the unit's digit. (Eg. The first term is 15, so the second term is 11, 3, 6, 12,... ) If no ten's digit, place 0.
3.) Do number step 2 repeatedly to get the following terms.

Now after several explorations, the sequence seems to go back to the first term and repeat again.

Any proof? Is there really a sequence like this? Haven't encountered this one.

Thanks

[quote="t0rajir0u"]Every term past the first can be at most $9+2 \cdot 9 = 27$, so there are only a finite number of possibilities for each term, which means that it is guaranteed to repeat after at most $27$ steps.[/quote]

A more interesting question: what number creates the longest sequence before a term is repeated?

[quote="t0rajir0u"]Every term past the first can be at most $9+2 \cdot 9 = 27$, so there are only a finite number of possibilities for each term, which means that it is guaranteed to repeat after at most $27$ steps.[/quote]

Can you please explain, i don't get what you mean.

I mean, consider what would happen if the sequence [i]didn't[/i] repeat.  That means each term would be different from all the other terms that came before it (because the moment you have the same term again, the sequence starts repeating).  But what happens when you go through $27$ terms (after the first)?

You can only have, at best, the numbers $1, 2, 3, ... 27$ as those first $27$ terms.  The $28^{th}$ term cannot be larger than $27$ (the two digits of a two-digit number $ab$ can be at most $9$, so $a+2b$ can be at most $27$), so it has to be one of the other numbers - but that means that we will have to repeat a term.

[quote="guile"]Now after several explorations, the sequence seems to go back to [i]the first term[/i] and repeat again.[/quote]

This is inaccurate.  The sequence does repeat eventually, but if you start the sequence with a number larger than $27$, you will not be able to return to it.


Prove that the real numbers $x_1,x_2,...,x_n$ exist such that at least $n-1$ of them are irrational (for every natural number like $n$ which is bigger than one)and:

$0<\frac{\sqrt{2}}{2}(x_1+x_2+x_3+...+x_n-n)<2\sqrt{n}-1$

1. I don't understand the question.
2. you double posted it here: http://www.mathlinks.ro/Forum/viewtopic.php?t=64499
3. at the moment I don't see the combinatorics ;)


superior algebra
superior algebra unsolved

what's the difference between centralizer and normalizer?  I looked at the definitions and they looked the same.  I must have missed something..

Have a closer look, they're not the same.

When posting a vague question like this you will get more detailed replies if you say exactly what definitions you are using and say a little more about the nature of your confusion beyond "I don't get it."

The centralizer fixes each element. The normalizer fixes the set, but can permute elements within it. The centralizer and normalizer of a 1-element set are equal, while the normalizer is usually larger for a larger set.

For a group $ G$, the centralizer of $ G$ is its center, and the normalizer is $ G$.


[quote="orl"][quote="darij grinberg"]BTW, Orl, have you already noted that you are behind me in the members ranking by post numbers? So you'll have to post more problems  :D  :D

  Darij[/quote]

Hi Darij,

funny as always.  :D  Actually it is not my goal to be the member with the highest number of postings. And have you already noted that I have same number of postings as you right now. Did you also realize that you are still 24th in this ranking ?  :D  A tactic might be: Do NOT post as Isotomion and you would be almost number 20.  :) Regarding problems I can post some more.  :)[/quote]

post count can take weird leaps. I just topped both of you :D

[quote="orl"]funny as always.  :D  Actually it is not my goal to be the member with the highest number of postings.[/quote]

Strangely enough, this is not my goal, too  :D  Else, I would never post under two different names...

I just wanted some more problems, and I thought this could be a nice provocation...

[quote="orl"]Regarding problems I can post some more.  :)[/quote]

Thanks!

  darij

19 year bump
i don't post that much on official forums, so i have 856 posts

Wow mad props to @above how'd you manage to dig this up???

[quote name="K124659" url="/community/p29828533"]
He probs spent 20 minutes scrolling to the bottom of fun factory lol
[/quote]
no why would i do that
i just used search


Let $n$ be a positive integer. Show that there exist three distinct integers
between $n^{2}$ and $n^{2}+n+3\sqrt{n}$, such that one of them divides the product of the other two.


let a, b, c be integers such that a+b+c=0, how to prove that 2a^4+2b^4+2c^4 is a perfect square?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group with $ 520$ elements. Proove that there exists a non-trivial normal subgroup of $ G$.

$ 520\equal{}2^3\cdot5\cdot13$, and thus the only possibilities for the number of Sylow $ 13$-groups are $ 1$ and $ 40$.  If there are $ 40$, that accounts for $ 12\cdot40\equal{}480$ elements of the group.  The number of Sylow $ 5$-groups is either $ 1$ or $ 26$.  But then there are $ 4\cdot26>40$ elements of order $ 5$, a contradiction.  So either the Sylow $ 13$-group or Sylow $ 5$-group is normal.

Steve


I am interested in taking the Intro Independent Study, however I do not have enough money to take it (my parents make me pay for everything  :mad: ).

What I want to know is if there is any possible way for me to get a more cost-effective way to do a self study course, such as an outline of things to read and problems to solve in say, AOPS volume 1 or Mathematical Circles.

If it helps any, I am preparing for MATHCOUNTS. I made nationals last year, but am not good enough to do very well.

Thanks in advance.

Start from the beginning in AoPS Vol 1.  You'll whiz through the early chapters.  Don't worry about Mathematical Circles until you are finished Vol 1.  (Though it would be a fine thing to pick up and play with every once in a while during your study of Volume 1.)

There are also some free resources on this site.  In the contests area, most of the problems are well above MATHCOUNTS level, but many problems on the first halves of AMC/AHSME exams make good study for a student who has been to nationals in MATHCOUNTS.  Since each posted problem includes a discussion of a solution, there is plenty to learn from and you can always ask questions in the problem threads.

Ok thanks guys  :)
You never know, I may be able to convince my parents. :P


Suppose tan A = 2,
would A be rational or irrational?

(In other words, for $A$ measured in radians, the question is whether $\frac{A}{\pi}$ is rational.)

This is really number theory, not analysis. Here's one way to approach it: Consider the complex number $z=\frac{1+2i}{\sqrt{5}}.$ Does there exist a natural number $n$ such that $\text{Im}(z^{n})=0?$


hello everybody :D
I doubt anyone can answer this exactly, but how have you all done?  :)

i did ok only.. first paper, not much time for chem.

IMHO, second paper was easier than the first.. especially chem..

PS: pardesi, if this thread should rather be in the 'General Discussion' sub-forum, pls move it :)

I ve already started a thread there.  :D

I think this shd be here only :) As for my performance, is decent though i could've done better :)

tougher........easy are relative terms so, let's speak a little statistically.....

considering last year's highest was 418/486....wat do u think it wud be for this year?????????

im betting it'll be around the same. though the paper was slightly tougher than last yrs, the standard of the students also increases every yr..  :|

wtf. if that is the highest, ill have to kiss my chances of getting in goodbye. the chem paper was absolute sh** but math was screwing. :o pardon my bad language but im really pissed coz i prepared more for chem than for math and phy.
im sure nobody has crossed 400.

[quote="anandghegde"][quote="hash_include"]im betting it'll be around the same. though the paper was slightly tougher than last yrs, [b]the standard of the students also increases every yr[/b]..  :|[/quote]

how?? :|[/quote]

It's a general trend. Not some thing which HAS to be true or anything. maybe the difference won't be seen in consecutive years, but over a long time, it sure as hell shows.

and besides, competition increases every year (at least till now) and so, obviously, the quality of students should too.  :roll:

wat do you think will be the overall cutoff?

I did decently good..... But I am so glad that its finally over!!!! Some questions were way too easy, and some werent. But it was a decent paper overall. Neither too easy, nor too difficult...

I think over 500 people will get above 300 and another 500 will get from 270-300 :)

[hide][img]http://www.fiitjee.com/test/images/JEE-2008-Solutions-04.gif[/img]. Just saw this in FIITJEE website. Maybe I shouldn't sleep. :D:D But I think I will. tired.
Knowing FIITJEE's solutions, it will only depress people with the patheticness.
[/hide]

I think it was a bit (quite) tougher than JEE 2007. I believe the top mark was near 400, last year. Should come down to 375 odd this year. The standard was quite close to FT11. Infact it was only 25-35 marks easier..

I believe that 250-270 odd should be the cutoff for 1000 rank. Might be higher, because FIITJEE is a very unreliable tool. I didn't find mathematics too difficult, nor did I find chemistry too easy.  :rotfl:  :rotfl:  But thats just me.

I too did second paper really well. Good split up. First paper, I wasn't able to finish, especially math comprehensions. :( But physics more than took care of that.

I think I could have got 20-30 marks more very easily, but no use whining now. Lets all hope for the best. Anybody getting 300+???

[quote="hash_include"]http://www.prernaclasses.com/IITJEE2008Solutions.htm
!!
looks like fiitjee's second best.. again  :P[/quote]

Wow. Awesome~! But I'm scared to check it out.

[quote="VARUNARASU"][quote="anitsahu"]i dont think the crowding will start so early
or else cut off would be somewhere around 260
the pouplated zone this time is 230-250
which was 270-290 last time
where students get 1500-2500 ranks
this time i feel 270 onwards one mark may dash your hopes
u never know may be this time lasy years 270-290 becomes 300+
and we end up getting 3k,4k ranks :o[/quote]
anit, u r a big hypocrite da. phoenix(wherever he is rite now) was rite. i know u r expecting nothing less than 350.
& madness is gonna be AIR 1. :D[/quote]

@ANIT 230 250 may be d peak of crowding da but 290's d startin pt and dat's wat matters ppl in aops i find none claimin here a below 250 marks (maybe wit my exception)
also it sounds really strange but u know wen u get activ in aops phoenix gets silent and vice versa however u too must be communicatin  wit each oder as he knew your score details of evry fiitjee test    :!:

note : ankur gupta claims to know dis yr topper from bhansal institute scorin 450  :D  think he isn't aware of anit  :P

what lies varun and anand :o
No chance, ill get around 2000 :(

And ofc anand, i said ill be happy if i get below 1000, that hasn't happened has it? Im not happy am I? :(

zeroes r insignificant!!!! :D  :D  :rotfl:

so many of u feel me to be phoenix
well hes also appearing for pmt
he has qualified pmt prelims
hes appearing nearly 15 odd entrances
so u can understand why he doesnt come here
u see he just lives pretty near to me
as well as knows my fiitjee enrollment no and etc
he loves to compare others with everybody :mad:  :mad:
if the topper has checked his solutions from bansal s solutions i would suggest him to deduct 40 odd marks from it
because bansals solution was most hillarious
i couldnt help myself from laughing some reaLLy stupid answers :rotfl:  :rotfl:

naa, phoenix cant be u. i know u wud never be online days b4 jee, but boy, was he angry with u or what? :rotfl:

thats pretty sure anit cant be online days b4 jee and yet score so brilliantly :lol:
regardin phoenix its sad to know he was makin illegitimate use of others reg.no  :huh: neways wat happened to those twins dude, phoenix was a gr8 fan of theirs  :rotfl:  :rotfl:

by the way has anyone recieved any response from fiitjee regardin their projected rank i didn't but a few in our region have so...

FIITJEE called a person (Chappli) here to ask for Roll No. When he told he's getting 316, they said he would get from 200 - 500 rank.

is that they who check the papers :rotfl:  :rotfl:
over 300 shud b <200   ~just guessing~

@madness--2222????????that confirms AIR 2 'cause u cud've written 2398 OR...... :rotfl:

[quote="achu_182"]anyone writing VITEEE tomorrow?? (i dont know if this shd be in this thread)

if yes, have u seen the semiconductors syllabus of VITEEE!?!?[/quote]

You are writing VIT? Wow! I thought you were pretty confident of getting through the top 3. ;)

those who gave VIT said that the paper was cool!!!!!!!is that so????????? :oops:

ya it was quite easy. but i m getting only 96-99/120 (i have a doubt in a few questions in the solutions)

i was told that people easily get 100.

my guessing theory works. the probability of a answer being C is very high, when u dont know it!!


superior algebra
superior algebra solved

I am sorry, I am not sure this is the right place to post that :
let $f: E\rightarrow F$ be an application
show that : $\left( \forall y\in F\right) \left( \exists !x\in E\right) : y=f(x)\Leftrightarrow \left( \forall X\sqsubset E\right) : f\left( \overline{X}\right) =\overline{f\left( X\right) }$

Both sides say $f$ is bijective(or, in saxon terminology one-to-one and onto).

yeah !!!!!
but how to show that :  Right Side $\Rightarrow$ Left Side ??   :?

ok, maybe you didn't understand my question,
How to show that : $( \forall X\sqsubset E) : f( \overline{X}) =\overline{f( X) }$ $\Rightarrow$ $f$ is bijective

Consider $X=\phi$, f is surjective.
Suppose $f(x)=f(y)=z$ for distinct elements $x,y \in E$.
Take $X=E-\{x\}$, then f(x) has no image. Contradiction.


inequalities
function
induction
algebra proposed
algebra

Find all functions $f,$ defined on the positive real numbers and taking real numbers such that
\[f(x)+f(y)\leq \frac{f(x+y)}{2},\ \ \frac{f(x)}{x}+\frac{f(y)}{y}\geq \frac{f(x+y)}{x+y}\] for all $x,\ y>0.$

Let $a_{n}=f(nx)$. We have (y=x) $a_{2}\ge 2a_{1}$ and $a_{2}\le 2a_{1}$, therefore $a_{2}=2a_{1}$. By induction (y=(n-1)x) $a_{n}=na_{1}$. Therefore $f(\frac{x}{n})=\frac{1}{n}f(x)$. It give $f(rx)=rf(x) \ \forall r\in Q_{+}$.
If f(x) is continiosly, then $f(x)=ax$.
It is true for all functions Let $f(x)\not =0, a=\frac{y}{x}$. Let $\frac{m_{k}}{n_{k}}\to a$.
We have ${2f(x)+2\frac{n_{k}}{m_{k}}f(y)\le \frac{1}{m_{k}}f(m_{k}+n_{k}y}\le f(x)(1+\frac{n_{k}y}{m_{k}y})+f(y)(\frac{x}{y}+\frac{n_{k}y}{m_{k}x}.$
It give $f(x)=const x, \ \forall x$.

[quote="Rust"].
We have ${2f(x)+2\frac{n_{k}}{m_{k}}f(y)\le \frac{1}{m_{k}}f(m_{k}+n_{k}y}\le f(x)(1+\frac{n_{k}y}{m_{k}y})+f(y)(\frac{x}{y}+\frac{n_{k}y}{m_{k}x}.$
It give $f(x)=const x, \ \forall x$.[/quote]

Why do you deduce this agrement?

(*) $f(x)+f(y)\leq  \frac{f(x+y)}{2}$
(**) $\frac{f(x)}{x}+\frac{f(y)}{y}\geq  \frac{f(x+y)}{x+y}$
Then for $g(x)={f(x)\over x},\ x\in\mathbb{R}_{+}$ we have
(***) $g(x)+g(y)\geq g(x+y)$ for all $x,y>0$.

1) $g(nx)\leq ng(x)$ and $f(nx)\leq n^{2}f(x)$ for all $x\in\mathbb{R}_{+}$ and $n\in\mathbb{N}$:
This follows easily from (***).

2) $f(2^{n}x)=4^{n}f(x)$ and $g(2^{n}x)=2^{n}g(x)$ for all $x\in\mathbb{R}_{+}$ and $n\in\mathbb{N}$:
From (*) follows $f(x)+f(x)\leq  \frac{f(2x)}{2}$ and $f(2x)\geq 4f(x)$.
Then 1) gives $f(2x)=4f(x)$, and the rest follows by induction on $n$.

3) $g(nx)=ng(x)$ for all $x\in\mathbb{R}_{+}$ and $n\in\mathbb{N}$:
Otherwise $g(nx)<ng(x)$ by 1), and for $r\in\mathbb{N}$ with $2^{r}>n$ we get
$2^{r}g(x)=g(2^{r}x)\leq g(nx)+g((2^{r}-n)x)<ng(x)+(2^{r}-n)g(x)=2^{r}g(x)$
using (***) and 1) and 2). Contradiction!

4) $f(x)\leq 0$ and $g(x)\leq 0$ for all $x\in\mathbb{R}_{+}$:
From (*), 1) and 2) folllows
$10f(x)=2(f(x)+f(2x))\leq f(3x)\leq 9f(x)$ and $f(x)\leq 0$ and $g(x)\leq 0$.

5) $g$ is monotonically decreasing:
From (***) and 4) follows $g(x)\geq g(x)+g(y)\geq g(x+y)$ for all $x,y>0$.

And now for $a: =g(1)=f(1)\leq 0$ we get
6) $g(x)=ax$ and $f(x)=ax^{2}$ for all $x\in\mathbb{R}_{+}$:
From 3) follows $g(r)=rg(1)=ar$ for all $r\in\mathbb{Q}_{+}$ and 5) gives the rest.

Conversely $f(x)=ax^{2}\ \ \forall x\in\mathbb{R}_{+}$ is a solution for all $a\in\mathbb{R},a\leq 0$,
as $x^{2}+y^{2}={(x+y)^{2}\over 2}+{(x-y)^{2}\over 2}\geq{(x+y)^{2}\over 2}$ for all $x,y\in\mathbb{R}$.

Congralution [b]Olorin[/b]. Thanks for your nice solution, honestly!

Is my answer correct?

Substitute $y \rightarrow 1$, we have:
$\frac{f(x)}{x}+f(1) \geq \frac{f(x+1)}{x+1} \geq \frac{2f(x)+2}{x+1} \Leftrightarrow (x-1)(xf(1)-f(x)) \geq 0$
We choose $x>1$, hence we have $f(x) \leq kx, k=f(1)$
$\Rightarrow f(x)+f(y) \leq \frac{f(x+y)}{2} \leq \frac{k(x+y)}{2}$
Substitute $y \rightarrow x$, we have $f(x) \leq \frac{kx}{2}$
By induction, we have $f(x) \leq \frac{kx}{2^n}$ for arbitrary large $n$, so $f(x) \rightarrow 0 \Rightarrow f(x)=0 \forall x>1$
For $x \leq 1$, we have: $2f(x)=f(2x) \Rightarrow f(2^{n}x)=2^nf(x)$, and because there exists $n$ such that $2^{n}x > 1$, we have $f(x)=0$ for every $x<1$.
Hence, $f(x)=0 \forall x$

@Nguyenhuyhoang
Your solution is incorrect, since there exist non-zero functions.
You forgot, that $f(x)$ can be $<0$ and thus $f(x)\rightarrow 0$ may not be true.

Rust's solution is incorrect as well, since $f(nx)=nf(x)$ cannot be proved so easily.

See also my solution here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3380139#p3380139

could you please clarify the sixth part of ur answer ?


trigonometry
inequalities proposed
inequalities

Let $ x,\ y$ be postive real numbers such that $ (x^2\plus{}y^2)^2\equal{}x^2\minus{}y^2$. Find the maximum value of $ x\plus{}y.$

[quote="kunny"]Let $ x,\ y$ be postive real numbers such that $ (x^2 \plus{} y^2)^2 \equal{} x^2 \minus{} y^2$. Find the maximum value of $ x \plus{} y.$[/quote]
It's very known problem.
My ugly solution:
let $ x\equal{}r\cos\alpha$ and $ y\equal{}r\sin\alpha,$ where $ 0<\alpha<\frac{\pi}{4}.$
Hence, $ r\equal{}\sqrt{cos2\alpha}$ and $ x\plus{}y\equal{}f(\alpha),$ where $ f(\alpha)\equal{}\sqrt{\cos2\alpha}(\sin\alpha\plus{}\cos\alpha)\leq f\left(\frac{\pi}{12}\right).$

[hide]
Let $ p\equal{}x\plus{}y,q\equal{}x\minus{}y$. Then $ p$ positive and $ (p^2\plus{}q^2)^2\equal{}4pq$. This means $ q$ positive as well.

But AM-GM gives
\[ \left( 3\cdot \frac{p^2}{3}\plus{}q^2\right)^2 \geq \frac{16}{\sqrt{27}}p^3q\]
That gives $ p^2\leq \frac{\sqrt{27}}{4}$, or $ p\leq \frac{\sqrt[4]{27}}{2}$.

There should be equality for $ p\equal{}\frac{\sqrt[4]{27}}{2}$ and $ q\equal{}\frac{\sqrt[4]{3}}{2}$. That gives
\[ x\equal{}\frac{\sqrt[4]{3}}{4}\left( \sqrt{3}\plus{}1\right) ,y\equal{}\frac{\sqrt[4]{3}}{4} \left( \sqrt{3}\minus{}1\right)\]
so that $ x^2\equal{}\frac{\sqrt{3}}{8}\left( 2\plus{}\sqrt{3}\right) ,y^2\equal{}\frac{\sqrt{3}}{8}\left( 2\minus{}\sqrt{3}\right)$ which works, making $ \boxed{\frac{\sqrt[4]{27}}{2}}$ the maximum.
[/hide]

[quote="arqady"][quote="kunny"]Let $ x,\ y$ be postive real numbers such that $ (x^2 \plus{} y^2)^2 \equal{} x^2 \minus{} y^2$. Find the maximum value of $ x \plus{} y.$[/quote]
It's very known problem.
My ugly solution:
let $ x \equal{} r\cos\alpha$ and $ y \equal{} r\sin\alpha,$ where $ 0 < \alpha < \frac {\pi}{4}.$
Hence, $ r \equal{} \sqrt {cos2\alpha}$ and $ x \plus{} y \equal{} f(\alpha),$ where $ f(\alpha) \equal{} \sqrt {\cos2\alpha}(\sin\alpha \plus{} \cos\alpha)\leq f\left(\frac {\pi}{12}\right).$[/quote]

$ f'(\alpha) \equal{} \frac {\cos 3\alpha \minus{} \sin 3\alpha}{\sqrt {\cos 2\alpha}}\ \left(0 < \alpha < \frac {\pi}{4}\right)$ gives $ f(\alpha)\leq f\left(\frac {\pi}{12}\right) \equal{} \sqrt {\frac {\sqrt {3}}{2}}\cdot \sqrt {2}\sin \left(\frac {\pi}{12} \plus{} \frac {\pi}{4}\right) \equal{} \frac {\sqrt [4]{27}}{4}$


This question may seem simple, but i have never seen sucha question out of all the database questions i have done!
Question: Mane and Janani are playing a coin toss game with a fair penny. Manu gets a point if the penny lands on heads, and Janani gets a pointif the penny lands on tails.The score is Janani 9, Manu 7, in a game to 10 points. What is the probabilty that Janani will win hte game?
Please help, i would apprecieate it.

basically, the only way janani can lose is if manu gets three straight heads. the chance of this is $ \frac12\cdot\frac12\cdot\frac12=\frac18$. thus, the probability that janani wins is $ 1$ minus this, or ${ 1-\frac18=\boxed{\frac78}}$.

Or, you could think of it like this:

FIRST TOSS
There is a $ \frac{1}{2}$ chance of flipping a tails and Janani winning.
There is a $ \frac{1}{2}$ chance of flipping a heads and the game continuing to toss 2.

SECOND TOSS
There is a $ \frac{1}{2}$ chance of flipping a tails and Janani winning.
There is a $ \frac{1}{2}$ chance of flipping a heads and the game continuing to toss 3.

THIRD TOSS
There is a $ \frac{1}{2}$ chance of flipping a tails and Janani winning.
There is a $ \frac{1}{2}$ chance of flipping a heads and Manu winning.

So the chance of Janani winning is:
$ \frac{1}{2}\plus{}\frac{1}{2}\cdot\frac{1}{2}\plus{}\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\equal{}\frac{1}{2}\plus{}\frac{1}{4}\plus{}\frac{1}{8}\equal{}\frac{4}{8}\plus{}\frac{2}{8}\plus{}\frac{1}{8}\equal{}\boxed{\frac{7}{8}}$


function
real analysis
real analysis unsolved

Let $X$ and $Y$ be metric spaces, with $Y$ complete, let $W$ be a dense subset of $X$ and let $f: W\rightarrow Y$ be a Lipschitz-continuous function.  Prove that $f$ extends to a continuous function $f_{X}: X\rightarrow Y$.  (Prove that there is a continuous function $f_{X}: X\rightarrow Y$ such that $f_{X}\left(w\right)=f\left(w\right)$ for $w\in W$.)

Define $f_{X}(x)=y$ where $\{y\}=\bigcap_{n=1}^{\infty}\overline{f(W\cap B(x,1/n))}$. (The intersection of shrinking closed sets in a complete space consists of precisely one point.) This $f_{X}$ is actually Lipschitz, not just continuous.


there are 3 altitudes in a triangle.... (obviously)

if the lengths of two of the altitudes are $h$ and $k$ with $h$ does not equal $k$, find the MAXIMUM and MINIMUM length of the thrid altitude, $p$, in terms of $h$ and $k$...

hint: use the traingle inequality, if a,b,c are sides of a tirangle, a < b+c

Don't make double posts! :o http://www.mathlinks.ro/Forum/viewtopic.php?highlight=there+are+3+altitudes+in+a+triangle&t=32438


Find the perimeter of this figure, which consists of six consecutive squares and has an area of 294 square units.

[asy]size(3cm,3cm);
pair a,b,c,d,e,f,g,h,i,j,k;

a=(0,3);
b=(0,2);
c=(1,1);
d=(2,1);
e=(3,0);
f=(4,0);
g=(4,1);
h=(4,2);
i=(3,2);
j=(2,3);
k=(1,3);

draw(a--b--h);
draw(c--d--g);
draw(e--f--g--h);
draw(e--i);
draw(a--j);
draw(k--c);
draw(j--d);[/asy]

The area of one square is $ \frac {294}{6} \equal{} 49$, so each unit side length is $ 7$.
The perimeter of the shape consists of $ 14$ lengths, so our answer is $ 14 \cdot 7 \equal{} \boxed{98}$.


Consider pairs of sequences of positive real numbers:
$a_1 \geq a_2 \geq a_3 \geq ... $ ,  $b_1 \geq b_2 \geq b_3 \geq ... $
and the sums:
$A_n=a_1+a_2+...+a_n$ , $B_n=b_1+b_2+...+b_n$ , $n=1,2,...$
For any pair define $c_i=min(a_i,b_i)$ and $C_n=c_1+c_2+...+c_n$ , $n=1,2,...$
(1) Does there exist a pair  $(a_i)_{i\geq1}$,  $(b_i)_{i\geq1}$ such that the sequences $(A_n)$,  $(B_n)$ are unbounded while  the sequence $(C_n)$ is bounded?
(2) Does the answer to part (1) change by considering $b_i=\frac{1}{i}$ , $i=1,2,...$
Justify your answer.

Just additional well known problem, which is equivalent to (1) (I think).

Let $\{a_n\}_1^{+\infty}$, $\{b_n\}_1^{+\infty}$ be increasing sequences of positive numbers. It is known that series $\sum_{i=1}^{+\infty}\frac{1}{a_i}$ and $\sum_{i=1}^{+\infty}\frac{1}{b_i}$ disverge. Is it true that $\sum_{i=1}^{+\infty}\frac{1}{a_i+b_i}$ disverges?

[quote="grobber"]I remember it being posted and I think Darij solved it (?).[/quote]

It's the second time I'm hearing that I solved the problem, but this doesn't imply that I did.  :D  It was solved by Jmerry:

http://mathlinks.ro/Forum/viewtopic.php?p=100351

  Darij


complex analysis
complex analysis unsolved

let f be continuous in $ \mathbb{C}$ and holomorphic outside the rays $ {re^{\frac{i\pi}{n}}}$  where n=1,2...

show that f is actually entire

i have done a similar problem where just a line is excluded and in that case i used morera's theorem and continuity. i wonder if the same will work here.


Find the shortest distance between $y^2=x$ and $(x-3)^2+y^2=1$.

This isn't hard, just use a bit of calculus.

First, we begin by using the distance formula, then replace the x is the square root with y^2.

After that, take the derivative, and take the second derivative to check if it's a max or a min. Once you're done, plug the number back into the original distance formula, and voila.


When adding 4 consecutive numbers, they will always equal an even number.  Is this statement true and explain your answer? I understand that the answer is true but why?

Let the first number be x.
Then x+(x+1)+(x+2)+(x+3)=4x+6=2(2x+3).
So it's divisible by 2.


[quote="randomdragoon"]Four twenty times five, and nine, is thirty. True or false?

A man drives up to a hotel and immediately knows that he is bankrupt. How?[/quote]
[hide="1"]True. $4.20\cdot5+9=30$[/hide]

[hide="2"]He is playing Monopoly.[/hide]

for 2.

The man drives up to the hotel, but realizes he's banrupt, right?

That's cuz he lost his wallet.

For 2, I was thinking of something more like
[hide]He knows that he has buried his retirement nest egg underground at an exact location. When he drives up to the hotel, he realizes that the hotel was built over that spot. Therefore, he has lost his money. Kinda far-fetched, don't you think?[/hide]

i_like_pie,

No, the answer is "false."

You do not read "4.20" in mathematics as "four twenty."

The decimal point MUST be read.

If the poser meant to get away with that, then this question
itself must be discarded.

[quote="Summation"]The answer is still "false."

The fact that you read that amount of money that way is irrelevant.  You can't treat a phrase in two totally different
ways here.

If it's "four twenty," then it CANNOT be multipled by 5 because
there is no sense to this expresssion multiplying any number.

However, if it's "4.20" then it COULD.

Also, there are no units, i.e. dollar signs, attached to the "9" and the "30."  You can't assume dollar units with the "4.20" and then automatically do the same with the second two numbers in the problem.

Again, this problem is flawed.[/quote]
Honestly, I agree that the problem is flawed, but then again, ALL puzzles and brainteasers are flawed, because there are many possible answers.
Also, all other stuff you say, imo, is total bs. It's supposed to be a BRAINTEASER for god's sake.

[quote="randomdragoon"]However, strangely enough, you read $\$4.20$ as "four twenty" in real life.[/quote]

and yet some people read 4040 as four thousand and fourty

umm for the first one,

[hide]if it says four twenty, then, it would be $20*4$ that is $80$, and that multiplied by 5, is $400$, and then the word 'and' represents a decimal point, so would it be $400.9$? and that is most certainly not equal to $30$[/hide]
does this seem like a possibility?


Seven fairies don't make friends with each other, if they aren't located within the distance of 1m .
Determine the arrangement such that any three of the fairies will make friends.

Are there any restrictions to the fairies?
For example, the minimum distance between any two fairies, the space that a fairy occupies, etc.


[color=blue][i]Σε έναν ρουμάνικο διαγωνισμό , όχι στην ολυμπιάδα , τέθηκε το παρακάτω πρόβλημα :[/i][/color]

Οι παρεγγεγραμμένοι κύκλοι ενός τριγώνου $ ABC$ εφάπτονται των  πλευρών $ BC,CA,AB$ στα σημεία $ A_1,B_1,C_1$.

 Να αποδείξετε ότι οι περιγεγραμμένοι κύκλοι των τριγώνων $ ABB_1 , ACC_1$  τέμνονται  στη διχοτόμο της γωνίας $ A$ του τριγώνου $ ABC$.

[u]ΣΧΟΛΙΟ[/u]

 Παρακάλω τον κύριο Βήττα να μας στείλλει το σχήμα.Τους παρεγγεγραμμένους κύκλους ας τους κρύψει , να αφήσει όμως τα σημεία επαφής $ A_1,B_1,C_1$

Τα σημεία $ A_{1},$ $ B_{1},$ $ C_{1}$ ορίζονται εύκολα, γιατί ως γνωστό, τα σημεία επαφής των παρεγεγραμμένων κύκλων στις πλευρές ενός τριγώνου, είναι συμμετρικά των σημείων επαφής του εγγεγραμμένου κύκλου στις ίδιες πλευρές, ως προς τα μέσα τους.

Κώστας Βήττας.

Γράφω μια λύση αν και δεν είμαι τόσο σίγουρος αν είναι σωστή.

Κατ' αρχάς είναι γνωστό ότι $ BC_1\equal{}CB_1\equal{}p\minus{}a$, όπου $ p$ η ημιπερίμετρος.
Εστω τώρα $ K$ η τομή των $ BB_1$, $ CC_1$. Εστω $ M$ το σημείο τομής των περιγεγραμμένων κύκλων των τριγώνων $ ABB_1, ACC_1$. Τότε το $ M$ θα αποτελεί σημείο τομής και των περιγεγραμμένων κύκλων των τριγώνων $ BKC_1, CKB_1$, αφού θα είναι το σημείο Miquel του αντίστοιχου πλήρους τετραπλέυρου. Εστω τέλος $ R_1, R_2$ οι ακτίνες των περιγεγραμμένων περί τα τρίγωνα $ BKC_1, CKB_1$ κύκλων, αντίστοιχα.

Θα έχουμε (από νόμο ημιτόνων) $ 2R_1\equal{}\frac{BC_1}{\sin{BKC_1}}\equal{}\frac{CB_1}{\sin{CKB_1}}\equal{}2R_2\iff R_1\equal{}R_2$

Αρα: $ \frac{MK}{\sin{MBK}}\equal{}R_1\equal{}R_2\equal{}\frac{MK}{\sin{MCK}}\iff \angle{MBK}\equal{}\angle{MCK}$  $ (1)$

Ομως απο τα εγγράψιμα $ BB_1$, $ CC_1$ προκύπτει αντίστοιχα:
$ \angle{MAC_1}\equal{}MCC_1\equal{}MCK$ και $ \angle{MAB_1}\equal{}MBB_1\equal{}MBK$ $ (2)$

Από $ (1), (2)$ έπεται ότι:
$ \angle{MAB}\equal{}\angle{MAC}$ κι άρα το ζητούμενο :)

[b]
Σημαντική παρατήρηση[/b]: Νομίζω ότι το πρόβλημα θα ίσχυε για οποιαδήποτε 2 σημεία $ B_1, C_1$ για τα οποία $ BC_1\equal{}CB_1$

[quote="bilot4"][b]Σημαντική παρατήρηση[/b]: Νομίζω ότι το πρόβλημα θα ίσχυε για οποιαδήποτε 2 σημεία $ B_1, C_1$ για τα οποία $ BC_1 \equal{} CB_1$[/quote]
Πολύ σωστή παρατήρηση, αυτό είναι το κύριο και αυτό χρησιμοποιείται για τη λύση.

Αφαιρώντας τα περιττά, ομορφαίνει και το σχήμα. :)

Κώστας Βήττας.

[quote="bilot4"]
..............................................
Σημαντική παρατήρηση[/b]: Νομίζω ότι το πρόβλημα θα ίσχυε για οποιαδήποτε 2 σημεία $ B_1, C_1$ για τα οποία $ BC_1 \equal{} CB_1$[/quote]

Επομένως , όπως δείχνει και το σχήμα του Κώστα , μετά την παρατήρηση ότι τα δύο τμήματα είναι ίσα (αυτό είναι το κλειδί
της άσκησης και το συναντάμε σε πολλά θέματα διαγωνισμών ), η ισότητα των δύο τριγώνων τελειώνει το πρόβλημα  :) .

 Μπάμπης


Euler
number theory
least common multiple
induction
geometry
perpendicular bisector
number theory proposed

Find all $ n$ with the following property:
Consider a regular $ n$-gon, with $ n$ odd. Consider the sequence of vertices: $ P_{0}$ and $ P_{1}$ are adjacent vertices. For $ k \geq 2$, $ P_{k}$ is the vertice that lies on the perpendicular bisector of $ P_{k \minus{} 1}P_{k \minus{} 2}$. This sequence, covers all vertices of the polygon.

I think you want the $ n$ from the last sentence to be a $ k$. At least it shouldn't be the same variable as $ n$, the number of vertices.

Yes, you're right ZetaX.  :blush:
I've edited it now.

It is a very nice problem but, unless you have a better solution than the following, I'm not sure it is so simple :wink:

We will prove that the desired $ n$ are those of the form $ n \equal{} 3^t$, for some positive integer $ t$.

Let $ n \equal{} 2k \plus{} 1$.
Note that $ n$ and $ k$ are coprimes.

Number the vertices consecutively $ V_0,V_1,\cdots,V_{n \minus{} 1}$ so that the first two chosen vertices are $ V_0$ and $ V_1$.

Let $ U_i$ be the subscript of the $ i^{th}$ chosen vertex, with $ U_0 \equal{} 1,U_1 \equal{} 1.$

Since $ n$ is odd, we note that $ 2$ is inversible modulo $ n$ and $ k \equal{} \minus{} \frac 1 2 \mod [n]$.

In all the following the $ U_i$ will be considered modulo $ n$.

It is straightforward to see that $ U_{i \plus{} 1} \equal{} \frac {U_i \plus{} U_{i \minus{} 1}} {2}$ for each $ i > 1$.

Thus $ U_p \equal{} \sum_{i \equal{} 0}^{p \minus{} 1} ( \minus{} \frac 1 2)^i \equal{} \sum_{i \equal{} 0}^{p \minus{} 1} k^i$.

Moreover $ U_p \equal{} U_q$ if and only if $ U_{p \minus{} q} \equal{} 0$.

Thus, we are looking for the $ n$ such that $ U_0, \cdots,U_{n \minus{} 1}$ are pairwise distinct, which is the same that none of the $ U_p$ is $ 0$ (except $ U_0$).

- If $ n$ is not divisible by $ 3$ :

Then $ k \minus{} 1 \equal{} \frac {n \minus{} 3} 2$ and $ n$ are coprime, which ensures that $ k \minus{} 1$ is inversible modulo $ n$.

We then have $ (k \minus{} 1)U_p \equal{} k^p \minus{} 1$ for each $ p$.
But, using Fermat-Euler theorem, we deduce that $ U_{\varphi(n) }\equal{} 0$, where $ \varphi$ is the Euler totient, and $ 0 < \varphi(n) < n$.
Thus, $ n$ is not a solution.

- If $ n \equal{} 3^tb$ for some positive integers $ t,b$, where $ b$ is odd and not divisible by $ 3$.
Note that $ k$ is not divisible by $ 3$ and that $ \varphi (b)$ is even.
Let $ p \equal{} lcm (\varphi (b), \varphi (3^t)).$
It is well-known that $ \varphi (3^t) \equal{} 2 \cdot 3^{t \minus{} 1}$. Therefore
$ p \leq 3^{t \minus{} 1} \varphi(b) \leq 3^{t \minus{} 1}(b \minus{} 1) \equal{} \frac n 3 \minus{} 3^{t \minus{} 1} < \frac n 3$.
Thus $ 0 < 3p < n.$

In another hand, we also have $ \frac {k \minus{} 1} 3 U_i \equal{} \frac {k^i \minus{} 1}{3}$ for each $ i$, where $ \frac {k \minus{} 1}3$ and $ n$ are coprime.
Using Fermat-Euler theorem again, we deduce that there are integers $ a_1,a_2,a_3$ such that
$ k^p \minus{} 1 \equal{} a_1n$ and $ k^{2p} \minus{} 1 \equal{} a_2n$ and $ k^{3p} \minus{} 1 \equal{} a_3n.$

If one of $ a_1,a_2,a_3$ is divisible by $ 3$, we deduce that one of the numbers $ \frac {k \minus{} 1} 3 U_p ,\frac {k \minus{} 1} 3 U_{2p} ,\frac {k \minus{} 1} 3 U_{3p}$ is divisible by $ n$, so that one of the numbers $ U_p,U_{2p},U_{3p}$ is $ 0$. In that case $ n$ is not a solution.
If none of the $ a_i$'s is divisible by $ 3$ then, from the pigeon-hole principle, we deduce that two of them are equal modulo $ 3$.
If $ a_1 \equal{} a_2 \mod[3]$ then $ k^{2p} \minus{} 1 \minus{} (k^p \minus{} 1) \equal{} 0 \mod[3n]$ from which we deduce that $ k^p \minus{} 1 \equal{} 0 \mod [3n]$, a contradiction.
We get similar contradiction in the case $ a_1 \equal{} a_3$ and in the case $ a_2 \equal{} a_3$.

Thus, $ n$ is not a solution.

It follows that if $ n$ is a solution then $ n \equal{} 3^t$ for some positive integer $ t$.

By an easy induction, we deduce that for $ t \geq 2$, we have $ ( \minus{} \frac 1 2)^{3^{t \minus{} 1}} \equal{} 1 \mod [3^t]$ and $ ( \minus{} \frac 1 2)^{3^{t \minus{} 1}} \neq 1 \mod [3^{t \plus{} 1}]$  (recall that $ \minus{} 2$ is inversible modulo $ 3$).

It follows that for each positive integer $ t$, the order of $ \minus{} \frac 1 2$ modulo $ 3^t$ is $ \omega_{3^t} ( \minus{} \frac 1 2) \equal{} 3^{t \minus{} 1}$ :
This is clearly true for $ t \equal{} 1,2$. Assume it is true for some $ t > 1$.
From above, we have $ ( \minus{} \frac 1 2)^{3^t} \equal{} 1 \mod[3^{t \plus{} 1}]$. Therefore $ \omega_{3^{t \plus{} 1}} ( \minus{} \frac 1 2)$ divides $ 3^t$. But, it cannot divides $ 3^{t \minus{} 1}$ otherwise, we would have $ ( \minus{} \frac 1 2)^{3^{t \minus{} 1}} \equal{} 1 \mod[3^{t \plus{} 1}]$, contradicting the result above. Hence conclusion.

Now, let's consider $ n \equal{} 3^t \equal{} 2k \plus{} 1$ for some positive integer $ t$.
Clearly, $ n \equal{} 3$ is a solution, so let's assume that $ t \geq 2$.

Recall that $ k \equal{} \minus{} \frac 1 2 \mod [n]$.

If there exists $ p \in \{1, \cdots, n \minus{} 1 \}$ such that $ U_p \equal{} 0$ :
Then $ \frac {k \minus{} 1} 3 U_p \equal{} \frac {k^p \minus{} 1} 3 \equal{} 0 \mod [n]$, so that $ k^p \minus{} 1 \equal{} 0 \mod[3n]$.
Thus $ ( \minus{} \frac 1 2 )^p \equal{} 1 \mod [3^{t \plus{} 1}]$.
Since $ \omega_{3^{t \plus{} 1}} ( \minus{} \frac 1 2) \equal{} 3^t$, it follows that $ n \equal{} 3^t$ divides $ p$. A contradiction.

Therefore, $ n \equal{} 3^t$ is a solution, and we are done.

Pierre.

this is problem 6 from the iberoamerican olympiad 2006.


I took the AMC 10B this year and scored a 109.5. I'm currently taking Algebra 2/Trig at school and doing fine.

Thanks for the input.

I don't see why they would be too hard.  I didn't take the classes but you seem to have plenty of pre-requisites.


calculus
derivative
calculus computations

let$ f(x) \equal{} (x^n \minus{} 1)^n .$
 caculate : $ f^{(n)}(x)$

Using the chain rule,
$ f'(x)\equal{}n(x^{n}\minus{}1)^{n\minus{}1}(nx^{n\minus{}1})$
I put this into Mathematica and it gets real ugly real fast.

im guessing we'll be able to get a pattern pretty quickly but im too lazy to solve it  :D

It is n^n n!. This cane be got by looking at coefficient of (x-1)^n in taylor expansion of ((x-1)^n-1)^n.  Please see http://www.goiit.com/posts/list/0/differenciation-the-solution-46803.htm

it can also b solved thru pure logic.. refer 2..http://www.goiit.com/posts/list/differenciation-nth-derivative-46692.htm..posts of comp001


topology
real analysis
real analysis unsolved

Can anyone help me on this problem?

Let $S$ be a subset of the metric space $E$. A point $p \in S$ is called an $interior point$ of $S$ if there is an open ball in $E$ of center $p$ which is contained in $S$. Prove that the set of interior points of $S$ is an open subset of $E$ that contains all other open subsets of $E$ that are contained in $S$.

Here's what I have so far:

Let $p \in S$. Since $p$ is not on the boundary of $S$, the minimum distance $d(p, border) > 0$. So let this distance $d = \epsilon$. By Archimedian Principle, we can always find $n \in N$ such that $0 < \frac{1}{n}< \epsilon$. Obviously we can select the radius of the open ball centered at $p$ to have radius $\frac{1}{n}$. So this open ball is contained entirely within $S$. Clearly, we can construct some open ball using this logic for all the points like $p$ that are contained within $S$, that are not on the boundary. So this set of points make up the open subset S (without the boundary).

Is this logic correct? It seems a bit too simple for an advanced problem. Also, how do I go on to show that this open subset contains all other open subsets of $E$ that are contained in $S$?

Thanx

The interior of a set is actually commonly used and is usually encounted early on, so yeah, don't expect the reasoning to be too tricky. I'm not sure what you mean by d(p, border) though. How do you define this thing? Try working pretty much straight from the definitions of an "open set" and "interior point" instead.

For the second part, just get past the wording - you want to show that any open subset of E is a subset of S, i.e. if you have a point p that is an element of an open subset of E, it is an interior point. Again, just work from the definitions, pretty much.

Ok, since $p$ is located inside the subset S, we know that the distance from $p$ to the border of the subset $\ne 0$. But we can't just choose any point on the border b/c we might get part of the open ball on the outside of $S$ if the figure is concave (or convex, whatever). But by picking the minimum distance from $P$ to any border point, we can ensure such as overlap does not happen. So that's what the $d(p, border)$ means. Clearly, I should have written $min\{d(p,q)\}$, where $p, q \in S$ and $p$ is interior of $S$, while $q$ is on the border of $S$. Let us then elaborate such that a point on a border means a point $k \in S$ such that it is impossible to find an open ball with positive radius centered at $k$ that is wholly contained in $S$.

Anyone else?

What is a border?

How do you know the min exists?

Don't try defining other expressions when you really don't need them...

[quote="bythecliff"]Do we have to prove each statement such as the existence of the min and the boundary? See, what confuses me is what are the parts we need to prove and what the parts we don't need to prove are.[/quote]

The parts you need to prove are everything that you haven't already proven (as theorems in class, for instance, or on earlier homework problems).  In particular, if you're going to start using some object, you should be able to define what that object is and show that it exists.

In this particular problem, you've gotten off to a terrible start with this line:

[quote]Let $p \in S$. Since $p$ is not on the boundary of $S$, the minimum distance $d(p, border) > 0$. [/quote]
The problems with this that immediately jump to mind are:
1) What is the border of a set?  (You may perhaps have defined such a thing, in which case this question is null.)
2) Why is $p$ not on the border?
3) Why should $p$ not being on the border imply the existence of a minimal distance from $p$ to the border?

From a problem-solving standpoint, there is another serious flaw: you were given a definition of what an interior point is, and you've never actually used that definition in your proof.  This strongly suggests that something is wrong.  Although there are exceptions, a good principle to keep in mind is that you will probably need all the information you're given, so if you haven't used some of the givens, you're probably doing something wrong.


Can Someone give me the solution for:

    If  (x-y)^3  +  (y-z)^3  +  (z-x)^3  = 48

    Then maximum value of |x-y| + |y-z| + |z-x| is ?

     (Please reply with complete solution)

This surely isn't the correct forum for this kind of question.


[hide]
We want to show $ 2005*1003$ divides this.  But both $ 2005$ and $ 1003$ divide this because

$ 2005|i^{2005}+(2005-i)^{2005}$
and
$ 1003|i^{2005}+(2006-i)^{2005}$

(in fact, we could have used any odd exponent, not just $ 2005$ on the right.)
[/hide]


can anyone kindly provide me the easiest proof of the converse of the alternate segment theorem :(

If you tell us what the theorem is...

  darij

I think it is:

$ A,B$ are two points on a circle. Line $ CD$ tangent to the circle at $ C$ and $ A,D$ is on different sides of line $ CB$. Then $ \angle BAC \equal{} \angle BCD$

so the converse is

$ A,B,C$ are three points on a circle and $ D$ is a point on the plane such that $ A,D$ is on different sides of line $ CB$. If $ \angle BAC \equal{} \angle BCD$. Then $ CD$ is a tangent

Try connect the points to the centre and trace the angles  :D


[quote="Arvind_sn"]What is the sum of the first ten perfect squares starting with one (1, 4, 9,...)?[/quote]

[hide]$1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}+7^{2}+8^{9}+10^{2}=385$[/hide]

[size=75][color=green]Mod edit: please hide your answers.[/color][/size]

[hide="my cooler way"]
The sums are like: $1+4+9+.....+n^{2}=\frac{n*(n+1)*(2n+1)}{6}$
Using it we get  $\frac{10*11*21}{6}=5*11*7=385$[/hide]

edit:what is "derive"?  :blush:

[quote="alex__ib"][hide="my cooler way"]
The sums are like: $1+4+9+.....+n^{2}=\frac{n*(n+1)*(2n+1)}{6}$
Using it we get  $\frac{10*11*21}{6}=5*11*7=385$[/hide][/quote]
Can you derive your formula :P.

it seems like something you could do with induction... I'm going to try to do that

alex_ib, derive is basically "discover/prove"

edit:  I figured out how to prove it using induction  :D :

for the base case, $1^{2}=\frac{1\cdot2\cdot3}{6}$.  For the inductive step, we prove that, if it works for $k$, it works for $k+1$.
$1^{2}+2^{2}...+k^{2}+(k+1)^{2}$
$=\frac{k(k+1)(2k+1)}{6}+k^{2}+2k+1$
adding these together and expanding gives $\frac{2k^{3}+9k^{2}+14k+6}{6}$.  What we are trying to prove is that this sum is equal to $\frac{[k+1]([k+1]+1)(2[k+1]+1)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}$.  Expanding the desired product gives the above sum, so our induction is complete :coolspeak:


Am I missing something? By the principle of unsmoothing, you need as much negatives as you can get, and the rest (when possible, if not all but 1) maximized.

So 1736 times $-\frac{1}{\sqrt{3}}$, 260 times $\sqrt{3}$ and one time $\frac{2}{\sqrt{3}}$.

That gives an awful maximum of $189548$ for the requested expression...  :maybe:

Does it help to choose $a_{i}$'s which are $\tan^{-1}(x_{i})$'s?

Perhaps that's the right solution Peter.
Unfortunately I'm not familiar with the principle of unsmoothing, nor can I find a solution on the net, so I can not say if that is the right answer.  :)

But I think there is another solution, since I found this problem as an exercise after the first chapter concerning trivial inequality... No word of unsmoothing in chapter 1  :wink:

:P Trivial...
why is the 1! there? Also I think it is a good problem to prove this...

What's your problem? I said the maximum was there iff we had  $1736$ times $-\frac{1}{\sqrt{3}}$, $260$ times $\sqrt{3}$ and one time $\frac{2}{\sqrt{3}}$. So as the $x_{i}$ are numbered, in how many ways can this be attiained?

Call them 1736 red balls, 260 blue balls and 1 green ball if you like, in how many different orders can we pull them from a box? ... :maybe:

Sorry didn't scroll the page up to your previous post... that was stupid of me..


This seems too easy, so I'm suspicious.

Suppose a set $ S$ is an open subset of a metric space. Then for each point in $ S$, we can choose some radius $ r_p$ such that the neighborhood $ N_p(r_p)\subset{S}$.

Claim: Let $ G\equal{}\displaystyle\bigcup_{p\in{S}}{N_p}(r_p)$; then $ S\equal{}G$. That is, $ S$ equals the union over all points $ p\in{S}$ of some neighborhood of each point contained entirely in $ S$.

Proof of Claim: since some neighborhood of every point in S is in the union, $ S\subset{G}$. Since each neighborhood in the union is contained entirely in S, $ G\subset{S}$. Therefore $ G\equal{}S$.

I ask since I think this would allow me to simplify somewhat the proof of Baby Rudin Theorem 2.30.

Well, then isn't the theorem almost trivial?...

2.30 Theorem. Suppose Y is a subset of X, a metric space. A subset E of Y is open relative to Y if and only if E=(Y intersect G) for some open subset G of X.

Proof:

Part I: If a subset E of Y is open relative to Y, then E=(Y intersect G) for some open subset G of X.

Proof of Part I: Let E=G, or let G be defined as in my first post. Then part I follows (trivially?).

Part II: If E=(Y intersect G) for some open subset G of X, and for a subset Y of X, then E is an open subset of Y.

Proof of Part II: (This part is quoted from Rudin) Since E is a subset of G, then every p in E has a neighborhood V_p, a subset of G. Then (V_p intersect Y) is a subset of E for all p, so that E is open relative to Y.

----------

So actually I'm pretty sure Rudin's proof of Part I implicitly utilizes the claim in my first post, so I haven't actually simplified anything. It just took a few minutes to get my head around the relative topology in part II.

This is certainly not the hardest theorem in the book. I do not have the book at hand now but my usual way to prove part 1 when I teach this course is the following:

If $ E\subset Y$ is open relative to $ Y$, then for each $ p\in E$, there exists $ r_p>0$ such that $ N_p(r_p)\cap Y\subset E$. Let now $ G\equal{}\bigcup_{p\in E}N_p(r_p)$. It is an open set (as a union of open sets), and, since $ \{p\}\subset N_p(r_p)\cap Y\subset E$ for all $ p\in E$, we can pass to the union over $ p\in E$ in this double-sided inclusion to get $ E\subset G\cap Y\subset E$.

I also usually draw a picture on the board with $ X$ being the entire plane, $ Y$ being a line, and $ E$ being an open interval.


calculus
integration
function
AMC
AMC 8
HMMT
geometry

Well

Mine is 98 on AP Calculus :mad:

I studied it 4 years ago, I am still failing it!

Darn it.

I actually will have to face a ridiculous circumstance that I actually will have to study for Calculus exam! Duh.

I want Linear Algebra or Multivariable.
 :(  :(  :(

I am suck at math.

Why does it say that your post count is 0?

My recent failure...

I forgot the song in the middle of a piano concert with 500 people staring at me.

cuz he doesnt have any posts outside of FF

____________
i left a certain electronic device inside my desk at the end of school  :o  :o  :o  i got it back quickly teh nxt day. whew.

Oh where to start... I've had about 27 Epic Fails in the past week.

The funniest I think,
is when a guy squirted a lemon in my eye and I somehow danced in pain.
Fail.

My most epic fail so far was doing the limbo on skates

my most epic fail was stubbing my toe under the door, WHEN I WAS LOOKING WHERE I WAS GOING
epic failiure

I tripped on a rock...
in the middle of Target :|

I walked into a volleyball net... It hurt...  :stretcher:

lollllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll
mine is
not buttoning up my shirt properly, in front of my whole school..AH

I said $ 2009 \times 3 \equal{} 8017$.  :blush:

"The integral of a piecewise function is the combination of the integrals of the pieces."

"Which of the following answers are FALSE (choose both answers)"
I chose A, B, C
Correct: D, E

*facepalm*

Epic fail.

I am an eighth grader and now we are studying heredity,

AND NO ONE KNOWS ANYTHING ABOUT DOWN'S SYNDROME.

EPIC FAIL.

Somebody said that one with the Down's Syndrome has three sex chronosomes :rotfl:  :rotfl:  :rotfl:

My recent fail...

I was trying to solve this geometry problem that nobody else in my class could solve. I got the right idea of making the triangle turn into a rectangle then subtracting that from the other polygon. I got the answer so close. I figured out that the last step, I did 40-16=32... That's just fail. I just would've killed myself just then.

When on my super easy Algebra 1 stuff, I misread the question and got 12/13.
I'm so PISSED!

My grade is now a 96.10 :(
CRYING.

My most recent epic fail,

I was solving a problem, and I was determined to solve it.
So I used about an hour or so to finish that problem.
It was a geometry problem, and I used very weird analytic geometry to solve it.
A teacher came by and solved it within 2 minutes.

And the method was extremely simple all right.

I was transferring an answer from my work to my answer section. I got the problem wrong because I put the wrong answer on the answer section!!! I was so  :censored: pissed!

NO 1 CAN BEAT THIS!
I had a multiple choice question write and they drew WXYZ and W'X'Y'Z' right.
Well for some reason I drew my own W'X'Y'Z' and found the degree change in that. I got it wrong and corrected in 1 sec.................. I spent like 30 min checking over that test too. Man and they gave me the picture, but for some reason I drew my own picture and it was wrong anyway!!!!!  :oops_sign:   :cursing:  :help: I keep making mistakes like this.

Also, Chuthecommie you probably shouldn't write p***** out that's a bad word too.

Check this out.

My friend: "Hey, I suddenly forgot how to spell the word 'FAILURE'"
Me: "lol, are you serious? It spells P-H-A-I-L-U-R-E!"

.
.
.

Phail much?

My most epic fail recently?

Saying $6x^2/2x^-3$=$3x$.  So I got 55.5/54 on the quiz.

most recent epic fail: 142 on Gauss 7

that's a true EPIC FAIL

Most recent epic fail:
I was at school taking a test, and I accidently put down fail for one of the answers. It was a spelling test. It brought me down to 95%
sob.

We had a test on the reproductive system in science and I got a B+.  That was an epic fail. :P

wrote an answer and it was confusing so i got a bad grade. this is what it looked like:
$\boxed{\emph{\frac{4} {3}}}$
epic fail.

most recent epic fail: losing 10 points cuz i have no color printer

now have A- in SS

i mis-transferred my answers on a bubble sheet (for science) and got a 20 (\pm 5) on a test (100 max).  I went back and corrected it to find I would have gotten a 98...

FaiL

I did 2005 Mathcounts Chapter countdown round.  I was on the final step of Q19, which was something like 60x=79200.  I accidentally did 79200/6 and ended up with x=13200.  FAIL...

Another one is when I got a 99/100 on a Algebra 1 Accl Chapter 1 test on the last question.  I ended up saying w=7,l=11.  When my test came back, I noticed that my teacher wrote:"w=7 [b]feet[/b], l=11 [b]feet[/b]"...


[quote="shinwoo"]Insert parentheses to make each statement true...

4 + 16 / 2 + 3 * 5 = 20

This is pretty interesting

this is the problem that we were struggling in Middle School section

is there any way to solve it beside guess and check?[/quote]How about $(4+16)/(2+3)\cdot5$?

[quote="shinwoo"]well, we already have the answer and I put it up actually, but I don't know how to solve it without using guess and check... That's what we are looking for.[/quote]I couldn't imagine any systematic method for solving such problems. You just have to find it. If you like stuff like that, try [url=http://www.ridiculopathy.com/crappy_flash_games.php?gamename=numbers]this game[/url]. I have the number 11, 12, and 14 high scores.

OK, "Big Sexy", let me see if I can beat your high score.


I've been trying to do problem 6.2.8 from the book
"The Art and Craft of Problem Solving" by Paul Zeitz

Here it goes:

Sal the Magician asks you to pick any five cards from a standard deck
(13 denominations in each of 4 suits).Yo do so, and then show them to
Sal's assistant Pat, who places one of the five cards back in the deck
and then puts the remaining four cards into a pile. Sal is blindfolded,
and does not witness any of this. Then Sal takes off the blindfold, takes the pile of four cards, reads the four cards that Pat has arranged, and is able find the fifth card in the deck (even if you shuffle the deck after Pat
puts the card in the deck). Assume that neither Sal nor Pat has supernatural powers, and the deck of cards is not marked.
How is the trick done?

Harder version: you pick which of the the five cards goes into the deck
(instead of Pat).

...Now, this is a well known trick whose solution can be sketched as
follows:

Since there are five cards and only four suits, two of them must be of the
same suit. The assistant can choose one of these to be the hidden card,
choosing the one which is (cyclically) 6 or fewer cards above the other.
For instance, if the assistant had the queen and 3 of a suit,
the assistant would hide the 3 since it is the fourth card above the queen (king, ace, 2, 3).
Then the assistant would place the other card of the same suit as the first card and the remaining three cards in a permutation (relative to their sorted order, based on some assignment of numbers 1-52 to cards) to indicate whether the hidden card is one, two, three, four, five, or six ranks above the first card.
Since a card that is seven above another card is six below it,
the assistant can always choose one of the two cards that is six or
fewer above the other.

But, notice that this solution does not solve the second part of the problem,
that is, when YOU (instead of the assistant) get to pick the card that goes into the deck.

Can anyone devise a solution to this variation?

[quote="Ragingg"]How bout if Sal does other tricks like putting the first card face down if its hearts, 2nd if its clubs, etc?   Are they looking for something creative like that? Heh[/quote]

No...The problem is asking to devise an encoding of 48 cards
based only on the values (demominations and suits) and the order
(relative to each other) of the 4 remaining cards (The ones which the assistant shows the magician).
For example, look at the encoding of the cards which I explained above.

However, that particular encoding only works if the assistant
is allowed to pick the card which goes into the deck.

Also...Sal is only allowed to read the 4 cards that Pat hands to him.

[quote="Peter VDD"]So... you can pick the card... but what can the assistant still do? Can he decide on the order of the card? Or what?[/quote]

OK...The trick goes like this:  You choose five cards at random
from a standard deck. Then YOU pick one card from
the five chosen cards to be the hidden card. The assistant
has been witnessing all this (but not the magician).
That is, the assistant has seen  the five cards and knows
which card will be the hidden card. Now, the assistant
puts the hidden card back into the deck and shows the other
four cards to the magician. Finally, the magician names the
fifth card correctly.

Of course, the assistant can show the four cards in
some previously established order that will allow
the magician to decode that ordering and name
the fifth card.

Obviusly...the order of the four cards will not be enough
since the ordering of 4 cards give only 24 permutations
and we need to encode 48 cards.

To encode 48 cards we will need the values and
also the order of the four cards.

PS: I get the feeling my explanations are not
very clear. Please tell me if that is the case.

I understand how his assistant would tell him which card it is.

Here's my method for when YOU pick which one to put back.
First, you assign each of the 4 remaining cards a number 1-4 based on it's relative standing to the others (smaller numbers get a lower encoding, suits go clubs, diamonds, hearts, spades so if you have both 5 of clubs and 5 of diamonds, the clubs would be ranked lower).  Now, this only gives him 24 numbers to deal with.  However, you can then use the card's orientation.  Say that the first 2 cards will be either vertical, vertical=clubs, (v,h)=diamonds, (h,v)=hearts, (h,h)=spades.  That would provide enough information.

If orientation isn't allowed, you could use something else like how the assistant picks up the cards after he shows them to the magician.

Oh, I just got an idea!  Since there's only 48 cards left (24*2), you could do the 24 encoding, but either place all the cards horizontal or all vertical.  That would then give him 48 possibilities to deal with, so he could figure out which one you picked.  Although memorizing this encoding would take some time.

[quote="Wumbate"]
If orientation isn't allowed, you could use something else like how the assistant picks up the cards after he shows them to the magician.
[/quote]

You are using lexicographical ordering which is OK,
but orientation is not allowed.
The method can ONLY rely on the values of the cards
and their ordering (e.g. lexicographical ordering).

For example, the encoding which I explained as an example
was only based on the values and some ordering of the cards.
The method worked because of the pigeonhole principle.
However, I don't know how to generalize it when
you (instead of the assistant) select the hidden card.

[quote="dts"]I haven't been able to solve the harder one, but I have found a much simpler solution to the easier one.  The assistant picks a card, and then puts it 22 (or some other predetermined number) places below the top card.  Then, the magician must simply count down the deck until he reaches the correct card.[/quote]

I think you didn't read the problem carefully enough.
The problem says that the magician finds the fifth card:

"even if you shuffle the deck after Pat puts the card in the deck"

"The method can ONLY rely on the values of the cards
 and their ordering"

If that is true, and you get to pick which card gets put back, then the best you can do is 4*3*2*1=24, I would suppose.  The fact that 24*2=48, and there's 48 cards to pick from, makes it very tempting to use something else in addition to the relative encoding.

Are you sure it can only rely on the value and their orderings?  And do you know the answer?

[quote="Wumbate"]
Are you sure it can only rely on the value and their orderings?  And do you know the answer?[/quote]

The fact is that four cards convey much more information
than it is needed to encode 48 cards. For example,
it can be shown that if the assistant picks the hidden
card you can encode any deck up to 124 cards (Amazing as it is!!!!)

If you are interested in this problem I will refer you to
the following article:

http://people.brandeis.edu/~kleber/Papers/card.pdf

Finally....I don't know the answer. All I can tell you
is that I found the problem in the book
"The Art and Craft of Problem Solving" by Paul Zeitz
(Problem 6.2.8, page 222)

It's a pretty good book if you like problem solving.  ;)

I got it, I think.
This method is basically a modified form of the answer when the assistant picks which one to kick out.

First, know that I am employing a card ranking system, where each card gets it's numerical value, plus 13 if it's diamonds, 26 if it's hearts, and 39 if it's spades.  This allows every card to be assigned a number 1-52.  Next, the assistant takes the highest ranked card of the 4 and puts it in slot 1, 2, 3, or 4.  Then the other 3 cards are ordered in the remaining 3 slots in such a way that their relative ranking provides a number 1-12 (3*2*1=12).  Combined with the highest ranking part, this gives you 48 different encodings.  Since you only have 48 cards left, you can go through them and pretend that they're sorted by their ranking.  Then the number corresponds to the card in the cards-left pile.

PS I'll check out that website.

OK...the ranking method seems OK to me.

But there are two things which I fail to see:

You mentioned that :

[quote="Wumbate"]
Then the other 3 cards are ordered in the remaining 3 slots in such a way
that their relative ranking provides a number 1-12 (3*2*1=12).
[/quote]

The thing is that 3*2*1=6 (Not 12)
You can only produce 6 encodings using 3 cards.

Second: How exactly do you combine the highest ranking
card to produce 48 encodings?

Yea, that's something I realized late in the night while thinking about this problem.  Both those objections are really the same problem (that 3*2*1=6 not 12).  I'm currently looking at a way where you do it one way if all four suits are there, and another way if you have 2 cards of a same suit, but I don't have anything yet.  I realized another problem, too, that since you get to pick which card to go out, the going from a base and counting up to 6 won't work since it could be actually down 5, or up 7, and also there is the possibility you have 13 cards to deal with, not 12.  So I don't have an answer but I'm thinking about it.

One way to deal with the 13 cards, not 12 possibility would be to use the fact that you're COULD encode one of the cards that are in the 4 selected, and then switch that to the 7th cyclic one (which otherwise would be impossible to encode).  Otherwise you would not be able to reach the 7th card.  But that's assuming a method similar to the original solution exists.

Yes..The original method relies cleverly on the pigeonhole
principle. In fact, it uses the pigeonhole principle twice.
First, by assuring that there will be two cards of the same suit
and secondly by assuring that their distance will
never be greater than 6 ranks from each other.

I suspect that the original method can not be
adapted as it is when you get to pick the fifth card.
Since the pigeonhole principle doesn't apply in this case.
Basically, because your choice will be random
which is not the case if the assistant is allowed
to choose the fifth card.

I read the document, and it only confirms my initial thoughts: the first problem is correct (and can be extended to 124), the second is wrong.

Just put all the non-sense aside. The assistents gets the lowest 4 cards (those with indices 1,2,3,4) to be ordered. No tricks here, the indices are just {1,2,3,4} and can be ordered in 24 ways.

You, as magician, can recieve those in only 4! ways, thus there are only 24 options you can decide on. Simple as that.

[quote="Peter VDD"]the second is wrong.[/quote]

But there are many other ways to produce encodings
than by looking at the ordering of the cards.

For instance... you could do something like
defining a function of the values of the cards.

Maybe a combination of the ordering of the cards
and a function (like parity) will work out....

I think your example only proves that by
relying on the ordering of the cards
it's impossible to get 48 encodings.

Certainly ordering is not enough... but how can
you be certain that there doesn't exist any other
method?

I came to the conclusion that it was impossible, too.  The best you can do is narrow it down to 2 cards.
I think ordering is all you get.  If you had something else to go by than relative ordering of the 4 cards (which produces 4!=24 permutations), then I could pick different cards to ruin what you had to go by.  Since the 4 "code" cards and the fifth picked card have no possible relationship, I think it is impossible.  In the original solution, where the assistant picks which one to throw back, he gets around the fact that he can only encode 24 cards by picking one of the two cards he can encode (if he picked the other card of the same suit, he would not be able to encode it).  This just isn't possible in the 2nd question.

I'm also beginning to think that it's impossible
to perform the trick if someone besides
the assistant picks the fifth card.

Clearly once the pile of four cards is chosen they will
need to encode 48 cards which is impossible since
in this case we can only rely on their ordering.

When I first read this problem I thought that
I might be overlooking something. But now
I also believe that it's impossible to perform
the trick without any sort of trickery (like
orienting the cards or something)

I guess the author of the book where
I got this problem from is asking
to show something which is impossible.  :?

Yes, but that happens sometimes. If I consider the amount of errors in my Engel book (and still it's great), I can't imagine how other books would have none :)