The MOEMS competition went public in the fall of 1979, the brainchild of Dr. George Lencher. There are 25 problems per year, with 5 problems on each of several monthly competitions in the fall. How many monthly competitions have there been? Assume that this year's MOEMS season has not started yet.

start of competition: 1979
last: 2004
problems per year: 25
problems per monthly competition: 5

.: 5 competitions per year; 26 years of competition

26 * 5 = 130 monthly competitions all in all

...[hide]it started in 1779 and not counting this year so 2004...thats 26 yrs...5 problems each compettion for 25 problems 25/5=5...so 26 x 5 which is....130[/hide]


trigonometry
algebra
polynomial
geometry unsolved
geometry

Let points A1, A2, ... , An form a regular polygon inscribed in a circumference with radius 1.

Prove that A1A2*A1A3*...*A1An = n

An elementary method would be interesting =)

By sin law we have $A_{1}A_{2}=2\sin{\frac{\pi}{n}},...$ etc. Also we need prove $2^{n-1}\prod_{i=1}^{n-1}\sin{\frac{i\pi}{n}}=n$.

That's right... I came across the same solution. But how can we prove it? And I wonder if there's any other solution.

Nice solution !  :)

[quote="mathmanman"]It remains to compute $\prod_{k=1}^{n-1}(1-e^{2ik\pi/n}).$[/quote]

However, we could directly see on the unit circle that $A_{1}A_{2}\times A_{1}A_{3}\times ... \times A_{1}A_{n}= \prod_{k=1}^{n-1}(1-e^{2ik\pi/n})$.

Benjamin

[quote="Virgil Nicula"][b]Similar exercise[/b] : $\prod_{k=1}^{n}\sin\frac {(2k-1)\pi}{2n}=\frac{1}{2^{n-1}}$ .[/quote]

Tchebycheff polynomials :D


geometry
circumcircle
trigonometry
geometry proposed

In an triangle $ ABC$ with circumradius $ R$,altitudes $ AD,BE,CF$ have lengths $ h_1,h_2,h_3$ respectively.If $ t_1,t_2,t_3$ are the lengths of the tangents from$ A,B,C$ to the circumcircle of triangle $ DEF$ respectively,prove that
\[ \frac {{t_1}^2}{h_1} \plus{} \frac {{t_2}^2}{h_2} \plus{} \frac {{t_3}^2}{h_3} \leq \frac {3}{2}R
\]

[b][color=darkblue]Denote $ \| \begin{array}{cc} L \in AD \cap (O_9) \\
J \in BE \cap (O_9) \\
K \in CF \cap (O_9) \end{array} \|$

Then by power of the point in $ (O_9)$:

$ \| \begin{array}{cc} t_1^2 = AL \cdot h_1 \\
 \\
t_2^2 = BJ \cdot h_2 \\
 \\
t_3^2 = CK \cdot h_3 \end{array} \| \Rightarrow \| \begin{array}{cc} \frac {t_1^2}{h_1} = AL = \frac {1}{2} AH \\
 \\
\frac {t_2^2}{h_2} = BJ = \frac {1}{2} BH \\
 \\
\frac {t_3^2}{h_3} = CK = \frac {1}{2} CH \end{array}$

Then we should prove:

$ \begin{array}{cc} AH + BH + CH \le 3R \\
 \\
2R(\cos{\alpha} + \cos{\beta} + \cos{\gamma}) \le 3R \\
 \\
\cos{\alpha} + \cos{\beta} + \cos{\gamma} \le \frac {3}{2} \\
 \\
\frac {R}{r} + 1 \le \frac {3}{2} \\
 \\
R \ge 2r \end{array}$
$ \mathrm{Q.E.D.}$
[/color][/b]
[hide="Image"][img]http://ipicture.ru/uploads/080605/GQdzG7OELn.png[/img][/hide]


VASC problem:
Let $ a,b,c >0, a^{2}+b^{2}+c^{2}=3$. Prove that:
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{9}{a+b+c}$
Stronger:
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1 \ge \frac{12}{a+b+c}$
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{6}{a+b+c}+\frac{3}{\sqrt{3(ab+bc+ca)}}$


number theory
greatest common divisor
Putnam

Prove that $\frac{gcd(m,n)}{n}C(n,m)$ is an integer.

Yes, $C(n,m) \equiv C_n^m \equiv {n \choose m}$

[hide="it was an AOPS challenge problem"]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7009&postorder=asc

You prove that $n$ divides both $m\binom{n}{m}$ and $n\binom{n}{m}$ and use the fact that $\gcd(km,kn)=k\gcd(m,n)$.
[/hide]

[quote="fuzzylogic"]use the fact that $\gcd(km,kn)=k\gcd(m,n)$.[/quote]

Or use the fact that the gcd of two numbers is a linear combination of those numbers.

This is a Putnam question, by the way.


number theory
relatively prime
number theory solved

does anyone know an ellegant proof for:

totient of 2^n-1 is divisible with n, for any integer n>0 ?

best regards

can someone post a solution, at least, pls?
(number of all positive integers less than 2^n-1 and relatively prime with 2^n-1 is divided by n, for any positive integer n)

Do you think someone will post a solution in half an hour? Not everyone reads the forum every second. And it takes time to solve these problems. Please be patient.

I think it is right not only for [tex]2^n-1[/tex],but for [tex]a^n-1[/tex] for any natural [tex]a[/tex],and it seems to me I can prove it.

That is exactly what I've done in the solution given in the link above. :D

Pierre.

[quote="pbornsztein"]Well, anyone who knwos that it has already been proposed and solved on ML can answer in less than half an hour :D

http://www.mathlinks.ro/viewtopic.php?t=14167

Pierre.[/quote]

What did you mean Pierre? This is still a conjecture?

If it has been solved I do think we can say it is a conjcture....
Please cleck on the links I gave, and you will find a proof of your task.

Pierre.

[quote="pbornsztein"]Please cleck on the links I gave, and you will find a proof of your task.

Pierre.[/quote]

i tried many times that link without any result (topic does not exists)
finally i succed to watch at that topic
thanks

p.s. it was so easy!  :blush:


function
limit
number theory proposed
number theory

[quote="TTsphn"][quote="Max D.R."]Prove that there are innitely many positive integers such that $ \pi(n)|n$.[/quote]
It is only a consequence of this problem  :wink:[/quote]
Which problem?  :huh:

Better, for every $ m\in\mathbb{N}$ there exist $ n \in \mathbb{N}_0$ such that $ m\pi(n) \equal{} n$. For $ m \equal{} 0$ take $ n \equal{} 1$. If $ m > 0$, since $ \lim_{i\to\infty}{\pi(i)i^{ \minus{} 1}} \equal{} 0$, define $ n: \equal{} \min\{x \in \mathbb{N}_0: \pi(x)x^{ \minus{} 1}\le m^{ \minus{} 1}\}$: by construction we have $ \pi(n \minus{} 1)(n \minus{} 1)^{ \minus{} 1} > m^{ \minus{} 1}$, it means that $ \pi(n \minus{} 1)\pi(n)^{ \minus{} 1} > (n \minus{} 1)n^{\minus{}1}$. But $ f(x): \equal{} (x \minus{} 1)x^{ \minus{} 1}$ is a strictly increasing function, so we [i]must[/i] have $ \pi(n) \equal{} \pi(n \minus{} 1)$. Summing up we have $ \pi(n)n^{ \minus{} 1}\le m^{ \minus{} 1} < \pi(n)(n \minus{} 1)^{ \minus{} 1}$, that is $ n \equal{} m\pi(n)$. [] :D


ah`, de thi chon doi tuyen toan quoc gia thi toan Quoc te kha lau roi
                  va chac chan bay gio thi co danh sach chinh thuc doi tuyen toan Quoc te roi chu ? ban nao biet thi Post len de cho than dan " Vietnam " xem ten biet mat :6 anh hao dai dien cho toan hoc Vietnam nam nay nao ?

bạn thanhnam xem cái này xem
[url]http://mathnfriend.net/index.php?showtopic=3085[/url]

ah`, co the cxho ten chinh xac + tinh thanh cac thanh vien do khong ?  de lam wen thoi
                      chi biet moi Le Hong Qui thoi .

         con may nguoi kia wen nhung ma khong biet ten that su + tinh thanh nao nua ?  hi vong la som nhe

Lê Hồng Quý
Lê Nam Trường
Nguyễn Xuân Thọ
Hoàng Mạnh Hùng
Nguyễn Duy Mạnh
Đặng Bảo Đức

Hix!!!cho em biết lớp luôn đi!!!em muốn biết xem các bác í già hơn mình bao nhiêu tuổi á mà!!! :blush:  :blush:

5 bạn 1988,một bạn 1989.
  có thể xem thêm profile của mấy người đó thông qua bảng kết quả thi quốc gia trên THTT...

Thanh tich doi tuyen Vietnam trong kithi toan quoc te  kha on :
 2 vang
2 bac va 2 dong
  minh chi biet : Le Nam Truong -Hatinh : HCB
                        Le Hong Qui -SP Toan DHVinh :HCD
 Doan Vietnam dung thu 13 toan doan ( TQ van la so 1 )


geometry
geometric transformation
rotation
analytic geometry
graphing lines
slope
trigonometry

Assuming you mean rotating around the origin, your method doesn't work. For example, take a horizontal line and an almost vertical line both passing through the origin. Then the average is really really high, but the slope of the angle bisector is barely below $ 1$.

Edit: Sorry, silly mistake.

Slope should be $ 1$.

What if the line isn't $ x \equal{} 1$?

A pretty straightforward way to do this is using polar coordinates; this at least makes the computation obvious.

You could do it with complex numbers too.
\[ e^{i\frac{\pi}{4}}\equal{}\cos\frac{\pi}{4}\plus{}i\sin\frac{\pi}{4}\equal{}\frac{1}{\sqrt{2}}\plus{}i\frac{1}{\sqrt{2}}\]
And the slope would be
\[ \frac{\text{Im}(e^{i\frac{\pi}{4}})}{\text{Re}(e^{i\frac{\pi}{4}})}\equal{}\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\equal{}1\]

Though I think both complex numbers and polar coordinates entail trigonometry.

Well, they're all really closely related...

[quote="iYOA"]is there a way to find the slope of a line rotated 45 degrees without having to use trig?

my method was to use the line and find the  slope of its perpendicular part
i was then trying to find the slope of the angle bisector of these 2 perpendicular lines.  can that be done by taking the average of the slopes?[/quote]

I don't know about the without using trig bit, in as much as the operation of rotation requires multiplication of a matrix whose entries are trig functions by a vector whose components are x and y. That said, applying a bit of linear algebra and differential calculus, neither of which I intend to show here, the slope of a line rotated by an angle $ \theta$ is given by:

\[ m' \equal{} \frac{m\cos\theta \plus{} \sin\theta}{\minus{}m\sin\theta \plus{}\cos\theta}\]

where $ m$ is the slope of the original un-rotated line.

Let's assume the line goes through the origin. Suppose the point $ A(1,m)$ is on the line. This point rotated 90 degrees is $ C( \minus{} m,1)$. Add them to get $ B(1 \minus{} m,1 \plus{} m)$. Thinking in terms of the square defined by these three points and the origin, we find that this fourth point is on the line which is a 45-degree rotation of the first line. The slope of the line through the origin and this point is $ \frac {1 \plus{} m}{1 \minus{} m}$.

It is crucial that we are dealing with a square (or really, a rhombus). We had to verify that $ A$ and $ C$ are the same distance from the origin first before we could assert that the diagonal $ B$ was the angle bisector of $ \angle AOC$.

Dr. No's result can also be derived from complex numbers.
If $ (x,y)$ is a point which must be rotated $ \theta$ radians, we have:
$ (x\plus{}yi)\cdot e^{i\theta}\equal{}(x\plus{}yi)(cos\theta\plus{}i\sin\theta)\equal{}x\cos\theta\plus{}xi\sin\theta\plus{}yi\cos\theta\minus{}y\sin\theta\equal{}z$.
Our slope is given by the traditional rise/run formula, which translates to $ \frac{\text{Im}(z)}{\text{Re}(z)}$ in the complex plane:
\[ \frac{\text{Im}(z)}{\text{Re}(z)}\equal{}\frac{x\sin\theta\plus{}y\cos\theta}{x\cos\theta\minus{}y\sin\theta}
\]

Which is Dr. No's result, without the use of differential calculus or linear algebra.

It's all pretty much the same.  Complex numbers and $ 2 \times 2$ rotation matrices are [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=216343]essentially the same thing[/url], so there's not a reason to prefer one over the other in this context except for notational reasons.

That having been said, any answer to this problem is at least equivalent to a special case of an angle addition formula, which can be phrased without the use of trigonometry but is really more natural phrased trigonometrically (and even more natural phrased with rotation matrices).

ok thanks.

after giving it much thought, it seems trig is the best way to approach this

fortunatley, since it is 45 degrees we can use right special right triangles to identify the coordinates of a point on the rotated line
we know the other coordinates of one other point, the origin.
and then we can find the slope  using theses 2 points

its a bit tedious and sometimes slopes can be irrational.

iYOA, I urge you not to ignore this solution:

[quote="TZF"]Let's assume the line goes through the origin. Suppose the point $ A(1,m)$ is on the line. This point rotated 90 degrees is $ C( \minus{} m,1)$. Add them to get $ B(1 \minus{} m,1 \plus{} m)$. Thinking in terms of the square defined by these three points and the origin, we find that this fourth point is on the line which is a 45-degree rotation of the first line. The slope of the line through the origin and this point is $ \frac {1 \plus{} m}{1 \minus{} m}$.[/quote]

Given a line with a positive slope $ m$, TZF has basically outlined a proof that the slope of the line rotated 45 degrees CCW is $ \frac {1 \plus{} m}{1 \minus{} m}$.  I think plugging a slope into this fraction is simpler than using special triangles and actually finding another point as you've suggested.


[quote="236factorial"]How many such two digit numbers are there where when you reverse the digits, the number becomes twice the original number?[/quote]

[hide]

  Lets say our orginal number is:

$10x +y$

  By reversing the digits we yield:

$10y +x$

  From the conditions of the problem we obtain:

$10y+x= 2(10x+y)$

$10y+x=20x + 2y$

$8y=19x$

  Thus, $x$ and $y$ must be in the ratio of $8:19$ meaning $y$ a single digit, must be a 2-digit number or greater for the expression to hold, a contridiction, thus no such numbers exist.
[/hide]

[quote="pianoforte"][hide]
12
24
36
48

4?
[/hide][/quote]

Uh.... 21 is not twice 12, 42 isn't twice of 24, etc.


function
quadratics
modular arithmetic
superior algebra
superior algebra unsolved

here is a easy one:
let p be a prime number and k>=2 an integer. In Z/pZ,
what is the cardinal of {x^k where x belongs to Z/pZ}.

and in Z/nZ i don't find the solution ..

Is it $\frac {p-1}{(k,p-1)}$? Or that number $+1$, if we include $0$.

This is because given $x^k\ne 0$, there are as many $y$ in the set $\{y\in\mathbb Z_p|y^k=x^k\}$ as there are solutions to the equation $a^k-1=0$, and it's well-known that there are $(k,p-1)$ such solutions.

I'm not sure, but this seems to be it.

yes, I found like you. Here's my solution:

consider the morphism f(x)=x^k from (Z/pZ)* to itself.
Then Card(E)=Card((Z/pZ)*)/card(ker f) where E={y s.t. y=x^k, x<>0}.
But ker f={x s.t. x^k=1}. Hence x is in (ker f) iff ord(x) is a divisor of
m=GCD((p-1),k) . Thus (ker f)={x s.t. x^m=1}. But Z/pZ is a field so x^m-1=0
has at most m solution and we know there are at least m solutions by Cauchy
theorem. So we are done.

but it can't be generalized for arbitrary n with the same idea because
we used the fact that Z/pZ is a field.. what the solution in Z/nZ ??

yes, I found like you. Here's my solution:

consider the morphism f(x)=x^k from (Z/pZ)* to itself.
Then Card(E)=Card((Z/pZ)*)/card(ker f) where E={y s.t. y=x^k, x<>0}.
But ker f={x s.t. x^k=1}. Hence x is in (ker f) iff ord(x) is a divisor of
m=GCD((p-1),k) . Thus (ker f)={x s.t. x^m=1}. But Z/pZ is a field so x^m-1=0
has at most m solutions and we know there are at least m solutions by Cauchy
theorem. So we are done.

but it can't be generalized for arbitrary n with the same idea because
we used the fact that Z/pZ is a field.. what the solution in Z/nZ ??

Maybe $\frac{\phi(n)}{(k,\phi(n))}$, where $\phi(n)$ is the Euler's Phi (or Totient) function?

Not by a long shot.. In particular, this would imply that there are $\frac{\phi(n)}2$ quadratic residues $\pmod n,\ \forall n$, and we can find lots of counterexamples ($n=10$, for exmple, but there are more :)).


algebra polynomial calculus integration induction

Prove that the nth degree polynomial P(x) has integral values for x=0,1,2,...,n then it has integral values of x

b) Prove that if a polynomial P(x) of degree n has integral values n+1 succesive integers x, then it has integral valued for all integers x.

c) Prove that if the polynomial P(x) of degree n has integral values for x=0,1,4,9,16,...,n^2, then it has integral values for all integers x which are perfect squares (but this does not necessarily follow for all integers x)
Give an example of a polynomial which assumes integral values for all integers x which are perfect squares, but which for some other value of x yields a rational (not whole) number.

[quote="mdk"]Prove that the nth degree polynomial P(x) has integral values for x=0,1,2,...,n then it has integral values of x[/quote]

[hide="Humongous hint"] Write $ P(x)$ in the form

$ P(x) \equal{} a_{n}{x\choose n}\plus{}a_{n\minus{}1}{x\choose n\minus{}1}\plus{}...\plus{}a_{0}$

And induct. [/hide]

[hide="or just do it"]
a) Induction on degree:

Base case: constant polynomials. Duh
Inductive step: $ P(x)\minus{}P(x\minus{}1)$ is a polynomial of degree $ n\minus{}1$, if $ P$ is of degree $ n$. This not hard to show. This polynomial has integral values for the $ n$ differences given. Thus, by the inductive hypothesis, it has integral values everywhere. Thus, P(x) also has integral values everywhere.

(everywhere means every integral input)

b is very similar. c is more interesting.
[/hide]

c isn't that much more interesting.  [hide="Very big hint for the first part"]Consider the polynomial $ P(x^{2})$.[/hide]

Well, all that's left is to give the example, then.

[hide="Here is one"] $ P(x) \equal{}\frac{x(x\minus{}1)}{4}$ [/hide]


algebra
polynomial
induction
modular arithmetic
function
calculus

This is indeed a difficult problem, which has been proposed as ex. 10674 in the A.M.M. It seems that only the proposer provided a solution (see. vol.107, Aug.-Sept. 2000, p. 654), which is probably impossible to find....
It is not easy to write it here, so I will give it only if you cannot reach the above reference.

Pierre.

No, I cannot find that reference and I would be most glad if you could post a solution.

Ok, let's go (I have interchanged a and b, but it is too late....i will not write it again...) :

Define a sequence (Q_k) of polynomials with deg(Q_k)  \leq  k by Q_0 = -1, and
Q_(k+1)(x) = b k+1 (x-1)Q_k(bx) - a(b k+1x - 1)Q_k(x)  for k  \geq  0.
Observe that Q_(k+1)(0) = (a - b k+1 )Q_k(0). Iterating this and employing Q_0 = -1 leads to
Q_k(0) = -(a - b k )(a - b k-1 )...(a - b).
Assume that a = b j  for every non-negative integer j, so that Q_k(0) =/=0.
We will obtain a contradiction by identifying a k such that Q_k is identically 0.

Let r_(0,n) = (a n  - 1)/(b n  - 1) for n > 0.
By assumption r_(0,n) is an integer.
For k  \geq  0 define r_(k+1,n) recursively by :
r_(k+1,n) = b k+1 r_(k,n+1) - a*r_(k,n).
Let p_0 = 1 and p_(k+1) = a(1 - b k+1 )p_k.

By induction on k it follows for n  \geq  1 and k  \geq  0 that :
r_(k,n) = [a n p_k + Q_k(b n )]/[(b n+k  - 1)(b n+k-1  - 1)...(b n  - )]

Now fix k so that b k  < a < b k+1 . Since :
(b n  - 1)(b n+1  - 1)...(b n+k  - 1) = b n(k+1) (1 - b -n )(b - b -n )...(b k  - b -n )  \geq  b n(k+1) /2,
we see that
|r_(k,n)|  \leq  |a n p_k + Q_k(b n )|/[b n(k+1) /2]  \leq  2( |p_k|(a/b k+1 ) n  + |Q_k(b n )|/(b n ) k+1  ).  (*)

Since a < b k+1  and deg(Q_k) < k+1, the rightmost expression in (*) is bounded above by 1 when n is sufficiently large. Thus r_(k,n) = 0 for large n, since it is an integer. For all large n, this yields
a n p_k + Q_k(b n ) = 0 and thus p_k(a/b k ) n  + Q_k(b n )/(b n ) k  = 0.
This forces p_k = 0, since otherwise the lefts side is unbounded as n - > +oo. We now conclude that Q_k(b n ) = 0 for all n and thus Q_k is the zero polynomial, and we are done.

Pierre.

Oh, my God! From where did he take those? My ideas ar the following:
  Suppose that b is not a power f a and take k=[log(base a) b]. Then a^k<b<a^(k+1). Put b^n-1=x_n(a^n-1). I've manged to show that we can find some non-zero constants C-k,...,C-0 such that for all sufficiently large n we have C_k*x_(n+k)+C_(k-1)*x_(n+k-1)+...+C_1*x_n+1+C_0=0. But I don't know how to continue. Help!

Who was the proposer of this question in the A.M.M.?

Wow! I didn't know that. I know Marius Cavachi has extremely difficult problems, but this one is the hardest of them all. He is really the teacher with the hardest proposed problems in our country.

Look what I did. Maybe someone knows how to continue.
   Let us choose k such that a^k<b<a^(k+1). Define x_n=(b^n-1)/(a^n-1) and r_1=b/a, r_2=b/a^2,...,r_k=b/a^k. Then we have the identity:
  x_n=[b^n-a^kn]/[a^kn(a^n-1)]+ \sum (r_i)^n. Take the sequence a_n= \sum (r_i)^n and consider also the polynomial P(X)=(X-r_1)...(x-r_k). Then if we denote C=k(k+1)/2, we find that:
    a_n-( \sum r_i)a_(n-1)+....+(-1)^kr_1*...*r_k*a_(n-k)=0. But we can easily see looking at the polynomial that  \sum r_i,..., r_1*...*r_k are of the form T/C, with T an integer. Thus, if we write S_1=C*(- \sum r_i),...
S_k=C(-1)^kr_1*...*r_k, we find that S_1,...,S_k are integers and also that Ca_n+S_1a_(n-1)+...+...+S_ka_(n-k). But from the first identity we can see that the sequence x_n-a_n goes to 0. Then so does the sequence
C(a_n-x_n)+....+S_k(a_n-a_(n-k)). Thus the sequence Cx_n+...+S_k*x_(n-k) goes to 0 and it's formed by integers. Therefore we can find n_0 such that whenever n>n_0 this last sequence is 0. But I cannot find a contradiction from this assertion.

Does everybody think that this idea cannot lead to a solution? I still hope, but I still can't prove that. Any ideas?

It can also be solved in an entirely different way. Specifically, pick an arbitrary prime $p>a$, and let $n$ be the order of $a$ modulo $p$. Since $p$ divides $a^n-1$, it must also divide $b^n - 1$. But as $a$ has order $n$ modulo $p$, $b$ has to be a power of $a$ modulo $p$ (that is, the congruence $b \equiv a^x \pmod{p}$ is solvable). We conclude that $b$ is a power of $a$ modulo all sufficiently large primes. It is then a well-known result of A. Schinzel (proven easily with the Chebotarev density theorem) that $b$ must be a power of $a$, which is the assertion of our problem.

In fact, this argument proves a stronger claim: [i]Let $S \subset \mathbb{N}$ be a set that intersects all (infinite) integer arithmetic progressions. If, for all $n \in S$, the set of prime divisors of $a^n - 1$ is a subset of those of $b^n-1$, then $b$ is a power of $a$[/i].

--Vesselin

Yes, vess, if we throw the atomic bomb on it, it works even if we suppose that the divisibility relation holds for infinitely many values of n, but who can prove this result? I've read in an article that this is true, but they say the proof is monstruous. Probably for you is easy, since you are expert in this field, but do not ask anyone to be.

[quote="harazi"]Someone gave me the following extraordinary problem:
  Let a,b>1 be integers such that for all n>0 we have a^n-1|b^n-1. Then b is a natural power of a. I can't find a solution.[/quote]
Well, I think using Cyclotomical polynomials, Eulerphi function and modular theory this problem is TRIVIAL   :D  :D  :D  so read my problem in the FORUM, and try to use it!

Best Regards.

RX TCM

Excuse me, great genius of number theory, for such a trivial problem. Probably, since it seems there is no problem enough difficult for you you can start thinking about great unsolved problems in nulber theory. Probably they are not so trivial. It seems that all readers of AMM and many mathlinkers are too idiot.

Guys, guys, guys settle down. Untill we see proof, disregard RobertuX statement of the problem above being trivial. Maybe he mis-understood the problem ;) Let's not be harsh with each other here, shall we? :)

[quote="harazi"]I've manged to show that we can find some non-zero constants C-k,...,C-0 such that for all sufficiently large n we have C_k*x_(n+k)+C_(k-1)*x_(n+k-1)+...+C_1*x_n+1+C_0=0. But I don't know how to continue. Help![/quote]

But you have already done most of the work!  :?  Perhaps you have already finished the proof from here, but anyway, here it goes:

You have proven that, for $n>n_0$, the $x_n$ take the form
\[
x_n = c_1\alpha_1^n + c_2\alpha_2^n + \cdots + c_s\alpha_s^n,
\]
where $\alpha_1,\ldots,\alpha_s \in \mathbb{C}^*$ are pairwise different non-zero complex numbers, and the $c_i$ are (non-zero) constants. Hence, for every $n > n_0$, we have the identity
\[
b^n -  1 = (a^n - 1) x_n = \sum c_i(a\alpha_i)^n - \sum c_i \alpha_i^n,
\]
i.e.
\[ (*) \quad
b^n - 1^n - \sum c_i(a\alpha_i)^n + \sum c_i \alpha_i^n = 0 \quad \forall \, n>n_0.
\]
Now, the following simple assertion, which is an immediate corollary of the properties of the Vandermonde determinant, has received at least two other solutions in this forum:

[i]If $y_1,\ldots,y_r \in \mathbb{C}^*$ are pairwise different non-zero complex numbers and $b_1,\ldots,b_r$ are constants satisfying $b_1y_1^n + \cdots + b_ry_r^n = 0$ for all $n>n_0$, then $b_1 =\cdots =b_r = 0$. [/i]

Now, as the constants $c_i$ are assumed to be non-zero and $a>1$, (*) implies that there is an $i_0$ such that $\alpha_{i_0} = 1$, and there is a $j_0$ such that $a\alpha_{j_0} = b$ (otherwise, the coefficients of $1^n$ and  of $b^n$ are $-1 \neq 0$ and $1 \neq 0$, respectively, and we get a contradiction with the above assertion). Furthermore, because the $\alpha_i$ are pairwise different, every $\alpha_i$ with $i \neq i_0$ must appear as some $a\alpha_{i'}$, and also conversely: each $a\alpha_j$ with $j \neq j_0$ must appear as some $\alpha_{j'}$. We conclude that the following sets coincide:
\[
\big\{ \alpha_i \mid i \neq i_0 \big\} \equiv \big\{ a\alpha_j \mid j \neq j_0 \big\}.
\]
As $\alpha_{i_0} = 1$, it is easy to conclude from here that $\{ \alpha_1,\ldots,\alpha_s \} = \{1,a,a^2,\ldots,a^{s-1}\}$. Hence, $b = a\alpha_{j_0}$ is a power of $a$ (actually, it is $a^s$), as claimed. QED

--Vesselin

To Valentin: I consider a little lack of respect considering a problem that did not received solutions from the readers of AMM as being trivial. Anyway, I apologise for the tone.
   To vess: yes, vess, I have already found the continuation of the solution. Thus, we have another solution to this magnificient problem.

[quote="harazi"]To Valentin: I consider a little lack of respect considering a problem that did not received solutions from the readers of AMM as being trivial. Anyway, I apologise for the tone. [/quote]Of course Harazi. It is possible for this to be just a language / problem misunderstanding (which I think is the case here), in which case there is no need to argue one with each other :). With this post, let us end the off-topic, and continue posting regarding this indeed marvelous problem ;)

This is a really nice and hard problem, with an excellent solution. But i wonder that we could have an arithmetical solution ? We can see that every prime divisors of b is divide a and if a is divide b then b is a power of a. Is it can be continued ?

Yes, the problem does admit a solution with algebraic number theory. The key is that it suffices to show that, for any prime $\ell$, there exist $j = j(\ell),c =c(\ell)$ such that $b = a^jc^{\ell}$. For $\ell = 2$, for example,  it is easily seen that if $b$ does not have the form $a^jc^2$, then there exists a prime $p$ such that $\Big( \frac{a}{p} \Big)=1, \Big( \frac{b}{p}  \Big) = -1$, and taking $n = \frac{p-1}{2}$ contradicts $a^n - 1 \mid b^n - 1$. Try generalizing this idea for any $\ell$.

Yeah,  when i solved this problem, i see that if $b^n-1$ is divisible by $a^n+1$ then $b$ is a even power of $a$. That means  $b^n-1$ is divisible by $a^n+1$ with every n then $b^n-1$ is divisible by $a^n-1$ with every $n$. What do you think ?

[quote="zscool"]proposed by Marius Cavachi of romania.[/quote]
Is he really the author of this problem?

Actually this problem was published in russian book

[i]В.А.Садовничий, А.А.Григорьян, С.В.Конягин "Задачи студенческих математических олимпиад"[/i]
("Problems of mathematical olympiads for university students")

back in 1987 as a problem 27 in the section 4.1. But there is nothing said about who is its author. BTW, solution from that book is quoted [url=http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/solutions/April2005.html/%24FILE/opuz0405.pdf]here[/url].

Well, after the problem about $a^n+n|b^n+n$ was given at ROM TST /Taiwan TST 2006 (and you can guess that also somewhere else), it's natural to ask for the following generalisation, which is done by the same method:

If $p,q \in \mathbb{Z}[X]$ with $a^n-p(n)|b^n-p(n)$ for all sufficiently big $n$, then $b=a^k, q=p^k$,

[quote="maxal"]Actually this problem was published in russian book

В.А.Садовничий, А.А.Григорьян, С.В.Конягин "Задачи студенческих математических олимпиад"
("Problems of mathematical olympiads for university students")

back in 1987 as a problem 27 in the section 4.1. But there is nothing said about who is its author. BTW, solution from that book is quoted here.[/quote]

  I would be very appreciate if someone could explain for me this argument:
" In the case $k=1$ we have that for n sufficiently large $qm_{n+1}-pm_{n}=0$, $m_{n} <>0$ this implies that $q=1$"
  Please help me! :(

[quote="Gibbenergy"]I would be very appreciate if someone could explain for me this argument:
" In the case $k=1$ we have that for n sufficiently large $qm_{n+1}-pm_{n}=0$, $m_{n} <>0$ this implies that $q=1$"
  Please help me! :([/quote]
Let $t$ be any positive integer. We have
$qm_{n+1} = p m_n$
$qm_{n+2} = p m_{n+1}$
...
$qm_{n+t+1} = p m_{n+t}$
Multiplying them up we have $q^t m_{n+t+1} = p^t m_n$ meaning that $q^t\mid m_n$.
Since $t$ is arbitrary and $m_n\ne 0$, it must be $q=1$.

in the opuz0405.pdf

In the induction, we can get $x_{1},x_{2},\ldots,x_{k-1}\in\mathbb{Z}$ but how can we get $x_{k}\in\mathbb{Z}$? And where do we need $x_{1},x_{2},\ldots,x_{k}\geq 1$? (Actually we need it I think)

[quote="shalex"]in the opuz0405.pdf
In the induction, we can get $x_{1},x_{2},\ldots,x_{k-1}\in\mathbb{Z}$ but how can we get $x_{k}\in\mathbb{Z}$?[/quote]
Choice of $x_{k}=p/q$ was rather arbitrary.
Applying the same arguments to $x_{1}=p/q,$ one can get that $x_{2},\ldots,x_{k}\in\mathbb{Z}.$
[quote="shalex"]And where do we need $x_{1},x_{2},\ldots,x_{k}\geq 1$?[/quote]
To get $m_{n}\ne 0.$

A very similar problem but about integral functions instead of integers:
[url]https://artofproblemsolving.com/community/c6h136198[/url]

[quote="vess"]The congruence $b \equiv a^{x}\pmod{p}$ is solvable. We conclude that $b$ is a power of $a$ modulo all sufficiently large primes. It is then a well-known result of A. Schinzel (proven easily with the Chebotarev density theorem) that $b$ must be a power of $a$.[/quote]

I'd like to see if the following elementary proof of this "well-known result" is correct, just to have a completely elementary proof of the whole problem:

The equation is solvable for every prime such that $(a,p)=1$, so it is solvable for every positive integer such that $(a,n)=1$

Let $n=a^{c}-1$ for a positive integer $c$ such that $a^{c}-1>b$

Obviously $c=ord_{n}(a)$ so there is a nonnegative integer $x<c$ such that $n|a^{x}-b$ but since $-n<a^{x}-b<n$, then we have $a^{x}-b=0$ that is our thesis.

EDIT: ops Vess, you are right, it works only if $(ord_{q}(a),ord_{p}(a))=1$

[quote="piever"]
The equation is solvable for every prime such that $(a,p)=1$, so it is solvable for every positive integer such that $(a,n)=1$
[/quote]

Not so obvious! How, for instance, the existence of $x,y$ with $b \equiv a^{x}\mod{p}$ and $b \equiv a^{y}\mod{q}$ yields a $z$ with $b \equiv a^{z}\mod{pq}$? An example such as $b = 4, a =-1, p = 3, q = 5$ show that the first two propositions do not imply the third: we have $4 \equiv (-1)^{0}\pmod{3}, 4 \equiv (-1)^{1}\pmod{5}$, but $4 \equiv (-1)^{z}\pmod{15}$ is not solvable.

What you do show is this weaker statement: if the congruence $b \equiv a^{x}\pmod{n}$ is solvable for all $n$ coprime to any given $D$, then $b$ is a power of $a$.

For generalisations and analogues for abelian schmes and odd-dimensional Quillen $ K$-theory groups of algebraic number fields see http://arxiv.org/abs/0809.1991 and the references given there.

A more general statement: [i]"if $a^n-1$ divides $b^n-1$ for [b]infinitely many[/b] integers $n$, then $b$ is a power of $a$"[/i] is proved in the paper http://www.emis.de/journals/JTNB/2005-1/pages423-435.pdf

I think they have a proof in the last example of chapter 17 of PFTB.
It is amazing how a number theory problem has such a not number theory solution! (See post #28) How in the world would anyone come up with such a solution?

[quote=vess]It is then a well-known result of A. Schinzel (proven easily with the Chebotarev density theorem) that $b$ must be a power of $a$, which is the assertion of our problem.
[/quote]

Sorry, but can someone post a solution to this using Chebotarev density theorem? Thank you very much!

[hide=Solution from 1 year ago?] Step 1: Show $rad(b)\mid rad(a)$. This implies $b\mid a^m$ for all sufficiently large integral $m$.

Step 2: Show$$a^n-1 \mid \prod\limits_{e=0}^{n-1} (b-a^e)$$
This implies$$(a^t-1)^{\lfloor \frac nt \rfloor} \mid \prod\limits_{e=0}^{n-1} (b-a^e)$$
Now some degree bounding would suffice, as $\gcd(b,LHS)=1$ so we can take out a bunch of $b$'s from the RHS. We also employ the identity $a^t-1 = \prod\limits_{n\mid t} \Phi_n(a)$

We have $P(a)=\prod\limits_{t=1}^n \Phi_t(a)^{\lfloor \frac nt \rfloor} $ actually has degree$$\sum\limits_{t=1}^n \varphi(t) \lfloor \frac nt \rfloor = \sum\limits_{t=1}^n \varphi(t) \sum\limits_{kt\le n} 1 $$
Let $u=kt$ then this sum is $\sum\limits_{u=1}^n 1 \sum\limits_{t\mid u} \varphi(t) = \sum\limits_{u=1}^n u$ which is equal to the degree on RHS. However, observe that RHS is divided by $b^{n-C}$ for some constant C, while the left hand side generally has smaller growth (sometimes you take out prime factors, but they seem rather small) so we are done. [/hide]

Here's an elaboration of the above solution (for my own benefit, since I suck at NT)

Step 1: Prove that $\text{rad}(b)\mid \text{rad}(a)$.
Proof: If $p\mid b$ then $p\nmid b^n-1\implies p\nmid a^n-1\implies p\mid a$. So (maybe a hard skill?) $b\mid a^m$ for large $m$.

Step 2: Prove that for an odd prime power $p^k$ all solutions to
\[b^{\text{ord}_{p^k}(a)}\equiv 1\pmod {p^k}\]
are exponents of $a$.
Proof: I probably overcomplicated this (not good at NT intuition, rip). Let's set $a=g^e$ for primitive root $g$ and $b=g^f$, then
\[f\cdot \text{ord}_{p^k}(a)=f\cdot \frac{p^k-1}{\gcd(e,p^k-1)}\]
is an integer, hence $\gcd(e,p^k-1)\mid f$ is necessary and sufficient, and yet this is equivalent to having $(g^e)^c\equiv g^f\pmod {p^k}$ for some $c$.

Step 3: Use [b]polynomials[/b], a useful hard skill.
Proof: Well, since $b$ satisfies the equation in Step 2, there's always some $b-a^e$ which is divisible by $p^k$. I find this hard to motivate, but it anyway implies
\[a^n-1\mid \prod_{e=0}^{n-1}(b-a^e)\]
where $\nu_2$ analysis is trivial (since $a,b$ are both odd). (Possible motivation: since $b-a^e$ will be divisible by $p^k$ and the product of the prime powers forms the odd part of $a^n-1$, we have the LHS. But $e$ varies, so we should [b]multiply together[/b] to get the RHS. I think you just have to push it and hope for the best.)

Step 4: [b]Bound degree[/b], another useful hard skill. [b]Bounding[/b] in general is nice.
Okay, at this point we want to bound degree (heuristically, size arguments should work). First realize that [b]distinct irreducible polynomials are relatively prime[/b]. Since the LHS has too small degree as a polynomial of $a$, and the RHS is fixed if we decrease $n$, we might as well try to multiply together a bunch of irreducible polynomials (hopefully this step becomes more natural as I improve). So let's take the cyclotomic polynomials $\Phi_t(a)$ for $1\le t\le n$. We should get
\[\prod_{t=1}^n \Phi_t(a)^{\lfloor \frac{n}{t}\rfloor}\mid \prod_{e=0}^{n-1}(b-a^e)\]
and now the degrees match, and yet the RHS should be divided further by $b^{n-C}\ge a^{n-C}$ from Step 1. Hence the LHS has "larger" degree (grows faster) so the expression in the RHS is $0$, meaning $b$ is a power of $a$.

Wow that's kind of hard. Probably beyond my skill level right now, most of the ideas are totally new to me and I have no experience with intpoly. :(


Prove that the interior angles of a quadrilateral add up to 360 degrees.

Also try using vectors.

Isn't that the definition of a quadrilateral?

I guess we can choose four points for a given quadrilateral such that no three are collinear, find component vectors, use $ \cos\theta\equal{}u\cdot{v}/|uv|$, and sum these angles to be 360 degrees.

Or we can combine two triangles to form any quadrilateral.  2 times 180 degrees is 360 degrees.

im just wondering, why did you want us to use vectors?

By the definition of polygons, the exterior angles of a quadrilateral must sum to $ 360^\circ$. Then the sum of the interior angles is $ 720 \minus{} 360 \equal{} \boxed {360^\circ}$. :D


Euler
function
superior algebra
superior algebra unsolved

Prove that if a and b are elements of a finite group G, then

ord(ab) = ord(ba).

trivial, and does not have anything to do with the phi-function or the finiteness of G. check the definition of ord.


hi.

How many palindromes $ N$ exist, such that all of its five digits are odd ?

[hide]$ 5\cdot5\cdot5\cdot1\cdot1$, which equals $ 125$.[/hide]


Yeah, I never really understood this issue: when is r in (r, theta) negative and why? From the formula r = Sqrt(x^2 + y^2), it would seem that r is always positive...but its not. Any clarifications?

Fierytycoon

[color=cyan]Heading along the ray a negative distance is just going to be the same as heading backwards.  In other words, (-r, :theta:) = (r, :theta: + :pi:).  Remember that the formulae for r and :theta: in terms of x and y will only give you one possible name for the point, while in polar coordinates points have infinitely many names.  So the fact that you can get a few more names for any given point shouldn't really cause any problems.[/color]

Yeah it can be written in a few ways like that. However the r cis theta form always is supposed to have a positive r, you change the angle bit to make it correct. So even though it can equal r cis theta with a negative r, this isn't really the proper way to write it.

In general, it's not going to be very useful to have a negative value for r.  (Although there must be some counterexample out there!)

a negative r value just means you go backward.  You probably knew that already, but I'm just clarifying...so like JBL said, if you go 180 degrees (pi) around the circle, you'll be where you want.

Thanks. Hmm, yeah, I saw this in my school precalc book. It was introducing polar coordinates, and had a table going like

r          1   2   2   1   -2   -2    -1
theta     0  30 60  90  120  150   180

Or something like. I don't have it with me right now, but it had negative values here and there, and, since the theta values sorta ascended in order, it didn't really agree with what most ppl have said. Hmm...do any of you know why the textbook would have something like this?

Fierytycoon

[color=cyan]What exactly is your problem with the table the book had?

As for complex numbers, yes, usually the r in r cis :theta: form is positive.  The reason for that, however, is that r cis :theta: is e^(x + iy) = e^x * e^iy, and e^x is always positive for real x.  But there isn't any particular reason otherwise that it should be so.[/color]

I will get back to you once I get hold off the book later today, so I can tell you exactly how the book had it.

Fierytycoon

Hmm, actually, i think what you guys said does seem right. The table in the textbook was of r = 2cos(theta), which is sorta only in the positive x quadrants (-90 to 90 degrees). The table goes like:

theta  0  30    45    60 90  120   135     150     180
r____  2  :rt3:  :rt2: 1   0   -1    -:rt2:    -:rt3:  -2

Fierytycoon

[quote="mathfanatic"]In general, it's not going to be very useful to have a negative value for r.  (Although there must be some counterexample out there!)  [/quote]

If you allow r to take any real value, it helps in graphing. Take :theta: = :pi:/4 - if  r  :ge: 0, you'll get half a line. There's probably a better example out there, though.


algebra
polynomial
superior algebra
superior algebra unsolved

E is the extension of field F.and $ a \in E$ is algebraic on F.suppose b$ \in F(a)$.if n= degree of  minimal polynmial b on F,and m=degree of minimal polynomial a on F.proof $ n|m$.


Just curious about this...Personally, I think it is fair that they hire based on the content on your website- you don't want someone who does illegal activities working for you/someone who is an excessive drinker, etc.....I don't really know...

[color=green]Note from moderator: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]

Of course its fair, why wouldn't it be?

[color=green]Note from moderator: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]

Because some people don't have websites and don't want to make any.

[color=green]Note from moderator: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]

And so they wont be discriminated against for no reason.

I think an employer should be able to hire or fire whomever they want for whatever reason.
It's up to us as a society to model our social employment safeguards as we see fit (for example, making it unlawful for someone to fire an employee just because they're gay, or making it so someone with a single felony conviction can have it expunged after a certain period of time so they can find employment later in life, etc, etc) but as far as peripheral issues like "websites", appearance, etc, I do believe that employers should have a wide range of discretion on things like that.

Contrary to popular belief, "discrimination" in and of itself isn't unlawful.
Discrimination against certain protected classes of individuals based on their 'protected status' is illegal (discriminating against blacks because they're black, women because they're women, etc), but the rights of employers to make discriminatory decisions as they see fit, for the benefit of their own business, is the cornerstone of a free economy and in my opinion, should be left largely untouched.

We had an issue here locally where an individual who weighed 500 pounds was denied employment at a  furniture store because the employer didn't feel that person was who he wanted representing his business and openly told him that his weight was the reason why. He sued and lost, resued and won on appeal and it is presently pending another appeal.
It made for a very robust debate.

In a Massachusetts court, plaintiff has the burden to prove that:
1. the principle reason of denial of employment was their race/gender/etc
2. that they were otherwise qualified for the job
3. that the reasons presented by the defense comprise a subterfuge
4. that the race/gender/etc issue was not protected by a bona-fide occupational requirement
[i]- ex1. a man would have difficulty suing for being denied the position of frisking women at a security point[/i]
[i]- ex2. a particularly large woman would have difficulty suing Hooters for employment denial[/i]

That being said, I absolutely think that one should be able to fire an employee on the basis of a website if it shows or highly suggests that the employee would have been ineligible for the job, such as:
- great moral turpitude
- that the employee does not meet/no longer meets a particular bona-fide occupational qualification
- that the employee is a danger
- that the employee misrepresented his/her qualifications, abilities, characteristics, or history
etc.

Beyond severe things such as the above, I think the employer should take a look at the website/myspace/etc before hiring. I don't think the employer should have as much latitude when terminating a job as when denying a job.

[quote="WantToGetBetter"]I think an employer should be able to hire or fire whomever they want for whatever reason.[/quote]

So if I open a business, you think it's right for me to not hire any black or hispanic people?

[quote="Ignite168"][quote="WantToGetBetter"]I think an employer should be able to hire or fire whomever they want for whatever reason.[/quote]

So if I open a business, you think it's right for me to not hire any Black or Hispanic people?[/quote]

Absolutely. You hire whoever is best qualified and suited for the position. So (hypothetically), if no Black people that apply to you for jobs are best qualified or best suited for the job, then why would you go out of your way to hire one?

And who would fire someone for their ethnicity? In most cases, that doesn't make sense. If the employer had a thing against some ethnicity, they probably wouldn't have hired.

Really?

In the book freakonomics, I remember seeing an experiment where the same man with the same qualifications sent different names to different employers - some names common among the coloured people and some among the white. The white names had a much higher success rate.

You can make all the laws against racism - but you can't change basic human instincts.

I am of the opinion that the employer must lay down rules for selection and there be a neutral arbitrator to judge whether a candidate qualifies or not. If he wants white people or people with clean records or pretty girls, he must announce it without hypocricy.

Theoretically, if all people of all ethnicities, races, genders, etc. were equally qualified, then discrimination would be an economically suicidal move.

Say Company A won't hire hispanic people, and Company B hires all people. The demand for Hispanic laborers is lower than that of others, so they command a lower wage. Company B, being a profit maximizing firm and knowing that all people are equally qualified, obviously hires the hispanic people. Their costs are now lower than Company A and are able to undercut them, driving Company A out of business. Other companies may try to take advantage of the same fact, also hiring Hispanic people, pushing their demand back up and equalizing wages. By this time, Company A is long gone and there is no more wage differential.

Of course, there are several assumptions that are made here, some of which do not apply to real life. First, not all people from all groups are equally qualified, especially if you consider that many customers are bigoted and will receive lower satisfaction from being served by a certain group of people regardless of how good the actual service is.

As to the original question, the employer has every right to look at materials you voluntarily placed online and not hire you if there is a legitimate reason.


Let $x,y$ be integer number with $x,y\neq-1$ so that $\frac{x^{4}-1}{y+1}+\frac{y^{4}-1}{x+1}\in\mathbb{Z}$. Prove that $x^{4}y^{44}-1$ is divisble by $x+1$

[hide="hint"]This is your hint :D
From hypothesis we have $a=\frac{x^{4}-1}{y+1}$ and $b=\frac{y^{4}-1}{x+1}$ are rational roots of integer equation $t^{2}-(a+b)t+ab=0$ therefore $a\in\mathbb{Z}$ and $b\in\mathbb{Z}$.[/hide]

If $a,b \in Q$ and $a.b \in Z$ ,$a+b \in Z$ then $a,b$ is integer.

We set $\large \frac{x^{4}-1}{y+1}$ = $\large \frac{a}{b}$ and $\large \frac{y^{4}-1}{x+1}$ = $\large \frac{c}{d}$, where $b,d$ > $0$,$(a,b) = (c,d) = 1$.
From the hypothesis, we have : $\large\frac{ad+bc}{bd}$ is an integer. So $ad$ is divisible by $b$, since $(a,b) = 1$, then $d$ is divisible by $b$.    $(*)$
On other hand, $\large\frac{ac}{bd}$ = $(x+1)(x^{2}+1)(y+1) (y^{2}+1)$ is an integer, so $ac$ is divisible by $d$, and $a$ is divisible by $d$                         $(**)$
From $(*)$ and $(**)$ we have $a$ is divisible by $b$ and $b = 1$. Similarly, $c$ is divisible by $d$, and $d = 1$ as required.

I'm sorry about my poor English.

I don't understand why $ \frac{ad\plus{}bc}{bd}\in Z$,So $ ad$ is divisible by $ b$.What about $ bc$ is divisible by $ d$.Can someone help me??

[quote="toanIneq"]I don't understand why $ \frac {ad \plus{} bc}{bd}\in Z$,So $ ad$ is divisible by $ b$.What about $ bc$ is divisible by $ d$.Can someone help me??[/quote]

$ \frac{ad\plus{}bc}{bd}\in\mathbb{Z}\Rightarrow bd\mid ad\plus{}bc\Rightarrow b\mid ad\plus{}bc\Rightarrow b\mid ad$
I think it's clear now...

First it's very easy to see both are integers.Rest is easy. :)


Hey! :)
Let z1 and z2 be two complex numbers. Knowing that their modulus is : 1.
Show that : $\frac {z_1 + z_2}{1+z_1z_2}$ is real.

I've proved it showing that  : A - $\bar{A}$ equaled 0 (if A represent the above fraction).

But, I don't know how to prove it using the exponential form. (See, like : z1 = e^(i*theta1) and z2 = e^(i*theta2))

Thanks!  :D

Huh?  :?
Sorry, I don't see how it helps  :blush:

The easier way is: show that $w=\bar{w}$ such that $w = \frac {z_1 + z_2}{1+z_1z_2}$...

$\bar{w} = \overline{ \left(\frac {z_1 + z_2}{1+z_1z_2}\right) }=\frac {\overline{z_1 + z_2}}{\overline{1+z_1z_2}}$

$\bar{w} = \frac {\bar{z_1} + \bar{z_2}}{\bar{1}+\overline{z_1z_2}} = \frac {\bar{z_1} + \bar{z_2}}{\bar{1}+\bar{z_1}\bar{z_2}}$

For every $x$ such that $x \in \mathbb{C}$ we have $\lvert x \rvert = x\bar{x}$.
If $\lvert x \rvert = 1$ then $\bar{x} = \frac 1x$.

... resuming

$\bar{w} = \frac {\bar{z_1} + \bar{z_2}}{1+\bar{z_1}\bar{z_2}}$

$\bar{w} = \frac {\frac 1{z_1} + \frac{1}{z_2}}{1+\frac1{z_1z_2}}$

$\bar{w} = \frac {\frac{z_1 + z_2}{z_1z_2}}{\frac{1 + z_1z_2}{z_1z_2}}$

$\bar{w} = \frac {z_1 + z_2}{1+z_1z_2}$

$\bar{w} = w \quad \therefore \quad \boxed{\mathfrak{R}(w) = 0}$

Yeah, I know.. that's what I did (read my first post)

But, I want to solve it using exponentials..

$\overline{e^{\theta i}} = e^{-\theta i}$

 :what?:

lol...
Nvm, I got it ;)

Here goes my "proof" :
|z1| = 1 <==> $\exists \theta_1 \in \mathbb{R}$ : $z_1 = e^{i\theta_1}$
|z2| = 1 <==> $\exists \theta_2 \in \mathbb{R}$ : $z_2 = e^{i\theta_2}$

$\frac {z_1 + z_2}{1+z_1z_2} = \frac {e^{i\theta_1}+e^{i\theta_2}}{1+e^{i\theta_1}e^{i\theta_2}}$ $= \frac {{e^{i(\frac {\theta_1+\theta_2}{2})}}2cos\frac {\theta_1-\theta_2}{2}}}{e^{i(\frac {\theta_1+\theta_2}{2})2cos\frac {\theta_1+\theta_2}{2}}$
$=\frac {cos(\frac {\theta_1-\theta_2}{2})}{cos(\frac {\theta_1+\theta_2}{2})}$

Therefore, it is real! YAY!  :coolspeak:


function
analytic geometry
Functional Analysis
real analysis
real analysis unsolved

Show in l_1,the norm N(x)=sum absx_i is nowhere edifferentiable.

Actually, a Lipschitz function defined on a separable Banach space is Gateaux differentiable a.e. (if "a.e." is properly understood).

But it's true that $ x\mapsto \|x\|$ is not Frechet differentiable at any $ x\in\ell_1$. Indeed, for any $ \epsilon > 0$ there is $ i$ such that $ |x_i| < \epsilon/2$. The norm as a function of $ i$th coordinate does not look like a linear function on the interval $ (x_i\minus{}\epsilon,x_i\plus{}\epsilon)$.


arithmetic sequence
number theory
prime numbers
number theory unsolved

let  $ P_n$ be (ensemble) the integre number positive and lower of n and (premier avec )  n

find all n such as the element of $ P_n$ constitute an arithmetic progression

sorry about using the world in french , and I hope if somebodye can translate this probleme in english

[quote="mohamed-01-01"] ... I hope if somebodye can translate this probleme in english[/quote]

Let $ P_n\equal{}\{x\in\mathbb N$ such that $ 0<x<n$ and $ \gcd(x,n)\equal{}1\}$. Find all $ n\in\mathbb N$ such that all elements of $ P_n$ make an arithmetic progression

[quote="mohamed-01-01"] Let $ P_n \equal{} \{x\in\mathbb N$ such that $ 0 < x < n$ and $ \gcd(x,n) \equal{} 1\}$. Find all $ n\in\mathbb N$ such that all elements of $ P_n$ make an arithmetic progression[/quote]

If $ n$ is odd, $ 1\in P_n$ and $ 2\in P_n$ and so the common difference of the progression is $ 1$ and $ P_n\equal{}\{1,2,3,...,n\minus{}1\}$ and $ n$ is an odd prime.

If $ n$ is even, let $ n\equal{}2^au$ with odd $ u$. If $ u\equal{}1$, $ P_n\equal{}\{1, 3, 5, 7, ..., n\minus{}1\}$ and is an arihmetic progression.
If $ n$ is even, $ n\equal{}2^au$ with odd $ u\geq 3$, then $ u\minus{}2\in P_n$ and $ u\plus{}2\in P_n$ and common difference of the sequence must be $ 2$ or $ 4$.
$ 2$ is impossible since then $ u$ would be in $ P_n$
$ 4$ would imply $ P_n\equal{}\{1, 5, 9, ...\}$ but $ 9$ cant be in $ P_n$ if $ 3\notin P_n$, and so either $ P_n\equal{}\{1\}$ and $ n\equal{}2$, either $ P_n\equal{}\{1,5\}$ and $ n\equal{}6$

So the required set is $ \{1,6\}\cup\{$ set of prime numbers $ \}\cup\{2^n$ $ \forall n>1\in\mathbb N\}$

Also, see 1991 IMO, #2:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60725[/url]


function
induction
real analysis
real analysis solved

we consider the functions f,g,h:N*->N* one to one mapping (bijective) s.t.
f^3(n)+g^3(n)+h^3(n)=3.n.g(n).h(n)

This is almost trivial:
   We have by AM-GM that f(n)<=n for any n>=1. Thus f(1)=1, f(2)<=2 and f(2) is not 1, thus f(2)=2. By induction f(n)=n for any n. THus we have equality in the AM-GM which shows that f(n)=g(n)=h(n)=n.


inequalities
ratio
geometry
Ross Mathematics Program
geometry proposed

In Triangle ABC cevians AD,BE and CF meet at P (D is on BC,E on CA and F is on AB).

Prove that the ratios AP/PD, BP/PE, CP/PF have sum greater than or equal to 6 and product greater than or equal to 8.

As trivial as it is, still please don't forget to say that the point P has to lie inside the triangle ABC - this is not automatically assumed, at least not in the "Geometry Section" (in the "Inequalities Section" maybe).

Just to give a newer reference to this theorem, see Chapter 12 paragraph 2 of

[url=http://www.amazon.com/exec/obidos/tg/detail/-/0883856395/qid=1092737678/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/104-6043065-0271943?v=glance&s=books&n=507846]Ross Honsberger, [i]Episodes in Nineteenth and Twentieth Century Euclidean Geometry (New Mathematical Library)[/i], Washington 1995[/url].

  Darij


Art of Problem Solving shirts are now availabe [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Comm_TShirt.php]here[/url].  There is also a tab to get there at the top of the forum area.

[quote="djimenez"]For how long are they going to be available? I am a little too much overbudged this month, but I would like to order one for next month.

Is there more than one color?

Best,[/quote]

I would like to have one in blue or red.  :)  Having a ML and AoPS T-Shirt is nice. The best would be a ML/AoPS T-Shirt. Not sure why you are still making a difference between ML and AoPS.

For some colour advice with respect to [url=http://www.inf.ed.ac.uk/resources/tshirts.html]red, blue and white look here.[/url]  :)

[quote="sundma"]If I order the T-shirt how long before they get sent out?  Should I get them in 6 weeks or less?[/quote]

Almost certainly.  We ship books and shirts almost every weekday.

[quote="orl"]
Not sure why you are still making a difference between ML and AoPS.
[/quote]
well, i think the main difference between ML and AoPS is the books!
I don't have the AoPS books so i think am only a Mathlinker, not an AoPSer...


geometry
parallelogram
similar triangles

In parallelogram $ABCD$, $E$ is on $\overline{BC}$. $\overline{AE}$ cuts diagonal $\overline{BD}$ at $G$ and  line $CD$ at $F$. Show that $AG$ is one-half the harmonic mean between $AF$ and $AE$.

Prove the theorem when $E$ is on the extension of $\overline{CB}$ through $B$.

[hide="Part 1 Solution"]Draw a diagram first. It's clear that $\triangle ABG \sim \triangle FDG$ and $\triangle ADG\sim \triangle EBG.$

From the first pair of similar triangles, $\frac{GF}{AG}=\frac{GD}{BG}.$ It follows that
\[\frac{GF+AG}{AG}=\frac{GD+BG}{BG}\implies \frac{AF}{AG}=\frac{BD}{BG}.\]
From the second pair of similar triangles, $\frac{GE}{AG}=\frac{BG}{GD},$ so
\[\frac{GE+AG}{AG}=\frac{BG+GD}{GD}\implies \frac{AE}{AG}=\frac{BD}{BG}.\]
Simplifying, $\frac{1}{AF}=\frac{BG}{(BD)(AG)}$ and $\frac{1}{AE}=\frac{GD}{(AG)(BD)}.$ Therefore,
\[\frac{1}{AF}+\frac{1}{AE}=\frac{1}{AG}\left(\frac{BG}{BD}+\frac{GD}{BD}\right)=\frac{1}{AG},\]
and $AG=\cfrac{1}{\frac{1}{AF}+\frac{1}{AE}},$ as desired.
[/hide]

Anyone have a solution for the extension?

$\frac{1}{AG}=\frac{1}{AF}+\frac{1}{AE}$
$1=\frac{AG}{AF}+\frac{AG}{AE}$
we just have to prove that $\frac{GF}{AF}= \frac{AG}{AE}$.
\begin{eqnarray*}\frac{AG+GE}{AG}&=& 1+\frac{BE}{AD}\text{ }(\triangle ADG \sim \triangle GBE)\\ &=& 1+\frac{AB}{DF}\text{ }(\triangle FDA \sim \triangle ABE)\\ &=& 1+\frac{AG}{GF}\text{ }(\triangle ABG \sim \triangle FDG)\\ \frac{AE}{AG}&=&\frac{AF}{GF}\end{eqnarray*}


Show that for all $a,b,c,d\leq k$ we have \[ \frac{abcd}{\prod(2k-a)}\leq\frac{a^4+b^4+c^4+d^4}{\sum(2k-a)^4}. \]

This one is a problem of the Taiwan TST 2002.

This problem has appeared in other threads, four of which are listed by Darij at http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2002+Taiwan&t=46398

I haven't worked through all of the others carefully; a purported solution from 2003 hasn't been refereed either yay or nay by a second individual. I was only recently able to do this problem after I was tipped off that it involved an unusual sum of squares argument.

[i]Solution.[/i] Without loss of generality, suppose that $a \ge b \ge c \ge d.$ The desired is equivalent to
\begin{eqnarray*} & & \frac{(a^2-b^2)^2 + (c^2-d^2)^2 + 2(a^2b^2+c^2d^2)}{abcd} \ge \\ & & \frac{\left((2k-a)^2-(2k-b)^2\right)^2 + \left((2k-c)^2+(2k-d)^2\right)^2 + 2\left((2k-a)^2(2k-b)^2+(2k-c)^2(2k-d)^2\right)}{(2k-a)(2k-b)(2k-c)(2k-d)} \end{eqnarray*}
We will realize this equivalent inequality as a sum of three simpler inequalities. We have
\begin{eqnarray*} \frac{\left(a^2-b^2\right)^2}{abcd} & \ge & \frac{\left((2k-a)^2-(2k-b)^2\right)^2}{(2k-a)(2k-b)(2k-c)(2k-d)} \qquad (*) \\ \Longleftarrow (a-b)^2\left[(a+b)^2(2k-a)(2k-b)\right] & \ge & (a-b)^2\left[(4k-a-b)^2ab\right] \\ \Longleftarrow \left(\frac{a+b}{2}\right)^2(2k-a)(2k-b) & \ge & \left(2k-\frac{a+b}{2}\right)^2ab \\ \iff \left(\frac{a+b}{2}\right)^2\left(\left(2k-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2\right) & \ge & \left(2k - \frac{a+b}{2}\right)^2\left(\left(\frac{a+b}{2}\right)^2 - \left(\frac{a-b}{2}\right)^2\right) \\ \iff \left(\left(2k - \frac{a+b}{2}\right)^2 - \left(\frac{a+b}{2}\right)^2 \right)\left(\frac{a-b}{2}\right)^2 & \ge & 0 \end{eqnarray*}
As a consequence of $a, b \le k$, it is evident that the last line holds, thus validating our inequality $(*)$. Substituting $c$ and $d$ for $a$ and $b$ shows that the related inequality
\[ \frac{(c^2-d^2)^2}{abcd} \ge \frac{\left((2k-c)^2-(2k-d)^2\right)^2}{(2k-a)(2k-b)(2k-c)(2k-d)} \qquad (**)  \]
holds as well.

We now verify our third inequality. Writing $f(x) = x + 1/x$, we have that $f(x) = f(1/x)$ and that $f(x)$ is increasing for $x \ge 1$. Thus,
\begin{eqnarray*} \frac{2(a^2b^2+c^2d^2)}{abcd} & \ge & \frac{2\left((2k-a)^2(2k-b)^2 + (2k-c)^2(2k-d)^2\right)}{(2k-a)(2k-b)(2k-c)(2k-d)} \qquad (***) \\ \iff f\left(\frac{ab}{cd}\right) & \ge & f\left(\frac{(2k-c)(2k-d)}{(2k-a)(2k-b)}\right) \\ \iff \frac{a(2k-a)}{c(2k-c)} \cdot \frac{b(2k-b)}{d(2k-d)} & \ge & 1 \\ \iff \left(\frac{k^2-(k-a)^2}{k^2-(k-c)^2}\right) \cdot \left(\frac{k^2-(k-b)^2}{k^2-(k-d)^2}\right) & \ge & 1 \end{eqnarray*}
It follows that $(***)$ holds as a consequence of $k \ge a \ge b \ge c \ge d > 0$, and that the desired inequality is realized by adding the corresponding sides of $(*)$, $(**)$, and $(***)$. $_\Box$

sorry,Mildorf,I don't think your solution is very good,but it's indeed a solution.I know a solution which is more symmetrical than yours,I'll post it latter.
by the way, this problem was originally in China TST 2001.

[quote="Hawk Tiger"]I know a solution which is more symmetrical than yours,I'll post it latter.[/quote]

I have found solutions to this problem rather hard to find. Please share yours, I am very interested!


[img]http://i3.photobucket.com/albums/y57/lilshantypanty/prob1.jpg[/img]

--
Thanks!

I'd like to try proving this:

If $ a_0$ is the limit point of singularities, then $ f(z)$ cannot be analytic in any punctured disc $ 0 < |z \minus{} a_0| < \delta$ and thus cannot have a Laurent expansion about $ a_0$.  Hence, $ a_0$ cannot be a regular point or an isolated singularity.   Using the same argument for the proof of Casorati-Weierstrass, assume the converse is true:  There exists some $ c\in \mathbb{C}$ such that $ |f(z) \minus{} c|\geq \epsilon$ which implies the function $ F(z) \equal{} 1/(f(z) \minus{} c)$ is analytic in $ 0 < |z \minus{} a_0| < \delta$ for some delta.  This implies that $ f(z) \equal{} c \plus{} 1/F(z)$ is regular at $ z \equal{} a_0$ if $ F(a_0)\ne 0$ or a pole of order m.  In either case a contradiction is reached.  Since $ \epsilon$ was arbitrary, this shows the values $ w \equal{} f(z)$ are dense in any $ \delta$-neighborhood of $ a_0$.

This one's talking to us! Can you hear it?

Let $ \mathbb{D}=\{x,y\}: Re[Sec(1/z)] = 1$.  Now just raise this set 1 unit above the x-y plane.  The plot below shows this set as $ \{\infty\}=\{\mathbb{D},1\}$ in blue superimposed over the yellow sheet $ Im[Sec(1/z)]$.  Now, drop the set $ \{\infty\}$ directly onto the yellow sheet.  For each $ \infty$, this translates to not one $ \mathbb{R}$ but $ 2\mathbb{R}$ because the loops cut directly through the spindle poles.  And this happens infinitely often in any $ \delta$-neighborhood of the origin as shown by the infinitely regressing $ \infty$'s.  For me, this graphically shows (intuitively, I know, not proving)  that $ Sec(1/z)$ assumes the values $ 1 + iy, y\in\mathbb{R}$ infinitely often about the origin.  I'm sure it will turn out that ANY set ${ \mathbb{D} = \{x,y}\}: Re[Sec(1/z) = s, s\in\mathbb{R}$ will likewise cut through the spindle poles in such a way to assure $ Im[Sec(1/z)(\mathbb{D})] = \mathbb{R}$ thereby allowing $ Sec(1/z) = \mathbb{C}$ in any $ \delta$-neighborhood of the origin.  I've attempted to make the sheet transparent so try and interpret it as such with the lower component of each spindle showing through the image.  I realize I'm abusing the meaning usually given to the symbol $ \infty$ but in this case I think it's nicely appropriate. :)

[img]http://img249.imageshack.us/img249/7158/secsaysvj6.jpg[/img]


algebra
polynomial
combinatorics proposed
combinatorics

Given the recurrence $ a_n \equal{} \frac {a_{n \minus{} 1}a_{n \minus{} 3} \plus{} a_{n \minus{} 2}}{a_{n \minus{} 4}}$.
Show that if $ a_0 \equal{} a_1 \equal{} a_2 \equal{} a_3 \equal{} 1$ than this sequence contains only integers.

More generally show that if $ a_i \equal{} x_i$ for $ 0\le i \le 3$ then the sequence consists of Laurent polynomials.
(Polynomials in $ \mathbb{R}\left[x_0,x_1,x_2,x_3,x_0^{ \minus{} 1},x_1^{ \minus{} 1},x_2^{ \minus{} 1},x_3^{ \minus{} 1}\right]$)

If $ a_0\equal{}a_1\equal{}a_2\equal{}a_3\equal{}1$, you have $ a_4\equal{}0$. Then you cannot define $ a_8$. I guess you can change the condition to $ a_0\equal{}a_1\equal{}a_2\equal{}a_3\equal{}\minus{}1$.

I think it's also true if you add integer exponents to each term in the numerator, yes?  One can show this using the Caterpillar Lemma of Fomin and Zelevinsky, from their work on cluster algebras.  See, for example, their paper on the "Laurent phenomenon": [url]http://www.math.washington.edu/~curtis/M308/sdarticle.pdf[/url].  (Of course, this particular example (something related to the Somos sequences?) is almost certainly easier than the general case -- it's also [url=http://www.research.att.com/~njas/sequences/A048736]A048736[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].)

Nice. Thanks for the paper. It's the first time I see the Caterpillar lemma and its consequences. This indeed is a Somos sequence, and I thought it was a little more interesting because it is non-homogeneous. It doesn't have an obvious pattern.


Simplfy as much as you can, you can use factorial

$\prod_{k=2}^{2n+1}(1-\frac{1}{4k^2})^\frac{2n+2-k}{2}$

I can't find an expression of your value which length doesn't depend on n.
Can you? Please can you post your solution? I'd like to see it. Thanx.


[quote="mr. math"]A plane is headed in a direction of 130 degrees with an air speed of 180 knots. The course is 105 degrees with a ground speed of 110 knots. Find the direction of the wind.

North = 0 degrees, East = 90 degrees, South = 180 degrees, West = 270 degrees[/quote]

[hide="My thoughts"]

This is an application of the concept of relative velocity. That is

\[ \vec{v}_A \equal{} \vec{v}_B \plus{} \vec{v}_{A/B}\]

In this case $ \vec{v}_A$ is the velocity of the aircraft, $ \vec{v}_{A/B}$ is velocity of the aircraft relative to the wind. This means,

\[ 110\angle 105^{\circ} \equal{} \vec{v}_B \plus{} 180\angle 130^{\circ} \\ \implies \vec{v}_B \equal{} 110\angle 105^{\circ} \minus{} 180\angle 130^{\circ} \equal{} 92.79\angle 340^{\circ}\]

The direction of the wind then is $ 340^{\circ}$.[/hide]


geometry
geometric transformation
linear algebra
matrix
vector
analytic geometry
combinatorics proposed

I have been asked to post some rather hard problems from Kvant, so here I do it. (Sorry for my bad translation.)

$n$ teams took part in a hockey tournament; every team played against every other team. Each time two teams played against each other, each of these teams got $0$ points for defeat, $1$ point for stalemate and $2$ points for victory. After the end of the tournament, it turned out that for every nonempty set $S$ of teams, there exists another team $Q$ such that the sum of all points which the team $Q$ got in its matches against all teams from $S$ is odd. Prove that $n$ is even.

  Darij

[hide]suppose n is odd,

Lets look at the sets of all the teams except one, that team scored an odd number of points against all the other teams.  For an odd number of teams, this means an odd number of points were scored.  However, there are 2 points awarded in each game, the total number of points must be even. [/hide]

Sorry - my translation was wrong! (And if you look at it closely, you see that it doesn't make much sense - it requires that "for every set $S$ of teams, there exists another team $Q$ such that the sum of all points which the team $Q$ got in its matches against all teams from $S$ is odd"; however, this condition can never be satisfied, because if $S$ is the set of all teams, then there is no "another team $Q$" at all.)

The right problem is:

$n$ teams took part in a hockey tournament; every team played against every other team. Each time two teams played against each other, each of these teams got $0$ points for defeat, $1$ point for stalemate and $2$ points for victory. After the end of the tournament, it turned out that for every nonempty set $S$ of teams, there exists some team $Q$ (which may and may not belong to $S$) such that the sum of all points which the team $Q$ got in its matches against all teams from $S$ is odd. (If the team $Q$ is in the set $S$, then we say that $Q$ got $0$ points against itself  :) .) Prove that $n$ is even.

[b]EDIT:[/b] This problem has already been posted at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5143 ...

  Darij

nice one. first, a lemma:

let $A$ be a symmetric integer $n$x$n$-matrix such that on the main diagonal, all entries of $A$ are zero. suppose $n$ is odd. then $\det A\equiv 0 \mod 2$.
proof: the determinant is $\sum_{\sigma} (-1)^{%Error. "sign" is a bad command.
\sigma} a_{1\sigma_1}\cdots a_{n\sigma_n}$ over all permutations $\sigma$. if $\sigma$ has a fixed point, then term is zero. if it has not, then $\sigma^{-1}\neq \sigma$, so we can group all terms, grouping the terms for $\sigma$ and $\sigma^{-1}$ together, which are the same. that gives the result.

consider the graph on $n$ vertices, where two vertices are connected iff the two teams played stalemate. for every vertex $i$, consider the vector $a_i$, where the $j$-th entry of $a_i$ is 1 if the vertices $i$ and $j$ are connected, else 0. let $A$ be the matrix formed by the $n$ vectors $a_1, ..., a_n$. by the lemma, $\det A\equiv 0\mod 2$. that gives that the $n$ vectors $a_1, ..., a_n$ are linearly dependent over $\mathbb{F}_2$. now, we know that after choosing $k\neq 0$ vectors(the indices of whom are the teams chosen in S), their sum may not be zero(consider them as vectors over $\mathbb{F}_2$), else there would be no team Q which got an odd number of points against them(Q must have played an odd number of times stalemate against the teams of S, thus the corresponding coordinate of the sum vector has to be 1). that shows that there is no nontrivial linear combination of $a_1, ..., a_n$ over $\mathbb{F}_2$ which is nonzero. this contradicts the linear dependence of the vectors.

Peter


Princeton college Stanford MIT Support Harvard

argh "the party college"(penn state) and UGA are on the rarely study list! *takes UGA and penn state off list*
UGA looked good until now.

Yeah I really don't have a list of colleges I plan to attend it's kind of early for that stuff-I was just pointing out colleges I know more people who went there are in that one list. :lol:

[quote="tokenadult"]If you've taken a good AP statistics course, you know that the data-gathering method (voluntary response surveys) used for these rankings is not to be trusted for important decisions like choosing a college to attend.[/quote]

but, on the other hand msn is quite reliable :P

...really

-jorian

AoPS is quite reliable but I hardly ever give much consideration to the polls on here.

[quote="funcia"]... takes UGA and penn state off list ...[/quote]
domt kmow about UGA, but pen steta was a god schol went i when (go Rolling Rock! go JoePa! :D)

Well, Penn State is not a "dry" campus. However, that has never stopped the majority who can balance the act from earning a top motch education. At least there will be fun things there for one to balance the act with. Some bigger brand names may not offer such balance :wink:).

AoPSers should look into these two Penn State programs:

Shreyer Honors College - Ivy education without the Ivy price tag
http://www.scholars.psu.edu/

Accelerated Premedical-Medical Program (6 or 7 years) - for the Doogie Howsers
http://www.science.psu.edu/premed/6yrpremedprogram.html

[quote="Kent Merryfield"]Well, I've got to say that I appreciate the #3 listing on the "Best Value Colleges." :thumbup:[/quote]
Yes! And to me that certainly adds credibility to the Princeton Review list.

[quote="MithsApprentice"]I think there's something wrong with this list. Princeton and Stanford are on almost all the "good" lists...beautiful campus, academic experience, happy kids...but there's no MIT!

And MIT isn't on the list for best professors either - in fact, none of HYPMS is. I just got through watching a Walter Lewin lecture on OCW. I don't think Princeton Review knew about him.[/quote]
I found MIT in three categories:

#1 - Top 10 Toughest Colleges to Get Into
#2 - Top 10 Schools Where Students Always Study
#6 - Top 10 Diverse Student Bodies

Yeah, MIT probably should have been on a few more lists although it doesn't really bother me.  I wouldn't expect MIT to make the happiest students list even though I know many people love this place (i.e. MIT) more than anything else.  It's that "IHTFP" attitude.  The students are happy yet they are not.  It's like that feeling after you've gone 5 days with an average of 4 hours of sleep a night resulting in you finishing your psets, writing your paper, and studying for your tests.  You're incredibly happy yet you're miserable.  It's that sick, demented, nerdy kind of bliss.

I do think that MIT should probably have been a little higher on the diversity list.  I mean...

"DePaul University is a very diverse campus." Undergraduates tell us that "everyone gets along with each other remarkably. There are gays/lesbians, African Americans, Caucasians, Asians, and people of all different races who all accept each other for who they are." The university works hard to create this open atmosphere. It "strongly supports multiculturalism; we even have to take a class on it."

Gays/lesbians eh?  MIT has [url=http://web.mit.edu/housing/undergrad/senior.html]Senior Haus[/url] (among other things).

You have a class on multiculturalism?  amazing  :roll:  Here's just a few classes I randomly pulled from the online catalog at MIT: Understanding Culture; Identity and Difference; Gender, Sexuality, and Society.

Anyway, there were many discrepancies all over.  Harvard should be higher up there on many of them, for example.  The poor was poorly done and I'm not sure you can get much good information from it.  At most you can see that Princeton's an excellent school which you probably already knew.

Coming in second to Caltech for how much studying goes on is probably wrong (it's a stinging blow at least...or at least it would be if this poll were credible).

By the way, Lewin is pretty awesome.  But Sadoway for 3.091 (solid-state chem) is also very, very good.  Today's lecture was awesome (see [url=http://web.mit.edu/smcs/3.091/fall2006/mit-3.091-lec-mit-10250-06dec2006-1100-220k.ram]here[/url]).  I recommend skipping ahead to 34 minutes into it unless you really want to see the whole lecture (your choice).

[quote="jhredsox"]I saw this as I was looking through msn.

[url]http://encarta.msn.com/encnet/departments/college/default.aspx?page=Rankings[/url]

it is quite interesting

-jorian[/quote]

Is it really necessary to sign your name after every post? (hint: it's not)

By the way, some of the rankings are pretty dumb inasmuch as they try to list each school as few times as possible, and some things are pretty arbitrary (for example, every Harvard/Princeton cross-admit I know got a virtually identical, and very generous, financial aid package from each institution).  Also, the only reason WUSTL is on the most selective list is because they try to increase their number of junk applications as much as possible e.g. by spamming admissions materials.

whats wrong with me putting my name at the end of my posts?

-jorian

It's annoying, adds nothing to your posts, and you already have it in your signature.

Yeah, repeating something that occurs in your signature looks redundant. But I'm not going to moderate it out or anything--I'll let peer pressure take care of that kind of netiquette issue for a teenage participant here.

off-topic, but i'm so used to doing it that its hard to stop :P
-jorian

(i like it as well)


How many positive divisors does the number $ 2\times 3^2\times 5^3$ have?

Note that we can pick $ 0$ or $ 1$ factors of $ 2$, $ 0$, $ 1$, or $ 2$ factors of $ 3$, and $ 0$, $ 1$, $ 2$, or $ 3$ factors of $ 5$.  Thus, we multiply $ 2 \times 3 \times 4\equal{}\boxed{24}$ positive divisors.


function
Asymptote
geometry
algebra
rational function
complex analysis
complex analysis unsolved

Find $\bf{the}$ rational function $f$ that maps the plane as shown on the diagram.  The regions labeled $+/-$ are mapped 1:1 onto the upper/lower half-plane and $1,i,-1,-i,0$ are mapped to $1,\infty,1,\infty,0$.

(Now here is where I ask for forgiveness in advance -- I do not know how to include an image in this post, so I will describe it in words.)

Imagine the x and y axes with the unit circle drawn.  This creates 8 regions.  Let us number them as follows:

$I= \{|z|<1\}\bigcap \{Im z>0\}\bigcap \{Re z>0\}$
$II= \{|z|<1\}\bigcap \{Im z>0\}\bigcap \{Re z<0\}$
$III= \{|z|<1\}\bigcap \{Im z<0\}\bigcap \{Re z<0\}$
$IV= \{|z|<1\}\bigcap \{Im z<0\}\bigcap \{Re z>0\}$
$V= \{|z|>1\}\bigcap \{Im z>0\}\bigcap \{Re z>0\}$
$VI= \{|z|>1\}\bigcap \{Im z>0\}\bigcap \{Re z<0\}$
$VII= \{|z|>1\}\bigcap \{Im z<0\}\bigcap \{Re z<0\}$
$VIII= \{|z|>1\}\bigcap \{Im z<0\}\bigcap \{Re z>0\}$

Then we have $I,III,VI,VIII$ are $+$ and $II,IV,V,VII$ are $-$.

I just wanted to write that the figure is correct.  It looks much better than a set of $8$ equalities too!

As a follow-up to the would-be problem solver, if you have seen a problem like this before or if this problem relates to some higher or deeper concept in complex analysis, please let me know.  It is so unusual to me, I would like to learn more about it if there is more to learn.

The function is $\frac{4z^{2}}{(1+z^{2})^{2}}$.

We get there by a sequence of transformations. First, $z\to z^{2}$ gives this:
[asy]import cse5; size(200); DPA((-2,0)--(2,0)^^CR((0,0),1)); MP("1",D((1,0)),SE); MP("0",D((0,0)),SE); MP("i",D((-1,0)),SW); void f(string s1, string s2, real a) { MP(s1,0.5*dir(a),(0,0)); MP(s2,1.5*dir(a),(0,0)); } f("+","-",90); f("-","+",270);[/asy]
(Points are labeled by where they came from in the original)

Next, $z\to\frac{1-z}{1+z}$ gives this:
[asy]import cse5; size(200); DPA((-2,0)--(2,0)^^(0,-1)--(0,1)); MP("0",D((1,0)),SE); MP("1",D((0,0)),SE); void g(string s1, real a) { MP(s1,1*dir(a),(0,0)); } g("-",45); g("+",135); g("-",225); g("+",315);[/asy]

Now, $z\to-z^{2}$ gives:
[asy]import cse5; size(200); DPA((-2,0)--(2,0)); MP("0",D((-1,0)),SE); MP("1",D((0,0)),SE); void g(string s1, real a) { MP(s1,1*dir(a),(0,0)); } g("+",90); g("-",270);[/asy]

Finally, adding 1 puts $0$ and $1$ in the right place. Composing our transformations, the function is $z\to 1-\left(\frac{1-z^{2}}{1+z^{2}}\right)^{2}$.

This isn't a terribly deep problem- it's all about the relationship of geometry and complex algebra. These transformations are my thought process for it; the reasons for each step shouldn't be hard to see.

The pictures were created by reverse-engineering fedja's code. I can't say I know Asymptote yet, but it's not hard to understand what a piece of code does and tweak its parameters.


Express $y$ as a function of $x$, given that the constant $C$ is a positive number.

$ln(y-3)=-4x+lnC$

The answer confused me.  I'd really appreciate an explanation,

Step-by-step tutorial:

[hide="Step 1"]What is an inverse function to logarithm?
[hide="Answer"]Right, exponential.[/hide][/hide]
[hide="Step 2"]So we apply exponential function to both sides in order to get rid of the logarithms. How the equation should look like then?
[hide="Answer"]That's right, $e^{\ln(y-3)}=e^{-4x+\ln C}$[/hide][/hide]
[hide="Step 3"]How do we simplify exponents?
[hide="Answer"]Right, summing within an exponent becomes a product of powers.[/hide][/hide]
[hide="Step 4"]How does that affect our equation?
[hide="Answer"]Exactly: $e^{\ln(y-3)}=e^{-4x}\cdot e^{\ln C}$[/hide][/hide]
[hide="Step 5"]Are we ready now to simplify our exponential expressions?
[hide="Answer"]I should think so: $y-3=Ce^{-4x}$[/hide][/hide]
[hide="Step 6"]What's left to do?
[hide="Answer"]You're right: $y=3+Ce^{-4x}$[/hide][/hide]
[hide="Step 7"]Just kidding :)[/hide]

[quote="inom"]Shortest way
[hide]$y=Ce^{-4x}+3$
[/hide][/quote]

The poster clearly stated that he already had an answer and that he desired an explanation.  You gave only an answer with no explanation, which is of no help.


email
summer program
PROMYS
LaTeX
HCSSiM

What is the deadline for registration this year? I could not find it on the site. thanks

[quote="yjneb"]from http://hcssim.org/app1.php:

"We would prefer to receive it by May 15th (but that's a preference, not a deadline)"

That said, getting it in earlier is not a bad idea. (You don't have to start rushing yet though--the IT doesn't get sent out to applicants until March, if I remember correctly)[/quote]
yo when is the IT coming??

Soon(tm).

From http://hcssim.org/apply.php:
[quote]The 2009 Interesting Test will be emailed to applicants beginning in March[/quote]

I stilll have not recieved it. WHAT IS GOING ON??

[quote="Smartguy"]mathemonster, i would try again if i were you as in emailing them[/quote]
have you got it?

nope, that reminds me, i need to email them heh

I got the IT a while ago and just finished it. Does anyone know how many people they accept? I'm just afraid I won't get in . . . :(

Last year there were forty of us. I'm not entirely sure what the acceptance rate is, but if you feel comfortable with your solutions to most of the IT, I think you'll be fine.

[quote="mathemonster"]When should we expect to hear back?[/quote]

Did you get your Interesting Test yet?

I sent my application on Monday (Apr 20) and got it on Tuesday.

Yes I recieved the IT and mailed my [b]finished[/b] app ~ 2 weeks ago

[size=150][color=green]Yes, I sent in my application 2 weeks ago, and I haven't heard back either. Supoosedly it's in about 17 days, but I'm getting nervous....[/color][/size]

are we talking about the one you have to send before they send you the interesting test?

no i already did the IT a while ago. But I'm going to PROMYS so I don't care anyways really

I sent my IT in a few days ago. Has anyone heard back yet? How long will it be?

No, I mean I sent the IT in about 2 weeks ago. Has anyone heard back yet?

Did you all print the test out and write your solutions on the test or typed your solutions using latex?

for PROMYS I printed out, solved them thoroughly and then $ \text{\LaTeX}$ed them beautifully. For HCSSiM I did none of those.

I just wrote all my solutions on the test + back of test.
Speaking of which, I mailed them my test exactly 3 weeks ago. My sponsor sent in the reference 19 days go.
This is not good. 19>17.
If I don't hear back from them, does that mean I got rejected?

Well, I finally heard from David Kelly through email yesterday. It just said that I had been accpeted and to await other paperwork and such through snail mail.
Whoopee!!

I haven't got it yet but i wouldn't be surprised if i didn't get in b/c i did it in like 2 hours

If I didn't get the e-mail yet, does that mean I didn't get in? Should I send them an email? Help me guys!

I don't really know. I would wait a little while, though. I know that I sent in an application online very early and finished the IT in two weeks before sending it in. From what I understand, they do enrollment on a rolling basis, which could mean that you just sent in your IT later.

On a different note:   I'M SO EXITED!!!

I know, me too!
Woo!

Now I just need to find a way to wait until June 28th . . .

Congratulations to those who got in. I went last year, and can tell you that you're in for an amazing summer.

Yay I got in.
Interestingly enough, they sent it by snail mail rather than email.
Well, that explains the delay.

I got in but I declined.
$ \mathbb{HCSSIM} \leq  \mathbb{PROMYS}$

Does this program ask for an recommendation?  If so, is there a specific format that we need to follow?


calculus
integration
trigonometry
calculus computations

Find the integral representing the arc length for intersection of $ x^2\plus{}y^2\plus{}z^2\equal{}4$ and the cylinder $ x^2\plus{}y^2\minus{}2y\equal{}0$ in the first octant.

[hide]
First parametrize the intersection. The second curve can be represented as
\[ x^2 \plus{} (y \minus{} 1)^2 \equal{} 1\Rightarrow x \equal{} \cos \theta,\ y \equal{} 1 \plus{} \sin \theta
\]
and for $ x\geq 0$, we require $ |\theta |\leq \frac {\pi}{2}$.

Substituting into the first equation gives
\[ z^2 \equal{} 4 \minus{} x^2 \minus{} y^2 \equal{} 4 \minus{} \cos^2 \theta \minus{} (1 \plus{} \sin \theta)^2 \equal{} 2 \minus{} 2\sin \theta
\]

\[ \equal{} 2 \minus{} 2\cos \left( \theta \plus{} \frac {\pi}{2}\right) \equal{} 4\sin^2 \left( \frac {\theta}{2} \plus{} \frac {\pi}{4}\right)
\]

\[ \Rightarrow z \equal{} 2\sin \left( \frac {\theta}{2} \plus{} \frac {\pi}{4}\right)
\]
(We choose this square root to make $ z$ positive.)

Now integrate over the required interval for $ \theta$:
\[ \int_{ \minus{} \frac {\pi}{2}}^\frac {\pi}{2} \sqrt {x^\prime (\theta)^2 \plus{} y^\prime (\theta)^2 \plus{} z^\prime (\theta)^2}\ d\theta
\]

\[ \equal{} \int_{ \minus{} \frac {\pi}{2}}^\frac {\pi}{2}\sqrt {1 \plus{} \cos^2 \left( \frac {\theta}{2} \plus{} \frac {\pi}{4}\right)}\ d\theta
\]
[/hide]


[quote="mustafa"]Using each of the numbers 1-9 exactly once, find a way to make these three fractions equal

$a/b=c/de=fg/hi$

If $d=3$ and $e=4$, then $de=34$ NOT $12$

Is there a nice way to do this problem? Brute force seems so bad...[/quote]
To do fractions in LaTeX, use \frac{a}{b}.

$\frac{a}{b}=\frac{c}{de}=\frac{fg}{hi}$

[hide="Answer"]$\frac{3}{6}=\frac{9}{18}=\frac{27}{54}$[/hide]

i believe he did it by educated brute force, if you do it logically brute force isn't as bad as one might think.

[quote="mustafa"]How did you get your answer?[/quote]
Guess and check. Took a few minutes. I'm probably going to make a program to find all the solutions. :D

I tried guess and check, and it took me 15 minutes! Is there any better way to do it?

a/b = c/de = fg/hi

So, b(c/de) = b(fg/hi) = a

1. Now, we know that hi and de must be multiples or factors of b.
However, because b is in the ones, we know hi and fg are multiples of b.

2. Because c<de, a<b.

3. hi>de since fg>c.

I guess use brute force from here by substituting b multiple times.

Can't the answer be:

2/4 = 8/16 = 16/32???

This is a very vague question. Too many possible solutions.

[quote="neelnal"]Can't the answer be:

2/4 = 8/16 = 16/32???[/quote]
No, because you used 1, 2, and 6 twice each.


[quote="Ji Chen"]Prove that for all nonegative real  numbers $x,y,z$ the following inequality always holds:

$\sqrt{578x^{2}+1143yz}+\sqrt{578y^{2}+1143zx}+\sqrt{578z^{2}+1143xy}$
$\leq{\frac{253}{140}}\sqrt{527}(x+y+z).$

When does the equality occur?[/quote]
i think it can be proved by Cauchy-Schwarz

[quote=Crazy_LittleBoy]I proved the interesting inequality about 4 years ago, now I notice there is maybe no solution for it, so I push my solution here.
It just by Cauchy-Schwarz：
$
{\left(\sum_{cyc}\sqrt {{\frac {578\,{x}^{2}+1143\,yz}{ \left( x+y+z \right) ^{2}}}}\right)}^{2}
$

$
\leq
$

$
{\left(\sum_{cyc}{\frac {1}{8636\,{y}^{2}+8636\,{z}^{2}+17065\,x \left( y+z \right) }}\right) \left(\sum_{cyc}{\frac { \left( 578\,{x}^{2}+1143\,yz \right)  \left( 8636\,{y}^{2}+8636\,{z}^{2}+17065\,x \left( y+z \right)  \right) }{ \left( x+y+z \right) ^{2}}}\right)}
$

how to continue？
just see here:[url]https://artofproblemsolving.com/community/u248534h1167255p5587138[/url][/quote]

Crazy_LittleBoy, a long time to see you :-D

[quote=dragonheart6][quote=Crazy_LittleBoy]I proved the interesting inequality about 4 years ago, now I notice there is maybe no solution for it, so I push my solution here.
It just by Cauchy-Schwarz：
$
{\left(\sum_{cyc}\sqrt {{\frac {578\,{x}^{2}+1143\,yz}{ \left( x+y+z \right) ^{2}}}}\right)}^{2}
$

$
\leq
$

$
{\left(\sum_{cyc}{\frac {1}{8636\,{y}^{2}+8636\,{z}^{2}+17065\,x \left( y+z \right) }}\right) \left(\sum_{cyc}{\frac { \left( 578\,{x}^{2}+1143\,yz \right)  \left( 8636\,{y}^{2}+8636\,{z}^{2}+17065\,x \left( y+z \right)  \right) }{ \left( x+y+z \right) ^{2}}}\right)}
$

how to continue？
just see here:[url]https://artofproblemsolving.com/community/u248534h1167255p5587138[/url][/quote]

It is a very nice application of C-S to eliminate the root signs. I proceed along this to get a polynomial with huge coefficients. Have you gotten a simple solution for it?[/quote]

Based on Crazy_LittleBoy's result, one may use the pqr method. Let $p = a + b + c = 140, q = ab+bc+ca, r = abc$. We need to prove that
$$-127214133858750120000q^4+2486302107244169360854400q^3-847435180689084960000q^2r-29843798259735207188443520000q^2+1062870209453886603346248000qr-11225672892515471432842800r^2+106051470451652695507614720000000q-4591727971548467094471526400000r \ge 0.$$
It is not difficult.


superior algebra
superior algebra unsolved

Let $ K$ be a field such that for every element $ a$ in $ K$ there exists a nonzero natural number $ n$ such that $ a^{n}$ belongs to $ Z(K)$. Prove that $ K$ is commutative.

In standard terminology, a field is already assumed to be commutative; you can add, subtract, multiply, and divide, just as with $ \mathbb{Q}$ or $ \mathbb{R}$. Can you give the relevant definitions for your version?

[quote="mon"]Sorry, I mean $ K$ is a ring with $ (K$\{0}$ ,*)$ group (I thought this was what is called "field") and then $ K$ is commutative iff $ (K$\{0}$ ,*)$ is commutative...[/quote]

A ring $ K$ in with $ \forall g \in K, \exists g^{ \minus{} 1} \in K$ is called a [u][i]division ring[/i][/u] or [u][i]skew field[/i][/u]. A commutative division ring is called a field.

So:

“Let $ K$ be a division ring such that every element $ a \in K, \exists n \in \mathbb{Z}$\ $ \{ 0 \}$ such that $ a^n \in Z(K)$. Prove that $ K$ is a field”


Let be a,b,c real numbers such that |a*x*x+b*x+c|<=1 for any x in [-1,1]. Show that if q in [0,1], then
           q(1+q)|b|+(1-q*q)|c|<=1+q*q.
When the equality establish?

[b]Hint:[/b] Try to use $b=\frac{f(1)-f(-1)}{2}\; ,\; \; c=f(0)\; ,$ where $|f(-1)|\le 1\; ,\; |f(0)|\le 1\; ,\; |f(1)|\le 1\; .$
However the equality case holds (in your proposed inequality)  for $a=2, b=0,c=-1, q=0 .$

It is not so simple but thankyou for your answer anyway. If |b|=8/9 and |c|=7/9 the equality establishes when q=1/2. Any ideeas? :?:

Do not quadrupelpost! Be patient! Don't do ever again what you did the last three posts!

Ooops. There is a problem like this. You have to prove that |b|+|c| \le \frac {5} {3} , which is true. I've tried to prove the inequality when q=0, q=1 and q= \frac {1} {2} . There is an interesting result:
   b^2 + c^2 \le \frac {4(29+6 \sqrt {6} )} {125} . I hope that these things are useful.


conics
parabola
analytic geometry
geometry unsolved
geometry

This didn't get any reply in the Pre-olympiad section, so I'm moving it here

There is a parabola $ y \equal{} x^2$ drawn on the coordinate plane. The
axes are deleted. Can you restore them with the help of compass and ruler?

There are a number of ways. For example, pick 5 arbitrary parabola points $ A, B, C, D, E.$

Draw parallel $ k \parallel AD$ through $ B,$ cutting the parabola again at $ F$; this point has to be constructed. By Pascal theorem, intersections $ P \equiv AD \cap BC$, $ Q \equiv AE \cap BF,$ $ R \equiv CE \cap DF$ are collinear. $ P, Q$ can be constructed; $ PQ$ cuts $ CE$ at $ R$ and $ DR$ cuts $ k$ at $ F.$  Find midpoints $ K, L$ of the parabola chords $ AD \parallel BF$; the line $ KL$ is parallel to the parabola axis $ y.$

Construct perpendicular $ n$ to $ KL$ through $ B,$ intersecting the parabola again at $ G$; this point has to be constructed. Again, intersections $ P \equiv AD \cap BC$, $ S \equiv AE \cap BG,$ $ T \equiv CE \cap DG$ are collinear. $ P, S$ can be constructed; $ PS$ cuts $ CE$ at $ T$ and $ DT$ cuts $ n$ at $ G.$ Find midpoint $ M$ of $ BG$ and construct parallel $ y \parallel KL$ through $ M.$ This is the parabola axis.

The y-axis cuts the parabola at its vertex $ O$ (to be constructed) and at a point $ Z$ at infinity; this infinite point is known, it is identical with the direction of the parabola axis. As before, intersections $ U \equiv AD \cap ZC,$ $ V \equiv AE \cap ZO,$ $ W \equiv CE \cap DO$ are collinear. $ U, V$ can be constructed ($ ZO \equiv y,$ $ ZC$ is parallel to $ y$ through $ C$); $ UV$ cuts $ CE$ at $ W$ and $ DW$ cuts $ y$ at $ O.$ Perpendicular to $ y$ through $ O$ is the parabola vertex tangent and the x-axis.


Can someone help me with this problem?

AH is an altitude on base of an isoceles triangle ABC. From point P on BC, draw two lines PD perpendicular to AB and PB perpendicular to AC respectively. If AH = PD + PE, prove angle C = 60°.

Hints would be the best because I still want to try to do this by myself first. I tried looking for angles but since the question doesn't give anything about angles, I have a feeling I should use sides and similar triangles. I'm not sure how to progress from there though.

Thanks in advance for your help.

Would a trigonometry solution suffice?
If so,
[hide="a hint:"]
Label the altitude AH = h, the base angles $ \theta$, and PD = x. Express all other lengths in terms of these variables.[/hide]

i will give some hints and finally a total solution open as more as you want to

1)[hide]remember that its enough to prove that AB=AC=BC[/hide]
2)[hide]remember that $ (ABC)\equal{}1/2 {AH} {BC}$[/hide]
3)[hide]notice that $ (ABC)\equal{}(APB)\plus{}(APC)(1)$[/hide]
4)[hide]$ (APB)\plus{}(APC)\equal{}1/2{AB}{PD}\plus{}1/2{AC}{PE}\equal{}1/2{AB}{AH}$(because AH=PD+PE,AB=AC)(2)[/hide]
finally,
solution:[hide]use (1) and (2) and you get that AB=AC=BC[/hide]


algebra
binomial theorem
number theory theorems
number theory

Help Prove $ a^p\equiv b^p(\mod p^n)\Rightarrow a\equiv b(\mod p^{n\minus{}1})$
or
$ a^{p\minus{}1}\plus{}a^{p\minus{}2}b\plus{}...\plus{}ab^{p\minus{}2}\plus{}b^{p\minus{}1}\not\equiv0(\mod p^2)$

p just prime number or for all integer???

I will assume that $ p$ is prime :D ,especially I will take the case when $ p$ is an odd prime  .

We have to prove that if $ a^{p} \equiv b^{p}\bmod{p^{n}} \implies a \equiv b \bmod{p^{n - 1}}$ .
I will prove the contrapositive of the statement ,which will in turn prove the original question .

I will prove if $ a \not\equiv b \bmod{p^{n - 1}}$ then $ a^{p} \not\equiv b^{p} \bmod{p^{n}}$

Let $ a \equiv l \bmod{p^{n - 1}}$ where $ l < p^{n - 1}$ and $ l$ not congruent to $ b$ mod $ p^{n - 1}$. Let $ a = p^{n - 1}k + l$ . Now raise both sides to power $ p$ and expand by binomial theorem , now we can easily prove that $ a^{p} \equiv l^{p} \bmod{p^{n}}$.By fermats little theorem ,$ l^{p} \equiv l\bmod{p}$. Since $ l \not\equiv b\bmod{p}$ ,$ l^{p} \not\equiv b \bmod{p} \implies l^{p} \not\equiv b\bmod{p^{n}}$  .Since $ a^{p} \equiv l^{p}\bmod{p^{n}}$,it implies that $ a^{p} \not\equiv b\bmod{p^{n}}$.This implies that $ a^{p} \not\equiv b^{p}\bmod{p^{n}}$ because if $ a^{p}\equiv b^{p}\bmod{p^{n}}$ then $ a^{p} \equiv b^{p}\bmod{p}$ as well . Since$ b^{p} \equiv b\bmod{p}$,it implies $ a^{p} \equiv b\bmod{p}$ ,which is a contradiction,because we have already proved $ l^{p} \not\equiv b\bmod{p}$ and $ a^{p} \equiv l^{p}\bmod{p}$. So $ a^{p} \not \equiv b^{p} \bmod{p^{n}}$. Hence proved


probability
linear algebra
matrix
expected value
probability and stats

Suppose that Alice and Bob repeatedly play some game G. The game is actually determined and the first player should win, but Alice and Bob don't know this. They play with randomized strategies that get successively better and better as they are more experienced, so that the probability that the first player wins the $ n^{th}$ game is $ p \equal{} 1 \minus{} \frac{1}{n\plus{}1}$.

In the absence of knowledge about the game, Alice and Bob decided before they started that whoever wins the $ n^{th}$ game gets to go first in the $ (n\plus{}1)^{th}$ game.

Suppose that Alice wins the first game. What is the expected proportion of games that Alice wins, in the limit as the number of games they play goes to infinity?

How about the same question, but with $ p \equal{} 1 \minus{} \frac{1}{(n\plus{}1)^2}$?

Let $ p_n$ be the probability that the second player wins the $ n$th game. The transition from game $ n$ to game $ n\plus{}1$ can be represented by $ M_n\equal{}\begin{bmatrix}1\minus{}p_n&p_n\\p_n&1\minus{}p_n\end{bmatrix}$; Alice's expected number of wins after $ n$ games is the first component of $ \begin{bmatrix}1\\0\end{bmatrix}\plus{} M_2\begin{bmatrix}1\\0\end{bmatrix}\plus{} M_3M_2\begin{bmatrix}1\\0\end{bmatrix}\plus{} M_4M_3M_2\begin{bmatrix}1\\0\end{bmatrix}\plus{}\cdots\plus{} M_nM_{n\minus{}1}\cdots M_2\begin{bmatrix}1\\0\end{bmatrix}$.

Write $ M_n\equal{}I\minus{}2p_nA$, where $ A\equal{}\begin{bmatrix}\frac12&\minus{}\frac12\\\minus{}\frac12&\frac12\end{bmatrix}$; we have $ A^2\equal{}A$. The $ M_n$ commute, and $ (I\minus{}aA)(I\minus{}bA)\equal{}I\minus{}(a\plus{}b\minus{}ab)A\equal{}I\plus{}((1\minus{}a)(1\minus{}b)\minus{}1)A$
The product $ M_nM_{n\minus{}1}\cdots M_2$ is thus $ I\plus{}\left(\prod_{k\equal{}2}^n (1\minus{}2p_n)\minus{}1\right)A$

In the first case, the product diverges to zero as $ n\to\infty$, and the probability of the first player winning the $ n$th game tends to $ \frac12$. The limit in this case is $ \frac12$; we could calculate values exactly as well.

In the second case, the product converges to some nonzero $ L$, and the limit is $ \frac12\plus{}\frac12L$. I'm not going to try to calculate $ L$ right now.


In very complex LaTeX equations, clicking on the image will open up an empty asymptote code window.  :|

I know the pop ups are buggy and I'm trying to find a way to permanently fix them unfortunately IE has me very limited on this one.

IE? What's wrong with it's newWindow support? Can't you just alter the URL in the links?

It's the limit for the GET statement length that generates the problems.

[quote="#H34N1"]Can't it be possible just to get PHP to "read" the latex image and get the title for it from there?[/quote]If you found a miraculous method to decode a sha1 encoding please submit it to the NSA. Otherwise there is no way to do that :) (without creating a new table in the database with each formula and its sha1 encoding which would make the whole thing useless in the first place).

The thing that I will probably do is not support MSIE on the code popup. Not sure yet though.

No. I mean like right now, a html code is like this right?

[code]
<img src="insertthinghere.gif" title="EQUATION GOES HERE" />
[/code]

or something along the lines of that.

Can't PHP use str_pos to get the text between title=" and the "?

Bleh why not  just get the onClicks to send the code or give the window like:
[code]
... onClick="w=window.open('', 'code', 'toolbar=no,location=no,scrollbar=yes,resizable=yes,status=no,left=200,top=100, width=400,height=250');w.document.write('<code>'+this.alt+'</code>');" ...
[/code]

[quote="Temperal"]That's not the problem. The problem is that if th string is too long, IE messes up the javascript command.[/quote]

eh no. Its the PHP get statement. IE can't let the GET go over (700) chars methinks or something like that.

Oh, I see. URL limits, then basically.

The IE users should just be redirected to mozilla.org.  :P

[quote="Temperal"]Oh, I see. URL limits, then basically.[/quote]Exactly. Nothing that I can think of can right now that is both practical and useful will work.

Just a thought that may not be practical or useful (and I haven't tried it either :)), but how about using onClick to save the latex code as a cookie, which is read again when the popup window opens? At least the maximum size for a cookie at 4KB is large enough.

Then it would be impossible to link to the window, though, without resetting the cookie. Also, you would either have to have multiple randomly generated cookie names, which would be messy, or give up multiple windows at a time.

The cookie would only need to exist for a few seconds so could be set to expire very quickly. The name is not a problem since each image has such a randomly generated name. But, since I haven't tested it, I agree that it may not be as good as using a URL.

ccy's way works: http://www.mathprob.com/test/ccy.php

Meh, couldn't you bypass the URL by writing the code into the window from the original document with javascript after it opens? Then you wouldn't need code.php at all.

I will investigate ccy's idea which might be the solution indeed.

I've modified (and thus fixed) the latex popup tool based on ccy's idea (though not the way he suggested entirely). Very soon the same code will be applied to the asymptote formulas.