algebra
polynomial
number theory unsolved
number theory

Given nonconstant polynomials $ W_1(x),W_2(x),...,W_n(x)$ with integer coefficients. Prove that for some integer $ a$ all the numbers $ W_1(a),W_2(a),...,W_n(a)$ are composite.

This is an easy problem .
[u]Lemma [/u]
$ P(x)$ is a nonconstant polynomial then exist an integer $ m$ such that $ P(m)$ has at least M  prime divisor  .
Proof :
Exist an integer $ x_o$ such that $ |P(x_0) > 2$
Consider the sequence
$ x_{n \plus{} 1} \equal{} P(x_n)^2 \plus{} x_n$
Then $ P(x_{n \plus{} 1})\equiv P(x_n) (\mod P(x_n)^2)$
So $ P(x_{n \plus{} 1}) \equal{} P(x_n)(t_nP(x_n) \plus{} 1)$
Because $ \gcd(P (x_n),t_nP(x_n) \plus{} 1) \equal{} 1$ so exist an integer prime $ p$ such that
$ p|P(x_{n \plus{} 1})$ and $ p\not | P(x_n)$
This is an well know lemma .
Apply this exist $ m_i ,\forall i \equal{} 1,..,n$ such that $ W_i (m_i)$ is composite .
It means that for each $ i \equal{} 1,..,n$ exist pairs $ (p_i,q_i)$ such that
i)$ p_i\geq q_i$
ii) $ q_{i \plus{} 1} > p_i$
Now consider the system
$ a\equiv m_i (\mod p_iq_i)$
Apply chinese remainder theorem ,this system has a solution a.
So $ p_iq_i|W_i (a)$ follows that $ W_i(a)$ are  composite .
Problem claim .

let $ S \equal{} \{j: \ w_j(0)\neq 0\}$
we take $ B \equal{} \prod_{k\in S}w_k(0)\neq 0$
for some $ A$ large.
$ \forall j\in S: \ w_j(0)|w_j(AB)$ and $ |w_j(AB)| > \max_{j\in S}(|w_j(0)|)$
and $ \forall j\not\in S: \ B|w_j(AB)$ and $ |w_j(AB)| > |B|$

[quote]$ A$ exists because:
 $ \lim_{A\to\plus{}\infty}\frac{|w_j(AB)|}{max(|B|,\max_{j\in S}(|w_j(0)|))}\equal{}\plus{}\infty$
so for some $ A$, $ \frac{|w_j(AB)|}{max(|B|,\max_{j\in S}(|w_j(0)|))}>1$[/quote]

More general as I posted . For each $ M$ there exist an integer $ a$ such that
$ W_i(a)$ have at least $ M$ prime divisor for any $ i\equal{}1,..,n$

Remark that for every polynomial P(x) we have

P(x) | P(k. P(x) + x) for all integer k

Let 's take a number a such that Wi(a) >1 for all i

Put  M= K.W1(a).W2(a)....Wn(a) + a

Then Wi(M) is divisible by Wi(a) for all i

.....


How are biology and problem solving related? I mean, there's no numbers or anything you can manipulate to get a correct answer, is there? i have absolutely no background in biology so please don't get angry at me.

Being a doctor or a research requires extensive problem solving skills.  You need to be able to analyze data and draw conclusions based on things that may or may not be there.  Diagnosing an illness requires analysis of symptoms and the patient's history, and doing research requires analysis of data to determine the success of the drug or treatment.

Going along with what JRav said, extensive problems in genetics require a lot of problem solving. Maybe not as obvious as mathematics, but [advanced] biology does require understanding simple concepts and applying them to specific and difficult problems. (For example, given an extensive pedigree, and only a pedigree, what are some possible common genetic diseases and what is the percent of person A of having this disease?)

I participated in the Science Olympiad, particularly the genetics event.  I can tell you, you must be able to see something and draw a conclusion quickly.  You need to be able to look at a pedigree and trace family history, or look at a gel map and match DNA.  Pretty cool stuff.

And that's just classical biology.  There's a new and rather exciting field going by the name of [url=http://en.wikipedia.org/wiki/Computational_biology]computational biology[/url] that is deals explicitly with large-scale problems in the mathematical (or computer-scientific) sense.

Problem-solving is a skill that pervades all of the scientific disciplines - in a sense, it's all scientists do.  I'm not sure what you think scientists do if they don't problem-solve, even biologists.

genetics and gene mapping is greatly computational based

and not too forget the plant water relations :D

wow! those are some really good explanations!

can someone please show me an example for a biology Olympiad that requires problem solving?

ofc see this_____

<though this si easy>

In a family both husband and wife are heterozygous 4 infantile blindness.They hav atleast 3 children
wats the %probablity that all r normal??  :D

Here is a link that contains many old IBO problems: http://www.ibo-info.org/questions.html

Most of the biological material you encounter is descriptive and not quantitative. Biology still has a long way to go to become like other sciences. One of the areas in which biology does excel is the area of structural analysis of biological materials. For example, suppose that drugs are being investigated for their ability to mitigate the effects of osteoporosis. Suppose we want to determine if a disease or diet causes loss of surface area in the gut. How can we measure the damage to lungs due to exposure to different materials?

These problems are not simple. Many papers published today get the wrong answers. They get the wrong answer and publish bad numbers because the obvious methods to get a number are wrong.

Fortunately, researchers are learning that these problems are not trivial and require special methods to obtain the correct answer.

Okay. That's actually pretty cool. I'll try and get some more knowledge about biology, I seem to be liking it, especially microbiology.

if u take more interest  in it u'll like it more... :P  :rotfl:


[quote="13375P34K43V312"]Given an arbitrary quadrilateral $ ABCD$, we draw the angle bisectors $ AZ,BY,CX,DW$. Let $ AZ\cap BY=M$, $ BY\cap CX=N$, $ CX\cap DW=O$, and $ DW\cap AZ=P$. Prove that quadrilateral $ MNOP$ is cyclic.[/quote]

[hide]Would Angle Chasing work for this problem, I believe it can.[/hide]

Call the two angles at A a, the two angles at B b, and so on. Then 2a+2b+2c+2d=360. a+b+c+d=180. So angle M is c+d because in triangle AMB, you have angles a and b, and then angle M, which must sum to 180. Then by the same process, angle O=a+b. Opposite angles O and M are a+b and c+d respectively. Since O+M=a+b+c+d=180, ABCD is cyclic.


geometry
parallelogram
geometric transformation
reflection
rotation
3D geometry

Several times I've noted a word I'm unfamiliar with -- the word is "image" such as a question about a  parallelogram that  says "this is done such [b]that D' is the image of D, and D" is the image of D' [/b]--  first, what does "image" MEAN mathematically? And secondly, what do the one or two quotes mean?

"image" means a reflected point. just say you reflect a point (2,3) about the x axis, the reflected point would be the called the image, which would be (2,-3). the quotes are just there to help clarify which reflection of the point u're talking about if u do multiple things to the point (i don't know if that makes sense to u, cuz it seems really vague)...

Actually, mwpl11, it doesn't have to be a reflection.  It can be anything (rotation, translation, etc).
Basically, when they say  "Do XYZ to this polygon", the image is the result.

The X' (read as X prime) is basically there so that the image of point X doesn't have to have a completely different name.  When you see X' you know it's the point corresponding to the image of X after the transformation.

A way to think about this is "primes" ,like B', are places some point is going to move to or was originally. another example of this is if a cone  (any geometric shape. This is just an example) is cut into a sector of a circle at point B. point B and b' are both at the "corners" where the cone was originally connected.

An image of some shape is in a way a "prime" for a geometric point.


Let $(a_{1}, a_{2}, ... , a_{n})$ be a permutation of $(1, 2, ... , n)$. Find the permutation which maximises $a_{1}a_{2}+a_{2}a_{3}+...+a_{n-1}a_{n}+a_{n}a_{1}$ and the permutation which minimises it.


geometry
cyclic quadrilateral
geometry solved

The diagonals AC and BD of a cyclic quadrilateral ABCD intersect at E. Let O be the
circumcentre of ABCD.
If the midpoints of AD, BC and OE are collinear, prove that AB = CD.

Bomb

[quote="bomb"]The diagonals AC and BD of a cyclic quadrilateral ABCD intersect at E. Let O be the
circumcentre of ABCD.
If the midpoints of AD, BC and OE are collinear, prove that AB = CD.[/quote]

Bomb, do you think renaming the points can make a wrong problem correct?

http://www.mathlinks.ro/Forum/viewtopic.php?t=38712

  Darij

This is the correct version of the problem. I typed it wrongly last time. :blush:

Bomb

No, it's absolutely the same problem, you just permuted the points B and D.

  Darij


search
quadratics
number theory unsolved
number theory

Posted before . You can search on the forum with the general problems.

It is clear for $ p\equal{}2$, thus assume that $ p$ is an odd prime.
The squares of the numbers $ 0,1,\cdots, \frac {p\minus{}1}{2}$ give $ \frac {p\plus{}1}{2}$ pairwise distinct classes modulo $ p$.
- If one of them is $ \frac {p\minus{}1}{2}$ then choose $ x,y$ such that their squares belong to that classe, and we are done.
- Otherwise, for at least one of the pairs $ (0,p\minus{}1),(1,p\minus{}2),\cdots, (\frac {p\minus{}3}{2}, \frac {p\plus{}1}{2})$, the two members of this pairs are quadratic residues, and we are done.

Pierre.

[quote="Γιώργος"]prove that for any prime p it is possible to find integers x and y such that x^{2}+y^{2}+1 is divisible by p[/quote]
but for $ p\equal{}5$ $ p|1\plus{}2^2\plus{}5^2$

[quote="aviateurpilot"][quote="Γιώργος"]prove that for any prime p it is possible to find integers x and y such that x^{2}+y^{2}+1 is divisible by p[/quote]
but for $ p \equal{} 5$ $ p|1 \plus{} 2^2 \plus{} 5^2$[/quote]
I don't understand because he post can find.
other solution is consider the set
$ A\equal{}\{1^2,..,(\frac{p\minus{}1}{2})^2\}$
$ B\equal{}\{\minus{}1^2\minus{}1,..,\minus{}(\frac{p\minus{}1}{2})^2\minus{}1\}$


number theory
greatest common divisor
modular arithmetic
relatively prime
number theory proposed

I'd like to know how to attack the following problem.

Find $ GCD( 11^n \plus{} 5^n , 11^m \plus{} 5^m)$ for $ (m,n) \equal{} 1$ i guess the answer is 1

Let d be a common divisor of both. Assume m>n. Then
$ d | 11^{m \minus{} n}(11^n \plus{} 5^n) \minus{} (11^m \plus{} 5^m) \equal{} 5^n(11^{m \minus{} n} \minus{} 5^{m \minus{} n})$
But d can't have any factor of 5 in it since neither of the original two numbers is divisible by 5. So now we have the original problem again, but where the exponents are n and m-n. We can keep repeating this. Since m and n are relatively prime, one of the exponents will eventually be 1 (Euclidean algorithm). So d | 16. 16 divides each of them if m and n are odd (just use mod 16), otherwise, only 2 divides each of then, because $ 11^{2k}\plus{}5^{2k}\equal{}121^k\plus{}25^k \equiv 1^k\plus{}1^k \equal{} 2 \pmod 4$.

Lemma:- If $ a^{x} \equal{} b^{x}(modk)$ and $ a^{y} \equal{} b^{y}(modk)$ and a and b are relatively prime to k then $ a^{(x,y)} \equal{} b^{(x,y)}(modk)$

From this, if n and m were odd, take a=11, b=-5 and x=n, y=m, we have k divides 16 and 16 indeed divides both and hence gcd=16

If one of them were even, say m, take a=11, b=5, x=2n, y=2m and we have k divides 96
3 cannot be a divisor of $ 11^{n} \plus{} 5^{n}$ for any integer n, and m is even would imply $ 11^{m} \plus{} 5^{m} \equal{} 2(mod4)$
So, in this case gcd=2

A different approach:
Let $ a_n\equal{}11^n\plus{}5^n$. We have $ a_{n\plus{}2}\equal{}16a_{n\plus{}1}\plus{}55a_n$, with $ a_0\equal{}2$ and $ a_1\equal{}16$. As long as $ m$ is not divisible by either $ 5$ or $ 11$, we can solve this recurrence mod $ m$ for $ a_n$ in terms of $ a_{n\plus{}1}$ and $ a_{n\plus{}2}$. We also note that $ a_n$ is never divisible by either $ 5$ or $ 11$; since we look for divisibility, we henceforth ignore all multiples of $ 5$ and $ 11$. In this light, we can interpret $ a_k$ for negative $ k$ as an integer mod $ m$.
Any linearly recursive sequence mod $ m$ which is continued infinitely in both directions is periodic. For a second order sequence such as this, the zeros are simply periodic; if $ a_m\equiv 0,a_{m\plus{}1}\equiv b,a_{m\plus{}k}\equiv 0$, and $ a_{m\plus{}k\plus{}1}\equiv c$, we have $ a_{n\plus{}k}\equiv cb^{\minus{}1}a_n$ for all $ n$, and $ a_{n\pm k}\equiv 0$ if $ a_n\equiv 0$.
Now, what if $ a_n\equiv 0\mod d$? We have $ d|11^n\plus{}5^n$, so $ 11^{\minus{}n}\plus{}5^{\minus{}n}\equal{}55^{\minus{}n}(11^n\plus{}5^n)\equiv 0\mod d$, and $ a_{\minus{}n}$ is also divisible by $ d$. Suppose WLOG that this is the minimal positive $ n$ with this property. The period for zeros mod $ d$ must divide $ 2n$, and we have two cases:
Case 1: $ d$ divides $ a_0\equal{}2$. In this case, the period is exactly $ n$, and by observation $ n$ is $ 1$.
Case 2: $ d$ does not divide $ a_0$. In this case, the period is greater than $ n$, and must be $ 2n$. We have $ d$ dividing $ a_m$ exactly when $ \frac{m}{n}$ is an odd integer.

From this, we can find the greatest common divisor of $ a_m$ and $ a_n$ in all cases. If $ \frac{m}{n}$ has either even numerator or even denominator in lowest terms, the gcd of $ a_m$ and $ a_n$ is $ 2$. If $ \frac{m}{n}$ has odd numerator and odd denominator in lowest terms, $ a_{\text{gcd}(m,n)}$ is the gcd of $ a_m$ and $ a_n$.

This generalizes pretty easily.


abstract algebra
vector
Ring Theory
group theory
superior algebra
superior algebra unsolved

I need an example for a commutative semigroup that we could not put any $ \plus{}$ operation to be a ring .

i.e. if $ (S,\cdot)$ is a commutative semigroup , then $ (S,\plus{},\cdot)$ is not a ring for any operation $ \plus{}$.

Hint : find a commutative band which it has an element that is not a zero divisor .

I found this answer somewhere if $ (N,\plus{},\cdot)$ is not a ring for any $ \plus{}$ , then how we could prove this if that argument true ?

In fact , I need example as in the hint.

Any help ?  :)

Simple version: anything without an absorbing element $ 0$ ($ 0\cdot a\equal{}0$ for all $ a$) or anything without an identity element $ 1$ ($ 1\cdot a\equal{}a$ for all $ a$).

For something that has these pieces, I'll go with $ R\equal{}\{0,1,x,x^2,\cdots\}$; this is really a simplification of the natural number example. For this to be a ring, we must have $ \minus{}1\equal{}x^k$ for some $ k$, since additive inverses exist, and the inverse of a nonzero element can't be zero. But then $ 1\equal{}\minus{}x^k\equal{}(\minus{}1)\cdot x^k\equal{}x^k\cdot x^k\equal{}x^{2k}$, which is impossible unless $ k\equal{}0$. From $ 1\plus{}1\equal{}0$, addition has characteristic 2, much like Boolean rings.
Now, what's $ 1\plus{}x$? It can't be $ 0,$ $ 1,$ or $ x$ by cancellation laws, so it's some $ x^j$. From that, we have a ring homomorphism from $ (\mathbb{Z}/2)[x]$ onto $ R$, which sends $ 1\plus{}x\plus{}x^j$ to zero. Thus, it factors through the quotient $ ((\mathbb{Z}/2)[x])/(1\plus{}x\plus{}x^j)$. That quotient is finite and $ R$ is infinite, so this is impossible.
For a finite version, make $ R\equal{}\{0,1,x,x^2,x^3,x^4\}$, with the relation that $ x^5\equal{}1$. It comes out as characteristic $ 2$ for the same reasons, and an abelian group with characteristic $ 2$ can't have order $ 6$.

Thx for the answer
Could you give more details for these points :

1)
[quote="jmerry"] Thus, it factors through the quotient $ ((\mathbb{Z}/2)[x])/(1 \plus{} x \plus{} x^j)$. That quotient is finite and $ R$ is infinite, so this is impossible.
[/quote]

2)
[quote="jmerry"]
 and an abelian group with characteristic $ 2$ can't have order $ 6$.[/quote]
 :)

What's an abelian group with characteristic 2? It's an abelian group in which every element has order $ 1$ or $ 2$. The number of elements in such a group must be a power of $ 2$ (if it's finite), and in fact the group must be isomorphic to the vector space $ (\mathbb{Z}/2)^k$ for some $ k$.

Factoring through the quotient? That's the universal property of quotient objects. Suppose we have an object (group, ring, etc.) $ A$ and a subobject $ I$, such that the quotient object $ A/I$ makes sense. This quotient comes with a quotient map $ \pi$ from $ A$ to $ A/I$. If $ f$ is a homomorphism from $ A$ to $ B$ which becomes trivial when restricted to $ I$, there is a unique homomorphism $ g$ from $ A/I$ to $ B$ satisfying $ f\equal{}g\cdot \pi$.
So, for the example we're looking at, $ A$ is the infinite ring $ (\mathbb{Z}/2)[x]$, $ I$ is the ideal generated by $ 1\plus{}x\plus{}x^2$, and $ B$ is the ring $ R$. $ g$ is a ring homomorphism from the finite ring $ ((\mathbb{Z}/2)[x])/(1\plus{}x\plus{}x^2)$ to the infinite ring $ R$. Since $ f$ is surjective and $ f\equal{}g\cdot\pi$, $ g$ is surjective, and this is impossible.


How many cubes are there in a 3 by 3 by 4 rectangular prism made up of 36 unit cubes and the cubes have vertices that are also vertices of any of the unit cubes vertices and edges also on the unit cubes (so no slanted cubes and stuff)?


algebra
polynomial
linear algebra
matrix
complex numbers
superior algebra
superior algebra solved

Prove there is no matrices A,B,C \in M_3(R) such that
AB=BC=CA=M where
M=
(1 2 3)
(4 5 6)
(7 8 10)
 By Cristinel Mortici
------------------------------------------------------------------------
If there exist three such matrices,
detA.detB=detB.detC=detC.detA=detM=-3
A,B,C are invertible
detA=detB=detC
det(AB)=(detA)^2 =-3 wich is a contradiction

Remark: Let T \in Mn(R) with detT<0 there is no three A,B,C \in M_3(R) such that AB=BC=CA=M.
Suppose there are such matrices se have 0 \leq (detA)^2=detT<0 contradiction.

Hi Valentin another solution,

Suppose there exist three matrices A,B,C such that AB=BC=CA=M
Consider the product of matrices  define by blocks

(A 0 0)(B 0 0)
(0 B 0)(0 C 0)
(0 0 C)(0 0  A)
=
(AB 0 0)
(0  BC 0)
(0 0 CA)
=
(M 0 0)
(0 M 0)
(0 0 M)

take the determinant
LHS = (detA.detB.detC)^2=
RHS = (detM)^3 = (-3)^3<0 again a contradiction

Another remark Valentin:

But there exist A,B,C \in M_3(R) such that
ABC=BCA=CAB=M

Answer A=B=I_3,  C=M so obvious.

Prove there exist  3 matrices A,B,C \in  M_3(C) such that
AB=BC=CA=M where
M=
(1 2 3)
(4 5 6)
(7 8 10)
-----------------------------------------------------------------------
Characteristic polynomial of M is Q(x)=x^3-16x^2-12x+3 and the
canonical form of Q(x) (change x in y-b/(3a))
is R(y)=y^3-292/3.y-9839/27
its discriminant D=4p^3 +27q^2 = 4( -292/3)^3 +27(-9839/27 )^2=
-103053<0 this means R,and Q has three distincts complex roots a,b,c
M is diagonalisable in M_3(C)
M=P.diag(a,b,c)P^(-1) where P invertible matrix in M_3(C)
A=B=C=P.diag( \sqrt a, \sqrt b, \sqrt c)P^(-1) is a solution

[quote="Moubinool"]...
 has three distincts complex roots a,b,c...[/quote]
oops, should read one real root a, and two complex roots b, c=b*(complex conjugate of b)..


quadratics
algebra
polynomial
number theory
relatively prime
number theory unsolved

What are some good general techniques that can be used to prove the infinitude (or finitude) of primes of a certain type? Also, same question but for composites.

As a more specific question, a type that I'm currently interested in (as part of a problem I'm working on actually) are those of the form:
$ an^2\plus{}b$ (a,b fixed, positive, and coprime)
$ an^4\plus{}bn^2\plus{}c$ (a,b,c fixed, a,c positive, and a,b,c coprime (if b is 0 then a,c coprime).
Is there a good way to go about finding conditions on a,b(,c) s.t. there are infinitely many primes of that form? Even better, can we prove that there are infinitely many primes of one or both of those forms for all a,b,c?

The question even for $ a \equal{} b \equal{} 1$ is an open problem (in other words, no results of this type are currently known).  The best result known is [url=http://mathworld.wolfram.com/DirichletsTheorem.html]Dirichlet's Theorem[/url], although some special cases are straightforward to prove using quadratic reciprocity.

The general conjecture (for an arbitrary irreducible polynomial with relatively prime coefficients) is called the Buniakowski conjecture.

Edit:  So much for the question of primeness.  It is easy to show that a polynomial with integer coefficients takes composite values infinitely often using the fact that $ a \minus{} b | P(a) \minus{} P(b)$.


[quote="ashrafmod"]Prove that for all n greater than or equal to 2,
$\frac{ (5)(7)(9) . . . (2n+1)}{ (4)(6)(8) . . . (2n) }< \sqrt n$[/quote]

This actually looks to me like it could be an interesting result on binomial coefficients -- it's actually vaguely (just vaguely) reminiscent of Stirling's formula, but it must be much weaker because this is solvable with [hide]induction; you just have to [hide]compare the ratios of successive terms[/hide] unless I made an error.[/hide]


In how many ways can you fill $ 2\times2\times5$ box with $ 1\times1\times2$ blocks?


Given a finite set of genes $G$, every gene is represented by a sequence of $0$s and $1$s of lenght $l$. The result of the crossing-over of 2 genes $(a_1,a_2,...,a_l)$ and $(b_1,b_2,...,b_l)$ is one of the genes $(a_1,...,a_k,b_{k+1},...,b_l)$ (for some $k$, $0\le k\le l$). $g_1,g_2\in G$ are called a protopair iff the exists such an ordering of $G$:
$g_1,g_2,g_3,...,g_n$, so that $\forall i\ge 3\ \exists j,k<i$, such that $g_i$ is the result of the crossing-over between $g_j$ and $g_k$. It is given that for $G$ exists at least a protopair and that $\forall i$, there exist 2 genes $(a_1,...,a_l)$ and $(b_1,...,b_l)$ with $a_i\ne b_i$.
Prove or disprove that $(a_1,...,a_l)$ and $(b_1,...,b_l)$ are a protopair iff $a_i\ne b_i$ $\forall i$.

[i]Edited by Myth[/i]

Sasha told me that he has some reason to think that the statement is false, i.e. non-equality $a_i\neq b_i$ doesn't imply that $a$ and $b$ is a protopair.
Hmm... Did somebody think on it?

[quote="Sasha"]Given a finite set of genes $G$, every gene is represented by a sequence of $0$s and $1$s of lenght $l$. The result of the crossing-over of 2 genes $(a_1,a_2,...,a_l)$ and $(b_1,b_2,...,b_l)$ is one of the genes $(a_1,...,a_k,b_{k+1},...,b_l)$ (for some $k$, $0\le k\le l$). $g_1,g_2\in G$ are called a protopair iff the exists such an ordering of $G$:
$g_1,g_2,g_3,...,g_n$, so that $\forall i\ge 3\ \exists j,k<i$, such that $g_i$ is the result of the crossing-over between $g_j$ and $g_k$. It is given that for $G$ exists at least a protopair and that $\forall i$, there exist 2 genes $(a_1,...,a_l)$ and $(b_1,...,b_l)$ with $a_i\ne b_i$.
Prove or disprove that $(a_1,...,a_l)$ and $(b_1,...,b_l)$ are a protopair iff $a_i\ne b_i$ $\forall i$.
[/quote]
Yes, did someone think on this problem? :)


calculus
integration
logarithms
limit
search
real analysis
real analysis unsolved

[img]http://bbs.pep.com.cn/attachments/123_juRrwAfh76oF.jpg[/img]

It seem that not converge

You are right. The sum diverges. As a matter of fact the series behaves the same as the improper integral:

$\int_{a}^{\infty}\frac{1}{x\ln x}dx=\ln(\ln(x))\biggm|_{a}^{\infty}=\infty$.

I think that series diverges showly than $n$→[size=150]∞[/size] , $1+1/2+1/3+1/4+...+1/n$
How do you think ?

Oh , and one more thing
the series $n$→[size=150]∞[/size] , $1/n*(1+1/2+1/3+1/4+...+1/n)$diverges or converges

If $H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$ is the $n^{th}$ harmonic number then observe that

the sequence $\frac{H_{n}}{n}$ decreases, hence the series $\sum_{n=1}^{\infty}\frac{H_{n}}{n}$ diverges, since

$\lim n \frac{H_{n}}{n}=\lim H_{n}=\infty$.

I want to know what series diverges very very showly
Could you give me some examples ?

We can continue modifying your example to find series that diverge even more slowly.  e.g.  $\sum_{n = 1}^\infty \frac{1}{(n+1)\cdot(1+\log(n+1)) \cdot \log(1+\log(n+1))}$.  (The +1's are just there so that we can continue to begin our sum at 1.  Divergence is by the integral test.)

In general series of the form $\sum \frac{1}{an+b}$ diverge slowly, namely harmonic series and another type of series which diverge slowly are the series involving logarithms, as for instance:

Define $\log_{0}(x)=x$, $\log_{1}(x)=\log x$ and $\log_{k+1}(x)=\log \log_{k}(x)$,

then

it is known that

$\sum_{n=N}^{\infty}\frac{1}{n\log_{1}(n)\log_{2}(n)\cdots \log_{k}(n)\left(\log_{k+1}n\right)^{\alpha}}$ diverges if $\alpha\leq 1$.

You can also search the net, I think you should find examples of slowly divergent series.

A somehow more natural approach, we can group the terms of the sum into blocks of dimension $2^{k}$:
$\sum_{k=2}^{\infty}\frac{1}{n\cdot \ln n}=\sum_{k=1}^{\infty}\sum_{i=2^{k}}^{i=2^{k+1}}\frac{1}{n\cdot \ln n}$
But $\sum_{i=2^{k}}^{i=2^{k+1}}\frac{1}{n\cdot \ln n}\geq \frac{2^{k+1}-2^{k}}{2^{k+1}\cdot \ln (2^{k+1})}=\frac{1}{2\ln(2)\cdot (k+1)}$
Thus $\sum_{k=2}^{\infty}\frac{1}{n\cdot \ln n}=\sum_{k=1}^{\infty}\sum_{i=2^{k}}^{i=2^{k+1}}\frac{1}{n\cdot \ln n}\geq\sum_{k=1}^{\infty}\frac{1}{2\ln(2)\cdot (k+1)}=\frac{1}{2\ln(2)}\cdot \sum_{k=1}^{\infty}\frac{1}{k+1}$ from where it can be seen that the series diverges.


[quote="Jumbler"]Let $ p > 2007$ is a prime number and $ n$ is an integer exceeding $ 2006p$, Prove that
$ \binom{n}{2006p} \minus{} \binom{k}{2006p}$ is a multiple of p, where $ k \equal{} [\frac {n}{p}]$ is the largest integer which is not larger than $ \frac {n}{p}$[/quote]

The source is an obvious fake, it's a Maths&Youth problem.
Anyway, I gave a good hint (in my opinion) at http://www.mathlinks.ro/Forum/viewtopic.php?t=164079 . Let's continue the discussion there.

  darij


calculus
integration
calculus computations


topology
linear algebra
matrix
function
vector
linear algebra unsolved

Hi all

I am taking a differential topology course but my linear algebra is a little rusty, this problem came up in one of my lectures, and since it is of a linear algebra nature, I was hoping someone here could help me out.

In the lecture we were talking about the construction of Grassmanians as quotient spaces, and so we were considering when two "k by n" matrices of maximal rank k (k<=n), say A and B, represent the same subspace in R^n. It is clear that this occurs iff there is an invertible k by k matrix G such that G*A = B. Now it was mentioned in class that you can construct a function f, such that given A and B of rank k, you can prove such a G exists iff f(A,B) = 0. Now I know construction of this f has to do with sums of determinants of the minors of A and B (or possibly sums of their squares), but I can't figure out exactly what f is supposed to be. I would greatly appreciate it if anyone out there who is familiar with this sort of thing could tell me what the function f is precisely, as I have been unable to figure it out myself.

Thanks for the help.

I think of it this way: the columns of $ A$ are vectors spanning the subspace determined by $ A$. Therefore, two matrices $ A$ and $ B$ determine the same subspace iff all of their columns, put together, have a $ k$-dimensional span. In other words, let $ C$ be a $ 2k$ by $ n$ matrix obtained by putting $ A$ and $ B$ together; the condition is that $ C$ has rank $ k$. We already know it has rank at least $ k$, so it remains to check that all minors of size $ (k \plus{} 1)$ are zero. You can take $ f$ to be the sum of their squares.

I'm not entirely sure why introducing this function is useful. If you really understand why such a $ G$ exists, then that is all that should matter. The "it is clear" part that you wrote about seems "straightforward," but making sure you can actually fill in the details would be good linear algebra practice if it is rusty.

My only guess at such a function would be something along the following lines. If $ A$ and $ B$ are of rank k, then each has an invertible $ k\times k$ minor. Let's denote that $ A_k$ and $ B_k$ respectively. Then $ AA_k^{\minus{}1}$ will be a matrix with the identity for that $ k\times k$ minor. Note that you said the Grassmannian is a quotient space. The quotient map being the span of a particular k-frame. So in essence what this multiplication by the inverse does is chooses a canonical form for the subspace.

Now my guess is that $ f(X,Y)\equal{}det(XX_k^{\minus{}1}\minus{}YY_k^{\minus{}1})$ ... or something like that. We do get that there is a $ k\times k$ minor of all zeros in there, but it also is not a square matrix that we are taking the determinant of. So maybe there is a step missing where you extend $ XX_k^{\minus{}1}$ to a basis of $ \mathbb{R}^n$?

Edit: Ignore or get what you want from this, it appears someone else has posted during my writing this out.


You should post this in the lower sections. Seems like a school problem or homework problem.

"Getting started" is somewhat of a misnomer. The problems are possible for people who haven't learned trigonometry at a deep level, but they are still supposed to be problem solving problems, rather than excersizes, such as the problem in question.

[quote="ryan22"]i'm sorry i ment to post it in getting started. What do you mean its a hw problem? It just review? What kind of problmes are the ones for intermediate leavel could it be HW?? Cause the problems I see in the intermeidtae leavel form dont look like math they look like puzzles......???[/quote]This is one issue a lot of people are confused about. Instead of explaining it to you, I'll give you examples for you to see what is meant by homework problems and what are the type of problems that intermediate levels look like (reminder: please DO NOT SOLVE these problems here)

Howework problem: Simplify $\frac{x^3-y^3}{x-y}$

Getting Started problem: What is the value of $r^2+s^2$ if the $r,s$ are the roots of $x^2-3x+1=0$?

Intermediate Topics problem: Prove that you cannot use 31 dominoes to successfully cover an 8x8 chessboard with two opposite corners removed.

Now, do you see the increasing difficulty in the problems? I hope you understand, if not, send me a PM :)

Getting Started forum is for like the easy AMC questions or AHSME  problems or some county math league problems like WOCOMAL(Worcester County Math League). You can see some problems here http://www.wocomal.org

The difference between simple exercise problems(often called "homework" problems) and problem-solving problems:

"Exercise" problems (usually inappropriate for this forum or higher-level forums) are relatively straight-forward problems that you can apply some theorems/formulas to get the answers.
On the other hand, "Problem-solving" problems cannot be solved simply by applying theorems/formulas; they require some thinking/problem solving skills.

[quote="Apprentice"]would the answer be $\frac{3+\sqrt{5}}{2}^2 + \frac{3-\sqrt{5}}{2}^2$
$\rightarrow \frac{28}{4} = \boxed{7}$ ? I have some free time...[/quote]This is a great example of why I ponder sometimes the intellectual level of our readers. Yes I am not happy because I SPECIFICALLY said DO NOT SOLVE. And here comes an amazing reader with free time and answers it... :furious:


function
analytic geometry
graphing lines
slope
calculus
calculus computations

how do I prove that given a differentiable function f and point (x_1, y_1) that the shortest path from (x_1,y_1) to the curve is a line segment perpendicular to the tangent line of the graph?

-this is from a set of problems given before a calc. I exam from previous years.

You want to find a point on the graph of $ f$ that is closest to $ (x_{1},y_{1})$. This means minimizing the expression $ g(x)\equal{}(x\minus{}x_{1})^{2}\plus{}(f(x)\minus{}y_{1})^{2}$. Once you apply some witch magic to the equation $ g'(x)\equal{}0$, it  will show that the slope of the tangent line at $ x$ is $ \minus{}\frac{x\minus{}x_{1}}{f(x)\minus{}y_{1}}$.


Every sequence is really a function defined on N (natural numbers).  In fact, {a_n} for all n in N is just shorthand for f: N -----> R where f(n) = a_n . Prove, using episilon, delta that every sequence ( every function defined on N) is a continous function.

Well...i know the episilon delta definition...and when i try to apply it to the sequence...it doesnt really make sense...perhaps there is some other trick?

whoops..i didnt mean uniform..i just meant continous..sorry..but its somewhat similiar to the uniform continuity isnt it? and oh yea sorry about the double repost...didnt know it was gonna get move by itself..thanks..

That definition makes sense for functions between any two metric spaces, which $ \mathbb{N}$ and $ \mathbb{R}$ are. What can we say about two natural numbers $ m,n$ with $ |m\minus{}n|<1$?
The generalized version of this: Suppose $ X$ is a discrete topological space, and $ Y$ is any topological space. Any function from $ X$ to $ Y$ is continuous.

By the way, this belongs in the calculus/real analysis forum.

There is no trick. It is just one pecularity of the definition of continuity. If you are talking about uniform continuity, as the subject of your post suggests, you should get something like

[color=red]For every $ \varepsilon > 0$, there exists $ \delta > 0$ such that $ |a_m \minus{} a_n| < \varepsilon$ whenever $ m,n\in\mathbb N$ and $ |m \minus{} n| < \delta$. [/color]

(if you get something drastically different, let me know what it is).

Now the whole point is that if you take $ \delta < 1$, then the conditions $ m,n\in\mathbb N$ and $ |m \minus{} n| < \delta$ imply that $ m \equal{} n$, so you are down to checking that $ |a_n \minus{} a_n| \equal{} 0 < \varepsilon$, which is obvious.

Moderator remark: [color=red]Please, do not double post![/color] (especially after you have been answered; if you posted in the wrong subforum, we'll move your post, do not worry :) ).


symmetry
combinatorics unsolved
combinatorics

Let $ m,n$ be positive integers, and let $ F$ be a family of $ m$-element subsets of $ \{1,2,...,n\}$ satisfying $ A\cap B \not \equal{} \emptyset$ for all $ A,B\in F$. Determine the maximum possible number of elements in $ F$.

[quote="N.T.TUAN"]Let $m,n$ be positive integers, and let $F$ be a family of $m$-element subsets of $\{1,2,...,n\}$ satisfying $A\cap B \not = \emptyset$ for all $A,B\in F$. Determine the maximum possible number of elements in $F$.[/quote]

After some calculation I got $C_{n}^{m}\cdot \sum_{i=1}^{m-1}C_{n-m}^{m-i}$. Is this correct?

There are $C_{n}^{m}$ ways to choose $m$ elements from $\{1,2,...,n\}$ . Supose that we have chosen $m$ elements for the set $A$, and now we want tho choose $m$ elements for the set $B$ in that way that  $card(A\cap B )=i$. Then we need $m-i$ elements completing the set $B$, and these elements we choose from the $n-m$ elements remained in the set $\{1,2,...,n\}$. This  can be made in $C_{n-m}^{m-i}$ ways. So therefore $cardF=C_{n}^{m}\cdot \sum_{i=1}^{m-1}C_{n-m}^{m-i}$. Is this ok?
Forgive my bad english.

Assuming that $m<n$, the answer is: $\binom{n}{m}$ for $2m>n$, and $\binom{n-1}{m-1}$ for $2m \leq n$.

If $2m>n$, then $F$ can simply contain all possible $m$-element subsets of $\{1, 2, ..., n\}$: for any two such $A, B$, since they have $2m$ elements in total and this is more than $n$, they must share some element and thus have non-empty intersection. Then we get $\binom{n}{m}$.

If $2m \leq n$, first note that it is trivial to get a family $F$ containing $\binom{n-1}{m-1}$ subsets: let all the subsets contain the element $1$, and then pick $m-1$ numbers from the remaining $n-1$ in all possible ways.

Now, is it possible to get more? Note that $\binom{n-1}{m-1}=\frac{m}{n}\binom{n}{m}$. If $n=2m$, it's easy to see why $F$ can't contain more than half of all possible subsets: for any $A$, let $A'$ be the only set such that $A \cap A' = \emptyset$. Then the family of all possible subsets is broken up into pairs, and we can take at most one subset from each pair.

If $n=lm$, a similar argument should prove that we can take at most one out of $l$ subsets: partition the family of all possible subsets into disjoint sub-families each containing $l$ disjoint subsets, and the result follows. However, it's not obvious that such a partition exists. We can sidestep that problem: first take a sub-family consisting of some $l$ disjoint subsets which form a partition of $\{1, 2, ..., n\}$. Now, rename the elements $1, 2, ..., n$ in all $n!$ possible ways; for each of these ways we will get a new sub-family with $l$ disjoint subsets. In the resulting huge family, each possible subset $A$ is represented many times, but by symmetry arguments every $A$ is represented the same number of times. Now, given any valid $F$, for each of these sub-families each having $l$ subsets, at most one of them can be in $F$. Since all subsets are equally represented, this means that at most one $l$th of all possible subsets are in $F$.

If $n$ is not a multiple of $m$, we would like create a sub-family containing $n$ subsets such that any valid $F$ can contain at most $m$ of them. This is easy: take $\{1, 2, ..., m \}, \{2, 3, ..., m+1 \}, ..., \{n, 1, ..., m-1 \}$. This clearly works. Now, as above, rename the elements $1, 2, ..., n$ in all $n!$ possible ways, etc. Then at most $m$ out of every $n$ possible subsets are in $F$.


I study it for the tingling feeling you get in your tummy when you finish a hard problem :P
And, I also like how its practical and abstract at the same time.

I study math because it is fun and it is useful for many things. I also study math because we don't know everything there is to know about math yet there are many theorums and patterns out there waiting for you to discover them.

I'm not sure why I study math.... hmmm....

I study math because 6 out of 5 people think that 1=0 is OK

LOL

Lemme see...  I study math for the fun of it.  I like the feeling you get after answering a really tough problem.


A light-year is $ 9.5 \times 10^{12}$ km. A star in the Andromeda Galaxy is estimated to be $ 2.0 \times 10^6$ light-years away. How many kilometers away is that star? Express your answer in scientific notation.

[quote="KingJamesMvP"]2*9.5 is 19.
10^6*10^12 is 10^18.
So the answer is 1.9 * 10^19[/quote]

But you have to use LaTeX.

Okay, To find the answer. (9.5*10^12)*(2.0*10^6) This equals [u][b]1.9*10^19[/b][/u]. You can do this beacuse to find how to find the distance you multiply the rate by the time.

[quote="pi31415926"]It says you're wrong if you type 1.9*10^19
[/quote]

You have to type 1.9times10^19.

They really should make the wording a lot more lenient. I mean, I probably got 5% of my problems wrong due to wording!


ARML email AMC AIME MATHCOUNTS USA(J)MO USAMO

Does anyone know who is confirmed going to ARML?

I just want to know because I know a lot of good people are not going for a variety of reasons.

Also, who do you think will be the 15 on the Gold team?

I'm going...

Which people are going anyways? And if there are only like 20 people, are we going like 12-8 or 10-10 or 15-5 on each team?

Top people...

Phil
Sachin
Nathan
E.G.
Tyler
Shawn
Lauren
Jeffery
Lyndon
Richard
Joseph
Yuqing
Andrew
Mark
Jessica

That's 15, but there's probably some people I don't know about...

problaby 15-5, they gotta have 15 on their first team

what about udayan?  is he going..i dont talk to him in school taht often these days(no classes together) and he doesnt respond pms........

oh well, I bet indianamath knows............he should be with him tonight................................................................................ :wink:

yea I know like some people from terre haute that go there to practices but I dont know if they will actually go (like mathcountsfreek, and some girl named ashley? (sorry)..and like these indian poeple (no offense) like pardha and raj, maybe matthew croop and defiently sam dittmer-he went to that practice 2 weeks ago)....yea...and i just know alot of good people arent going....including about like 10 people from carmel that went last year........

undefined are you going?

hurrary no sleep, now im getting stupider by the minutessssssssssssssss......
o well....i get to take my finals on the half day on thursday so I can go to arml..lucky me..and I cant even skip french..........................................

umm apparently mihika and anirudh prahbu got emails from Mrs. Benjamin inviting them to come. I didn't get one, but should I come anyway?

idk; if it's during the weekend, my dad might be able to drive me. Otherwise, I can't go or will have to carpool

I'm guessing most (if not all) the practices are over now?

teehee andrew keiser is coming??!! :rotfl:

And uh... i dun see David on this list any where why?? or the other two people from north central? All the people who passed AIME are going on the A team. 11 people
So phil, sachin, nathan, tyler, none of the other people on ur list.  In other words, the other 4 people will be E.G., and 3 others.  This narrows down the possibilities... And they might take top 4 MC people, and also depending on ur practice scores. So. Me and richard (possibly). Then, Jeffrey, lauren, or shawn

Phil
Sachin
Nathan
E.G.
Tyler
Shawn
Lauren
Jeffery
Lyndon
Richard
Joseph
Yuqing
Andrew
Mark
Jessica

[quote="mathcrazed"]And uh... i dun see David on this list any where why?? or the other two people from north central? All the people who passed AIME are going on the A team. 11 people
So phil, sachin, nathan, tyler, none of the other people on ur list. In other words, the other 4 people will be E.G., and 3 others. This narrows down the possibilities... And they might take top 4 MC people, and also depending on ur practice scores. So. Me and richard (possibly). Then, Jeffrey, lauren, or shawn [/quote]

nathan said david isnt going.  there arent 2 people from NC, theres onlyu one, the other is sam dittmer frmo zionsville.  Furthermore, undefined passed AIME, but he said he might not be able to go.  ALso, there are about 3 other peopel that passed AIME that I dont even know about (never showed up to practices).  If they want to take llike best 5 that didnt pass AIME, itll probalby be E.G. (since he passed last year), some kid name Pardha (he passed last year), lauren (since she has finished all of high school math in school), probalby jeffery over me (uh...because jeffery likes math better than me), and probalby richard and lydnon if they like mathcounts nationalists...

I dont really care what team they put me on, as long as I go.....

undefined you know yixin wang?
another usamo qualifier, michael harris, whos heard of him? (it said he went to arml as a 9th grader, but not as 10th o indidna dr. rickers arml page)
Mohit agrawal?  whos that?
sorry, but im just trying to find out whos who.

Pardah's gonna be on the gold team for sure. Meh, this post rating thing is bs, mine is 1.1...wtf?

[quote="Fanatic"]Uhh yuqing, you obviously weren't paying attention at ARML practice on saturday

Andrew Keiser was sitting like 20 feet from you across the room horizontally...[/quote]

wait that was him?  yuqing told me he was some kid that didnt make state.  oh well, problaby heard wrong

[quote="undefined117"]I think I'll go now.. mihika's parents agreed to give me a ride (thank you!) on thursday for the practice..
I know Yixin Wang, but not personally. I think he's going, but I might be wrong.
Mohit Agrawal is a senior at our school. He passed the AIME, but only got a 3 on the USAMO cause he missed the first day of it.[/quote]

hehe im going on thursday for the practice.....i just wana see how many people show up (probalby way less than last year).
3 on USAMO is actually beating like 1/6 of the people (guess?)
yeah ill be on the second team for sure now.....now i cant screw up the first team..............................

lets make a list (i dont know last names......why am i listing random people on the internet? :ninja: )

Phil Thomax
Sachin Shinde
Nathan Benjamin
E.G. Wright
Tyler Zou
Soo-Hyun Yoo
Yixin Wang
Sam Dittmer
Matthew Croop
Pardha Pongouti

I dont know about the other people like Mohit Agrawal, or Michael Harris..so im going to leave them out first

 I dont know anyone else that used to pass AIME (this is directed at Lyundon saying if you pass AIME, youll be on team A)
so
5 more
uh..
Lyndon Ji
Richard Ni
Jeffery Shen
Lauren Stevens
ummm..................one more perrson
this is biased.........im forgetting lots of great people frm west lafayette and terre haute and other regions of indiana.................sorry.

ehh but if you put me in the A team I might screw it up

and mihika and anirudh prahbu are going too :)

[quote="stupidkid"][quote="undefined117"]I think I'll go now.. mihika's parents agreed to give me a ride (thank you!) on thursday for the practice..[/quote]
Dude man, Terre Haute on thurs? Man, math is alright, but don't lose sleep over it... I <3 sleep. (would be sleeping if it weren't for the spurs game)[/quote]

wait your not going on thursday (this is thrusday next week)/..wtf, of couse no one would go on thrusday as in two days from now........................i thoguht its required to go.............thats why im skipping 2 of my finals................you shouldnt lose any sleep watching the spurs demolish the jazz .

and wtf portland won the lottery...too bad they cant draft oden because they already hvae a PF and they used a pick to draft ALdridge......

[quote="rebirth"][quote="indianamath"]uhh..do you not have any1 who wants to carpool with you???? we're going to terrre haute...it'd be stupid if every1 went by himself...

we're just meeting there, since it's sort of in the middle of everything and every1 knows where it is.[/quote]

can i please please plase carpool with you guys..please!!! my phone number is 849-1127[/quote]

he was being sarcastic. he's not going to arml

[quote="rebirth"]and how do you know im not going to arml

richard, seeing that your nice, DONT CALL THAT PHONE NUMBER, tis got 911 in it?[/quote]

I told him (or I think I did, implied)

911, that's if you only have one of those dialing phones that dials one number at a time instead of total one where you put in the whole number first.  I think thats why you cant have 911.

McDonalds.  Umm, I'll be late (like 3:00) because I have to take some tests by myself with my French teacher monitoring.  Hope you don't get her as your French teacher.  She is the only reason why I can't even skip the final when every other teacher allows foreign language skips.

Also, what is this rooming thing.  I mean, is it some college dormatory?  or some normal hotel?  Also, someone bring some cards, and/or ill just borrow a DS.

it's at Rose Hulman dorm.

i thought we were meeting at around 4:30.

? Dr. Chi told me it was 7:00 (in Rose Hulman)

we're meeting at my house at... whatever time my mom told u guys.

what? i thought we were meeting at the McDonalds on 96th at 4:30.

can i PLEASE go with you guys.  flying isnt safe.

indianamath,

you're right it might be the McDonalds... just do whatever my mom told you lol.

[quote="indianamath"][quote="Proud Asian"]Im going to ARML.

Calvin White.
Carmel High School.[/quote]

wtf?? proud asian is a white guy??? lol.[/quote]

No actually hes semi asaian....and half  black

I dont find both of your conversations very amusing.

I thought we were meeting at McDonalds around 2ish (after I get done with my finals), like my dad will pick me up from Carmel High and pick Yuqing up from his house (or wait, you dont have half days do you... :?: )...o well, my parents know this more than me.  I'll just ask them.

Dont worry about me, I like wasting myself before finals...its good luck for me.

EDIT: Richard, why are you up so late.  Its bad for you, its 1:00 almost

I get to sleep in tomorrow before graduation, and I'm fixing the crap out of my graduation speech.

Oh wow, private schools take graduation seriously...
Considering it is a middle school as when I was in middle school we never even had a graduation where names were called and you guys have to make speeches on top of that.


[hide]For a start, $ \cos \alpha \equal{} \frac{b^2\plus{}c^2 \minus{} \sqrt{b^4\plus{}c^4}}{2bc}$.[/hide]

[quote="Altunshukurlu"]If there is a triangle with sides (a,b,c) and a^4 = b^4 + c^4
Then find the measure of angles[/quote]
the existence of such a triangle is evident fact,take,for example $ a \equal{} \sqrt [4]{2},b \equal{} c \equal{} 1$.
what makes you think that its angels must be fixed?obviously the provided triangle and a triangle with sides $ a \equal{} \sqrt [4]{2},b \equal{} \frac {1}{2}$ and $ c \equal{} \sqrt [4]{\frac {31}{16}}$ are not similar,hence these two triangles have different angles...
of course,by using Carcul's hint you can find some relation between angles,but nothing more.


I just got confirmation of this news from ETS and my ARML coach a few hours ago. ETS claims that the reason for not giving an alternate test date is because students from too many different states want to do it. What do people think about this and its impact on ARML?

(Sorry, I put this in the wrong forum, someone move this please. Thanks.)

Does ARML [i]have[/i] to be on the SAT date?  I mean, can't they move it to another week?

I think people should boycott ETS and not take their tests since they are working to prevent students from attending ARML.

[quote="ComplexZeta"]I think people should boycott going to college and not take their tests since they are working to prevent students from attending ARML.[/quote]

Fixed the quotation.

I think ARML is probably actually at fault here, because I imagine, possibly mistakenly, that ETS announced their date first.  Though, even if this is true, ETS is still wrong not to offer an alternate date.  Oh well, doesn't affect me, I'm not going to ARML.

You ``fixed'' the quote so that it doesn't make sense. Why should I blame ARML which has been the cause of some of my fondest memories when I can blame ETS which only ever causes trouble?

Yeah, ETS does suck.  I'm especially opposed to standarized testing in general, so I dislike ETS even more.  But only a very widespread ETS boycott would accomplish anything, and that simply won't happen.  The layperson doesn't care about ETS and ARML having the same date, and plenty of mathy people don't care that much anyway.  ARML participants are much more likely to be affected by the SAT than SAT people are by ARML, which is why I don't think ETS really did anything wrong.  It would be unreasonable for them to care about every single conflict.

Hm we can arrange for thousands of students to go to various ETS testing sites on June 5th and hold up picket signs that say ``Let students do math!'' Then they'll take notice of us.

It's a bit unfair for them to do it at this time, the final day of (early?) registration.  Teams who take the alternate date yearly are probablly counting on this date, and consequently many people have already registered.  So now if people want to go to ARML and take the SAT at another date, they will lose their registration money (or have to pay an extra test-date changing fee).

Thousands?  I don't think many people would be willing to sacrifice both the SAT and ARML.

I'm going to stop being so humorless now.

Well, we could have math majors who don't want to go to ARML and parents of ARML people doing it. I don't know.

[quote="Ravi B"]Is there any chance that the ARML folks can administer the SAT I and II at the ARML site, maybe the day before the contest?[/quote]

I doubt it -- besides, people going to ARML would almost certainly not want to take the SAT there.

It's a pointless exercize to "blame" one or the other of the two organizations.  They have no responsibility to schedule around each other.  I would blame ETS for waiting until it was this late to tell people, though.  Why does it matter so much to take the SAT2 now?  You can always take it some other time.  (Yeah, yeah, I know, class just ended and whatnot.  Your mind will melt if you have to take the test after summer.)

[quote]Why does it matter so much to take the SAT2 now? You can always take it some other time. (Yeah, yeah, I know, class just ended and whatnot. Your mind will melt if you have to take the test after summer.)[/quote]

Unless you're a junior...

I took my one and only SAT 1 test when I was a senior.

That works for you Simon. Unfortunately, for me, my parents will ultimately determine my location June 6th, and sadly it might be taking some tests at a local high school :(.

I think ARML should just move up or back a week.

I also think the ETS should be more flexible-- after all, ARML is an educational thing. Isn't promoting education one of their goals or something?

Grr, this is annoying.  Hopefully my brain won't "melt" over the summer ;)

One of the two programs had to have posted their dates first, and the other one could have had the foresight... oh well.

[quote="mathnovice"]I also think the ETS should be more flexible-- after all, ARML is an educational thing. Isn't promoting education one of their goals or something?[/quote]

But if the ETS is more flexible and allows students to take the SAT on a different date, it becomes much easier for someone to cheat.  I'm not saying that it will happen, but in general, the ETS likes to keep everything synchronized to keep the tests more secure.

And promoting education isn't really one of their goals.  All they care about is determining what people know (and by some accounts, their tests don't even measure that.)

But they give a different test if you take it on an alternate date.

Oh.  Well, in that case, they are just being uncooperative.

But I think the ARML people are the ones at fault.  They should try and schedule ARML far, far away from whenever they think the SAT date will occur.

ARML and SAT are ALWAYS on the SAME weekend. At least for the past three years.

Don't know why. Makes no sense. But it happens anyway. And it's probably not going to change anytime soon.

[quote="Nukular"]ARML and SAT are ALWAYS on the SAME weekend. At least for the past three years.[/quote]

I'm sure this has everything to do with the unchangeable fact that there are only so many weekends near the end of the usual United States school year. There are [url=http://www.collegeboard.com/student/testing/sat/calenfees.html]other SAT dates[/url], and always have been, and SAT dates next autumn are still timely for early action admission applications by next year's high school seniors. Personally, I think the "solution" is to spring kids from more regular school days for both of these purposes  :)  , but as long as the College Board and ARML both have to accommodate school schedules, something will conflict with something.

(After edit:) Oh, yeah, and this whole thread really ought to be over on the Other Contests and Programs forum, not on this Other Problem Solving Topics forum.

Is anyone else here going to have to miss ARML because of this?

I think if you're <junior, then you should definitely go...

Who runs ARML, i.e. who makes all the decisions? Is ARML a company, an organization, or what?


logarithms
inequalities
inequalities unsolved

Prove that for all positive numbers $x \neq 1$  (where $\ln x$ is the natural logarithm $\log_{\textrm{e}} x$) we have
\[x^{1.5}\dfrac {\ln x} {x^{1.5}-1} < \ln(x+1) < x(2x+1)\dfrac {\ln x} {2(x^2-1)}.\]


[quote="t0rajir0u"]Sometimes this is how the Catalan numbers are defined.  What definition are you using?[/quote]

Errrrrrrrrr...Sorry. I mean C_n=1/(n+1)*2nCn

[quote="Stephen"]If a_1=1, a_2=1, a_(n+1)=a_0a_n+a_1a_(n-1)+...+a_na_0, prove that a_n are Catalan Numbers.[/quote]
Exactly $ a_0\equal{}a_1\equal{}1, a_{n\plus{}1}\equal{}\sum_{k\equal{}0}^na_ka_{n\minus{}k}$. It give $ a_2\equal{}2\equal{}\frac{1}{2\plus{}1}C_4^2$.
Consider $ f(x)\equal{}\sum_{k\equal{}0}^{\infty} a_kx^k$, then $ f^2(x\sum_{k\equal{}0}^{\infty} c_kx^k,$ were $ c_k\equal{}\sum_{i\equal{}0}^k a_ia_{k\minus{}i}\equal{}a_{k\plus{}1}$. Therefore $ f^2(x)\equal{}\frac{f(x)\minus{}1}{x}$ or $ f(x)\equal{}\frac{1\minus{}\sqrt{1\minus{}4x}}{2x}\equal{}\sum_{n\equal{}0}^{\infty}Ca_nx^n$, were
$ Ca_n\equal{}\frac{4^n*\frac 12 *\frac 32 *...\frac{2n\minus{}1}{2}}{(n\plus{}1)!}\equal{}\frac{1}{n\plus{}1}\frac{2^n(2n\minus{}1)!!}{n!}\equal{}\frac{1}{n\plus{}1}\frac{2n!}{n!n!}\equal{}\frac{1}{n\plus{}1}C_{2n}^n.$


I have posted this your ineq but perhaps you have not noticed it, so I post it again

[quote="manlio"]For $a,b,c$ sidelenghts of a triangle prove that

$\sum \frac{a}{b}+5 \geq \frac{8}{3}\sum \frac{a+b}{b+c}$

I write it as

$[\frac{1}{ca}-\frac{8}{3(b+c)(a+b)}](a-c)(b-c)+[\frac{1}{ab}-\frac{8}{3(b+c)(a+c)}](a-b)^{2}\geq 0$

WLOG we can assume $a\geq b \geq c$

It is easy to prove that $[\frac{1}{ca}-\frac{8}{3(b+c)(a+b)}]$ is positive but I cannot prove that $[\frac{1}{ab}-\frac{8}{3(b+c)(a+c)}]$ is positive.

[/quote]

I am trying to solve this one from a week and I asked kindly you if you have a solution. Also I sent you two private msgs to ask you if you have a solution, but you never answer me. Can you please tell me if you have a solution and give me a hint, please, to solve it. Thank you very much, 10maths_tp0690. :)

Sorry you, I was very busy that time, so I couldn't send you the solution for this inequality. I've just sent you a solution to apologize.


I found the test to be easy this year. Did anyone solve question #3?

Q3:

Paul Martin is racing Steven Harper, Steven has acceleration, a, and top speed, v.
Paul have half of Steven's acceleration, but 50% more of Steven's top speed. Since Steven has a faster acceleration, he will be ahead of Paul in the beginning part of the race, but Paul will eventually catch up to Steven. At what distance will Paul catch up to Steven?

$(A) \frac{v^2}a (B) \frac{2v^2}a (C) \frac{3v^2}a (D) \frac{4v^2}a (E) \frac{5v^2}a$

it was 3v^2/a

btw, did anyone actually understand that question about Bush's new skating rink? It was totally non physics and so badly worded.

I didn't find it to be bad at all, we were suppose to find the difference between the length of the striaght for the metric track and the mile track with the information given. The problem that is totally non physics was the IPod question. Btw, what do u think the cutoff for top 100 will be?

I find it really amazing to talk about SIN exam. 8th Army and Liu , you guys sound confident which is a good sign that you did the exam well. How many problems did you solve?
I think I spent quite long on presenting the solutions for my answers.
“The problem that is totally non physics was the IPod question.” You can say that again, Liu. What was your choice for this Ipod question ( question number 7)?
About your question “ cutoff for top 100”, personally, I think that it depends on how the students did the test. For example, the best one could solve 10/12 problems, and you solved 9/12 ( just for example) , you would expect the good result. But if the best one had solved all, bad luck.  I know that there were 14 problems previous years.
“Remember, this is a prize exam, not an entrance test. If you have than half the answers right, you will likely be in the top 5%”. Honestly, I don’t believe it. Think about it, there are over 15,000 students taking part in this test. 5% which is 750 and they only mark 200 top paper. So, I don’t feel confident at all.
Any approval?

[quote="satyam"]I probably got 3/12 on the SIN exam. I didn't do that good this year, but I expect to get a better mark next year when I am in grade 12. Btw, does anyone know any good physics book to prepare from for the SIN exam. Also, when do the results come out.

[quote]btw, does anyone know when the CAP usually comes out? its getting quiet close to June, if I calculated correctly...[/quote]
8th Army, if you mean the 'CAP Art of Physics Competition', then the results already came out.[/quote]

i recommend a "decade of sin +14" whatever the title is now. being sold by uw physics department. sin is relatively easy in comparison though...

[quote="winter97"]“decade of sin+14”, is that the book which contains all the tests from 1969 until now with solutions ? I am not quite sure if that was the book you recommended.
Satyam, you meet my approval. It would be a good preparation if you could have that book and you tried to solve all the previous tests. You will find it easy.Here is my experience: at first, I solved problems of each year but it was not a good way. Let me explain, once I met a tough one which I could not solve, I looked at the solution and I thought that I would remember it. Then, I tried test of another year and I met the same one , I didn’t remember it at all. Until one month before the test, I found an effective way: there are a lot of different parts in the test of one year like : acceleration, freely falling objects, collisions, relationship between conservative forces and potential energy, circular motion……I collected all the problems having the same topic from all years and solved them, then I changed to another topic. It was a good way to review. Try it. Good luck.[/quote]

yup, thats the book i'm talking about. if u want to go further, ask elites on this forum like Yufei, Peng, and Elyot who did amazing in sin 2005.

The way I studied for the SIN exam was like this: I looked at the sample questions and the answers, then covered the answers and tried to solve that problem without looking at the answer. Plus, since I didn't know any Grade 12 Physics, I took the grade 12 physics textbook home and memorized as many equations as I can. This process worked quite well for me. I got 6 questions right and ended up in the 86th percentile.
Btw, does anyone know what is the highest mark of a Grade 11 student in SIN exam. One of my friends told me that during 2005 the highest mark of a grade 11 student was 5/14, but I am not sure if this is the fact. Also, I will really appreciate if you guys can post your studying stragies, mark and grade.

I wrote the SIN exam and i got pwned...mainly because i'm in gr. 11 and at that point had not completed gr. 11 physics. I would like to see the results though could someone please give me the site..

Thank you

The names of the top 100 contestants are at http://www.sin.uwaterloo.ca in early June. If you were not in the top 100, then you can ask your teacher for the result, or simply compare your answers with the official answers. :lol:

ya, i definitely didnt make the top 100 but thx for letting me know

SIN this year wasn't that bad, the last question was surprisingly easier than some of the questions before that.

[quote="liudaveliu"]After checking the result, I realized that I only got 6/12. I got the first question wrong by forgeting the kinetic energy formula was  $1/2 mv^2$ :blush:[/quote]

i probably got 4/12..if i'm lucky

[quote="satyam"]Does anyone know what is the highest mark of a Grade 11 student in SIN exam?
Also, when do the OAPT physics contest results come out?[/quote]

the oapt results are already out, you can find them here:

http://www.oapt.ca/grade_11_contest/winners.html

I'm not sure about the highest mark for a Grade 11 in a SIN exam, but last year a few grade 11's placed in the top 10 so it must be fairly high.

Hope i helped!

did anyone found the computer mark was wrong?
my teacher told me mymark is 58%....i was soooo shocked,
i have my answers and i'm sure i got 10/12....
and he checked again this week....and it became 69%....but still not my answers onthe web.... :(  :(  :(  and a lot of other people in our school got mark changed....
they had computer problem this yr?!


Find the exact values of $ x$ which satisfy $ \sqrt [3]{x} \plus{} 2\sqrt [9]{\frac {x^2}2} \minus{} \frac {1}{\sqrt [3]{2}} \equal{} 0$.

[hide="Comment/hint"]
Apparently nobody got this question right; I left my graphing calculator at home so I couldn't solve it. The solution suggests
\[ \sqrt [3]{x} \plus{} 2\sqrt [9]{\frac {x^2}2} \minus{} \frac {1}{\sqrt [3]{2}} \equal{} \left(\sqrt [9]{x} \plus{} \frac {1}{\sqrt [9]{2}}\right)\left(\sqrt [9]{x^2} \plus{} \sqrt [9]{\frac x2} \minus{} \frac {1}{\sqrt [9]{4}}\right) \equal{} 0
\]
which looks a bit arbitrary to me, thoughts?
[/hide]

Edit: [hide="Better hint"]
In fact this question was a lot easier than I thought. Make the substitution $ y \equal{} (2x)^{1/9}$, and the root $ x \equal{} \minus{}1/2$ becomes obvious, we have a quadratic left over.
[/hide]

The suggestion from the solution seems at bit [i]out of ordinary[/i].  The substitution method would seem to be the quickest.

Is the CJML, Central Jersey ML?

[quote="AstroPhys"]Is the CJML, Central Jersey ML?[/quote]

Yep, it is. Its used to decide the Central Jersey ARML team.

I almost got this one. But I simplified wrong somewhere, and ended up with a cubic, with a missing $ x$ term. How stupid.

Oh, and congratulations on your 9. I'm from UC Magnet, and I made like a bunch of dumb mistakes and got a 6.

When I use azjps' substitution hint, I get $ y^3 \plus{} 2y^2 \minus{} 1 \equal{} 0$. However, how the heck would somebody think of the substitution $ y\equal{}\left(2x\right)^{1/9}$? $ y\equal{}x^{1/9}$ is reasonable, but the other substitution??

It's not really as unmotivated as I thought it seemed; $ \sqrt [3]{x} \plus{} 2\sqrt [9]{\frac {x^2}2} \minus{} \frac {1}{\sqrt [3]{2}} \equal{} (2x)^{(1/3)} \plus{} 2(4x^2)^{(1/9)} \minus{} 1 \equal{} (2x)^{3/9} \plus{} 2(2x)^{2/9} \minus{} 1 \equal{} 0$. Hmm but even if I got it right they probably wouldn't have taken it due to the weird answer format they requested.

Oh btw hope you're ready [to get owned].  :P Edit: We'll see.


Putnam
function
integration
limit
real analysis
real analysis solved

Let f:]0,1[->]0,1[ a strictly decreasing function s.t.  \int {0->x}f(x)dx has a finite limit as x->00. Prove that xf(x)->0 as x->00.

[i]Putnam/forgot the year, maybe someone else knows, so I can edit the post[/i]

f is decreasing so
0 \leq x/2.f(x) \leq  \int _[x/2;x]f(t)dt= \int _[0,x]f(t)dt- \int _[0,x/2]f(t)dt-->L-L=0 when x-->+oo

THat's SO COOL? My solution is horrible. I can't even remember it now that I've seen yours! :D

Oh, my God, this can't be true. I also had a monster of solution to this one, but this is really incredible. This is really memorable.

Another one wich is the discrete case
a(n)  \geq 0 sequence of positive number decreasing a(n+1) \leq a(n)
such that \sum a(n) is convergent, prove that  \lim n.a(n)=0

well, this one is 2571st problem in the Demidovich's problem set.

denote r(m) = \sum {i=m}^\infty ai
For any m and n>m,  we have
(n-m)a(n) < \sum {i=m+1}^n a_i < r(m)
n a(n) < n/(n-m) r(m).

for a given  \varepsilon > 0, we can find m0 such that r(m0) <  \varepsilon.

since n/(n-m_0) \lim 1, there exists n0>m0 s.t. n/(n-m0)<2 for all n\geq n0.

hence, we have 0 < n a_n < 2\varepsilon for n\geq n0, which implies n a_n  \lim 0


inequalities
function
inequalities solved

Let x,y,z \geq 0 such that x+y+z=x 2 +y 2 +z 2 . Prove that:
    \sum x(y^3+z^3-2) \leq 1/8*[ \sum (x-y) 2 ] 2 .

after plenty of troubles i think the inequality reduces to proving
2+xyz \geq xy+yz+zx
but i aint sure whether this is true or not.so harazi (or ne1 else for that matter who is reading this) check if it is true(and tell me otherwise i'd be wasting time to prove sumthing wrong :) .i think it's true after trying out some values :? )
and harazi u need not give the solution right now.i am trying hard at this one. :D

Now, I won't post a solution now.  And yes, the inequality at which you arrived is valid.

This inequality is quite a hellish and yet beautiful problem  :cool:  ;)  :blush:

ok another step forward (i'm doing this in bits and pieces :D )
i'm replacing the variables x,y,z by a,b,c
let
a+b+c=A
ab+bc+ca=B
abc=C
P(X)=X^3-AX 2 +BX-C
The condition is equivalent to A 2 =A+2B
And the inequality to be proved is 2 \geq B-C
NOTE that A \leq 3 and B \leq 3
and now the only case i considered is P(1) \leq 0
that is equvalent to B-C \leq A-1 \leq 2
as for the case P(1) \geq 0 I'm still searching :(

I bet you will lough when you'll see the solution.

now here's another thought
since we have completed the case where two of the numbers are less than 1,we set
a=1+l,b=1+m,c=1-n with l,m,n  \varepsilon [0,1]
then the condition is l 2 +m 2 +n 2 +l+m-n=0
and the inequalty to be proved (i.e 2+abc \geq ab+bc+ca) becomes
0 \geq lmn+l+m-n
now here's what i think ,this function is linear in each variable so we just check the endpoints,am i correct?? :?
and if this is not,GIVE THE SOLUTION,HARAZI (This is SOS)
coz if i dont get the solution ,i wont have my mind relaxed and can't study chemistry(chemistry xam on monday) :D  :D
and BTW,ur inequality coupled with my failure to solve it has given me new signature food :)

The function is linear in each variable, but the condition that links the variables is very messy. I'm not sure you can apply directly this argument. Anyway, my solution to xy+yz+zx<=2+xyz si the following:
   In the topic "A proposal" I proved the inequality a^2+b^2+c^2+abc+2>=a+b+c+ab+bc+ca. And this solves the problem.

Topic "A proposal" doesn't exist. Dou you remember the exact title.

Thank you.


I am solving a problem in which there is a hint to use the fact that  trace(BAB^-1)=trace(A) but I cannot prove it. Can anyone help me?

[quote="mlok"]There is a slightly more general identity: $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ for any square matrices $A$ and $B$. Just compute both traces: $\sum_{i}(AB)_{ii}=\sum_{i,j}A_{ij}B_{ji}$ and $\sum_{i}(BA)_{ii}=\sum_{i,j}B_{ij}A_{ji}$. The double sums coincide up to relabeling of indices.[/quote]

Do you mean that trace(BAB^-1)=trace(A) is not right?

[quote="mlok"]There is a slightly more general identity: $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ for any square matrices $A$ and $B$.[/quote]
You can strike the word "square." All you really need is for both $AB$ and $BA$ to be defined, so that $A$ is $m\times n$ and $B$ is $n\times m.$ There's no need to change the proof - reversing the order of summation, or "relabeling the indices" as you put it, works just fine.

Sayeh: that's what I meant on the other topic when I said that similar matrices have the same trace. Study what Xevarion said.


prove that sqrt(m+1)^sqrtm<(sqrtm)^sqrt(m+1) when m>7 for integers m

Have a look at
http://www.math.uiuc.edu/~hildebr/putnam/problems/mock04-2sol.pdf

square both sides
m+1^(sqrtm)>m^sqrt(m+1)
then multiply both radicals by sqrt m
m+1^m>m^sqrt(m^2+m) and then from there we can
then we take the mth root of both sides
m+1>m^sqrt(1+1/m)
when m is larger, sqrt (m+1)/m is smaller.
so if it works for any m, it will work for integers
since it works for 7, it works for all integers above 7

tell me if this doesnt work

No. Unfortunately it doesn't work.

There is a problem with your reasoning here:

[quote="Scrambled"]
m+1>m^sqrt(1+1/m)
[b]when m is larger, sqrt (m+1)/m is smaller.
so if it works for any m, it will work for integers
since it works for 7, it works for all integers above 7[/b]
[/quote]

ok then, but i am quite certain there is a non-calculus or log using solution

Is the text of the problem correct?!  :?

$\sqrt{m+1}^\sqrt{m}$ > $\sqrt{m}^\sqrt{m+1}$ or $\sqrt{m}^\sqrt{m+1}$ > $ \sqrt{m+1}^\sqrt{m}$  :P ?

Similar to: witch is bigger between $e^\pi$ and $\pi^e$. (a classic problem) :)


Solve $ \frac{1}{y}\frac{dy}{dx} \equal{} x^{2}y \plus{} \frac{x^{2}}{1\plus{}x^{3}}$ where $ y\equal{}1$ when $ x\equal{}0$.

$ y' \equal{} x^{2}y^2\plus{}\frac{x^{2}}{1\plus{}x^{3}}y$ --its a Bernulli equation.
You can make subst. $ z\equal{}y^{\minus{}1}$

hello, after the hint we get the following solution
$ y(x)\equal{}\minus{}\frac{4\sqrt[3]{1\plus{}x^3}}{\sqrt[3]{1\plus{}x^3}\plus{}x^3\sqrt[3]{1\plus{}x^3}\minus{}5}$
Sonnhard.


geometry
parallelogram
geometry unsolved

I centered incircle of triangle $ABC$ $(m(\hat{B})=90^\circ)$ touches $\left[AB\right], \left[BC\right], \left[AC\right]$ respectively at $F, D, E$. $\left[CI\right]\cap\left[EF\right]={L}$ and $\left[DL\right]\cap\left[AB\right]=N$. Prove that $\left[AI\right]=\left[ND\right]$.

A beautiful problem!
Its easy observe AB parallel ID. Now, AI parallel ND because ILBD is cyclic (¿why?). Conclusion, AIDN is paralelogram!!!  :D

Let's say $\angle A = 2a$, $\angle C = 2c$. So $\angle AIC = 135$, $a + c = 45$, $\angle AFE = \angle AEF = 90 - a$. $\triangle CDL$ congruent to $\triangle CEL$. Since $AI$ perpendicular to $EF$, $\angle ELC = \angle DLC = 45$. Hence $ND//AI$. Also $AN // ID$. Furthermore, $ADIN$ is a paralellogram and $AN = DI$


Some that happened
(I'm too lazy to watch  the fights from the front row, i rely on other people telling me about them)

october- near locker rooms, a gangfight, Surenos versus Nortenos. It started out as a verbal confrontation, but then fists were flying

February- Near the gym and near the vending machines- people FLOCKED to see  the fight. Therewere two people, who were picking each other up and bashing each other against the vending machines and walls and poles. No one was hurt.

February- Girl fight!!! Two girls broke up, and someone tried to stop the fights, but the two girls ended up slapping that person. By then lunchtime brawls had almost ended as the principal threatened to confine us to the gym.

So far, only two major fights--saw these both firsthand. I didn't hang around for the second half of the second one, though.
One involved some guys taking a few swings at each other in the cafeteria--where they were then promptly swarmed by teachers. The kids didn't swarm around them, as we were all quite comfy sitting at our tables.
The other one involved two girls in a nasty cat fight. Let's just say it wasn't pretty...

[quote="helenamei"]By then lunchtime brawls had almost ended as the principal threatened to confine us to the gym.[/quote]

WTH that is a great punishment! You can exercise for the rest of the day!

There was this one fight last year in the cafeteria, apparently one guy insulted this other guy's sister and that other guy drop-kicked him from the table...

Some kids tried to break it up, and it seemed alright for a second but the other guy went back and tried to drop-kick him again.

Apparently the comments were pretty mean though.

yesterday, a kid tried to strangle another kid by the vending machine in the cafeteria (during lunch) because that kid had tried to sit in his seat.   :o

He ended up getting in-school-suspension, but luckily nobody was hurt.

no offense to anyone, but only americans hate school suspensions... dont u get it? its a relief, not punishment

Its not really in school, but we're all in teh bathroom and we look out of the window and we see two gangs come together

now these are rough gangs, not like school people but like dropouts etc

so one gang has like five ppl and hteo ther like three. the pplwith five all pull out knives and get at the three person gangs. two police cars swipe by but don't stop. so one guy from the five person gang stabs a guy from teh three person gang. Then, one of the people from the two remaining people from teh 3 person gang (presumably the leader), pulls out a knife and goes back at the other gang. He takes down two people.

Meh I had to leave after that but my friend provided coverage. He claimed that the two gangs flew at each other and the three person guy who went with the offensive got killed. When we left school, police were at the scene. This was like a year ago.

usually, its just one person trips onother person, and they start talking each other.

[quote="ZhangPeijin"]My school is so weird.
Basically, if you're late to class, nothing happens.
But if you're late to school you get an automatic detention.
...
:|[/quote]

I was late because i did not get an admit slip the day after the aime. I got to skip running, though!

[quote="CatalystOfNostalgia"][quote="Nexmus"][quote="CatalystOfNostalgia"]haha, this reminds me in the middle of the usamo last year, there was a shouting match between two girls in the next room over, we pretty much heard every word

i forget what was said though, there was a lot of swearing, and i'm not sure if it got physical or not.[/quote]

Sounds awesome, 4.5 hrs of pwnage yelling.[/quote]

nah it was like 2 minutes[/quote]

wow your school has pretty thin walls :P

Eh...My friend and his friend KO'd(litterally) a kid for cutting in the lunch line. Umm...Oh and two girls got really mad at each other and started screaming vandettas at each other and slapping each other, I got a front row seat. The last one this year was when I hit this dork and he went sprawling. That was like last month though and I didn't get in trouble because he threw the first punch but I dared him because he said he could beat me up anytime, apparently not. And no I do not have anger management issues

[quote="#H34N1"][/quote][quote="CatalystOfNostalgia"][quote="Nexmus"][quote="CatalystOfNostalgia"]haha, this reminds me in the middle of the usamo last year, there was a shouting match between two girls in the next room over, we pretty much heard every word

i forget what was said though, there was a lot of swearing, and i'm not sure if it got physical or not.[/quote]

Sounds awesome, 4.5 hrs of pwnage yelling.[/quote]

nah it was like 2 minutes[/quote][quote="#H34N1"]

wow your school has pretty thin walls :P[/quote]
In order to save budget, and give the teachers more coffee and cake to eat, our school has turned each classroom into two classrooms by putting artificial walls 5 inches thick and covered with some sort of condensed paper diagonally across each classroom!
Yay!
Now we can get $ \approx$ 30 children squeezed into a trianglular classroom.

I've saw maybe 4 or 5 fights at lunch in my lifetime, all up close. In two instances, when an administrator/teacher intervened they both got hit and the fighters were expelled.


number theory
greatest common divisor
modular arithmetic
prime numbers

If $a$ and $b$ $(a > b)$ are prime numbers, consisting of at least two digits, then $a^4 - b^4$ is divisible by $240$. Moreover, $240$ is the $gcd$ of all numbers $a^4 - b^4$ arising in this way. Prove this.

the first part is rather easy with the knowlegde that $a^4-b^4=(a^2+b^2)(a^2-b^2)$ and a few mod-tables:
$240=2^4\cdot3\cdot5$
we know that every prime greater than two is odd. and $(2n+1)^2=4(n^2+n)+1$ with $n^2+n$, being even for every $n$. ($n=2m$: $4m^2+2m=2\cdot(2m^2+m)$; $n=2m+1$: $4m^2+4m+1+2m+1=2\cdot(2m^2+3m+1)$)
so we get that $a^2+b^2$ is divisible by $2$ and $a^2-b^2$ by 8. and $2\cdot 8= 2^4$

also we can conclude that for all $a\not\equiv 0 \pmod 3$ or $\pmod 5$ $\Rightarrow$ $a\equiv 1 \pmod 3$ or $\pmod 5$:

modulo 3:
\begin{tabular}{c|c}
a & a^4\\\hline
1 & 1 \\
2 & 16\equiv 1
\end{tabular}

modulo 5:
\begin{tabular}{c|c}
a & a^4\\\hline
1 & 1 \\
2 & 16\equiv 1\\
3 & 81\equiv 1\\
4 & 256\equiv 1
\end{tabular}

since $1-1=0$ we proved for every two numbers not dividible by 2,3, and 5 the given "theorem". and $\gcd(13^4-11^4;17^4-13^4)=240$ that is true for all prime numbers.


Let a and b by distinct real numbers for which a/b + a+10b/b+10a = 2. Find a/b. I saw this somewhere, but I forget how to solve it.
Thanks to anyone that can help.

10 bucks says that there's a typo in there -- look at it again.

Sorry if there was any confusion, but here is a better restatement of the problem - if that helps. a/b + (a+10b)/(b+10a) = 2. Find a/b. This is only one equation - but it's possible to manipulate it somehow. That process, I forget. This appeared on an AMC test, so I don't think there are typos other than that clarification.

If so, that makes a/b equal to 1.  But, are a and b not supposed to be equal to each other?

ya - the problem states a and b are distinct

they are not supposed to be equal to each other...

i multiplied the whole thing out and got [hide]4/5[/hide]

And here is the manipulation you were looking for:

[hide]

We are solving for a/b right?

So lets set k = a/b and see what we get.

Then a = k*b.

Plug that into our equation and you get a quadratic equation in k:

specifically (5k-4)(k-1) = 0. Since k!=1 k = 4/5.

[/hide]

oh that works perfectly fine in the end; its just thats really not a manipulation and I figured he was looking for one.

Luckily multiplying the whole thing out WORKED in this case. Sometimes, what you get in the end is SO convoluted its ridiculous.

That's not a manipulation? Hmm. It could have fooled me.

well i dont think it is in the traditional sense ...

depends how you define it, since otherwise almost all algebraic equations require manipulation

Um -- your two methods are identical.  The entire process is exactly the same.  "Manipulation" doesn't mean anything special as a mathematical word.

I guess thats true. :)

By "manipulation" I was trying to emphasize substitution as a way of simplifying algebra. Although, you only save a couple of multiplications by substitution here.


[size=150][color=red]MATHCOUNTS-AMC Contest[/color][/size]

This is going to be a great contest that is a super fast one. Each round will be in 2 days. Suppose you miss a round on a day, then what happens is that you can finish 2 rounds the next days.

[hide="Directions"]

1. We recommend you to do the contest on the first day that is there. Each question will be asked in an html format. This is for the first round.

2. After teh first roudnd, all rounds are single elimination. There will be a pre selected number of questions, and the order will be decided on the spot.

3. If you get 2 questions wrong in the elimination, you are out.

4. All decisioins are final by the judge.

5. People who can work from 7:00 am-9:00pm ET are the only ones who can participate, becasue the webiste has downtime at all otehr times.

6. The questions will be asked  in an html format, so you click on next to get your question.

7. The questions can be asked from MATHCOUNTS level to AMC-10.

8. The questions will be hosted [url=http://www.4aunlimited.co.nr]here[/url].

9. IF you have any doubts ask here.

10. We hope you join. To signup, post something here.[/hide]

[hide="people who signed up"]236Factorial
h_s_potter_2002
dakyru
math92[/hide]

Good. You can preview the site where the contest will take place [url=http://4aunlimited.co.nr]here[/url].

Why do you refer to yourself in the plural, anirudh? I'll try. Why don't you just post the problems here? :?

okay, count me in.  but what's with the other thread, is there one going on already?  and when is this going to start?

It's probably a contest going on, but run by someone else.

[quote="10000th User"]I see anirudh copied my answer and posted it on his forum for 'Geometric Progressions' problem... :D

Please give me some credit like put my login name in it.  :)[/quote]

if you siggn up, i will make you the author.

Do we need to register at your forum to be able to participate?

Well, yes or no

Everyone, now registration is required. You can access the registration form [url=http://4aunlimited.co.nr/AoPS/com/registration.pdf]here.[/url]. I am requiring registration, because it is then easier to manage everything. Thanks. Also, if you register here, you automatically get registratied for my forum, so you can access the problmes.

[quote="jhollenbeck"]:? might that be a lot off the topic?[/quote]

Must double post?  ;)

When is this going to start?

[quote="236factorial"][quote="chess64"]I'm unsigning up, even though you never put my name on in the first place. :roll:[/quote]

Unsigning up?  :rotfl:

I think it's actually called... removing self.

Why are you "unsigning up"?[/quote]

All the cool people are always unsigning up.

:D :D jk, jk. I don't want to sign up on his site. :ninja:

I don't see the registration form. It says the file is damaged.

btw, do you think the purpose of this contest is to challenge us or to advertise his site  ;)

what? I thought it was fine. Oh welel, how about no more registration forms, and no  more signing up somewhere else. Here are all the people who i have that signed up:

[hide="clcik here"] 236Factorial
h_s_potter_2002
dakyru
math92
danne89
omg-i think(he signed up on my site that's why)
varun-i think(he signed up on my site that's why)
gobraves[/hide]

[quote="anirudh"]what? I thought it was fine. Oh welel, how about no more registration forms, and no  more signing up somewhere else. Here are all the people who i have that signed up:

[hide="clcik here"] 236Factorial
h_s_potter_2002
dakyru
math92
danne89
omg-i think(he signed up on my site that's why)
varun-i think(he signed up on my site that's why)[/hide][/quote]

what about me?

You rwebpage isn't working. Can yo plz fix it.

Plz don't sign me up as yet. I want to see the questions before entering!!1

Well, my question is that when it is going to start. Is he even online anymore (he's hidden I think)?

well, i am online sometimes. for the website, click [url=http://aplethora.forumsplace.com]here.[/url]The other one doesn't work.

OK guys. I'll create the accounts, and i'll pm the passwords. Okay?

of course not. just go to http://aplethora.forumsplace.com and sign up yourself(it saves me al ot of time). the signups don't end until Sept. 5

Is this contest going on?

I don't see a problem but this thread has been stickied for long time. I'm not sure who made it stickied but if the contest is not going anytime soon, I'll remove it.

well, currently i am working on making the questions :)

Contest questions made. Now just pm me the answers on aplethora.forumsplace.com or here. Thanks. Download the attachment for questions. Show steps please. Partial credit may be given if you answer the question wrong, but show steps. You can discuss the topic (or maybe find answers if some of teh contestants decided to) at http://aplethora.forumsplace.com . All you have to do is that you signup for the usergroup competition to get access to the competition discussion place.

Wait, for Number 2, do you find the products of the roots of each individual equation, and then add them up?
 :(

yep

EDIT: THE DEADLINES FOR THE CONTEST IS TOMORROW AT 12:00 ET. I HAVE ONLY 2 PEOPLE SO FAR.

How about this is unstickied and locked. It is of practically no use, since i only got 2 people.  :D

Perhaps you should try under a different forum section or change the difficultly level of the contest.

Anyways, relocked for now.  ;)


the diagonal is 16.

so the side is 18\6/(sqrt2)

that squared is 16^2 / 2 = 128

128 is ans

Just a note --

Consider a square is side length $ s$.

[asy]draw((0,0)--(5,0));
draw((5,0)--(5,5));
draw((5,5)--(0,5));
draw((0,5)--(0,0));
draw((0,0)--(5,5),red);
label("$s$",(0,2.5),W);
label("A",(0,5),NW);
label("B",(5,5),NE);
label("C",(5,0),SE);
label("D",(0,0),SW);[/asy]

The diagonal is shown in red.

Obviously, BD = BD. Since ABCD is a square, that means all the sides are equal, so AB = BC = CD = AD. By SSS, $ \triangle ABD \cong \triangle CBD$. That implies $ \angle ADB \equal{} \angle BDC$ and $ \angle ABD \equal{} \angle BCD$, and so those angles are $ 90 \div 2 \equal{} 45^\circ$.

Therefore, the diagonal is $ \sqrt{s^2 \plus{} s^2} \equal{} \sqrt{2s^2} \equal{} s\sqrt{2}$.

That's why the diagonal of any square is the side length multiplied by $ \sqrt{2}$.

The area of the square is also $ \frac{d^2}{2}$, since a square is a rhombus.


I have decided to accept both of the possible answers for the COP the second round 1.

By that, I have included more winners.

whitehorseking and RC-7th.

When was #5 contested? I feel that #4 should be changed. :D

Well, I'm pretty sure that's what he meant. Apparently, I misunderstood the question.

Such, I got that wrong. But, It isn't Silverfalcon's fault, so I guess he's being pretty generous.

Fixed.

Yeah, I should have included the term, "arithmetic sequence."

 :)

I just got a private message from h_s_potter 2002.

I knew there was one more contestant who only got 4 wrong but couldn't remember.

Hopefully, I'll have all people passed to next round in the list soon.

I think it's like 6 or 7 people (more than last time :))


geometry
geometric transformation
rotation
trigonometry
inequalities
USAMTS
ratio

I did very strange things...  :| how about ya'll?

Haha - I guess you beat me to the punch.  Can a mod merge our thread somehow?

#1: Not much to say on this solution
#2: The one solution I managed to cram into one page without really cramming.  Fairly straightforward problem, not hard.
#3: I tried to make this one as clear as possible.  I never realized that there was a much simpler area solution.
#4: Lots of casework: overall, a pretty messy solution, I suppose...that Lemma I used was probably a little bit cheap...
#5: Part a was easy, Part b: I realized that Lisa first has to get to an even number of lights each way, and try to get a symmetric pattern.  However, I soon got lazy with all of the pairing and parity and patterns, and just listed out all possibilities and gave solutions.

Very impressive work on #5, Zanttrang.  That was the problem giving me far and away the greatest difficulty.  I wound up brute-forcing it in much the same method that k81o7 used (though with diagrams showing each rotation, instead of just declaring the valid moves), which, needless to say, was painful and unsatisfying.  In fact, k81o7's solutions are interesting for me, since he used a trig approach to simplify all of the fractions over 2R on #3, which I spent a good amount of time trying myself, before seeing the method given in the math jam.

In any case, my solutions are massive in both filesize and page length (14 pages... boo), far in excess of any of my other solutions (6 and 7 pages for rounds 1 and 2, respectively), mostly due to #s 4 and 5, in both of which I used a large amount of casework.  I'm expecting a 24 this round, since I proved #5, but in such an inelegant (and typo-riddled) way, that the grader is bound to remove a point somewhere for spite.  Anyway, it's giving me an upload quota reached when I attempt to attach the file (It's 3.6 MB, but I'm getting the Upload Quota Limit error at 2MB, though it says that PDFs are allowed up to 5MB on http://www.artofproblemsolving.com/Forum/attach_rules.php?f=164), but that doesn't matter much, since my solutions aren't worth reading anyway (#s 1-3 are the same as those from the math jam, #4 is workable, but more complex than that given in the math jam, and my #5 is disgusting).  I'm curious to see how many people manage to make elegant solutions for #s 4 and 5, since those were killers this round.

I forgot to do the problems until last minute, so I have only #3.

Grr. I was finally going to get a perfect twenty five, but for some odd reason I converted 7/2 into 27/3 in problem number 2. It was very add, at first my answer was six less than it should have been(because for some reason I only had 12+7/2+11/3 instead of a 13 there) As it turns out, I had made another error which cancled that one :P. Of course, I only caught one of those errors.  :D Oh well. A small mistake like that should only cost me one point, right? A 24 is nothing to laugh at, unless they deduct me more for semi brute forcing problem 4.  :blush: .(I am on a new computer right now, and thus dont have the problems to post. But they all appear good.  :) )

I solved the first four, but didn't start the 5th until the day they were due. Solved part a but couldn't get part b and finally went to bed. Was hoping for a 25  :(

#1: Not too hard. I also found a generlization, ie I found every single prime number that works.

#2: This was insanely easy, took about 15 mins to solve.

#3: I solved a geometry problem, yay! I used normal geo (similar triangles, etc) at first, then used non-messy trig at the end.

#4: I ordered them, then did out the cases with inequalities

#5: Got part a, didn't get part b...

[quote="solafidefarms"]Is there a reason fewer people than usual have posted their solutions? Is it an indicator of the relative number of people who entered this round?[/quote]

For me, I didn't post solutions because number 1-3 is the same as in math jam (or almost the same), the number 4 and 5 I just brute force.  so there's really no point to post my solutions

[quote="solafidefarms"]Is there a reason fewer people than usual have posted their solutions? Is it an indicator of the relative number of people who entered this round?[/quote]
It's also possible that a lot of people are preparing for final exams right now - my school has them next week, and I know a lot of other schools have them soon, so that might account for the lackluster response up to this point.

[quote="solafidefarms"]Is there a reason fewer people than usual have posted their solutions? Is it an indicator of the relative number of people who entered this round?[/quote]
Another reason is that less people are posting in general. Compared to the last to rounds, nobody is posting at all. There also seems to be much less talk about random stuff.

i was lazy and did not type up my solutions...but i have a [b]really[/b] simple solution for #3
Lemma 1:
in a triangle nonright triangle ABC, $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ (easy to show with a+b+c=180 and expanding)

Proof:
i note that $D$ lies on segment $BC$ because the triangle is acute, then by power of a point, $BD*DC=AD*DX \implies \frac{DX}{AD}=\frac{BD*DC}{AD^2}=\cot B*\cot C$, then adding 1, we get $\frac{DX}{AD}+1=\frac{AX}{AD}= \cot B*\cot C+1$ using since there is nothing special about the ratio I chose, the same logic follows for the other ones, adding the relations up,

$\frac{AX}{AD}+\frac{BY}{BE}+\frac{CZ}{CF}=\cot A \cot B+\cot A \cot C+\cot B\cot C+3$
$=\frac{\tan A +\tan B+\tan C}{\tan A\tan B\tan C}+3=\boxed{4}$...i wonder if i could have gotten a commended solution for this?

[quote="dts"][quote="eyefragment"]Salmon, if your work is well documented, then you may well still get full credit for #2, in spite of the error.  In round 1, I simplified $\sqrt{4r^2}$ to $4r,$ but my solution was otherwise correct.  The grader saw that the error was minor, and still awarded a 5.[/quote]

I don't know... last year, I made an addition error in the last line of one of my solutions, and I was docked a point.  I ended up getting a commended 4.[/quote]
I didn't know that was possible

[quote="eyefragment"][quote="joml88"]It should also be mentioned that for the first two rounds they allowed discussion of the USAMTS [i]before[/i] the Math Jam but this time they did not allow any posting about Round 3 until after the Math Jam.[/quote]

I'm fairly sure that the same policy (Do NOT discuss rounds 1/2 before the math jam) was in place for rounds 1 and 2, though I have nothing but my word to back that up.[/quote]

Round 2:

Date [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62784&postorder=asc]this[/url] topic started: Nov. 23.
Date of the [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=126]Math Jam[/url].

I think you're right about Round 1 though.

[quote="eyefragment"][quote="joml88"]It should also be mentioned that for the first two rounds they allowed discussion of the USAMTS [i]before[/i] the Math Jam but this time they did not allow any posting about Round 3 until after the Math Jam.[/quote]

I'm fairly sure that the same policy (Do NOT discuss rounds 1/2 before the math jam) was in place for rounds 1 and 2, though I have nothing but my word to back that up.[/quote]
You're half right - the "Math Jam barrier" was in place for the first round, but for the second round a different cut-off was put before that Math Jam, possibly because they weren't able to host the Math Jam as early as they wanted.  If I remember, generally last year the Math Jam did serve as the barrier after which discussion could begin.


More general question:

$ (m, n, x)$ denotes: In $ \triangle ABC, \angle BAC \equal{} m^\circ, \angle ADC \equal{} n^\circ$ and $ \angle ABC \equal{} x^\circ$ where  $ D\in [BC]$ such that $ DC \equal{} AB$

[b]a)[/b]
Find $ x$s in the triples $ (m, n, x)$.

1. $ (m, 60 \plus{} 2m/3, x)$
2. $ (m, 45 \plus{} m/4, x)$

3. $ (30, 150, x)$
4. $ (54, 126, x)$
5. $ (60, 150, x)$
6. $ (102, 30, x)$
7. $ (105, 45, x)$
8. $ (110, 100, x)$
9. $ (120, 30, x)$
10. $ (127.5, 112.5, x)$
11. $ (138, 18, x)$

[b]b)[/b]

Prove that if $ n+x > 180$ and $ (m, n, x)$ is a solution, then ordered triple $ (2x\plus{}m\minus{}180, n, 180\minus{}x)$ is also a solution.

[b]c)[/b]

Prove that if  $ 90 \minus{} m/2 > x$ and $ (m, n, x)$ is a solution, then ordered triple $ (180\minus{}x\minus{}n, 180\minus{}m\minus{}x, x)$ is also a solution.