Diagonal $ DB$ of rectangle $ ABCD$ is divided into $ 3$ segments of length $ 1$ by parallel lines $ L$ and $ L'$ that pass through $ A$ and $ C$ and are perpendicular to $ DB$.  The area of $ ABCD$, rounded to the nearest tenth, is

[asy]size(200);
defaultpen(linewidth(0.7)+fontsize(10));
real x=sqrt(6), y=sqrt(3), a=0.4;
pair D=origin, A=(0,y), B=(x,y), C=(x,0), E=foot(C,B,D), F=foot(A,B,D);
real r=degrees(B);
pair M1=F+3*dir(r)*dir(90), M2=F+3*dir(r)*dir(-90), N1=E+3*dir(r)*dir(90), N2=E+3*dir(r)*dir(-90);
markscalefactor=0.02;
draw(B--C--D--A--B--D^^M1--M2^^N1--N2^^rightanglemark(A,F,B,6)^^rightanglemark(N1,E,B,6));
pair W=A+a*dir(135), X=B+a*dir(45), Y=C+a*dir(-45), Z=D+a*dir(-135);

label("A", A, NE);
label("B", B, NE);
label("C", C, dir(0));
label("D", D, dir(180));
label("$L$", (x/2,0), SW);
label("$L^\prime$", C, SW);

label("1", D--F, NW);
label("1", F--E, SE);
label("1", E--B, SE);
clip(W--X--Y--Z--cycle);
[/asy]

$ \textbf{(A)}\ 4.1 \qquad \textbf{(B)}\ 4.2 \qquad \textbf{(C)}\ 4.3 \qquad \textbf{(D)}\ 4.4 \qquad \textbf{(E)}\ 4.5$

Call the first bottom point of intersection with the diagonal M and the top N. $ \triangle ADB \sim \triangle AMB$. Let $ \overline {AB} \equal{} x$. We have $ \frac {2}{x} \equal{} \frac {x}{3}$. $ x \equal{} \sqrt 6 \overline {AD} \equal{} \sqrt 3$ Area is $ \sqrt {(3)(6)} \equal{} 3\sqrt 2 \equal{} 4.24... \rightarrow \boxed B$

Let's first label the points.  On the diagonal from D going up, the points are M & N.  Let's believe that NC & MA is y, and the ends of the perpendiculars (the stub after y) is x.  Finally, let's say that AB is a and BC is b.

Because you drop an altitude from B to a perpendicular, the corresponding ratio is x/1=1/y, or xy=1

Furthermore, it is easily reasoned that y^2+4=a^2 & y^2+1=b^2 (use Pythagorus accordingly).

The area of this rectangle is ab, and therefore what we try to solve for.  The areas of triangles AMB, CND, and the two remaining small triangles in the rectangle have areas: y,y,y/2,&y/2, which add up to 3y.
Therefore,

3y=ab
9y^2=a^2*b^2
9y^2=(y^2+4)(y^2+1)        Substitute from paragraph 3.
9y^2=y^4+5y^2+4
0=y^4-4y^2+4
0=(y^2-2)^2                         Factor
y=sqrt.2

Therefore, dc=3*sqrt.2 (substitute)
or 2* 1.414..., or 4.242...

Answer is 4.2, or B.

Stick to shorter methods, this is just the longer, yet less knowledgeable way.


function algebra domain calculus derivative integration geometry

Let $ (\Omega)$ is a bounded domain in $ \mathbb{R}^3$. A function $ u \in C^2(\Omega)$ which satisfies the Laplace equation is called a harmonic function i.e:
$ \Delta u \equal{} u_{xx} \plus{} u_{yy} \plus{} u_{zz} \equal{} 0$
Prove that if $ u \in C^2(\Omega)$ be harmonic in $ \Omega$ then $ u \in C^\infty(\Omega)$ and every partial derivative of $ u$ is a hamonic  function in $ \Omega$.

[b]Definition (Mean value property).[/b] For $ u \in C(\Omega)$ we define
[list]
$ \star$ $ u$ satisfies the first mean value property if \[ u\left( x \right) \equal{} \frac{1}
{{{\omega _n}{r^{n \minus{} 1}}}}\int\limits_{\partial {B_r}\left( x \right)} {u\left( y \right)d{S_y}} \]for any $ B_r(x) \subset \Omega$

$ \star$ $ u$ satisfies the second mean value property if \[ u\left( x \right) \equal{} \frac{n}
{{{\omega _n}{r^n}}}\int\limits_{{B_r}\left( x \right)} {u\left( y \right)dy} \]for any $ B_r(x) \subset \Omega$
[/list]
where $ \omega_n$ denotes the surface area of the unit sphere in $ \mathbb R^n$.

[b]Lemma 1.[/b] Let $ u \in C^2(\Omega)$ be harmonic in $ \Omega$. Then $ u$ satisfies the mean value property in $ \Omega$.

The proof relies on the so-called Divergence Theorem.

[b]Lemma 2.[/b] If $ u \in C(\Omega)$ has mean value property in $ \Omega$ then $ u$ is smooth and harmonic in $ \Omega$.

The proof is a little bit complicated since it deals with convolution, mollifiers,..

You may find these fundamental stuffs through some advanced PDEs book.


A point inside a square is positioned so that the distances to the four vertices are 27, 21, 6, and x. If x is a whole number, what is the value of x?

[hide="Answer, no solution"]
18
(Too lazy to make a diagram. Just make a square with the distances given and use Pythagorean. There are multiple possibilities, but only one will give an integer answer.)
[/hide]

British Flag Theorem. [url]http://www.artofproblemsolving.com/Wiki/index.php/British_Flag_Theorem[/url]


I posted this in the inequality forum about 4 months ago and did not get a response, so I think I can be forgiven posting it here and now...

Prove:
$ \frac {a \plus{} abc}{1 \plus{} ab \plus{} abcd} \plus{} \frac {1 \plus{} ab}{a \plus{} abc \plus{} abcd} \plus{} \frac {a \plus{} abcd}{1 \plus{} ab \plus{} abc} \plus{} \frac {1 \plus{} abc}{a \plus{} ab \plus{} abcd} \plus{} \frac {ab \plus{} abcd}{1 \plus{} a \plus{} abc}\ge$ $ 10/3$

Now my original inequality was five variable and used $ abcde \equal{} 1$, but I put it down to this for some reason. Maybe you guys can do it either way...

EDIT: If it is five variable with $ abcde \equal{}1$, the inequality has some interesting properties when you plug $ abcde \equal{} 1$. Not sure how this helps either.

EDIT2: Okay I figured it out, but you can try it anyway :P

Is there a typo? This doesn't appear very symmetric...maybe I'm just not looking hard enough.

err dude you should read his comments

let e=1/abcd and do blah stuff then it is pretty symmetric-ish

[quote="Altheman"]Just add one to each term and cauchy it.[/quote]

But is it given that the variables are positive?

um usually when given an inequality, the variables are postive

also his method was my method

[quote="Anaxerzia"]um usually when given an inequality, the variables are postive[/quote]
That's not really true.  Some inequalities are true regardless of sign.  You need to specify what values the variables are able to take (especially if $ 0$ is allowed; this ruins some inequalities).


Let $S$ be an infinite set of real numbers such that $|x_1+x_2+\cdots + x_n | \leq 1 $ for all finite subsets $\{x_1,x_2,\ldots,x_n\} \subset S$. Show that $S$ is countable.

It's clear that the only point which could, eventually, be an accumulation point of $S$ is $0$, so $S$ can be written as $\displaystyle \cap_{n\ge 1} (S\setminus [\frac {-1}n,\frac 1n])$, and each $S\setminus [\frac {-1}n,\frac 1n]$ is a discrete set, so I guess we're done.

A slightly more interesting problem is to extend the result to complex numbers.


[quote="spix"]Prove that $a_{n}=\{ n\sqrt{2}\}+\{ n\sqrt{3}\}$ diverges.[/quote]

What does that mean?

$n\sqrt{2}\to+\infty$ and $n\sqrt{3}\to+\infty$
So definitely thier sum.

$\{x\}$ stands for the fractional part of $x$

Diverges as a sequence, or as a series?

Hint: $\{(n+1)\sqrt{2}\}-\{n\sqrt{2}\}$ is equal to either $\sqrt{2}-1$ or $\sqrt{2}-2$.


(1) The fractions $ \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}$, and $ \frac{6}{7}$ are all represented by the same six digits in their repeating decimal equivalents. Only the order of the digits is changed.  In which of the decimals do the third and fourth digit form a two- digit number which is one-half of the number formed by the fifth and sixth digit?  Write the fraction corresponding to this decimal.

[hide]$ \frac{4}{7}$[/hide]

(2) When the three integers $ 618, 343,$ and $ 277$ are divided by a positive integer, $ d$, where $ d > 1$, the remainders are the same.  What is the smallest possible value of $ d$?

[hide]$ 11$[/hide]

(3) How many digits are there in the expanded form of 2 to the 40th power?

[hide]$ 13$[/hide]

[hide="1."]
The digits are 142857 in some order.
The number in which the 3rd and 4th digits are 1/2 the 5th and 6th digits will have these digits 1428, which means it starts with 5 and it is $ \boxed{\frac{4}{7}}$[/hide]

For prob 3
all you need to do is take your calculator
type in 2^40= ??????????
 ----- that is the easiest method i can think of

but while i was trying to do this problem i have noticed that
2^0 = 1 digit
2^10 = 4 digit
2^20 = 7 digit
2^30 = 10 digit
2^40 = 13 digit
Notice that as you increase the power by 10, you add 3 new digits

ANS. 13

i always rember $ 2^{10k}\equal{}1024....$ 1024 are the first 4 digits. i think i got it from learning about computers or something :wink:

#2 all i can think of is list the remainders when divided odd numbers(evens wont work b/c 618 will leave a even remainder and everything else will leave an odd one)

***These three problems are from a National Sprint Round.  Therefore, could***
***someone please find a way to do them WITHOUT a calculator? Thanks.***

[quote="Hustla25"]but while i was trying to do this problem i have noticed that
2^0 = 1 digit
2^10 = 4 digit
2^20 = 7 digit
2^30 = 10 digit
2^40 = 13 digit
Notice that as you increase the power by 10, you add 3 new digits

ANS. 13[/quote]

Hustla25 - Does this pattern always hold true?  Thanks.

I'm pretty sure it doesn't for very large powers

about 3, you could just go down about 7 topics and find this:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180641

For #1, notice the pattern and see which order you need it to be in.

For #2, find (618-343,343-277,618-277), which =11

For #3, you need to find a power of 2 that is very close to a power of 10, and you see 2^10=1024, which is near 10^3


logarithms
geometry
trapezoid
integration
calculus
derivative
real analysis

let $ p_{n}\sim q_{n}$ mean that $ \frac{p_{n}}{q_{n}}\to 1$ as $ n$ tends to infinity

prove
\[ 1^{1}\cdot 2^{2}\cdot\ldots\cdot n^{n}\ \sim\ A\cdot n^{\frac{n^{2}\plus{}n}{2}\plus{}\frac{1}{12}}\cdot e^{\minus{}\frac{n^{2}}{4}}\]
for some constant $ A$

and
\[ 1^{n\minus{}1}\cdot 2^{n\minus{}2}\cdot\ldots\cdot (n\minus{}1)^{1}\ \sim\ \frac{1}{A}\cdot e^{\frac{1}{12}}\cdot n^{\frac{n^{2}}{2}\minus{}\frac{1}{12}}\cdot (2\pi)^{\frac{n}{2}}\cdot e^{\minus{}\frac{3}{4}n^{2}}\]
where $ A$ is from the previous one

I'm convinced Stirling should be used here

It's not directly Stirling, but it's the same technique. Take logarithms: we want to estimate $ \sum_{k = 1}^{n}k\ln k$ in the first case, and $ \sum_{k = 1}^{n}(n-k)\ln k$ in the second. I'll focus on the first sum:
Consider the trapezoid rule estimate $ \int_{n}^{n+1}f(x)\,dx =\frac{1}{2}(f(n)+f(n+1))-\frac{1}{12}f''(c)$ for some $ c\in (n,n+1)$ (assuming $ f$ is $ C^{2}$). Applying this to $ \int_{1}^{n}x\ln x\,dx$, we get
$ \int_{1}^{n}x\ln x\,dx =\frac{1}{2}\cdot 1\cdot\ln 1+\sum_{2}^{n-1}k\ln k+\frac{1}{2}n\ln n-\frac{1}{12}\sum_{k = 1}^{n-1}\frac{1}{c_{k}}$ where $ k < c_{k}< c_{k+1}$, since the second derivative of $ x\ln x$ is $ \frac{1}{x}$. Solving for the sum we care about, this becomes
$ \sum_{k = 1}^{n}k\ln k =\int_{1}^{n}x\ln x\,dx+\frac{1}{2}n\ln n+\frac{1}{12}\sum_{k = 1}^{n-1}\frac{1}{c_{k}}$
That integral is $ \int_{1}^{n}x\ln x\,dx =[\frac{x^{2}}{2}\ln x-\frac{x^{2}}{4}]_{1}^{n}=\frac{n^{2}}{2}\ln n-\frac{n^{2}}{4}+\frac{1}{4}$
Now, we deal with the $ c_{k}$ term. Here, note that $ |\ln(\frac{k+1}{k})-\frac{1}{c_{k}}| <\frac{1}{k}-\frac{1}{k+1}$; the sum of these errors converges, and $ \sum_{k = 1}^{n-1}\frac{1}{c_{k}}=\ln n+a+o(1)$ as $ n\to\infty$ for some $ |a| < 1$.

Putting the pieces together,
$ \sum_{k = 1}^{n}k\ln k =\frac{n^{2}}{2}\ln n-\frac{n^{2}}{4}+\frac{1}{4}+\frac{n}{2}\ln n+\frac{1}{12}\ln n+\frac{1}{12}a+o(1)$
Exponentiate to get your result. $ A =\exp(\frac{1}{4}+\frac{1}{12}a)$.

[quote="jmerry"]

Putting the pieces together,
$ \sum_{k \equal{} 1}^{n}k\ln k \equal{}\frac{n^{2}}{2}\ln n\minus{}\frac{n^{2}}{4}\plus{}\frac{1}{4}\plus{}\frac{n}{2}\ln n\minus{}\frac{1}{12}\ln n\minus{}\frac{1}{12}a\plus{}o(1)$
[/quote]
Yes jmerry is right. I'm not sure, but I think there must be $ \plus{}\frac{1}{12}\ln n$ instead $ \minus{}\frac{1}{12}\ln n$
this can be cheked by Stolt's Theorem also.

I lost track of the sign changes- fixed now.

[quote="jmerry"]
Now, we deal with the $ c_{k}$ term. Here, note that $ |\ln(\frac {k + 1}{k}) - \frac {1}{c_{k}}| < \frac {1}{k} - \frac {1}{k + 1}$[/quote]

could you explain this estimation?
thanks

Both quantities are between $ \frac1n$ and $ \frac1{n\plus{}1}$. It's a crude overestimate, but it's good enough.


Evaluate the expression:

$\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}k^{n}$ for $n=10$

[hide]
Prove that:

$\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}k^{n}= (-1)^{n}\cdot n!$

[/hide]

[hide="Further hint"]If the desired sum is $a_{n}$, prove the recursion $a_{n}=(-n)a_{n-1}$[/hide]

right, i saw the pattern by trying out smaller values of $n$, but i wasnt sure how to prove it

i tried using induction but its .. difficult i think

[hide="Another_Hint"]Let $f(k) = k^{n}$, which is a polynomial of degree $n$.  This problem is asking about the $n$th difference of this polynomial.[/hide]


Given continuous functions $ f,g: \mathbb{R} \to \mathbb{R}$ such that $ f(g(x))\equal{}g(f(x))$, $ \forall x \in \mathbb{R}$.
Prove that exists $ x \in \mathbb{R}$ such that $ f(x)\equal{}g(x)$.

[quote="MellowMelon"]$ f(x) \equal{} x \plus{} 1, g(x) \equal{} x \minus{} 1$ seems to be a counterexample.[/quote]

yes, seems so..i'll check my source, thanks for your observation..


graph theory
combinatorics proposed
combinatorics

Hi every one,
I am going to teach discrete mathematics(combinatorics and graph theory) to the students of mathematics and
computer science, and I want your opinion about the text books in that concept.Please let me know which text books are used in your country, with a brief discription, if possible.
(Have you seen the books by Anderson, Grimaldi?)
Thanks in advance

combinatorics- principles and techniques in combinatorics        (chen chun chong and koh khee-meng)
                    - titu andrescu has a good book i think its called a path to combinatorics for undergraduates.

                - Theres another one really good that is called A walk through combinatorics (i think)
graph theory- Douglas West's Graph Theory and also Reinhard Diestel's Graph theory.

By the way, Diestel's text is available online too:

http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/

 :)

[quote="blahblahblah"]What kind of students are you aiming at?

Grimaldi is a decent introductory textbook for not especially bright students, but there are better options for students at the olympiad level.[/quote]

You know this is not going to be an advanced class like IMC preparation classes, but rather a normal university class.
Has anyone experienced "First course in combinatorial mathematics "by Anderson or
"Aspects of combinatorics "by Bryant ?

We, at MNNIT India follow Tremblay-Manohar. Personally I find Kenneth H Rosen's text for discrete mathematics as a very enlightening source...


The altitude from one of the vertices of an acute-angled triangle $ABC$ meets the opposite side at $D$. From $D$ perpendiculars $DE$ and $DF$ are drawn to the other two sides. Prove that the length of $EF$ is the same whichever vertex is chosen.

WLOG assume $CD\perp AB, DE\perp AC, DF\perp BC$. Standard markings $a,b,c,\alpha,\beta,\gamma$. Denote $h=CD$

Then

$\angle DCA={\pi\over 2}-\alpha\implies DE=h\cos\alpha$
$\angle DCB={\pi\over 2}-\beta\implies DF=h\cos\beta$
$\angle DEC=\angle DFC={\pi\over 2}\implies \angle EDF=\pi-\gamma$

By Cosine Law in $\triangle DEF$

$EF^{2}=h^{2}(\cos^{2}\alpha+\cos^{2}\beta+2\cos\alpha\cos\beta\cos\gamma)$

Then we have

\begin{eqnarray*}\cos^{2}\alpha+\cos^{2}\beta+2\cos\alpha\cos\beta\cos\gamma &=& \cos^{2}\alpha+\cos^{2}\beta-2\cos\alpha\cos\beta\cos(\alpha+\beta)\\ &=& \cos^{2}\alpha+\cos^{2}\beta-2\cos^{2}\alpha\cos^{2}\beta-2\sin\alpha\sin\beta\cos\alpha\cos\beta\\ &=& \cos^{2}\alpha(1-\cos^{2}\beta)+\cos^{2}\beta(1-\cos^{2}\alpha)+2\sin\alpha\sin\beta\cos\alpha\cos\beta\\ &=& \cos^{2}\alpha\sin^{2}\beta+\cos^{2}\beta\sin^{2}\alpha+2\sin\alpha\sin\beta\cos\alpha\cos\beta\\ &=& (\sin\alpha\cos\beta+\cos\alpha\sin\beta)^{2}\\ &=& \sin^{2}(\alpha+\beta)\\ &=& \sin^{2}(\pi-\gamma)\\ &=& \sin^{2}\gamma\end{eqnarray*}

Hence $EF=h\sin\gamma=\frac{2[ABC]}{c}\cdot\frac{c}{2R}=\frac{[ABC]}{R}$, where $[ABC]$ is the area and $R$ the circumradius of $\triangle ABC$. Obviously, $EF$ doesn't depend on the choice of the vertex.

$\triangle ABC\ : D\in BC\ ,\ AD\perp BC$ $\mathrm{\ ;\ }$ $E\in AC\ ,\ DE\perp AC$ $\mathrm{\ ;\ }F\in AB\ ,\ DF\perp AB\ .$
The quadrilateral $AEDF$ is cyclically with the diameter $[AD]$  $\Longrightarrow$ $EF=AD\cdot \sin A=h_{a}\cdot\frac{a}{2R}=\frac{2S}{2R}$ $\Longrightarrow$ $\boxed{\ EF=\frac{S}{R}\ }\ .$


Let $ X\ne \emptyset$ be compact and Hausdorff space and $ f: X\to X$ continuous. Prove that there is a closed non-empty set $ A\subset X$ such that $ fA\equal{}A$. I guess we can take $ A\equal{}f\circ f\circ\cdots\circ f X$ but I don't see how to make that argument precise.

You can make that argument precise by transfinite induction; equivalently, use [url=http://en.wikipedia.org/wiki/Zorn%27s_lemma]Zorn's lemma[/url].

This seems to work:

Set $ A_0 : \equal{} X, A_n : \equal{} f(A_{n\minus{}1}), n \in \mathbb{N}$ and $ A : \equal{} \bigcap_{n \in \mathbb{N}} A_n$. Then $ A$ is nonempty and closed.
Clearly $ f(A) \subset A$. Let $ y \in A$ then $ y \in A_{n\plus{}1} \equal{} f(A_n)$ for all $ n \in \mathbb{N}$.
Thus $ f^{\minus{}1}(y) \cap A_n \not\equal{} \emptyset$ and $ f^{\minus{}1}(y) \cap A_n$ form a closed, nested sequence in $ X$. Hence $ \bigcap_{n \in \mathbb{N}} f^{\minus{}1}(y) \cap A_n \not\equal{} \emptyset$ and $ x \in \bigcap_{n \in \mathbb{N}} f^{\minus{}1}(y) \cap A_n \Rightarrow f(x) \equal{} y$ it follows $ A \subset f(A)$.


trigonometry
inequalities unsolved
inequalities

[quote="ashrafmod"]prove
$(\frac{\sin^{2}x \cos^{2}x+1}{\sin^{2}x \cos^{2}x}) \geq \frac{25}{2}$[/quote]
There might be some typing mistakes
try $x=\frac{\pi}{4}$


geometry
algebra
polynomial
area of a triangle
Heron\u0027s formula

what is (.5A+.5B+.5C)(-.5A+.5B+.5C)(.5A-.5B+.5C)(.5A+.5B-.5C)??
I know how to do it with the distributive, but there has to be a shortcut. If there is no other way, what is the answer??

Where did you get this? I'd start by factoring out all the .5s, and making it A+B... all over 16

i got it off a website (don't remember name). Ill try factoring

this is a half-guess, but is is (-a^4-b^4-c^4)/16?

everything looks like it cancels out to me
1/16?
guessing

[hide]If a, b, and c can represent the lengths of the sides of a triangle, then by Heron's Formula the expression above equals the square of the triangle's Area.

Other than that, I don't think that it simplifies to anything nice algebraically.[/hide]

Well I guess it could be the area of the triangle, but I need a polynomial answer.


Find $\lim_{x\to +\infty}\frac{\pi(x).\ln(x)}{x} $
Denote $\pi(x)$ be the number of primes which is not exceed $x$

Please do not post directly into the soved problems section... ;)

By the way the desired limit is $1$ and the result is called 'the prime number theorem'. Considering its proofs, I don't think it is quite fair to propose it as an olympiad problem  ;)

Pierre.

Yes and i'm sorry for my mistake!I read it from a Lhvbao's post but there no any idea!
It's really nice and hard.Anybody have solution??

Maybe you should ask Hadamard or Vallée Poussin (a Belgian mathematician, if I am not mistaken) for a proof of this funny little theorem :D

http://mathworld.wolfram.com/PrimeCountingFunction.html

see also

http://en.wikipedia.org/wiki/Prime_counting_function


Which compound is expected to be the most soluble in water at 25 degrees Celsius?

a) $ N_{2}(g)$
b) $ O_{2}(g)$
c) $ (C_{2}H_{5})_{2}NH(l)$
d) $ C_{2}H_{5}OC_{2}H_{5}(l)$.

Both nitrogen and oxygen are sparingly soluble in water, because they have non-polar molecules. Both diethyl ether and diethyl amine can form hydrogen bonds with water molecules. However, diethyl amine can form more of those bonds (via the nitrogen and via the amine hydrogen) and so it is expected to be more soluble. This is indeed the case: diethyl amine is completely soluble in water, whereas diethylether has a solubility of 6.9g/100mL.


Support search probability geometry inequalities

Can discrimination be justified? Explain.

Just curious, all views are welcome. Just try to avoid using real examples, in case it offends anyone.

I believe all people are individuals...so no, I can't think of a time where discrimination is ever justified.

[quote="probability1.01"]Discrimination really is a broad term.  Please be more specific.  For example, I could argue that selective admissions in college is discrimination... :-/[/quote]

Yes, you could ... so do you think that selective admissions is justified?

I'm referring to the generalization of a characteristic to a certain group - the separation. It doesn't necessarily have to be in a negative sense.

You really are going to have to be more specific on the definition of the word "discriminate" if you want to ask questions like this.  The first definition of the word I came across on a quick search was "dis·crim·i·nate :  To make a clear distinction; distinguish: [i]discriminate among the options available.[/i]"  This, I imagine, is most likely not the sense of the word about which you are curious.

Providing context is also very helpful.

[quote="JBL"]"dis·crim·i·nate :  To make a clear distinction; distinguish[/quote]

This defintion of discriminate is the one I intended.

Then asking a question like that would be akin to asking someone, "Do you believe in inciting?"

What do you expect people to respond with?

[quote="probability1.01"]I would suspect that for the vast majority of people, discrimination is permissible to a certain extent.  Therefore it would be nearly impossible to give a straight yes/no answer.  Anyways in the general case I do support discrimination, in the sense that people have the right to discriminate.  I did say [b]general[/b] case ;).[/quote]

^Something like that is fine...  :)

[quote="banblink182"]Would you consider a statement such as, the mean height of a NBA player is greater than the mean height of a non NBA player, given that it has been statistically shown?[/quote]

Technically speaking, that's disinguishing the two groups of people. But it is clearly independent of the prejudice connotation that is generally associated with "discrimination".

So you feel that [quote="MithsApprentice"]it is clearly independent of the prejudice connotation that is generally associated with "discrimination"[/quote], even though [quote="MithsApprentice"]Technically speaking[/quote] it is separation.

But do you feel that such a statement makes one discriminatory?

Discrimination: Treating a person or group differently (usually worse) than others.

[quote="shobber"]Discrimination: Treating a person or group differently than others.[/quote]

 :yup:  That works too.

Discrimination can be justified from the point-of-view of the "discriminators" -- well that's precisely why they are doing it. But I think discrimination can be justified in general as long as it doesn't go to an extreme (yet to be defined). You could say that nationalism is discrimination because people might think their country is better than others. Interestingly, this brings up the concept of cosmopolitanism (which we discussed recently in AP World History) and if it is possible to implement a cosmopolitan education upon the world in an effort to get rid of tensions between groups. Basically, can we become a world community and thus effectively eliminate discrimination (on a large scale, at least)?

But anyway, all I have to say is that discrimination, whether or not it is justified or not, will exist and the only thing we can do is try to control it. Justification is up to whatever point-of-view you are coming from.

I think a distinction needs to be made between discrimination and prejudice. Discrimination, where I have seen it used, implies(necessitates) action, prejudice does not. On that note, I think prejudice is OK(though not particularly justified via logic), while discrimination is not.

To say something is justified depends on what standards you are weighing it on. E.G. murder of jews is justified if you use hitler's moral calculus. So, figuring out whether something is "justified" is pretty blurry unless the top level post specifies exactly by what standards constitute a justified action.

But in my own opinion, I don't think discrimination is good. Because people are generally discriminated based on some aspect of their life which they cannot control: nationality, race, religion, sex, etc.

[quote="banblink182"]Thanks for the responses.  :)

Do you feel that we should not be judged on anything that we did not have a choice over, such as race, sex, etc.?

If so, what about intelligence and the idea of "Honors" classes.[/quote]I don't feel that we should be judged over anything we don't have a choice over, but we should be judged over things that we have control over.

Being judged only on things one can control is not only impossible, it'd also be bad for society.

Unfortunately, genes matter.  Now, I'm not saying we go back to the days of natural selection for society where if you broke your leg you starved to death.

However, if you and another guy are competing for a job, and you are more qualified than the other guy, don't you want the job?  What if he's an idiot who's worked harder than you his whole life, but because of his genes is still and idot?  Do you want him to get the job?  Of course not!  You want the job--you know you can do better.  The fact that he's an idiot is out of his control, but in the end, you have to admit that you don't really care--you want the job anyway.

I'd say, we want a world where the most qualified people get things.

[quote="bubala"]I'd say, we want a world where the most qualified people get things.[/quote]

Do you think you'd say that if you weren't well-qualified?  Also, how can you bring such a thought into line with an "anti-discrimination" view?  (I think this is what banblink was getting at, after all.)

Rule of thumb for designing an ideal society: you have to remember that you have an equal chance of being born as each member; accordingly, you must provide for each member's happiness, not just your own.

JBL, I'm not sure that's true.  First of all, what does it mean to have an equal probability of being each member?  I assume you are imagining a scenario in which souls are randomly inserted into bodies.  Thus, we are in a sense morally bound to help those less fortunate than us because they are not there by choice; we could have been born into their situation, too.  But who knows that this process is random?  Maybe some souls are destined to get better bodies (whatever that means) than others.  I'm not saying that then people are not obligated to help each other.  They may still be bound by feelings such as compassion or love.  If I recall, Buddhists don't think this process is random at all (but I'm not all that familiar with Buddhism, so I might be wrong).
If you don't believe in souls...then I don't really know what that statement means.  Since you are just you (your physical body), the probability that you could have been anyone else is 0.

I agree with bubala, who is essentially advocating a meritocracy.  In this sense, I'm all for discrimination.  It is impossible to have a world without discrimination.  Everyday, we have to discriminate over the most mundane things such as what we are going to wear, what we are going to eat, what we are going to do.  If someone asks me who I think is going to win the NBA playoffs, I discriminate between the teams and say the SA Spurs because they are obviously the best team  :) .  A world without discrimination is a world of indecision and nothing good will come of it.  The problem is that there are different ways of discriminating.  Choosing to eat eggs scrambled instead of fried because they taste better that way is a legitimate form of discrimination.  Deciding that the SA Spurs will win the playoffs because their overall play is stronger and more consistent than that of any other team is also a legitimate form.
All the "ok" or PC forms of discrimination discriminate based on things that actually matter.  Notice that I didn't say that scrambled eggs are better than fried eggs because "scrambled" has more letters than "fried."  This fact is utterly irrelevant and thus would not be a proper way of deciding which to eat.  So in my opinion, discrimination is bad when the logic behind it is flawed.  So all discrimination is by definition irrational!  This may be the reason that a lot of mathy people I know all tend to be very PC...but that's another discussion.  The problem is complicated because sometimes it may become difficult deciding whether the discrimination is legitimate (logical) or not.  For instance, racial profiling.  Can statistics be used to discriminate?  Certainly in sports and in many areas, this is done all the time.  But in others, it seems to be very controversial.  In this PC age, it is becoming harder and harder to talk politely and intelligently about this issue.

Perhaps I wasn't clear.  Let's say I wanted to design an ideal society or to judge which of two historical societies I would have prefered to live in.  To design a meritocratic utopia or decide if Peter the Great's Russia was a nice place to live, you have to not just imagine yourself as a highly qualified individual or as Peter the Great.  Everyone in those societies is an individual and all of their experiences in society matter.  In Russia, Peter the Great may have lived really well, but his experience is outweighed by the millions of serfs who shared that society with him.  In order to decide if a meritocratic society would be a "good" society, you must examine it not just from your position as someone with a lot of merit.

How you judge a society good under this system will vary.  A communist might only accept a society as good if the poorest were given a standard of living on a par with everyone else.  A capitalist would be much more likely to accept a society with inequality and a moderate number of poverty-stricken in exchange for ceating opportunities for immense wealth.  Ayn Rand would accept something much more extreme than that, I'd imagine.  So your conclusions might vary, but the method of investigation should be the same.
(There are of course other things you might use to help you decide -- Americans seem to have a preference for democratic societies, for example, regardless of other factors.)

Your last two paragraphs kind of wander:  what it sounds like you say is, "I like the idea of a meritocracy, and I like it because some times discrimination is okay but it's hard to talk about it."  I don't see what PC has to do with it -- after all, here's a thread devoted to the discussion of discrimination right here, and no moderator or administrator is telling anyone to tone it down.

*Will write more later -- must go to dentist appointment.

No moderator is telling anyone to tone it down because there has been no need to.  The effects of PC are felt implicitly.

[quote="gr8sk8r"]i wonder how the american court system defines discrimination[/quote]
That's the point.

As moderator:  So what did you want to say that you felt was implicitly being suppressed?  "I think discrimination is good; I should be able to not sell my house to someone just because of the color of her skin."?  I mean, come on -- if you say something unacceptable, I'll let you know.  There's no point in complaining about being suppressed if you aren't even trying the boundaries.

Not as moderator: As for the final paragraph of your previous post:  "logical discrimination" is in the eye of the beholder.  Give me a standard that I can apply to a case of discrimination to determine whether or not it is "logical."

[quote="banblink182"]Thanks for the responses.  :)

Do you feel that we should not be judged on anything that we did not have a choice over, such as race, sex, etc.?

If so, what about intelligence and the idea of "Honors" classes.[/quote]

There's quite a difference between honors classes and discrimination against people based on some random variable such as race, sex, etc.

First, honors classes aren't purely based on intelligence. You can sign up for them yourself, so it's a choice.
Second, intelligence differs from race/sex in that you can make yourself more intelligent/knowledgable etc. Unless you're michael jackson, you can't really change your race.
Third, there's a difference between been judged on and being discriminated against. Discrimination has to do with categorical labels(in the philosophical, legal sense) and action. Judging someone based on their individual characteristics is fine. In fact, I'd argue that judging a group(prejudice) is fine(it's not good, but no one can stop you, so oh well) but acting upon that prejudice is bad.

btw, Jbl, are you familiar with John Rawls? Your ideas sound a lot like his.

[quote="agolsme"]There's quite a difference between honors classes and discrimination against people based on some random variable such as race, sex, etc.

First, honors classes aren't purely based on intelligence. You can sign up for them yourself, so it's a choice.
Second, intelligence differs from race/sex in that you can make yourself more intelligent/knowledgable etc. Unless you're michael jackson, you can't really change your race.
Third, there's a difference between been judged on and being discriminated against. Discrimination has to do with categorical labels(in the philosophical, legal sense) and action. Judging someone based on their individual characteristics is fine. In fact, I'd argue that judging a group(prejudice) is fine(it's not good, but no one can stop you, so oh well) but acting upon that prejudice is bad.

btw, Jbl, are you familiar with John Rawls? Your ideas sound a lot like his.[/quote]

Although people can always make themselves more knowledgable, I do not believe that one can make themselves more intelligent. Intelligence, or the quality which IQ so desperately tries to measure (but fails quite miserably), cannot be [i]quantified[/i], but can be sensed by other people. Even though Honors Classes aren't purely based on intelligence, intelligence has a great impact. A student who is born with a "low IQ" will not be able to learn the information, no matter how hard they try, [i]even if they sign up[/i].

As for you third point, since that is the topic of discussion in this thread, I would just like to point out that separating students based on ability can be considered "acting upon" your sense of people's skills and capabilities. After all, Honors Classes learn much more and much more quickly than non-Honors, so it truly is separate and unequal as well.

Thanks for all the replies!  :lol:

[quote="polymorphic"] ... I'm all for discrimination.  It is impossible to have a world without discrimination.  Everyday, we have to discriminate over the most mundane things such as what we are going to wear, what we are going to eat, what we are going to do ... The problem is that there are different ways of discriminating.  Choosing to eat eggs scrambled instead of fried because they taste better that way is a legitimate form of discrimination.[/quote]

How do you feel about a certain man or woman who only considers getting married with a person of a certain race, due to culture, appearance, etc.

[quote="polymorphic"]In this PC age, it is becoming harder and harder to talk politely and intelligently about this issue.[/quote]

Agreed, completely. You nailed exactly why I started this thread.  :)

[quote="JBL"](I think this is what banblink was getting at, after all.)[/quote]

Nah, I don't really hold any real views on this topic, I just find it interesting. Everyone is free to say whatever he or she wants.

[quote="JBL"]Rule of thumb for designing an ideal society: you have to remember that you have an equal chance of being born as each member; accordingly, you must provide for each member's happiness, not just your own.[/quote]

I don't know about "chance"; that's a little too statistical for something like birth, but I agree that those who are born with certain talents are more "lucky" than "deserving". Nevertheless, for the common good, for example, wouldn't it be better to have an [i]undeserving, lucky[/i], highly intelligent president, than a [i]undeserving, unlucky[/i], unintelligent one?

I'm not trying to prove you wrong or anything, I'm just posting these questions. Anyone can answer them.

[quote="eryaman"]I don't feel that we should be judged over anything we don't have a choice over, but we should be judged over things that we have control over.[/quote]

Let's make a (partial) list of "things we can control over" and "things we don't". I think that is the source of much disagreement. Obviously we must not interpret these two categories too literally, for obviously you could purposely give yourself brain damage and "lower" your IQ, etc.  ;)

I'll begin: "Things we don't have control over" - IQ, EQ, Race, Appearance, Maximum Physical Ability (you can definitely improve, but there are things that are just "humanly impossible", and each person's line of physical impossibility is drawn at a slightly different place)...

I agree with judging based on what we have control over being permissible. On the topic of honors - as, no matter how much you know, you may not be able to grasp concepts as fast as others who are just 'lucky' (i mean this by the right genes, too), that is also justified discrimination, because it only helps you succeed. By the basketball players question, I'm not sure what you are asking - to say that people who dont play basketball are short? If thats the case - well its not really discrimination - it might be stereotyping. Prejudice is like discrimination, but without acting, prejudice can influence how you act to a certain person, and can cause discrimination. Stereotyping can cause prejudice. Therefore Stereotyping (can) cause prejudice, which (can) cause discrimination, which is acting on a decision based on a prejudice. By the basketball players - well i dont see how you can act on that. If you see a guy in the street who doesnt play basketball, you cant call him short if hes 6"5. You can, however, say that the mean of professional basketball players is higher (if it is), because statistically its true, and its not based on any prejudice. Notice prejudice is made up of 'pre' and 'judice' , to judge someone before hand, e.g. before meeting the person.

You can also ask, is choosing an apple over an orange prejudice? Simple, but explain why it isnt.

Just a quick sentence (turned into a paragraph *shrugs* heh)- People place too much emphasis on one's IQ. I've never seen anyone try their best continuously at a subject and not eventually succeed in it (yes there are exceptions with ESE kids and stuff like that, but this is in general). The fact is that when people see that they have low IQs they DON"T TRY as hard to get the higher grades because they think "Ah, I'm too stupid" which is not true- many highschool classes require little use of any intelligence, and if you focus on it you can master it. That's why I hate "IQ"s - they just give numbers that people can flaunt over others and have others create mental blocks because now they have a reason to slack off - because they aren't smart enough.

Where there's a will there's a way.

What I meant by my "control over" and "no control over" quote I don't think I expressed very well.  I meant that, say, someone decides to wear all black, sticks his hair up, wears chains.  Obviously this person is trying to convey something by his appearance and clothing, somthing he has control over, in this case.  So if you see someone like that, I'm saying it's justified to discriminate him over his choice of clothes.


The finite sets $A_1$, $A_2$, $\ldots$, $A_n$ are given and we denote by $d(n)$ the number of elements which appear exactly in an odd number of sets chosen from $A_1$, $A_2$, $\ldots$, $A_n$. Prove that for any $k$, $1\leq k\leq n$ the number \[{ d(n) - \sum\limits^n_{i=1} |A_i| + 2\sum\limits_{ i<j} |A_i \cap A_j | - \cdots + (-1)^k2^{k-1} \sum\limits_{i_1 <i_2 <\cdots < i_k} | A_{i_1} \cap A_{i_2} \cap \cdots \cap A_{i_k}}|  \] is divisible by $2^k$.

[i]Ioan Tomescu, Dragos Popescu[/i]

Firstly note that the $r^{th}$ summand (without the $2^r$) is precisely #$\{(x,A{}_i{}_1,A{}_i{}_2,\ldots,A{}_i{}_r)|x \in A{}_{i{}_1} \cap A{}_{i{}_2} \cap \cdots \cap A{}_{i{}_k}\}$. Counting from the perspective of the points instead, we get that this sum is the same as $\sum_{x\in S} \binom{|A_x|}{r}$ where $A_x=\{i|x\in A_i\}$ and $S$ is the superset of the $A_i's$.
 what we want to show is that for each $1\leq k\leq n$ we have that $d(n)-\sum_x \binom{|A_x|}{1} +2\sum_x \binom{|A_x|}{2}-\cdots+(-1){}^k2{}^{k-1} \sum_x \binom{|A_x|}{k}$ =
\[ d(n)-\sum_x\big{(} \binom{|A_x|}{1} +2\binom{|A_x|}{2}-\cdots+(-1){}^k2{}^{k-1}\binom{|A_x|}{k}\big{)} \] is divisible by $2^k$.
Note that to prove this it suffices to show that $d(n)-\sum_x\big{(} \binom{|A_x|}{1} +2\binom{|A_x|}{2}-\cdots+(-1){}^n2{}^{n-1}\binom{|A_x|}{n}\big{)}$ is divisible by $2^n$. but this summand (for each $x$ ) is simply $\dfrac{(1-\theta)^{|A_x|}-1}{\theta}|_{\theta=2}$ which is  $-1$ if $|A_x|$ is odd and zero if it is even. Hence the result.


calculus Putnam Harvard college MIT HMMT derivative \/closed

I know AoPS likes to use interesting problems and creative techniques, and I'm sure this is one of the goals of the calculus course.  Just out of curiosity, could an instructor please post a couple of problems that demonstrate the kind of non-routine thinking encouraged by the calculus course?

(Sorry I didn't reply sooner...I've been out sick the last couple of days.)

We do a lot of the harder calculus problems from contests such as the Putnam and the Harvard-MIT Math Tournament.  (Take a look at some of the past calculus tests on the HMMT website to see what sorts of problems these are.)

We also do non-trivial applications.  For example, in next week's class, we are going to prove that
$ 1 \plus{} \frac14 \plus{} \frac19 \plus{} \frac{1}{16} \plus{} \cdots \equal{} \frac{\pi^2}{6}$

Also, two problems on the most recent Challenge Problem set were to prove the formula for the volume of a torus and to prove certain basic facts about the Gamma function.

[quote="benzi455"]That sounds great.

Unrelated question: Are you (or other readers) familiar with Michael Spivak's "Calculus"?  If so, do you recommend it?[/quote]I very highly recommend it.  It is very well written.  In fact, I use the occasional problem from Spivak in our class.  But it is a [b]hard[/b] book.  Our calculus class does not cover calculus at the theoretical level that Spivak does.

I would recommend learning the mechanistic aspects of calculus first (e.g. computing derivatives and integrals and basic applications) before attempting to work through Spivak.

Thanks.  I think I'm ready for Spivak now, since I've picked up many of the basics from school.  Now the problem is figuring out how to get it >_<


inequalities
algebra
polynomial
function
inequalities open

Let $ a,$ $ b$ and $ c$ are non-negative numbers, no two of which are zeros. Prove that:
 \[ \frac{a}{b^{3}\plus{}c^{3}}\plus{}\frac{b}{a^{3}\plus{}c^{3}}\plus{}\frac{c}{a^{3}\plus{}b^{3}}\geq\frac{18}{5(a^{2}\plus{}b^{2}\plus{}c^{2})\minus{}ab\minus{}ac\minus{}bc}.\]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167491
The first inequality here is easy enough, but I have found a mistake in my proof for the second inequality.

[quote="ehku"]Very nice inequality!
Do you have any proof, Arqady ? Because, I found one.[/quote]
No! I still don't have a proof. :(

I used the following lemma:
Lemma. If we want to prove an inequality in the form $ A(abc)^2\plus{}B(abc)\plus{}C \le 0$ for all $ a,b,c \ge 0$ where $ A, B, C$ are fonctions of $ (a\plus{}b\plus{}c)$ and $ (ab\plus{}bc\plus{}ca)$, moreover, $ A \ge 0$, then if suffits to prove it in two case $ a\equal{}b$ or $ c\equal{}0$.
I think you can find a proof for it here http://reflections.awesomemath.org/2006_5/2006_5_breakingpoint.pdf

[quote="ehku"]I used the following lemma:
Lemma. If we want to prove an inequality in the form $ A(abc)^2 \plus{} B(abc) \plus{} C \le 0$ for all $ a,b,c \ge 0$ where $ A, B, C$ are fonctions of $ (a \plus{} b \plus{} c)$ and $ (ab \plus{} bc \plus{} ca)$, moreover, $ A \ge 0$, then if suffits to prove it in two case $ a \equal{} b$ or $ c \equal{} 0$.
I think you can find a proof for it here http://reflections.awesomemath.org/2006_5/2006_5_breakingpoint.pdf[/quote]
I think it is ABC theorem,just one of the cases of this theorem.For more information see here:
http://www.mathlinks.ro/viewtopic.php?t=165132

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers, no two of which are zeros. Prove that:
\[ \frac {a}{b^{3} \plus{} c^{3}} \plus{} \frac {b}{a^{3} \plus{} c^{3}} \plus{} \frac {c}{a^{3} \plus{} b^{3}}\geq\frac {18}{5(a^{2} \plus{} b^{2} \plus{} c^{2}) \minus{} ab \minus{} ac \minus{} bc}.
\]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167491
The first inequality here is easy enough, but I have found a mistake in my proof for the second inequality.[/quote]
Does any one have another way ? It is so nice  :(

Yes, in theory it is solved using $ p,q,r$ method. That is, we express it in terms of $ p\equal{}a\plus{}b\plus{}c$, $ q\equal{}ab\plus{}bc\plus{}ca$, and $ r\equal{}abc$. Then we use a lemma. However, this is computationally expensive.

[quote="pvthuan"]Yes, in theory it is solved using $ p,q,r$ method. That is, we express it in terms of $ p \equal{} a \plus{} b \plus{} c$, $ q \equal{} ab \plus{} bc \plus{} ca$, and $ r \equal{} abc$. Then we use a lemma. However, this is computationally expensive.[/quote]

I agree with you about that. But it seems to be very hard to do. Very complicated !!!

[quote="pvthuan"]Yes, in theory it is solved using $ p,q,r$ method. That is, we express it in terms of $ p \equal{} a \plus{} b \plus{} c$, $ q \equal{} ab \plus{} bc \plus{} ca$, and $ r \equal{} abc$. Then we use a lemma. However, this is computationally expensive.[/quote]

But what’s the use of extensive computations? I claim that [b] any symmetric homogenous polynomial in $ a,b,c$ can be generated by [/b] $ p,q,r$. :) Let’s prove it!
For $ n\in\mathbb N$, denote by $ p_3(n)$ the number of partitions of $ n$ with three or less parts. Note that by conjugation of the partitions, this is equal to the number of partitions of $ n$ consisting only of 1’s, 2’s and 3’s. (We’ll use that.) :D
Let $ H_n$ the space of all symmetric homogenous polynomials of total degree $ n$ in 3 variables with real coefficients. It is well-known that $ H_n$ is the span of the elementary symmetric functions $ [\alpha]$ (the ones used in Muirhead) for all partitions $ \alpha\vdash n$ with three or less parts, and they form a basis, hence the dimension of $ H_n$ is $ p_3(n)$.
On the other hand, by taking products of powers of $ p,q,r$ with exponents $ \in\mathbb N_0$, we can generate exactly $ p_3(n)$ different ones with total degree $ n$. (Each partition of $ n$ consisting only of 1’s, 2’s and 3’s yields one.) Hence these polynomials, being linearly independent, must also form a basis of $ H_n$. :) :)

In the given case, we only need to make sure there is no term $ (abc)^3$ before we can apply the lemma. But – oops – it doesn’t vanish. One can see without heavy computation that expanding will yield $ 15(abc)^3$ on LHS but $ 36(abc)^3$ on RHS. :( :?:

Did you come to this arqady?

\[ [5(a^2\plus{}b^2\plus{}c^2)\minus{}ab\minus{}bc\minus{}ca](a^2\plus{}b^2\plus{}c^2)^2\geq 36(a^3b^3\plus{}b^3c^3\plus{}c^3a^3).\]

@pvthuan,zaiai-hoang,spanferkel: that is not what I mean. Try your way, you will see your problem.
Using the above lemma, my solution is very simple !

[quote="pvthuan"]OK, it shouldn't be an open problem. I tried it by Cauchy
\[ {\rm LHS} \equal{} \sum\frac {a^4}{a^3b^3 \plus{} a^3b^3}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{2(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3)}.
\]
Now we need to show that
$ [5(a^2 \plus{} b^2 \plus{} c^2) \minus{} ab \minus{} bc \minus{} ca](a^2 \plus{} b^2 \plus{} c^2)^2\geq 36(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3).$
Let $ a \plus{} b \plus{} c \equal{} 1$, and $ q \equal{} ab \plus{} bc \plus{} ca$, $ r \equal{} abc$, the the above reads
\[ (5 \minus{} 11q)(1 \minus{} 2q)^2\geq 36[q^3 \minus{} 8r(q \minus{} r)].
\]
[/quote]
Try for $ (a,b,c)\equal{}(0.1,1,1)$.

It wasn't an open problem, because I solved it as soon as seeing it.  :(
By Cauchy-Schwarz inequality we have
\[ (\sum \frac{a}{b^3\plus{}c^3})(\sum \frac{a(b^2\minus{}bc\plus{}c^2)}{b\plus{}c}) \ge (\sum \frac{a}{b\plus{}c})^2
\]
It suffits to prove
\[ (\sum \frac{a}{b\plus{}c})^2(5(a^2\plus{}b^2\plus{}c^2)\minus{}ab\minus{}bc\minus{}ca) \ge 18 \sum \frac{a(b^2\minus{}bc\plus{}c^2)}{b\plus{}c}
\]
or
\[ (\sum \frac{a}{b\plus{}c})^2(5(a^2\plus{}b^2\plus{}c^2)\minus{}ab\minus{}bc\minus{}ca) \ge 18 (2\sum ab\minus{} 3abc \sum \frac{1}{a\plus{}b}) \ \ (*)
\]
Put $ p\equal{}a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$ and notice that
$ \sum \frac{1}{a\plus{}b}\equal{}\frac{p^2\plus{}q}{pq\minus{}r}$
$ \sum \frac{a}{b\plus{}c}\equal{}(a\plus{}b\plus{}c)(\sum \frac{1}{a\plus{}b})\minus{}3\equal{}p.\frac{p^2\plus{}q}{pq\minus{}r}\minus{}3\equal{}\frac{p^3\minus{}2pq\plus{}3r}{pq\minus{}r}$
, the last inequality is equivalent to
\[ (\frac{p^3\minus{}2pq\plus{}3r}{pq\minus{}r})^2(5p^2\minus{}11q) \ge 18(2q\minus{}3r.\frac{p^2\plus{}q}{pq\minus{}r})
\]
or
\[ (p^3\minus{}2pq\plus{}3r)^2(5p^2\minus{}11q) \ge 18(pq\minus{}r)(2pq^2\minus{}3p^2r\minus{}5qr)
\]
Notice that $ 9(5p^2\minus{}11q) \le 18(3p^2\plus{}5q)$, the last inequality can be rewritten in the form of the lemma, then we have only to prove (*) in two case $ b\equal{}c\equal{}1$ or $ c\equal{}0$.
If $ c\equal{}0$, (*) becomes
\[ (\frac{a}{b}\plus{}\frac{b}{a})^2(5a^2\plus{}5b^2\minus{}ab) \ge 36ab
\]
, which is clearly true by AM-GM inequality.
If $ b\equal{}c\equal{}1$, (*) becomes
\[ (\frac{a}{2}\plus{}\frac{2}{a\plus{}1})^2(5a^2\minus{}2a\plus{}9) \ge 18(4a\plus{}2\minus{}3a(\frac{2}{a\plus{}1}\plus{}\frac{1}{2}))
\]
or
\[ (a^2\plus{}a\plus{}4)^2(5a^2\minus{}2a\plus{}9) \ge 36(a\plus{}1)(5a^2\minus{}3a\plus{}4)
\]
or
\[ a(a\minus{}1)^2(5a^3\plus{}18a^2\plus{}81a\plus{}4) \ge 0
\]
, which is obvious.
The inequality is proved.

(Sorry if I made some mistakes in calculating).

[quote="arqady"]Thanks  all!
I have proved it by expanding and SOS. :D
It's ugly but transparent enough. :wink:[/quote]
Could you post it, please ? Thank you.

[quote="ehku"]Then, Arqady, what is your solution ?  :wink:[/quote]
Oh, ehku, I didn't see your posts.
$ \frac {a}{b^{3} \plus{} c^{3}} \plus{} \frac {b}{a^{3} \plus{} c^{3}} \plus{} \frac {c}{a^{3} \plus{} b^{3}}\geq\frac {18}{5(a^{2} \plus{} b^{2} \plus{} c^{2}) \minus{} ab \minus{} ac \minus{} bc}\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(5a^9 \minus{} 2a^8b \plus{} 10a^7b^2 \minus{} a^7bc \minus{} 26a^6b^3 \plus{} 8a^5b^4 \plus{} 8a^5b^3c \plus{} 8a^4b^3c^2 \minus{}$
$ \minus{} 3a^4b^4c \minus{} 7a^3b^3c^3)\geq0\Leftrightarrow\sum_{cyc}(a \minus{} b)^2(c^7 \minus{} 6(a^2 \plus{} ab \plus{} b^2)c^5 \minus{} (a \plus{} b)abc^4 \plus{}$
$ \plus{} 7a^2b^2c^3 \plus{} 2a^3b^3c \plus{} 5a^7 \plus{} 8a^6b \plus{} 20a^5b^2 \plus{} 6a^4b^3 \plus{} 6a^3b^4 \plus{} 20a^2b^5 \plus{} 8ab^6 \plus{} 5b^7)\geq0,$
which obviously true because if $ a\geq b\geq c$ then $ S_c\geq0,$ $ S_b\geq0$ and $ S_a \plus{} S_b\geq0.$

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers, no two of which are zeros. Prove that:
\[ \frac {a}{b^{3} + c^{3}} + \frac {b}{a^{3} + c^{3}} + \frac {c}{a^{3} + b^{3}}\geq\frac {18}{5(a^{2} + b^{2} + c^{2}) - ab - ac - bc}.
\]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167491
The first inequality here is easy enough, but I have found a mistake in my proof for the second inequality.[/quote]
A simpler solution is
Rewrite the inequality as
\[ \sum \frac {a(a + b + c)}{b^3+c^3} \ge \frac {18(a + b + c)}{5(a^2 + b^2 + c^2) - ab - bc - ca}
\]
or
\[ \sum \frac {a^2}{b^3 + c^3} + \frac {a}{b^2 + c^2 - bc} \ge \frac {18(a + b + c)}{5(a^2 + b^2 + c^2) - ab - bc - ca}
\]
By the Cauchy Schwarz, we have
\[ \sum \frac {a^2}{b^3 + c^3} \ge \frac {(a^2 + b^2 + c^2)^2}{\sum a^2(b^3 + c^3)}
\]

\[ \sum \frac {a}{b^2 + c^2 - bc} \ge \frac {(a + b + c)^2}{\sum a(b^2 + c^2 - bc)}
\]
It suffices to prove
\[ \frac {(a^2 + b^2 + c^2)^2}{\sum a^2(b^3 + c^3)} + \frac {(a + b + c)^2}{\sum a(b^2 + c^2 - bc)} \ge \frac {18(a + b + c)}{5(a^2 + b^2 + c^2) - ab - bc - ca}
\]
Assume $ a + b + c = 1$ and denote $ ab + bc + ca = q,abc = r \Rightarrow r \ge \max { 0,\frac {(4q - 1)(1 - q)}{6}}$. We have to prove
\[ \frac {(1 - 2q)^2}{q^2 - (q + 2)r} + \frac {1}{q - 6r} \ge \frac {18}{5 - 11q}
\]
The last inequality can be proved easily by consider two cases $ 1 \ge 4q$ and $ 4q \ge 1$ :)
Equality holds for $ a = b = c$ and $ a = b,c = 0$. :)


geometry
geometric transformation
reflection
real analysis
real analysis unsolved

prove that C+C=[0,2].(C means cantor set)

I think the following problem was posted recently (can't find it :(), and I made an attempt to solve it, but only posted a sketch:

Show that every number in $[0,1]$ is the differrence between two Cantor numbers.

The same idea would work here: we have to show that every line parallel to the diagonal of the unit square has a point of [url=http://mathworld.wolfram.com/CantorDust.html]Cantor's dust[/url] on it. That's all I wrote there, but I'll try to prove it now.

Consider the way Cantor''s dust is formed, taking out open regions at each step. Now consider our (closed) segment parallel to a diagonal in the square. At each step, we take out of that segment either an open interval symmetric wrt the midpoint of our segment, or two open intervals, one being the reflection of the other one in the midpoint of our segment. Another important property is that the set we take out at a certain step is disjoint from the sets taken out in other steps. From here it's easy to show, by using Cantor's theorem (a decreasing sequence of compact intervals has non-void intersection), that there's always at least a point left on each segment parallel to a diagonal (here we include the down-left and up-right corners of the square as being degenerate closed segments parallel to the main diagonal).

I hope it works. Does it?

reasonable solution .my firend's solution is over my knowledge he construct
C+C as union of some contraction and then conclude that C+C is fractal and is unique so it just [0,2].

Maybe I am being too simplistic and not understanding the point of your proof, but what about this:

(1) Prove that $C$ is a closed set.
(2) Prove that if $x\in C$ then $1-x\in C$.
(3) Prove that $C-C=[-1,1]$.
(4) If $\alpha\in[0,2]\Rightarrow\alpha-1\in[-1,1]$, $1-\alpha=c_1-c_2$, $c_i\in C\Rightarrow\alpha=c_1+(1-c_2)$.
(5) Conclude thus that $[0,2]\subseteq C+C$.
(6) Trivially conclude that $C+C\subseteq[0,2]$.

To poof that $C-C=[-1,1]$,

(7) Note that the $C-C$ has central symetry.
(8) Prove that every triadic  number in $[0,1]$ is in $C-C$ (a triadic number is a number of the form $\frac{n}{3^k}$) where $n\mathbb{Z}$ and $k\in\mathbb{Z}^+$.
(9) Prove that if $A$ and $B$ are closed, then so is $A-B$.
(A) Conclude that $C-C=[-1,1]$.

I hope it helps. Nice problem, I hadn't seen it before. Best regards,


analytic geometry
geometry proposed
geometry

\[A,B,C,D,P_1\in d_1,\ M,N,P_2\in d_2;\ P_1P_2\perp d_1\parallel d_2$\] \[\ [M\longleftrightarrow (A,B),\ N\longleftrightarrow (C,D)]\] \[A_1\in MP_1\cap AP_2,\ B_1\in MP_1\cap BP_2,\ C_1\in NP_1\cap CP_2,\ D_1\in NP_1\cap DP_2.\] \[************************************************\] \[\frac{\overline {AB}}{\overline {CD}}+\frac{\overline {P_1 A}}{\overline{P_1 C}}\cdot \frac{\overline {P_1 B}}{\overline {P_1 D}}=0\Longleftrightarrow A_1 D_1\cap B_1 C_1\cap P_1P_2\ne\emptyset\] [b]See:[/b]
http://www.mathlinks.ro/Forum/viewtopic.php?t=46358
http://www.mathlinks.ro/Forum/viewtopic.php?t=47140

What does

$\displaystyle \ [M\longleftrightarrow (A,B),\ N\longleftrightarrow (C,D)]$

mean?

For the four mentioned intersections: $MP_1$ with $P_2 A,\  P_2 B$ (analoguos N). It is a correspondence. It isn't essential.

[quote="levi"]\[\frac{\overline {AB}}{\overline {CD}}+\frac{\overline {P_1 A}}{\overline{P_1 C}}\cdot \frac{\overline {P_1 B}}{\overline {P_1 D}}=0\Longleftrightarrow A_1 D_1\cap B_1 C_1\cap P_1P_2\ne\emptyset\] [/quote]
I believe this is not true, until you put
\[\frac{\overrightarrow {AB}}{\overrightarrow {CD}}+\frac{\overrightarrow {P_1 A}}{\overrightarrow{P_1 C}}\cdot \frac{\overrightarrow {P_1 B}}{\overrightarrow {P_1 D}}=0\Longleftrightarrow A_1 D_1\cap B_1 C_1\cap P_1P_2\ne\emptyset\]

Let $d$ a orientated line with the origin $O$, i.e. $X=X(x)\in d$,
where the coordinate $x\in R$ (the axiom of rule).
Thus, for A(a) and B(b) $,\overline {AB}=b-a\in R$.
But, for two points A, B of one plane $\vec {AB}\ $is a vector.


calculus
integration
trigonometry
calculus computations

kemon achen apnara shobai ?  :D

$ \int_0^1\;\sin\,(\pi\,x)\cdot x^x\cdot (1\minus{}x)^{1\minus{}x}\;\;\textbf dx$

answer below:

[hide]
did you at least try it out  first ?  :(
[hide]
giving up that soon..... OK !  :D
[hide]
$ \text{answer: }\qquad\boxed{ \dfrac{\pi\,e}{24}}$
[/hide][/hide][/hide]


Calculate

$ \sum_{n \equal{} 1}^{N} C_{n} \sum_{k \equal{} 1}^{N} X_{k} e^{{j 2 \pi (n\minus{}k)/N}}$

Here $ C$ and $ X$ are complex variables.

I know its upperbound would be $ N^{2}$ but you know that is a quite big upperbound. Some practical solution or upperbound perhaps?

Hmmm. I don't think this should be in intermediate topics. Try the college forum.

You'd need more information. For example, you could have $ X_1 \equal{} a, X_{i \neq 1} \equal{} 0$, and similarly $ C_1 \equal{} b, \; C_{i \neq 1} \equal{} 0$ and your sum would be $ ab$. Then, depending on what $ a$ and $ b$ are the sum can be arbitrarily large...

I assume $ j$ is the engineer's $ \imath$.
[hide="Discussion"] Rewrite as

$ \sum_{a \equal{} 1\minus{}N}^{N\minus{}1} e^{\frac{2 \pi \imath a}{N} } \left( \sum_{n\minus{}k\equal{}a} C_n X_k \right)$.

The rest depends on the nature of $ C_n, X_k$.  Note that the inside summation has $ N \minus{} |a|$ terms, so if a bound on $ |C_n|$ and $ |X_k|$ is known we have some upper bounds available. [/hide]


linear algebra
matrix
linear algebra unsolved

\[ \begin{array}{l}
 {\bf{A}} \equal{} {({a_{ij}})_{n \times n}}, \\
 {a_{ij}} \equal{} \sqrt {\frac{{\min \{ i,j\} }}{{\max \{ i,j\} }}} , \\
 \end{array}\]

Try to calculate
\[ \det ({\bf{A}})\]

Calculating the determinant for $ n\equal{}1,2,3,4$ we get $ 1,2,6,24$, so we can suppose det is $ n!$.
There is a interesting pattern in the inverse, only 3 diagonals, but don't know how to use it to verify the formula

LU decomposition works here:

$ A\equal{}(b_{ij})_{n\times n}(c_{ij})_{n\times n}$
where $ b_{ij}\equal{}\left\{\begin{matrix}
0,i<j\\
\sqrt{\frac{i}{j}}, i\geq j
\end{matrix}\right.$ and $ c_{ij}\equal{}\left\{\begin{matrix}
0,i>j\\
\frac{1}{\sqrt{ij}}, i\leq j
\end{matrix}\right.$

And then $ det A\equal{}det (b_{ij})\times det (c_{ij})\equal{}\frac{1}{n!}$


a) If S is the subspace of M6(R) consisting of all upper triangular matrices, find the dimension of S.

b) If S is the subspace of M8(R) consisting of all diagonal matrices, find the dimension of S.

[hide]The dimension of $Upp_n(\mathbb{R})$ is $n(n+1)/2$
The dimension of $Diag_n(\mathbb{R})$ is $n$[/hide]


least common multiple
number theory
relatively prime
number theory unsolved

Let $ n$ be a positive integer, let $ A$ be a subset of $ \{1, 2, \cdots, n\}$, satisfying for any two numbers $ x, y\in A$, the least common multiple of $ x$, $ y$ not more than $ n$. Show that $ |A|\leq 1.9\sqrt {n} \plus{} 5$.

This is the official solution:

For $ a\in (\sqrt{n},\sqrt{2n}]$, we have $ [a,a\plus{}1]\equal{}a(a\plus{}1)>n$.So $ |A\cap (\sqrt{n},\sqrt{2n})|\le \frac{1}{2}(\sqrt{2}\minus{}1)\sqrt{n}\plus{}1$.
For $ a\in (\sqrt{2n},\sqrt{3n}]$, we have $ [a,a\plus{}1]\equal{}a(a\plus{}1)>n$, $ [a\plus{}1,a\plus{}2]\equal{}(a\plus{}1)(a\plus{}2)>n$, $ [a,a\plus{}2]\ge \frac{1}{2}a(a\plus{}2)>n$. So $ |A\cap (\sqrt{2n},\sqrt{3n}]|\le \frac{1}{3}(\sqrt{3}\minus{}\sqrt{2})\sqrt{n}\plus{}1$.
For the same reason $ |A\cap (\sqrt{3n},2\sqrt{n}]|\le \frac{1}{4}(\sqrt{4}\minus{}\sqrt{3})\sqrt{n}\plus{}1$.

So
$ |A\cap [1,2\sqrt{n}]|\le \sqrt{n}\plus{}\frac{1}{2}(\sqrt{2}\minus{}1)\sqrt{n}\plus{}\frac{1}{3}(\sqrt{3}\minus{}\sqrt{2})\sqrt{n}\plus{}\frac{1}{4}(\sqrt{4}\minus{}\sqrt{3})\sqrt{n}\plus{}3\equal{}(1\plus{}\frac{1}{6}\sqrt{2}\plus{}\frac{1}{12}\sqrt{3})\sqrt{n}\plus{}3$

For positive integer $ k$, assume $ a,b\in (\frac {n}{k\plus{}1},\frac {n}{k}]$, $ a>b$, and assume $ [a,b]\equal{}as\equal{}bt$, then $ \frac{a}{(a,b)}s\equal{}\frac{b}{(a,b)}t$
Because $ \frac{a}{(a,b)}$ and $ \frac{b}{(a,b)}$ are relatively prime, so $ s$ is a multiple of $ \frac{b}{(a,b)}$. Therefore $ [a,b]\equal{}as\ge \frac{ab}{(a,b)}\ge \frac{ab}{a\minus{}b}\equal{}b\plus{}\frac{b^2}{a\minus{}b}>\frac{n}{k\plus{}1}\plus{}\frac{(\frac{n}{k\plus{}1})^2}{\frac{n}{k}\minus{}\frac{n}{k\plus{}1}}\equal{}n$

So $ |A\cap (\frac{n}{k\plus{}1},\frac{n}{k}]|\le 1$.
Take positive integer $ T$ such that $ \frac{n}{T\plus{}1}\le 2\sqrt{n}<\frac{n}{T}$, then $ |A\cap (2\sqrt{n},n]|\le \sum_{k\equal{}1}^T |A\cap (\frac{n}{k\plus{}1},\frac{n}{k}]|\le T<\frac{1}{2}\sqrt{n}$

Combined all above we get  $ |A|\le (\frac{3}{2}\plus{}\frac{1}{6}\sqrt{2}\plus{}\frac{1}{12}\sqrt{3})\sqrt{n}\plus{}3<1.9\sqrt{n}\plus{}5$.
Q.E.D

For $n\le64$ it is verry easy, try  it yourself.
In the last part i meant it has at least 1.9.... divisors smaller that $\sqrt{n+1}$


algebra
polynomial
quadratics
Rational Root Theorem

Factorise $ x^4 \plus{} x^3 \plus{} 6x^2 \plus{} 5x \plus{} 5$ completely.

Factor Theorem doesn't seem to work. Haha.

A quartic polynomial reducible over the integers (equivalently, the rationals) either has a rational linear factor or two rational quadratic factors.  The Rational Root Theorem makes it easy to rule out the first case; the only question is then ruling out the second.

[hide="From there..."] There are only two cases:  a factorization of the form $ (x^2 \plus{} ax \plus{} 1)(x^2 \plus{} bx \plus{} 5)$ or $ (x^2 \plus{} ax \minus{} 1)(x^2 \plus{} bx \minus{} 5)$ for integers $ a, b$.  Expanding the first factorization gives $ a \equal{} 1, b \equal{} 0$ and the factorization

$ (x^2 \plus{} x \plus{} 1)(x^2 \plus{} 5)$.

In hindsight, inspection says we should've written the original polynomial as $ (x^4 \plus{} x^3 \plus{} x^2) \plus{} 5(x^2 \plus{} x \plus{} 1)$.[/hide]


For all non-negative $ a,$ $ b$ and $ c$ prove that:
\[ (ab\minus{}c^2)(a\plus{}b\minus{}c)^3\plus{}(ac\minus{}b^2)(a\plus{}c\minus{}b)^3\plus{}(bc\minus{}a^2)(b\plus{}c\minus{}a)^3\geq0\]

[hide="Ugly proof"] it is given by summing up
$ a^5\plus{}b^5\plus{}c^5\geq abc(a^2\plus{}b^2\plus{}c^2)$ and $ 4\sum_{sym}a^4b \geq \sum_{sym}a^3b^2\plus{}6abc(ab\plus{}bc\plus{}ca)$[/hide]

Stupid calculation mistake sorry... :oops:

[quote="arqady"]For all non-negative $ a,$ $ b$ and $ c$ prove that:
\[ (ab \minus{} c^2)(a \plus{} b \minus{} c)^3 \plus{} (ac \minus{} b^2)(a \plus{} c \minus{} b)^3 \plus{} (bc \minus{} a^2)(b \plus{} c \minus{} a)^3\geq0
\]
[/quote]

I like this inequality, very beautiful one. I would like to see your solution, Arqady. I hope the following is also true.
\[ (ab\minus{}c^2)(a\plus{}b\minus{}c)^5\plus{}(bc\minus{}a^2)(b\plus{}c\minus{}a)^5\plus{}(ca\minus{}b^2)(c\plus{}a\minus{}b)^5\geq0.\]

[quote="arqady"]For all non-negative $ a,$ $ b$ and $ c$ prove that:
\[ (ab \minus{} c^2)(a \plus{} b \minus{} c)^3 \plus{} (ac \minus{} b^2)(a \plus{} c \minus{} b)^3 \plus{} (bc \minus{} a^2)(b \plus{} c \minus{} a)^3\geq0
\]
[/quote]

My solution :).

By expanding, assume $ a \plus{} b \plus{} c \equal{} 1$ and setting $ q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$ the inequality becomes
\[ \left( {ab \minus{} c^2 } \right)\left( {a \plus{} b \minus{} c} \right)^3 \plus{} \left( {bc \minus{} a^2 } \right)\left( {b \plus{} c \minus{} a} \right)^3 \plus{} \left( {ca \minus{} b^2 } \right)\left( {c \plus{} a \minus{} b} \right)^3 \ge 0
\]

\[ \Leftrightarrow \minus{} \left( {24q \plus{} 4} \right)r \plus{} 16q^2 \minus{} 7q \plus{} 1 \ge 0
\]
Besides that we have $ r \le \frac {{q^2 \left( {1 \minus{} q} \right)}}{{2\left( {2 \minus{} 3q} \right)}}$ - [b]canhang_2007[/b]

So it suffices us show that
\[ \minus{} \left( {24q \plus{} 4} \right)\left[ {\frac {{q^2 \left( {1 \minus{} q} \right)}}{{2\left( {2 \minus{} 3q} \right)}}} \right] \plus{} 16q^2 \minus{} 7q \plus{} 1 \ge 0
\]

\[ \Leftrightarrow \frac {{12q^4 \minus{} 58q^3 \plus{} 51q^2 \minus{} 17q \plus{} 2}}{{2 \minus{} 3q}} \ge 0
\]

\[ \Leftrightarrow \frac {{\left( {q \minus{} \frac {1}{3}} \right)\left( {12q^3 \minus{} 54q^2 \plus{} 33q \minus{} 6} \right)}}{{2 \minus{} 3q}} \ge 0
\]
which is clearly true. Equality holds if $ a \equal{} b \equal{} c$.
We have done :).

[quote="pvthuan"][quote="arqady"]For all non-negative $ a,$ $ b$ and $ c$ prove that:
\[ (ab \minus{} c^2)(a \plus{} b \minus{} c)^3 \plus{} (ac \minus{} b^2)(a \plus{} c \minus{} b)^3 \plus{} (bc \minus{} a^2)(b \plus{} c \minus{} a)^3\geq0
\]
[/quote]

I like this inequality, very beautiful one. I would like to see your solution, Arqady. I hope the following is also true.
\[ (ab \minus{} c^2)(a \plus{} b \minus{} c)^5 \plus{} (bc \minus{} a^2)(b \plus{} c \minus{} a)^5 \plus{} (ca \minus{} b^2)(c \plus{} a \minus{} b)^5\geq0.
\]
[/quote]
I think it is true too and I espect that the followed is true also
\[ \left( {ab \minus{} c^2 } \right)\left( {a \plus{} b \minus{} c} \right)^7 \plus{} \left( {bc \minus{} a^2 } \right)\left( {b \plus{} c \minus{} a} \right)^7 \plus{} \left( {ca \minus{} b^2 } \right)\left( {c \plus{} a \minus{} b} \right)^7 \ge 0
\]
for all $ a,b,c \ge 0$.

More general will we have
\[ \left( {ab \minus{} c^2 } \right)\left( {a \plus{} b \minus{} c} \right)^k \plus{} \left( {bc \minus{} a^2 } \right)\left( {b \plus{} c \minus{} a} \right)^k \plus{} \left( {ca \minus{} b^2 } \right)\left( {c \plus{} a \minus{} b} \right)^k \ge 0
\]
for all $ a,b,c \ge 0$ and $ k$ is an given natural odd number.  :?:

I think if it true, it is not easy.

Did you think about them arquady? :)

[quote="Honey_S"]
More general will we have
\[ \left( {ab \minus{} c^2 } \right)\left( {a \plus{} b \minus{} c} \right)^k \plus{} \left( {bc \minus{} a^2 } \right)\left( {b \plus{} c \minus{} a} \right)^k \plus{} \left( {ca \minus{} b^2 } \right)\left( {c \plus{} a \minus{} b} \right)^k \ge 0
\]
for all $ a,b,c \ge 0$ and $ k$ is an given natural odd number.  :?:

I think if it true, it is not easy.

Did you think about them arquady? :)[/quote]
I'll think about it.
I have proved it by SOS for $ k \equal{} 5.$
By the way, if $ a,$ $ b$ and $ c$ are side lengths of triangle then it's true for all real $ k\geq1.$  :wink:

The following inequality is true too:
Let $ a,$ $ b$  and  $ c$ are side lengths of triangle. Prove that:
\[ (ab \minus{} c^2)\sqrt [3]{a \plus{} b \minus{} c} \plus{} (ac \minus{} b^2)\sqrt [3]{a \plus{} c \minus{} b} \plus{} (bc \minus{} a^2)\sqrt [3]{b \plus{} c \minus{} a}\geq0
\]
but the following
\[ (ab \minus{} c^2)\sqrt [4]{a \plus{} b \minus{} c} \plus{} (ac \minus{} b^2)\sqrt [4]{a \plus{} c \minus{} b} \plus{} (bc \minus{} a^2)\sqrt [4]{b \plus{} c \minus{} a}\geq0
\]
is wrong already.


Julian wants to label the five fingers on his left hand with the integers 1 through 5, such that no two adjacent fingers have consecutive integers, and each integer is used exactly once. In how many ways can Julian label his fingers?

Labeling fingers...ah, such a great recreational activity.

Case 1: First Finger is 1

We can have second number 3,4, or 5.

If 3, then we have to have 5 for the third number, we have to have 2 for the fourth number, and 4 must be the last number.

[b]1[/b] possibility.

If 4, then the third number has to be 2, the fourth has to be 5, and the fifth has to be 3.

[b]1[/b] possibility.

If 5, then the third number has to be 2 or 3.

- If 2, then the fourth has to be 4, and the fifth has to be 3.

- If 3, then the fourth has to be 5, and the fifth has to be 2.

[b]2[/b] possibilities.

Case 2: First Finger is 2

We can have the second finger be 4 or 5.

If 4, then the third finger has to be 1, and the fourth has to be 3 or 5.

- If 3, then the fifth has to be 5.

- If 5, then the fifth has to be 3.

[b]2[/b] possibilities.

If 5, then the third finger has to be either 3 or 1.

- If 3, then the fourth finger has to be 1, and the fifth has to be 4.

- It doesn't work for 1.

[b]1[/b] possibility.

Case 3: First Finger is 3

We can have the second finger be 1 or 5.

If 1, then the third finger has to be 4 or 5.

- If 4, then the fourth finger has to be 2, and the fifth has to be 5.

- If 5, then the fourth finger has to be 2, and the fifth has to be 4.

[b]2[/b] possibilities.

If 5, then the third finger has to be 1 or 2.

- If 1, then the fourth finger has to be 4, and the fifth has to be 2.

- If 2, then the fourth finger has to be 4, and the fifth has to be 1.

[b]2[/b] possibilities.

Case 4: The First Finger is 4

The second finger can be 1 or 2.

If 1, then the third finger has to be 3 or 5.

- If 3, then the fourth finger has to be 5 and the fifth has to be 2.

- If 5, then it doesn't work.

[b]1[/b] possibility.

If 2, then the third finger has to be 5, and the fourth finger has to be either 3 or 1.

- If 3, then the fifth finger has to be 1.

- If 1, then the fifth finger has to be 2.

[b]2[/b] possibilities.

Case 5: First Finger is 5

The second finger has to be 3, 2, or 1.

If 3, then the third finger has to be 1, the fourth finger has to be 4, and the fifth finger has to be 2.

[b]1[/b] possibility.

If 2, then the third finger has to be 4, the fourth finger has to be 1, and the fifth finger has to be 3.

[b]1[/b] possibility.

If 1, then it doesn't work.

Add up the [b]bold[/b] numbers to get $ \boxed{16}$.

No, the answer is actually 14 (based on the National Sprint answer key)

Case 1: Suppose that the middle finger is labeled 1. (or 5)
Then the 2nd and 4th fingers must be 5 and 4, giving the 1st and 5th fingers 2 and 3.
So the cases possible here are 25143, 24153, 34152, 35142. There are 4 cases. By symmetry, there are also 4 cases if the middle finger is 5.

Case 2: Suppose that the middle finger is 2 (or 4).
If the middle finger is 2, then the 2nd and 4th fingers must be 4 and 5. Then the 1st and 5th ones must be 1 and 3.
So the cases possible here are 14253 and 35241. There are 2 cases. By symmetry, there are also 2 cases if the middle finger is 4.

Case 3: The middle finger is 3. So the 2nd and 4th fingers must be 1 and 5, and the 1st and 5th must be 2 and 4.
The only possibilities are 41352 and 25314, so there are 2 ways.

So there are a total of 2(4)+2(2)+2=14 ways for Julian to label his fingers.

Of course, if you had to, you could mem it. :P


Find the minimum and the maximum values of $A=x^2+y^2+z^2$ if

$\frac{x^2}{4}+\frac{y^2}{5}+\frac{z^2}{25}=1$ and $z=x+y$

Your maximum is correct but I think in the minimum you have a typo .
Your solution is without using Langrange multipliers arquady ?

[quote="silouan"]
Your solution is without using Langrange multipliers arquady ?[/quote]
No. Middle school!
arqady :)

Could you post your solution arquady? Please....


for function $ f:\mathbb{N}\rightarrow\mathbb{N}$ we have:
$ I)\ \forall x,y\in\mathbb{N}$ we have : $ f(xy)\equal{}f(x)\plus{}f(y)\minus{}1$
$ II)$ there are finite x that $ f(x)\equal{}1$
$ III)\ \ f(30)\equal{}4$

find $ f(2)$.

Put $ x\equal{}y\equal{}1$ then $ f(1)\equal{}1$
If there exist $ a\in\mathbb{N},a\not \equal{} 1$ so that :$ f(a)\equal{}1$.

Put $ x \equal{} y\equal{}a$ then $ f(a^{2}) \equal{} 1$ and $ a^{2}\not \equal{} a$ then the set $ \mathbb{A}\equal{}\{x\in\mathbb{N}|f(x) \equal{} 1\}$ has infity elements, contracdiction .

So there is only natural $ a\equal{}1$ so that $ f(a) \equal{} 1$.

We have $ f(30) \equal{} f(2)\plus{}f(3)\plus{}f(5)\minus{}2 \equal{} 6$ then $ f(2)\plus{}f(3)\plus{}f(5) \equal{} 6$.

Because $ f(a)\ge 2,\forall x\in\mathbb{N},a\ge 2$ so we get :$ f(2)\equal{}1$

[code]edited [/code]

This problem quite easy but Quangpbc remember $ N$ is set of positive integer.You can not let $ x\equal{}0$

note that if $ a\not\equal{}1$ and $ f(a)\equal{}1$ then put $ x\equal{}y\equal{}a$ we get that:
$ f(a^{2})\equal{}2f(a)\minus{}1\equal{}1$ then put $ x\equal{}a^{2},y\equal{}a$ we ge that $ f(a^{3})\equal{}1$ and so on,we get infinitely many numbers $ t$ s.t $ f(t)\equal{}1$ which is a contradiction,so for $ a\not\equal{}1$ we get that $ f(a)\not\equal{}1$...
now we have:
$ f(30)\equal{}f(2.15)\equal{}f(2)\plus{}f(15)\minus{}1\equal{}f(2)\plus{}f(3)\plus{}f(5)\minus{}2$
so we get that:
$ 6\equal{}f(2)\plus{}f(3)\plus{}f(5)$ but as we said at the begining of the solution,$ f(2),f(3),f(5)>1$ so we get that:
$ f(2)\equal{}f(3)\equal{}f(5)\equal{}2$...

[quote="TTsphn"]This problem quite easy but Quangpbc remember $ N$ is set of positive integer.You can not let $ x \equal{} 0$[/quote]

Oh, i am sorry  :blush: . I don't know it  :(


Να βρεθεί η ελάχιστη τιμή του $M$ έτσι ώστε να ισχύει

$ab(a^4+b^4)\leq M(a+b)^6$

για κάθε $a,b\geq 0$ .

Stelios...

Βρήκα ότι  $M=\frac{5\sqrt{10}-14}{27}$ .
Έλυσα το πρόβλημα με τη μέθοδο των συμμετρικών παραστάσεων πολυωνύμων .
Έστω $a+b=s$  και  $ab=p$ .Επίσης λόγω ομοιογένειας(μπορούμε εύκολα να διαπιστώσουμε ότι η παράσταση είναι ομογενης αν θέσουμε $a=kx$ $b=ky$  $c=kz$ τότε το $k$ απαλοίφεται ;) και η παράσταση παραμένει αναλοίωτη) θέτουμε $a+b=s=1$ .(Και χωρίς να το υποθέσουμε λύνεται ομοίως αλλά με λίγες περισσότερες πράξεις ;) )
Είναι $a^4+b^4=s^4-4ps^2+2p^2$ .Άρα θέλουμε να δείξουμε ότι

$p(s^4-4ps+2p^2)\leq Ms^6$  ή $p(1-4p+2p^2)\leq M$  αφού $a+b=1$ .
Άρα αρκεί να βρούμε το μέγιστο της συνάρτησης $f(x)=2x^3-4x^2+x$  με το x να παίρνει τιμές  $0<x\leq\frac{1}{4}$  από ΑΜ-ΓΜ .$f'(x)=6x^2-8x+1$  και έχει ρίζες τις
$\frac{4-\sqrt{10}}{6}$ kai $\frac{4+\sqrt{10}}{6}$ . Όμως  $0<x\leq\frac{1}{4}$ έχουμε ότι f γνησίως αύξουσα στο $\left(-\infty,\frac{4-\sqrt{10}}{6}\right]$  και f γνησίως φθίνουσα στο $\left[\frac{4-\sqrt{10}}{6},\frac{1}{4}\right]$  άρα η f παίρνει τη max τιμή για $x=\frac{4-\sqrt{10}}{6}$  και
$f(\frac{4-\sqrt{10}}{6})=\frac{5\sqrt{10}-14}{27}$

Τι θα λέγατε για τη γενίκευση ??

$ab(a^{n-2}+b^{n-2})\leq M(a+b)^n$     :D

[quote="RAKAR"]Sιlουανε νομιζω οτι υπαρχει λαθος στη λυση σου.
Εγω διαιρεσα με β^6 ,εθεσα α/β=x και βγαινει Μ περιπου0,067.(βρισκω ακροτατα και μονοτονια της συναρτησης που προκυπτει με το x)
Ως αντιπαραδειγμα α=1/6 β=5/6 δινει ΑΤΟΠΟ!

ΡΑΛΛΗΣ :D[/quote]

Ναι βγαίνει $M= \frac{5 \sqrt{10}-14}{27} = 0.06708...$ .

Για να το λέτε και οι δύο  :blush:  .Κάπου θα έκανα λάθος.
Πού όμως ??Μπορεί να το βρει κάποιος ??

Λοιπόν το βρήκα το λάθος . Έκανα λάθος σε μια δευτεροβάθμια στις ρίζες  :blush:
Δείτε παραπάνω τη διόρθωση και πείτε μου αν πλέον είναι σωστό.

Το ίδιο αποτέλεσμα βρήκα και εγώ.Είναι πράγματι $M= \frac{5 \sqrt{10}-14}{27}$ και η ισότητα ισχύει αν και μόνο αν
$ab=\frac{4-\sqrt{10}}{6}(a+b)^2$ !

Παρατήρηση:Η ανισότητα $ab(a^4+b^4)\leq \frac{5 \sqrt{10}-14}{27}(a+b)^6$ ισχύει για κάθε $a,b \in R$ και όχι μόνο για θετικούς.

Stelios... :yup:

Και τώρα η Γενίκευση:

Να βρεθεί ,για κάθε $n \in N$, η ελάχιστη τιμή του $M$ έτσι ώστε να ισχύει

$ab(a^{n-1}+b^{n-1})\leq M(a+b)^{n+1}$

για κάθε  $a,b\geq 0$ .

Stelios...

[quote="stedes"]

Να βρεθεί ,για κάθε $n \in N$, η ελάχιστη τιμή του $M$ έτσι ώστε να ισχύει

$ab(a^{n-1}+b^{n-1})\leq M(a+b)^{n+1}$

για κάθε  $a,b\geq 0$ .

[/quote]

Έχει κανείς καμιά ιδέα ή να γράψω τη λύση μου και να βάλω την [b]πιο γενικευμένη[/b] μορφή;

Στέλιο αν θα μπορούσες βάλε την εντελώς γενικευμένη μορφή αλλά μην βάζεις λύση σε αυτή γιατί δεν εχω προλάβει ακόμα να την προσπαθήσω . ;)

[u][b]Η Γενίκευση της Γενίκευσης[/b][/u]

Αν  $a,b\geq 0$  με  $a^k+b^k=c$(σταθ.)   για κάποιο  $k \in N^{*}$ ,τέτοιο ώστε  $k<\frac{n^2}{n+2}$ ,
να βρεθεί η μέγιστη τιμή  της παράστασης:

$A=ab(a^n+b^n)$   ,συναρτήσει των $n,k \in N^{*}$

Stelios...

Λοιπόν η παραπάνω γενίκευση είναι ειδική περίπτωση της παρακάτω γενικοτέρης  για τρεις μεταβλητές την οποία είχα παραθέσει σε άλλο τόπικ.... ;)  . Η ανισότητά σου προκύπτει για c=0.
Έστω $n,a,b,c>0$ με   $a+b+c=1$.
$F_{n}(a,b,c)=a^{n}(b+c)+b^{n}(c+a)+c^{n}(a+b)$
Τότε έχουμε
a)με $n \le 3$   $maxF_{n}(a,b,c)=max(\frac{2}{3^{n}},\ \frac{1}{2^{n}})$

b)με  $n >3$ τότε  $max F_{n}(a,b,c)=\frac{t^{n}+t}{(t+1)^{n+1}}$ με το  $t$ να είναι η ρίζα της εξίσωσης  $t^{n}-n\cdot t^{n-1}+n\cdot t-1=0$ που βρίσκεται στο $(0,1)$.

Εν μέρει σωστός ,αφού για $n\leq 3$ είναι $max=\frac{1}{2^n}$  ;)  (μιλάω για τις δυο μεταβλητές).

Στη γενικότερη περίπτωση όμως με το $k$ το να βρεις το μέγιστο είναι πιο δύσκολο ,αφού υπάρχει μια σημαντική διαφορά σε σχέση με την περίπτωση των τριών μεταβλητών  :!:

Μπορούμε να δείξουμε ότι το μέγιστο επιτυγχανεται όταν η μία μεταβλητή είναι μηδέν .Αυτό αποδεικνύεται με SMV (αρκετές πράξεις όμως..... :( )


Set A consists of 4 consecutive integers whose sum is 18. Set B consists of 6 consecutive integers whose sum is 9. How many integers are members of both A and B?

We know 1+2+3+4=10, so it is obvious that Set A won't start with 1. Going farther, we know 1+2+3+4+5+6=21, and we notice that we are three away from 18 and that 1+2=3. Thus, Set A is made up of the numbers 3,4,5,6.

Like we did above, we know that 1+2+3+4=10, so we want to get rid of the first 1. Since the numbers have to be consecutive, Set B is made up of the numbers -1,0,1,2,3,4.

Thus, the numbers 3 and 4 are both members in set A and set B.

Or we can do this algebraically.

Set $A$:

Let $n$= the first number of the consecutive sequence

$n+(n+1)+(n+2)+(n+3)=18$

$n=3$

Thus the consecutive integers are $3$, $4$, $5$, $6$.

Set $B$:

$n+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=9$

$n=-1$

$-1$, $0$, $1$, $2$, $3$, $4$

It's evident that $A$ and $B$ only share $\boxed{2}$ integers.


I dont know I post this problem in right topic or not :?
we put 22 points in the interval[0,1], we can replace two points and put their midpoint in return,prove that we are able to  do this operation 20 times then we get the situation that the distance between two remain points doesnt exceed 0.001?

There's something I don't quite get: do you mean to say that there is one such sequence of moves? From what you've written I understand that we must show that no matter whaich $20$ moves we make, we get the conclusion. However, this isn't true if we put $11$ points really close to $0$ and the other $11$ really close to $1$ and make $10$ moves among the ones close to $0$ and the other $10$ among those close to $1$.


[quote="airrafer"]A(1) = m

A(n+1) = 3*A(n) - 3*A(n)^2

find A(n)[/quote]

I'm very surprised since I think there is no closed form for this sequence (see below). Where is this problem coming from ? (what contest / book / training session ?).

Setting $ b_n\equal{}3a_n\minus{}\frac 32$ and so $ a_n\equal{}\frac{b_n}3\plus{}\frac 12$, we get $ b_{n\plus{}1}\equal{}\frac 34\minus{}b_n^2$

And, in my kowledge (but maybe i'm wrong), the sequence $ u_{n\plus{}1}\equal{}a\minus{}u_n^2$ has a closed form only for $ a\equal{}0$ and $ a\equal{}2$

So ....  :(


i know i have to use the distance formula but i am super confuse with settng x=0

find all of the points on the y axis whose distance from the point (-3,5) is equal to 6

then i need to discribe what a graph would look like if it was symmetric about the  x axis

im so confused

[quote="kali"]i know i have to use the distance formula but i am super confuse with settng x=0

find all of the points on the y axis whose distance from the point (-3,5) is equal to 6

[/quote]

Do not think at first about the distance formula. Try to make a figure of what's going on.  Then probably you may use just the Pythagorean theorem. You will see that is easy. By the way there are going to be just two points.

[quote="kali"]

then i need to discribe what a graph would look like if it was symmetric about the  x axis

im so confused[/quote]

Still you need to draw a picture and observe how the values of y are.

[quote="kali"]find all of the points on the y axis whose distance from the point (-3,5) is equal to 6[/quote]

This sounds like a circle. $ ( \minus{} 3,5)$ is the center and $ 6$ is the radius.

EDIT: The graph of a circle will most likely cross the Y-axis 2 times.


Jack's end of the year exam came. He found that he hadn't studied at all, and couldn't do one single question. Since no points would be taken off if he gets it wrong, he decided to guess all of his answers. There are 25 multiple choice questions, and each question has 5 choices. what is the probability that jack is going to get 5 questions right?

[hide]There are $\binom{25}{5}$ ways in which 5 questions can be right.

The probability that a selected 5 is right and the rest are wrong is:

$\left(\frac15\right)^5 \times \left(\frac45\right)^{20}$

After you mulitiply this be $\binom{25}{5}$, you get a small number in which I have no time to figure out right now.[/hide]

This type of probability is an extension of the binomial theorem.

(25C5)*(1/5)^5*(4/5)^20

approx. equal to [b]1/5[/b]


Let $ a\geq 1,\ b\geq 1\ c\geq 1$ such that $ (a^2\minus{}a\plus{}1)|a\minus{}2|\plus{}(b^2\minus{}b\plus{}1)|b\minus{}2|\plus{}(c^2\minus{}c\plus{}1)|c\minus{}2|\equal{}3.$

Prove that $ a^3\plus{}b^3\plus{}c^3\plus{}a\plus{}b\plus{}c\geq a^2\plus{}b^2\plus{}c^2\plus{}3.$

It seems the second condition is useless but that is a way that can make solvers dizzy  :)

$ \sum_{cyc} a^3 \ge \sum_{cyc}a^2$

$ a\plus{}b\plus{}c \ge 3$

You are right. :blush:

Here is my solution:

$ \sum (a\minus{}|a\minus{}2|)(a^2\minus{}a\plus{}1)\geq 0$ for $ a,\ b,\ c\geq 1.$

Hi kunny! you're problem is obvious without the condition,
we have $ a^3\geq a^2$ and $ b^3\geq b^2$ and $ c^3\geq c^2$  because $ a,b,c\geq 1$ then $ a^3\plus{}b^3\plus{}c^3\geq a^2\plus{}b^2\plus{}c^2$ and from Am-Gm $ a\plus{}b\plus{}c\geq 3\sqrt[3]{abc}\geq 3$
with summation we find you're inequality, equality holds when $ a\equal{}b\equal{}c\equal{}1$


inequalities
geometry
3D geometry
octahedron

[color=darkblue]Find $ (x;y;z)$ of this system: $ \begin {cases} x\plus{}y\plus{}z\equal{}1\\x^4\plus{}y^4\plus{}z^4\equal{}xyz \end {cases}$[/color]

[hide="Answer"]

$ \frac {1}{3} \equal{} \frac {x \plus{} y \plus{} z}{3} \geq \sqrt [3] {xyz} \implies \frac {1}{27} \geq xyz$

$ \sqrt [4]{\frac {x^4 \plus{} ^4 \plus{} z^4}{3}} \geq \frac {x \plus{} y \plus{} z}{3} \equal{} \frac {1}{3} \implies x^4 \plus{} y^4 \plus{} z^4 \geq \frac {1}{27}$

Transitively, $ x^4 \plus{} y^4 \plus{} z^4 \geq xyz$, with equality iff. $ x \equal{} y \equal{} z$.

$ \therefore (x,y,z) \equal{} \left( \frac {1}{3}, \; \frac {1}{3}, \; \frac {1}{3}\right)$[/hide]

[hide="Or..."] You can prove directly that $ x^4 \plus{} y^4 \plus{} z^4 \ge xyz(x \plus{} y \plus{} z)$ by AM-GM, Rearrangement, or even Muirhead (in order of amount of work).

[hide="The AM-GM solution"] Sum the inequalities

$ x^4 \plus{} x^4 \plus{} y^4 \plus{} z^4 \ge 4x^2yz$
$ x^4 \plus{} y^4 \plus{} y^4 \plus{} z^4 \ge 4xy^2z$
$ x^4 \plus{} y^4 \plus{} z^4 \plus{} z^4 \ge 4xyz^2$.[/hide] [/hide]

Before using AM-GM, one should check that the equation system has no roots in negative reals.

Consider those solutions in the first octant. The second equation is symmetric across the planes x,y,z=0, so we can bound the graph of the second equation in the octahedron of points |x|+|y|+|z| = 1. However, one easily sees that the octahedron is strictly under the plane x+y+z=1 in the other 7 octants so there can be no solutions.

[color=darkblue]$ \bullet$ Aply AM-GM we get:

$ x^4 \plus{} y^4 \plus{} z^4 \equal{} \frac {x^4 \plus{} y^4}{2} \plus{} \frac {y^4 \plus{} z^4}{2} \plus{} \frac {z^4 \plus{} x^4}{2}$

$ \ge x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2\equal{} \frac {x^2y^2 \plus{} y^2z^2}{2} \plus{} \frac {y^2z^2 \plus{} z^2x^2}{2} \plus{} \frac {z^2x^2 \plus{} x^2y^2}{2}$

$ \ge xy^2z \plus{} yz^2x \plus{} zx^2y \equal{} xyz(x \plus{} y \plus{} z)$

$ \bullet$ But $ x \plus{} y \plus{} z \equal{} 1$ $ \Longrightarrow x^4 \plus{} y^4 \plus{} z^4 \ge xyz$

$ \bullet$ So that equation $ \Longleftrightarrow \begin {cases} x \equal{} y \equal{} z \\
x \plus{} y \plus{} z \equal{} 1\end {cases}\Longleftrightarrow x \equal{} y \equal{} z \equal{} \frac {1}{3}$[/color]


An $ n\times n$ complex matrix $ A$ is irreducible if and only if its directed graph $ G(A)$ is strongly connected.

You may want to see section 29.5 of [i]Handbook of Linear Algebra, CRC Press[/i], Edited by Prof. Hogben for the definitions you need and reference to the proof of your statement.


A runner is jogging at a steady x=5km/hr.  When the runner is 5 km from the finish line, a bird begins flying from the runner to the finish line at x=10km/hr(2 time as fast as the runner).  When the bird reaches the finish line, it turns around and flies back to the runner. Even though the bird is a dodo, we will assume that it occupies only one point in space, i.e. a zero length bird.  How far does the bird travel? Answer in units of km.

[hide="dodo?"]

Okay, so we have by the time the bird get's to the finish line ($ \frac{1\;\text{hour}}{2}\equal{} 30\;\text{mins}$)
which will happen 30 minutes into the 10km run, the runner will be at $ 5\text{km/hr}\cdot\frac{1}{2}\;\text{hr}\equal{} 2.5\text{kms}$. Next, the bird flies back at 10kms/hr and the runner runs from 2.5kms toward the bird at 5km/hr:

$ \frac{2.5}{10\plus{}5}\equal{} .1\overline{66}$
$ 10\cdot.1\overline{66}\equal{} 1.\overline{66}$

Finally, we add:

$ 5\text{km}\plus{}1.\overline{66}\text{km}\equal{} 6.\overline{66}\text{km}$

Hence our answer is $ 6.\overline{66}$

[/hide]

[quote="Nexmus"][hide="dodo?"]

Okay, so we have by the time the bird get's to the finish line ($ \frac{1\;\text{hour}}{2}= 30\;\text{mins}$)
which will happen 30 minutes into the 10km run, the runner will be at $ 5\text{km/hr}\cdot\frac{1}{2}\;\text{hr}= 2.5\text{kms}$. Next, the bird flies back at 10kms/hr and the runner runs from 2.5kms toward the bird at 5km/hr:

$ \frac{2.5}{10+5}= .1\overline{66}$
$ 10\cdot.1\overline{66}= 1.\overline{66}$

Finally, we add:

$ 5\text{km}+1.\overline{66}\text{km}= 6.\overline{66}\text{km}$

Hence, our answer is $ \boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{6.\overline{66}\text{km}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

[/hide][/quote]

Please refrain from making extremely large boxes like that.

This problem is ambiguous. Are you telling us that the bird does that process infinitely (which is the classic problem)?

Or, does the bird only do the process once?

[hide="Solution for first case"]The runner will reach the finish line in 1 hour, so the bird has one hour to be squished. So, it will fly for a total time of 1 hour. Hence, it flies 10 kilometers.[/hide]

[hide="Solution for second case"]First, the bird flies from the runner to the finish line, which is a distance of 5 km. By this time, the runner has ran $ 2.5$ kilometers. So, the bird and runner are $ 2.5$ kilometers apart, and they are approaching each other at a speed of $ 15$ km/hr (combined speed). So, the reach each other in $ 1/6$ hour. By this time, the bird has traveled $ 10/6 = 5/3$ km. So, the total length the bird has ran is $ 5+5/3 = 20/3$ km.[/hide]


There are cards labeled from 1 to 2000. The cards are shuffled and placed in a pile. The top card is placed on the table, then the next card at the bottom of the pile. Then the next card is placed on the table to the right of the first card, and the next card is placed at the bottom of the pile. This process is continued until all the cards are on the table. The final order (from left to right) is 1, 2, 3, ... , 2000. In the original pile, how many cards were above card 1999?


linear algebra
matrix
vector
induction
superior algebra
superior algebra solved

A_1,A_2,....,A_n matrix in Mn(C) nilpotent

A_i.A_j = A_j.A_i for any i,j

Show that A_1.A_2....A_n  = 0

Oral examination
--------------------------------------------------
A matrix B is nilpotent if there is an positive integer q such that

B^q = 0 and B^(q-1) =/=0

Is n-the number of matrices the same as n from Mn(C)?

Yes n is the same for A_1,...,A_n and  for Mn(C)

I am sure there is an easier solution then this, but here we are:

Consider the matrix as vectorial space endomorfisms.
We have f1,f2,...,fn : C^n -> C^n such that fi o fj = fj o fi and fi^n=0
We want to pruve that f1of2o...ofn=0

Lema: if f,h are two endomorfisms of a vectorial space V such that foh=hof and f is nilpotent then dim(foh(V))<=dim(h(V)) and if they are equal they are equal to 0

let S1=h(V) and S2=(foh)(V)
S2=h(f(V)) \leq h(V) =S1  meas S2 is a subspace of S1
Let see that if S1<>0 then S2 can't be S1

Take x from S1, x<>0
and i>0 such that  f^j(x)<>0 any j < i and f^ i(x)=0;
Notice that x, f(x), f^2(x)... f^(i-1)(x) are liniar independent.

(take  a1x+a2f(x)+..+aif^(i-1)(x)=0 then
         a1f^(i-1)(x)+a2f^ i(x)+...+aif^2(i-1)(x)=0  => a1f^(i-1)(x)=0 so
a1=0 and so on)

complet {x, f(x),...,f^(i-1)(x)} to a base in S1:
{y1,...,yl,x,f(x)...f^(i-1)(x)}
S2 will be generated by f(y1),f(y2)..... f^ i(x) which is 0 so dim S2<dim S1

So we pruve that lema
Suppose that (f1of2o..ofn)(C^n)<>0
We have:

n>dim(f1(C^n))>dim((f1of2)(C^n))>...>dim((f1o..ofn)(C^n))>0 which is imposible.

Here another solution

Let E an C-vector space with dim_(C)E = n

Lemma 1: Let u :E--->E, C-linear map nilpotent, and F subspace of E
such that u(F)  included in F, F=/={0}
Then dim u(F) < fim F

Proof:If dim u(F) = dimF then u_|F is bijectif, so
(x in F x=/=0) ==> u^n(x) =/=0 we get a contradiction that u is nilpotent

Lemma 2: Let g_1,...g_n linear maps E-->E ,g_iog_j = g_jog_i
and g = (g_1^k_1)o(g_2^k_2)o...o(g_n^k_n)
Then for any i, Im g is stable by g

Proof:
For any x in E , g_i(g(x)) = g(g_i(x)) is in Im g so
g_i(Img) is included in Im g

Now let U_1,...,U_n (linear maps associate to A_i)
by the lemma 2 for any k in {1,2,...,n-1}

U_koU_(k+1)o...oU_n(E) is included in  U_(k+1)o...oU_n(E)

with U_n(E) incuded in E taking the dimension we get

dim U_koU_(k+1)o...oU_n(E) <= dim U_(k+1)o...oU_n(E)

the inegality is strict, and by the lemma 1 if the dimension od the RHS
is =/=0, by induction we have

dim U_(k+1)o...oU_n(E) <= k

So dim U_1o...oU_n(E) = 0

U_1o...oU_n = 0


If $w$ is one of the imaginary roots of the equation $x^3 = 1$, then the product
\[ (1 - w + w^2)(1 + w - w^2) \]
is equal to
\[ \text{(A)}\ 4\ \ \text{(B)}\ w\ \ \text{(C)}\ 2\ \ \text{(D)}\ w^2\ \ \text{(E)}\ 1 \]

[hide="My answer and explanation"]I got the answer of 4

I solved my answer by first factoring the given equation

$x^2 - 1 = (x-1)(x^2 + x +1)$

Combining $(1 - w + w^2)(1 + w - w^2)$ gives $(-w^4 + 2w^3 - x^2 +1)$

The roots for $(x-1)(x^2 + x +1)$ are $1, -.5 + 3i/\sqrt{2}, and -.5 + -3i/\sqrt{2}$

Plugging in either of those into $(-w^4 + 2w^3 - x^2 +1)$

Gives $4$

I had to use a calculator for the last part. I never was good with imaginary numbers. Perhaps someone could post the correct way of solving it.[/hide]

Heres how I did it:

[hide]We see that the factors are really a difference of two perfect squares.  (1-(w-w^2))(1+(w-w^2)).  So expanding it we get 1-(w-w^2)^2.  Expanding this we get 1-(w^2-2w^3+w^4).  Now since w^3=1, we can rewrite this as 1-(w^2-2+w^4).  So our final equation is equal to -w^2+-w^4+3.  Since w^3=1 then (w^3)w=w=w^4.  So -w^2-w+3.  Now if we call our solution z, -z=w^2+w-3=w^2+w+1-4.  If w^3=1  then w^2 +w +1 must equal 0.  So -z=-4 so z=4.[/hide]

[hide="peek if you like"]If you know $1+\omega+\omega^2 =0$ then first  factor is $(-2\omega)$ and second factor is $(-2\omega^2)$ giving the product $4$ as $\omega^3 = 1$ :) [/hide]

[quote="ankur87"][hide="My answer and explanation"]I got the answer of 4[/hide][/quote]After I read that it took about 20 seconds to find the solution that I spent 10 minutes not seeing at all. Thanks.

[hide]By the conditions of the problem we have $w^3=1\Rightarrow{w}^3-1=0\Rightarrow(w-1)(w^2+w+1)=0\Rightarrow{w}^2+w+1=0,$ since $w\ne1$ (it is one of the two complex solutions). Thus \[(1-w+w^2)(1+w-w^2)=1-(w-w^2)^2=1-(w^2-2w^3+w^4)=3+(-w^2-w^4)=3+(-w^2-w)=3+1=4.\]Edit: Or, for a much simpler solution see the post above^  :D. [/hide]


What is the typical level of math knowledge of a person doing a math project at RSI? Would Multivariable Calculus be enough?

Also, on the application, it lists Technical Skills, and then Beginning, Intermediate, and Advanced. What do those mean? (I'll use Mathematica as an example). In other words, if I was checking that I knew Mathematica, what would constitute as Beginning skill, Intermediate skill, and Advanced skill?

Most of the math you'll end up using at RSI is things outside the standard curriculum (so if your question is "do I need to know complex analysis?" the answer is [b]of course not![/b]).  Fortunately, you have some opportunity to learn it (although not much time).  A brief overview of the projects I remember in my year:

- Me:  [url=http://en.wikipedia.org/wiki/Elliptic_curve]Elliptic curves[/url] in three dimensions.  I spent the first two weeks learning the background I needed.  (What I ended up proving didn't relate directly to most of the background I learned, but that's okay :P )
- David Ng:  [url=http://en.wikipedia.org/wiki/Tropical_geometry]Tropical geometry[/url].  He didn't get any results, but not for lack of trying.
- Alex Chen:  [url=http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6WKD-4HYN0FG-3&_user=10&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=55d285ec28cfd6fbb32b9f4deb3ded51]Generalized base representations[/url].  He ended up improving on the results in that paper.
- Paul Kominers:  [url=http://en.wikipedia.org/wiki/Combinatorial_game_theory]Combinatorial game theory[/url].  Doesn't require much background; fun to play with, from what I saw.
- Philip Hu:  The [url=http://www.math.ubc.ca/~ilaba/kakeya.html]Kakeya Problem[/url] over finite fields.  Also no results; again, not for lack of trying.
- Sana Raoof:  [url=http://en.wikipedia.org/wiki/Knot_theory]Knot theory[/url].

If I am making a point here (aside from the fact that the standard curriculum doesn't play heavily into this), it is that what is more important is a willingness to learn and an ability to learn quickly.  :)

As to your second question:  I don't really know.  I sort of guessed.

Only on an individual basis (and most of them, as far as I know, have not chosen to share their papers).  Some Rickoids end up publishing their papers, and a lot of them get submitted to Siemens and Intel, so having these papers available online would be problematic.

oh right...but they'll turn up eventually then...


Is there a way to prove geometry without using the statement vs reason chart?

For example, how can I prove that two triangles are congruent using numbers in place of statement and reason charts?

You can write a paragraph proof (the most common style of proof by far) or a coordinate proof.

[hide="paragraph proof example"]Given two triangles, $\triangle ABC$ and $\triangle DEF$, and $m\angle A=m\angle D$, $AB=DE$, and $m\angle B=m\angle E$, prove that $\triangle ABC\cong\triangle DEF$.

We have a side and two angles given, thus by $ASA$ congruence, $\triangle ABC\cong\triangle DEF$.[/hide]

of course, that still has statements and reasons. the only difference is that it isn't a chart.

Yes, the different proof techniques are just different ways of saying the same thing, but paragraph proofs are best when explanation is needed.  The chart-like proofs are only useful to someone else who has taken Geometry, because only something like "Substitution" is given as a reason.

How can you prove that two triangles are congruent using numbers or coordinate geometry?

What I learned from my previous teacher is that he said one of the reasons for doing line statement type of proofs is to emphasize deductive reasoning, which is essentially how mathematics is built, an indispensable concept to learn.

You can apply analytical skills like the distance formula, trigonometry, etc...