trigonometry
inequalities
inequalities proposed

Let $ a,b,c,d>0$, s.t. $ \dfrac{1}{1\plus{}a^4}\plus{}\dfrac{1}{1\plus{}b^4}\plus{}\dfrac{1}{1\plus{}c^4}\plus{}\dfrac{1}{1\plus{}d^4}\equal{}1$. Prove that:

$ abcd\ge 3$

A trigonometric substitution can help here.
Let $ a\equal{}\sqrt{\tan{\alpha}},b\equal{}\dots$
so $ \boxed{\sum\cos^2{\alpha}\equal{}1}$.
and we need to prove that
$ \prod\sin{\alpha}\geq 9\prod\cos{\alpha}$.
Due to the boxed relation it follows that $ \sin^2{\alpha}\equal{}\sum_{\beta,\gamma,\delta}cos^2{\beta}\geq 3\sqrt[3]{\prod_{\beta,\gamma,\delta} \cos^2{\beta}}$
multiplying all similar inequalities,with respect to $ \beta,\gamma,\delta$, we get the desired inequality.

[quote="Inequalities Master"]Let $ a,b,c,d > 0$, s.t. $ \dfrac{1}{1 \plus{} a^4} \plus{} \dfrac{1}{1 \plus{} b^4} \plus{} \dfrac{1}{1 \plus{} c^4} \plus{} \dfrac{1}{1 \plus{} d^4} \equal{} 1$. Prove that:

$ abcd\ge 3$[/quote]
$ \dfrac{1}{1 \plus{} a^4} \plus{} \dfrac{1}{1 \plus{} b^4} \plus{} \dfrac{1}{1 \plus{} c^4} \equal{} \frac {d^4}{1 \plus{} d^4}\geq 3\sqrt [3]{\frac {1}{(1 \plus{} a^4).(1 \plus{} b^4).(1 \plus{} c^4)}}$
product of 4 similar inequalitis we have
$ abcd\geq 3$

[quote="Erken"]A trigonometric substitution can help here.
Let $ a \equal{} \sqrt {\tan{\alpha}},b \equal{} \dots$
so $ \boxed{\sum\cos^2{\alpha} \equal{} 1}$.
and we need to prove that
$ \prod\sin{\alpha}\geq 9\prod\cos{\alpha}$.
Due to the boxed relation it follows that $ \sin^2{\alpha} \equal{} \sum_{\beta,\gamma,\delta}cos^2{\beta}\geq 3\sqrt [3]{\prod_{\beta,\gamma,\delta} \cos^2{\beta}}$
multiplying all similar inequalities,with respect to $ \beta,\gamma,\delta$, we get the desired inequality.[/quote]
Nice substitution!!
In the same way: If $ \frac{1}{1\plus{}(a_1)^n} \plus{} ... \plus{} \frac{1}{1\plus{}(a_r)^n} \equal{} 1$ with $ a_1,...,a_r > 0$ then the minimum of $ a_1a_2...a_r$ is $ (r\minus{}1)^{\frac{r}{n}}$. (So the minimum $ abcd$ in this problem is $ (4\minus{}1)^{\frac{4}{4}} \equal{} 3$ :) )

[quote="Inequalities Master"]Let $ a,b,c,d > 0$ and $ \frac {1}{1 \plus{} a^4} \plus{} \frac {1}{1 \plus{} b^4} \plus{} \frac {1}{1 \plus{} c^4} \plus{} \frac {1}{1 \plus{} d^4} \equal{} 1,$ prove that $ abcd\ge 3.$[/quote]This problem is introduced as Example 2.2.7 on page 43 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://www.ewen.cc/cache/books/view/15/view-0202010000973315.htm]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)

With the same conditions, we have the stronger inequality

$ \left(\frac {1}{a} \plus{} \frac {1}{c}\right)\left(\frac {1}{b} \plus{} \frac {1}{d}\right)\leq\frac {4}{\sqrt {3}},$

with equality if and only if $ a \equal{} b \equal{} c \equal{} d \equal{} \sqrt [4]{3}.$

[hide=Solution]One may proof the above inequality $\left(\text{call }S=\left(\frac{1}{a}+\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{d}\right)\right)$.

First, by homogenization, let $w,x,y,z>0$ be real numbers such that $w+x+y+z=1$ and :
\[\begin{cases}a=\sqrt[4]{\frac{x+y+z}{w}}=\sqrt[4]{\frac{1-w}{w}}\\b=\sqrt[4]{\frac{w+y+z}{x}}=\sqrt[4]{\frac{1-x}{x}}\\c=\sqrt[4]{\frac{w+x+z}{y}}=\sqrt[4]{\frac{1-y}{y}}\\d=\sqrt[4]{\frac{w+x+y}{z}}=\sqrt[4]{\frac{1-z}{z}}\end{cases}\]

Then, expanding $S$ and using Cauchy–Schwarz we get :
\[S^2=\left(\frac{1}{ab}+\frac{1}{ad}+\frac{1}{bc}+\frac{1}{cd}\right)^2\leqslant4\left(\frac{1}{a^2}+\frac{1}{c^2}\right)\left(\frac{1}{b^2}+\frac{1}{d^2}\right)=P\]

And applying a second time Cauchy–Schwarz we obtain :
\[S^4\leqslant P^2=16\left(\frac{1}{a^2b^2}+\frac{1}{a^2d^2}+\frac{1}{b^2c^2}+\frac{1}{c^2d^2}\right)^2\leqslant64\left(\frac{1}{a^4}+\frac{1}{c^4}\right)\left(\frac{1}{b^4}+\frac{1}{d^4}\right)=Q\]

Thus :
\[Q=64\left(\frac{w}{1-w}+\frac{y}{1-y}\right)\left(\frac{x}{1-x}+\frac{z}{1-z}\right)\]

Using Lagrange multiplier, defined $L=f-\lambda g=Q-\lambda\left(w+x+y+z-1\right)$ hence :
\[\begin{cases}\frac{\partial L}{\partial w}=\left(\frac{x}{1-x}+\frac{z}{1-z}\right) \frac{1}{\left(1-w\right)^2}=\lambda\\\frac{\partial L}{\partial x}=\left(\frac{w}{1-w}+\frac{y}{1-y}\right) \frac{1}{\left(1-x\right)^2}=\lambda\\\frac{\partial L}{\partial y}=\left(\frac{x}{1-x}+\frac{z}{1-z}\right) \frac{1}{\left(1-y\right)^2}=\lambda\\\frac{\partial L}{\partial x}=\left(\frac{w}{1-w}+\frac{y}{1-y}\right) \frac{1}{\left(1-z\right)^2}=\lambda\end{cases}\]

From where we deduce $\boxed{w=y}\Longrightarrow\boxed{a=c}$ and $\boxed{x=z}\Longrightarrow\boxed{b=d}$. So :
\[S^4\leqslant64\times4\frac{1}{a^4b^4}\leqslant\frac{16^2}{9}\]

Since from above $abcd\geqslant3$.

Thus :
\[S^4\leqslant\frac{4^4}{\sqrt{3^4}}\Longleftrightarrow S\leqslant\frac{4}{\sqrt{3}}\]
[/hide].

Let $x=\frac{1}{1+a^4} \rightarrow a=\sqrt[4]{\frac{1-x}{x}}$
define $y,z,m$ similarly ,so $x+y+z+m=1$ and we need to prove that:
$ \frac{y+z+m}{x}\frac{x+z+m}{y}\frac{y+x+m}{z}\frac{y+z+x}{m}\ge 81$ which is just AM-GM .

Make the subtitution $ a=\sqrt[4]{\frac{y+z+t}{x}}$

the same thing for $b,c,d$ ,so we need only to prove that:

$ (y+z+t)(x+z+t)(y+x+t)(y+z+x) \ge 81xyzt$ which is just AM-GM .


Anybody know anything about it? I've never heard about it, but one of my friends was talking about it. I've never seen it on the college board website.

I would guess that he's talking about one of [url=http://www.collegeboard.com/student/testing/ap/scholarawards.html]these[/url].


Define the number :a(n)=[1+(2/n)]^n . Using any computer package,investigate the values of a(n),as n gets very large.

But what if you're smarter than a computer?

There's potential trouble here if you use floating-point approximate arithmetic.

I tried this in a spreadsheet with $n=3^{k}.$ Of course, we know (since we are smarter than the computer) that the limit is $e^{2}.$ The error in our expression looks like $\frac{C}{n}$ were $C$ sems to be somewhere between $-14$ and $-15.$ At least that's what's happening out to about $3^{15}.$ After that, the error starts bouncing around. The absolute error reaches a minimum (in this case) of about $9\times 10^{-8}$ for $n=3^{18}\approx 4\times 10^{9}.$ After that, the error starts increasing. For $n=3^{3}3\approx 6\times 10^{15},$ the original expression is about $11.8.$ For $n=3^{34}$ the original expressin is about $40.6$ And for larger $n$ the original expression is $1.$

All of that stuff at the end is an artifact of doing approximate arithmetic - our accuracy is being destroyed by roundoff.


induction
vector
linear algebra
linear algebra theorems

HI all, I am quite confused with the following statement:

Let T be a linear operator on V, and let k belong to the field F. Then, the sum of E_k is direct.
E_k denotes the eigenspace associated with the eigenvalue k.

From most of the proof I have seen, induction is always used. However, by using induction, we have assumed the countability of the field itself. Hence, how if the field itself is uncountable? To me, by assuming its countability through induction seems not right. Any idea?

Isn't your book implicitly assuming that vector space $ V$ is finite dimensional [i]when discussing eigenvalues[/i]? If $ V$ is finite dimensional, then $ T$ has only finitely many eigenvalues, and hence we can use induction.

It does not mention whether the vector space is finite-dimensional, so in this context, it may include infinite dimensional vector space as well.

It technically works in infinite-dimensional spaces, because we're actually only working with the span of finitely many eigenvectors at any one time.


A coin is tossed ten times, what is the likelihood of 3 or more heads in a row?

How would one solve this problem theoretically (sans simulations)?

i saw this one on the 2003 team scramble. IT was the exact same problem. I hope someone answers thie one .  :)

[hide="The general idea"] Consider the likelihood of $2$ heads, then $1$ head, when you've had $n$ tosses, and form a recursion. [/hide]

I get a probability of 520/1024.
Can someone try this problem and see if my answer is correct?


trigonometry
geometry
trig identities
Law of Sines
geometry proposed

[quote="Virgil Nicula"]Given is $ \triangle ABC$ for which $ C = 60^{\circ} + A$ and exists $ x\in(0,60^{\circ})$ so that $ B = 60^{\circ} + x$.

Consider the points $ \|\begin{array}{c} M\in (AC)\ ,\ m(\widehat {ABM}) = \frac x2 \\
 \\
N\in (AB)\ ,\ m(\widehat {ACN}) = 30^{\circ}\end{array}$. Prove that $ m(\widehat {ANM}) = x\ .$[/quote]
Nice problem!
[hide="Solution"]
Take point $ R$ on $ AC$ so that $ \angle RBC = 60$ and $ Q$ on $ AB$ so that $ \angle QCB = 60$.  Allow $ QC$ and $ BR$ to intersect at $ P$.  Now, we notice that $ \angle ABR = x$.  Thus, $ \angle ANM = x$ if and only if $ NM\parallel BR$.  Thus, $ \angle ANM = x\iff \frac {AN}{NB} = \frac {AM}{MR}$.  Now, we notice that $ \frac {AM}{MR} = \frac {AB}{BR}$ since $ BM$ bisects $ \angle ABR$ and thus $ \angle ANM = x\iff \frac {AN}{NB} = \frac {AB}{BR}$.  Now, we name $ \angle A = \alpha$.  Thus, $ \angle ACB = 60 + \alpha$.  Now, $ \angle ABC = 60 + x = 180 - \alpha - 60 - \alpha = 120 - 2\alpha\implies x = 60 - 2\alpha$.  Now, this implies that $ \angle ARB=\angle ACB+\angle RBC=60+60+\alpha=120+\alpha$, $ \angle ABR = 60 - 2\alpha$, $ \angle QCN = 30 - \alpha$, $ \angle NCB = 30 + \alpha$, $ \angle NBC = 120 - 2\alpha$, and $ \angle BNC = 30 + \alpha = \angle NCB$.  From here, we apply Law of Sines a few times.  We notice that
\[ \frac {AB}{BR} = \frac {\sin \angle ARB}{\sin \angle A} = \frac {\sin (120 + \alpha)}{\sin \alpha} = \frac {\sin (60 - \alpha)}{\sin \alpha}
\]
Also,
\[ \frac {AN}{NC} = \frac {\sin \angle ACN}{\sin \angle A} = \frac {\sin 30}{\sin \alpha} = \frac {1}{2\sin \alpha}
\]

\[ \frac {BN}{NC} = \frac {\sin (30 + \alpha)}{\sin (120 - 2\alpha)} = \frac {\sin (30 + \alpha)}{\sin (60 + 2\alpha)} = \frac {1}{2\cos (30 + \alpha)}
\]

\[ \frac {AN}{NB} = \frac {\frac {AN}{NC}}{\frac {NB}{NC}} = \frac {\cos (30 + \alpha)}{\sin \alpha} = \frac {\sin (60 - \alpha)}{\sin \alpha} = \frac {AB}{BR}
\]
and we are done.
[/hide]


In a triangle $ABC$ , the bisector of $B$ intersect $AC$ at $M$ while the bisector of $C$ intersect $AB$ at $N$ . If $BM=CN$ , prove that $ABC$ is isosceles .

Well , I have a trigo solution for this one , but im finding a geometry solution . Anyone ?

The Chinese language in the document does not make trouble for you to understand the geometrical proof.

thats a beatiful geometry proof . Thanks shobber  :lol:


trigonometry
algebra
polynomial
binomial theorem
absolute value
complex numbers

What is the value of,

$ 2 \plus{} 8 ( \minus{}\textonehalf \plus{} \frac{(\imath \sqrt{3})}{2} )^8 \plus{} 2 ( \minus{}\textonehalf \plus{} \frac{(\imath \sqrt{3})}{2} )^9$.

Remember expansion is not an option because this is supposed to be solved in less than 2 mins.... and no advanced math....

Its a question from sample entry exam paper for LUMS SSE and is the last question from basic math section. [url]http://sse.lums.edu.pk/documents/admissions/Fall_2009/SSE%20Admission%20Test%20-%20Sample%20Questions.pdf[/url]

Use [url=http://www.artofproblemsolving.com/Wiki/index.php/De_Moivre%27s_Theorem]DeMoivre's Theorem[/url] after simplifying to $ 2 \plus{} 8(\cos{120^\circ} \plus{} i\sin{120^\circ})^8 \plus{} 2(\cos{120^\circ} \plus{} i\sin{120^\circ})^9$

$ \equal{} 2 \plus{} 8\cos{960^\circ} \plus{} 8i\sin{960^\circ} \plus{} 2\cos{1080^\circ} \plus{} 2i\sin{1080^\circ}$

$ \equal{} 2 \plus{} 8\left( \minus{} \frac12\right) \plus{} 8i\left( \minus{} \frac {\sqrt3}2\right) \plus{} 2(1) \plus{} 2i(0) \equal{} 2 \minus{} 4 \minus{} 4i\sqrt3 \plus{} 2 \equal{} \boxed{ \minus{} 4i\sqrt3}$.

I think this is taught in Algebra II but I'm not sure?

i'm sorry i forgot to mention.... de movire's theorem is not supposed to be used either because basic math section is supposed to be solved even by students of biology and other subjects.... infact $ \imath$ (iota) is defined before the question as $ \sqrt{\minus{}1}$ so the students solving the question are not even supposed to know what a complex number is....  :wink:

I think demoivre's theorem is completely appropriate here (but not necessary).  It is often taught in high school (I learned it 1st semester in Honors Precalculus.  Not sure if it is part of the regular precalc curriculum though).  Anyway, keep in mind that this is a science/engineering university - they will expect you to be good at math.

Anyway, let $ \omega \equal{} \minus{} 1/2 \plus{} i/2\sqrt {3}$. Note that $ \omega^3 \equal{} 1$ (that is, it is a root of $ x^3 \minus{} 1 \equal{} 0$, which can be factored as $ (x \minus{} 1)(x^2 \plus{} x \plus{} 1) \equal{} 0$).  Then the question simplifies to:

$ 2 \plus{} 8\omega^2 \plus{} 2$

and $ \omega^2 \equal{} \minus{} (1 \plus{} \omega)$ (either by expansion or from the polynomial above) so we have
$ 2 \minus{} 8 (1/2 \plus{} i/2 \sqrt {3}) \plus{} 2 \equal{} \minus{} 4i\sqrt {3}$

ok so that probably is the hardest question on the basic math test.  Anyway, chances are if it is supposed to be solved without much advanced math, you aren't going to use binomial theorem and then you would probably go ahead and compute the powers of $ \omega$ recursively, and run into $ \omega^3 \equal{} 1$.  At which point you have already calculated $ \omega^2$ and can just plug it all in.  So really it can be solved with easy algebra.  and if you try to speed up the exponentiation by squaring, you'll ge $ (\omega^2)^2 \equal{} \omega^4 \equal{} \omega$, so clearly $ \omega^3 \equal{} 1$.  So there are few ways to do this that will actually make it take too long since you'll probably stumble into this time-saving fact.

Mentioning that $ i\equal{}\sqrt{\minus{}1}$ is generally done as a reminder that the problem is talking about complex numbers.  Besides, if the problem mentions $ \sqrt{\minus{}1}$, it must assume that you understand what a complex number is.

facis: thanks! very nice method. i guess we cannot go more elementary than this.

just to clarify, here is the syllabus,
[url]http://sse.lums.edu.pk/documents/admissions/Fall_2009/SSE%20Admission%20Test%20-%20Syllabi.pdf[/url]
no mention whatsoever of complex numbers.... not even binomial.... so i guess facis's is the only way...

It really is a problem you should be able to tackle in under 2 minutes, we used to solve dozens of problems of this type.

First notice that the absolute value of $ \omega$ is one. Now $ \omega$ to any (integer) power is a complex number on a unit circles, i.e. its absolute value doesn't change. Since $ \omega$ subtends an angle of 120 degrees, $ \omega^7$ subtends the very same angle. $ \omega^8$ equals $ \minus{}1/2 \minus{} i\sqrt {3}$. $ \omega^9$ equals 1.

Now we can see that the value is:
$ \minus{} 4i\sqrt {3}$

i hate to be a ball buster, but again u r assuming the person solving the question knows about complex numbers, polar form, and to some extent argand diagrams.... :oops: facis's method is actually very short once you realize $ \omega^3 \equal{} 1$


linear algebra
matrix
inequalities
linear algebra unsolved

Let $A\in\mathcal l M_{3}(R)$ with $Tr(A)=Tr(A^{*})=0$. Show that : $\det(A^{9}+B^{3})\det(A^3+B)\geq 0$ for any $B\in\mathcal l M_{3}(R)$.

[i]$A^{*}$ is the adjoint matrix of $A$ [/i]

If $tr(A)=tr(A^*)=0$ then by Hamilton Cayley we get that $A^3=\det(A) I_3$. Let`s denote $\alpha=\det(A)$ then $A^3=\alpha I_3, A^9= \alpha^3 I_3$. It remains to prove  that $\det(B^3+\alpha^3 I_3)\det(B+\alpha I_3)\geq 0$. Let $\beta_1,\beta_2,\beta_3$ be the eigenvalues of $B$ then the eigenvalues of $B^3$ are $\beta_1^3,\beta_2^3,\beta_3^3$. So we must prove that:
$(\beta_1^3 +\alpha^3)(\beta_2^3+\alpha^3)(\beta_3^3+\alpha^3)(\beta_1+\alpha)(\beta_2+\alpha)(\beta_3+\alpha)\geq 0$.
Now to prove the inequality $(x^3+y^3)(x+y)\geq 0 , \forall x,y \in\Re$. We have that:
$(x^3+y^3)(x+y)=x^4+y^4+xy(x^2+y^2)=(x^2-y^2)^2 +xy(x+y)^2=(x+y)^2[(x-y)^2+xy]=(x+y)^2(x^2-xy+y^2)\geq 0$
so $(x^3+y^3)(x+y)\geq 0 , \forall x,y \in\Re$.(1)
Now assuming that all the eigenvalues of $B$ are real the conclusion is immediate.
Suppose that $\beta_2,\beta_3$ are non real then they are conjugated then $(\beta_2^3+\alpha^3),(\beta_3^3+\alpha^3)$ are also conjugated as well as $(\beta_2+\alpha),(\beta_3+\alpha)$ so their product is positive and using (1) for $\alpha,\beta_1$ which are of course real numbers the conclusion follows.  :)

actually we only need $\det(B^2-\alpha B +\alpha^2 I_3)>0$ which is quite obvious


Dear Chinese Mathlinkers.
I'd like problems and solutions in English of Chinese National Olympiad and Chinese Team Selection Tests.I like to solve problem from contests especially from your country.If any of you have it,I would like to exchange with you.Any links are also welcome.I'll give you many links to download books and articles from all over the world and every thing if I can.Thank you very much.
Respectfully.

iam from Morocco and i would like to have some problems to work too

Thanks Superman I'm very interested in sharing information.


prove that every natural  number can be wrriten as sum of for sguares .

This is Lagrange's theorem... :wink:

Pierre.

For the case of four squares, see [url=http://www.maths.mq.edu.au/~wchen/lnentfolder/lnent.html]this page,  chapter 5[/url]. For a general Waring's theorem, there is a proof for example in a book A. Y. Khinchin, [url=http://www.amazon.com/Three-Pearls-Number-Theory-Khinchin/dp/0486400263/sr=1-1/qid=1163445681/ref=pd_bbs_sr_1/102-3058430-0351349?ie=UTF8&s=books]Three pearls of number theory.[/url]

Well, if you haven't already tried this proof, try Minkowski's theorem.


Do you mean something like $ 10^y$?  This is just a computation of the solutions to $ x^3 \plus{} x \equiv 0 \bmod 10$.

Wait but if you mod out the 10, wouldn't it be congruent to 8 mod 10?


Well i dont think that we can actually get a value for this.
The max. which can be done is Taking series expansion of sina and then considering only the first two possibilities or using Descartes rule of signs  :maybe:

Please don't post the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180959]same problem[/url] twice.


Prove that if m,n,k are the positive integers satisfying $ 1\plus{}m\plus{}n\sqrt2\equal{}(2\plus{}\sqrt3)^{2k\minus{}1}$,then m is a perfect square.

You probably meant 2 instead of 3, or vice versa. Otherwise there are clearly no solutions.
The key seems to be Pell Equations - $ (2,1)$ supposed to be the fundamental solution of $ x^2\minus{}3y^2\equal{}1$, and the LHS is the $ 2k\minus{}1$th solution. But I need to see the correct version...

$ \left(m\plus{}1\right)^{2}\minus{}3n^{2}\equal{}1\Leftrightarrow m\left(m\plus{}2\right)\equal{}3n^{2}$. $ m\equiv 2^{2k\minus{}1}\minus{}1\equiv 1\bmod 3$ yields, that $ m$ is a perfect square.


Prove that $(k-2)(k-1)k(k+1)(k+1)$ is not a perfect square, when $k$ is a positive integer.

Is it (k-2)(k-1)k(k+1)(k+[u]2[/u])?

I think it's obvious because (x,x+1)=1.

I guess we'll assume $k>2$...

Well, if it's correct as posted, then it's sufficient to show $k(k-1)(k-2)$ is not a perfect square.  But $k-1$ is relatively prime to $k$ and $k-2$, so it then would have to be a perfect square.  So then $k=a^{2}+1$.  But then $k(k-1)=a^{4}-1$ which is not a perfect square unless $a^{4}=1 \implies k=2$...

Why would it be sufficient to show that the product of any 3 consecutive numbers is not a perfect square?

EDIT: Ok I see why you're saying that, but I think the original poster made a mistake. It should be the product of 5 consecutive numbers.

Actually it has been proven for $n>1$ that product of $n$ consecutive numbers is not a perfect power.

This problem has been discussed a  few times before

For example
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=kabootar&t=30932]here[/url]

I am cutting and pasting  my post from that thread, because I think it is one
of the funnest proof.

[hide]Pigeon hole principle:

Solution inspired by a famous hindi song which goes "kabootar kabootar ").. (kabootar  = pigeon BTW).

Get five kabootars and tie one number to each of them. $(n, n+1, n+2, n+3, n+4)$
Get 4 kabootar-pijaras and number them "1", "2","3", "6"

Now each kabootar would know that his number can be written as $(a \cdot x^{2})$  where $a$ is sqaure free. furthermore, as pointed out before ( none of the numbers have a common prime factor greater than 3) "$a$" must be one of the kabootar-pijara number.  (That is $a$  must be one of  the numbers of the set{1,2,3,6} )

Let every kabootar fly and get to the corrrosponding pijara.

4 kabootar-pijara's, 5 kabootars .. at least one pijara would contain two (or more) kabootars...

For $n>1$  , this is obviously a contradiction ( difference $a (x_{1}^{2}-x_{2}^{2})$  would be greater than 4). (one can easily check n=1 case (120 is not a  perfect square) .

QED!
P.S. (The above solution will work even if you do not know that kabootar is pigeon in hindi and even if kabootars can't do math)

PPS - Using similar logic you can prove the more general case of any number (say k) of consecutive integers. (With somve work one can prove that all a's would be distinct..etc) ( Narumi proved that this is true for k's less than 200 and Erdos proved for all the numbers .. in fact in 1975 he proved, as pointed out by pbornsztein above, that the product is not equal to any power too.[/hide]

You may have to look up in the previous thread, if some steps above need
further explanations.

The  problem is discussed also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6337&highlight=kabootar&start=20]here[/url]

The funny terminology in that proof was hard to understand  :wink:

Yeah, the problem is a folklore/classic one and is not considered  that "simple"  ...you may like to look at the previous thread(s)  to understand the proof fully and may also looks for some  alternative proof(s) (and there are  also some erroneous proofs and false starts in those threads.. :) ).

There is a proof by Myth in that thread that is compelete.
The pigeon-hole proof (which I posted) is rather simple .


Let $ABC$ be a triangle with centroid $G$,inscribed in circle $(O)$.The lines $AG$;$BG$:$CG$ meet the circle at $D;E;F$ respectively.Let $a;b;c$ denote the lengths of sides $BC;CA: AB$ of the triangle.Prove that:
$\frac{1}{GD}+\frac{1}{GE}+\frac{1}{GF}\leq \sqrt{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

This is a hard problem,very hard! :rotfl:

[quote="Jonhs"][color=darkred]Let $ABC$ be a triangle with centroid $G,$inscribed in circle $(O).$ The lines $AG,\,BG,\,CG$ meet the circle $(O)$ at $D,\,E,\,F$ respectively. Let $a,\,b,\,c$ denote the lengths of sides $BC,\,CA,\, AB$ of the triangle. Prove that:
$\frac{1}{GD}+\frac{1}{GE}+\frac{1}{GF}\leq \sqrt{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ :rotfl:  [size=150]Very easy problem![/size][/color] [/quote]
[color=indigo][b]Proof.[/b] Let $M=AG\cap BC,\,N=BG\cap CA,\,P=CG\cap AB$
We have: $MA\cdot MD=MB\cdot MC=\frac{a^{2}}{4}\Longrightarrow MD=\frac{a^{2}}{4m_{a}}$
Thus, $GD=GM+MD=\frac{m_{a}}{3}+\frac{a^{2}}{4m_{a}}\ge\frac{a}{\sqrt{3}}\Longrightarrow\frac{1}{GD}\le\frac{\sqrt 3}{a}$
Similarly, we also have: $\frac{1}{GE}\le\frac{\sqrt 3}{b},\,\frac{1}{GF}\le\frac{\sqrt 3}{c}$
And hence: $\frac{1}{GD}+\frac{1}{GE}+\frac{1}{GF}\leq \sqrt{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$[/color]


Let $ a,b,c\geq0$ and $ abc\equal{}1$. Prove that

$ (a\plus{}b\plus{}c)\plus{}3(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\minus{}\frac{1}{3})(\frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c})\geq15$

---------------
 :blush:  Panagiote LIGOURAS

Nice problem!
We have

$ 3 \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right) \minus{} 1 \ge \frac {4}{3} \left( \frac {1 \plus{} a}{a} \plus{} \frac {1 \plus{} b}{b} \plus{} \frac {1 \plus{} c}{c} \right)$

It suffices to prove that

$ \frac {4}{3} \left( \frac {1 \plus{} a}{a} \plus{} \frac {1 \plus{} b}{b} \plus{} \frac {1 \plus{} c}{c} \right)\left(\frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \right) \ge \frac{4}{3}\left( \frac {1}{\sqrt {a}} \plus{} \frac {1}{\sqrt {b}} \plus{} \frac {1}{\sqrt {c}} \right)^2 \ge 12$

It is trivial to prove that
$ a \plus{} b \plus{} c \ge 3$

We derive our inequality straightaway.


How do u prove the following using Intermediate value theorem

Let f[0,1]->[0,1] be a continuous function. show that there exists at least one x in [0,1] such that f(x)=x

Hint:look at g(x)=f(x)-x and apply intermediate value theorem

The problem statement is true if $ f(0) \equal{} 0$ or $ f(1) \equal{} 1$.  Suppose neither of these are true.  Then what must be true about $ g(0)$ and $ g(1)$?

One has $ g(0) \equal{} f(0) \minus{} 0 \geq 0$ and $ g(1) \equal{} f(1) \minus{} 1 \leq 0$, since $ 0 \leq f(x) \leq 1$ for all $ x$. So the conclusion is right (and trivial: the simplest fixed point result), no matter the values of $ f(0)$ and $ f(1)$. [color=red]Probably a careless read of the previous poster.[/color]
EDIT. [color=blue]Sorry, the ambiguity of plain language ... Yes, at second reading it's a hint.[/color]

tora wasnt questioning the result. He was giving a hint there.


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ \Phi_{n}$ be the $ n$-th ciclotomic polynomial, $ \Phi_{n}\in\mathbb{Q}[X]$, $ n\geq 2, n\in\mathbb{N}$.
Find $ Res_{X}(\Phi_{n}(X), X^{m} \minus{} 1)$.

Could you tell me you e-mail, I want to communicate with you directly. My e-mail:
shenrulin@hotmail.com


I've sure that I've already seen the following pb on the forum:
prove that there exists a countable set $E$ of points in $R^3$ such that:
1/$E$ is dense
2/no 4 of them are on a same plane

prove that the conclusion also holds with uncoutably many points.

Consider the collection $\mathcal C$ of all the dense subsets of $\mathbb R^3$ with no four coplanar points. We now from the previous problem (I also remember it) that it's non-void. Every chain (wrt inclusion) of elements from $\mathcal C$ is bounded from above: just consider the union of all the sets in the chain. It's also dense, and if it had four coplanar points, then these points would belong to one of the sets in the chain, which would yield a contradiction. Zorn's Lemma says that, in this case, $\mathcal C$ must have a maximal element.

Assume now that our maximal element is a countable set. Its points yield countably many planes, so we can add a point to this set which does not belong to any of those planes, thus getting a strictly larger set. Since our set was maximal, we have a contradiction. This means that the maximal element of $\mathcal C$ is uncountable.

nice ! and could we find such a set of point such that there is a bijection between R and this set ? (without the contium hypothesis  :D  )

Actually, grobber's maximal set is automatically of cardinality $c$. To prove it, one needs two statements. The first is the well-known statement that the cardinality of all triplets of elements of a given infinite set $A$ has the same cardinality as $A$. The second statement, whose proof I prefer to leave to someone else, is that you cannot fill the space with fewer than continuum planes. I also wonder whether there exists a construction independent not only of continuum hypothesis, but also of axiom of choice.

[quote="fedja"]you cannot fill the space with fewer than continuum planes. [/quote]

Wonder why I missed that when I initially posted.. :? :)

After we've got a family of planes with union $\mathbb R^3$, we can take another plane (if all the planes were in the initial family, then we would already have $c$ of them, and we would be done). This plane is covered by the lines formed by intersecting it with the planes in our family. Since these lines are at most as many as the planes in the family, it would suffice to show that a plane cannot be covered with $<c$ lines. We now repeat the procedure to decrease the dimension again, and reach the conclusion that it suffices to show that no line can be covered by $<c$ points, which is, of course, obvious :).


Prove that $ \frac{2}{9}\leq x^4\plus{}y^4\leq 8$ where $ 1\leq x^2\minus{}xy\plus{}y^2\leq 2$, $ x,y\in R$.

We have $ x^2\plus{}y^2\le 2\plus{}xy$, square both sides of this inequality, we obtain:
$ (x^2 \plus{} y^2)^2\le (2 \plus{} xy)^2\implies x^4 \plus{} y^4\le 8 \minus{} (xy \minus{} 2)^2\le 8$

[hide="see my solution"]
$ x^4\plus{}y^4\minus{}8\equal{}(x^2\plus{}y^2)\minus{}2(xy)^2\minus{}8\leq (2\plus{}xy)^2\minus{}2(xy)^2\minus{}8\equal{}\minus{}(xy)^2\plus{}4xy\minus{}4\equal{}\minus{}(xy\minus{}2)^2\leq0$ (1)
$ x^2\plus{}y^2\minus{}xy\leq\frac{3}{2}(x^2\plus{}y^2)$
$ \Leftrightarrow x^2\plus{}y^2\geq\frac{2}{3}$
with caushy: $ x^4\plus{}y^4\geq\frac{1}{2}(x^2\plus{}y^2)^2\geq\frac{1}{2}\times{4}{9}\equal{}\frac{2}{9}$ (2)
(1) and (2) $ \Leftarrow \frac{2}{9}\leq x^4\plus{}y^4 \leq 8$ [/hide]
I add this.
prove that: $ x^{2n}\plus{}y^{2n}\geq\frac{2}{3^n}$


I am not very comfortable with proofs (along with some other rookies who are going to take the USAMO pretty soon) and was wondering if there was somewhere I could find model solutions.

I found that the official USAMO solutions sometimes took quite a bit of effort for me to follow (which would mean either that I am really stupid or the official solutions are not very good => I would like to believe in the later lol).

The official solutions are very, very concise, to they point that they leave out minor details.


This question is from Aops vol. 2. I read the solution, but it's a little confusing. :?

Point D is chosen on side BC of ∆ABC, such that the incircles of ∆ACD and ∆ABD are tangent at G.
Let line l be the angle bisector of angle ABC, line m be the angle bisector of angle ACB, and line n be perpendiculer to BC at pt D. Prove that line l, m, & n are concurrent.

Any solutions would be appreciated. :idea:

It isn't? Then could one of the mods move this to intermediate. My solution is much different than the book's. Thanks PenguinIntegral

2 things:

1. if you want to know if your solution is valid, then post it

2. is G the centroid?

[quote="Altheman"] 2. is G the centroid?[/quote]
I just drew a quick diagram.
EDIT: B and C should be switched  :oops:

[quote="Altheman"]1. if you want to know if your solution is valid, then post it[/quote]
I'm kind of stuck, so my solution is incomplete. I'll post a new diagram and my solution in a little while.

Solution:
AE=AG=AF=x
JD=GD=BI=y
FB=KB=s
CE=CJ=z
Now we find BD in terms of the sides of the triangle (:idea:).
BD=s+y
AB=c=x+s
BC=a=2y+z+s
AC=b=x+z
So, using a little common sense, I got
(a+c-b)/2=(2y+2s)/2=s+y=BD
Then... here's where i got confused. Any help would be appreciated. :)

[quote="PenguinIntegral"][quote="Inspired By Nature"]It isn't? Then could one of the mods move this to intermediate. [/quote]
Sir, I *am* a mod. ;)[/quote]
oh, sorry I didn't realize :blush:. Can someone please finish the proof?

[quote="Inspired By Nature"][quote="Inspired By Nature"]Can someone please finish the proof?[/quote]
 :?: anyone[/quote]
You have completed it yourself so what we need to do is to let you know  you have done!

The problem is actually asking you to show that the perpendicular line from $D$ to side $BC$ passes through the incentre of the triangle $ABC$. Thus if we let the incircle of $\triangle{ABC}$ touches side $BC$ at $D'$, then $ID' \perp BC$ and $BD' = p-b$. Since you have shown that $BD= p-b$, thus $D=D'$, thus $ID \perp BC$. Hence the perpendicular line from $D$ to $BC$ goes through the incentre of the triangle. Therefore the angle bisectors and line $n$ are concurrent.


email
number theory theorems
number theory

please help me find the proof of the unseen proofs of andrew potter's article found on this site:

http://www.scieng.ed.ac.uk/students/awards/Potter.pdf

please help me secure the proofs of theorem 3.3 up to 3.7 especially theorem thm3.3 nad 3.4.
I badly need the proofs for my undergraduate thesis... My theses depends on that proofs.. thank you,

I'm looking for the book entitled Algebraic Numbers and Fourier Analysis by raphael Salem. i want to see the proofs of some theorems on page 4 to 21 because i working on my undergraduate theses but i don't hav the book and our library also don't hav that book... I just want to see the proofs there.. Please email me at queue_zniv16@yahoo.com.. Thanks a lot..

thanks

You can find some things on the forum, especially here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=100067
http://www.mathlinks.ro/Forum/viewtopic.php?t=110608

Also, the book [url=http://books.google.fr/books?vid=ISBN3764326484&id=yxeRXLG19cwC&pg=RA3-PA153&lpg=RA3-PA153&ots=Sxz-aROMWL&dq=pisot+and+salem+numbers&sig=r5zkfi-KV08Gixs7vX1tvBfoSQ0#PPP11,M1]Pisot and Salem numbers[/url] might be interesting to you.


abstract algebra
superior algebra
superior algebra unsolved

Let A and B be two finite abelian groups. Suppose that for all n in N,
#{a in A : (p^n) a=0} =#{ b in B (p^n)b=0}
Prove that A is isomorphic to B.

Thanks

it's an immediate consequence of the structure theorem.

I am sure about the statement of the Structure theorem. Is there any reference that I can look at it?

Thanks.


$\sqrt{\frac{x-8}{2}}+\sqrt{\frac{x-6}{4}}+\sqrt{\frac{x-7}{3}}=\sqrt{\frac{x-2}{8}}+\sqrt{\frac{x-4}{6}}+\sqrt{\frac{x-3}{7}}$

Put $y=x-10$ to get

$\sqrt{{y\over 2}+1}+\sqrt{{y\over 4}+1}+\sqrt{{y\over 3}+1}-\sqrt{{y\over 8}+1}-\sqrt{{y\over 6}+1}-\sqrt{{y\over 7}+1}=0$

That's equivalent to

$\frac{y({1\over 2}-{1\over 8})}{\sqrt{{y\over 2}+1}+\sqrt{{y\over 8}+1}}+\frac{y({1\over 4}-{1\over 6})}{\sqrt{{y\over 4}+1}+\sqrt{{y\over 6}+1}}+\frac{y({1\over 3}-{1\over 7})}{\sqrt{{y\over 3}+1}+\sqrt{{y\over 7}+1}}=0$

After extracting $y$ as a common factor, what remains is always positive, hence the equation is equivalent to $y=0\iff x=10$


1. We can start from the last daughter. Since half of what was left plus half a dollar was all that he had left, he gave his last daughter a dollar. Now we proceed to the middle daughter. She got half of what he had plus a half a dollar, and he was left with a dollar. So she got $2. We could calculate how much the first daughter got, but that won't be necessary. Since the second daughter has $2, should needs another $4.25 to buy the MATHCOUNTS T-shirt.

2. Pick any two points. A line can be drawn through these two points. Thus there are 10 choose 2 = 45 possible lines.

1.  call what was left for the last girl L (in cents)

L/2 + 50 = L --> L = 100

add 50 and double to get the amount that was left before 2nd girl.

thats 300.  subtract 3rd girls --> 200.

she needs 425 more cents

2. 10c2


Prove that $\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{7}$ is not rational.

The solution depends on if you want to do ugly computations or not ;)
See http://www.mathlinks.ro/Forum/viewtopic.php?t=80213 , especially the links there.


Two boxes, A and B, are connected by a lightweight cord and are resting on a frictionless table. The boxes have masses of 12 kg and 10 kg. A horizontal force $ F_p$ of 40.0 N is applied to a 10 kg box. What is the acceleration of each box, and what is the tension of the cord connecting the boxes?

Oh oops wrong forum.

Can a mod please move this to the physics forum?
Sorry

Clearly the two boxes will have the same acceleration.
So we want to F=ma like crazy
We have:

$ 40 \equal{} (10 \plus{} 12)a$,
$ a \equal{} \frac{20}{11}$

Now, drawing a free-body diagram for the first box, we get

$ 40 \minus{} T \equal{} \frac{20}{11} \cdot {10}$,
$ T \equal{} \frac{240}{11}$.

Use a free-body diagram on the second body to check:

$ T \equal{} \frac{20}{11} \cdot {12} \equal{} \frac{240}{11}$


college Putnam calculus Harvard MIT ARML conics

Well folks, I'm going to be a freshmen at UC Berkeley next semester and I was just wondering if anyone could give me a link to some competitions for undergraduates. I'm looking for something like the equivalent of the olympiads for undergraduates.

Appreciate the help.

[quote="texas137"]Putnam for math, ACM for computer science. I think MAA has something for undergrads also.[/quote]

putnam: http://math.scu.edu/putnam/

ACM (is berkeley socal?): http://socalcontest.acm.org/       http://www.acm.org/

Actually Berkeley is in northern California. You'll find universities such as UCSD, UCLA, USC, and CSU SLO in SoCal.

here's the MAA one:
http://www.maa.org/awards/morgan_nom.html

Many colleges have courses (usually 1 unit) or clubs devoted to preparing for the Putnam competition, which is on the first Saturday in December. At UC Berkeley, the course number is MATH H90 - you can find it in the schedule of classes.

Does preparing for high school math competitions help you prepare for undergraduate competitions like the Putnam, or do you have to learn a ton of new material? What is covered on the Putnam? Are there a lot of people who are into math competitions when they are undergraduates like there are for high school competitions?

[tex]\displaystyle\{\text{Putnam material}\}=\{\text{high school contest material}\}\cup\{\text{calculus}\}.[/tex]

How much calculus? Up to AP Calculus BC curriculum? Up to Multivariable Calculus? I'm surprised that it doens't include undergraduate material like Linear Algebra, Abstract Algebra, Differential Equations, Real/ Complex Analysis, etc.

The calculus-using questions are sometimes among the easier ones. You do see questions that might be considered abstract algebra, in that they involve some operation definined on some set, but you don't often need the specific technicalities of an abstract algebra course. High school contest skills, especially with proof-oriented contests, should be very good preparation for the Putnam.

Last year, about 3600 students at over 400 colleges took the Putnam. Large groups - 50 or 100 or so - take it at Harvard and MIT and maybe one or two other places; everywhere else, it's a pretty small (self-selected) group of students taking it.

As an ARML coach who also teaches the Putnam preparation course at my college, I will admit that there is often some overlap between some materials I use for the two groups, although the college students do see, on average, more sophisticated material.

Just as a sample, I'll show you two problems. They were A5 and A6 on the 2001 Putnam - among the more difficult problems that year. A5 is the finest example of a "year" problem I've ever seen, one in which it very much matters exactly what the year was, and A6 is an example of a problem that may use calculus.

A5: Find, with proof, all solutions in positive integers a and n to the equation
a^(n+1) - (a+1)^n = 2001.

A6: A parabola intersects a circle of radius 1. Can the arclength of the portion of the parabola that lies inside the circle ever exceed 4? Prove or disprove.

George Berzsenyi recently pointed me at this:

[url]http://www.imc-math.org/[/url]

I know nothing about it besides what's on that page.

Hm, that's quite interesting. I don't see many from the US. There was MIT.

If anyone has any other info on this I'd appreciate it.

For anyone intending to spend 1-2 years at a community college, AMATYC sponsors a 2-part Student Mathematics League competition: precalculus, roughly AMC-12 level.  Nice scholarship for national winner.

While searching for more info on 2004 IMO, I found this website which accesses at least 50 national and intenational math competitions, often including old tests:

http://www.mathpropress.com/competitions.html


There is a sequence of $ n$ integers such that the sum of $ 3$ consecutive members in this sequence is positive and the sum of $ 4$ consecutive members in this sequence is negative. What is the largest possible $ n$?

A convex quadrilateral has consecutive side lengths of $ 3$, $ 9$, and $ 19$ and its diagonals are perpendicular. What is the length of the fourth side?

Is this an ongoing competition?
I know the answer to #2 and am ready to give it when appropriate.

These are from NJML.  Though I do not believe the competition is still ongoing, since the test date has past.  :wink:
If the competition is still ongoing, then the thread can wait.

Alright then
[hide="Hint for #2"]
Look to the diagonals and use the Pathagorean Theorem. Recognize the relation of the diagonals to the sides.
[/hide]
[hide="Solution"]
I have yet to know how to create a diagram.

The diagonals can be divided into four lengths: $ a$, $ b$, $ c$, and $ d$ such that:
[list]$ a^2\plus{}b^2\equal{}(9^2) \rightarrow 81$
$ a^2\plus{}c^2\equal{}(3^2) \rightarrow 9$
$ \iff b^2\minus{}c^2\equal{}72$
Similarly with the other sides:
$ d^2\plus{}b^2\equal{}361$
$ d^2\plus{}c^2\equal{}x^2$
$ b^2\minus{}c^2\equal{}361\minus{}x^2$
$ \iff 72\equal{}361\minus{}x^2$
$ 289\equal{}x^2$
The length of the fourth side is $ \boxed{17}$[/list][/hide]

No it shouldn't be. They're actually from all the math leagues from [url]mathleague.org[/url]

[hide="1"]OK, let's start by finding some sequences. For $ n\equal{}4$, we can pretty easily see some, like $ \minus{}3,2,2,\minus{}3$. For $ n\equal{}5$, it's a bit tougher. We see that $ a_1\plus{}a_2\plus{}a_3,a_2\plus{}a_3\plus{}a_4,a_3\plus{}a_4\plus{}a_5>0$ and $ a_1\plus{}a_2\plus{}a_3\plus{}a_4,a_2\plus{}a_3\plus{}a_4\plus{}a_5<0$. Now WLOG let all entries in the sequences be integers. Thus, we can determine that $ a_1,a_2,a_4,a_5\leq\minus{}2$, and thus $ a_3\geq5$. We then see a sequence $ \minus{}2,\minus{}2,5,\minus{}2,\minus{}2$. Now, let $ n\equal{}6$. Then, $ a_1\plus{}a_2\plus{}a_3,a_2\plus{}a_3\plus{}a_4,a_3\plus{}a_4\plus{}a_5,a_4\plus{}a_5\plus{}a_6>0$ and $ a_1\plus{}a_2\plus{}a_3\plus{}a_4,a_2\plus{}a_3\plus{}a_4\plus{}a_5,a_3\plus{}a_4\plus{}a_5\plus{}a_6<0$ Adding all in each group yields a contradiction(see yourself), so $ n\neq6$. This accounts for all $ n$, as a sequence of 6 must be inside one of $ n>6$. So, the maximum possible value of n is $ \boxed{5}$.[/hide]

2 i don't feel like making a picture, otherwise I'd post it.


geometry
analytic geometry
graphing lines
slope
algebra
function
domain

Here Few Problems I Need Help With:

1.
[img]http://i6.photobucket.com/albums/y208/outoflimitz/1.jpg[/img]
2.
[img]http://i6.photobucket.com/albums/y208/outoflimitz/2.jpg[/img]
3.
[img]http://i6.photobucket.com/albums/y208/outoflimitz/3.jpg[/img]

[hide="3"]To find the surface area, just find the area of all the sides and the base. The area of one side, because it's a triangle, is $\frac{13*10}{2}=65$. Multiply this by 4 to get 260. The area of the base is 10*10=100. 260+100=360. For volume, multiply the area of the base times the height then divide by 3.$\frac{12*100}{3}=400.$[/hide]
It's the only one I know how to do...

[hide="1"]$\frac{xyz^{-1}-xz^{-1}}{yz^{-1}-1}=\frac{\frac{1}{xyz}-\frac{1}{xz}}{\frac{1}{yz}-1}\times{\frac{xyz}{xyz}}=\frac{1-y}{x-xyz}$[/hide]
2. I don't have any easy way to make a graph, sorry.
[hide="4"]I'm going to assume that each mark is one unit. We need an equation of the form $y=mx+b$. From this graph, we can get two points: (-3,0) and (0,2). The slope ($m$) of the line is: $\frac{y_1-y_2}{x_1-x_2}$ or with these numbers:2/3. Now we have $y=\frac{2}{3}x+b$. Plug in one of the coordinates (I used (0,2)): $2=\frac{2}{3}\times{0}+b$. $b$ is clearly 2. The equation is $y=\frac{2}{3}x+2$[/hide]
[hide="g)"] Again, I'm going to assume that each mark is one unit. The domain is the set of possible x coordinates. A solid point means it is included; an open point means that it is not included. So, $-2\le{x}<4$[/hide]

[hide="#1"]$(\frac{xy}{z}-\frac{x}{z})$ divided by $(\frac{y}{z}-1)$

The rest is like homework.[/hide]


rotation
geometry
geometric transformation
reflection

there are two discs rotating with angular speed $ \omega$ in oppsite direction.Both of them have some number of mirrors attached to them in such a way that a light ray coming from a source kept in between them  gets reflected $ n$ times at each mirror in one second .Find the angular speed $ \Omega$ of such a ray.

also assume that when the ray strikes the mirror each time the mirror is vertical and the ray started horizontally

Hmm, That is an interesting one Let’s see
Assume that the mirror rotates through an angle theta (say) then the angle of  incidence increases by the same angle and so does the angle of reflection. Since the incident ray is stationary the reflected ray has turned an angle of 2*theta. Therefore the angular velocity of the reflected wave is 2*omega. But since the ray undergoes n-1 reflection before coming out of the so-mentioned setup the angular speed of the way is 2*omega*(n-1)

PS: sorry that I couldn’t Insert the symbols
By the way Can you tell me the source of the problem since such type of problems are going to help me in Inpho.

Cheers

Hope that helps!

also i couldn't get this right the answer is $ 2\omega(2n\minus{}1)$

The ray undergoes 2n-1 reflection and not n-1 reflection. This is because the ray is being reflected from two mirrors and for the ray to come out at last it shouldn't undergo reflection therefore it is 2n-1
therefore the answer is 2*omega*(2n-1)

A slight variation is to calcuate the angular velocity of the ray when the disks are rotating in opp.. direction. try it out

yes the other one is eay u have just two cases even and odd reflections
why should that be i got $ 4\omega n$
why can't the ray come out??
where is it trapped
remember it has to undergo n times reflection on eac mirror
also visit the IPho site and also the sticky column of this forum in the physics resources section

for u you to seen an image the light ray has to hit the retine of your eye. if ht e ray according to you goes like that it is never going to come out. Well it undegoes n reflection at the first disk and n-1 at the second disk. So total is 2n-1 and not 4n

see the question.it specifically asks u the angular speed after n reflections on eac mirror

YOu should try to draw the picture and you will get a headway.

PS: I am very bad at using paint

pinnaka i have already tried a lot on this
read the question again it asks for the angular speed of a ray that is reflected n times on 'each' mirror.
so how can u say it does so n times on one and n-1 times on other
ur specifically given it does so n times on each mirro

I have come up with a better way to explain the solution of this problem. Now the disks are  rotating in the opposite direction (forget that variation which I wrote about it should have been in the same direction). Now as per my first post the reflected ray from is  rotated about 2*omega*t. Now this ray goes and falls on the second disk where the reflected ray becomes the incident ray. But the mirror is not stationary and it rotates therefore the angle of incidence becomes 3*omega*t+alpha (alpha is the angle of incidence on the first disk). The angle of reflection will be the same (think about it!!).So after 2 reflection the beam has rotated about 3*omega*t. then for 3 reflection the angle is 5*omega*t.
So the angular velocity of the beam is (2n-1)*omega*t

i don't get t why the distinction between two mirrors
why don't u post the answer for mirrors in same direction and u will know whether tha's right or wrong
also  pinnaka i have thee USSR book not an e-link

ok here it goes the beam from the first disc will fall on the second disk rotated about an angle 2*omega*t. Now since they are rotating in the same direction the angle of reflection decreases by the same amount. The same reasoning can be adopted for any two conse.. reflections, the angular velocity of the beam will be 0 if the no.of reflection is even and 2*omega if it is odd

yes that's the point....
see no matter how it hots which mirror it is turned an $ 2\theta$...so don't diff. between mirrors


function
Ring Theory
superior algebra
superior algebra unsolved

Background: Let $ K$ be a field. A discrete valuation on $ K$ is a function $ \nu$ on $ K$ is a function $ \nu$ $ : K^* \rightarrow %Error. "texbb" is a bad command.
{Z}$ such that
(i) $ \nu(ab)=\nu(a)+\nu(b)$ for all $ a, b \in K^*.$
(ii) $ \nu$ is surjective
(iii) For all $ x, y$ $ \in K^*$ such that $ y \neq -x$, $ \nu(x+y)\geq$ min$ \{ \nu(x), \nu(y) \}$. The set $ R=$$ \{ x \in K^* | \nu(x) \geq 0 \} \cup \{ 0 \}$ is called the valuation ring of $ \nu$.

Prove:
(a) Prove that $ R$ is a subring of $ K$.
(b) Prove that for each nonzero element $ x \in K$, $ x \in R$, or $ x^{-1} \in R$.
(c) Prove that $ x \in R$ is a unit if and only if $ \nu(x)=0$.
(d) Now let $ K = %Error. "texbb" is a bad command.
{Q}$ and let $ p$ be a prime. Given a nonzero element $ x \in %Error. "texbb" is a bad command.
{Q}$, write $ x=p^e \cdot \frac{c}{d}$ where $ e$ is an integer and $ p$ divides neither $ c$ nor $ d$; then define $ \nu_p=e$. Prove $ e$ is uniquely defined and that $ \nu_p$ is a discrete valuation on $ %Error. "texbb" is a bad command.
{Q}.$

Most of these are straightforward applications of the definitions.  What are you getting stuck on?

Could you help me get started on part (a), for example.  I guess I just need to know how to start these parts.

For part a), you need to show: (this is [i]nothing[/i] but the definition of a subring!)

a1) $ R$ contains $ 0$.
a2) $ R$ contains $ 1$.
a3) If $ a,b\in R$, then $ a\plus{}b\in R$.
a4) If $ a,b\in R$, then $ ab\in R$.

When asking for help with a question like this it's usually a good idea to say which of these things you have been able to show, and which ones have caused you trouble.


I suggest that the moderator (b-flat) email Mr. Archie Benton and ask him to send the following link of this forum to every ARML participant in his emailing list.

http://www.artofproblemsolving.com/Forum/index.php?f=432

I have informed Archie of this forum previously.  However, I will email.


Let x,y,z be positive variables and a,b,c be positive constants. A rectangular box in the first octant has one vertex at the origin and the opposite vertex on the surface
a/x + b/y + c/z =1. Determine the maximum volume of the box.

That is, maximize $ xyz$ subject to $ \frac ax \plus{} \frac by \plus{} \frac cz \equal{} 1.$

Set this up as a Lagrange multipliers problem:

$ yz \equal{} \minus{} \frac {\lambda a}{x^2}\\ xz \equal{} \minus{} \frac {\lambda b}{y^2}\\ xy \equal{} \minus{} \frac {\lambda c}{z^2}$

Multiply the first equation by $ x,$ the second by $ y$ and the third by $ z,$ and equate.

$ \minus{} \frac {\lambda a}{x} \equal{} \minus{} \frac {\lambda b}{y} \equal{} \minus{} \frac {\lambda c}{z}$

which becomes $ \frac xa \equal{} \frac yb \equal{} \frac zc$ or $ (x,y,z) \equal{} (ta,tb,tc).$ Putting this into the constraint gets us that $ t \equal{} 3$ and the point $ (3a,3b,3c)$ is the only critical point, and at that point, the volume of the box is $ 27abc.$

So, we're done?

[hide="No, we are not done."]We said it was the only critical point. We didn't say it was a maximum. For instance, try the point $ (2a,4b,4c).$ At that point, $ xyz \equal{} 32abc.$ Since that's bigger than $ 27abc,$ we've got a problem. In fact, we could take the point $ \left(na,\frac {2nb}{n \minus{} 1},\frac {2nc}{n \minus{} 1}\right)$ for which $ xyz \equal{} \frac {4n^3}{(n \minus{} 1)^2}\,abc.$ Let $ n\to\infty$ and that is unbounded above. So it is clear that there is no maximum volume at all, and the one critical point we found is the [i]minimum[/i] volume instead.[/hide]

Or, there's the non-calculus argument that uses AM-GM:

$ \left(\frac{abc}{xyz}\right)^{\frac13}\le\frac13\left(\frac ax\plus{}\frac by\plus{} \frac cz\right)$

so if $ \frac ax\plus{}\frac by\plus{} \frac cz\equal{}1,$ then

$ \frac{abc}{xyz}\le\frac1{27}$

$ xyz\ge27abc.$

With equality when $ \frac ax\equal{}\frac by\equal{}\frac cz$ or $ (x,y,z)\equal{}(3a,3b,3c).$


A tunnel in the shape of a semi-ellipse is sixty feet wide and twelve yards high in the center. Find its height (in feet) six feet from the edge.

We'll assume (for simplicity) that the ellipse is centered at the origin so it's equation is
$ \frac{x^{2}}{a^{2}}\plus{}\frac{y^{2}}{b^{2}}\equal{}1$
where $ a$ is the semi-major axis and $ b$ is the semi-minor axis.  In your problem, $ a\equal{}60/2\equal{}30$ and $ b\equal{}36$.  Now it's just a matter of plugging in and solving.

Did you mean 6 [u]yards[/u] from the edge?

[hide="Solution"]
Stretch horizontally by a factor of $ \frac{2}{5}$ to get a circle of radius $ 12$ yards.

Then, what was perviously $ 6$ feet ($ 2$ yards) from the edge, or $ 28$ yards from the center, now becomes $ \frac{56}{5}$ yards from the center of the new circle under the stretch.

Then, by Pythag, we have $ h \equal{} \sqrt{12^2 \minus{} \left( \frac{56}{5} \right)^2} \equal{} \frac{4 \sqrt{29}}{5}$ yards, or about $ 12.94$ feet.

[b]Note:[/b] If the problem had said 6 [u]yards[/u] from the edge, we would've had a nice 3-4-5 right triangle, giving $ \frac{36}{5}$ yards, or exactly $ 21.6$ feet.[/hide]


I think Temperal and I made one, but it's awful. And I still need to write up the team problems.

I have written four. The first three are too hard, and frost13 is meshing them into an easier one.

Hm, if you want to see my half of our horrible Mock MC, see [url=http://www.artofproblemsolving.com/Wiki/index.php?title=User:Temperal/sandbox&oldid=21129]this[/url].
Siddhant's sprint is like

[quote]1. Let a@b=a^b+b^a. Find (5@2)(-1@1)
2.Consider a right triangle with a leg of length 17. What is the area of the triangle?
3.The ratio of girls to boys at Timber high school is 10 to 11. After 50 girls and 30 boys leave, the ratio of girls to total kids is 5 to 13.How many kids were there originally?
4.Nate finds that the area of a regular hexagon is 9/4. What is the side length?
5.A sequence of numbers is 2,5,8...101. Another is 7,11,15...107. What is the sum of the 17th numbers in both sequences?
6. Out of the 120 kids at the local middle school, 56 take algebra, 47 take geometry, and 17 take both. How many kids take neither?
7.Apotato and an onion cost 89 cents, an onion and a mango cost 60 cents, and all 3 cost 1 dollar. How much does a potato and a mango cost?
8. Set A is the set of primes less than 23, set B is the set of factors of 168, and set C is the set of numbers less than 12. How many numbers are in both sets?
9.Define *x to be the numbers of even factors of x. What is *(842)?
10. How many triangles are in a 4 by 4 grid?
11. Some numbers are picked from the set of {1,2,3,4,5,6,7,8}. Mary wants to kmow what the numbers are. Joe tells her the product of the numbers is 120, and the sum  of the remaining numbers is odd. What's the largest the sum of the chosen numbers could be?
12. How many cubes are factors of 7!*4!*3!?
13. Let m be in {1,2,4,6} and n be in {6,7,9,11}. Find the max value of (m+n)^2/mn.
14. If y is the square of the last two digits of (2662)^10001 and x is the product of the last two digits in 7^2008, find xy.
15. Find the number of solutions, in positive integers, too x^2+625^2=y^2.
Target-
1. A triangle has sides 7,21, and x. Find the max value of x divided by the min value of x. Express your answer as a common fraction.
2. If 1/x+1/y+1/z=24, and xyz=3, find xy+yz+xz.
3. What is the largest prime factor of 17!-15!.
4. Find 1/{3*6}+1/{6*9}+1/{9*12}+...+1/{33*36}.
Yeah, my targets are a bit trivial.[/quote]

question: what approximate level are these supposed to be?
they're way too hard for chapter and school, i think even too much for states. these have to be nationals.

The problem with #10 is that is it a grid of squares, or a bunch of dots?

Also,
@Temperal: According to my answer, #1 on the countdown is irrational.

The dot kind.
And the set one is supposed to be all 3, not both.

Fixed, Alan.

@mihail: Well, there's some pretty hard national ones, but there's also really easy ones suitable for chapter, like target #4 on mine.


Consider P is a point in ABC triangle that <APB-<C=<APC-<B. D and E are centers of scribed circles of APB and APC respectively..prove that lines AP,BD and CE intersect each other in one point.

[hide="Solution"]Apply an inversion with center at A and radius r. The given condition becomes $ \angle B'C'P'\equal{}\angle C'B'P'$ thus P'B'=P'C'. $ P'B'\equal{}\frac{r^2}{AP\cdot AB}\cdot PB$ and $ P'C'\equal{}\frac{r^2}{AP\cdot AC}\cdot PC$. Hence $ \frac{PB}{AB}\equal{}\frac{PC}{AC}$ which solves the problem.[/hide]

[quote="behdad.math.math"]your solution is really good dear limes123.but is there another solution?[/quote]

See my solution here, pre-last post in the thread:

http://www.mathlinks.ro/viewtopic.php?p=3459#3459


Hey everybody.

It's only a little over two weeks until the Canadian Open.  I'm wondering...what have some of you been doing (or have done, if your high school days are over)?.  I think one of the most obvious strategies is taking every single past COMC test there is however, I'd like to know what else (books?)  you guys are doing to help maximize that score.

Also, another concern is during the test.  How fast should one pace themselves???  The Part A questions of the COMC aren't difficult at all, however, I'm just worried I might make some stupid mistake or something.  Then there's part B.  No comment there, but I think the difference in points comes down to the last one or two questions in that section.  How should one do part B?

Part B also deals a lot with solution writing style.  What is and what is not an acceptable solution?

Thanks!

UW has a series of books called [i]Problems[/i]. Other than that, I'm not too sure.. Trying Euclid might help though.

As for pacing, I try to finish Part A within first 30 minutes. That way, you have about half an hour per each problem in Part B, and you don't have to be stressed out about time. But then, everyone's different, so it probably doesn't help anyways..

Good luck in COMC everyone!


1.$Find all continuous functions , such that
f (x+y+f(xy))=xy+f(x+y).
for all x,y$


Does anyone know how compilers (in my case, Free Pascal) evaluate boolean expressions? For example, suppose I have an array a[1..100]. If I write
"if (x>100) or (a[x]=3)" , will the program work if x>100?

Most languages that I know of stop evaluating an expression as soon as the final value is determined so that you can do stuff like what you have there. I wouldn't have a clue personally about Free Pascal, but it was the first link I found in google:

[quote]Remark: In Free Pascal, boolean expressions are always evaluated in such a way that when the result is known, the rest of the expression will no longer be evaluated (Called short-cut evaluation).[/quote]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hey all
here's the problem statement:

Suppose $ H$ is a subgroup of $ G$ such that whenever $ Ha \neq Hb$ ,then
$ aH \neq bH$ . Prove that $ gHg^{\minus{}1} \subset H$ for all $ g \in G$

I think the condition basically states that when two right cosets differ then the above mentioned cosets also differ. Since the left and right cosets are in one-to-one correspondence, would i be wrong if i say

$ G \equiv \bigcup Ha_i$ implies $ G \equiv \bigcup a_{i}H$ for the same set of $ a_i$

How do i go about doing the problem now?

thanks

Note that $ Ha\equal{}Hb$ if and only if $ b\in Ha$, and similarly for left cosets. We have that $ G\setminus aH\supset G\setminus Ha$, so $ aH\subset Ha$. Multiplying on the right by $ a^{\minus{}1}$, $ aHa^{\minus{}1}\subset H$ and we are done.
This containment is actually equality; we also have $ H\subset a^{\minus{}1}Ha$ for the same reason, and $ H$ is normal.

Ah thanks. I was thinking that it should be equality  :)

pardon me
but is this obvious or I'm plain stupid
$ G\setminus aH\supset G\setminus Ha$
in the night i thought it was making sense ... alas  :lol:

$ G\setminus Ha\equal{}\{b: Hb\neq Ha\}$.


probability
function
probability and stats

A particular very large college with kN students assigns dorms via a housing lottery. The procedure is as follows:
- Students must form groups of N
- Each student is randomly assigned a unique integer rank from 1 to kN
- All students within a group adopt the rank of the group member with the best rank
- A student's normalized rank is his/her rank divided by kN

Given N, what is the expected distribution of normalized ranks as k tends to infinity?

[b]Edit:[/b] Rephrased, variables changed, to address Tyl's concern.
(The original problem had groups of size M and a student body of size N, with N tending to infinity)

What do you do when $ M$ does not divide $ N$? Also, what do you mean by "expected distribution"?

Assume $ M$ is small compared to $ N$, so the remainder doesn't matter. Expected distribution? If $ M\equal{}1$, the ranks are a discrete uniform distribution on $ \frac1N,\frac2N,\cdots,1$. As $ N\to\infty$, the fraction in any particular slice $ [x,x\plus{}dx]$ tends to $ dx$, so we can say that the distribution tends to the uniform distribution on $ [0,1]$ with density $ 1$ everywhere.
This is convergence in distribution: as $ N\to\infty$, the probability the rank is between $ x$ and $ y$ tends to that of some fixed distribution.

[quote="TZF"]Does anyone have any ideas on what to do with this problem?[/quote]

the phrase 'expected distribution' does not make sense.  do you really want to determine the expected rank of a given person?  here's an idea.  say that a person belongs to a group of $ N$ people.  Then it's easy enough to figure out combinatorially how many permutations of $ kN$ rank all $ N$ of those people below $ t$, in fact, it's just a binomial coefficient.  But if you know these, you can recover the expected rank fairly easily.

No, it does make sense. There's a full distribution there.

Within a group of $ N$ students, the correlations tend to zero and the original normalized ranks look like a uniform distribution on $ [0,1]$. The minimum $ r$ of $ N$ independent uniform distributions has cumulative distribution function $ P(r\ge x)\equal{}(1\minus{}x)^N$ and density $ N(1\minus{}x)^{N\minus{}1}$. That's the limiting distribution.


[quote="A. S. K."]Find seven distinct primes $p_{1}, p_{2}, ..., p_{7}$ less than 1000 satisfying
\[p_{2}-p_{1}=p_{3}-p_{2}=...=p_{7}-p_{6}. \]
[/quote]

i also discovered the problem. but i didn't have any solution but i found six such primes. one of my friends gave me the two solutions: ${7,157,...907}$ and ${47,257,...1307}$. at a book of mine i found ten such primes(which are in an A.P.). i wonder if there always exists a set of $n$ such primes $\forall n\in \mathbb N$. :P

[quote="nayel"]i wonder if there always exists a set of $n$ such primes $\forall n\in \mathbb N$. :P[/quote]

Yes, this has been proved in 2004 by Green and Tao with a non-constructive proof.
Have a look here :
http://mathworld.wolfram.com/PrimeArithmeticProgression.html

Pierre.

[quote="A. S. K."]Please, can you tell me how did you get that solution? Thanks[/quote]

i invented a theory and i use it to solve these kinds of problems and then use trial and error. but it's a quite boring system actually.

"if $a_{1}, a_{2},...a_{n}$ is an A.P. with common difference $d$ and $k$ is a positive integer prime to $d$ then of any $k$ consecutive terms of the A.P. there exists an $a_{i}$ such that $k|a_{i}$." :D

Please, can someone post a complete solution of the problem I asked? Thanks a lot.


Express the ratio of $ \frac{2}{7}$ to 1.$ \overline{09}$ as a common fraction.


$ 1$. A finite noncyclic group $ G$  in which every proper subgroup is abelian possesses a normal subgroup $ N$ which is distinct from $ G$ and $ 1$.

$ 2$. If $ G$ is a finite group such that for each abelian subgroup $ A$, the normalizer of $ A$ in $ G$ is equal to the centralizer of $ A$ in $ G$, then $ G$ is abelian.

$ 3$. If $ H$ is a normal subgroup of the finite group $ G$, and $ p$ divides $ |H|$, then $ H$ contains every Sylow $ p$-subgroup of $ G$.

[size=117]1.[/size]
Assume $ G$ is a non-cyclic simple group with every proper subgroup abelian
and $ P$ some non-trivial Sylow subgroup of $ G$.
Then $ G\not \equal{} P$, as $ P$ is nilpotent, and also $ P$ can't be normal in $ G$, so $ N_G(P)\lneq G$.
This gives $ P\le N_G(P)$ abelian, and therefore $ N_G(P) \equal{} C_G(P)$.
Then [url=http://www.matheplanet.com/default3.html?call=fav.php?op=view&fav_id=23296&uname=Gockel&select=ddddff&ref=http%3A%2F%2Fwww.google.com%2Fsearch%3Fie%3DUTF-8%26oe%3DUTF-8%26gfns%3D1%26sourceid%3Dnavclient%26rls%3Dcom.google%3Ade%3Aofficial%26q%3Dn_g%2528p%2529%253Dc_g%2528p%2529]Burnside's transfer theorem[/url] gives a normal complement of $ P$ in $ G$,
which contradicts $ G$ simple and $ 1\not \equal{} P\not \equal{} G$.
[size=117]2.[/size]
Let $ G$ be a minimal counterexample.
Then every proper subgroup of $ G$ also satisfies the condition and is therefore abelain.
If $ N$ is a maximal normal subgroup of $ G$, then $ G/N$ is simple with every proper subgroup also abelain.
Then 1. gives $ G/N$ cyclic, and $ N$ abelian gives $ G \equal{} N_G(N) \equal{} C_G(N)$, so $ N\le Z(G)$.
This gives $ G/Z(G)$ cyclic and $ G$ abelain. Contradiction!
[size=117]3.[/size]
This is nonsense.
I assume you mean $ p\nmid |G: H|$, in which case it follows immediately from Sylow:
$ H$ must contain some Sylow $ p$-subgroup of $ G$,
the Sylow $ p$-subgroups of $ G$ are all conjugate in $ G$
and $ H$ is closed under $ G$-conjugation.

Dear Olorin,
Thank you very much for your help.
With #1, I'm a complete novice in group theory but I'll try my best to understand the Burnside Transfer Theorem better (sorry can't read German fluently  :( ). I thought your solution to #2 was quite clever. I'm very sorry about the typo in q3, the question is exactly as you suggested ( $ (p, [G: H])\equal{}1$), I was trying the case when $ p^2$ does not divide $ [G: H]$ and forgot to change the condition to suit the general case.  :oops:  :oops:    :|

Ok, here is a more elementary solution for 1). (though I like the [url=http://en.wikipedia.org/wiki/Transfer_%28group_theory%29]transfer[/url] argument much more)

For any maximal subgroup $ M$ of $ G$ we have $ M\le N_G(M)\lneq G$, as $ M$ is not normal in $ G$.
So $ M\equal{}N_G(M)$ and $ M$ has $ |G: N_G(M)|\equal{}|G: M|$ conjugates in $ |G|$.
Also for any two maximal subgroups $ M,N$ of $ G$ (which are abelian) we have $ M\cap N\equal{}1$,
for if $ 1\not\equal{}x\in M\cap N$, then $ x\not\in Z(G)\equal{}1$, so $ M,N\le C_G(x)\lneq G$ gives the contradiction $ M\equal{}C_G(x)\equal{}N$.

So now if $ M_1,\ldots,M_n$ are representatives of the conjugacy classes of maximal subgroups of $ G$,
we get the disjoint union $ G\setminus\{1\}\equal{}\bigcup_{i\equal{}1}^n\bigcup_{gM_i\in G/M_i}(gM_ig^{\minus{}1}\setminus\{1\})$, which gives
$ |G|\minus{}1\equal{} \sum_{i\equal{}1}^n|G: M_i|(|M_i|\minus{}1)$ and $ (n\minus{}1)|G|\plus{}1\equal{}\sum_{i\equal{}1}^n|G: M_i|\le n{|G|\over 2}$.
This gives the contradiction $ n\equal{}1$ and $ 1\equal{}|G: M_1|$.

P.S.
Also $ M\cap gMg^{\minus{}1}\equal{}1$ for all $ g\in G\setminus M$ gives $ G$ a [url=http://en.wikipedia.org/wiki/Frobenius_group]Frobenious group[/url] with Frobenius complement $ M$,
which has the Frobenius kernel as normal complement. (this proof uses [url=http://en.wikipedia.org/wiki/Character_theory]character theory[/url])
So $ G$ can't be simple.

P.P.S.
Another interesting way to prove 1. would be the following.
For any two involutions $ x,y\in G$ we have $ \langle x,y\rangle$ a [url=http://en.wikipedia.org/wiki/Dihedral_group]dihedral group[/url], and therefore solvable.
This gives $ G\gneq \langle x,y\rangle$ abelian for all involutions $ x,y\in G$.
So the subgroup of $ G$ generated by all involutions is abelian and normal in $ G$, and so $ \equal{}1$.
So $ G$ has no involution, and is therefore a [url=http://en.wikipedia.org/wiki/CA_group]CA-group[/url] of odd order, which is solvable by Suzuki.


A manager wants to hire a good problem solver to work at her computer store for just four weeks.  When Terry applies, the manager offers him, "I will pay you  0.01 dollars on Day 1, 0.02 dollars on Day 2,
0.04 dollarson Day 3, 0.08 dollars on Day 4, and so on, doubling your pay each day."  Terry laughed and walked out.  How much would Terry have earned on Day 20 alone?

A.    0.40 dollars	 B.10.24 dollars	 C.1,024	dollars   D.    5,242.88 dollars    E.10,485.76dollars

[hide] We see that if it is Day x, Terry will recieve $ \$\frac {2^{x \minus{} 1}}{100}$ that day. Thus, on Day $ 20$ he received $ \frac {2^{19}}{100} \equal{} 5242.88$. Hence, the answer is D.[/hide]


I would like an algebraic proof of

$\sum_{i=1}^{n}\left( \begin{array}{c}n-i\\j-k\end{array} \right) \left( \begin{array}{c}i-1\\k-1\end{array} \right) = \left( \begin{array}{c}n\\j\end{array} \right) $.

I have spent two days to find it  :blush:  :blush:

Thank you very much.

What do you mean by 'algebraic proof'. You do accept a combinatorial one or not? If no, I think it has to be moved into the Algebra section... ;)

Pierre.

Dear Pierre,

this is my problem.

[quote]To prove Vandermonde identity it is necessary to use the binomial theorem for the expression $(1+x)^n(1+x)^m=(1+x)^{m+n}$[/quote]

So

 [quote]I would like to find an expression whose binomial expansion gives the formula I proposed.[/quote]

Sorry, but I cannot move it in algebraic section as only Administrator can do that.

If you prefer, I can post the same topic in Algebraic Section.

I found combinatorical solution, but it is the same as in mentioned paper.

Dear Myth,

do you think it is possible to find an algebraic solution ?

Dear Manlio!
Only for you!

Consider $S=\sum_{i=1}^n(1+x)^{n-i}(1+y)^{i-1}$. We are to calculate coefficient of $x^{j-k}y^{k-1}$ in $S$. It is easy to get closed form for $S$:
\[S=\frac{(1+x)^n-(1+y)^n}{x-y}.\]
Suppose $S=\sum_{j=1}^{n}S_j$, where $S_j=a_{j0}x^0y^{j-1}+a_{j1}x^1y^{j-2}+...+a_{j,j-1}x^{j-1}y^0$. Then $S_j\cdot (x-y)$ is all members of degree $j$ in expression $(1+x)^n-(1+y)^n$, it follows that $S_j\cdot (x-y)=C_n^jx^{j}-C_n^jy^j$. And we immediately obtain $a_{j0}=a_{j_1}=...=a_{j,j-1}=C_n^j$.

That's all.

You are really a great mathematician, Myth.

Thank you very much.

I also love this insightful solution.

Thanks, Myth!

  Darij

:blush:  :blush:  :blush:  :blush:  :blush:  :blush:

Ok, I move it into the algebra section before Myth collapse... :D

Pierre.


I think it is a very nice problem.The whole solution is not shown here,just the steps:
1/ Times both side by (a^2+b^2+c^2) inequality is equivalent to:
 a^4*(b+c)/(b^2+c^2)+b^4*(c+a)/(c^2+a^2)+c^4*(a+b)/(a^2+b^2) >= a^3+b^3+c^3. (1)
2/AM-GM applies then:
  a^4*(b+c)/(b^2+c^2) + b^4*(c+a)/(c^2+a^2) + c^4*(a+b)/(a^2+b^2) +
+a^2*(b^2+c^2)/(b+c) + b^2*(c^2+a^2)/(c+a) + c^4*(a^2+b^2)/(a+b) >= 2*(a^3+b^3+c^3)
3/It is final step to prove that:
  a^2*(b^2+c^2)/(b+c) + b^2*(c^2+a^2)/(c+a) + c^4*(a^2+b^2)/(a+b) <=  a^3+b^3+c^3
 This is done by means of algebraic manipulation and the use of two inequalities:
  a/(b+c) + b/(c+a) + c/(a+b) >= 3/2 .
  a^3 + b^3 + c^3 + 3abc >= a^2*(b+c) + b^2*(c+a) + c^2*(a+b) .
 Long and complicated :blush:  :rotfl: ,some one may present a better one????

Use [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX]LaTeX[/url], please. It's really hard to read otherwise.


calculus
integration
LaTeX
algebra
partial fractions
calculus computations

Evaluate $ \int_0^{\infty} \frac{|x\minus{}1|}{(x\plus{}1)(x^2\plus{}1)}\ dx$.

split the integral from 0->1 and then from 1->infinityt...

and then split it as partial fractions and then solve....

PS-don know to use latex

didnt work out...
sorry.. studying chemistry.. will work out later


αν $ abc \equal{} 1$ ,kai $ a,b,c>0$,ν.δ.ο.
 $ \sum \frac {1}{a^3(b \plus{} c)} \geq \frac {3}{2}$   :wink:

[quote="Athinaios"]αν $ abc \equal{} 1$ ,kai $ a,b,c > 0$,ν.δ.ο.
$ \sum \frac {1}{a^3(b \plus{} c)} \geq \frac {3}{2}$   :wink:[/quote]

Η συγκεκριμενη ειναι αρκετα γνωστη :P

[hide]
Απο Cauchy-Swartz
$ 2(ab \plus{} bc \plus{} ca)LHS \equal{} [(ab \plus{} ac) \plus{} (bc \plus{} ba) \plus{} (bc \plus{} ca)]\left(\sum_{cyclic}{ \frac {(ab)^2}{ab \plus{} bc}}\right) \geq (ab \plus{} bc \plus{} ca)^2 \Leftrightarrow LHS \geq \frac{ab \plus{} bc \plus{} ca}{2} \geq \frac{3\sqrt[3]{a^2b^2c^2}}{2} \equal{} \frac{3}{2}$
Απο AM-GM και η αποδειξη ολοκληρωθηκε.[/hide]

Simply put $ a \equal{} \frac{x}{y}$ κλπ. και είναι η ευκολότερη λύση και αυτή που θα σκεφτόταν κανείς πρώτη. :wink:


algebra
polynomial
inequalities
inequalities proposed

I use the notation
$[a, b, c] = \sum_{sym}x^{a}y^{b}z^{c}$

Prove that for nonnegative real numbers $x, y, z$ we have

$[4, 1, 1]+[3, 3, 0]+4[2, 2, 2] \ge 6[3, 2, 1]$

Note that
$[4, 1, 1]+4[2, 2, 2] \ge 5[3, 2, 1] \iff [3, 0, 0]+4[1, 1, 1] \ge 5[2, 1, 0]$
is false.
;)

edited: I don't know why I titled it "cyclic." Now it says symmetric as it should.

it's equivalent to:
$\sum_{cyc}c(a-b)^{2}\left((a+b)(ab+c^{2})-4abc\right)\geq 0$
but $(a+b)(ab+c^{2})\geq 2\sqrt{ab}\cdot 2\sqrt{abc^{2}}=4abc$ so our inequality is true!
What's your proof Xevarion?

It is also equivalent to

$\sum_{cyc}2bc(a-b)^{2}(a-c)^{2}\ge 0$

which we can get to from yours by

$(a+b)(ab+c^{2})-4abc = b(a-c)^{2}+a(b-c)^{2}$

;)

But by checking on my computer I found it is also equivalent to

$2(a^{2}c+b^{2}a+c^{2}b-3abc)(a^{2}b+b^{2}c+c^{2}a-3abc) \ge 0$   :o

which is obvious by AM-GM!!


The degree measures of the interior angles of a pentagon form an arithmetic sequence. What is the middle term of this sequence?

It has to be the sum of the interior angles divided by the number of sides, thus, 540/5=[b]108[/b]


Prove that there exist precisely two non-zero function $f: \mathbb{R} \to \mathbb{R}$ which satisfies :
$f(f(x)) = f(x) + x$ for all $x \in \mathbb{R}$.

I haven't completely solved it;but I found these solutions.
[hide]
If f(x)=ax we have a^2*x = (a+1)x;so,if x is not 0,
a^2=a+1.(if it's 0,it's true for every a)
Solving this,we obtain a=(1+sqrt(5))/2 and a=(1-sqrt(5))/2

And the function are f(x)=x*(1+sqrt(5))/2 and f(x)=(1-sqrt(5))/2[/hide]

Thanks ;)
but how prove that there is no more solutions ?

Well, are you sure that there is no hypothesis missing? Continuity or something like that, because otherwise, at first sight I'm not sure that the conclusion holds... ;)

Pierre.

sorry, I forgot to say that function is countinous ;p

In that case, I think the proof I gave here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=continuous+injective&t=15554
can be easily adapted to your problem.

Pierre.


There are 16 non-overlapping equilateral triangles (unit triangles) in the figure shown. Each new number to be written in an empty unit triangle is the product of the three closest numbers in the row directly below it. What number will be written in the shaded unit triangle at the top?

[asy]draw((0,0)--(8,0), linewidth(1));
draw(2dir(60)--(2dir(60)+(6,0)), linewidth(1));
draw(4dir(60)--(4dir(60)+(4,0)), linewidth(1));
draw(6dir(60)--(6dir(60)+(2,0)), linewidth(1));

draw((0,0)--8dir(60),linewidth(1));
draw((2,0)--(6dir(60)+(2,0)),linewidth(1));
draw((4,0)--(4dir(60)+(4,0)),linewidth(1));
draw((6,0)--(2dir(60)+(6,0)),linewidth(1));

draw(2dir(60)--(2,0),linewidth(1));
draw(4dir(60)--(4,0),linewidth(1));
draw(6dir(60)--(6,0),linewidth(1));
draw(8dir(60)--(8,0),linewidth(1));

label("1",(1,2/3));
label("4",(3,2/3));
label("2",(5,2/3));
label("1",(7,2/3));

label("0",(2,1));
label("1",(4,1));
label("3",(6,1));

label("8",(4,sqrt(3)+2/3));
label("6",(5,sqrt(3)+1));

filldraw(6dir(60)--(6dir(60)+(2,0))--8dir(60)--cycle,gray);[/asy]

Instead of working through everything, notice how the triplet 1,0,4 produces 0, and this affects everything above it. Therefore, the answer is $ \boxed{0}.$


Betty goes to the store to get flour and sugar.  The amount of flour she buys, in pounds, is at least 6 pounds more than half the amount of sugar, and is no more than twice the amount of sugar.  Find the least number of pounds of sugar that Betty could buy.

soooo $ 6\plus{}\frac{s}{2}<f<2s \implies 12\plus{}s<2f<4s \implies 12\plus{}s<4s \implies 12<3s \implies 4<s$

lots of arrows

so 4<s

answer : 5

Wouldn't the answer be 4?
The question says "at least" and "no more than" so the inequalities would be less than or equal to, instead of less than. Then, s would equal 4.

Two things:

One, the problem never stated the amount had to be an integer. So 4<s would not give a minimum value.

Second, the question does say "at least" and "no more than", so it should be 4<=s, which of course gives s = [b]4[/b]

Yes,

The correct signs are less than/equal to or greater than/equal to.

$ s\equal{}\boxed{4}$


Let $ P\in\mathbb{C}[x]$ be s.t. $ P$ has no single roots. Let $ A\in\mathcal{M}_n(\mathbb{C})$ be s.t. $ A^{n-1}\not={0},A^n=0$. Find the $ X\in\mathcal{M}_n(\mathbb{C})$ s.t. $ P(X)=A$.

What about the following:
If $ n\geqslant 2$, $ P\in\mathbb{C}[x]$ be s.t. $ P$ has no single roots, $ A\in\mathcal{M}_n(\mathbb{C})$ be s.t. $ A^{n - 1}\not = {0},A^n = 0$.
There is no $ X\in\mathcal{M}_n(\mathbb{C})$ s.t. $ P(X) = A$.

Let $ P(t)=\prod_{i=1}^{m}(t-\lambda_i)^{n_i}$, with $ n_i\geqslant 2$.
Assume there is ${ X \in \mathcal{M}_n(\mathbb C}$ s.t. $ P(X)=A$.
We have $ \oplus_{i=1}^{m} \ker ((X-\lambda_iI_n)^{n_i n}) =\mathbb{C}^n$, and $ \oplus_{i=1}^{m} \ker ((X-\lambda_iI_n)^{n_i (n-1)}) \neq \mathbb{C}^n$. So there is $ i$ s.t. $ \ker ((X-\lambda_iI_n)^{n_i (n-1)})  \subsetneq \ker ((X-\lambda_iI_n)^{n_i n})$, which is impossible since $ n_i(n-1) \geqslant n$.


[quote="Mathpro"]Let $ \{a_n \} _{n \equal{} 1}^{\infty}$ be sequence of real numbers such that $ a_1 \equal{} 2$ and $ a_{n \plus{} 1} \equal{} {a_n}^2 \minus{} {a_n} \plus{} 1$, for $ n \equal{} 1,2,...$ Prove that
$ 1 \minus{} \frac {1}{2005^{2005}} < \frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} \frac {1}{a_3} \plus{} ... \plus{} \frac {1}{a_{2005}} < 1$[/quote]
We can prove that for all $ n \geq 2$ then
$ 1\minus{}\frac{1}{2^{2^{n\minus{}1}}} <\sum_{k\equal{}1}^n \frac{1}{u_k}<1\minus{}\frac{1}{2^{2^n}}$


function
induction
algebra unsolved
algebra

Find all the function
$ f(k,n): N^2 \minus{} > N (1 < \equal{} k < \equal{} n)$ $ f(0,n) \equal{} f(n,n) \equal{} 1 ; f(k,n \plus{} 1) \equal{} f(k,n) \plus{} f(k \minus{} 1,n)$

This seems a bit too easy for the olympiad section.

[hide="Hint"]
Show by induction on $ n$ that $ f(n,k)$ must be $ \binom{n}{k}$.
[/hide]


calculus
integration
logarithms
calculus computations

Evaluate

\[ \int_e^{e^{2009}} \frac{1}{x}\left\{1\plus{}\frac{1\minus{}\ln x}{\ln x\cdot \ln \frac{x}{\ln (\ln x)}}\right\}\ dx\]

Substituting $ \ln{x}\equal{}t$, we get

$ \int_{e}^{e^{2009}}\frac{1}{x}\left\{1\plus{}\frac{1\minus{}\ln x}{\ln x\cdot\ln\frac{x}{\ln (\ln x)}}\right\}\ dx$
$ \equal{}\int_{1}^{2009}\left(1\plus{}\frac{1\minus{}t}{t(t\minus{}\ln{t})}\right)dt$
$ \equal{}2008\plus{}\int_{1}^{2009\minus{}\ln{2009}}\frac{\minus{}1}{u}du$($ u\equal{}t\minus{}\ln{t}$)
$ \equal{}\boxed{2008\minus{}\ln\left(2009\minus{}\ln{2009}\right)}$.

Yes, I think there was a typo in the original problem statement as Kunny obviously intended for the above solution.

is there a way to solve it with the mistake?


The first term and last term of an A.p are a and b. There are altogether (2n+1) terms. A new series is formed by multiplying each of the first 2n terms by the next term. Find the sum of the new series.

$ a_1\equal{}a$
$ a_{2n\plus{}1}\equal{}b\equal{}a_1\plus{}2nr\Rightarrow r\equal{}\frac{b\minus{}a}{2n}$

We have to find the sum
$ \sum_{k\equal{}1}^{2n}a_ka_{k\plus{}1}\equal{}\sum_{k\equal{}1}^{2n}[a\plus{}(k\minus{}1)r)](a\plus{}kr)\equal{}$
$ \equal{}\sum_{k\equal{}1}^{2n}(a^2\plus{}ar(2k\minus{}1)\plus{}r^2k(k\minus{}1))\equal{}2na^2\plus{}ar\sum_{k\equal{}1}^{2n}(2k\minus{}1)\plus{}r^2\sum_{k\equal{}1}^{2n}k(k\minus{}1)$

$ \sum_{k\equal{}1}^{2n}(2k\minus{}1)\equal{}4n^2$

$ \sum_{k\equal{}1}^{2n}k(k\minus{}1)\equal{}\sum_{k\equal{}1}^{2n}k^2\minus{}\sum_{k\equal{}1}^{2n}k$

Now use $ \sum_{k\equal{}1}^nk^2\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}, \ \sum_{k\equal{}1}^nk\equal{}\frac{n(n\plus{}1)}{2}$ replacing n with 2n


[quote="indybar"]Let $a_1,...,a_n\in Z^+$, prove that
$a_1+\frac{2}{\frac{1}{a_1}+\frac{1}{a_2}}+\frac{3}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}}+...+\frac{n}{\frac{1}{a_1}+...+\frac{1}{a_n}}<2(a_1+...+a_n)$[/quote]
I have seen this problem at komal and have a long time thought on it long ago.The solution is also interesting to me.


How many odd perfect square factors does $2^4 \times 3^6 \times 5^{10} \times 7^9$ have?

Hint: even factors, 1 is a perfect square, and try combining the primes ($2^4\cdot3^2$), etc.

[quote="JesusFreak197"][hide="Answer"]Taking even powers of the odd factors only: $(3^2)^3\cdot (5^2)^5\cdot (7^2)^4$
$(3+1)(5+1)(4+1)=\boxed{120}$.[/hide][/quote]

Whoa, that's a nice solution  :D , using prime factorization-like things.

[hide]

to have odd perfect square , all the prime factor must be odd . So we can only choose from $3,5,7$ . Since it is perfect square , the power of the odd number must be EVEN.

So by letting the perfect square having the form

$(3^2)^l(5^2)^m(7^2)^n$ , where $l,m,n$ are integer .

We notice that $l$ can take the range $0\leq l\leq 3$ similarly $0\leq m\leq 5$ and $0\leq n\leq 4$ . Here we have $4$ choice to choose from $l$ ( since $l$ can be either $0,1,2,3$ .) Similarly $6$ choice for $m$ and $5$ choice for $n$ .

Hence the number of odd perfect is $4\times 6\times 5=120$ [/hide]

[quote="anirudh"] 29+1/2=30?[/quote]
um that doesn't equal 30 :?

[hide="answer"]

Only even powers will be perfect squares. You can only use the odd numbers.
So: $ (3^2)^3 * (5^2)^5 * (7^2)^4 $ because there are 4 squares in 3^6, 6 in 5^10, and 5 in 7^9.
Since you can combine two numbers together to get a square of the product of both, (5*3^2=225 square root is 15) you have to multiply them. So the answer is $4 * 6 * 5 = \framebox{120}$ [/hide]


Prove that Cos(CosX) > SinX for all real X.

that should be a $\geq$ sign :) (check $\frac\pi2$ for example)

I think I have proved that $\mid\cos(cosx)\mid\geq\mid\sin(x)\mid$ for all $x\epsilon\ R$

Which proof did you use??!!! ;)   :)


Who's going? Palmer is definitely going once I can get a team together.

Woodward's going. I'll go, assuming I qualify for our team.

Dodgen is going, and maybe Dickerson. Not sure about that yet. It's time for ONE of the Cobb schools to win that thing. Cobb County has never won the GCTM State Middle School tournament.

Fulton

Oh; I didn't realize that you were talking about the middle school one.

Wow, seriously? Cobb's never won. If Dickerson goes, they'll win easy. My memory's a little sketchy, but from what I remember the GCTM's easier than most tournaments. We got 5th last year and we only had 7th graders and a 6th grader on that team.

[quote="Jenn7"]Wow, seriously? Cobb's never won.[/quote]

Probably since the tournament is too far for Middle Schools to be interested.

Hmmh...middle school eh? can I register as an individual? when is it?

Yeah, it is far: 2 hours.

I have no clue if you can register for individual, arvind. It's on April 21st.


linear algebra
matrix
vector
linear algebra unsolved

Let A an m *n Matrix .
[b]Prove that Rank A <=1 [/b]if Only if A =BC  where B is an m*1 Matrix and C 1*n Matrix

Note that if A is expressible of that form, that means it looks like

b_1 c_1   b_1 c_2   b_1 c_3  ... b_1 c_m
...
b_n c_1   b_n c_2 ...                  b_n c_m

Obviously the rows span the subspace spanned by the vector <c_1 ... c_m> and thus the matrix has rank 1 (or rank 0 if actually everything is 0). On the other hand, if the matrix has rank 1 (rank 0 is trivial), that means the rows span a 1-dimensional subspace, i.e. they are all multiples of some row vector (we get our row vector here); the multiples form the column vector (i.e. if the first row is twice the row vector, the first entry of the column vector would be 2).


I am trying to build a robot
i hav a few questions:
How do i program the robot?
What lang?
Where do i get the circuit board?  :?:

There are many different microcontroller manufacturers, you should be able to buy a microcontroller in any electronics shop. The ones I've seen were PIC Microchip ones but there are [url=http://en.wikipedia.org/wiki/List_of_common_microcontrollers]many others[/url]. I think the circuit board for programming them can be bought as well, but you can also make one yourself, you'll only need a serial port interface. Most of the microcontrollers (all of them?) can be programmed in C and you can usually download the compiler from manufacturer's website.

Can someone provide me some links or a book to learn about the basics of building a robot. Or can somebody answer about what all I need to do.

This is [url=http://www.robotbuilder.org/index.php/Main/HomePage?userlang=en]an example[/url] of a robot created by a frien of me with details.

what exactly is a microcontroller ?
and how do i plug into it?
will C++ work for the most microcontrollers?

Microcontroller is something that looks like this: [url]http://www.intertexelectronics.com/images/-MC68701S.jpg[/url]. It's really just a tiny version of a regular PC - it usually contains its own CPU, RAM, persistent memory and clock. The "legs" that you see on the picture are the input/output pins through which you can communicate with the controller (by setting the appropriate voltage, you send 1s and 0s). You should be able to get an instruction manual on how exactly they work from the manufacturer's website. To program the microcontroller, you'll have to use a special compiler supplied by the manufacturer, most of the compilers are C or assembler compilers. (You probably won't have all the C functionality though - only what's the microcontroller capable of.) I doubt that there are any microcontrollers with C++ compilers but that's just my guess.

So, about how much does a microcontroller cost?
Can i input a C++ compiler into the small memory?
And, how do i plug into the thing with my computer?

thanks a lot for all the help,
Poincare

The cost really depends on the features you need, the basic ones are very cheap (less than a dollar). I don't know what you mean by "input a C++ compiler into ...". As I'm saying, you'll have to download the compiler from the manufacturer's website, you can't use any other compiler. To program the microcontroller, you'll need a special circuit borad. You can either build it yourself or buy it from somewhere. I'm sorry but I've never actually done this so I can't help you with any more details, but maybe I could ask my dad - he knows quite a bit about these things and he's been recently programming some microchips for his RC airplanes...


[color=darkblue]Let $ a > 0, b > 0, c > 0$. Prove that: $ 8(a^3 \plus{} b^3 \plus{} c^3)^2\ge 9(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)$[/color]

first of all WLOG assume that $ abc \equal{} 1$,then rewrite the inequlity as:

$ 8(a^3 \plus{} b^3 \plus{} c^3)^2\geq 9(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)$

now let $ x \equal{} a^3,y \equal{} b^3,z \equal{} c^3$,so $ xyz \equal{} 1$ and we have to prove that:

$ 8(x \plus{} y \plus{} z)^2\geq 9(x \plus{} 1)(y \plus{} 1)(z \plus{} 1)$

which is easy to prove,for example after expanding,we are left with:

$ 7\sum yz \plus{} 8\sum x^2\geq 18 \plus{} 9\sum x$

which is true,because by AM-GM we have:

$ 4\sum yz \plus{} 2\sum x^2\geq 18$

and

$ 3\sum yz \plus{} 6\sum x^2\geq 9\sum x$

[color=darkblue]Thank you very much.[/color]

We hold the strong inequality. This is proved by BCS inquality:
$ 3(a^3\plus{}b^3\plus{}c^3)^2 \geq  (a^2\plus{}b^2\plus{}c^2)^3$


Are there exist rational numbers like $x,y,z$ such that $x^2+y^2+z^2=x+y+z+1$? :)

No, since $7$  is not the sum of squares of 3 rational numbers.  ;)


probability
function
parameterization
algebra
polynomial
symmetry
expected value

There are 100 numbered cowboys. At the same time each cowboy randomly chooses one of the others as his principal ennemy. Then the first cowboy shoots his ennemy. Then the second cowboy (if he is alive) shoots his ennemy (if the ennemy is alive). If his ennemy is already dead the second cowboy passes. Then the third, etc. We assume that cowboys never miss.
What's the probability that the first cowboy will be alive after all?
The expected number of killed cowboys?

[hide="answer"]$\frac{1}{2}$ and 50[/hide]

[hide="addition"]With computer experiments i've found:
If there n cowboys, and k of them have no guns then for large n the probability of survival would be $\frac{1}{(2-y)}$, where y=k/n
[/hide]

[quote="roah"][hide="answer"]$\frac{1}{2}$ and 50[/hide][/quote]

I think that $\frac{1}{2}$ is wrong.

I wrote a program wich calculates [u]all[/u] the [b]1024[/b] possible situations for $N=5$

I got that the expected number of killed cowboys is [b]2.5[/b]. But:

In the [b]1024[/b] iterations the cowboy #1,2,3,4,5 stays alive [b]544, 528, 512, 496, [/b] and [b]480[/b] times respectively

So the first cowboy has more than 50% chance of survive.
Notice that the middle cowboy has exactly 50%

I tried a mathematical solution but it's a little difficult, because we have events that are not independent and not exclusive ...

This is a very interesting problem! Unfortunately so far I haven't got very much.

Let $N$ be the number of cowboys, $p_{N}$ the probability of the first cowboy to survive
and $e_{N}$ the expected number of killed cowboys. Then by direct computation

$p_{2}= 1$, $e_{2}= 1$
$p_{3}= 1/2$, $e_{3}= 3/2$
$p_{4}= 5/9$, $e_{4}= 2$

So $e_{N}= N/2$ could be true, and I suspect there is a short proof, but so far I couldn't find it.
However $p_{N}= 1/2$ cannot be true, and I can't see a reason, why it should hold for $N=100$.
Maybe $p_{N}\to 1/2$ for $N \to \infty$?

I agree with Pontios -- with 3 cowboys, it seems that each has an equal chance of survival (and this chance is $\frac{1}{2}$), but it does not appear to be the case for larger numbers of cowboys.  (I also agree with solyaris' computations.)

It seems that one should be able to build a recursion based on $n$ cowboys, $k$ without guns, for the chances of survival of the $i$-th cowboy.  An easier question to start, maybe, is, "What is the probability that the $i$-th cowboy stays alive long enough to take his turn?"

An interesting variation is what happens if the cowboys choose enemies so that none is chosen by more than one other (i.e., we have a permutation rather than a function with no fixed points).  I suppose one could also consider the possibility where each cowboy can choose himself as his own enemy.

A much better recursion is the one that considers the situation of a fixed number $n+1$ of cowboys out of which $k$ (for kind :)) are without guns but alive and $d$ are dead. Let $E(k,d)$ be the expected number of survivors in this situation. Then, if $k+d=n+1$, we, obviously, have $E(k,d)=k$. Let $k+d<n+1$. Looking at what the first shooter can possibly do, we get the relation
\[E(k,d)=\frac{n-k-d}{n}E(k+1,d+1)+\frac{k}{n}E(k,d+1)+\frac{d}{n}E(k+1,d) \]
which is symmetric in $k$ and $d$ and has the sum of coefficients on the right equal to $1$. Thus, if $E(k,d)$ is a solution, then $(n+1)-E(d,k)$ is a solution as well. But the boundary condition determines the solution uniquely. Hence, for $d=k$, we have $E(k,d)=\frac{n+1}2$ and we are interested in $E(0,0)$. Of course, for large $n$, we can replace the recurrence relation by the first order PDE, which can be then solved by the method of characteristics. I leave this routine exercise to someone else. The probability of surviving of the first shooter seems, indeed, to be slightly different from $1/2$ but I do not know how to compute it exactly yet.

Edit: the probability of survival for the first shooter in the original problem is $\frac{1}{2}+\frac 1{8n}+O(n^{-2})$ for large $n$. The higher order terms can also be found but I'm too lazy to attempt that. There seems to be no nice explicit formula though :(. The idea is exactly that of  JBL: build a recursion, replace it with a second order ODE with a small parameter $1/n$, solve the arising system of first order linear equations and get the result. I'll post the details if somebody is interested :).

nice idea, fedja!

Well, when I first read your solution I was confused about the usage of $k$

Of course you mean that every man who has already shooted is now considered a man without gun.

So I understand now the recursive equation, but I don't know how we apply "the method of characteristics".

I'm interested about this. Could you post more details, please?

[hide="What I΄ve done so far"]
My method has not mathematics at all (I didn't find the numbers by hand, the program did it for me).
 I'll post my work only because I found a function which is a polynomial, and I think that it's interesting

I run a simulation in JavaScript with [b]N+k[/b] cowboys.
[b]N[/b] of them are have gun and [b]k [/b] have not.
The total cases are $\boxed{T(N,k) = (k+N-1)^{N}}$
(Because each of the $N$ shooters has $k+N-1$ choices)

Let $f(N,k)$ be the function which indicates how many times (in total [b]T[/b]) the first shooter is [b]killed[/b]. Then $f$ is a polynomial of [b]k[/b] (for [b]N[/b] constant). And the degree of $f$ is $N-1$.

It is

$f(N,k) = (N-1)\cdot\left((k+N-2)^{N-1}+g(N,k)\right)$, for some polynomial $g$

$\star \text{ }$ Notice that $(k+N-2)^{N-1}=T(N-1,k)$

I found the polynomial $g$ for $N=3,4,5,6$. For larger $N$ the computer needs much time

$g(3,k) = 1$
$g(4,k) = 3k+4$
$g(5,k) = 6k^{2}+28k+39$
$g(6,k) = 10k^{3}+100k^{2}+365k+456$

I can't find any relation between these polynomials, except the fact that the first coefficient seems to follow the sequence [b]1,3,6,10,15,21,...[/b] the triangular numbers

[color=green][b][u]An example[/u][/b][/color]

For $N=3$ I got

[code]k	f(3,k)
-----------
0	4
1	10
2	20
3	34
4	52
5	74
6	100
7	130
8	164
[/code]

You can see that $f(3,k) = 2[(k+1)^{2}+1]$
[/hide]

[quote="pontios"]
So I understand now the recursive equation, but I don't know how we apply "the method of characteristics".
[/quote]
Here goes: Assume that $E$ is actually a smooth function in the triangle $x,y\ge 0$,$x+y\le n+1$. Then, replacing $E(k+1,d)-E(k,d)$ by $E_{x}(x,y)$ and such, we get the first order PDE
\[(1-y)E_{x}+(1-x)E_{y}=0\]
Let $(x(t),y(t))$ be such a curve that $\frac{dx}{dt}=1-y$, $\frac{dy}{dt}=1-x$. Then the equation essentially says that $E$ is constant on that curve. Thus, given a point $x,y$, one can look at the curve of this type passing through the point and find where it intersects the line $x+y=n+1$ on which the values are known to determine $E(x,y)$. I hope you know how to solve systems of linear ODE with constant coefficients, so I'll let you proceed from here.

Indeed, there was a mistake: not the first cowboy, but randomly chosen.

Elementary proof for expected value:
At each moment of time there cowboys of 3 types Dead, Active (wich are alive and are waiting for their turn) and Passive (alive, but they have already done their shoot).
A+P+D=100 (always)
There may be 3 type of change after one shoot:
1) D:=D+1
2) P:=P+1
3) D:=D+1 and P:=P+1
Each trajectory starts at (P,D)=(0,0) and ends somewhere on P+D=100.
Each trajectory has symmetric one (symmetry about P=D) with equal probability.
So, the resulting distribution of A is symmetric about 50. QED.

Source of problem and solution:
http://community.livejournal.com/ru_math/511593.html


limit
calculus
algorithm
analytic geometry
graphing lines
slope
algebra

why does maxima not calculate the following limit?:
$\lim \frac{1-cosx+3sinx-(asinx)^{3}-x^{2}+4x^{4}}{(atanx)^{3}-6(sinx)^{2}-5x^{3}}$
for $x\longrightarrow0+$

Because the limit does not exist.

If you approach from the right, it diverges to infinity, and if you approach from the left, it diverges to negative infinity.

Try using series or L'Hopital's (or just graphing) to see why.

FYI, you're better off posting calculus questions in the calculus forum  :) From the forum index:
1. select the College Playground Section
2. then select "Calculus-Real Analysis"
3. Go to "Calculus Computations and Tutorials"

ok, but when I make, e.g., the left-sided limit ($x\longrightarrow0-$), maxima doesn't give a result, while I can plot a graph with derive; and derive gives me, as a result for the left-sided limit $\pm\infty$ and not just the positive infinity;
ps: in the graph I have plotted with derive, if you approach from the left, it diverges to infinity (and from right to neg inf), just the opposite to what you said
pps: I think there's some trouble with both the algorithms to calculate limits of maxima and derive, or there must be something wrong with the graph derive plots to me.

You're write, I did write them backwards.

How exactly are you trying to use maxima to solve this? I don't know exactly what you're trying to do.

Take a look at the numerator and denominators though.

Plot each separately. Notice that the numerator can be approximated near $x=0$ by a line with positive slope (I believe $y=3x$.

The denominator has a horizontal tangent there, and can be approximated with the line $y=0$, but is essentially negative.

So when you're sufficiently close to zero, the denominator is a much "stronger" zero than the numerator.

Coming from the positive side, you're dividing $3x$, which is positive, by a very small negative number, so it diverges to negative infinity. From the negative side, $3x$ is now negative, and it diverges to positive infinity.

What exactly do you mean by "maxima"?

And I take it that $asin$ and $atan$ mean $\arcsin$ and $\arctan$, respectively.

I agree with you (The Zuton Force), and I can plot them both separately and together;
but when I ask maxima (I mean "wxMaxima 0.7.0a http://wxmaxima.sourceforge.net; Maxima 5.10.0 http://maxima.sourceforge.net" or more recent versions) to make the limits (both left-sided and right-sided) it doesn't work (to be exact it says: " maxima is calculating").
I think maxima has some trouble with taylor polynomials or something similar or maybe its algorithm doesn't work well.
ps: with asin(x) I mean $arcsin(x)$, or $sin^{-1}(x)$.
pps: maybe we have to ask a moderator to move this post in another part of the forum..

:rotfl: I thought you meant you were trying to use relative extrema to solve it... my bad...

So yeah, I'm having the same problem with your program. I can't help you though because I have no idea how it works  :( sorry.

A bunch of other programs seem to do it just fine though, including Mathematica, Matlab, Geometer's Sketchpad, and a bunch of online Java plotters.


inequalities
trigonometry
inequalities open

If $ ABCDE$ is a convex pentagon with angles: $ \angle \alpha_1\equal{}\angle EAB$, $ \angle \alpha_2\equal{}\angle ABC$, $ \angle \alpha_3\equal{}\angle BCD$, $ \angle \alpha_4\equal{}\angle CDE$, $ \angle \alpha_5\equal{}\angle DEA$(apparently angles are consecutive), determine if the following inequality true:

\[ \sin \alpha_1 \sin \alpha_2 \sin \alpha_3 \sin \alpha_4 \sin \alpha_5 \geq |\sin (\alpha_1\plus{}\alpha_2)\sin (\alpha_2\plus{}\alpha_3)\sin (\alpha_3\plus{}\alpha_4)\sin (\alpha_4\plus{}\alpha_5)\sin (\alpha_5\plus{}\alpha_1)|\]

Is anybody familiar with this problem? Do u have any ideas how to solve it if its true? I've checked it for like 10 pentagons and it seems true, but i cant find a proof for it... If its true, than probably the general version is correct too ;) :

Let $ \alpha_1, \alpha_2,...,\alpha_n$ be consecutive angles of a convex $ n$-gon. Then the following inequality stands: (of course $ \alpha_{n\plus{}1}\equal{}\alpha_1$)
\[ \prod_{i\equal{}1}^n \sin \alpha_i \geq |\prod_{i\equal{}1}^n \sin(\alpha_i\plus{}\alpha_{i\plus{}1})|\]