Chapter Mathcounts coming up, just curious, but who else is doing mathcounts?

I am!

Darn, my message is too small.  Whatever, this line will make it longer.

Disclaimer. Please don't talk about anything but your placement. No scores. No problems. Not even if you think it was hard or not.

So how did everyone do? (That is, if you are done.)

I saw on the national board that Sitan took first in his chapter and River Trail won.

In Cobb, Dickerson won with Hightower trail 2nd. (Dodgen was third, but didn't make it as a team.)

Edward Park was first in the written, followed by Claus Zheng and Slok Carpenter. Dickerson had 7 of the top 12. Dodgen had 3 make state. Claus won the countdown over Edward in the final. Edward reminds me a lot of Sitan. First of all he's short....  :lol: Edward is in 6th grade, REALLY quick, and pretty intuitive. (I think I'm starting to figure out the 2012 Ga. ARML team!)

Tom

Hi, Mr. Fulton!  Um, yeah, I just wanted to remind you about the private message about nats problems from last year.  Sorry if I sound impatient. :oops:
Hope to see you at state MC! :lol:

PS: I'm not that short...  I'm 5 ft 1.25 in., so ha! jkjk

Um, Mr. Fulton, does Edward Park have a username on the forum, cuz I might know him via AoPS?

What did Claus (or is it Klaus? I've read it both ways) get on written? How abt. Edward?

OOOOOOOOOOOooooooooo....  Arvind, you told us your score, which is forbidden.

Arvind, please edit your post and remove the score. We can't talk about those until later. And Claus (that's the correct spelling unless he takes both) beat you!  :lol:

Sorry, I edited it... :blush:

When can we discuss?

Probably not until the end of the month. I think that chapters have the end of February to administer the test.

There's a different competition that I am eager to discuss, but I will not name it. I will just wait until Sunday.

Well, Mathcounts State's tomorrow...HA JUST KIDDING IT'S SATURDAY.

So, is Claus coming? What about Edward and Slok? I am coming, and so are sitan, yubo, and alan.

River Trail VS Dickerson, Round 1 Results pending...

Yes, they are all coming. As well as 3 other Dickerson kids. They qualified 7 total for State (and 4 for the AIME.)

Hmm, well we have me, Sitan, Alan, and Yubo. We're the team, but we have other people (I forgot, but I think we have 6 or 7 total). I think we have 3 or four who made AIME. this shud be good, we shud bet on it.  :P

But wait, we can't bet in skool (or so they say...) :roll:

[quote="Arvind_sn"]I think we have 3 or four who made AIME[/quote]

Um, actually, Arvind, we have only 2 people who made AIME.

PS: Arvind, I like your goals for 2008. :wink:

Don't worry; you'll all meet each other during practices.  Nationals practices were the first time I really got to know people my age who were into math and weren't from my school.  Also, in my day, they posted a growing pdf of the national competitors on the national mathcounts website.

Hm, getting 5th must be pretty painful.  The good news is that it gets much better in high school, and plus, if you're that good now and you work hard through next year, you will probably make MOP.  (The "working hard" part is very important.)  Unless you're in seventh grade, in which case you have nationals next year again.

Congrats to the winners and good luck at nationals! :)

Um, when are practices? :|

Or do we organize those ourselves?

Umm, the state coach is supposed to organize them, so I guess Mrs. Harris will be setting up a meeting some time soon.  idk when though.

For those that are interested, the Countdown round will be webcast live tomorrow (Friday) at 5pm. Go to the MATHCOUNTS home page. http://www.mathcounts.org

Why is one post on Apr 10, but the next on May 10? Did people forget about this topic?

This isn't of much importance, but I'm just wondering  :D

P.S. Hey Sitan! You changed your signature. When did you do that? I'm not very attentive

Um, what's your real name?  I don't seem to remember seeing you on this forum before, but my memory is probably bad.

perhaps the guy who sat behind me at practice this sunday? he seemed to cough a lot.

[quote="1077400"]Your'e right, I haven't posted on this topic before that. As to who I am, I'm... drum roll please....  SOMEBODY....... MUAHAHAHAHAHAHAHAHAHAH *cough cough cough cough cough*
Sorry, I have coughing problems

You still didn't answer my question about the signature...  :D

I gave you a clue as to who I am. Find out yourself[/quote]

Edward Park?  But isn't your username edwardpark97?  Or did you make another account?

I didn't care about this site before Nats, but I'm was the 4th place person in Georgia.

Can we comment about Nats now?

OMG, it's John!  Wow, I didn't know that you had an AoPS account...

I just made one a few days ago.

BTW, it's spelled Jiyang.

Hi Stan!


Suppose a problem says "so-and-so holds for all triangles." May we assume they hold for degenerate triangles as well?

EDIT: By the way, sorry if this is in the wrong forum.

EDIT2: If you're asking why, I was practicing on [url=http://www.unl.edu/amc/a-activities/a7-problems/USAMO-IMO/q-usamco/pdf/usamo2001.pdf]this[/url], and assuming it holds for degenerate triangles trivializes the problem. I just need a clarification on this.

I'm pretty sure it shouldn't include degenerate triangles...

For example, a degenerate triangle can't have a circumcenter, since all the bisectors are parallel. Similarly, it can't have an orthocenter, since the perpendiculars from the "vertices" to the "opposite sides" would also be parallel to one another. Moreover, you can't define the centroid for it, since the "intersection point" of the "medians" would actually be a segment.

[quote="Zhero"]Suppose a problem says "so-and-so holds for all triangles."
 May we assume they hold for degenerate triangles as well?
[/quote]

Also:

Asking this is partly analogous to stating "so-and-so holds
for all polygons."
And then the question asked is "May we assume they hold for
self-intersecting polygons?"  No, not necessarily, because they
are deviants.

In my opinion, I would make sure your triangles are in the
Euclidean Plane, have sides of finite length, and have positive,
but finite area.

Alright, thanks.

Dang, that #6 would've been so trivial T__T


There's no elementary antiderivative- but there is symmetry. Try substituting $ x\mapsto \frac1x$, and then adding the two integrals.

thanks

only maybe give me complete solution please

Not without see an effort from you.

Look at jmerry's hint.


Prove that $\forall x,y,z \in R$ we have:
$x(x+y)^{3}+y(y+z)^{3}+z(z+x)^{3} \geq 0$.

cheers! :D :D :cool:

This one is a very difficult problem. I spent a lot on it, but I didn't manage to find a satisfactory solution. So, any suggestion? You say you have a solution Lagrangia. I'm interested.

This is a very interesting and very difficult inequality.

well, I have a solution, but I think we should also give the other members of the forum a chance to think about it! :D:D

When do you post a solution for this VERY VERY NICE inequality, Lagrangia?
I am very curious to know it.

Thank you very much.

yea, I think by now enough time's passed?

I came with a solution for this one (dont know if its correct... maybe you can tell me :D )
Here it is:
$[x(x+y)^3+y(y+z)^3+z(z+x)^3][ \frac{1}{x(x+y)}+\frac{1}{y(y+z)}+\frac{1}{z(z+x)}]\geq 4(x+y+z)^2$
so

$x(x+y)^3+y(y+z)^3+z(z+x)^3\geq \frac{4(x+y+z)^2}{\frac{1}{x(x+y)}+\frac{1}{y(y+z)}+\frac{1}{z(z+x)}}$
Now we have to prove that:
$\sum{\frac{1}{x(x+y)}}\geq0$
Since
$x+(x+y)\geq 2\sqrt{x(x+y)}$ or
$\frac{(2x+y)^2}{4}\geq x(x+y)$
Hence
$\sum{\frac{1}{x(x+y)}}\geq\sum{\frac{4}{(2x+y)^2}}\geq0$

Gohan, your solution is wrong. I see that you are applying AM-GM for real numbers? :o

See http://www.mathlinks.ro/Forum/viewtopic.php?t=30448 for a solution and http://www.mathlinks.ro/Forum/viewtopic.php?t=25451 for a similar problem.

  darij

I will show that $x(x+y)^3\geq0$
We have that $x(x+y)^3=x^4+3x^3y+3x^2y^2+y^4\geq x^4+3x^2y^2+3x^2y^2+y^4\geq0$
This from Muirhead. Similar we work at the others and we were done.

This is not true.

Take $x=-1$ and $y=2$ ....

This problem was put in M&Y long time ago .
   This problem can also be solved by using Vasc's ineq : $ (x^2+y^2+z^2)^2 \geq 3(x^3y+y^3z+z^3x) $ . I tried it , it's OK ;)


function
inequalities
inequalities proposed

After transformations we must prove $ F(8,0, \minus{} 2) \plus{} 2F(4,3, \minus{} 1)\ge F(6,0,0)$ where $ F$ Murchad function, but for any $ x,y > 0$
$ \frac {x^8}{y^2}\ge 4x^6 \plus{} 4x^2y^4 \minus{} 6x^4y^2 \minus{} y^6$ so $ F(8,0, \minus{} 2)\ge 3F(6,0,0) \minus{} 2F(4,2,0)$ but $ F(4,3, \minus{} 1)\ge F(4,2,0)$

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
\[ \left( \frac {a^4}{b} \plus{} \frac {b^4}{c} \plus{} \frac {c^4}{a}\right)^2 \ge 3(a^6 \plus{} b^6 \plus{} c^6)
\]
[/quote]
Dear Can,that you knew,i am quite busy now,so i only post my idel,sorry if it is wrong (because i am a stupid student)
Let $ Let a\equal{}e^x,b\equal{}e^y,c\equal{}e^z$
Then I think we can prove by Karamata Ineq with convex $ e^x$ (the hard step is proving  majorizes)
Who can help me

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
\[ \left( \frac{a^4}{b}\plus{}\frac{b^4}{c}\plus{}\frac{c^4}{a}\right)^2 \ge 3(a^6\plus{}b^6\plus{}c^6)\][/quote]
Buffalo Way ($b=a+u$, $c=a+v$) helps here.

[quote="Mirus"]Arqady, what do you mean ? After buffalo substitutions we need to expand all ? If so, it is almost impossible by hand !!![/quote]
Why? The hour of the work. :wink:  But I agree with you that it's very ugly.

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
\[ \left( \frac{a^4}{b}\plus{}\frac{b^4}{c}\plus{}\frac{c^4}{a}\right)^2 \ge 3(a^6\plus{}b^6\plus{}c^6)\][/quote]

Very beautiful inequality! :)
I really like it, so I hope I will find a proof using only standard inequalities for the last step.

Using substitution: $\frac{a^4}{b}=x^{21}, \; \frac{b^4}{c}=y^{21}, \; \frac{c^4}{a} = z^{21}$ , inequality becomes:

$\left( \sum \frac{x^{20}}{yz} \right)^2 \ge 3\sum x^{30}y^6$

Using Cauchy-Schwarz inequality, we have:

$\left( \sum x^{30}y^6 \right)^2 \leq  \left( \sum x^{24} \right) \left( \sum x^{36}y^{12} \right)$

Now, because of well known Vasc's inequality: $(p^2+q^2+r^2)^2 \ge 3(p^3q+q^3r+r^3p)$
we have:
$ \sum x^{36}y^{12} \leq \frac{1}{3}  \left( \sum x^{24} \right)^2$

Therefore, it remains to prove:

$ \left( \sum \frac{x^{20}}{yz} \right)^4 \ge 3  \left( \sum x^{24} \right)^3$

What can be proved using EV theorem. $\blacksquare$

Cauchy schwarz gives us $(\frac{a^4}{b}+\frac{b^4}{c}+\frac{c^4}{a})^2 \ge (a^8+b^8+c^8)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}). Expanding this gives a^6+b^6+c^6+\frac{a^8+b^8}{c^2}+\frac{a^8+c^8}{b^2}+\frac{b^8+c^8}{a^2}. This is clearly greater than or equal to the RHS of the problem statement by rearrangement.

Is this ok, or did i do my cauchy wrong?

EDIT: oops.

[quote="filetmignon821"]
Is this ok, or did i do my cauchy wrong?[/quote]
The Cauchy application is indeed wrong. The statement of Cauchy-Schwarz is $(a_1^2+a_2^2+\cdots+a_n^2)(b_1^2+b_2^2+\cdots+b_n^2)\ge (a_1b_1+a_2b_2+\cdots+a_nb_n)^2$, you used the reverse of this.


A "perfect power" is a numer that can be written as a positive integer raised to an integer power greater than 1. For example, 125 is a perfect power because it is equal to $5^{3}$. The list 2,3,5,6,7,10,11,12,13,14,15,17... contains every positive integer less than 1000 that is not a perfect power. How many integers are in the list.

[hide]Use complementary counting.
So we are trying to count all of the perfect powers. Since powers get very big very fast, this will not take too much time.

Perfect Squares:

There are 31 perfect squares, as $31^{2}= 961$.

Perfect Cubes:

There are 9 perfect cubes, because $10^{3}=1000$ is not included.

Now we have overcounted all of the powers of 6, and there are 3 of these.

Perfect fourths were already counted when we counted perfect squares.

Perfect fifths (sounds like music):

There are 2 perfect fifths (other than 1).

Perfect sixths were counted.

Perfect sevenths: there is 1 perfect seventh (other than 1).

Eighths were counted.

Ninths were counted.

Tenths are too big.

So there are $31+9-3+2+1=\boxed{40}$ perfect powers.

EDIT: Forgot to subtract. $999-40=\boxed{959}$ nonpowers.
[/hide]


calculus
integration
real analysis
real analysis unsolved

my friends recommended to me this question
$\int-2x^{3}\cdot e^{ \int{-2x dx}}\cdot dx$


Steven is making a scale drawing of his living room. He is using a scale of one-quarter inch to one foot. In the drawing, how many inches will represent the length of a 72-inch long couch? Express your answer as a decimal to the nearest tenth.

72 inch in real life = 6 feet in real life = 6*(1/4) inches in scale drawing = [b]1.5[/b] inches.

since one foot is one quarter inch, we have to convert the length of the couch into feet

$ 72 \text {inches} \equal{} 6 \text{feet}$
so now we set up a proportion

$ \frac {0.25\text{inch}} {1\text{foot}} \equal{} \frac {x\text{inches}} {6\text{feet}}$

$ x \equal{} 1.5$

our answer is $ \boxed{1.5}$

edit: beaten...
BTW, how would you make a space between the number and the unit?
pm me the answer to reduce spam plz


linear algebra
matrix
vector
linear algebra unsolved

$E$  is an euclidian space. $G$ is the gramian matrix of
$v_1,..,v_n$. Suppose that all eigenvalues of $G$ are either
$0$ or $1$. prove then that there exists an orthogonal projector
$p$ such that $u_i=p(e_i)$ for all $i$ where $(e_1,..,e_n)$ is the
canonic basis.

Since the gramian is symmetric, it's orthogonally diagonalizable, so there is an orthogonal matrix $A=(a_{ij})_{i,j}$ s.t. $AGA^t$ ahs $1$ as its first $m\le n$ diagonal entries, and $0$ everywhere else. However, $AGA^t$ is the gramian matrix of the system of vectors $(\tilde v_i)_{1\le i\le n}$, with $\tilde v_i=\sum_{j=1}^n a_{ij}v_j,\ \forall i\in\overline{1,n}$, and the conclusion is easy to draw from here (all we need to do is express the $v_i$'s in terms of the $\tilde v_j$'s, and we'll obtain exactly what we want).

I hope it's ok.


calculus
integration
rotation
LaTeX
geometry
geometric transformation
calculus computations

Okay, so I ran into another "tough" problem...

You have the region enclosed by y=1-x^2  and y=0

How would I set up this integral if this region were rotated around the axis x=2?  I can do it if it were rotated around the x-axis just fine:

Integral from -1 to 1 of pi(1-x^2)^2

But I can't seem to figure out how to set up the integral for the washer method going around x=2....

PS- Does anyone know any good LaTex tutorials?  I'm sorry this is probably hard to read, but I haven't been able to find any good latex tutorials yet.  Thanks again

Try setting it up with respect to $y$.  (The integral will involve some expression in terms of $y$ and $dy$.)

For help with $\LaTeX$:

http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php
http://test.izyba.com/equationeditor/equationeditor.php

To use $\LaTeX$ on these forums just include the code between two dollar signs ($\$$).  No downloading is necessary to use $\LaTeX$ on these forums.

So I would get x = sqrt(-y-1)....  The problem is I don't know what to do to take into account the rotation around x=3...  Would I have to find new bounds?  Thanks


[quote="April"]Solve the system of equations:
\[\begin{cases}2^{x}-2^{y}=\left(y-x\right)(xy+2)\\ x^{2}+y^{2}=2\end{cases}\]
[/quote][hide]$2^{x}-2^{y}=\left(y-x\right)(xy+x^{2}+y^{2})=y^{3}-x^{3}$
$x^{3}+2^{x}=y^{3}+2^{y}$, and $f(x)=x^{3}+2^{x}$ is increasing, so $x=y$, and the only solutions are $(1,1)$ and $(-1,-1)$
[/hide]


Consider a non conducting thread of length $ 2l$ attached to whose ends are two like charges of magnitude $ q$ and mass $ m$.Suddenly the mid point of the thread starts moving perpendicular to the thread at  constant speed $ v$ whcih is alsway perpendicular to the thread.Find the minimum distance $ d_{0}$ between the charges.

Consider frame of reference of the midpoint. The charges have initial kinetic energy $ mv^2$ and final potential energy $ \frac{q^2}{4\pi\epsilon_0 d_0}$
As tension in thread is always perpendicular to motion, energy is conserved.
Hence, $ d_0\equal{}\frac{q^2}{4\pi\epsilon_0 mv^2}$

$ 2 \frac{mv^2}{2}\plus{}\frac{q^2}{8\pi \epsilon_0 l}\equal{}0\plus{}0\plus{}\frac{q^2}{4\pi \epsilon_0 a}$

$ a\equal{}\frac{kq^2}{mv^2 \plus{}\frac{kq^2}{2l}}$

$ k\equal{}\frac{1}{4\pi \epsilon_0}$
Thus,

$ d\equal{}\frac{2lq^2}{q^2 \plus{}8\pi \epsilon_0 mv^2 l}$

[quote="pardesi"]yes that was ob...but importantly what's the reason :D  u conserved energy especially what aout the work done by the rope[/quote]

As My namesake posted. F Perpendicular to V ==> F Perpendicular to X =? W.D by Thread =0/

Nice question man.

thanks....
but the mid point of the thread moves perpendicular to it ...what about the charges :)
.

Since he has considered it as one system. Thread becomes internal force na?


Prove that the solutions (n, x, y)of equation   u^x -u^y =n!  are finite.


function
logarithms
inequalities
number theory
prime numbers
number theory open

Here are two problems
The first one is trivial while second is hard(seems Open)
1.For every $n \in \mathbb{N}$ ,prove that there exist a $k \in \mathbb{N}$ s.t
$k$,$k+1$,$k+2$,...,$k+(n-1)$ are not primes.

2.let $P_k$ be the $k_{th}$ prime number.
For any even natural number like $\mathcal F$, prove that there exist a $\mathcal R \in \mathbb{N}$ such that
$P_{\mathcal R+1}-P_{\mathcal R}=\mathcal F$

PS.In another word,These problems are saying that There exist any big enough space between the numbers without any prime  and
In the second part ,there is a big enough Gap between exactly two primes.

Yes Ali ,This is right for the fist part cause any number with the form of
$2 \le i \le n+1, (n+1)!+i$ is not prime cause $i(\frac{(n+1)!}{i}\in \mathbb{N} +1)$.

Now the real problem ,solve the Second one.

Maybe Rankin proved that $g_n > cF(n)$ for infinitely many $n$?  Because the inequality is not true for all $n$.

By the way, my daughter asked me Problem 2 when she was 4 years old!  I had to confess to her that I didn't know the answer, and suspected that it was an open problem.

:o  :o  :o  :o Ravi are you serious ,4 years old.Gosh.
She is definitely, like this father, a genuis .
I am sure ,I couldnt understand the meaning of prime numbers when i was 4!

Thanks, but neither of us is a genius.  I was showing her the Sieve of Eratosthenes, and we were looking for prime gaps of length 2, 4, 6, ....  That's when she asked the question.  At first, I said the answer is yes, thinking she meant is there a prime gap of at least n (which is true for all n by the Prime Number Theorem).  But then I realized what she was asking, and I was stumped.

I wouldn't have guessed that a 4-year-old could stump me with a math question!

P.S. I like your signature.  The funny thing is that, even though I proved the inequality, I have trouble remembering the statement of it!  All those indices...

[quote="silouan"]Let the function $F(n)=\frac{(\log n)(\log (\log n))(\log(\log (\log (\log n))))}{\log (\log (\log n)))^2}$
Rankin prove that $g_n>cF(n)$ where c a constant positive number c<1,31256.
The $g_n$ is the gap between two numbers prime p and p+2.[/quote]

Rankin proved that the gap between consecutive primes  $p_{n+1} - p_n > \exp(\gamma) F(n)$, where $\gamma$ is Euler's number, [i]infinitely often[/i] in Proc. Edinburgh Math. Soc [b]13[/b] (1962/3) 331-332.

[quote="Ravi B"]Maybe Rankin proved that $g_n > cF(n)$ for infinitely many $n$?  Because the inequality is not true for all $n$.

[/quote]
My book says that it is true for infinitely many $n$


symmetry
inequalities
inequalities solved

So equality occurs if a=b=c=d=1. Now we slightly decrease a and slightly increase b. In other words, we keep c^2+d^2 constant at 2, having a^2+b^2=2, now AM-GM gives 1>=ab. So however we change a and b, we only will decrease the product ab. Now since c,d, must equal 1, we must show that a+b+2>=ab+a+b+1, or a+b+1>=ab+a+b, which we just showed. By symmetry, it follows that if any two variable are kept constant, the inequality holds. I think you could finish up the proof here, but I'm not sure...
Other attempt: we have $a^2+b^2+c^2+d^2=8$, add $ab+bc+cd+da$ gives $(a+b)^2+(b+c)^2+(c+d)^2+(d+a)^2=8+2(ab+bc+cd+da)\geq 16(a+b)(b+c)(c+d)(d+a)$

[hide]
Cauchy: $(ab+bc+cd+da)^2\le (a^2+b^2+c^2+d^2)(b^2+c^2+d^2+a^2)$

$\iff ab+bc+cd+da\le 4$

ISTS $a+b+c+d\ge 4$.

AM-GM: $\frac{1}{4}(a^2+b^2+c^2+d^2)=1\ge \sqrt{abcd}\iff abcd\le 1$

AM-GM: $a+b+c+d\ge 4\sqrt[4]{abcd}$

Eh, I can't finish it.
[/hide]

[quote="chess64"][hide]
ISTS $a+b+c+d\ge 4$.

[/hide][/quote]
I do not think this is always true. Take $a$ approaches $2$ from the bottom, and $b,c,d$ are really small.
I find this problem quite difficult.

Maybe something like this:

By Cauchy:
4=sqrt(4(a^2+b^2+c^2+d^2))>=(a+b+c+d);
So, 2(a+b+c+d)>=?2(ab+bc+cd+da)
8+2(a+b+c+d)>=?(a+b+c+d)^2
(a+b+c+d-4)(a+b+c+d+2)<=? 0
Since a+b+c+d-4 <=0, a+b+c+d+2 >0, the above inequality holds, so the given inequalitiy do holds, too.

a possible sol [hide]
$ab+bc+cd+da \leq 4$
$4(ab+bc+cd+da) \leq (a+b+c+d)^2 \Leftrightarrow (a+c-b-d)^2 \ge 0$[/hide]

[quote="pvthuan"]chess64, you are WRONG. :?[/quote]

How can I be wrong if I didn't even finish? :? I just suggested a possible beginning of a solution.

Here is the solution:

We rewrite the inequality as
\[ a+b+c+d\geq (a+c)(b+d),  \]
or
\[ \frac1{a+c}+\frac1{b+d}\geq1.  \]
It remains to prove that
\[ a+b+c+d\leq4,  \]
since $1/(a+c)+1/(b+d)\geq4/(a+b+c+d)$.
I think now you can do the rest.

[quote="pvthuan"]
It remains to prove that
\[ a+b+c+d\leq4,  \]
since $1/(a+c)+1/(b+d)\geq4/(a+b+c+d)$.
I think now you can do the rest.[/quote]

\[ a+b+c+d\leq 4 \Leftrightarrow \frac{a+b+c+d}{4}\leq 1 \]

\[ \frac{a+b+c+d}{4}\leq \sqrt{\frac{a^2+b^2+c^2+d^2}{4}}=1 \]

[quote="pvthuan"]Let $a,b,c,d$ be positive real numbers such that \[a^{2}+b^{2}+c^{2}+d^{2}=4.\] Prove that \[a+b+c+d\geq ab+bc+cd+da.\] [/quote]
I am biginner. :lol:
$a+b+c+d\geq ab+bc+cd+da\Leftrightarrow$
$\Leftrightarrow(a^{2}+b^{2}+c^{2}+d^{2})(a+b+c+d)^{2}\geq4(ab+bc+cd+da)^{2}.$
But $(a+b+c+d)^{2}\geq4(a+c)(b+d)=4(ab+bc+cd+da)$ and
$a^{2}+b^{2}+c^{2}+d^{2}\geq ab+bc+cd+da\Leftrightarrow(a-b)^{2}+(b-c)^{2}+(c-d)^{2}+(d-a)^{2}\geq0.$

Wow, this problem has so many solutions... also go to the topic I started about this problem arqady, it is called "easy" inequality... also, I didn't understand some of the steps in your amazing proof, could you explain how to go from line 1 to line 2?

[quote="me@home"] could you explain how to go from line 1 to line 2?[/quote]

raelly simple :wink:

$a+b+c+d\geq ab+bc+cd+da$
$\Rightarrow (a+b+c+d)^{2}\geq (ab+bc+cd+da)^{2}$
multiply both sides by 4:
$\Leftrightarrow(a^{2}+b^{2}+c^{2}+d^{2})(a+b+c+d)^{2}\geq4(ab+bc+cd+da)^{2}.$

inequality is almost trivial. :)

yeah i figured that out about a minute after posting it... except by then i had to leave the library and didn't get computer access till i got home a half hour ago today...

I was just looking at this problem 5 minutes before you posted it... what a coincidence! (it is in Hojoo Lee's pdf)
It seems like a mess to me, first substitute $a^{2}\mapsto_{cyc}a$
Then homogenize:
it becomes \[(a^{2}+b^{2}+c^{2})^{3}\left(a^{2}+b^{2}+c^{2}+2\left( \sum_{cyc}{ab}\right) \right) \geq (27)\left(\left(\sum_{cyc}a^{4}b^{4}\right)+2\left(\sum_{cyc}a^{2}b^{4}c^{2}\right)\right)\] This doesn't seem a good way to go, now I'm thinking Newton symmetric inequalities...

i know this has nice solution, let me thik :wink:

Well, let me try :)

We can write it as $(a+c)+(b+d) \geq (a+c)(b+d)$, but, for AM-GM, we have $(a+c)+(b+d) \geq 2\sqrt{(a+c)(b+d)}$, so we need $2\sqrt{(a+c)(b+d)}\geq (a+c)(b+d) \Longleftrightarrow 4 \geq (a+c)(b+d)$

By Cauchy: $(ab+bc+cd+da)^{2}\leq (a^{2}+b^{2}+c^{2}+d^{2})^{2}\Leftrightarrow ab+bc+cd+da \leq 4$ (on $(a,b,c,d)$ and $(b,c,d,a)$)

Wow I've improved so much from the beginning of this year :P [quote="Do Tien Dang"]I'm beginner too  :lol:  :lol:
$a+b+c+d-ab-bc-cd-da = \frac{1}{2}((a+b+c+d)-(a+b+c+d)^{2}+a^{2}+b^{2}+c^{2}+d^{2})$
$\text{ Let }t = a+b+c+d \leq 2 \sqrt{3}\text{ Then }-t^{2}+t+4 \geq 0$
$\text{ ( As desire )}$  :)

$\text{ We have a similar little problem }$
$a+b+c= 3 \text{ prove that }\sqrt{a}+\sqrt{b}+\sqrt{c}\geq bc+ca+ab \text{( Russia 2000 )}$ :blush:[/quote]
[hide="Okay this is rather tight but still only moderate level (basically, its not trivial)."]
Note that $2\ge (b+c)\sqrt{a}\iff 1+1+a^{3/2}\ge 3\sqrt{a}$ which follows by Am-Gm. It follows that: \[\sum_{cyc}\sqrt{a}\left(2-(b+c)\sqrt{a}\right)\ge 0 \iff \sum_{cyc}\sqrt{a}\ge \sum_{cyc}ab\][/hide]

[quote="Je_nie_dieu"][quote="arqady"]
I am biginner. :lol:
$a+b+c+d\geq ab+bc+cd+da\Leftrightarrow$
$\Leftrightarrow(a^{2}+b^{2}+c^{2}+d^{2})(a+b+c+d)^{2}\geq4(ab+bc+cd+da)^{2}.$
But $(a+b+c+d)^{2}\geq4(a+c)(b+d)=4(ab+bc+cd+da)$ and
$a^{2}+b^{2}+c^{2}+d^{2}\geq ab+bc+cd+da\Leftrightarrow(a-b)^{2}+(b-c)^{2}+(c-d)^{2}+(d-a)^{2}\geq0.$[/quote]

Could somebody explain why the last but one line is true?[/quote]
By Am-Gm, $(b+c)^{2}>4bc$

[quote=pvthuan]Let $a,b,c,d$ be positive real numbers such that
\[ a^2+b^2+c^2+d^2=4. \]
Prove that
\[ a+b+c+d\geq ab+bc+cd+da. \]

That's not very hard inequality.[/quote]

By rewriting the inequality 2 times we get:

$a(1-b)+b(1-c) + c(1-d)+ d(1-a) \geq 0$  and
$a(1-d)+b(1-a) + c(1-b)+ d(1-c) \geq 0 $

Then
 $ (2-a-c)(b+d) +(a+c)(2-b-d) \geq 0 $  But
   $  (2-a-c)(b+d) +(a+c)(2-b-d) \geq (b+d)(4-a-b-c-d) \geq 0 $

If we consider  WLOG, $ b+d \geq a+c $  and also  by CS :

 $$ 4(a^2 + b^2 +c^2 + d^2)= 4^2 \geq (a+b+c+d)^2 $$
The equality case is easy.

after proving (a+c)(b+d)<or=4 using GM >or=HM our desired result is obtained.

Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=3$. Prove that
$$a+b+c\ge ab+bc+ca.$$


show that In any triangle, orthcenter,insogonal conjugate of nagel point,feuerbachpoint are collinear

Isogonal conjugate of nagel point is external similitude center of $ (O)$ and $ (I)$, Feurebach point is external similitude center of $ (I)$ and $ (E)$, orthocenter is external similitude center of $ (O)$ and $ (E)$. We have the result.


If $ \omega ^ 2 \plus{}\omega \plus{}1\equal{}0$, then find all pairs of integers $ a,\ b$ such that $ |a\omega \plus{}b|\equal{}1.$

Just a tip that might help:
The roots of the quadratic are the primitive third roots of unity.
Maybe some manipulation may be involved.


Hi manlio. Could you solve it? Mu solution is very nice.

Thank you very much for this nice inequality. I hope to solve it by myself. I am trying :huh:

[quote="manlio"]This is very like to Vasc ineq

[quote="Vasc"]My solution is a little simpler. Assume that $a\le b \le c$, then write the ineq as
$\sum\frac{a(b+c-a)}{a^2+bc} \ge 0$.
Since $b+c-a>0$, it suffices to show that
$\frac{b(c+a-b)}{b^2+ca}+\frac{c(a+b-c)}{c^2+ab} \ge 0$.
This is equivalent to
$(b-c)^2(a-b)(a-c)+abc(2a+b+c) \ge 0$,
which is clearly true. ;)[/quote]

but I cannot apply the same method :(

[quote="ductrung"]It is true that
$\frac{a(b+c-a)}{a^2+2bc}+\frac{b(c+a-b)}{b^2+2ca}+\frac{c(a+b-c)}{c^2+2ab}\ge\frac{81abc(ab+bc+ca)}{(a+b+c)^5}$[/quote]

How did you find this very interesting fact?

Thank you very much.[/quote]

The solution for this problem is very nice and hard. It's not easy, I think.

[quote="hungkhtn"]As a long time my friend Manlio hasn't come to Mathlinks. Now I'm very glad because of his returning. The following inequality is to him with my best compliment... Others mathlinks member could try proving it.

Prove that for all non-negative real numbers $a,b,c$ we always have
\[ \frac{a(b+c-a)}{a^2+2bc}+\frac{b(c+a-b)}{b^2+2ca}+\frac{c(a+b-c)}{c^2+2ab} \ge 0.  \]

I think this problem is not as easy as its appearence. Gook luck.[/quote]

$\frac{a(b+c-a)}{a^2+2bc}+\frac{b(c+a-b)}{b^2+2ca}+\frac{c(a+b-c)}{c^2+2ab}$
=$\frac{2(a-b)^2(b-c)^2(c-a)^2+3abc(ab+bc+ca)(a+b+c)}{(a^2+2bc)(b^2+2ac)(c^2+2ab)} \geq 0$ :)

Yes, Gibbenergy. $(a-b)^2(b-c)^2(c-a)^2$ is the key.

My initial problem is that

Prove that if $ab+bc+ca=1$ then
\[ \frac{1+ab}{c^2+2ab}+\frac{1+bc}{a^2+2bc}+\frac{1+ca}{c^2+2ca} \ge 3. \]
Anh in this problem, we can't replace $2$ by a greater number. Even the solution for this problem, as Gibbenergy show is not easy. But I conjecture a harder one

Prove that if $ab+bc+ca=1$ then
\[ \sqrt{\frac{1+ab}{c^2+2ab}}+\sqrt{\frac{1+bc}{a^2+2bc}}+\sqrt{\frac{1+ca}{c^2+2ca}} \ge 3. \]
I hope someone can kill it.  ;)

[quote="ductrung"]It is true that
$\frac{a(b+c-a)}{a^2+2bc}+\frac{b(c+a-b)}{b^2+2ca}+\frac{c(a+b-c)}{c^2+2ab}\ge\frac{81abc(ab+bc+ca)}{(a+b+c)^5}$[/quote]

  Can anyone prove this one? It is too hard for me to manipulate by hand.
  Ductrung, do you use indentity again?

[quote="Simo_the_Wolf"][quote="Gibbenergy"]
$\frac{a(b+c-a)}{a^2+2bc}+\frac{b(c+a-b)}{b^2+2ca}+\frac{c(a+b-c)}{c^2+2ab}$
=$\frac{2(a-b)^2(b-c)^2(c-a)^2+3abc(ab+bc+ca)(a+b+c)}{(a^2+2bc)(b^2+2ac)(c^2+2ab)} \geq 0$ :)[/quote]

You have almost done!

We know: $2(a-b)^2(b-c)^2(c-a)^2 \geq 0$ and $27(a^2+2bc)(b^2+2ac)(c^2+2ab) \leq (a^2+2bc+b^2+2ac+c^2+2ab)^3= (a+b+c)^6$.

So:

$\frac{2(a-b)^2(b-c)^2(c-a)^2+3abc(ab+bc+ca)(a+b+c)}{(a^2+2bc)(b^2+2ac)(c^2+2ab)} \geq$ $\frac{81abc(ab+bc+ca)(a+b+c)}{(a+b+c)^6}$

 :lol:[/quote]

  You are right! Grr... I am sometimes so dump indeed! Spending some hours with SCP to find the indentity of this one is totally disaster. Probably classic inequality is the best!
   :lol:

[quote="shyong"][quote="hungkhtn"]
Prove that if $ab+bc+ca=1$ then
\[ \sqrt{\frac{1+ab}{c^2+2ab}}+\sqrt{\frac{1+bc}{a^2+2bc}}+\sqrt{\frac{1+ca}{c^2+2ca}} \ge 3.  \]
I hope someone can kill it.  ;)[/quote]

I check with computer and when $a=b=1/20$ , $c=399/40$ we have counter example LHS = $2.54884$ .......[/quote]

Thank you, shyong. This problem costed me too much time, but I failed.


The letters $ A$, $ B$, $ C$ and $ D$ represent digits. If $ \begin{tabular}{ccc} &A&B \\ \plus{}&C&A \\ \hline &D&A \end{tabular}$ and $ \begin{tabular}{ccc} &A&B \\ \minus{}&C&A \\ \hline &&A \end{tabular}$, what digit does $ D$ represent?

$ \textbf{(A)}\ 5 \qquad
\textbf{(B)}\ 6 \qquad
\textbf{(C)}\ 7 \qquad
\textbf{(D)}\ 8 \qquad
\textbf{(E)}\ 9$

[hide="Solution"]

Since $ B\plus{}A\equal{}A$, obviously $ B\equal{}0$. Next, since $ B\minus{}A\equal{}A\implies0\minus{}A\equal{}A$, we have $ A\equal{}5$.

Now, we rewrite this as $ \begin{tabular}{ccc}&5&0\\ \plus{}&C&5\\ \hline &D&5\end{tabular}$. Hence, $ D\equal{}5\plus{}C$ since no carrying occurs. Finally, we have $ A\minus{}1\minus{}C\equal{}0\implies{A\equal{}C\plus{}1}\implies{C\equal{}4}$. By our equation above, we have $ D\equal{}5\plus{}C\implies{D\equal{}5\plus{}4}\equal{}\boxed{9}$.

[/hide]

since it is urrggg forgot the problem ........ oh yeah it is d

Ehh I don't know if I did it right but I got 9.  Basically, I put B as 0.   Then, I wrote A + C = D.  So then I put A0 - CA=A.  A has to be 5.  So 50 - X5 = 5.  I got C as 4 and hence 5 +4 = 9.  I don't have a definite, amazing proof.


Tony is 50 years older than brett. twenty years ago tony was at least three times as old as brett was. At most, how old was brett now?
Please show work for this

[hide="Solution"]Let $ t$ be Tony's current age and $ b$ be Brett's current age. We are given $ t \equal{} b \plus{} 50$ and $ (t \minus{} 20)\ge3(b \minus{} 20)$.

Substituting the first equation into the second yields $ (b \plus{} 50 \minus{} 20)\ge3(b \minus{} 20)$, or $ (b \plus{} 30)\ge3(b \minus{} 20)$.

Expanding, we get $ b \plus{} 30\ge3b \minus{} 60$. Simplifying, $ 90\ge2b$, or $ b\le45$. Thus, Brett is at most $ \boxed{45}$ years old now.[/hide]


Is this problem too difficult? :wallbash:  :(

[quote="James Potter"]Denote that [x](x is a real number) is the largest integer number  x.
Let be given two sequence of numbers
$a_1,a_2,...$:$a_n=3n^2-2n\forall n\ge 1$ and $b_1,b_2,...$:$b_n=[n+\sqrt{\frac{n}{3}}+\frac{1}{2}]\forall n\ge 1$
Prove that {$a_n|n\ge 1$}$\bigcup${$b_n|n\ge 1$}$=N^*$ and {$a_n|n\ge 1$}$\bigcap${$b_n|n\ge 1$}$=$ empty[/quote]
[b]James[/b] what mean $N^*$?

P.S.Nice emoticons James!!, :wallbash:  :rotfl:

I think that's the Romanian-style notation - $N$ includes zero, and $N^{*}$ starts from $1$ (in the rest of the world $\mathbb{N}$ starts from $1$, and $\mathbb{N}_0$ or $\mathbb{Z}_{0}^{+}$ denote the extended set).


advanced fields
advanced fields unsolved

Let $ A$ be an infinite set. Prove:

Every subset $ G$ of the $ P(A)$ that satisfies the conditions:
(i) $ B$ is a finite set for all element $ B\in G$
(ii) $ \bigcup_{B\in G}B \equal{} A$
is equipotent to $ A$.

G is obviously infinite.

$ |A|\equal{}\left|\bigcup_{B\in G}B\right|\leq\sum_{B\in G}|\mathbb{N}|\equal{}|G\parallel \mathbb{N}|\equal{}|G|$

On the other hand we have $ G\subseteq P_{fin}(A)\Rightarrow |G|\leq|P_{fin}(A)|\equal{}|A|$ and hence $ |A|\equal{}|G|$.


$ p$ is prime number,$ p^{2}|2^{p\minus{}1}\minus{}1$ ,$ p\minus{}1\equal{}2^{\alpha}$  find all $ (p,\alpha)$

[quote="shoki"]obviously if $ ord_p(2) \equal{} d$ then if $ p^2|2^{p \minus{} 1} \minus{} 1 \implies p^2|2^d \minus{} 1$.
obviously $ p | 2^{2 \alpha} \minus{} 1$ so we must have $ p^2|2^{2 \alpha} \minus{} 1$.
the rest is trivial (inequality).[/quote]

Why $ p^2|2^{p \minus{} 1} \minus{} 1 \implies p^2|2^d \minus{} 1$ ?   :huh:

assume $ \gcd(a,p)\equal{}1$ where $ a$ is an integer and $ p$ is an odd prime.
suppose that $ ord_p(a)\equal{}d$ then $ ord_{p^2}(a)\equal{}pd$ or $ ord_{p^2}(a)\equal{}d$.
u can prove it by using the fact that $ p^2 | a^{pd}\minus{}1$.so we must have $ ord_{p^2}(a) | pd$.
so we have two cases:either $ \gcd(ord_{p^2}(a),p)\equal{}1$ or $ \gcd(ord_{p^2}(a),p)\equal{}p$.
and both cases imply that
1-$ ord_{p^2}(a) | d$.but we know that $ ord_p(a)\equal{}d \implies ord_{p^2}(a)\equal{}d$.
2-$ ord_{p^2}(a) \equal{}p.k,k| d$ but we know that if $ p^2 | a^{pk}\minus{}1 \implies p | a^{k}\minus{}1 \implies k\equal{}d$.
now put $ a\equal{}2$.

[quote="shoki"]assume $ \gcd(a,p) \equal{} 1$ where $ a$ is an integer and $ p$ is an odd prime.
suppose that $ ord_p(a) \equal{} d$ then $ ord_{p^2}(a) \equal{} pd$ or $ ord_{p^2}(a) \equal{} d$.
u can prove it by using the fact that $ p^2 | a^{pd} \minus{} 1$.so we must have $ ord_{p^2}(a) | pd$.
so we have two cases:either $ \gcd(ord_{p^2}(a),p) \equal{} 1$ or $ \gcd(ord_{p^2}(a),p) \equal{} p$.
and both cases imply that
1-$ ord_{p^2}(a) | d$.but we know that $ ord_p(a) \equal{} d \implies ord_{p^2}(a) \equal{} d$.
2-$ ord_{p^2}(a) \equal{} p.k,k| d$ but we know that if $ p^2 | a^{pk} \minus{} 1 \implies p | a^{k} \minus{} 1 \implies k \equal{} d$.
now put $ a \equal{} 2$.[/quote]

Oh sorry i understand now  :wink:


geometry
3D geometry
tetrahedron
combinatorics unsolved
combinatorics

Consider a regular tetrahedon where 4 insects are placed on the corners .  Determine the number of ways in which the insects can move along the sides of the tetrahedron such that no two insects ever meet.
Consider this argument- If the insects  dont meet at any point they can be assumed to be moving in a cyclic path following each other thee number of ways to do this are only SIX .(though we acn find $6$ cyclic paths i.e two at each edge actually there are effectively only $3$ paths in effect since if we consider a single circu;lar path , the cyclic path obtained by removing this cyclic path i.e the complementary path is the same e.g For a tetrahedron $VABC$ with vertex $V$ the paths $VABCV$ and $VCBAV$ are the same..hence there are $6$ ways considering both clockwise and anticlockwise arrangements

Now consider this argument- any one insect has three paths to follow and  the next has 2 and the next also has two and the last has only one option.this gives $3*2*2 = 12$ ways
which is the correct one.

The third insect only has 1 choice in the second argument (unless the insects are allowed not to move at all, in which case there are further possibilities).


let $ f(k)$ be the number of such numbers satisying the following conditions,
when the number is written in its decimal representation,
(1) the number has exactly $ k$ digits, and we can choose any number we want for each digit,for $ k \equal{} 5$ we can have $ 00211$
(2) after rearranging the digits of the number we can get a new number which is divisble by $ 11$. in the previous example we can arrange $ 00211$ into $ 00121$ which is divisble by $ 11$
prove that
$ f(2m) \equal{} 10f(2m \minus{} 1)$ for all $ m$ that is a positive integer

does anybody have any good ideas on this problem?


logarithms
trigonometry
calculus
calculus computations

and here is the last problem for today....

find a nice [b]closed[/b] form for the following sum:

$\sum_{n=0}^{\infty}\;\frac{(-1)^{n}}{4^{n}}\left( \frac{3}{4n+1}\;-\;\frac{\frac{1}{2}}{4n+3}\right)$

(answer hidden below...)

[hide]

$\ln{5}\;+\;\tan^{-1}{2}$

[/hide]

[hide="Solution less a few calculations"]When $x$ comes from the appropriate region (which it will), everything will converge absolutely and so the following calculations are valid:
we have $-\log(1-x) = \sum_{n=1}^\infty \frac{x^{n}}{n}$.  Thus $-\frac{1}{2}\log(1-x^{2}) = \sum_{n = 1}^\infty \frac{x^{2n}}{2n}$ and thus $\sum_{n = 0}^{\infty}\frac{x^{2n+1}}{2n+1}$ is their difference.  We also have $\arctan x = \sum_{n = 0}^\infty (-1)^{n}\frac{x^{2n+1}}{2n+1}$; adding this to the previous gives us $2\sum_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1}$, while subtracting gives us the $4n+3$ terms.  Taking the appropriate difference and substituting $x =-\frac{1}{4}$ gives the result.[/hide]


function
algebra
domain
real analysis
real analysis unsolved

Find a bijection from (0,infinity) onto (0,1)

So, does it mean that f(x) = e^(-x) is good as well?

The range of the function is (0,1)
But in these two functions, the domain of the function is (-infinity, infinity), and not (0,1)......

I wasn't sure about that.
Any explanation would be appreciated.

[quote="allrighty"]So, does it mean that f(x) = e^(-x) is good as well?

The range of the function is (0,1)
But in these two functions, the domain of the function is (-infinity, infinity), and not (0,1)......[/quote]

Yes, $ e^{\minus{}x}$ works, and so does, for example, $ \frac{2}{\pi}\arctan x$.  We are restricting all of these functions to the domain $ (0,\infty)$.  Because they still map to $ (0, 1)$ in that domain, they are all valid bijections.

I meant

"But in these two functions, the domain of the function is (-infinity, infinity), and not (0,infinity)...... "

Now, I got it, thank you :)


A standard deck of 52 cards has 13 ranks (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King) and 4 suits ($ \spadesuit$, $ \heartsuit$, $ \diamondsuit$, and $ \clubsuit$), such that there is exactly one card for any given rank and suit.  Two of the suits ($ \spadesuit$ and $ \clubsuit$) are black and the other two suits ($ \heartsuit$ and $ \diamondsuit$) are red.  The deck is randomly arranged. What is the probability that the top card is a $ \heartsuit$?

There are $ 13$ cards that are in the $ \heartsuit$ suit and $ 52$ cards total.  Thus, the probability that we want is $ \frac{13}{52}\equal{}\boxed{\frac{1}{4}}$.

there are 4 suits. hearts is on top. therefore the probability is 1/4


geometry
calculus
integration
real analysis
real analysis unsolved

Two discs, each of radius=a are placed one over the other. The disc on the top begins to move with a velocity=v such that v is inversely proportional to the area of the stationary disc underneath exposed. Find time taken by the moving disc to expose the disc underneath.

Please, do not post twice. Also, this should belong to one of the Physics forums.

Agreed on the double posting, but I'm going to disagree with Carcul about the location: this is a differential equation, and a differential equations are calculus. Perhaps this could go to the calculus computations and tutorials section, but for the time being, I'll leave it here.

i get something nasty with a square root and a tangent inverse and some x terms all on the denominator ?  eesh

$ x'(0)\equal{}v_0$  $ f(x)\equal{}a^2\pi\minus{} 2[a^2 \arctan (\frac{ \sqrt{a^2\minus{}x^2}}{x})\minus{}x\sqrt{a^2\minus{}x^2}]\equal{}$ exposed area of bottom circle when $ x(t)\equal{}2x$

$ \to f(\frac{x(t)}{2})x'(t)\equal{}2K \to \int_0^{\frac{x(T)}{2}} f(u)du \equal{}KT$

$ \int_0^T f(x)dx \equal{} a^2\pi T\minus{} 2a^2 \int_0^T \arctan (\frac{ \sqrt{a^2\minus{}x^2}}{x})dx\plus{} \int_0^T 2x\sqrt{a^2\minus{}x^2}dx$

$ \equal{}a^2\pi T \plus{}\frac{2a^3}{3}\minus{} \frac{2}{3}(a^2\minus{}T^2)^{\frac{3}{2}}\minus{}2a^2  [T\arctan (\frac{ \sqrt{a^2\minus{}T^2}}{T}) \minus{} \int_0^T \frac{dx}{ \sqrt{a^2\minus{}x^2}}]$

$ \equal{}a^2\pi T \plus{}\frac{2(a^3\minus{}(a^2\minus{}T^2)^{\frac{3}{2}})}{3}\minus{}2a^2 T\arctan (\frac{ \sqrt{a^2\minus{}T^2}}{T}) \plus{}2a^2 \arcsin \frac{T}{a}$

so

$ \frac{KT}{2a^2}\minus{}\frac{a}{3}\plus{}\frac{(a^2\minus{}x(T)^2)^{\frac{3}{2}}}{6a^2}\equal{}x(T)(\pi\minus{}2\arctan (\frac{ \sqrt{a^2\minus{}x(T)^2}}{x(T)}) \plus{} \arcsin \frac{x(T)}{a}$

and at this point i give up?


inequalities
geometric sequence
number theory unsolved
number theory

1.  Verify that no power of prime can be a perfect number

2.  Show that no proper divisor of a perfect number can be perfect

First part:
Let the number be $n=p^q$. The divisors of $n$ are:
$1,p,p^2,...,p^(q-1)$
Their sum can be find using the sum of a geometric progression formula.Hence the sum is:
$(p^q-1)/(p-1)$
This is obviously less than $p^q=n$ and the desired result follows.

Rem's solution is on the right track. But the sum of divisors of p^q has to always be greater than p^q, since p^q divides itself. The exponent of p after summing as a geometric progression should be q+1, not q. You can use an inequality to see what values it will be less than 2p^q for (for the required perfectness), and playing around with that will eventually show that it is true for all values.

As for the second one, you can show it for even n by using the fact that n is an even perfect number if and only n=(2^p-1)(2^(p-1)), where 2^p-1 is prime, since its only odd divisor is a prime, which is not perfect, and all other even divisors are not of the required form. All that remains to show is that there are no odd perfect numbers. Just kidding everyone, a little math joke.

Forgive me for my inability to use Latex.


I have a basic installation of GeoGebra on my computer.  Under Options -> Right angle style, the $ \Box$ option is marked.  I've created a pair of perpendicular lines, but how do I get the $ \Box$ to show up?  Thanks!

[quote="herefishyfishy1"]Measure the angle, and the arc that usually appears should automatically be a right angle mark instead.

 [geogebra]3296c568fd34e0578f7065d28fc218dfb678dc3d[/geogebra][/quote]

How did you do that?

I measured the angle using the angle measure tool.

[geogebra]f781135d7fef780ec96a8587e67bbdd510a004ce[/geogebra]

I typed 90 in the dialog box that came up, once with counter clockwise and once with clockwise.  But it isn't working.  The angle is not 90 degrees and I only get an angle, not two perpendicular lines.

Thanks in advance!


function
calculus
derivative
complex analysis
complex analysis unsolved

Let $ \mathbb D \equal{} \{ z \in \mathbb C : |z|<1\}$, and suppose $ f : \mathbb D \to \mathbb C$ is an analytic function such that its derivative $ f'$ is injective on $ \mathbb D$. Show that $ f''(z) \ne 0$ for all $ z \in \mathbb D$.

Thank you  :( .

Very stange formulation of standart problem :) .Just forget about $ f$ and prove that if you have an analytic function $ g$ from unit disc and $ g'(z_0)\equal{}0$ then $ g$ is not injective in the neighbourhood of the point $ z_0$.To prove this use Rouche theorem :wink:

[quote="leshik"]Very stange formulation of standart problem :) .Just forget about $ f$ and prove that if you have an analytic function $ g$ from unit disc and $ g'(z_0) \equal{} 0$ then $ g$ is not injective in the neighbourhood of the point $ z_0$.To prove this use Rouche theorem :wink:[/quote]

Leshik : to  prove that if $ g$is one-to-one than $ g'(z)\neq 0$ for all $ z$ in $ G$ I use the following fact :
 if $ f$ is analytic in $ B(a,R)$ and $ f(a)\equal{}\alpha$. If $ f(z)\minus{}\alpha$ has a zeros of oder $ m$  at $ a$ then there is $ \epsilon>0$ and $ \delta>0$  such that $ 0<|\eta\minus{}\alpha|<\delta$ the equation $ f(z)\equal{}\eta$ has exactly $ m$ simple solutions in $ B(a,\epsilon)$

how can you use  Rouche theorem to prove that if $ g$ is one-to-one then $ g'(z)\neq 0$ for all $ z$ in $ G$?

I think he means that we can set $ g(z) \equal{} a_0 \plus{} a_kz^k \plus{} ...$, where $ a_k$ is the first nonzero coefficient after $ a_1$. Then use Rouche's theorem to compare the number of zero of $ g(z)\minus{}a_0$ and $ z^k$

My viewpoint: If $ f(0) \equal{} 0$ and $ f'(0) \equal{} 0$, then let $ n > 1$ be the first $ n$ such that $ f^{(n)}(0) \neq 0$. Write $ f(z) \equal{} z^{n} g(z)$ where $ g$ is zero-free around 0. $ f \equal{} h^n$ where $ h(z) \equal{} z(g(z))^{1/n}$. Note that $ h'(z) \equal{} (g(z))^{1/n} \plus{} \frac {1}{n}zg'(z)(g(z))^{1/n \minus{} 1}$, so that $ h'$ is also zero-free around 0. Hence $ h$ maps a neighborhood of $ 0$ biconformally to another neighborhood of $ 0$, by the inverse function theorem. It follows that that $ f$ maps a neighborhood $ U$ of 0 to another neighborhood $ V$ of 0 such that the image of $ U$ covers $ V$ exactly $ n$ times.

Loosely stated, when $ f'(0) \equal{} 0$, $ f$ behaves like $ z^n$ (plus a constant if $ f(0) \neq 0$) for some $ n > 1$, making all angles at zero $ n$ times as large. So it certainly isn't injective.


algebra
polynomial
quadratics
probability
expected value

Let a, b, c, be random integers 1-9.  What is the expected value of the zeros of the quadratic f(x) with coefficients a, b, and c?

[hide="Solution"]\begin{eqnarray*}E &=& \sum_{a,b,c}(x_{1}(a,b,c)P(a,b,c)+x_{2}(a,b,c)P(a,b,c))\\ &=& \sum_{a,b,c}\left(-{b\over a}\cdot{1\over 9^{3}}\right)\\ &=&-{1\over 9^{3}}\sum_{a,b,c}{b\over a}\\ &=&-{1\over 9^{3}}\sum_{a,b}\left(9\cdot{b\over a}\right)\\ &=&-{1\over 9^{2}}\sum_{a}\left(\sum_{b}{b\over a}\right)\\ &=&-{1\over 9^{2}}\sum_{a}\left({1\over a}\cdot{9\cdot 10\over 2}\right)\\ &=&-{5\over 9}\sum_{a}{1\over a}\\ &=&-{5\over 9}\left(1+{1\over 2}+\dots+{1\over 9}\right)\end{eqnarray*}

The last expression can be reduced to a fraction, giving $E=-{7129\over 4536}$[/hide]


calculus symmetry inequalities derivative algebra function domain

Let $a,b$ be positive reals such that $a+b \ge 4$. Find minimum value of:
\[ 3a+2b+\frac{11}{a}+\frac{6}{b}  \]

[quote="stancioiu sorin"]Just Cauchy.[/quote]

Why not AM-GM? It seems trivial.

I think we need to check whether the condition $a+b\geq 4$ is satisfied!
Sometimes, a lower bound is not accessible.  :)

Some ideas:
Let $f(x,y)=3x+\frac{11}x+2y+\frac 6 y$ for $x,y>0$.
The minimum value of $f$ will be attained at some points of the boundary set $S=\{(x,y)|x+y=4,x>0,y>0\}$. ( Need to be clarified with Calculus of 2 variables.)
Therefore, we may substitute $y=4-x$.
Then $g(x)=f(x,y)=3x+\frac{11}x+2(4-x)+\frac 6{4-x}$, which is symmetric about $x=2$ for $x \in (0,4]$.
$g'(x)=1+\frac 6 {(4-x)^2}-\frac{11}{x^2} <0$ for all $x \in (0,2)$.
By the symmetry of $g$,the decreasing property of $g$ on (0,2), and the continuity of $g$, the absolute minimum of $g$ is attained at $x = 2$.
In conclusion, the absolute  minimum of $f$ = $f(2,2) = 3(2)+ \frac{11}{(2)}+2(2)+\frac 6{(2)}=18.5$

Wish someone can give a complete solution to this problem and please point out my mistake if I have written something incorrect.  :)
[b]
Edit[/b]: Sorry, $g$ is in fact not symmetric about $x = 2$. Therefore, the minimum is not necessarily attained at $x=2$.

The question has no nice solution; the minimum is just a shade smaller than 18.5, with expressible in terms of the irrational solution of a quartic.  If the 6 were replaced by 7, there would be a clean solution using AM-GM.

I've tried to find a solution with Cauchy that makes use of the initial problem condition, but I am at a loss.  Lovasz, do you have a solution?

I really, really promise that there is no nice solution.

[quote="stancioiu sorin"]Just Cauchy.[/quote]

I'm allergic to people making this kind of posts. Please don't do this.

I don't have solution. I luimichael's one, we cannot replace $b$ with $4-a$. I's quite hard.
Is there any proof here?

[quote="rem"][quote="stancioiu sorin"]2 $\sqrt 33$ + 4$\sqrt 3$[/quote]
The lower bound is satisfied and we are done. ;)[/quote]
*********************************************************************
It will be perfect if you can find such a solution.
Can you show us the values of a and b in such a situation?

[quote="rem"][quote="sen"]Im not sure, but $a=0 b=4$ yields the minimum $=9.5$

any $(a,b)$ majorises $(0, 4)$ so anyone have a slightly stronger argument?sorry  :blush:  didnt read has to be positive.[/quote]
wtf? you can't divide by 0! :!:[/quote]

He meant $a = 0, b = 4$.

Isn't there a quartic equation which we could use to find the exact solution?

[quote="luimichael"]Some ideas:
Let $f(x,y)=3x+\frac{11}x+2y+\frac 6 y$ for $x,y>0$.
The minimum value of $f$ will be attained at some points of the boundary set $S=\{(x,y)|x+y=4,x>0,y>0\}$. ( Need to be clarified with Calculus of 2 variables.)
Therefore, we may substitute $y=4-x$.[/quote]Can you explain your justification for this?

[quote="JBL"][quote="t0rajir0u"]He meant $a = 0, b = 4$.[/quote]

In which case, in the original expression, you get division by zero.

If you want a nice solution, replace the 6 in the original question by 7; then you get the following solution:
[hide]$3a+2b+\frac{11}{a}+\frac{7}{b} = (\frac{11a}{4} + \frac{11}{a}) + (\frac{7b}{4} + \frac 7b) + (\frac{a}{4} + \frac b4) \geq 11 + 7 + 1 = 19$ where the inequalities are by AM-GM, AM-GM and the given.  Equality holds at (2, 2).[/hide]

Now, for the problem as written, this nice method doesn't work.  The solutions proposed in the first 4 posts on this thread fail because the minimum they give can't be achieved with $a + b \geq 4$.  A full, ugly solution of the question as posed is basically going to be brute force; you create a fourth degree equation by taking a derivative, it doesn't have any nice roots, and you get a numerical result.  The details are all due to Jae Bae: [hide]In particular, if $k = 0.406602\ldots$, then $\sqrt{\frac{11}{3-k}} + \sqrt{\frac{6}{2-k}} = 4$ and setting $a = \sqrt{\frac{11}{3-k}}$ and $b = \sqrt{\frac{6}{2-k}}$ gives
$3a+2b+\frac{11}{a}+\frac{6}{b} = k(a + b) + [(3-k)a + \frac{11}a] + [(2-k)b + \frac6b] \geq 4k + 2\sqrt{11(3-k)} + 2\sqrt{6(2-k)} = 18.49258\ldots$
which is the minimum value.  [/hide][/quote]

I don't understand why you would use an inequality if you already computed numerical values of $a$ and $b$ from your given value of $k$ ...

Let $f(x,y)=3x+\frac{11}x+2y+\frac 6 y$ for $x,y>0$.
$\frac{\partial f}{\partial x}=3-\frac{11}{x^2} = 0$ if and only if $x=\sqrt{\frac{11}3}$.
$\frac{\partial f}{\partial y}=2-\frac 6{y^2}=0$ if and only if $y=\sqrt{3}$.
When $x=\sqrt{\frac{11}3}$ and $y=\sqrt{3}$, we have $x+y=3.6469....$, showing that this critical point does not belong to $A=\{(x,y)|x+y\geq 4,x>0,y>0\}$.
Therefore, $f$ has no minimum points or maximum points in the interior set $B=\{(x,y)|x+y>4,x>0,y>0 \}$.
In other words, if the minimum point or maximum point of $f$ does exist, it has to be found in the boundary set $S={(x,y)|x+y=4,x>0,y>0\}}$.
Luckily, the minimum can be found at $g'(x)=0$......
I didn't complete the solution in my  post because I had obtained an ulgy solution by means of Mathematica.
Probably JBL got the same solution in a similar fashion.  :)

hmm... luimichael, $\min f_x(x+y)=m_1,\ \min f_y(x+y)=m_2 \implies \min f(x,y)=m_1+m_2\ \boxed{?}$

Sorry, I don't know what you are going to imply.  :)

In my last post, the main idea is to point out that $f(x,y)$ has no maximum or minimum in the interior set {(x,y)|x+y>4,x>0,y>0}. Therefore, if we want to find any maximum or minimum of $f$, we only need to consider the values of $f$ on the boundary set {(x,y)|x+y=4,x>0,y>0}.

Please let me know the mistake if I have made any wrong assumptions or calculations. :)

[quote="t0rajir0u"]I don't understand why you would use an inequality if you already computed numerical values of $a$ and $b$ from your given value of $k$ ...[/quote]

It is just to illustrate where the value of $k$ came from.  Really, we choose $k$ so that the expressions for $a$ and $b$ work out nicely.  Alternatively, you can just plug in $a = 4 - b$, take a derivative and solve the quartic that results and sub back in.  Technically, you need steps to verify that 4 gives the minimum, which is what luimichael appears to be doing for us.

[quote="luimichael"]Let $f(x,y)=3x+\frac{11}x+2y+\frac 6 y$ for $x,y>0$.
$\frac{\partial f}{\partial x}=3-\frac{11}{x^2} = 0$ if and only if $x=\sqrt{\frac{11}3}$.
$\frac{\partial f}{\partial y}=2-\frac 6{y^2}=0$ if and only if $y=\sqrt{3}$.
When $x=\sqrt{\frac{11}3}$ and $y=\sqrt{3}$, we have $x+y=3.6469....$, showing that this critical point does not belong to $A=\{(x,y)|x+y\geq 4,x>0,y>0\}$.[/quote]It doesn't belong to $S=\{(x,y)|x+y=4,x>0,y>0\}$ either, does it?  I'm a little confused...

Let me make it clear then.
I mean the point $(x,y)=(\sqrt{\frac{11}3},\sqrt{3})$ is the only critical point for  x >0 and y >0. Besides, this point  will correspond to the  absolute minimum as quoted in some other posts ( without the constraint $x+y\geq 4$).

Note that  the above mentioned point does not belong to $\{(x,y)|x+y>4, x>0,y>0\}$ because by direct checking, $x+y<4$ is true.

So, the only possible candidate for a maximum or minimum of $f$ on the interior  set {(x,y)|x+y>4, x>0, >0} is kicked away. This proves that $f$ has no minimum points or minimum points on the interior set, so the possible candidates for an extreme point can only go to the boundary set $\{(x,y)|x+y=4,x>0,y>0\}$.

I think the ambiguity may come from the domain of $f$.
To find the extreme points, we extend the domian to {(x,y)|x>0,y>0} for convenience.
Yet, on discussion of our original problem, we come back to the set {(x,y)|x+y>=4,x>0,y>0}.

One more thing: The derivative method only apply to the interior set. Normally we have to deal with the boundary set separately. Think about the same type of question in one variable you may get the idea.

Sorry. Here's my question:
If the minimum value of $f_x(x,y)$ happens if $x=x_1$, and the minimum of $f_y(x,y)$ occurs when $y=y_1$, then it follows $\min f(x,y)$ holds if $(x,y)=(x_1,y_1)$. Are you  [b]sure[/b]?

Yeah, it's the definition of critical points for more than one variable.
[quote=Wikipedia]For a function of several real variables, the condition of being a critical point is equivalent to all of its partial derivatives being zero. ....In one or several variables, the maxima and minima of a function (if they exist) can occur either at its critical points or at points on its boundary, or points where the function is not differentiable.[/quote]

[url=http://en.wikipedia.org/wiki/Critical_point_%28mathematics%29]Check it out here[/url]

Ok, but what will happen if someone replaces $\frac{6}{b}$ with $\frac{13}{b}$, then:
\[ \frac{df(x,y)}{dx}=3-\frac{11}{a^2} \newline \frac{df(x,y)}{dx}=0 \iff \boxed{a=\sqrt{\frac{11}{3}}} \newline \frac{df(x,y)}{dy}=2-\fra{13}{b^2} \newline \frac{df(x,y)}{dy}=0 \iff \boxed{b=\sqrt{\frac{13}{2}}} \newline \implies a+b=2.345+2.549=4.894 \boxed{>4} \newline \boxed{?}  \]

Hi! Lovasz,

It seems that we don't need to apply Calculus in your new problem.
But I guess you want to see what happens if we insist on using Calculus?
In this case, we observe that there is [b]only one critical point[/b] and it is in the interior of the domain, so we proceed to apply the [b]second derivative test[/b].Here is a link for you reference:
http://mathworld.wolfram.com/SecondDerivativeTest.html

The critical point you found should correspond to a [b]local minimum[/b] by  the test. But remember that there is only one minimum point for the entire domain, and our function is [b]continuous[/b], therefore this minimum is in fact a [b]global minimum[/b]. Besides, we don't even need to check the values of $f$ on the boundary set of the domain.

I hope the info is useful.  :)


logarithms
algebra
polynomial
number theory
algebra unsolved

it dont have a algebraic solution.

a good approximation would be 1.20399

[quote="mornik"]If you ment that x must be natural, then it's equivalent to:

$4x^{2}+2$=$(x^{6}+x^{2}+1)*2007^{x^{6}-3}$. Now, if x is natural (or integer at all), we have 2 cases:

1. $|x|>1$

$(x^{6}+x^{2}+1)*2007^{x^{6}-3}$ is odd, while $4x^{2}+2$ is even. Contradiction.

2. $|x|=1$

$6=3*2007^{-2}$, contradiction.

3. $|x|=0$

$2=2007^{-3}$, contradiction.

Therefore, there are no integer solutions.[/quote]

Haha mornik, this remids me of your sentece today "i added one more condition and solved the problem..."

:D Yeah, right... Totally. (That was one of my better solutions.  :D )
Nah... I'm not really sure what this problem is all about. Since it's not a polynomial and since yagaron already gave the aproximation, I thought maybe nobody1 wanted an integer solution (even though that should go to number theory then).


[quote]
How many 6 character lists can you make, where A-Z can be used for the first 3 positions and the digits 0-9 can be used for the last 3 positions. Important: The letter O and the digit 0 cannot be used in the same list. [/quote]

I think I need to use inclusion/exclusion... so something like:

number of all possible permutations (26^3 * 10^3) minus number of bad lists (lists with both O and 0).

I'm not sure how to calculate the number of bad lists, tho.

Thanks for your help!

I've been thinking about it more...

Let X be the set of all possible permutations.
That is, |X| = (26)^3 * (10)^3

Let A be the set of all permutations containing at least one 0.
That is, 26^3 * 1*10^2

Let B be the set of all permutations containing at least one O.
That is, 1*26^2 * 10^3

IS the answer simply |X| - intersection of A and B?

That is:

(26)^3 * (10)^3 - 26^2 * 10^2?


Let $x,y,z$ be positive reals such that $xyz=1$ and let a real $p>0$. Prove that

$\frac{1}{x^p+y+z}+\frac{1}{x+y^p+z}+\frac{1}{x+y+z^p} \le 1$.

We have

$\sum \frac{1}{x+y+z^p} \le \sum \frac{2 + z^{1-p}}{(\sqrt{x} +\sqrt{y} + \sqrt{z})^2} \le 1$

which is evident since

$\sum 2\sqrt{xy} \ge 6$ and $x+y+z \ge x^{1-p} + y^{1-p} + z^{1-p}$ for $3/2 \ge p \ge 0$. (Edit/Correction:  Vasc pointed out that this doesn't work for $p > 3/2$.)

[quote="Mildorf"]We have

$\sum \frac{1}{x+y+z^p} \le \sum \frac{2 + z^{1-p}}{(\sqrt{x} +\sqrt{y} + \sqrt{z})^2} \le 1$

which is evident since

$\sum 2\sqrt{xy} \ge 6$ and $x+y+z \ge x^{1-p} + y^{1-p} + z^{1-p}$ for $p \ge 0$.[/quote]
Mildorf, the inequality $x+y+z \ge x^{1-p} + y^{1-p} + z^{1-p}$ is true for $-1/2 \le 1-p \le 1$, that is $0\le p\le 3/2$.
It remains to solve the case $p>3/2$. :?:

[quote="Vasc"]Let $ x,y,z$ be positive reals such that $ xyz \equal{} 1$ and let a real $ p > 0$. Prove that

$ \frac {1}{x^p \plus{} y \plus{} z} \plus{} \frac {1}{x \plus{} y^p \plus{} z} \plus{} \frac {1}{x \plus{} y \plus{} z^p} \le 1$.[/quote]

$ \frac {1}{x^p \plus{} y \plus{} z} \le \frac {x^{\frac {1 \minus{} p}{3}}}{x^{\frac {1 \plus{} 2p}{3}} \plus{} y^{\frac {1 \plus{} 2p}{3}} \plus{} z^{\frac {1 \plus{} 2p}{3}} }$

[b]We are done , Because :[/b]

$ x^a \plus{} y^a \plus{} z^a\ge (xyz)^b(x^{a \minus{} 3b} \plus{} y^{a \minus{} 3b} \plus{} z^{a \minus{} 3b})$

$ \rightarrow LHS\le (abc)^{\frac { \minus{} 4p}{9}} \equal{} 1$

$ p>1$  :oops:  :oops:

[quote="khanhsy"]
$ \frac {1}{x^p \plus{} y \plus{} z} \le \frac {x^{\frac {1 \minus{} p}{3}}}{x^{\frac {1 \plus{} 2p}{3}} \plus{} y^{\frac {1 \plus{} 2p}{3}} \plus{} z^{\frac {1 \plus{} 2p}{3}} }$

[/quote]
What if $ p>1$ :?:  :wink:

For $ p\ge \frac 1{2}$, we can prove this inequality using
$ \frac 2{x^p\plus{}y\plus{}z}\le \frac {y^q\plus{}z^q}{x^q\plus{}y^q\plus{}z^q}$,
where $ q\equal{}\frac {2(p\minus{}1)}{3}.$
Let $ t\equal{}\sqrt{yz}$. Since $ y\plus{}z\ge 2t$ and $ y^q\plus{}z^q\ge 2t^q$, it suffices to show that
$ \frac 1{x^p\plus{}2t}\le \frac{t^q}{x^q\plus{}2t^q}$,
which is equivalent to
$ t(t\minus{}1)(t^{2p\minus{}1}\minus{}1)\plus{}(t\minus{}1)(t^{2p}\minus{}1)\ge 0$.


Hi! Im new here and I have a question... What exactly is Moems  :?:

Check out their website:

http://www.moems.org

Hey BuckeyeFan! Glad to see your papers went through at last. :)

MOEMS stands for Math Olympiads for Elementary and Middle Schools. It's for grades 4-8, open to homeschoolers too, and it's a great introduction to problem solving. It is extraordinarily fun!

If you can find a teacher/parent to sponsor your math club doing it, you can have a lot of fun working the problems. The top people get special awards :winner_first: .

As DrunkenMath said, http://www.moems.org is the official website.

Billy


Show how to divide a circle into 9 regions with equal area with compass and straightedge.

This problem just requires some ingenuity. See how many different ways you can do it!

first solution: find the center of the circle(easy), find its radius r(easy) and draw a circle around the center of the circle with radius $\frac{r}{3}$. it has one ninth of the area of the original circle. draw eight rays from the center such that the angle between two neighboring is 45 degree(easy). the area outside the small circle but inside the larger is divided into 8 regions with equal area, so all have one ninth of the area of the circle.

Peter


function
algebra
domain
calculus
calculus computations

Given $ f(x) \equal{} \cfrac{3}{x}$, find a function $ b(x)$ which gives the y-intercept of a line tangent to $ f(x)$ for any value of $ x$ in the domain of $ f(x)$.

I tried to do this by intuition alone, but I'm not sure if my method is the best one:

$ b(x) \equal{} f(x) \minus{} f'(x)\cdot{x} \equal{} \boxed{\cfrac{6}{x}}$

is there a clear way to do this, maybe using equations?

But this is a clear way! If you define a line that is tangent to a function as:

$ y \minus{} f(x_0) \equal{} f'(x_0)(x \minus{} x_0)$

then all you need to do is set $ x \equal{} 0$ because your trying to find the interceptions with the $ y$ axis. This gives you:

$ y \equal{} f(x_0) \minus{} x_0 f'(x_0)$

and that's it - just as you've done it.

Ohh of course! hehe idk why I never thought to set $ x \equal{} 0$.. No wonder I was getting weird results.. Thanks! :)


If $ (x \plus{} y)^2 \equal{} 1$ and $ xy \equal{} \minus{}4$, what is the value of $ x^2 \plus{} y^2$?


How do you do this one.....

Consider a 4x4 square with a circle of radius 1 centered in the square. A point P is randomly chosen inside the square but not on the circle. A new circle is drawn with point P as the center such that the new circle does not intersect the original circle. Wihat is the probability that there exists a line that is tangent to both circles? (Note: A line is tangent to a circle if it intersects the circle in exactly one point. )

[/hide]

It depends.  What is the probability distribution for the radius of the new circle?

There are infinitely many new circles whih can be drawn which don't intersect the old circle and have no common tangents with it, but none of these circles lie entirely inside the square...


Im kinda confused why i cant find some usernames on the memberlist when i know they exist..

i think this is the wrong forum but im kinda confused why this is.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160965, and yes, a mod should move this to Support

can someone tell me how they are deactivated?

i dont get why youdonotknowme and yalldonotknowme both got banned or deactivated or that..

i never spammed or did any of that so can someone tlel me?

a admin?

huh my accounts got banned too.

Im vikas indianamath

[quote="indianamath"]you don'tknowme. how have you never spammed? you crated the accounts who aint the man, the great one...and about 5 others.

I'm pretty sure they banned all the account made so far from that ip address.[/quote]

How did you guys find out it was all the same person?

We are aggressively banning multiple accounts now.  We spend too much time having to clean the messes formed by these users and figuring out which accounts are 'real' or 'fake'.  It is clear there is a small number of very troublesome users.  We are now aggressively banning them from the forum.  If your account has magically disappeared, chances are that's because you are one of the offending parties.  If you think your account has been wrongfully banned, then you can email us at webmaster@artofproblemsolving.com.  Do not create a new account and complain on the message board.


algebra
polynomial
group theory
abstract algebra
Galois Theory
Ring Theory
superior algebra

Show that the polynomial $ x^{6}\plus{}22x^{5}\minus{}9x^{4}\plus{}12x^{3}\minus{}37x^{2}\minus{}29x\minus{}15$ has galois group $ S_{6}$ over the rationals.

This is part of a larger exercise in Hungerford.  The previous part was to show that if $ \overline{f}$ is the reduction of $ f$ modulo a prime $ p$, then $ \overline{G}$, the galois group of $ \overline{f}$ over $ \mathbb{Z}_{p}$, is a subgroup of $ G$, the galois group of $ f$.

A hint suggests to apply this fact with $ p\equal{}2,3,5$.

Thanks for any help.

this first part (a theorem of dedekind) is an exercise?! this is very difficult, or have you found an easy proof?

the fact that galois groups over finite fields are cyclic and the fact that the degree of irreducible factors correspond to the lengths of the orbits, yields the corollary: if $ f\in\mathbb{Z}$ is separable mod a prime $ p$ and has a factorization of degrees $ j_{1}, j_{2}, ...$ over $ \mathbb{F}_{p}$, then the galois group of $ f$ over $ \mathbb{Q}$ (as a permutation group) has a permutation of the cycle type $ (j_{1},j_{2},...)$.

in your case, the galois group is a subgroup of $ S_{6}$ consisting of a $ 6$-cycle, a $ 5$-cycle and a $ 2$-cycle. try to show that these already generate $ S_{6}$.

thanks -00-, i just had a similar idea, and it's good to know i'm on the right track.  this problem is from a part of the book prior to the introduction of galois theory for finite fields, which made things harder.  the proof of the initial result is based on this polynomial which they introduce.
\[ \prod_{\sigma\in S_{n}}(x\minus{}(u_{1}y_{\sigma(1)}\plus{}\cdots\plus{}u_{n}y_{\sigma(n)}))\]
Where $ u_{1},\ldots,u_{n}$ are the distinct roots of a polynomial $ f\in K[x]$, and the $ y_{i}$ are indeterminates, as is $ x$.  You show that the above polynomial is actually in $ K[y_{1},\ldots,y_{n},x]$, which isnt too hard, just a lot of symmetric polynomials, and then it factors in $ K[y_{1},\ldots,y_{n},x]$ into polynomials of the form
\[ \prod_{\tau\in G}\left (x\minus{}\Sigma_{i}\left (u_{\tau\sigma(i)}y_{i}\right )\right )\]
where $ G$ is the galois group of $ f$.  You get the result by noting that a factorization over $ \mathbb{Z}[x]$ always extends to one in $ \mathbb{Z}_{p}[x]$.  Well that's how I did it in any case.

ok that's exactly the proof I know for this. but to get the result, that the galois group over Fp lies in the one over Q, I used "lying over" (evere prime ideal of a ring comes from a prime ideal in a integrally extension) to construct a splitting field over Fp which is related to the one over Q.


Evaluate without calculator.

\[\sqrt{7 + 4 \sqrt 3} + \sqrt{7 - 4 \sqrt 3}\]

$7+4\sqrt{3}$ can be written as $4+3+4\sqrt{3}=(2+\sqrt{3})^2$
Similarly, simplifying with the other one, we get:

$(2+\sqrt{3})+(2-\sqrt{3})=4$

[color=green][size=75]mod edit: if you want people to read your posts, LEARN LATEX!!!![/size][/color]

[quote="shadysaysurspammed"][color=green][size=75]mod edit: if you want people to read your posts, LEARN LATEX!!!![/size][/color][/quote]

WILL YOU TEACH ME!!!!cuz i cant mug up the formula's n stuff for katex,nevertheless chess,i'll try but my xams are comin so i wont learn till the 10th of october. :D

$\LaTeX$ is easy.

Go here:

http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

When I learned my Latex, it was before AoPS had this nice section devoted to Latex and it was hard for me to learn it by internet.  :mad: But you guys are lucky and should have no problem since AoPS gives nice explanation on how to use Latex and etc.

Don't ask any more questions in here. Ask on Latex forum.

I don't memorize formulas like that, I find it a better idea to just use the algorithm.

[quote="ifai"]Here is a formula:
$\sqrt{a\pm\sqrt b}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$[/quote]
Or you could say $\sqrt{p+q\pm2\sqrt{pq}}=\sqrt{p}\pm\sqrt{q}$ (generalization of shadysaysurspammed's solution).

I agree with chess64, but I think it's nice to remember formulas as long as you know how you derive it. (That way, it's easier to use the algorithm anyway.)


linear algebra
matrix
linear algebra unsolved

Let $A \in M_4(Z), det(A) = 1, n \in N, n \geq 2$
In the sequence: $A^k - I_4, k \geq 1$ there exists a matrix with all elements divisible by n.

Obvious. The sequence $A^k$ mod $n$ is finite (since $M_4(Z/nZ)$ is) and clearly if $AX=AY (mod n)$ then $X=Y$ (since $A$ is in $GL_4(Z)$). Thus conclusion.

this was a 5 minutes contest problem, but i found it cute, so... what can i say more... :)


One receipt have some number missing, only can read 3 of the 5 digits in the total, the bill show:

72 gadgets at $-. each =$ _74.5_.

How many dollars did one gadget cost?

Any one please help out and explain how to get the result?

[hide]Notice that the result has to be divisible by 72. This means that it must be divisible by both 8 and 9. To be divisible by 8, the last 3 digits must be divisible by 8. We see that the only combination of three that works with the given 2 digits is 456. For it to be divisible by 9, the sum of digits must be divisible by 9. Adding them up, we have x+7+4+5+6= a multiple of 9, so x+22= a multiple of 9, and the only one that works is 27, so x+22=27, so x=5. Now we have 574.56 for 72 gadgets, so when we divide, we get $7.98$ for one[/hide]

No problem! It gives me good practice too!

Ya thats a good solution, I wouldve kept on brute forcing it and getting no where.. never thought of ur way


algebra function domain logarithms limit

Hello there,

I am having trouble with the following exercises and I would appreciate any help or hints.

Thank you very much.

---
[b]
Give the domain of the following functions: [/b]

1. $ b(x) \equal{} (3x)^{x \plus{} 2}$

My original answer was $ D \equal{} ( \minus{}\infty, \infty)$, but the provided answer is $ D \equal{} (3, \infty)$. Could anyone please explain why? When I graphed this function on my graphing calculator, the function starts at $ (0,0)$.

2. $ c(x) \equal{} log_3 (x^2 \minus{} 1)$

[u] My Work : [/u]

According to the definition of logarithms, $ (x^2 \minus{} 1) > 0$:

$ x^2 \minus{} 1 > 0$:

$ x^2  > 1$:

$ x \equal{} \minus{}1, 1$

When I graphed this function, the function intersected the x axis at -1.5 and at 1.5. The provided answer for this question was $ D \equal{} (\minus{}1.5, \infty)$. Could this be explained?[/u]

[quote="catalan"]Hello there,

I am having trouble with the following exercises and I would appreciate any help or hints.

Thank you very much.

---
[b]
Give the domain of the following functions: [/b]

1. $ b(x) \equal{} (3x)^{x \plus{} 2}$

My original answer was $ D \equal{} ( \minus{} \infty, \infty)$, but the provided answer is $ D \equal{} (3, \infty)$. Could anyone please explain why? When I graphed this function on my graphing calculator, the function starts at $ (0,0)$.

2. $ c(x) \equal{} log_3 (x^2 \minus{} 1)$

[u] My Work : [/u]

According to the definition of logarithms, $ (x^2 \minus{} 1) > 0$:

$ x^2 \minus{} 1 > 0$:

$ x^2 > 1$:

$ x \equal{} \minus{} 1, 1$

When I graphed this function, the function intersected the x axis at -1.5 and at 1.5. The provided answer for this question was $ D \equal{} ( \minus{} 1.5, \infty)$. Could this be explained?[/u][/quote] For (2), would the fact it's log base 3 have anything to do with it?

I would have given the same domain as yourself for part (1)!

Thank you for your reply, beans.

About #2, I am not sure what further restrictions there are. Also, I am unsure of how the base of 3 is significant. Since 3 > 0 and does not equal to 1, wouldn't I just focus on $ x^2 \minus{} 1$?

For #1: $ (3x)^{x\plus{}2} \equal{} 9x^2 \cdot 3^x \cdot x^x$

The first two terms are defined over the reals and the third is defined for all positive numbers, so the domain is $ (0,\infty)$ - that is, strictly positive numbers. The range, too, $ (0, \infty)$ because it $ \lim_{x \to 0} 9x^2 \cdot 3^x \cdot x^x \equal{} 0$, and the function increases with $ x$, so that couldn't have been the confusion.

I don't know where the given answer came from, but it's wrong.

For #2:

$ 3^{c(x)} \equal{} x^2 \minus{} 1 \implies x^2 \minus{} 1 > 0 \iff |x| > 1$ as you derived. Again, the given answer is wrong - there's nothing to be confused about.

Where the graph intersects the $ x$ axis is irrelevant. To find that, you just use $ log_k 1 \equal{} 0$ and be like, $ x^2 \minus{} 1 \equal{} 1 \iff x \equal{} \pm \sqrt{2} \approx \pm 1.41$. That doesn't really mean much though.

2)  The base of the logarithm is irrelevant; for any $ b$,

$ \log_b(x^2 \minus{} 1) \equal{} \frac{\ln (x^2 \minus{} 1)}{\ln b}$

so only the range is affected, not the domain.

Just a note, you shouldn't just be posting homework problems on this site. It's kind of like cheating because you're waiting for people to do the work for you.

Anyways, but the provided answers probably are incorrect. It's quite obvious that there are no restrictions for exponential functions, and you were right with the restrictions for the logarithmic function as well. So I think you were right on both.

[quote="gauss1181"]It's quite obvious that there are no restrictions for exponential functions[/quote]

Of the form $ b^x$, maybe.  The function $ x^x$ is more complicated; it is not definable (as a real function) for negative irrational numbers or negative rational numbers with even denominator in lowest terms.  It's convenient to define its domain as $ (0, \infty)$ (or, since $ x \equal{} 0$ is a removable singularity, we can define $ 0^0 \equal{} 1$ if we like and extend a little to $ [0, \infty)$).


geometry
ratio
trigonometry
circumcircle
trapezoid
radical axis
projective geometry

Let $M, N$ be points inside the angle $\angle BAC$ usch that $\angle MAB\equiv \angle NAC$. If $M_{1}, M_{2}$ and $N_{1}, N_{2}$ are the projections of $M$ and $N$ on $AB, AC$ respectively then prove that $M, N$ and $P$ the intersection of $M_{1}N_{2}$ with $N_{1}M_{2}$ are collinear.

It's just an outline of a proof: Let $E$ be intersection of $M_{1}M$ and $N_{2}N$ and $F$ be intersection of $MM_{2}$ and $NN_{1}$. Angle chase to get that $M_{1}M_{2}N_{1}N_{2}$ is cyclic then prove that $E$ and $F$ are isogonally conjugated. To prove that $M,N,P$ are collinear use Menelaos for triangle $EM_{1}N_{2}$. To count wanted ratios use Menelaos for triangle $M_{1}N_{2}A$ and line $N_{1}M_{2}P$, power of point ($AN_{2}\cdot AM_{2}=AN_{1}\cdot AM_{1}$) and sine theorem to get $\frac{EM}{MM_{1}}=\frac{AE sin \angle{EAM}}{AM_{1}sin \angle{MAM_{1}}}$ and $\frac{EN}{NN_{2}}=\frac{AE sin \angle{EAN}}{AN_{2}sin \angle{NAN_{2}}}$.

First of all let us prove that
$X_{1}X_{2}Y_{1}Y_{2}$ is concyclic. Observe that $AX_{1}UX_{2}$ and $AY_{1}VY_{2}$
are concyclic, it follows that

$\angle X_{2}X_{1}A = \angle X_{2}UA = 90^{\circ}-\angle BAU ; \angle AY_{2}Y_{1}=\angle AVY_{1}=90^{\circ}-\angle CAV.$

Therefore $\angle X_{2}X_{1}Y_{1}+\angle X_{2}Y_{2}Y_{1}= 180^{\circ}, \text{ }\Rightarrow$ $X_{1}X_{2}Y_{1}Y_{2}$ is
concyclic.

Second thing that we want to prove is that $O$, the
circumcenter of $X_{1}X_{2}Y_{1}Y_{2}$ lies on $UV$. Indeed, circumcenter
lies on the intersections of perpendiculars to the angle sides in
the midpoints of $X_{1}Y_{1}$ and $X_{2}Y_{2}$. As $UX_{1}Y_{1}V$ and $UX_{2}Y_{2}V$
are trapezoids, we get these perpendiculars intersect in the
midpoint of $UV$.

 Let $M$ be point of intersection $X_{1}X_{2}$ and $Y_{1}Y_{2}$.
Observe that $M$ is the radical center of $AX_{1}UX_{2}$, $AY_{1}VY_{2}$,
and $X_{1}X_{2}Y_{1}Y_{2}$. It follows that $AM$ is the radical axe for
$AX_{1}UX_{2}$, $AY_{1}VY_{2}$ and is perpendicular to the center line. As
$AU, AV$ are diameters, we get that $AM$ is perpendicular to the
midline of $AUV$. Hence $AM \perp UV$.

Using the properties of projective geometry we have

(a) $AW$ is a polar of $M$ with respect to the circle
$X_{1}X_{2}Y_{1}Y_{2}$.

 (b) $MW$ s a polar of $A$ with respect to the circle
$X_{1}X_{2}Y_{1}Y_{2}$. \newline  Thus $AM$ is polar of $W$ with respect to
the circle $X_{1}X_{2}Y_{1}Y_{2}, \text{ }\Rightarrow AM \text{ }\perp WO$.

Finally, adding the above facts we get that $W \in UV$.

[quote="Megus"] then prove that $E$ and $F$ are isogonally conjugated. [/quote]

How to prove it?

Also, in the end I'm left to prove $\frac{M_{2}N_{2}}{M_{1}N_{1}}=\frac{\sin FAN}{\sin EAN}$. But I don't know how it is true.

[quote="TomciO"]Hey, I think that after proving the concyclity you can end the problem easily with using Pascal two times directly (proving that $U, W, O$ and $W, O, V$ are collinear). You've proved Pascal Theorem for this special case probably :).[/quote]

Pascal how? I see only four points on the circle, with no tangents.

Ok, it's almost direct application of Pascal. Just extend the lines $UX_{1}$ and $UX_{2}$ to meet the circle again at $X_{1}'$ and $X_{2}'$ respectively. Then notice that $O$ is a intersection of $X_{1}'Y_{1}$ and $X_{2}'Y_{2}$. Is it clear now :)?

[color=darkred][b][u]An interesting generalization.[/u][/b]

In $\triangle ABC$ let $\{M,N\}\subset (BC)$ so that $M\in (BN)$, $\widehat{BAM}\equiv\widehat{CAN}$ and

$\{\begin{array}{ccc}\{M_{1},N_{1}\}\subset (AB) & , & \frac{BM_{1}}{BM}=\frac{M_{1}N_{1}}{MN}=m\\\\ \{M_{2},N_{2}\}\subset (AC) & , & \frac{CN_{2}}{CN}=\frac{N_{2}M_{2}}{NM}=n\end{array}$.

For $P\in M_{1}N_{2}\cap M_{2}N_{1}$ prove that $\boxed{P\in BC\Longleftrightarrow nb-mc=\frac{b^{2}-c^{2}}{a}}$.

[b]Particularly[/b], for $\{\begin{array}{c}m=\cos B\\\ n=\cos C\end{array}$, i.e. $\{\begin{array}{ccc}MM_{1}\perp AB & , & MM_{2}\perp AC\\\ NN_{1}\perp AB & , & NN_{2}\perp AC\end{array}$ obtain

the proposed problem : $P\in BC$ $\Longleftrightarrow$ $b^{2}-c^{2}=a(b\cos C-c\cos B)$ (true !). [/color]

Because of the quadrilateral $X_{1}X_{2}Y_{1}X_{2}$ is cyclic $($ [b][size=109]''prowler’s''[/size][/b] wording $),$ we can also complete the solution, based on the well known theorem which it has already been proved in the last replies at

http://www.mathlinks.ro/Forum/viewtopic.php?t=110887

Kostas Vittas.

Your ideas are all nice.
Sincerely I proposed this problem for Moldavian TST.
I stole it from the fact that Pascal's line for projections of two
isogonal points on triangle's sides passes through these two points.

Actually, we had something related to this on ML three years ago:

[quote="[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=13816#13816]http://www.mathlinks.ro/Forum/viewtopic.php?t=4322 post #5[/url]"]Let the lines YZ' and ZY' meet at R. [...] so that the point R lies on PQ.[/quote]

Here, P and Q are two isogonally conjugate points with respect to a triangle ABC, but the idea of the proof works for any angle BAC and two points M and N satisfying < MAB = < CAN (with directed angles).

  darij

[color=darkred][b][u]An interesting generalization.[/u][/b]

In $\triangle ABC$ let $\{M,N\}\subset (BC)$ so that $M\in (BN)$, $\widehat{BAM}\equiv\widehat{CAN}$, $E\in (AB)$, $F\in (AC)$ and

$\{\begin{array}{ccc}\{M_{1},N_{1}\}\subset (AB) & , & MM_{1}\parallel NN_{1}\parallel CE\\\\ \{M_{2},N_{2}\}\subset (AC) & , & MM_{2}\parallel NN_{2}\parallel BF\end{array}$.

For $P\in M_{1}N_{2}\cap M_{2}N_{1}$ prove that $\boxed{P\in BC\Longleftrightarrow AE\cdot AB=AF\cdot AC}$.

[b]Particular cases.[/b] Denote the intersection $X\in BF\cap CE$.

$1.\blacktriangleright$ For $X: =H$- the orthocenter of $\triangle ABC$  obtain the proposed problem : $P\in BC$ $\Longleftrightarrow$ $c\cdot b\cos A=b\cdot c\cos A$ (true !).

$2.\blacktriangleright$ For $X: =A$ obtain a new problem : $\{\begin{array}{c}MM_{1}\parallel NN_{1}\parallel AC\\\\ NN_{2}\parallel MM_{2}\parallel AB\\\\ P\in M_{1}N_{2}\cap M_{2}N_{1}\end{array}\|$ $\Longrightarrow$ $P\in BC$. [/color]


Let T_n be the right triangle with sides (4n^2,4n^4-1,4n^4+1), n a positive integer, and let \gamma _n be the angle opposite the sides of length 4n^2. Prove that, as n runs through the positive integers, the sum S of all the angles \gamma _n is a right angle: S = \gamma _1 + \gamma _2 +... = 90.

It is easy to show \gamman=2.atan(1/2n2).
So we must determine \sum 2.atan(1/2n2). This is concluded from identity atan(1-1/n)+atan(1/2n2) = atan(1-1/(n+1)).

and by atan I assume that you mean the inverse function of the tangent (tan) function, right? :)

I prefer use russian notation tg for tangent and arctg for inverse function, and I don't know exact foreign notations. I supposed tan and atan (like in programming languages) are clear notations.

Actually it's tan and arctan.  So you weren't that far off actually. :)


Let $ ABC $ be an acute triangle . Prove that :
  $ (\frac{sinA.sinB}{sinC})^2 + (\frac{sinB.sinC}{sinA})^2 + (\frac{sinA.sinC}{sinB})^2 \geq \frac{9}{4} $

Do you have a nice proof for this?...

http://www.mathlinks.ro/Forum/viewtopic.php?t=5311
http://www.mathlinks.ro/Forum/viewtopic.php?t=357

  darij

Have you seen a " mixing variables " solution , Darij ?  ;)
 This solution is nicer than that using Muirhead


Determine the maximum size of a subset $S\subseteq \{1,2,\ldots,2002\}$ such that
$(i)$ no element of $S$ divides any other element of $S$, and
$(ii)$ no two consecutive integers are both in $S$.


Let \[\begin{array}{cccc}a_{1,1}& a_{1,2}& a_{1,3}& \dots \\ a_{2,1}& a_{2,2}& a_{2,3}& \dots \\ a_{3,1}& a_{3,2}& a_{3,3}& \dots \\ \vdots & \vdots & \vdots & \ddots \end{array}\] be a doubly infinite array of positive integers, and suppose each positive integer appears exactly eight times in the array. Prove that $a_{m,n}> mn$ for some pair of positive integers $(m,n)$.

[quote="Robert Gerbicz"]For a given positive integer n consider the numbers $ {a}_{k,l}$ where $ k*l\leq n$. There are $ \sum_{k=1}^{n}\lfloor \frac{n}{k}\rfloor$ such pair of $ (k,l)$ indexes. And this is at least $ \sum_{k=1}^{n}(\frac{n}{k}-1)=n*(-1+\sum_{k=1}^{n}\frac{1}{k})$, but it is well known that $ \sum_{k=1}^{n}\frac{1}{k}$ is divergent so for large $ n$ value there are more than $ 8*n$ pair of $ (k,l)$ indexes meaning that there is at least one $ (k,l)$ pair of indexes for that $ k*l\leq n$ and $ a_{k,l}>n$ (because there are $ 8*n$ pairs of $ (k,l)$ for that $ a_{k,l}\leq n$ is true) from this $ a_{k,l}>k*l$ is also true what is needed.[/quote]


[quote="sanbittu"]problem 1

$ x \equal{} \sqrt {2 \plus{} \sqrt {2 \plus{} \sqrt {2 \plus{} .....\infty } } }$

problem2
\[ \sqrt {2x^{\rm{2}} \minus{} 9x \plus{} 4} \plus{} 3\sqrt {4x \minus{} 1} \equal{} \sqrt {2x^{\rm{2}} \minus{} 21x \minus{} 11}
\]
members please submit more sums of same type with solutions
STUDENTS FROM INDIA
ONLINE COACHING FOR SSC CBSE ICSE BOARD MATHS SITTING AT YOUR HOUSE
FOR DETAILS MAIL sanbittu@hotmail.com[/quote]

[hide="1"][b]Problem 1[/b]

We have $ x \equal{} \sqrt{2\plus{}x}$

$ x^2 \equal{} 2\plus{}x$

$ x^2\minus{}x\minus{}2\equal{}0$

$ (x\minus{}2)(x\plus{}1)\equal{}0$

$ x \equal{} 2$ or $ x \equal{} \minus{}1$

But since the RHS must be positive, we can only have $ \boxed{x \equal{} 2}$[/hide]

For problem 2:

Note that $ \sqrt {2x^{\rm{2}} \minus{} 9x \plus{} 4} \plus{} 3\sqrt {4x \minus{} 1} \equal{} \sqrt {2x^{\rm{2}} \minus{} 21x \minus{} 11}$
$ \implies \sqrt {2x^{\rm{2}} \minus{} 9x \plus{} 4} \leq \sqrt {2x^{\rm{2}} \minus{} 21x \minus{} 11}$
$ \implies 2 x^2 \minus{} 9x \plus{} 4 \leq 2 x^2 \minus{} 21 x \minus{} 11$
$ \implies 15 \leq \minus{} 12 x$
$ \implies x \leq \minus{} \frac {5}{4}$

However, $ 4x \minus{} 1 \geq 0 \iff x \geq \frac {1}{4}$

We have reached an internal contradiction, and there are thus no real solutions.

There are actually no complex solutions either, but that is well beyond the scope of your intended audience.

Oh, and [i]please don't advertise directly in your posts [/i] :)  it would be better to advertise in your signature (go to "Profile") in the horizontal button bar near the top of the page. Not like I see how you could tutor all across India, anyway  :P Oh, and please don't name all your posts the same thing


calculus
integration
LaTeX
logarithms
real analysis
function
limit

Evaluate integral((ln(x)ln(1-x))/(1+x)^2). The limits of integration are 0 and 1.

Here is the integral in $\LaTeX$, so maybe more people will attempt it. \[\int_{0}^{1}\frac{\ln x\ln(1-x)}{(1+x)^{2}}\ dx\]

Ok. I have given this problem some thought. I believe I have found a solution (edit: it turns out I have not). I will consider the integral a Lebesgue integral over $(0,1)$.

Let $I(a) = \int_{0}^{1}\frac{\ln(ax)\ln(1-x)}{(1+x)^{2}}dx$ for $a\in E$, where $E=[1/2,1]$. Also define the set $\omega = (0,1)$

Define $f(x,a) = \frac{\ln(ax)\ln(1-x)}{(1+x)^{2}}$ in $\Omega = \omega \times E$. We make the observation that $f$ is continuous in $\Omega$. We also have that $f$ can be differentiated with respect to $a$ so that $f'_{a}(x,a) = \frac{\ln(1-x)}{a(1+x)^{2}}$ and this function is also continuous in $\Omega$. Furthermore $\left| f'_{a}(x,a) \right| \le \frac{-2\ln(1-x)}{(1+x)^{2}}$ for all $x\in \omega$ and $a \in E$. It is clear that the right hand side here is a continuous function that it has a convergent integral over $w$ (I will show the latter soon). That is, we have what in swedish is called dominated convergence of $f'_{a}$. It follows that $I(a)\in C^{1}(E)$ and that

$I'(a) = \int_{w}\frac{\ln(1-x)}{a(1+x)^{2}}dx= \lim_{b\to1}\left[-\frac{1}{2}\,\ln \left(-1-x \right)-\frac{1}{2}\,{\frac{\ln \left( 1-x \right) \left( 1-x \right) }{x+1}}\right]_{0}^{b}=-\frac{\ln2}{2a}$

$\ln$ means the principal branch in the complex plane.

It follows that $I(a) =-\frac{\ln2}{2}\ln(a)+C$ for some constant $C$. Unfortunately this is actually immediately obvious from the first integral and $C = I(1)$, so this is really no help. I will post this anyway because it took so much time to write it :P

Sorry about this, I suppose the technique only works in excercises specifically defined for it.


A vector space $K^{n\times n}$ over $K$. Prove the following subsets of $K^{n\times n}$ are subspaces:

$1.$ the set of all diagonal matrices
$2.$ the set of all matrices with trace zero
$3.$ the set of upper triangular matrices
$4.$ the set of all symmetric matrices
$5.$ the set of all skew-symmtric matrices

All of these are fairly straightforward. You have to show two things (or you could bundle them together as one thing): is this set closed under addition, and is this set closed under scalar multiplication?

In the case of #2, #4, and #5, the set in question can be defined as the null space of an easily-described linear operator.

For #2, the trace of a matrix is a linear operator.

For #4, that would be the operator $A\mapsto A-A^{T}.$

For #5, that would be the operator $A\mapsto A+A^{T}.$


geometry
geometric transformation
reflection
puzzles

A man is pronounced dead. His wife tells the police that the reason he died was because a butterfly died first. What happened?

Huh Butterfly Effect? Wife is retarded, kills him and tries to find a really awesome excuse?

[hide="Random guess"]The butterfly got stuck in the man's throat and died, causing the man to choke to death?[/hide]

The man was being carried over a gaping chasm by a swarm of butterflies; together, they were barely able to sustain his weight.  However, a butterfly died and he plummeted to his demise.  Had the butterfly died later, he would have been safely transported and thus avoided his sad fate.

[hide]The man was in a butterfly museum of some sort, and there was a glass wall that was very clean.  The museum noticed that people would not be able to see the glass wall, so they put a fake butterfly on it to show people that it was there.  The butterfly fell off the wall, or "died", and the man ran into the glass wall and got hurt and bled to death internally or something of the sort.[/hide]

[quote="serialk11r"]I don't see how you can die from running into a glass wall. Also glass reflects 4% of the light and transmits 96% I think, so unless it was dark, you would be able to see the glass wall (not to mention whatever it is holding the glass wall). If it was dark, then the butterfly wouldn't have helped in the first place.[/quote]

the majority of people don't seem to notice :wink:

Other than the INTENDED answer already mentioned from the OP,

"A Sound Of Thunder"  -- science fiction story by Ray Bradbury

People went back into time of the dinosaurs for killing certain dinosaurs (but not capturing them), but to stay on a pathway.
A traveler went off the pathway and stepped on a butterfly and killed it.

That "knocked down a bunch of dominoes" and changed the future.
That traveler managed to get back to the present day.  Certain politics
had changed drastically, and the time traveler who had strayed off the
pathway back in the dinosaur time was murdered by a weapon by an angry co-traveler for changing the future to its degree, which was evidenced by
"the sound of thunder."

[quote="123456789"]Does this have something to do with that ancient Chinese philosopher's dream?[/quote]

What ancient Chinese philosopher's dream?!? :huh:

*digs out chinese story book*

The Chinese philosopher's dream went something like this:

He fell asleep and had an extremely vivid dream that he was a butterfly. When he woke up, he couldn't tell if the butterfly had turned into him or if he had turned into a butterfly.

Deep, I know.

Hehe, if I were the person carried by butterflies in Begoner's response, I'd probably have fallen off from ticklishness, never mind the death of a butterfly.

[quote="hm29168"]*digs out chinese story book*

The Chinese philosopher's dream went something like this:

He fell asleep and had an extremely vivid dream that he was a butterfly. When he woke up, he couldn't tell if the butterfly had turned into him or if he had turned into a butterfly.
[/quote]

Alas, I tried to find this story once, but I had to dig out [i]Sophie's World[/i].

So, uh....anyone going to tell us the answer?

[quote="perfect628"]Tennis is basically right. There's no reason for it to be a butterfly museum though...[/quote]

...abcd


Assume $(I)$ is incircle of triangle $ABC$. $(I)$ meet $BC,CA,AB$ at $A_1,B_1,C_1$. Denote $A_2,B_2,C_2$ are midpoints of $B_1C_1,C_1A_1,A_1B_1$. Pover that circles $(AA_1A_2),(BB_1B_2),(CC_1C_2)$ has two common points.

Hmm... GSP shows that there is one common point. :?

Hmm... Euklid DynaGeo shows that there is no common point at all, although the circles are quite near to have one...

  darij

I can't post a large picture here, but I see one common point pretty well for various configurations. It is hard to believe these are three distinct points. :?

Well, Myth, that's a very nice problem for you: Show that if the circles $\left(AA_1A_2\right)$, $\left(BB_1B_2\right)$, $\left(CC_1C_2\right)$ have one common point, then they also have a second common point (or are tangent to each other). I'm serious here, it's a nice problem and you don't need a computer for that.

  darij


Find all solutions $x,y \in \mathbb{Z}$, $x,y \geq 0$, to the equation \[ 1 + 3^x = 2^y. \]

This is really easy since $1+3^y = 2 \ or \ 4 \mod 8$

So $y < 3$, checking cases manually gives only $(1,2)$ as solution.

Your method works, but (0,1) is a solution as well.
There is thus two solutions:
(0,1) and (1,2)

With Zigmony theorem it is obvius that $y \leq 1$ since $2|3^{1}+1^{1}$

[quote="gouthamphilomath"]It is direct Catalan Mihalescu's theorem. Can we use such theorems in TSTs and Olympiads?[/quote]
My God! Goutham, Dirichlet's Theorem was proved more than a century ago, but we still don't use it to prove that there are infinitely many primes of the form $ 4n \plus{} 1$. What you're talking about is analogous to saying that "Can Fermat's Last Theorem be used to prove that the equation $ x^3\plus{}y^3 \equal{} 1$ has no solution in positive integers?". Catalan's Conjecture was only proved a few Years ago, after 1996, when this MO was held, so thus or otherwise, there's no question that it can't be used.
I hope you understand! :)


geometry
circumcircle
geometric transformation
reflection
incenter
parallelogram
geometry proposed

Given triangle $ ABC$, its circumcircle $ (O)$ and its incircle $ (I) (BC>CA>AB)$. Denote $ D$ the midpoint of arc $ BC$ which does not contain $ A, F$ the midpoint $ BC$. Let $ G$ be the reflection of $ I$ wrt $ F$. $ DG$ intersects $ (O)$ again at $ N$. Prove that $ NB\equal{}NA\plus{}NC$.

I'm changing the enunciation according to my notations.

[b]Proposition:[/b] In $ \triangle ABC,$ let $ D$ and $ M$ be the midpoints of $\overline{BC}$ and the arc $ BC$ of its circumcircle $ (O).$ $ J$ is the reflection of the incenter $ I$ about $ D.$ If  $ MJ$ cuts  $ (O)$ again at $ F,$ then one of the segments $ FA,FB,FC$ equals the sum of the other two.

Let $ \triangle A_0B_0C_0$ be the anticomplementary triangle of $ \triangle ABC,$ whose vertices are the reflections of $ A,B,C$ about the midpoints of $ BC,CA,AB.$ We prove that $ F$ is the Feuerbach point of $ \triangle A_0B_0C_0.$

Let $ L$ denote the midpoint of the arc $ BC$ containing $ A,$ $ E$ the tangency point of the incircle $ (I')$ of $ \triangle A_0B_0C_0$ with $ B_0C_0$ and $ A'$ the foot of the $ A_0$-altitude. Ray $ EL$ cuts $ (O)$ at the Feuerbach point $ F'$ of $ \triangle A_0B_0C_0$ and since $ F'E$ bisects $ \angle AFA',$ it follows that $ MF'$ is the external bisector of $ \angle AF'A'$ $ \Longrightarrow$ $ MF' \perp F'E.$ Let $ J' \equiv MF' \cap I'A_0.$ Note that $ \triangle A_0CB$ and $ \triangle ABC$ are symmetric about $ D,$ thus $ J$ is the incenter of $ \triangle A_0CB,$ moreover the midpoint of $ A_0I'.$  Then the $ A_0$-Garitte's circle with diameter $ EJ$ cuts $ (O)$ at $ F'$ $ \Longrightarrow$ $ J \equiv J',$ $F \equiv F'.$ Thus, one of the segments $FA,FB,FC$ equals the sum of the other two.

Thanks you dear [b]Luis[/b] for your attention. But we can solve it by another simple way.
[b]My proof:[/b]
Denote $ J$ a point on $ BN$ such that $ NA\equal{}NJ$. We have $ \angle BJA\equal{}90^o\plus{}\angle ANB\equal{}90^o\plus{}\angle ACB\equal{}\angle AIB$.
Then $ AIJB$ is a cyclic quadrilateral. We obtain $ \angle GNC\equal{}\angle BAI\equal{}180^o\minus{}\angle BJI$. But $ BICG$ is a parallelogram thus $ BI\equal{}CG$, which follows that $ R_{BJI}\equal{}R_{CGN}. (1)$
On the other side, $ \angle BNG\equal{}\angle BAI\equal{}\angle IJN$ then $ IJ//NG$. Therefore $ \angle BIJ\equal{}\angle NGC (2)$
From $ (1)$ and $ (2)$ we claim $ BJ\equal{}NC$ or $ NB\equal{}NA\plus{}NC$.


function
trigonometry
calculus
calculus computations

For $ 0 < \theta < \frac {\pi}{2},$ find the minimum value of the function $ \frac {\sin ^ 3 \theta \plus{} \cos ^ 3 \theta}{\sin \theta \plus{} \cos \theta}$

[hide="Possible solution"]$ \frac{\sin^3x \plus{} \cos^3x}{\sin x \plus{} \cos x} \equal{} \frac{(\sin x \plus{} \cos x)(\sin^2x \minus{} \sin x \cos x \plus{} \cos^2x)}{\sin x \plus{} \cos x} \equal{} 1 \minus{} \frac{\sin(2x)}{2} \equal{} f(x)$

Making f'(x) = 0 we get x = pi/4, which is a minimum, since f'(x) < 0 for 0 < x < pi/4 and f'(x) > 0 for pi/4 < x < pi/2. Therefore, the minimum value is f(pi/4) = 1/2.[/hide]

Your answer is correct, but you didn't need to differentiate the function.
This problem should have been posted in Interdimate Topics, I think.

Well, you can minimize the numerator and maximize the denominator separately and then note that both these extrema are attained at the same point. Another option is just to simplify the expression to $ 1\minus{}\frac 12\sin 2\theta$, which makes everything obvious. :)

You are right, fedja.

Here is my solution.

Let $ t \equal{} \sin \theta \plus{} \cos \theta$, two graphs; the unit circle $ x^2 \plus{} y^2 \equal{} 1$ and the line $ y \plus{} x \equal{} t$ gives $ 1 < t\leq \sqrt {2}.$

$ \therefore \frac {\sin ^ 3 \theta \plus{} \cos ^ 3 \theta}{\sin \theta \plus{} \cos \theta} \equal{} 1 \minus{} \sin \theta \cos \theta \equal{} 1 \minus{} \frac {t^2 \minus{} 1}{2} \equal{} \minus{} \frac {1}{2}t^2 \plus{}\frac {3}{2}.$ Since $ 1 < t^2 \leq 2$, we have the minimum value is $ \frac {1}{2}$ when $ t \equal{} \sqrt {2}\Longleftrightarrow \theta \equal{} \frac {\pi}{4}.$


[quote="José"]Prove that the following system of equations: $x^m+y^m=z^m$ ; $x^n+y^n=z^n$ , with $m$ and $n$ different numbers do not have solutions in natural numbers...[/quote]
Well, by Fermat's Last Theorem, if $m$ or $n$ is greater than $2$, then there is no solution in the natural numbers.  The only thing now is to try $m=1$ and $n=2$, and it is obvious that there aren't solutions because

\[ z^{2}=x^{2}+y^{2}\neq \left(x+y\right)^{2},  \]

for each $x,y,z\in\mathbb{N}$.

Right. But who can prove it without using Fermat??? Because this problem was in a contest which was held before Wiles prove Fermat...


calculus
integration
function
calculus computations

Evaluate \[ \int_0^1(\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3})dx \]

I've tried using riemann sums, substitutions, and expressing this as a function $g(k)$ where $k$ is the upper bound, but I can't really find a way to evaluate this integral.

If $y=\sqrt[3]{1-x^7}$, $y^3=1-x^7$ and $x=\sqrt[7]{1-y^3}$. The two functions are inverses of each other.

Let $f(x)=\sqrt[3]{1-x^7}$. We have $f(0)=1,f(1)=0$. $-\int_0^1 f^{-1}(t)\,dt=-\int_1^0 xf'(x)\,dx=\left.xf(x)\right|_0^1-\int_0^1 f(x)\,dx=-\int_0^1 f(x)\,dx$ by substitution and parts, and the integral of the difference is zero.

Ah, wonderful, I didn't see that they were inverses.

What exactly was the substitution you used in the first line?