Determine all trple of real numbers $ x;y;z$ such that
$ x^6 \plus{} y^6 \plus{} z^6 \minus{} 6(x^4 \plus{} y^4 \plus{} z^4) \plus{} 10(x^2 \plus{} y^2 \plus{} z^2) \minus{} 2(x^3y \plus{} y^3z \plus{} z^3x) \plus{} 6(xy \plus{} yz \plus{} xz) \equal{} 0$

See here:
 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168836,my

We will prove that:
$ \sum{x^6} \minus{} 6\sum{x^4} \plus{} 12\sum{x^2} \minus{} 2\sum{x^3y} \plus{} 6\sum{xy} \geq 0$
The inequality is equivalent to:
$ \sum{3(x^3 \minus{} 4x)^2} \plus{} \sum{(x^3 \minus{} 4y)^2} \geq 12 \sum{(x \minus{} y)^2}$
By Cauchy-Schwarz, we get:
$ (1 \plus{} 3)(3(x^3 \minus{} 4x)^2 \plus{} (x^3 \minus{} 4y)^2) \geq 3(4x \minus{} 4y)^2$
which helps us complete the proof.
Equality holds when $ x\equal{}y\equal{}z\equal{}0$ or $ x^2\plus{}2\frac{y}{x}\equal{}y^2\plus{}2\frac{z}{y}\equal{}z^2\plus{}2\frac{x}{z}\equal{}6(1)$
Someone please solves (1) :(


Is there a standard way to show that f(x)+f(y)=f(x+y) for all real numbers implies that f(x)=kx?

That is false. It's only true if you assume f is a continuous function.

But to answer your original question, a proof of this must be on the forum somewhere. I don't have time to type one up right now, but I'm sure it has been posted before.

As the Wikipedia article states, the solutions to [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy's functional equation[/url] can be made to take the form $ f(x) \equal{} kx$ using one of several surprisingly weak conditions such as being continuous at a single point.  Conversely, the nonlinear solutions are highly pathological and cannot be constructed without the axiom of choice.

The functional equation
$f(x+y)=f(x)+f(y)$ is called as [b]Cauchy Equation.[/b]

About [b]Cauchy Equation[/b],the following facts are all equivalent.

-$f$ is linear
-$f$ is continuous at some point
-$f$ is monotone
-$f$ is either lower-bounded or upper-bounded on some open interval

Furthermore if $f$ is non-linear,the graph of $f$ is dense on $\mathbb R^2$ :agent:


[quote="Treething"]In Treethingland, telephones are four-digit numbers. Each digit can be either 0, 2, 5, 8, or 9. How many possible phone numbers are there in Treethingland?[/quote]

[hide="Answer"]
4 places. First spot, 5 choices. Second spot, 5 choices and so on to fourth so 5^4
5^4=[b]625[/b][/hide]

[hide]1st digit-5 possibilities
2nd digit-5 possibilities
3rd digit-5 possibilities
4th digit-5 possibilities

5x5x5x5= 5:^4: =625[/hide]

OK...

If each digit must be different, how many ways are there? (It's still easy this way, but it adds a little twist)

Oooh I know! I know!
[hide="Click to reveal hidden content...if you dare"]There can be five digits in the first space, then four in the second, then three, and then two, since all must be different. 5*4*3*2=120 possibilities.[/hide]

[hide]
The little extension would make it 5!. So 5*4*3*2*1=120
[/hide]

Huh?  If each has a four digit number shouldn't it be
ABBB
where A > 0
so 4*5^3 is 500
500 telephone numbers?

[hide]jimli: Treething said that ALL numbers have to be different, not 1 number different from the rest.
5 possibilities for 1st digit
4 possibilities for 2nd digit
3 possibilities for 3rd digit
2 possibilities for 4th digit
5*4*3*2 = 120 possiblities of phone numbers.[/hide]

the extension one is more interesting ,it would be 120


trigonometry
inequalities proposed
inequalities


trigonometry
algebra
polynomial
Trigonometric Equations
Trigonometric Identities
IMO
trig identities

[quote="dashmiz"]i think the sum of these three cosin is equal to 1/2.[/quote]
 :o  the question is to prove it

Does anyone know how to find the polynomial in which those values are the roots?

Sorry for reviving this very old topic but I was looking through old problems and saw [b]pkothari13's[/b] question.

What I did was to consider the roots of unity of $ z^{7}=1$ so $ z= cis \frac{2 \pi k}{7}$ where$ k=-3,-2,-1,0,1,2,3$. Now consider the sum of roots:

$ 0 = 1+2( \cos{\frac{2 \pi}{7}}+\cos{\frac{4 \pi}{7}}+\cos{\frac{6 \pi}{7}})$

Now since: $ \cos{\frac{6 \pi}{7}}=-\cos{\frac{\pi}{7}}$ and $ \cos{\frac{4 \pi}{7}}=-\cos{\frac{3 \pi}{7}}$

We have $ \cos{\frac{\pi}{7}}-\cos{\frac{2 \pi}{7}}+\cos{\frac{3 \pi}{7}}= \frac{1}{2}$.

On Oct 30th, I took the 1st grade of Japanese Suugaku Kentei (The Mathematics Certification).
http://www.su-gaku.net/
In Question [2] of the first exam, I encountered the similar problem.

Find the polynomial $P(x)$ of the lowest degree such that
(1) the coefficient of the highest degree is 1
(2) all coefficients are rational
(3)$P \left( 2 \cos \frac{2\pi}{7}  \right)=0$

[quote=gary2003][quote=tc1729]The $\text{LHS}$ of the desired identity is $K=\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}$. Now observe that \[K\sin\frac{\pi}{7}=\frac{\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}}{2}=\frac{\sin\frac{6\pi}{7}}{2}\implies K=\frac{1}{2}\] as desired. $\Box$[/quote]

Could someone explain this to me in more detail? Is it a result of the product sum formulas?[/quote]

Yes.

Let $z = \cos \frac{\pi}{7} + i \sin \frac{\pi}{7}$. Problem is to prove that the real part of the number $z-z^2+z^3$ equals to $\frac{1}{2}$. Note that $z^7 = -1$ and $z \neq -1$. Introducing $w = z^3-z^2+z$, we have $z^3w-w+1=0$ or
\[ w = \frac{1}{1-z^3} \]
Then the proof is completed by the following general observation;
lemma: For every complex number $\alpha \neq 1$ with absolute value equals to $1$, the real part of $\frac{1}{1-\alpha}$ equals to $\frac{1}{2}$.
<Proof> Let $\beta = \frac{1}{1-\alpha}$. It suffices to show that $\beta + \overline{\beta} =1$. Actually
\[ \beta + \overline{\beta} = \frac{1-\overline{\alpha} + 1- \alpha}{(1-\alpha)(1-\overline{\alpha})} =1 \]
$\square$


Where can you find the average GPAs at undergraduates? Or average GPAs at particular departments of undergraduates?

I'm not sure what you want this information for, but be aware that grading standards vary widely between colleges and this is not a meaningful measure of anything - either the quality of the department or the quality of the students.

I am still interested in if anyone happens to know of such a place to find the average GPAs of particular departments or of undergraduates.

There's got to be a balance between difficulty and quality in choosing which undergraduate programs or undergraduates to apply to.

I would look at dropout rates before grades to determine how stressful a program is.  Again, grades vary too widely between colleges.

If you have a particular school in mind, you could try calling up the department and asking.


Ahh... THere we go. Accidentally signed up for COP 4th. Please cross me off and Sign me Up here. I'm in.

Uhh yeah I did that too. I'll do the COPSD instead hahahaha..

I think i did that 2. Put me here instead

[quote="Silverfalcon"]Please sign up for COPSD second if you're interested.

Basically, COPSD is a middle-school only contest for COP where you'll be given 10 questions to do and send the solutions (meaning answer with explanation) and if you get at least 7 right, you're promoted to next round.

Please show by "I'm in," or "I am interested", etc..[/quote]

Do you know when the questions will be posted? Sorry if I'm rushing you.

I don't know exactly when.

Probably around in next two or three weeks.

Not sure.

You can do both of COP and COPSD so I'm not going to remove your name.

I'm interested.

But is it possible for you to email me when e questions are up, and 1 day before the answers are due?  THat way, I won't forget, unlike last time.

Can a moderator make this a sticky so when Silverfalcon makes an announcement it will be easy to find?

If this is possible, I'll be grateful as well since it would gain more attention to middle-schoolers.

I try to make the contest so that not only excelled middle-schoolers can participate but average middle-schoolers should be at least make the border line of qualifying (7 out of 10).

uh....

put me in
( though try to notify me somehow because I'm not really on as much as I did before...)

and how did you sticky it????

and also,

could none-AoPS user take this?? one of my friend is interested in this round

Hmm, do you know yet when these questions will be up? Thanks.

[quote="Silverfalcon"]I think it's around 29th or something.

It's in COP 4th Announcement Thread. Same date that is will be the same date that COPSD will be posted.

I'm in no hurry.[/quote]

Oh, ok thanks. I'm eager to get this competition started.

[quote="YSJeong"]could i do both COPSD and COP??? I am a middle schooler.[/quote]

Yes, you can do both COPSD and COP. I'm also in middle school and I'm doing both.

I am interested in participating but where will I find when the tests are up? Here?

They will be up on the 29th like it says at the top of this page. The questions will be posted in a different topic but in this forum most likely.

I want to join but i am in 9th grade  :?

[quote="ExterAY"]I want to join but i am in 9th grade  :?[/quote]

If you are in 9th grade, there is the COP in the Getting Started forum.

[quote="ExterAY"]I want to join but i am in 9th grade  :?[/quote]

Sorry, but you may join the COP.

Please click "Getting Started Forum" and read the stickied thread.

Thank you.

[quote="shobber"][quote="DarkKnight"][quote="shobber"]what is the age bound? Doesn't 9th grade belong to middle school in USA?[/quote]

No. In the USA middle schools are 6th-8th grade.[/quote]
then what about high school?
9th-11th?[/quote]

High school is 9th-12th.

[quote="Silverfalcon"]Yes.

1-5 (or 6. It varies from school to school) grade = Elementary School

6-8 = Middle School or Junior High School

9-12 = High School

Just few words:

In high school:

9th grade = freshman
10th grade = sophomore
11th grade = junior
12th grade = senior

-----------------------------

For COP, COPSD is only for middle schoolers. But regular COP is for high school students although middle school students do attend it. (Middle schoolers can do both)

I think COPSD is a lot easier than COP, however.[/quote]

THis isn't true everywhere, just to let you know.  Some schools just skip middle school and make junior high 7-8, or make middle schoole 4-5, or middle school 7-9.  But for here. middle school = 6-8, and high school = 9-12, no matter what your school system does.

Please unsticky this threada and sticky the new COPSD sign up thread.

Thanks.

This is the COPSD sign up from a long time ago. Check the stickied thread.


function
inequalities
inequalities proposed

you are right .[b]arqady[/b]  . 'do you have any idea?

[quote="zaya_yc"] [b]arqady[/b]  . 'do you have any idea?[/quote]
By Zhaobin's method we can obtain a function of one variable and to estimate the extremum of him, but it's very ugly.

[quote="zaizai-hoang"][quote="dduclam"]Given $ a,b,c\geq0$ such that $ a \plus{} b \plus{} c \equal{} 3$.Prove that
a) $ (4 \plus{} 11a^2)(4 \plus{} 11b^2)(4 \plus{} 11c^2)\leq3375$
b)$ (11 \plus{} 4a^2)(11 \plus{} 4b^2)(11 \plus{} 4c^2)\geq3375$[/quote]

I think it should be:
\[ (4 \plus{} 11a^2)(4 \plus{} 11b^2)(4 \plus{} 11c^2)\geq3375
\]
And I think the following inequality hold (I solved it):
Let $ a_1,a_2,...,a_n$ be positive real numbers such that $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} n$. Prove that:
\[ (4 \plus{} 11a_1^2)(4 \plus{} 11a_2^2)...(4 \plus{} 11a_n^2)\ge 15^n
\]
Am I wrong?  :roll:[/quote]

Yes,sorry,you are wrong,Hoang. Try $ a\equal{}2,b\equal{}1,c\equal{}0$  :)

[quote="arqady"]
By Zhaobin's method we can obtain a function of one variable and to estimate the extremum of him, but it's very ugly.[/quote]
what is the method

pardesi, it's here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64933


Let $ p\geq 3$ be a prime number. Show that there exist $ p$ positive integers $ a_{1},a_{2},...,a_{p}\leq 2p^{2}$ such that the $ \frac{p(p-1)}{2}$ sums $ a_{i}+a_{j}(i<j)$ are all distinct

Let $ r_{k}$ be the remainder of $ k^{2}$ mod p. Then take the numbers $ i+2p.r_{i}$ for i = 0,1,2..p-1. All of this numbers are less than $ 2p^{2}$ and if $ a+2pr_{a}+b+2pr_{b}=c+2pr_{c}+d+2pr_{d}$ then we have mod p $ a+b=c+d$ and $ a^{2}+b^{2}=c^{2}+d^{2}$. From here it is easy to see that pairs (a,b) and (c,d) coincide.

So the problem is solved.


$ a_{1}^{2} \plus{} a_{2}^{2} \ge100$ is the equation of the exterior of a circle, centre at the origin, radius $ 10$.

$ a_{1} \plus{} a_{2} \equal{} c$ is the equation of a straight line passing through the points $ (0, c)$ and $ (c, 0)$ and so the least value of $ c$ which satisfies the inequalities is $ 10\sqrt {2}$, when the line is a tangent to the circle and $ a_{1} \equal{} a_{2} \equal{} \sqrt {\frac {100}{2}} \equal{} 5\sqrt {2}$.

Similarly, for the other inequality, it seems that we require two of $ a_{3}, a_{4}, ..., a_{100}$ to be $ 5\sqrt{2}$ and the others to be $ 0$.

Therefore the minimum value of $ S$ appears to be $ 20\sqrt {2}$.

(There is no maximum.)


When do your star-ranky-thingies change?

I know this much:

0 posts - 0 stars
20 posts (?) - 1/2 a star
50 posts - 1 star

???

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13045&postdays=0&postorder=asc&&start=0

You can get some info there. Not all the information is there though. Anyways, posting rankings do not matter much at all. They're just to judge how active people are on the forums. But don't spam just to get your post number up.

[quote="yif man12"] Anyways, posting rankings do not matter much at all. They're just to judge how active people are on the forums. But don't spam just to get your post number up.[/quote]

Too true...it's highly annoying...and u don't make too many friends this way...


algebra
polynomial
quadratics
quadratic formula

Given that the polynomial $ x^2\minus{}5x\plus{}b$ has only positive integer roots, find the average of all distinct possible values of $ b$.

since the roots are positive
then using Viete theorem
the value of b is also positive
now using quadratic formula
we get values of b as 0,4 ,6
so average is 10/3 :rotfl:

If $ b\equal{}0$, then you have $ x^2\minus{}5x\equal{}0$, whose roots are 5 and 0, but the question only asked for positive integer roots, so the answer is $ (4\plus{}6)/2$, or $ \boxed{5}$.


a,b.c,d>0 with abcd=1. than we have to prove that : $ (a^4\plus{}b^3\plus{}c^2\plus{}d)(b^4\plus{}c^3\plus{}d^2\plus{}a)(c^4\plus{}d^3\plus{}a^2\plus{}b)(d^4\plus{}a^3\plus{}b^2\plus{}c)>\equal{}256$

we can take $ AM\minus{}GM$ inequality in the following
$ \frac{a^4\plus{}b^3\plus{}c^2\plus{}d}{4}>\equal{}(a^4b^3c^2d)^{\frac{1}{4}}$
also
$ \frac{b^4\plus{}c^3\plus{}d^2\plus{}a}{4}>\equal{}(b^4c^3d^2a)^{\frac{1}{4}}$
also
$ \frac{c^4\plus{}d^3\plus{}a^2\plus{}b}{4}>\equal{}(c^4d^3a^2b)^{\frac{1}{4}}$
also
$ \frac{d^4\plus{}a^3\plus{}b^2\plus{}c}{4}>\equal{}(d^4a^3b^2c)^{\frac{1}{4}}$

multiply all above equations we get

$ (a^4\plus{}b^3\plus{}c^2\plus{}d)(b^4\plus{}c^3\plus{}d^2\plus{}a)(c^4\plus{}d^3\plus{}a^2\plus{}b)(d^4\plus{}a^3\plus{}b^2\plus{}c)>\equal{}256(abcd)^{\frac{1}{10}}$
and hence
$ (a^4\plus{}b^3\plus{}c^2\plus{}d)(b^4\plus{}c^3\plus{}d^2\plus{}a)(c^4\plus{}d^3\plus{}a^2\plus{}b)(d^4\plus{}a^3\plus{}b^2\plus{}c)>\equal{}256$ because abcd=1

[quote="petraru"]a,b.c,d>0 with abcd=1. than we have to prove that : $ (a^4 \plus{} b^3 \plus{} c^2 \plus{} d)(b^4 \plus{} c^3 \plus{} d^2 \plus{} a)(c^4 \plus{} d^3 \plus{} a^2 \plus{} b)(d^4 \plus{} a^3 \plus{} b^2 \plus{} c) > \equal{} 256$[/quote]
Using Holder inequality,we have:
$ (a^4 \plus{} b^3 \plus{} c^2 \plus{} d)(b^4 \plus{} c^3 \plus{} d^2 \plus{} a)(c^4 \plus{} d^3 \plus{} a^2 \plus{} b)(d^4 \plus{} a^3 \plus{} b^2 \plus{} c) \ge (abcd\plus{}\sqrt[4]{a^3b^3c^3d^3}\plus{}\sqrt[4]{a^2b^2c^2d^2}\plus{}\sqrt[4]{abcd})^4\equal{}256$

You've already posted this problem:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169817[/url]

Do not double post.


Show that if $D\in\mathcal M_{2}(C)$ and $rank(D)=rank(D^{2})$ then $rank(D)=rank(D^{n})$ for any natural $n>0$.

Since $rank(D)\leq 2$ when he is $0$ or $2$ is preatty clear. But what's happening when $rank(D)=1?$ Maybe we need to use the information that is given in the hypotesis but i can't figure out how..

The general case was discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=term&t=73922]here[/url]. But I don't see how to avoid the college stuff. With them, it's rather easy.

Also, the case $n=2$ is easily solved. Just use Cayley-Hamilton $\left( A^{2}-\textrm{tr}\, A+\left( \det A \right) I_{2}= 0, \, \forall A \in \mathcal M_{2}\left( \mathbb C \right) \right)$.

Let's do everything over $F^{k}$, where $F$ is an arbitrary field and $k$ is an arbitrary positive integer.

The rank of $D$ is the dimension of $U=D\cdot F^{k}$. The rank of $D^{2}$ is the dimension of $V=D^{2}\cdot F^{k}=D\cdot U$. Since $D\cdot U\subset D\cdot F^{k}$, $V\subset U$. If the two ranks are equal, we must have $V=U$, and $D$ is surjective from $U$ onto itself. The composition of any number of surjective maps is surjective, and therefore $D^{n}\cdot F^{k}=D^{n-1}\cdot U=U$ for all $n\ge1$. Now $\text{rank }D^{n}=\dim U=\text{rank }D$ and we are done.
[This is essentially grobber's argument in the linked thread]

I don't think this special case is really any easier than the generally case; the tools for dimension 2 don't really make things any easier.
For anyone who isn't Romanian, all linear algebra is college material anyway, and this argument works in the context of a first course.

Here, I used this problem for a beautiful problem. You could find a proof [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97845]here[/url].

Actually, things are much easier in dimension 2 for this very particular case, since if $A$ has rank 1 then $A$ verifies a relation of the form $A^{2}=aA$ and everything is finished.

That's better :). jmerry's solution is too complicated and requires advanced knowledge that a student who just finished 12 grades isn't aware of :P.

I like natural approaches and Jmerry's one is the best! And since it's quite a long time since I'm no longer in 12-th grade, all I can say is: if you are to find a solution for a problem, try a conceptual one and not the most tricky.


Let $ ABCD$ be a quadrilateral such that$ AD \equal{} CD$and $ \angle DAB \equal{}
\angle ABC < 90^\circ$ . The line passing through D and the midpoint of $ BC$ intersects $ AB$ at point E. Prove that $ \angle BEC \equal{} \angle DAC.$

Construct $ (ADC)$, it cuts AB at G.
Let E' be a point such that $ \angle CE'B\equal{}\angle CAD$, we will prove that DE passes through midpoint M of CB.
Since $ ADCG$ is cyclic we get $ \angle AGD\equal{}\angle ACD\equal{}\angle DAC\equal{}\angle DGC\equal{}\angle GEC\equal{}\angle GCE$
Let $ I\equal{}CG\cap DE, H\equal{}DG\cap CB$.
Applying [b]Menelaus[/b]'s theorem, $ D.M.E$ are collinear iff $ \frac{MB}{MC}.\frac{IC}{IG}.\frac{EG}{EB}\equal{}1$
$ \Leftrightarrow \frac{IC}{IG}\equal{}\frac{EB}{EG}\equal{}\frac{CE}{DG} (*)$
But $ \angle DAB\equal{}\angle ABC$ then $ \Delta DCG\sim \Delta CBE$
$ \Rightarrow \frac{CE}{DG}\equal{}\frac{DC}{CB}\equal{}\frac{CH}{CB}\equal{}\frac{EB}{EG}$
So $ (*)$ is true.
We are done.


It is a well-known fact that the harmonic series
\[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}... \]
diverges. Let $S$ be the set of positive integers without a $0$ in their base-10 representation. Does the sum of the reciprocals of $S$ converge or diverge?

At least the third time I have seen this problem.

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=harmonic+representation+base&t=115768


I don't understand, How to find the number 16 ?
31 is a prime number, why 16^14 ?


calculus
algebra
polynomial
integration
function
domain
Ring Theory

Let $ p$ be a prime. Prove that
\[ p \mid \sum_{\frac {j}{p^{k \minus{} 1}} \le i \le p \minus{} 1} \binom{ip^{k \minus{} 1}}{j}\]
when $ 0 \le j < p^k \minus{} p^{k \minus{} 1}$ is an integer.

Edit: Nevermind, I believe the problem is trivial.

[quote="calc rulz"]Let $ p$ be a prime. Prove that
\[ p \mid \sum_{\frac {j}{p^{k \minus{} 1}} \le i \le p \minus{} 1} \binom{ip^{k \minus{} 1}}{j}\]
when $ 0 \le j < p^k \minus{} p^{k \minus{} 1}$ is an integer
Edit: Nevermind, I believe the problem is trivial.[/quote]
This is an application of  theorem :[url]http://en.wikipedia.org/wiki/Lucas%27_theorem[/url]. By this theorem we have two facts here  1) If $ j$ is not a multiplication of $ p^{k \minus{} 1}$ then $ {i.p^{k \minus{} 1}}\choose {j}$ is divisible by $ p$ .
2) If $ j \equal{} t.p^{k \minus{} 1}$ where $ 0\leq t\leq p \minus{} 2$ (by the condition of this problem ) then $ {{i.p^{k \minus{} 1}}\choose {t.p^{k \minus{} 1}}}\equiv {{i}\choose {t}} (\mod p)$ .
Then it is easy to show that :
\[ S\equiv \sum_{i \equal{} 0}^{p \minus{} 1}\sum_{j \equal{} 0}^i {{i}\choose {j}} \minus{} 1 (\mod p )\]
or it is equivalent to the following congruence after some calculus:
\[ S\equiv 2^p \minus{} 2 (\mod p )\]
By Fermat theorem we have $ p|S$ and that's what we need .

Nice! [hide="Here is the motivation for it."]
Consider the $ p^k$th cyclotomic polynomial $ \Phi_{p^k} = \sum_{i = 0}^{p - 1} x^{ip^{k - 1}}$. To show this is irreducible, we apply Eisenstein's criterion to
\begin{eqnarray*} \Phi_{p^k}(x + 1) & = & \sum_{i = 0}^{p - 1} (x + 1)^{ip^{k - 1}} \\
& = & \sum_{i = 0}^{p - 1} \sum_{j = 0}^{ip^{k - 1}} \binom{ip^{k - 1}}{j} x^j \\
& = & \sum_{j = 0}^{p^k - p^{k - 1}} x^j \sum_{\frac {j}{p^{k - 1}}\le i\le p - 1}\binom{ip^{k - 1}}{j} \end{eqnarray*}
To apply Eisenstein's criterion, we need to show that the $ j$th coefficient is divisible by $ p$ for $ 0 \le j < p^k - p^{k - 1}$. We can show this directly, as TTsphn did.

Alternatively, taking $ (x^{p^{k - 1}} - 1)\Phi_{p^k}(x) = x^{p^k} - 1$ and reducing both sides of this equation mod $ p$, we get $ (x - 1)^{p^{k - 1}}\Phi_{p^k}(x) \equiv (x - 1)^{p^k} \mod{p}$ since the Frobenius map is a homomorphism in characteristic $ p$. Since $ (\mathbb{Z}/p)[x]$ is an integral domain, we can conclude $ \Phi_{p^k}(x) \equiv (x - 1)^{p^k - p^{k - 1}} \mod{p}$, so $ \Phi_{p^k}(x + 1) \equiv x^{p^k - p^{k - 1}} \mod{p}$. This means that all the coefficients other than the leading coefficient of $ \Phi_{p^k}(x + 1)$ are divisible by $ p$.
[/hide]


Found on an other forum  :

Find the unit (= right most) digit of :

[10^1992/(10^83+7)]

where [x] is as usual the greatest integer less than or equal to x.

Left ! I still have problem with my left and my right ....

Errr, you mean right? The units digit is on the right. You're reading from the left hand side of the screen to the right.

12345
1 is on the left.
5 is the units digit. It is on the right.

[10^1992/(10^83+7)]

[hide]So, my first suspicion involved factoring 1992, and, lo and behold, 1992 = 24*83.  Thus, we want [x^24/(x + 7)], where x = 10^83.  Just carrying out long division, we see that

x^24/(x + 7) = x^23 - 7x^22 + 49x^21 - 343x^20 + . . . - 7^23*x^0 + 7^24*x^(-1) - . . .

Now, 7^24 has significantly fewer than 83 digits, so we can basically take all of the terms with negative powers of x and just treat them as a small positive constant.  Everything with powers of x greater than or equal to 1 end in many zeroes.  Thus, we can treat all that as a large positive constant with 0 units digit.  Thus our expression reduces to

E - 7^23 + e.  7^23 ends in a 3, which means E - 7^23 ends in a 7, which means that the thing we're looking for is 7.

[/hide]


$ \tan (45^\circ \plus{}k^\circ)\equal{}\frac{1\plus{}\tan k^\circ}{1\minus{}\tan k^\circ}\ (k\equal{}1,\ 2,\ \cdots 45)$

[hide][quote="AndrewTom"]Evaluate $ (1 \plus{} \tan 1^{\circ})(1 \plus{} \tan 2^{\circ})(1 \plus{} \tan 3^{\circ})...(1 \plus{} \tan 44^{\circ})(1 \plus{} \tan 45^{\circ})$.[/quote]
$ (1 \plus{} \tan x)(1 \plus{} \tan(45 \minus{} x)) \equal{} (1 \plus{} \tan x)[1 \plus{} (1 \minus{} \tan x)/(1 \plus{} \tan x)]$
$ \equal{} (1 \plus{} \tan x)(2/(1 \plus{} \tan x)) \equal{} 2$
So the answer is $ 2^{22}\cdot2 \equal{} 2^{23}$[/hide]

$ \equal{}(1\plus{}\tan x)\plus{}(1\minus{}\tan x))\equal{}2$
I got a little confused there

Thanks, mathwizarddude. That's more or less what I had. But I still don't get what kunny's getting at.


search
inequalities proposed
inequalities

[quote="arash nk"]let $ a,b,c$ be positive numbers.find $ k$ such that:
$ (k \plus{} \frac {a}{b \plus{} c})(k \plus{} \frac {b}{c \plus{} a})(k \plus{} \frac {c}{a \plus{} b})\ge (k \plus{} \frac {1}{2})^{3}$[/quote]
I have got $ k\geq\frac{\sqrt5\minus{}1}{4}$ or $ k\leq\frac{\minus{}1\minus{}\sqrt5}{4}.$ :maybe:

Take a look here http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1533960969&t=272187 :wink:


calculus
integration
derivative
advanced fields
advanced fields unsolved

How do you prove the following:

    if g(t) = t f(t) then g(w) = i df(w) / dw

with g(w) and f(w) representing Fourier Transforms of g and f.
The only way I can think of is using integration by parts, but even that doesn't seem to get me anywhere. Any help would be greatly appreciated.

For at least a formal argument, try writing down the integral for the Fourier transform (I'm not quite sure what convention you're using in terms of placemen of $2\pi$ and the signs of the exponents) and differentiate with respect to $w$ (or is that $\omega$), bringing the derivative inside the integral sign.


linear algebra
matrix
linear algebra unsolved

Solve the system equation matrix :
[color=red]1)[/color]$ \{\begin{array}{c}{1}A^6 + B^6 = I_n \
2A + B = \sqrt [6]{(\sqrt [5]{64} + 1)^5}I_n \end{array}$
[color=red]2)[/color]$ \{\begin{array}{c}{1}A^6 + B^6 + C^6 = I_n \
3A + B + C = \sqrt [6]{(\sqrt [5]{729} + 2)^5}I_n \end{array}$
 Here $ A,B,C \in M_n{\mathbb{(C)}}$ are Hermitian matrices and semipositive define.
 $ I_n$ is $ n$x$ n$ identity matrix.
[hide="Answer of me"]For 1)$ A = \sqrt {\frac {\sqrt [5]{4}}{\sqrt [3]{\sqrt [5]{64} + 1}}}I_n$ ,${ B = \sqrt {\frac {1}{\sqrt [3]{\sqrt [5]{64} + 1}}}}I_n$,
For2) [b]Implies if you known[/b]  :P  [/hide]

Hi Quybac, do you force yourselves to post all the exercises that you imagine ?
For 1) $ B\equal{}\alpha{I}\minus{}2A$ and $ A,B$ commute. Then we may assume $ A,B$ are diagonal matrices with non negative entries. After it is easy.
You have posted many similar exercises.
The ii) is much more difficult and even (I think) unfeasible.
This is not serious at all !

I will post lemma of me last (English of me so bad  :blush: ) . I try for shorter lemma .


Points $P$ and $Q$ are both in the line segment $AB$ and on the same side of its midpoint. $P$ divides $AB$ in the ratio 2:3, and $Q$ divides $AB$ in the ratio 3:4. If $PQ=2$, then the length of $AB$ is:
(A) $60$
(B) $70$
(C) $75$
(D) $80$
(E) $85$

[hide="Answer"]Given that the midpoint of segment $AB$ is point $M$, and that $P$ and $Q$ are both on segment $AM$. Therefore, $AP=\frac{2}{5} AB$, and $AQ=\frac{3}{7} AB$. $PQ=2=AQ-AP=\frac{1}{35} AB \Rightarrow AB=70$, and the answer is $B$.[/hide]

I just kinda did it in a really easy, way, you know that the line is divided in $5$ regions and $7$ regions, so just find the common multiple which is $35$, now it says $PQ$ is $2$ and so it implies that $35(2)=70$. Or just look for a multiple of $35$ in one of the answers.

Another way to do it, is that you can set equations, $3z+4z=y,  2x+3x=y,  3z-2x=2$, I solved for $z$ and got that as being $10$ so the segment is $70$.

Please edit your post rather than double posting.

Your first solution is basically the same thing that I did, except I wrote out a little more of the thought behind it.

[hide]
Let $x$ be the length of $\overline{AB}$. Therefore
$\overline{AQ}-\overline{AP}=2=\frac{3}{7}x-\frac{2}{5}x=\frac{15}{35}x-\frac{14}{35}x=\frac{1}{35}x, x=70$
[/hide]

Do not post an answer with no solution, it is complete spam, especially when a complete solution has already been posted - more than one, in fact. If you want to post an answer with solution, do so, but you should never just post an answer.

[hide]  Either moving from $B$ to $A$ or $A$ to $B$.  $Q$ and $P$ must be $\frac{3}{7}$ and $\frac{2}{5}$ repectively, for anything else would be greater than $\frac{1}{2}$.  Thus, $\frac{3}{7} - \frac{2}{5}=\frac{15}{35} - \frac{14}{35} = \frac{1}{35}x = 2$  Hence, $x=\boxed{70}$[/hide]


$ xy\equal{}56$ and $ \frac{7}{x}\plus{}\frac{14}{y}\equal{}4$.  Then $ y\equal{}\frac{56}{x}\Rightarrow \frac{7}{x}\plus{}\frac{14}{\frac{56}{x}}\equal{}\frac{7}{x}\plus{}\frac{14x}{56}\equal{}4\Rightarrow 14x^{2}\minus{}224x\plus{}392\equal{}0$, which comes from some algebra and re-arranging.  14 is a common factor so this reduces to $ x^{2}\minus{}16x\plus{}28\equal{}(x\minus{}14)(x\minus{}2)\Rightarrow x\equal{}2, 14$.  But recall that $ x<y$ so we must have that $ x\equal{}2$.


If you just want the Greek letter, it's \Sigma for uppercase $ \Sigma$ and \sigma for lowercase $ \sigma$. \sum is an uppercase sigma, but enlarged and set up to take subscripts and superscripts nicely in display mode.

Yes, although the second is nicer, and first isn't supported everywhere (at least for me)

[quote="AwesomeToad"]There is more use for summation, so most use $ \sum$ (\sum)
Also there are 2 ways to do combination? $ \displaystyle{n}\choose{r}$ and $ \binom{n}{r}$[/quote]
You're doing the first one wrong. It should be "{n\choose r}" : $ {n\choose r}$

He was saying that there are two ways to write it, and i think i like the second one betta but there's nothing wrong with the first.


Let the polynomial $P(x)=x^3+19x^2+94x+a$ where $a\in\mathbb{N}$. If $p$ a prime number, prove that no more than  three numbers of the numbers $P(0), P(1),\ldots, P(p-1)$  are divisible by $p$.

Well, note that $\mathbb{Z}_p$ is a field as one can always find an inverse for a non-zero element in it and consider the given polynomial in this field. It is of degree $3$ so it has no more than $3$ roots in the above field. Q.E.D.

Suppose 0<a1<...<ai<p and p|P(a1),.....,p|P(ai). I will prove that i<=3.

  If i=1 or 2 or 3 ok

 If i>=4 then

 P(a1)=k*p , so suppose Q(x)=P(x)-k*p= (x-a1)*Q1(x) with degQ1(x)=2.
 We have p| Q1(a2) because p| Q(a2) and p doesn't divide a2-a1.

   So suppose Q1(a2)=l*p and Q1'(x)=Q1(x)-l*p. Then Q1'(x)=(x-a2)Q2(x) with degQ2(x)=1.
Again we have that p|Q2(a3) because p| Q1'(a3) and p doesn't divide a3-a2. So
Q2(a3)=m*p, so suppose Q2'(x)=Q2(x)-m*p. Q2'(x)=(x-a3).   So if p| P(a4) then
p| a4-a3 (contradiction).

We suppose that $p|P(b),P(c),P(d),P(e) (0\le b<c<d<e\le p-1)$.

$P(c)-P(b)=(c-b)(c^2+cb+b^2+19c+19b+94)$.
Since $0<c-b\le p-1$, $p|c^2+cb+b^2+19(c+b)+94$---①
Similarly we get $p|d^2+db+b^2+19(d+b)+94$---②
From ①,②, $p|(d-c)(d+c+b+19)$.Since $0<d-c\le p-1$, $p|b+c+d+19$
Similarly we get $p|b+c+e+19$.Then $p|e-d$ which is absurd.Therefore the conclusion holds.

Q.E.D.


Here is an interesting article: http://news.bbc.co.uk/2/hi/science/nature/3430481.stm

How do you feel about animal intelligence? Do you think that they can use language, though maybe not int he way we use it?

I also conducted a search on Google of N'kisi. Before I read this article, I thought that Alex (an African grey subject of Irene Pepperberg who had a vocabulary of around 150 words and knew what he was saying) was amazing. Alex could make requests such as requests for more nuts or to go back into his cage. But N'kisi is so much more amazing. He has been suspected of having telepathy but that still cannot make one doubt his intelligence. Many of his notable quotes were notable because they were unexpected and the owner wasn't thinking about them.

This conversation is especially interesting:

N'Kisi: "Remember, we went in a car"
Aimee: "Yes! Did you like it?"
N'Kisi: "I like that - wanna go out in the car"
Aimee: "We can't, wo don't have a car now"
N'Kisi: "Wanna go in a car right now"
Aimee: "I'm sorry, we can't right now - maybe we can go again later"
N'kisi: "Why can't I go in a car now?"
Aimee: "Because we don't have one"
N'kisi: "Let's get a car"
Aimee: "No Kisi, we can't get a car now"
N'kisi: "I want a car"
Aimee: "I'm sorry, baby, not today"
N'kisi: "Hurry up, wanna go in a car. Remember? We were in a car"

I believe that all life is intelligent, even if it cannot think for itself. All life can evolve to adapt to its surroundings and that, that is intelligence.


In the figure below, isosceles $\triangle ABC$ with base $\overline{AB}$ has altitude $CH=24\text{ cm}$. $DE=GF$, $HF=12\text{ cm}$, and $FB=6\text{ cm}$. What is the number of square centimeters in the area of pentagon $CDEFG$?

[img]7232[/img]

use similarity for the are of the small triangles and subtract the triangle area from the area of the largest triangle.

[hide="Solution"]
We know that $\angle B$ is common for both $\triangle CHB$ and $\triangle GFB$. We also get equal angles $\angle GFB=\angle CHB$. By AA similarity, $\triangle CHB\sim\triangle GFB$. We also know that $BF: FH$ is in a $1: 3$ ratio. Therefore, $GF=8$. We use $bh/2$ for triangle area to find the area of $[GFB]=24$. We also know that by $bh/2$ for triangle area that $[CHB]=216$. Therefore, $[CHFG]=[CHB]-[GFB]=216-24=192$. However, this is only half of pentagon $CGFED$. We double $216$ to get the final answer $\implies\boxed{384\text{ square centimeters}}$.[/hide]

[hide]
you can also realize that the triangles are 3,4,5 triangles, so that may help.
the big ones are 18,24,30 or area of 216 and the smaller ones 6,8,10 with area of 24, so subtract to get 192 the multiply by 2 to get 384.[/hide]


pigeonhole principle
combinatorics unsolved
combinatorics

1.A 2004 x 2004 array of points is drawn.Find the largest interger n such that it is possible to draw a convex n-sided polygon whose vertices lie on the points of the array
2. The transportation ministry has just decided to pay 80 companies to repair 2400 roads. These roads connects 100 cities.Each road is between two cities and there is at most one road between any two cities.Each company must repair exactly 30 roads,and each road is repaired by exactly one company.For a company to repair a road,it is necessary for the company to set up stations at the both cities on its endpoints.Prove that there are at least 8 companies stationed at one city.

2. The transportation ministry has just decided to pay 80 companies to repair 2400 roads. These roads connects 100 cities.Each road is between two cities and there is at most one road between any two cities.Each company must repair exactly 30 roads,and each road is repaired by exactly one company.For a company to repair a road,it is necessary for the company to set up stations at the both cities on its endpoints.Prove that there are at least 8 companies stationed at one city.

Solution: For each company, it is easy to see that it repairs at least 9 cities as K8 contains only 28 lines. So there are at least 720 different stations. By the PigeonHole Theory, it is easy to see that at least 8 companies stationed at one city. We are done.


calculus
search
SFFT
special factorizations

well i am trying to solve $abc-2=a+b+c$ over positive integers.
so i tried to make it easier and solve $abc=a+b+c$ is there any way to do this? (all i have so far is $(1,2,3)$ and that was a guess) i have tried making $c$ constant.

[hide="A hint"]
Is it possible for $a, b, c \ge 2$?  If not, then what does this tell you about the smallest of $a, b, c$?
[/hide]

[hide="Extended to the original problem"]
Can you show that $a, b, c \ge 2 \implies abc-2 \ge a+b+c$?[/hide]

[quote="me@home"]$abc=a+b+c$ let $x=ab,y=bc,z=ca$ $1=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
and solve this...[/quote]

to do this would u do

WLOG $x\geq{y}\geq{z}$ so $\frac{3}{z}\geq{1}$ which implies that
$z\leq{3}$ and then just solve all cases.

this gives $(x,y,z)=(3,3,3),(2,3,6),(2,4,4)$
then solving for $(a,b,c)$ we see that the only solution is $(1,2,3)$.

is this right?

[b]zgorkster[/b] how would you solve $c=\frac{a+b+2}{ab-1}$ for the other values of $c$?

Take a look.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108621[/url]

[quote="zgorkster"]These problems aren't hard, they are all similiar so I'll just solve your original question.  First note that if $a=b=1$, there are no solutions.  Now, after moving terms and factoring, we have

$c=\frac{a+b+2}{ab-1}$.

Now, using calculus, it is not hard to show that the RHS of the equation above is always decreasing, but you can show this without calculus (you can just use long division and see the expression is the sum of decreasing functions).  Now, since both $a$ and $b$ cannot be $1$, the next best thing is $a=1$, $b=2$, which implies $c=5$.  Hence $c\le 5$.

Now there are just a finite number of cases to check, either by hand or by Diophantine equations.  These methods yield the solutions to be

$(a,b,c)=(1,2,5),(1,3,3),(2,2,2)$ and their permutations.[/quote]
after i got to $c=\frac{a+b+2}{ab-1}$ i did WLOG  $a\geq{b}$ so i now had $c\leq\frac{2}{b-1}$ and the only possible values for $b$ were $3$ and $2$ so doing this and all possible values for $c$ i ended up with the same solutions.
thanks for the help

well since there are so few solutions a good way to attack this problem is by showing they are less than 3; I was shooting more towards something like Simon's favorite factoring trick, i'm not sure if you've heard of it... do a search :)


Suppose that $a,b,c$ are positive real numbers. Prove the inequality
\[\sqrt[5]{\frac{a^{5}+b^{5}+c^{5}+1}{3}}\geq\frac{a+b+c+1}{3}-\frac{5/9}{\sqrt{a/bc+b/ca+c/ab+1/abc}}. \]

Another similar problem is: Prove the inequality \[\sqrt[5]{\frac{a^{15}+b^{15}+c^{15}+(abc)^{5}}{3}}\geq \frac{a^{3}+b^{3}+c^{3}+abc}{3}-\frac{10/9}{a/(bc)^{2}+b/(ca)^{2}+c/(ab)^{2}+1/abc}\] for all positive real numbers $a,b,c.$ This problem follows from the previous one.

Related to the first post: If $a,b,c$ are positive real numbers prove the inequality
\[\sqrt[7]{\frac{a^{7}+b^{7}+c^{7}+1}{3}}\geq \frac{a+b+c+1}{3}-\frac{7/12}{\sqrt{a/bc+b/ca+c/ab+1/abc}}. \]


geometry
parallelogram
geometric transformation

In a parallelogram $ ABCD$ let $ M \in AB$ and $ N \in AC$ and $ P \in AD$ such that $ AMNP$ is also a parallelogram.

If parallelogram $ AMNP$ is now allowed to "slide" down the side $ AB$ until $ M$ coincides with $ B$, prove that $ P$ lies on $ BD$.

Not sure if this is OK, but:

If $ N$ moves to $ N'$ and $ P'$ is the point wher $ PNN'$ meets $ DB$, we want to show that $ P'$ is the image of $ P$ under the translation that takes $ M$ to $ B$, that is we want to show that $ PP' \equal{} NN'$.

Now triangles $ ABD$ and $ CDB$ are congruent as are triangles $ ABD$ and $ BAC$ and so are $ PP'D$ and $ N'NC$, so $ PP' \equal{} NN'$.


An integer between 000 and 999 is [i]multiple[/i] choice. :D (and Kalva's problems don't always have integer answers between 000 and 999).


2 selected harder problems from the Greater Atlanta Mathematics Examination, some help would be greatly appreciated.

1) Given the system:
3x-2y+az=4
4x+y-z=10
5x+4y+(a+5)z= -4
Each equation of the system describes a plane in 3D space and these three planes have exactly one point in common. What is the only value that [i]a[/i] cannot have?

2) In triangle ABC, let A, B, and C be the dergee measures of the three angles and let 36, 82, and [i]b[/i] be the lengths of the sides of AB, BC, and CA respectively. Given that the determinant of K is zero, where K=
|82  -b||sinB  82|
|-01  1||sinA  b   |,
Find the largest possible area of triangle ABC. (Note: the -01 has the 0 just so the lines would match up)

$\left[\begin{array}{ccc|c}3&-2&a&4\\4&1&-1&10\\5&4&a+5&-4\end{array}\right]$

Row-reduce and look at the bottom row.  You should have $f\left(a\right)$ as your 3,3 entry and some constant to the right.  Then you know that $f\left(a\right)\neq 0$.


calculus
integration
complex analysis
complex analysis unsolved

$ \mbox{res}_0 \frac{1}{(1\minus{}e^{\minus{}z})^n}$ for all $ n \in \mathbb N$

This is computationally hard for me. Any trick to do this?

Here $ \oint$ denotes integral along a sufficiently small simple closed curve around zero:
\[ [z^{\minus{}1}]\frac{1}{(1\minus{}e^{\minus{}z})^{\minus{}n}}\equal{}\frac{1}{2\pi i} \oint \frac{dz}{(1\minus{}e^{\minus{}z})^n}\equal{}\frac{1}{2\pi i} \oint\frac{dt}{t^n(1\minus{}t)}\equal{}
[t^{n\minus{}1}]\frac{1}{1\minus{}t}\equal{}1.\]
Here I changed the variable $ t\equal{}(1\minus{}e^{\minus{}z})$, and this maps a small simple closed curve around zero into a similar curve. Hopefully I did no mistake.


solve the equation
\[ \log_2^2(x^2\minus{}x)\log_2\left(\frac{x\minus{}1}{x}\right)\plus{}\log_2^2 x^2 \equal{} 4\]

[quote="anik_andrew"]solve the equation
\[ \log_2^2(x^2 \minus{} x)\log_2\left(\frac {x \minus{} 1}{x}\right) \plus{} \log_2^2 x^2 \equal{} 4\]
[/quote]
If it was supposed to be $ \log_2(x^2 \minus{} x)\log_2\left(\frac {x \minus{} 1}{x}\right) \plus{} \log_2^2 x^2 \equal{} 4$, my solution is:

$ \log_2(x^2\minus{}x) \equal{} \log_2 (x\minus{}1) \plus{} \log_2 x$
$ \log_2\left(\frac {x \minus{} 1}{x}\right) \equal{} \log_2(x\minus{}1) \minus{} \log_2 x$.
So $ LHS \equal{} (\log_2^2(x\minus{}1) \minus{} \log_2^2x) \plus{} \log_2^2 x \equal{} 4 \iff \log_2^2 (x\minus{}1) \equal{} 4$ $ \iff \log_2(x\minus{}1) \equal{} \pm 2 \iff x\minus{}1 \equal{} 2^{\pm 2} \iff x \equal{} 2^{\pm 2}\plus{}1$. So we have the solutions $ x\equal{}5,x\equal{}\frac{5}{4}$

If not, then I'll try again.

[quote="anik_andrew"][quote="Mathias_DK"][quote="anik_andrew"]solve the equation
\[ \log_2^2(x^2 \minus{} x)\log_2\left(\frac {x \minus{} 1}{x}\right) \plus{} \log_2^2 x^2 \equal{} 4\]
[/quote]
If it was supposed to be $ \log_2(x^2 \minus{} x)\log_2\left(\frac {x \minus{} 1}{x}\right) \plus{} \log_2^2 {\bf x^2} \equal{} 4$, my solution is:

$ \log_2(x^2 \minus{} x) \equal{} \log_2 (x \minus{} 1) \plus{} \log_2 x$
$ \log_2\left(\frac {x \minus{} 1}{x}\right) \equal{} \log_2(x \minus{} 1) \minus{} \log_2 x$.
So $ LHS \equal{} (\log_2^2(x \minus{} 1) \minus{} \log_2^2x) \plus{} \log_2^2 {\bf x} \equal{} 4 ......$[/quote]
1) you lost the square (((
2) in my book the equation is exactly as in my first post  :([/quote]
Oh. I see. Forget my comment then.

[quote="anik_andrew"]maybe there is mistake in book...[/quote]

what book might that be ?

its russian book for 10-11 forms by М.Л.Галицкий, А.М.Гольдман, Л.И.Звавич


Find the volume of the set of points at most 1 unit away from a 1*1*1 cube, including the cube itself.

1 unit away from the center or at any point?  Because the set of points at most some distance away is a sphere.  Although in this case it is somewhat different.

From any point of the cube.

If we draw this out, we see that our solid looks like a 3 by 3 cube except the edges and vertices are rounded out (sorry, bad description..). From each face (6 of them) of the original cube, we form another 1x1x1 cube, and counting the original, the volume is $ 6 \plus{} 1 \equal{} 7$. From each edge, we form a quarter cylinder with height 1 and radius 1; there are twelve such quarter cylinders with total volume $ \frac {12}{4} \left(1\right)^2 \pi \cdot 1 \equal{} 3\pi$. From each corner we form an eighth sphere, and there are eight of a vertices, with total volume $ \frac {4}{3}\pi 1^3 \equal{} \frac {4}{3}\pi$. These add up to $ \frac {13\pi}{3} \plus{} 7$.

[quote="The Zuton Force"][b]Off Topic:[/b] Colts, you may want to specify a city before "oregon" in your location. This would bump "Rating" to the next line and making it look more realistic  :)[/quote]

ALso to make it more realistic gocolts should not make up his location :lol:

anyways heres a similar AIME problem that you can practice with
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=508863#508863


A non-zero digit is chosen in such a way that the probability of choosing digit $ d$ is $ \log_{10}(d\plus{}1) \minus{} \log_{10} d$.  The probability that the digit $ 2$ is chosen is exactly $ \frac12$ the probability that the digit chosen is in the set

$ \textbf{(A)}\ \{2,3\} \qquad \textbf{(B)}\ \{3,4\} \qquad \textbf{(C)}\ \{4,5,6,7,8\} \qquad \textbf{(D)}\ \{5,6,7,8,9\} \qquad \textbf{(E)}\ \{4,5,6,7,8,9\}$

bleh that was confusing to understand at first:
[hide]
probability 2 is chosen = log 3/2 = 1/2 probability of something else.

probability of something else = 2log3/2 = log 9/4.

checking the answers, we see that C works[/hide]

Let $ P(d) \equal{} log \frac {d \plus{} 1}{d}$ be the probability that digit $ d$ is chosen and $ d_1, d_2, ..., d_k$ be the digits in the set .Then $ P(the \; set) \equal{} 2log \frac {3}{2} \equal{} log \frac {9}{4}$, $ 10^{P(the \; set)} \equal{} \frac {d_1 \plus{} 1}{d_1} \times \frac {d_2 \plus{} 1}{d_2} \times ... \times \frac {d_k \plus{} 1}{d_k}\equal{}\frac{9}{4}$ .As  the sets in the choices contains consecutive numbers , $ d_1 \equal{} 4$ and $ d_k \equal{} 8$ . So the answer is $ C$


integration
real analysis
real analysis unsolved

Find the volume enclosed inside of the surface $ x^2\plus{}y^2\equal{}2x,$ and between surfaces $ z\equal{}\minus{}x^2\minus{}y^2$ and $ z\equal{}9\minus{}x^2\minus{}y^2.$

Let $ D \equal{} \{(x,y) \in \mathbb{R}^2 \colon (x \minus{} 1)^2 \plus{} y^2 \leq 1\}$, that is, the intersection between the plane $ z \equal{} 0$ and locus of the points in $ \mathbb{R}^3$ $ x^2 \plus{} y^2 \plus{} 2x \leq 0$. Now the volume you have to calculate is $ \int_D \int_{ \minus{} x^2 \minus{} y^2}^{9 \minus{} x^2 \minus{} y^2}dz dx dy \equal{} 9\int_D dx dy \equal{} 9\pi$. You can also easily draw those surfaces, cause $ (x \minus{} 1)^2 \plus{} y^2 \equal{} 1$ is a cylinder, and the other surfaces are two "parallel" paraboloids whose "distance" is 9 so the volume you have to calculate is the same of the one of a cylinder with height 9 and base $ \pi$.


symmetry
geometry
geometric transformation
reflection
trapezoid
conics
ellipse

Inside a circle of center $ O$ and radius $ r$, take two points $ A$ and $ B$ symmetrical about $ O$. We consider a variable point $ P$ on the circle and draw the chord $ \overline{PP'}\perp \overline{AP}$. Let $ C$ is the symmetric of $ B$ about $ \overline{PP'}$  ($ \overline{PP}'$ is the axis of symmetry) . Find the locus of point $ Q \equal{} \overline{PP'}\cap\overline{AC}$ when we change $ P$ in the circle.

I do believe you meant that $ C$ is the reflection of $ B$ across $ PP'.$ Let $ D$ be the reflection of $ A$ across $ PP'.$ It's easy to see that $ ABCD$ is an isosceles trapezoid with symmetry axis $ PP'$ $ \Longrightarrow$ $ Q \equiv DB \cap AC \cap PP'$  and  $ AC \equal{} BD \equal{}2PO \equal{} 2r.$ Since $ \triangle QCB$ is isosceles with apex $ Q,$ we get $ QB \equal{} QC$ $ \Longrightarrow$ $ QA \plus{} QB \equal{} QA \plus{} QC \equal{} AC \equal{} 2r \equal{} \text{const}.$ Locus of $ Q$ is an ellipse $ \mathcal{E}$ with foci $ A,B$ and pedal circle  $ (O,r)$ and the straight line $ PP'$ is always tangent to $ \mathcal{E}.$

Sorry, I edited the proposition.
Many thanks.


describe all positive integers which can be written as

a) $ x^2\plus{}xy\plus{}y^2$

b) $ x^2\plus{}xy\plus{}6y^2$

where $ x,y\in\mathbb{N}$

You are looking for the norms of the numbers of type $ a\plus{}\frac{1\plus{}\sqrt{d}}{2} b$ where a) $ d\equal{}\minus{}3$, b) $ d\equal{}\minus{}23$. So you should work in the corresponding rings of algebraic integers.
Alternatively, there is very much known about number of type $ x^2\plus{}ny^2$.


Three fair, standard six-faced dice of different colors are rolled.
In how many ways can the dice be rolled such that the sum of
the numbers rolled is 10?

List the partitions of $ 10$ that have three numbers, none of which are greater than $ 6$:

$ \{6,3,1\}$ -> 6 ways

$ \{6,2,2\}$ -> 3 ways

$ \{5,4,1\}$ -> 6 ways

$ \{5,3,2\}$ -> 6 ways

$ \{4,4,2\}$ -> 3 ways

$ \{4,3,3\}$ -> 3 ways

This gives a total of $ \boxed{27}$ ways.

[hide="Solution to this question"]This question is basically asking you to find all the combinations for $(x,y,z)$ such that $x+y+z=10$ and $1\le x,y,z \le 6$.

We can solve this with the divider method. First let's account for the fact that each dice rolls at least a $1$. Set each dice to an initial value of $1$. So the sum of the dice is currently $3$. For the sum to equal $10$, the three dice rolls must increase by a total value of $7$.

So what we have to do is break $7$ into three chunks and distribute them to each of the dice, and we can do that with the divider method. Solving, we can do that in $\binom{9}{2} = 36$ ways.

Now we have counted all dice combinations that add up to $10$ where $0$ is not rolled, and of these, we must subtract the scenarios in which a number greater than $6$ is rolled.

The two possible cases are:
$8-1-1$
$7-2-1$

The first case can be ordered in $3!/2!$ different ways and the second case can be ordered in $3!$ different ways. $3!/2!=3$, and $3! = 6$, and we subtract $6+3=9$ from $36$ to get $\boxed{27}$, our answer. [/hide]

Use generating functions:
(x+x^2+x^3+...+x^6)^3, find the coefficient of x^10


Let $ABC$ be a triangle with side lengths $a,b,c$, such that $a$ is the longest side. Prove that $\angle BAC = 90^\circ$ if and only if
\[ (\sqrt { a+b } + \sqrt { a-b} )(\sqrt {a+c } + \sqrt { a-c } ) = (a+b+c) \sqrt 2. \]

[quote="spider_boy"]THIS CAN BE EASILY TREATED WITH INEQUALITIES!!!
THERE ARE TWO CASES.
   A 2 <B 2 +C 2
  A 2 >B 2 +C 2
          SQUARING BOTH SIDES THEN APPLYING THE INEQUALITIES INTHE CASES
GIVES THE ANSWER!!!

[quote=huashiliao2020]Nice @above! I did the same, any hints for the converse?[/quote]

I doubt they are actually active lmaooo


You know in the part of the forum index, it shows whose online??? What does hidden mean, and can you become hidden?????

This and much more is all in the FAQ.
Hidden just means other people can't see that you are online. Theres an option in your profile.

OH, and THANKS!!!! I can't wait until I try to be hidden...


For all positive integers $ n$ less than $ 2002$, let
\[ a_n \equal{} \begin{cases} 11 & \text{if }n\text{ is divisible by }13\text{ and }14 \\
13 & \text{if }n\text{ is divisible by }11\text{ and }14 \\
14 & \text{if }n\text{ is divisible by }11\text{ and }13 \\
0 & \text{otherwise} \end{cases}
\]Calculate $ \sum_{n \equal{} 1}^{2001} a_n$.

$ \textbf{(A)}\ 448 \qquad \textbf{(B)}\ 486 \qquad \textbf{(C)}\ 1560 \qquad \textbf{(D)}\ 2001 \qquad \textbf{(E)}\ 2002$

Are the conditions to be taken to mean "$ n$ is divisible by $ 13$ and $ 14$, but not by $ 11$?"

2002=11*13*14^

well anyways, this means that you cannot have a condition with all 3

this makes it even easier to notice that there are 10 numbers divisible by 13 and 14, 12 numbers divisible by 11 and 14, and 13 numbers divisible by 11 and 13 (because 11*13*14=2002, if you just use two of those divisors, you will notice that for less than 2002, there will be 1 less than the other divisor, so you want number of ones divisible by 11*13, its just 14-1 of them less than 2002).

So its just 10*11+12*13+13*14=448


geometry
rhombus
trigonometry
algebra
system of equations
power of a point
trig identities

ABCD is a rhombus. The circumradii of ABD, ACD are 12.5, 25. Find the area of the rhombus. (http://www.kalva.demon.co.uk/aime/aime03b.html)

[hide]

using the formula $R = \frac{abc}{4A}$ where $A$ is the area of the triangle, we begin by labeling the point of intersection as $E$, and letting $AE = CE = a$ and $BE = DE = b$

hence, since $AE$ is the altitude to $BD$, the area of triangle $ABD$ is $\frac{1}{2} \cdot a \cdot 2b$ and similarly for triangle $ACD$

so putting this all into our formula, we get the two equations:

$25 = \frac{(s)(s)(2a)}{(4)(1/2)(2a)(b)} = \frac{s^2}{2b}$

$12.5 = \frac{(s)(s)(2b)}{(4)(1/2)(2b)(a)} = \frac{s^2}{2a}$

dividing the two, we get:

$2 = \frac{a}{b} \implies 2b = a$

now, from pythagoreans, we get:

$s^2 = (2b)^2 + b^2 = 5b^2$

so substituting this into our first equation above, we have:

$25 = \frac{5b^2}{2b} \implies b =10$

hence $a = 20$, and the area of each of the four triangles is $100$, so the area of the rhombus is $\boxed{400}$

[/hide]

Using the formula $R=\frac{abc}{4A}$, we know that, after labeling the triangle in which we let $X$ be the midpoint of $AC$ and $BD$ and calling $AX$ and $BX$ as $a$ and $b$ respectively we need to find $2ab$ and we know that  $12.5=\frac{a^2+b^2}{2a}$ and that $25=\frac{a^2+b^2}{2b}$ we can get a system of equations and solve for $2ab$

We should get $2ab=\boxed{400}$

EDIT: Looks like gotztabeazen beat me to it :P

Does anyone know another way to solve this...or in other words, if we didn't know $R=\frac{abc}{4A}$?  ;)

[hide]if you draw out the circles that lends to a solution by power of a point theorem. such that if you label the shorter diagonal $2a$ and the longer diagonal $2b$ you have two equations:
\[ a^2 = (25- b)\cdot b \]
and
\[ b^2 = a\cdot(50-a) \]

so that $a^2 = 25b - b^2$ and $b^2 = 50a - a^2$ which means that $25b = 50a \implies b = 2a$
and when we substitute back in $5a^2 = 50a \implies a =10$ which means $b = 20$
thus, the total area is $2ab = 400$
[/hide]

ah yes...why didn't I think of power of point?? Good solution agolsme

another easy solution lies in using the extended law of sines

[quote="Adunakhor"]How do you derive that formula, $R=\frac{abc}{4A}$?[/quote]

Use extended law of cosines and $[ABC] = 1/2\sin{C}ab$

some info can be found here
http://mathworld.wolfram.com/Circumradius.html

[quote="agolsme"][quote="Adunakhor"]How do you derive that formula, $R=\frac{abc}{4A}$?[/quote]

Use extended law of [i]cosines[/i] and $[ABC] = 1/2\sin{C}ab$[/quote]
I believe you meant "extended law of sines"... ;)

Or...how about the proof?
$A=\frac{1}{2}ab\sin C=\frac{abc}{4R}\implies R=\frac{abc}{4A}$

[quote="chess64"][quote="boxedexe"]$A=\frac{abc}{4R}$[/quote]

Prove it.[/quote]
[hide]$2R=\frac{c}{\sin{C}}$, so $\text{Area}=\frac12ab\sin{C} =\frac12ab \frac{c}{2R} =\frac{abc}{4R}$
[/hide]


Let $ ABCD$ be a convex quadrtilateral such that $ AB$ is not parallel with $ CD$. Let $ \Gamma_1$ be a circle that passes through $ A$ and $ B$ and is tangent to $ CD$ at $ P$. Also, let $ \Gamma_2$ be a circle that passes through $ C$ and $ D$ and is tangent to $ AB$ at $ Q$. Let the circles $ \Gamma_1$ and $ \Gamma_2$ intersect at $ E$ and $ F$. Prove that $ EF$ passes through the midpoint of $ PQ$ iff $ BC \parallel AD$.

Previously discussed at http://www.mathlinks.ro/viewtopic.php?t=212500.


Έστω τρίγωνο ΑΒΓ και (ε) ευθεία του επιπέδου του.Είναι Α΄,Β΄, Γ΄οι προβολές των Α,Β,Γ αντίστοιχα στην (ε).Από το Α΄φέρμουμε κάθετη (χ) στην ΒΓ και κυκλικά (ψ) και (ω).
1) Δείξτε ότι οι ευθείες (χ),(ψ),(ω) διέρχονται δια του αυτού σημείου Ρ
2) Αν Κ,Λ,Μ τα μέσα των πλευρών ΒΓ ,ΓΑ ,ΑΒ και η ευθεία (ε) διέρχεται από το περίκεντρο Ο του ΑΒΓ τότε οι κύκλοι (Κ,ΚΒ΄). (Λ,ΛΓ΄) και (Μ,ΜΑ΄) διέρχονται δια του αυτού σημείου Σ
3) Ποιός ο γεωμετρικός τόπος του Σ αν ισχύει το ερώτημα 2?
(ΠΑΛΙΑ ΑΣΚΗΣΗ ΠΟΥ ΕΚΑΝΑ ΣΤΟ ΦΡΟΝΤΙΣΤΗΡΙΟ ΟΤΑΝ ΗΜΟΥΝ ΜΑΘΗΤΗΣ .ΔΥΣΤΥΧΩΣ ΔΕΝ ΕΧΩ ΟΛΟΚΛΗΡΗ ΤΗΝ ΛΎΣΗ ΤΟΥ (2) ΔΣΤΕ ΝΑ ΤΗΝ ΣΤΕΙΛΩ)

[quote="r_boris"](ΠΑΛΙΑ ΑΣΚΗΣΗ ΠΟΥ ΕΚΑΝΑ ΣΤΟ ΦΡΟΝΤΙΣΤΗΡΙΟ ΟΤΑΝ ΗΜΟΥΝ ΜΑΘΗΤΗΣ .ΔΥΣΤΥΧΩΣ ΔΕΝ ΕΧΩ ΟΛΟΚΛΗΡΗ ΤΗΝ ΛΎΣΗ ΤΟΥ (2) ΔΣΤΕ ΝΑ ΤΗΝ ΣΤΕΙΛΩ)[/quote][color=red]ΠΑΡΑΚΑΛΩ [/color]Αν έχει κάποιος χρόνο ας στείλει την λύση του 2 και θα στείλω την λύση των υπολοίπων

Στέλνω την λύση στο attachment

Η πλήρης Λύση είναι εδώ


ARML email calculus AMC AIME MATHCOUNTS trigonometry

anybody from mo or a bordering state, especially if you're trying out for mo arml should post here. we should make this forum more active! c'mon ppl

so anyways, im from ks

(maybe i shoulda wait until it's atleast closer to next year's competition, but ok, whatever...)

[b]S[/b]am Lite? Is he from [b]K[/b]S? Who is h[b]e[/b]? Is he good a[b]t[/b] math? How [b]c[/b]an I contact [b]h[/b]im or some other person who has been involved in ARML recentl[b]y[/b]?

:rotfl:   So funny!!!!

 :| I don't know...because you have his email, I think you could email him.

Blah.

Calculus question: In general, how do you find the volume of a 3-D object? :maybe:

You guess. What a stupid question... gosh.... :rotfl:

OH-EHM-GEE!!!!!!!!!! Thank you so much!!!! :D  :D  :D

Do you know him? If you don't that's a pretty sketchy comment.

actually yeah...

but to makes things sketchy again

SAM LITE WAS LIKE MY ROOMMATE FOR ARML *wink* *wink*

Did you guys have fun? (interpret "fun" whichever way you like) :wink:

Yeah, weren't 50% of this years nationals counting/probability?

Yeah, depending on how you define a "count/probability" problem, there were 15-19 of them. So 40-50%. :rotfl:

I hate MC. MC hates me. When I study geo, they load up on C&P. When I study C&P, we get a geometry test.

Who studies?

Nobody studies, people just "study".

If people liked math, they wouldn't "like" math.

It is all an "illusion".

Or illusion.

:ewpu: Im so CONFUSED(or is it "confused")

Now my confusion is worse! (not pokemon confusion xD)

Um no. I dont think you get it yet, ask your mom.

No, I think he gets it and "gets it" but doesn't [i]get it[/i]. :wink:


prove that every positive integer can write in this form:
[b]a,b,c,d,e[/b] is natural number and [b]a 2 +b 2 +c 2 +d 2 +e 2 [/b] divisible by [b]1+abcde[/b].

In what form? What do you mean?

And again. Please, read http://www.mathlinks.ro/viewtopic.php?p=140098#140098
Title "I know you don't know" is out of law ;)

[b]I am very sad excuse me... sorry......[/b]

Please, reformulate your problem.
I can't understand what actually you ask to prove.

It seems that he means every positive integer $N\ne 4^k(8t+7)$ can be written in the form :

\[N=\frac{a^2+b^2+c^2+d^2+e^2}{abcde+1}\]

This's a problem in Vietnam TST 1994 but i am sure that it appeared somewhere earlier!


How to proof, that
$tg \frac{3 \pi}{36} \cdot tg \frac{ 5 \pi}{36} \cdot tg \frac{ 7 \pi}{36}= tg \frac{ \pi}{36}$

$tg x=\tan x= \frac{ \sin x}{ \cos x}$ :)

[hide]$\tan15\cdot\tan25\cdot\tan35\cdot=\tan5$

$\implies \cos5\cdot\sin15\cdot\sin25\cdot\sin35=\sin5\cdot\cos15\cdot\cos25\cdot\cos35$

$\implies (\sin20 + \sin10)(\cos60-\cos10)=(\sin10 -\sin20)(\cos60+\cos10)$

after simplification; $2\sin20\cos60=2\sin10\cos10$ which is true. :D [/hide]

[quote="lingomaniac88"]Then why not just say $\tan \frac{3 \pi}{36} \cdot \tan \frac{ 5 \pi}{36} \cdot \tan \frac{ 7 \pi}{36}= \tan \frac{ \pi}{36}$?[/quote]
Some countries use $tg$ instead of $\tan$.  ;)

Thanks [b] sen [/b].
PS: in my country your nick means " a dream" :)


Prove that 27!/(3!16!4!4!) is an integer.

Then, generalize

[hide]that is the coefficient of a^3*b^16*c^4*d^4 in (a+b+c+d)^27. That's gonna be an integer.[/hide]

Hmm, that is correct

You can also think of it as the [hide]permutation of 27 objects, with 3 of the same kind, 16 of a different kind, etc...[/hide]

Now, can you generalize?

[quote="MysticTerminator"]sure...multinomial theorem...or permutations with repetitions....anything of the form A!/a_1!a_2!a_3!a_4!...a_n! is an integer if the sum of the little a's is A.[/quote]

Or actually less than A.. unless you usually write something out with lots of 1!s in it :)

[quote="mathfanatic"]I will give anyone a dollar if they can think of a real 27-letter word with exactly four different kinds of letters: 16 of one kind, 4 of another, 4 of a third, and 3 of a fourth.   :wink:

That expression you put up would be the number of distinct permutations.[/quote]
$1000? :)
Oh well, a dollar's good enough for me. I'm sure it's somewhere in [url=http://www.scrabble-assoc.com/boards/dictionary/10-15-20030401.txt]here[/url].

Not fare, tare.  You have to find it yourself...

Ok, so maybe we should change the wording to the permutation of a 27 string of letters instead of a 27 letter word.


[quote="AYMANE"]lets $ a$,$ b$ & $ c$ reals >0 such that $ a+b+c=3$ then
$ (a^{2}-ab+b^{2})(b^{2}-bc+c^{2})(c^{2}-ac+a^{2})\leq 12$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62544

[quote="AYMANE"]Lets reals $ x,y,z\geq 0$ such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ \left(y^{2} \minus{} yz \plus{} z^{2}\right)\left(z^{2} \minus{} zx \plus{} x^{2}\right)\left(x^{2} \minus{} xy \plus{} y^{2}\right)\leq 12$.[/quote]$ 4(x \plus{} y \plus{} z)^6 \minus{} 243\left(y^{2} \minus{} yz \plus{} z^{2}\right)\left(z^{2} \minus{} zx \plus{} x^{2}\right)\left(x^{2} \minus{} xy \plus{} y^{2}\right)$

$ \equiv F(x,y,z) \equal{} F(x,x \plus{} s,x \plus{} t)$

$ \equal{} 2673x^6\plus{}5346(s\plus{}t)x^5\plus{}243\left(17s^2\plus{}38st\plus{}17t^2\right)x^4$

$ \plus{}27(s\plus{}t)\left(62s^2\plus{}151st\plus{}62t^2\right)x^3\plus{}27\left(11s^4\plus{}71s^3t\plus{}84s^2t^2\plus{}71st^3\plus{}11t^4\right)x^2$

$ \plus{}9(s\plus{}t)\left(8s^4\plus{}5s^3t\plus{}48s^2t^2\plus{}5st^3\plus{}8t^4\right)x$

$ \plus{}(s\minus{}2t)^2(2s\minus{}t)^2\left(s^2\plus{}11st\plus{}t^2\right)\geq0,$

which is clearly true for $ x \equal{} \min\{x,y,z\}.$


trigonometry
function
trig identities
Law of Cosines

A man always drives at the same speed at all times. He makes it from A direct to C in 30 minutes; from A through B to C in 35 minutes; and from A through D to C in 40 minutes. Distance from point B to C is 10 miles, same distance from C to D. By the way angle C is right angled. Points A, B, C and D are 4 corners of a quadrilateral and labeled in a counterclockwise manner. How fast does he drive? (Popular science, September 1939)

my ugly solution went this way, I labeled triangle ABD with corresponding angles A, B, and D. Then i use the "law of cosines" for each angle in terms of the length of the diagonal AC. Then i brute force it, angle A + angle B + angle C = 180 degrees. What an ugly solution! Any comment? I'm not still sure if the answer is true.

I find this problem VERY difficult!

The September 1939 issue of the magazine Popular Science addressed this problem, 2 months later an answer was given by a READER, but the process was not shown. I know, Virgil Nicula and YETTi will play with this without difficulty. but for us, it might pose a great challenge. If only a synthetic solution can be obtained, much better to my delight. It seems that this problem is an adventitious angle problem, sort of.

after constructing the quadrilateral ABCD, my computation runs in circle . . . i cann't have a solution other than brute forcing it. Any TAKERS? May it be synthetic, metric, analytic or whatever method is acceptable . . .

[b]Whew! No takers?[/b] I used the triangle ABC as reference, then taking the line segment AC = y,

and line segment AB = x and line segment Bc = 10.

Since they have the same velocity directly from A to C and from A to C through B,

we have v = (x + 10)/35 = y/30; thus x = (7y-60)/6. Triangle ACD seems to me not needed, a flowery expression.

From triangle ABC, we have angle A + angle B + angle C = 180 degrees.

Use the Law of Cosines to determine angle A as a function of y, the distance of the diagonal AC.

Angles B and C follows the same method. Substituting it, we have . . . .

cos^(-1) ((x^2 +y^2-10^2)/(2xy)) + cos^(-1) ((10^2+x^2-y^2)/(2(10)y)) + cos^(-1)((10^2+y^2-x^2)/(2(10)y)) = 180 degrees.

where x = (7y-60)/6, simplifying

 cos^(-1) [(85y-840)/(12(7y-60))] + cos^(-1) [(13y^2-840y+7200)/(120(7y-60))] + cos^(-1) [(1/720)(840-13y)] = 180 degrees.

it is now unknown in y right? Where i made a mistake? My calculator returns an error message.

Maybe i have to convert the minutes to hours first . . . NO ONE cares to solve this problem in this Forum. Only sightseers

please help . . .

What an ugly expression, devoid of beauty. [b]Kunny and Virgil[/b], help!

Until now, i could not arrive at the claimed answer which is 38.843 miles/hour. [b]My previous conceived solution was a DUD, whew![/b] Any takers? It seems, that this problem remains a problem to me until now. No one dared venturing into this chaotic maze . . . .

angle A is an adventitious angle, right? Pretty hard, harder than the Langley problem


Sorry for my double posting, but no one answer me in the Intermediate sub.

Find real $ x$
$ 2001^{x}\plus{}2004^{x}\equal{}2002^{x}\plus{}2003^{x}$

0 and 1 are both real number solutions.  I'm not sure if there are any non-trivial solutions.

Rearrange the equation to get
\[ 2004^x\minus{}2003^x\equal{}2002^x\minus{}2001^x\]
Now by the mean value theorem on $ f(t)\equal{}t^x$, we can find $ p\in [2003,2004],q\in [2001,2002]$ such that
\[ xp^{x\minus{}1}\equal{}2004^x\minus{}2003^x\equal{}2002^x\minus{}2001^x\equal{}xq^{x\minus{}1}\]
In particular, $ p>q>0$ and $ x(p^{x\minus{}1}\minus{}q^{x\minus{}1})\equal{}0$, so the only possible solutions for $ x$ are 0 and 1, and it's easy to see they both work.

Can you explain your application of the Mean Value Theorem more?

For using the mean value theorem on $ 2004^x\minus{}2003^x$, we have
\[ 2004^x\minus{}2003^x\equal{}\frac{f(2004)\minus{}f(2003)}{2004\minus{}2003}\equal{}f^\prime (p)\equal{}xp^{x\minus{}1}\]
Note that $ x$ is fixed and the differentiation is with respect to $ t$.

Thanks. I do not know how I did not see the before.  :|


Given a,b,c are 3 possitive real numbers prove that
$\frac{a^3}{a^2+b^2+a^2b}+\frac{b^3}{b^2+c^2+abc}+\frac{c^3}{c^2+a^2+a^2c}\ge \frac{a+b+c}{a+2}$

Oh it's my stupid to post one problem twice.
Some one help me to delete one  :D

OH,Your ineq look not beautiful and quite hard.But I find the solution is quite nice.

$a+b+c-LHS \le a+b+c -RHS$
$\leftrightarrow \frac{ab(b+a^2)}{a^2+b^2+a^2b}+\frac{bc(c+ab)}{b^2+c^2+abc}+\frac{ac(a+ac)}{c^2+a^2+a^2c} \le \frac{(a+b+c)(a+1)}{a+2}$
Use Cauchy
$a^2+b^2 \ge 2ab$
$b^2+c^2 \ge 2bc$
$c^2+a^2 \ge 2ca$
And $LHS \le \frac{b+a^2+c+ab+a+ac}{a+2}=RHS$

Here is my solution
It has the same way with yours but it's more naturalize  :D
$\frac{a^3}{a^2+b^2+a^2b}+\frac{b^3}{b^2+c^2+abc}+\frac{c^3}{c^2+a^2+a^2c}\ge \frac{a^3}{a^2+b^2+a(\frac{a^2+b^2}{2})}+\frac{b^3}{b^2+c^2+a(\frac{b^2+c^2}{2})}+\frac{c^3}{c^2+a^2+a(\frac{c^2+a^2}{2})}$
Now we have to prove the ineq
$\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2} \ge (\frac{a+b+c}{2}$
which is simply and I have already post before :D
NOTE: This solution belong to my friend


inequalities
function
inequalities proposed

For all $ a,b,c$ be nonnegative real numbers, show that
\[ \frac{3(ab\plus{}bc\plus{}ca)}{2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)}\le\frac1{a^2\plus{}bc}\plus{}\frac1{b^2\plus{}ca}\plus{}\frac1{c^2\plus{}ab}\le\frac{3(a^2\plus{}b^2\plus{}c^2)}{2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)}\]

[quote="dduclam"]For all $ a,b,c$ be nonnegative real numbers, show that
\[ \frac {3(ab \plus{} bc \plus{} ca)}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\le\frac1{a^2 \plus{} bc} \plus{} \frac1{b^2 \plus{} ca} \plus{} \frac1{c^2 \plus{} ab}\le\frac {3(a^2 \plus{} b^2 \plus{} c^2)}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\]
[/quote]

assume $ a \equal{} max{(a,b,c)}$

$ \frac1{a^2 \plus{} bc} \plus{} \frac1{b^2 \plus{} ca} \plus{} \frac1{c^2 \plus{} ab}\le\frac {3(a^2 \plus{} b^2 \plus{} c^2)}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}$

is equivalent to

$ (3a^4bc \plus{} (b \plus{} c)(b^2 \plus{} c^2)a^3 \plus{} ( \minus{} b^4 \minus{} c^4 \plus{} 3b^3c \plus{} b^2c^2 \plus{} 3bc^3)a^2 \plus{} 2bc(b \plus{} c)(b^2 \minus{} bc \plus{} c^2)a \plus{} bc(b^2 \plus{} 3bc \plus{} c^2)(b \minus{} c)^2)(a \minus{} b)(a \minus{} c)$
$ \plus{} (4b^4ca \plus{} 6b^3c^2a \plus{} 6b^2c^3a \plus{} 4bc^4a \minus{} b^4c^2 \minus{} 2b^3c^3 \minus{} c^4b^2)(b \minus{} c)^2$

$ 3a^4bc \plus{} (b \plus{} c)(b^2 \plus{} c^2)a^3 \plus{} ( \minus{} b^4 \minus{} c^4 \plus{} 3b^3c \plus{} b^2c^2 \plus{} 3bc^3)a^2 \plus{} 2bc(b \plus{} c)(b^2 \minus{} bc \plus{} c^2)a \plus{} bc(b^2 \plus{} 3bc \plus{} c^2)(b \minus{} c)^2\geq (b \plus{} c)(b^2 \plus{} c^2)a^3 \plus{} ( \minus{} b^4 \minus{} c^4 )a^2\geq a^2((b^3 \plus{} c^3)a \minus{} (b^4 \plus{} c^4))\geq 0$
$ 4b^4ca \plus{} 6b^3c^2a \plus{} 6b^2c^3a \plus{} 4bc^4a \minus{} b^4c^2 \minus{} 2b^3c^3 \minus{} c^4b^2\geq 2b^3c^2a \plus{} 2b^2c^3a \minus{} b^4c^2 \minus{} 2b^3c^3 \minus{} c^4b^2\geq 0$

so..............

$ \frac {3(ab \plus{} bc \plus{} ca)}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\le\frac1{a^2 \plus{} bc} \plus{} \frac1{b^2 \plus{} ca} \plus{} \frac1{c^2 \plus{} ab}$

is equivalent to

$ ((2b^3 \minus{} b^2c \minus{} c^2b \plus{} 2c^3)a^3 \plus{} ( \minus{} b^2c^2 \plus{} b^4 \plus{} c^4)a^2 \plus{} ( \minus{} 2b^4c \minus{} 2c^4b \plus{} 3c^5 \plus{} 3b^5)a$
$ \minus{} c^5b \minus{} b^5c \minus{} b^4c^2 \minus{} c^4b^2 \plus{} b^3c^3 \plus{} 3c^6 \plus{} 3b^6)(a \minus{} b)(a \minus{} c) \plus{} (b \plus{} c)((3a \minus{} 3c)b^4$
$ \plus{} 2c(a \minus{} c)b^3 \plus{} 2c^2(2a \minus{} c)b^2 \plus{} c^3( \minus{} 3c \plus{} 2a)b \plus{} 3c^4a)(b \minus{} c)^2\geq 0$

$ 2b^3 \minus{} b^2c \minus{} c^2b \plus{} 2c^3\geq b^3\plus{}c^3\minus{}c^2b\minus{}b^2c\equal{}(b\plus{}c)(b\minus{}c)^2\geq 0$
$ \minus{} b^2c^2 \plus{} b^4 \plus{} c^4\geq 0$
$ \minus{} 2b^4c \minus{} 2c^4b \plus{} 3c^5 \plus{} 3b^5\geq \minus{} 2b^4c \minus{} 2c^4b \plus{} 2c^5 \plus{} 2b^5\equal{}2(c\plus{}b)(b^2\plus{}c^2)(\minus{}c\plus{}b)^2\geq 0$
$ \minus{} c^5b \minus{} b^5c \minus{} b^4c^2 \minus{} c^4b^2 \plus{} b^3c^3 \plus{} 3c^6 \plus{} 3b^6$
$ \geq (b^6\plus{}c^6\minus{}c^5b\minus{}b^5c)\plus{}(b^6\plus{}c^6\minus{}b^4c^2\minus{}c^4b^2)$
$ \equal{}(b^4\plus{}b^3c\plus{}b^2c^2\plus{}c^3b\plus{}c^4)(\minus{}c\plus{}b)^2\plus{}(b^2\plus{}c^2)(\minus{}c\plus{}b)^2(c\plus{}b)^2\geq 0$

$ (3a \minus{} 3c)b^4 \plus{} 2c(a \minus{} c)b^3 \plus{} 2c^2(2a \minus{} c)b^2 \plus{} c^3( \minus{} 3c \plus{} 2a)b \plus{} 3c^4a\geq 0$

so..........

[quote="dduclam"]For all $ a,b,c$ be nonnegative real numbers, show that
\[ \frac {3(ab + bc + ca)}{2(a^2b^2 + b^2c^2 + c^2a^2)}\le\frac1{a^2 + bc} + \frac1{b^2 + ca} + \frac1{c^2 + ab}\le\frac {3(a^2 + b^2 + c^2)}{2(a^2b^2 + b^2c^2 + c^2a^2)}\]
[/quote]
Let $ a+b+c=3u,$ $ ab+ac+bc=3v^2$ and $ abc=w^3.$
Hence, the left inequality is equivalent to $ f(w^3)\geq0,$ where $ f$ is decreasing function.
Thus, for the proof enough to check only one case: $ b=c=1,$ which gives $ (a-1)^2(2a^2-a+3)\geq0.$
The right inequality is equivalent to $ g(w^3)\geq0,$ where $ f$ is increasing function.
Thus, here enough to check two cases:
1) $ b=1$ and $ c=0$ ;
2) $ b=c=1.$

[quote="MeKnowsNothing"]I can't do the case $ a \equal{} b \equal{} c \equal{} 0$. Is there something wrong in the problem?[/quote]
It means $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0.$ :wink:

$ (\sum a^2b^2)( \sum \frac{1}{a^2\plus{}bc} ) \le \frac{3}{2}(\sum a^2)$
$ (\sum a^2)( \sum \frac{ a^2}{a^2\plus{}bc})\minus{}\sum a^2\plus{}\sum bc \le \frac{3}{2}(\sum a^2)$
$ \frac{1}{2}.( \sum a)^2 \le (\sum a^2)( \sum \frac{bc}{a^2\plus{}bc})$
We have :
$ \sum \frac{bc}{a^2\plus{}bc} \ge \frac{ (\sum ab)^2}{ \sum a^2bc\plus{}\sum b^2c^2} \ge \frac{ (\sum a)^2}{2(\sum a^2)}$

[quote="Vincent Gilbert"]$ (\sum a^2b^2)( \sum \frac {1}{a^2 \plus{} bc} ) \le \frac {3}{2}(\sum a^2)$
$ (\sum a^2)( \sum \frac { a^2}{a^2 \plus{} bc}) \minus{} \sum a^2 \plus{} \sum bc \le \frac {3}{2}(\sum a^2)$
$ \frac {1}{2}.( \sum a)^2 \le (\sum a^2)( \sum \frac {bc}{a^2 \plus{} bc})$
We have :
$ \sum \frac {bc}{a^2 \plus{} bc} \ge \frac { (\sum ab)^2}{\sum a^2bc \plus{} \sum b^2c^2} \ge \frac { (\sum a)^2}{2(\sum a^2)}$[/quote]

Nice, Vincent Gilbert, the first part of your solution is similar to me. After that, I use the lemma: $ \sum_{cyc}\frac {bc(b^2 \plus{} c^2)}{a^2 \plus{} bc}\ge a^2 \plus{} b^2 \plus{} c^2$  to completed the proof.

[quote="Rose_joker"]For elementary proof...
$ \sum_{cyc} \frac {1}{a^2 \plus{} bc}\geq \frac {3(ab \plus{} bc \plus{} ca)}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}$
from cauchy
$ \sum_{cyc} \frac {(b \plus{} c)^2}{(a^2 \plus{} bc)(b \plus{} c)^2}\geq \frac {4(\sum_{cyc} a)^2}{\sum_{sym} a^3b \plus{} 2\sum_{cyc} a^2bc \plus{} 4\sum_{cyc} a^2b^2}\geq \frac {3(ab \plus{} bc \plus{} ca)}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}$
which is equivalent to
$ 8\sum_{sym} a^4b^2 \plus{} 24a^2b^2c^2 \plus{} 16\sum_{cyc} a^3b^3 \plus{} 16\sum_{sym} a^3b^2c\geq 3\sum_{sym} a^4b^2 \plus{} 21\sum_{sym} a^3b^2c \plus{} 6\sum_{cyc} a^4bc \plus{} 18a^2b^2c^2 \plus{} 12\sum_{cyc} a^3b^3$
or
$ 5\sum_{sym} a^4b^2 \plus{} 6a^2b^2c^2 \plus{} 4\sum_{cyc} a^3b^3\geq 5\sum_{sym} a^3b^2c \plus{} 6\sum_{cyc} a^4bc$
which is obviously true from schur and muirhead... ($ \sum_{cyc} a^3b^3 \plus{} 3a^2b^2c^2\geq \sum_{sym} a^3b^2c$)[/quote]

You are right, Rose_joker. But we have a shorter proof only with Cauchy-Schwarz.

[quote="Materazzi"][quote="dduclam"]For all $ a,b,c$ be nonnegative real numbers, show that
\[ \frac {3(ab \plus{} bc \plus{} ca)}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\le\frac1{a^2 \plus{} bc} \plus{} \frac1{b^2 \plus{} ca} \plus{} \frac1{c^2 \plus{} ab}\le\frac {3(a^2 \plus{} b^2 \plus{} c^2)}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\]
[/quote]

I think the following inequality is true also .

Let $ a,b,c$ be nonnegative real numbers .

Prove that :
\[ \ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\]
:)[/quote]

You are right, Materazzi. Infact, we have the following inequality for all $ a,b,c$ be nonnegative relnumbers
\[ \sum_{cyc}\frac1{a^2 \plus{} bc}\ge\sum_{cyc}\frac1{a^2 \plus{} 2bc} \plus{} \frac {ab \plus{} bc \plus{} ca}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\]

Dear dduclam,

I proved $ \sum_{cyc}\frac {bc(b^2 + c^2)}{a^2 + bc}\ge a^2 + b^2 + c^2$

this way: By BCS $ LHS \geq \frac{(\sum ab\sqrt{a^2+b^2})^2}{\sum bc(a^2+bc)}$

so it suffices to prove:

$ (\sum bc\sqrt{b^2+c^2})^2 \geq (\sum a^2)(abc\sum a+\sum a^2b^2)$

Opening LHS and using $ \sqrt{a^2+b^2}{\sqrt{a^2+c^2} \geq a^2 +bc}$

we get shur of degree 3. Using this LEMMA to prove your inequality I obtained:

$ 2abc\sum \frac{bc(b^2+c^2)}{a^2+bc} +\sum a^2 \geq 2\sum ab$

that I cannot prove. :blush: Can you please help me to end your proof?

Thank you very much.

Can you also post your proof for second inequality by only Cauchy? I am very intereste in it. Thank you very much again for thes nice inequalities. :)

[quote="dduclam"][quote="Vincent Gilbert"]$ (\sum a^2b^2)( \sum \frac {1}{a^2 + bc} ) \le \frac {3}{2}(\sum a^2)$
$ (\sum a^2)( \sum \frac { a^2}{a^2 + bc}) - \sum a^2 + \sum bc \le \frac {3}{2}(\sum a^2)$
$ \frac {1}{2}.( \sum a)^2 \le (\sum a^2)( \sum \frac {bc}{a^2 + bc})$
We have :
$ \sum \frac {bc}{a^2 + bc} \ge \frac { (\sum ab)^2}{\sum a^2bc + \sum b^2c^2} \ge \frac { (\sum a)^2}{2(\sum a^2)}$[/quote]

Nice, Vincent Gilbert, the first part of your solution is similar to me. After that, I use the lemma: $ \sum_{cyc}\frac {bc(b^2 + c^2)}{a^2 + bc}\ge a^2 + b^2 + c^2$  to completed the proof.
[/quote]

[quote="manlio"]Dear dduclam,

I proved $ \sum_{cyc}\frac {bc(b^2 + c^2)}{a^2 + bc}\ge a^2 + b^2 + c^2$

this way: By BCS $ LHS \geq \frac {(\sum ab\sqrt {a^2 + b^2})^2}{\sum bc(a^2 + bc)}$

so it suffices to prove:

$ (\sum bc\sqrt {b^2 + c^2})^2 \geq (\sum a^2)(abc\sum a + \sum a^2b^2)$

Opening LHS and using $ \sqrt {a^2 + b^2}{\sqrt {a^2 + c^2} \geq a^2 + bc}$

we get shur of degree 3. Using this LEMMA to prove your inequality I obtained:

$ 2abc\sum \frac {bc(b^2 + c^2)}{a^2 + bc} + \sum a^2 \geq 2\sum ab$

that I cannot prove. :blush: Can you please help me to end your proof?

Thank you very much.
[/quote]

Very nice your proof, manlio  :lol:

Using this lemma to prove my Inequality as follow

We have
\[ (a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2})\left(\sum_{cyc}\frac {1}{a^{2} + bc}\right) = \sum_{cyc}\left(b^2 + c^2 - bc\frac {b^2 + c^2 - bc}{a^2 + bc}\right)\]
The Inequality is equivalent to
\[ 2(a^2 + b^2 + c^2) - \sum_{cyc}bc\frac {b^2 + c^2 - bc}{a^2 + bc}\le\frac3{2}(a^2 + b^2 + c^2)\]
or
\[ 2\sum_{cyc}bc\frac {b^2 + c^2 - bc}{a^2 + bc}\ge a^2 + b^2 + c^2\]
By AM-GM Inequality, we have $ 2(b^2 + c^2 - bc)\ge b^2 + c^2$, then using above lemma, we have done!

[quote="manlio"]
Can you also post your proof for second inequality by only Cauchy? I am very intereste in it. Thank you very much again for thes nice inequalities. :)
 [/quote]

The proof for second Inequality. We have
\[ (a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2})\left(\sum_{cyc}\frac {1}{a^{2} + bc}\right) = \sum_{cyc}\left(bc + \frac {a^2(b^2 + c^2 - bc)}{a^2 + bc}\right)\]
The Inequality is equivalent to
\[ 2\sum_{cyc}\frac {a^2(b^2 - bc + c^2)}{a^2 + bc}\ge ab + bc + ca\]
or
\[ 2\sum_{cyc}a^2\left(1+\frac {b^2 - bc + c^2}{a^2 + bc} \right)\ge 2(a^2 + b^2 + c^2)+ab + bc + ca\]
or
\[ \sum_{cyc}\frac {a^2}{a^2 + bc}\ge1 + \frac {ab + bc + ca}{2(a^2 + b^2 + c^2)}\]
By Cauchy-Schwarz Inequality, we get
\[ \sum_{cyc}\frac {a^2}{a^2 + bc}\ge\frac {(a + b + c)^2}{\sum_{cyc}a^2 + \sum_{cyc}bc} = 1 + \frac {ab + bc + ca}{\sum_{cyc}a^2 + \sum_{cyc}bc}\ge 1 + \frac {ab + bc + ca}{2(a^2 + b^2 + c^2)}\]
We completed the proof.

Very nice your proof, dduclam. :)

Thank you very much. :)

[quote="manlio"]Dear dduclam,

did you prove these very nice inequalities by SOS or have you a neat proof by Cauchy-Schwarz?

Thank you very much. :)
\[ \ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\]
:)
\[ \sum_{cyc}\frac1{a^2 \plus{} bc}\ge\sum_{cyc}\frac1{a^2 \plus{} 2bc} \plus{} \frac {ab \plus{} bc \plus{} ca}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\]
For the first inequality I found this solution on VIMF:
http://ddbdt.co.cc/forum/showthread.php?t=238

It is nice but it uses SOS method, so I would like to know if you have a solution by only BCS.

Thank you very much[/quote]

Hint for you, manlio:
\[ 2\sum_{cyc}\frac{a^2(b^2\minus{}bc\plus{}c^2)}{a^2\plus{}bc}\ge\sum_{cyc}\frac{a^2(2b^2\minus{}bc\plus{}2c^2)}{a^2\plus{}2bc}\ge ab\plus{}bc\plus{}ca\]

Dear dduclam,

thank you very much for your hint.

\[ 2\sum_{cyc}\frac {a^2(b^2 \minus{} bc \plus{} c^2)}{a^2 \plus{} bc}\ge\sum_{cyc}\frac {a^2(2b^2 \minus{} bc \plus{} 2c^2)}{a^2 \plus{} 2bc}\ge ab \plus{} bc \plus{} ca\]

$ 1)$     $ 2\sum_{cyc}\frac {a^2(b^2 \minus{} bc \plus{} c^2)}{a^2 \plus{} bc}\ge\sum_{cyc}\frac {a^2(2b^2 \minus{} bc \plus{} 2c^2)}{a^2 \plus{} 2bc}$

$ 2)$     $ \sum_{cyc}\frac {a^2(2b^2 \minus{} bc \plus{} 2c^2)}{a^2 \plus{} 2bc}\ge ab \plus{} bc \plus{} ca$

I tried the first inequality, but I failed :blush: For the second I wrote it as

$ 2\sum a^2 \sum \frac{a^2}{a^2\plus{}2bc} \plus{}3abc \sum\frac{a}{a^2\plus{}2bc} \geq \sum ab\plus{}2\sum a^2$

But using $ \sum \frac{a^2}{a^2\plus{}2bc} \geq 1$ it is too much weak to prove it. :blush:

Can you please help me?  :oops:

I don't understand how your hint can be used to prove your  inequalities?

Thank you very much. :)

[quote="arqady"]
Hence, the left inequality is equivalent to $ f(w^3)\geq0,$ where $ f$ is decreasing function.
The right inequality is equivalent to $ g(w^3)\geq0,$ where $ f$ is increasing function.
[/quote]

Can you post the whole form of $ f$ and $ g$ , or the  only thing you  need to know is that it is a function of $ w$ ,whatever is it .

[quote="enndb0x"][quote="arqady"]
Hence, the left inequality is equivalent to $ f(w^3)\geq0,$ where $ f$ is decreasing function.
The right inequality is equivalent to $ g(w^3)\geq0,$ where $ f$ is increasing function.
[/quote]

Can you post the whole form of $ f$ and $ g$ [/quote]
Certainly!
$ f(w^3)=8(u^2-v^2)w^6-(63u^3v^2-54uv^4)w^3+54u^2v^6-45v^8.$
$ g(w^3)=16(u^2-v^2)w^6+(81u^5-180u^3v^2+108uv^4)w^3+27u^2v^6-36v^8.$

[quote="manlio"]Dear dduclam,

thank you very much for your hint.
\[ 2\sum_{cyc}\frac {a^2(b^2 \minus{} bc \plus{} c^2)}{a^2 \plus{} bc}\ge\sum_{cyc}\frac {a^2(2b^2 \minus{} bc \plus{} 2c^2)}{a^2 \plus{} 2bc}\ge ab \plus{} bc \plus{} ca\]
$ 1)$     $ 2\sum_{cyc}\frac {a^2(b^2 \minus{} bc \plus{} c^2)}{a^2 \plus{} bc}\ge\sum_{cyc}\frac {a^2(2b^2 \minus{} bc \plus{} 2c^2)}{a^2 \plus{} 2bc}$

$ 2)$     $ \sum_{cyc}\frac {a^2(2b^2 \minus{} bc \plus{} 2c^2)}{a^2 \plus{} 2bc}\ge ab \plus{} bc \plus{} ca$

I tried the first inequality, but I failed :blush: For the second I wrote it as

$ 2\sum a^2 \sum \frac {a^2}{a^2 \plus{} 2bc} \plus{} 3abc \sum\frac {a}{a^2 \plus{} 2bc} \geq \sum ab \plus{} 2\sum a^2$

But using $ \sum \frac {a^2}{a^2 \plus{} 2bc} \geq 1$ it is too much weak to prove it. :blush:

Can you please help me?  :oops:

I don't understand how your hint can be used to prove your  inequalities?

Thank you very much. :)[/quote]

Dear Manlio, I will explain carefully about my post.
The first Inequality,
\[ \ \frac {1}{a^{2} \plus{} 2bc} \plus{} \frac {1}{b^{2} \plus{} 2ca} \plus{} \frac {1}{c^{2} \plus{} 2ab}\geq\frac {ab \plus{} bc \plus{} ca}{a^{2}b^{2} \plus{} b^{2}c^{2} \plus{} c^{2}a^{2}}\]
we use the equality  $ 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \equal{} bc(a^2 \plus{} 2bc) \plus{} a^2(2b^2 \plus{} 2c^2 \minus{} a^2)$ and rewrite the inequality as
\[ \sum_{cyc}\frac {a^2(2b^2 \minus{} bc \plus{} 2c^2)}{a^2 \plus{} 2bc}\ge ab \plus{} bc \plus{} ca\]
or
\[ \left(2\sum_{cyc}\frac {a^2(2b^2 \minus{} bc \plus{} 2c^2)}{a^2 \plus{} 2bc} \plus{} a^2\right)\ge (a \plus{} b \plus{} c)^2\]
or
\[ \sum_{cyc}\frac {a^2(a^2 \plus{} 4b^2 \plus{} 4c^2)}{a^2 \plus{} 2bc}\ge (a \plus{} b \plus{} c)^2\]
By Cauchy-Chwarz Inequality, we have
\[ \sum_{cyc}\frac {a^2(a^2 \plus{} 4b^2 \plus{} 4c^2)}{a^2 \plus{} 2bc}\ge \frac {\left(\sum_{cyc}a\sqrt {a^2 \plus{} 4b^2 \plus{} 4c^2}\right)^2}{(a \plus{} b \plus{} c)^2}\]
Therefore, it suffices to prove that
\[ \sum_{cyc}a\sqrt {a^2 \plus{} 4b^2 \plus{} 4c^2}\ge (a \plus{} b \plus{} c)^2\]
This Inequality was proven at [hide="here"]http://www.mathlinks.ro/viewtopic.php?t=313454[/hide].

The second Inequality
\[ \sum_{cyc}\frac {1}{a^{2} \plus{} bc}\ge\sum_{cyc}\frac {1}{a^{2} \plus{} 2bc} \plus{} \frac {ab \plus{} bc \plus{} ca}{2(a^{2}b^{2} \plus{} b^{2}c^{2} \plus{} c^{2}a^{2})}\]
by also the equality  $ 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \equal{} bc(a^2 \plus{} 2bc) \plus{} a^2(2b^2 \plus{} 2c^2 \minus{} a^2)$ and $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \equal{} bc(a^2 \plus{} bc) \plus{} a^2(b^2 \plus{} c^2 \minus{} bc)$, we rewrite the inequality as
\[ 2\sum_{cyc}\frac {a^2(b^2 \minus{} bc \plus{} c^2)}{a^2 \plus{} bc}\ge\sum_{cyc}\frac {a^2(2b^2 \minus{} bc \plus{} 2c^2)}{a^2 \plus{} 2bc}\]
or
\[ 2\sum\frac {b^2 \minus{} bc \plus{} c^2}{1 \plus{} x}\ge\sum\frac {2b^2 \minus{} bc \plus{} 2c^2}{1 \plus{} 2x}\]
where $ x \equal{} \frac {bc}{a^2}, y \equal{} \frac {ca}{b^2}, z \equal{} \frac {ab}{c^2}.$
After that, we use expansion to complete the proof, but this way is not meaningful (althought the first Inequality is also use this way).  :blush:

Thank you very much, dduclam. I will study your solution very carefully. :)

I tried to solve this

$ 2\sum\frac {b^2 \minus{} bc \plus{} c^2}{1 \plus{} x}\ge\sum\frac {2b^2 \minus{} bc \plus{} 2c^2}{1 \plus{} 2x}$
where $ x \equal{} \frac {bc}{a^2}, y \equal{} \frac {ca}{b^2}, z \equal{} \frac {ab}{c^2}.$

by full expansion, but it is terribly long. :(  Has anyone a trick to simplify it, please? Thank you very much.

[quote=Materazzi]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2} \quad (1)\][/quote]
Put $p=a+b+c,\,q=ab+bc+ca$ and $r=abc$ we have lemma
\[2p(2p^2-3q)r \leqslant q^2(p^2-q). \quad (2)\]
Write inequality $(1)$ as
\[\frac{\displaystyle \sum (a^2+2bc)(b^2+2ca)}{(a^2+2bc)(b^2+2ca)(c^2+2ab)} \geqslant \frac{ab+bc+ca}{a^2b^2+b^2c^2+c^2a^2},\]
equivalent to
\[\frac{q(2p^2-3q)}{27r^2+2p(2p^2-9q)r+2q^3} \geqslant \frac{q}{q^2-2pr},\]
or
\[q(2p^2-3q)(q^2-2pr) \geqslant q\left[27r^2+2p(2p^2-9q)r+2q^3\right],\]
\[\left[q^2(p^2-q)-2p(2p^2-3q)r\right]+\left[p^2q^2-4q^3+2p(9q-2p^2)r-27r^2\right] \geqslant 0.\]
Whicch is true because lemma $(2)$ and
\[p^2q^2-4q^3+2p(9q-2p^2)r-27r^2=(a-b)^2(b-c)^2(c-a)^2 \ge 0.\]
We are done.


vector
search
algebra
polynomial
linear algebra
matrix
induction

i need the following result:
"Let V be and n-dymensional vector space and T an linear operator in V. Let's supóse that there's and positive integer k that T^k=0. Demonstrate that T^n=0."

Thanks b4 anything for the attention! :lol:

Topics like this really should be in the College Playground Linear Algebra section. I suggest that a moderator move this topic.

This exact question has been posted there several times.

humm.. I haven't seen that this was Olympiad section... my bad...

I've been trying to search for this; so far [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=259998755&t=234130]this[/url], which is the problem in the case $ n\equal{}2,$ is what I've found. I think there's more out there.

Anyway, one possible outline goes like this: if $ T$ is nilpotent (that is, $ T^k\equal{}0$ for some $ k$), then its only eigenvalue is zero and it's minimal polynomial is $ T^j$ for some $ j.$ That means that the characteristic polynomial for $ T$ is $ x^n.$ By Cayley-Hamilton, $ T$ satisfies its characteristic polynomial, so $ T^n\equal{}0.$

For another approach: we prove that $ T$ is similar to an upper triangular matrix with zeros on the main diagonal. Once we have that, matrix multiplication will get us that $ T^n\equal{}0.$

We prove that by induction on $ n.$ Assume the theorem is true for $ n\minus{}1.$

We find a nonzero vector such that $ Tx\equal{}0.$ All we have to to is take some arbitrary $ y$ and look at $ y, Ty, T^2y, \dots.$ That sequence is eventually zero, so we take the last nonzero entry in that sequence as our $ x.$

Extend $ x$ to a basis; that is, construct an invertible $ P$ that has $ x$ as its first column. Then $ TP$ is a matrix with first column zero, as is also $ P^{\minus{}1}TP.$

Write $ P^{\minus{}1}TP$ as $ \begin{pmatrix}0&*\\0&T'\end{pmatrix}.$

Since $ T^k\equal{}0,$ we have $ (P^{\minus{}1}TP)^k\equal{}P^{\minus{}1}T^kP\equal{}0.$ But from the block-structured matrix, that implies that $ (T')^k\equal{}0.$ By the inductive hypothesis, $ T'$ is similar to an upper triangular matrix with zeros on the diagonal, and we are done.


analytic geometry
geometry
3D geometry
sphere

So say I told you that:

$ \Psi_{4f_{xyz}} \equal{} \frac {1}{96\sqrt {35}} \times \frac {1}{8} \times e^{\frac { \minus{} r}{4}} \times \sqrt {\frac {1}{4 \pi}} \times \sqrt {\frac {105}{4}} xyz$

How would you deduce what it looks like?

P.S. This is for a hydrogen atom.

What would be the nodal planes of that orbital?

Okay, I am going to be honest here. I have no idea whatsoever. I have no idea how to even start a question like this. :blush:

By the way, what does "r" stands for in your orbital?

It should be easy to see that the planes x = 0, y = 0, z = 0 are nodal planes, and that gives a rough information about the orbital.

$ r$ comes in when you write $ \psi$ is spherical coordinates.

But you have xyz, which are not spherical coordinates.

Which is why I am confused. I mean we can write everything in spherical coordinates but I don't know if that had be easier.

So for this one there are nodal planes $ xy, yz, zx$ and then I picture a sphere and "slice" it using the nodal planes?

That should give a rough picture of the orbital. However, the precise shape can only be obtained by drawing it with a computer program.

Yep thanks. By the way here is a link to some pictures of atomic orbitals if anyone is interested:

http://www.uky.edu/~holler/html/orbitals__1.html


2005 Vestavia Hills math tournament, Geometry division

1.The Statement "A zoop is a goop if a zoop is a boop," is given. What is the inverse of the converse of the inverse of the contrapositive of the given statement?

A. If a zoop is not a boop, then a zoop is not a goop.
B. If a zoop is a boob, then a zoop is a goop.
C. If a zoop is not a goop, then a zoop is not a boop.
D. If a zoop is a goop, then a zoop is a boop.
E. None of the Above(Nota)

The two inverses cancel, so there is the convese. The answer is A

Patrick jkust because you know the answer doesn't mean yiou should answer it unless you know how.......... and in this case you don't but the answer is [hide]a[/hide]

After inverse and inverse cancels out, you take the converse of the contrapositive, which is essentially the inverse.

Therefore, p -> q, then the inverse is ~p -> ~q.

Keep in mind that the second part of the statement is actually the condition (p)


I'd like to ask you: I need to include two jpeg pictures, one below the other, let's say p1 and the p2 are the names of the files. They should be centered on the page and their size should be like 9 cm height and 10 cm width, each. How could I do this?

Thanks,
Jean

\usepackage{graphicx}

\begin{center}
\includegraphics[height=9cm,width=10cm]{p1.jpg}\\
\includegraphics[height=9cm,width=10cm]{p2.jpg}
\end{center}

It worked. Thanks and wish you the best!


[This is related to a problem posted by harazi. I hope you'll find it interesting as i did.]

If $a_n$ is the number of digits equal to 5 or greater in digits representation of n, then find the limit of:
$x_n = (1/10^k)\sum_{i=0}^{10^k - 1} a_i$.

[Also you may change the 5 digit to other digit and check what's happening with the above sum  :roll: ]

I hope I am wrong :P .Because If I am right,then I  find the answer is $+\infty$,not good.
because:\[ \sum_{i=0}^{10^k - 1} a_i =10^{k-1}\cdot 5k \]
am I right?

I've got $\frac{10}{9}$.
If we denote by $x_k=\sum_{i=10^{k-1}}^{10^k-1}a_i$, then $x_{k+1}-x_k=9\cdot 10^k$ and $x_k=x_1+9(10^k+10^{k-1}+...+10)=10^{k+1}-9$ and now we want to find the limit of $y_k=\frac{x_1+x_2+...+x_k}{10^k}$ when $k \to \infty$. Thus computing $y_k=\frac{\frac{10}{9}(10^k-1)-9k}{10^k} \to \frac{10}{9}$.
I've not checked it carefully- i'm sorry if i've made stupid mistakes.


A finite collection of squares has total area $4$. Show that they can be arranged to cover a square of side $1$.

This problem i sreally old, it appeared in Russia MO for Grade 9 in 1979.
Anyways, here is the solution:
Let the sides of the squares be equal to $a_{i}$ for $i=1,2,...,N$($N$ is the number of squares).
If some $a_{k}>1$ then $k$th square will cover the unit square. Now let's assume the case when $a_{i}<1$ for all $i$
Every number $a_{i}$ must belong to some interval $(2^{-k_{i}}, 2^{-k_{i}+1})$, ie $2^{-k_{i}}\le a_{i}< 2^{-k_{i}+1}$, where $k_{i}$ are  positive integers for all $i$. Now let's decrease every $i$th square to the square with side $b_{i}=\frac{1}{2^{k_{i}}}$. Then its area would decrease by at most $4$ times, because $1\le \frac{a_{i}}{b_{i}}< 2$. Therefore the area of all squares will be greater than $1$.
Now let's prove we can tile the unit square fully with the new squares. Let's divide the unit square into $4$ squares of side $\frac{1}{2}$. First place the squares with side $\frac{1}{2}$(if they exist). Then on the non-tiled squares with side $\frac{1}{2}$(if they exist) place the squares with side $\frac{1}{4}$(if they exist), with dividing each non-tiled square with side $\frac{1}{2}$ into $4$ equal squares. We will continue this procedure for $k=3,4,...$ by placing squares with side $\frac{1}{2^{k}}$ on the on-tiled squares with side $\frac{1}{2^{k-1}}$, in turn dividing them into $4$ equal squares.
Finally, because the sum of areas of squares is greater than $1$, then on some step, we will cover the square. Then increasing the $i$th square with side $b_{i}$ to the square with side $a_{i}$, we will get the tiling of the unit square with the given squares.
[size=59]Yay post numebr 600![/size]

are moscow oympiads the same as russian olympiads

No. Moscow olympiad is the olympiad at city level(maybe region level, I am not sure), while All-Russian olympiads are national level, one level higher than moscow olympiads.


induction
number theory
relatively prime
prime numbers

Let $ x,\ y$ be relatively prime numbers such that $ xy\neq 1$. For positive even integer $ n$, prove that $ x \plus{} y$ isn't a divisor of $ x^n \plus{} y^n$.

[quote="Mewto55555"]If x+y is a factor, then its (x+y)(stuff) so if x+y=0, then x^n+y^n=0

However, if x=-y then its equal to 2x^n which is not 0.[/quote]

Sorry, if x=-y , then x,y are not relatively prime, contradicts with the given condition. :P

Consider

$ x^n \plus{} y^n \equal{} (x \plus{} y)(x^{n \minus{} 1} \plus{} y^{n \minus{} 1}) \minus{} xy(x^{n \minus{} 2} \plus{} y^{n \minus{} 2})$

$ x \plus{} y$ is a divisor of $ x^n \plus{} y^n \iff x \plus{} y$ is a divisor of $ x^{n \minus{} 2} \plus{} y^{n \minus{} 2}$

But,
$ x \plus{} y$ is a not divisor of $ x^2 \plus{} y^2$

By induction, the proposition is true when $ n$ is even.

Consider $ x\equiv \minus{}y\: mod\: (x\plus{}y)$

So, for even $ n, x^n \equiv y^n\: \mod(x\plus{}y)$

Suppose $ x^n\equiv \minus{}y^n\: mod\: (x\plus{}y)$

So, $ 2y^n \equiv 0\: mod\: (x\plus{}y)$

This is not always true.

gouthamphilomath: Your proof is incorrect. You have to prove that $ x^n \equiv \minus{}y^n$ modulo $ x\plus{}y$ can never be true. There is a difference between "not always" and never; "not always" is equivalent to sometimes, which is not the desired result.

[quote="darkdieuguerre"]gouthamphilomath: Your proof is incorrect. You have to prove that $ x^n \equiv \minus{} y^n$ modulo $ x \plus{} y$ can never be true. There is a difference between "not always" and never; "not always" is equivalent to sometimes, which is not the desired result.[/quote]
Yeah, you are right. I will complete my proof.

Suppose $ x\plus{}y|\left(2y^n\right)$

Let $ 2y^n\equal{}(x\plus{}y)y_1$ and so, $ y|xy_1$ but $ (x,y)\equal{}1 \Longrightarrow y|y_1$ and so, let $ y_1\equal{}yy_2$

So, $ 2y^{n\minus{}1}\equal{}(x\plus{}y)y_2$

Similarly arguing and defining $ y_3, y_4, \cdots y_n,$ we get

$ 2y\equal{}(x\plus{}y)y_n$ and now let $ y_n\equal{}yk$

So, $ 2\equal{}(x\plus{}y)k$ which is false as $ (x\plus{}y)>2$

It is well known that the followings hold:

[b]Lemma 1[/b]: if $a\in \mathbb{N}^{*}$ is odd then  $x^a +y^a=(x+y)\cdot A(x,y)$ with $A(x,y)\in \mathbb{Z}[x,y]$;
[b]Lemma 2[/b]: if $a\in \mathbb{N}^{*}$ is even then $x^a -y^a=(x+y)\cdot A(x,y)$ with $A(x,y)\in \mathbb{Z}[x,y]$.

Now we will prove the following claim, which suffices to show the statement of the problem.

[b]Claim[/b]: $x+y\, \vert \,x^n +y^n$ for even $n$ if and only if $x+y\,\vert \,2x^{n/2}y^{n/2}$.

[b]Proof[/b]: if $n=2k$ with $k$ odd then $$x^{2k}+2x^{k}y^{k}+y^{2k}=(x^{k}+y^{k})^2=(x+y)^2\cdot A(x,y)^2$$ due to Lemma 1, while if $n=2k$ with $k$ even then $$x^{2k}-2x^k y^k +y^{2k}=(x^k-y^k)^2=(x+y)^2 A(x,y)^2$$ due to Lemma 2. So the claim is proved.

But $\gcd(x+y,x)=1$ and $\gcd(x+y,y)=1$ implies  that $x+y\,\vert \,2$ due to Generalized Euclid Lemma, which cannot be true for coprime positive integers $x,y$ such that $xy\neq 1$.