geometry
3D geometry
sphere
incenter
quadratics
pyramid
modular arithmetic

Recently, I've noticed an increase in number of people questioning, "How can I study for AMC?" or "My score is ### and I can't seem to improve," and so on. So, I'm bringing back the old idea that I used sometime ago.  Hopefully, this can help others and I know it will help me as well (I don't get to practice math as I used to nowadays).

Everyday (this is tentative), I'm going to post two problems that I wrote on specific category.  These problems will have ideas taken from actual AMC questions.  They will be [b]very[/b] similar to the original one, and I'll cite the original ones.  But, problems are purely MINE (not copied off from some internet site) so please respect them and if there's an error, please correct or let me know.  :)

Another thing... To make this more fun, you'll be able to gain points on answering the questions.  I'm going to take FIRST TWO CORRECT responses.  Each question will worth 1 point so maximum you can get on each day is 2 points. Here are brief rules (on hidden text):

[hide]Question Answering and Score Rule

There are just some rules to follow.

1. When posting your answer, answer itself does NOT count.  You must have your own solution typed in your own words so that I can see that you have solved the problem on your own. If you take previous poster's answer and change it, I may or may not know it.  But it is disgrace to yourself.

2. You cannot win everyday. If you win on one day, you cannot win the next day but the day after. It's only possible to win every other day.  This is to even out the scores and give other people some chance.  So, if you have gotten the question right, hold your answers until at least two people with correct answers post. Then post or elaborate or so on.

3. I'm not native speaker so don't argue with me about English rule in these problems. You're here to do math, not English.  :D

4. When you have questions to past problems, contact me FIRST.  I'll probably make a new thread dedicated just for that problem so new problems don't get mixed up with old ones.

5. There is no prize folks.  Just the fact that you might be one step closer to 150. Who knows.  :roll:

6. I'll try to do this everyday. I will really try but I have millions of things to do now. I don't usually get home till 7 so if I'm not posting problems, just wait until I post new one. Don't send millions of PMs about this.

[/hide]

PLEASE READ THOSE RULES!

First Set of Problems

1. (Modified from AMC 12A 2004 #22)  Three mutually tangent unit spheres (radius 1) are placed on the surface.  A soccer ball of unknown radius is placed on top of them.  The distance from the surface to top of a soccer ball is in form $ \displaystyle 8 \plus{} \frac{2 \sqrt{m}}{3}$ where $ m$ is not divisible by any perfect square greater than 1.  What is the remainder when $ m$ is divided by 23?
A) 3
B) 11
C) 17
D) 19
E) 22

2. (Modified from AMC 12B 2004 #19) Roskolnikov makes a slice in a cone-shaped lumber so that it is now a truncated cone with two different radii.  Inside the lumber, he positions a sphere so that it touches top, bottom, and lateral surfaces of it.  If one of the radii is 2008, what is the minimum length of other radii so that the sphere has an integer radius?
A) 411
B) 502
C) 636
D) 675
E) 1004

*Just for fun, where is this character from?*
[hide][i]Crime and Punishment[/i][/hide]

Still working on solution 1, but here's #2.

[hide="Solution 2"]
Taking a radial cross-section of the cone before the cut, it is obvious that the center of the sphere lies at the incenter of this radial cross-section.  Drawing in the angle bisectors and the radii of the sphere to the tangential points in this cross-section, we notice that we have determined several right triangles. Those that are of interest to us:

Triangle 1:
Side a = r
Side b = r_1
Angle b = x/2

Triangle 2:
Side a = r
Side b = 2008
Angle a = x/2

Now, since r/2008 must equal r_1/r, we end with the quadratic r^2 = 2008*r_1, which leads us to the obvious conclusion that r_1=502 at minimum. (2008=2^3*251, for a perf. square, we need another 2 and another 251.)

Answer: B, 502.[/hide]

[hide="1?"]Connect the centers of the balls to form a pyramid. Let $ x\equal{}1\plus{}r$ where $ r$ is the radius of the soccerball. The pyramid has a 2 x 2 x 2 base and a vertex that has a distance of x from the other three points. Then $ h^2\plus{}\frac{4}{3}\equal{}(x)^2$

The total distance is $ x\plus{}h\equal{}8\plus{}\frac{2}{3}\cdot \sqrt{m}$

So okay...lets analyze $ \sqrt{h^2\plus{}\frac{4}{3}}\plus{}h$...taking h=0, it is root 4/3...taking h arbtrarily large, that expression is arbitrarily large. Hence for any prime p=m, we have a solution for h. But these all satisfy the problem (to be extra ridiculous, the sequence 23n+b where n is an integer has infinitely many primes by dirchlet's theorem so there cannot be a unique answer...)

If I missed something, my bad. But I read the problem like ten times.[/hide]

@Altheman: Try to use number 8. It's actually a lot simpler than what you're thinking (honestly, I think you've gotten the hard part).

The question for #1 needs to be slightly clarified as Altheman noted (its sort of confusing whether or not you want $ m$ to always have a certain residue $ \pmod{23}$ or for a certain specific case).

[hide="1"]
[img]http://www.artofproblemsolving.com/Wiki/images/7/76/2004_AMC12A-22a.png[/img]

(Ignore those 2s and pretend they are $ r$s, because the original problem statement had 2 instead of unknown. Its from the wiki [I made it])

The heights of the bottom parts (eg below the center) of the bottom spheres and the top part of the top sphere is $ 1 \plus{} r$. The height of the centers of the bottom sphere to that of the top sphere can be calculated by drawing a triangular pyramid between those four points. The distance from any of the three bottom centers to the center of the equilateral triangle face of the pyramid is $ \frac {2}{\sqrt {3}}$ using $ 30 \minus{} 60 \minus{} 90 \triangle$s. The other edges are of length $ r \plus{} 1$. Thus by Pythagorean thm $ h \equal{} \sqrt {(r \plus{} 1)^2 \minus{} 4/3}$, and the total height is $ 1 \plus{} r \plus{} \sqrt {(r \plus{} 1)^2 \minus{} 4/3}$.

Here comes the ambiguous part, we assume $ r \equal{} 7$ just to make our answer nice. Then the height becomes $ 8 \plus{} \sqrt {64 \minus{} 4/3} \equal{} 8 \plus{} \frac {2\sqrt {141}}{3}$, and the residue upon division by $ 23$ is $ 3\ \mathrm{(A)}$.
[/hide]

Edit: h_s_potter2002, I noticed that and fixed it before I saw your post. Thanks anyway!

[hide="1"]Draw lines from the center of the soccer ball to each of the three unit spheres and connect the centers of the three spheres. Thus, we have a tetrahedron with an equilateral triangle base of side length 2. Let the radius of the soccer ball be $ r$ and the height of the tetrahedron be $ h$. The distance to the top of the soccer ball is just $ 1 \plus{} r \plus{} h$.

Now, we find that $ h^2 \plus{} \left(\frac {2\sqrt {3}}{3}\right)^2 \equal{} (r \plus{} 1)^2$. Hence, $ h \equal{} \sqrt {r^2 \plus{} 2r \minus{} \frac {1}{3}}$. The distance to the top of the soccer ball is just $ 1 \plus{} r \plus{} \sqrt {r^2 \plus{} 2r \minus{} \frac {1}{3}}$. Since there is an 8, let $ r \equal{} 7$. Then, we have $ 8 \plus{} \sqrt {63 \minus{} \frac {1}{3}} \equal{} 8 \plus{} \sqrt {\frac {564}{9}} \equal{} 8 \plus{} \frac {2\sqrt {141}}{3}$. Thus, the residue is 3, or A.[/hide]

edit: too late  :roll:

My bad on #1. I didn't really consider the problem in full case. Thanks for pointing out this.

Anyway:

[hide="Correct Answer"]
1. A
The expression was indeed $ \displaystyle 8 \plus{} \frac{2 \sqrt{141}}{3}$.

2. B
I solved it using trapezoid and circle though. It's probably similar logic as other person solved.
[/hide]

Second Set of Problems (Just one problem; Lack of time)

3. (Inspired by AMC 12B 2002 #23) Quadrilateral DACB is drawn such that AD = AC, and BC = 20. A line with length of 20 was drawn from A so that it touches BC at its midpoint, E.  If ABD is a right triangle, then find BD.
A) 28
B) $ 21 \sqrt{2}$
C) 30
D) $ 10 \sqrt{10}$
E) $ 15 \sqrt{7}$

[hide="Solution #3"]
By the Law of Cosines on $ \triangle AEC, AEB$, we have
\begin{eqnarray*} AC^2 & = & 20^2 + 10^2 - 2\cdot 20 \cdot 10 \cos \angle AEC \\
AB^2 & = & 20^2 + 10^2 - 2\cdot 20 \cdot 10 \cos (180 - \angle AEC) \end{eqnarray*}
Summing these give $ AB^2 + AC^2 = AB^2 + AD^2 = 1000$ (alternatively, use Stewarts). In triangle $ DAB$, Pythagorean Thm gives $ AB^2 + AD^2 = BD^2$, so $ BD = \sqrt {1000} = 10\sqrt {10}\ \mathrm{(D)}$.
[/hide]

[hide="Edit history"]*edits* I just read the rules ... sorry. (I'm not so much interested in winning than in practicing though)

Request for clarification: Which angle in $ \triangle ABD$ is right? I don't think it was specified in the problem statement, but I believe its necessary for the answer. Below is how I drew my diagram.

Edit again: Thanks

Edit again: Nobody's posting a full solution, so I did.
[/hide]

Yes, I remembered early this morning that I didn't specify that BD was the hypotenuse.

So yes, your left diagram is correct.

is it just me, or does 1 look like an old AIME problem... one with balls aligned so that their centers form octagon, etc... if so, would a similar approach work?

I think silverfalcon mentioned in the rules to post questions regarding old problems on a separate thread, but one hasn't been created yet so I guess its safe to answer here for now.

I'm guessing the AIME problem in question is [url=http://www.artofproblemsolving.com/Wiki/index.php/1998_AIME_Problems/Problem_10]1998 #10[/url]; and the answer to your question is to a degree, yes (since both solutions involve connecting the radii and then Pythagorean).

[hide="Correct Answer"]
3. D
Nice job, azjps, especially with neat drawing. [/hide]

I'll keep total scores once we hit like 10 problems. I don't want to do it every set.

Again, lack of time (I have a big test tomorrow and Tuesday) so here's one.

4. (My own) Orthodiagonal trapezoid PQRS (PQ and RS are bases) has median MN, and it is given that ON = 19 where O is the midpoint of MN. If RS = 59, then the sum of the squares of diagonals is $ \displaystyle S_d$. What is the remainder when $ \displaystyle S_d$ is divided by 2008?
A) 1760
B) 1761
C) 1762
D) 1763
E) 1764

what i have so far: we can find PQ because of what we know about SR and ON (and thus MN). i was wondering if you could use ptolemy's theorem.
here's my general question: are orthodiagonal trapezoids cyclic?

[quote="ghjk"]You meant the solution of "The Zuton Force" is wrong, Teneroez?? :o . Can you show me your solution?? Maybe you also get it wrong  :lol:.Anyway, I think solution of Zuton Force is right! Brute force won't work in this case! :wink:[/quote]

No I also agree that his solution is wrong.  You cannot set one plate because some plates have multiples (more than just 1).

Like firecricket91 said, try the case with 2 blue plates and 2 red plates.

According to Zuton Force, he did it with the idea of $ \frac {(4 - 1)!}{2!2!}$ (generically of $ \frac {(n - 1)!}{k!l!m!...}$ where $ k + l + m + .... = n$, and each is just how many times each different item appears which is $ \frac {6}{4}$ in this case, which isnt even an integer so clearly his idea is wrong.
He just did the normal $ (n - 1)!$ in a circle then put it over how many of each appears to account for overcounts.
${ (n - 1)}!$ only works for if all things are different, but not in this case.  It is very hard to account for overcounts using this case (since objects can be the same).

Thats why in my earlier post, I said this problem is a lot harder than AMC--this can be like 10+ on AIME.

Heres an old topic about the exact same type of question.  You will see the only way to do this [i]is[/i] brute force.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1662713791&t=57731
but even that "simplification" does not work when there are like 4 different colored ones

Wait, but the original problem said that there was 1 white plate, so isn't Zuton Force correct?

Okay, try it with 1 white plate 2 red plates and 2 blue plates.

List them out.  you wont get the same answer.

Why does Zuton only set the 1 white plate?

WHy cant he set one of the two red plates.

Setting one of the two red plates would give a completely differnt answer than setting one of the 1 white plates.

That shouldnt be the case because both answers should be the same.

Uh, can you list them out for me?

I have W B B R R, W R B B R, W R R B B, W B R R B, W B R B R , W R B R B.

isn't that it?

Isn't WBRBR different from WRBRB because one is a mirror image of the other?

Mirror images are not superimposable, right?

Uh...to answer your question, white is chosen over the other colors because it has  only $ \emph{one}$.  If you choose the red, you have to consider the case where where the rotation makes it indistinguishable. (IE, RRBBW, RBBWR)

Uh, here's a problem (not sure if it's AMC level)

$ \cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7} \cos \frac{4\pi}{7} \cos \frac{5\pi}{7} \cos \frac{6\pi}{7} \equal{} ?$

[quote="Zellex"]Uh, here's a problem (not sure if it's AMC level)

$ \cos \frac {\pi}{7} \cos \frac {2\pi}{7} \cos \frac {3\pi}{7} \cos \frac {4\pi}{7} \cos \frac {5\pi}{7} \cos \frac {6\pi}{7} \equal{} ?$[/quote]
[hide="Solution"]
Let $ k \equal{} \frac {\pi}{7}$.  Therefore, we wish to find the value of $ \cos k\cos 2k\cos 3k\cos 4k\cos 5k\cos 6k$.  Notice that $ \cos ik \equal{} \minus{} \cos k(7 \minus{} i)$.  Hence, the value that we seek is $ \minus{} \cos^2k\cos^22k\cos^23k \equal{} \minus{} \cos^2k\cos^22k\cos^24k$.  Let $ P \equal{} \cos k\cos 2k\cos 4k\implies \minus{} P^2 \equal{} \cos k\cos 2k\cos 3k\cos 4k\cos 5k\cos 6k$.  Thus,
\[ P\sin k \equal{} \cos k\sin k\cos 2k\cos 4k
\]
Noticing that $ \cos k\sin k \equal{} \frac {\sin 2k}{2}$, we have that
\[ P\sin k \equal{} \frac {\sin 2k\cos 2k\cos 4k}{2}
\]
Again, we notice that $ \cos 2k\sin 2k \equal{} \frac {\sin 4k}{2}$, so
\[ P\sin k \equal{} \frac {\sin 4k\cos 4k}{4} \equal{} \frac {\sin 8k}{8} \equal{} \frac { \minus{} \sin k}{8}\implies P \equal{} \frac { \minus{} 1}{8}\implies \minus{} P^2 \equal{} \boxed{\frac { \minus{} 1}{64}}
\]
[/hide]

Let $ 1 < log_{10}({log_{10}{a}}) < 100$.  There are $ N$ integer solutions for $ a$.  Let $ S$ be the sum of the digits of $ N$.  Let $ S'$ be the sum of the digits of $ S$.  Find $ S'$

[quote="123456789"][quote="firecricket91"]Let $ 1 < log_{10}({log_{10}{a}}) < 100$.  There are N integer solutions for $ a$.  Let $ S$ be the sum of $ N$.  Let $ S'$ be the sum of the digits of $ S$.  Find $ S'$[/quote]

When you say "Let $ S$ be the sum of $ N$", do you mean digits of $ N$, or all possible integer solutions for $ a$?[/quote]
yea sorry.

Solving the inequality, we have $ 10^{10} < a < 10^{10^{100}}$

Thus, $ N \equal{} 10^{10^{100}} \minus{} 10^{10} \minus{} 1$
The number $ 10^{10^{100}}$ in base-10 form is a single $ 1$ followed by $ 10^{100}$ zeros.
The number $ 10^{10}$ in base-10 form is a single $ 1$ followed by $ 10$ zeros.
The number $ 1$ in base-10 form is a single $ 1$ followed by $ 0$ zeros.

Thus, $ N$ consists (in digits) of $ (10^{100} \minus{} 1)$ nines and $ 1$ eight.

So $ S \equal{} 9(10^{100}) \minus{} 1$.

Then, $ S$ consists of $ 100$ nines and $ 1$ eight.
So $ S' \equal{} 100(9) \plus{} 8 \equal{} \boxed{908}$.

[quote="firecricket91"][quote="123456789"][hide="Solution"]Solving the inequality, we have $ 10^{10} < a < 10^{100^{10}}$ or $ 10^{10} < a < 10^{10^{20}}$.

Thus, $ N \equal{} 10^{10^{20}} \minus{} 10^{10} \minus{} 1$
The number $ 10^{10^{20}}$ in base-10 form is a single $ 1$ followed by $ 10^{20}$ zeros.
The number $ 10^{10}$ in base-10 form is a single $ 1$ followed by $ 10$ zeros.
The number $ 1$ in base-10 form is a single $ 1$ followed by $ 0$ zeros.

Thus, $ N$ consists (in digits) of $ 10^{20} \minus{} 1$ nines and $ 1$ eight.

So $ S \equal{} 9(10^{20}) \minus{} 1$.

Then, $ S'$ consists of $ 20$ nines and $ 1$ eight.
So $ S' \equal{} 20(9) \plus{} 8 \equal{} \boxed{188}$.[/hide][/quote]

Its not 10^{10^{20}} , it would be 10^{10^{100}}, also you cannot go from

[quote]Thus, $ N$ consists (in digits) of $ 10^{20} \minus{} 1$ nines and $ 1$ eight.

So $ S \equal{} 9(10^{20}) \minus{} 1$.[/quote]

because some digits can carry forming a new number.[/quote]

I edited my post. Is it right now?

yea thats what I got.

you can post next one if you want.

lol. I don't know any problems. You can post one if you want.


do any of u have any recommendation on honor chemistry text book? cuz i wanna learn i honor chemistry at home, so i can take chemistry AP in 10th grade

as a homeschooler, i use A Beka for chemistry I...very good book in my opinion, short, precise, and to-the-point...

Our school uses Zhumdahl (sp?), but my school is extremely inferior compared to many of those on the forums.

but what text book does most high school use?

prentice hall and mcgraw hill are typically very common and popular among public schools...

however, they are not necessarily always the best...


inequalities
function
Cauchy Inequality
inequalities unsolved

Let $ x,y,z>0$ satisfy $ xyz\equal{}1$. Solve that:
$ \sqrt{4\plus{}9x^{2}}\plus{}\sqrt{4\plus{}9y^{2}}\plus{}\sqrt{4\plus{}9z^{2}}\leq\sqrt{13}(x\plus{}y\plus{}z)$

It's equivalent to:
$ \sum(\sqrt{4+9x^{2}}-\sqrt{13}x+(\sqrt{13}-\frac{9}{\sqrt{13}})\cdot ln(x))\leq 0$
Let $ f: (0,\infty)\to\mathbb R$ ,$ f(x)=\sqrt{4+9x^{2}}-\sqrt{13}x+(\sqrt{13}-\frac{9}{\sqrt{13}})\cdot ln(x)$
Because $ f''(x)<0$ and $ f'(1)=0$ we are done.

[hide="PS"]
I've seen this technique used by Arqady.Very nice
[/hide]

Who can solve this problems by the Cauchy-Schwarz inequality?

[quote="bibobeo"]How do you choose $ \sqrt{13}\minus{}\frac{9}{\sqrt{13}}$[/quote]

Consider the function $ f(x) \equal{}\sqrt{4\plus{}9x^{2}}\minus{}\sqrt{13}\plus{}C\cdot ln(x)$
Because our inequality has the equality case when $ x \equal{} y \equal{} z \equal{} 1$,we must find  $ C$ in such a way that $ f(1)\geq f(x),\forall x\in (0,\infty)$ If we managed to find such a constant it would mean that the inequality is true ,since $ 0 \equal{} 3f(1)\geq f(x)\plus{}f(y)\plus{}f(z)$ ,which is equivalent to the initial inequality.
If $ f(1)\geq f(x),\forall x\in (0,\infty)$  it means that $ f(1)$ must be the maximum value of $ f$ $ \Leftrightarrow$ $ f'(1) \equal{} 0$ $ (*)$
From $ (*)$ we get the value of $ C$.
It remains only to check if $ f$ really has a maximum point (check if $ f$ is concave ).

Sorry, but I don't see why $ f''(x)<0$ with $ x\geq 0$

I got $ f''(x)\equal{}\frac{36}{(4\plus{}9x^{2})\sqrt{4\plus{}9x^{2}}}\minus{}\frac{4}{\sqrt{13}x^{2}}$
I don't see why $ f''(x)<0$ when $ x>0$
It's easy to see when $ x\geq 1$, what about $ 0<x<1$?

This is my homework and my teacher say that it can be solved by Cauchy-Schwarz inequality but he don't say clearly!  But I can't do it by that way

[quote="nktp"]Let $ x,y,z$ are positive real numbers such that  $ xyz = 1$. Prove that

$ \sqrt {4 + 9x^{2}} + \sqrt {4 + 9y^{2}} + \sqrt {4 + 9z^{2}}\leq\sqrt {13}(x + y + z)$.[/quote]Using Cauchy inequality,

$ \sum\sqrt {4 + 9x^{2}} = \sum\sqrt {4\sqrt [3]{x^{2}y^{2}z^{2}} + 9x^{2}}\equiv\sum\sqrt {4u^{2}v^{2}w^{2} + 9u^{6}}$

$ = \sum{u\sqrt {4v^{2}w^{2} + 9u^{4}}\leq\sqrt {\sum{u^2}}\sqrt {\sum{\left(4v^{2}w^{2} + 9u^{4}\right)}}\leq\sqrt {13}\sum{u^3}\equiv\sqrt {13}\sum{x}}$,

the last inequality holds for

$ 13\left(\sum{u^3}\right)^2 - \sum{u^2}\sum{\left(4v^{2}w^{2} + 9u^{4}\right)}$

$ = 4\sum{u^6} - 13\sum{u^4\left(v^2 + w^2\right)} + 26\sum{v^3w^3} - 12u^2v^2w^2$

$ = 2\sum{\left(v^6 + w^6 - v^4w^2 - v^2w^4\right)} - 11\sum{\left(v^4w^2 + v^2w^4 - 2v^3w^3\right)}$

$ + 4\left(\sum{v^3w^3} - 3u^2v^2w^2\right)$

$ = 2\sum{(v - w)^2(v + w)^2\left(v^2 + w^2\right)} - 11\sum{v^2w^2(v - w)^2}$

$ + 2\sum{vw}\sum{u^2(v - w)^2}$

$ = \sum{(v - w)^2\left[2v^4 + 4v^3w - 7v^2w^2 + 4vw^3 + 2w^4 + 2u^2(vw + wu + uv)\right]}\geq 0$.

The generalization see: http://www.mathlinks.ro/Forum/viewtopic.php?t=165627

By the way, if $ x,y,z$ be nonnegative real numbers such that $ 2xyz + yz + zx + xy = 5$, then

$ \sqrt {4 + 9x^{2}} + \sqrt {4 + 9y^{2}} + \sqrt {4 + 9z^{2}}\leq\sqrt {13}(x + y + z)$,

with equality if and only if $ x = y = z = 1$.


I noticed that everyone seems to expect the AMC B cutoff score to drop after the A cutoff score did.  That means people don't really have a score of 120 to aim for any more, they just do the best they can, and maybe it will be enough.

This seems to show a way of satisfying the original goal of the scoring changes in a different way than the one originally given.

What if the target score of 120 was not given, and they just take the top 5%.  That way people would not have a score to shoot for, and would be forced to do their best and try all the problems.

Perhaps they should get rid of a number cutoff and go to the percentages directly. Doesn't that make more sense? Or, better yet, go by strictly number of people qualifying - like how they do it for USAMO qualification.

LOL, I just edited that idea into my original post, and then I saw your post.  So perhaps it is a good idea.

This year was weird because of the scoring change. Usually having an absolute cutoff helps people rather than hurts them, when using the 5% rule might push the cutoff up to 106 or higher.

If they were to go by a percentage cutoff - is there enough of a difference between the skill levels between the A testers and the B testers to potentially skew the results?

Right, and also it sucks when you can't even tell what score you're shooting for to qualify. Having an absolute cutoff just makes it so much easier, especially for the people who don't care or aren't savvy enough to dig up statistics to estimate a relative cutoff. Besides, AMC at the lower levels should be (and I think is) more about individual achievement than competition, and a relative cutoff sort of violates that spirit.


Let $ \lambda_1 \leq \lambda_2 \leq...$ be a positive sequence and let $ K$ be a constant such that \[  \sum_{k=1}^{n-1} \lambda^2_k < K \lambda^2_n \;(n=1,2,...).\] Prove that there exists a constant $ K'$ such that \[  \sum_{k=1}^{n-1} \lambda_k < K' \lambda_n \;(n=1,2,...).\]

[i]L. Leindler[/i]

fourth problem ,here : [url]http://www.mathematik.uni-muenchen.de/~yarotsky/jarnik/Jarnik_daniel_harrer.pdf[/url]


What is conduction, convection and radiation and what states of matter does it occur in?

I also need to know how to draw a diagram of each type of heat transfer.

thanks  :)

Conduction is direct transfer of heat due to physical contact.  It can occur in all states of matter.  Convection is heat transfer due to convection currents.  It is based on the idea that things at higher temperature tend to be less dense than the same thing at a lower temperature.  Therefore, when something is heated, the heated substance tends to move in convection currents throughout the substance and transfer heat to the rest of the substance.  This is typical of liquid and gaseous substances.  Radiation is simply an electromagnetic wave emitted by a substance.  I believe radiation exists for all states of matter.

Dragon was right as far as he went.

Allow me to add a bit.

Conduction is energy transfer between molecules in contact.  Think of a stove in contact with a pot.  (It is part of heat transfer in all phases.)

Convection is energy transfer by conduction, when supplemented by movement of molecules to bring these moving molecules in contact to conduct energy.  Think of the water inside the pot, particularly when boiling.  (This occurs in mobile phases -- liquid, gas, plasma.)

Radiation is energy transfer by emitting/absorbing photons.  Think of the pot inside a normal oven.  (They don't sell as well if they are called "radiation ovens.")  (This occurs in all phases.)


conics
hyperbola
ellipse
parabola
geometry
3D geometry
trigonometry

How is the eccentricity values determined for the conic sections?

ie: if e>1 it is a hyperbola
     if 1>e>0 it is an ellipse

I understand why a parabola has an eccentricity of 1 and why a circle has one of 0, but I don't understand how they derived the e for hyperbolas and ellipses.

Also, how do you derive the directrix for the conic sections?

Edit:  http://mathworld.wolfram.com/Eccentricity.html

I have actually derived an expression for the eccentricity in terms of the angle of the plane that cuts the conic section and the angle of the cone itself :D

$e = 1 - \cos \theta + \cot \alpha \sin \theta$

Where $\theta$ is the tilt of the plane away from the plane perpendicular to the axis of the cone (that is, $\theta = 0$ is a circle) and $\alpha$ is the tilt of the cone away from the plane perpendicular to the axis of the cone (that is, small $\alpha$ is a wide cone, and $\alpha \approx \frac{\pi}{2}$ is almost a stick).

For a parabola, $\theta = \alpha$ (can you see why?) and the expression becomes $e = 1 - \cos \theta + \cos \theta = 1$.

For a circle, $\theta = 0$ and the expression becomes $e = 1 - 1 + 0 = 0$.

For a hyperbola, $\theta > \alpha$ (can you see why?) and it can be shown either graphically or with angle summation formulas that $e > 1$.

Finally, for an ellipse, $0 < \theta < \alpha$ and it can again be shown that $0 < e < 1$.

I can tell you that the eccentricity of an ellipse is a measure of how far it deviates from being a circle.  In fact, the polar equation for an ellipse is $r = \frac{k}{1 - e \cos t}$ (I have avoided $\theta$ because of my use of it above) and you can clearly see that $e = 0$ describes a circle, and $0 < e < 1$ describes a wobbly sort of circle - in fact, an ellipse!

(I worked for awhile on describing the polar equation of a conic section solely in terms of the geometry of the cone and the plane that cuts it, so that is why I have derived such an odd equation for $e$ - I could also tell you my expression for $k$ if you would like  :D )

Well, e=c/a (c=distance from center to focus, a=distance from center to vertex).  If you think about the ellipse, the vertex is farther away from the center than is the focus.  It is the opposite on the hyperbola.  This makes e<1 is an ellipse and e>1 a hyperbola.

If $e = c/a$, why is e also equal to the ratio $PF: PD$ where $PF$ is the distance from focus to a point and PF is distance from directrix to focus?

This is what Wikipedia has to say about your question:

"The four defining conditions above can be combined into one condition that depends on a fixed point F (the focus), a line L (the directrix) not containing F and a positive number e (the eccentricity). The corresponding conic section consists of all points whose distance to F equals e times their distance to L. For 0 < e < 1 we obtain an ellipse, for e = 1 a parabola, and for e > 1 a hyperbola.

"For an ellipse and a hyperbola, two focus-directrix combinations can be taken, each giving the same full ellipse or hyperbola. The distance from the center to the directrix is a/e, where a is the semi-major axis of the ellipse, or the distance from the center to the tops of the hyperbola. The distance from the center to a focus is ae.

"In the case of a circle e = 0 and one imagines the directrix to be infinitely far removed from the center. However, the statement that the circle consists of all points whose distance is e times the distance to L is not useful, because we get zero times infinity.

"The eccentricity of a conic section is thus a measure of how far it deviates from being circular."

I think it answers your question perfectly.  You can read the entire article at http://en.wikipedia.org/wiki/Conic_sections.

That's a good question.  I don't have the answer as to how to prove it, but those two distances you mentioned are used to distinguish conics.  If the sum of distances from a given point on the conic to the two foci is equal to a constant, it is an ellipse.  If the difference of the distances is equal to a constant, it is a hyperbola. (The constant is always equal to 2a, or the length of major/transverse axis in a ellipse/hyperbola, respectively).

I don't get why the formulas are the way they are. I'm looking for a proof or derivation of the eccentricity formulas of the conic shapes.

My textbook says eccentricity is $PF: PD$ where P is a point on the section, F is focus, and D is directrix.

The wikipedia article says it is equal to$c: a = F_1F_2 : 2A$

I think you are looking for a simple explaination?  I'm not exactly sure.  As weiyan said earlier, eccentricity measures how much the graph deviates from being a circle.  I don't know much about trig/polar but think of it this way.  A cirlce is made by intersecting the two cones with a plane, and having the plane parallel to the base of the cone.  Now slightly tilt the plane.  The intersection is an ellipse.  Tilt a little more so that exactly one of the bases of the cone is interected.  This creates a parabola.  Tilt still more, such that both bases are intersected by and the plane, and it has been nearly a 90 degree tilt from the way you formed the circle.  The intersection is now a hyperbola.  Notice, the eccentricity is increasing as the new conic sections are formed.  This also relates the conic's graph to its eccentricity.


Let $ a\ge b\ge c>0$ and $ x,y,z$ be real numbers such that $ x\plus{}y\plus{}z\equal{}0,\,x\ge 0,\,z\le 0$. Show that
\[ a^xb^yc^z\ge 1.\]

[quote="nayel"]Let $ a\ge b\ge c > 0$ and $ x,y,z$ be real numbers such that $ x \plus{} y \plus{} z \equal{} 0,\,x\ge 0,\,z\le 0$. Show that
\[ a^xb^yc^z\ge 1.
\]
[/quote]

1)if $ y\ge0$
$ a^xb^yc^z\ge a^x c^{y \plus{} z} \equal{} a^xc^{ \minus{} x} \equal{} (\frac {a}{c})^x\ge1$.

2)if $ y\le0$
$ a^xb^yc^z\ge a^{x\plus{}y}c^z\equal{}a^{\minus{}z}c^z\equal{}(\frac {a}{c})^{\minus{}z}\ge1$

Very nice NextPeace! :)

An interesting corollary of the above inequality: let $ a\ge b\ge c>0$ and $ x,y,z$ be real numbers such that $ x\ge \frac{x\plus{}y\plus{}z}{3}\ge z$. Then the following inequality holds
\[ a^xb^yc^z\ge (abc)^\frac{x\plus{}y\plus{}z}{3}.\]


function
induction
number theory unsolved
number theory

[i]A given funtion $ f \ : \ N \  \to \ N$ satisfies three conditions :

 $ 1 / f(1) \ \equal{} \ 1$

$ 2/ 3f(n). f( 2n\plus{}1 ) \ \equal{} \ f(2n) \left( 1 \plus{} 3f(n) \right)  \ \forall \  n \ \in \ N$

 $ 3/ f(2n) < 6f(n) \  \forall  \ n \ \in \ N$

Find all pairs of positive integers $ \left( k ; m \right)$ such that :

   $ f(k) \ \plus{} \ f(m) \ \equal{} \ 293$[/i]

From 2) we conclude $ 3f(n)|f(2n)$ and this yields a stronger version of 3) given by $ f(2n)=3f(n)$. After substitution into 2) we get $ f(2n+1)=1+3f(n)$. Switching to the dual system we observe that for all $ N>0$ and $ a_k\in\{\pm 1\}, a_N=1$ we get the recursion $ f(\sum_{k=0}^Na_k2^k)=a_0+3f(\sum_{k=0}^{N-1}a_{k+1}2^k)$. This together with 1) can be used in a straightforward manner to prove (by induction on $ N\geq 0$) that $ f(\sum_{k=0}^Na_k2^k)=\sum_{k=0}^Na_k3^k$ holds.

For $ N\leq M$ assume that $ f(\sum_{k=0}^Na_k2^k)+f(\sum_{k=0}^Mb_k2^k)=293$ with $ a_k,b_k\in\{0,1\}$ and $ a_N=b_M=1$ holds. This is equivalent to $ 1\cdot 3^5+1\cdot 3^3 +2\cdot 3^2+1\cdot 3^1 +2\cdot 3^0=\sum_{k=0}^M(a_k+b_k)3^k$, where we have formally set $ a_k=0$, if $ k>N$. Since $ 0\leq a_k+b_k<3$ holds, this is equivalent to the equations $ a_0=b_0=a_2=b_2=1$ and $ a_1+b_1=a_4+b_4=a_5+b_5=1$. Therefore, all pairs $ (A,B)$ with $ f(A)+f(B)=293$ are given by the set $ \{(5+2a+16b+32c\,,\,55-2a-16b-32c)\,|\,a,b,c\in\{0,1\}\}$


floor function
inequalities
combinatorics unsolved
combinatorics

find the minimum value of k>1 that: for any k subsets of X={1,2,..,n} ,there are 2 set A,B that A C B

Your minimum value is $\binom{n}{\left\lfloor\frac{n}{2}\right\rfloor}+1$. In fact, Sperner's theorem states that if we have k subsets $X_1$, $X_2$, ..., $X_k$ of an n-element set Y such that $X_i\not\subseteq X_j$ for any $i\neq j$, then $k\leq\binom{n}{\left\lfloor\frac{n}{2}\right\rfloor}$; and indeed, this inequality is strict, since we can set $k=\binom{n}{\left\lfloor\frac{n}{2}\right\rfloor}$ and take as $X_1$, $X_2$, ..., $X_k$ all possible $\left\lfloor\frac{n}{2}\right\rfloor$-element subsets of Y, and these k subsets will obviously satisfy $X_i\not\subseteq X_j$ for any $i\neq j$.

A proof of Sperner's theorem can be found at http://www.mathlinks.ro/Forum/viewtopic.php?t=14689 .

[Edited to make it more clear.]

  Darij


In the game of Bing Bong, there are two ways to score. A smash is worth 7 points, and a bash is worth 9 points. What is the greatest score that cannot be attained in a game of Bing Bong?

The question is not clear. Please state the exact rules of the game.

[hide] By the Chicken McNugget Theorem, $ 7\times{9}\minus{}(7\plus{}9)\equal{}\boxed{47}$.[/hide]


function
number theory solved
number theory

Let $f$ be a bijective function from $A = \{ 1, 2, \ldots, n \}$ to itself. Show that there is a positive integer $M$ such that $f^{M}(i) = f(i)$ for each $i$ in $A$, where $f^{M}$ denotes the composition $f \circ f \circ \cdots \circ f$ $M$ times.

Let $k \in A$ be given.
Since $A$ is finite the set $\{k,f(k), \cdots, f^p(k), \cdots, \}$ is finite, thus there exist $a,b$ such that $b>a$ and $f^a(k) = f^b(k).$
Thus $f^a(f^{b-a}(k)) = f^a(k)$. Since $f$ is bijective it follows that $f^{b-a}(k) = k.$
Therefore, for each $k \in A$ there exists a positive integer $M_k$ such that $f^{M_k}(k) = k.$
It is easy to deduce that for every positive integer $p$, we have $f^{pM_k} (k) =k.$

Now, to conlude, just choose $M = \text{lcm} (M_1, \cdots, M_n).$

Pierre.

Honestly, this is a pretty cool problem.

Essentially we must find a $M$ so that $f^{M-1}(k)=k$.
Since we have that $f$ is a bijection that means that after some turns, there exists some $B_k \in \mathbb{N}$ so that $f^{B_k}(k) = k$.
So now let's have a graph $G_k$, with $n$ vertices. We connect the pairs of vertices of the form $(v_{f^{t}(k)},v_{f^{t+1}(k)})$, where $t$ is a non-negative integer number less than $B_k$ (note that $f^{0}(k)=k$), but now notice that we must have a cycle in the graph $G_k$, and essentially we must have that $B_k$ is the length of the cycle.

So we must have that $B_t \mid M-1$, where $t$ is a positive integer number less than or equal to $n$.
Thus we can set $M-1=B_1B_2\dots B_n$, this implies that $M=B_1B_2\dots B_n +1$.


analytic geometry
vector
trigonometry
Pythagorean Theorem
geometry

How fast something is traveling south and/or east when you're given how fast it is when you're traveling southeast?  Example:  A car is traveling southeast at $ \frac{\sqrt{3}}{3}$.  How fast is the car traveling going south?

Is it any different for northeast, northwest, or southwest?

Thanks!!

Take a Cartesian plane with a right angled coordinate system in it, and let the positive direction of y axis denote north. Now, positive direction of x axis denotes east, negative y south, and negative x west.
Assign a vector to the car's velocity using this system-if you put the car in the origin of the system, then the given vector is a radius vector whose modulus is $ \frac {\sqrt {3}}{3}$, in fourth quadrant with a 45 degrees angle between it and the positive x (hence, the angle between negative y and the vector is also 45 degrees).
The south component is its projection on the y axis, i.e. $ \frac {\sqrt {3}}{3}\frac {\sqrt {2}}{2} \equal{} \frac {\sqrt {6}}{6}$.

This was a complicated manner of showing something rather simple, but when put in this manner, it will be easier for you to move to harder problems of this type (i.e. Car is traveling 60 degrees southeast at the given velocity)

Simpler way would consist of taking the vector $ v$ into its components $ v_S$ and $ v_E$ denoting the speed toward south and east, respectively (they're orthogonal). Knowing that SE is "right in the middle of it", you know that your task is reduced to a simple application of Pythagorean theorem in an isosceles right triangle with a know hypotenuse.

Could you explain how you got $ \frac {\sqrt {2}}{2}$?  Thanks again!

Basic trigonometry: $ \cos 45^0\equal{}\frac{\sqrt{2}}{2}$. How did it appear in our calculations? Well, in the first approach, you have a vector (whose modulus is $ v\equal{}\frac{\sqrt{3}}{3}$) and want to know the modulus of its orthogonal projection on an axis. If the angle between the vector and the axis is $ \alpha\equal{}45^0$, then the projection's modulus is (it's pretty much obvious, just draw a sketch) $ v_S\equal{}v\cos\alpha\equal{}\frac{\sqrt{3}}{3}\frac{\sqrt{2}}{2}$.

In the other approach, the same formula is obtained, applying Pythagoras' theorem:

$ v^2\equal{}v_S^2\plus{}v_E^2$

noting that $ v_S\equal{}v_E$, we have $ 2v_S^2\equal{}v^2$

or $ v_S\equal{}\frac{\sqrt{2}}{2}v\equal{}\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{3}$.


What number is $ 10\%$ of $ 20\%$ of $ 30\%$ of 40? Express your answer as a decimal to the nearest hundredth.


geometry
geometric transformation
homothety
ratio
combinatorics proposed
combinatorics

$A$ is a compact convex set in plane. Prove that there exists a point $O \in A$, such that for every line $XX'$  passing through $O$, where $X$ and $X'$ are boundary points of $A$, then
\[ \frac12 \leq \frac {OX}{OX'} \leq 2.\]

For all $M$ on the boundary of $A$, let $A_M$ be the homothetic of $A$ with respect to the homothty $h_M$ of pole $M$ and ratio $\frac 2 3 $.
Using the convexity of $A$, we deduce that $A_M$ is convex and contained in $A$ for all such point $M$.

Using the convexity of $A$ and $A_M$, it is easy to verify that is $M,N,P$ are three points on the boundary of $A$ and $G$ is the center of gravity of $MNP$ then $G$ belongs to each of the sets $A_M,A_N,A_P$.
It follows that the family of the sets $A_M$ is a set of bounded convex sets such that any three have a common point.
From Helly's theorem, we deduce that there exists a point $O$ which belongs to all the sets $A_M$ where $M$ is a point of the boundary of $A$.

Now, let's consider any line $d$ which passes through $O$. This line meet the boundary of $A$ in two points $X,X'$.
Then, from above $O$ belongs to the segment $XX''$ where $X'' = h_X(X')$, and to the segment $XX'$. Thus, $OX \leq XX'' = \frac 2 3 XX'$ so that $OX' \geq \frac 1 3 XX'$. Therefore $\frac {OX } {OX'} \leq 2$.
The other inequality is obtained by interchanging $X$ and $X'$.

Pierre.

It's very similar to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16474]this[/url] problem. In fact, in order to prove that, I proved a very similar lemma (although weaker).

Given a collection of convex sets in $\mathbb R^{n}$ s.t. each $n+1$ have non-void intersection, all of them have non-void intersection.

Another solution that is very beautiful.
If A and B are two convex figures. Then there is atranslation that takes A in B if and only if Every 3 points of A could be taken in B with a translation.
Proof of this is easily with Helly theorem.

A is a convex set in plane so it is compact. :)
So, we can set a triangle which the size is maximum. If we set  $ O $ the centroid of it, we are done. :)

Here is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=520132]another thread dedicated to this problem[/url]. One can see that the center of gravity(the centroid) of $A$ satisfies the requirement.

a nice problem
 here's my solution:
we will use $S$ to refer to the convex set
take a triangle $ABC$ with vertices in the boundary of $S$ with maximum area
note that if we construct the parallel from $ A$ to $BC$ let it $\ell_a$ similarly  $\ell_c$ and  $\ell_b$
then the triangle from  $\ell_a$  $\ell_c$  $\ell_b$ contains $S$ (let it $T$)
take $O$ the centroid of $ABC$
take a line throw $O$ cuts $S$ in $X,Y$,  $ABC$ at $X',Y'$ and $T$ at $X'',Y''$ (with $X',X''$ closer to $X$)
$2 \ge \frac{OX''}{OY'} \ge \frac{OX}{OY} \ge \frac{OX'}{OY''} \ge \frac{1}{2}$
and we win :D

It can be proved immediately from Minkovski's theorem. It states that for convex sets in $R^d$ we can find a point $O$ such that for every line $XOX'$ we have $\dfrac{XO}{X'O} \leq d$


Source: Rocket City Math League Round 1, Apollo Division

9. If $\overline{AB}$ and $\overline{CD}$ are perpendicular chords to a circle that intersect at E where $\overline{AE}=6$, $\overline{EB}=2$,$\overline{CE}=4$, and $\overline{ED}=3$. What is the area of the circle?

(BTW, no calculators are allowed on RCML's)

[quote="tarquin"][hide]An inscribed triangle in this circle has sides of $8$, $2 \sqrt {13}$, and $2 \sqrt 5$. Since $\frac {abc}{4A}$ gives us the radius of the circumscribed circle of a triangle with sides $a$, $b$, $c$, and area $A$, we plug in the values to get $\frac {8 \cdot 2 \sqrt {13} \cdot 2 \sqrt 5}{\frac 12 \cdot 4 \cdot 8 \cdot 4} = \frac {\sqrt {65}}{2}$. The area of the circle is then $\left(\frac {\sqrt {65}}{2}\right)^2 \cdot \pi = \frac {65\pi}{4}$.[/hide][/quote]

Thanks, but how did you figure out the lengths of the sides?

The chords are perpendicular so they make right triangles if you connect different points.

There is a simple formula for perpendicular chords.
Let
$AE=a$
$BE=b$
$CE=c$
$DE=d$

$\text{Radius}=\displaystyle \frac{\sqrt{a^2+b^2+c^2+d^2}}{2}$

In this case

$\text{Radius}=\displaystyle \frac{\sqrt{2^2+3^2+4^2+6^2}}{2}=\displaystyle \frac{\sqrt{65}}{2}$

$\text{Area}=\left({\displaystyle \frac{\sqrt{65}}{2}\right)}^2\cdot{\pi}=\displaystyle \frac{65\pi}{4}}$

I believe that that's called the root-mean-square

[quote="NoSoupForYou"]If you don't like formulas, here is a graphical approach to the problem.[/quote]

wow, did you make that using paint?


If p and q are different odd primes, then, if p - 1|q-1, a is an integer, an (a,pq) = 1, prove that $a^{q-1}$ is congruent to 1 mod pq.

If $q-1=k(p-1)$, then $a^{q-1}=(a^{p-1})^k\equiv1\bmod p$. Since also $a^{q-1}\equiv1\bmod q$, the number $a^{q-1}-1$ is divisible by p and q, i.e. $a^{q-1}\equiv1\bmod pq$.


Can people who've taken classes in something you have not share information from their transcripts and what they have learned or is that against the rules?


limit
trigonometry
calculus
calculus computations

(1) Calculate $ \lim_{n\to\infty}n\left(\sec^{n}\frac{a}{n}\minus{}1\right)$

(2) Let $ \theta \equal{}\theta(n) \equal{}\frac{2\pi}{n}$. Calculate $ \lim_{n\to\infty}\frac{\tan\theta (\sec^{n}\theta\minus{}1)}{\sec\theta\minus{}1}$. This limit has a simple geometrical interpretation.

[quote="Dr. Doe"]$ \sec x \equal{} 1\plus{}\frac{1}{2} x^{2}\plus{}O(x^{4})\quad ,\quad\left(1\plus{}\frac{a^{2}}{2}x^{2}\right)^{\frac{1}{x}}\equal{} 1\plus{}\frac{a^{2}}{2}x\plus{}O(x^{2})$

Therefore $ \sec^{n}\left(\frac{a}{n}\right) \equal{}\left(1\plus{}\frac{a^{2}}{2}\frac{1}{n^{2}}\plus{}O\left(\frac{1}{n^{4}}\right)\right)^{n}\equal{} 1\plus{}\frac{a^{2}}{2}\frac{1}{n}\plus{}O\left(\frac{1}{n}\right)$

$ \Rightarrow n\left(\sec^{n}\left(\frac{a}{n}\right)\minus{}1\right)\to\frac{a^{2}}{2}$

$ \frac{\tan\theta\left(\sec^{n}\theta\minus{}1\right)}{\sec\theta\minus{}1}\equal{}\frac{\tan\frac{2\pi}{n}}{n\left(\sec\frac{2\pi}{n}\minus{}1\right)}n\left(\sec^{n}\frac{2\pi}{n}\minus{}1\right)\sim\frac{\frac{2\pi}{n}}{n\frac{1}{2}\left(\frac{2\pi}{n}\right)^{2}}\frac{(2\pi)^{2}}{2}\equal{} 2\pi$[/quote]
Hi everyone.
Some correction to Dr.Doe

[quote="Dr. Doe"]$ \quad\left(1\plus{}\frac{a^{2}}{2}x^{2}\right)^{\frac{1}{x}}\equal{} 1\plus{}\frac{a^{2}}{2}x\plus{}O(x^{2})$[/quote]
here must be
$ \quad\left(1\plus{}\frac{a^{2}}{2}x^{2}\right)^{\frac{1}{x}}\equal{} 1\plus{}\frac{a^{2}}{2}x\plus{}O(x^{3})$
and also here
[quote="Dr. Doe"]
Therefore $ \sec^{n}\left(\frac{a}{n}\right) \equal{}\left(1\plus{}\frac{a^{2}}{2}\frac{1}{n^{2}}\plus{}O\left(\frac{1}{n^{4}}\right)\right)^{n}\equal{} 1\plus{}\frac{a^{2}}{2}\frac{1}{n}\plus{}O\left(\frac{1}{n}\right)$[/quote]
 here must be
$ ...\equal{}\left(1\plus{}\frac{a^{2}}{2}\frac{1}{n^{2}}\plus{}O\left(\frac{1}{n^{4}}\right)\right)^{n}\equal{} 1\plus{}\frac{a^{2}}{2}\frac{1}{n}\plus{}O\left(\frac{1}{n^{3}}\right)$

I don't understand your first correction.

$ \left(1\plus{}\frac{a^{2}}{2}x^{2}\right)^{\frac{1}{x}}\equal{}1\plus{}\frac{a^{2}}{2}x\plus{}\frac{a^{4}}{8}x^{2}\plus{}...$ and that's not $ 1\plus{}\frac{a^{2}}{2}x\plus{}O(x^{3})$ .

But in your second correction actually a small Landau symbol $ o\left(\frac{1}{n}\right)$ would be better than a big one.


GIVEN x,y,z are positive reals such that x+y+z>=xyz . prove that x^2+y^2+z^2>= xyz :blush:

hmm.. it's a bit weird
setting $ a\equal{}\frac{1}{x} ; b\equal{}\frac{1}{y} ; c\equal{} \frac{1}{z}$
we have $ \sum ab \geq 1$
we need to prove that $ \sum \frac{ab}{c} \geq 1$
which is equivalent to $ abc (\sum \frac{1}{c^2} ) \geq 1$
note that $ abc (\sum \frac{1}{c^2}) \geq abc (\sum \frac{1}{ab}) \equal{} \sum a$
so it's suffice to prove that $ \sum a \geq 1$
on the other hand , $ \sum a \geq \sqrt{3(\sum ab)} \geq \sqrt{3} > 1$  ..... a bit weird

$ \sum {x^2} \ge  \sum {xy}$
$ \sum\frac{1}{x} \sum {x^2} \ge  \sum {x} \ge xyz$

$ \text {since }\sum\frac{1}{x}\equal{}\frac{\sum{xy}}{xyz}$

$ \implies \sum {x^2}  \ge \frac{( xyz)^2}{\sum {xy}} \ge \frac{( xyz)^2}{\sum {x^2}}$

$ \implies \sum {x^2}  \ge xyz$


for m;n is two integer number such that : (m;n)=d
          prove that : $m^{phi(n)}+n^{phi(m)}=1+kmn$

It is not true. For example n=m=2.
Obviosly $d=(m,n)=1$. If d=1, then for $x=m^{\phi(n)}+n^{\phi(m)}-1$ we have $n|x$ and $m|x$, therefore $x=kmn$.

thanks you
firt ; i think it true  :lol:


function
calculus
derivative
inequalities theorems
inequalities

how to use it?

can somebody show me how to find the minimum of

this simple example $ x \plus{} 1/{x}$

by AM-GM its minimum is 2

but how by lagrange. can we use lagrange?
please :)

Lagrange multipliers are used when we have some multi-variable function subjected to some constraints, and we want to find the extrema.

I don't think we need to use Lagrange for this problem.  If you use the derivative tests, we get
$ 1\minus{}\frac{1}{x^{2}}\equal{}0\Rightarrow 1\equal{}\frac{1}{x^{2}}\Rightarrow x\equal{}\pm1$
Plugging in,
$ 1\plus{}1\equal{}2$ and $ \minus{}1\minus{}1\equal{}\minus{}2$
Then 2 is a local minimum and -2 is a local maximum.

[quote="JRav"]Lagrange multipliers are used when we have some multi-variable function subjected to some constraints, and we want to find the extrema.

I don't think we need to use Lagrange for this problem.  If you use the derivative tests, we get
$ 1 \minus{} \frac {1}{x^{2}} \equal{} 0\Rightarrow 1 \equal{} \frac {1}{x^{2}}\Rightarrow x \equal{} \pm1$
Plugging in,
$ 1 \plus{} 1 \equal{} 2$ and $ \minus{} 1 \minus{} 1 \equal{} \minus{} 2$
Then 2 is a local minimum and -2 is a local maximum.[/quote]

thanks,
actually i dont know alot about calculus. :blush:
 :D
why $ 1 \minus{} \frac{1}{x^{2}} \equal{} 0$

Take the derivative of each term and set it equal to zero.

oh ya thanks

the gradient equal to $ 0$


Please differentiate these:

Differential Geometry, Algebraic Geometry and Projective Geometry. It is my elective, and I do not know which is which.

Thanks in advance

Information on all of these topics is easy to find on [url=http://www.math.niu.edu/~rusin/known-math/welcome.html]Dave Rusin's site[/url] or wikipedia.  However, this is the wrong subforum to be asking this question in (advanced fields in the college playground would be better, for example)


function
advanced fields
advanced fields unsolved

Let $ \mathcal{A}$ unital $ C^*$-algebra, $ x\in\mathcal{A}$ normal element and $ f\in C(\sigma(x))$. Prove that \[ \sigma(f(x))\equal{}f(\sigma(x))\equal{}\{f(\lambda): \lambda\in \sigma(x)\}\].

I would be grateful for any useful hints.

everything takes place in the commutative, unital C*-algebra which is generated by $ x$, and the spectrum does not change when we make the C*-algebra smaller. then, by gelfand-neymark, the claim may be translated to ordinary functions where the spectrum is just the image.


Given that the midpoints of the sides of a triangle are connected, how many triangles are in the resulting diagram?

[asy]pair A=(-5,0), B=(5,0), C=(0, 5sqrt(3)), D=(-2.5, 2.5sqrt(3)), E=(2.5, 2.5sqrt(3)), F=(0,0);
draw(A--B--C--cycle);
draw(D--E--F--cycle);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, NW);
label("$E$", E, NE);
label("$F$", F, S);[/asy]
We can see, from the diagram, that the triangles are $ \triangle ADF,\triangle DEF,\triangle EFB,\triangle DEC,$ and $ \triangle ABC$.  Therefore, we have a total of $ \boxed{5}$ triangles.


For real numbers $ x_1, x_2, \cdots x_n$, we have $ x_1 > 1, x_2 > 2, \cdots x_n > n$, prove that

$ \frac {(x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n)^2}{\sqrt {x_1^2 \minus{} 1^2} \plus{} \sqrt {x_2^2 \minus{} 2^2} \plus{} \cdots \plus{} \sqrt {x_n^2 \minus{} n^2}}$.

---

Positive integers $ x_1, x_2, \cdots x_n$ satisfy $ x_1\plus{}x_2\plus{}\cdots\plus{}x_n\equal{}1$.

Prove that

$ \frac{1}{1\plus{}x_1}\plus{}\frac{1}{1\plus{}x_1\plus{}x_2}\plus{}\cdots\plus{}\frac{1}{1\plus{}x_1\plus{}x_2\plus{}\cdots\plus{}x_n}<$

$ \sqrt{\frac{2}{3}\left(\frac{1}{x_1}\plus{}\frac{1}{x_2}\plus{}\cdots\plus{}\frac{1}{x_n}\right)}$.

---

Real numbers $ x_1, x_2, \cdots x_n$ satisfy $ x_1^2\plus{}x_2^2\plus{}\cdots\plus{}x_n^2\equal{}1$.

Prove that $ \frac{x_1}{1\plus{}x_1^2}\plus{}\frac{x_2}{1\plus{}x_1^2\plus{}x_2^2}\plus{}\cdots\plus{}\frac{x_n}{1\plus{}x_1^2\plus{}x_2^2\plus{}\cdots\plus{}x_n^2}<\sqrt{\frac{n}{2}}$.

What are we supposed to prove in no 1? :huh:

For some reason I can't edit anymore...

#1 should be:

For real numbers $ x_1, x_2, \cdots x_n$, we have $ x_1 > 1, x_2 > 2, \cdots x_n > n$, find the minimal value of

$ \frac {(x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n)^2}{\sqrt {x_1^2 \minus{} 1^2} \plus{} \sqrt {x_2^2 \minus{} 2^2} \plus{} \cdots \plus{} \sqrt {x_n^2 \minus{} n^2}}$.

Thanks for pointing out.  :)


We will be holding a Math Jam on Wednesday, October 5, at 9:30 p.m. Eastern / 6:30 p.m. Pacific to discuss the solutions to the Round 1 problems.

[b]Please do not discuss solutions to the Round 1 problems (on this board or anywhere else) until the Math Jam has concluded.[/b]

We will announce here when it is OK to discuss the Round 1 problems and solutions.

[b]It's now OK to discuss Round 1 solutions![/b]

how do we enter a math jam (I know this sounds dumb...)???

They are also accessible through the "Math Jams" tab at the top of this page.

[quote="jb05"]Err... Where's the transcript? Am I missing something? It's not where it usually is, but it claims that there was something else yesterday.[/quote]We had some technical problems getting the transcript up last night, but it's up now.  Click on "Math Jams" at the top of this screen, then click on the "Transcripts" button.

Er, 6508/3465 [b]is[/b] 1.87821... I don't quite understand what you're asking.

[quote="l156steven"]uh oh... for number 2 i put 6508/3465  = 1.87821 when its supposed to be 6508/3465  = 2.87821...   :oops: do u think they're gonna take off points  :|[/quote]

I think he put a "2" in the second one instead of a "1". This is probably what he means:

[quote="l156steven"]uh oh... for number 2 i put 6508/3465  = 1.87821 when its supposed to be 6508/3465  = 1.87821...   :oops: do u think they're gonna take off points  :|[/quote]

He's probably saying that he forgot to put $\ldots$ ;)


[FACT: If $n\in N$ can not be written as $4^s (8t+7)$ for nonnegative integer $s, t$,

$n$ can be written as sum of three squares of integers. ]

Using FACT, prove that for odd integer $a,b\in N$ satisfying $\dfrac{b^2}{4}<a<\dfrac{b^2}{3}$

we can find four integer $x, y, z, w\in N$ s.t $a=x^2+y^2+z^2+w^2, b=x+y+z+w$.


geometry
3D geometry
number theory
relatively prime
number theory proposed

find all $ n \in \mathbb{N}$ such that $ n^4\plus{}n^2\plus{}1$ is a cube.

[quote="shoki"]find all $ n \in \mathbb{N}$ such that $ n^4 \plus{} n^2 \plus{} 1$ is a cube.[/quote]
$ n^4 \plus{} n^2 \plus{}1 \equal{} (n^2 \plus{} n \plus{}1 )(n^2\minus{}n\plus{}1)\equal{}c^3$ we have $ (n^2 \plus{} n \plus{}1 , n^2\minus{}n\plus{}1)\equal{} 1$ . so

$ n^2 \plus{} n \plus{}1 \equal{} a^3$ and $ n^2 \minus{} n \plus{}1 \equal{} b^3$

$ a\geq b\plus{}1 \Rightarrow a^3\minus{} b^3 \equal{} 2n \geq 3b^2 \plus{} 3b \plus{}1$

$ \Rightarrow n^2\minus{}n\plus{}1 \equal{} b^3 \geq \frac{(3b^2\plus{}3b)^2 \minus{}1}{4}$  contradiction !! so it can`t be cube .

[hide]***Shoki please check your PM***[/hide]

The only solution is n = 0.

w.l.g assume $ n \ge 0$

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

suppose $ d \mid  n^2 \minus{} n \plus{} 1$ and $ d \mid n^2 \plus{} n \plus{} 1$
then $ d \mid 2n$ but $ n^2 \plus{} n \plus{} 1$ is odd, so $ d \mid n$ but $ d \mid n^2 \plus{} n \plus{} 1 \equal{} n(n\plus{}1) \plus{}1$ so $ d \mid 1$

therefore the two factors are relatively prime and must each be a cube

$ a^3 \equal{} n^2 \plus{} n \plus{} 1$ and $ b^3 \equal{} n^2 \minus{} n \plus{} 1$

$ 2n \equal{} a^3 \minus{} b^3 \equal{} (a \minus{} b)(a^2 \plus{} ab \plus{} b^2) > a^2$

$ a^3 > n^2 > (1/4) a^4$ so $ a < 4$ and $ n < 2$

n = 1 does not work, so n = 0 is only solution


modular arithmetic
number theory unsolved
number theory

[hide="idea"]use pell equation.note that $ 17^2 \minus{} 12^2.2 \equal{} 1$.[/hide]
if u want a full solution...
[hide="solution"]
define these sequences:
$ X_k \equal{} 17X_{k \minus{} 1} \plus{} 24Y_{k \minus{} 1},Y_k \equal{} 12X_{k \minus{} 1} \plus{} 17Y_{k \minus{} 1}$.we can easily prove that $ X_k \plus{} Y_k\sqrt2 \equal{} (17 \plus{} 12\sqrt2)^k$.
and also we can prove that $ A_k \equal{} \frac {X_k \minus{} 1}{16}$.note that $ X_k \equiv 1 \pmod {16}$.
but we have
$ X_k^2 \minus{} 2.Y_k^2 \equal{} 1 \implies (X_k \minus{} 1)(X_k \plus{} 1) \equal{} 2.Y_k^2,16 \mid X_k \minus{} 1 \implies \gcd(\frac {X_k \minus{} 1}{16},\frac {X_k \plus{} 1}{2}) \equal{} 1 \implies X_k \equal{} 2.l_k^2 \minus{} 1,\frac {X_k \minus{} 1}{16} \equal{} m_k^2 \implies A_k \equal{} \frac {l_k^2 \minus{} 1}{8},l_k \equal{} 2.t_k \minus{} 1 \implies A_k \equal{} \frac {t_k(t_k \minus{} 1)}{2}$
DONE![/hide]

define $ x \equal{} \sqrt{2} \minus{} 1$ and $ y \equal{} \sqrt{2} \plus{} 1$

  notice that $ xy \equal{} 1$ and $ x^4 \equal{} 17 \minus{} 12\sqrt{2}$ and $ y^4 \equal{} 17 \plus{} 12\sqrt{2}$

  $ \implies 32A_k \equal{} x^{4k}  \plus{} y^{4k} \minus{} 2 (xy)^{4k}$
          $ \equal{} (y^{2k} \minus{} x^{2k})^2$
          $ \equal{} (y^k \minus{} x^k)^2 (y^k \plus{} x^k)^2$
          $ \equal{} (y^{2k} \plus{} x^{2k} \minus{} 2)( y^{2k} \plus{} x^{2k} \plus{} 2)$

  so if $ N_k \equal{} \frac{1}{4}(y^{2k} \plus{} x^{2k} \minus{} 2)$
  then $ A_k \equal{} \frac{1}{2}N_k(N_k \plus{} 1)$

  so it remains to show that $ N_k$ is an integer which can be done in the usual way either by expanding the powers using the binomial expression or by forming a linear recurrence relation for $ N_k$

[quote="Lomisedu"]Give a sequence
$ A_k \equal{} \frac {1}{32}( (17 \plus{} 12\sqrt {2})^k \plus{} (17 \minus{} 12\sqrt {2})^k \minus{} 2 )$

$ k$ is a natural number.
Prove that $ A_k$ is triangular number.

thanks![/quote]
If $ A_k\equal{}\frac{x_k(x_k\plus{}1)}2$ then $ x_k\equal{}\frac{\sqrt{1\plus{}8A_k}\minus{}1}2$

$ 8A_k\plus{}1\equal{}\frac 14((3\plus{}2\sqrt 2)^{2k}\plus{}\frac 1{(3\plus{}2\sqrt 2)^{2k}}\plus{}2)$ $ \equal{}\frac 14((3\plus{}2\sqrt 2)^k\plus{}\frac 1{(3\plus{}2\sqrt 2)^k})^2$

So $ x_k\equal{}\frac 14(2\plus{}(3\plus{}2\sqrt 2)^k\plus{}\frac 1{(3\plus{}2\sqrt 2)^k})$

And we just have to show that this $ x_k\equal{}\frac {2\plus{}(3\plus{}2\sqrt 2)^k\plus{}(3\minus{}2\sqrt 2)^k}4$ is an integer, which is easy :

$ x_0\equal{}1$
$ x_1\equal{}2$
$ x_{k\plus{}2}\equal{}6x_{k\plus{}1}\minus{}x_{k}\minus{}2$

And $ A_k\equal{}\frac{x_k(x_k\plus{}1)}2$

(a bit too late :) )


The vertices of a regular 2005- gon are colored red, white and blue. When ever two vertices of different colors stand next to each other, we are allowed to recolor them into the third color.
(a) Prove that there is a ﬁnite sequence of allowed recolorings after which all the vertices are of the same color.
(b) Is that color uniquely determined by the initial coloring?

Hi. Now I'm at the first variant which is [i]a[/i]. Let's begin with I and II vertices. The first is blue. Anyway, we must choose one. And we chosen. Now we have two options. First is that we can select [i]same[/i] or [i] different[/i] colors. ... After some operations such that we're already doing, will establish the same situation we'd had at the outset. It's time now to use advantages of amateaurity. We can easily assume that in doing the requested operation, quite same thing will happen and wherever in a place, the all will go replaying.
I will then think of the [i]b[/i] variant of the problem.
Thanks.

Drive an injection one of the corner to another i.e., make an injection from the first to the last number (and these numbers are what are used in counting the operations while you're solving the problem). Then you'll see whether or not there exists unique way to do.
Once thanks![/i]

[quote="K81o7"][hide="b"] Yes.  Instead of red, white, and blue, put 0's, 1's and 2's.  Under the given operation, the sum of all numbers around this 2005-gon does not change modulo 3.  In order for us to have all numbers around the circle to be 0, 1 or 2, then, we would have to have an initial sum congruent to 0, 1 or 2, respectively, modulo 3.  Thus, the initial configuration determines a final color.[/hide]

For part a, I think you can create a process by which you can strictly increase the number of occurrences of whichever color you want.  But I'm too lazy to outline it.[/quote]

Yes.. I'm not sure if there is an easier way.. but here's mine:
Let $ \{a_{i}\}$ be the vertices' colour and $ b$ be the final colour
Let $ a_{i+kn}=a_{i}$ $ \forall k\in\mathbb{Z}$
Let the operation on $ (a_{i},a_{i+1})$ be $ O(i)$
If $ \exists a_{i}\neq a_{i+1}$, either keep or do $ O(i)$ so that $ a_{i}=b$ or $ a_{i+1}=b$
Start with any contiguous block of $ \{a_{i}=b| j\leq i\leq k\}$, $ a_{k+1}\neq b$
Let $ x\geq k+2$ be the smallest such that $ a_{x}\neq a_{k+1}$
If $ a_{x}\neq a_{k}$ then do $ O(x-1)$ ($ x$ decreases by 1)
If $ a_{x}=a_{k}$ and $ x=k+2$ do $ O(k),O(k+1),O(k)$ ($ k$ increases)
If $ a_{x}=a_{k}$ and $ x>k+2$ do $ O(k+1),O(k)$ ($ x$ decreases by 2)
Since $ x$ cannot decrease indefinitely without $ k$ increasing, the process will terminate with the desired outcome.

(In a more easy-to-see form, WLOG 0 is the final colour so I use the operations 01[color=red]0[/color] -> 220 -> 211 -> 0[color=red]0[/color]1 and 11[color=red]0[/color] -> 122 -> [color=red]0[/color]02 to bunch zeros together to the left)


analytic geometry
probability
calculus
integration
function
symmetry
algebra

For [i]n [/i]distinct points chosen randomly and independently in the unit interval, what is the expected length of the largest sub-interval.

Where "randomly" means uniform distribution and the "largest subinterval" is the maximum of {$x_1,x_2-x_1,x_3-x_2,\dots,x_n-x_{n-1},1-x_n$} where $0\leq x_1\leq x_2\leq \dots\leq x_n\leq 1$.

The differences are distributed uniformly on the simplex $u_i>0,\sum u_i=1$. This is not obvious, but it can be routinely proven by induction.

Now, the largest coordinate- that's ugly. It comes down to finding the center of mass of an irregular box-like region (quadrilateral when $n=3$, hexahedral with no two faces parallel when $n=4$, etc.)

[quote="jmerry"]The differences are distributed uniformly on the simplex $u_i>0,\sum u_i=1$. This is not obvious, but it can be routinely proven by induction. [/quote]

All that is needed is that the probability distribution on the differences is cyclically invariant.
Thus, one can assume the longest interval is the last one, $[x,1]$.  The probability density of such configurations (among all $n$-point configurations) is proportional to $x^{n-1}$.   The integral of that density is $1/n$ so that the correctly normalized probability density is $n x^{n-1}$.  The answer to the problem is therefore $\int_0^1 (1-x) n x^{n-1} dx = \frac{1}{n+1}$.

[quote="fleeting_guest"]
All that is needed is that the probability distribution on the differences is cyclically invariant.
[/quote]
You mean that for example $1-x_n$ and $x_{n}-x_{n-1}$ has the same distribution??? ($x_{n-1}<x_n<1$)

[quote="fleeting_guest"]The answer to the problem is therefore $\int_0^1 (1-x) n x^{n-1} dx = \frac{1}{n+1}$[/quote]
There are $n+1$ intervals, so the largest is always greater than $\frac1{n+1}$. You've answered the wrong question.

[quote="anik_andrew"]You mean that for example $1-x_n$ and $x_{n}-x_{n-1}$ has the same distribution??? $(x_{n-1}<x_n<1)$[/quote]
That's true- as is the stronger statement I posted. Note that these two intervals are negatively correlated with each other- we can't look at any one interval in a vacuum.

Ok. I define more exactly our conditions.
We have $\xi_1,\xi_2,\ldots,\xi_n$ which are totaly independent and have the same distribution.
$\xi_i \simeq U[0,1].$ Then we define new random quantitis $x_1,x_2,\ldots,x_n$ in such way:
$\forall \omega \; x_i(\omega) = \xi_{k_i}(\omega)$ if $\xi_{k_1}\le \xi_{k_2}\le \ldots \le \xi_{k_n}$. where $k_1,k_2,\ldots,k_n$ is a permutation of $1,2,\ldots,n.$ Then $\zeta(\omega) = \max_{k=0,\ldots,n} x_{k+1}-x_k$
And we want to find ${\bf M} \zeta.$
Is it the right conditions of our problem?

Yes.

A few calculations: For $n=1$, the answer is $\frac34$. For $n=2$, the answer is $\frac{11}{18}$.

Let $f_n$ be the density function of $\zeta$, as you call it, when there are $n$ $x_i$. Let $F_n(x)=\int_0^x f_n(t)\,dt$, the probability that $\zeta$ is less than $x$. We have the recursive formula
$f_n(x)=n(n+1)(1-x)^{n-1}F_{n-1}\left(\frac{x}{1-x}\right)$ for $0<x<1$.
Also, $f_1(x)=2$ if $\frac12<x<1$ and zero otherwise.

An explanation: $n(1-x)^{n-1}$ is the probability density of a particular length. $F_{n-1}\left(\frac{x}{1-x}\right)$ is the probability that all of the others are shorter than $x$, using the symmetry of the overall distribution. We multiply by $n+1$ since there are $n$ possible intervals that could be largest.
If you evaluate this, you get a piecewise polynomial function of degree $n-1$ with $n+1$ jump discontinuities in the $(n-1)$st derivative.


search
geometry
circumcircle
analytic geometry

A ball of mass $ m$ attached to a string is whirled clockwise in a [i]vertical[/i] circle of radius $ R$. At one instant, it is right at the top of the circle, and its speed then is $ \sqrt{Rg}$ (the minimum speed needed here to keep the ball moving in a circle), where $ g$ is the acceleration due to gravity. After the ball goes through a little more than 270 degrees of whirling, the string can be cut so that the ball goes through a parabolic trajectory and passes through the centre of the circle. At what angle above the horizontal (passing through the circle's centre) is the string cut?

(Assume that no energy is added to the system throughout)

You'll have to deal with the parabolic phase, finding the initial velocity and position in terms of the "launching" angle, by Mechanical Energy Conservation or (equivalently) Torricelli's Equation.

Knowing the final position, the center of  the circle, you will search for the initial one, getting to the required angle.

1.The x axis is the circle diameter and y axis is tangent to the circle at the left point of the diameter
2.The point P is the point on the circle that the ball does begin the parabolic voyage
3.The velocity is (gr*(3-2*sin(teta))^.5 and the launching angle with x axis is (pi/2-teta)
4.The point P has coordenates [r-rcos(teta),rsin(teta)]
5.The parabolic way goes through point O (circumcenter) with coordinates [r,0]
6.For obey item 5 angle teta=10.57389...degrees

As usual in parabolic launching (with some exceptions) there are two solutions: 5.26439° and 65.26439° .


inequalities
function
algebra
domain
inequalities proposed

i hope this is original and more importantly correct
abc=1
a,b,c all greater than 0
prove that
[(a^3-1)(b^3-1)/(b^3)]+[(c^3-1)(b^3-1)/(c^3)]+[(a^3-1)(c^3-1)/(a^3)] <=0

Denote :
\[ x=a^3 , y=b^3 , c=z^3 \]
The inequality becomes:

Let $ x,y,z>0 , xyz=1$
Prove that :
\[ x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leq \frac{x}{y}+\frac{y}{z}+\frac{z}{x} +3 \]

I try so hard and i only solve that :
\[ x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leq \frac{x^n}{y^n}+\frac{y^n}{z^n}+\frac{z^n}{x^n} +3 \]
\[ n \geq 2 \]
I think it's so hard and interesting .
Anybody help me ???

Yes, Galois, your inequality is true. Here is my proof:

In fact, just as Tung Lam, I substitute $x=a^3$, $y=b^3$ and $z=c^3$. Then your problem can be rewritten as follows:

[i]If x, y, z are arbitrary positive numbers with xyz = 1, then prove that

  $\displaystyle \frac{\left( x-1\right) \left( y-1\right) }{y}+\frac{\left( y-1\right) \left( z-1\right) }{z}+\frac{\left( z-1\right) \left( x-1\right) }{x}\leq 0$.[/i]

Now here is my proof of this inequality: We can rewrite the inequality in the form

  $\displaystyle \left( x-1\right) \left( 1-\frac{1}{y}\right) +\left( y-1\right) \left( 1-\frac{1}{z}\right) +\left( z-1\right) \left( 1-\frac{1}{x}\right) \leq 0$,

or, equivalently,

  $\displaystyle \left( x-1\right) \left( \frac{1}{y}-1\right) +\left( y-1\right) \left( \frac{1}{z}-1\right) +\left( z-1\right) \left( \frac{1}{x}-1\right) \geq 0$.

Since xyz = 1, we can use Harazi's substitution $\displaystyle x=\frac{u}{v}$, $\displaystyle y=\frac{v}{w}$, $\displaystyle z=\frac{w}{u}$, where u, v, w are three positive reals. Then, the inequality we have to prove simplifies to

  $\displaystyle \left( \frac{u}{v}-1\right) \left( \frac{w}{v}-1\right) +\left( \frac{v}{w}-1\right) \left( \frac{u}{w}-1\right) +\left( \frac{w}{u}-1\right) \left( \frac{v}{u}-1\right) \geq 0$.

We expand this and get

  $\displaystyle \frac{v^{3}w^{3}+w^{3}u^{3}+u^{3}v^{3}+3u^{2}v^{2}w^{2}-vu^{3}w^{2}-u^{3}wv^{2}-v^{3}u^{2}w-uv^{3}w^{2}-uv^{2}w^{3}-w^{3}vu^{2}}{u^{2}v^{2}w^{2}}\geq 0$.

Thus, it remains to prove that

$v^{3}w^{3}+w^{3}u^{3}+u^{3}v^{3}+3u^{2}v^{2}w^{2}-vu^{3}w^{2}-u^{3}wv^{2}-v^{3}u^{2}w-uv^{3}w^{2}-uv^{2}w^{3}-w^{3}vu^{2}\geq 0$.

But if we now define X = vw, Y = wu and Z = uv, then we see that

$v^{3}w^{3}+w^{3}u^{3}+u^{3}v^{3}+3u^{2}v^{2}w^{2}-vu^{3}w^{2}-u^{3}wv^{2}-v^{3}u^{2}w-uv^{3}w^{2}-uv^{2}w^{3}-w^{3}vu^{2}=X^{3}+Y^{3}+Z^{3}+3XYZ-\left( YZ\left( Y+Z\right) +ZX\left( Z+X\right) +XY\left( X+Y\right) \right)$,

and this is $\geq 0$ after Schur. Hence, the inequality is proven.

Nice inequality, Galois! Although the "^3"'s were a bit needless...

  Darij

Schur's Again!!!  :)  :)

[quote="darij grinberg"]
 $\displaystyle \left( \frac{u}{v}-1\right) \left( \frac{w}{v}-1\right) +\left( \frac{v}{w}-1\right) \left( \frac{u}{w}-1\right) +\left( \frac{w}{u}-1\right) \left( \frac{v}{u}-1\right) \geq 0$.
[/quote]

May we declare victory here since it's equivalent to

$\displaystyle v^{-2}(v-u)(v-w)  + w^{-2}(w-u)(w-v) + u^{-2}(u-v)(u-w) \geq 0$

which is the Schur's inequality for $r=-2$.

See (and pay attention to the first reply by Moubinool)
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1166
[quote="Moubinool"]
For any t \in R, (a,b,c)\in R+3

at(a-b)(a-c) + bt(b-c)(b-a)+
ct(c-a)(c-b) \geq 0

with equality iff a=b=c
[/quote]

Or you can prove a more generalized Schur's inequality:

If $f(x)$ is a monotone function (increasing or decresing, not necessary strict) and $f(x)\geq 0$ in it's domain, then for any $a,b,c$ in it's domain:

$f(a) (a-b)(a-c) + f(b)(b-c)(b-a) + f(c)(c-a)(c-b) \geq 0$.

Make sure you prove this as a lemma if you ever use it in a Olympiad contest and don't forget to quote my name if you use it in the case $f(x)$ is a decresing function. :D  :D  :D

[quote="fuzzylogic"]Or you can prove a more generalized Schur's inequality:

If $f(x)$ is a monotone function (increasing or decresing, not necessary strict) and $f(x)\geq 0$ in it's domain, then for any $a,b,c$ in it's domain:

$f(a) (a-b)(a-c) + f(b)(b-c)(b-a) + f(c)(c-a)(c-b) \geq 0$.[/quote]

This is so cool! And the proof is really the same as for the "standard" Schur inequality, just with a little modification. Thanks for this generalization!

  Darij

hmm... I knew that it's for increasing or convex functions:

if $f:\mathbb{R}\to\mathbb{R}_+$ and its an increasing or a convex function we have
$f(a)(a-b)(a-c)+f(b)(b-a)(b-c)+f(c)(c-a)(c-b)\ge 0$

[quote="Tung Lam"]
Let $ x,y,z>0 , xyz=1$
Prove that :
\[ x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leq \frac{x}{y}+\frac{y}{z}+\frac{z}{x} +3 \]
[/quote]

Hehe, in his post  "Inequality - abc = 1", [b]Arne[/b] noticed that this is actually equivalent to IMO2000/2.  :D

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=848


limit
real analysis
real analysis unsolved

0.
First method: Stirling's approximation. The expression is just $ \frac {\binom{2n}{n}}{4^n}$, and from Stirling's approximation, we have: $ \lim \frac {n!}{(n/e)^n\sqrt {2\pi n}} \equal{} 1$, thus $ \lim \frac {\binom{2n}{n}\sqrt {\pi n}}{4^n} \equal{} 1$, and together with $ \lim \frac {1}{\sqrt {\pi n}} \equal{} 0$  it implies that $ \lim \frac {\binom{2n}{n}}{4^n} \equal{} 0$

A more elementary approach: Let $ a_{n} \equal{} \frac {1}{2}\cdot\frac {3}{4}\cdots\frac {2n \minus{} 1}{2n}$. We can show that $ a_n \le \frac {1}{\sqrt {3n\plus{}1}}$, and this is enough (this problem was found in Titu's "50 Introductory Problems in Algebra"). It is just induction.


rotation
Support
geometry
geometric transformation

Can anybody explain the direction of friction on the front and rear wheels of a bicycle with a proper argument. I was actually analyzing the way my cycle chain rotates today so let me check if my analysis was correct. (Proper Reasoning obviously required from your side)

it is towards the front(the direction u r moving) on the back wheel and toward back on the front wheel.
well basically the reason is the construstion of bicycle...wehen u push the pedals of the bicycle u apply both torque and force on the back bicycle wheels but the impulse due to torque is more than due to force hence the wheel tends to rotate witha speed $ \omega$ such that $ \omega r > v$ hence a frictional force acts wich supports translation that is forward motion...
In case of the front wheel due to the additional condition of the cycle being a rigid system as u pedal along with the back wheel sthe front ones are bound to move forward but not rotate...hence friecction acts backward to provide torque and support rotation and oppose translation

[quote="pardesi"] the impulse due to torque is more than due to force [/quote]

This is what I was pondering about. Any specific reason or is just standard theory. :D

let me assume the biscycle tyres to be discs...
So $ \omega r \equal{} \frac {2Fdt}{M} > \frac {Fdt}{M} \equal{} v$ ...and we are done..
and yes i don't beleive in standard theories..i beleive in change( i hate hillary :D )


I'm trying to find one that does not use java (flash preferred), has free chat enabled, and isn't a "club" that you have to join.

EDIT: And my dad doesn't let me install anything on his computer (his is the one with internet access.)

http://www.jagex.com has all of these, but java doesn't work on my compy for whatever reason.

[quote="Fermatprime"][quote="TripleM"]You might want to look at [url=http://www.freechess.org]FICS[/url]. It has all the things you mention. You will need to download some graphical chess program though - eg winboard for windows, which is free, as FICS is entirely textbased. Its easy to set up something like winboard to run your games though.[/quote]

Problem with FICS is that it won't let you on if you have a free email. Which kinda sucks.[/quote]

They used to let u sign (i'm not sure if they still do) in as a guest and then talk to a moderator and he'll set up an account with your free email. That's what I did, but for some reason my email wasn't free it was @msn.com which costs money.

Playchess is free and pretty good, but I think you have to get fritz.

Yeah, I play there too. Although people seem to get annoyed at me because I have dialup. But I suppose that happens everywhere. I don't think you need Fritz - I'm pretty sure theres something else (free) that you can download that lets you play there, but Fritz has apparently a nicer interface.

[quote="Fermatprime"]Problem with FICS is that it won't let you on if you have a free email. Which kinda sucks.[/quote]
Be clever. Go to http://www.mailinator.com and read the blurb on their front page. The site is rather useful, and FICS accepts a @mailinator.com "address" for your email.

[url]http://www.chesskids.com[/url] has some nice links to chess-playing sites.

Thanx for the info on Mailinator. I may sign up for FICS that way later on. It's probably worth it, from what I've heard...

Of course, I just played an absolutely horrible game on Yahoo!, so I'm not in the mood right now. I got to a point where I was four pawns up, and then basically went insane and ended up losing...first a pawn, then the exchange, then a rook, then my queen...

[quote="Kyptonite"]A number of people on Yahoo use chess programs to cheat.  A word to the wise. :D[/quote]

Yeah, but I doubt he did. Look at this position:

White: Kd2, Qb7, Ra1, Re1, Ne3, Pa2, Pc5, Pe2

Black: Kg8, Qg5, Ra8, Re8, Bc4, Pa6, Pc7, Pd6, Pf3, Pg6, Ph5

The game continued: 1.exf3 Rxe3! (a sound exchange sac) 2. Qxa8+?? (a line I had nearly overlooked in my calculations) 2...Kg7????

2... Re8+! would have won the queen and the game.

Im not a chess addict, but ill play someone if they want to play. If you go to funorb.com and make an account, add me to your friends list, my name is hyperhavoc5. they have lots of games, one including chess

http://www.chessvariants.org doesnt have [i]normal[/i] chess but it has many very fun chess variants free on the internet that dont require download

unfortunately, the computer in the chess games is very stupid and has very little sense of strategy

[quote=mathleticguyyy]nowadays we have chess.com[/quote]

(-_-)
I broke chess.com computer mode by letting 2 level 1 computers play each other

I think that you should go with chesskid.com

chesskids has very funny animations for the videos


inequalities
pigeonhole principle
geometry
perpendicular bisector
combinatorics unsolved
combinatorics

a) is classic: Consider the pair of boys standing closest to each other; they will throw to each other. If none of the other boys throw to them either of them, the result follows by induction. Otherwise, there will not be enough throws among the remaining boys for everyone to get a ball, so the result follows.

b) is pretty interesting: Suppose that a boy gets more than five balls. Then there will be at least six boys who are closer to him than anyone else. Consider the boy he is closest to, and draw a circle with the original boy as center and the radius as the distance between the two boys. Now note that any of the remaining five boys must lie closer to the original boy than the other one, so they must lie on the original boy's side of the perpendicular bisector of the segment connecting the two boy, which cuts off a 2pi/3 arc of the circle - we'll drop the line restriction for this weaker restriction, because it turns out this suffices.

Repeat this logic for the second-closest boy, the third, etc. up to the fifth, and note that each arc extends to 2pi/3 on the largest circle of the fifth boy because the circles are concentric. Now note that the sixth boy has to be outside the fifth boy's circle by assumption, but also that because the 2pi/3 arcs cannot contain any other boys, the projections of the boys onto the largest circle are at least pi/3 apart, so they take up at least 5pi/6 of the circle. If they take up more than 5pi/6, then the sixth boy cannot possibly be closer to the first boy than each of the other boys by the logic. If equality holds, however, we bring back the perpendicular bisector restriction and note that this requires all of the first five boys to be equidistant from the first boy; contradiction.

That was rather longwinded for a fairly simple and intuitive problem; does anyone have a better solution?

[hide="Solutions"]
(a) Suppose the opposite case. If all the boys get a ball, each one would get just one ball (obvious). Now, take the pair whose distance has the shortest length. It's easy to see that each boy will throw his ball in the other, because this is the least distance in all the configuration. So these boys will get a ball. Analogously, taking the other $ 13$ boys, and since no boy throw his ball in that "isolated pair", we can apply the same reasoning. Thus only one boy will remain, a contradiction.
(b) If there exist a boy which get at least six balls (call this boy "$ A$"), there would be two balls' trajectories which form a $ \pi /3$ angle, and using some inequalities in the triangle, you can notice that there's two boys which throw his ball in $ A$ but the distance between them is smaller than the distance to $ A$, contradiction.
[/hide]

[quote="raf92"]
(b) If there exist a boy which get at least six balls (call this boy "$ A$"), there would be two balls' trajectories which form a $ \pi /3$ angle, and using some inequalities in the triangle, you can notice that there's two boys which throw his ball in $ A$ but the distance between them is smaller than the distance to $ A$, contradiction.
[/quote]

This seems more like handwaving than an actual proof. It's the same idea I used, but you can't just say "using some inequalities in the triangle".

[quote="fwolth"]
This seems more like handwaving than an actual proof. [/quote]
Ok, forgive me. I wanted to be brief, and maybe my solution lost clarity. The inequality I mean is, using the conventional notation, given a triangle $ \Delta ABC$, if $ \angle BAC>\angle ABC$, therefore $ a>b$.

Oh, I understand now. That's indeed a very nice proof. By Pigeonhole, there must be an angle of at most $ \pi/3$ which must be the largest angle in that triangle, contradiction.

Also another problem is to show that the lines formed by balls travelling don't form a closed polygon.

Suppose $ A_1,A_2,\ldots,A_n$ are arranged so that $ A_i$ throws to $ A_{i\plus{}1}$ where $ A_{n\plus{}1}\equal{}A_1$. Then we must have $ A_1A_2>A_2A_3>\ldots>A_{n\minus{}1}A_n>A_nA_1>A_1A_2$, contradiction.

Alternately, we can apply pure logic, assuming that this process continues finitely....

Suppose $ C$ is the closest person to both $ A$ and $ B$, Suppose A gets the ball, passes it on to C, and finally the ball reaches $ B$ , the cycle repeats, the process continues infinitely, thus to make the opp. happen $ B$ should not receive the ball.....thus proved.[hide][/hide]

[quote="reza"]Fifteen boys are standing on a field, and each of them has a ball. No two distances between two of the boys are equal. Each boy throws his ball to the boy standing closest to him.

(a) Show that one of the boys does not get any ball.
(b) Prove that none of the boys gets more than five balls.[/quote]

a)suppose that all the boys stay with 1 ball.Look at then as a graph.consider the boy $ b_1$ whose shortest edge is the longest edge, when compared to other boys.This boy $ b_1$ throws to boy $ b_2$, and so on.But boy $ b_{15}$ throws to $ b_1$. But due to the fact that $ b_1$ has the propertie that his shortest edge is thr longest edge when compared to others, $ b_{15}$ couldn´t possibly throw to him and the result follows.

[quote]You appear to have badly misunderstood the question. (Each person has their own ball, and each person throws once and only once. If everyone threw whenever they got a ball, the process would continue indefinitely no matter what: the two closest-together people would toss back-and-forth forever.)
[/quote]

OOps...thanx JBL...


$a,b,c$ are positive reals that are not equal to 1. And satisfy: $a^{x}=b^{y}=c^{z}$, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$.

Find $abc$.


Who,in your opinion, is more likely to become the next winner of the Champions League 2008-2009. As for me, it is Liverpool, though, I hope Barca will bash them in the semi-final.

P.S: for those who are still unaware of what champions league is:

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

My favourite is LIverpool too.They show fascinating games nowadays(Especially their game with MU).
But apparently Barca has the best staff in League.So it would be interesting to watch the match between these 2 teams :D

Talking about soccer, you should've seen the Corinthians-Santos match for the Brazilian Cup the other day. There were tons of amazing plays, and the Corinthians scored a goal with 15 seconds left in extra time to win.

[quote="Erken"]So eventually Barcelona and Chelsea qualified for the semi-finals. I'd say their chances are about 50-50.
By the way the game between Liverpool - Chelsea was awesome, probably, the best match ever, well, at least it deserves to be among the most memorable matches of Champions League.[/quote]
I think Barca will own Chelsea easily!
I'm quite disappointed with Liverpool's defense on that "crazy" match.Steven Gerrald should be on that match, not in the audience.

[quote="ghjk"]
I think Barca will own Chelsea easily!
I'm quite disappointed with Liverpool's defense on that "crazy" match.Steven Gerrald should be on that match, not in the audience.[/quote]

I sincerely hope so.

btw, goalkeepers quite sucked that night as well, both Cech and Reina.

So finally, all semi-finalists are known. MU is not as impressive as it was last year, so I am more inclined to think that the winner of the couple Barcelona - Chelsea will win the CS cup.

[color=red]Manchester United [/color]WILL WIN THE CL

btw, barca 0-0 chelsea.

btw, Manchester United 1-0 Arsenal. hope we get the away goal against arsenal.

1-1 my prediction at the emirates

So the second day of the semi-finals is coming. I'd like to see MU and Barca in the final, and the Red Devils, in my opinion, are already in the final. But Barcelona will have to do a harder job tomorrow.

My prediction for today is $ 1\minus{}1$.

[quote="Erken"]So eventually Barcelona and Chelsea qualified for the semi-finals. I'd say their chances are about 50-50.
By the way the game between Liverpool - Chelsea was awesome, probably, the best match ever, well, at least it deserves to be among the most memorable matches of Champions League.[/quote]

I agree about Barcelona and Chelsea. The game tonight is pretty much unpredicatble, having in mind the key players that won't play for Barcelona (including the injured T. Henry) and Chelsea, with their new coach. :P

About Arsenal and Manchester, I believe that Manchester will qualify for the final, but they are not as good as they were in the beginning of the year, so I am not sure if they are going to be able to win CL.  :huh:

My personal favourite is Chelsea, but all I want is to watch some nice soccer, as most of us watched during the game Liverpool - Chelsea.  :)

Manchester won 1 - 3:) It was a nice game
I am looking forward to the match tomorrow, to see the other finalist :)

arsenal 1-3 manchester united aka CHAMPIONS OF ENGLAND, EUROPE, and WORLD

here we come rome!!!!!!!!!!!!!!!

[b]Can Barcelona find a way past a Premier League defence?[/b]

Barcelona may be dominant domestically, but defeating Premier League opposition in Europe is another matter

For a club with the St George's cross on its badge, in a city where Sant Jordi's day is celebrated with thousands of improvised stalls draped in the Catalan flag selling roses for her and books for him, FC Barcelona must be sick of the sight of the English. Last season the roses were yet to wilt when Barça's Champions League hopes withered at the hands of Manchester United; this season another English visitor to Camp Nou and another 0-0 draw threatens a familiar end.

When Gerard Piqué squeezed the ball past Real Madrid's Iker Casillas on Saturday night, it completed a 6-2 hammering that has been declared the greatest performance in the club's history. It also took Barcelona's total to a staggering 100 league goals. In all competitions their front three alone have 94. Barcelona have scored 146 times in 55 games and are the Champions League's top scorers, with 10 goals more than Chelsea.

The thing is, of course, that only one of their matches has been against an English team. Dominating domestically is one thing; defeating Premier League opposition another. Last Tuesday was the first time Barcelona had failed to score at home this season.

The last time they had been held in Camp Nou was a year earlier – against Manchester United at the same stage. A 0-0 draw then, followed by a 1-0 defeat at Old Trafford, put them out. Barcelona's bitter reaction to Chelsea's tactics spoke of frustration – the English again the spanner in Barça's works.

Barcelona have won only 20 of their 50 games against English clubs in European competition. In knockout ties they have emerged defeated as often as victorious and they have won only one of their last five European semi-finals against English teams – against Tottenham in the 1981-82 Cup Winners' Cup. United, Liverpool and Leeds all saw them off. United beat them in the 1991 Cup Winners' Cup final, Mark Hughes scoring the winner.

Recent Champions League campaigns have been more striking still. It is six years since Barcelona were knocked out by a non-English side, when an extra-time breakaway saw Juventus beat 10 men at Camp Nou. Last season they were knocked out by United, the season before by Liverpool and two seasons before that by Chelsea. In between they knocked out Chelsea and defeated Arsenal in the 2006 final but there were extenuating circumstances: Arsenal had their goalkeeper sent off and Chelsea, too, saw red. Barcelona have not beaten a Chelsea side with 11 men.

Barcelona's record over the last five games against English teams at Camp Nou reads: drawn three, lost two, won none. It is these last two draws that most stick in the throat, chiefly for the manner in which they were achieved.

Manchester United arrived feted as a devastating attacking force but spent 90 minutes defending, with Wayne Rooney employed as an auxiliary full-back. Chelsea came with Guus Hiddink promising attack but they parked the bus, the goalkeeper Petr Cech providing more "passes" than any of his team-mates.

The style is not unusual but its success is. Most teams have attempted to pack players behind the ball this season. "Chelsea did the same as we did at the Camp Nou," said the Real Madrid coach, Juande Ramos. But Barcelona eventually found a way through their defences, as they have against other sides who have sought to frustrate them.

They should have scored against Chelsea too, Samuel Eto'o, Bojan Krkic and Alexander Hleb missing excellent chances that would have put a different spin on the supposed tactical "genius" of Hiddink's approach.

But breaking down a team that comes to defend is difficult; breaking down an English team that has come to defend is harder still. They are physically stronger, better organised and faster. Beating them requires qualities that some Spanish sides do not have.

As Robert Pires admitted after Villarreal were defeated by Arsenal: "They played with an intensity La Liga teams cannot live with." Barcelona's deep commitment to a footballing philosophy is laudable but some criticise the lack of a plan B.

Chelsea assumed Barcelona's technical superiority but not the inevitability of defeat. Barcelona did not like it but it was licit – especially with European referees. The Spanish game permits far less contact and, as the Athletic Bilbao coach, Joaquin Caparros, says, it is causing teams problems beyond the country's borders. Daniel Alves described Chelsea as employing "excessive force" in the first leg, whereas John Terry said he saw "only one bad foul".

In Europe Barcelona have discovered that the referees are not the same as they are in Spain. And nor are the opponents, especially the ones from the other land of St George.

[i]- written by Sid Lowe[/i]

A fun trivia fact concerning the article Erken quoted: today's Saint George's day in the old (Ortodox) calendar  :lol:

And concerning the tonight's semifinal: I believe Barça has what it takes to get through. After the Saturday Real demolition at Bernabeu, they're confident, and Guardiola has probably prepared a tactical surprise for tonight. We'll just have to wait and see the final outcome (it's the match between Total Football (something like the Dutch in the past) and Destructive football (something like Italians in the past) )

Chelsea 1 - 1 Barcelona! It's not fair!  :(  The referee is not good enough. I am not satisfied with his decisions  :mad:  Anyway, congratulations, Barca!  :|

2day the ref was really, really bad. abidal shouldnt had a red card (what a dive by anelka!), maluda foul should have been a pk, not a fk, pique handball should be a pk (he even admitted!!!), and the last minute hand ball by etoo should have been a pk.

though the reaction of ballack and drogba to the ref was priceless.

Yep, you are right. We could understand Michael Ballack and Didier Drogba's disappointment and their reaction to the decisions of the ref. After Barcelona's performance today, I think that Manchester United will defend their Euro crown successful.

That was really some bad refereeing. I honestly believe that the last 20 minutes the referee was 'compensating' Abidal for Barça by taking their side in some dubious situations, but it still doesn't approve his actions (and what about first 70 minutes and all those dives in Barça's penalty box, wasn't at least one a penalty?). Two beautiful goals, and total domination of science (tactics) over art (technique)- but Barça goes to the finals. I was rooting for Barça last night, so I'm satisfied  :lol:

Probably tomorrow, despite the result, some people will say that Hiddink totally overplayed Guardiola, and only some crazy case prevented him from the triumph. Maybe they are right, maybe not. Maybe only this crazy case helps Essien to score an unbelievable goal in the very beginning of the match. Afterwards, the match went under the scenario of Hiddink, who knows what would've happened, if Barcelona opened the score in the beginning? There was a moment, when Xavi shoted from the distance, and the series of ricochets might've led to goal, who knows?
Then, Chelsea would need to play open football, and I am pretty sure - Barcelona would've just bashed them. And what would you say then?

That Guardiola is a genius? And that Barcelona is the best fotball team and that they play celestial football?

It is not about the bad refereeing , it's not about the tactics, it's not about who desreved it more. It is football, and the case plays a great role in it.

[quote="Erken"]no, guys, you are wrong.

Fristly, probably, we'll see Barcelona in the final, because they've been showing a great perfomance every time they were on a field, well maybe except for this match.

Yes, someone might argue that Barcelona didn't  deserve to tie the match, but that's not the case, as long as, they are the only team, in my opinion, who is still trying to play entertainment football.

And we'll see two strongest teams of the world in the final, and that's the only thing matters.[/quote]

Yes, I agree that Barca always performs great performance and indeed they are a strong team. But you see when they meet an English FC, they didn't (or can't?) show their strength. This is why I guess and believe that Man Utd will defend their Euro crown successful, not only because of the way Barca played last night.  :)

As you said, it's football, and we don't know what will happen. We will have to wait to see who is the new king of Europe.  :wink:

may the best one win, but where was messi or etoo at the 2 games (barc-chels)? i didn't know either of them was playing untill the game ended in standford bridge (i probably spelled that wrong), celebrating

[quote="hjoon0125"]so, any predictions who's going to win the final?

heres mine: Man Utd 2-1 Barca

its going to be an great match. both teams r best know for their goals. attack vs attack.[/quote]

are u kidding? There is no chance MU will play an attacking football. I expect somewhat a repetition of the last Camp Nou game. But since Barcelona is way stronger now, my prediction is MU 1 - 2 Barcelona. Besides, Barcelona has scored more goals than MU and managed to beat more serious opponents than MU did.

I hope Manchester United wins in a similar manner like they had won in 1999.

Hopes: Barca wins

Reality: MU ALWAYS scores in last minute (especially in Premier League), and I'm afraid that something similar will happen now...

Predict: Barca-MU 1:2

Reality is that MU last year played way better than now, while Barca was way weaker. Nevertheless, MU had to do they best to overcome Barcelona in the semi - final last year. So if Iniesta and Henry will play - I believe Barcelona will win. If english clubs played open attacking football, nobody would be able to confront the Barcelona.

im pretty sure iniesta and henry will play. no abidal nor alves 4 the final, but captain puyol's back so that doesnt hurt there chances.

imo, we need to play attacking football. big teams like barca know how to score even if the opponet puts 10 or 11 men behind the ball.

Barca 2-0 ManUtd

gah, now i look like a fool. barca were way better than manutd. deserved at the least well done. congratulations barcelona fans. they were just the better team. congratulations once again. that was not the manutd i expected.

also congratulations to erken and bugi. man i watched the game and none, yes NONE of our players showed up. though give credit to puyol, OUTSTANDING performance.

It takes a special team to make United look ordinary, let alone ugly, but Barcelona, a truly beautiful team, had done so and no one, not least Ferguson, denied the best side had won.


Find all primes $ p,q$ such that $ pq | 2^p \plus{} 2^q$.

If $ p \equal{} q$ then $ p^2 | 2^{p \plus{} 1}$, then $ p \equal{} q \equal{} 2$.

$ p \equal{} 2$ implies $ 2^{q} \plus{} 4 \equiv 0 (mod 2q) \Rightarrow 2^q \plus{} 4 \equiv 0 (mod q)$

$ \Leftrightarrow 2 \plus{} 4 \equiv 0 (mod q) \Leftrightarrow 6 \equiv 0 (mod q)$ and, since $ (q,2) \equal{} 1$, $ q \equal{} 3$.

If $ (p,2) \equal{} 1$ and $ (q,2) \equal{} 1$ then $ 2^p \plus{} 2^q \equiv 0 (mod pq) \Rightarrow 2 \plus{} 2^q \equiv 0 (mod p) \Rightarrow 2^q$

$ \equiv \minus{} 2 (mod p) \Leftrightarrow 2^{q \minus{} 1} \equiv \minus{} 1 (mod p)$     $ (1)$  and $ 2^{2(q \minus{} 1)} \equal{} 1 (mod p)$    $ (2)$.

Let $ K$ be the order of $ 2$ mód $ p$. By $ (1)$ $ K$ doesn't divide $ (q \minus{} 1)$, but by $ (2)$ $ K$ divides  $ 2(q \minus{} 1)$. Then the greatest

power of $ 2$ that divides $ K$ is the gratest power of $ 2$ that divides $ 2(q \minus{} 1)$. Let $ 2^{m \plus{} 1}$ be such power.Then $ 2^{m \plus{} 1}$

divides $ p \minus{} 1$ (because it divides the order and the order divides $ p \minus{} 1$).

Similarly, let $ 2^n$ be the greatest power that divides $ p \minus{} 1$, then $ 2^{n \plus{} 1}$ divides $ q \minus{} 1$.

Then $ 2^{n \plus{} 1} | 2^m$ and $ 2^{m \plus{} 1} | 2^n$. Contradiction, so there is not solution for $ p$ and $ q$ coprimes with $ 2$.


Let $ a,b,c$ be integers such that none is zero and $ abc$ divides both $ a^2b\plus{}b^2c\plus{}c^2a$ and $ ab^2\plus{}bc^2\plus{}ca^2$. Prove that $ |a|\equal{}|b|\equal{}|c|$.

We can resume that $ \gcd(a,b,c) \equal{} 1$
Now we will prove that $ \gcd(a,b) \equal{} 1$ ,if not there exist a prime p such that $ p^n\parallel{}\gcd(a,b)$
Therefore exist $ a$ or $ b$ such that $ p^n\parallel{}a$ or $ p^n\parallel{}b$
Suppose $ p^n\parallel{}a$( other case is similar )
Because $ ab|ac^2 \plus{} b^2c \plus{} a^2b$ so $ p^{2a}|ac^2 \plus{} b^2c \plus{} a^2b$
From condition we have $ p^{2n}|a^2b \plus{} b^2c$ therefore $ p^{2n}|c^2a$
Because $ p^n\parallel{}a$ so $ p|c$ (contradiction )
So $ \gcd(a,b) \equal{} gcd(b,c) \equal{} \gcd(c,a) \equal{} 1$
But from condition then $ a|b^2c$ so $ |a| \equal{} 1$
It is similar to prove that $ |b| \equal{} |c| \equal{} 1$
Problem claim.


calculus
integration
function
topology
calculus computations

Suppose that $\vec{F}(\vec{r})$ is continue in a connected and open set $D$. Then line integral $\int_{C} \vec{F}(\vec{r}) . d \vec{r}$ independent of path If and only if $\vec{F}(\vec{r}) = \nabla f(\vec{r})$ (conservative) .
any proof?

Say the integral is independent of path.  You need to define a function $\phi$ such that $F=\nabla \phi$.

The idea is that we pick some $a\in D$ and define

$\phi(x)=\int^x_a F\cdot dr$

It remains to show that $\phi$ has the property we want of it.  Restrict consideration to a path along the $x-$axis and show that the $x-$component of $F$ is $\frac{\partial\phi}{\partial x}$.  You'll need the mean value theorem there.  Proceeding similarly, you get $F=\nabla\phi$.

Now for the other direction.  Say that $F=\nabla\phi$.  Since $C$ is smooth, we can parametrize it with respect to $t$, using the relations $dx=\frac {dx}{dt}dt$ via the chain rule and so on.  Hence you should get that

$\int_C F\cdot dr=\int_C \frac {d\phi}{dt}dt=\phi(b)-\phi(a)$, where $a,b$ are the endpoints of $C$.  But this is just independence of path.


Determine the number of ordered pairs $ (a,b)$ of integers such that $ \log_a b \plus{} 6\log_b a\equal{}5, 2 \leq a \leq 2005$, and $ 2 \leq b \leq 2005$.

Why is the below method incorrect?

[b]Incorrect Method:[/b]
$ \log_a b\equal{}\frac{1}{\log_b a}$
Substituting $ x\equal{}\log_a b$, we have $ x \plus{} 6\times\frac{1}{x}\equal{}5$. Multiplying both sides by $ x$ and factoring, we get $ (x\minus{}6)(x\plus{}1)\equal{}0$ so either

$ 1) x\equal{}\log_a b\equal{}6$
So these ordered pairs are of the form $ (a,a^6)$.

$ 1) x\equal{}\log_a b\equal{}\minus{}1$
This doesn't yield integer solutions within the given range.

Plugging in 2, 3, and 4, we find that 2 and 3 works so our answer is 2.

Except not...

when you multiply through by x, you should get $ x^2 \minus{} 5x \plus{} 6 \equal{} 0$, not $ (x\minus{}6)(x\plus{}1) \equal{} x^2 \minus{} 5x \minus{} 6 \equal{} 0$.

Anyways to finish off what max said.

It would be a^3=b and a^2=b,

since the solutions are 2 and 3.

For a^3=b, we find that a can any integer from 2 to 12 inclusive
For a^2=b, we find that a can be any integer from 2 to 44 inclusive

It doesnt matter what b is because there will be no reapeating (a,b) since a and b are greater than 1.

So first one has 11, second has 43, 11+43=054


[quote="jin"]Maybe $x^2+xz+z^2=16$?[/quote]
    oh you are right,I am very sorry,I will edit it

[quote="quangtrung"]find $xy+2yz+3zx$ given that
                 $x^{2}+xy+ \frac{y^{2}}{3}=25$
                 $\frac{y^{2}}{3}+ z^{2}=9$
                 $x^{2} +xz+ z^{2}=16$[/quote]

It is an old USSR math olympiad problem.
You can use geometrical interpretation.It is easy.

$3^2=x^2+(\frac{y}{\sqrt3})^2+2x(\frac{y}{\sqrt3})sin{\frac{5\pi}{6}}$
$4^2=z^2+(\frac{y}{\sqrt3})^2$
$5^2=x^2+z^2+2xzsin{\frac{2\pi}{3}}$
And $\frac{5\pi}{6}+\frac{2\pi}{3}+\frac{\pi}{2}=2\pi$
So we can found a triangle ABC,in which contains a point O.And $OC=x,OB=\frac{y}{\sqrt3},OA=z,\angle{AOB}=\frac{\pi}{2},\angle{AOC}=\frac{2\pi}{3}$,then we can have:$AB=3,AC=4,BC=5$,so $\angle{BAC}=\frac{\pi}{2}$
For one thing, $S_{\triangle{ABC}}=6$
For another thing,$S_{\triangle{ABC}}=\frac{1}{4\sqrt3}(xy+2yz+3xz)$
so $xy+2yz+3xz=24\sqrt3$