vector
algebra
polynomial
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

I'm struggling with the topic of Tensor Products.  I would appreciate some help on these questions:

Let k/K = k(a) be a purely inseparable field extension, let μ_a be the k-linear map on K given by μ_a(x) = ax, and let f = id_K@μ_a (tensor product) be the induced K-linear map on the K-vector space V = K@K (tensor over k). Show that f which is in End_K(V) has a Jordan canonical form and determine this canonical form.

Let R = k[x, y] be a polynomial ring over a field k and I = (x, y). Find a free R-module F of finite rank and a submodule U of F so that I@I is isomorphic to F/U, where @ is tensor product over R.

1) let $ p$ be the characteristic and $ x^{p^n} \minus{} b$ the minimal polynomial of $ a$ over $ k$. Then we have canonical isomorphisms

$ K \equal{} k(a) \equal{} k[x]/( x^{p^n}\minus{} b )$

$ K \otimes_k K \equal{} K[x]/(x^{p^n} \minus{} b) \equal{} K[x]/((x \minus{} a)^{p^n})$

then $ f$ acts as multiplication with $ x$, and we get a simple jordan block of size $ p^n$ with eigenvalue $ a$.

2) first observe that $ R^2 \to I, (a,b) \to ax \minus{} by$ is an epimorphism whose kernel is generated by $ (y, x)$ as $ R$-module. thus $ I \cong R^2 / \langle (y, x) \rangle$. now, using the general isomorphism $ F/U \otimes G/V \equal{} (F \otimes G)/(U \otimes G \plus{} F \otimes V)$, we get $ I \otimes_R I \cong (R^2 \otimes_R R^2) / \langle R^2 \otimes \{(y, x)\} \cup \{(y, x)\} \otimes R^2 \rangle \cong R^4 / \langle (y,0, x,0),(0,y,0, x),(y, x,0,0),(0,0,y,x) \rangle$.

What is the rank of $ \langle(y,0,x,0),(0,y,0,x),(y,x,0,0),(0,0,y,x)\rangle$ over $ k[x,y]$?

the rank is $ 4$ (since it's bounded by the rank of $ R^4$, and these $ 4$ generators are linearly independant).


geometry
parallelogram
circumcircle
geometric transformation
reflection
geometry proposed
U sviti matematyky

Point $ M$ is chosen at the diagonal $ BD$ of parallelogram $ ABCD.$ The straight line $ AM$ intersects the side $ CD$ and the straight line $ BC$ at points $ K$ and $ N$ respectively. Let $ \omega_1$ be the circle with centre $ M$ and radius $ MA$ and $ \omega_2$ be the circumcircle of triangle $ KNC.$ Denote by $ P$ and $ Q$ the intersection points of circles $ \omega_1$ and $ \omega_2.$ Prove that the circle $ \omega_2$ is inscribed into the angle $ QMP.$

Let $ O$ be the center of the parallelogram and $ A'$ the reflection of $ A$ about $ M.$ Since $ O$ is midpoint of $ AC$ $ \Longrightarrow$ $ DB \parallel CA'.$ Let $ T_{\infty}$ be the infinity point of this direction. Since $ (D,B,O,T_{\infty}) \equal{} \minus{} 1,$ it follows that the pencil $ C(D,B,O,A')$ is harmonic $\Longrightarrow$ $ (K,N,A,A')=-1.$ By Newton's theorem, we have then  $ MA^2 \equal{} MN \cdot MK$ $ \Longrightarrow$ Circle $ \omega_1$ centered at $ M$ with radius $ MA$ is orthogonal to $ \omega_2.$


Calculate the sodium ion concentration when 70.0 mL of 3.0 M sodium carbonate is added to 30.0 mL of 1.0 M sodum bicarbonate.

I have no clue how to work this problem.  Please help me again.

The total amount of solution after you've mixed the solutions is 100mL.  Now, you need to find how many moles of sodium ion are in the solutions.  For the sodium carbonate, you have 3.0 M and 70 mL, meaning 3M *0.07 L= 0.21 mol of the compound, however, by stoich, there are 2 sodium atoms per one molecule of sodium carbonate, so the total amount of moles of Na contributed by sodium carbonate is 0.21 * 2 = 0.42 mol.  The other compound... 1.0M sodium bicarbonate you have 30 mL of. 1.0M * 0.03 L = 0.03 mol.  Since the ratio of sodium ion to sodium bicarbonate is 1:1 in this compound, this compound contributes a total of 0.03 mol Na to the solution.  Now, you add the two up so you have a total of 0.45 mol of Na in the solution.  You know you have 100 mL of solution total... so molarity = 0.45 mol / 0.1 L = 4.5M

yea just write out the compounds for both, so you have:

$Na_2(CO_3)$ and $NaHCO_3$

hence you have 0.42 mol + 0.03 mol = 0.45 mol of $Na^{+}$ total, dividing by $0.1 L$ gives $4.5$ molarity


$1989$ equal circles are arbitrarily placed on the table without overlap. What is the least number of colors are needed such that all the circles can be painted with any two tangential circles colored differently.

This is a very old problem of Paul Erdos. There are two more spin-off questions of it;

1. Same for an infinite number of equal circles in the plane;
[color=blue]Both the finite and infinite problems have answer $ 4$.[/color]
Hint. [hide]Pick a point $ P$ in the plane, such that its distances to the centers of the circles are all distinct (there exists such a point, since the perpendicular bisectors of the segments determined by the centers are at most countable), and order the circles by the distance of their centres from $ P$. Show that there always is a fourth color available for the next ...[/hide]
2. Find the configuration with the least circles that requires $ 4$ colors.

Contruction for 4 colours is :https://youtu.be/-3-t9NQcevQ i think it can be found here .Next we show by induction that we need at most 4 colours .We got some contruction with n+1 circles and we know that we can colour any number of circles up to n with 4 colours.Then make coordinate system and take circle with highest y coordinate and if some of them have same y coordinates take one with lowest x coordinate.Then that circle is tangent to at most 3 circles(can be proved ) and take by inductive step we can colour all other n circles and that circle can have fourth colour .


modular arithmetic
number theory
relatively prime

If you write out what $ a\equiv b\pmod{c}$ actually means, you should have an easier time doing it.  Also, recall that if $ a\mid bc$ and $ a$ and $ b$ are relatively prime, then $ a\mid c$.

[quote="veezbo"]Prove that $ ak \equal{} bk \pmod{p}$ implies $ a \equal{} b \pmod{p}$ for all positive integer values of $ k$ iff $ p$ is a prime or $ k$ is not a factor of $ p$.[/quote]

I think it should be "iff $ gcd(k,p)\equal{}1$" for example if $ p$ is a prime number and $ k\equal{}2p$ then $ ak\equiv bk\pmod{p}\implies 2pa\equiv 2pb \pmod{p}\implies 0\equiv 0\pmod{p}$ regardless of what $ a$ and $ b$ are.

Solution:

$ \dfrac{ak}{k} \equiv \dfrac{bk}{k}\pmod{\dfrac{p}{gcd(k,p)}}$

$ a\equiv b\pmod{p}$ iff $ gcd(k,p)\equal{}1$.

[hide="Another Solution"]
Using the definition of a congruence, we have, for an integer $ j$,

$ \frac {ak \minus{} bk}{p} \equal{} j \implies \frac {k(a \minus{} b)}{p} \equal{} j.$

Because we are given that $ p$ is prime or $ k$ is not a factor of $ p$, we must have $ p | (a \minus{} b)$, which implies that $ a \equiv b \pmod{p}$.

Now we must prove that converse. If $ a \equiv b \pmod{p}$, we know that $ \frac {a \minus{} b}{p} \equal{} j$ for some integer $ j$. Multiplying both sides by $ k$ still yields an integer, so we have

$ \frac {ak \minus{} bk}{p} \equal{} kj \implies ak \equiv bk \pmod{p}$. $ \Box$
[/hide]

Just to comment, sunehra's solution the one I was hoping the OP would discover based on the hint I gave in the first post.


If $ a,b,c \ge 0$ then prove

$ \frac{a}{1\plus{}a\plus{}ab}\plus{}\frac{b}{1\plus{}b\plus{}bc}\plus{}\frac{a}{1\plus{}c\plus{}ca} \le 1$

the inequality is equivalent to:

\begin{eqnarray*}a(1 + b + bc)(1 + c + ca) + b(1 + c + ca)(1 + a + ab) + c(1 + a + ab)(1 + b + bc)\leq (1 + a + ab)(1 + b + bc)(1 + c + ca)\end{eqnarray*}

now expand,and simplify the equal terms,then we are left with:

$ 2abc\leq 1 + a^2b^2c^2$

which is equivalent to:

$ (1 - abc)^2\geq 0$

Other problem:
Let $ a;b;c$ be positive real numbers satisfying $ 0 <a;b;c \leq 1$
Prove that
$ \frac{a}{1\plus{}a\plus{}ab}\plus{}\frac{b}{1\plus{}b\plus{}bc}\plus{}\frac{1}{1\plus{}c\plus{}ca}\plus{}(1\minus{}a)(1\minus{}b)(1\minus{}c) \leq 1$


Let A be a subset of S = {1, 2, 3, ..., 100} such that |A| = 55.
Prove that A contains two numbers a and b such that |a - b| = 9.

[color=cyan]Okay, this is only semi-rigorous.

[size=125]<Warning, Solution Follows> [/size]

Let us assume the contrary, and attempt to construct a solution.  Every time we pick a number for our list, it makes two others unavailable.  However, each of those made unavailable can also be made unavailable by one other point.  So all of our points make, in the best case scenario (in which every point made unavailable is also made unavailable by a different point) 1 other point unavailable.  There is an exception, however:  the points 1 through 9 and 92 through 100 only make 1/2 of a point unavailable (in the best-case scenario).  So, at best our 55 points will make
18(1/2) + (55-18) = 46 points unavailable.  However, 55 + 46 > 100, so there is at least 1 point which we have chosen that is also unavailable.  In this case, it is a distance of 9 from some other point, and we are all done.[/color]

This is my own solution :

Let a1 < a2 < ... < a55 be the 55 elements of A.
Consider the 55 numbers bk =  ak + 9. Then b55 :le: 109.
So we have 110 numbers ak, bk, all in [1,109].
By the Pigeonhole principle, two of them must be equal.
This means that there exist indices i and j such that bi = aj.
(We can't have two equal a's or b's - this is obvious.)
In that case indeed |ai - aj| = 9. So we're done.

But your solution seems to be OK also. There are even some other solutions.

Too late  :evil:

My solution looks very much like Arne's.


A row of 8 light switches controls different banks of lights in a school gymnasium. In preparation for a school dance, students decide that they want either one or two banks of lights on during the dance. How many possible ways are there of setting these 8 switches so that either one or two banks of lights are on?

In this situation, the answer is 8 + 8 C 2, or 8 + 28, which is 36.


New version of the site began to work.
http://olympiad.sanjesh.org/en/index.asp
There is a notice that Chairman of 2008 Jury will be Professor Nezam Mahdavi-Amiri from Sharif University (who was the Chairman at 2007 Olympiad also).
The 2008 Olympiad will be in Tehran, Iran in July.

The 2008 Olympiad will be during 15-18 July.
The days of competition are 16 and 17 July.
This year olympiad in math will be firstly accompanied with separate olympiad in chemistry.

[quote="Imp"]Is it possible to take part in competition if you are below third year of study?[/quote]

Well, you should better ask organizers.
But I think if Ministry of Education or Mathematical Society in your country agrees to send you then no problem should be.

P.S. We hope to see team of Poland. :wink:
Professor Michał Krych from Warsaw University told me at IMC-2007 that in that year there were some troubles with sending a team to Iran.


A series with sum of first n terms S(n) is such that S(n) = n^3. Compute the 10th term of the series.

[hide="Solution"]Let $ a_{n}$ denote the $ n$-th term of the sequence. Then $ a_{n}=S(n+1)-S(n)=(n+1)^{3}-n^{3}=3n^{2}+3n+1$
$ \Longrightarrow a_{10}=3\cdot 10^{2}+3\cdot 10+1=331$.[/hide]

yes you are right. my stupidity :oops: $ a_{n}=S(n)-S(n-1)\Longrightarrow a_{10}=271$.


Let f be analytic and $ f'(z) \neq 0$ on a region A. Let $ z_0 \in A$ ans assume that $ f(z_0) \neq 0$. Given $ \epsilon > 0$, show that there exists a $ z \in A$ and a $ \xi\in A$ such that $ |z\minus{}z_0| < \epsilon, |\xi\minus{} z_0| < \epsilon$ and
$ |f(z)| > |f(z_0)|$ and $ |f(\xi)|<|f(z_0)|$

the existence of such a $ z$ is just restating the conclusion of the maximum modulus theorem, isn't it? similarly, the existence of $ \xi$ is restating the conclusion of the minimum modulus theorem, which is the maximum principle applied to $ \frac{1}{f}$ (restrict $ A$ to be some disc about $ z_0$ on which $ f$ does not vanish to ensure $ \frac{1}{f}$ stays analytic rather than developing some random poles)


Hello Canada forum,

I am a highschool student planning on writing the Euclid exam for the University of Waterloo in April 19. Unlike a lot of the members on this board, I have literally no experience in competitions such as the Euclid. As a prospective math student at UW, this test is important to me in terms of scholarships, etc. Any suggestions as to what would be useful to me in terms of preparing for the exam will be appreciated. Thus far, I've been working through some books published by the University, namely "Mathematics Preparation Manual for Post Secondary Studies", and "Problems, Problems, Problems Volume 8". What worries me is that there are some questions in the books (namely the former), that involve concepts that I have learned some time ago and are somewhat hazy. With only roughly a month left, are there any things I should be looking at in general? Any tips for an inexperienced problem solver? Thanks.

While I doubt I could do it, some people at this bored could : get a hold of aops vol 1, and push through all 500 problems in the month before.

Otherwise, do Old Euclids, don't peak at solutions untill you have done at least a half hour of 'trying' ( not staring at the page ).

Yes.  Try the [url=http://cemc.uwaterloo.ca/english/contests/euclid_prep/index.shtml]Euclid E-Workshop[/url] here.

i am sorry for saying that :blush:  :blush:
btw, the "problems, problems, problems' books are really good, and not just volume 8.
Another important thing is timing. You need to make an outline of how much you'll spend on each problem.

ok, I ordered the problem solving book online. It should be here in a week or so. Until then, I guess I'll look over old Euclid exams and such. Thanks for the suggestions

Dude do you read Camus too? You should try COMC 2004 Part B #4 and tell me why you think they named the sequence that way. It's been bothering me for months!

I guess my intentions for writing the thing aren't purely for scholarships though. I mean, scholarships are definitely an asset when it comes to going into university and all in financial terms, but my understanding is that your Euclid score does has some bearing on whether or not you will be able to take certain advanced math courses in your first year, which is something that I am definitely interested in. On top of that of course, there is personal satisfaction and so on, which is normally assumed.

After some thinking though, I think that my chances of doing well on this thing are pretty slim =\

Edit:  and yes, I'm a pretty avid reader of Camus  :)

HEY!
Spend a little time each day preparing and you may surprise yourself  :)
You dont need to be a grade 9 euclid winner, or whatever he is, to be good at math.
Im sure if you practice the right things, come contest day, youll get a great score!  :lol:
Good luck!

it all depends on how ur definition of good is. i dunno how ur level of problem solving is like right now, but try doing a past contest right now and see where u stand. then just practice practice practice and u'll be surprised on how much u can improve! it's quite easy to improve from like a 50 to 70 on euclid if u got a sound basis in a few months. just don't expect to jump from like 50 to winning euclid. but just practice, and u'll be fine.

I'd be pretty content with a 70, I think. I've tried some of the Euclid problems on the website, and the results kind of worry me. It's not that I get questions wrong, but that when I look at a question, I either know it for sure (usually the case with "lightbulb" problems), or I have no idea as to how to even get started (the long answer problems give me the most grief in this regard)! I think that this has got to do with the fact that I am pretty unfamiliar with problem solving strategies, and that I have forgotten certain concepts that have been taught to me in my previous years. I don't know though; I mean, since I'm writing the test either way, I might as well make an attempt at doing well on it.

definitly review the topics though. as posted above, go thru the euclid workshop. they would help with the concepts. i never did well in euclid, although thats the one i actually cared about. i've been writing it from gr 10 to gr 12 but no accomplishment/increase in marks from grade to grade. one advice, don't expect too high, or u'll get nervous. the first 7 questions r straight forward, as long u get those right and do #8a and u'll be fine. definitly go through the waterloo books and past contests. good luck!

Argh, I'm working through the workshop, and I'm having quite a bit of trouble with some of the problems, which worries me quite a bit. Anyhow, when I'm working through this, should I go question-by-question checking my answers, or try and complete as much of it as possible and then go back to see what I did wrong?

i would check answer by answer. thats just my personal preference, cause i tend to forget the small details.

[quote="haunter"]WTF?
Some guy wants help with getting a strong Euclid score so he can get a scholarship.
Why are you using this as an opportunity to show off about winning Euclid? :huh:
You got a lot to learn kid.[/quote]

You should make "Kid you got a lot to learn" your signature on the forum  :D . Because it's true,  we all have a lot  to learn. In math and in life.

[quote="J-I-M-B-O"] one advice, don't expect too high, or u'll get nervous.[/quote]

Yup, this is exactly how it works for me - when I wrote Fermat my only aspiration was to get to the Waterloo Seminar, which didn't require a particularly high score, so I relaxed and did pretty well. If I was aiming, for example, to win the Fermat, I would've spent half the time nervously counting up the remaining problems and reminding myself of my great mission. Some people say that nervousness helps them concentrate, but in my case it fills my mind as opposed to letting the problem fill my mind, and I can't just relax and explore the problem without a shuddered glance at the watch every five seconds. Even on a speed contest relaxation is often necessary.

Thankfully, anxiety comes [i]after[/i] writing tests for me. When I'm writing tests, I can usually focus pretty well. But who knows, tests usually don't mean this much to me, and this is my first math contest, so I might find myself in cold sweat come exam time.

don't worry, just remember, contests r extra to that of ur school work, unless ur olympian or working towards in becoming an olympian.  :)

[quote="Sisyphus"]With the contest coming up in less than a week, I cannot believe how underprepared I am.  :D

(I am screwed  =\ )

ahh crap  :([/quote]

Aww no worries.  Relax, and Good luck! To everyone!

Thanks for the encouragement.

I didn't mean to come off as if I were fishing for support though, just coming to terms with hard facts.   :|


Given a triangle $ ABC$ with $ BC\equal{}a,\ CA\equal{}b,\ AB\equal{}c$ such that $ \frac{\tan A\minus{}\tan B}{\tan A\plus{}\tan B}\equal{}\frac{b\plus{}c}{c}$.

Find the angle of $ A$.

$ \frac{\tan A\minus{}\tan B}{\tan A\plus{}\tan B} \equal{}\frac{b\plus{}c}{c}$

$ \Longrightarrow\frac{\sin A \cos B\minus{}\cos A \sin B}{\sin A \cos B\plus{}\cos A \sin B}\equal{}\frac{\sin B\plus{}\sin C}{\sin C}$

$ \sin (A\minus{}B)\equal{}\sin B\plus{} \sin C$

$ \sin (A\minus{}B)\equal{}\sin B\plus{}\sin (A\plus{}B)$

$ \sin (A\minus{}B)\minus{}\sin (A\plus{}B)\equal{}\sin B$

$ \minus{}2 \cos A \sin B\equal{}\sin B$

$ \sin B(1\plus{}2\cos A)\equal{}0$

$ \cos A\equal{}\frac{\minus{}1}{2}$

$ \boxed{A\equal{}\frac{2 \pi}{3}}$


8th
yes
4-6
i come here often....you should see the first 3 pages.(pg. 4, 3, 2) :D

[quote="sirorange"]Just so I get to know who visits this forum, I'd like you to post three pieces of information.

What grade are you in?

Do you compete in MOEMS?

If so, what division(s) do you compete in? 4-6, 7-8 or both.

If you don't feel comfortable answering any, I understand, it would just give me the oppurtunity to post more problems suitable for what you guys are doing.[/quote]

7th
Yes I do since 5th grade
7-8....no choice
Post harder MOEMS problems please!

Well, todays problem of the day is from a 7-8 competition.

Grade: 8th
MOEMS? No.  :(

I'll try to leave the problems to other students first, and then help out if needed :P

Grade: 6th
I did MOEMS since I was in 4th grade.
Now, though, I have to do 7-8th :(

I think they could be a little harder.

Grade:8th, I do MOEMS, i compete in the 7-8th grade division....

Grade 8.

Moems: 7-8th division, also did it in 6th grade, but not 7th

8th
MOEMS No

Does anybody mind explaning MOEMS or telling me where I can get info about it?
Other than knowing that it is another math comp. I have no idea what it is

Nevermind....I found the topic that talked about the  MOEMS compitition


Solve the equation
$\cos(\cos(\cos(\cos(x))))=\sin(\sin(\sin(\sin(x))))$
Could someone check my solution?
[hide="solution"]$\cos(\theta)=\sin(\theta)$ iff $\theta$ is $\pi/4$ or $5\pi/4$ so $\cos(\cos(\cos(x)))=\sin(\sin(\sin(x)))=\pi/4,5\pi/4$(*) but cos and sin of something is always less than 1 so the expression is equal to $\pi/4$.
By the same reasoning, $\cos(\cos(x))=\sin(\sin(x))=\pi/4$
substituting this into (*) we get that $\cos(\cos(\cos(x)))=\sin(\sin(\sin(x)))=\pm\frac{\sqrt{2}}{2}\neq \pi/4$ so we get a contradiction and therfore no solutions.[/hide]
edited to fix mistakes

[quote="junggi"]Solve the equation
$\cos(\cos(\cos(\cos(x))))=\sin(\sin(\sin(\sin(x))))$
Could someone check my solution?
[hide="solution"]$\cos(\theta)=\sin(\theta)$ iff $\theta$ is $45^\circ$ or $225^\circ$ so $\cos(\cos(\cos(x)))=\sin(\sin(\sin(x)))=45^\circ,225^\circ$(*)
By the same reasoning, $\cos(\cos(x))=\sin(\sin(x))=45^\circ,225^\circ$
substituting this into (*) we get that $\cos(\cos(\cos(x)))=\sin(\sin(\sin(x)))=\pm\frac{\sqrt{2}}{2}\neq45^\circ,225^\circ$ so we get a contradiction and therfore no solutions.[/hide][/quote]

I think you meant make use of the fact that $sin \theta=cos (\frac{\pi}{2}-\theta)$. And plus the values of $sin \theta$ or $cos \theta$ can never exceed $1$. So I think you might have made some errors.

[quote="Franz Joseph"]I think you meant make use of the fact that $sin \theta=cos (\frac{\pi}{2}-\theta)$[/quote]
well I meant that the graph of cos and sin intersects only when theta is 45 or 225...
[quote]And plus the values of $sin \theta$ or $cos \theta$ can never exceed $1$.[/quote]
er...in that case, would it help if I just replace the degrees with radians for the 45 degree case and drop the 225 degree case?

It doesn't matter if you use degrees or radians the thing is that the value of $sin \theta$ and $cos \theta$ is always between $-1$ and $1$. During one stage of your proof you had reasoning,
$cos (cos (x))=sin (sin(x))= 45^\circ, 225^\circ$. Which is impossible.

[quote="junggi"]:maybe: [hide="how about this"]
since $\cos(\theta)=\sin(\theta)$ only when they are $\pi/4,5\pi/4$,
$\cos(\cos(\cos(x)))=\sin(\sin(\sin(x)))=\pi/4,5\pi/4$
but since sin or cos of something is <1, we can drop the $5\pi/4$ case.
similarly, $\cos(\cos(x))=\sin(\sin(x))=\pi/4$ which leads to a contradiction... :maybe: [/hide][/quote]

Okay I am sorry if I didn't make myself clear when I mentioned the fact that they must be smaller than or equal to one. $sin \theta$ and $cos \theta$ are equal when $\theta= 45^\circ. 225^\circ.........$ etc. What I meant by smaller than or equal to $1$ is that $sin \theta$ and $cos \theta$ must be smaller than or equal to one for any value of $\theta$.

Then doesnt the problem lead to a sort of obvious contradiction(I'm not saying that you're wrong)?
the stuff inside the first cos and sin are smaller than 1 but the cos and sin are equal only when they are greater than 1....actually $45^\circ$ is less then 1 when written as radians...and I should have written all the angle measures in radians from the beginning since its cos inside of cos...

graaahh maybe i'm just not thinking straight...its 2 am here o.o

[quote="junggi"]Solve the equation
$\cos(\cos(\cos(\cos(x))))=\sin(\sin(\sin(\sin(x))))$
Could someone check my solution?
[hide="solution"]$\cos(\theta)=\sin(\theta)$ iff $\theta$ is $45^\circ$ or $225^\circ$ so $\cos(\cos(\cos(x)))=\sin(\sin(\sin(x)))=45^\circ,225^\circ$(*)
By the same reasoning, $\cos(\cos(x))=\sin(\sin(x))=45^\circ,225^\circ$
substituting this into (*) we get that $\cos(\cos(\cos(x)))=\sin(\sin(\sin(x)))=\pm\frac{\sqrt{2}}{2}\neq45^\circ,225^\circ$ so we get a contradiction and therfore no solutions.[/hide][/quote]
Oops, sorry I think I see what you are doing. :blush:

does that mean my solution is sort of ok, if I fix the some of the mistakes?

Posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105367]here[/url] and solved [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=91594]here[/url].


geometry
geometric transformation
invariant
function
abstract algebra
algebra
polynomial

Fix a bounded sequence {$ k_n$}$ _{n\equal{}\minus{}\infty}^{\infty}$ of complex numbers, and define an operator $ Tf$ by
 $ Tf(x)\sim \sum_{n\equal{}\minus{}\infty}^{\infty}$ $ k_n a_n$ $ e^{inx}$ whenever $ f(x)\sim \sum_{n\equal{}\minus{}\infty}^{\infty}$ $ a_n e^{inx}$.
Such an operator is called a Fourier multiplier operator.

Prove that if $ T$ is any bounded operator on $ \L^2 ([\minus{}\pi,\pi])$ that commutes with translations (that is , if we define $ f_h (x)\equal{}f(x\minus{}h)$ then $ T(f_h)\equal{}f_h (T)$), then $ T$ is a Fourier multiplier operator.

Let $ T$ be such a translation invariant linear operator, and consider its action on the function $ f(x)\equal{}e^{inx}$ for some $ n\in\mathbb{Z}.$ Suppose $ Tf\equal{}g.$

Use the notation $ f_y(x)\equal{}f(x\minus{}y)$ for translation. We are told that $ (Tf)_y\equal{}T(f_y).$ But for this particular $ f,$ $ f_y(x)\equal{}e^{inx}e^{\minus{}iny}\equal{}e^{\minus{}iny}f(x).$ Since $ T$ is linear, we have that $ g_y\equal{}(Tf)_y\equal{}T(e^{\minus{}iny}f)\equal{}e^{\minus{}iny}Tf\equal{}e^{\minus{}iny}g.$

So, $ g$ must be a function that satisfies $ g(x\minus{}y)\equal{}e^{\minus{}iny}g(x)$ for every $ y$ and almost every $ x.$ I'll leave the details to you, but such a function must be equal (almost everywhere) to $ g(x)\equal{}ke^{inx}$ for some constant $ k.$

This argument has made essential use of the fact that $ e^{inx}$ is a [i]character[/i] of the group $ \mathbb{T}\equal{}\mathbb{R}/2\pi\mathbb{Z}.$ A character of a locally compact abelian group is a continuous homomorphism of the group into the group of complex numbers of modulus $ 1.$

So for each $ n,$ we have that $ T(e^{\minus{}inx})\equal{}k_ne^{inx}$ for some constant $ k_n$ that depends on $ n.$ Hence if $ f$ is a trigonometric polynomial (a finite linear combination of function of the form $ e^{inx},$ then $ T$ acts as a multiplier (as in the the equation in Byun's post). The trigonometric polynomials are dense in $ L^2,$ which we will eventually use.

Just as an aside, at this point I'd prefer to write the $ L^2$ norm as
\[ \|f\|_2\equal{}\left[\frac1{2\pi}\int_{\minus{}\pi}^{\pi}|f|^2\right]^{\frac12}\]
With the $ 2\pi$ placed there, the group $ \mathbb{T}$ is a probability space and the functions $ e^{inx}$ form an orthonormal set.

Let $ f$ be a trigonometric polynomial: $ f(x)\equal{}\sum_na_ne^{inx}$ where that is a finite sum. Then we know that

$ \|f\|_2^2\equal{}\sum_n|a_n|^2.$

Then, $ Tf(x)\equal{}\sum_nk_na_ne^{inx}$ and $ \|Tf\|_2^2\equal{}\sum_n|k_na_n|^2.$

The inequality $ \|Tf\|_2\le\sup_n|k_n|\,\|f\|_2$ should be clear from that, as well as the fact that for any $ n$, I can make $ \|Tf\|_n\ge |k_n|\,\|f\|_2$ just by choosing $ f(x)\equal{}e^{inx}.$

A famous example: Define the operator $ H$ as a multiplier operator by setting its sequence of multipliers equal to $ k_n\equal{}\minus{}i\,\text{sgn}(n).$ That is, $ k_n\equal{}\minus{}i$ for $ n>0,$ $ k_n\equal{}i$ for $ n<0,$ and $ k_0\equal{}0.$

Can you find other characterizations of this operator?


[quote="iamhe"]There is a sphere inscribed in a frustrum. The sphere has radius 3 and the frustrum has slant height of 10. What is the volume of the frustrum?[/quote]Hard to visualize, but since everything is very symmetrical the problem can be reduced to plane (and very plain) geometry by considering a section sliced along the altitude of the cone. After that, sketching on paper that section, we see a bunch of nice similar right triangles.[hide]The height, slant height, and base radius of the cone involved form the largest of those triangles. The intersection of the sphere and the cone is a circle parallel to the base of the cone, but in the section sliced before they are parallel lines, both of them perpendicular to the altitude of the cone. Connect the center of the sphere to the intersection with the cone and you have another right triangle. Draw a line passing through that intersection and parallel to the altitude (call it vertical), and see the right triangle it forms with the base and the slant height of the cone. Another similar triangle is similarly formed using the slant height of the frustrum, its base and the vertical line passing through the top edge of the frustrum. We know two sides of that right triangle, and we can calculate the third one. Since all those right triangles are similar, we can calculate the height of the whole cone the frustrum came from, the radius of the base of the cone and the radius of the top of the frustrum.[/hide]


Hello.
Does anyone have any links to problems involing matrices on a level above solving intersections of planes using them,
Thx.

I couldn't directly help you but it sounds like you would want to look into Linear Algebra.  You can look in the Linear Algebra forum, search on google for it, or try and find a good textbook on it.

Regards

I would like to recommend the Linear Algebra book by Serge Lang.


Putnam
AMC
USA(J)MO
USAMO
college contests

With the Putnam coming in a week and a half, what is your personal goal? Here's a table showing how the scoring has looked in recent years. The column marked $ N$ is the number of competitors. The columns marked "top 100", "top 200" and "top 500" are the minimum score needed to appear in those parts of the published list of accomplishments. (Bear in mind that the "top 500" won't be exactly 500 - in might be 479, it might be 513, wherever it breaks.) The column labeled "median" is just that, the median score, and the column labeled "% zero" is the percentage of all competitors who scored zero points.

[hide="Updated below"]$ \begin{array}{ccccccc}\text{Year}&N&\text{top }100&\text{top }200&\text{top 500}&\text{median}&\%\text{ zero}\\1998&2573&58&42&29&10&30\%\\ 1999&2900&38&21&11&0&69\%\\ 2000&2818&39&21&11&0&58\%\\ 2001&2954&50&36&20&1&45\%\\ 2002&3349&56&40&24&2&35\%\\ 2003&3615&42&30&19&1&28\%\\ 2004&3733&49&39&22&0&54\%\\ 2005&3545&44&32&20&1&47\%\end{array}$[/hide]

I suppose I'll shoot for a 20...I haven't really done much preparation.  Hopefully I can get A1 and B1 though (or the two 'easiest' problems).

[quote="joml88"]I suppose I'll shoot for a 20[/quote]
The grading is relatively harsh- if you think you have two solutions, there's a good chance one is flawed and will lose most of the points. Try to spend enough time on A1/A2/B1/B2 to polish your solutions and ensure the credit.

There's an interesting link between the difficulty of the test and participation the next year- it looks like 1999 and 2004 scared some students away. I'd guess that about 3700-3800 students will participate this year, based on the long-term trend.

[quote="jmerry"]... it looks like 1999 and 2004 scared some students away.[/quote]
But the only sense in which 2004 was "harder" than 2003 and 2005 was the zero percentage. It may have been relatively more difficult to get  1 to 3 points in 2004 but it was relatively easier to get 10 to 30 points.

The easiest exam in these years was 1998. Having worked all the way through it with a class, I can see why. And the the hardest was 1999, with the second hardest being 2000. The other years are in between those extremes.

I'd really like to make top 200 this year, but I probably don't deserve it, considering I haven't prepared at all since last year, my classes are mostly beyond the level of Putnam math now, and I wasn't that close to top 200 last year... Oh well, I've started to feel that contest math isn't that important after all in the grand scheme of things, although it's still fun. And it's good to be able to solve nontrivial problems quickly, as a general principle.

I'd like a 20 too - which seems to be around the level for the top 500 students in the nation, which seems like a good goal for a freshman. I haven't done any preparation and I feel very out of practice, but I believe I could get oneof the first two on each set (well with a little luck.)

What can I expect from the experience of taking the Putnam, especially compared to other contests like the USAMO? What about the grading, which everyone says is very tough?

I was somewhere in the middle of the top 200 last year, with 'about' four questions right.  I'm aiming for 45-60 points and an honorable mention this year.  This is my third, and last time, writing the Putnam (I graduate in the spring).

I feel like my problem solving ability has come a long, long ways since last spring, but I haven't done many Putnam type problems lately, so it could be interesting.  I certainly don't think I'll do worse than last year, though.

20. I may/may not take it officially (high school senior here).

Dude if you take it officially as a high schooler, you can't take the IMO if qualified.

That's nonsense. Gabriel Carroll was a Putnam Fellow in 2000, and a member of the IMO team for the USA in 2001. The Putnam has nothing to do with high-school contest eligibility.

The Putnam does have a restriction which discourages high school students- you only get at most four years of eligibility.

[quote="hmmm"]Is there any way to find a more detailed breakdown of scores from previous years?  I couldn't find it on any relatively official Putnam website.  Thanks![/quote]
The AMC-hosted [url=http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml]site[/url] maintained by Kiran Kedlaya has the publicly released score distribution. Click in the column marked "scores." That's the information that is mailed to exam supervisors (including me) and the information on which I'm basing the chart above. This also includes a problem-by-problem breakdown, but only for the top 200, not for the whole population.

An update of the first post on this topic. To repeat what I said above:

[quote]The column marked $ N$ is the number of competitors. The columns marked "top 100", "top 200" and "top 500" are the minimum score needed to appear in those parts of the published list of accomplishments. (Bear in mind that the "top 500" won't be exactly 500 - in might be 479, it might be 513, wherever it breaks.) The column labeled "median" is just that, the median score, and the column labeled "% zero" is the percentage of all competitors who scored zero points.[/quote]

$ \begin{array}{ccccccc}\text{Year}&N&\text{top }100&\text{top }200&\text{top 500}&\text{median}&\%\text{ zero}\\1998&2573&58&42&29&10&30\%\\ 1999&2900&38&21&11&0&69\%\\ 2000&2818&39&21&11&0&58\%\\ 2001&2954&50&36&20&1&45\%\\ 2002&3349&56&40&24&2&35\%\\ 2003&3615&42&30&19&1&28\%\\ 2004&3733&49&39&22&0&54\%\\ 2005&3545&44&32&20&1&47\%\\ 2006&3640&50&31&14&0&63\% \end{array}$


linear algebra
matrix
combinatorics unsolved
combinatorics

In a group of people, every two people have exactly one friend in common. Prove that there is a person who is a friend of everyone else.

It has been discused before, but I can't remember if there are any posted proofs. I think this should work:

First of all, assum that the graph (I think it's obvious which graph I'm talking about :)) has at least five vertices (otherwise it's a triangle, and there's nothing to prove). Take any two vertices $x,y$  which are not connected to each other. Every neighbour of $x$ is connected to precisely one neighbour of $y$ and vice-versa, so every two vertices which are not connected have the same degree. There are now two cases:

[b](a) The graph is not regular[/b]

Take a vertex with maximal degree. If we assume that it's not connected to all the other vertices, there is at least one other vertex of maximal degree. Now consider a vertex of minimal degree (different from the maximal degree, by assumption). Since it's not connected to all the other vertices, there is at least another vertex of minimal degree. The two vertices of minimal degree are connected to the two vertices of maximal degree, so we have a contradiction. This means that when the graph is not regular, the conclusion holds.

[b](b) The graph is regular[/b]

Let $2<r<n-1$ be the common degree of the vertices of the graph, and let $A$ be the adjacency matrix. We have $A^2=AA^t=(r-1)I+J$, where $I$ is the $n\times n$ identity matrix, and $J$ is the $n\times n$ matrix with all entries equal to $1$. This means that the eigenvalues of $A$ are some $\sqrt{r-1}$, some $-\sqrt{r-1}$, and an $r$ or $-r$. Since the trace of $A$ is zero, we get something of the form $k\sqrt{r-1}=r$ for some non-zero $k\in\mathbb Z$. This means that $k=tr,\ t\in\mathbb Z$, thus yielding $t\sqrt{r-1}=1$. This is impossible, since $r>2\Rightarrow |t\sqrt{r-1}|>1$.

here is my solution :
assume the graph (the graph that these people are the vertices of it and two vertices are connected to each other when  two people are friend to each other.)
then take the vertix with maximum degree and call it $v_1$ .(and $v_3,...,v_k$ are it neghbours)
suppose that there is a vertix such that it is not neighbour to $v_1$ and call it $v_2$ $\rightarrow$ $v_1$ and $v_2$ have exactly one friend in common $\rightarrow$ $v_2$ is connected to one of $v_1$ neighbours like $v_3$ and it is not connected to the other $v_1$ friends.(?)
for each $v_1$ neighbours  like $v_i , 3 \leq i \leq k$
$v_i$ and $v_2$ have exactly one friend in common $\rightarrow$ and we know that two vertices of $v_1$'s neighbours have no more friend in common.(?)
$\rightarrow$ for each $v_i,3 \leq i \leq k$ $v_2$ and $v_i$ have exacly one friend in common like $v_q,k \leq q$ and none of these vertices ($v_q$) are same.(?)
$rightarrow$ the degree of $v_2$ is $k-1+1=k>k-1$ and it is bigger that the degree of $v_1$ $\rightarrow$ $v_1$ is friend of every one else.


every time i login, i enter my information and then it goes to the forum, but i'm not logged in

I need to go back to login and login once more

can this be fixed??

Posted by MrMcGeek

Check "Automatically log me in each time I visit" or something like that and that should be fixed.

Probable cause for this is that phpBB recognizes the links to the top and the left of the screen as a "new visit" thus automatically considers it logging off, then visiting this forum again.

McGeek - when does this occur?  When you use the bottom to login?  When you press the login button?  Specifically let me know what screens you're using/seeing, as I can't recreate this problem.

You shouldn't have to use the 'Log me on automatically button' - I don't.


Source: Furman University Wylie Math Tournament Senior Exam 2001 #16

How many times does the digit "9" appear in the representation of:
999,999,999,999,999,999,999:^2:?

A guess but does it appear [hide]20[/hide] times? My solution:

[hide]

9^2 = 81

99^2 = 9801

999^2 = 998001

9999^2 = 99980001

So the amount 9 appears is the total number of digits minus 1.

21 - 1 = 20.[/hide]

Solution (in spoiler):

[hide]n = (9...<21 9's>...9)^2

= (10^21 - 1)^2

= 10^42 - 2*10^21 + 1

= 10...<42 0's>...0 - 2*10...<21 0's>...0 + 1

100....0000...1

- 00....2000...0

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

n

Notice that the column with 0 at the top and 2 at the bottom is the 22nd place. Due to the necessary carrying needed in the subtraction, every digit in n to the left of that column has a 9. Thus, there are 20 9's in the answer.[/hide]

Can you explain your solution, I don't quite get it.

Hmm, where do you start to not understand?

Yes, you're both right.  Nice use of patterns to make a long problem quick, Ripper1 - importatn for the AMC!  Fierytycoon's method is more rigorous...


geometry Gauss function calculus integration

Let $n$ be positive numbers. Consider the following region on $xy$ plane.

\[ D_n=\left\{(x,y) |\ \frac{x}{n+\frac{1}{2}}\leq y\leq [x+1]-x,\ x\geq 0 \right\}. \]

Here $\ [x]$ is the greatest integer $\ \leq x.$ Find the area of $D_n.$

[quote="me@home"]Can you explain what the | means?
Also are you looking for a formula in N?[/quote]

| = "such that"

...and yes, I believe the formula will be in terms of $n$.

By the way, are we allowed to use calculus?  I think this could be set up nicely using an iterated integral, but since you posted it in the Intermediate forum I wasn't sure if you wanted it done otherwise.

[quote="The Zuton Force"][b]Curiosity question @ kunny[/b] - what does the term "humanities" mean in Japan? Because here in America humanities has nothing to do with math  :?[/quote]

This is a generic name, in case of Kyoto university Humanities are composed of the following faculties.

Facaulty of

Integrated Human studies

Letters

Education

Law

Economics

kunny

[quote="kokosai"]kunnyさん、京都大学の試験をとりましたか。京都大学はいい大学ですね。高校生ですか。東大の試験もとりましたか。^_^[/quote]

いいえ。最近, 京都大学も以前に比べましてずいぶんやさしくなりました。特に、今年は、本学独自路線ですすんでらしく、単なる話題づくりの感が否めません。この10年のゆとり教育が原因と思われます。はじめて、このサイトに日本の大学入試を紹介したわけですがおそらく、MathLinkers はこんなもんかと思っていると思います。今度、本来の京大の問題を紹介したいとおもいます。kokosai  さん、是非、日本語フォーラムにもお越しください。

To Mathlinkers.

Sorry for that I wrote this statement in Japanese.

Japanese communities Moderator

kunny


inequalities
calculus
inequalities proposed

Let $ a,b,c$ be positive numbers.Prove that:
$ \frac{a^3}{a^2\plus{}b^2}\plus{}\frac{b^3}{b^2\plus{}c^2}\plus{}\frac{c^3}{c^2\plus{}a^2} \ge \sqrt[4]{\frac{27(a^4\plus{}b^4\plus{}c^4\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2)}{32}}$
I hope you will like it :) .

It's weaker than my old result:
$ \frac{3(a\plus{}b\plus{}c)}{2(ab\plus{}bc\plus{}ca)} \geq\ \frac{a}{a^2\plus{}b^2}  \plus{}\frac{b}{b^2\plus{}c^2} \plus{} \frac{c}{c^2\plus{}a^2}$
 :)

[quote="nguoivn"]It's weaker than my old result:
$ \frac {3(a \plus{} b \plus{} c)}{2(ab \plus{} bc \plus{} ca)} \geq\ \frac {a}{a^2 \plus{} b^2} \plus{} \frac {b}{b^2 \plus{} c^2} \plus{} \frac {c}{c^2 \plus{} a^2}$
 :)[/quote]
Why,my friend? :)
[hide] lau lam moi thay anh len?   :(   :)    [/hide]

[quote="nguoivn"]It's weaker than my old result:
$ \frac {3(a \plus{} b \plus{} c)}{2(ab \plus{} bc \plus{} ca)} \geq\ \frac {a}{a^2 \plus{} b^2} \plus{} \frac {b}{b^2 \plus{} c^2} \plus{} \frac {c}{c^2 \plus{} a^2}$
 :)[/quote]
Dear nguoivn,
Your old result is nice but I can't prove this.
Could you tell me where I can find a full solution for this result ?
Thanks.

[quote="nguoivn"]In my result, let $ a \equal{} \frac {1}{x}$,... we obtain:
$ \sum\ \frac {x^3}{x^2 \plus{} y^2} \geq\ \frac {2\sum\ x^2 \plus{} \sum\ xy}{2(x \plus{} y \plus{} z)}$
And by pqr, easily to prove that: $ \frac {2\sum\ x^2 \plus{} \sum\ xy}{2(x \plus{} y \plus{} z)} \geq\ \sqrt [4]{\frac {27(x^{4} \plus{} y^{4} \plus{} z^{4} \plus{} x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2})}{32}}$
 :)
@quykhtn-qa1: [hide]Uh, bay gio` cong viec chinh' cua? anh la` kinh doanh ma`. Thinh? thoang? vao` lam` ti' cho vui thoi  :) [/hide][/quote]
Oh,very nice :lol: .But I think $ p,q,r$ very strong to solve it.
My inequality 's weaker but I have a nice solution for it :)

[quote="nguoivn"]It's weaker than my old result:
$ \frac {3(a \plus{} b \plus{} c)}{2(ab \plus{} bc \plus{} ca)} \geq\ \frac {a}{a^2 \plus{} b^2} \plus{} \frac {b}{b^2 \plus{} c^2} \plus{} \frac {c}{c^2 \plus{} a^2}$
 :)[/quote]
Nguoivn,please post full your solutions? :)

Here is can_hang's proof for the stronger:
http://www.mathlinks.ro/Forum/viewtopic.php?t=264418
And my another hint but not also not nice  :(
http://www.maths.vn/forums/showthread.php?t=9244&page=8
[quote="nguoivn"]And by pqr, easily to prove that: $ \frac {2\sum\ x^2 \plus{} \sum\ xy}{2(x \plus{} y \plus{} z)} \geq\ \sqrt [4]{\frac {27(x^{4} \plus{} y^{4} \plus{} z^{4} \plus{} x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2})}{32}}$
 :)
[/quote]
Indeed, it'll take us a long time to calculus  :oops:
@quykhtn-qa1 [hide]Anh rat to` mo` ve` loi giai? cua? em  :) [/hide]

[quote="nguoivn"]Here is can_hang's proof for the stronger:
http://www.mathlinks.ro/Forum/viewtopic.php?t=264418
[quote="nguoivn"]And by pqr, easily to prove that: $ \frac {2\sum\ x^2 + \sum\ xy}{2(x + y + z)} \geq\ \sqrt [4]{\frac {27(x^{4} + y^{4} + z^{4} + x^{2}y^{2} + y^{2}z^{2} + z^{2}x^{2})}{32}}$
 :)
[/quote]
Indeed, it'll take us a long time to calculus  :oops:
@quykhtn-qa1 [hide]Anh rat to` mo` ve` loi giai? cua? em  :) [/hide][/quote]
The inequality of can_hang2007 's weaker than my inequality
Here 's my proof,nguoivn:
By Holder inequality,we have:
$ LHS^2 \ge \frac{(a^2(a^2+c^2)+b^2(b^2+a^2)+c^2(c^2+b^2))^3}{(a^2+b^2)^2(a^2+c^2)^3+(b^2+c^2)^2(b^2+a^2)^3+(c^2+a^2)^2(c^2+b^2)^3}$
Put $ \sqrt{x}= b^2+c^2;\sqrt{y}=c^2+a^2;\sqrt{z}=a^2+b^2$,we only need to prove that:
$ (x+y+z)^{\frac{5}{2}} \ge 3\sqrt{3}({xy\sqrt{x}+yz\sqrt{z}+zx\sqrt{x})}$
Assume,$ x+y+z=3$,we must prove that:
$ xy\sqrt{y}+yz\sqrt{z}+zx\sqrt{x} \le 3$
Which 's obvious true by Vasc 'inequality  :)

Nice proof, quykhtn-qa1  :)
PS: In can_hang's proof, he proved my stronger. I also had read not carefully. But now, I realize this thing  :oops:

[quote="quykhtn-qa1"][quote="nguoivn"]Here is can_hang's proof for the stronger:
http://www.mathlinks.ro/Forum/viewtopic.php?t=264418
[quote="nguoivn"]And by pqr, easily to prove that: $ \frac {2\sum\ x^2 + \sum\ xy}{2(x + y + z)} \geq\ \sqrt [4]{\frac {27(x^{4} + y^{4} + z^{4} + x^{2}y^{2} + y^{2}z^{2} + z^{2}x^{2})}{32}}$
 :)
[/quote]
Indeed, it'll take us a long time to calculus  :oops:
@quykhtn-qa1 [hide]Anh rat to` mo` ve` loi giai? cua? em  :) [/hide][/quote]
The inequality of can_hang2007 's weaker than my inequality
Here 's my proof,nguoivn:
By Holder inequality,we have:
$ LHS^2 \ge \frac {(a^2(a^2 + c^2) + b^2(b^2 + a^2) + c^2(c^2 + b^2))^3}{(a^2 + b^2)^2(a^2 + c^2)^3 + (b^2 + c^2)^2(b^2 + a^2)^3 + (c^2 + a^2)^2(c^2 + b^2)^3}$
Put $ \sqrt {x} = b^2 + c^2;\sqrt {y} = c^2 + a^2;\sqrt {z} = a^2 + b^2$,we only need to prove that:
$ (x + y + z)^{\frac {5}{2}} \ge 3\sqrt {3}({xy\sqrt {x} + yz\sqrt {z} + zx\sqrt {x})}$
Assume,$ x + y + z = 3$,we must prove that:
$ xy\sqrt {y} + yz\sqrt {z} + zx\sqrt {x} \le 3$
Which 's obvious true by Vasc 'inequality  :)[/quote]
It is also my proof in Old and New Inequalities 2, you can find that book and read it ;)


Let a,b,c be positive real numbers satisfying:
$abc(a+b+c)=1$
Prove that: $(a+b)(a+c)\geq2$

[quote="chien than"][color=darkred]Let $a,\,b,\,c$ be positive real numbers satisfying: $abc(a+b+c)=1.$ Prove that:[/color]
\[(a+b)(a+c)\geq 2 \]
[/quote]
[color=darkblue][b]Proof.[/b] We have:
\[\left(a+b\right)\left(a+c\right)=a^{2}+ab+bc+ca=a(a+b+c)+bc\ge 2\sqrt{abc\left(a+b+c\right)}=2 \]
[/color]

Another solution:
$(a+b)(a+c)=a^{2}+ab+ac+bc=a(a+b+c)+bc=\frac{1}{bc}+bc\geq 2$


geometry
inradius
circumcircle
trigonometry
symmetry
Pythagorean Theorem

If $ \triangle ABC$ has inradius $ r$ and [[circumradius]] $ R$, then $ \cos{A} \plus{} \cos{B} \plus{} \cos{C} \equal{} \frac {r \plus{} R}{R}$.

Also prove Euler's theorem for a triangle in the link to "circumradius". Thanks.

I don't see how this is possible. In any given 30-60-90 triangle cos(30)+cos(60)+cos(90) is equal to a negative number, and since the inradius and circumradius are positive I don't see how this can happen. Furthermore, the lengths of inradii and circumradii can change while angles remain constant, which makes the formular impossible. Also I'm sure I would've heard of it if such an elegant theorem existed.

[quote="Captain_P"]Also I'm sure I would've heard of it if such an elegant theorem existed.[/quote]
That's kind of arrogant. :huh:

The equality is actually true. In your example none of the terms $ \cos 30, \cos 60, \cos 90$ are negative. You can prove this by drawing the perpendiculars from the circumcenter to the sides and then using the area formulas and pythagorean theorem a lot, but I'll let someone else better at geometry post a solution that isn't brutally computational.

I wouldn't say the computation is brutal. The symmetry simplifies them a lot. Maybe you're thinking of a variant of the proof I give, but the basic idea is the same as you suggest.
[hide]Let $ a=BC, b=CA, c=AB, p=a+b+c,s=\frac{p}{2}$ and $ O$ be the circumcenter. Then the area of $ \triangle OBC = \frac12(R\cos A)a$, since the height from $ O$ is $ R\cos A$. Similarly for $ \triangle OCA$ and $ \triangle OAB$. Then the area of $ \triangle ABC = \frac12(R\cos A)a  + \frac12(R\cos B)b + \frac12(R\cos C)c = rs$, so $ a\cos A + b\cos B  + c\cos C = \frac{rp}{R}$.

Let $ H$ be the foot of the altitude to $ BC$ from $ A$. Then $ BH=c\cos B$ and $ HC=b\cos C$. Thus $ c\cos B + b\cos C = a$. Obtain similar equations for the other sides.

\begin{align*}
a\cos A + b\cos B + c\cos C &= \frac{rp}{R}\\
b\cos A + a\cos B &= c\\
c\cos A + a\cos C &= b\\
c\cos B + b\cos C &= a\\\hline
(a+b+c)(\cos A + \cos B + \cos C) &= \frac{rp}{R} + p\\
\cos A + \cos B + \cos C &= \frac{r}{R} + 1.
\end{align*}[/hide]


calculus
integration
trigonometry
calculus computations

If you rationalize the denominator, this turns into
$ \frac{1}{2}\int\sqrt{x^{2}\plus{}1}\minus{}\sqrt{x^{2}\minus{}1}\,dx$, and each of these can be handled by a trig substitution (or by looking it up in a table).

Is this a joke?
Just rationalize and apply the respective formula!

It's no joke. Rather than using a formula (which doesn't seem like much fun), I've done as JRav suggests but instead of using a trig substitution (presumably $ x\equal{}\tan \theta$), I used integration by parts. For example, for $ \int \sqrt{x^{2} \plus{}1} \ dx$, I let $ u\equal{} \sqrt{x^{2} \plus{}1}$ and $ dv \equal{}dx$.

Your idea of fun seems to be a bit too morbid. You could do a problem any way you want as long as you don't wanna reach the answer rather that waste sheets after sheets trying to extract FUN out of it!


This could either be my making Algebra mistakes, or not understand the notation of parametric equations.

Let a curve be given parametrically in terms of $ t$ by the equations $ x\equal{}4t$ and $ y\equal{}2t^2$ where $ t$ takes positive and negative values.

$ P$ is the point on the curve with parameter $ t$.

Show that the gradient at $ P$ is $ t$.
$ \frac{dy}{{dx}} \equal{} \frac{4t}{4} \equal{} t$

Find and simplify the equation of the tangent at $ P$.
Just that $ y\minus{}y_1\equal{}m(x\equal{}x_1)$ right?
so $ y\minus{}2t^2\equal{}t(x\minus{}4t)$
which simplifies to $ y\equal{}tx\minus{}2t^2$

So far so good, but then the question asks:
The tangents at two points $ Q$ (with parameter $ t_1$) and $ R$ (with parameter $ t_2$) meet at $ S$.
Find the co-ordinates of $ S$.

I figure this is just a matter of substituting $ t_1$ and $ t_2$ respectively for $ t$ in the general tangent equation I found for $ P$ above, and then setting the two equal to one another, however, after simplifying the equation, I don't get the right answer.

$ t_1x\minus{}2t_1^2\equal{}t_2x\minus{}2t_2^2$
Then grouping like terms and factoring, I can arrive at:
$ x\equal{}\frac{2(t_1^2\minus{}t_2^2)}{(t_1\minus{}t_2)}$

The co-ordinate pair I should get is $ [2(t_1\plus{}t_2),2t_1t_2]$, but I don't come close. I can't see how I simplify that equation for $ x$ above into $ 2(t_1\plus{}t_2)$. I'm either being daft with my Algebra, or I've misunderstood how to find the co-ordinate pair.

Thanks very much for reading.

[hide="Hint"]$ 2(a^2\minus{}b^2) \equal{} 2(a\plus{}b)(a\minus{}b)$.[/hide]

Hi cHt.

I also get $ x\equal{} \frac{2(t_{1}^{2} \minus{} t_{2}^{2})}{t_{1}\minus{}t_{2}}$.

The numerator is the difference of two squares so this cancels down to
$ x\equal{} 2(t_{1} \plus{} t_{2})$

Substituting, I get $ y \equal{} \frac{(t_{1} \plus{} t_{2})^{2}}{2}$.

I also get this value of $ y$ by eliminating $ t$ and using $ y\equal{} \frac{x^{2}}{8}$.

[quote="AndrewTom"]
Substituting, I get $ y \equal{} \frac {(t_{1} \plus{} t_{2})^{2}}{2}$.
[/quote]

Thanks both, I see that was silly mistake, but I'm still having a hard time getting that $ y$ value.

When I substitute in, I get:

$ y \equal{} 2t(t_1 \plus{} t_2) \minus{} 2t^2$ and then what?  :maybe:

AndrewTom, I'm sorry, but I don't understand how your value of $ y$ is the same as $ 2t_1t_2$
I see that expanding that bracket gets me $ t_1^2 \plus{} 2t_1t_2 \plus{} t_2^2$, but where does the $ t_1^2$ and $ t_2^2$ go?

Sorry again, thanks for your time.


As we all know the major trig identity is Sin+Cos=1. But thats not the way its supposed to be written if following the proper rules it should be written Cos+Sin=1. Can anyone answer this questions.

It doesn't really matter how you write it.  They both mean the same thing.  Additionally, I've never heard of these 'proper rules'.


On the RHS, between lines 2 and 3, you lost a positive x.

On another note, though, I'm a little worried by this problem: normally, one defines an infinite expression such as the one shown as a limit of its finite sub-expessions.  This seems somewhat problematic to me, as the partial expressions for the given expression seems to only be defined only some of the time, and may not in fact have a limit.  In fact, by its construction, it appears that every even partial term has lots and lots of $\frac{2}{0}$ terms, something that doesn't bode well for this having a limit.  If you want it to have a chance of actually working out right, I suppose you could (1) define it as a limit of only the even terms or (2) always have the - signs above.  That is,

$\frac {1-\frac{1-\frac{1-..}{1+..}}{1+\frac{1-..}{1+..}}}{1+\frac{1-\frac{1-..}{1+..}}{1+\frac{1-..}{1+..}}} $

but I don't promise that this doesn't have the same problems.

Oh yeah JBL is right.  x could very well be 1.  And yes I somehow don't think this infinite fraction actually converges.  I'm troubled too  :blush: .

Actually, there is something interesting. Depending on how you interpret the expression, you get to something different. You could express it as

$ \frac {1+\frac1x}{1-\frac1x} = x $

and in that case, $x=1\pm\sqrt2$, that is different to the other solution probability 1.01 gave!  :?

More evidence that it doesn't actually converge !

And I got $ 1 $ and $ -1 $ as solutions.

This is weird...

Do you guys know any other expressions that behave like this?

[quote="djimenez"]and in that case, $x=1\pm\sqrt2$, that is different to the other solution probability 1.01 gave!  :?[/quote]

Well, probability1.01 was wrong, so there's no conflict with his thing.  But doing it his way (without losing the x) yields

\[x = \frac{1+\frac{1-x}{1+x}}{1-\frac{1-x}{1+x}} = \frac{\frac{2}{1+x}}{\frac{2x}{1+x}} = \frac{1}{x},\]

so x^2 = 1, or $x = \pm 1$.  But $x \neq -1$ since there are a bunch of $x+1$ terms in there.  I think x = 1 is okay.

For your interpretation, I think that $x = \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}$ is different.  This is because

\[\frac{1}{x} = \frac{1 - \frac{1}{x}}{1 + \frac{1}{x}} = \frac{x-1}{x+1},\]

so

\[x = \frac{1 + \frac{x-1}{x+1}}{1 - \frac{x-1}{x+1}}.\]

That is, there are $x-1$s in place of $1-x$s.

Maybe I should ask: do you think that the $\frac{1+\ldots}{1-\ldots}$ things (the smallest things shown in Hex's initial problem statement) are equal to the entire expression?  That is, is it okay to replace those with $x$s or not?

call x the top expression, and y the bottom. then

x/y = (1 + y/x)/(1 - y/x)

Cross multiplication yields

x^2 - xy = xy + y^2

x^2 - 2xy - y^2 = 0

(x-y)^2 = 0

x-y = 0

x = y

x/y = 1

so what do u think? Did I assume anything incorrect?

[quote="drunner2007"]djimenez, there is an error in your setup. I noticed that 1/x in the numerator is not equal to what it is replacing.
1/x flips the total expression. The expression is 1 + and then a (1 - etc.) in the numerator and a (1+ etc.) in the denominator. This is not equivalent to the (1 + etc.) originally in the numerator and (1 - etc.) originally in the denominator.
This is very hard to explain or see. If you do not see what I am getting at look at the expression for yourself.[/quote]

Yes, you are right!  :blush: Sorry!


geometry
incenter
circumcircle
search
geometry proposed

Given is triangle ABC(it's not said that it's acute). Let Q be an interior point and AQ meets BC at $ A_{1}$, BQ meets AC at $ B_{1}$ and CQ meets AB at $ C_{1}$. Let $ QA_{1} \equal{} QB_{1} \equal{} QC_{1}$. Is it true that the triangle is equilateral if:

a) Q is the centroid
b) Q is the incenter
c) Q is the circumcenter
d) Q is arbitrary point

d) No, see http://tech.groups.yahoo.com/group/Hyacinthos/message/6589 (search for "Verkaart point").
c) Yes (very easy).
a) Yes (very easy).
b) Such triangles satisfy $ \frac{\left|b\minus{}c\right|\left(b\plus{}c\minus{}a\right)}{b\plus{}c}\equal{}\frac{\left|c\minus{}a\right|\left(c\plus{}a\minus{}b\right)}{c\plus{}a}\equal{}\frac{\left|a\minus{}b\right|\left(a\plus{}b\minus{}c\right)}{a\plus{}b}$. Now someone with more time can check whether this yields $ a\equal{}b\equal{}c$...

  darij

I have forgotten  :|
 e) Q is the orthocenter(it's not said that the triangle is acute)


geometry
3D geometry
sphere
geometry proposed

Let S be aset of 2003 points in 3D space such that each of its subsets consisting of 78 points contains at least 2 points that have distance at most 1 from each other. Prove there exists a sphere of radius 1 such that at least 27 points of S lie on or inside it.

Indeed it is easy, and it reminds me another one which has already been proposed here (I will look after it) :

Suppose for a contradiction that no unit sphere contains (interior or boundary) at least 27 of the points.
Let $A_1$ be any of the points, and $S_1$ be the unit sphere with center $A_1$. Then, no more than 25 other points belongs to $S_1$. Thus, at least 1977 of the points are exterior of $S_1$. Let $A_2$ be any of these 1977 points and $S_2$ be the unit sphere with centre $A_2$. Then at most 25 of the 1976 remaining points belong to $S_2$. Then at least 1951of the points do not belong to $S_1$ nor $S_2$. Let $A_3$ be any of these 1951 points and so on...
Since $2003 = 26 \cdot 77 + 1$, we may construct $A_1, \cdots, A_{78}$ such that for each $i,j$ the point $A_i$ does not belong to $S_j$. That means that $A_iA_j > 1$ for all $i,j$ contradicting the statement of the problem.

Pierre.

Ok, I do not not have totally loose my mind :
http://www.mathlinks.ro/Forum/viewtopic.php?t=4886&highlight=sphere

Pierre.


geometry
geometric transformation
rotation
calculus
derivative
3D geometry
sphere

Well, as you might have noticed, I haven't been around for some time, but I'm back now and have some interesting problems ;) Let's start with a fairly easy (but tricky) one:

Two copper wires of equal length are connected in parallel and then  connected to a source of direct current whose internal resistance is much greater than the resistance of either wire. When a stationary regime is reached, the temperature of one wire is $\Delta T_1 = 16 \text K$ higher than the room temperature while the temperature of the other wire is $\Delta T_2 = 8\text K$ higher than the room temperature.  Determine the temperature difference between each wire and the room temperature when the parallel connection of the wires is replaced with a serial one. Neglect the expanding of the wires due to heating and consider that the resistance of the wires does not depend on the temperature.

(The result might surprise you...)

Angular momentum of the system is
$L=\frac{2}{5} M_E R_E^2 \omega_E+\frac{2}{5} M_M R_M^2 \omega_M+\frac{M_M M_E}{M_E+M_M} \omega r^2$
Since spin-angular momentum of the moon is too less
$L=\frac{2}{5} M_E R_E^2 \omega_E+\frac{M_M M_E}{M_E+M_M} \omega r^2$
From the Kepler's third law
$\omega^2 r^3=G (M_E+M_M)$
$\omega=\sqrt{\frac{G (M_E+M_M)}{r^3}}$
$L=\frac{2}{5} M_E R_E^2 \omega_E+M_M M_E \sqrt{\frac{G r}{M_E+M_M}}=const$
so it's derivative must be zero
$dL=-\frac{2}{5} M_E R_E^2.2\pi \frac{\Delta T}{T^2}+M_M M_E \sqrt{\frac{G}{M_E+M_M}} \frac{\Delta r}{2\sqrt{r}}=0$
And at last
$\Delta r=\frac{8\pi}{5} \sqrt{\frac{(M_E+M_M) r}{G}} \frac{R_E^2}{M_M} \frac{\Delta T}{T^2}=9.65\cdot{10^{-4}}\text{m}$
Is this one true???

[quote="azereyvazov"]
$I_1+I_2=\frac{U}{R}$
$R$ is the internal resistance of the source
[/quote]

This is true because R is very high, isn't it?

Curious to be seen, if R=0 (I had solved the problem in this way because i didn't carefully read the text :blush: ) i get:

$T'=0,64 K$ , $T''=5,12 K$

@ [b]Bacco[/b]: Yes, the high $R$ makes for the wires not to affect the total current. About your result: possible... In my experience, whatever mistake you make in your reasoning, you'll get a more sensible result like the one you obtained -- having a wire heat up by 200K and still neglecting thermal expansion and such things is rather... unbelievable ;)

@ [b]azereyvazov[/b]: Can you come up with an approximation which will allow you to express $\Delta r$ in terms of the variables given explicitly in the problem? As far as I see, this approximation isn't valid, but still gives a result of the correct order of magnitude, which isn't so bad... I'll wait a little longer to see whether someone will get the idea...

[quote="Djole"]Yup ;) But it's interesting to note that in vacuum the results would be quite different because there would be no convection.

Here's another one, maybe a bit harder but still not very difficult:

Due to global warming, the period of Earth's rotation decreases by $\Delta T = 4.4 \cdot 10^{-8}\text s$ every day, ie. every $T = 24\text h$. This effect causes the distance between the Moon and the Earth to change by $\Delta r$ daily. Assuming that the Earth and the Moon form an isolated system and that both can be approximated as homogenous balls with constant masses and radii, determine $\Delta r$. The 'innitial' distance between the Earth and the Moon is $r = 3.84 \cdot 10^8\text m$. The revolution of the Moon around the Earth and the rotation of the Earth around its axis have the same direction.[/quote]
I am very sure that a very question was given at the APhO2001.  Check it out.

Hmmm, according to what I gathered at first glance, the problem deals with the same system, but there is a difference... And as far as I can see the idea that was used in the solution of my problem wasn't used on the APhO...

@ [b]AndersoN[/b]: It's good that you've reminded me, I almost forgot to do the APhO's before Singapore ;) I understand you'll be there as well?

[quote="azereyvazov"]Angular momentum of the system is
$L=\frac{2}{5} M_E R_E^2 \omega_E+\frac{2}{5} M_M R_M^2 \omega_M+\frac{M_M M_E}{M_E+M_M} \omega r^2$
[/quote]

Is it true? The dimensions of the two spheres aren't negligible, so i think that the angular momentum should be $L=\frac{2}{5} M_E R_E^2 \omega_E+\frac{2}{5} M_M R_M^2 \omega_M+\frac{M_M M_E}{M_E+M_M} \omega r^2+\frac{2}{5}(M_E R_E^2+M_M R_M^2)\omega$

I tried to calculate those quantities: i think that the only one negligible is the second guy, not the fourth.

I don't see where does that fourth term come from :huh: The first two are ok -- due to the rotation of both bodies about their axes; indeed, the second term is negligible because it's approx. five orders of magnitude lesser than the first. And the third term is ok -- its due to the orbital movement of the Earth and Moon. But that forth term :?:

Let's calculate the angular momentum respect to an axis that passes through the center of mass of the earth+moon system.

$L_E=\frac{2}{5}M_E R_E^2 \omega_E$ due to earth rotation along its axis
$L_M=\frac{2}{5}M_M R_M^2 \omega_M$ due to moon rotation along its axis
$L_S=(I_1+I_2)\omega$ due to the rotation of the two body along the axis through the center of mass of the earth-moon system

Because of Huygens-Steiner theorem:
$I_1=\frac{2}{5}M_E R_E^2 + M_E d^2$
$I_2=\frac{2}{5}M_M R_M^2 + M_M (r-d)^2$

where $d=\frac{M_M r}{M_M+M_E}$

hence the formula I wrote. Perhaps it's wrong... where is the error?

The 4th term would disappear if we considered earth's and moon's dimensions much smaller than $r$... i think.

No, $L_S$ (as you labeled it) is [i]not[/i] $(I_1+I_2) \omega$. Since every motion of a rigid body may be broken into the motion of its center of mass and the rotation about an axis passing through the center of mass (actually this works for any point, not only the center of mass). The angular momentum due to the rotation about the axis is $I_E\omega_E$ and the ang. mom. due to the motion of the CM is $\frac{M_EM_M}{M_E+M_M}\omega^2r$...

Excuse me,
Could you write where is my mistake, please.

@Djole: Sorry, I have not understood. Let's consider a sphere (radius $r$) with a very very light bar attached on it in a radial direction. The lenght of the bar is $l$. The end of the bar is fixed and is the centre of rotation. I want to make the body rotate: how much is I? In this case, the sphere can't rotate itself. I, by Steiner, is $\frac{2}{5}mr^2 + m(l+r)^2$, and $L=I\omega$. In tha case of our earth-moon problem, there is an additional angular momentum because of the rotation about the axis. I think your formula is correct only if we're talking about material particles.

@ [b]Bacco[/b]: I understand what you mean, sorry for the misunderstanding. Yes, you're right, but as far as I know it's a custom that in problems involving planetary motion all planets are treated as material points.

@ [b]azereyvazov[/b]: No, as far as I can see you haven't made a mistake -- I was referring to the approximation in the solution I have. Basically, it states that $\frac{\Delta L_S}{L_S} = - \frac{\Delta L}{L}$, where $L_S$ is due to the Earth rotation and $L$ is due to the revolution of both the Earth and the Moon around their CM. As far as I can see, this is incorrect, but has led to a result which is, if I remeber correctly, about $5 \cdot 10^{-4}\text m$... Now, is this an error, or do you think there is a 'hidden sense' within? (I vote for the former!)

[quote="Djole"]@ [b]AndersoN[/b]: It's good that you've reminded me, I almost forgot to do the APhO's before Singapore ;) I understand you'll be there as well?[/quote]

You bet!  I gather there'll be at least 3 people from here, us and azereyvazov. Cool, the more the better.

Btw, how are you preparing?

@ [b]AndersoN[/b]: Well, I'm not preparing as vigorously as I'd expected I would... I'm mainly meditating, thinking about some past IPhO experiments, generally organizing my thoughts (I've finished a great lot of problems, and I somehow feel that there isn't much that I can do within the remaining week...) But I'm still open to new problems; as I said, I should go over the APhO's this weekend and that would be that. What about you?

This entire post of mine is off-topic... I'll be away tomorrow, so if anybody wishes to discuss the upcoming Olympiad it would be best to open a new thread, lest we want the entire conversation deleted (although I haven't seen Rushil active for a while...) I'll open a new thread myself the day after tomorrow if somebody doesn't open it before.


compute :  $ A\equal{}\sum_{k\equal{}0}^n k.\cos(x)$

                $ B\equal{}\sum_{k\equal{}0}^n k.\sin(x)$

                 $ C\equal{}\sum_{k\equal{}0}^n\cos^{k}(x)$

                 $ D\equal{}\sum_{k\equal{}0}^n\sin^{k}(x)$

:)

Are they just ... so easy ?

$ A \equal{} \sum{k} \cos(x) \equal{} \frac {n(n \plus{} 1) \cos(x)}{2}$
$ B \equal{} \frac {n(n \plus{} 1) \sin(x)}{2}$
$ C \equal{} \frac {1 \minus{} \cos^{n \plus{} 1}(x)}{1 \minus{} \cos(x)}$
$ D \equal{} \frac {1 \minus{} \sin^{n \plus{} 1}(x)}{1 \minus{} \sin(x)}$

oh i'm sorry :

$ A\equal{}\sum k\cos(kx)$

$ B\equal{}\sum k\sin(kx)$

$ C\equal{}\sum k\cos^k(x)$

$ D\equal{}\sum k\sin^k(x)$

Start by asking yourself if they even converge.


$ Let x\equal{}y\equal{}0$

$ f(0)\equal{}0$

Let $ x\equal{}0$ to get:

$ 0\equal{}yf(y)$

Then, if $ y \ne 0$ we have $ f(y)\equal{}0$

Therefore, $ f(x)\equal{}0$ is the only solution.

sorry rofel my problem have some mistakes
I have edited it :blush:

[quote="Math pro"]find $ f: (0,1) \minus{} > R$ such that:
$ f({xy}) \equal{} x.f({x}) \plus{} y.f({y})$[/quote]
Let $ x \equal{} y \equal{} \frac{1}{2}$. Then $ f(\frac{1}{4}) \equal{} f(\frac{1}{2}) \equal{} a \in \mathbb{R}$.
Then $ f(\frac{1}{8}) \equal{} \frac{1}{2}f(\frac{1}{2}) \plus{} \frac{1}{4}f(\frac{1}{4}) \equal{} \frac{3}{4}a$.
So $ f(\frac{1}{16}) \equal{} \frac{1}{2}f(\frac{1}{2}) \plus{} \frac{1}{8}f(\frac{1}{8}) \equal{} \frac{5}{8}a$. On the other hand $ f(\frac{1}{16}) \equal{} \frac{1}{4}f(\frac{1}{4})\plus{}\frac{1}{4}f(\frac{1}{4}) \equal{} \frac{1}{2}a$. So $ a\equal{}0$. And by induction: $ f(\frac{1}{2^n}) \equal{} 0$.

Let $ x \equal{} \frac{1}{2^n}$ and $ y \equal{} k \in (0;1)$.
Then $ f(\frac{1}{2^n}k) \equal{} \frac{1}{2^n}f(k)$
So $ f(\frac{1}{8}k) \equal{} \frac{1}{8}f(k)$.
On the other hand $ f(\frac{1}{8}k) \equal{} \frac{1}{4}f(\frac{1}{4}k) \plus{} \frac{1}{2}f(\frac{1}{2}k) \equal{} \frac{5}{16}f(k)$. Then $ f(k) \equal{} 0 \forall k \in (0;1)$ is the only solution. Inserting gives $ 0 \equal{} x \cdot 0 \plus{}  y \cdot 0$, which is true, so it is a solution. QED.


calculus
derivative
calculus computations

For what values of $x\in\mathbb{R}$ does the following series converge?

$\sum_{n=0}^{\infty}ne^{-nx}$

I took the derivative of this sum: $\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$ and multiplied by $x$ to get:

$\sum_{n=0}^{\infty}nx^{n}=\frac{x}{\left(1-x\right)^{2}}$

Then I plugged in $x=e^{-x}$ to get:

$\sum_{n=0}^{\infty}ne^{-nx}=\frac{e^{-x}}{\left(1-e^{-x}\right)^{2}}$

From this I can say that all values will work except $x=0$.  So it converges for $x\in\left(-\infty,0\right)\cup\left(0,\infty\right)$.

Is my answer / reasoning correct?

Thank you.

Nope. Take for example $x=-1$. Then you are calculating terms of the form $ne^n$. But $ne^n>n$ if $n>0$ so the series is divergent at the point $x=-1$.

$ne^{-nx} \rightarrow +\infty$ when $x \leq 0$ is a major problem I think.

OK I see what you mean.  But why does my formula say otherwise?  I mean, I am pretty sure that $\sum_{n=0}^{\infty}nx^{n}=\frac{x}{\left(1-x\right)^{2}}$.  Any ideas where I went wrong?

[quote="amcavoy"][quote="jmerry"][quote="apmcavoy"]$\sum_{n=0}^{\infty}nx^{n}=\frac{x}{\left(1-x\right)^{2}}$[/quote]
Only when $|x|<1$. Check what happens to that condition when you substitute $x=e^t$.[/quote]
Right, $e^{-x}<1$ so $x>0$ would be the final answer, correct?

Thank you.[/quote]
yes, i think.
The geometic series converges if $|x|<1$, after taking the derivative. You need to deal with the endpoints (The Abel's theorem tell us about endpoints). In your problem we can see that the endpoind $x =0$ give divergen result.

Edit: silly me  :rotfl: I thought it was term by term integration. but i said derivative.
Of course we know that power series $f(x)= \sum_{k=0}^{\infty} a_k (x-x_0)^k$ with interval of convergence I, with $f$ has continous derivative of all orders throughout the interval I, term by term differentiation is allowed and the series of derivative has the same radius of convergence. ;)


[quote="nsato"]There are no mock AIMEs, just weekly message board problems.  All problems in the lectures will be drawn from past AIME contests.

[quote="bpms"]In the AIME problem series will there be a mock AIME? Also, questions from old AIMEs will obviously be used, but will AIME like questions be used or just AIME questions?[/quote][/quote]
Where will the weekly message board problems be posted?

[quote="rrusczyk"][quote="bpms"][quote="Phelpedo"]On the message board?  :wink:[/quote]
I meant, is the message board the regular AMC forum here, or is there a special one for the problem series?[/quote]

There's a special one just for the class.  We pack it with extra problems.[/quote]
About how many a week? Also, how many of each difficulty (1-5 / 6-10 / 11-15)? Lastly, will the problems be old AIME ones or will they be original problems you guys make up? I should be done with questions now unless I think of more...

[quote="bpms"][/quote][quote="rrusczyk"][quote="bpms"][quote="Phelpedo"]On the message board?  :wink:[/quote]
I meant, is the message board the regular AMC forum here, or is there a special one for the problem series?[/quote]

There's a special one just for the class.  We pack it with extra problems.[/quote][quote="bpms"]
About how many a week? Also, how many of each difficulty (1-5 / 6-10 / 11-15)? Lastly, will the problems be old AIME ones or will they be original problems you guys make up? I should be done with questions now unless I think of more...[/quote]
Does anyone know?

[quote="SplashD"][quote="bpms"][quote="robinhe"]There are about 12 problems each week and the difficulty is medium (so 6-10) except I think there are some harder ones.

You know, you could always just try the class and then many of your questions will be answered.[/quote]
Well I thought it might be too hard for me, and then it would be a waste of money. I'm going to take it though unless I see something which really turns me off of it...[/quote]

You do know you can always drop the class before the third week.[/quote]
I didn't know that :|


I hope that Carpo and Mathgod did them, and personally I think this contest are really easy.
Question 1-9, all jokes.
Question 10, typical question, it asks us to find roots of this equation $x= \sqrt{2+\sqrt{2+\sqrt{2+...}}}$
so that $x= \sqrt{2+x} \Rightarrow x = 2$
Question 11, 12 towns. $21 *2 - (4*4 + 3*6) = 8$
Question 12, 1534/4602 = 1/3

i did it

yea the whole contest was wayy too easy, should have gotten perfect

BUT THE STUPID NUMBER 8 AHJHVNTMFVHDO;SF

still kinda confused about what the hell its asking  :?

I didn't get the problem's wording at first, either
Mathematically, a set with [b]n[/b] elements can have[b] m[/b](express as $m_n$) different subsets, where m can be expressed as
$m_n = 1 + n + (^{n}_{2}) +(^{n}_{3}) + ... + (^{n}_{n})$
In this case, set S can have 24 more subsets then set T.
Plug in different values of n, like 1,2,3,4,5,6,... we get $m_3 = 7, m_5 = 31, m_6 = 63$ Since $31-7 = 24$ and for $m_k$ where $k \ge 6$, $m_k - m_j \ge m_5 - m_3 = 24$, so that the answer is [b]5[/b] (set S has 5 elements)

i know this has got to be the noobiest question, and this is like really basic stuff.... but...

whats the 2 numbers one on top of each other with the bracket around both of them mean? i never know what they mean... or how to calculate them... lol sorry  :P

$\binom{n}{k} = \frac{n!}{k!(n-k!)}$

This is a very basic counting identity...the number of ways to choose k objects in an n object set.  I would dare say this is something you would have to know.

See if you can prove it.

I guess my math teacher doesn't know about this contest, so I didn't write it. Though I don't think I'm missing much . . .

[quote="Kalimdor"]It's okay, Steven
Coz I missed Fermat...
 :P
Walking along on the "Boulevard of Broken Dreams"[/quote]Wow how did you manage to miss Fermat? COOL

Well I feel that the preliminary rounds of that BC Colleges contest are so much harder than its final rounds.... I remember getting so seriously owned last year  :blush:

There is a bc college contest!? How come noe one told me?
 :(

[quote="d.lam.86"]Are these contests posted online like the CMC ones?[/quote]Yes.


Βρείτε όλες τις συναρτήσεις από το R\ {-1, 1}$ \rightarrow$R ώστε:

$ f(\frac {x\minus{}3} {x\plus{}1})\plus{}f(\frac {3\plus{}x} {1\minus{}x})\equal{}x$

Απλή αλλά όμορφη....

Πραγματι ειναι ωραια,

Θετουμε $ g(x)\equal{}\frac{x\minus{}3}{x\plus{}1}$

και παρατηρουμε οτι $ g(g(x))\equal{}\frac{x\plus{}3}{1\minus{}x}$ και οτι $ g(g(g(x)))\equal{}x$

Οποτε η δοσμενη γινεται:$ f(g(x))\plus{}f(g(g(x)))\equal{}x$    $ (1)$
Θετουμε οπου $ x$ το $ g(x)$ στην    $ (1)$

$ f(g(g(x)))\plus{}f(x)\equal{}g(x)$    $ (2)$

και οπου $ x$ το $ g(x)$ στην    $ (2)$

$ f(x)\plus{}f(g(x))\equal{}g(g(x))$    $ (3)$

Οποτε απο τις σχεσεις $ (1)$, $ (2)$, $ (3)$ φαινεται οτι

$ f(x)\equal{}\frac{g(x)\plus{}g(g(x))\minus{}x}{2}$


function
integration
calculus
calculus computations

Why does not the following function have [b]any extreme points [/b]?

$F(x) = \int_0^x\frac{t^2-5t+4}{2+e^t}dt$

I also got the same answer:

$\frac{d}{dx}\int_0^x\frac{t^2-5t+4}{2+e^t}dt = \frac{x^2-5x+4}{2+e^x}=\frac{(x-1)(x-4)}{2+e^x}$

Because $F(x)$ is increasing in $(-\infty,1)$ and in $(4,\infty)$, and decreasing in $(1,4)$
$\therefore F(x)$ has a local maximum at $x=1$ and a local minimum at $x=4$.

 But because [b]Maple[/b] returned no extreme points for this function, I thought my answer is wrong and the function has no extreme points. Anyway, then it is an error from Maple.

Thank you!

And you're letting an anomalous (and probably notation-dependent) result from a computer program shake your own confidence that you know what you're talking about because ... ?


[hide]$ f(x)\equal{}f(2\cdot\frac{x}{2})\equal{}\left(\frac{x}{2}\right)^{2}$. Hence $ f(a)\equal{}\frac{a^{2}}{4}\equal{}1\Longrightarrow a^{2}\equal{}4$. So $ a\equal{}\pm 2$.[/hide]

Is this correct?

I am not too sure what you did to find that a = -2 and =2.

Can you explain further?

Thanks

$ f(2x) \equal{} x^{2}$ says that, when $ x$ is multiplied by $ 2$, the function $ f(x)$ produces $ x^{2}$.

To find $ a$ when $ f(a) \equal{} 1$ you obviously need the equation $ f(x)$ in order to do this.

To get the original equation you divide the $ x$, [i]not the $ x^{2}$[/i], by two. In other words, $ f\left(2\cdot\frac{x}{2}\right) \equal{}\left(\frac{x}{2}\right)^{2}$ and $ 2\cdot\frac{x}{2}\equal{} x$ as well as $ \left(\frac{x}{2}\right)^{2}\equal{}\frac{x^{2}}{4}$.

So, $ f(x) \equal{}\frac{x^{2}}{4}$ and $ f(a) \equal{}\frac{a^{2}}{4}\equal{} 1$. Multiply by $ 4$, $ a^{2}\equal{}4$, and take the square root $ \boxed{a \equal{}\pm2}$.


quadratics
algebra
quadratic formula
algebra unsolved

Find the condition so that all the roots of the equation

$x^4 - 2ax + x + a^2 - a = 0$ are real.

I am getting $a \geq  0$

[quote="Andreas"]Indian RMO 2000 Problem 7
Posted on Pre-Olympiad[/quote]

Can You send me the link to it. I could not find it.

According to the correct version of RMO 2000 , it should be

$x^4-2ax^2+x+a^2-a=0$

Now let us change it into a quadratic equation in term of $a$ ,

$a^2-(2x^2+1)a+(x^4+x)=0$ . Notice that $a$ is real and this means that the determinant is $\geq 0$ , hence we have

$(2x^2+1)^2-4(x^4+x)\geq 0$ $\implies 4x^2-4x+1\geq 0$ $\implies (2x-1)^2\geq 0$ which is always true for any $x$ .

Hence using the quadratic formula , we should obtain

$a=\frac{2x^2+1\pm (2x-1)}{2}$ so $a=x^2-x+1$ or $a=x^2+x$

this mean we can factorise it into

$(a-x^2+x-1)(a-x^2-x)=0$ or $(x^2-x+1-a)(x^2+x-a)=0$

But since $x$ has all real roots , means the determinant of the both the quadratic above ( which in term of $x$ ) must be $\geq 0$

Hence we get $1-4(1-a)\geq 0$ $\implies a\geq \frac{3}{4}$ and
$1+4a\geq 0$ $\implies a\geq -\frac{1}{4}$

Combine both and get $a\geq \frac{3}{4}$  :)


a,b in R
$a-b=1$
prove that
$a^{3}-b^{3}\geq  \frac{1}{4}$


Find n such that

$(x + k)^n - x^n - k^n = 0$

$n \in {\rm N},k \in \Re ^*$

and
$x = ke^{\dfrac{{2\pi i}}{3}}$

divide by k which is nonzero

we get for r = cis 120deg:

$(r+1)^n = r^n + 1$

Now $(r+1)^2 = r$ and $r^3 = 1$, implying $(r+1)^6 = 1 = r^6$ so that both sides are periodic functions of n with period 6.  We just need to try n = 0, ... 5.

For 0 it dosnt work.
For 1 it works.
For 2 it doesn't.
For 3 it doesn't.
For 4 it doesn't.
For 5 it works.

So it works for n = 1,5 mod 6.  Done.


Source: 2005 AMC-10B #25

A subset $B$ of the set of integers from $1$ to $100$, inclusive, has the property that no two elements of $B$ sum to $125$. What is the maximum possible number of elements in $B$?

$\text{(A)}\ 50 \qquad\text{(B)}\ 51 \qquad\text{(C)}\ 62 \qquad\text{(D)}\ 65 \qquad\text{(E)}\ 68$

[hide]
\begin{eqnarray*}125&=&100+25\\ &=&99+26\\ &\vdots\\ &=&64+61\\ &=&63+62 \end{eqnarray*}

So we conclude that $\{1,2,\ldots,24\}\subset B$.
From every partition given above, we may pick exactly one number, so we can add $38$ more numbers to $B$ having totaly $24+38=62$ elements.
[/hide]


Is there a commonly used notation to say "W is a subspace of V"? Or is this something that must always be written out in words?  Thanks!


geometry
incenter
search
function
perpendicular bisector
geometry unsolved

The altitude $CH$ of a right triangle $ABC$, with $\angle{C}=90$, cut the angles bisectors $AM$ and $BN$ at $P$ and $Q$, and let $R$ and $S$ be the midpoints of $PM$ and $QN$. Prove that $RS$ is parallel to the hypotenuse of $ABC$

I'm pretty sure it was posted before but I can't find it right now (can someone find the "missing" link? :)),
so here is my solution:

An easy angle chase shows that $\Delta CNQ$ and $\Delta CMP$ are isosceles, so $CS \perp BN$ and $CR \perp AM$. Let $I$ be the incentre of $\Delta ABC$ (the intersection of $AM$ and $BN$), then $CRIS$ is obviously cyclic and $\angle ISR = \angle ICR = \angle ICB - \angle RCM = 45^{\circ} - \angle A/2 = \angle B/2 = \angle IBA$. Hence $AB \parallel RS$.

Here's another way to see this. Notice that $CRHA$ is cyclic, so $\angle CHR = \angle CAR = \angle A/2 = \angle HCR$. So $CR = HR$ and $R$ lies on the perpendicular bisector of $CH$. $S$ also lies on this perpendicular bisector, so $SR$ *IS* in fact the perpendicular bisector of $CH$, and therefore it is parallel to $AB$.

Well, I use the "search" function, but I can't find the problem. So, I post it here  :)


geometry
circumcircle
trigonometry
LaTeX
geometry proposed

Let $ ABC$ be a triangle. The point $ L$ belongs to the side $ [BC]$ . Prolong the segment $ [AB]$ beyond the point $ B$ up to the point $ M$ so that  $ m(\widehat {ALC})\equal{}2\cdot m(\widehat {AMC})$ . As well prolong the segment  $ [AC]$ beyong the point  $ C$ up to the point  $ N$ so that  $ m(\widehat {ALB})\equal{}2\cdot m(\widehat {ANB})$ . Denote the circumcenter $ O$ of the triangle $ AMN$ . Prove that $ OL\perp BC$ .

[quote="minsoens"][quote="mr.danh"]Thanks Virgil for your nice problem.
The circle (L;LA) meets BC at D,E with D,B,C,E in order.
1) Two qudrilateral ADML, AENL are cyclic.
2) Denote I=DM∩EN. Triangle IDE is isosceles, the quadrilateral OMIN is cyclic.
3) OM=ON, IO is the bisector of angle MIN, hence result.[/quote]mr.danh, I think in your first step [b]ADMC[/b] and [b]AENB[/b] are cyclic, but everything else looks fine. Thank you for a nice solution![/quote]
Thank you for pointing out my type mistakes. You were right.

By Stewart Theorem;

$ |OB|^2=R^2-|AB|.|BM|$ and $ |OC|^2=R^2-|AC|.|CN|$ where $ R$ is radius of circumcircle of triangle $ AMN$.

We will prove that $ |OB|^2-|OC|^2=|BL|^2-|CL|^2$

Let $ m(AMC)=\alpha$ , $ m(LAC)=\beta$ , $ m(LAB)=\theta$ then $ m(ALC)=2\alpha$ , $ m(ANB)=90-\alpha$

$ |AB|=x.sin(2\alpha+\beta)$ , $ |AC|=x.sin(2\alpha-\theta)$ , $ |BL|=x.\frac{sin(2\alpha+\beta)sin{\theta}}{sin{2\alpha}}$ , $ |CL|=x.\frac{sin(2\alpha-\theta)sin{\beta}}{sin{2\alpha}}$ , $ |BM|=x.\frac{sin(\beta+\theta)sin(\alpha-\theta)}{sin\alpha}$ , $ |CN|=x.\frac{sin(\beta+\theta)cos(\alpha+\beta)}{cos\alpha}$

$ |OB|^2-|OC|^2=|AC|.|CN|-|AB|.|BM|=|BL|^2-|CL|^2$    if and only if

$ \frac{sin(2\alpha-\theta)sin(\beta+\theta)cos(\alpha+\beta)}{cos\alpha}-\frac{sin(\beta+\theta)sin(2\alpha+\beta)sin(\alpha-\theta)}{sin\alpha}=(\frac{sin(2\alpha+\beta)sin\theta)}{sin{2\alpha}})^2-(\frac{sin(2\alpha-\theta)sin\beta}{sin{2\alpha}})^2$

$ \frac{sin(2\alpha+\beta)sin\theta+sin(2\alpha-\theta)sin\beta}{sin{2\alpha}}=\frac{1}{2}.\frac{cos(2\alpha+\beta-\theta)-cos(2\alpha+\beta+\theta)+cos(2\alpha-\beta-\theta)-cos(2\alpha+\beta-\theta)}{sin{2\alpha}}=sin(\beta+\theta)$

$ |OB|^2-|OC|^2=|AC|.|CN|-|AB|.|BM|=|BL|^2-|CL|^2$    if and only if

${ \frac{sin(2\alpha-\theta)sin(\beta+\theta)cos(\alpha+\beta)}{cos\alpha}-\frac{sin(\beta+\theta)sin(2\alpha+\beta)sin(\alpha-\theta)}{sin\alpha}}=(\frac{sin(2\alpha+\beta)sin\theta}{sin{2\alpha}}-\frac{sin(2\alpha-\theta)sin\beta}{sin{2\alpha}}).sin(\beta+\theta)$ $ \Longleftrightarrow$

$ 2sin\alpha\cos(\alpha+\beta)sin(2\alpha-\theta)-2cos\alpha\sin(2\alpha+\beta)sin(\alpha-\theta)=sin(2\alpha+\beta)sin\theta-sin(2\alpha-\theta)sin\beta$ $ \Longleftrightarrow$

$ sin(2\alpha+\beta)[2sin(\alpha-\theta)cos\alpha+sin\theta]=sin(2\alpha-\theta)[2cos(\alpha+\beta)sin\alpha+sin\beta]$ $ \Longleftrightarrow$

$ sin(2\alpha+\beta)[sin(2\alpha-\theta)-sin\theta+sin\theta]=sin(2\alpha-\theta)[sin(2\alpha+\beta)-sin\beta+sin\beta]$   $ \Longleftrightarrow$

$ sin(2\alpha+\beta)sin(2\alpha-\theta)=sin(2\alpha-\theta)sin(2\alpha+\beta)$ which is obviously true. :)

A $ \text{\LaTeX}$ remark: use \cdot ($ x\cdot y$) for the multiplication sign; it looks much nicer. Also use \sin and \cos ($ \cos x)$ for the same reason :)


group theory
abstract algebra
function
superior algebra
superior algebra theorems

A,B are normal subgroups of C with no common elements besides the identity. I want to show that C is isomorphic to a subgroup of C/A + C/B (direct product).

I'm having a very hard time constructing a subgroup with order |G|.

I defined a function from C to a subgroup as: f(x) = (xA, B). I am pretty sure this is onto, but I do not think it is 1-1. (f(x) = f(y) doesn't imply x=y).

Any hints? Am I even on the right track?


[quote="epsilonist"]For $ a,b,c,d \geq 0$. Prove that
\[ a^{4} \plus{} b^{4} \plus{} c^{4} \plus{} d^{4} \plus{} 2abcd \geq a^{2}b^{2} \plus{} a^{2}c^{2} \plus{} a^{2}d^{2} \plus{} b^{2}c^{2} \plus{} b^{2}d^{2} \plus{} c^{2}d^{2}
\]
.
 :|[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6397
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204175


Let $M=\{\frac{m+n}{\sqrt{m^{2}+n^{2}}}|m,n\in\mathbb{N}\}$. Prove that for every pair $x\in M,y\in M$ with $x<y$, there exists $z\in M$ such that $x<z<y$.

[hide="Solution"] It's enough to prove the result for the [i]square[/i] of every element in $M$.  Call this new set $N = \{ 1+\frac{2mn}{m^{2}+n^{2}}| m, n \in \mathbb{N}\}$.

Now $\frac{2mn}{m^{2}+n^{2}}= \cos 2\theta$ for the relevant angle $\theta$ in the triangle $m, n, \sqrt{m^{2}+n^{2}}$.

Hence two values $x, y \in N$ correspond to two right triangles with integer legs $a, b$ and $c, d$.  Then $a+c, b+d$ is another right triangle with integer legs, and moreover it has an acute angle between that of the first two (obvious geometrically).

Then let $z = 1+\frac{2(a+c)(b+d)}{(a+c)^{2}+(b+d)^{2}}$.  Because $\cos$ is monotonic on the relevant interval the conclusion follows.  [/hide]

[quote="t0rajir0u"]

Now $\frac{2mn}{m^{2}+n^{2}}= \cos 2\theta$ for the relevant angle $\theta$ in the triangle $m, n, \sqrt{m^{2}+n^{2}}$.

[/quote]
Why? I think it must is $\sin{2\theta}$. Do you agree?

Hmm.  True.

[hide="And then..."] We need the WLOG assumptions $m \le n, a \le b, c \le d$.  Then $\theta < \frac{\pi}{4}$ so that $\sin 2 \theta$ is still monotonic on the relevant interval. [/hide]

I will rewrite your solution, use algebra.
[hide]Suppose $a,b,c,d\in\mathbb{N}$ such that $\frac{ab}{a^{2}+b^{2}}<\frac{cd}{c^{2}+d^{2}}$(*). If $a=b$ or $c=d$ then easy else suppose WLOG $\frac{a}{b}<1$ and $\frac{c}{d}<1$. From (*) $f(\frac{a}{b})<f(\frac{c}{d})(f(x)=\frac{x}{1+x^{2}})$ therefore $\frac{a}{b}<\frac{c}{d}$ and we have also $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}<1$, or $\frac{ab}{a^{2}+b^{2}}<\frac{(a+c)(b+d)}{(a+c)^{2}+(b+d)^{2}}<\frac{cd}{c^{2}+d^{2}}$. Done !

Note: $f'(x)=\frac{1-x^{2}}{(1+x^{2})^{2}}>0\forall x\in (0,1)$.[/hide]


inequalities
trigonometry
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers such that $ a^2\plus{}b^2\plus{}c^2\plus{}2abc \equal{}1$. Prove that
\[ a\plus{}b\plus{}c \le 4\left( 3\minus{}2\sqrt{2}\right) abc\plus{}\sqrt{2}.\]
This inequality is slightly stronger than the following: http://www.mathlinks.ro/viewtopic.php?t=240117 with equality holds when $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$ and $ a\equal{}b\equal{}\frac{1}{\sqrt{2}},c\equal{}0$ and its cyclic.
:)

Maybe the following can help us prove this one  :)
Let
\[ f(a,b,c) \equal{} kabc \plus{} \sqrt {2} \minus{} a \minus{} b \minus{} c
\]
where is $ k \equal{} 4\left( 3 \minus{} 2\sqrt {2}\right)$
So we need prove that:
\[ f(a,b,c)\ge \max \left \{f\left(a,\sqrt {\frac {1 \minus{} a}{2}},\sqrt {\frac {1 \minus{} a}{2}}\right);f \left(a,\sqrt {1 \minus{} a^2},0\right) \right \}
\]
And we also have:
\[ f(a,b,c) \minus{} f\left(a,\sqrt {\frac {1 \minus{} a}{2}},\sqrt {\frac {1 \minus{} a}{2}}\right)
\]

\[ \equal{} \frac {(b \minus{} c)^2}{1 \plus{} a}\left[\frac {1 \minus{} a}{\sqrt {2 \minus{} 2a} \plus{} b \plus{} c} \minus{} \frac{ka}{2}\right]
\]
Let find the relationship between
\[ f(a,b,c) \minus{} f \left(a,\sqrt {1 \minus{} a^2},0\right)
\]
And
\[ f(a,b,c) \minus{} f\left(a,\sqrt {\frac {1 \minus{} a}{2}},\sqrt {\frac {1 \minus{} a}{2}}\right)
\]
The last things are really easy when we can solve that:
\[ f\left(a,\sqrt {\frac {1 \minus{} a}{2}},\sqrt {\frac {1 \minus{} a}{2}}\right)\ge 0
\]
Of course
\[ f \left(a,\sqrt {1 \minus{} a^2},0\right)\ge 0
\]
The equality as same as canhang's post  :wink:

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 1$. Prove that
\[ a \plus{} b \plus{} c \le 4\left( 3 \minus{} 2\sqrt {2}\right) abc \plus{} \sqrt {2}.
\]
[/quote]
It may be viewed as a generalization of the Jack Garfunkel's inequality, when restricting $ x,y,z\le 1/\sqrt {2}$,
\[ a \plus{} b \plus{} c\ge \frac {4}{3}(1 \plus{} abc)
\]
The Jack Garfunkel's inequality is in fact still true when $ x,y,z\le \sqrt {3} \minus{} 1$, and $ \sqrt {3} \minus{} 1$ is the best constant.

Some years ago (maybe in 2002), my friends also considered the problem of finding the best constant k satisfying
\[ a \plus{} b \plus{} c\ge k \plus{} (4/3 \minus{} k)abc
\]
without any upper bound of variables, and they got the same result (i.e. $ k \equal{} \sqrt {2}$). However, this is a so long solution using the trinomial polynomial. Now this problem can be solved simply by mixing variable  :D

[quote="Hatucdao"][
Some years ago (maybe in 2002), my friends also considered the problem of finding the best constant k satisfying
\[ a \plus{} b \plus{} c\ge k \plus{} (4/3 \minus{} k)abc
\]
without any upper bound of variables, and they got the same result (i.e. $ k \equal{} \sqrt {2}$). However, this is a so long solution using the trinomial polynomial. Now this problem can be solved simply by mixing variable  :D[/quote]
I've got the following result (but it is very easy): If $ a,b,c \ge 0,a^2\plus{}b^2\plus{}c^2\plus{}abc\equal{}4$, then $ a\plus{}b\plus{}c \ge 2\plus{}abc$ ;) :)
See it here: http://canhang2007.wordpress.com/2008/11/04/an-easy-inequality-vo-quoc-ba-can/ :)

[quote="zaizai-hoang"]Maybe the following can help us prove this one  :)
Let
\[ f(a,b,c) \equal{} kabc \plus{} \sqrt {2} \minus{} a \minus{} b \minus{} c
\]
where is $ k \equal{} 4\left( 3 \minus{} 2\sqrt {2}\right)$
So we need prove that:
\[ f(a,b,c)\ge \max \left \{f\left(a,\sqrt {\frac {1 \minus{} a}{2}},\sqrt {\frac {1 \minus{} a}{2}}\right);f \left(a,\sqrt {1 \minus{} a^2},0\right) \right \}
\]
And we also have:
\[ f(a,b,c) \minus{} f\left(a,\sqrt {\frac {1 \minus{} a}{2}},\sqrt {\frac {1 \minus{} a}{2}}\right)
\]

\[ \equal{} \frac {(b \minus{} c)^2}{1 \plus{} a}\left[\frac {1 \minus{} a}{\sqrt {2 \minus{} 2a} \plus{} b \plus{} c} \minus{} \frac {ka}{2}\right]
\]
Let find the relationship between
\[ f(a,b,c) \minus{} f \left(a,\sqrt {1 \minus{} a^2},0\right)
\]
And
\[ f(a,b,c) \minus{} f\left(a,\sqrt {\frac {1 \minus{} a}{2}},\sqrt {\frac {1 \minus{} a}{2}}\right)
\]
The last things are really easy when we can solve that:
\[ f\left(a,\sqrt {\frac {1 \minus{} a}{2}},\sqrt {\frac {1 \minus{} a}{2}}\right)\ge 0
\]
Of course
\[ f \left(a,\sqrt {1 \minus{} a^2},0\right)\ge 0
\]
The equality as same as canhang's post  :wink:[/quote]
Have you thought of an easier proof? There exists a nice inequality for my inequality ;) :)

[quote="Hatucdao"]However, it is a really weak inequality. By a change of variables, i.e. a->2a, this problem of letting $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 1$ ($ a,b,c\ge 0$) and proving $ a \plus{} b \plus{} c\ge 1 \plus{} 4abc$, in which the number $ k \equal{} 1$ is low far from the best constant $ k \equal{} \sqrt {2}$. In trigonometry view point, it is equivalent to the inequality
[/quote]
Did you mean that: If $ a,b,c \ge 0,a^2\plus{}b^2\plus{}c^2\plus{}2abc\equal{}1$, then $ a\plus{}b\plus{}c \ge \sqrt{2} \plus{}\left( \frac{4}{3} \minus{}\sqrt{2}\right) abc$? If so, try $ a\equal{}1,b\equal{}c\equal{}0$ ;)

[quote="Hatucdao"][quote="can_hang2007"] I've got the following result (but it is very easy): If $ a,b,c \ge 0,a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4$, then $ a \plus{} b \plus{} c \ge 2 \plus{} abc$ ;) :)
[/quote]
It looks like an Olympiad problems appeared in 2002 (maybe of England or Australia, I don;t remember exactly). This prolem can be solve easily as follows:
\[ (2 \minus{} a)(2 \plus{} a) \equal{} 4 \minus{} a^2 \equal{} b^2 \plus{} c^2 \plus{} abc\ge 2bc \plus{} abc \equal{} bc(2 \plus{} a)
\]
implies $ 2\ge a \plus{} bc$ and hence $ 2a\ge a^2 \plus{} abc$. Adding the latter inequality and two similar ones, we obtain the result.

However, it is a really weak inequality. By a change of variables, i.e. a->2a, this problem is of letting $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 1$ ($ a,b,c\ge 0$) and proving $ a \plus{} b \plus{} c\ge 1 \plus{} 4abc$, in which the number $ k \equal{} 1$ is low far from the best constant $ k \equal{} \sqrt {2}$. In trigonometry view point, it is equivalent to the inequality
\[ sin(A/2) \plus{} sin(B/2) \plus{} sin(C/2)\ge cos(A) \plus{} cos(B) \plus{} cos(C).
\]
A more interesting inequality is that $ 2 \plus{} abc\ge ab \plus{} bc \plus{} ca$ (when $ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4$, $ a,b,c\ge 0$). This is the Olympiad problem I mentioned above. The solution is also simple: Because $ 2\ge a \plus{} bc$, we just need to prove that $ abc \plus{} a\ge ab \plus{} ac$, which is equivalent to $ a(b \minus{} 1)(c \minus{} 1)\ge 0$. To make the latter inequality true, we can simply assume that $ (b \minus{} 1)(c \minus{} 1)\ge 0$ (answer by yourself why? - this interesting technique makes me still remembered this inequality although 6 years passed  :D ).[/quote]
Also see my proof here, I used to Mixing Variable method too
[url]http://www.mathlinks.ro/viewtopic.php?t=236977[/url]
I haven't thought about simple way to prove this one but I think Mixing Variables method is a nutural way when you want to solve this kind of inequality (with a special condition without $ \sum a \equal{} I,abc \equal{} LOVE ,\sum ab \equal{} U,$ etc...  :D)  It also help me solve another one, Aqady's problem  :)
[quote]
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 5.$ Prove that
\[ 2(ab \plus{} ac \plus{} bc)\leq5 \plus{} abc
\]
[/quote]
Of course, it is not the best way but it is really useful for inequality with conditon type:
\[ k(a^2\plus{}b^2\plus{}c^2)\plus{} pabc\equal{}r\]
Or some other...

[quote="Hatucdao"][quote="can_hang2007"] Did you mean that: If $ a,b,c \ge 0,a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 1$, then $ a \plus{} b \plus{} c \ge \sqrt {2} \plus{} \left( \frac {4}{3} \minus{} \sqrt {2}\right) abc$? If so, try $ a \equal{} 1,b \equal{} c \equal{} 0$ ;)[/quote]
Oh, I am sorry about this typo. It is of course $ a \plus{} b \plus{} c \ge k \plus{} 4(3 \minus{} 2k)abc$.[/quote]
If you meant $ a \plus{} b \plus{} c \ge k \plus{} 4(3 \minus{} 2k)abc$, then $ k \equal{} 1$ ;) and it is equivalent to my "easy" inequality "If $ a,b,c \ge 0 ,a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4$, then $ a \plus{} b \plus{} c \ge 2 \plus{} abc$"! ;)
And if you meant $ a \plus{} b \plus{} c \le k \plus{} 4(3 \minus{} 2k)abc$, then $ k \equal{} \sqrt {2}$ and it is equivalent to the first one posted by me at this topic. ;)

[quote="can_hang2007"] And if you meant $ a \plus{} b \plus{} c \le k \plus{} 4(3 \minus{} 2k)abc$, then $ k \equal{} \sqrt {2}$ and it is equivalent to the first one posted by me at this topic. ;)[/quote]
Yes, I just restate your first problem at all. In the previous post, I want to recall some inequality related (in fact weeker than) this general one. This inequality is not so difficult because some "regular methods" such as Mixing Variable work. However, it is not easy to give a nicer solution.

The following inequality is also true
[img]http://s6.sinaimg.cn/middle/0018zOAxgy70TX9F9w995&690[/img]


geometry
geometric transformation
reflection
geometry proposed

A circle centered at $O$ is tangent to all sides of the convex quadrilateral $ABCD$.The rays $BA$ and $CD$ intersect
at $K$,the rays $AD$ and $BC$ intersect at $L$.The points $X,Y$ are considered on the line segments $OA,OC$,respectively.Prove that $\angle {XKY}=\frac{1}{2}\angle {AKC}$ if and only if $\angle {XLY}=\frac{1}{2}\angle {ALC}$   :)  :)

This is from currect issue of Mathematical Reflections - I suppose solutions shouldn't be posted till 1st March.


Let $KL$ and $KN$ be tangent to the circle $C$ (with$L,N$ on $C$), and let$M$
be a point on the extension of $KN$ beyond $N$. The circumcircle of triangle
$KLM$ meets$C$ again at $P$. Point$Q$is the foot of the perpendicular from
$N$ to$ML$. Prove that $\angle MPQ = 2\angle KML$.