in 1995 :
90= MAR1 Mounssif Aderkaoui 7 5 3 7 7 0 29 Silver Medal
193= MAR2 Charaf Tahiri 7 7 1 4 0 0 19 Bronze Medal
137= MAR3 Rachid Rabih 7 4 1 6 7 0 25 Bronze Medal
245= MAR4 Tarik Belhaj Soulami 7 0 0 6 0 1 14
114= MAR5 Abdellatif Badidi 7 5 7 7 0 1 27 Bronze Medal
148= MAR6 Abdelmajid Youssfi 7 7 3 7 0 0 24 Bronze Medal
 MAR Morocco Totals 42 28 15 37 14 2 138

1 year later :

353= MAR1 As-Salih Ahmed 1 0 2 0 0 0 3
180= MAR2 Belhaj Soulami Tarik 2 7 2 1 0 0 12 Bronze Medal
392= MAR3 Hadja Abdel Kader 0 0 1 0 0 0 1
392= MAR4 Oueld Lebled Hassan 0 0 1 0 0 0 1
392= MAR5 Riffi Hicham 0 0 1 0 0 0 1
392= MAR6 Zeouita Ahmed 0 0 1 0 0 0 1
 MAR Morocco Totals 3 7 8 1 0 0 19

9 years later(2004):

200= MAR1 Armel, Kamal 7 1 0 7 3 0 18 Bronze Medal
351= MAR2 Belabess, Ali Jadallah 0 1 0 7 0 0 8
169= MAR3 Ben Moussa, Hamid 6 7 0 7 0 0 20 Bronze Medal
159= MAR4 Elharti, Yassine 7 4 0 7 3 0 21 Bronze Medal
316= MAR5 Kamil, El Mehdi 0 0 0 7 3 0 10
295= MAR6 Maach, Saifeddine 7 1 0 0 3 0 11
 MAR Morocco Totals 27 14 0 35 12 0 88

10 years later(2005):

MOROCCO
Score P1 P2 P3 P4 P5 P6 Final Medal
MAR1 0  0  0  2  0  0  2
MAR2 0  1  0  1  0  0  2
MAR3 0  0  0  1  0  0  1
MAR4 0  1  0  0  0  0  1
MAR5 0  0  0  1  0  0  1
MAR6 2  0  0  7  2  0  11  Honor
TOTAL 18

 :(  :(  :(  :(  :(  :(  :( so any comments ? :blush:

i am MAR6,(i needed 1 point to get the bronze), no problem ;), we couldn't focus the first day  :sleeping:,and we were disappointed the second...even MAR3 and 4 who scored last year...all this is just excuses,it's true we lacked rest,but we mainly lacked training,we usually rely on luck,but when we lack luck:................... :blush:  :blush:

[quote="persian"][quote="Al-Khawarizmi"]i want just to know who is MAR6[/quote]
Nice screen name! He's my favourite [url=http://en.wikipedia.org/wiki/Al-Khwarizmi]persian mathematician[/url].[/quote]
 yes, he is my favourite too  :)

[quote="senouy"]1995 moroccan results werent that bad.
No one of the 1995 contestant appears in the following year.
Is it age constraint.[/quote]

No this is not true. Tarik Belhaj Soulami took part in IMOs 95 and 96.

Concerning the Moroccan results in IMO, they are very changing from year to year, this is because there is not a real mathematics competition tradition in Morocco, and one has to wait a year where some clever students represent their country to win a bronze medal or at most a silver one (as Mahmoud Chilali in 1986, Youssef El Hajjam in 1990, and Mounssif Aderkaoui in 1995).

[quote="Guest forever"]Nice comments Mr Assila, and from now on i'll call gold : "Le défi 301"  ;)[/quote]

thank you  :lol:  :lol:  :lol:

Hi everybody,

I have participated in the IMO 95. The results were very good for the moroccan team that year.

Are you interested in creating an association (Jam3ya) in morooco to help the moroccan team in its preparation for the next IMOs?

Abdellatif


A white plane is partitioned onto cells (in a usual way). A finite number of cells are coloured black. Each black cell has an even (0, 2 or 4) adjacent (by the side) white cells. Prove that one may colour each white cell in green or red such that every black cell will have equal number of red and green adjacent cells.

This is a triple-post.  Moderator, please delete this thread.


group theory
abstract algebra
superior algebra
superior algebra unsolved

(i) Show that the group of inner automorphisms is normal in the group of all automorphisms of a group $ G$.

(ii) Let $ H_1, H_2$ be two normal subgroups of $ G$. Show that $ H_1\cap H_2$ is normal.

(i) if $ c_g$ denotes the inner automorphism $ x \mapsto g^{\minus{}1} x g$, show that $ \phi^{\minus{}1} c_g \phi \equal{} c_{\phi^{\minus{}1}(g)}$.
(ii) look at the kernel of the natural map $ G \to G/H_1 \times G/H_2$ (or use the definition of normal).

Well, you need the specific definition of normal: $ H$ is normal iff $ aha^{\minus{}1} \in H$ for all $ a \in G, h \in H$.

[quote="JSteinhardt"]Well, you need the specific definition of normal: $ H$ is normal iff $ aha^{ - 1} \in H$ for all $ a \in G, h \in H$.[/quote]

If $ H_1 \cap H_2$ is normal then $ aha^{ - 1} \in H_1 \cap H_2$ for all $ a \in G, h \in H_1 \cap H_2$ and $ \exists h' \in H_1 \cap H_2$ such that $ ah = h'a$ for all $ a \in G, h \in H_1 \cap H_2$

$ H_1$ and $ H_2$ are normal so... for all $ h \in H_1 \cap H_2, g \in G$ exists $ h'_1 \in H_1, h'_2 \in H_2$ such that:

$ \{ \begin{array}{l} ah = h'_1a\\ ah = h'_2a \end{array}$


I need to know how to find the best fit equation on a TI-86 calculator. When I enter the data then go to STAT, CALC, LinR, and then hit enter I get an error that reads ERROR 12 DIM MISMATCH. I think after I press LinR I am supposed to enter something with an x and a y and a comma or something because somebody showed me this two years ago but I forgot.

Many thanks in advance

I dun use ti-86 but you probably have the lists set up wrong, read this:

http://www.saintmarys.edu/~cpeltier/calcforstat/StatTI-86.html


[quote="frankvista"]Now we know that
$ a^5 \plus{} b^5 \equal{} 2a^2b^2,$
$ a,b \in \mathbb Q,$
$ 1 \minus{} ab \equal{} t^2,$
Prove $ t \in \mathbb Q$.[/quote]

1) If $ a\equal{}\minus{}b$ then $ 2a^2b^2\equal{}0$ and so $ a\equal{}b\equal{}0$ and so $ t^2\equal{}1$ and so $ t\in\mathbb Q$

2) If $ a\equal{}0$ then $ b\equal{}0$ and so $ t^2\equal{}1$ and so $ t\in\mathbb Q$

3) If $ a\neq \minus{}b$ and $ a\neq 0$, then let $ c\equal{}\frac ba\neq \minus{}1\in\mathbb Q$ We have :

$ a^5(1\plus{}c^5)\equal{}2c^2a^4$ and so $ a\equal{}\frac{2c^2}{1\plus{}c^5}$ and $ b\equal{}ca\equal{}\frac{2c^3}{1\plus{}c^5}$

So $ ab\equal{}\frac{4c^5}{(1\plus{}c^5)^2}$ and $ 1\minus{}ab\equal{}\left(\frac{1\minus{}c^5}{1\plus{}c^5}\right)^2$

And so $ t\equal{}\pm\frac{1\minus{}c^5}{1\plus{}c^5}\in\mathbb Q$

[quote="frankvista"]Now we know that
$ a^5 \plus{} b^5 \equal{} 2a^2b^2,$
$ a,b \in Q,$
$ 1 \minus{} ab \equal{} t^2,$
Prove $ t \in Q$.[/quote]

Sometimes,
$ 1 \minus{} ab \equal{} (\frac {a^5 \minus{} b^5}{2a^2b^2})^2$

[quote="frankvista"]Sometimes,
$ 1 \minus{} ab \equal{} {a^5 \minus{} b^5}/{2a^2b^2}$[/quote]

 :huuh:
??


function
limit
calculus
derivative
real analysis
real analysis unsolved

Let $ \alpha \geq 1$ and consider
\[ f_\alpha (x) \equal{} \begin{cases} \frac {1}{q^{\alpha}} & \text{if } x \equal{} \frac {p}{q} \neq 0, gcd(p,q) \equal{} 1, q > 0 \\
0 & \text{if } x \equal{} 0 \,\,\mbox{or x is irrational}\end{cases}\]
I'm trying to determine the values of $ \alpha$ for which $ f_\alpha$ is differentiable at 0. As a consequence of Dirichlet's approximation theorem, a number c is rational if and only if there exist only a finite number of rational numbers p/q such that $ |c \minus{} \frac {p}{q}| \leq \frac {1}{q^2}.$ I think this implies that choosing x = m/n (x rational) in a sufficiently small interval around 0 gives
\[ \left|\frac {f_\alpha (x) \minus{} f_\alpha (0)}{x \minus{} 0}\right| \equal{} \frac {|f_\alpha(x)|}{|x|} < \frac {1}{n^{\alpha}}\cdot n^2 \equal{} n^{2 \minus{} \alpha}.\]
Thus I'm inclined to say that $ \alpha > 2$ works. According to wikipedia, the irrationality measure of a rational number is 1, and if I understood the discussion, this means that $ \alpha > 1$ is potentially suitable as well. Can someone verify this? Thanks in advance.

Any function $ f$ such that $ \lim_{x\to 0}\frac {|f(x)|}{|x|} \equal{} 0$ is differentiable at $ 0$. And this is the case for your function whenever $ \alpha > 0$, because $ \frac {1}{q^{\alpha}}$ is much smaller than $ \frac {1}{q}$ which itself does not exceed $ |x|$.

But what you claim doesn't seem to hold for $ 0 < \alpha < 1.$

That was probably a typo, I imagine mlok meant $ \alpha > 1$.

$ \alpha > 1$ is also the best bound, because for $ \alpha \leq 1$ taking the difference quotient limit over the sequence $ \lbrace x_k\rbrace$ with $ x_k \equal{} \begin{cases} \frac {1}{k} & k \text{ even} \\
\frac {2}{k} & k \text{ odd}\end{cases}$ demonstrates non-differentiability of $ f_\alpha$ at $ 0$.

Thanks JoeBlow and mlok. I understand now that the simpler approach is to use the estimates given in the definition of the limit instead of trying rational approximation methods. I just have one question.

The claim in the first sentence of mlok's post is essentially a slight reformulation of the definition of the derivative, with the slight exception that f(0) might not be 0. But if we write out the definition, we end up with $ |f(x)| \leq \varepsilon\cdot|x|,$ so this implies f(0) = 0, right?

No: mlok should have said, "any function $ f(x)$ such that $ f(0) \equal{} 0$ and $ \lim_{x \to 0} \frac {|f(x)|}{|x|} \equal{} 0$ is differentiable at $ 0$," or alternatively "any function $ f(x)$ that is continuous at $ 0$ and satisfies $ \lim_{x \to 0} \frac {|f(x)|}{|x|} \equal{} 0$ is differentiable at $ 0$."  If you defined the value of your function differently at $ 0$ then it wouldn't be continuous, so certainly not differentiable.


summer program
MathPath
email
geometry
geometric transformation
reflection
Harvard

Hi, everyone,

Almost three years ago I heard, on an email list for parents, about the then brand-new MathPath summer program for young would-be mathematicians.

http://www.mathpath.org/

Since then I have been in email correspondence with a member of MathPath's board of directors (herself a parent of math-liking children), with the program director, and with various parents or friends of parents of MathPath alumni. Everything I have heard about the program sounds wonderful. MathPath has now updated its Web site to reflect the details of its plans for the summer of 2005, the fourth year of its summer program. The coming location will at Colorado College in Colorado Springs, CO

http://www.coloradocollege.edu/index.asp

The application quiz for 2005 is already posted on the MathPath Web site.

http://www.mathpath.org/Quiz/quiz.2005.pdf

As in previous years, MathPath expects to have an outstanding faculty of Ph.D. mathematicians to lead courses on a variety of topics outside the standard curriculum. In my humble opinion, meeting John Horton Conway alone would be almost worth the full cost of the program. Professor Conway is a strong supporter of MathPath, and suggested the name of the program.

http://www.mathpath.org/faculty.htm

One of the big changes in MathPath this year is a change in the financial aid policy that should make the program more affordable for more middle-class families.

http://www.mathpath.org/fees/schol.htm

The MathPath staff (= the program director) is pretty good about answering email inquiries about the program, and I encourage you to contact MathPath if you have questions. See you in Colorado Springs in June.

Cool. I hear one of my son's recommendation letters is about to be mailed to MathPath.

This reply is really just a cheap way to bump the thread, but I am happy about this news. I hope I'll see some of you AoPSers at MathPath this year when I take my son there, which I also hope to do.

A question in the Parent Forum asks about summer math camps.  I felt my response should be to a broader audience.

[quote]As a parent, one of our tasks right now is to pick summer activities for our kids.  Any suggestions for good summer math camps that are available for mid and high school students?[/quote]
For Middle School Students, who are exceptionally gifted, mathematically, I strongly endorse [url=http://www.mathpath.org]MathPath[/url].

The Application Quiz is quite challenging.  The Camp Founder, Dr. Thomas, personally evaluates every application.  If your child doesn't enjoy working on the Application Quiz, I would recommend that they not apply to MathPath.  Their days at MathPath will be loaded with questions and problems of this nature.  If your child doesn't have the emotional maturity to be away from home for 4 weeks, I would recommend that they not apply to MathPath.

If your child enjoys the Application Quiz, and if they can solve some of the problems, and if they have the maturity to be away from home for 4 weeks, you should consider MathPath.  Dr. Thomas has a knack for identifying exceptionally gifted math students.  If your child's Application Quiz shows that gift, your child will probably be accepted to MathPath.  There were approximately 70 attendees last summer.

My daughter attended the camp last summer, and she intends to attend the camp again, this summer.  This summer the camp will be held at Colorado College in Colorado Springs.

Last summer, the camp was 4 weeks in length.  Due to my travel arrangements, I was there for the first couple of days of camp and for the last couple of days of camp.

The only real negative was the facilities.  Roger Williams University pulled the switch-a-roo on the facilities they had promised MathPath.  JHU's larger CTY program got the first-rate facilities, while MathPath was forced into second-rate, run-down facilities.  The rustic cabins that Roger Williams U called "dorms" were marginally fit for human occupation.

Luckily, the kids barely noticed the facilities.  This is a testament to the skill and dedication of the MathPath staff and counselors, who were sharing the same or similar facilities.  The MathPath Director arranged for evening use of the Student Union, which meant the kids didn't need to use their dorms for much more than sleeping and showering.

With the change of venue to Colorado College, I don't expect the facilities to be a problem this year.  :)

The positives of MathPath completely overwhelmed the negatives...

Weekend non-math activities included:  Whale-watching and a trip to the beach; a visit to Harvard and MIT; a visit to the Mayflower, the USS Constitution, and Bunker Hill; a day of Kayaking; watching the oldest 4th of July Parade in the country in Bristol, RI; a volleyball tournament; and church, for those who wanted to attend.

Daily non-math activities included:  morning jogging; evening swimming; and ping-pong.

The reason the kids were there was to study mathematics in a high-level, high-energy, high-expectation manner.  And that's where MathPath really shined.

The mathematics were high-level, but not beyond the skill of the students.  In the morning, there were 3 lectures a day, one hour per lecture.  Topics included:  "History of Mathematics" by Thomas Drucker and Mike Molinsky, "Spirals" by ???, "Logic and Proofs" by Stephen Maurer, "Problem Solving" by Paul Zeitz, "Hyperbolic Geometry" by George R. Thomas, "Hyperspace" (aka Dimensions beyond the 3rd) by Philip Davis, and Seminars by John Conway.

After lunch, there was an hour of recreation -- library, musical intrument practice, card games, etc.

Then an afternoon of problem solving.  Students were split into three groups, depending at least partially, on ability-level.  These groups were led by Al Lippert, Stephen Maurer, and Paul Zeitz.  One of the groups was large and split in half a bit later in the camp.  My daughter reports there was quite a bit of mobility between groups, as students found one group too hard or too easy or not enough friends or whatever.

There was a daily "Problem of the Day".  A very challenging problem, requiring out-of-the-box thinking.

After dinner, there was a period of "Relatively Silent Mathematics", where students could work on the Problem of the Day or "homework" assignments.  Following that was more recreation -- swimming, ping-pong, musical instrument practice, card games, basketball, wall-ball, capture-the-flag, etc.

There were also special Saturday morning competitive math activities, which included "Capture the Flag -- Math Style"; "Math Relays -- ARML-style"; and a "Math Quiz Bowl".

Unlike MathCamp, where students can opt-in/opt-out of nearly every activity, this is a highly structured camp.  Aside from the recreational options, all students were expected to attend all of the planned activities.

This is the perspective of one parent and one child.  I'm sure I missed many of the experiences other campers found important.  But I hope this gives an idea of the rich variety of experiences the students enjoyed.

RCV,

Thanks for sharing your experience of  mathpath. I was debating between mathpath and mathcamp for my 8th-grader son, and your story helped me make up my mind now.

Tokenadult:

Thanks for bring this up.

MathCamp is also run by Dr. Thomas isn't it? But it's mostly for high-school student's and maybe a couple of middle school students of Greg Gauthier's caliber (who by the way went to MP2003 and became National Champion the next year :D )

He started it but is no longer affiliated with it?

As the younger AoPSers might say,

\/\/007!

My son A+MATH has been accepted to MathPath, so I will be making the drive out to Colorado with him this summer. I hope to see some other AoPS parents and young people when I drop him off and pick him up. We'll wear our AoPS T-shirts to be recognizable.  :)

The hard postmark deadline for applications is next week, so apply if you are interested. It looks like a great summer of intense math.

 :first:

Cool, I'll be wearing an AoPs T-shirt too then. See you there!

[quote="tokenadult"]As the younger AoPSers might say,

\/\/007!
[/quote]

I'm not young... ;)

[quote="xxreddevilzxx"]i heard that Dr. Thomas is not affiliated with Canada/USA MathCamp anymore, although it was he who started it, before he started MathPath.[/quote]

This is correct.

[quote="tokenadult"]... My son A+MATH has been accepted to MathPath ...[/quote] Was that supposed to be a surprise?  :D

Congratulation just the same. He'll have lots of fun there.

[quote="tokenadult"]Who has received the MathPath acceptance packet? Has anyone received it by postal mail (snail mail), or have you only received it by email?[/quote]

I've received the registration packet a few days after I was accepted, and the briefing sheets just yesterday.

How many of you are driving, and how many flying? I'm driving, because my brother lives in Colorado Springs (what an amazing coincidence) and I will visit him for a few days while I am there.

I'm going!  :) I don't know if I'll be able to pay 3500 dollars :( , but I got accepted at least  :!:

I haven't gotten my registration packet in "snail mail" yet, mainly because I sent my application quiz pretty late.  :blush:

How are you guys paying the 3500 dollars? The need-based scholarship is only about 800 dollars for me, and my dad says he'll pay 1000 dollars, but he doesn't want to pay the rest. Any ideas (for fundraising)?

Also, are we allowed to talk about the application quiz now?

DON'T discuss the application quiz until you arrive at camp. There will still be some late applicants.

Do you have the fundraising kit? Most of the best fundraising sources don't apply to us. (Camper's grandparents are poor and camper's parents are self-employed and thus have no employer.) But we are racking our brains figuring out if we can do some fundraising.

[quote="tokenadult"]Do you have the fundraising kit?[/quote]
No, you get it when you send your registration, w/ 750 dollars, but I need to fundraise to get that. My dad says that if I raise the 750, he'll give me 1000 dollars in addition. He doesn't think I'll be able to raise the rest of the money once he gives me 1000, so it would go to waste :( It sucks.

My son has also been accepted to MathPath, so I will be making the flight out to Colorado with him this summer. I also hope to see some other AoPS parents and young people when I drop him off and pick him up. We'll also be wearing our AoPS/Nasa T-shirts to be recognizable.

My son has been a member of AoPS since last year, but I logged on recently to find more information about MathPath and MathCamp. Although I am not a mathematician the breadth of posts on the various threads are quite fascinating.

I have no relation with the original 'tokenadult' (although I do hope to meet him in Colorado) but chose this name as the base for my username since I think it is appropriate to identify oneself on such a site as being an adult.

If any other parents will be going to drop their kids off at MathPath, please let me know -- I would greatly enjoy meeting you.

If folks are still looking for a summer camp, I understand MathPath still has a few openings.  Their [url=http://www.mathpath.org/camp.htm]Web-Site[/url] says they are still accepting applications, but if you plan to postmark your application after June 10, you should call them.

Nevermind, I'm not going  :( My parents changed their mind and said Colorado was too far for 1 month and they would miss me and it would be unsafe  :(  :wallbash_red:  :wallbash_red:  :wallbash_red: darn

I will be there. But I don't think it's a very good idea to reveal my name online.  :D  If I meet any AoPSer's, one of us will eventually let the other know!!!

I hope to be wearing my AoPS T-shirt the day I drive up with my son. He may be wearing his AoPS T-shirt also.

See you there. Just remember, the air is THIN in the Rocky Mountains, so plan on some acclimation if you usually live nearer to sea level.

I'll probably be wearing an ordinary black shirt and no pants.

>_>

We're getting back from MathPath to various places we call home now. I met Treething at MathPath and several other AoPSers. Let's post our impressions of MathPath here in this thread.


Let $ a, b, c > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
prove: $ \frac{1}{1\minus{}ab} \plus{} \frac{1}{1 \minus{} bc} \plus{} \frac{1}{1 \minus{} ca} \le \frac{9}{2}$

$ \sum a^2\geq a^2\plus{}b^2\geq 2ab\Rightarrow ab\leq\frac{1}{2}$
Similar:$ bc;ca\leq\frac{1}{2}$
$ \frac{1}{1\minus{}ab}\leq\frac{9}{2}a^2b^2\minus{}\frac{3}{4}ab\plus{}\frac{5}{4}\Leftrightarrow (3ab\minus{}1)^2(2ab\minus{}1)\leq 0,Right$
$ \Rightarrow ineq\Leftrightarrow\frac{9}{2}\sum a^2b^2\minus{}\frac{3}{4}\sum ab\plus{}\frac{15}{4}\leq\frac{9}{2}$
$ \Leftrightarrow 6\sum a^2b^2\minus{}\sum ab\leq 1\Leftrightarrow 6\sum a^2b^2\minus{}\sum ab\sum a^2\leq(\sum a^2)^2$
$ \Leftrightarrow \sum a^4\plus{}abc\sum a\plus{}\sum(a^3b\plus{}ab^3)\geq 4\sum a^2b^2$
that $ LHS\geq 2\sum(a^3b\plus{}ab^3)\geq 4\sum a^2b^2$

It 's here
http://www.mathlinks.ro/Forum/viewtopic.php?t=109029


A tank contains 200 L of fluid in which 30 grams of salt is dissolved. Pure water without salt is pumped into the tank at a rate of 4L/min; the well-mixed solution is pumped out at the same rate. Find the number $ A(t)$ of grams of salt in the tank at time $ t$.

The way I understand the problem, the rate of salt coming in at time $ t$, where $ A$ is the amount of salt at time $ t$ is:

$ \frac{dA}{dt}\equal{} R_{in}\minus{}R_{out}$.

Note that no salt comes in, so $ R_{in}\equal{} 0$. Note also that $ R_{out}\equal{}\frac{4 L}{min}(\frac{30 g}{200L}) \equal{} 0.6$.

So this means that $ \frac{dA}{dt}\equal{}\minus{}0.6$. But this seems wrong b/c it is too simple. Am I doing something wrong?

Thanks.

What you're missing is that the amount of salt in the tank does not stay at $ 30$ grams. That's only the concentration for an instant. The quantity you're calling $ R_{\text{out}}$ is not constant but is rather proportional to $ A(t).$

I'll put it another way: the differential equation $ \frac{dA}{dt}\equal{}\minus{}0.6,\ A(0)\equal{}30$ has as its solution $ A(t)\equal{}30\minus{}0.6t.$ In that model the concentration reaches zero in $ 50$ minutes and is negative for $ t>50.$. That makes no sense; this can't be the right model.


limit
real analysis
real analysis unsolved

If f:[1,oo)--> (0,oo) is strictly growing, so that

\lim oo  [f(n)/ln(n)]=l, then \lim oo {1/ln(n)* [sum f(k)^(1/n),k=1,n)-n]}=l

also I think that in this case  stolz cesaro should also work.. pls help!!

cheers! :D

There seems to be sth wrong. I'm not sure, but maybe l=lim of LN(f(n))/lnn instead of f(n)/lnn? I think the method from 'theorem limit 1' should work here too, but I'm not sure.

the limit is for  [f(n)/ln(n)]=l and you should try to solve it.. it's not so simple...  :(

cheers!

Then the problem is really simple if f(n)/lnn tends to L. All you need is that e^x-1<ex for 0<x<1. Then you observe that for a certain n_o we have that the sequence is smaller than  e* \sum lnf(k)/nlnn and this tends to 0 by Cesaro-Stolz.
   The problem is more interestng when f(n)/n tends to L. But it's not very hard. Use the definition of the limit e^x-1/x =1 , the fact that f(k)^1/n=e^[lnf(k)/n] and Cesaro-Stolz.


function
real analysis
real analysis unsolved

I'm just wondering if this law is true or not  :

if $ E_1,E_2, and E_3$ are measurable sets , then :

$ \mu(E_1\cup E_2\cup E_3) \equal{}$

$ \mu(E_1) \plus{} \mu(E_2) \plus{} \mu(E_3) \minus{} \mu(E1\cap E_2) \minus{} \mu(E_1\cap E_3) \minus{} \mu(E_2\cap E_3) \plus{} \mu (E_1\cap E_2 \cap E_3)$

If it's true , how we can prove it  :?:

I can prove it using the law:

$ \mu(E_1 \cup E_2)\equal{}\mu(E_1) \plus{} \mu(E_2) \minus{}\mu(E_1\cap E_2)$

but how we can prove it without using it ?

 :)
 :)

[quote="Hidden Scofield"]

but how we can prove it without using it ?

[/quote]
We can not.  :lol: Because that is a condition of the definition.

prove the corresponding identity for the characteristic functions pointwise and then get the equality by integrating the characteristic functions.


trigonometry
inequalities
inequalities proposed

[color=purple]A long time ago hungkhtn posted the following problem
   " Let triangle ABC and denote its element practice.Prove or disprove that
  $m_{a}r_{a}+m_{b}r_{b}+m_{c}r_{c}\ge p^{2}$(*)
  And the nice answer of dariji is disprovation it mean in an acute triangle ABC in plane we have
    $m_{a}r_{a}+m_{b}r_{b}+m_{c}r_{c}\le p^{2}\longleftrightarrow \frac{m_{a}r_{a}+m_{b}r_{b}+m_{c}r_{c}}{p^{2}}\le 1$(1)
  Now prove or disprove that
  $\frac{m_{a}r_{a}+m_{b}r_{b}+m_{c}r_{c}}{p^{2}}\le\frac{2}{3}(\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2})$(2)
    We always know $\sin\frac{A}+\sin\frac{B}{2}+\sin\frac{C}{2}\le \frac{3}{2}$ thus (2) is sharper than (1)
     If you can show (2)is correct please try with the following problem
$\bigstar \frac{m_{a}r_{a}+m_{b}r_{b}+m_{c}r_{c}}{p^{2}}\le\frac{2}{3}(cosA+cosB+cosC)$
   These problem is my present of me for PKH and DG . I hope you'll like them but this topic is welcome other mathlinkers :lol: [/color]

[quote="evarist"][color=purple] gemath i hope you can help me to prove or disprove all problem in this topic.
   :wink: [/color][/quote]

I am curious of when I post the problem you memtioned there?

My sketches show that for acute triangles $ABC$, we probably have

$\frac23\left(\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\right)\geq\frac{m_{a}r_{a}+m_{b}r_{b}+m_{c}r_{c}}{s^{2}}\geq\frac23\left(\cos A+\cos B+\cos C\right)$.

For non-acute triangles $ABC$, both inequality signs can be arbitrary. Here, my $s$ is your $p$.

Another question is the proof...

  darij

My cabri show

$\frac23\left(\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\right)\geq\frac{m_{a}r_{a}+m_{b}r_{b}+m_{c}r_{c}}{s^{2}}$

is true for acute triangle and the inequality

$\frac{m_{a}r_{a}+m_{b}r_{b}+m_{c}r_{c}}{s^{2}}\geq\frac23\left(\cos A+\cos B+\cos C\right)$

is false when triangle is in "neighbourhood" of equilateral triangle.

[color=purple]Thanks Darij and gemath  :D .Very good but it only " sketches" and " my cabri" not a proff.Who can post a complete solution ? Thanks alot
@hungkhtn: May be i don't know your post , sorry my English is bad so i can't translate " memtioned" :( [/color]
[hide="To hungkhtn"][color=purple]Em tam dich the' nay` dc ko a : "Khi anh post bai` do' chu' da tham gia o do' chua" co' phai ko anh[/color][/hide]


probability
combinatorics unsolved
combinatorics

Hi

ten marries couples are sitting around a round table.

what is the probability the non of husband and wife will sit together ?


Hey,
[quote="Arachnotron"]
So my question is: What exactly does the W.L. Putnam exam measure? Speed? Creativity? Both? Some people I've talked to tell me that the Putnam is primarily a speed test at the higher levels, while others tell me that the Putnam measures the ability to think very creatively in addition to forcing that thought to be fast.

Thoughts?
[/quote]
The Putnam exam more than anything measures your Problem Solving skills. While many of the problems that appear on the Putnam exam are indeed very difficult, many also have elegeant/trivial solutions - once you "see" the "crux move" as Paul Zeitz refers to it in "The Art and Craft of Problem Solving."

In short, to really have a good understanding of the Putnam exam one needs to have a solid foundation in "problemsolvingology"-so to speak.

Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]

Is it an innate talent? Or is it a "studyable" or "practicable" skill?

"innate" talent can only get you so far... most of the proficiency has to come through [b]lots[/b] of hard work over time.


limit
integration
calculus
calculus computations

$ \lim_{n\rightarrow \infty} n\int^a_{ \minus{} a}|(x^2 \minus{} a^2)^n| dx$

edit: I suppose I should add these, and a is a constant.

If $ 0<a<1,$ this limit is zero, and if $ a\ge1,$ this limit is infinity.

First, we can eliminate the need for the absolute values by writing $ n\int_{\minus{}a}^a(a^2\minus{}x^2)\,dx.$

Next we factor out $ a$ to get $ na^{2n}\int_{\minus{}a}^{a}\left(1\minus{}\left(\frac xa\right)^2\right)^n\,dx.$

Next, make the substitution $ \frac xa\equal{}\frac u{\sqrt{n}}:$

$ \lim_{n\to\infty}\sqrt{n}\,a^{2n\plus{}1}\int_{\minus{}\sqrt{n}}^{\sqrt{n}}\left(1\minus{}\frac{u^2}n\right)^n\,du.$

But $ \lim_{n\to\infty}\left(1\minus{}\frac{u^2}n\right)^n\equal{}e^{\minus{}u^2}.$ From that (and some reasonable use of some convergence theorem) we get

$ \lim_{n\to\infty}\int_{\minus{}\sqrt{n}}^{\sqrt{n}}\left(1\minus{}\frac{u^2}n\right)^n\,du\equal{}\int_{\minus{}\infty}^{\infty}e^{\minus{}u^2}\,du\equal{}\sqrt{\pi}.$

The conclusions in my first line are then just statements about the behavior of $ \sqrt{n}\,a^{2n\plus{}1}.$

I have a nagging suspicion that you might have meant a slightly different problem, but I'm not going to guess what.

[quote="Kent Merryfield"]
I have a nagging suspicion that you might have meant a slightly different problem, but I'm not going to guess what.[/quote]

Actually no, not really. I just thought of this off the top of my head, and it looked cool enough so I posted it. Although now that I think about it, it didn't turn out exactly as planned. Oh well

[quote="Kent Merryfield"]As you can see from what I did, the interesting problem would have been
\[ \lim_{n\to\infty}\frac {\sqrt {n}}{a^{2n \plus{} 1}}\int_{ \minus{} a}^a(a^2 \minus{} x^2)^n\,dx
\]
Or, for simplicity, the same thing but with $ a$ set equal to $ 1.$[/quote]

The funny thing about that is I was going to post it with a=1, but I thought throwing in a constant would be more fun.


Let $ p$ be an odd prime and let $ n\equal{}\frac{2^{2p}\minus{}1}{3}$.
Prove that $ \phi (n)|n$, where $ \phi (n)$ is the phi function.

If $ p\equal{}3$, then $ n\equal{}21$, and $ \phi(n)\equal{}12$. But $ 12\not |\,21$..

Actually $ \phi(n)\mid n$ iff $ n\equal{}2^a3^b$ for some $ a,b\ge 0$.


Please, no discussion of the problems and solutions of the AMC 8 until after Tuesday, November 25, 2008.  Schools may administer the AMC 8 until that date,

On the contest covers, it says:  "The publication, reproduction or communication of the problems or solutions of the AMC 8 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via copier, telephone, e-mail, World Wide Web or media of any type during this period is a violation of the competition rules."

In the Teachers Manual, page 5, Section VI, it says:  "After you have delivered all of the Answer Forms
to the school office to be mailed, you may discuss the contest and solutions with your students. Remember that there will be schools taking the contest in other locations at different times, and some will be taking the contest on other “window” dates.   Inform the students that the contest and solutions may not be  discussed with anyone outside of their contest administration either orally, via email, www, copier or media of any type until after November 25.
The contest booklets must be collected from the students. They may be returned after November 25."

In order to keep the contest fresh and interesting for everyone, I ask that you not discuss the problems until November 25,
Enjoy the AMC 8 and thank you for your cooperation.

Steven Dunbar
MAA Director,  American Mathematics Competitions

So about how long does it take for the AMC8 to get checked and the scores sent out? Also, will the answers be posted on the AMC website after the 25th?

I just want to say, good job with the AMC 8; it was really fun, and I'm looking forward to the AMC 10 and 12 and the AIME. Being my last AMC 8, I tried my best, and the test didn't disappoint me. Great job!

The AMC 8 was exhilrating. It was my first AMC 8 and it was a lot of fun.

Chances are you won't be able to say those either; you're not allowed to say scores for mostly any other competition until you're given the okay to. At least wait until the AMC director or someone else tells if you may or may not.

Why don't all of you just keep quiet about the competition?

Just say NOTHING about it until the period ends.
You know that people don't want to discuss/break any rules or anything so just STOP posting until this time period is over!!

Yes, for example, the students in Bedford (I'm co-proctoring :D) haven't taken it yet! Shush!

[quote="Smartguy"]Just say NOTHING about it until the period ends.[/quote]

Just say nothing about what? :wink:

[quote="AMCDirector"]From the Teachers' Manual, page 2, paragraph 5:
" Most schools will receive their results before the holiday break."

From the Teachers' Manual, page 4, Section II, items  7:
" Your contest package contains
7. Contest Manager’s Envelope (contains a complimentary copy of
   the AMC 8 contest and solution),"

From the Teachers' Manual, page 5, Section VI.
""After you have delivered all of the Answer Forms
to the school office to be mailed, you may discuss the contest and solutions with your students."

Steve Dunbar
MAA Director, American Mathematics Competitions[/quote]

Oh, shoulda read first. Sorry  :oops:


parameterization
modular arithmetic
floor function
number theory unsolved
number theory

Let f(n) denote the sum of (all) digits of natural number n. Prove that for each natural n we can choose convenient value of natural parameter p such that the equation f(npx)=f(x) has solution in natural numbers x that doesn't contain any "9" in its notation.

Does anybody have any idea? I don't... but I hope that you do... Actually I can solve a lot of special cases on a lot of pages... but I can't solve it generally. Is there any trick or only hard work? Thank you.

I'm late for over four years, but "better late than never". :)

Let $ n$ be given. Among the numbers $ 9,99,999,9999,\dots,\overbrace{999\dots 999}^{n\plus{}1}$ there are $ \overbrace{999\dots 999}^i\equiv \overbrace{999\dots 999}^j\pmod n$, $ i>j$. Therefore, $ n\mid \overbrace{999\dots 999}^i\minus{}\overbrace{999\dots 999}^j\equal{}\overbrace{99\dots 99}^{i\minus{}j}\overbrace{00\dots 00}^j$. We choose \[ p\equal{}\frac{\overbrace{99\dots 99}^{i\minus{}j}\overbrace{00\dots 00}^j}n.\]
Since the trailing zeroes obviously make no difference, we have just reduced the problem to the following: prove that for each $ k\in\mathbb{N}$ the equation $ f(\underbrace{99\dots 99}_kx)\equal{}f(x)$ has solution $ x\in\mathbb{N}$ that doesn't contain any $ 9$ in its notation.

We shall now prove that the required solution is \[ x\equal{}\overline{\underbrace{88\dots 88}_{\left\lfloor\frac k 8\right\rfloor}(k\text{ mod }8)\underbrace{888\dots 888}_k}.\]
Notice that $ f(x)\equal{}8\left\lfloor\frac k 8\right\rfloor\plus{}k\text{ mod }8\plus{}8k\equal{}9k$. We now need to evaluate $ f(\underbrace{99\dots 99}_kx)$. We have $ \begin{array}{rl}
\underbrace{99\dots 99}_kx&\equal{}(10^k\minus{}1)x\equal{}10^kx\minus{}x\equal{}\\
&\equal{}\overline{\underbrace{88\dots 88}_{\left\lfloor\frac k 8\right\rfloor}(k\text{ mod }8)\underbrace{888\dots 888}_k\underbrace{000\dots 000}_k}\minus{}\overline{\underbrace{88\dots 88}_{\left\lfloor\frac k 8\right\rfloor}(k\text{ mod }8)\underbrace{888\dots 888}_k}\end{array}$.

Subtracting:
\[ \begin{array}{ccccc}
&88\dots 88(k\text{ mod }8)8\dots 8&88\dots 88&8&000\dots 000\\
\minus{}&&88\dots 88&(k\text{ mod }8)&888\dots 888\\\hline
&88\dots 88(k\text{ mod }8)8\dots 8&00\dots 00&(7\minus{}k\text{ mod }8)&111\dots 112\end{array}\]
Finally,
\[ \begin{array}{rl}
f(\underbrace{99\dots 99}_kx)&\equal{}f(\overline{\underbrace{88\dots 88}_{\left\lfloor\frac k 8\right\rfloor}(k\text{ mod }8)\underbrace{8\dots 8}_{k\minus{}\left\lfloor\frac k 8\right\rfloor\minus{}1}\underbrace{00\dots 00}_{\left\lfloor\frac k 8\right\rfloor}(7\minus{}k\text{ mod }8)\underbrace{111\dots 11}_{k\minus{}1}2})\equal{}\\
\noalign{\bigskip}
&\equal{}8\left\lfloor\frac k 8\right\rfloor\plus{}k\text{ mod }8\plus{}8\left(k\minus{}\left\lfloor\frac k 8\right\rfloor\minus{}1\right)\plus{}(7\minus{}k\text{ mod }8)\plus{}(k\minus{}1)\plus{}2\equal{}9k\end{array}\]
The problem is solved.


Given three distinct points $P_1(x_1,y_1),P_2(x_2,y_2),P_3(x_3,y_3)$.
Assumed that $P_2,P_3$ lie in the line $x_1x+y_1y=1$, and $P_3$ lies in the line $x_2x+y_2y=1$.
Let $O$ be the origin,prove the following statements.

(1)  $OP_1<1 \Longrightarrow OP_2>1$.
(2)  $OP_2>1\ and\ OP_3>1\Longrightarrow OP_1<1$.

The equations given are that \[x_1x_2+y_1y_2=1\]
\[x_1x_3+y_1y_3=1\]
\[x_2x_3+y_2y_3=1\]

For the first part, $OP_1.OP_2\geq x_1x_2+y_1y_2$ from Cauchy, and the result follows.

I'm working on the second part...


If you do look at this, could you please find me a right picture of a ferret to be my avatar and plz post the site on here?

First,
a heads-up.
If you want someone to make you a avatar,
PM instead of making a new topic.
To know who's experienced...
read signatures. Blogs. Look at avatars.
Anyways...
here's a couple of random ones made out of boredom,
 [img]http://img003.picture2life.net/7275206/untitled_animation_web-large_medium.gif[/img]
and
[img]http://img003.picture2life.net/7275213/untitled_animation_web-large_medium.gif[/img]

@meewhee: I believe Butler wishes to ask in general so he doesn't know who to PM yet.

Just a few looks around the site will lead you to a signature that can guide you to a Professional Avatarsmith. Or use Google Images.


A traffic cone has slant height 20 inches and a base radius 5 inches. The cone is then supported with a square base 12 inches to a side. Assuming the cone has no thickness to it, how many square inches is the exterior surface area of the cone and base? Answer to the nearest square inch.

[quote="janieluvsmusic"]How is the answer possibly 380??[/quote]

Wow, I just figured it out and got 380... :ninja:
[hide="Hint"]
SA of a cone is $\pi sr+\pi r^{2}$ where $s$ is slant height and $r$ is radius... :!:
Then you've got your square...[/hide]


Bart and Lisa hiked to a point $\frac{3}{4}$ the way up the mountain in $43$ minutes and $12$ seconds. They took the same amount of time to hike each half of the distance from bottom to top. It took them $1$ minute and $36$ seconds longer to hike the third quarter than the fourth quarter. How many minutes did it take them to hike the entire mountain from bottom to top?

Were they really Bart and Lisa???
[hide]
WOW.

Let all times be in seconds
Call the amount that they biked the third part $x$.
Thus in the fourth part they biked $x-96$
The first $\frac 34$ took them 2592.
So the first half was
$2592-x$
The second half was $2x-96$
Solve the equation
$2592-x=2x-96$
$3x=2688$
$x=896$

Thus for the second half they traveled for
$2(896)-96=1696$
THe total is
$3392\approx 56.5 \text{min}$
Since the answer was meant to be to the nearest tenth of a minute[/hide]


Cat de bine ai spus (ca raspuns la ultimul post al lui pureblack2003 de la topicul  "Rezultate Finale - ONM 2006")...Pacat ca n-a fost o organizare mai adecvata, cel putin la premiere...Sa se puna o ordine clara de luare a premiilor. Eu, de exemplu, pendulam de la un profesor la altul ca sa-mi iau ba premiu, ba carti, ba medalie, ba suma penibila de bani-750000 (mentionez ca la Moisil unde am luat locul III am primit 1000000).
Alta chestie...mi s-a parut total de lipsit de respect sa se apuce toti sa aplaude in batjocura, chipurile, ca sa mearga mai repede. Oamenii aia aveau ceva de spus si nu se face sa intrerupi niste profesori cu niste aplauze in zeflemea. Ar fi trebuit ca profesorii insotitori sa stea cu ei sa-i mai tempereze, ca fac de ras toti matematicienii. Si nu vorbesc doar de cei care au inceput, vorbesc si despre cei care s-au luat dupa ei ca oile. Poti sa fii cat de genial, daca nu ai putin bun simt, totul se duce de rapa. E rusinos ca niste non-valori (nu-i cunosc pe respectivii, dar banuiesc ca nu sunt cine stie ce, avand in vedere ca au facut ce au facut) sa-i faca de ras pe ceilalti. In ziarele iesene a aparut: "Matematicienii (?!?) s-au purtat ca pe stadion". NU se poate asa ceva! Nu e de mirare ca ministerul da niste premii penibile, daca, chipurile, noi, matematicienii, suntem niste nesimtiti si niste prost crescuti. Bine au spus ziarele...totusi poate ar trebui sa fie mai stricta calificarea la ONM. Sa nu ajunga toti cretinii (iertati-mi, va rog, limbajul) sa ne faca de ras in fata intregii tari, cand noi ar trebui sa fim niste modele pentru colegii nostri mai putin inzestrati. Parerea mea e ca o injumatatire a locurilor fiecarui judet ar fi absolut necesara. Sau daca nu, sa fie lasati pe dinafara cel putin o treime din cei care ar fi fost calificati. Atat am avut de spus, sper sa nu va supere tonul meu prea radical si, daca am jignit pe cineva pe nedrept, il rog sa ma ierte.

[quote="Stormraven"]Eu, de exemplu, pendulam de la un profesor la altul ca sa-mi iau ba premiu, ba carti, ba medalie, ba suma penibila de bani-750000 (mentionez ca la Moisil unde am luat locul III am primit 1000000).[/quote]

750.000 ??? Pentru ce loc? Anul trecut am luat 1 milion pentru argint. Oricum, nu cred ca au prea multi bani sa ofere. Intrebarea mea: cine da banii?

[quote="Stormraven"]Alta chestie...mi s-a parut total de lipsit de respect sa se apuce toti sa aplaude in batjocura, chipurile, ca sa mearga mai repede. Oamenii aia aveau ceva de spus si nu se face sa intrerupi niste profesori cu niste aplauze in zeflemea.[/quote]

Asta cred ca se intampla in fiecare an... si nu numai la mate. E un fel de traditie. Eu nu cred ca ar trebui sa te enervezi pentru atata.

750000 pentru locul II pe ţară ...Cât despre aşa zisa tradiţie eu n-am văzut la ONM 2005 aşa ceva. Probabil că atunci nu s-au calificat persoanele in cauză sau poate au avut o criză de conştiinţă (bine ar fi!). Totuşi, trebuie să fii de acord că asemenea lucruri nu sunt normale, mai ales la acest nivel!

Era si pe vremea mea chestia asta. La deschidere se facea, de obicei cand vorbeau persoanele aflate acolo din cauze politice: prefect, primar, etc. Nu am vazut vreodata sa fie "impins mai departe" prof. Gologan de exemplu cu aplauze :)

Da stiu si eu. Mai mult au dat premiu la propunatoriii de probleme. Eu am luat 1 milion 200 de mii premiu pentru problemele date la clasa a-12-a. :)

Felicitari!
Desi s-ar putea spune ca multe probleme sunt uneori mai usor de compus decat de rezolvat  :D  :D

Bineînteles că nu îşi permit asta cu d-l Gologan, doar dânsul e "marele şef"  :D  , dar totuşi acei primari sau prefecţi dau niste bani pentru chestia asta şi nu e frumos ceea ce se face. Aţi văzut şi cu domnul academician Radu Miron ce au făcut şi, vorba aia, e şi el om bătrân şi ai zice ca ar trebui sa aibă măcar respect pentru dânsul...

Mersi. Da sa stii ca unele probleme chiar sunt mai usor de compus decat de rezolvat,dar doar unele. ;)

matematicienii mi-au lasat impresia ca ar fi niste oameni lipsiti de un substrat filosofic :) adica nu par genul de oameni cu care ai putea aborda niste teme filozofice, sa discuti literatura, arta, sa iti dai cu parerea despre lucrurile ce ne inconjoara ( excludem informatica, gazetele si jocurile pe calculator, ca toata lumea acolo se da mare cu astea ) . totusi , iti trebuie un minim de bun-simt, si un minim de cultura care nu tine de matematica . nu e cazul pt toti, fireste, sunt multi oameni deosebiti acolo . dar acestia au de suferit din cauza celorlalti ... eu ma simteam penibil la inceput, ca aplaudau jumate la deschidere ( cealalta jumatate invata  :| ) ce era amuzant in asta, ca eu nu m-am prins ? si eu le am cu chestiile amuzante, pot sa afirm cu tarie . Domnul academician Radu Miron merita tot respectul , si in nici un caz nu trebuie tratat cu acele aplauze nici macar ironice, nici macar amuzante, nici macar ... nimic. eu mi-am luat inima in dinti si l-am rugat sa dea noroc cu mine, probabil ca nu trebuia sa imi permit asa ceva, dar cine stie cand o sa mai ajung vreodata premiat  ... iiiii, ce deosebit ma simteam . chiar respect pe acesti oameni care si-au dedicat intreaga viata acestei stiinte, si care au muncit cu seriozitate ani de-a randul pentru ei , si pentru altii . cum poti sa aplauzi ironic un asemenea monstru sacru, ca sa zic asa ?

la mentiune a dat 400 k looooooooooooooooool , iar la premiu III 550 k loooooooooooooooooool . l-am vazut pe Zahariuc cum discuta cu un baiat care luase premiu III, si ii zicea ca a luat 750 k , cu un chip ce-i drept mai degraba amuzat ... ditai olimpicu international ii dai restu de la semintze.

acuma banii primiti drept premiu nu ar trebui sa fie influentati de comportamentul la deschidere . asta e vrajeala . am vazut la sponsori acolo pe brosurica aia ... unde or fi oare toti banii ? anu asta a fost ceva de genu : premiu I : 17 milioane, premiu II : 800 de mii . pai ce-i asta ? eh... in sfarsit, daca banii erau prioritatea, ne ocupam de altceva ... evident . banii trebuiesc sa fie un stimul , si nu ceva de genu : hmmmmm , a luat premiu , hai sa-i dam si noi ceva ! stai , stai , ca parca am putin marunt ! na 400 de mii !

la premiere era o asa lipsa de interes ... mai lipsea sa ne dea la fiecare cate un numar . in loc de Zamorzaev , premiu I , sa zica : numarul 070707 cod MDV sa pofteasca sa-si ia premiul I . s-or fi suparat si d-aia , eu stiu ...

Am o intrebare: unde pot gasi solutiile de la baraje si myller? :) Sa inteleg ca prima problema(geometria) era pt cei de a 9-a ;)

Si perfect_radio avea dreptate, aplauzele sunt o traditie. Desi la bistrita n-a iesit mare lucru la deva a fost la fel ca la iasi.;)

Pentru pureblack: Stimabile, dumneata ai o perceptie complet gresita despre matematicieni. NU suntem niste incuiati, niste unilateral dezvoltati, cum dai dumneata de inteles. Poate nu putem discuta despre Hegel, Marx, Spinoza si altii (unii dintre noi poate pot, nu e cazul meu  :D ), dar multi dintre noi sunt muzicieni talentati (sau mai putin, dar tot muzicieni), iar alti sunt pasionati cititori. Daca iti vine sa crezi, eu am un tabel in care mi-am scris toate cartile si romanele citite de cand aveam 8 ani si pana in prezent am ajuns la numarul de 122. Poate parea mult, poate parea putin, dar tot am trecut de stadiul de incuiat...
In celelalte privinte ai avut perfecta dreptate si ai punctat totul asa cum trebuie punctat.
In opinia mea nu poti sa-i numesti pe aceia matematicieni sau, daca ii numesti, ei nu sunt incuiati, sunt pur si simplu, iremediabil si ireversibil, prosti.

Bun, sa zicem ca sunt o traditie acele aplauze...dar dom'le, aplauzi cand e nevoie, cand pretuiesti un om sau ce a spus acesta, nu aiurea in tramvai doar ca sa arati ca esti acolo si ai tupeu. E de neiertat.

Scuzati-ma ca am deviat putin cam tare de la discutia initiala si ar fi trebuit sa fac un topic pentru asta, dar a trebuit sa-mi vars naduful undeva... :D

mai raven :) scuza-ma daca te-am jignit . dar nu vorbeam DOAR despre tine . am zis : nu toti ! din cate am vazut p-acolo , sunt niste oameni f tristi . si goi pe dinauntru . NU TOTI ! si care nu au un cuvant de spus in viata de zi cu zi, orientandu-si timiditatea inspre matematica, ca de ... din cauza agitatiei brawniene, omu tre sa se ocupe cu ceva nu ? NU TOTI ! te respect pt muzica si tot ceea ce te pricepi , si eu ma mai pricep la alte chestii ( in afara de gazete , matematica , ca la astea nu le am , la info , si la jocuri pe calculatoare , ca asta e ocupatia la 70% dintre matematicieni ) NU TOTI deci. haideti ca sunt offtopic, ma scuzati !  :!:

Doamne iarta-ma! Nu m-ai jignit absolut deloc...probabil ca eu am interpretat gresit spusele tale. Tonul meu e putin cam prea dur acum pentru ca am rememorat aceste intamplari regretabile. Te rog sa nu crezi ca ma cert cu tine sau asa ceva.

P.S. In legatura cu muzica, nu ma pricep absolut deloc  :) , dar, dat fiind ca am vazut ca exista un astfel de supratopic pe acest forum am facut niste conexiuni, plus ca am vazut ca exista si un sondaj acolo cu instrumente muzicale, deci am presupus automat ca acolo exista niste oameni care au ceva aptitudini in acest minunat domeniu :-)

Hai ca offtopicul e atat de mare incat isi merita propriul topic, asadar iata-l :D Sa nu intelegeti gresit: aplauzele nu erau (in vremea mea cel putin) la intamplare: apareau intotdeauna cand discursul era plictisitor si nu spunea nimic (da, exista oameni care vorbesc 15 minute fara sa spuna [b]absolut[/b] nimic). Si atunci ideea probabil ca a pornit (nu de la persoanele cu 10p sau mai putin asa cum credeti, ci chiar de la cei cu mai multe puncte) ca cei care au discursuri goale sa isi primeasca aplauzele la inceput, si sa scuteasca elevii de pierderea timpului. Lucrurile aparent au degenerat un pic de atunci ...

Nu am fost la deschidere. Eram pe E85 pe la Urziceni la ora aia :)

Deschiderea a fost cu mult mai nerusinat batjocorita decat premierea. Gandeste-te, la fiecare cuvant al doamnei ce prezenta izbucneau rafale de aplauze.
Un lucru n-am inteles totusi: Singurul care a facut un timid gest de dezaprobare a fost domnul Branzei, in rest, absolut nimic. Iar la incheiere, nici atat.
Oare de ce n-au luat domnii profesori putina atitudine, sa le faca putine observatii la oamenii astia? Nu spun asta cu repros, bineinteles...

[quote="Stormraven"]Oare de ce n-au luat domnii profesori putina atitudine, sa le faca putine observatii la oamenii astia? Nu spun asta cu repros, bineinteles...[/quote]S-a mai incercat si in anii trecuti si raspunsul a fost acelasi: mai multe aplauze :D

[quote="Stormraven"]De aceea zic ca ar trebui o metoda mai radicala: Sa se puna un punctaj minim: sa zicem 20 de puncte (depinde de subiecte) si cei de sub acest punctaj sa nu fie admisi la ONM.[/quote]Gandesti prea radical pentru varsta ta :) Uite-te si la jumatatea plina a paharului: numarul de medalii  (si premii) depinde de numarul de participanti in mod proportional. Mai multi participanti, mai multe sanse ca cineva care e chiar bun sa nu plece cu mana goala acasa ;)

Si mai multe sanse ca o sleahta de needucati sa ne faca de ras in ziare... :)

aplauzele sunt o traditie la onm...chiar nu are rost sa va enervati fiindca asta se repeta in fiecare an. si, dupa cum spunea si Valentin, de cele mai multe ori cei care isi permit sa fie "nesimtiti" sunt cei cu punctajele mari...tocmai poate si pt ca simt ca-si irosec timpul cu disursuri inutile....

Din partea mea pot sa aiba 29,5 puncte din 28, daca n-au bun simt...pentru mine nu inseamna nimic. Ideea e ca eu n-am auzit UN SINGUR discurs inutil de-a lungul carierei mele (vorba vine, cariera...). Discursurile sunt tinute de niste oameni inteligenti, care stiu ce spun (de obicei).

Eu am aplaudat frenetic pe tipii care cantau la bucium ( a zis tipa care prezenta fagot sau mi se parea mie?) si pe aia mica cu flautul ei care ziceai ca mai are putin si cade pe jos.... canta cu sentiment !
Insa cand au vorbit domnul Radu Miron si domnul Gologan m-am abtinut... btw de ce apare cu numele de G.Rene ?  :ninja:

Foarte bine ai facut...asta inseamna sa aplauzi cand trebuie, nu aiurea cum fac unii...
Simpatica saraca fetita, nu-mi dau seama totusi de ce era ea mai stresata cu un flaut decat cei cu fagotul cat ei  :D , dar, asta e, nu ma pricep...

[quote="Stormraven"]Ideea e ca eu n-am auzit UN SINGUR discurs inutil de-a lungul carierei mele (vorba vine, cariera...). Discursurile sunt tinute de niste oameni inteligenti, care stiu ce spun (de obicei).[/quote]E clar: cariera ta este foarte scurta deocamdata :D

Crede-ma cand o sa fii la a 5-a nationala o sa apara o groaza de discursuri inutile :)

Fara indoiala ca ai mult mai multa experienta decat mine deci ma inclin in fata parerii tale... :)
Bun, se face putin cam tarziu si mai am de facut ceva la geografie ( :mad: ) asa ca am de gand sa ies de pe forum. Va spun somn usor tuturor si va urez o noapte buna.

[quote="Stormraven"]Intr-adevar...poate ai observat, Valentin, ca la deschidere la ONM 2006 i-au luat peste picior si pe copiii aceia talentati cu instrumentele muzicale.
Poate pe vremea ta mai erau din astia care tineau discursuri aiuristice si chiar meritau sa li se taie macaroana putin, dar acum a ajuns chiar nesimtire curata.[/quote]
Daca vreau sa vad copii talentati care canta la bucium si alte chestii ma duc la filarmonica sau cine stie pe unde.... daca vreau sa vad un film ... ma duc la cinema.....
Ce vreau sa zic e ca: avand in vedere caracterul nerabdator al participantilor la ONM ar trebui evitate chestii artistice care pot fi caracterizate de majoritatea ca fiind plictisitoare.(la concurs la Focsani a inceput inspectoarea de romana sa recite poezii , la o sesiune de comunicari unde fiecare pregatise ceva specific dintr-un domeniu : analiza, combinatorica etc . 5 fete de la filologie fac despre fractali : aduc 2 poze cu fractali si incep sa recite Eminescu  :wallbash_red: si unde pui ca unii chiar se chinuisera sa faca ceva frumos).
Ideea mea:
  - presedintele de onoare sa zica : "Este o onoare sa fiu presedinte de onoare ." si inca ceva avand in vedere ca e o personalitate distinsa in lumea matematicii .
  - d-l Gologan niciodata nu vorbeste pe alaturi asa ca stie el ce sa zica
  - din cadrul administratiei orasului sa zica: "Este o onoare pentru noi sa gazduim olimpiada si sa va dam multi bani la premiu.Succes."

P.S.: cred ca am exagerat :blush:

Mai ales partea cu "multi bani"  :D . E improbabila  :)  Administratia nu da bani de bunavoie...Sunt totusi simpatice sugestiile tale...

Orice lucru bun mai are si partea lui rea...Stormraven, nu te mai stresa cu <nesimtitii> aia de la nationala, tu continua-ti treaba de olimpic...

Bun, intr-adevar nu prea merita sa discutam prea mult pe subiectul asta. Daca mai vrea cineva sa comenteze, e binevenit...

spix, ce faci ? :D m-am plictisit teribil la deschidere ... am vorbit cu altii ... si unii in spate imi faceau semn sa tac ... le-am zis vreo 2 ...deschiderea aia artistica a avut loc si acuma 2 ani , cand eram la chimie. exact aceeasi baieti cu hoarnele alea mari , cam aceleasi fetite la pian parca ... se dau si ei mari cu ceva, ce vrei ... dar mai bine le baga cu o alta ocazie, nu la deschidere ... cand toata lumea e nervoasa si concentrata si alea alea . aplauzele alea mi s-a parut glume de oameni care nu au o limita ... e ca si cum ai zice aceeasi gluma de 20 de ori : prima oara e amuzanta, si a doua oara, a treia oara mai zambesti ... dupa aia nu isi mai are rostu . eu nu stiam traditia, si mi se pare o traditie naspa ... daca era traditia ca ala care iese primu sa fie aruncat in garla sau ceva de genu ... era mai interesant :P

cum adica sa fie punctaj minin 20 ??? ce sa spun ... te-ai vazut cu premiu II si acu deja vrei sa scapi de astia mai prosti , carevasazica ... nu mai te crede asa de mare si tare , pt ca esti mica ! olimpiada e bine organizata asa cum e ... sa fie sanse pt oricine ... ca sunt multi care la judeteana au clacat rau de tot , si la nationala iau auru ... dar si viceversa ...   asta-i spectaculos . baieti care castiga interjudetene , fac legea peste tot , sunt siguri pe ei , fac pregatire luni si luni de zile, si cand ajung la olimpiada nationala ... ce sa este oare ??? tre sa fie aceeasi sansa pt toti , orice este posibil ... de la un anumit loc in jos evident . pe primele locuri mereu vor fi Boreico si Zamorzaev :lol:

Ba copile tu vorbesti de spam? Ia mai uita-te in oglinda.

Stimabile pureblack, te rog sa nu-ti faci pareri gresite despre mine. Eu spusesem in cu totul alt context, pentru o imbunatatire a disciplinei la ceremoniile de deschidere, respectiv premiere.
Crede-ma, nu sunt genul sa ma bucur de necazul altuia si nici genul invidios ca "haha, sa nu am concurenti".
Gandeste-te logic ca daca intrec pe mai multi ma simt mult mai increzator, si deci nationala imi aduce mult mai multa satisfactie...

Sa revenim la aplauze (acuma erau bune cateva nu? :P) ...

A fost si eu la toate nationalele posibile ( adica la 6) si recunosc faptul ca pe la ultimele doua (cel putin) faza cu aplaudatul deviase si isi pierduse umorul,desi aplaudam si eu....
 La prima mea nationala (Brasov 2000) se aplauda la citirea unor sponsori (ex : Vincon Vrancea :)) .
 Legat de premiere ....am luat si eu de trei ori premiu :) (adik am fost in primii 5) si recunosc ca la ultimul premiu (pe a XI-a la Deva)a fost cea mai fugitiva premiere (next...next...next...). In general ,parca nimeni nu mai avea rabdare la sfarsit ,nici elevii si nici profesorii. Poate e si o chestie de buna crestere (ma refer la elevi...).
 Si mai este ceva : o viziune coerenta asupra vietii nu ti-o poti forma doar cantand la un instrument (cant si eu la pian) ,ci citind si rafinadu-ti     permanent gustul lecturii...cel putin asa cred.
 Succes tuturor!!! Si nu uitati ca matematica , oricat ati iubi-o,nu e decat un capriciu al vietii. Unul nobil....

anul asta am ratat deschiderea, dar mi se spusese ca parca aplauzele nu au mai fost in ceilalti ani  :? , in sensul ca 'copiii' inca nu au deprins vechea traditie  :D..  parerea mea e ca aceste deschideri sunt considerate in general inutile de catre participanti si in orice caz, mult prea lungi, deci e normal ca ei sa incerce prin metode care le gasesc amuzante sa le scurteze! dupa atata matematica, normal ca avem chef sa radem putin de niste manifestari artistice care nu le solicitasem si de niste discursuri fara cap, fara coada si evident, fara continut!
poate ca ar trebui ca la deschideri sa se prezinte doar cei interesati [eventual numai profesorii, care sa transmita informatiile vitale elevilor].

eventual toate discursurile de bun-venit sa fie transferate pentru premiere  :diablo:      acolo, prezentandu-se numai premiantii, ori se vor supune si vor asculta povestile nemuritoare, ori vor continua traditia aplauzelor  :D     si atunci nu vor mai putea fi acuzati ca sunt incapabili.. poate doar niste genii nerabdatoare  :wow:      totusi ar trebui motivati si cu niste premii substantiale.. poate vor fi mai intelegatori stiind ca respectivele oficialitati s-au zbatut ca sa le obtina premii pe masura valorii lor..

ma mir totusi ca a fost aplaudat academicianul Radu Miron.. acum cativa ani la Myller tinuse un domn un discurs despre viata lui Myller, toate descoperirile mici, mari, de toate culorile, de sala fierbea ca nu mai termina.. aplauzele erau bineinteles prezente si la un moment dat a aplaudat si Radu Miron!!, incat cel ce isi citea discursul ii spusese jenat: "imediat termin, domnule academician." :read:

Ce e prea mult strica..."Gluma" n-are in orice caz pic de umor si denota proasta crestere. Oricat de plictisitor, enervant, agasant ar fi discursul, trebuie aratat macar un minim respect celui ce isi pierde timpul sa il scrie si sa-l prezinte.

Dar ce tot aveti cu discursurile astea?Am mai spus o data, ceea ce e important este sa dai impresie buna la ONM.
Nu trebuie sa va stresati cu asta... :?

Chiar daca nu am participat la onm (am fost la onf), vreau sa zic ca si acolo a fost exact la fel. Si acest fapt (elevii aplaudau dupa fiacera 2 cuv ale profesorilor d p scena) ar fi putut fi un motiv pt. care presedintele comisiei a dat un subiect suplimentar, destul de valoros, pe care insa nimeni nu l-a rezolvat... :lol:

Aplauzele sunt acolo de mult timp...eu unul am aplaudat timp de 5 ani la nationale si n-am facut-o neaparat cu intentia de a-l batjocori pe vorbitor...in plus e vorba si de faptul ca discursurile lungi nu sunt pe sufletul elevilor iar adultii care vorbesc tind sa uite uneori lucrul asta, fiind de obicei destul de departe de varsta adolescentei. Eu iti sugerez sa fii mai ingaduitor(are) (iarta-ma, n-am reusit sa citesc chiar totul...) cu colegii tai si sa uiti de injumatatirea loturilor pentru ca nu stii niciodata cand ai o zi proasta si nu faci pe cat de bine poti, si atunci ai sa privesti altfel lucrurile. In plus, incearca mai mult sa te bucuri de experienta de la nationala (si loturi viitoare) decat sa critici si sa te uiti urat la altii, pentru ca numai asa ai sa-ti amintesti mai incolo cu placere de aceasta perioada din viata ta.

Mult succes in continuare!

[color=darkblue][b]Frumos spus, Stgmihnea ![/b] Se vede consecinta fireasca de a face frontiera benefica cu matematica in orice imprejurare, inclusiv si in acest dialog. Ai gandit ca la sah, cel putin 5 mutari inainte de a pune mana pe piesa. Va puteti inchipui ce deranjeaza cel ce da mutarea inapoi. Ca analogie, si scrierea presupune o gandire ca la sah : cel care foloseste frecvent guma (buretele) nu gandeste inainte de a pune mana pe condei (creta). Imprimanta este cel mai lenes terminal intr-un sistem informatic. Cand am apucat s-o folosesc, inseamna ca se apropie sfirsitul sau am incheiat ... cu cerebelul ! De exemplu, foarte des in urma unei operatii puneti imediat $+$ pentru a constata ca de fapt este $-$. Dar iata un exemplu mai "complex".
Doresc sa adun doua fractii : $\frac{7}{12}$ si $\frac{11}{30}$.
Etapa 1. Determin $d\equiv (12,30)=60\ .$
Etapa 2. Trag linia de fractie a sumei si la numitorul fractiei-suma pun $d\ .$
Etapa 3. Vad la fiecare fractie-termen ce-i lipseste numitorului ei pentru a deveni $d\ .$ In cazul nostru, $5$ si $2\ .$
Etapa 4. Amplific fractiile corespunzator si la numaratorul fratiei-suma (ca pe un pat al lui Procust) asez frumos rezultatul !
Tot timpul se pleaca de la ultima operatie spre prima (ca intr-o compunere de functii). Aici ultima operatie este impartirea, deci trag linie de fractie mai intai etc.
$\frac{7}{12}+\frac{11}{30}=\frac{}{60}$
$\frac{7^{(5}}{12}+\frac{11^{(2}}{30}=\frac{}{60}$
$\frac{35}{60}+\frac{22}{60}=\frac{57}{60}$
$\frac{7}{12}+\frac{11}{30}=\frac{57}{60}\ ,$ raspuns final.
Altfel, linia de fractie va apare pe undeva, mai jos, si nu este nici logic si nici estetic. Frumoase aceste cuvinte, logic, estetic, dar folositi-le si in practica !
Va doresc din toata inima mult succes la anul, pentru unii ... si la anul ![/color]

E...eu am propus un lucru...Daca cumva asa ceva se va aplica si voi avea de suferit din cauza asta asta e...Eu tot imi pastrez parerea si nu sunt genul ipocrit sa spun ca injumatatirea ar fi doar buna pentru altii...Daca nu m-as califica, asta e...as face-o pe propria mea prostie...
Nu mai repet ce am spus in posturile anterioare pentru ca am latrat destul...Daca am spus ce trebuia, foarte bine...Daca nu, asta e...


[hide="hint"]make the right side 0.[/hide]

[quote="Glen Mann"][hide="answer"]
$\frac{x-3}{x+3}> 2; x \not=-3$
$x-3 > 2x+6; x \not=-3$
$-x > 9; x \not=-3$

$x >-9; x \not=-3$

so $-9 < x <-3 < x$

im not too sure if this is right or not.. dont know what you meant by the red herring thing :wink:
[/hide][/quote]

Two things:

1) When you divide by a negative number (such as in the third and second to last steps), the inequality sign switches.

2) When you multiplied both sides by x+3, you forgot to account for if x+3 is negative, which leads back to the first point.

Could you explain the problem with our math (it's really just the same, but I did it a little faster and neglected denominator=0)? It's kind of hard, staring at my math and saying "what is wrong with that?"

[hide]1st cross multiply:
$x-3>2x+6$ Now solve.
$x>2x+9$
$-x>9$
$\boxed{x<-9}$. There's no domain restriction since none of these values include $x=-3$[/hide]

[quote="Ignite168"]Solve for $x: \frac{x-3}{x+3}>2$.

The topic title may or may not be a red herring.[/quote]

[hide]
If the numerator and denominator are different signs, then the quotient is less than 2, therefore either $x<-3$ or $x>3$

Case 1: $x<-3 \implies x+3<0$.

$\frac{x-3}{x+3}> 2$
$x-3 < 2x+6$ (remember $x+3$ is negative).
$-9 < x$, but remember that $x<-3$
$-9 < x <-3$

Case 2: $x>3 \implies x-3, x+3 > 0$

$\frac{x-3}{x+3}>2$
$x-3 > 2x+6$ ($x+3>0$)
$-9 > x$, but $-9>x>3 \implies-9>3$, so no solutions.

Therefore, $-9 < x <-3$ or, equivalently, $x \in (-9,-3)$.
[/hide]

Which is, of course, what you do when you face such problems.
Our precalc class spent a month on these (on and off) during our first semester. It was pretty boring.

Stupid cross-multiply error. $x+3$ can be less then $1$ and throw the whole equation off.

You mean less than $0$, so the inequality sign may need to flip. This is why you can't cross multiply in this case.

easier way without cases

[hide="the proof"]
(x - 3)/(x + 3) > 2
(x - 3)/(x + 3) - 2([i](x + 3)/(x + 3)) [/i]> 0
(x - 3 - 2x - 6)/(x + 3) > 0
(-x - 9)/(x + 3) > 0
-(x + 9)/(x + 3) > 0
(x + 9)/(x + 3) < 0
so, either x+9 is negative or x+3 is negative, but not both. if x+9 is negative, then x+3 will always be negative. go to x+3. this will be negative with x<-3. when x reaches 9, then this inequality is false. so, {x: -9 < x < -3}[/hide]

sorry, i don't know  how to make the text fancy like all you other guys.


[quote="chien than"]Nobody can do it![/quote]

Take $x,y>1$ and it's obviously wrong.


probability
inequalities
inequalities unsolved

p+q=1
m,n,p,q>=0
Prove that:
(1 - q^n)^m + (1 - p^m)^n  >= 1
Can U help me, I think just Holder but I stuck! Thanks for reading!

Hint: [hide]Use propabilistic interpretation :)[/hide]

[quote]
_________________
A cat has nine tails. Why?
No cat has eight tails. One cat has one more tail than no cat. So one cat has nine tails
 :D
[/quote]

It's very interesting
We can prove A cat (dog,...) has $ n $ tails for every positive integer $ n $

 :P  :P  :P

Yes, but I don't like that general form so much  ;)  :P  :D

What do U mean??? I don't understand (because of my bad English!!!) Please help with the provement!

I've written jsut a hint because I know this problem and also how to solve it using propabilistic interpretation but I'm not so good in english to explain it properly. But I can try :)

Let's have a table $m\times n$ ($m$ columns and $n$ rows) filled with just 0's and 1's. Let say we construct such a table at random in such way that at each place there's propability $p$ of $0$ being there and $1-p=q$ of $1$ being there. Now look at the sum in the inequality:

1) $(1-q^n)^m$. It's propability of fact that in each column there are not all $1$'s or that in each column there is at least one $0$.

2) $(1-p^m)^n$. It's propability of fact that in each row there are not all $0$'s or that in each row there is at least one $1$

But it's easy to find out that each random table is in group 1) or in group 2) => sum of these propabilities is at least 1 and it's what we wanted to show.

This problem can be easily generalized:

Let's have real numbers $0\leq y_1,y_2,\ldots,y_n\leq 1$ Prove that (if $m$ is positive integer):
\[(1-y_1y_2\cdots y_n)^m+(1-y_1^m)(1-y_2^m)\cdots(1-y_n^m)\geq 1\]


I know how to do
What is the maximum value of --
$%Error. "display" is a bad command.
\frac{1}{a} + \frac{1}{b} = x$
but how do you do
What is the maximum value of --
$%Error. "display" is a bad command.
\frac{1}{a} + \frac{1}{b} + \frac{1}{c}= x$
?
Is there a way to do
What is the maximum value of --
$%Error. "display" is a bad command.
\frac{1}{a} + \frac{1}{b} + \frac{1}{c}+ \frac{1}{d}= x$

The maximum value of what? x? Wouldn't that just be 3 or 4 or whatever, assuming a, b, c, etc. are positive integers? Or is it the maximum value of a+b+c or something like that?

What exactly is the question?  Couldn't you just make a+b+c as large as you want?  I'm not following...

I don't get it either.  Just make a,b,and c equal to $\frac{1}{(really big number)}$.  Numbers can be infinitely large, so there is no maximum value.

He is trying to say:

"Given that $\frac{1}{a}+\frac{1}{b} = x > 0$, it is possible to find the maximum value for a + b, assuming that they are positive.  If instead we are given that $\frac{1}{a}+\frac{1}{b} +\frac{1}{c}= x >0$, is it possible to find a maximum for a + b + c?  What about for larger numbers of variables?"

...i still don't see the problem...if a, b and c are positive, then the first equation holds and thus you can make them as large as you want to.  For the second one, if you're talking about negative numbers as well, and you want to find the minimum, then i can see a problem...but otherwise i just don't see it

By the way, if you multiply out AM-HM, it boils down to the trivial inequality, in case you want to prove it.

[quote="Singular"]Let a' = 1/a.  Now the sum of a' is fixed, call this x.

Notice $ x \geq n/(a+b+...)$ by AM HM.  So the sum a+b+... is >= n/x, so it has a minimum.

How do you show there is no maximum?

Let all the numbers except one of them tend to infinity.  Then the reciprocals of those numbers tend to zero.  Then set the last number to 1/x.[/quote]
Usually when this type of question appears on a contest, a, b, and c are specified to be positive integers. How would the maximum value of a+b+c be calculated then?

Sorry about the confusion, the thing I was asking is if x is a constant and a, b, and c are all positive intergers than what is the maximum value of a+b+c if
$%Error. "display" is a bad command.
\frac{1}{a} + \frac{1}{b}+ \frac{1}{c}=x$
an example would be
$%Error. "display" is a bad command.
\frac{1}{a} + \frac{1}{b} + \frac{1}{c}=7/10$ what is the maximum value of a+b+c. Which was from an AMC.
Somewhere I saw some divide it into casework somehow and
I hope that clarifies it.
and is there a way to find the maximum value for 4 numbers with the same resrictions?

By AM-HM, $\displaystyle\frac{a+b+c}{3}\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$
$\displaystyle a+b+c\ge\frac{9}{\left(\frac{7}{10}\right)}$
$a+b+c\ge\frac{90}{7}$

Equality holds when $a=b=c=\frac{30}{7}$.

In general, try playing around with your inequalities.

Yet I specified a,b,c had to be integers so they cannot equal 30/7, and I think you found the minimum value of a+b+c not the maximum.

Oh, I have issues then.  :D

I remember seeing that problem from the AMC, and I think I just used guess and check.

Denote a'=1/a. a'+b'+c'=7/10. What is the maximum of a+b+c?

(ab+bc+ac)/(abc)=7/10
10(ab+bc+ac)=7abc
343abc-490ab-490bc-490ac=0
(7a-10)(7b-10)(7c-10)=700a+700b+700c-1000

Since a,b,c are integers and 700(a+b+c)-1000 is divisible by 100, 100|(7a-10), 100|(7b-10), or 100|(7c-10). Therefore, 7n :equiv: 10 (mod 100) (n=a, b, or c). Therefore n :equiv: 30 (mod 100). We can assume that this is true for n=a. What if a=30?

(7b-10)(7c-10)=200+7b+7c -> 49bc-70b-70c+100=200+7b+7c -> 49bc-77b-77c=100 -> (7b-11)(7c-11)=221 -> no solutions

I'm having trouble beyond this, but I think you can just test everything and see if it works. The first solution you come upon for a that leads to a solution for b and c will (probably) be the maximum. I'm really not great with stuff like this, so maybe I'm wrong but I suggest you try it and see.


Hello everyone,

I am having some trouble with the following two logarithmic equations and I would appreciate any help or hints. For the first equation, I am not sure how to solve it fully. For the second equation, my answer was different from the given answer.

Thank you once again!

---

1. $ \log_3 (x^2 \minus{} 9) \minus{} \log(x \plus{} 3) \equal{} 1$

$ \log_3 [(x \plus{} 3)(x \minus{} 3)] \minus{} \log(x \plus{} 3) \equal{} 1$

I am unsure of what to do. Should I change the base from log 3 to log 10?

2. $ 2\log(y \plus{} 2) \equal{} 1 \plus{} \log(y^2 \minus{} 4)$

[u] My Work : [/u]

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

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

$ \log[\frac {(y \plus{} 2)^2}{(y \plus{} 2)(y \minus{} 2)}] \equal{} 1$

$ \log(y \plus{} 2)(y \minus{} 2) \equal{} 1$

$ (y \plus{} 2)(y \minus{} 2) \equal{} 10$

$ y^2 \minus{} 4 \equal{} 10$

$ y \equal{} \pm {14}$

$ y \equal{} \plus{} {14}$

However, the given answer is $ \frac {22}{9}$.

Yeah, you should use the change of base formula so that the logarithms all have the same base, and you should be able to solve the resulting equation (if you know how)

[quote="catalan"]$ \log[\frac {(y \plus{} 2)^2}{(y \plus{} 2)(y \minus{} 2)}] \equal{} 1$
$ \log(y \plus{} 2)(y \minus{} 2) \equal{} 1$[/quote]I think you canceled incorrectly here :)

2: When you did

$ \log[\frac {(y \plus{} 2)^2}{(y \plus{} 2)(y \minus{} 2)}] \equal{} 1$

$ \log(y \plus{} 2)(y \minus{} 2) \equal{} 1$

it should have been $ \log\frac{(y \plus{} 2)}{(y \minus{} 2)} \equal{} 1$

[hide="2"] We can re-write this equation as $ 2\log_{10}(y\plus{}2)\equal{}\log_{10}(10)\plus{}\log_{10}(y^{2}\minus{}4)$ This gives us $ \log_{10}(y\plus{}2)^{2}\equal{}\log_{10}10(y^{2}\minus{}4)$. This equals $ 0\equal{}\log_{10}10(y^{2}\minus{}4)\minus{}\log_{10}(y\plus{}2)^{2}$, which then equals $ \log_{10}\left(\frac{10(y\minus{}2)}{(y\plus{}2)}\right)\equal{}0$. Thus, $ \frac{10(y\minus{}2)}{(y\plus{}2)}\equal{}1$. Solving for y gives us $ \boxed{\frac{22}{9}}$.[/hide]

Thank you for all the replies. I solved it and the mistake was that I didn't simplify $ \frac{(y \plus{} 2)^2}{(y \plus{} 2)(y \minus{} 2)}$ correctly.

For #1, here is what I have done:

1. $ \log_3 (x^2 \minus{} 9) \minus{} \log(x \plus{} 3) \equal{} 1$

$ \log_3 [(x \plus{} 3)(x \minus{} 3)] \minus{} \log(x \plus{} 3) \equal{} 1$

$ \frac{\log[(x \minus{} 3)(x \plus{} 3)] \minus{} \log(x \plus{} 3)(\log 3)}{\log 3} \equal{} 1$

Without solving the equation for me, could someone please tell me how I can simplify $ \log(x \plus{} 3)(\log 3)$? Would it just be $ \log(x \plus{} 3)^{\log 3}$?

Thanks again.

I don't think this is solvable by means of elementary methods:
$ \log_3{(x^2 \minus{} 9)} \equal{} \log{(x \plus{} 3)} \plus{} 1$
$ \log_3{(x^2 \minus{} 9)} \equal{} \log{10(x \plus{} 3)}$
$ \log{(x^2 \minus{} 9)} \equal{} \log{\left(10(x \plus{} 3)\right)^{\log{3}}}$
$ x^2 \minus{} 9 \equal{} 3(x \plus{} 3)^{\log{3}}$
Since $ x \plus{} 3\neq 0$, divide it at both sides
$ x \minus{} 3 \equal{} 3(x \plus{} 3)^{\log{3} \minus{} 1}$

I don't see the next step...

I used a calculator and cheated and got around 4.0782353, but like timwu said, I can't solve it algebraically...


geometry
circumcircle
ratio
geometry proposed

Let a line $ \ell$ intersect the sides $ BC$, $ CA$, $ AB$ of a triangle $ ABC$ at $ A_{1}$, $ B_{1}$, $ C_{1}$, respectively. Let $ P$ be an arbitrary point. Let the lines $ AP$, $ BP$, $ CP$ intersect $ \ell$ at $ A_{2}$, $ B_{2}$, $ C_{2}$, respectively. Prove that the three circumcircles of triangles $ \triangle A_{1}A_{2}P$, $ \triangle B_{1}B_{2}P$, $ \triangle C_{1}C_{2}P$ are coaxal.

[quote="Fang-jh"]Who can prove this problem?[/quote]
Yetti can prove this problem  :P. In any case, Amir is a dangerous boy  :!:  To make the solution easier to follow, I change notation to get rid of indices:

Let a line $l$ intersect the side lines $BC,\ CA,\ AB$ of a triangle $\triangle ABC$ at $D,\ E,\ F,$ respectively, and let $P$ be an arbitrary point (not on the sidelines or on the line $l$, but it does not have to be inside the triangle). Let the concurrent cevians $AP,\ BP,\ CP$ intersect the line $l$ at $X,\ Y,\ Z,$ respectively. Prove that the 3 circumcircles $\mathcal C_{1},\ \mathcal C_{2},\ \mathcal C_{3}$ of the triangles $\triangle PDX,\ \triangle PEY,\ \triangle PFZ$ are coaxal.

The main idea is this: The points $F,\ Z$ lie on the same circle $\mathcal C_{3}$ in the pencil defined by the circles $\mathcal C_{1},\ \mathcal C_{2}$ (intersecting at P), iff powers of $F,\ Z$ to these 2 circles are in the same ratio (see the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=121062]Constant value[/url]). The circle $\mathcal C_{3}$ then also passes through $P.$ Thus we have to show that

$(?)\ \ \frac{FY \cdot FE}{FX \cdot FD}= \frac{ZY \cdot ZE}{ZX \cdot ZD}$

Equivalently, we have to show that the following 2 cross ratios are equal:

$(?)\ \ \frac{FX}{ZX}\cdot \frac{ZY}{FY}= \frac{FE}{ZE}\cdot \frac{ZD}{FD}$

To do this, we project any line through the point $P$ to infinity, which makes the projected cevians $A'P' \parallel B'P' \parallel C'P'$ parallel (because $P'$ is at infinity). Let $A'P'$ meet the line $B'C'$ at $K'$ and let parallels to $A'P' \parallel B'P' \parallel C'P'$ through $E',\ F'$ (i.e., the lines $E'P',\ F'P'$) meet the line $B'C'$ at $M',\ N'.$ By Menealus theorem for the triangle $\triangle A'B'C'$ cut by the transversal $l' \equiv D'E'F',$

$\frac{F'B'}{F'A'}\cdot \frac{E'A'}{E'C'}\cdot \frac{D'C'}{D'B'}= 1$

which parallel-projects on the line B'C' as

$\frac{N'B'}{N'K'}\cdot \frac{M'K'}{M'C'}\cdot \frac{D'C'}{D'B'}= 1$

WLOG, assume that the points $B',\ N',\ K',\ M',\ C',\ D'$ follow on the line $B'C'$ in this order and denote the segments

$t = B'N',\ u = N'K',\ v = K'M',\ x = M'C',\ y = C'D',\ z = B'D'$

We have 2 equations for these segments:

$\frac{t}{u}\cdot \frac{v}{x}\cdot \frac{y}{z}= 1,\ \ t+u+v+x+y = z$

The cross ratios in question are preserved in our central projection, hence we have to show

$(?)\ \ \frac{F'X'}{Z'X'}\cdot \frac{Z'Y'}{F'Y'}= \frac{F'E'}{Z'E'}\cdot \frac{Z'D'}{F'D'}$

which parallel-projects on the line $B'C'$ as

$(?)\ \ \frac{N'K'}{C'K'}\cdot \frac{C'B'}{N'B'}= \frac{N'M'}{C'M'}\cdot \frac{C'D'}{N'D'}$

Rewriting this using the segments $t,\ u,\ v,\ x,\ y,\ z:$

$(?)\ \ \frac{u}{v+x}\cdot \frac{z-y}{t}= \frac{u+v}{x}\cdot \frac{y}{z-t}$

Rearranging and substituting Menelaus theorem:

$(?)\ \ \frac{(z-t)(z-y)}{(v+x)(v+u)}= \frac{ty}{ux}= \frac{z}{v}$

Multiplying through:

$(?)\ \ vz(z-t-y)+tyv = zv(v+x+u)+uxz$

Since $z$ is sum of the remaining segments, $z-t-y = v+x+u.$ Canceling equal terms on the LHS, RHS:

$(?)\ \ tyv = uxz$

which is obviously true (Menelaus theorem). The central projection of P to infinity just makes the proof easier to follow, otherwise we could do without it.

I remember someone asking, what IMO means. Go to the website [url=http://www.acronymfinder.com]Acronym Finder[/url] and type in WLOG or wrt.

Can you prove (ZY*ZE)/(ZX*ZD)=(FY+ZE)/(FX+ZD) without project geometry?

The main idea is to show $ \frac {p(F,\ \mathcal C_{1})}{p(Z,\ \mathcal C_{1})} \equal{} \frac {p(F,\ \mathcal C_{2})}{p(Z,\ \mathcal C_{2})},$ based on Menelaus theorem. The algebra to do this is simple, it can be done in various ways, and I said that it can be done without projecting the point P to infinity. In case there are any doubts, using Menelaus theorem for the triangles $ \triangle AFX,\ \triangle AEX$ cut by the transversals BYP, CZP, respectively, we have:

$ \frac {BF}{BA}\cdot \frac {PA}{PX}\cdot \frac {YX}{YF} \equal{} 1,\ \ \ \frac {CE}{CA}\cdot \frac {PA}{PX}\cdot \frac {ZX}{ZE} \equal{} 1,$

yielding

$ \frac {BF}{BA}\cdot \frac {YX}{YF} \equal{} \frac {CE}{CA}\cdot \frac {ZX}{ZE},\ \ \ \frac {BF}{BA}\cdot \frac {CA}{CE} \equal{} \frac {FY}{YX}\cdot \frac {XZ}{ZE}.$

Using Menelaus theorem for the triangle $ \triangle AFE$ cut by the transversal DCB,

$ \frac {BF}{BA}\cdot \frac {CA}{CE}\cdot \frac {DE}{DF} \equal{} 1,$

yielding

$ \frac {FY}{YX}\cdot \frac {XZ}{ZE}\cdot \frac {ED}{FD} \equal{} 1,\ \ \ \frac {t}{u}\cdot \frac {v}{x}\cdot \frac {y}{z} \equal{} 1,$

where $ t \equal{} FY,\ u \equal{} YX,\ v \equal{} XZ,\ x \equal{} ZE,\ y \equal{} ED,\ z \equal{} FD.$ Assuming (WLOG) that E, F are inside the segments CA, AB, that D is on the ray (BC outside the segment BC, and that P is inside the triangle $ \triangle AFE,$ we have $ t \plus{} u \plus{} v \plus{} x \plus{} y \equal{} z.$ Now we can write (even though the following chain of equivalences in actually done in reverse):

$ uy(t \plus{} u \plus{} v) \plus{} tvy \equal{} uy(z \minus{} x \minus{} y) \plus{} uxz,$

because the left terms on LHS, RHS are equal and the right terms on LHS, RHS are also equal. Factoring this, we get

$ y(t \plus{} u)(u \plus{} v) \equal{} u(x \plus{} y)(z \minus{} y),\ \ \ \frac {(t \plus{} u)(u \plus{} v)}{(x \plus{} y)(z \minus{} y)} \equal{} \frac {u}{y} \equal{} \frac {tv}{xz}.$

Substituting back for $ t,\ u,\ v,\ x,\ y,\ z,$ we arrive to

$ \frac {FX \cdot YZ}{ZD \cdot FE} \equal{} \frac {FY \cdot XZ}{ZE \cdot FD},\ \ \ \frac {FX \cdot FD}{ZD \cdot ZX} \equal{} \frac {FY \cdot FE}{ZE \cdot ZY},$

which is exactly what we need.

Thank yetti very much! I have learned much knowledge from you and Darij.
Let circle FZP intersect again circle XDP at Q. Let PQ intersect FD at K.
FK*KZ=PK*KQ=XK*KD. KZ=(ZX*ZD)/(FX+ZD). If YK*KE=FK*KZ KZ=(ZY*ZE)/(FY+ZE)  This mean that circle YEP passes through two point P,Q.
Because vxz=vx(t+u+v+x+y)   -uvx+vxz-xv^2=tux+vx^2+vxy
uxz-uvx+vxz-xv^2=tvy+tux+vx^2+vxy
(u+v)x/v(x+y)=(t+x)/(z-v)  (ZY*ZE)/(ZX*ZD)=(FY+ZE)/(FX+ZD)
The circle passes through two point P,Q
Dear yetti! Is my solution right?

You can combine
http://www.mathlinks.ro/Forum/viewtopic.php?t=5860 post #2 (this shows that the problem is equivalent to Romanian ROM TST 2004, problem 14)
and post #6 of the same topic (this solves that TST problem)
to get a different solution.

  Darij

Corrected:
Let circle PFZ intersect again circle PDX at Q. Let PQ intersect FD at K.
FK*KZ=PK*KQ=XK*KD.
yeilding KZ=(ZX*ZD)/(FX+ZD).
If YK*KE=FK*KZ=PK*KQ
yeilding KZ=(ZY*ZE)/(FY+ZE)
This mean that the circle PEY passes through two point P,Q.
Because vxz=vx(t+u+v+x+y)
-uvx+vxz-xv^2=tvx+vx^2+vxy    uxz=tvy
uxz-uvx+vxz-xv^2=tvy+tvx+vx^2+vxy
(u+v)x/v(x+y)=(t+x)/(z-v)  This is equivalent to
(ZY*ZE)/(ZX*ZD)=(FY+ZE)/(FX+ZD)
As a result, the circle PEY passes through the two points P,Q.  Q.E.D

as yetti said we have to show that $\frac{p(D,\ \mathcal C_{3})}{p(D,\ \mathcal C_{2})}=\frac{p(X,\ \mathcal C_{3})}{p(X,\ \mathcal C_{2})}$
note that here $\mathcal C_{2},\ \mathcal C_{3}$ are the circumcicles of $\triangle PEY$ , $\triangle PFZ$ respectively...
so as yetti said we have to show that
$(?)\ \ \frac{DF \cdot DZ}{DE \cdot DY}=\frac{XF \cdot XZ}{XE \cdot XY}$
with some rearangements we have to show that:
$(?)\ \ \frac{\frac{FD}{ED}}{\frac{FX}{EX}}=\frac{\frac{ZX}{YX}}{\frac{ZD}{YD}}$
or we have to show that:
$(?)\ \ (ZY,XD)=(FE,DX)$
now if we project line $EDF$ on the line $BC$ with point $P$,we get that $(ZY,XD)=(CB,X'D)$ where $X'$ is the intersection of $AP$ and $BC$
now if we project the line $BC$ to the line $EDF$ agaian but with the point $A$ we get that $(CB,X'D)=(EF,XD)=(FE,DX)$ and we are done...!


integration
logarithms
induction
real analysis
real analysis unsolved

Another way : by induction prove that : $ \frac{n}{3^{n^2}}<\frac{1}{3^n}$


Four kids (Allie, Bobby, Callie, and David) meet four teletubbies  (Dipsy, Laa Laa, Po, and Tinky Winky) and Barney's crew  (Barney, Riff, Baby Bop, and BJ). Each person/being shakes hands with every other one once except for those of his/her same species (i.e. a human doesn't shake hands with a human, a teletubby doesn't shake with teletubby, a dinosaur doesn't shake with dinosaur) How many handshakes would take place if Callie wants to shake hands with Laa Laa twice, Bobby refuses to shake hands with Baby Bop, and Allie (being so friendly) wants to shake hands with everyone no matter the species?

Hard but silly problem! Enjoy!

each person shakes hadns with 8 others and we just have to divide by two and aplay all those special conditions

we have $ \frac {12*8}{2} \minus{} 1 \plus{} 1 \plus{} 3 \equal{} 51$

No, that can't be it. You forgot to consider the exceptions that Allie shakes hands with everyone and Bobby doesn't want to shake hands with Baby Bop. Also Callie wants to shake hands with Laa Laa twice.

Btw, mods/admins, why did this get moved? I thought this is AMC level.

1. Because the AMC forum is for AMC test problems or issues about the test. Problems on their own go into the high school forums.

2. Bobby doesn't want to shake hands once: $ \minus{}1$. Callie wants two handshakes once: $ \plus{}1$. Allie wants to shake with the three other humans: $ \plus{}3$. These are accounted for. Why is his solution incorrect?

oh wait didn't see the last few steps in his equation...sorry :oops:

This is a public shaming of myself. TYVM


This inequality is equivalent to:
$\sum_{cyc}\frac{a^{4}}{a^{3}+b^{3}}\geq \frac{a+b+c}{2}$
it's indeed very hard. it's problem 93 of chapter 8 in Vasc's book
 it has been discussed also before in the site
[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sos&t=77613]here[/url] post #6
also in some other threads but i can't find them now.


inequalities
integration
trigonometry
absolute value
inequalities solved

maybe  I am wrong, but a2k=2k and a2k+1=1 ? then the inequality becomes something like

[n/2] + \sum k=1[n/2] 1/(2k) > 2\sqrt n -1 which is true for n >9 ?!

upppss :) sorry I wrote a2 = 2 on the scrath paper and iterating I've got that ugly result upthere :D:D

Well, if it makes you feel any better, at first I saw it the same way you did  :D  :D

[quote="liyi"]1.
Sequence {a_n} satisfies
a_1 = 1
a_n * a_{n+1} = n
show that 1/a_1 + 1/a_2 + ... + 1/a_n > 2sqrt(n) - 1

2.
x, y, z are reals and x^2 + y^2 + z^2 = 2.
show that |xyz - (x+y+z)| <= 2.
(|x| means the absolute value of x)[/quote]

============
The second one is exactly the Poland 1991
"x, y, z are reals and x^2 + y^2 + z^2 = 2 show that x + y + z <= 2 + xyz."

We can prove it as following:
WLOG  |x| <= |y| <= |z| => |xy| <=1
By Cauchy-Schwartz
(xyz - (x+y+z))^2= (z(xy-1)-x-y)^2 <= [z^2+(x+y)^2][(xy-1)^2+1]
= (2+2xy)(2-2xy+(xy)^2) = 4 - 2(xy)^2(1-xy) <= 4

:)

[quote="liyi"]Sequence $ {a_n}$ satisfies $ a_1 \equal{} 1,a_{n}a_{n \plus{} 1} \equal{} n,$ show that

$ \frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} \cdots \plus{} \frac {1}{a_n} \geq 2\sqrt {n} \minus{} 1.$[/quote]$ a_{2k} \equal{} \frac {3\cdot 5\cdot \cdots \cdot (2k \minus{} 1)}{2\cdot 4\cdot \cdots \cdot (2k \minus{} 2)} \equal{} \frac {4k}{\pi}\int_0^{\frac {\pi}{2}}\sin^{2k}{x}d{x},$

$ a_{2k \plus{} 1} \equal{} \frac {2\cdot 4\cdot \cdots \cdot (2k)}{1\cdot 3\cdot \cdots \cdot (2k \minus{} 1)} \equal{} (2k \plus{} 1)\int_0^{\frac {\pi}{2}}\sin^{2k \plus{} 1}{x}d{x},$

$ \sum_{k \equal{} 1}^{n}\frac {1}{a_k} > (\pi \plus{} 2)\sqrt {\frac {n}{2\pi}} \minus{} 1 \plus{} \frac {(\pi \minus{} 2)( \minus{} 1)^n}{4\sqrt {2\pi n}}.$


function
induction
algebra proposed
algebra

Suppose f is a function on Q such that f(x) is 0 or 1 for every x and such that if f(x)=f(y) then f(x)=f((x+y)/2)=f(y) for every x,y in Q. If  f(0)=0 and f(1)=1 show that f(q)=1 for every q in Q greater than or equal to 1.

Suppose we have such function $ f $. Then let us prove by induction
that for every natural $ n $, $ f(n) = 1 $. First of all, if $ f(2) = 0 $ then
$ f(0) = f(2) = 0 $, so $ f(1) = 0 $ ,but $ f(1) = 1 $. Therefore
$ f(1) = f(2) = 1 $. Now suppose that we know that for some $ m \geqslant 1 $,
for every $ 1 \leqslant k \leqslant 2m $ we have $ f(k) = 1 $. Then
if $ f(2m+2) = 0 $ then $ f(0) = f(2m+2) = 0 $ so we must get that $ f(m+1) = 0 $,
but $ f(m+1) = 1 $. So we get that $ f(2m+2) = 1 $, but also $ f(2m) = 1 $,
so $ f(2m+1) = 1 $. Therefore for every $ 1 \leqslant k \leqslant 2m+2 $,
$ f(k) = 1 $. Now suppose the for some $ r \in \mathbb{Q} $, $ r > 1 $
we have $ f(r) = 0 $. Define new function $ g(x) := 1 - f(1 + x(r-1)) $.
Then of course $ g $ satisfies the conditions of the problem, therefore
we get that for every $ n \in \mathbb{N} $, $ g(n) = 1 $,
so for every $ n \in \mathbb{N} $ we have $ f(1 + n(r-1)) = 0 $.
But if we take $ n $ to be a denominator of $ r-1 > 0 $, then
$ 1 + n(r-1) \in \mathbb{N} $, but as we have seen before, there must be
$ f(1 + n(r-1)) = 1 $, and we come to contradiction. Therefore for every
rational $ r > 1 $ we have $ f(r) = 1 $.

Nice solution Leva1980 :D In my solution i used the same ideea :D


So...I encountered this interesting problem today:
Let n be an integer. Of all fractions 1/n, the fractional part of √123456789 is closest to one such fraction. Compute that value n.
(2003 ARML Individual Round #6)
Does this belong here, or in the Getting Started forum?  :?

Is this one of the ARML problems where you get to use a calculator or no?

[hide="if it is"]
$\sqrt{123456789}= 11111.111060555555440541666143353 \rightarrow n \approx \frac{1}{0.111060555555440541666143353} \approx \boxed{9}$[/hide]

Of course you're not allowed to use a calculator! :roll:

Of course there are no caluclators... :)
Hint: Infinite geometric series...

well, first you want to find the perfect square that it is close to:

$11111^2=123454321$ (note the cool identity  :P )

then the difference is $2468$

then you want to do
$\frac{1}{\sqrt{123456789}-11111}=n$, rationalize it to get

$\frac{\sqrt{123456789}+11111}{2468}=n$, and we have those 2 squares are approximately equal, so $n \approx \frac{11111}{1234}\approx 9$


trigonometry
limit
calculus
calculus computations

Define $ f(x,y)=\frac{\sin{x^3}}{x^2+y^2}$ in $ (0,0)$ to get $ F(x,y)$, which will be continuous on $ R^2$ and study its differentiability.

[hide]
[b]Continuity in (0,0):[/b]

$ f(0,0)=\lim_{\begin{array}{c}
x\rightarrow 0\\
y\rightarrow 0
\end{array}}f(x,y)=\lim_{\begin{array}{c}
x\rightarrow 0\\
y\rightarrow 0
\end{array}}\frac{\sin{x^3}}{x^2+y^2}=\lim_{\begin{array}{c}
x\rightarrow 0\\
y\rightarrow 0
\end{array}}\frac{\sin{x^3}}{x^3}\frac{x^3}{x^2+y^2}=\lim_{\begin{array}{c}
x\rightarrow 0\\
y\rightarrow 0
\end{array}}\frac{x^3}{x^2+y^2}\leq \lim_{\begin{array}{c}
x\rightarrow 0
\end{array}}\frac{x^3}{x^2}=\boxed{0}$

[b]Differentiability:[/b]

$ f_x=\frac{3x^2\cos{x^3}(x^2+y^2)-\sin{x^3}2x}{x^2+y^2} \Rightarrow f_x \; \mathrm{continuous \; on \; R}^2 \setminus \{(0,0)\}$

$ (0,0): \lim_{h \to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h \to 0}=\frac{\sin{h^3}}{h^3}=\boxed{1}$

But $ \lim_{y\to 0}\lim_{x\to 0}f_x=0$, therefore $ F_x$ is not continuous in $ (0,0)$

Similarly $ f_y \; \mathrm{continuous \; on \; R}^2-\{(0,0)\}$ and $ F_y{(0,0)}=0$.

Since $ F(x,y)$ is continouos on $ \mathrm{R}^2$ and $ F_x, F_y$ are continuous on $ \mathrm{R}^2 \setminus \{(0,0)\}$, it is differentiable on $ \mathrm{R}^2 \setminus \{(0,0)\}$.

[b]Differentiability in $ (0,0)$:[/b]

$ F(h,k)-F(0,0)=\frac{\sin{h^3}}{h^2+k^2}=F_x (0,0) h+F_y(0,0)k+\epsilon\sqrt{h^2+k^2}=h+\epsilon\sqrt{h^2+k^2} \Rightarrow
\epsilon=\frac{\sin{h^3}-h(h^2+k^2)}{\sqrt{(h^2+k^2)^3} }
\Rightarrow
\lim_{\begin{array}{c}
h\rightarrow 0\\
k\rightarrow 0
\end{array}}\epsilon=\lim_{h\to 0}\frac{\sin{h^3}-2h^3}{2\sqrt{2}h^3}=-\frac{1}{2\sqrt{2}}.$

Since $ \epsilon$ does not go to $ 0$ with $ h,k \rightarrow 0$, $ F$ is not differentiable in $ (0,0)$.

Is the upper reasoning correct?
[/hide]

This is fine except the notation $ {\lim_\substack{h\to 0\\k\to 0}} \varepsilon$ near the end is terrible. It would be much nicer and more formally correct to write $ \varepsilon(h,k)$ and $ \lim_{h\to 0}\varepsilon(h,h)$


The 2 digit integers from 19 to 92 are written consecutively to form the large integer N=19202122...909192. If 3^k is the highest power of 3 that is a factor of N, find k.

The sum of the digits of $ 19202122\ldots909192$ is congruent to its $ \bmod3$ residue.

We find that the sum of the unit's digits is $ 9\plus{}8(0\plus{}1\plus{}2\plus{}\cdots\plus{}8\plus{}9)\plus{}0\plus{}1\plus{}2\equal{}372$, and the sum of the ten's digits is

$ 1\plus{}10(2\plus{}3\plus{}\cdots\plus{}8)\plus{}9\cdot3\equal{}378$, so the total sum is

$ 750\equal{}2\cdot3^{\boxed1}\cdot5^3$...

I don't think that's correct. The question asks about the largest 3^k that's a factor of N, not the sum of the digits of N.

[quote="math154"]The sum of the digits of $ 19202122\ldots909192$ is congruent to its $ \bmod3$ residue.

We find that the sum of the unit's digits is $ 9 \plus{} 8(0 \plus{} 1 \plus{} 2 \plus{} \cdots \plus{} 8 \plus{} 9) \plus{} 0 \plus{} 1 \plus{} 2 \equal{} 372$, and the sum of the ten's digits is

$ 1 \plus{} 10(2 \plus{} 3 \plus{} \cdots \plus{} 8) \plus{} 9\cdot3 \equal{} 378$, so the total sum is

$ 750 \equal{} 2\cdot3^{\boxed1}\cdot5^3$...[/quote]

OK, sorry, my solution wasn't really finished.

The fact that $ 750$ is not divisible by $ 9$ means that $ 19202122\ldots909192$ isn't divisible by $ 9$.

However, it is divisible by $ 3$, so our answer is $ \boxed1$.

@MrBob: Recall that if a number is congruent to some value mod 3, the sum of its digits is congruent to that same value mod 3. (this also applies mod 9).

Proof:
$ \overline{a_0a_1a_2\cdots a_n}\equal{}\sum_{k\equal{}0}^{n}10^ka_k\equiv \sum_{k\equal{}0}^{n}1^ka_k\equal{}\sum_{k\equal{}0}^{n}a_k\pmod{3}$, because $ 10\equiv 1\pmod 3$.

This applies modulo 9 too, because $ 10\equiv 1\pmod 9$ as well.


We call an integer as "good number" if for all the prime divisors of that integer, square of the prime divisor also divides it.
Prove that there are infinitely many consecutive good number twins.
For example 8 and 9 are good integers and they are consecutive.

If $(x,x+1)$ is a solution, then $ 4x(x+1), (2x+1)^2$ seems to be a solution.

thank you,
It id easy but I couldn't see that point.

i think i have the solution
if $(m,m+1)$ satisfy the condition then $(4m(m+1),(2*m+1)^2)$ satisfy the condition :D  :D  :D  :D


Let A1,A2,...An be real positive numbers.

Prove that  \sum [k=1,..,n] (Ak)^3 >  \sum [k=1,..,n] A[k+1](Ak)^2

where A[n+1]=A1

The inequality is large (there is equality when the A5k] are all equal), and it follows directly from the rearrangement inequality.

Pierre.


Hi. I tried to use $ \binom{n}{k}$ in TeXnicCenter and It came out as $ nk$.

What happened??

You must have the line \usepackage{amsmath} in the preamble (between \documentclass and \begin{document}).


Suppose for any three distinct members $ a,b$ and $ c$ from an infinite set $ A \subset \mathbb{N}$ we have $ \gcd(a,b)\plus{}\gcd(b,c) > \gcd(a,c)$. Prove that there exists a $ d \in \mathbb{N}$ such that $ \gcd(a,b)\equal{}d$ for all $ a\not\equal{} b \in A$.

Assume that there exist $ a\neq b\in A$ such that $ \gcd(a,b)\equal{}1$. For all $ c\in A$ ($ c\neq a,c\neq b$), we have $ \gcd(a,b)\plus{}\gcd(b,c)>\gcd(a,c)$ and $ \gcd(a,b)\plus{}\gcd(a,c)>\gcd(b,c)$. So $ 1\plus{}\gcd(b,c)>\gcd(a,c)>\gcd(b,c)\minus{}1$, which implies $ \gcd(a,c)\equal{}\gcd(b,c)$. But $ \gcd(a,c)$ divides $ a$ and $ \gcd(b,c)$ divides $ b$. Since $ \gcd(a,b)\equal{}1$, we must have $ \gcd(a,c)\equal{}\gcd(b,c)\equal{}1$. So $ \gcd(m,n)\equal{}1$ for all $ m\neq n\in A$.

Now assume that $ \gcd(m,n)>1$ for all $ m\neq n\in A$. Let $ a$ be an arbitrary element of $ A$ and let $ S\equal{}\{d_1,d_2,\ldots,d_k\}$ ($ d_1<d_2<\ldots<d_k$) be the set of positive integers which divide $ a$. Also let $ S_i\equal{}\{b: d_i|b,b\in A\}$ for all $ 1\le i\le k$. Since $ A$ is infinite, there exists $ i$ such that the set $ S_i$ is infinite. Take the smallest $ i$ such that there exist $ m\neq n\in S_i$ with $ \gcd(m,n)\equal{}d_i$. For all $ p\in S_i$ ($ p\neq m,n$), we have $ \gcd(m,n)\plus{}\gcd(p,n)>\gcd(p,m)$ and $ \gcd(m,n)\plus{}\gcd(p,m)>\gcd(p,n)$. Therefore $ d_i\plus{}\gcd(p,n)>\gcd(p,m)>\gcd(p,n)\minus{}d_i$. Since $ d_i|\gcd(p,m),\gcd(p,n)$, we must have $ \gcd(p,m)\equal{}\gcd(p,n)\equal{}d_i$. Thus $ \gcd(x,y)\equal{}d_i$ for all $ x\neq y\in S_i$.

We shall prove that $ S_i\equal{}A$. Assume, by way of contradiction, that there exists an integer $ b$ such that $ b\in A$ and $ b\not\in S_i$. If $ \gcd(b,d_i)\equal{}d_k>1$ ($ d_k$ is a divisor of $ d_i$ and hence is also a divisor of $ a$), then the set $ S_i$ is a subset of $ S_k$. But $ S_k$ has the property that there exist $ m\neq n\in S_k$ with $ \gcd(m,n)\equal{}d_k$, contradicting the minimality of $ i$. Thus $ \gcd(b,d_i)\equal{}1$.

Let $ S_i\equal{}\{a_1,a_2,\ldots\}$. Since $ \gcd(b,d_i)\equal{}1$ and $ a_1,a_2,\ldots$ are all divisible by $ d_i$, then $ \gcd(b,a_m)\equal{}\gcd(b,a_m/d_i)$ for all $ m\ge1$. Note that $ a_1/d_i,a_2/d_i,\ldots$ are pairwise relatively prime. But we know that $ \gcd(b,a_m/d_i)>1$ for all $ m\ge1$. So $ \gcd(b,a_1/d_i),\gcd(b,a_2/d_i),\ldots$ are pairwise distinct integers. This implies that $ b$ has infinitely many distinct divisors, which is impossible.

Therefore, our assumption that there exists $ b$ such that such that $ b\in A,b\not\in S_i$ is wrong. So $ S_i\equal{}A$ and $ \gcd(a,b)\equal{}d_i$ for all $ a\neq b\in A$. This ends our proof.


find all rational non-zero numbers such that (x+6)/x is an integer.

[hide]First of all, this expression will never come to 1, so the smallest positive integer it can be is 2. 1, 2, 3, and 6 work. -1, -2, -3, and -6 also work.

I think you assume integers for x however, because there are infinite values for x. Just set up a proportion, and you see that for any integer we want to find as the ratio of x+6 and x, there is a rational solution to x.  [/hide]

yes, you're correct. i know that there is an infinite amount of solutions for x if you want the fraction to be rational, but, i was only asking for integer solutions to the fraction and rational x values.


[hide]
My solution has nothing to do with the title's suggestion but here it goes:
Knowing
\[\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}\]
We can show
\[\displaystyle \sum_{n=1}^\infty \frac{1}{(2n-1)^2}=\sum_{n=1}^\infty \frac{1}{n^2}-\sum_{n=1}^\infty \frac{1}{(2n)^2}=\frac{3}{4}\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{8}\]
Which means
\[\displaystyle \frac{1}{3^2}+\frac{1}{5^2}+...=\frac{\pi^2}{8}-1<\frac{1}{4}\]
This proves the case for all n because the sum is always increasing.
[/hide]
I don't know how induction would fit in since the upper bound doesn't change with increasing $n$.  I'm sure someone has a more elegant solution than mine.


The math journals titles are abbreviated in articles. I have hard time sometimes to figure out the full name and search for it at library.
One way to do is to google them and hope I can find the full name. Is there any faster way?
Thank you

[url=http://www.ams.org/msnhtml/serials.pdf]This[/url] is a reasonably full alphabetical list of abbreviated and full math. journal names. I'm not sure that scrolling works faster than typing in the Google search line but if Google doesn't help, this list, most likely, will.