MATHCOUNTS
summer program
MathPath
geometry
probability
AMC
AIME

I got a question.

I'll be coaching my younger sister on her Mathcounts next year. But before I do, I wondered if any coach who were successful at making their students to national or high place at students had any tips to give me.

I remember frost commenting about his students going to national.. There are more except I can't remember..

Thanks for the good tips.

Silverfalcon

[quote="Silverfalcon"]I got a question.

I'll be coaching my younger sister on her Mathcounts next year.
Silverfalcon[/quote]
Do make full use of the summer break.

All of summer my son worked on AMC 8s and MathCounts handbooks and old tests.  He put in about 15 hours a week.  During the last month of summer I got him the AoPs books too.  He enjoyed reading book 1.

He improved from 23rd in chapter of 6th grade to second in state of 7th grade.

He could not have covered that much ground during the school year.  I now repeat, [b]do not let the summer break go by[/b].  Last summer (after 7th grade) he reviewed AoPs vol. 1, covered AoPs vol 2. and also went to MathPath.

Have an incentive plan - I can't remember where I read but one of the coaches/parents here gave a nickel for every problem done correctly and the kid had collected 200 dollars while becoming stronger in Math.

I also have formulas in sheet protecters.  Before I used sheet protectors we lost several copies of them.  You will be surprised how handy they are.  You can have them in the car, living room or dining room - no problem.

Practice countdown with her.

In my coaching at school, the first year we practiced once a week during the school year.  We were third in state.  We started working in summer the next year.  We practiced 2-3 times a week during the school year.  We also practiced full length tests on Sundays.  We(the team) were first in state.  The early bird will catch the worm.

Keep track of her progress on tests.  Let her repeat the tests every 3-4 months.  You could also collect problems she missed and have her do them independently a week or two after the review.  It's not enough for a student to watch you do the problem.  Unless they apply themselves, it (the review) does not really soak in.  This kind of review took one of our school kids from the top 25% in chapter to 11th in state (top 10%).

One thing missing in our school curriculums is construction in geometry.  They do not do anything until the year of high school geometry.  You will be surprised how much she will understand if she knows how to bisect a segment and angle, construct equilateral triangles, circumscribed and inscribed circles for triangles, etc.  I think that kids get  fascinated with this and learning is fun this way.  It is more like art.  It also helps them focus.

If you have the scores for your chapter and state - let her know what was required in the last few years to make it.  When they see how far (or near) they are, it helps.

Lastly, remember that she should want to go to Nats.  After a certain point, they should be motivated.   You cannot goad them forever.

Coaching is fun.  Enjoy!

Naga

Thankyou! I wanted to coach my little sister too.

What my coach did was make a student go through every problem of every handbook he had, and every time the student missed a question, he was to find out HOW to do the problem by ASKING someone.  The idea was that you memorize formulas and ideas much better when they are explained to you by a human being instead of a book.  And judging by how well that student did, I would say this is a great way of coaching!

Also,
[quote="Naga"]
I also have formulas in sheet protecters.  Before I used sheet protectors we lost several copies of them.  You will be surprised how handy they are.  You can have them in the car, living room or dining room - no problem.[/quote]

I'd say that is a great idea.  Here is a starter list of formulas, but a few are fuzzy because of the Latex.

Falcon,

Two words : Early and Often  :)

Start early(now) and practice as frequently as you can.  For simple volume of practice problems you want to get your hands on as many old MATHCOUNTS handbooks as you can.  If your sister is in 6th grade(or going to be next year), I woud start with the handbooks.  Another good source of problems for younger kids can be the old math league tests.  They are multiple choice, but you can always photocopy and white the answers out, or just go with the multiple choice.  You probably want top use the 8th grade tests even with a younger kids, but you can buy books with 7th and 8th grade old tests at http://www.mathleague.com

Also try to get her to memorize things like square, triangular, cubic, etc numbers that don't require great understanding at first.  I'll attach the list I give my kids, it is pretty similar to Mr Diaz Bible, but with some things removed and others added.

Good luck,

frost

What would you recommend for a Gifted and Talented 4th grader working on 5th grade math? Also, what are your opinions on the JHU "Elementary Math Olympiad" course? (http://cty.jhu.edu/math/scope/olympiad.html) Does anyone know if it helps?

Thanks again.

[quote="Naga"]

He could not have covered that much ground during the school year.  I now repeat, [b]do not let the summer break go by[/b].  Last summer (after 7th grade) he reviewed AoPs vol. 1, covered AoPs vol 2. and also went to MathPath.

[/quote]

Your son went to MathPath? What's his name?

The MATHCOUNTS Toolbox is very comprehensive and easy-to-use (in my opinion): http://www.mathcounts.org/webarticles/articlefiles/105-toolbox.pdf

[quote="chess64"][quote="frost13"]
Also try to get her to memorize things like square, triangular, cubic, etc numbers that don't require great understanding at first.[/quote]

Why should you have triangular numbers memorized? They don't come up [i]that [/i] often (in my experience) and you can just do $\frac{n(n+1)}{2}$ when you do need to know one.[/quote]

I have found there are a lot of problems where you have to count up how many ways something can occur.  Often I find that as I count up the possibilities I recoginize the triangular numbers appearing in the pattern of how many ways something can occur.  By recognizing the triangular numbers you can often stop actually counting or listing then out and just follow the triangular numbers.

Also if you have a lot of them memorized, it can help you factor numbers more quickly.

Of course it's just my opinion.  It certainly can't hurt you to memorize them though.

[quote="mysmartmouth"][quote="tarquin"]My coach has us:
Practice every morning, have a class for practice, and practice everyday after school. Practice during spring break, Christmas break, and Thanksgiving break. Practice over the summer. Also, be sure to practice at home. Do lots of tests. When you run out, find problems from other contests.[/quote]

Note that he said [b]when[/b] you run out, not [b]if[/b] you run out. :)[/quote]

I don't worry about it too much because I know all math sources in various level such that some are not only in US but in other countries around the Europe and foreign countries.

Also, I'm active problem writer (maybe not that expert as some hard problem maker but I write my contest) so new problems can be produced any time.

An independent study during school can be extremely helpful.  Also, don't forget to practice a lot under time pressure.  Sometimes, though, it's just plain experience that's most helpful.

doing problems a lot can help you develop ways to solve each problem, so next time you see a problem that's similar, you know the solution immediately. It can be very helpful in the timed competitions.

Not doing Mathcounts for several months can be painful though... All those skills you built up will slip away in time if you don't practice.

These are all great tips (especially yours, naga) but if I can bring up a question from the third person's perspective: What about coaching a younger sibling when math competitions are not the premier extracirricular activity in the household? Obviously preparing for nationals is top priority this week, but I still will force myself to spend some time at the piano bench, I will be playing in several tennis matches (I play third singles this Tuesday! Yay!), and my sister has a similar schedule. How does one coach when time is a valuable asset that there just isn't enough of? Your strategies all seem to include hours of continual practice daily. That is simply impossible for many of us.

[quote="09husbri"]What about coaching a younger sibling when math competitions are not the premier extracirricular activity in the household? Obviously preparing for nationals is top priority this week, but I still will force myself to spend some time at the piano bench, I will be playing in several tennis matches (I play third singles this Tuesday! Yay!), and my sister has a similar schedule. How does one coach when time is a valuable asset that there just isn't enough of? Your strategies all seem to include hours of continual practice daily. That is simply impossible for many of us.[/quote]

Do what you can :)   I would say to be consistent -- try to set aside a minimum block of time for mathcounts, and do as much practice as you can within that time.  (Much of the time your sister spends working on mathcounts doesn't require you to be right there at the same time.  She cna work problems alone and then discuss any issues that come up with you later.)   If you sister has a feel for a certain type of problem that is a weakness for her, work on that.  Do try a couple of full competitions under the appropriate time constraints as well.

Having a variety of interests and extracurriculars is great.  I would imagine that few kids can make it to the very top without letting mathcounts take over their lives for some time, but it doesn't have to be everyone's goal to win nationals.  Just getting there is awesome.

Good luck!

The Aops books are good but taking the actual classes helps a WHOLE LOT MORE!  I took Intro to Geometry and a couple of classes of Intro to Counting and Probablity and advanced from not even on the top 50 (we won't go there) to 7th in the chapter.  It really helps but work more and practice more than i did :D.

elvenchamp777

Personally, I don't get much from text books or text-book-like resources. The AoPS books are some of the best I've seen for those who learn that way, and I think some of the best methods are:

1. Read materials like AoPS.
2. Take classes.
3. Work problems. If you can't figure them out look in math encyclopedias, textbooks, etc.

If you can't spend tons of time, it doesn't matter. Make sure that you set off some time each day or every other day (even every third day). For instance, you don't want to say, "Oh, I think we'll practice for three hours today, an hour three times next week, etc. We'll be working three hours a week." Instead, maybe spend an hour on Tuesday, Thursday, and Saturday of each week. Remember, slow and steady wins the race.

Something else that may be less conventional (I like unconventional methods as opposed to hearing something over and over again from different teachers) is to encourage active participation in math circles (real or virtual), including forums like this, and let the student post problems that they need help with, solve problems that others need help with, etc. I believe that socializing builds interest and capability.

[quote="09husbri"] How does one coach when time is a valuable asset that there just isn't enough of? Your strategies all seem to include hours of continual practice daily. That is simply impossible for many of us.[/quote]

Most of the students these days have hectic schedules like yours.  We do tennis and music too.  Use the week-ends to plan your next week.  Even if your Saturday is gone, you still have a few extra hours in which you can do 'known' homework.  Over a week during the school days, she should be able to get a couple of hours.  Like mathmom said, you don't have to be there all the time.  Let her stretch - Unless she has made an attempt, do not teach.  The young coach likes to show the solution.  It is interesting to see how the students think.

I will give you a very simple example.  In one of the school competition questions, (can't remember which as I do not have it right here) a student multiplied out the whole numerator and then the whole denominator.  Finally she divided the N by the D and started reducing the fraction.   After the test, when I did a one-on-one review, she realized how she could have canceled out factors and taken about fourth of the time.   I do not think that kid will ever forget what she learned from this experience.

Let them take long ways or incorrect ways and help them analyze what's wrong.  It's great that there are so many mathletes on this forum who want to do this one-on-one coaching for their siblings.  Good for you.  Do not be surprised if you learn something along the way - you will.  If her desire to win the Nationals becomes very strong, she will find the time to practice for several hours.

One of our third graders was so determined to win the spelling bee, that he mastered the 3000 word list given by the Topeka Capital journal.  While preparing for this, he was also swimming and doing gymnastics.  Such is the spurt of energy and motivation that comes when the desire is strong.

Last and most important - it's not the winning that counts but the experience and fun of it.  Less stress on winning - more on doing their best.

Naga

Hehe, I'm lucky. My 5th grade sister has no interest whatsoever in doing MathCounts. She can do some State problems, but she just isn't interested.

If you're the sadistic bad older sibling (like me) who enjoys watching their younger sibling squirm in frustration, boredom, and semi-agony, by all means push your younger sibling to practice lots. Your parents will say, "Great job, you're teaching your sibling!", as an added bonus.

Formulas = good if you know how to derive them.

I'm on only child :) .
However, that appears to just make things worse. Next year, I will be coaching the entire mathcounts team from my school! It's not a big public school, but it doesn't matter anyway. Plus I'm helping to write a competition that started this year and I will be working constantly on AIME and USAMO, so as to go to MOSP.
What a fun life! (no that's not being sarcastic)

Thanks for the advice, everybody. I appreciate your input. :)

What is more important for mathcounts, practicing a lot of questions after you know the basic middle school skills, or practice problems not as much but learn stuff up to the early high school level?

I tend to do both, but does the state level or below really involve that much high school material?

I think at nationals, some harder problems pop up.

For Mathcounts: practice tons of middle school problems. If you want to stay active and successful in competitions, you should start practicing harder material (high school level) early, but not give up on mathcounts. I am really looking forward to next summer (after mathcounts, before high school) because I will be able to practice just the AIME, USAMO, etc. If you are in 8th grade, you should try to take advantage of the summer (in fact, it is 1 month longer than usual in a lot of NC).


integration
real analysis
inequalities
calculus
derivative
real analysis unsolved

Let $ (X,\mathcal{B})$ be a measurable space. Let $ \mu$ and $ \nu$ be measures on the sigma-algebra $ \mathcal{B}$ with $ \mu \geq \nu$. Show that there is a measure $ \lambda$ on $ \mathcal{B}$ such that $ \mu\equal{}\nu\plus{}\lambda$.

[I would define $ \lambda\equal{}\mu\minus{}\nu$ where it takes the value 0 when both $ \mu$ and $ \nu$ are infinite. Does that work?]

Show that the measure $ \lambda$ is unique if $ \nu$ is $ \sigma$-finite.

Show that in general the measure $ \lambda$ need not be unique but there is always a smallest such $ \lambda$.

[quote] I would define $ \lambda\equal{}\mu\minus{}\nu$ where it takes the value 0 when both $ \mu$ and $ \nu$ are infinite. [/quote]

I figured this will not work because of the monotonicity or additivity required by $ \lambda$ as a measure. However, I certainly will want $ \lambda$ to reduce to that difference in the case where $ \nu$ is finite.

The $ \sigma$-finite case is typically done by Radon-Nikodym. Take $ \lambda(E) \equal{} \int_E 1 \minus{} \frac {d\nu}{d\mu} \, d\mu$. You will need to verify some technical details, but it's standard stuff (cf. a proof of radon nikodym or lebesgue decomposition or something like that).

In general I guess you can use Zorn's lemma. Consider all measures $ \lambda$ on script B such that $ \nu \plus{} \lambda \leq \mu$, partially ordered by $ \leq$. If you have a chain of $ \lambda_j$'s then take $ \lambda(E) \equal{} \sup \lambda_j(E)$ as an upper bound. So hence there exists a maximal element $ \lambda$. If there were a set such that $ \nu(E) \plus{} \lambda(E) < \mu(E)$ we reduce this to a $ \sigma$-finite case (i.e. by extracting a $ sigma$-finite subset of $ E$ where the inequality still holds), and then apply Radon-Nikodym to show that $ \lambda$ would not be maximal.

I'm just writing what I think will happen intuitively, so I haven't checked. And again there would be a bunch of technical details. Maybe the problem is easier than this, but it seems like zorn's lemma is necessary since defining such a measure seems like a very transcendental problem when we don't have countable finiteness

I would think that it can be done using ideas simpler than Radon-Nikodym because, for the record, this problem appears in the very first section (on measure spaces) where Royden begins the part on general measure and integration theory.

Well the Radon-Nikodym part is maybe not necessary. You can always write a sigma finite space as $ X \equal{} \bigcup_1^\infty E_n$ where $ E_n$ is an [i]increasing[/i] sequence of sets with finite measure. So you can probably easily define $ \lambda$ via that. The thing with Radon-Nikodym is just that we get the verification that what we define is a measure for free, because it's been done already in integration theory (all you need verify is that the derivative lies in [0,1] a.e. but that is easy)

I happened to see a suggestion elsewhere to define $ \lambda(E) \equal{} \sup_{F \subset E, \nu(F) < \infty}(\mu(F) \minus{} \nu(F))$. Certainly it reduces to just the difference when $ \nu(E) < \infty$.

To show that $ \lambda$ is a measure, here's what I did:
First, $ \mu\minus{}\nu$ when restricted to measurable sets with finite $ \nu$-measure is a measure. Using this and some arguments taking supremum, I showed that $ \lambda$ is finitely additive, after which I used this to show that $ \lambda$ is countably additive.

To show that $ \lambda$ satisfies the requirement:
If $ \nu(E)\equal{}\infty$, then $ \mu(E)\equal{}\infty$ so $ \nu(E)\plus{}\lambda(E) \equal{} \mu(E)$ and if $ \nu(E)<\infty$, then $ \nu(E)\plus{}\lambda(E)\equal{}\nu(E)\plus{}(\mu(E)\minus{}\nu(E))\equal{}\mu(E)$. Hence $ \mu\equal{}\nu\plus{}\lambda$.

If $ \mu\equal{}\nu\plus{}\eta$ for some other measure $ \eta$, and if $ \nu$ is finite, then $ \eta\equal{}\lambda$.

If $ \nu$ is $ \sigma$-finite, then $ X\equal{}\bigcup X_{n}$ where $ \nu(X_{n})<\infty$ for each $ n$. We may assume the $ X_{n}$ are disjoint. For any measurable set $ E$, we have $ \eta(E \cap X_{n})\equal{}\lambda(E \cap X_{n})$ for each $ n$. Thus $ \eta(E)\equal{}\sum \eta(E \cap X_{n})\equal{}\sum \lambda(E \cap X_{n})\equal{}\lambda(E)$. i.e. $ \eta\equal{}\lambda$.

I'll appreciate any comments on what I've done, especially if I've made some mistake.
I'll also need some help with the last part of the problem:
Show that in general the measure $ \lambda$ need not be unique but there is always a smallest such $ \lambda$.

[quote="chungyc"]I happened to see a suggestion elsewhere to define $ \lambda(E) \equal{} \sup_{F \subset E, \nu(F) < \infty}(\mu(F) \minus{} \nu(F))$. Certainly it reduces to just the difference when $ \nu(E) < \infty$.[/quote]
How is $ \mu(E) \minus{} \nu(E)\equal{}\lambda(E)$ when $ \nu(E)<\infty$? Can you please justify this?

I was thinking along the lines that $ \mu\minus{}\nu$ would be a measure so when $ \nu(E)<\infty$, then by monotonicity I would get the equality. Perhaps I got that wrong?

But  $ \mu \minus{} \nu$ need not be a  measure on $ (X,\mathcal{B})$. But even if we restrict $ \mu \minus{} \nu$  to the collection of all sets $ E$ for which $ \nu(E)<\infty$ then this collection need not be a sigma-algebra, isn't it?

Oh yes $ \mu\minus{}\nu$ need not be a measure on $ (X,\mathcal{B})$. I realised that in one of the earlier posts. Rather what I should have wrote was that for this definition of $ \lambda$, I could show that it is a measure and perhaps add that those sets $ F$ should be measurable subsets of $ E$, then use monotonicity.

[quote]Show that in general the measure  need not be unique but there is always a smallest such $ \gamma$[/quote]

there are so many notations :lol:  but the idea is followed like this :
suppose that $ \mu \equal{} \nu \plus{} \eta \equal{} \nu \plus{} \eta'$, then we have $ \eta(F) \equal{} \eta'(F)$ for any set $ F$ with $ \nu(F) < \infty$
let $ E\in B$ ( script B), take $ F\subset E$ with $ \nu(F) < \infty$, then we have $ \mu(F) \minus{} \nu(F) \equal{} \eta(F) \equal{} \eta'(F)\leq\eta'(E)$


[color=blue][size=150]PERU TST IMO - 2006[/size]
Sabado, 20 de mayo.[/color]

[b]Pregunta 03[/b]
En cada casilla de un tablero cuadriculado de $2^n$
 filas y $n$ columnas $(n\geq1)$ se escribe un 1 o un -1, de tal
 modo que las filas del tablero constituyen todas las posibles
 secuencias de longitud $n$ que se pueden formar con los numeros 1
 y -1. A continuacion, algunos de los numeros son reemplazados por
 ceros. Probar que es posible elegir algunas de las filas del
 tablero (podria ser una fila solamente) de manera que en las
 filas elegidas, se cumpla que la suma de los numeros en cada
 columna sea cero.

$\text{\LaTeX}{}$ed by Carlos Bravo 2006 - MathLinks AoPS

Se puede reemplazar por cero cualquier número que nosotros elijamos?

Bueno lo que hacen es darte un tablero en donde se han reemplazados [b]varios[/b] numeros por ceros(es decir tienes que analizar [b]todos[/b] los casos).


As you may (or should) know, AoPS stands for Art of Problem Solving.
Try to make up a phrase also named "AoPS" (keep this appropriate).
1) Attacks over Postal Service

1) Attacks over Postal Service
2) Ambush on Pig Street
3) Anaconda on Popcorn Shrimp
4) After one Popular Setting
5) Alice Owned Peter's Sister


Let be $ a,b,t\in \mathbb{R}_\plus{}$ . Prove that :

i) If $ a\equal{}b$ then : $ \frac{1}{a\plus{}bt}\plus{}\frac{t}{at\plus{}b}\equal{}\frac{2}{a\plus{}b}$

ii)If $ a<b$ then : $ \frac{1}{a\plus{}bt}\plus{}\frac{t}{at\plus{}b}\ge \frac{2}{a\plus{}b}$

iii)If $ a>b$ then : $ \frac{1}{a\plus{}bt}\plus{}\frac{t}{at\plus{}b}\le \frac{2}{a\plus{}b}$

For the second and third:
after expansion we got $ 2abt \plus{} b^2 \plus{} t^2b^2 \geq 2b^2t \plus{} ab \plus{} abt^2 \Leftrightarrow 2bt(a \minus{} b) \minus{} b(a \minus{} b) \minus{} t^2b(a \minus{} b) \geq 0 \Leftrightarrow b(a \minus{} b)(2t \minus{} 1 \minus{} t^2) \geq 0 \Leftrightarrow \minus{} b(a \minus{} b)(t \minus{} 1)^2 \geq 0$
From this follows directly the second and the third  :)


AB and AC are two equal chords in a circle.  PA and PB are tangent segments to the circle. Prove that \angleAPB = \angleBAC

this is just an angle chasing problem...use the tangent secant angle theorem, and the inscribed angle theorem, and 180 degrees in a triangle to show that they are equal

Thanks for bringing this up:

The tangent-secant angle theorem was in chapter 14 of my Geometry book. Unfortunately, that was last year, and I forgot almost everything I learned in chapter 14 (but nothing of any other chapter). Please remind me of that theorem.

using the picture that you have, lets say that arc AB = x, then angle PBA = x/2


I was reading about Logarithms and there was a section of problems.
What are the answers to the problems?

Evaluate $ (\log_{50}2.5)(\log_{2.5}e)(\ln2500)$. - I know that one is 2

Evaluate $ (\log_{2}3)(\log_{3}4)(\log_{4}5)\cdots(\log_{2005}2006)$

Simplify $ \frac{1}{\log_{2}N}\plus{}\frac{1}{\log_{3}N}\plus{}\frac{1}{\log_{4}N}\plus{}\cdots\plus{}\frac{1}{\log_{100}N}$ where $ N\equal{}(100!)^{3}$.

[quote="Quevvy"]I was reading about Logarithms and there was a section of problems.
What are the answers to the problems?

Evaluate $ (\log_{50}2.5)(\log_{2.5}e)(\ln2500)$. - I know that one is 2

Evaluate $ (\log_{2}3)(\log_{3}4)(\log_{4}5)\cdots(\log_{2005}2006)$

Simplify $ \frac{1}{\log_{2}N}\plus{}\frac{1}{\log_{3}N}\plus{}\frac{1}{\log_{4}N}\plus{}\cdots\plus{}\frac{1}{\log_{100}N}$ where $ N \equal{} (100!)^{3}$.[/quote]

[hide="hint"]
use change of base for 2 and 3. Then use the chain rule.[/hide]

[quote="Quevvy"]I was reading about Logarithms and there was a section of problems.
What are the answers to the problems?

Evaluate $ (\log_{50}2.5)(\log_{2.5}e)(\ln2500)$. - I know that one is 2

Evaluate $ (\log_{2}3)(\log_{3}4)(\log_{4}5)\cdots(\log_{2005}2006)$

Simplify $ \frac{1}{\log_{2}N}\plus{}\frac{1}{\log_{3}N}\plus{}\frac{1}{\log_{4}N}\plus{}\cdots\plus{}\frac{1}{\log_{100}N}$ where $ N \equal{} (100!)^{3}$.[/quote]

Changes of base isn't needed for 1 and 2, though it is useful.
[hide="Hint"]Try the mutliplication rule and the fact taht $ b^{\log_{b}{x}}\equal{} x$
[/hide]

Edit: fixed now  :oops:
[hide="Answers"]
[hide="Problem 2"]
$ \log_{2}{2006}$
[/hide]
[hide="Problem 3"]
$ \frac{1}{3}$
[/hide]
[/hide]

[quote="Quevvy"]Simplify $ \frac{1}{\log_{2}N}\plus{}\frac{1}{\log_{3}N}\plus{}\frac{1}{\log_{4}N}\plus{}\cdots\plus{}\frac{1}{\log_{100}N}$ where $ N \equal{} (100!)^{3}$.[/quote]

[hide="Solution to 3 with work"]Um...replacing one with the respective logs....ie $ \frac{\log_{2}2}{\log_{2}N}\plus{}\frac{\log_{3}3}{\log_{3}N}\plus{}...\plus{}\frac{\log_{100}100}{\log_{100}N}$,
and then using the change of base rule...you get:
$ \log_{N}2\plus{}\log_{N}3\plus{}...\log_{N}100$, and then that simplifies to $ \log_{N}100!$

Substituting in $ (100!)^{3}$ for $ N$, we get:
$ \log_{(100!)^{3}}100!$, which is equal to simply $ \frac{1}{3}$.[/hide]

Question...
[quote="Anaxerzia"]
[hide="Hint"]Try the mutliplication rule and the fact taht $ b^{\log_{b}{x}}\equal{} b^{x}$
[/hide][/quote]

Were you talking about 1 and 2 or 3? If 1 and 2, why not use the chain rule, which is way faster and simpler? If you were talking about 3, could you show me how that would apply to the question?

[quote="Anaxerzia"]Try the mutliplication rule and the fact taht $ b^{\log_{b}{x}}\equal{} b^{x}$[/quote]
Your hint is incorrect, it should be $ b^{\log_{b}x}\equal{}x$.


topology
advanced fields
advanced fields unsolved

Question :

[b]Find a Separation axioms on a Space $ X$ that's necessary and Sufficient for the uniqueness of the limit of a sequence.?[/b]

please ,Need very useful details.

Is there anybody know the book about [u][b]TOPOLOGY its Author [/b][b]Shukla[/b] :help: [/u].

I appreciate your help.

Are you sure that limits of [i]sequences[/i] are considered? If one uses filters or nets instead limits are unique iff X is T_2. But there are T_1 spaces that are not T_2 where the limit of a convergent sequence is unique but the limits of filters are not.

[b]Hey Gockel

Yes I'm pretty sure about the limit of the sequence... Seldom to see this , since the only book has this question is the book by Shukla , But I couldn't find that book.

So, the only hint I knew about that question is to define a new space between $ T_1$ and $ T_2$.

Thx , I'm waiting your response.

 :) [/b]

[b]Where are the Clevers people here.... :|

I don't think this is very hard on people like you.

Anyway , what if I define new separation axiom as $ T_*$ , if $ x\in U$ where $ U$ is open and it contains $ x$, then $ \forall V_x$,   [s.t $ V_x$ is open containing x], $ V_x$ must meet $ U$.

Is this satisfies the uniqueness of the limits of the sequence...

Need your Opinion .. :) [/b]

[b] :roll:  :roll:  :roll:

I don't think that space satisfies the uniqueness .

Any suggests  please.

[/b]

Hey guys :

What about if we define the following space :

for any distinct $ x$ and  $ y$ in $ X$ , there are an open set $ U$ contains $ x$ not $ y$, and open set $ V$ contains $ y$ not $ x$ s.t. If $ E$ is any countable set in $ X$ , then $ U \cap V \cap E$ has finite elements.

Is this space satisfies my condition for uniqueness?

Need your opinion.

 :D

It js just a very fancy way to state $ T_2$. Indeed, the last condition is equivalent to $ U\cap V$ being finite and you can easily get rid of finitely many points in the intersection using $ T_1$.

O.k. , what about if we make the intersection to be countable set of them.

Is it work?

 :maybe:


For a real number $x$ let $[x]$ be the greatest integer less than or equal to $x$ and let $<x>$ be equal to $x-[x]$. If $a$, $b$ and $c$ are distinct real numbers, prove that $\frac{a^{3}}{(a-b)(a-c)}+\frac{b^{3}}{(b-a)(b-c)}+\frac{c^{3}}{(c-a)(c-b)}$ is an integer if and only if $<a>+<b>+<c>$ is an integer.

[hide="Hint"]The given expression in $a,b,c$ reduces to $a+b+c.$[/hide]


Maybe I'm understanding the problem incorrectly but...

Isn't C on AD?...so isn't ABCD a triangle and shouldn't it be ABDC?

[quote="Carbon57"]Isn't C on AD?...so isn't ABCD a triangle[/quote]\

Yes, I'm sorry. It's fixed now.

[quote="jvenezuela716"]Consider a circle with perpendicular diameters $ AB$ and $ MN$. Let $ C$ be any point on $ MN$ and $ D$ be the point where the line $ AC$ extended intersects the circle. Prove that quadrilateral $ OBDC$ is cyclic.[/quote]
[hide="Solution"]
Since $ AB$ is a diameter of the circle, we have that $ 90\equal{}\angle ADB\equal{}\angle CDB$.  Since $ AB$ and $ MN$ are diameters of the circle, we have that they both have $ O$ in common.  Therefore, $ AB$ and $ MN$ intersect at $ O$.  Furthermore, $ AB\perp MN$, so $ 90\equal{}\angle NOB\equal{}\angle COB$.  Therefore, $ \angle COB\plus{}\angle CDB\equal{}90\plus{}90\equal{}180$, so $ COBD$ is cyclic, as desired. [/hide]

[quote="jvenezuela716"]Correct! Now for the more general case:

Consider a circle with perpendicular diameters $ AB$ and $ MN$. Let $ C$ be any point outside the circle on the line $ MN$ extended and $ D$ be the intersection of $ AC$ and the circle. Prove that quadrilateral $ OBCD$ is cyclic.[/quote]

[hide="Solution"]Draw segment $ BD$. Notice that $ \angle BDC \equal{} 180^{\circ} \minus{} \angle ADB \equal{} 180^{\circ} \minus{} 90^{\circ} \equal{} 90^{\circ}$. Also, $ m\angle COB \equal{} 90^{\circ}$. So $ \angle BDC \cong \angle COB$, so $ OBCD$ is cyclic.[/hide]

vishalarul - how do you go from $ \angle BDC \cong \angle COB$ to the fact that the quadrilateral is cyclic?  These angles are not angles of the quadrilateral.

So we know that $ \angle BDC$ is right.  Since $ \overline{OD}$ and $ \overline{OB}$ are both radii of the circle, we have $ \angle OBD \cong \angle ODB$, so lets define $ x \equal{} m\angle OBD$.  Since $ \triangle ADB$ is right, $ m\angle A \equal{} 90 \minus{} x$.  Since $ \triangle AOC \cong \triangle BOC$, $ m\angle A \equal{} m\angle B \equal{} 90 \minus{} x$.  But $ m\angle ODC \equal{} 90 \plus{} x$, so we have $ m\angle ODC \plus{} m\angle B \equal{} 180$.  Thus quadrilateral $ OBCD$ is cyclic.

[quote="facis"]vishalarul - how do you go from $ \angle BDC \cong \angle COB$ to the fact that the quadrilateral is cyclic?  These angles are not angles of the quadrilateral.[/quote]Concyclic points generate a cyclic quadrilateral, his proof is correct.

10000th User, I probably should have realized that.  By concyclic, you mean that the two right angles opposite the same segment have to lie on the same circle as the endpoints of that segment?

Perhaps this wiki article can illustrate the idea
[url]http://en.wikipedia.org/wiki/Concyclic_points[/url]

Aha.. so the 90 degree angles are just a coincidence - the only thing that matters is that have the same measure.  Thanks.


I have seen a non-standar theorem
Given $\triangle ABC$ and the points $A'$, $B'$, $C'$. The lines $AA'$, $BB'$, $CC'$ are concurrent iff there exist three vectors $u//AA'$, $v//BB'$, $w//CC'$ with $u+v+w=0$ such that \[\overrightarrow{OA'}\times u+\overrightarrow{OB'}\times v+\overrightarrow{OC'}\times w=0\] for any $O$.

I have seen it a page of Daniel Pedoe's book (I just have 3 pages of this book because here in Peru is too much dificult to get the book)

Sincerely

I think that Darij can help you. He mentioned that theorem [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=criterion&t=17902]here[/url]. Note that he knows it from the same book as you.

PS: I always considered that criterion a bit too exotic to try to prove it.

ah a nice problem
if you put $AA',BB',CC'$ be concurrent at $I(x,y,z)$(barycentric) $\Rightarrow x\vec{IA}+y\vec{IB}+z\vec{IC}=\vec{0}$
you can choose $\vec{u}=x\vec{IA}//AA',\vec{v}=y\vec{IB}//BB',\vec{w}=z\vec{IC}//CC'$ we have
$\sum\vec{OA'}\times\vec{u}=\sum (\vec{OI}+\vec{IA'})\times (x\vec{IA})=\vec{OI}\times(\sum x\vec{IA})+\sum\vec{IA'}\times\vec{IA}=O$
note:$\sum x\vec{IA}=\vec{0}$ and $(I,A,A'),(I,B,B'),(I,C,C')$ are colinear!
if exist $\vec{u}+\vec{v}+\vec{w}=\vec{0}$ and $\sum\vec{u}\times \vec{OA'}=\vec{0},\forall O$and $\vec{u}//AA',\vec{u}//BB',\vec{u}//CC'$ we will prove $AA',BB',CC'$ are concurrent indeed:
Called $\{I\}=BB'\cap CC'$ choose $O\equiv I\Rightarrow \vec{u}\times\vec{IA'}+\vec{v}\times\vec{IB'}+\vec{w}\times\vec{IC'}=0(1)$
because $\vec{v}//BB',\vec{w}//CC'\Rightarrow \vec{v}\times\vec{IB'}=0,\vec{w}\times\vec{IC'}=0(2)$
From $(1),(2) \Rightarrow \vec{u}\times\vec{IA'}=0 \Rightarrow IA'//\vec{u}\Rightarrow I\in AA'$(follow Euclide's axiom :D)


Gauss
number theory proposed
number theory

Prove [b]without using Gauss Three-square-representation [/b]Theorem:
If a nature number can be represented as three-square of rational numbers,then it can be represented as three-square of integers.

p.s.One book mentioned that it can be proved by Minkowski Theorem.Can anyone give a solution using Minkowski's by computing the volume?

Thank you very much :) This is also the proof I knew
But can anyone use Minkowski Theorem to give an alternative proof?


logarithms
AMC
AIME
algebra
system of equations

[hide]Look at the first equation.
log(2000xy) = log(2xy) + log(1000) = log(2xy)+3
With this, the first equation may be expressed as...
log(2xy)+3 - (log x)(log y) = 4
log(2xy) - (log x)(log y) = 1
This looks like our 2nd equation! In fact, we can infer that x=z.
Using this fact on the third equation:
log (x^2) - (log x)^2 = 0
2log x - (log x)^2 = 0
(log x)(2 - log x) = 0
log x =0 or log x = 2
x,z=1 or x,z=100
If z=1, then from the second equation:
log(2y) - (log y)0=1
2y=10 [b]y=5[/b]
If z=100, then from the second equation:
log(200y) - 2log(y)=1
(log2y) + 2 - 2log(y) = 1
2log y - log(2y) = 1
2log y - log 2 - log(y) = 1
log(y) = 1+ log(2) = log(10) + log(2) = log(20)
[b]y=20 [/b]
Adding up the 2 values of y, the answer is [b][u]25[/u][/b][/hide]

Now that was a HARD problem!!! :D

[quote="Farnak"]Thanks a lot Vishalarul, that was bugging me all morning. Something odd though, I got the problem from an AIME kit at my school, but #9 on the AoPS AIME resources is different from this problem, why is that?[/quote]
Which year was it from? You could also check Kalva.

[quote="Farnak"] Something odd though, I got the problem from an AIME kit at my school, but #9 on the AoPS AIME resources is different from this problem, why is that?[/quote]

That's because the AoPS Resources section only has the Alternate AIME from that year. This problem came from the 1st AIME of that year.


A 20-inch by 12-inch picture is centered within a 28-inch by 18-inch frame.  What is the number of inches in the distance from a corner of the frame to a corresponding corner of the picture?

[hide]since the picture is in the center, then each side would either have a distance of 3 or 4 to the edge of the picture fram. the, but it ask to find the number of inches from a corner of the frame to a corresponding corner of the picture so it would be the square root of (4 square + 3 square ) which is [b]5 inches.[/b]
[/hide]


Stanford
college
email
geometry
AMC
AIME
3D geometry

Announcement:

The 2009 Stanford Math Tournament will be held on Saturday, February 28th.  Info and registration will be at sumo.stanford.edu/smt soon, and I'll also be sending out paper invites once we get a few wrinkles ironed out.  Until then, feel free to direct any questions to me either in this thread or at my email, which you can find on the SUMO page.

Hope to see you there!

So it'll be the same time as Science Bowl? That's kind of bad for us :( And on the same campus, too...

Oh dang it is the same day as Science Bowl Regionals...

@bookaholic: You're in science bowl too? What team? :D

Guys guys guys, Science Bowl is a load of BS from what I heard, why don't you just drop it and come over to SMT? You always have a chance to do Physics or Chem or Bio Olympiads or whatever.

...
I'm not going to say anything...

...In public.

serialk11r, some time in the near future you are going to get a very...colorful...IM from me...when im not feeling too lazy...then we can debate the relative importance of these things to your heart's content...and a bit more. :dry:  :furious:  :spider:

tl;dr: I may not be able to go to SMT this year either :(

ditto here, for the same reason as everyone else

Gah, Kevin Yang, Jeff Han and Joseph Swernofsky are going to be missing cuz of Science Bowl crap probably. Lynnelle you should join us and ditch Science Bowl!

BTW guys, I don't have anything against science (I would not be able to imagine that), its just that SMT happens to seem more important.

No!  Go to Science Bowl!  Then my school can win xD

I'm going to SMT.

http://www.scied.science.doe.gov/nsb/default.htm

Science Bowl is kinda like MC Countdown (complete with buzzers), except with questions from several different branches of science (including math :D )

[quote] Can someone explain to me what Science Bowl is and what advantages it might have over USAPhO/USABO/USAChO/etc.? [/quote]

There are no advantages to participating in the Science Bowl over USAPho/BO/Cho/MO...but it seems like fun =)

[quote="bookaholic"]I'm not actually going to Science Bowl, it's the rest of our SMT team I'm worried about. I would never choose any kind of countdown-style contest over SMT...come on.[/quote]
I counted up our team's guaranteed membership, and it seems to be around 6 (Boyle, me, Brian, another soph, Mr. Hahn and ccy)...so pull your non-Science Bowl paly folk and we can form a Palo Alto team.

The MC Chapter is on Feb 14 and the regional middle school Science Bowl is on Feb. 21, 2009.
So I can still go to SMT. Yay!

Registration forms and all that are now up.  You can register online again this year.

http://sumo.stanford.edu/smt/

I think that AAST (from New Jersey) flew their math team out here, and will be competing against all of us.

They have 40-50 people, I think.

I lost my calculator yay!
The tests were pretty ridiculous, seeing how many perfects there were and how good people didn't win stuff like geometry...
Even though I would lose my tied for first status, I think it would've been more fair to have the pentagon problem be $ \frac{1\plus{}\sqrt5}{2}$ instead of 2cos36.

Well... so it seemed Team/Alg/Advanced/Geo were way way way too easy. I won't whine about my tests, since I'm dumb and like making careless mistakes. I'm pretty disappointed about the Alg and Geo tests, because #4 on Alg was a legitimate (albeit stupid) question, and I can't believe that 2cos36 was accepted on Geo #3, which is just retarded. Plus, it was a very well known problem... like a lot of the other ones.

Why is it that every time I actually go to a contest (last year I didn't go to SMT), it sucks?

[quote="Ubemaya"]I can't believe that 2cos36 was accepted on Geo #3, which is just retarded[/quote]

Only because you were too dumb to realize that golden ratio = whatever the hell you were trying to do with sin18! [hide=":P"][/sarcasm][/hide]

I felt that calc was pretty easy too... none of it required much more ingenuity than what you learn in school.
ah... z=0! and the gear problem from adv. was just wtf?

Sorry about the absurdly long wait for grading.  We were having some computer bugs in the scoring room and wanted to make sure we got all the scores added up correctly.  And we're quite glad we did, as the scores we initially tabulated gave significantly different rankings from the correct ones.

And yes, looking at the scores, this was an easier year.  Personally, I'm happy to see that; I never like seeing tons of zeros on the scoresheet, and 6's and 7's winning awards.  I also would rather not have multiple perfect scores, but I'd rather err on the side of being too easy than too hard.  The part I like best, though, was the balance in difficulty between subjects.  At least looking at the top scores, it seems like most of the tests were about at the same level, with geometry maybe being a little easier.  I think general was also at about the right level of difficulty.  Curiously enough, power turned out to be the hardest event--our counterparts at Rice were worried about that one being too easy.  I hope it gave you all a good challenge.

I'll probably post solutions and rankings over the next couple of days.  Until then, I suggest you take another look at Power #7.  If you spend enough time playing with construction of nontransitive dice, you might discover the trick.

The gear problem was a quarter stepper problem. I say quarter stepper because it only required one computation: 7X21.
wtf was up with all the complex numbers problems that I've seen on school tests before?

Albert, I don't know what's up with that. Last year's SMT was a lot better...maybe contest writers hate you so they replace all the problems with crappy ones as soon as they find out you're going :P
Although I wouldn't say team was too easy, seeing that last year's was not that much harder (maybe it was, but not much more). The power round had lots of silly questions though, like the ones that wanted to find the sum of the visible faces or whatever.

Silas, I think the top scores don't really represent the difficulty well. For geometry, many of the top people finished before time was up, while Stephen was the only person who finished it last year. Advanced topics had a lot of problems that you see in school textbooks like I said, which I think is pretty ridiculous. Last year's test was very very hard compared to this year's. I got 7 this year, from computation mistakes and the likes, while last year I got 4 (and no I didn't improve that much) or a 3. I'd much rather take last year's test however, because it had much more interesting problems that were not completely trivial.

When a test gets so easy that people who clearly do not have as much geometry skill (me) can get higher scores than people who do (like Amol/Quatto, or Taylor), I think there is something wrong. A lot of people can't get anything correct on a lot of contests, but their lackings should not interfere with people at the competitive level.

Holy smokes. I got (9,9) on Adv/Geo instead of the projected (6,7) (practiced 2008).
[b]And z=0... ARGH.[/b]

Goals for next year: don't epic fail tiebreakers, get perfect on at least 1 subject test...

I thought the power round was pretty interesting.
Wait so #10 on algebra had an undefined term in the denominator? What's up with that?

lawl I ended up missing SMT for a swim meet and a mock SAT...wtvr

WHAT? 2cos36 was accepted???
Well that was a waste of like a minute of simplifying...

I did pretty well on alg (felt like 7ish). It was WAY easier than last year's.
I left early, so I have no idea what our scores were (anyone from Combo Beiiige here who got them?)

The power round was cool, as archimedes1 mentioned. As archimedes1 didn't mention, we Beiiigers failed it. (Our #7 proof was "Jesus divined it to be so" and our #8 was "...IMPOSSIBLE!!!! QED").

Geo was cool. I bashed #5 with complex numbers  :lol:

And I got the right values of $ x$ and $ y$ on the minimization problem on algebra by cheating with multivariable calculus, and then I plugged them in wrong and got $ \minus{}7/4$ instead of $ \minus{}5/4$.  :mad:

Person from the Rice Math Tournament here.

#10 on the algebra left me with a feeling of "Jesus Christ". The solution did nothing to the feeling. Was pulling out the implicit differentation on minimization thing, but got lazy. Everything else was pretty normal, used to seeing.

grn, the proof works on number 7, but methinks your grader was atheist.

I looked at the Geo test and realized how incompetent in Geo I really am.

[quote="serialk11r"]Silas, I think the top scores don't really represent the difficulty well.[/quote]

Well, top scores were all I had to go on yesterday.  Now that I've compiled some more data, try these:
On algebra, 35% of students got 5 or more.  5% got 1.  No one got 0.
On topics, 57% got 5 or more.  11% got 1 or less.  (Note that this has a much smaller sample size.)
On geometry, 44% got 5 or more.  21% got 1 or less.
On calc, 35% got 5 or more.  11% got 1 or less.
On general, 45% got 15 or more.  6% got less than 5.

I'm satisfied with all but one of those numbers.  The percentages scoring 5 or more could maybe be a little lower, but with the exception of topics, they're all well within what I would consider to be an acceptable range.  (For topics, I'm inclined to chalk it up to self-selection, since so few students take that event.)  The one number I don't like is that over 1/5 of the geometry students got 0 or 1.  I would like to see that number go way down.  I would also like to point out that at the very top, algebra and calculus differentiated very well.  Advanced topics didn't do too badly (at least only one person got a perfect); geometry obviously wasn't so good.

[quote]A lot of people can't get anything correct on a lot of contests, but their lackings should not interfere with people at the competitive level.[/quote]

This brings me to my main point.  I'm sure this year's SMT was easier than most.  However, I think the usual SMT is harder than it should be.  Like any contest, SMT is supposed to provide a good experience for all competitors, not just the best of them.  Ideally, this means having a range of easy questions at the beginning of tests, transitioning into very challenging ones at the end.  In the past, SMT has lacked the former.  The result was that average students from less well-established schools were scoring 1's, 2's, and 3's on individual rounds, while people won with 7's and 8's.  Judging by the scores, I think the difficulty got dialed down a little this year, to the point where an average student was capable of a 4 or 5, and the very top students scored 9's and 10's.  (Geometry, of course, is the exception; if 7 people got perfect scores, it's obvious there weren't any killer problems.)

That said, you'd have to ask Rice if any of this was intentional.  They assembled the tests this year, so our part in it was merely to write the best problems we could.

[quote="grn_trtle"]Our #7 proof was "Jesus divined it to be so"[/quote]

But #7 is so cool!  Like I said, everyone should play with it a little more.  Nathan graded it, and from what he told me, it sounded like the total number of points he gave out was a one-digit number.  By the way, a more general result is that as the number of dice increases, the maximum possible advantage approaches a limit of 3/4.  (From Martin Gardner's [u]The Colossal Book of Mathematics[/u], which was the source for that particular problem.)

By the way, my favorite non-answer of the day was on an individual test, someone wrote "...and here's a drawing of a hampster!" with an accompanying drawing.  I told the grader they should write "We would have given you a point for this, but you misspelled 'hamster'".  That was when I discovered that half the SMT graders couldn't spell "hamster" either.

[quote="Sly Si"]Yeah, we're not doing tiebreakers or a super relay this year.  They just weren't all that popular last year.  And I expect more volunteers this year, so I think we'll be able to finish up the grading quicker.  No promises, but we can hope.

AIME15, believe it or not, you'd only tie for that award.  There's another group bringing five full teams.[/quote]

I was on AIME15's team - We ended up sending exactly 42 people, and counting the official results, there seems to be also exactly 42 teams.

I liked the power round; maybe it was because I didn't do the "bad" problems, but to say the least, it was an interesting topic.

I found the tests fairly easy. Algebra, I got an 8/10, tying me for fourth. I actually got #9 for algebra--I got the roots of the polynomial, and found the coefficient using Vieta's formula. Except I forgot to multiply by -1.

I just failed calculus. I have a feeling that I got #5 correct (the one with the acceleration, but it was in answer that was not accepted (I fed in the problem to my TI-89, and it gave me an answer different from what I got, though they were numerically the same.)

I failed the Riemann sums one. I really liked the one with the floor functions--it was quite original and got me thinking. Unfortunately, it had a bit too much computation, and I ended up getting something completely wrong.

#9 was a total joke. If I recall correctly, that was #10 from last year's algebra section. We don't even need calculus for that one. Even with calculus, it's completely trivial.

I liked #10 somewhat. Perhaps it's because I evaluate all of my infinite sums (including the Riemann sums one) by bounding them with integrals. Bad habit, I know. It's a shame that I thought that 200*200 = 4,000,000.

I ended up with a 5/10. I hate myself.


The representation of a prime in decimal expansion has the property that each permutation of its digits is a prime in decimal exansion again. Prove that no matter how many digits are in the prime's decimal expansion there are at most n distinct digits in the prime's decimal expansion  !

a.) n = 3
b.) n = 2

Hi,

I dont understand this problem. What do you mean "at least 2 different digits" or "at least 2 different digits". What about 11 for n=2 and 13 for n=3? Can you explain clearly? Thanks.

Namdung

Hi Namdung,

do you understand the problem now ? Or is there anything unclear yet ?


function
linear algebra
real analysis
real analysis unsolved

Prove that $ card A \equal{} card B$ there $ A \equal{} [a,b)$ and $ B \equal{} (a,b)$

[quote="Opalg"][quote="Friendly1"]Prove that $ card A \equal{} card B$ where $ A \equal{} [a,b)$ and $ B \equal{} (a,b)$[/quote]
Let $ A \equal{} (a,b)\cap\mathbb Q$ and let $ A \equal{} \{a_1,a_2,\ldots\}$ be an enumeration of the points in A. Let $ a_0 \equal{} a$. Then the map given by $ f(a_n) \equal{} a_{n \plus{} 1}$ for n≥0, and $ f(x) \equal{} x$ for $ x\in[a,b)\setminus A$, is a bijection from [a,b) to (a,b).[/quote]
Ok,thank you,but can you follow a function satisfying it,please!thank again

This isn't linear algebra and shouldn't have been posted in that section.

[quote="Friendly1"]Ok,thank you,but can you follow a function satisfying it,please!thank again[/quote]
I don't understand this comment. Opaig has given a complete answer.

Here's a slight variation of the same idea, that is a tiny bit more explicit.

Let $ f(x)\equal{}\begin{cases}a\plus{}\frac{b\minus{}a}2&\text{ if }x\equal{}a\\a\plus{}\frac{b\minus{}a}{n\plus{}1}&\text{ if }x\equal{}a\plus{}\frac{b\minus{}a}{n},n\equal{}2,3,4,\dots\\x&\text{ otherwise}\end{cases}.$

Then $ f$ is a bijection from $ [a,b)$ onto $ (a,b).$


i think people who do problems often would certainly have come accross this problem. I just started so this is my first time.

prove that $\log_{10} 2 + \log_{10} 5 = 1$

From your question... would you like for us to prove the log addition rule?   Because the problem is just its application+the definition of a logarithm

$\log_{10}2$
$10^x=2$
Second part,
$\log_{10}5$
$10^y=5$
Multiplying the 2,
$10^x*10^y=10$
$10^{x+y}=10$
$x+y=1$
which is what we wanted to prove


analytic geometry
graphing lines
slope
calculus

Hello,
I am having a problem with the following problem:

"Find an equation of the tangent line to the circle $x^{2}+y^{2}=25$ at the point $(4,-3)$."

Does anyone have any pointers? Thank you.

(Sorry for the crosspost - someone said this was too hard for Classroom Math)

[hide="Hint/Start of Solution"] Let the line be $y=mx+b$. We know it passes through $(4,-3)$, so $-3=m(4)+b\Longrightarrow b=-4m-3$. $y=mx-4m-3$. Now plug this into the circle equation and find when the determinant, with respect to $x$, is $0$, which means it intersects at only one point and is tangent. From there, you can find the slope of the line. [/hide]

[hide="Easier approach"]Find the equation of the radius in point $(4,-3)$ - i.e. of the line passing through the origin and $(4,-3)$ - and from that you can find the slope of the tangent, knowing that radius and tangent are always perpendicular and for the slopes of perpendicular lines it holds $s_{1}s_{2}=-1$.[/hide]

because the circles center is at (0,0), you can draw the line that is perpendicular using points (0,0) and the point of tangency. Then we know the slope of the line using negative reciprocal. Plug in the point of that we know (the point tangent to the circle) in to the slope intercept form and solve for b.

The equation of a line that is tangent to $x^{2}+y^{2}=r^{2}$ at $(x_{0}, y_{0})$ is $xx_{0}+yy_{0}= r^{2}$.

[hide]You know that the slope from (0,0) to (4,-3) is $-\frac{3}{4}$.

Now by law the tangent is perpendicular to the radius, so what we have to do is find the line with slope $\frac{4}{3}$, that pasess through (4,-3).

Which is $y=\frac{4x}{3}-\frac{25}{3}$.[/hide]

I will try to solve by Derivative.

$\frac{d}{dx}(x^{2}+y^{2})=\frac{d}{dx}(25)$
$2x+2yy'=0$
$8-6y'=0$
$y'=\frac{4}{3}$   so slope of tangeent line is also $m=\frac{4}{3}$
At point (4,-3)
Equation of tangent line is
$4x-3y-25=0$

I don't think you are allowed to used Calculus on this forum, if I am not mistaken.


vector
function
linear algebra
matrix
real analysis
real analysis unsolved

In the problems for section 4.4 (inner product spaces) of Herstein's [i]Topics in Algebra[/i], there is a question asking to show that the set of real-valued functions $ y \equal{} f(x)$ satisfying
\[ \frac {d^3y}{dx^3} \minus{} 6\frac {d^2 y}{dx^2} \plus{} 11\frac {dy}{dx} \minus{} 6y \equal{} 0
\]
is a three-dimensional real vector space. I believe that $ e^{x}$, $ e^{2x}$, and $ e^{3x}$ span all solutions, but I'm wondering how one would go about proving it (and how much of a rigorous proof did Herstein actually intend?)

If I assume that $ f(x)$ is analytic, then I can compare Taylor coefficients to show that the vector space has dimension $ 3$. (Specifically, we can choose any values for the first three Taylor coefficients of $ y$ and the rest follow from the differential equation.)

I'm thinking one might be able to prove this under the assumption that $ f$ is bounded by an exponential function by using a Laplace transform.

Though I'm not sure whether there might be other pathological examples. Any thoughts?

It's the existence-uniqueness theorem for ordinary differential equations. A sufficiently nice $ k$th order differential equation (linear with nonzero $ y^{(n)}$ coefficient is more than enough) has exactly one solution for each vector of initial coefficients $ \begin{bmatrix}y^{(n\minus{}1)}(c)&\cdots&y'(c)&y(c)\end{bmatrix}^T$ in a neighborhood of $ c$. If the equation is nice enough (again, linear with never-zero $ y^{(n)}$ coefficient is enough), these solutions can be extended to all of $ \mathbb{R}$.

Actually proving the theorem is fairly hard, requiring quite a bit of analysis. Just find a version from a differential equations textbook and quote it.

The rest is linear algebra, showing that the span of the three linearly independent solutions you have does in fact cover all initial conditions. With the exponentials, you get Vandermonde matrices if you do this explicitly.


A person sells paintings. He makes a 2.50 profit on every small painting and a 4.00 dollar profit on every large painting. He generally sells 5 large paintings for every 2 small paintings. This year, he wants to make a profit of 10,000 dollars from both large and small paintings.

a) How many of each painting does he need to sell in order to reach his goal?

b) If the person can paint a painting in 12 minutes, how many hours would he need to make enough to reach his profit target if he sells them all?

c) What is the painters hourly rate of pay?

[hide="a"]

We have the two equations:
\[ 2.5s + 4b = 10000 \\
 \\
\frac {5s}{2} = b\]
Where $ s = \text{ small painting}$ and $ b = \text{big painting}$.  Substituting $ b$ into the first equation we get:
\[ 2.5s + 4\left(\frac {5s}{2}\right) = 10000 \\
 \\
2.5s + 10s = 10000 \\
 \\
12.5s = 10000 \\
 \\
s = 800 \\
 \\
\therefore b = 2000\]
[/hide]
[hide="b"]
Assuming it takes the same amount of time to paint a small and big painting, we have $ 800 + 2000 = 2800$ paintings, and since each takes $ 12$ minutes, it will take a total of $ 33600$ minutes, or $ 560 \text{ hours}$[/hide]
[hide="c"]
The painter is making $ \$10,000$ and it takes him $ 560$ hours (from (b)), making his rate of pay $ %Error. "about" is a bad command.
\$17.86 \text{ per hour}$[/hide]

Are the above answers correct?

Yes, I get exacty the same answers., the exact value in c being [hide=" click to reveal hidden content"] $ 17\frac{6}{7}$ dollars. [/hide]


modular arithmetic
algebra
polynomial
advanced fields
advanced fields theorems

[quote]Integer n is prime if and only if (x +1)^n ≡ x^n +1 (mod n) in Z[x].[/quote]

I don't really understand how this works...
for x= 4 and n=5
(4+1)^5=3125 but 4^5+1 mod 5 = 0
how does it (4+1)^5=3125  ≡  4^5+1 mod 5 holds?

[quote="JBL"]$(4+1)^{5}\equiv 4^{5}+1 \equiv 0 \pmod5$.[/quote]

I still not sure how it's works... assuming logical $\equiv$  how can it by equivalent?

[quote="ZetaX"]Do you knwo what $\equiv$ is¿ Then just look at the definition. Otherwise read the definition.[/quote]

well $\neg \neg p \equiv p$ but it doesn't clear things out..

Not logic, arithmetics!
See http://en.wikipedia.org/wiki/Modular_arithmetic .

It's a result on the (coefficient-wise) equality of two polynomials.  (That's what the "$x$" is for.)  Two different polynomials can agree at some values without being the same polynomial.

[quote="matumba"]ok, thanx. but this theorem still doesn't make seance...
$(4+1)^{4}\equiv (4^{4}+1) (mod 4) = 1$
but 4 is not prime..[/quote]

But you're concerned with 5, not 4. 5 is prime.

No, he was (I think) complaining about something else, namely that $(4+1)^{4}\equiv 4^{4}+1^{4}\equiv 1 \pmod 4$, while the statement is an if-and-only-if.


find the nth term for this sequence
1, 2, 10, 67, 467, 3268, 22876

I do not understand why you are posting the same problem in 5 places. It has been solved here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98457[/url]

[quote="shadysaysurspammed"][quote="Inspired By Nature"][quote="climskywell"]you mean pokemon?[/quote]
 :?: :huh:[/quote][/quote]

Wow shady.   :spam: At least add a smiley face or something.  Wow.

[quote="shadysaysurspammed"][quote="Inspired By Nature"][quote="climskywell"]you mean pokemon?[/quote]
 :?: :huh:[/quote][/quote]
what's that supposed to mean?

[quote="rem"]It means he wants to increase his post count, and can't think of any other way but spamming.[/quote]
I see :|


[quote="Walk Around The River"][quote="Quevvy"][quote="#H34N1"]On a four-by-four grid (see below), the points are  1  unit apart horizontally and vertically. Two distints points  P  and  Q are randomly selected from this grid. What is the probability that the distance between  P  and  Q  is less than  2  ? Express your answer as a common fraction.

          . . . .
          . . . .
          . . . .
          . . . .[/quote]
Can P and Q be on the same point?[/quote]

it says two distinct points.[/quote]
Oh, I missed that  :blush:
[hide="Anyways..............................................................."]
At the corners, there are 3 points that are less than 2 units away. (3 points * 4 corners=12)
On the edges (not on a corner), there are 5 points. (5 points * 8 on edge=40)
In the middle, there are 9 points that are less than 2 units away. (9 points * 4 in the middle=36).
Add up the numbers $12+40+36=88$
And the denominator is $\binom{16}{2}=120$
$\frac{88}{120}=\boxed{\frac{11}{15}}$[/hide]

[quote="usaha"]Wait I have 2 questions

How are there 9 points for the middle?(isnt it 8)
And why is the denominator 16C2?, isnt it 16P2, the way  u are counting?[/quote]
there are four points in the middle:
....
.[color=red]..[/color].
.[color=red]..[/color].
....
and nine points around each of those points that are less that 2 units away.

The denominator is 16C2 because point 1 and and point 2 is the same as point 2 and point 1.

thers 8 points and 16C2 doesnt account for the fact u repeated lines..

[hide]I get 3/5

[/hide]

ur answer is too big quevvy

[quote="Quevvy"][/quote][quote="Walk Around The River"][quote="Quevvy"][quote="#H34N1"]On a four-by-four grid (see below), the points are  1  unit apart horizontally and vertically. Two distints points  P  and  Q are randomly selected from this grid. What is the probability that the distance between  P  and  Q  is less than  2  ? Express your answer as a common fraction.

          . . . .
          . . . .
          . . . .
          . . . .[/quote]
Can P and Q be on the same point?[/quote]

it says two distinct points.[/quote][quote="Quevvy"]
Oh, I missed that  :blush:
[hide="Anyways..............................................................."]
At the corners, there are 3 points that are less than 2 units away. (3 points * 4 corners=12)
On the edges (not on a corner), there are 5 points. (5 points * 8 on edge=40)
In the middle, there are 9 points that are less than 2 units away. (9 points * 4 in the middle=36).
Add up the numbers $12+40+36=88$
And the denominator is $\binom{16}{2}=120$
$\frac{88}{120}=\boxed{\frac{11}{15}}$[/hide][/quote]

You are almost right except there are 8 points that are less than 2 units away in the middle. That makes 12+40+32=84. Divide by 2 to account for counting them twice, total 42. That's 42/120 which is 7/20.

davidli, thats what I was tryin to tell him.. but apparently he doesnt listen (both of ur points)

So ya, (12+40+32)/(16P2) because we are implyin order matters because we have repeated the same line, just from P1-P2 instead of frm P2-P1 where p is a point, so that is 7/20

You have to look at 3 sceniros- corner point, non-corner edge point, and 4 middle points

[hide] If 1 of the 4 corners, which is 1/4th of the time then the odds of the other point being within 2 are 5/15 if one of the edges which is 1/2 of the time then odds of other point are 7/15
If one of the centers which is 1/4th then the other points odds are  10/15

1/4(5/15)+1/2(7/15)+1/4(10/15)
5/60+14/60+10/60=29/60[/hide]

the distance is less than 2, zserf did u do less or equal to 2?


Someone found this answer in about 5 seconds.

 How Many of the 1st 500 positive integers are divisble by 3,4,and,5.

I don't know the answer but do I just find the LCD of 3, 4, and 5? in this case 3 times 4 time 5, which i s 60, and see how many times 60 goes into 500 and round that number down? 500/60=8.33333, so 8????


I'd like to inform mathlinkers that the international congress of mathematicians will be held this year in Madrid from 22 to 30 august ,the official website is :http://www.icm2006.org .
I guess it might even be a chance for mathlinkers to get to know each other ,at least for those who are in Europe ,so what do u think ??

MoHaMeD: Can you or anyone explain what is it about???

The site is great ! Thank you MoHaMed !
It's [b]HUGE![/b] there will be the most concentrate maths because the great mathematiciens of the moment (about ...a few... [b]thousands[/b]! )are there!
What about 7th junior Math Congress in Romania from 25 June to 2 July ?
web adress : http://www.uni-miskolc.hu/matjun7
I think is more convenient for some people of Eastern Europe...

sorry José my answer comes l8 a bit .Different activities are scheduled :
-Lectures in different topics.
-Mathematical prestigious prizes awarded (for instance Fields(mathematical Nobel) and for the first time Gauss prize).
and also some special activities:http://www.icm2006.org/scientificprogram/specialactivities/
plz check the link for further information .If u have specific questions i'll ask a spaniard friend who was the one who's told me about the congress .
Regarding the junior math congress i do agree mstoenescu it sounds interesting too ,but its website doesn't contain sufficient informations yet i wonder if there is a Romanian mathlinker who might tell us more about it .

Hi MoHaMeD !
I'm Romanian (and French too !) and I shall post the question in Romanian Forum for some details about 7e junior congress of maths !
Personnally i can't go to Romania because work but  if someone give me details I'll put then [b]HERE ![/b] :roll:


I have the following doubt about some specific definition, please help me, thanks.

1.When we say that two numbers are coprime, does it count numbers like 1 and 6(are they coprime?).
2.When we talk about divisor of a number n, does 1 and n itself always count?

2. yes, unless the problem specifies otherwise.

the answer to both is yes

for more information check this
[url]http://en.wikipedia.org/wiki/Coprime[/url]

2 nos are said to be co prime when they don't have any common factor greater than 1
1 is co prime to every no.

You can talk about proper divisors, which are the positive divisors of $ n$ excluding $ n$ itself. Unfortunately this term is among the ambiguously defined in mathematics (the same way some countries include $ 0$ in the natural numbers and others don't). Sometimes it may refer to all divisors of $ n$ except $ 1$, or all divisors of $ n$ except $ 1$ and $ n$, or include negatives.

http://mathworld.wolfram.com/ProperDivisor.html

Thanks to all of you.

Also, since mention the word "proper", if we say "a circle properly contains another circle, what is it actually mean. (can they be internally tangent or can they be the same circle)

That usage of proper probably has multiple meanings as well, though I'm not very familiar with it.


Karen drove continuously from 9:40 a.m. until 1:20 p.m. of the same day and covered a distance of 165 miles. What was her average speed in miles per hour?

in 3 hours and 40 minutes, he went 165 miles. $ \frac{165}{3\frac23}\equal{}165\cdot\frac3{11}\equal{}15\cdot3\equal{}\boxed{45}$


听说前几年江苏高考出现了一道错题,能不能介绍一下?

Sorry,but could you translate this? :huh:  :huh:  :huh:  :?:  :!:

[quote="shobber"]应该是$R^+$吧?[/quote]
不要受對數函數影響, 還有很多可能.

[quote="ifai"]說點言外話. 我校高一級的這次數學考試也出了一道錯題, 老師和學生對此都不以為然,
但我就覺得難以接受:
設函數 $f$ 的定義域是 $\mathbb R$, 且 $f$ 不恆為零. 對於任意實數 $x,\ y$, 都有
$f(xy)=f(x)+f(y)$... (還有一些條件, 叫你去證明一些結論)
已知條件跟本是矛盾的, 函數根本不存在, 題目還有甚麼意義?[/quote]

高中挺流行出这样的题，我见过好几回了。
从来没有质疑过其存在性（考试时间紧迫，还有闲工夫管那个），要是不存在就好玩了，因为不管他存不存在，我们这样研究它也毫无应用价值，抽象的很（声明：我不是韩寒论者）。现在发现他不存在也好，存在也罢，根本就是毫无感觉，毫无影响，只是让这种缺乏应用的数学教育自己砸自己的脚而已。至于题目的意义，仍然是有，那就是考你那些方法会不会用--这是和函数存不存在无关的，反正是叫你证明，不可能得出多个结论的；你能证出来，题目意义就有了，即使是个错的证明；由于题目错误使你证不出来，只能自认倒霉。
你说如果就是对数函数，倒也还能应用。现在这个R上的函数真不知道是啥了。或许可以说它在R+上能够应用。“近一半能够使用”。
题目矛不矛盾不知道，但你的签名"impossible=nothing"，不也跟你发的内容矛盾？


A finite graph is such that there are no subsets of $5$ points with all $15$ edges and every two triangles have at least one point in common. Show that there are at most two points $X$ such that removing $X$ leaves no triangles.

I'm pretty sure this has been posted before by orl.


Let a sequence $ \{F_n\}_{n\equal{}0}^{\infty}$ be defined by $ F_0\equal{}0, F_1\equal{}1, F_{n\plus{}1}\equal{}F_n\plus{}F_{n\minus{}1}$ for all $ n\geq 1$. Let $ k$ be any positive integer.

Prove that we can find a positive integer $ a$, with at least $ k$ distinct prime divisors, such that $ a$ divides $ F_a$.

[quote="Albanian Eagle"]Simply use the property that $ F_m|F_{mn}$[/quote]

But how do you guarantee the relationship between $ F_m$ and $ m$ itself? And how to find $ F_m$ with a lot of divisors?


When I search for someone's posts, I always get the number of posts found was less than the number of posts he or she has posted.

For example, Athena has posted 94 times.  But when I search for hear messages by clicking the "Find all posts by Athena" link, it says that no posts have been found.

[quote="Osiris"]The forum has been through several clean-ups - old posts were removed.[/quote]
Fun and Games also auto-prunes.

[quote="Peter VDD"]I've remarked this too.
You know, Tare (no#1 poster) shows a number of 4471 while his real post count is only 2414.

I've asked valentin about this and sent him a tool to fix it though I haven't heard from him since then.[/quote]I am looking into it. :)

Ok, problem solved. Thanks for the script Peter.

[quote="Peter VDD"]You know, Tare (no#1 poster) shows a number of 4471 while his real post count is only 2414.

I've asked valentin about this and sent him a tool to fix it . . .[/quote]

What?! You want to see MORE posts from Tare than you can already see?  ;)

I thought one more reason  post counts vary when you look up people's profiles is that some people have posts to online class forums that aren't accessible to people who haven't registered for that class.

No actually, he wanted exactly the opposite. The difference in posting (and especially that one in Tare) is that he posted tens of hundreds of posts into the games and fun factory, which was auto-pruning. But it seems that phpbb2 does not decrease a users posts in some cases (auto-pruning for example).

[quote="tokenadult"]
I thought one more reason  post counts vary when you look up people's profiles is that some people have posts to online class forums that aren't accessible to people who haven't registered for that class.[/quote]

This is definitely another reason for post counts to vary.

The difference seen in, say, Tare's posts (and others) is that we used to have auto-pruning in the Fun & Games section.

[quote="darij grinberg"][b]1.[/b] What exactly is auto-pruning?

[b]2.[/b] Do you delete maths postings? (I hope not...)

  darij[/quote] 1) pruning basically means getting rid of, deleting. So, on the old AoPS board some forums had this feature enabled, which ment that each time a week passed and no replies were made in that specific topic, the topic would be deleted (automatically). This is not valid anymore. The moderators only delete topics.

2) Of course not. :)

In fact, we don't delete anything. We just make the correct number appear. The posts are gone for a long time yet, but the number the forum remembers as "post count" is still the number from before the deletings. All we did is correcting those mistakes by adjusting the numbers.

[quote="Valentin Vornicu"]1) pruning basically means getting rid of, deleting. So, on the old AoPS board some forums had this feature enabled, which ment that each time a week passed and no replies were made in that specific topic, the topic would be deleted (automatically). This is not valid anymore. The moderators only delete topics.

2) Of course not. :)[/quote]

Aha, thanks, now I can sleep well...

What concerns deleting messages, I think one should do it only if the message is really unnecessary and contains no information. For instance: Double postings caused by clicking on "Submit" two times; Spams (unless they are funny or, what happens more often, the answers are funny); Postings like "sorry, I posted in the wrong forum, please move it" (of course, after the topic is moved to the right forum) and similar things. Postings with mathematical errors should not be deleted (people often learn from others' errors, and also a wrong solution can include a part of a correct solution).

  Darij


For any real numbers $ A,B,C$ prove that \begin{eqnarray*}&&\sin(B+C-A)+\sin(C+A-B)+\sin(A+B-C)\\
&=&\sin(A+B+C)+4\sin A\sin B\sin C\end{eqnarray*}

[quote="nayel"]For any real numbers $ A,B,C$ prove that
\begin{eqnarray*} & & \sin(B + C - A) + \sin(C + A - B) + \sin(A + B - C) \\
& = & \sin(A + B + C) + 4\sin A\sin B\sin C\end{eqnarray*}
[/quote]

First, let's denote the angles by lowercase greek letters because it looks extremely intellectual. Next, application of the addition theorems yields
\begin{align*} \sin(\alpha + \beta + \gamma) & = \sin(\alpha + \beta)\cos\gamma + \cos(\alpha + \beta)\sin\gamma \\
& = \sin\alpha\cos\beta\cos\gamma + \cos\alpha\sin\beta\cos\gamma + \cos\alpha\cos\beta\sin\gamma - \sin\alpha\sin\beta\sin\gamma \end{align*}
Let's collect terms and try to prove the statement.
$ \sin\alpha\sin\beta\sin\gamma$ occurs 3 times on the LHS (one of the angles is always negative and cancels out the - sign). It also occurs -1+4 = 3 times on the RHS.
$ \cos\alpha\cos\beta\sin\gamma$ occurs 3 times on the LHS but one time with a negative sign, so that makes 1 time in total. It also occurs once on the RHS.
The other terms are cyclic permutations of the second one and cancel out the same way.

Remark: The statement can also be written as $ \sin\alpha + \sin\beta + \sin\gamma = \sin(\alpha + \beta + \gamma) + 4\sin\left(\frac {\alpha + \beta}{2}\right)\sin\left(\frac {\beta + \gamma}{2}\right)\sin\left(\frac {\gamma + \alpha}{2}\right).$ This doesn't look as nice as the other one though :P


calculus
integration
trigonometry
complex analysis

Define
\[ h(z)=\oint_{C}\frac{w\sin^{2}{w}}{(w-z)^{2}}dw, \qquad |z-1|<3 \]
where $ C$ is the positively oriented circle $ |z-1|=3$ (1 turn).

a) Compute $ h(i)$.

b) Compute $ \oint_\gamma \frac{h(z)}{z-i}dz$, where $ \gamma$ is the positively oriented circle $ |z-i|=1$ (1 turn).

The first one is a pretty straight-forward application of Cauchy's Integral formula (since $ i\in C$):
$ f^{n}(z_{0}) = \frac{n!}{2\pi i}\oint_{C}\frac{f(z)dz}{(z-z_{0})^{n+1}}$
So differentiate once:
$ f(w) = w\sin^{2}w$
$ f'(w) = \sin^{2}w+2w \sin w \cos w$
Thus the integral reduces to:
$ h(i) = \frac{2\pi i}{1!}f'(i) = 2\pi i [\sin^{2}i+2 i \sin i \cos i]$
Of course there are hundreds of forms this answer could take, but from here you should be able to manipulate to whatever is most useful.


The On-Line Encyclopedia of Integer Sequences says  :D

Number of letters in n (in Danish) -> 9
Numbers of letters in n (in the Norwegian language Bokmal) -> 9   ;)


At a party there are only single wome and married men with their wives. The probability that a randomly selected women is single is 2/5. What fraction of people in the room are married men?

[size=75]This from AMC 8 2004[/size]

Single Women are 2 of 5. That means 3 of 5 Women are married. --> The same number of married Men

---> If they were 8 People, there were 2 of them Singles, 3 of them married Woman, und also 3 married Men. So, number of men is in my opinion 3/8.

BTW: Sorry for my bad english, but I am german ;)


Euler
number theory proposed
number theory

Find all positive intergers $n$ such that $2^{\phi(n)}-1$ divides $n^n$, where $\phi$ is the Euler $\phi$-function.

i guess none, since there is a prime dividing it not dividing $n$, I posted this problem long time ago!

well except for four small cases 1, 2, 3 and 6.


[quote=Inequalities Master]Let $ a,b,c\geq0$, no two of which are zero. Prove that:
$ \frac{a(b\plus{}c)}{a^2\plus{}2bc}\plus{}\frac{b(c\plus{}a)}{b^2\plus{}2ca}\plus{}\frac{c(a\plus{}b)}{c^2\plus{}2ab}\geq 1\plus{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}$.[/quote]

We have
\[LHS-RHS=\frac{1}{2}\sum{\frac{c^2(a-b)^2}{(2ac+b^2)(a^2+2bc)}}+\frac{(a-b)^2(b-c)^2(c-a)^2\sum{(3a^2+2bc)}}{2(a^2+b^2+c^2)\prod{(a^2+2bc)}}\ge{0}\]

[quote=Inequalities Master]Let $ a,b,c\geq0$, no two of which are zero. Prove that:
$ \frac{a(b\plus{}c)}{a^2\plus{}2bc}\plus{}\frac{b(c\plus{}a)}{b^2\plus{}2ca}\plus{}\frac{c(a\plus{}b)}{c^2\plus{}2ab}\geq 1\plus{}\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}$.[/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, it's obvious that our inequality is equivalent to $f(w^3)\geq0$, where $f$ is a concave function and [url=http://www.artofproblemsolving.com/community/c6h278791]uvw[/url] kills it.


geometry
geometric transformation
rotation
symmetry
analytic geometry

hi,
Show that there are exactly 6 symmetries of an equailateral triangle.

I have found the 6 symmetries , R120, R240, e(rotation 360), and the 3 medians lines of symmetries.  Now I have to prove why there are only 6? I am thinking of matrices.  Any help please?

Consider the vertices.  A symmetry of the triangle restricts to a symmetry of the vertices.

thanks for the hint, I thought of that and since there are  vertices then the symmetry will sent one two either other two vertices therefore using permutations there are 6 symmetries, but my question is how can we prove it using isometry?
ans since I am to this topic, I have another question: :blush:
find the matrices of an equilateral triangle with its center at the origin of  x,y,z coordinative system, when the vertex coordiantes are (1,0,0), and (-1/2,sqrt3/2,0) and (-1/2,-sqrt3/2,0),

thanks a lot


a certain number of people are standing in a circle holding numbers in consecutive order. if person 8 is standing across from person 177, how many people are standing in the circle?

how would u do these kind of problems?

[hide]Think of a clock.

1 is opposite 7
2 is opposite 8
3 is opposite 9
etc.

Since there are 12 numbers on a clock, the formula is $x_{1}-x_{2}=0.5n$ where $x_{1},x_{2}$ are diametrically opposite numbers and $n$ is the total number of numbers.

Plugging 177 and 8 in gives $n=338$.[/hide]

Person $0$ is across from person $169$. So $169$ is half the answer. The answer is $338$. Generally, use this procedure:
1) Find who is across person $0$.
2) Multiply that result by $2$.

I don't think there would really be a person $0$, that would be person $338$ actually, but that method will give you the right answer...

[hide="Answer"]Like others have said you just do $2(177-8)$ $=$ $2(169)$ $=$ $338$ people. This is since they are equally spaced.[/hide]

[hide="Problem Extension"]In a circle where $300$ people stand equidistant from one another, what is the measure of the smaller angle formed when a segment is drawn from Person $107$ to the center and another segment is drawn from the center to Person $212$?[/hide]

[hide]The arc between two of the people is $\frac{360}{300}=\frac{6}{5}^\circ$.

$212-107=105$

$105\times\frac{6}{5}=\boxed{156^\circ}$[/hide]


For 7th and 8th grader, how should one decide whether to participate in AMC8 competition vs MOEMS.

Thanks,
Khem
kchande@yahoo.com

AMC 8 is better. After AMC's, you have the AIME if you qualify, the USAMO, if you qualify, and the IMO if you qualify.  :)


I have two quickly questions:

1. Let ABC be a triangle, let P be a point in AC sucha that the incircles of ABP (X,r) and1. PBC (Y,r) are congruent. give tha area of triangle ABC using the length of the internal common tangent to (X,r) and (Y,r) and m<ABC.

2. let x be a line, all paralel lines to x meet x and meet all of them just in a point.

Thanks a lot

José Carlos

What do You mean by "m less than ABC" ? :(

"savoir attendre et douter" -D'Alembert


First and second inequality can be obtained by Cauchy-Schwarz's inequality and third one gives the condition of the existence of the triangle with side-length $R,\ r$ and $d$ such that $|R-r|<d<R+r,$ where denote $R, r$ by radii of two circles and the distance of the centers between them.

Ok ,

One  difficult now :

[img]http://img443.imageshack.us/img443/3394/8014vm0.png[/img]


geometry
3D geometry
sphere
tetrahedron
Pythagorean Theorem
geometry solved

Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of the larger sphere?

The tetrahedron formed by the centers of the spheres has a face parallel to the plane, which is an equilateral triangle of sidelength 2, and the other 3 faces isosceles triangles with base 2 and the other 2 sides equal to 3, so it's easy to calculate its height (with the aid of Pythagoras' thm). Call this height h. Then the distance we're looking for is h+2+1=h+3.

h=(23/3)1/2 (I think; I might have made some mistakes).

Hmmm...grobber are you sure? When I asked my friend about it he said he thought the answer was the square root of 25, which is 5... :?  Can anyone else give this problem a try, or grobber can you see if you made a mistake somewhere?

well the actual distance would be 1 + \sqrt 23/3 :D, but I guess it's ok.

1+\sqrt (23/3) is the distance from the center of the large sphere to the plane, so the distance we are looking for, the one from the top of the sphere to the plane, is 3+\sqrt (23/3). So, the final answer is

3+\sqrt (23/3).

ok...i guess that seems valid. Anyway, if you guys were wondering here is what my friend had to say about it:

If 3 spheres are mutually tangent means touching each other, from any one side you will be able to see only two, so draw two spheres, no. 1 and 2, third one is hiding behind them. Now draw the fourth sphere no. 4 on top of these two touching both of them. Now draw a straight line from the top of the sphere no. 4 upto the table top. This line will pass through the tangent point of spheres no. 1 and 2. Draw another line joining the centers of the sphere 1 and 4. Draw a perpendicular from the center of sphere 1 to the tangent point of 1 and 2.

Now label the various parts of the figure. Use Pythagorean theorem to find other unknown parts, and you will get the total distance from the plane to the top of the larger sphere equal to (1+ sqrt 8 +2)= 3+ sqrt 8

thats why i was wondering...feel free to comment

I might be rushing, so this may very well be wrong, but it seems to me like the first sentence is false: it's not true that from a "side" (I don't think I even know what that means :D) the 3'rd sphere is invisible; you can still see some of it..

actually, yeah now that I take a closer look it at, i don't think that can be possible, because although most of the 3rd sphere is hidden a little portion is visible, and besides 3+ \sqrt 8 is not 5 (earlier post), it is approximatley 5.828...hmm both answers seem to be ver close to each other 3+  \sqrt (23/3) is about 5.768 and 3+ \sqrt 8 is about 5.828.  :D  Oh well, by now im convinced that the answer is 3+  \sqrt (23/3)

actually, grobber could you draw a diagram or something to show what you mean- because for some reason i can not see what values you plugged in to pythagoras's thm. If you drew something, then it would be clear.

[quote="grobber"] Then the distance we're looking for is h+2+1=h+3.

h=(23/3)1/2 (I think; I might have made some mistakes).[/quote]

if h= 1+ \sqrt (23/3), and the distance we need is h+3, wouldn't the total distance be 1+ \sqrt (23/3) +3 which equals to 4+ \sqrt (23/3)?

no :) h is not 1 + \sqrt 23/3. I just hurried and added only the distance from the lower spheres' centers to the plane, and forgot to add 2, the distance from the larger sphere's center to the plane above. so it's 3+ \sqrt 23/3 the final answer ;)

:D okay, thanks for clearing that up-but do you know what values that you plugged into the pythagoras thm? just makin sure im doin the right thing

the following picture should finish your doubts :D


So I will put all the romanian Tst-s here, going backwards in time.
[color=red][UPDATED: Romanian TSTs 1,2 and 3 in 2004. Two more to come at the end of May, so stay tuned :)][/color]

Just added Rom Sel tests 2004, test 2. Test 3 added!


Find all polynomials $ P(x)$ with real coefficients such that $ P(x^2)P(x^3)\equal{}(P(x))^5$ holds for every real number $ x$.

[quote="Aumn"]if $ x\neq0$ is one of p(x)'s roots(x can be complex number)
$ x = r(cos\alpha + isin\alpha),0\le \alpha < 2\pi$
so it's easy to show that $ \sqrt [3]{r}(cos\frac {\alpha}{3} + isin\frac {\alpha}{3})$ is it's root
then for all n,$ \sqrt [3^{n}]{r}(cos\frac {\alpha}{3^{n}} + isin\frac {\alpha}{3^{n}})$ is root too
it's a contradiction,so p(x) only has zero root
$ p(x) = x^{n}$or$ p(x) = 0$[/quote]

including $ n=0$ leading to $ P(x)=1$

[quote="Aumn"]if $ x\neq0$ is one of p(x)'s roots(x can be complex number)
$ x = r(cos\alpha + isin\alpha),0\le \alpha < 2\pi$
so it's easy to show that $ \sqrt [3]{r}(cos\frac {\alpha}{3} + isin\frac {\alpha}{3})$ is it's root
then for all n,$ \sqrt [3^{n}]{r}(cos\frac {\alpha}{3^{n}} + isin\frac {\alpha}{3^{n}})$ is root too
it's a contradiction,so p(x) only has zero root
$ p(x) = x^{n}$or$ p(x) = 0$[/quote]
A little detail: You should decide $ 0 < \alpha \le 2\pi$ instead, so that $ x = 1 \Rightarrow \alpha = 0, r = 1$ doesn't occur :)

Denoting $ \deg P(x)\equal{}n$ and
\[ P\left( x \right) \equal{} {x^n} \plus{} Q\left( x \right)\]
then we have $ \deg Q\left( x \right) \equal{} k < n$
So
\[ \left( {Q\left( {{x^2}} \right) \plus{} {x^{2n}}} \right)\left( {Q\left( {{x^3}} \right) \plus{} {x^{3n}}} \right) \equal{} {\left( {Q\left( x \right) \plus{} {x^n}} \right)^5}\]
Simplify the equation and compare the degrees of two hand sides, we reach the contradiction


Hey everyone...lets count prime numbers

I'll start:

2
 :trampoline:  :trampoline:

[quote="Chompy"][hide="41"]HEY WHY'D YOU CLICK IT???[/hide][/quote]

Next time make a rickroll :D

47

[hide][list][size=200] :arrow:  :arrow:  :arrow: [color=white]191 [/color] <--- HIGHLIGHT[/size][/list][/hide]

$ 14^2\plus{}1^2$

mod it, whatever. prime.

197!

[hide="The last prime under 200"][hide="guess what it is"][hide="guess it yet?"] :spider: [color=cyan][size=150]SPIDER SURPRISE!!![/size][/color][size=200]199[/size][/hide][/hide][/hide]

[hide="made up of prime numbers"][hide="2"]2[/hide][hide="3"]3[/hide][hide="3"]3[/hide][/hide]

[hide="ERROR: 301 is not a prime..."][hide="i will post the next two after 239..."][hide="ready?"] :welcomeani:  :welcomeani: to the primes! I'm [size=42]not[/size] going to show them to you!
[color=white]241[/color]and[color=white]251[/color][/hide][/hide][/hide]


Contruct triangle $ABC$ if we are given the points $A',B',C'$ are intersections of altitude with circumcircle $(O)$.
Similar questions with median, bisector.

we begin with the triangle $XYZ$.

We make a circumscribed circle on it.

And after we make the bisectors for the triangle $XYZ$

the intersection of the bisector of the triangle $XYZ$ with the circumscribed circle is the triangle $ABC$

It's just too look at the angles.


A rectangular flower bed has a perimeter of 96m. Given that its length it twice its breadth, calculate its length and its area.
The flower bed is surrounded by a concrete path of width 40cm and height 15 cm. Calculate the volume of concrete used for the path, giving your answer in m3.

Set the length to $ l$. We thus have $ 2(l\plus{}\frac{l}{2})\equal{}96$.
Simplify

$ l\plus{}\frac{l}{2}\equal{}48$

$ \frac{3}{2} \cdot l\equal{}48$

$ l\equal{}48 \cdot \frac{2}{3}$

$ l\equal{}\boxed{32}$ meters.

Area=$ 32 \cdot \frac{32}{2}\equal{}\boxed{512}$ square meters.

Volume:

(Picture is not accurate, wanted it to be visible.)
[asy]draw ((2,2)--(34,2)--(34,18)--(2,18)--cycle);
draw ((0,0)--(36,0)--(36,20)--(0,20)--cycle);[/asy]

Basically, we take the difference of those two rectangles and subtract them, and then multiply it by the width. (note that the 0.8 extra means that 0.4 were added on each side, note that all measurements are in meters)

Larger rectangle: Length =32.8, Width=16.8
Smaller rectangle, Length =36, Width=16

$ \left(32.8\cdot 16.8\minus{}36 \cdot16 \right)0.15\equal{}\boxed{5.856 \: \text{m}^3}$.

Let $ b$ be the breadth .

Then $ l\equal{}2b$

Perimeter= $ 2(l\plus{}b)$ $ \equal{}6b$

According to the question: $ 6b\equal{}96$ or $ b\equal{}16$

So $ l\equal{}32$

Area: $ l\times b$=$ 32\times 16$=$ 512$


hello everybody!
do you know problems of CWMO2006?
Please, post them!
how about rezults?
where can I find any information about it?
thanks!

Please find it carefully  :P
It's just below your post!


quadratics
inequalities
number theory unsolved
number theory

Let $ m$ and $ n$ be positive integers such that $ x^2 \minus{} mx \plus{}n \equal{} 0$ has real roots $ \alpha$ and $ \beta$.

Prove that $ \alpha$ and $ \beta$ are integers [b]if and only if[/b] $ [m\alpha] \plus{} [m\beta]$ is the square of an integer.

(Here $ [x]$ denotes the largest integer not exceeding $ x$)

Note that $ m \equal{} \alpha \plus{} \beta$, $ n \equal{} \alpha\beta$. If $ \alpha$ and $ \beta$ are integers, then $ [m\alpha] \plus{} [m\beta] \equal{} m(\alpha \plus{} \beta) \equal{} m^2$, which is a perfect square.

Now assume that $ [m\alpha] \plus{} [m\beta] \equal{} k^2$, where $ k \ge 0$ is an integer. We have, $ [x \plus{} y]\ge [x] \plus{} [y]\ge[x \plus{} y] \minus{} 1$, hence $ m^2\ge k^2\ge m^2 \minus{} 1$. Therefore $ k \equal{} m$ or $ k \equal{} 0,m \equal{} 1$. The latter implies, $ \alpha \plus{} \beta \equal{} 1$ and $ [\alpha] \plus{} [\beta] \equal{} 0$. But then
\[ 1\le n \equal{} \alpha\beta\le\frac {m^2}{4} \equal{} \frac 14,\]
a contradiction. So we must have $ k \equal{} m$, and $ [m\alpha] \plus{} [m\beta] \equal{} m^2 \equal{} m(\alpha \plus{} \beta)$. But $ [m\alpha]\le m\alpha$ and $ [m\beta]\le m\beta$. Hence both of the inequalities must be equalities and so both $ m\alpha$ and $ m\beta$ must be integers. This forces $ \alpha$ and $ \beta$ to be rational. Hence by the rational roots theorem we conclude that $ \alpha$ and $ \beta$ are both integers.

let $ \alpha, \beta $ integers
then$ [m\alpha]=m\alpha,[m\beta]=m\beta $
so $ \  [m\alpha]+[m\beta]=m(\alpha+\beta)=m^{2} $
so this part is done.

other part.let $ \ [m\alpha]+[m\beta]$ is a square number.
see
$ \  m^{2}-2=m(\alpha+\beta)-2=<[m\alpha]+[m\beta]=<m(\alpha+\beta)=m^{2}$
so the only option is $ \ [m\alpha]+[m\beta]=m^{2} $
that is $ \ [m\alpha]=m\alpha,[m\beta]=m\beta $
the discriminant  $ \ D= m^{2}-4n$
let if m is odd then w is the largest odd integer not surpassing $ \sqrt(D) $
if m is even then w is the largest even integer not surpassing $ \sqrt(D) $
$ \sqrt(D)=w+z $
then $ \frac{z}{2}<1 $
now easy to see that if $ \ [m\alpha]=m\alpha,[m\beta]=m\beta $
$ \ [\frac{z}{2}]=0,[1-\frac{z}{2}]=0 $
which is possible iff $ \frac{z}{2}=0$
but then $ \sqrt(D)=w $
so roots are integers

[hide]$[m\alpha]+[m \beta] =m^2$ or $m^2-1$ Also m>1 for real roots. So, if the expression is a perfect square $m \alpha$ and $m \beta$ have to be integers and so $\alpha, \beta$ are rational and since leading coefficient is 1, they are integers[/hide]

[quote="hsbhatt"]$[m\alpha]+[m \beta] =m^2$ or $m^2-1$ Also m>1 for real roots. So, if the expression is a perfect square $m \alpha$ and $m \beta$ have to be integers and so $\alpha, \beta$ are rational and since leading coefficient is 1, they are integers[/quote]
I didn't understand how you concluded from here $\alpha ,\beta $ rational
[hide="Alternative Finishing"]
We already deduced that $[m\alpha ]+[n\beta ]=m^2\ \text {or} m^2-1$
Now we first see the case $m=1,$then from $\text {Vietta's Rule,}$we get $m^2-4n\ge 0\implies 4n\le 1\implies n=0\implies (\alpha ,\beta )=(1,0),(0,1)$
Now note that when it is $m^2,$then $[m\alpha ]+[n\beta ]=m^2=(\alpha +\beta )^2=(\alpha -\beta )^2+4\alpha \beta $
And $\alpha \beta =n$.Hence $[m\alpha ]+[n\beta ]=(\alpha -\beta )^2+4n\implies (\alpha -\beta )$ integer and therefore, $\alpha ,\beta $ rational,as desired.

[/hide]

That is a consequence of the Rational Roots Theorem. (nayel too refers to this in his post above)

[quote=INMO 2007 P3]Let $ m$ and $ n$ be positive integers such that $ x^2 \minus{} mx \plus{}n \equal{} 0$ has real roots $ \alpha$ and $ \beta$.

Prove that $ \alpha$ and $ \beta$ are integers [b]if and only if[/b] $ [m\alpha] \plus{} [m\beta]$ is the square of an integer. [/quote]
[b][color=#000]Solution:[/color][/b] Since, $x^2-mx+n=x^2-(\alpha+\beta)x+\alpha \beta \implies \alpha , \beta > 0$. Suppose, if $\alpha , \beta$ are integers,
$$\lfloor m\alpha \rfloor + \lfloor m\beta \rfloor = m\alpha +m \beta =m(\alpha + \beta ) =m^2$$
Now, Suppose, $\lfloor m\alpha \rfloor + \lfloor m\beta \rfloor =k^2$ for some $k \in \mathbb{Z}$
$$\lfloor m\alpha \rfloor + \lfloor m\beta \rfloor=2\alpha \beta +\lfloor \alpha^2 \rfloor + \lfloor \beta ^2 \rfloor=k^2$$
Note: $\lfloor m\alpha \rfloor + \lfloor -m\alpha \rfloor =0,1$ $\implies$ $k^2=\lfloor m\alpha \rfloor + \lfloor m\beta \rfloor$ $=$ $m^2 + \lfloor m\alpha \rfloor + \lfloor -m\alpha \rfloor$ $\implies$ $k^2=m^2$ or $k^2=m^2-1$, But since, $k,m \in \mathbb{Z}$ $\implies$ Either, $m=1$ or $\lfloor m\alpha \rfloor + \lfloor m\beta \rfloor$ $=$ $m^2$. But if, $m=1$ $\implies$ $\lfloor \alpha \rfloor + \lfloor \beta \rfloor$ $=$ $0$ $\implies$ $\alpha \beta$ $\cancel{\in}$ $\mathbb{Z}$ $\implies$ $\lfloor m\alpha \rfloor + \lfloor m\beta \rfloor$ $=$ $m^2$
$$\therefore ~ \lfloor \alpha^2 \rfloor + \lfloor \beta^2 \rfloor = \alpha^2 + \beta^2$$
Hence, $\alpha^2, \beta^2$ are integers $\implies$ $\alpha + \beta$ and $\alpha - \beta$ are rational numbers $\implies$ $\alpha, \beta$ are integers $\qquad \blacksquare$

[hide=Nice and easy...]
One side is clear, so suppose \([m\alpha] + [n\alpha] = k^2\) for some positive integer \(k\). Then we must have
\[\{m\alpha\} + \{n\alpha\} = m^2 - k^2 \in \mathbb{Z}_{\geq 0}\]
As \(m > k > 0\) we must have \(m=k\), and we are done.
[/hide]


Let ABCD be a convex quadrilateral with area is "S".Let construck  new quadrilateral which vertices midpoints of [AC],[AD],[BD],[BC] line segments.Show that  area of new quadrilateral area is not bigger than  $\frac{S}{2}$

I am waiting for a proof, please.Any response?


Could some one explain to me the following wierd observations reg the amorphous form and crystalline forms of Si:   :huh:
1. Preparation: For Amorphous: Heat Potassium Silicoflouride with K
Crystalline: The same with Zn or Al..
2. Why does Amorphous form of silicon burn Brilliantly when heated in O2, while Crystalline one not even on strong heating?

Consider the reactivity of both of them: obviously K>Zn or Al. So that means there is rapid reduction taking place that the atoms do not get enough time to settle into their lattice points. Whereas in case of the latter the Si atoms formed get time sufficient to form a crystalline solid. :)

[quote="Euclidean Geometer"]2. Why does Amorphous form of silicon burn Brilliantly when heated in O2, while Crystalline one not even on strong heating?[/quote]

In which form of Silicon are the bonds stronger? Or, in general, where are stronger the bonds: in an amorphous solid or on a crystaline one?

i have heard that in computer chips they now super oxidize the silicon inside

maybe, but how does answer the question?


Positive integer x, y, and n are each less than 10. If x^2-y^3=n, find the median of all possible values for the sum x + y + n.

Basically, it's first asking you to find all perfect squares and cubes with base $(x,y)$ less than $10$ where the square is less than $10$ larger than the cube, then sum the corresponding triples and find the median value.

I hope that made sense. Just ask if you need any clarification.

I think i understand it. What did you get for the answer? I need to verify if i got it correct.

[quote="Jenniohmy"]The answer says its 13, but I got 12 as the answer.[/quote]

13? The list of values I got are 4, 6, 12, 14, 18. If the median was 13, they would have to have another integer number in the set $\ge13$.  :?


Exactly,my solution is the same as mathfanatic's one.

I have another problem.

Find all integers $l$ , $m$ and $n$ which satisfy

\[lmn-lm-mn-nl+l+m+n=15\]

such that  $1\leqq{l}\leqq{m}\leqq{n}$.

kunny

Mod note: This is split from a thread in the MATHCOUNTS section since it is way too advanced for MATHCOUNTS.

Spoiler:

[hide]l, m, and n look like symmetric sums, which suggests polynomials.  So let l, m, and n be the roots of a cubic equation f(x) = x^3 + ax^2 + bx + c.  (We can let the leading coefficient be 1 because if it were some number k, we'd just divide everything by k anyway.)  Then by Viete's formula,

a = - l - m - n

b = lm + mn + ln

c = -lmn

Thus the given equation becomes

- a - b - c = 15

a + b + c = -15

Thus f(1) = 1 + a + b + c  = 1 - 15 = -14.

But since l, m, and n are roots of f(x), we can rewrite f(x) as (x-l)(x-m)(x-n).  Thus f(1) = (1-l)(1-m)(1-n).

So (1-l)(1-m)(1-n) = -14

(l-1)(m-1)(n-1) = 14

The only (unordered) triples of integers that multiply to 14 are (1,1,14) and (1,2,7).  Thus (l-1,m-1,n-1) = (1,1,14) or (1,2,7), or (l,m,n) = (2,2,15) or (2,3,8).

...arguably tidier than T&E with modular arithmetic.[/hide]

Exactly, my solution is the same as WarpedKlown1335'one.

Thank you.

kunny


inequalities
real analysis
real analysis solved

sorry for my incompetence ,but wots stolz-cezaro :?  :(

Lemma (Stolz-Cesaro):

Let x_n and y_n be two series with the following properties:

1) y_n>0 for any n;
2) y_n <y_(n+1) for any n;
3)y_n is not bounded (does not have an upper bound)
4) there exists the limit a=lim (x_(n+1)-x_n)/(y_(n+1)-y_n)

then there exists the limit lim (x_n/y_n) and more x_n/y_n=a

cheers!

Has somebody actually tried this?

SOmeone sugested me to use something with Riemann sums.. but I don't know what's that!

Is there a solution at the level of the 11th grade?
I tried Stolz Cesaro but got nowhere :(

cheers! :D

[quote="Vialli"]Prove that lim n-->oo ln(1+2e+4e^4+6e^9+..+2ne^(n^2))/n^2=1.[/quote]

you must minorate and majorate all terms of sum relative to last term:

2n e^(n^2) <= sum (k=1, k<=n, 2k e^(k^2)) <= n x 2n e^(n^2)

=>

ln(2n) / n^2 + 1 <= ln (sum ( ... )) / n^2 <= ln (2n^2) / n^2 + 1

the left and right member of our double inequality have limit 1.

tibi
cluj


Hi.
   I have noticed that there are two sub-forums of the new album function. One is IMO 2005, the other one is Geometry pictures. I wonder what the geo pics used for?

[quote="yif man12"]Olympiad Geo probably. Mr. Rusczyk mentioned it once, though I can't see any album function right now.[/quote]
It is on top of the ML skin, beside the Blogs button.

[quote="yif man12"]Olympiad Geo probably. Mr. Rusczyk mentioned it once, though I can't see any album function right now.[/quote]The link is http://www.mathlinks.ro/Forum/album.php

It will be put up on the AoPS skin as well, but we got so much stuff to do today at the office, that we forgot this detail :D

Please continue the discussion in the following thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=43978 as this thread is closed :)