Two distinct numbers are selected simultaneously and at
random from the set $ \{1, 2, 3, 4, 5\}$. What is the probability that
the smaller one divides the larger one? Express your answer as a
common fraction.

I think the only way to do this is if you actually do it out.

1/2, no
1/3, no
1/4, no
1/5, no
2/1, yes
2/3, no
2/4, no
2/5, no
3/1, yes
3/2, no
3/4, no
3/5, no
4/1, yes
4/2, yes
4/3, no
4/5, no
5/1, yes
5/2, no
5/3, no
5/4, no

We have 5 yes-es out of 20 possible, so the answer is $ \boxed{\frac{1}{4}}$.

A little trick to go faster:

Rather than list them all out like ernie did, which is very time-consuming, you can start with 5 and work your way down.

Since 5 is prime, there is only one integer less than 5 that divides - 1.

For the number 4, there are two integers - 2 and 1.

For 3, we have only one integer - 1.

For 2, we have one integer - 1.

For 1, we have none.

Counting these up, we have 5 ways out of a possible $ 5 \times 4 \equal{} 20$ ways, so $ \boxed{\frac{1}{4}}$.

I did it by thinking the smaller divides INTO the larger...

So 1 divides into everything else, thats 4.
2 divides into only 4.
(Note that choosing 2 then 4 has the same result as when you choose 4 and then 2)
3 and 5 are prime, and 4 is too big, so they don't divide into anything there...

And there are 5x4=20 possibilities, so $ \frac{5}{20}\equal{} \boxed{1/4}$.

There are $ 5$ pairs which satisfy the condition ($ (1,2)$, $ (1,3)$, $ (1,4)$, $ (1,5)$, $ (2,4)$), and the number of ways we can pick two integers is $ \binom{5}{2}\equal{}10$, so $ \frac{5}{10}\equal{}\boxed{\frac12}$.

Yeah, two years ago, I made the mistake using permutations.  Stupid me. :P

The only way is if the denominator is 1 or 2 which there are 5 ways out of 5choose2=10 so 1/2


I wrote this down. I'll try to verify it doing the day.

[quote="shobber"]If $\sin{\beta}=2\sin{(2\alpha+\beta)}$, then prove:
\[\tan{(\alpha+\beta)}+3\tan{\alpha}=0\][/quote]

First I Prove $\tan A + \tan B = \frac{{\sin (A + B)}}{{\cos A\cos B}}$

by writing $TanA$  as $\frac{{\sin A}}{{\cos A}}$ and $TanB$  as $\frac{{\sin B}}{{\cos B}}$

Given $\frac{{\sin (2\alpha  + \beta )}}{{\sin \beta }} = \frac{1}{2}$

writing it as

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

$ \Rightarrow \frac{{\tan (\alpha  + \beta ) + \tan \alpha }}{{\tan (\alpha  + \beta ) - \tan \alpha }} = \frac{1}{2}$

$ \Rightarrow \frac{{\tan (\alpha  + \beta ) + \tan \alpha }}{{\tan (\alpha  + \beta ) - \tan \alpha }} = \frac{{3 - 1}}{{3 + 1}}$
applying componendo-dividendo, we get

$\tan{(\alpha+\beta)}+3\tan{\alpha}=0$


calculus integration limit function geometry rectangle derivative

Please excuse me if this the wrong board to post this in.

I am currently taking AP Calc AB in High School, and the book introduces terrible equations for the limit definition of the integral. They are:
$ \lim_{\parallel{}P\parallel{}\rightarrow0}\sum_{k \equal{} 1}^{n}f(c_{k})\Delta x_{k}$
and
$ \lim_{n\rightarrow\infty}\sum_{k \equal{} 1}^{n}f(c_{k})\Delta x$
also, I don't know the code the little triangle delta, but that's what those $ \text{delta}$s are. (EDIT: fixed)

As such, I have decided to develop my own, however it has issues and doesn't evaluate the integrals properly. It looks like it should work, but it doesn't, incorrectly evaluating $ \int_{0}^{2}[3x^{2}]dx$ as 16. It is, in its two variations:

$ \lim_{i\rightarrow\infty}\sum_{n \equal{} a}^{b\cdot i}\frac {f\left(a \plus{} \frac {n \minus{} a}{i}\right)(b \minus{} a)}{i}$
or
$ \lim_{i\rightarrow\infty}\left(\frac {b \minus{} a}{i}\sum_{n \equal{} a}^{b\cdot i}f\left(a \plus{} \frac {n \minus{} a}{i}\right)\right)$

The idea is for these to equal $ \int_{a}^{b}[f(x)]dx$, $ a$ and $ b$ have the same definitions, $ n$ is the sum variable, and $ i$ is how many times the interval $ [a,b]$ is divided.

How it should work is it takes the initial function value ($ f(a)$) and multiplies it against the width ($ \lim_{i\rightarrow\infty}\frac {b \minus{} a}{i}$), and then takes $ \lim_{i\rightarrow\infty}f\left(a \plus{} \frac {1}{i}\right)$ multiplies that against the width, it takes all these values and adds them together. The second functions in the same way, except that is multiplies the constant width afterwards.

I thought I'd put it up so that you could completely annihilate it, and tell me what's wrong. Thanks in advance.

[quote="JRav"]The definition of the integral given (the real one) is rigorous and was proven so that is how you should define it.  Basically, this definition selects some test point in an infinitesimal square (or rectangle) in the region, finds its area, and then distributes it throughout the entire region.  I'm not sure you can define the integral to be in terms of the way you defined it.  If you (or no one) can prove it, then it is wrong.  Anyway, the definition itself is not particularly important in AP.  And the code is $ \Delta$ is \Delta.[/quote]

Thanks, JRav. It's not that I think that the definitions given by the book are [i]wrong[/i] per say, just that they're aren't functional (least ways, I don't get them). For instance, they don't limit the function to just $ [a,b]$ in their definitions, and good luck getting a calculator to process them without some extreme coding.

The derivative limit definition works splendidly in the calculator, these do not. I guess what I want is something evaluable.

The way I defined the integral is obviously incorrect as it produces incorrect values. I'd like someone to help me figure how to make it work. Or just explain the book's definitions, I'm not picky.

It looks like you don't understand the notation- your sums are nonsense. The standard definition of $ \sum$ notation includes that the indices must be integers, and yours aren't.

Now, if you want to  fix your ideas- it looks like you're trying to restrict to the case of an evenly spaced partition. A way to express this correctly:
$ \lim_{N\to\infty}\sum_{j \equal{} 1}^N\frac {b \minus{} a}{N}f\left(a \plus{} j\frac {b \minus{} a}{N}\right)$

That works- we don't actually need unevenly spaced partitions. In practice, it's better to make the definition broad- those other partitions have their uses. Two examples:

Use a geometrically spaced partition to calculate $ \int_a^b x^r\,dx$ for some real $ r\neq \minus{} 1$ and fixed $ a,b > 0$. What happens if $ r \equal{} \minus{} 1?$
Geometric sums are a lot easier to handle than the stuff that comes up with uniform partitions.

Show that $ \int_0^1 f(x)\,dx$ exists and is zero, where $ f(x) \equal{} \begin{cases}\frac1q & x \equal{} \frac pq\text{ in lowest terms} \\
0 & \text{otherwise}\end{cases}$.

There's also the upper and lower bounds approach you could consider.

I surmise you're looking for something you can develop for your TI-whatever number. I'm sure there are programs like that.

If you are interested in learning integration your best bet is to get an analysis book and start from scratch.

Oh, and you also want to check that your definition shows that continuous functions are Riemann integrable.

Honestly though AP doesn't require you to spend your time on the definition of Riemann integral.

Maybe it's because I never focused on the definition in AP, but once I learned the rigorous definition of double and triple integrals, I found it a lot easier to understand the single variable case.

Thanks very much, guys. I'll stop wasting my time trying to develop this. I would have gotten back sooner, but my e-mail wasn't (and isn't) receiving e-mails, so I had no idea that these replies were posted. Sorry.

On a tangent, during my time away, I realized that the $ b\minus{}a$ in my formula is incorrect.

Thanks, again.


AMC AIME calculus college articles geometry function

Do you ever get frustrated because your math teacher won't or can't answer your questions in class?  Do your classmates groan when you raise your hand because they want to get through class as quickly as possible whereas you actually want to understand what's being taught?

if that is the case, you gotta find another class, buddy...

What if it's the only Calc BC section in your school?

I ask on AoPS if I have any questions....which does happen often

My math teacher likes to rush through sections and assign homework as fast as possible. My classmates are helpful, but I'm the type that doesn't pay much attention in class, and I like to learn it at home. I feel I know it better if I can teach it to myself through a book.

I used to ask my math teacher questions about different types of math. I was greatly disappointed when he couldn't teach me basic mod functions. With the AP Test tomorrow, class is just a formality for me....I'm only going because I want to graduate high school....

If you do have any questions, I suggest that you bring them to AoPS...I have yet to encounter a question that someone here can't answer (barring milennium problems and what not.)

_peace

At the beginning of the semester I would ask my AlgII teacher how different formulas were derived, or things like that, but he said that it didn't matter as long as I knew how to use the formula. So I basically stopped asking questions like that in my class. I do, however, enjoy pointing out when he has made a mistake on a problem that he's doing in front of the class. :lol:

[quote="ankur87"]If you do have any questions, I suggest that you bring them to AoPS...I have yet to encounter a question that someone here can't answer (barring milennium problems and what not.)

_peace[/quote]

Give us a little more time ;)

you can PM me any questions.... not about BC, because i am not in yet. But anything else is okay

Welcome to American classroom math :P.  You should ask people here or try to figure things out yourself (it is sometimes within reach!).  Or ask your math teacher very nicely and cross your fingers and toes... :?

You ask questions in math class??? :P

I never really bothered to after like middle school, because I figured out that anything I don't understand (not that it happens often ;)), someone else will not either and he or she will ask and I'll get my explanation. Then again, I usually don't come across this very much, so maybe I shouldn't be talking.

Part of the problem is that so few (i.e. almost none) of the better math students want to go on to become teachers when they grow up. :(

The other problem is that too many American students know that they can go on and have comfortable middle-class lives in this country (if not become President of the US) by simply going through the motions of education without really becoming intellectually engaged in the process.

[quote="joml88"]A great skill to acquire is being able to look stuff up and learn it on your own.  Becoming an independent learner goes a long way in furthering your education.  On the other hand, do NOT be afraid to ask questions when you don't get something.  You won't (or at least shouldn't) lose any pride in the process.[/quote]

THAT helps a lot when you want to get better at AIME or USAMO. Alot of independent studying and studying other solutions... and other stuff helps.

In my BC calculus class, we rush in order to get through the material. We occasionally do a proof, so it's not a "memorize and regurgitate" class. I knew a good portion of what we did cover, so I didn't have any questions really. Most of my comments were just pointing out faster or significantly different ways of solving problems, for the benefit of the class, really.
I save theory and proof questions for after class because I know no one else is interested in that kind of stuff and because I feel that if we went through all of my questions, the curriculum would never be finished. So far, I've gotten relatively thorough answers with this approach and haven't annoyed any peers. I also learned that the point of the class is to get through the AP exam. Although I do not like that, I've accepted that it's what a majority of the students in the class care about and it's their loss if they're not interested in the theory.
I find independent study helpful as well, and AOPS has been great in terms of providing challenging questions as well as complete answers.

On the other hand, if you don't understand how to do the problem after following an example and making an honest attempt, you should ask for help. It does you no good to leave class confused. If your teacher can't/won't explain these types of questions, s/he's not much of a teacher, so get someone else. (Ignore your class when they groan.)

Agreed. Asking questions after class can do wonders. I am very fortunate this year and have an excellent math teacher who will make every effort to answer my questions. If he doesn't know, he'll look it up.

I've been less fortunate in my other classes, however, where asking questions in class can elicit groans and rolled eyes (yes, from the teacher) and sometimes even openly hostile reactions. Haha...ok, so it doesn't happen that much, but it does happen! So like joml88 says, it's a good thing it doesn't hurt my pride to ask questions.

So if you're stuck in one of the less fortunate circumstances, here are some of the things I've done:

-Give the teacher a chance and instead of asking questions while they're teaching, ask them afterwards. (i.e. First accept blindly and try to follow along as you jot down questions; later, once you've picked through the lesson and know what you can understand/accept, ask questions about what is left to be desired.)

-Select question-asking representatives. (No joke.) lol. My friends have been great in this respect. If I can interest them in a question enough that they want to know the answer, they'll be willing to ask the question for me. :) This diversifies (to some extent) the question-askers in class! (My friends also enjoy seeing me squirm when the teacher glances over at the true source of the question.)

-Self-study. As other posters have indicated, self-study is an EXCELLENT way to learn material. I cannot emphasize this more. I have taken up self-study in a certain unnamed and absolutely frustrating HL nightmare (I mean...class), and since then, it's become much more enjoyable. I've learned a lot, and because I also have to teach the material to my peers, I can now say that I really understand the material. :)

-And if all else fails, go look for help. Teachers, for the most part, WANT you to succeed. Hopefully that's why they went into the profession. If your classroom teacher doesn't want to help for whatever reason, go ask another one. Besides, if you ask a teacher that doesn't teach you, you're in no danger of playing teacher's pet, and it's a complement of that teacher's teaching ability.

Oh, and of course, I'm sure AoPS would be happy to answer any/most/all of your math questions!

Well, I haven't really been in a normal classroom setting since first semester of last year, but back in the day, I had the exact same problem.  Well, sometimes they would groan, and sometimes they would know that I was about to pwn my math teacher, because she was entirely incompetent.

yea... same here. in the MIddle school, i used to pwn my teacher. To keep her superiority, she would just try to shut my mouth... which didnt work. Then i would get in trouble... which was.... really ridiculous. School teachers just try to impress the students, although half of them are already smarter then them. ;)

[quote="xxreddevilzxx"]School teachers just try to impress the students, although half of them are already smarter then them. ;)[/quote]

Perhaps... (Sorry if this is getting slightly off-topic, but I do feel quite compelled to respond to xxreddevilzxx's comment.)

I'm a pretty hard person to impress (ok...so a lot of you on this site regularly impress me, but still!), but I've also had the good fortune of having excellent teachers.

I've heard many fantastic things about my math teacher: he's so smart, he's so good at math, he's so this...he's so that. Since finally attending his class, I find that it's not his intellect that impresses me (he is quite good), but the way he carries himself, treats others, and approaches life. I have seen so many students literally flunk his class (i.e. D or below), but NONE of them have ever hated him OR his class. How many teachers can pull that off? Not very many. What's more, he has a way of always being a student's biggest cheerleader. He fished me out of some obscure classroom when I was in 8th grade and has been watching over me since. His ability to inspire and give courage to those who lack it is absolutely phenomenal.

I guess the thing is that in life, you'll probably come across many brilliant people. And initially, it may be their brilliance that impresses you. But eventually, for most people you meet, the qualities that will sustain the "impress" are personal qualities.

Ok...so, how off-topic was THAT?

Thanks for all the responses.  Actually this is more my friend's situation than my own. I was wondering how common this problem is.  She started the year asking her calc teacher for proofs and derivations but soon gave up.  She says no one in her class asks questions.  Maybe the teacher doesn't know the proofs.  More likely he has so much material to cover, he can't stop to answer questions.  I will pass along your advice to self-study and make use of the vast knowledge available on AoPS.

It just seems too bad that our schools, which should be encouraging bright, curious students, are instead turning out passive robots.  If students can't ask questions in the classroom, wouldn't we be better off videotaping the best math teachers in the country and showing their lectures to students?  It would be more effective and much cheaper than the current system.

I think a lot of teachers don't like to be asked questions because they're afraid they won't know the answers.  One of the other forum threads mentioned Professor Hung-Hsi Wu of UC Berkeley, who has written some excellent articles on math education.  He says that many elementary school math teachers are terrified of fractions!!! That's not surprising. I remember a grade school teacher who didn't understand the commutative law of multiplication: "To find the area of a triangle, you MUST multiply 1/2 * base * height.  You CANNOT multiply base * height * 1/2."

Mel, your strategy of enlisting friends to ask questions for you is hilarious. :lol:

our Alg2 teacher can't answer a lot of my classmates' questions

Just for the record, I'd like to point out that it is very difficult to answer questions that you haven't prepared for.  Often, the really interesting questions are about proofs or derivations that a teacher may not have seen in a long, long time.  A teacher is a store of knowledge, but (eep! I'm going to slip into the first person) we don't always have all of our knowledge readily available because we may not have thought about the question that's being asked when preparing the lesson.  So, not all of your in-class questions will be answered on the spot.  But, that doesn't mean they shouldn't be asked -- coming up to a teacher after class is an excellent way to have questions answered that go beyond the scope of the lesson.  Beyond that, there're always other kinds of resources -- textbooks, other individuals than your teacher, and of course the internet.

Teachers should feel comfortable saying, "I don't know but I'll look that up for you later", especially if they're teaching a difficult subject.  We shouldn't expect teachers to be all-knowing but we should expect them to encourage and respect their students.  Mel mentioned her experience with teachers who roll their eyes when students ask questions.  Such teachers don't belong in the classroom.

i really disliked my BC teacher last term, his response to my questions were usually "i dont' know, what do you think" and if other peopel asked something, he occassionally said Maria, why is that true... hated him!!!
this term, my teacher was much nicer and he actually cares about the theroy stuff. if he felt the class got too bored he would stop with the proof and just talk to me about it after class. i did alot of learning by example and talking to people though, that was alright
also, at least in my school, a few very bright math people have come back to teach (2 just this year!) so don't give up on smart math people :)

I had a teacher similar to you'res last term...everytime I asked him a question that was not in his curriculum,he ran to another teacher to get an answer...The problem is that I don't see many teachers giving an honest effort anymore...But that's only half the equation...It seems that many students have lost interest in mathematics...I truly believe that a year or two after I graduate, there will no real interest in mathematics from students in my school...apart from my year, you can only find a handful of people who show interest...
                                                                   Stan

[quote="pendulum"]...She started the year asking her calc teacher for proofs and derivations but soon gave up...[/quote]

You have to understand that a high school calculus class is not the right place place for proving theorms. Its about building students intuition for the concepts of calculus and giving them some dexterity with their applications. A few of the proofs of even elementary things in calculus (like the Intermediate Value Theorem for instance) are fairly deep, and would take way too long to develop in a normal AP Calculus class. So a teacher has to pick and choose which theorems to prove and which theorems to "explain."

I tell my students right off the bat that my aim is not to teach them how to [i]prove[/i] every theorem of calculus. My aim is to build their intuition for the theorems and concepts, and to give them plausible arguments for why they are true. So I might explain the Intermediate Value Theorem and its consequencesm and then tell them that its proof hinges on the not-quite-as-obvious-as-you-might-think fact that the Real Numbers have no "gaps" in them -- but not go into the nitty gritty details of its proof. [This year I actually did go into a discussion about how to prove that the rational numbers have "gaps" in them, to which I got the usual question, "Is this going to be on the AB Exam?" sigh].

Students will learn about the proofs of the theorems of calculus when they take an advanced calculus class or a real analysis class in college. For students being exposed to calculus for the first time though, its not really the best place to expose them to too many proofs.

I'm sure most high school calculus teachers have at some point seen the proofs of elementary calculus theorems, they just may not be able to reproduce them on the spot. For example, if someone asked me for a proof of the Chain Rule, I probably couldn't do it right there on the spot. And even if I could, it would depend on if it were instructive for me to do so for the whole class.

That being said, on the [i]rare[/i] occasion when I do get a student who asks me such a question -- I tell them where they can find an answer to it or to come back and see me later. Sometimes that's just the best I can do at the moment.

It sounds like you don't already know what the teacher is teaching. If you are on AoPS, I would think that this would mean that the class is relatively advanced, so I don't see why the other kids would have the mentality that you mentioned. If you actually do already know what the teacher is teaching, then why are you in that class (unless it is the highest class that your school offers)?

[quote="gauss202"]You have to understand that a high school calculus class is not the right place place for proving theorms. Its about building students intuition for the concepts of calculus and giving them some dexterity with their applications. A few of the proofs of even elementary things in calculus (like the Intermediate Value Theorem for instance) are fairly deep, and would take way too long to develop in a normal AP Calculus class. So a teacher has to pick and choose which theorems to prove and which theorems to "explain."

For example, if someone asked me for a proof of the Chain Rule, I probably couldn't do it right there on the spot. And even if I could, it would depend on if it were instructive for me to do so for the whole class.[/quote]

Agreed. High school calculus for the most part isn't about rigorously proving all the theorems. (Try [i]Calculus: An Intuitive and Physical Approach[/i] by Kline.)

However, even though we don't do rigorous proofs in my math class, the teacher still shows the development of the theorems and rules to make them more intuitive and to show students how/why they should accept them. It's true that most students don't care about why something is true, but for me, if I am to understand something, I have to be able to accept it. And acceptance comes from being able to understand its development/proof. It's not always possible, but one more proof understood is more mathematical insight gained.

There's a lot to be said for how subtle calculus really is, and how crucially it depends on countless theorems which we would think are 'obvious' but are actually far harder to prove.

In a thread a few days ago, I asked someone to prove that for every continuous function $f$ on an interval $[a,b]$, that there exists $c\in [a,b]$ such that $f(c)=\frac {1}{b-a}\int^b_a f(t)dt$

This is a relatively simple problem.  Continuity implies integrability.  Continuous functions attain their maximum $M$ and minimum $m$ on compact sets (such as $[a,b]$).  We have $m(b-a)\leq \int^b_a f(t)dt\leq M(b-a)$ by a simple theorem, which gives us the result by the intermediate value theorem.

But this is not so easy - this hinges on four theorems, none of which are really suitable for a calculus class.  And this is really an easy proof.  If you ask me to prove, say, L'Hopital, I'd be using the mean value theorem until every student in the class had nodded off to sleep.

gauss202, thanks for the teacher's perspective.  I think this is not so much a calculus class problem as a general passive learning problem across all subject areas.  Because of large classes and a desire to get through the curriculum, there is little time for questions.  Maybe questions and discussions should be an integral part of the curriculum.  Curiosity and active questioning should be promoted.  At a minimum, teachers should encourage self-learning by pulling aside bright students and saying, "I'm glad you want to learn more about this topic.  Here's a list of excellent books written by experts in the field.  You will probably find the answers to your questions in these books."  I don't think that happens often enough.

well, my teacher always ignores my geometry question. it seems that i shall never get the answer from her!!

[quote="blahblahblah"]Anyways, I think that the important thing to get out of a first calculus class is not knowledge of proofs (the existence of, say, the Riemann integral is not exactly a useful thing in the immediate future for most students) but rather a good geometrical interpretation of why things work

They should be able to understand roughly why the chain rule takes the form it does, they should be able to understand why the derivative of the integral gives back the original function, they should be able to understand why the arclength formula takes that form, etc.  If you understand why this stuff works, then the proofs just become a matter of speaking in the necessary epsilon language later on.[/quote]

Ahhh...well said! That's basically what I was trying to say. Now...how is it that you managed to put it to better words?  :P

Hopefully its because I've had a year more to think about it than you. :)

At my school, the teacher isn't as smart as I am and it's difficult to get questions answered.


hola amiguitos saludo su conocimiento adquiridos y compartidos a sus projimos

que solucion le darian a estos 2 problemita para calcular dicho ángulo

Al primero ya lo propusieron acá en la comunidad española, aunque sin respuesta, pero sigamos la discusión allá.

x, 2x ,7x lo resolvi usando ley de senos y el problema se complica al resolver la ecuacion trigonometrica

chekalo eso me lo dio un profesor y todavia no me sale

El problema de x,2x,7x es un problema de construcción(no me gusta ese tipo de problemas), asi que si lo intentan por trigo no les va a salir o se van a demorar mucho. Aqui les presento esta solución, un poco forzada pero válida.

Sea H un punto sobre AC tal que DHC=2x, entonces ADH=x y AH=HD=DC. Sea Q el simétrico de H respecto a la recta AD, entonces AQ=QD=BD. Ademas QAD=QDA=x=DAC, entonces QD es paralelo a AC, BDQ=BCA=2x. Hemos construido una nueva figura, el caudrilátero no convexo (en Q) ABDQ, con BD=DQ=AQ y QAD=QDA=x, QDB=2x y ABD=7x. Sea M el punto medio del triángulo isósceles QDB y sea P el pie de la perpendicular a AB trazada desde Q y sea y=BAM. Pueden ocurrir tres casos:

2PQ<BQ entonces PBQ=7x-(90-x)<30, entonces x<15. Ademas PQ<MQ y AQ=QD entonces y<x. Sumando los angulos del triángulo ABD, tenemos
(7x)+(3x)+(y+x)=11x+y<12x<180, contradicción.

2PQ>BQ entonces PBQ=7x-(90-x)>30, entonces x>15. Ademas PQ>MQ y AQ=QD entonces y>x. Sumando los angulos del triángulo ABD, tenemos
(7x)+(3x)+(y+x)=11x+y>12x>180, contradicción.

Por lo tanto debe ocurrir que 2PQ=BQ, es decir 7x-(90-x)=30, esto es x=15.

[quote="DarthTenebrus"]Una solucion trigonometrica para la primera de x , 2x , 7x :

Aplicando ley de senos : sen(10x)/sen(7x)=sen(x)/sen(2x)
                                 2sen(10x)cos(x)=sen(7x)
                                 sen(11x)+sen(9x)=sen(7x)
                                 sen(11x)-sen(7x)=-sen(9x)
                                 2sen(2x)cos(9x)=-sen(9x)
                                 2sen(2x)=-tg(9x)

pero como la suma de x , 2x , 7x es menor que 180 entonces :
 x < 18 y analizando las graficas de :

2sen(2x) = -tg(9x) se observa que l primer termino es positivo
asi que 9x>90 entonces x > 10 ; de esta manera analizando las graficas de : 2sen(2x) y -tg(9x) en el intervalo <10,18> nos damos cuenta que la primera es continua estrictamente creciente y la segunda es continua estrictamente decreciente
asi que a lo mas existira una solucion . pero vemos que al tanteo x=15 cumple luego x=15 es la unica solucion .[/quote]

  :rotfl:  jejeje Edson si ubieras aceptado mi solucion no la ubieras recibido dos veces jejeje


linear algebra
matrix
vector
linear algebra solved

It is known that all columns of $A$ are pairwise orthogonal. Prove that $|\det A|$ equals to product of lengths of vector-columns.

[i](6 points)[/i]

$A^TA$ is a diagonal matrix, the $i$-th diagonal element is $\|\alpha_i\|^2$, where $\alpha_1,\dots,\alpha_n$ are the column vectors of $A$.
Thus
$(\det A)^2 = \det (A^T A) = \prod \|\alpha_i\|^2$

:first:

I think it was the easiest problem on the contest. Try other problems! I believe in you!


The US Open, a tennis Grand Slam event, begins August 31.

Please vote and discuss!

Sorry the poll is only for men's singles.

Until someone takes out Federer at the Open, I'm sticking with him.  Some people will point to Nadal's victory in Australia this year but Nadal has a history of falling in Flushing.

I would love to see Roddick win, but I think Federer will take his sixth in a row.

I'm hoping for Roddick, Murray or Nadal.  Basically, anyone but Federer  :D
I feel like it's someone else's turn.  With Federer winning the slam on "Nadal's Turf" maybe it's time for Nadal to reciprocate?  And Roddick and Murray both had really good seasons, and really want to win.

We'll have to see what happens!

Federer is the biggest favorite, and in my opinion, he will win. Although I hate him.

nobody votes for Nadal? I noticed quite a lot of people do not believe the last year favourite can win the title, but I do. Still I support Federer.

i hope Andy Roddick wins Go America
Also does anyone know a american channel coverring the us open.

[quote="yaofan"][b] GO NADAL!!!!!!!!!!![/b] Federer is overrated.[/quote]

Wait what? How can you be overrated if you have the most grandslams, most consecutive weeks at #1, most consecutive wimbledons, most consecutive US Opens, and have won a grandslam at every turf, something that only 5 others in the history of tennis were able to accomplish?

[quote="yaofan"][b] GO NADAL!!!!!!!!!!![/b] Federer is overrated.[/quote]

How is Federer overated???
15 grand slams, most wimbledons, most #1, etc. etc. ETC.

need i say more??

OK. Federer is not overrated. And I think he's going to win. Unfortunately.

[quote="JRav"]Federer more or less lost the match.  I give del Potro credit for getting his act together after the first set, but Federer really went downhill.  In the first set, del Potro's forehand was flat and erratic, and Federer was putting every shot (save for his serves) exactly where he wanted.  In the later sets, especially the fourth and fifth, you could sense that Federer was tired and agitated.  I noted several points in which Federer just gave up.  Granted, del Potro's forehands became more like Federer's in those sets, but Federer just lost it-just look at how many unforced errors he had.[/quote]

I don't know if it is more Del Potro Pwn or Federer fail :?: . Federer did make lot of errors but look at what Del Potro did against Nadal.

Although Nadal was not doing well either.

Again, I give del Potro credit for forcing the unforced errors  :P , but on most other days, Federer is getting to those shots and hitting winners.  Also, while del Potro's first serve percentage was terrible, Federer's was even worse, which is pretty rare.

Yea, I know.

I don't know what happened to him today, underestimated?

Great tournament! Great thread too!

Congrats Juan Martin!

Yea,

CBS got like 7 mil from that in PROFIT hosting that final (30 sec commercials are like 1 mil dollars).

Wow.

Urgghh I hate to see Federer lose but yeah, Juan Martin played a great game.

[quote="polkadotman1"]In Argentina they having a party right now (not really).[/quote]

Ah my dad said Argentina sent lots of people to the USOPen

I could tell...

lol yea (not really).

audience screaming: roger, roger, roger, roger...etc.

Unfortunately Federer is already 28 so we'll be saying bye in about 5-6 years

Of course DP is only 20, will pwn for years

AwesomeToad, you may be a bit optimistic.  It's very unlikely in today's game that a player can perform at such a consistently high level into their 30's.  I say he's got less time than that: 3-4 years.

Fine. But I am a naturally optimistic person. [hide="This is a little off topic, but..."]Didn't Djokovic strip to his underwear once because he was bored? Sorry :blush: [/hide]

idk.

But i do know that nadal has a tendency to pick wedgies.

and Djokovic imitates them.


Given $ n$ consecutive integers starting from $ 1$, is it possible to arrange them in such a way such that the sum of any two consecutive integers is a perfect square for all $ n$(repetition is not allowed)?
If not, then for what values of $ n$ is it possible ? If it is true for some number, then how many such combinations exist?

are you sure that you mean sum of 'any' 2 'consecutive' numbers is a square?
if so then the squares shall be seperated by a number, leading to no such possiblity except for 1.
if not so, then could you explain it a bit clearer?

consider:
$ 16,9,7,2,14,11,5,4,12,13,3,6,10,15,1,8,17$

It contains numbers from $ 1$-$ 17$. Sum of any two consecutive numbers is a square.


Considere un cuadrillatero $ABCD$ circunscrito a una circunferencia $\Gamma$ de centro $O$. Sean $P$ y $Q$ los puntos de contacto de $\Gamma$ con $AB$ y $BC$ ,respectivamente. Si $X$ es el punto de interseccion de $PQ$ y $AC$, pruebe que:
a) $XO$ es ortogonal a $BD$.
b) Los puntos donde las tangentes a $\Gamma$ que pasan por $X$ interceptan a $\Gamma$ estan sobre $BD$.

Obs:  Si $R$ y $S$ son los puntos de contacto de $\Gamma$ con $CD$ y $DA$ , "creo que" $P$, $B$, $Q$, $R$, $D$ y $S$ estan sobre una elipse.

[hide]Sugerencia: Usar polos y polares[/hide]

Bueno, aqui mi solucion:
Pirmero llamemos T a la interseccion de PR, QS, AC y BD(es conocido que concurren).
Utilizando polos y polares (o algun menelao por alla), observamos que PQ, AC y RS concurren en X. Analogamente observamos que PS, QR y BD concurren en un punto Y.
a) Bueno, ahora tenemos un cuadrilatero PQRS ciclico. Ahora usaremos el siguiente teoremita que dice:
"Sea PQRS un cuadilatero ciclico de centro O, donde PQ y RS se intersectan en X, PS y QR se intersectan en Y, PR y QS se intersectan en T. Entonces O es ortocentro del triangulo XTY" Vemos que aplicando este teorema al problema ya demostramos que XO es perpendicular a BD.

b)Ahora, usaremos el sig. teorema:
"Sea PQRS un cuadrilatero ciclico, donde PQ y RS se intersectan en X, PS y QR se intersectan en Y, PR y QS se intersectan en T. Trazamos la recta YT y a los puntos donde corte a la circunferencia circuscrita de PQRS les llamamos J y K. Entonces XJ y XK son las tangentes desde X a la circunferencia"
Vemos que aplicando esto al problema obtenemos un resultado rapido.

Aqui pongo un link donde se puede encontrar la demostracion del teorema usado para el inciso a).
http://www.mathlinks.ro/Forum/viewtopic.php?t=108432
Jeje

Definamos $ M \equiv AD \cap BC,$ $ N \equiv AB \cap DC,$ $ X \equiv PQ \cap SR,$ $Y \equiv PS \cap QR.$ Por teorema de Newton se tiene que $ SQ,PR,AC,DB$ concurren en un punto, llámese $ E$ y como $ SQ$ y $ PR$ son las polares de $ M$ y $ N$ respecto a $ (O),$ se sigue que $ MN$ es la polar de $ E$ respecto a $ (O).$ En el cuadrilátero cíclico completo $ PQRS$ la diagonal $ XY$ es polar de su circuncírculo $ (O)$ respecto a $ E \equiv PR \cap SQ,$ así pues $ M,N,X,Y$ estan alineados. Las rectas $ PQ$ y $ SR$ se cortan en el polo $ X$ de la recta $ DB,$ por consiguiente $ OX \perp DB$  y los puntos de contacto de las tangentes desde $ X$ a $ (O)$ están en $ DB.$

[quote="jonfv"]Obs:  Si $ R$ y $ S$ son los puntos de contacto de $ \Gamma$ con $ CD$ y $ DA,$  "creo que" $ P,$ $ B,$ $ Q,$ $ R,$ $ D$ y $ S$ estan sobre una elipse.[/quote]
Esta conjetura es cierta y facil de demostrar. Los puntos $ P,B,Q,R,D,S$ estan en una misma cónica ya que los lados opuestos del hexágono $ PBQRDS$ se cortan en los tres puntos alinedos $ M,N,Y.$ Por el teorema de Pascal $ PBQRDS$ está inscrito en una cónica en la cual $ MN$ es una recta de Pascal.


geometry
3D geometry
inequalities
inequalities proposed

(1) Let $a,b,c$ be positive reals such that $ab+bc+ca=1$.  Prove that

\[ \sqrt[3]{\displaystyle \frac{a^2}{5+12bc}} + \sqrt[3]{\frac{b^2}{5+12ca}} + \sqrt[3]{\frac{c^2}{5+12ab}} \geq 1 . \]

(2) Let $a,b,c$ be positive reals such that $a+b+c=1$.  Prove that

\[1 \leq \sqrt[3]{\displaystyle \frac{a^2}{1+18bc}} + \sqrt[3]{\frac{b^2}{1+18ca}} + \sqrt[3]{\frac{c^2}{1+18ab}} < \sqrt[3]{3}.\]

I spent 3 hours trying to solve it  :(  Can anybody post the solution? Sorry for my bad English :blush:

I solved second one's left ineq.

Let $f(x)= \frac {1}{x^{\frac{1}{3}}}$. It is self-evident that $f(x)$ is convex.

then $\sqrt[3]{\frac{a^2}{1+18bc}}= a\sqrt[3]{\frac{1}{a+18abc}}$

We know $a+b+c=1$ and $f(x)$ is convex.

Using Jensen, We get $\sqrt[3]{\frac{a^2}{1+18bc}} + \sqrt[3]{\frac{b^2}{1+18ca}} + \sqrt[3]{\frac{c^2}{1+18ab}} \geq  \sqrt[3]{\frac{1}{\sum_{cy}(a^2 + 18a^2bc)}} $

So, It is enough to prove that $\sqrt[3]{ \frac{1}{\sum_{cy}(a^2 + 18a^2bc)}} \geq 1$

$\Leftrightarrow 1\geq \sum_{cy}(a^2 + 18a^2bc)= \sum_{cy} a^2 + 18abc$

$\Leftrightarrow  2\sum_{cy} ab \geq 18abc$

$\Leftrightarrow  \sum_{cy} ab \geq 9 abc$

using AD-GD, We get $\sum_{cy}ab \geq 3\sqrt[3] {a^2b^2c^2} $ and $ 1=a+b+c\geq 3\sqrt[3]{abc}$

$Q.E.D.$

Okay, here are my solutions:

(1)  [hide]We rewrite the inequality as

\[\sum_{\text{cyc}} \frac{a}{\sqrt[3]{5a^2b+17abc+5a^2c}} \geq 1.\]

From Holder,

\[\left(\sum_{\text{cyc}} \frac{a}{\sqrt[3]{5a^2b+17abc+5a^2c}}\right)^{3/4} \left(\sum_{\text{cyc}}a(5a^2b+17abc+5a^2c) \right)^{1/4} \geq a+b+c,\]

\[\sum_{\text{cyc}} \frac{a}{\sqrt[3]{5a^2b+17abc+5a^2c}} \geq \sqrt[3]{\frac{(a+b+c)^4}{\sum_{\text{cyc}}a(5a^2b+17abc+5a^2c)}},\]

so we need

\[(a+b+c)^4 \geq \sum_{\text{cyc}}a(5a^2b+17abc+5a^2c)\]
\[\iff \sum_{\text{cyc}} a^4+4a^3b+4ab^3+12a^2bc+6a^2b^2 \geq \sum_{\text{cyc}}5a^3b+5ab^3+17a^2bc\]
\[\iff \sum_{\text{cyc}} a^4-a^3b-ab^3-5a^2bc+6a^2b^2 \geq 0,\]

but this follows from

$\sum_{\text{cyc}} 6a^2b^2-6a^2bc \geq 0 \iff \sum_{\text{cyc}} a^2(b-c)^2 \geq 0,$
$\sum_{\text{cyc}} a^4+a^2bc \geq \sum_{\text{cyc}} a^3b+ab^3$ (Schur, r=2)

since adding gives the result.[/hide]

(2)  [hide]The LHS is similar, i.e. after Holder, it reduces to

\[(a+b+c)^4 \geq (a^2+b^2+c^2)(a+b+c)^2+18abc(a+b+c),\]

but this follow from

$(a+b+c)^4 = (a^2+b^2+c^2)(a+b+c)^2+2(ab+bc+ca)(a+b+c)^2$
$\geq (a^2+b^2+c^2)(a+b+c)^2+6(ab+bc+ca)^2$
$\geq (a^2+b^2+c^2)(a+b+c)^2+18abc(a+b+c).$

Now, for the RHS, we apply Holder once again (in the other direction)

\[\sum_{\text{cyc}} \sqrt[3]{\displaystyle \frac{a^2}{1+18bc}} \leq \left(\sum_{\text{cyc}}  \frac{1}{1+18bc}\right)^{1/3} (a+b+c)^{2/3} < (1+1+1)^{1/3} = \sqrt[3]{3}.\][/hide]

Well, now that I look at it, it would probably be nicer with the condition $a,b,c \geq 0$ and the $<$ replaced with a $\leq$ (for RHS of 2).  But anyways, did you like these problems?  :lol:

[quote="Chang Woo-JIn"]Using Jensen, We get $\sqrt[3]{\frac{a^2}{1+18bc}} + \sqrt[3]{\frac{b^2}{1+18ca}} + \sqrt[3]{\frac{c^2}{1+18ab}} \geq  \sqrt[3]{\frac{1}{\sum_{cy}(a^2 + 18a^2bc)}} $[/quote]

I don't understand your idea..

He WLOG set $\sum a^2=1$; the rest is just the definition of Jensen's inequality.

I just used $\displaystyle \sqrt[3]{\frac{a^2}{1+18bc}}= a\sqrt[3]{\frac{1}{a+18abc}}$ and $a+b+c=1$

Here, $a, b, c$ are weights.

Yes ,You're right!
It's very interesting!


Let $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}1$. Prove that

$ \frac{1\minus{}a}{2a\plus{}b}\plus{}\frac{1\minus{}b}{2b\plus{}c}\plus{}\frac{1\minus{}c}{2c\plus{}a}\geq2$

[quote="Ligouras"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that

$ \frac {1 \minus{} a}{2a \plus{} b} \plus{} \frac {1 \minus{} b}{2b \plus{} c} \plus{} \frac {1 \minus{} c}{2c \plus{} a}\geq2$[/quote]

$ \sum \frac {1 \minus{} a}{2a \plus{} b}\geq2$
<=>
$ \sum\frac {(b \plus{} c)^2}{(2a \plus{} b)(b \plus{} c)}\geq2$
by Chausy
$ \sum \frac {(b \plus{} c)^2}{(2a \plus{} b)(b \plus{} c)}$
$ \ge \frac {(2\sum a)^2}{\sum (2a \plus{} b)(b \plus{} c)}$
$ \equal{} \frac {(2\sum a)^2}{\sum a^2 \plus{} 5\sum ab}$
$ \equal{} \frac {(2\sum a)^2}{(\sum a)^2 \plus{} 3\sum ab}$
$ \ge \frac {(2\sum a)^2}{(\sum a)^2 \plus{} (\sum a)^2}$
$ \equal{} 2$


abstract algebra
geometry
geometric transformation
reflection
group theory
superior algebra
superior algebra unsolved

Let $\displaystyle G$ be a group with $\displaystyle 10$ elements such that there are $\displaystyle a,b \in G \setminus \left\{ e \right\}$, distinct, which fulfill $\displaystyle a^2=b^2=e$.

Is $\displaystyle G$ abelian?

There are just two groups with ten elements: the cyclic one and the dihedral group. Both fulfill your property (and the last one is not abelian: take $a,b$ as any reflection here).

There's only one element of order two in the cyclic group. That makes the group dihedral.

[quote="ZetaX"]There are just two groups with ten elements: the cyclic one and the dihedral group.[/quote]

Is that true for all $2p$, where $p \geq 3$ is a prime (the only groups of order $2p$ are the cyclic one and the dihedral group)? I'm asking this because I looked at a list of small groups, and I noted this fact.

If it's true, can you give some hints to prove it? Thanks.

i think you can find it somewhere on the forum..
anyway, the question can be poses with a prime $q$ instead of $2$: there are at most two groups of order $pq$ for any pair of primes $p\le q$ (not necessarily different): if $p=q$, there are the cyclic one and the product of the two cyclic ones, if $p|q-1$ there are a semidirect product of cyclic groups, and the cyclic one, otherwise there's just the cyclic one...

as for the proof.. just a short sketch.
suppose $p<q$. using cauchy's theorem, you have a subgroup $P$ of order $p$, and a subgroup $Q$ of order $q$. it's easy to show that they're disjoint, (hence $PQ$ is the whole group), and that $Q$ is normal.
so the whole group is a semidirect product.. but the group of automorphism of a cyclic group is cyclic itself, so you can prove the isomorphism between two different products, given by non-identity automorphisms...
(i've not been clear in the last part, since i never actually wrote down this part of the proof..)

This is first write. please understand if i write wrong reply. :blush:

[b]ⅰ)[/b] using Sylow's Theorem, $G$ contains subgroup of order $5$, let the subgroup $H$, $H$ is normal in $G$. moreover, $H$ is cyclic group.

[b]ⅱ)[/b] using Sylow's Theorem, the order of other elements, not $H$'s elements, is order $2$.

[b]ⅲ)[/b] the element multiplied two elements of order $2$(example, $a,b$) is $H$'s element. because, if the element is order $2$(example $c$), $\{e,a,b,c\}$ is Klein's 4-group. but, it is contradiction because $G$ is order $10$.(by Lagrange's Theorem)

[b]ⅳ)[/b] Let elements $a,b$ order $2$ and $H=<c>$. then, $ab$ and $ba$ is $H$'s element.(by ⅲ)
$ab=a^{-1}b^{-1}=(ba)^{-1}$. since $ab$ is $H$'s element,
$ab$ is $c$ or $c^{2}$ or $c^{3}$ or $c^{4}$. and
$ba$ is $c^{4}$ or $c^{3}$ or $c^{2}$ or $c$.
therefore $ab\neq ba$.

not good reply.. :blush:

The official solution is very short: If $G$ is abelian, then $\left\{ e,a,b,ab \right\}$ is a subgroup. By Lagrange's Theorem, $4 | 10$, contradiction.

But the other problem (with $pq$) is much more interesting. I'll look into it later.

[quote="perfect_radio"]
But the other problem with $pq$) is much more interesting. I'll look into it later.[/quote]
There was a topic about it, about a year ago (at that time I didn't understand it as well as I do now) :

[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=semi&t=45051[/url]


How hard do you think MOEMS is?  I gave it a 1.

Very very easy. Of course, I'm not the slow and careful type, so I missed about 4 this year. But I beat in Mathcounts someone who got only 1 wrong, so...

nvr did it before...but the problems sound easy enough to be a 1.

I think such a poll on AoPS will be severely biased by the math levels of the "regulars" who are mostly older and certainly more experienced relative to the MOEMS target age groups.  Remember that mathematical maturity is only beginning to show itself during grades 4-8.

I think using any standard at which MATHCOUNTS is a 3 (where only a few students per year score 90% of the problems) is examining something other than the appropriate level of MOEMS.

MOEMS introduces students to problem solving and I think that is on the level that it should be examined.

Well yes, I did MOEMS and it was an amazing entry to problem solving. It really is the best competition for elementary schoolers. I'm not sure how good it is for real competitive middle-schoolers, but it is still great for the casual participater.

Moems is easier than mathcounts.  For one thing, there is no speed element - if you know how to solve the problems there is more than enough time.  Doing the elementary olympiads in 5th grade and the middle school olympiads in 6th grade would be a good warm-up for mathcounts in 7th grade.  I have seen students follow this pattern, with considerable success.

Can you give me a link where i can find the former tests of MOEM?

http://www.moems.org/sample.htm
that's where you can find sample problems

I don't think it's too easy, for example, try this:

Let $a_1a_2a_3...a_n$ be a natural number, where $a_1$, $a_2$...are digits. What must this number be, so that $12\cdot2a_1a_2a_3...a_n1 = 21\cdot1a_1a_2a_3...a_n2$?

Yeah yeah, it's not [i]that[/i] hard, but it is on mathcounts level.

I gave it a two also. Looks like chess64 and I are the lonely 2's (as of 6/5 at 1:53 GMT - 4 hrs.)

Some of the problems can get nasty. I believe a problem on the marathon I gave out was solved wrong, but was let loose by me anyway.

I used to have trouble with MOEMS, but after math counts it seems pretty easy. All the top scorers from our school did MC.

[quote="chess64"]It depends on whether you're talking about school, chapter, state, or national mathcounts. That's why a gave it a 2 instead of a 1.[/quote]

Lol, think ur in the wrong forum. This is math olympiads.


To this problem I do not like tha solution I have.I'd like to see something more elegant.
Thanks!

Dear Stergiu and Mathlinkers,
I don't know your solution.
My quikly approach : consider the Miquel's point and circles leading to this point od the triangle MAB and the transversal NCD, then a simple angle chasing and we are done...
Sincerely
Jea-Louis

Because ABCD are concyclic,you can find a point G , that A,G,N,D  A,G,M,B are concyclic
G,D,C,M  G,B,C,N are concyclic  => △BGC∽△ADG  => AG/BG=AD/BC=AD/AE ∠AGB=∠CGD=∠CMD=∠EAD  => △AGB∽△DAE  =>  AD/AG=DE/AB=DE/CE   ∠DEC=∠DEA+∠CEA=∠GBA+∠ABC=∠GBC=∠GAD   => △AGD∽△CED
But we have △AGD∽△CNF   => △CNF∽△CED  => E,D,N,F are concyclic

Thanks to both of you  :) !

 I'm leaving in the morning for vacation  for a month, so ... have a nice summer ! See you again in September with your nice solutions!

 Babis


geometry AMC AIME USA(J)MO USAMO algebra function

Just wanted to talk about the upcoming UGA tournament. who is coming? I am. Also here is a good site:[url]http://www.math.uga.edu/mathmeet/[/url]
I went on that site and i saw many previous years' tests, but they weren't divided into jv and varsity like most others. this is the first time i am going to UGA, so are there two different tests each for jv and varsity, or is it one test?

It's all one test.

I'm coming!  Everybody, come, especially because Vestavia is coming!

By the way, the registration deadline is tomorrow.

I come, hope I make as good or better than last year, down with Vestavia :P ;). I may actually have a team, plus there will (as far as I know) be another homeschool team there :), so it should be a lot of fun. :) I wonder, how many people think Harrison will win? or will Harrison even come?

I'll be there as an individual, not as a team due to Wheeler policy.

But, I'll still be there, that is the important thing.

Yes I am coming too with a team...thx for answering the question. has everyone registered? i have

i wanna come. is there a geometry section  :rotfl:

especially if Vestavia is coming, jk, who cares if vestavia comes or not.

No. As a matter of fact, most of UGA is not geometry. In fact, last year had more geometry than usual.

question: is uga more amc 10 question difficulty, amc 12 difficulty, or aime difficulty? (or USAMO??? i doubt that HEAVILY :D )

ok scratch that, i dont think i'm going. our math team hasnt met since cobb. i think its roughly amc 12 difficulty, maybe a tiny bit easier

Unfortunately, almost all local regional competitions don't even approach olympiad difficulty.

Yeah, I'm having some trouble getting people from our school to show up as well.  It's unfortunate.  But I will still be there!

[quote="Boy Soprano II"][quote="Cicatriz"]Conjecture: each Georgia person is weirder than the last.[/quote]
Well, I assume that by "the last" you meant Billy, but weirdness does not assume a local minimum at Billy unless there exists an open interval containing Billy consisting entirely of irrational ARML members, which is false, since rational people are dense. :P   (No offense intended, Billy—I think possibly $\omega(\mbox{Billy}) \leq \omega(\mbox{Miles})$, where $\omega$ is the weirdness function :wink: )[/quote]

I don't think there are many irrational or rational ARML members since most of them are complex.

But I'm sure that rational people are dense.

BTW: Is Howard a moderator? It doesn't list him as a moderator, but he might be a special moderator.

^^^boy soprano^^  after that, you guys can never accuse me of wasting time posting stuff.  almost all of the stuff i post i type as im thinking it, so i rarely spend more than one minute per post.  i dont even wanna imagine how much time you spent on that when you could have been sleeping

haha nice proof. you should enter the national proof competition

u wud win 1st place, with 2nd elemmnop

[quote="captcha000"]Excuse you, Billy. I actually do read most of what gets posted in this forum at some point or another. Most of the stuff being posted here is quite useless anyway and I never bothered reading much of it. I don't mind "fun" stuff being posted, but if you guys are going to throw around insults and stuff do that on private messages or stuff, not on a public forum for the UGA tournament.

I'll keep this up for a while, but please think before you post.

Thanks[/quote]

I didn't say you didn't read everything ;) and if I implied that I am sorry for the misunderstanding... [quote]Somehow I think that Howard will delete this whole thread. On the other hand, having some superficial though firsthand knowledge of Howard, he isn't :D.[/quote]

OK :maybe:  :huh: so I'm a contender for the Weirdest Georgian award. Meh, I wonder, do they give awards at ARML for the weirdest ARMLer? :D[/quote]

Hmm... Thinking before posting would be a good idea, but then again, sometimes thinking too much gets you wound around in circles...

Let's divide the worth of Miles's post by the time he took, and compare that to the worth of certain other people's posts (admittedly mine included) and divide by the time taken...

Okay, okay, that may be interpreted as an insult of some sort. If so, I apologize--just pointing out a technicality.

I pulled myself back together, but sorry if I bothered anyone. And... maybe some of you won't be so quick to assume things, if you realize that sometimes something that sounds absurd is the sincere belief of the person posting it. Forgive me for having difficulty distinguishing between jokes and seriousness--I tend to chat with people who must be unlike what you're used to, and there are so many gradations of meaning, so many different sorts of intentions, and so many different standards of normal. I can say something that is a joke, a nod to a memory, an expression of anger and uncertainty... And I'm sometimes poor at controlling myself and making sure not to say unclear things.

I may apologize excessively, too--it doesn't make things any better, I know, but it's a good alternative to the rage and irritability that I gave myself over to last year.

Okay, okay, my real sin is probably talking too much about myself and thinking that my posts have impact, even to annoy y'all, but I don't have much else to say. Except that math and mania make a mouthwatering mix.

Cantor would have understood. Or maybe it was not understanding that got him in so much trouble...

[quote="Xantos C. Guin"]I don't think there are many irrational or rational ARML members since most of them are complex.[/quote]
True, but many of them have irrational real parts, and the set of complexes with rational rectangular coefficients is dense over the complex plane.

[quote="Xantos C. Guin"]BTW: Is Howard a moderator? It doesn't list him as a moderator, but he might be a special moderator.[/quote]
He is a (the) moderator of the Georgia memberlist, therefore he is a moderator of this forum.

[quote="Cicatriz"]Proof by induction:

Assume the nth Georgian is weirder than the n-1th. The nth Georgian spreads his or her weirdness to the n+1th Georgian by HIV. This weirdness adds to an inbred genetic weirdness possessed by each Georgian.

Thus each Georgia person is weirder than the last.[/quote]
You have two sizable holes in your "proof" which I won't even bother to point out, and many smaller ones.  For instance, you assume that each Georgian has positive weirdness, whereas many are weird in a very negative sort of way.  Plus, your knowledge of genetics is clearly cursory at best.  Everybody knows HIV is a somatic bacterium.  Plus, you are not even proving that each Georgian is weirder than the last; you're merely using it as your inductive step for some other obscure intent which you are clearly not sharing with us because you do not support the dissemination of knowledge and would clearly prefer a return to the Dark Ages wherein everybody but you would be so ignorant that you wouldn't even have to bother with learning new things to maintain the intellectual superiority over the masses necessary to maintain your power.

[quote="solafidefarms"]so I'm a contender for the Weirdest Georgian award.[/quote]
Sorry, it was only because you were the last GA person who posted before Cicatriz's post, and that particular misinterpretation was a crucial element of my argument. :wink:

[quote="jhcreinhardt"]i dont even wanna imagine how much time you spent on that when you could have been sleeping[/quote]
Did I mention that I spent two months learning topology, hyperbolic geometry, and set theory just so I could make that post?  I even learned two new notations.  That was the high point of my life, and now you have just ruined it for me.  On the other hand, you have opened my eyes to the wonderful value of sleep.  In fact, I think that from now on, I will spend my days sleeping.  Forget this math stuff!  What is the point of living and being awake at the same time?  In dreams, you can make everything how you want.

Anyhow, I would like to iterate (or reiterate—it's all recursive, anyhow) that if I have posted something offensive, it was not meant to be taken seriously, but only as a rebuttal that was intended (however dismally it might have failed) to be mildly entertaining, or at the very least interesting.

[quote="Boy Soprano II"]What is the point of living and being awake at the same time?  In dreams, you can make everything how you want.[/quote]

All kidding aside, is there anything wrong with that? Life is alot better asleep than awake.

But can it really be said that you MAKE things the way you want in dreams? It's all rather subconscious and confused...

Too bad ARML members are only partially real...

And I don't believe it's possible to measure weirdness accurately when you consider whether it is positive or negative--someone possessed of great weirdness with both positive and negative (and are these the only directional choices?) impact could then be considered to have only a low level of weirdness, which probably would not be useful in applications. I propose we assign weirdness vectors for each behavior exhibited by a person and define weirdness as the mean of the magnitudes of all weirdness vectors.

And, whatever the biology of weirdness... Wouldn't the trend toward being deader (corpses have 0 weirdness, right?) that eventually results from HIV counteract the weirdness being spread?

dude wtf are you talking about

no one cares

everything everyone is saying is

edible poop  :)

okay

guys

wtf?

note how the topic of this thread is: "stuff for uga tournament" and not "talk about random crap which is so pointless that it is retarded"

howard, please lock this thread...


[quote="pkerichang"]so if we're registered already, when should we send in the paper works for year 18???[/quote]

Whenever you like, but we don't have the paperwork up on the site yet.

Most students send them in with the first round of solutions.


vector
inequalities
triangle inequality
inequalities unsolved

vectors a,b,c,d satisfy a+b+c+d=0

prove |a|+|b|+|c|+|d| is not less than |a+d|+|b+d|+|c+d|

Adding up the vectors $ \vec{AB}\equal{}\vec{a},\vec{BC}\equal{}\vec{b},\vec{CA}\equal{}\vec{c},\vec{DA}\equal{}\vec{d}$, we get a closed polygon ABCD. It is easily to prove.

If $ ACBD$ is convex quadrilateral. $ AB cut CD\equal{}E$ .
 We can easily see that at least one of the six possible arrangements yields such a polygon.
 Adding up
$ \vec{AB}\equal{}\vec{a},\vec{CD}\equal{}\vec{b},\vec{BC}\equal{}\vec{c},\vec{DA}\equal{}\vec{d}$
$ |\vec{AE}|\plus{}|\vec{CE}|\geq |\vec{AC}|$,
$ |\vec{BE}|\plus{}|\vec{DE}|\geq |\vec{BD}|$, we get
                  $ |\vec{AB}|\plus{}|\vec{CD}|\geq |\vec{AC}|\plus{}|\vec{BD}|$
or     $ |\vec{a}|\plus{}|\vec{b}|\geq |\vec{b}\plus{}\vec{d}|\plus{}|\vec{a}\plus{}\vec{d}|$
The triangle inequality yields
$ |\vec{c}|\plus{}|\vec{d}|\geq |\vec{c}\plus{}\vec{d}|$
Adding up the last two inequalites, we get
 $ |\vec{a}|\plus{}|\vec{b}|\plus{}|\vec{c}|\plus{}|\vec{d}|\geq |\vec{b}\plus{}\vec{d}|\plus{}|\vec{a}\plus{}\vec{d}|\plus{}|\vec{c}\plus{}\vec{d}|$


inequalities
geometry
rectangle
analytic geometry
induction
triangle inequality
inequalities unsolved

Prove that if $ a_1,...,a_k$ e $ b_1,...,b_k$ are reals we have the inequality
\[ {\sqrt{{({\sum_{k\equal{}1}^n a_k^2})}^2\plus{}{({\sum_{k\equal{}1}^n b_k^2})}^2} \leq \sum_{k\equal{}1}^n \sqrt{(a_k^2\plus{}b_k^2)}}\]

[quote="Damien Hannigan"]$ {\sqrt {{({\sum_{k = 1}^n a_k^2})}^2 + {({\sum_{k = 1}^n b_k^2})}^2} \leq \sum_{k = 1}^n \sqrt {(a_k^2 + b_k^2)}}$[/quote]

Are you sure that you have post the right question?

[quote="Damien Hannigan"]
How to prove
$ {\sqrt {{({\sum_{k = 1}^n a_k})}^2 + {({\sum_{k = 1}^n b_k})}^2} \leq \sum_{k = 1}^n \sqrt {(a_k^2 + b_k^2)}}$[/quote]

Actually I also didn't know how to proof it.But I used it as a special case of Minkowski's inequality. Anyway i googled and got some good links about your question.
[url=http://en.wikipedia.org/wiki/Minkowski_inequality]Wikipedia[/url]
[url=http://planetmath.org/encyclopedia/ProofOfMinkowskiInequality.html]PlanetMath[/url]
Both of these proofs uses Holder's Inequality and some very advanced math (to be honest I haven't understood the proofs :blush: )
[url=http://planetmath.org/encyclopedia/HolderInequality2.html]Proof of holder's inequality[/url]

to solve the ineq,we can assume some rectangles which length of the side is $ a_1,b_1$......
I just write an example about when $ n\equal{}2$,and for $ n\ge 3$,the method is almost same.
We consider:
$ A(0,0),B(a_1,0),C(a_1,b_1),D(0,b_1),E(a_1\plus{}a_2,b_1),F(a_1\plus{}a_2,b_1\plus{}b_2),G(a_1,b_1\plus{}b_2)$
$ LHS\equal{}|AC|\plus{}|CF|$
$ RHS\equal{}|AF|$
of course,$ LHS \ge RHS$

[quote="Damien Hannigan"]Sorry,

$ {\sqrt {{({\sum_{k = 1}^n a_k})}^2 + {({\sum_{k = 1}^n b_k})}^2} \leq \sum_{k = 1}^n \sqrt {(a_k^2 + b_k^2)}}$[/quote]
Consider Cartesian coordinate system.
Let $ O,X_1,X_2,\dots,X_n$ be the points with coordinates $ (0,0),(a_1,b_1),(a_1+a_2,b_1+b_2),\dots,(a_1+a_2+\dots+a_n,b_1+b_2+\dots+b_n)$,respectively.
Obviously $ {\sqrt {{({\sum_{k = 1}^n a_k})}^2 + {({\sum_{k = 1}^n b_k})}^2}=OX_n\leq OX_1+X_1X_2+\dots+X_{n-1}X_n=\sum_{k = 1}^n \sqrt {(a_k^2 + b_k^2)}}$

Suppose that $ z_n\equal{}a_n\plus{}i\cdot b_n$ for $ n\equal{}1,2,3,...,k$.

Then the inequality is just \[ |z_1\plus{}z_2\plus{}...\plus{}z_k|\le |z_1|\plus{}|z_2|\plus{}...\plus{}|z_n|\] This is obvious by the triangle inequality (or induction or whatever).

To prove the general form, we'd like to have \[ \sqrt[p]{(a\plus{}c)^p\plus{}(b\plus{}d)^p}\le \sqrt[p]{a^p\plus{}b^p}\plus{}\sqrt[p]{c^p\plus{}d^p}\] Now...I'm not sure what to do exactly...I'm sure we can do it with calculus...or holder's. Perhaps there is a nice proof...


geometry rectangle function conics parabola quadratics

A rectangular pen which uses the wall of a house as one side, is enclosed by 1000 m of fencing.

a. What is w in terms of L?

b. Express the area in terms of L only?

c. Use the eq to find the area of the pen for x=50,100,200,300 and 400m

d. Describe the area of the pen as the length w increases?

e. Estimate the maximum are of the pen and the dimensions od the rectangle that would give one maximum possible area?

It is a hard problem for pre-algebra or beginning algebra. At higher levels, you may say “been there, done that.”

a) It would be best to take advantage of the “free side” by making the longer side of the pen (the length, $ L$) along the house.
There will be a length of fence $ (L)$ parallel to the side of the house connected to the house by two shorter sides of length $ W$ (the width of the pen).
The total fence length is $ 2W \plus{} L \equal{} 1000$
You can solve for $ W$ to get $ W$ as a function of $ L$.
[hide]$ W \equal{} 500 \minus{} L/2$[/hide]

b) The area is the product of the length $ L$, and the width $ W$ $ ($which you have as a function of $ L)$. That will give you the area as a function of $ L$.
[hide]$ A \equal{} 500L \minus{} L^2$[/hide]

c) The $ x$ in this part of the problem is meant to be the same as $ W$. We could have called the length of the side along the house $ x$ and the length of the other side $ y$ instead of $ L$ and $ W$, and the area as a function of $ x$ would be the same function. [hide]$ A(x) \equal{} 500L \minus{} x^2$[/hide]
The only difference is that when we called it $ L$ we meant for it to be longer than the other side. Calling it $ x$ makes it more general.

d) Part c) must have shown you that as you increased the length of the side along the house $ (x)$ from 50 to 400 the area of the pen increased.

e) IF you had studied parabolas (or quadratic functions) you would know that the area as a function of the length along the side of the house, $ A(x)$ or $ A(L)$, has a maximum, and would know how to find where that maximum occurs. Not knowing that, you can still estimate the location of the maximum by calculating $ A(x)$ for various values of $ x$. Since you did it for x=50,100,200,300 and 400m in part c), try x=500, 600m now.


geometry
trigonometry
trig identities
Law of Sines

in triangle ABC, side AC is of length 1, angle A is 30 degrees, and angle B is 15 degrees. what is the length of side B?

[hide="My sorution"]
hmm side B...AB or BC? well we'll just have to do both i guess.  :maybe:

So Law of Sines gives us \[\frac{\text{sin}15}{1}=\frac{\text{sin}30}{BC}=\frac{\text{sin}135}{AB}\]
And so we gots
\[BC=\frac{\text{sin}30}{\text{sin}15}\]
and
\[AB=\frac{\text{sin}135}{\text{sin}15}\]

[/hide]

side B normally refers to the side opposite angle B a.k.a segment AC
but sides should be in lower case, not in capitals

Does the problem statement have a mistake? Or does it just give the answer away..

[hide="My Solution"]Use an image:

[img]http://i96.photobucket.com/albums/l167/adasarathy/triangle_marked.gif[/img]

The Law of Sines states: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$.

By using it, $\frac{1}{\sin 15^{\circ}}=\frac{BC}{\sin 30^{\circ}}=\frac{AB}{\sin 135^{\circ}}$

Therefore, $ BC=\frac{\sin 30^{\circ}}{\sin 15^{\circ}}=\frac{\frac{1}{2}}{\sqrt{\frac{1-\cos 30^{\circ}}{2}}}=\frac{\frac{1}{2}}{\frac{\sqrt{6}-\sqrt{2}}{4}}\implies \frac{\sqrt{6}+\sqrt{2}}{2}$

And, $AB=\frac{\sin 135^{\circ}}{\sin 15^{\circ}}\implies\sqrt{3}+1$.[/hide]

[hide="@NeverOddOrEven"]You inverted the Law of Sines, even though you still got the right answer.  :rotfl: [/hide]

[hide="@sapphyre571"]In that case, the answer is already given.  :wink: [/hide]

[quote="anirudh"][hide="@NeverOddOrEven"]You inverted the Law of Sines, even though you still got the right answer.  :rotfl: [/hide]
[/quote]
I do that frequently myself. Keep in mind that $a=b$ means $\frac{1}{a}=\frac{1}{b}$. It only makes a difference in the general form...

It's obviously 1 since AC=b.

I'll find all three sides.

[hide="solution"]By the Law of Sines(which means $\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}$),

$2a=\frac{\sqrt{6}+\sqrt{2}}{2}=\frac{c\sqrt{2}}{2}$

a happens to be $\frac{\sqrt{6}+\sqrt{2}}{4}$

and c happens to be $1+\sqrt{3}$[/hide]

@anirudh: how do you get those images????

@NeverOddOrEven: Contests usually want the sides in simplest radical form, and side b is always the side opposite of angle B in problems that don't give a diagram.

[quote="nutz_for2.718281828"]@anirudh: how do you get those images????[/quote]

Geometers Sketchpad. You can download a demo to try it out [url=http://www.chartwellyorke.com/sketchpad/gspdemos.html]here[/url]. It is very useful in calculating those stuff, and if I input data, it'll do the math for me. :D

And, I use [url=http://www.webtree.ca/newlife/printkey_info.htm]Printkey 2000[/url] and capture the image in sketchpad. That saves it as an image, which I upload to [url=http://photobucket.com]Photobucket[/url] and post it in Image tags here. :)


analytic geometry
conics
parabola
quadratics
graphing lines
slope
algebra

A circle of radius $4\sqrt{5}$ has its center on the y-axis and is tangent to the parabola $y= \frac{1}{8}x^2$. Point $(p,q)$ is their common point in the first quadrant.  Find $p+q$

The equation of the circle can be written as $x^2+(y-b)^2=(4\sqrt{5})^2$. Then, since the parabola $y=\frac{1}{8}x^2$ is tangent to the circle, we can substitute $x^2=8y$ into the equation of the circle (which leads to quadratic equation of $y$) and then use the fact that the discriminant ($D$) of the quadratic is 0.

Or maybe there's an easier way... In my method, I have to find the equation of the circle first. Is there a way that you can directly find $p$ and $q$?

I think your method is good  :)

Maybe another method is to see that the tangent line has slope $\frac{p}{4}$


1. In the expansion:
$ (3x^8\minus{}2x^6\plus{}x^5\plus{}2x^4\minus{}x^2\plus{}1)^5\equal{}a_0\plus{}a_1x\plus{}a_2x^2...a_{40}x^{40}$,
the sum $ a_0 \plus{}a_2\plus{}a_4...a_{38}\plus{}a_{40}\equal{}$

2. Suppose that we label each of 8 cards with a +1 on one side and -1 on the other. The cards are randomly dealt onto a table, so that either side of each card is equally likely to be facing up. Let $ x$ be the absolute value of the sum of the numbers that are facing up. The most likely value of $ x$ is

3. Find the product $ cos(20^\circ)cos(40^\circ)cos(80^\circ)$

4. A number $ m\equal{}111\dots 11$ consists entirely of 1's in its decimal representation, and is divisible by 13. Which number need not divide $ m$? 3, 7, 11, 17, 37

[hide="4"]
It is well know that $ 1001 \equal{} 7\cdot11\cdot13$ and $ 111 \equal{} 3\cdot37$

Multiplying, we have $ 111111 \equal{} 3\cdot7\cdot11\cdot13\cdot37$

Thus, 17 doesn't have to divide.[/hide]

Mewto, I think you mean Problem 4?

And number 3 is just a specific case of a more general formula.

[hide="2"]
If I understood well, expected value?
If so, 4a.[/hide]

For $ \#3$ there are $ 2^{8}$ possible trials. Let's consider five cases:
$ 1.$ The sum is $ + 2$.
Hence, we must have that there are 5 $ + 1's$ and 3 $ - 1's$.  This can happen in $ {8 \choose 5} = 56$ ways.  Thus, the probability of this happening is $ \frac {56}{2^{8}}$.
$ 2.$ The sum is $ + 4$.
Hence, we must have that there are 6 $ + 1's$ and 2 $ - 1's$.  This can happen in $ {8 \choose 6} = 28$ ways.  The probability of this happening is thus $ \frac {28}{2^{8}}$.
$ 3.$ The sum is $ + 6$.
Hence, we must have that there are 7 $ + 1's$ and 1 $ - 1$.  This can happen in $ {8 \choose 7} = 8$ ways.  The probability of this happening is $ \frac {8}{2^{8}}$.
$ 4.$ The sum is $ + 8$
Hence, we must have that there are 8 $ + 1's$ and 0 $ - 1's$.  This can happen in $ {8 \choose 8} = 1$ way.  The probability of this happening is $ \frac {1}{2^{8}}$.
$ 5.$ The sum is $ + 0$
Hence, we must have that there are 4$ + 1's$ and 4 $ - 1's$.  This can happen in $ {8 \choose 4} = 70$ ways. The probability of this happening is $ \frac {70}{2^{8}}$.

We do not even have to list when the the totals are negative, since the probability of those happening will be the same as when the totals are positive, but have the same magnitude (i.e. a total of $ - 4$ and $ + 4$ will have the same value).  Thus, since we're looking for the absolute values of $ x$, $ -2$ and $ +2$ will have the same value, so we can just multiply each value by 2, except the sum of $ 0$.  Hence, the answer is, as the greatest probability, $ \boxed { + 2}$.

EDIT: just reworded the last paragraph, it was hard to understand.

An easier way for 3:

Let $ \cos(20^\circ)cos(40^\circ)cos(80^\circ) \equal{} P$

$ \sin(20^\circ)P \equal{} \sin(20^\circ)cos(20^\circ)cos(40^\circ)cos(80^\circ)$

$ \equal{} \frac{1}{2} \sin(40^\circ)cos(40^\circ)cos(80^\circ) \equal{}\frac{1}{4} \sin(80^\circ)cos(80^\circ)$

$ \equal{}\frac{1}{8} \sin(160^\circ) \equal{} \frac{1}{8} \sin(20^\circ)$

So
$ P\equal{}\frac{1}{8}$


modular arithmetic
number theory proposed
number theory
Zsigmondy

[quote="Ahiles"]Solve the equation

$ 3^x \plus{} 5^y \equal{} z^2$

in positive integers.[/quote]
We denote that a perfect square must be 3k or 3k+1, which k is a interger.
So $ 5^y$ must be 3k+1, then y divise by 2.
We recall $ y\equal{}2a$, a is a postisive interger.
$ 3^x \plus{} 5^(2a) \equal{} z^2<\equal{}> 3^a \equal{}(z\minus{}5^a)(z\plus{}5^a)<\equal{}>3^m\equal{}z\minus{}5^a,3^n\equal{}z\plus{}5^a$
Then $ 3^n\minus{}3^m\equal{}2.5^a$ so 3 is divise by 2 or 5 that wrong if m,n is positive interger.
So m must be zero, n=a.
We have:$ 3^m\equal{}2.5^a\plus{}1$
We consider m with modulo 4. Then m is 4e+1 which e is apositive interger.
But like that so $ 3^m\minus{}1$ is divise by 4, but right hand is not. So there is no pair (x,y,z) satify for this problem.

Considering modulo 3, $ 5^y\equiv1\pmod3$, so $ y$ is even. Let $ y\equal{}2a$.

So $ 3^x \equal{} z^2 \minus{} 5^{2a} \equal{}(z\plus{}5^a)(z\minus{}5^a)$

Thus $ z\plus{}5^a\equal{}3^b,z\minus{}5^a\equal{}3^c$, where $ b\plus{}c\equal{}x$ and $ b,c\in\mathbb{N}_0$.

We have $ 2\cdot 5^a \equal{} 3^b\minus{}3^c$. LHS is not divisible by 3, so $ c\equal{}0$, $ b\equal{}x$.

$ z\minus{}5^a\equal{}1\implies z\equal{}5^a\plus{}1$

$ 5^a\plus{}1\plus{}5^a\equal{}3^b$

So $ 3^b \equal{} 2\cdot 5^a\plus{}1$

Consider modulo 5, $ b$ is even. Consider modulo 4, $ b$ is odd. Contradiction, so there are no solutions.

I'm sorry. The question was incorrect.
Now everything is correct.

Looking mod 4, x is even. Let $ x\equal{}2a$
$ 9^a\minus{}5^y\equal{}z^2$
$ (3^a\minus{}z)(3^a\plus{}z)\equal{}5^y$
So let $ 3^a\minus{}z\equal{}5^b$, $ 3^a\plus{}z\equal{}5^c$
Then $ 3^a\equal{}\frac{5^b\plus{}5^c}2$, which is divisible by 5 unless $ b\equal{}0$ (since $ b\leq c$)
So $ 2\cdot3^a\equal{}5^c\plus{}1$
Since the only primes dividing the LHS are 2 and 3, this has no solutions for $ c>1$ by Zsigmondy's theorem. So the only solution is $ x\equal{}2,y\equal{}1,z\equal{}2$

[quote="caffeineboy"]... by Zsigmondy's theorem.[/quote]

Wow...

A sledgehammer is not required here. :wink:
Just look modulo $ 9$ to see that if $ a>1$ then $ c$ is divisible by $ 3$. Then deduce that $ 1\plus{}5^c$ is divisible by $ 7$ which is not the case of the $ LHS$.

Pierre.

alternatively, you might note that $ z \equal{} 3^a \minus{} 5$, and plug this into the main equation, we get:

$ \minus{}5^{y}\equal{}\minus{}10\cdot 3^{a} \plus{} 25$, so $ y\equal{}1$ and $ a\equal{}1$.

Like [b]pbornsztein[/b] I look modulo 9. Then again I get c is divisible by 3. Let c=3m. Then $ 2.3^{a}\equal{}\left(5^{m}\plus{}1\right)\left(5^{2m}\plus{}5^{m}\plus{}1\right)$ so $ 5^{m}\plus{}1\equal{}2.3^{h}$ and using the method of the infinite descent we get $ c\equal{}0$. Contradiction! So $ a\equal{}1$ and $ c\equal{}1$.

[quote="SAPOSTO"]Like [b]pbornsztein[/b] I look modulo 9. Then again I get c is divisible by 3. Let c=3m. Then $ 2.3^{a} \equal{} \left(5^{m} \plus{} 1\right)\left(5^{2m} \plus{} 5^{m} \plus{} 1\right)$ so $ 5^{m} \plus{} 1 \equal{} 2.3^{h}$ and using the method of the infinite descent we get $ c \equal{} 0$. Contradiction! So $ a \equal{} 1$ and $ c \equal{} 1$.[/quote]

Another way to finish it is by noticing that
\[ 5^m\plus{}1\equal{}2\cdot 3^{\alpha} \textrm{ and } 5^{2m}\plus{}5^m\plus{}1\equal{}3^{\beta} \Longrightarrow 5^m\equal{}4\cdot3^{2\alpha}\minus{}3^{\beta},\]
which leads to a contradiction.

We can also find out that $ c$ is divisible by 3 in following way:

From $ 2.3^{a}\equal{}5^{c}\plus{}1$ we conclude that $ c$ is odd (modulo 3) so $ c\equal{}2m\plus{}1$ and we get:

$ 2.3^{a}\equal{}(5\plus{}1)(5^{2m}\minus{}5^{2m\minus{}1}\plus{}...\plus{}5^{2}\minus{}5\plus{}1)$

From last one it is obvious that (for $ a>1$ , case $ a\equal{}1$ is easy) we get that 3 divides $ 5^{2m}\minus{}5^{2m\minus{}1}\plus{}...\plus{}5^{2}\minus{}5\plus{}1$ and since each of these $ 2m\plus{}1$ summands is congruent 1 modulo  3 we get $ 2m\plus{}1$ is congruent 0 modulo 3 being equaivalent to $ c\equal{}3d$.

there are 100 solutions for this question and I got only 4 points in BMO. I shuold kill myself!

I think it is a very easy problem for BMO and it can be a good problem for JBMO :lol:

[quote="nickthegreek"]Hello dear mathlinkers.

Could someone PLEASE answer the following question.

Apparently,the choice of modulo 9 is not random. So,what should lead us to consider this equation modulo 9,and why not modulo any other integer?

I am preparing for a team selection test in Greece and, since I do not live in a city,I dont have the resources or any teacher to explain these things to me.

So I would strongly appreciate your help.......

Waiting for a response,
Nick[/quote]
But only $\mod 9$ does not work,others work too,that depends on the way you have proceeded.Specially here $3^x$ is divisible by $9$ for $x>1$,so it was taken for providing informations.

[quote]Just look modulo $ 9$ to see that if $ a>1$ then $ c$ is divisible by $ 3$. Then deduce that $ 1\plus{}5^c$ is divisible by $ 7$ which is not the case of the $ LHS$[/quote]

I'm confused..  :maybe:  Can you explain why??? I don't see... :!:

@fighter
Hey,brother! :w00t:
Could you teach me solution by Pythagorean triple?

This thread is $7$ years old.But I want to solve problem by using [b]Zsigmondy's theorem[/b]. :bruce: So I revive this thread. :D

Thanks,
Takeya.O

$$3^x-5^y=z^2$$
first we use $\pmod 4$ and get $x=2m$ and $y$ is odd.
so we can write $(3^m-z)(3^m+z)=5^y(1)$
let $d=\gcd(3^m-z,3^m+z)=\gcd(3^m-z,2z)$ so $d|2z$ but from $(1)$ we get $d$ is in the form $5^k$. Note that $5\nmid z$ because if $5|z$ then $5|3^x$ which is impossible so we can conclude that $d=1$

so $3^m-z=1,3^m+z=5^y$ and $2\cdot 3^m=5^y+1$ by zygmondy theorem we cannot find $y>1$ because if for all $y>1$ RHS will get e new primitive prime so we conclude $y=1$ and $m=1\implies z=2$. Only $(2,1,2)$ works

hey brother don't be rood please because I didn't read any post just solved it sorry

@fighter

Hey,brother.I want to know Pythagorean triple solution. :D

Could you teach me?

[quote=AMN300]Very standard problem
[hide=Solution]
Taking$\pmod{4}$, $(-1)^x-1=z^2 \implies x=2a$ for positive integer $a$.
By difference of squares,
\[ (3^a-z)(3^a+z)=5^y \]
Suppose $3^a-z=5^m, 3^a+z=5^n$ where $0 \le m < n \le y$. Adding the equations, $2 \cdot 3^a=5^m(5^{n-m}+1)$. This means $m=0$, so
\[ 2 \cdot 3^a=5^y+1\]
We can finish through LTE now but here is a more elementary approach. Suppose $a \ge 2$, so$\pmod{9}$ reveals $y=6w-3$ since the order of $5\pmod{9}$ is $6$. By sum of cubes, $2 \cdot 3^a = (5^{2w-1}+1)(5^{4w-2}-5^{2w-1}+1)$. $5^{2w-1}\equiv -1\pmod{3}$, so the right hand side isn't divisible by $3$, contradiction.
This means that $a=1$. From here, we recover the solution
\[ \boxed{(x, y, z)=(2, 1, 2)} \]
[/hide][/quote]

not entirely sure your contradiction is a contradiction?
[hide=My solution]
First, we take $\pmod 4$. This gives us $x$ is even; if $x$ is odd, it follows that $3^x \equiv 3 \pmod 4$, and as $5^y \equiv 1 \pmod 4$, then the LHS would be $2 \pmod 4$; $2$ is not a quadratic residue $\pmod 4$, contradiction.

Now consider the equation $\pmod 3$. Clearly the LHS is equivalent to $-2^y \pmod 3$; from this we get $y$ is odd, as if $y$ is even, $-2^y \equiv 2 \pmod 3$, and $2$ is not a quadratic residue $\pmod 3$.

Let $x=2k$. Rearranging the equation gives us $3^{2k} - z^2 = 5^y$, and as $3^{2k}$ is a square we can apply difference of squares. The equation becomes $$(3^k - z)(3^k+z) = 5^y.$$
It follows that both $3^k-z$ and $3^k+z$ are multiples of $5$.

From this it follows that $(x,y,z) = (2,1,2)$ is the only solution; basically what we want is for $3^k$ to be the average of two different powers of $5$ where the exponent is nonzero. If it is, then it follows that the last digit of $3^k$ is $5$, which is clearly impossible. Therefore $3^k - z = 1$, from which it follows that
$$2 \cdot 3^{k} = 5^y + 1.$$
$k=1$ yields the solution $(x,y,z) = (2,1,2)$; now assume $k \ge 2$. Taking the equation $\pmod 9$ yields $0 \equiv 5^y + 1 \pmod 9$, from which it follows that $y$ is of the form $6m+3$ as $5^6 \equiv 1 \pmod 9$ and $5^3 \equiv -1 \pmod 9$.

Substituting in $y = 6m+3$ gives us $2\cdot 3^k = 125^{2m+1} + 1$. However, taking this equation $\pmod 7$ gives us $2\cdot 3^k \equiv 0 \pmod 7$, contradiction. Therefore the only solution is $(x,y,z) = (2,1,2)$.

The only solution is $\boxed{(x,y,z)=(2,1,2)}$. Taking the given equation modulo $4$, we learn that
\[(-1)^x-1\equiv 0,1\pmod{4},\]
so the only possibility is that $x$ and $z$ are even. We actually don't need that $z$ is even, but we will et $x=2b$. Thus, we have
\[3^{2b}-z^2=5^y,\]
so
\[(3^b-z)(3^b+z)=5^y.\]
Let the first factor be $5^p$, the second $5^q$. We see that $5^p+5^q=2\cdot 3^b$, which means that we must have $p=0$. Thus, $3^b-z=1$ and $3^b+z=5^y$, so
\[2\cdot 3^b=5^y+1.\]
For the left side to be divisible by $3$, we need $y=2c+1$ to be odd. Then, we see that $3\nmid 5^y-1$, so
\[\nu_3(5^y+1)=\nu_3(25^y-1)=1+\nu_3(y)\]
by LTE. Thus, $b=1+\nu_3(y)$, so
\[6\cdot 3^{\nu_3(y)}=5^y+1.\]
But note that the LHS is at most $6y$, so we have $5^y+1\le 6y$, or $y\le 1$. When $y=1$, we get that $b=1$, so $x=2$. This means that $z^2=9-5=4$, so $z=2$, so the solution must be the one that we claimed. It is easy to see that it works, so we're done. $\blacksquare$

[hide=Solution][quote=Balkan MO 2009 P1]Solve the equation
\[ 3^x \minus{} 5^y \equal{} z^2.\]
in positive integers.
[/quote]
[b][color=#000]Solution:[/color][/b] Clearly, $z$ is even. Taking $\pmod{4}$, we get, $x$ must be even too. Let $z=2k$ & $x=2x'$
$$5^y=(3^{x'}-2k)(3^{x'}+2k) \implies \begin{cases} 3^{x'}-2k=5^a \\ 3^{x'}+2k=5^b \end{cases}$$
So, $\implies$ $2\cdot 3^{x'}=5^a(1+5^{b-a})$ Since, $\text{ LHS }$ is not divisible by $5$ $\implies$ $a=0$. So,
$$1+5^y=2 \cdot 3^{x'}$$
If $x'=1$ $\implies$ $(x,y,z)=(2,1,2)$. When, $x' \geq 2$, Taking, $\pmod{9}$ $\implies$ $y=6y'+3$. Now, Taking $\pmod{7}$ $\implies$ $5^y+1$ is $0 \pmod{7}$, but $7$ does not divide $\text{ RHS }$, a contradiction! So, we have $\boxed{(x,y,z)=(2,1,2)}$ $\qquad \blacksquare$
[/hide]

I claim that the only solution is $(x,y,z) = (2,1,2)$.
First, if we take $\mod 4$, since $z^{2} \equiv 0,1\mod 4$ and $5^{y}\equiv 1\mod4$, we must have $x$ as even, or else $3^{x} - 5^{y}\equiv 2\mod4$ which can't be equal to $z^{2}$. Therefore, we denote $x = 2x_{1}$.
Rearranging the terms, we get
$$3^{2x_{1}} - z^{2} = 5^{y} \Rightarrow (3^{x_{1}} - z)(3^{x_{1}} + z) = 5^{y}$$
Since the only factors of $5^{y}$ is powers of $5$, we must have $3^{x_{1}} - z$ and $3^{x_{1}} + z$ as powers of $5$. I claim that $3^{x_{1}} - z = 1$. If $3^{x_{1}} - z$ is a power of $5$ greater than $1$, then $z\equiv 3^{x_{1}}\mod 5$. However, this implies that $3^{x_{1}} + z \equiv 2\cdot 3^{x_{1}}\mod 5$, so $3^{x_{1}} \equiv 0\mod5$, which is clearly impossible.
Thus, we have $$3^{x_{1}} - z = 1 \Rightarrow z = 3^{x_{1}} - 1 \Rightarrow 3^{x_{1}}+z = 2\cdot 3^{x_{1}} -1 = 5^{k}$$ where $k$ is a positive integer. I claim that for all $x_{1} \geq 2$, there are no solutions. This is because then $2\cdot 3^{x_{1}} \equiv 0 \mod 9$, thus $5^{k} \equiv 8 \mod 9$, which implies $k$ is a multiple of $3$. Rearranging, we get $2\cdot 3^{x_{1}} = 5^{k} + 1$. However, then because $k$ is a multiple of $3$, we have $5^{k} + 1 \equiv 0 \mod 7$. Thus $2\cdot 3^{x_{1}}$ must be divisible by $7$, which is clearly impossible. Thus, the only solution is when $x_{1} = 1$, so our only solutions for $(x,y,z)$ is $(2,1,2)$.

Not hard but beautiful problem!
Let us consider $mod 4$. We get that $x$ is even and let's denote it as $x = 2k$. Then $5^y = (3^k - z)(3^k + z)$. Then it follows that $3^k - z = 5^a$ $(1)$ and $3^k + z = 5^b$ $(2)$ where $a+b=y$ and $a<b$. If we sum $(1), (2)$ we get that $2*3^k = 5^a + 5^b$ but left side is not divisible by $5$ so we obtain that $a = 0$. Now our equation becomes $2*3^k = 5^b + 1$.If $k>1$ by taking $mod 9$ we get that $b$ is divisible by $3$ and let $b=3l$. Now we get that $5^l + 1 = 2*3^c$ and that $5^2l + 5^l + 1 = 3^d$, then it follows that $5^2l = 3^d - 2*3^c$ which gives us contradiction when watching $mod 3$ so $k=1$. Then plugging back we get that $y=1$, $x=2k=2$ and $z=2$. With this we are done!

Take $\pmod{4}$ we get that $x = \text{even}$

$$\implies x = 2m$$

Substituting back this in our original equation we get that
$$3^{2m} - 5^y = z^2$$
$$\implies (3^m + z)(3^m - z) = 5^y$$
$$\implies 3^m + z = 5^a \quad \text{and} \quad 3^m - z = 5^b, \quad a+b=y, \quad a > b$$
$$\implies 3^m = 5^b \left(\frac{1 + 5^{a-b}}{2}\right) = 5^b \cdot q$$
$$\implies b = 0 \quad \text{because power of 3 can't contain 5 in it}$$
Putting $b=0$ we get that
$$3^m = \frac{5^y + 1}{2}$$
Note that taking $\pmod{3}$ in out original equation we get that $y = \text{odd}$
$$\implies m = v_3\left(\frac{5^y + 1}{2}\right) = v_3(6) + v_3(y) - v_3(2)$$
$$\implies m = 1 + v_3(y)$$
$$\implies y = 3^{m-1} \cdot k \quad \text{(k is co-prime to 3)}$$
$$\implies y \ge 3^{m-1}$$
Plugging it back in we get
$$2\cdot 3^m = 5^{k \cdot 3^{m-1}} + 1 \ge 5^{3^{m-1}} + 1$$
But notice that $RHS > LHS$ for $m-1 > 0$ (Rigorous proof : Induction)
$$\implies m-1 = 0 \implies m = 1$$
$$\implies 3^1 - z = 1 \implies z = 2$$
$$\implies y = 1$$
Hence the solutions are $\boxed{(x,y,z) = (2,1,2)}$

I have done this task with mod:
3"x=5"y+z"2
Euler's theorem:
3*2/3=2
z"2=1(mod3)
5"y must be:
5"y=2(mod3)
because z"2+5"y must divided by 3.
if y is 2k+1 then 5"y=2(mod3)
because of that y is single integer.
let k=0
then x=2 y=1 z=2
problem solved
and its the only solve we can have

[quote=Ahiles]Solve the equation
\[ 3^x \minus{} 5^y \equal{} z^2.\]
in positive integers.

[i]Greece[/i][/quote]
$3^x-5^y=z^2$
$(-1)^x-1\equiv z^2 \pmod{4}$
$\Rightarrow x\equiv 0 \pmod{2} \Rightarrow x=2a$
$\Rightarrow (3^a-z)(3^a+z)=5^y$
$\Rightarrow 3^a-z=5^m , 3^a+z=5^n$
$\Rightarrow 3^a=\frac{5^m+5^n}{2}$
$\Rightarrow m=0$
$\Rightarrow z=3^a-1, z=5^n-3^a$
$\Rightarrow 2\times 3^a=5^n+1$
If $n>1:$
$5^1+1=2\times 3$
By Zsigmondy´s Theorem:
$\exists$ prime $p\neq 2,3/ p|5^n+1 (\Rightarrow \Leftarrow)$
$\Rightarrow n\le 1$
If $n=1:$
$\Rightarrow a=1 \Rightarrow (x,y,z)=(2,1,2)$ is solution
If $n=0:$
$\Rightarrow a=0 \Rightarrow x=0 (\Rightarrow \Leftarrow)$
$\Rightarrow (x,y,z)=(2,1,2)$ is a unique solution $_\blacksquare$

[b][color=#6AA84F]Claim:[/color][/b]The only solution is $\boxed{(x,y,z)=(2,1,2)}$
[i]Proof:[/i]
Checking$\pmod 3$ gives us $z^2=3^x-5^y\equiv 1,2\pmod 3$ furthermore from the fact that $z^2\equiv 0,1\pmod 3$ we get that $5^y\equiv 1\pmod 3\Longrightarrow 2\nmid y$, $y$ is odd.
Furthermore, by taking $\pmod 4$, we get that $z^2=3^x-5^y\equiv (-1)^x-1\pmod 4$ furthermore $z^2\equiv 0,1\pmod 4$ implies that $x$ is even. ($x=2k$)
Now taking $\pmod 8$, we get that $z^2=3^{2k}-5^y\equiv 4\pmod 8$ which implies that $z^2\equiv 4\pmod 8\Longrightarrow z$ is even.

After these observations, the equation can be rewritten as: $\left(3^k\right)^2-z^2=5^y\Longleftrightarrow (3^k-z)(3^k+z)=5^{a+b}\text{ where } a+b=y\Longrightarrow 3^k-z=5^a\text{ and } 3^k+z=5^b$, now notice that if we sum the equations we get $2\cdot 3^k=5^a+5^b\Longrightarrow 3^k=\frac{5^a+5^b}{2}$ however notice that for $RHS\in \mathbb{Z^+}$, $a$ must be equal to $0$. Thus $a=0\text{ and } b=y$
Thus the equation transforms into $2\cdot3^k=5^y-(-1)^y$ so now we can use $LTE$ to finish up our problem.
$\nu_3(2\cdot3^k)=\nu_3(5^y-(-1))\Longrightarrow k=\nu_3(y)+1\Longrightarrow y\ge3^{k-1}$, furthermore by plugging in our result into our original equation we get $6\cdot 3^k-5^{3^{k-1}}-1\ge0$ now let $f(k)=6\cdot3^k-5^{3^{k-1}}-1$
[i]Claim:[/i] The previously stated inequality in integers if and only if, $k=1$
[i]Proof:[/i]
$f'(k)=6\cdot3^k\ln {3}-3^{k-1}\cdot 5^{3^{k-1}}\ln {3}\ln {5}<0,  \forall k\ge2$, thus $k=1$ $\square$

Thus $k=1\Longrightarrow x=2$, furthermore form the equation $3^k=z+1$ we get that $z=2$, thus by plugging into our original equation we get $9-4=5^y\Longrightarrow y=1$
So the only solutions are the ones previously stated in the claim$\blacksquare$.


[quote="Interval"]Translate each the sentence given into a matematical equation.  Be sure to identify the meaning of all symbols.

Here it is:

The total variable cost of manufacturing x dishwashers is 150 dollars per dishwasher times the number of dishwashers manufactured.[/quote]
Wouldn't the equation just be $150x$?

is it just me, or does it seem like we are doing intervals algebra 1 homework for him?

I am not in algebra 1 or 2.

I graduated high school over 20 years ago.

I am just preparing for a math state exam.

The questions come from that textbook.

If the questions are bothering you, then skip my post.


Hey, I just recently heard of the UNSCO from this forum and other people.  I was wondering, how can one participate in this competition and what materials and resources do you recommend to practice for it? I appreciate any comments. Thanks in advance.

i am going to be the first person in the history of my school to participate in it for next year.
all you have you to do is to talk to your chem teacher, then your chem teacher has to e-mail someone (the coordinator) and say that there are some interested people. then you'll take the local examination. if you pass you go to nationals. then you can qualify for summer camp
for more info go to
http://www.chemistry.org/portal/a/c/s/1/acsdisplay.html?DOC=education%5Cstudent%5Colympiad.html


calculus
integration
LaTeX
trigonometry
calculus computations

Can't seem to figure this one out, would appreciate some help.

Evaluate $ \int \frac{1}{x^4 \sqrt{a^2\minus{}x^2}} \, dx$

Try doing this substitution....

$ {x\equal{}acosy}$...

@Valeri
$ \text{\LaTeX}$ tip use [code]$\cos x$[/code] instead of [code]$ cos x$[/code] to get the more nice looking $ \cos x$ :wink:

THen pcz.... use integration by parts.... split ur integral as $ {sec^2x}$*$ {sec^2x}$....

this is split like this because $ {sec^2x}$ is an integrable function...

Okay, got it from there. Thanks for the help :)

[quote="pcz"]Can't seem to figure this one out, would appreciate some help.

Evaluate $ \int \frac {1}{x^4 \sqrt {a^2 \minus{} x^2}} \, dx$[/quote]
Substitute $ x\equal{}\frac1u,$ then you'll get an easy integral.

[quote="pcz"]I've tried that which gets it down to $ \int \frac {1}{\cos^4y}dy$ or $ \int \sec^4y dy$, but I'm not sure the best way to attack that either :/

I also received this problem before learning anything about trig substitutions so I'm wondering if I am supposed to take a different approach...[/quote]

You can also do $ \int \sec^2 y [1\plus{} \tan^2 y] dy$ which you can do pretty easily since $ (\tan\theta)' \equal{} \sec^2 \theta$

pcz the method mentioned by chickendude will be better than mine.....


function
inequalities theorems
inequalities

Does there exist a generalization of Karamata?
i do not mean the generalization with two function or...,i am seeking one that puts some extra conditions on sequences,which yields the enequlity,that is some conditions weaker than majorizing.

There is also [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23644[/url]


f(x) is twice differentiable and it's defined on (-\infty,\infty). It satisfies f(0)=f'(0)=0 and |f''(x)|\leq C|f(x)f'(x)| for any x\in [b]R[/b], where C is a constant. Prove that f(x)=0.

[i]year 1990[/i]

Furthermore, this problem can be generalized as follows:

f:[b]R[/b]->[b]R[/b] is [i]n[/i]-times(n is an positive integer) differentiable. It satisfies f(0)=f'(0)=...=f(n-1)(0)=0 and there exists positive constant c and j\in {0,...,n-1} s.t. |f n(x)|\leq c|f(j)(x)| for any x\in [b]R[/b]. Prove that f(x)=0.

I don't know it the following is good or not, but at least I try:
   Take any a>0. We show that for any x in [0,a] f(x)=0. I believe we can treat similarly the case x<0.
    Let M_1 the maximum value of |f(x)*f'(x)| when x is in [0,a] and M-2 the maximum value of |f'(x)| on [0,a]. For arbitrary x in [0,a] integrate the relation in the problem between a and x. We find that |f'(x)|<=cM_1*x for all x. Then integrate the same relation and use the previous result to find that for all x in [0,a] we have |f'(x)|<=c^2*m_1*M_2*x^2/2. Integrate again and use the previous result. It follows that for all x in [0,a] we have |f'(x)|<=c^3*M_1*(M_2)^2/3!*x^3. By induction, we find that |f'(x)|<=c^k*M_1*(M_2)^k-1*x^k/k! for all k and all x in [0,a]. For fixed x, take k very large. It follows that f'(x)=0 for all x in [0,a] and thus f is constant of [0,a]. So f(x)=f(0)=0 for x in [0,a]. Since a is arbitrary, f is zero for positive values of x. The same for the negative ones.

I study your proof Harazi it's good, it seems f'(0)=0 is useless.
What is your proof Liyi ? Do you really need f'(0) = 0 ?

suppose f(x) \neq 0, for a x \geq 0. WLOG f(x) \neq 0 on [0,\varepsilon]. Hence f'(x)\neq 0 [0,\varepsilon] (if not, we can make an translation).

Take x_1>0 s.t. 1/x_1 > C, f is continous, so we can suppose that
max_[0,x_1] |f(x)| = M_0 \in (0,1).

Let M_1 = max_[0,x_1] |f'(x)|, M_1 > 0.

Take x_2 \in [0,x_1] such that |f'(x_2)| = M_1. From Lagrange's mean value theorem, we know there is a \xi \in (0,x_2) such that

|f''(\xi)| = |f'(x_2) - f'(0)| / |x_2 - 0| = |f'(x_2)/x_2|

thus

CM_1 \geq C|f'(\xi)| > C|f(\xi)f'(x_i)| \geq |f''(\xi)| = |f'(x_2)/x_2| > CM_1

a contradiction....

hence for all x\geq 0, f(x) = 0.

in the similar way we get proof that for x\leq 0, f(x) = 0

the codes r not showing up, i have NO idea wat u guys r saying, lol...

[quote="boxedexe"]I believe some people here forgot teh dollar signs.[/quote]

Yeah, really....

First,we choose an interval $[a,b]$ containing $x=0$,then it is obvious that$|f''(x)|\leq C'|f'(x)|$,where $C'$ is a constant on $[a,b]$.So according to Gronwall's lemma,we have $f'(x)=0$ on $[a,b]$,that is, $f(x)$ is a constant.And then $f(x)=0$.Cover the real axis with such successive intervals and the desired result obtained.

I also dont know if the following is good or not but I requires the condition f'(0)=0!
Take abitrarily a and b>a.For all x in  [a,b] choose M as max{$ \frac{f(x)*f'(x)}{x}*\frac {(b \plus{} a)}{(b \minus{} a)}$}.So,we get:

$ |f''(x)|\leq M*x*\frac {(b \minus{} a)}{(b \plus{} a)}$.Now let's intergrate between a and b both two sides of the inequality we get:

$ |f'(a) \minus{} f'(b)|\leq M*(b \minus{} a)^2/2$ or $ |f'(a) \minus{} f'(b)/(a \minus{} b)|\leq M*(b \minus{} a)/2.$
Let b tend to a and by the defination of derivation we gain:$ f''(a) \equal{} 0$.Hence $ f'(a) \equal{} f'(0) \equal{} 0$ and $ f(a) \equal{} f(0) \equal{} 0$.

I still doubt on what I wrote in the above message.Any check???


geometry
function
rectangle
quadratics
parallelogram
real analysis
real analysis solved

You have a linear function $f(x)=-\frac{2}{3}x+4$ where $0\leq x\leq 6$ and $0\leq y\leq 4$. What is the maximum area of a rectangle that has one side on the line of the function?

I know how to optimize this, I am just having trouble finding the equation for area of such a rectangle in terms of the function.

I have included an image of what I am talking about. Thanks a lot.

Here's a slightly offbeat approach. The region that is the rectangle is the result of starting with a big triangle and deleting from it three nonoverlapping triangles. All of the triangles are similar, and similar to a right triangle with legs 2 and 3 and hypotenuse $\sqrt{13}.$

The large triangle has area $12.$

Such a triangle with longer leg $u$ has area $\frac13\,u^2.$

Such a triangle with hypotenuse $u$ has legs $\frac{2u}{\sqrt{13}}$ and $\frac{3u}{\sqrt{13}}$ and thus area $\frac3{13}\,u^2.$

Now suppose the point at which the rectangle intesects the $x$ axis is $(z,0).$ The three triangles we subtract from the big triangles have a longer leg of $z$, a hypotenuse of $6-z$ and a hypotenuse of $4-\frac{2z}3$.

[oops - original post was in error. Everything from here down, deleted, fixed in next post.]

Thus, the area of the rectangle is

$12-\frac13\,z^2-\frac3{13}\,(6-z)^2-\frac3{13}\,\left(4-\frac{2z}3\right)^2= 4z - \frac{2}{3}\,z^2$

This quadratic function can be maximized by precalculus means. The maximum occurs when $z=3$ and the maximum area is $6,$ which is half of the area of the triangle.

---

That was clumsier than it had to be. There has to be a nicer solution.

Here's an argument. I'll won't do the drawing myself - you'll have to make your own.

We have a right triangle $ABC$ with right angle at $C.$ We would like to fit a rectangle of maximal area with one side along $AB.$ Let this rectangle be $DEFG$ with $D$ in $AC,$ $E$ in $BC$ and $FG$ a subsegment of $AB.$

Shear the rectangle into a parallelogram by fixing segment $DE$ and sliding $FG$ along $AB$ until it becomes $F'A$. We now have parallelogram $DEF'A$ in which side $EF'$ is parallel to $AC.$ Since this was a parallel shear, the area of $DEF'A$ is the same as the area of $DEFG.$

Now shear this again, fixing segment $EF'$ and sliding $AD$ along $AC$ until it becomes $HC.$ We now have a rectangle $CEF'H$ which has the same area as the original rectangle $DEFG.$

So our original problem is equivalent to finding the rectangle of largest area with one corner at $C$ that fits inside the triangle. This is a much easier problem than the original problem, and leads readily to the condition that point $F'$ must be the midpoint of the hypotenuse $AB$ and that the area of the rectangle $CEF'H$ (or of $DEFG$) must be half of the area of triangle $ABC.$