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MATHCOUNTS
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group theory

[quote="gauss202"]I know that some people believes that questions asking for proofs should not be put on the Getting Started board, but I disagree. Everyone has to start somewhere with writing proofs, and these are good questions for beginners to try their hands at.

1. Prove that there is a rational number between any two rational numbers.

2. Prove that there is a rational number between any two real numbers.[/quote]

I'm willing to let this thread develop and see how participants fare in solving those problems.

(After edit:) I edited the thread title to make clear that this thread is a challenge problem rather than an expository post about proof.

Have at it, everybody who's getting started.

This isn't quite convincing for me - you're just showing that there is a rational number between p/q and (p+1)/q.  What about a situation such as 22/23 and 20/21 (just randomly chosen)?

That's a very good start, Rep -- [hide]but if I give you any two rational numbers, you can't necessarily write them as p/q and (p + 1)/q, right?  You've got a very good idea -- now try being slightly more genera[/hide]l.

[quote="Rep123max"]Im probably still wrong but maybe this will make it better:

Again, let one of the rational numbers=[tex]\frac{p}{q}[/tex] where p and q are integers. Then let n and z be two other integers creating the fraction [tex]\frac{p+n}{q+z}[/tex].

The average of [tex]\frac{p}{q}[/tex] and [tex]\frac{p+n}{q+z}[/tex]=
[tex]\frac{2pq+pz+qn}{2q^2+2qz}[/tex] which is also a rational number because as I said, the set of integers is closed under addition and multiplication so [tex]2pq+pz+qn[/tex] and [tex]2q^2+2qz[/tex] are integers making that fraction rational.

And since the average of two numbers is in between them, there is a rational number between two rational numbers.

I dont know if that made any sense.[/quote]

Here's what it should be (I just edited the [tex]\LaTeX[/tex], not the mathematical content). Feel free to copy+paste my code and then delete this post after you have fixed it.

Osiris wrote:Can't you [hide]let the rationals be a and b and argue that (a + b) / 2 is a rational[/hide]?

Yes, but that rather defeats the purpose. You'd first have to prove that (in spoiler to follow example of quote) [hide]if a and b are rationals, (a+b) is rational. Sure, you could argue that using group theory, but using group theory to solve this kind of problem? C'mon. That's just ridiculous.[/hide]

I must admit, however, #2 is a lot harder than it seems. I've never seen that problem before (which I find weird.)

I severely doubt that problem number 2 in this thread is at the Getting Started level, but I am willing to be proved wrong if someone can show me a reasonably rigorous proof that is quite elementary, and that is really within the reach of a beginner. Problem number 1 looks susceptible to such a proof.

It's a bad sign already that most of the people posting to this thread are people well beyond the Getting Started level. I'll "promote" the thread (maybe all the way to Advanced, this time) unless I see evidence that these problems are as easy as they were claimed to be. I sure wouldn't see a problem like this on an AMC 8, nor on any level of Mathcounts.

(sorry tokenadult)

1. Is the fact that rationals are closed under addition and multiplication THAT obscure a fact? I use it like it's common sense and noone seems to find fault with it.

Certainly someone (even a student of "getting started") level would -expect- that an addition of two rationals yields a third rational.

2. That is certainly beyond getting started material...

rep123max's proof for #1 looks fine to me. Does someone else see a problem with it? Ok, maybe you should prove that the sum, product, and quotient of two rationals is also rational, but that's not hard. (Group theory is DEFINITELY not required.) There is also an elementary proof for #2 which involves a clever idea but should be accessible to getting started people who think about it very hard. (There are surely many others, but the one I am thinking of should be accessible.)

[quote="Osiris"]1. Is the fact that rationals are closed under addition and multiplication THAT obscure a fact? I use it like it's common sense and noone seems to find fault with it.[/quote]

Yale Professor of Mathematics Serge Lang has written a very interesting textbook for students in elite prep schools called Basic Mathematics (Springer-Verlag), in which he takes the field properties of the real numbers and uses those as local axioms to set a number of problems along the lines of "Prove . . . " in which the proposition to be proved is some well-known fact of secondary mathematics. Every math teacher who wants to improve his teaching should read this book at his earliest opportunity.

BUT, the contests that I take as indisputably Getting Started level contests are contests based on middle school early secondary mathematics as that is known to public school students in the United States. Even Lang's admirable attempt to introduce proof to teenagers is more advanced both in curriculum content and in approach than are typical Mathcounts problems.

[quote="Osiris"]Certainly someone (even a student of "getting started") level would -expect- that an addition of two rationals yields a third rational.[/quote]

I'm sure there are participants on the AoPS site who are vague about what a rational number is. [url=http://math.berkeley.edu/~wu]Adding fractions is actually a notoriously poorly taught subject[/url] and while I don't think the closure under addition over the rationals is contrary to anyone's intuition, I don't think many middle schoolers can marshal their intuition to form even a heuristic proof of Problem 1--although I'm still willing to be proved wrong on that point, "by construction," if some middle schooler shows me that he or she can do it.

[quote="Osiris"]2. That is certainly beyond getting started material...[/quote]

Because I have recently read complaints that Intermediate forum is also too hard for most would-be participants who are really at that level, and because I have seen the site owner express the same concern, I will probably promote this thread to Advanced pretty soon. Meanwhile, the middle-school-age beginners are welcome to post what they can post here.

I'm also thinking of a solution for #2 which I would think is within Getting Started people. (Although I don't really know what kind of level 'Getting Started' is; we don't have 'middle school' or AIME or any of that stuff in New Zealand...)

[quote="tokenadult"]I will probably promote this thread to Advanced pretty soon[/quote]

Why in the world would you do that?  These two proof questions are nowhere near "Advanced"!

rep123max has already provided a fine proof of part 1.  [Nobody asked him to prove any subtheorems like closure of various operations, and there is no need to do so.]

I agree with other posters that there are proofs to part 2 that are well within the grasp of some middle school students.

And finally, I personally think it is a good idea to get younger students exposed to simple proofs.

Please give these kids a chance.  This post has only been up for 6 hours.

Agreed. This would be far too easy for the intermediate forum, let alone the advanced forum! If you are unable to prove something like this, then you will be unable to do any single problem about number theory, as these are very basic results. Please leave it here.

I woke up this morning and right away thought of a "proof" of the proposition that there is a rational number between any two real numbers. I think it will satisfy the standard of rigor expected here, and it definitely only requires a "fact" that should be known to any seventh grader.

I'm probably the oldest participant in this thread, but also one of the least experienced in mathematical problem-solving, so I'll post an answer to this one in a while if someone else doesn't jump in first (or second, or third). Give it some thought, and try to post your reasoning process in spoiler here.

I don't think it's fair to say that this is too easy for Intermediate -- I think it's fair to say that it's easier than many of the problems in Intermediate, but I think it's a legitimate hard getting started problem or easy intermediate problem.

That said, yes, there is a nice way of proving it for real numbers (probably the same way everyone else has gotten it, just took me longer to find :)).  Interesting how different it seems, actually.

2.

[hide]

We can write a real number in decimal form to any desired degree of accuracy.  Let the two numbers be a,b, with a > b

Since the decimal representation of these two numbers must differ at some point (because they are distinct), say that they diverge at the nth decimal place.  Then truncate the decimals at the n+1th place.

Let d1 and d2 be the two decimal representations.

Then d1 < a < d2 < b

So d2 is the desired number.  But d2 is a decimal representation, and by multiplying d2 by (10^(n+1))/(10^(n+1)), we obtain the desired rational number.

[/hide]

[quote="Rep123max"]Hmmm, on number 2, really all I need to do is prove there is a rational number inbetween to irrational numbers (or a rational number between an irrational number and rational number) since I already proved there was a rational number between to rational numbers. Its harder then the other one. Im gonna have to think about it.[/quote]

Unfortunately, the irrationals do not have a general form.

[hide]Alternate 2:

Lemma: There is at least one rational number between 0 and any number.

WLOG we need to only prove it for positive numbers, because if x is rational, then -x is also rational.

Suppose there is an x>0 s.t. there is no rational number between 0 and x, and x is the smallest number for which this is true.  However, if this is true, then there is also no rational number between 0 and x/2, because x/2<x.  Therefore, we have found a smaller number, and there is a contradiction.

Suppose that a and b are a pair for which this trait exists.

Case 1: a is negative, b is positive.

There is a rational between 0 and b by the lemma.

Case 2: a and b are both positive.  Assume a<b and that a and b are the smallest pair for which this trait is true.  Because a and b are the smallest pair, there is a rational between a/2 and b/2, which we'll call m.  Because a/2<m<b/2, a<2m<b.  2m must be rational, because if m can be expressed as r/t so 2m can be expressed as 2r/t.

Case 3: a and b are both negative.  There is a rational m between |a| and |b| by case 2.  -m is also rational.[/hide]

Lots of words, but it's all pretty simple.

[quote="JBL"]I don't think it's fair to say that this is too easy for Intermediate -- I think it's fair to say that it's easier than many of the problems in Intermediate, but I think it's a legitimate hard getting started problem or easy intermediate problem.[/quote]

The post by rcv convinced me that Intermediate is about the right level for this pair of problems, which is the forum I am about to send this thread to.

I am ruthlessly empirical about judging the level of challenge problems. I checked the profiles of the people who posted in this thread who weren't already familiar to me, and they are all people ready AT LEAST to solve problems in Intermediate, as most of them have done before. It amazes me how readily experienced problem-solvers forget what is easy, and what is hard. Just because a problem only has the math curriculum prerequisites required of an eighth-grader doesn't make it easy--otherwise, everyone who has ever taken a precalculus course would be able to qualify for IMO.

I'm gonna be really eager to send problems up to Intermediate until

a) I no longer see complaints that Intermediate has too many hard problems

and

b) I see lots of REAL BEGINNERS (e.g., BHorseMath) posting solutions to challenge problems here on Getting Started.

There will always be a range of difficulty on each of the three forums, but the boundaries haven't been set at the right level for at least the bottom two for a while. I will try to do my part by posting more easy problems here and by resolving all doubts in favor of posting challenge problems to the HIGHER of the two forums to which they might belong.

Mangara, I regret to inform you that your proof doesn't work as stated.  It's actually a simple reason it fails: what is the smallest number greater than 0?

(Anyone at MOP 2002 might remember a question about 2-coloring a plane which some people tried to solve by taking the point nearest to a given point.)

Ok, here's the proof I came up with for 2:

[hide]Let the two numbers be a and b with a<b. Assume they're both positive. Then since b-a>0, we may choose an integer n large enough so that n(b-a)>1. Let m be the ceiling of na. We then have

0<m-na<1<n(b-a),

i.e.

a<m/n<b

as desired.[/hide]

Fine.

[hide]Assume that there is a largest value >0 for which this is true.  We know that there must be one such value because there is a rational number between 0 and, oh say, 1.  So the value must be less than 1.  We'll call this largest value x.  By definition, there is a rational number 0<x<m/n<2x, because x is the largest value that exhibits that property.  However, upon dividing by two, we find 0<x/2<m/(2n)<x and we have found a rational number.[/hide]

Happy?

I also had a similar proof for #1 as rep123. OK for #2 I have a good idea that this may work, however i'm not sure how rigorous it will be though

[hide] OK, let the two irrational numbers be denoted as p and q where p<q

To show that this works for any irrational numbers let the two numbers be written out to the nth decimal point. At this point is when the numbers vary, for example in number q, the value of the nth decimal point would be 5 and in p it would be 4.

If we eliminate all decimal points past n from number q, this number would be greater than p, because it varies in the nth place point, but less than q. I'll call this number y

So a<y<b

Because y stops repeating after n places, it is therefore rational. Does this work? I had to do a similar proof for another problem, I hope this applies as well.

[/hide]

Mangara -- I don't think any proof that relies on a largest or smallest number in a set of real numbers will be successful without calling on higher mathematics and more technical and complicated relationships between the rational and real numbers to prove that such a number exists.  For example, assume (falsely) that 1/2 was the smallest rational number.  Then the set of numbers for which there is not a rational between them and 0 would be (0, 1/2), a set of numbers without a largest or smallest term.

Yes, Magnara, your second proof (unlike your first) can be fixed to make it a valid proof, but first you have to stop cavalierly assuming that a set of real numbers (or even a bounded set of real numbers) has a smallest or largest element.  A nonempty, bounded set of real numbers has an infimum and supremum, but that's not the same as a smallest and largest element.

[quote]your second proof (unlike your first) can be fixed to make it a valid proof[/quote]

How?

Well, he might [hide]pick any number x with that property.  Consider its multiples x, 2x, 3x, ....  Eventually, one of them is bigger than 1.  Proceed from there.[/hide]

[quote="Ravi B"]Yes, Joel, that works.  Here's what I had in mind to fix Magnara's proof.  Let S be the set of positive real numbers x such that there is no rational number between 0 and x.  Assume, for contradiction's sake, that S is not empty.  Let x0 be the supremum of S.  Then argue as Magnara did to reach a contradiction.

That's not how I would have proved the result, but at least it fits Magnara's theme.[/quote]

How is that different from my proof?  Except that it's fancier of course.

Because your proof assumed that S had a largest element.  My revision of your proof did not.

I must have been mistaken in reading it.  What's a supremum?

Supremum is the same as least upper bound.  See [url]http://mathworld.wolfram.com/Supremum.html[/url] for a definition.  It's different from the maximum of a set.

To use Joel's example, let S be the set of real numbers bigger than 0 and less than 1/2.  The supremum of S is 1/2.  But S has no largest element, largest value, or maximum.

Every nonempty set of real numbers that is bounded above has a supremum.  Depending on your definition of real numbers, that fact is either an axiom or a theorem.


topology
superior algebra
superior algebra unsolved

A point P in a topology for a set $ X$ is called dense if P is contained in every non-empty open set of the topology. Alternatively, the closure of $ \{P\}$ equals $ X$.

Find and prove a necessary and sufficient condition for $ \text{Spec}(R)$ to have a dense point. The condition should related to the nilradical. How many dense points can $ \text{Spec}(R)$ have?

I am not seeing how to approach this problem right now. Any helpful hints will be very greatly appreciated. Thank you.

The equivalent condition is nilradical is prime.

The reason is V(p) = closure of {P} = X = V(nil R)
iff $ p \equal{} \mathrm{nil}R$, i.e. $ \mathrm{nil}R$ is prime.

There can only be one dense point, because any dense point must be the nilradical.


Given: angle 1 is congruent to angle 4, angle ABC and angle CDA are right angles.

INSTRUCTION:
draw a rectangle and label A on the bottom left, B on the upper left, C on the upper right, and D on bottom right.
then draw a diagonal line inside the rectangle from B to D. then label angle 1 below the diagonal line on the top. Label 2 above angle 1 on the top. then label 3 on the bottom below. Label 4 on the bottom above angle 3

a. write a congruence statement for the triangles in the diagram.

b. write a two-column proof of your statement in part (a)

Thank You for helping me.

P.S. The question does not say what needs to be proved.

Ah, sharkman has graduated from Algebra II and now wants us to do his Geometry homework.

[quote="sharkman"]P.S. The question does not say what needs to be proved.[/quote]

Then why post anything?


a)How many odd six digit palindromic numbers are there?
b)How many odd seven digit palindromic numbers are there in which every digit appears at most twice?

Could someone give me the correct answer for both of them, because for
a) i got 205 numbers and for
b) i got 63 numbers

much appreciated!

If I understood the question correctly...

a) we look for numbers ABCCBA with A odd. So we have 5 possibilities for A (1,3,5,7,9), and 10 possibilities for B and C. That makes $5\cdot10\cdot10 = 500$ numbers.

b) we look for numbers ABCDCBA with A odd. For A we have 5 possibilities. B can be every digit except A, so we have 9 possibilities for B. C can be every digit exept A and B, so we have 8 possibilities. And for D we have 7 possbilities. That makes $5\cdot9\cdot8\cdot7 = 2520$ numbers.

[quote="infinity"]a)How many odd six digit palindromic numbers are there?
b)How many odd seven digit palindromic numbers are there in which every digit appears at most twice?[/quote]

a) First digit must be odd, so 5 choices. 2nd and third have 10 choices for both, so $5 \cdot 10^2=500$

b) Are you sure that's right, Stijn?? :huh:

[quote="chess64"] Are you sure that's right, Stijn?? :huh:[/quote]

Why do you think it's wrong?

[quote="chess64"]b) Are you sure that's right, Stijn?? :huh:[/quote]Yes chess, we are sure. :yup:

Maybe if you or infinity told us your reasoning to get the number, we could see the mistake more easily...

Sorry, I didn't think about it hard enough. :D


AMC
USA(J)MO
USAMO
MATHCOUNTS
geometry
geometric transformation
reflection

For many that I know, the negative impact on performance arising from "stupid mistakes" is devastating. I've known people who have made these careless mistakes on over half the problems that they do. Even smart people can test significantly below average due to this one fatal flaw.

For me, this issue is pretty serious, since I make quite a lot of these mistakes. I'm curious as to how you guys are affected by this, as well as the cause for stupid mistakes. Sometimes, stupid mistakes are the result of a lack of true understanding of a topic or a "Freudian slip" in the other direction (misreading a word for another) - but there are times when you just simply are dumb and make computational errors or think extremely illogically without realizing it.

Is there any purpose for these temporary lapses in the brain? Could it be from lack of sleep? I'm not sure, but it's interesting.

14 points of careless errors at Nats.

Next thing I know, I'll make 14 points of careless errors at the USAMO (provided I make it there)...

I work very slowly and carefully and often double check my work right after doing a computation.  I still make stupid mistakes (everyone does!) but not too often.

But because of my slow speed I suck at things like Mathcounts.

eryaman brought up another related (in my opinion) topic - speed. Often associated with accuracy, and therefore the frequency of stupid mistakes, speed is nevertheless highly important in many competitions such as MATHCOUNTS.

I would consider speed, accuracy, and depth to be the three primary factors for success in most math competitions. Depth would be a person's level of understanding, thinking, and logical reasoning (a lot like what IQ tests try to measure). Again many interesting questions arise as there are people who are lightning quick and make hardly any stupid mistakes, but cannot understand difficult concepts and extrapolate from ideas, or some combination of the 8 different "types", if you by my tri-factor idea.

Are you someone who is fast? Do you usually finish things in class (for example) before everyone else? Are you average in terms of speed? Or do you usually finish things after most people have finished?

[quote="banblink182"]
Are you someone who is fast? Do you usually finish things in class (for example) before everyone else? Are you average in terms of speed? Or do you usually finish things after most people have finished?[/quote]
Yes. I also make stupid mistake a lot. I think the two are unconciously correlated, me getting my work done faster because I have no unconcious checking going on.

It's very interesting how I always miss 7 or 8 problems on every single Mathcounts Sprint that I do because of stupid mistakes. No more, no less.

Yeah... stupid mistakes, ergo lack of sufficient practice. I write legibly and acceptably (most of the time), but when I'm really tired I can't do arithmetic properly. Once I even missed a problem because I wrote AM instead of PM.

I'm also pretty slow at these types of things...

[quote="joml88"]I don't agree with blaming bad performances on "stupid mistakes".  Yes, you missed a problem because of something easy (like arithemic) but ultimately it is a reflection on you.  I think organization plays a huge role.  A lot of the sources of these careless errors will be diminished if you write neater, organize your work, and organize your thoughts more efficiently.  Correctly interpreting a word problem is part of the difficulty of the problem.[/quote]

I agree, but I still think that there are some errors that are not related to anything that can be corrected - mental arithmetic errors, for example.

Yes, correctly interpreting a word problem is certainly part of the difficulty of the problem; I was referring to more of a mistake made in terms of solving the problem that you would have known to be a mistake, not where someone misinterprets a word problem and tries to solve a different problem altogether, and thinks that he/she is doing the right thing.

[quote="mathnerd314"]Once I even missed a problem because I wrote AM instead of PM.[/quote]

Yea, that's not really lack of sufficient practice as just lack of concentration/fatigue/etc.

These are the things that I was referring to, not anything that is conciously done with a purpose. Reading or writing AM instead of PM is an example - if someone pointed it out to you, you would have immediately recognized it and changed it.

I used to make a LOT of stupid mistakes - where I'd do a problem, and, (for example) copy it wrong into the answer box (that stupid even, yeah). I think it has to do a lot w/ speed, although sometimes people make mistakes because they are tired - even if they are doing the test slowly.

Usually I finish math tests a bit before most of the class, but lately I've been getting better at  well.. not making mistakes.

[quote="ToBe"]I used to make a LOT of stupid mistakes - where I'd do a problem, and, (for example) copy it wrong into the answer box (that stupid even, yeah).[/quote]

I've done that before as well  ;)

I occasionally make mistakes, but it doesn't really affect me on fast-paced competitions, because I have a good speed:accuracy ratio at high speeds, but at low speeds the ratio sucks, because I will usually make the same amount of mistakes whether I work slowly or quickly. The double check is the key to catching mistakes, not working slowly in my case.

[quote="banblink182"]
I agree, but I still think that there are some errors that are not related to anything that can be corrected - mental arithmetic errors, for example.[/quote]

I believe that anything can be corrected, or at least improved upon to the point that the stupid errors won't occur often.  If you cannot do mental arithmetic without having a high risk of messing up, then A. you need to practice mental arithmetic more if you wish to do calculations in your head or B. you might want to use paper to carry out the computations.  Arithmetic mistakes will still occur but they wont' be as often.

If anything, instead of blaming bad performances on tests on "stupid mistakes" I think that laziness is really the cause.  You may disagree with me, and I'm fine with that.  I just think that 99% of stupid mistakes can be eliminated by maybe working slower, double/triple checking, learning to focus better, getting adequate sleep, et cetera.

Blah, I'm insanely fast, but I make way too many careless mistakes...

At States, I almost always answer before my opponent has even finished reading the question, but... 1/3 the time it's wrong xD .

[quote="TZF"]
[i]I define a stupid mistake as an error in a direct addition, subtraction, multiplication, or division computation that is independent of the context or meaning of the particular problem.[/i]
[/quote]
Agreed.But I would add another point - [i]misreading of a statement[/i]. Due to that mistake, I messed a very important olympiad up...didn't even manage to made National Olympiad...in fact, I messed two problems up that time. :roll:
btw, passing score was 49 out of 56...

I make stupid mistakes under the following circumstances:

1. The topic is boring but easy.
2. Time constraint so I have to work fast.
3. What on Earth is this? I cannot begin to comprehend the question!
4. Sometimes I read something and think something completely different.
e.g. Now find the vertex of parabola blah, blah and blah and I end up find the roots :P
5. Wandering mind. e.g. worrying about runny nose, headache, (the girl in front of me has nice hair?), what's for dinner? etc.

Well I would say basically when I am not fully focussed and double reading everything (usually I don't need to double check my work, though I still do it).

I make stupid mistakes on tests, important tests, and really important tests.

Not on HW though.

Yes, those mistakes cost a lot .... :( so i have this question : can we prevent those ' stupide mistakes' ? is there a very good way to do that ?...........am just wondering :wink:

I like your selfconfidence , coz some times we can think that we are just 'too stupid" to make such mistakes...any way,i have only to mention that : very intelligent doesn't mean too fast,it only means that you can make things very simple  :)

[quote="gauss1181"][quote="aleph0"]I make stupid mistakes all the time. And I'm not fast either.[/quote]
This applies to me on Mathcounts CD round and AOPS FTW. :D
(lol, I'm not happy about that, just being sarcastic because of our avatars)[/quote]
I am SLOW on FTW.

lol, what is your rating? Do you ever make stupid mistakes eggy?

/Me has a rating of 1344 :D  :D  :D

[quote="eggylv999"]1197. :(
Reasons:
1. I tense up when I see the time.
2. I make a stupid arithmetic error.
3. I grab my calculator and make a typo.
4. I play against pwnage people.[/quote]
Same with me. Except I'm 1397.
I noticed poll is in a teardrop shape. A comment on mistakes?

[quote="eggylv999"]1197. :(
Reasons:
1. I tense up when I see the time.
2. I make a stupid arithmetic error.
3. I grab my calculator and make a typo.
4. I play against pwnage people.[/quote]

Cough cough [size=59]NO CALCS ON FTW!!![/size] cough cough.

oh and 1680 :D

[quote="TPATPATPATPATPATPAM"]I make stupid mistakes, but I'm also incredibly brilliant, so it about evens out.
I usually do things like reversing the signs, skipping a step, missing questions, not reading questions completely, thinking I've done something but I haven't, etc. etc. Mostly it's carelessness, not mind lapses.[/quote]

Same here.

[quote="ernie"][quote="eggylv999"]1197. :(
Reasons:
1. I tense up when I see the time.
2. I make a stupid arithmetic error.
3. I grab my calculator and make a typo.
4. I play against pwnage people.[/quote]

Cough cough [size=200]NO CALCS ON FTW!!![/size] cough cough.

oh and 1680 :D

[quote="TPATPATPATPATPATPAM"]I make stupid mistakes, but I'm also incredibly brilliant, so it about evens out.
I usually do things like reversing the signs, skipping a step, missing questions, not reading questions completely, thinking I've done something but I haven't, etc. etc. Mostly it's carelessness, not mind lapses.[/quote]

Same here.[/quote]

I know. I make typos, anyway.

I am really bad at double-checking on tests, because usually if I get something wrong once, I'll get it wrong the second time, whether its a stupid mistake or lack of understanding.

This thread reminds me of a time in 7th grade Geometry, when we had a group test, and the teacher said that I had to let my groupmates copy off of me, and she would get mad if they didn't have the same answers as me.

During the test, I let them copy, and they copied my answers wrong. !!!

And then the teacher still got mad at me for a while.

And I hate calculators.


I try to show that:
Suppose $ |t|\leq 1$ fixed and $ x$ real variable.
${ e^{ - \frac {x^2}{2}}*[e^{tx} + e^{ - tx}]}$ attains maximum when x=0.
Or in other words,
${ e^{ - \frac {x^2}{2}}*[e^{tx} + e^{ - tx}]}\leq 2$
This came up from my computations and there is a 5% chance it's not correct.

[hide]By power series,
\[ \frac{e^{tx} \plus{} e^{\minus{}tx}}{2} \equal{} \sum_{n \equal{} 0}^\infty \frac{t^{2n} x^{2n}}{(2n)!},\]
and
\[ e^{x^2/2} \equal{} \sum_{n \equal{} 0}^\infty \frac{(x^2/2)^n}{n!} \equal{} \sum_{n \equal{} 0}^\infty \frac{x^{2n}}{2^n n!}.\]

Since $ |t| \le 1$, it suffices to show that
\[ \frac{1}{(2n)!} \le \frac{1}{2^n n!}\]
for all positive integers $ n$.  But this is equivalent to
\[ 2^n \le \frac{(2n)!}{n!} \equal{} (n \plus{} 1)(n \plus{} 2) \dotsm (2n),\]
which is obvious.

Therefore,
\[ \frac{e^{tx} \plus{} e^{\minus{}tx}}{2} \le e^{x^2/2} \quad \Rightarrow \quad e^{\minus{}x^2/2} (e^{tx} \plus{} e^{\minus{}tx}) \le 2.\][/hide]


trigonometry
inequalities unsolved
inequalities

Given that $ x,y,z \in \mathbb{R}^{\plus{}}$ and $ x \plus{} y \plus{} z \equal{} 1$, determine using trigonometry the minimum value of $ \frac {1}{x} \plus{} \frac {4}{y} \plus{} \frac {9}{z}$.

[quote="knol"]Given that $ x,y,z \in \mathbb{N}$ and $ x \plus{} y \plus{} z \equal{} 1$, determine using trigonometry the minimum value of $ \frac {1}{x} \plus{} \frac {4}{y} \plus{} \frac {9}{z}$.[/quote]

$ x,y,z \in \mathbb{N}$ and $ x \plus{} y \plus{} z \equal{} 1$ give $ (x,y,z)\equal{}(0,0,1)$  :huh: ......

knol, please edit the problem.
That doesn't make sense. :(

Thank you for your quick reply.

By the way, Cauchy-Schwarz can solve the problem easily, why do we have to solve with trigonometry?

The question appeared in a trigonometry book so I guess you have to use it. I am aware of a Cauchy-Schwarz solution, though.

by  Cauchy-Schwarz its very easier.
isn't?

amparvardi, check other's post before you post your post.


4th place. :lol: Going to 2 nationals now.

Billy

Congrats!  Please come introduce yourself in Detroit - I'll be there, too!

good job. do us well and go show them how good we are....still embarrassed about my performance at mc...

[quote="solafidefarms"]Ah, well. SO MUCH FUN!

Xantos C Guin got 16th. My other team members got something like a 23 score, and then 130th. I was 150th. I was the 3rd lowest AoPS score.

Cutiepi, Piggypi, and Hexahedron were in the countdown round. Xantos was in the tiebreaker and lost.[/quote]

That's ok. You did well. and your the 1st one on the team to discover this forum.
BTW: A person who gets 6 target will probably lose the tiebreaker since it is determined by the Sprint round. :(


I would be grateful if some of u could clarify these querries:

1)  Is it possible to apply for MS in say CS thru GRE after bsc from cmi or bstat/bmath from isi.

2)  Is it possible to give gate after bsc frm cmi or bstat/bmath frm isi.

3)  What r some good books for sat preparation?

[quote="bajpaipreetish"]
3)  What r some good books for sat preparation?[/quote]
It depends on your aim and potential. How much are you expecting?
In any case, one book that always helps is the Official collegeboard book(Read the explanations for even your correct answers)
Online resource- sparknotes.com
Rest will depend on your answer.


Let $ N\equal{}\overline{a_{1}a_{2}...a_{k}}$ is a $ k$ digits number
Find minimum positive integers $ N$ such that $ 2009\times \overline{a_{4}a_{5}...a_{k}}\equal{}\overline{a_{1}a_{2}...a_{k}}$

[quote="bambang"]Let $ N \equal{} \overline{a_{1}a_{2}...a_{k}}$ is a $ k$ digits number
Find minimum positive integers $ N$ such that $ 2009\times \overline{a_{4}a_{5}...a_{k}} \equal{} \overline{a_{1}a_{2}...a_{k}}$[/quote]

Let $ p\equal{}k\minus{}3$, $ x\equal{}\overline{a_{4}a_{5}...a_{k}}$ and $ y\equal{}\overline{a_{1}a_{2}a_{3}}$. The properties are :

$ x<10^p$, $ 100\leq y<1000$ and $ 2009x\equal{}x\plus{}10^py$.

So $ 2008x\equal{}8\times 251x\equal{}10^py$. So $ y\equal{}251z$ and $ y\equal{}251$, $ y\equal{}501$ or $ y\equal{}753$ (since $ 100\leq y<1000$).

$ y\equal{}251$ implies $ 8x\equal{}10^p$ And so $ p\geq 3$ and $ x\equal{}5^p2^{p\minus{}3}$
$ y\equal{}502$ implies $ 8x\equal{}2\times 10^p$ And so $ p\geq 2$ and $ x\equal{}5^p2^{p\minus{}2}$
$ y\equal{}753$ implies $ 8x\equal{}3\times 10^p$ And so $ p\geq 3$ and $ x\equal{}3\times 5^p2^{p\minus{}3}$

So all the solutions of the equation are :

$ N\equal{}2009\times 5^{k\minus{}3}2^{k\minus{}6}$ for any $ k\geq 6$
$ N\equal{}2009\times 3\times 5^{k\minus{}3}2^{k\minus{}6}$ for any $ k\geq 6$
$ N\equal{}2009\times 5^{k\minus{}3}2^{k\minus{}5}$ for any $ k\geq 5$

For $ k\equal{}5$, the only N is $ N\equal{}2009\times 5^{2}\equal{}50225$
For any other $ k\geq 6$, the minimum $ N$ is $ N\equal{}2009\times 5^{k\minus{}3}2^{k\minus{}6}$


[img]http://srishivananda.com/sumit/untitled.JPG[/img]
[hide="The answer"]
$ \frac{11}{18}$
[/hide]

Well, to get to F, it can either go to B, C, or D. The probability of going from A to B to F is 1/3 x 1/2 =1/6. The probability of going from A to C to F is 1/3 x 1=1/3. Finally the probability of going from A to D to F is 1/3 x 1/3=1/9. Add all those up because it is A to B to F, [b]or[/b] A to C to F [b]or[/b] A to D to F, you get 1/6 +1/9+1/3 to get
  [hide="ANSWER"]11/18[/hide]

What some people do is that they don't realize that your choices of where you go have weighted probability.

Some people would think that since there are 3 ways to get to F and 9 ways total, the answer would be 1/3.  Always consider this when doing one of these problems.  They usually appear on old State Targets.


$f(x)=-2e^{\frac 13 x}$

Now use the existence/uniqueness theorems to conclude that this is the only such solution.

[quote="tranthanhnam"]find $f(x) $ satisfying :
                     $f(x)=3f'(x)$ and $f(0)=-2$ .[/quote]

\[\boxed{f'(x)=kf(x)\Longleftrightarrow f(x)=Ce^{kx}}\]

proof: $f'(x)=kf(x)\Longleftrightarrow e^{-kx}f'(x)-ke^{-kx}f(x)=0$

$\Longleftrightarrow \{e^{-kx}f(x)\}^\prime=0$

$\Longleftrightarrow  e^{-kx}f(x)=C$

Hence $f(x)=Ce^{kx}$

The answer for your problem is $f(x)=-2e^{\frac{x}{3}}$

[b]Practice:Solve the following differential equation.[/b]

[1] $f'(x)=f(x)+1$

[2] $f'(x)=2f(x)+1$

[quote="shyong"]Kunny , shouldnt it be $f(x)=Ce^{\frac{1}{k}x}$ instead ? :?[/quote]

Sorry for that I made a typo. :blush:  I have just edited.

kunny

[quote="tranthanhnam"]find $f(x) $ satisfying :
                     $f(x)=3f'(x)\forall x\in R$ and $f(0)=-2$ .[/quote]
[i]tranthanhnam[/i], stop spamming!!! You don't need 3 posts to post 1 problem. Use the edit function if you want to edit your post!!
 :spam:

To [i]kunny[/i], you still have some more typo errors in ur post, but the idea is there.  :P

Thank you for pointing out my mistake,dblues. :)

kunny


Given $ x_1,x_2...x_n$ are positive reals and $ \sum x_i \equal{} \sum\frac {1}{x_i}$
Prove $ \sum\frac {1}{n \minus{} 1 \plus{} x_i} \le 1$


ive already done that, but i need help with the vectors lol

I'm guessing you were trying to put this with the 'Geometry Proof' thread.  Eh,  I wish moderators could merge topics so I could fix this.

Well, if you read this, please  make your post again on that thread so I can delete this one.


function
calculus
derivative
inequalities unsolved
inequalities

let $x, y \in [0;\sqrt{2}/2]$
       find the maximal of expression $P= \frac{x}{1+y^2} +\frac{y}{1+x^2}$

[url=http://http://mathnfriend.net/index.php?act=UserCP&CODE=24]web about math[/url]

[hide]

Fix , and consider the expression as a function of . The second derivative is

So the function is convex in . Same process for . It's convex in both variables, so it attains it's maximum at one or both of the endpoints, and is less elsewhere. It's enough to test the four possible cases, and we can see the  yields the maximum value of .

[/hide]


1) Find all polynomials $p(x)$ that satisfy
\[(p(x))^2-2=2p(2x^2-1), \quad \forall x \in \mathbb R.\]
2) Find all polynomials $p(x)$ that satisfy
\[(p(x))^2-1=4p(x^2-4X+1), \quad \forall x \in \mathbb R.\]

is there a mathlinker that tink on this problem?

[quote="mhmhm"]Who can solve this problem?[/quote]

Ask MJ GEO to post his solution. Since he posted in "proposed and own" category, he has one. And if he's not too busy, he will give it to you if you ask kindly.

[quote="mhmhm"]Who can solve this problem?[/quote]

if $ p(x)$ polyminal $ n$ th degree then $ p(x)\equal{}a_nx^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}...\plus{}a_1x\plus{}a_0$
then yours is a quite toy with equating coefficient!

[quote="MJ GEO"]1)find all $ p(x)$ that satisfify $ (p(x))^2 \minus{} 2 \equal{} 2p(2x^2 \minus{} 1)$[/quote]
[u]@all [/u]: MJ GEO asks a lot for solutions and never accepts to post his own solutions ... So here is mine for the first of these two problems.

[u]@voong [/u]: You have quite interesting toys in your country  :wink: . I tried your hint but found ugly heavy systems of equation and did not succeed. Could you, please, give us your full solution, up to the end (not just "is a quite toy with equating coefficients") ?. Thanks in advance.

My solution :
Since $ 1\plus{}\sqrt 3$ and $ 1\minus{}\sqrt 3$ are solutions of $ x^2\minus{}2\equal{}2x$, $ P(x)\equal{}1\plus{}\sqrt 3$ and $ P(x)\equal{}1\minus{}\sqrt 3$ are solutions.
Suppose now there exists a solution with degree $ \ge 1$
Let $ P(x)$ such a solution with the fewest positive degree.

Obviously $ P(x)^2\equal{}P(\minus{}x)^2$ and so (since polynomials), either $ P(x)$ is odd, either $ P(x)$ is even

1) if $ P(x)$ is odd :
$ P(x)\equal{}xQ(x^2)$ and the equation becomes : $ x^2Q(x^2)^2\minus{}2\equal{}2(2x^2\minus{}1)Q((2x^2\minus{}1)^2)$
Setting $ x\equal{}0$ in this equality, we get $ Q(1)\equal{}1$
Setting $ x\equal{}1$ in this equality, we get $ Q(1)^2\minus{}2\equal{}2Q(1)$
Hence contradiction and $ P(x)$ is not odd.

2) If $ P(x)$ is even :
$ P(x)\equal{}Q(x^2)$ and the equation becomes : $ Q(x^2)^2\minus{}2\equal{}2Q((2x^2\minus{}1)^2)$
So $ Q(x)^2\minus{}2\equal{}2Q((2x\minus{}1)^2)$ (remember these are polynomials).

Let then $ R(x)\equal{}Q(\frac{x\plus{}1}2)$ and so $ Q(x)\equal{}R(2x\minus{}1)$. The equation becomes $ R(2x\minus{}1)^2\minus{}2\equal{}2R(2(2x\minus{}1)^2\minus{}1)$ and so $ R(x)^2\minus{}2\equal{}2R(2x^2\minus{}1)$

And so $ R(x)$ is a solution too, but degree($ R$)=degree($ Q$)=$ \frac 12$degree($ P$), hence a contradiction since $ P(x)$ was supposed to be with the fewest positive degree.

Hence the two unique solutions :

$ \boxed{P(x)\equal{}1\plus{}\sqrt 3}$ and $ \boxed{P(x)\equal{}1\minus{}\sqrt 3}$

[quote="MJ GEO"]2)find all polynomial that $ (p(x))^2 \minus{} 1 \equal{} 4p(x^2 \minus{} 4X \plus{} 1)$[/quote]
For this one, I found a solution nearly similar (see previous post), except for elimination of the case "odd", more complex :

My solution :
Since $ 2\plus{}\sqrt 5$ and $ 2\minus{}\sqrt 5$ are solutions of $ x^2\minus{}1\equal{}4x$, $ P(x)\equal{}2\minus{}\sqrt 5$ and $ P(x)\equal{}2\plus{}\sqrt 5$ are solutions.
Suppose now there exists a solution with degree $ \ge 1$
Let $ P(x)$ such a solution with the fewest positive degree.

We can write the equation $ P(x)^2\minus{}1\equal{}4P((x\minus{}2)^2\minus{}3)$, and so $ P(x\plus{}2)^2\minus{}1\equal{}4P(x^2\minus{}3)$ and so $ P_1(x)^2\minus{}1\equal{}4P_1(x^2\minus{}5)$ where $ P_1(x)\equal{}P(x\plus{}2)$

Obviously $ P_1(x)^2\equal{}P_1(\minus{}x)^2$ and so (since polynomials), either $ P_1(x)$ is odd, either $ P_1(x)$ is even

1) if $ P_1(x)$ is odd :
Let $ x\in[\minus{}5,\plus{}5]$ : $ P_1(\sqrt{x\plus{}5})^2\minus{}1\equal{}4P_1(x)$ and so $ P_1(x)\ge \minus{}1$
Since $ P_1$ is odd, we also get $ P_1(x)\le 1$

So $ \forall x\in[\minus{}5,\plus{}5]$, $ \minus{}1\le P_1(x)\le \plus{}1$ and so $ P_1(x)^2\le 1$ and so $ P_1(x)^2\minus{}1\equal{}P_1(x^2\minus{}5)\le 0$

And so $ P_1(x)\le 0$ $ \forall x\in[\minus{}5,20]$ but this implies, since $ P_1(x)$ is odd : $ P_1(x)\equal{}0$, impossible
Hence $ P(x)$ is not odd.

2) If $ P_1(x)$ is even :
$ P_1(x)\equal{}Q(x^2)$ and the equation becomes : $ Q(x^2)^2\minus{}1\equal{}4Q((x^2\minus{}5)^2)$
So $ Q(x)^2\minus{}1\equal{}2Q((x\minus{}5)^2)$ (remember these are polynomials).

Let then $ R(x)\equal{}Q(x\plus{}5)$ and so $ Q(x)\equal{}R(x\minus{}5)$. The equation becomes $ R(x\minus{}5)^2\minus{}1\equal{}4R((x\minus{}5)^2\minus{}5)$ and so $ R(x)^2\minus{}1\equal{}4R(x^2\minus{}5)$

And so $ R(x)$ is a solution too, but degree($ R$)=degree($ Q$)=$ \frac 12$degree($ P$), hence a contradiction since $ P(x)$ was supposed to be with the fewest positive degree.

Hence the two unique solutions :

$ \boxed{P(x)\equal{}2\minus{}\sqrt 5}$ and $ \boxed{P(x)\equal{}2\plus{}\sqrt 5}$

[quote="pco"][quote="MJ GEO"]2)find all polynomial that $ (p(x))^2 \minus{} 1 \equal{} 4p(x^2 \minus{} 4X \plus{} 1)$[/quote]
For this one, I found a solution nearly similar (see previous post), except for elimination of the case "odd", more complex :

My solution :
Since $ 2 \plus{} \sqrt 5$ and $ 2 \minus{} \sqrt 5$ are solutions of $ x^2 \minus{} 1 \equal{} 4x$, $ P(x) \equal{} 2 \minus{} \sqrt 5$ and $ P(x) \equal{} 2 \plus{} \sqrt 5$ are solutions.
Suppose now there exists a solution with degree $ \ge 1$
Let $ P(x)$ such a solution with the fewest positive degree.

We can write the equation $ P(x)^2 \minus{} 1 \equal{} 4P((x \minus{} 2)^2 \minus{} 3)$, and so $ P(x \plus{} 2)^2 \minus{} 1 \equal{} 4P(x^2 \minus{} 3)$ and so $ P_1(x)^2 \minus{} 1 \equal{} 4P_1(x^2 \minus{} 5)$ where $ P_1(x) \equal{} P(x \plus{} 2)$

Obviously $ P_1(x)^2 \equal{} P_1( \minus{} x)^2$ and so (since polynomials), either $ P_1(x)$ is odd, either $ P_1(x)$ is even

1) if $ P_1(x)$ is odd :
Let $ x\in[ \minus{} 5, \plus{} 5]$ : $ P_1(\sqrt {x \plus{} 5})^2 \minus{} 1 \equal{} 4P_1(x)$ and so $ P_1(x)\ge \minus{} 1$
Since $ P_1$ is odd, we also get $ P_1(x)\le 1$

So $ \forall x\in[ \minus{} 5, \plus{} 5]$, $ \minus{} 1\le P_1(x)\le \plus{} 1$ and so $ P_1(x)^2\le 1$ and so $ P_1(x)^2 \minus{} 1 \equal{} P_1(x^2 \minus{} 5)\le 0$

And so $ P_1(x)\le 0$ $ \forall x\in[ \minus{} 5,20]$ but this implies, since $ P_1(x)$ is odd : $ P_1(x) \equal{} 0$, impossible
Hence $ P(x)$ is not odd.

2) If $ P_1(x)$ is even :
$ P_1(x) \equal{} Q(x^2)$ and the equation becomes : $ Q(x^2)^2 \minus{} 1 \equal{} 4Q((x^2 \minus{} 5)^2)$
So $ Q(x)^2 \minus{} 1 \equal{} 2Q((x \minus{} 5)^2)$ (remember these are polynomials).

Let then $ R(x) \equal{} Q(x \plus{} 5)$ and so $ Q(x) \equal{} R(x \minus{} 5)$. The equation becomes $ R(x \minus{} 5)^2 \minus{} 1 \equal{} 4R((x \minus{} 5)^2 \minus{} 5)$ and so $ R(x)^2 \minus{} 1 \equal{} 4R(x^2 \minus{} 5)$

And so $ R(x)$ is a solution too, but degree($ R$)=degree($ Q$)=$ \frac 12$degree($ P$), hence a contradiction since $ P(x)$ was supposed to be with the fewest positive degree.

Hence the two unique solutions :

$ \boxed{P(x) \equal{} 2 \minus{} \sqrt 5}$ and $ \boxed{P(x) \equal{} 2 \plus{} \sqrt 5}$[/quote]
I think that $ (Q(x))^2\minus{}1\equal{}4Q((x\minus{}5)^2)$ is true for all $ x\geq 0.$ So if what about $ x<0$? So, we can conclude $ R(x)$ is a solution. :) Can u explain plz. Thanks so much !

[quote="ngtl"] I think that $ (Q(x))^2 \minus{} 1 \equal{} 4Q((x \minus{} 5)^2)$ is true for all $ x\geq 0.$ So if what about $ x < 0$?[/quote]

You are right. We got $ (Q(x))^2 \minus{} 1 \equal{} 4Q((x \minus{} 5)^2)$ is true for all $ x\geq 0.$

But, as I said, these are polynomials So, if polynomial $ U(x) \equal{} (Q(x))^2 \minus{} 1 \minus{} 4Q((x \minus{} 5)^2)$ is equal to $ 0$ for all $ x\ge 0$, then $ U(x) \equal{} 0$ $ \forall x$ and so $ (Q(x))^2 \minus{} 1 \equal{} 4Q((x \minus{} 5)^2)$ is true for all $ x$

[quote] So, we can conclude $ R(x)$ is a solution. :) Can u explain plz. Thanks so much ![/quote]

Setting $ P_1(x) \equal{} P(x \plus{} 2)$, we started with $ P_1(x)^2 \minus{} 1 \equal{} 4P_1(x^2 \minus{} 5)$

and we got $ R(x)^2 \minus{} 1 \equal{} 4R(x^2 \minus{} 5)$ and so, you are right : $ R(x)$ is not a solution. But $ R(x \minus{} 2)$ is :) and so the conclusion is OK.

Thanks for reading so carefully my solutions.


Henry starts a trip when the hands of the clock are together between $ 8$ a.m. and $ 9$ a.m.  He arrives at his destination between $ 2$ p.m. and $ 3$ p.m. when the hands of the clock are exactly $ 180^\circ$ apart.  The trip takes:

$ \textbf{(A)}\ \text{6 hr.} \qquad
\textbf{(B)}\ \text{6 hr. 43\minus{}7/11 min.} \qquad
\textbf{(C)}\ \text{5 hr. 16\minus{}4/11 min.} \qquad
\textbf{(D)}\ \text{6 hr. 30 min.} \qquad
\textbf{(E)}\ \text{none of these}$

[hide="My Approach"]Sounds $ \boxed A$.
Because when he start it was about $ 8: 43$ and when he arrive it was like $ 2: 43$ also. But you can't make sure that its takes exactly $ \text{6 hr.}$[/hide]

@hasan4444: Actually, you can make sure it takes exactly 6 hours.

[hide="Solution"]
We will consider the number of degrees the hands move to find out when each event occurs. Let $ m$ be the number of minutes passed. For the first event, the hour hand has a inital position of $ 240^\circ$ (starting from the 12 clockwise) and we add on $ 30\cdot\frac {m}{60} \equal{} \frac {m}{2}$ degrees to account for the turning of the hour hand based on the minute hand. Every time the minute hand goes to a new minute, it travels $ 6^\circ$ (because there are $ 360/60 \equal{}$6 degrees per tick), so we the position of the minute hand is modeled by $ 6m$. Equating the two (since he will be at the same position), we obtain

$ 240 \plus{} \frac {m}{2} \equal{} 6m\implies 480 \plus{} m \equal{} 12m\implies m \equal{} 43\frac {7}{11}.$

So the trip starts at $ 8: 43\frac {7}{11}$. Tackling the second event, we find that the initial position of the hour hand is at $ 60^\circ$ (from the 12 clockwise) and the position of the minute hand is $ 6m$. Noting that the the minute hand will have to be past the hour hand, we obtain

$ 6m \minus{} 60 \minus{} \frac {m}{2} \equal{} 180\implies 12m \minus{} 120 \minus{} m \equal{} 360\implies 11m \equal{} 480\implies m \equal{} 43\frac {7}{11}.$

Therefore, Henry starts a trip at $ 8: 43\frac {7}{11}$ and ends it at $ 14: 43\frac {7}{11}$ (military time for convenience), so the trip takes $ \boxed{\text{(A) 6 hr.}}$.
[/hide]

Thanks "sunehra" for your smart approach  :)


inequalities
function
LaTeX
inequalities unsolved

HowTo apply Karamata's ienquality?
I just use Cauchy-Schwarz, for $k=n$, we have \[\sum a_{i}\leq \frac{n(n+1)(5n+4)}{6}\] Therefore, \[\sum \frac{1}{a_{i}}\geq \frac{n^{2}}{\sum a_{i}}\geq \frac{6n}{(n+1)(5n+4)}\]

but if $a_{i}=i(i+1)$
Then the equality obviously can be hold. :)
$Full$ $Solution$
Use Karamata's inequality:
because:
$\sum_{i=1}^{k}a_{i}\le \sum_{i=1}^{k}i(i+1),(k=1,2,\cdots,n-1)$
$\sum_{i=1}^{n-1}a_{i}+\sum_{i=1}^{n}i(i+1)-\sum_{i=1}^{n-1}a_{i}=\sum_{i=1}^{n}i(i+1)$
thus:$(a_{1},a_{2},\cdots,a_{n-1},\sum_{i=1}^{n}i(i+1)-\sum_{i=1}^{n-1}a_{i}) \prec (1\cdot 2,2\cdot 3,\cdots,n (n+1))$
So use Karamata's inequality:(Because the function 1/x is concave)
we have:
$\sum_{i=1}^{n}\frac{1}{a_{i}}\ge \sum_{i=1}^{n-1}\frac{1}{a_{i}}+\frac{1}{\sum_{i=1}^{n}i(i+1)-\sum_{i=1}^{n-1}a_{i}}\ge \sum_{i=1}^{n}\frac 1{i(i+1)}=\frac{n}{n+1}$
 :)
(Notice here is not really $<$,I forget how to type that notion.)
And the second $\ge$ in the last inequality Use Karamate(The first $\ge$ is trivial)

Post edited by the advice of $Xevarion$,thank you

Could you expalin more ?? :blush:
Sorry for this but I can't understand for which numbers did you use Karamata's inequality

Thank you very much . Fantastic solution ........ :D

Thank you :lol:
And I guess the originial solution may use $Abel$'s transform(Like the way prove  Karamata) :)

[b]zhaobin[/b] I think you want \succ $\succ$ and \prec $\prec$.

[b]Please stop using Imageshack or Photobucket for your graphics! You can use LaTeX for writing problems, and if you have graphics you can attach them to your post via the "Add an attachment" option when posting - no need to use external services![/b]

Before the image self-destroys, here is the problem:

Let be given $n$ real positive numbers $a_{1},\ a_{2},\ ...,\ a_{n}$ satisfying the condition $\sum_{i=1}^{k}a_{i}\leq\sum_{i=1}^{k}i\left(i+1\right)$ for all $k=1,\ 2,\ ...,\ n$. Prove that $\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}\geq\frac{n}{n+1}$.

Please write your problems in the same style (i. e. plain text and LaTeX)!

  darij

thank you Darij,
I couldn' see the image and found it quite frustrating trying to 'reconstruct' from all the partial solutions what the given problem might have been :weightlift:


please reference material in this link: http://www.artofproblemsolving.com/Wiki/index.php/1997_AIME_Problems/Problem_12

In this part...

definition, we have $ {\frac {a\frac {ax \plus{} b}{cx \plus{} d} \plus{} b}{c\frac {ax \plus{} b}{cx \plus{} d} \plus{} d} \equal{} x}$, which reduces to $ {\frac {(a^2 \plus{} bc)x \plus{} b(a \plus{} d)}{c(a \plus{} d)x \plus{} (bc \plus{} d^2)} \equal{} \frac {ex \plus{} f}{gx \plus{} h} \equal{} x}$. In order for this fraction to reduce to x, we must have f = g = 0 and e = h\not = 0. From ...

[b]I do not really get how "in order for this fraction to reduce to x", f=g=0 and e=h and both are not zero. WHY?[/b]

    Any help or insight into this problem would be most appreciated.
    I thank you in advance.

[quote="theax"]please reference material in this link: http://www.artofproblemsolving.com/Wiki/index.php/1997_AIME_Problems/Problem_12

In this part...

definition, we have $ {\frac {a\frac {ax \plus{} b}{cx \plus{} d} \plus{} b}{c\frac {ax \plus{} b}{cx \plus{} d} \plus{} d} \equal{} x}$, which reduces to $ {\frac {(a^2 \plus{} bc)x \plus{} b(a \plus{} d)}{c(a \plus{} d)x \plus{} (bc \plus{} d^2)} \equal{} \frac {ex \plus{} f}{gx \plus{} h} \equal{} x}$. In order for this fraction to reduce to x, we must have f = g = 0 and e = h\not = 0. From ...

[b]I do not really get how "in order for this fraction to reduce to x", f=g=0 and e=h and both are not zero. WHY?[/b]
[/quote]

If you don't understand it, just expand it and then compare like terms.

$ \frac {ex \plus{} f}{gx \plus{} h} \equal{} x \Leftrightarrow ex\plus{}f\equal{}gx^2\plus{}hx$ with $ gx \plus{} h \neq 0$

By comparing like terms we get $ f\equal{}g\equal{}0$ and $ e\equal{}h \neq 0$ (or $ gx \plus{} h\equal{}0$, for which the fraction is undefined).

but only the x's are like terms...
the x^2 and the f term are like terms...
please help!

I have no clue what you're talking about. A polynomial is zero everywhere if and only if its coefficients are 0. Since $ ex \plus{} f \equal{} gx^2 \plus{} hx$, $ gx^2 \plus{} x(h \minus{} e) \minus{} f \equal{} 0$; therefore, $ g \equal{} f \equal{} 0$, and $ h \minus{} e \equal{} 0$, yielding $ h \equal{} e$.

Zhero is right.

Generically, the equation is cx^2 + ax + b = 0.
Then, according to you, c=0, a=0, and b=0 in order for it to be = 0 [u][i][b]EVERYWHERE[/b][/i][/u]. True.

[quote="theax"]not necessarily...

x^2 - 5x + 6 = 0.
Factoring gives (x-3)(x-2) = 0
x=2 and x=3

Generically, the equation is cx^2 + ax + b = 0.
Then, according to you, c=0, a=0, and b=0 in order for it to be = 0. Not true.[/quote]

I don't think they are talking about particular instances where the polynomial happens to be 0. I think they are talking about the conditions under which the polynomial is identically equal to 0. (i.e, equal to 0 for all x, or whatever the independent variable is)

In your example, $ x^2\minus{}5x\plus{}6\equal{}0$, it is true that x=2 and x=3 satisfy this, but notice that x=1 (nothing particular about this number) does not satisfy it.

THANKS!
I don't think I realized that it was everywhere.
thanks. :)


a.Given three positive numbers x,y,z such that $ xyz \equal{} 1$ .Find the maximum of :
$ P \equal{} \frac {1}{x^2 \plus{} 2y^2 \plus{} 3} \plus{} \frac {1}{y^2 \plus{} 2z^2 \plus{} 3} \plus{} \frac {1}{z^2 \plus{} 2x^2 \plus{} 3}$

b.Consider $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$ with $ a,b,c \in Z$ a>0 and $ f(x) \equal{} 0$ has two real roots in $ (0;1)$
.Find the minimum of the coefficient a :D

Transform the first problem into an inequality:
$ \sum_{cyc}\frac{(xyz)^{2}}{x^{2}\plus{}2y^{2}\plus{}3(xyz)^{2}}\leq \frac{1}{2}$
Now get rid of $ xyz\equal{}1$ by putting into the inequality $ x\equal{}\frac{a}{b} \ y\equal{}\frac{b}{c} \ z\equal{}\frac{c}{a}$
Then we have $ \sum_{cyc}\frac{b^{2}c^{2}}{a^{2}c^{2}\plus{}b^{4}\plus{}3b^{2}c^{2}}$
Now assume that $ c\equal{}1$ (the inequality is homogenous) and then we have some work with multiplying the denominators, but the rest should be easy :wink:


If $a>1$, $b>1$, and $p=\frac{\log_b(\log_ba)}{\log_ba}$, then $a^n$ equals

$\textbf {(A) } 1 \qquad \textbf {(B) } b \qquad \textbf {(C) } \log_ab \qquad \textbf {(D) } \log_ba \qquad \textbf {(E) } a^{\log_ba}$

[hide="Solution"]Apply the logarithm base change formula to get $p=\log_a(\log_ba)$ so that $a^p=\log_ba$.

Oh yeah, one sentence!  :P [/hide]

[quote]then $a^n$ equals [/quote]

should be $a^p$

[quote]If $a>1, b>1$ and $p=\frac{\log_b(\log_ba)}{\log_ba}$, then $a^n$ equals

$\text {(A)} 1 \qquad \text {(B)} b \qquad \text {(C)} \log_ab \qquad \text {(D)} \log_ba \qquad \text {(E)} a^{\log_ba}$[/quote]

use the base changing rule on p to yield

$p=\frac{\frac{\log_a(\log_ba)}{\log_ab}}{\frac{log_aa}{log_ab}}$

$p=\log_a(\log_ba)$

let $x=a^p$

$\log_ax=p$

$\log_ax=\log_a(\log_ba)$

$x=\log_ba$

${(D)}$


The Fibonacci sequence is defined by $F_{0}=0, F_{1}=1, F_{n+1}=F_{n}+F_{n-1}$.

Prove that for all $k$, there exists a Fibonacci number that ends in $k$ zeros.

[hide="hint"]Think in mod $10^{k}$....[/hide]

[quote="t0rajir0u"][quote="randomdragoon"][hide="solution"]Look at the Fibonacci sequence in $\mod 10^{k}$. 2 consecutive terms define all preceeding and proceeding terms of the sequence. Furthermore, because there are only a finite number of pairs of numbers $\mod 10^{k}$, the Fibonacci sequence must repeat in $\mod 10^{k}$.

The sequence $1, 1$ appears once, so it must appear an infinite number of times in $\mod 10^{k}$. The term immediately before it must be congruent to $0 \mod 10^{k}$. If we do not choose the first instance of $1, 1$ in the Fibonacci sequence, this term is nonzero and ends in $k$ consecutive zeros.[/hide][/quote]

It ends in [b]at least[/b] $k$ consecutive zeros.  You haven't shown that it can't end in $k+1$, etc. zeros.[/quote]
well it doesn't really matter, does it?
I mean, 100 ends in a 0 right?
So ending in $k+1$ zeros implies that it ends in $z$ zeros which is what the problem asked... :maybe:

The problem requires that there exists a Fibonacci number that ends in [b]exactly[/b] $k$ zeroes for every positive integer $k$.

In other words, randomdragoon has only shown that the set

$K = \{ k \, | \, k \in \mathbb{N}, \exists i : F_{i}\equiv 0 \bmod 10^{k}\}$

Is unbounded, whereas the problem statement wants you to prove that

$K \equiv \mathbb{N}$

This is what the book wrote (and the solution it wrote), so I'm not sure if it was asking for exactly or at least.

Does 50,000 end in 3 zeros? Hmm...


An old meter stick was found that had parts of both ends missing. One end began with the 4 cm mark and the other ended with the 89 cm mark. What is the millimeter mark of the midpoint of the old stick?

The midpoint of the old stick in millimeters is $ \frac{4\plus{}89}{2}100\equal{}50(93)\equal{}\boxed{4650}$. I think.

[quote="yaofan"][quote="$ LaTeX$"]The midpoint of the old stick in millimeters is $ \frac {4 \plus{} 89}{2}100 \equal{} 50(93) \equal{} \boxed{4650}$. I think.[/quote]

It's only multiplied by $ 10$ because the units go from centimeters to millimeters.[/quote]
Actually, he multiplied it by $100$ accidentally.  He should have multiplied by $10$ (since there are 10 millimeters in one centimeter) and get the midpoint in millimeters to be $\boxed{465}$.


geometry
circumcircle
geometric transformation
homothety
power of a point
radical axis
geometry proposed

Let $ABC$ be a triangle inscribed in a circle $(O)$. Let $(I_a)$ be the circle that extouches $(O)$ at $M$ and touches the rays $AB$ and $AC$. Define $N, P$ similarly. Denote by $A', B' C'$ be the midpoints of the arcs $BC, CA, AB$ that contain $A, B, C$ respectively. Prove that $MA', NB', PC'$ are concurrent.

It is definitely true. But what follows, juancarlos ?   :?

From the last fact, $MA',NB',PC'$ perform a triangle.

Sorry, I consider the points $M,N,P$ as midpoints of arcs $BC,CA,AB$ that no cointain $A,B,C$.

I think it should be $AX, BY, CZ$ are concurrent (consider an equilateral triangle)

It is a nice proof, juancarlos.
 By the way, if we call $U$ the concurrent point $AK, BM, CN$ in your last post in http://www.mathlinks.ro/Forum/viewtopic.php?t=36502 , and $V$ the concurrent point of our problem, then $UV$ passes through the internal center of similitude of the incircle and the circumcircle.

Anybody, please explain to me what  the internal center of similitude of the incircle and the circumcircle is !!!!

It is the point $K$ in $IO$ such that $\frac{KI}{KO} = -\frac{r}{R}$

Further $AM,BN,CP$ concur at internal similitude center of incircle and circumcircle. Here $U,V$ is on the line $IO$ of $ABC$ triangle. Also $V$ is the homothety center of excental triangle of $ABC$  and triangle $A'B'C'$, i.e. $V$ is centroid of excentral triangle of $ABC$.

Dear juancarlos,
   It is very interesting. I did not notice that $U, V$ lie on the line $OI$. Anyway, it is easy to prove $V$ is the centroid of the excentral triangle of $ABC$ by using homothety. This yields that $V$ lies on $OI$.
   I think I have a nice way to prove $U, I, O $ are collinear.
    [i]Lemma[/i] Let $J$ be the intersection of $AB'$ and $A'B$ then $O, I, J$ are collinear.
    It follows from Pascal 's theorem for 6 points $A, A', B'', A'', B', B$, here $A'', B''$ are the antipodes of $A'. B'$ with respect to $(O)$.
    Recall that $K, D, A'$ are collinear, $M, E, B'$ are collinear. Moreover, $KA'$ intersects $MB'$ at the internal center of similitude. By applying Pascal 's theorem again to six points $A, B, B', K, M, A'$ and the lemma, we obtain $U, I, O$ are collinear.
    Is your proof similar to that, juancarlos ?

[quote="juancarlos"]If we consider $X,Y,Z$ the cut points of $(MA',BC),(NB',CA),(PC',AB)$. Now $X,Y,Z$ are collinear.[/quote]

  How can you prove it ?

[quote="juancarlos"]$A'$ is on the radical axis of the extouch circles $(Ib,Rb),(Ic,Rc)$.
Denote $t_A_'Ib$= tangent from $A'$ to extouch circle $(Ib)$,also:$t_A_'Ib$=$t_A_'Ic$, so:
$t_A_'Ib$= $(A'N/R). (\sqrt {R(R+Rb)}) = t_A_'Ic$ = $(A'P/ R).(\sqrt {R(R+Rc))}$
$(A'N/A'P) = (\sqrt {R+Rc})/(\sqrt{R+Rb})$
Similarly:
$(B'P/B'M) = (\sqrt {R+Ra})/(\sqrt {R+Rc})$
$(C'M/C'N) = (\sqrt {R+Rb})/({\sqrt {R+Ra})}$
Then, we have:
$(A'N/A'P).(B'P/B'M).(C'M/C'N)=1$
Now wrt $MNP$ triangle: $MA',NB',PC'$ are concurrent by Ceva in circular form.[/quote]

juancarlos,
can you give more details here?  I have no idea where you got that $A'$ lies on the radical axis, or the formula for the tangent lengths.

If we consider the touch points $D,E$ of $(Ic),(Ib)$ with $BC$. Now $DP,EN$ are the external bisectors of triangles $BPC, BNC$ resp., for a similar case with two extouch circles, look my post #4: http://www.mathlinks.ro/Forum/viewtopic.php?t=16007
Hence the lines $DP$ and $EN$ concur at $A'$ the mid point of arc $CAB$.Also $\angle PNA'= arc A'P/2= (arc A'B - arc PB)/2=(arc CA'-arc PB)/2= \angle PDE$. Then $DPEN$ is cyclic, also $A'P.A'D=A'N.A'E$, i.e. $A'$ is on the radical axis of $(Ib)$ and $(Ic)$.
The formula for the tangent you can to find: http://www.math.fau.edu/yiu/Geometry.html
Notes on Euclidean Geometry by Paul Yiu, pp141-142.

Wow, that's a lot of steps actually.  Thanks!


[quote="nguoigo"]Let a,b,c>0.Pro that$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{3(a^{2}+b^{2}+c^{2})}{a^{3}+b^{3}+c^{3}}$ :lol:[/quote]

This is direct Chebishev inequality..


calculus
integration
function
calculus computations

Let $ f: [0,1] \to \mathbb{R}$ with $ f\in\mathcal C^1$ and $ f(0)\equal{}0.$

Prove that $ \mathop {\sup }\limits_{0\,\le u\,\le1} \left| {f(u)} \right|\le\sqrt {\int_0^1 {(f'(u))^2 \,du} }.$

What does mean $ f\in\mathcal C^1$?

What does mean $ \mathop {\sup }\limits_{0\,\le u\,\le1}$?

[quote="Field"]Let $ f: [0,1] \to \mathbb{R}$ with $ f\in\mathcal C^1$ and $ f(0) \equal{} 0.$

Prove that $ \mathop {\sup }\limits_{0\,\le u\,\le1} \left| {f(u)} \right|\le\sqrt {\int_0^1 {(f'(u))^2 \,du} }.$

What does mean $ f\in\mathcal C^1$?

What does mean $ \mathop {\sup }\limits_{0\,\le u\,\le1}$?[/quote]
The problem will be very hard if you aren't familiar with $ \sup$...

$ f\in\mathcal C^1$ means $ f$ is in class 1 (set of differentiable functions???). I'm not sure if this is a standard notation.

$ \mathop {\sup }$ is the [[supremum]]

$ \mathcal{C}^1$ is the set of functions with continuous derivative. And showing $ \sup_{0\leq u\leq 1} |f(u)|\leq A$ is equivalent to showing $ |f(u)|\leq A$ for every $ u\in [0,1]$.
[hide]
So we must show that for every $ u\in [0,1]$, $ f(u)^2\leq \int_0^1 f^\prime (t)^2\ dt$. This is so because $ f(u)\equal{}\int_0^u f^\prime (t)\ dt$ (using $ f(0)\equal{}0$). Using the integral form of Cauchy,
\[ \left( \int_0^u f^\prime (t)\ dt\right)^2 \leq \left( \int_0^u \ dt \right) \left( \int_0^u f^\prime (t)^2 \ dt\right)\]
\[ \equal{}u\left( \int_0^u f^\prime (t)^2 \ dt\right) \leq \left( \int_0^1 f^\prime (t)^2\ dt\right)\]
[/hide]


Let the incircle of triangle $ABC$ have the center $I$ and touch the sides $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. Now let $AI \cap EF=O$, $BI \cap FD=P$, $CI \cap DE=Q$. Let the lines $AQ$ and $AP$ intersect the line $EF$ at $R$ and $S$; the lines $BO$ and $BQ$ intersect the line $FD$ at $T$ and $U$; the lines $CP$ and $CO$ intersect the line $DE$ at $V$ and $W$.

[b]1)[/b] Show that $RU\parallel DE$, $SV\parallel FD$ and $TW\parallel EF$. (Proving one is enough... :D )
[b]2)[/b] Show that the lines $RU$, $SV$ and $TW$ are concurrent.

I'm in a very awkward situation: there are so many beautiful properties that I'm lost :). Really!

For the paralellism: By applying Menelaus in triangle $DEF$ with transversals $BO,CO$ we reduce $TW\|EF$ to the fact that $A\in MN,\ MN\|EF$, where $M=BO\cap DE,\ N=CO\cap FD$. We know for sure that $A\in MN$ (from Pappus's theorem), so we need to show $MN\|EF$. In fact, we don't even need to employ Pappus's theorem. We can show $MA\|EF,\ NA\|EF$. I'll show $MA\|EF$:

The line $ME$ (since it's $DE$) cuts $AB$ in the point harmonically associated to $A,B,F$, so the lines $MA,MF,MB,ME$ form a harmonic quadruple, so $EF\|MA$ since $EF$ cuts the lines s.t. $EO=OF$ ($E=MD\cap EF,\ O=MB\cap EF,\ F=MF\cap EF$).

Regarding the $RU \cap SV \cap TW$ point, it seems to be the symmedian center of $D E F$, i.e. the Gergonne point of $ABC$. You can deduce it from the same Menelaus application in triangle $D E F$ with transversal line $CO$. You'll get $\frac{EW}{WD}=\frac{FN}{ND}$, and since $A$ is the $d$-exsymmedian center of $D E F$, you'll find $TW$ to be the line parallel to $EF$ and passing through the symmedian center of $D E F$.

I realize this is far from clear, and I'll be happy to make it clearer..tomorrow. Or should I say today? (:); it's 6 AM here, and I've been staying up all night).

grobber, I will read your solution to the first part more carefully later. Realizing time is a factor, however, I will state my ORIGINAL conjecture:

Show that lines AD, BE, CF, RU, SV and TW are ALL concurrent!  :D

That's also what I said and, if I'm not wrong, proved :).

Sorry for the delay in reply, grobber, your solution is correct, and, indeed, beautiful.

I think I have somewhat trivialized the problem by asking the second part (the 6 lines concurrent), it's just 3 times Pappus' Theorem. I can't believe that I didn't see it last night. Thanks! :)

The triangle DEF is the cevian triangle of the Gergonne point of triangle ABC, and the points O, P, Q are the midpoints of the sides EF, FD, DE of this triangle DEF (the latter is because the lines AI, BI, CI bisect the segments EF, FD, DE, respectively, since e. g. the points E and F lie symmetrically to each other with respect to the line AI). So the problem will be solved once the following generalization will be proven:

[b]Theorem 1.[/b] Let ABC be a triangle, and DEF the cevian triangle of an arbitrary point K with respect to the triangle ABC. Let O, P, Q be the midpoints of the sides EF, FD, DE of this cevian triangle DEF, and let the lines AP, AQ, BQ, BO, CO, CP intersect the lines EF, EF, FD, FD, DE, DE at the points S, R, U, T, W, V, respectively. Then, the lines TW, VS, RU pass through the point K and are parallel to the lines EF, FD, DE, respectively.

[i]Proof.[/i] At first, using the notation g /\ h for the point of intersection of two lines g and h, we can apply the Pappos theorem to the collinear point triplets (A; F; B) and (E; Q; D) and get the collinearity of the points AQ /\ FE = R, FD /\ BQ = U and BE /\ AD = K. Thus, the line RU passes through the point K.

Now, we know that the lines AD, BE, CF concur at one point (namely the point K), and that the lines AR, BU, CQ concur at one point (namely the point Q). Hence, applying the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6579]cevian nests theorem[/url] (in its most general form), we see that the lines DR, EU, FQ also concur at one point. Consequently, applying the Ceva theorem in triangle DEF, we get $\displaystyle \frac{DQ}{QE}\cdot \frac{ER}{RF}\cdot \frac{FU}{UD}=1$. Since Q is the midpoint of the side DE, we have $\displaystyle \frac{DQ}{QE}=1$, so this becomes $\displaystyle \frac{ER}{RF}\cdot \frac{FU}{UD}=1$. In other words, $\displaystyle \frac{ER}{RF}=\frac{UD}{FU}$, so that $\displaystyle \frac{ER}{RF}=\frac{DU}{UF}$. Hence, RU || EF.

Thus, we have shown that the line RU passes through the point K and is parallel to the line EF. Similarly we proceed for the lines TW and VS. And Theorem 1 is proven.

  Darij

Just another way to see the things (equal to that of Grobber?!)

Th.
Let ABC be a triangle, and DEF the cevian triangle of an arbitrary point K with respect to the triangle ABC. Let H a point on EF and let I=DF^BH and J=EF^IK. Then J is armonically coniugated to H wrt(E,F).

Proof

If G=EF^BC, G is armonically coniugated to D wrt(C,B). If L=IK^AB, the quadruple (J,K,I,L) is proj. corr. by F to  (G,C,D,B) then J is armonically coniugated to I wrt(K,L). Therefore, as the quadruple (J,E,H,F) is proj. corr. by B to  (J,K,I,L) then J is armonically coniugated to H wrt (E,F).
We have that EF//KI (i.e. J=oo)  iff H is the midpoint of EF.

M=DE^CH stay on KI. Infact if M'=KI^DE and N=IK^AC we have that (I,N,M',K) correnspond to (G,C,D,B) by E then M' is the fourth harminic wrt I,N,K. If M"=CH^IK we have that  (I,N,M",K) correnspond to (J,E,H,F) by C then M" is the fourth harminic wrt I,N,K. Therefore M'=M"=M.


Problema se afla deja postata la adresa:
http://www.mathlinks.ro/Forum/viewtopic.php?p=198180#p198180
Tot aici este si o rezolvare, la care da raspunsul $\frac{mn-m+n}{n}$.
Am insa o mare nelamurire. De fapt $\bigcup_{x\in G} H^x$ nu acopera intreg G-ul?? Altfel zis, $\bigcup_{x\in G} H^x$ are chiar m elemente.
Deci ar trebui sa avem $\frac{mn-m+n}{n}=m\Rightarrow m=n\Rightarrow H=G$, ceea ce contrazice faptul ca H e un subgrup propriu al lui G. De aici, concluzia ca ipoteza acestei probleme nu poate avea niciodata loc.

Gresela mea. Am crezut ca $H^x$-urile partitioneaza G-ul, dar de fapt nu e asa. Rog un moderator sa stearga postul.


floor function
algebra
polynomial
product of roots
sum of roots

Let $ \lfloor x \rfloor$ be the greatest integer less than or equal to $ x$.  Then the number of real solutions to $ 4x^2 \minus{} 40 \lfloor x \rfloor \plus{} 51 \equal{} 0$ is

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

[hide="Solution"]
Clearly, if $ 4x^2 \minus{} 40x \plus{} 51 > 0$ for some $ x$, then $ x$ can't be a solution to $ 4x^2 \minus{} 40\lfloor x\rfloor \plus{} 51\ge4x^2 \minus{} 40x \plus{} 51 > 0$, either. Since $ 4x^2 \minus{} 40x \plus{} 51 \equal{} (2x \minus{} 3)(2x \minus{} 17)$, this means that $ \frac32\le x\le\frac {17}2\implies1\le\lfloor x\rfloor\le8$. But $ 4x^2 \equal{} 40\lfloor x\rfloor \minus{} 51$, $ \lfloor x\rfloor\ge2$, so $ 2\le\lfloor x\rfloor\le8$ as well. Now, note that if and only if $ 4\lfloor x\rfloor^2\le40\lfloor x\rfloor \minus{} 51\le4(\lfloor x\rfloor \plus{} 1)^2$ will there be one solution to the equation. Testing, we find that $ \lfloor x\rfloor \equal{} \{2,6,7,8\}\implies\boxed{\textbf{(E)}\ 4}$ real solutions exist.
[/hide]

For $ x \ge 0$ the discriminant $ \equal{} 40^2 \minus{} 4(4)(51) > 0$, so two distinct positive roots (because sum and product of roots $ > 0$)
For $ x < 0$ the discriminant $ \equal{} 40^2 \minus{} 4(4)(51) > 0$ again, so there are two distinct negative roots .(because sum of roots is negative and product of roots is positive).
So there are 4 distinct real solutions .The answer is $ \boxed{\textbf{(E)}\ 4}$

[quote="math154"]This isn't very rigorous...[/quote]

Sorry , I mistook $ \lfloor \;\;\; \rfloor$ for $ |   \;\;\; |$ . :rotfl:

So the solution is just like yours.


trigonometry
analytic geometry
function
geometry theorems
geometry

i'm a new user...
please help me...
can anyone please explain me ,why is cosine of any angle always positive.....

Cosine of any angle isn't always positive...
Cosine of 104 (Radians, Degrees, and Grad) is negative.

It's not  ALWAYS positive.. When the angle is between $ \frac {\pi}{2}$ and $ \frac {3\pi}{2}$, its NEGATIVE. In other words, think of it as the coordinate system.

This shows what trigonometric functions are positive and when. In the first quadrant, $ 0^\circ$ to $ 90^\circ$, all the funtions are positive. Second quadrant, $ 90^\circ$ to $ 180^\circ$, just the sine. Third quadrant, $ 180^\circ$ to $ 270^\circ$, just the tangent function is positive. For the fourth quadrant, $ 270^\circ$ to $ 360^\circ$, just the cosine function is positive.

I think this should be enough :)

* In case you don't know what radians are, it's another way to write angles using multiples of $ \pi$. Use $ \pi\equal{}180^\circ$ as a reference point to convert between radians to degrees and vice versa.

This is neither true, as we have $ \cos ( \minus{} 3\pi/2) \equal{} 0$.

[quote="shibin.george"]sorry,but i asked the wrong question...
why is the cosine of a negative angle always positive.....[/quote]

thats  again a wrong question .  see  u better  study  sl  loney  trigonometry  first chapter.  :)


[quote="tranvanluan"]Let $ a;b;c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$
Prove
$ \frac {a^2 \plus{} 1}{a^2 \plus{} bc} \plus{} \frac {b^2 \plus{} 1}{b^2 \plus{} ba} \plus{} \frac {c^2 \plus{} 1}{c^2 \plus{} ba}\ge 15$[/quote]

Try $ a \equal{} b \equal{} \frac{1}{2},c \equal{} 0$.

I think the ineq would be
\[ \frac {a^2 \plus{} 1}{a^2 \plus{} bc} \plus{} \frac {b^2 \plus{} 1}{b^2 \plus{} ca} \plus{} \frac {c^2 \plus{} 1}{c^2 \plus{} ba}\le 15\] :maybe:


Ok, for now, this will be the topic related to what the title says.
I also include a poll, that is much more democratic than the first one :)
I invite to vote anyone ho wants to express his opinion about this matter :D
Btw, Valentin, the last topic with the same name should be deleted, to avoid confusion.

[quote="PiDeltaPhi"]Dust,I am sorry,but Valentin clearly stated that this section would imply higher bandwith traffic and more costs.So,we have to patient and wait for a much better opportunity for this section.Thanks for understanding.[/quote] Well of course, I could be "corrupted" if Dust would donate a server :P ...

We could open a paypal account for donations...I'd donate...Valentin would buy the server...etc.,I will probably not be the only one helping out...

Hey Valentin, I'm back.

You could eliminate part of the "Non Math topics" sections to create room for a Computer related (not necessarily programming section).  In response to the elimination, you could then force people to enter PiDeltaPhi's hosted chat room.

Locked for the same reason as the other programming thread. -.-


superior algebra
superior algebra unsolved

It is a classic problem to prove

"If G is finite and commutative of odd order n, then every element is a square."

The easy proof is to just construct an automorphism f: g-->g^2.  It is an automorphism b/c of commutative and so every element is a square.

But I am wondering, is commutativity necessary?  Consider:

g=g^(n+1), so g^((n+1)/2) squared is g.  What is wrong with this argument (if anything)?  Does this argument somehow use commutativity?  I don't see how.

Thanks for your help.

Your argument is fine.  It is equivalent to your classic problem for the commutative group <g>.


Hi, does anyone know how to do this question?

[color=olive]Let x = 1 + 3j − k and y = 1 + 2i + k in H. Find x + y, xy, N(x), x (bar) and x−1.
Check that xx(to the power -1) = 1.[/color]

Thanks very much

You just plug in all the definitions concerning [url=http://en.wikipedia.org/wiki/Quaternion]quaternions.[/url]


inequalities
search
function
inequalities proposed

[quote="chien than"]Let $ a;b;c$ be positive real numbers.Prove that:
$ x \plus{} y \plus{} z \plus{} \sqrt {x^2 \plus{} y^2 \plus{} z^2} \geq \sum \sqrt {x^2 \plus{} y^2}$[/quote]
Let $ a \equal{} x\vec{i},$ $ b \equal{} y\vec{j}$ and $ c \equal{} z\vec{k},$ where $ \vec{i}\vec{j} \equal{} \vec{j}\vec{k} \equal{} \vec{k}\vec{i} \equal{} 0$ and $ |\vec{i}| \equal{} |\vec{j} \equal{} |\vec{k}| \equal{} 1.$
Thus, it's remains to prove that
$ |a| \plus{} |b| \plus{} |c| \plus{} |a \plus{} b \plus{} c|\geq |a \plus{} b| \plus{} |a \plus{} c| \plus{} |b \plus{} c|,$ which very known.
For the proof see here the same trick:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153527

[quote="silouan"][quote="arqady"]
Thus, it's remains to prove that
$ |a| \plus{} |b| \plus{} |c| \plus{} |a \plus{} b \plus{} c|\geq |a \plus{} b| \plus{} |a \plus{} c| \plus{} |b \plus{} c|,$ which very known.
[/quote]

If I am not wrong you can use Popoviviu for the |x| and find an immediate proof[/quote]

You are. ;)

  darij

[quote="arqady"]silouan, darij grinberg, $ a,$ $ b$ and $ c$ are vectors.
It does not interfere?[/quote]

It does. See Billzhao's post in http://www.mathlinks.ro/Forum/viewtopic.php?t=19097 (search for the word "metal").
You cannot apply Popoviciu's inequality to (convex) functions of several variables.

  darij


From a two-digit number $ N$ we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then:

$ \textbf{(A)}\ {N}\text{ cannot end in 5}\qquad\\
\textbf{(B)}\ {N}\text{ can end in any digit other than 5}\qquad \\
\textbf{(C)}\ {N}\text{ does not exist}\qquad \\
\textbf{(D)}\ \text{there are exactly 7 values for }{N}\qquad \\
\textbf{(E)}\ \text{there are exactly 10 values for }{N}$

Well $ n \equal{} 85$ is one solution so answers $ \textbf{(A),(B),(C)}$ are out.

Now let $ n \equal{} 10a \plus{} b$, such that when you subtract by the reverse you get $ 10a \plus{} b \minus{} (10b \plus{} a) \equal{} 9a \minus{} 9b \equal{} 9(a \minus{} b)$. Since $ 9(a \minus{} b)$ has to be a positive perfect cube, and $ a,b$ are one digit integers $ 9(a\minus{}b)\equal{}3^3\implies a \minus{} b \equal{} 3$. and the solutions to this are:
\[ (a,b) \equal{} (3,0),(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)
\]
Or in other words, the two digit integers $ 30,41,52,63,74,85,96$ satisfy the requirements.

In total there are $ \boxed{7}$ solutions, so the answer is $ \boxed{\textbf{(D)}}$.

Since the difference between the original number and its reversed digit number must be a perfect cube, it can be 1, 8, 27, or 64. But the smallest difference a reversed number can be with its original is 9, that is, when the tens digit is 1 greater than the units digit. So the only possiblities are 27 and 64.

If a number with tens digits is 2 more than the units digit, then the difference is 18.
3 more: 27
4 more: 36
5 more: 45
6 more: 54
7 more: 63
8 more: 72

So the number must have tens digit 3 more than units. Possible values of $ N$ are 30, 41, 52, 63, 74, 85, and 96, for a total of exactly 7 values for $ N$.

Answer: D

EDIT: Beaten


Triangle $ ABC$ is scalene with angle $ A$ having a measure greater than 90 degrees. Determine
the set of points $ D$ that lie on the extended line $ BC$, for which
\[ |AD|\equal{}\sqrt{|BD| \cdot |CD|}\]
where $ |BD|$ refers to the (positive) distance between $ B$ and $ D$.

The set of points D isthe intersection of the tangent to circumcircle of triangle ABC at point A

All the measures I am going to write from now on refer only to the magnitudes.

We have $DA^2=DB\cdot DC$

We have the above expression similar to writing the power of point $D$

The circle contains the points $A, B,$ and $C$ and so, it is the circumcircle of $ABC$

So, to construct $D$, we draw the perpendicular bisectors of any two sides of $\Delta ABC$ and from the intersection point, the circumcentre, we join to $A$ and construct a perpendicular. It meets $BC$ at a unique point $D$

$D$ will be at infinity, if $A$, the circumcentre and the midpoint of $BC$ are collinear.

I don't think a set of points may exist.

[quote=soumitra15rta]in my diagram I,J,E are the required points.some mistake in naming my triangle is BCD[/quote]

Bro it is clearly written points only on extended line BC


Prove that for all real x,y the value of $x^2 + 2xy + 3y^2 -6x-2y$ cannot be less than -11

Well , do you know the theory of trionymes?This is a standard method.

 The ineq. becomes :

     $\ x^2 + 2(y-3)x + 3y^2 -2y + 11 \geq 0$ But the discriminant is
$\Delta = ......-8(y+1) \ ^2 \leq 0$ and we are done.We have equality only if $x=4$ and $y = -1.$

 If you do not know this theory , try to write the first member as sum of non negative terms.

[u]Babis[/u]

basically what he is saying is to complete the square...


if we have  a,b,c>0 ,abc=1  prove that
a/b + b/c + c/a >= a+b+c

[quote="payman_pm"]is that solution?[/quote]

why not?
Similarly you obtain two more inequalities, sum them up and finish


Math teachers drink alcohol using the mathematical induction method.They follow the following principles:
1.Every person has the right to drink 1 glass of wine
2.After a person drinks a glass of wine he becomes another person.

So he can drink any amount of wine!!!! :)  :)  :)  :)  :)  :)

no he can't

he's limited to the amount of persons there are :D

if a math teacher sees a glass of wine, he or she drinks it.

if a math teacher sees an empty glass of wine, he or she fills it up and thus reduces it to a previously solved problem  :)

[quote="MysticTerminator"]why do you become another person[/quote]A basic rule in mathematics is that you cannot argue on the axioms. (unless they are inconsistent, of course)

However, axioms 1 and 2 imply that either an infinite number of people exist, or some person who has the right to drink a glass of wine does not do so.  So if we take axiom 3 to be "The set of all people is finite", then the mathematician may be able to drink arbitrarily much wine, but not infinite amounts of it.  Unless you have an infinitely large glass.

But I think arbitrarily many glasses is what we're going for here.

[quote="Sly Si"]But I think arbitrarily many glasses is what we're going for here.[/quote]

That's what the Banach-Tarski Paradox is for :lol:

i don't agree with number 2 but it does work out.. :lol:


If F(n) denotes the n-th Fibonacci number prove that the number of digits in F(n) is less than n/4

Isn't it trivial when using Binet's formula
$F_n= \frac{1}{\sqrt{5}} \left( \left( \frac{1+ \sqrt{5}}{2} \right) ^n - \left( \frac{1- \sqrt{5}}{2} \right) ^n \right) $

you can prove by induction that $f(4n) \leq n$.


inequalities
real analysis
real analysis solved

Let (a_n)_n be a sequence of strictly positive real numbers s.t.
lim{n->oo}(1/n (\sum{k=1,n}a_k)-(a_1.a_2.....a_n)^(1/n))=0.
Prove that a_n/n^3 has the limit 0.

We use the fact that (a_1+...+a_n)-(a_1*a_2*...*a_n)^1/n  \geq 1/n*( \sqrt a_(n-1)- \sqrt a_n) 2  for any n.  Thus, the sequence  (\sqrt a_n- \sqrt a_(n-1))/ \sqrt n tends to 0. From Cesaro-Stolz theorem we find that  \sqrt a_n/(n* \sqrt n) also tends to 0, so a_n/n^3 tends to 0.

[quote="harazi"]We use the fact that (a_1+...+a_n)-(a_1*a_2*...*a_n)^1/n  \geq 1/n*( \sqrt a_(n-1)- \sqrt a_n) 2  for any n.  [/quote]

Here the proof
http://www.mathlinks.ro/viewtopic.php?t=4425&sid=8f95815f060c26cff87aa6ef6d490511

actually moubi this is another problem i posted few days ago :D

a few topics below this one. It is called " inequality" by Cristinel Mortici


A diagnostic test for a disease is said to be 90% accurate in that, if a person has the disease, the test will detect it with probability 0.9. Also, if a person does not have the disease, the test will report that he does not have it with probability 0.9. Only 1% of the population has the disease in question. If a person is chosen at random from the population and diagnostic test indicates that he has the disease, what is the conditional probability that he does, in fact, have the disease? Would you call this diagnostic test reliable?

There probability that the person selected has the disease  is $ \frac{1}{100}$.  The probability that the test detects his having the disease is $ \frac{9}{10}$.  The overall probability, then, that the selected person actually has the disease AND the tests detects it is thus: $ \frac{1}{100} \bullet \frac{9}{10}\equal{}\boxed {\frac{9}{1000}}$.

Oh, and, yes I would think this diagnostic test is reliable to a certain extent.  Its chance of showing that the person has a disease is $ 90\%$.  That is pretty high, $ 1$ point higher than my English test grade today -_-

That's not correct. It's $ \frac {\frac9{10}\cdot\frac1{100}}{\frac9{10}\cdot\frac1{100} \plus{} \frac1{10}\cdot\frac {99}{100}} \equal{} \boxed{\frac1{12}}$.

Edit: Thanks. Yes, I misread the problem (i.e. missed a "not"), but the test itself is still inaccurate.

[quote="math154"]That's not correct. It's $ \frac {\frac9{10}\cdot\frac1{100}}{\frac9{10}\cdot\frac1{100} \plus{} \frac9{10}\cdot\frac {99}{100}} \equal{} \boxed{\frac1{100}}$.[/quote]

You might have misread the problem, but the denominator (specifically the second term) is incorrect. The test would be pointless if a positive tells you there's a 1% chance you have the disease; you know that 1% of the population has the disease.


i want to buy 8 dogs all of which are either cocker spaniels, irish setters or russian wolfhounds.  how many ways can i make the choice?
dd_ddd_ddd
i know the answer is choose 2 in 10, but why are there 10 possible postitions for _? i only see 9 different intervals in between or beside the d.

You are not choosing 2 dividers/intervals between the d's and/or _'s

The _'s ARE the dividers.

You have 10 positions: _ _ _ _ _ _ _ _ _ _

8 of the positions are filled with d's

2 of them are the 2 dividers

Thus $ \dbinom {10}{2}$ possibilities.

If you still do not get it, try this problem with smaller numbers, and count the hard way, and then check by doing it using this slick one to one correspondence.

for example: 3 dogs total: (using A=number of cocker spaniels, B=number of irish settlers, C=number of russian wolfhounds)

ddd_ _     3A 0B 0C
dd_d_      2A 1B 0C
d_dd_      1A 2B 0C
_ddd_      0A 3B 0C
dd_ _d     2A 0B 1C
d_d_d      1A 1B 1C
_dd_d      0A 2B 1C
d_ _dd     1A 0B 2C
_d_dd      0A 1B 2C
_ _ddd     0A 0B 3C

We count 10 possibilities the hard way

Checking, we have $ \dbinom 52\equal{}10$ possibilities.

Hope that helped.

I just want to comment on [b]cdymdcool[/b]'s counting method.

Basically, you've found $ \binom{9}{2}$ of the arrangements. The one's that you've missed are the ones that are like:

dddd_ _dddd

In other words, you've missed the cases in which the middle category has no members. We have to count these cases. We can treat "_ _" as one character, and there are 9 places to put it. So there are an additional $ \binom{9}{1}$ cases which you missed.

Then, the actual answer is $ \binom{9}{1} \plus{} \binom{9}{2} \equal{} \binom{10}{2}$, where the last equivalence comes from Pascal's triangle.


induction
limit
algebra unsolved
algebra

let a_n be a sequence of réel numbers such that :
a_1=1 ,a_{n+1}=1+a_1.a_2.....a_n ,let Sn=1/a_1+1/a_2+...1/a_n
evaluate lim(Sn)

It is easy (by induction) \[S_{n}=2-\frac{1}{\prod_{i=1}^{n}a_{i}}\to 2.\]


Find limit:

$ \lim_{x\to 0}\frac{\frac{1}{\sqrt{1\plus{}x}}\minus{}1}{x}$

$ 1\minus{}\sqrt{x\plus{}1}\equal{}\frac{\minus{}x}{\sqrt{x\plus{}1}\plus{}1}$
So $ \lim\equal{}\frac{\minus{}1}{2}$


Let $ a$ and $ b$ be two positive integers such that $ a\plus{}b\equal{}1092$ and $ \text{lcm} (a,b)\equal{}3528$. Find $ a$ and $ b$.

3528 factored is 2^3*3^2*7^2 So we have to split these around to find something that works. A quick check reveals that a and b are not relatively prime.
So now we try different factors. 1/3 and 1/2 of 3528 is too high, so 1/4 is 882.
b must now be 210, which is impossible.

So 1/6 of 3528 is 588. b must be 504. This factored is 2^3*3^2*7 Which fits in the existing bounds so the two solutions (a,b) are (588,504) and (504,588)

[quote="v235711"][quote="ifai"]Let $ a$ and $ b$ be two positive integers such that $ a \plus{} b \equal{} 1092$ and $ \text{lcm} (a,b) \equal{} 3528$. Find $ a$ and $ b$.[/quote]

just factorize 3258 and check teh cases.[/quote]

What brilliant insight! It's astounding that nobody at all thought of this method and posted it right above yours hours ago.

Oh wait, they did :D

Another approach:

$ 1092\equal{}2^2 \times 3 \times 91$, $ 3528\equal{}2^3 \times 3^2 \times 7^2$

As $ lcm(a,b)\equal{}3528$, one of $ a,b$ is divisible by $ 2^2$, also $ 2^2|a\plus{}b$, so $ 2^2|a,b$

Similarly $ 3|a,b$ , now let $ a\equal{}12x,b\equal{}12y$,

$ x\plus{}y\equal{}91$ and $ lcm(x,y)\equal{}2 \times 7^2$

As one of $ x,y$ is divisible by $ 7^2$ we have $ (x,y)\equal{}(49,42),(42,49)$

,which gives $ (a,b)\equal{}(588,504),(504,588)$


function
real analysis
real analysis unsolved

if a function $f$  is a laplacian and say it's value at the boundary is same everywhere in the boundary and is equal to that at a certain point inside the region of interest then the function is constant through out the region

[quote]a function f is a laplacian[/quote]
Do you mean $f=\Delta u$ for some $u$? This does not give you much information about $f$. For any $C^{1}$ function $f$ there is a $C^{3}$ function $u$ such that $f=\Delta u$.

The term for that: $f$ is harmonic.

Just apply the maximum principle; if $f$ is equal to $c$ on the boundary of the region, $f(x)\le c$ inside, and $-f(x)\le-c$ inside. It's constant.


In triangle ABC the values of  b and c are known.
The length of the altitude from A to to the side BC is equal to the length of the side BC.

Prove that [c{Sqrt5 -1}]/2 <b <[c{Sqrt5 +1}]/2

It is rather easy. I f you give the points A,B,C coordinates $A=(x,1)$, $B=(0,0)$ and $C=(1,0)$ then you have to find the maximum and minimum of

$\sqrt{\frac{x^2+1}{(x-1)^2+1}}$.

The square root is increasing so we can just consider $\frac{x^2+1}{(x-1)^2+1}$. After the calculation of the derivate you see that the zeroes of $x^2-x-1=0$ are the $x$ for which the minimum and the maximum are achieved. After a bit of computations you get the result.

BTW: both estimates are equivalen because of the symmetry of $b$ and $c$.

Misha


greatest common divisor
number theory
least common multiple

The GCF of two numbers is 18. Their LCM is 630. what are the two numbers.

[hide]
Let the two numbers be $ 18a$ and $ 18b$.
$ \text{lcm}(a, b) \equal{} \frac{630}{18} \equal{} 35 \equal{} 5 \times 7$.
Choices for a and b are 5 and 7 or 1 and 35; luckily all these pairs are coprime (coprime = gcd is 1), so that no extraneous solutions occur.
So the two numbers are 18 and 630 or 90 and 126.
[/hide]


a)  If a,b,c are positiv real numbers such that $ a \plus{} b \plus{} c \equal{} 1$ , then

  $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\leq \frac {b^2 \plus{} c^2}{(b \plus{} c)^2} \plus{} \frac {c^2 \plus{} a^2}{(c \plus{} a)^2} \plus{} \frac {a^2 \plus{} b^2}{(a \plus{} b)^2}$

  b)  But if a,b,c are sides of triangle, is true inequality?

[quote="Marius Mainea"]If a,b,c are positiv real numbers such that $ a \plus{} b \plus{} c \equal{} 1$ , then

  $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\leq \frac {b^2 \plus{} c^2}{(b \plus{} c)^2} \plus{} \frac {c^2 \plus{} a^2}{(c \plus{} a)^2} \plus{} \frac {a^2 \plus{} b^2}{(a \plus{} b)^2}$[/quote]
Why $ a \plus{} b \plus{} c \equal{} 1$ $ ?$ Why does not $ a \plus{} b \plus{} c \equal{} 2$ $ ?$:wink:

Yes ,is equivalent with $ a\plus{}b\plus{}c\equal{}k$,   $ k>0$


Let $ a_{1},a_{2},...a_{n},p\in\mathbb{R}$ and $ m,n\in\mathbb{N}$. Prove that if  $ \sum_{ \equal{} 1i}^n a_{i}^{2s} \equal{} p^{2s}$ then $ \sum_{i \equal{} 1}^n a_{i}^m \le p^{m}$ for $ \forall m\ge 2s$.

Aren't you interested in the problem? It isn't hard at all...

I can't believe you haven't solved this problem yet  :(


probability
trigonometry
geometry
angle bisector
geometry proposed

I want to show two problems composed by me.

1. In the triangle ABC on the side (between A and B) are chosen two points M and N such that AM=BN and angle ACM= angle BCN. Prove that the triangle is  isosceles.

2. It is given acute-angled triangle ABC. On the sides AB, BC, CA (not on lines) exist points M, N and P, such that AM=BN=CP and angle ACM=angle BAN=angle CBP. Prove that the triangle is equaliteral. Is it true if the triangle is not acute angled?

Problem 1 is not difficult - there are at least two solutions. Problem 2 also is not difficult If I don't err I found an easy solution to it.

Why are you sending these problems in Open section?
the "[url=http://www.mathlinks.ro/Forum/index.php?f=47]Proposed & owned problems[/url]" section is for this.

The reason the problems to be posted together is because i think they are too easy for mathlinkers. To be here - there is a small probability for an error in second one. I think there is not, but I want to see a second opinion. While I not see second opinion it is an open problem.

For the first problem.
Idea for Solution 1. Use excircle, angles and arcs. After some additional constructions ... you may find two еqual triangles.
Idea for Solution 2. Use Sine law and similar triangles.

Problem 2. Idea: Use the method of supposal and contradiction for side lenghts or
angle dimensions and Sine Law.

Sorry, can you please precise your solution to 2.? In fact the problem is the particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=99674 for acute-angled triangle ABC, and even this particular case is, in my opinion, very hard (I am not even sure I have a proof).

  Darij

Hi, Darij,
I'm sending the solution (in the attached picture). I had problems with my Internet connection. Please, check it carefully - i almost believe I'm not wrong. If it is not a correct solution - the same idea may be used for this problem. I even suppose it is possible to prove - the case when the triangle is not acute angled is impossible. Have you any ideas for this proof? I think it is a good problem for math competition - if my solution is correct - even not very difficult.

P.S. If you have some problems with the attachment - please - write.

Hmm. First, why can you assume that $ a > b > c$ ? The situation is not symmetric, but only cyclic.
Then, why does $ a > b > c$ imply $ \sin\left(\gamma\minus{}\varphi\right) >\sin\left(\beta\minus{}\varphi\right) >\sin\left(\alpha\minus{}\varphi\right)$ ? I think it implies exactly the opposite...

  Darij

I'll write all cases carefully till 1-2 days. In this post or in new one. There is a good "probability" my solution to be right.

[quote="borislav_mirchev"]I'll write all cases carefully till 1-2 days.[/quote]

Not necessary. The cases $ a > b > c$ and $ a > c > b$ are enough.

  Darij

You made my task easier. You are right it is not "symetric" but the things works for me ;). In the attached file is the second part of my solution. What do you think? Any ideas for the case the triangle is not acute angled? May we prove this case is impossible?

P.S. If you have problems with attached file, please, say me! I saw this idea from a Russian problem: If AL=BM=CH (angle bisector, median, altitude) then ABC is equaliteral.  It is solved on mathlinks.

Why the ... does $ a>b>c$ imply $ \gamma>\beta>\alpha$ ? Or are your angles named in a strange way?

  Darij

gama = angle ACB = C
alpha = angle BAC = A
beta = angle ABC = B

by Sine law

a/sinA = 2R
b/sinB = 2R
c/sinC = 2R

My error was instead of sinA = a/2R i wrote ... sinA = 2R/a .... and I "proved" the statement. "It's human to err", but twice... I'll think about different solution.

darij, please see at the link you gave me for a beautiful solution. It is the most beautiful thing I've seen recently.

Take a look at this: http://www.mathlinks.ro/viewtopic.php?p=1131812#1131812


[quote="nguoivn"]Let $ a, b, c$ be nonnegative numbers. Prove that:
\[ \frac {a}{(b + c)^2} + \frac {b}{(c + a)^2} + \frac {c}{(a + b)^2} + \frac {28(ab + bc + ca)}{(a + b + c)^3} \ge \frac {11}{a + b + c}\]
:)[/quote]

Generally ,

$ \frac {a}{(b + c)^2} + \frac {b}{(c + a)^2} + \frac {c}{(a + b)^2} + \frac {4k(ab + bc + ca)}{(a + b + c)^3} \ge \frac {4+k}{a + b + c}$

holds $ \forall (a,b,c)\in{R^3_+}$iff $ {\frac{11}{2}\le{k}\le{20}.}$

>f:=numer(csum(x/(y+z)^2+4*k*x*y/(x+y+z)^3-(4+k)/3*1/(x+y+z))):

>symprove(numer(subs(k=11/2+(20-11/2)/(1+t),f)));

                   it is symmetric for {x, y, z}

                              ITS = [t]

                            IS = 1, N = 1

                        The inequality holds.

                              ITS = [t]

                            IS = 1, N = 1

                        The inequality holds.

                               DEGU = 3

                               DEG = 3

                            ITS = [t, ee]

                            IS = 1, N = 1

                        The inequality holds.

                            ITS = [t, ee]

                            IS = 1, N = 1

                        The inequality holds.

                            ITS = [t, ee]

                            IS = 1, N = 1

                        The inequality holds.

                            ITS = [t, ee]

                            IS = 1, N = 1

                        The inequality holds.

                  The original form is psd!(SYM3(1))

                            它是平凡非负的!

                             TIME = 0.312

[quote="nguoivn"]Let $ a, b, c$ be nonnegative numbers. Prove that:
\[ \frac {a}{(b \plus{} c)^2} \plus{} \frac {b}{(c \plus{} a)^2} \plus{} \frac {c}{(a \plus{} b)^2} \plus{} \frac {28(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^3} \ge \frac {11}{a \plus{} b \plus{} c}\]
:)[/quote]
____
Where's the equality?

For example, $a=b=1$ and $c=0$.
$uvw$ kills it.
But for the generalization there is a problem.