quadratics
limit
algebra
binomial theorem

[quote="chess64"]a) Prove that the values of $x$ for which $x=(x^2+1)/198$ lie between $1/198$ and $197.99494949\cdots$.

b) Use the result of problem a) to prove that $\sqrt{2}<1.41\overline{421356}$.

c) Is it true that $\sqrt{2}<1.41421356$?[/quote]

[hide="a"]

$x=\frac{(x^2+1)}{198}$
$198x=x^2+1$
$0=x^2-198x+1$You continue then.[/hide]
[hide="c"]$\sqrt{2}\equiv 1,4142135623730950488016887242097...$. Then, the statement is false.[/hide]

um, do you honestly think that he would post problems you can just plug in on a calculator in Pre-Olympiad? ;)

[hide="hint"]Extended Binomial Theorem should do the trick.[/hide]

[quote]m, do you honestly think that he would post problems you can just plug in on a calculator in Pre-Olympiad?[/quote]

Huh?  Does  truthfulness of $1+1=2$ depend on "in which forum the problem is posted?" :)
 IOW:  The problem is VERY easy, easy enough for a simple calculator (or  few seconds of  simple arithmetic even without a calculator).

Sure,  there may be a fun or a clever way to do this  problem, but there is no reason to think  that one has to have a  particular method just because a problem is in any particular forum.

I hope it increases your self-confidence to go around trivializing olympiad problems by using your calculator. :)

[quote="José"][quote="t0rajir0u"][quote="Gyan"]but there is no reason to think  that one has to have a  particular method just because a problem is in any particular forum.[/quote]

The point, I think, is that both problems are trivialized if one bothers to actually calculate with a calculator![/quote]

Sorry, but I can solve problems in any way I wanted. The thing to solve it. I don't "bother" because I use methods you don't like.[/quote]
1. You have already been proven false on your calculator in another thread.
2. You don't have one e.g. at IMO.

Jose, sure you can solve it any way you want, but honestly, what high degree of happiness did you gain by solving this problem with a calculator?  It all comes down to solving a quadratic, which is not hard to do if you know the quadratic formula...it's almost as if chess64 should restate the problem saying that someone gave him the problem(s) posted, and since he doesn't have a calculator, he need someone else who has a calculator to solve it for him (oh yes, and first one to post the answer wins!)

Basically, you have contributed nothing of any mathematical significance whatsoever. Before posting, Jose, please make sure that your either (a) your post contains informative value or (b) your post will interest or be useful to someone.

In general, the assumption on the main AoPS boards (i.e. the ones for problem solving math like GS, Intermediate, Pre Olympiad, and all the Olympiad forums) is that your solution should not invoke the use of a calculator unless the person who posted the problem specifically said you can't.

When a problem is trivialized by the use of a calculator, you should think to look for a more elegant method.  Additionally, whenever you see the source of a problem as ---- Math Olympiad, chances are that calculators are not allowed.

As an aside, I'd like to note that historically mathematicians have been displeased by proofs which require calculators/computers (most famously...the four color theorem).

He's just saying that it is obvious you can do it with a calculator, so we want a better way. A monkey could use a calculator. So please stop.

[quote="José"][quote="ZetaX"]
1. You have already been proven false on your calculator in another thread.
2. You don't have one e.g. at IMO.[/quote]

1. are you perfect???? can't you make a mistake?
2. and? My life is not only an IMO

what's wrong with you? what do you have against me, man?[/quote]
ZetaX just wants ppl not to decrease the quality of the forum. Consider other ppl who want to see a proof of a problem. They will not learn much if they just see a solution which requires pressing keys on a calculator instead of using mathematical logic. And ZetaX is pretty ownage in my opinion.

[quote="José"]
1. are you perfect???? can't you make a mistake?
2. and? My life is not only an IMO

what's wrong with you? what do you have against me, man?[/quote]

He, I, and many other people are annoyed that you continue to mess up perfectly good threads by posting uselessly.

Consider NOT POSTING AT ALL if you don't have anything useful to contribute (for your purposes, consider 'something useful' to be a complete, well written solution).

[quote="José"]1. are you perfect???? can't you make a mistake?[/quote]
No, I'm very far away from being perfect.
[quote="José"]what's wrong with you? what do you have against me, man?[/quote]
Nothing against you, but only against your posting/problem solving habit.

[quote="rem"]ZetaX just wants ppl not to decrease the quality of the forum.[/quote]
Yes, but I failed  :(

Sorry for reviving an old post, I didnt know where else to put this. I found another solution for a. Could you guys see if it's good?

[hide]We can manipulate the equation to yield
$ x^2 \minus{} 198x \plus{} 1 \equal{} 0$

Since it's a quadratic, it must have two roots. Let's let
$ f(x) \equal{} x^2 \minus{} 198x \plus{} 1$

We know that for two numbers $ a$ and $ b$, $ a < b$, if $ f(a) < 0$, and $ f(b) > 0$, or if $ f(a) > 0$, and $ f(b) < 0$ then between $ a$ and $ b$ there must be a root.

$ f(\frac {1}{198}) > 0$. $ f(1) < 0$. So, we know there is a root of $ f(x)$ between $ \frac {1}{198}$ and $ 1$.

We find that $ 197.994949\cdots$ is $ 198 \minus{} \frac {1}{198}$. $ f(198 \minus{} \frac {1}{198}) < 0$.r

$ f(198) > 0$. There must be a zero between these two also.

QED
[/hide]

Any comments or suggestions would be greatly appreciated.
Thanks!


MIT college Harvard Princeton Stanford Support Gauss

What is the strongest math program in the nation?

According to US News Ranking it is...

1. MIT
2. Harvard, Princeton, Stanford, UC-Berkeley

But.. how legitimate is the provided ranking?

Although it might be different for each field of mathematics, here we are primarily concerned with 'general' reputation of school and 'overall' image of the school.

I'm just taking a reasonable guesss here, but I'm saying that there's a difference between having the overall best program for math in the country or having the best certain course for math in the country. I'm sure if people come to a consensus that MIT is the best, that doesn't mean that all of their classes are the best in the country...I mean, I'm almost positive Harvard would have probably a few up there.

-jorian

Hmm, I would not have expected Stanford's math program to be up there, but I'm very glad it is :P.

[quote="JSteinhardt"]It is very difficult to say which is the best. Each of these schools have different but very strong math programs. I know the most about MIT, Harvard, and Princeton (based on talking with friends who are there), so I will discuss those.

Harvard has Math 55. This class is basically in a class of its own (no pun intended). Most of the people I know are Freshmen, so beyond that I don't know what they have.

Princeton, from what I have heard, probably has the best staff of any that you listed, if you are going based on contributions to super-famous modern developments in mathematics (I believe everyone involved in the proof of Fermat's Last Theorem is part of their faculty, for one thing).

MIT seems to have a student culture that supports mathematics (the same can be said of any of the above schools, but from what I have heard, the culture at MIT is particularly intense among the mathy folks). They also let you take any classes you want.[/quote]

mmm, ribet is at berkeley

The math program at any school is only as good as the students, and the students are only as good as the effort they put in. Many schools offer accelerated classes, accomplished faculty, and so forth, and most, if not all, have a library. Unlike other sciences, high quality facilities don't mean as much.

In short, all those math programs are great and who cares which one is best. It's just between the student and the math, the math in the books, and his/her peers.

I wonder which people have voted here. MIT is great but I do not see why it should have a much better math program than, say Berkeley. It seems people have rather voted for the school they like best. Go to the respective course lists and compare.

I haven't voted yet since I don't really know...

orl, people can vote for a school without thinking that school is [i]much[/i] better than the others.  They may think the programs are really close yet one is slightly better.

On the other hand, there are a lot of factors which would probably need to be considered when figuring out which one is the best.  I guess mostly what should be considered are the calibre of the students in the math department, the quality of the teachers, and the classes and how well they are taught.

It seems to me that the strongest student body is at MIT.  The other factors I'm not so sure about.  But basically, I think the math programs at Harvard, MIT, and Princeton are about equivalent in terms of what math undergraduates can get out of them.  It doesn't seem to me that math majors at one of these schools have a huge advantage over math majors at another school.  If you're trying to decide between the top schools, I would place much more of an emphasis on other factors such as geography, surroundings, and the required curriculum (do you want a solid science background or an exceptional humanities one?).

What is the meaning of "strongest" in this context? I could imagine that different schools would be "best" for different students.

I think its human nature to try to rank things -- it makes things easier to analyze. But I don't think everything (including the strengths of different math departments) has a logical ordering. Every student is different, and every student will have a different experience a given school.

If I were forced to try to make some sense out of this question though, I would probably look at what recent graduates from those school's departments have gone on to do, and see which ones have accomplished things that seem appealing to you. For example, what graduate schools did they go to and where are they working now? That information might be hard to find though. The best way to find it would probably be to talk to a current student or recent graduate of the school.

For strongest student body I suppose you could look at the quantity and quality of the math students at each school.  This is also hard to determine (to the accuracy needed to declare a student body as the strongest).  So even though my statement in my last post that MIT has the strongest body was an opinion, it was perhaps a little naive.

What exactly is Harvard math 55? I read the topic called that but it didn't really explain it.

It's a freshman math course at Harvard.  [url=http://www.math.harvard.edu/undergrad/Pamphlets/freshmenguide.html]Here's[/url] one site that gives a brief description of it along with other freshman math options at Harvard.

This debate probably won't get anywhere. In order to find out how each experience is, you'd need to have some guy attend each of those schools and take a bunch of classes in each, but even so it'd be biased because some schools have better this, some schools better that. Who cares.

The US News Ranking you quote there is only a ranking of the graduate programs, not of the undergraduate ones.

The revival of this thread from last year shows that this is a topic of continued interest.

[quote]I believe everyone involved in the proof of Fermat's Last Theorem is part of their faculty, for one thing[/quote]

Yes but Andrew Wiles, the man who is most acredited with proving the Tanyama-Shimura conjecture and thus Fermat's last theorem attended Cambridge university. He also chose to announce the proof at the Sir Isaac Newton institute in Cambridge. I would say be adventurous and travel abroad to England and follow in the footsteps of Sir Isaac Newton and in the footsteps (perhaps "wheel-marks") of Stephen Hawking (who holds the post of Lucasian professor that Newton once held). However, I am rather biased as I will be going to read maths at Cambridge in a month. I would have applied to Harvard if i were living in America.

[quote="curl"]"These courses are not taught in any other university because no other university has the same caliber of first-year mathematicians"

I saw this in a description of math 25 and 55 on Harvard's math dept. website. Does anyone else find this a little pretentious?[/quote]

Where did you see this on the website and do you know whom it was written by?

I happen to disagree with the statement as its written for several reasons. First of all, because I can think of several other schools who's top undergraduate students have comparable mathematical ability to Harvard's students.

Secondly, I disagree with the supposition that such courses are the only ideal introduction to undergraduate mathematics for top students.

The quotation comes from

http://www.math.harvard.edu/pamphlets/courses.html

I strongly agree with both of gauss202's points. I think this kind of language, presented by an official body, is a serious issue. The elitist intellectual attitude is very easy to develop when you are a student at a top university, and these universities should be making efforts to counter that effect, not to encourage it.

I heard that Princeton has a really good math program and physics program.
I heard that one professor is considered one of the top 5 physicists of all time.

Which professor?

Newton, Maxwell, Einstein already have three of the top 5 pretty easily...

Schrodinger is also almost certainly in the top 5, and then there's people like Ampere, Gauss, and Planck who, if not in the top 5, are certainly contenders. And as pointed out, Stephen Hawking is at Cambridge unless that changed at some point.

If you mean a current professor, perhaps it's Witten (Fields medalist)? Or Taylor (Nobel laureate)?
http://en.wikipedia.org/wiki/Edward_Witten
http://en.wikipedia.org/wiki/Joseph_Hooton_Taylor,_Jr.

What is University of Chicago ranked in mathematics? I have heard good things about that school as well.

I wanna add that its not all about the university you go to. It has to more do with amount of work you put into your studies. Yes, Princeton is one of the best but if you are a lazy type of guy its not gonna do any good. Similarly if you go to an ordinary college but you are very serious about mathematics you can sometimes get departmental approval to let you study you own advanced stuff not taught in college. Going to a top university will help you get better but you can avoid it altogether (not that I am saying that is what you should do). That is why Karl Weierstrass is a very big inspiration to me because he never got his doctrate and yet earned the immortal title "father or analysis".

Unless Prof. Jesus Christ Superstar Einstein IV is going to be your personal mentor, why exactly is it critical to your undergraduate education if you might occasionally pass by him on campus or have him be doing research you could never understand or even teach one of your classes that any qualified math professor would grasp to the same extent?


Here is a problem that took me quite a while...

There are twelve pennies. Among the 12 pennies there is

one bad penny that either weighs less or more than the

normal pennies. You have a scale that tells you whether one

side is heavier than the other or not.

Find the ultimate solution to find the bad penny in 3 steps.

(meaning using the scale only 3 times...)

Consider all the possibilities...

If you have seen this problem before, and solved it

give some chance for other people.

[hide] And Don't cheat...[/hide]

[hide]Put 6 on each side.  Set aside the six on the side that weighs more.  With the six remaining, split them into two groups of three, and weigh those.  Set aside the three that weigh more.  Take the last three, and pick two from there.  Put one on each side of the scale.  If they are equal, the one that you set aside is the light one.  If they are different, the lighter one is the light one.[/hide]

But you don't know whether the bad penny is lighter or heavier.

Magnara, I think you misunderstood the problem. The "bad" coin isn't necessarily lighter than the others, and you don't know if it's lighter or heavier when you start the weighings. This makes the question soooo much more fun.... and soooo much harder!

I think I really need to start actually reading these problems.  And if I think it's too easy for advanced, it probably is....

Having Fun?

[hide]

There is more than 1 answer to this problem.

And don't cheat!

And have fun w/ it!

[/hide]

[hide]Divide 12 coins into 4 piles a,b,c,d, each pile has 3 coins[/hide]

Am I at least on the right track?

hmm... you might be...

I've done the problem w/o any specific algorithm
but i've heard that it is possible to approach the problem
mathematically.

This doesn't quite work out...but here's what I did.

[hide]Divide into three piles of four. Weigh two piles.

Case 1: If they are the same, the bad penny must be in the third pile. Weigh any two of the pennies in the last pile.

Subcase 1: Weights are equal. Remove one penny and replace with any other penny in the last pile. If they are still both the same, the other penny left is the bad penny. If they are different, the one replaced is the bad penny.

Subcase 2: Weights differ. Remove any one penny, and replace with another penny. If weights still differ, the penny not removed is the bad penny. Otherwise, the penny removed is the bad penny.

Case 2: Two piles are of unequal weight. Err...unless I'm mistaken, this requires four tries. Can anyone come up with fewer?

[/hide]

It might seem that unequal case needs 1 extra step but
it can be done within 3 steps as the problem is stated.

I got four tries too using my way, but the question asks for three.

Is any of us on the right track?

[hide]Divide 12 coins into 4 piles a,b,c,d, each pile has 3 coins

Take any a and b piles and weigh them.

  If they are balanced, then none of them have the bad one.

The bad one is in pile c or d.  Take pile a and compare it with pile c.

     If they are balanced, then bad one is in pile d. Take any two of the 3 in pile d and compare them. if they balance, then good, the other one is the bad. If they don't, we  have a problem.that takes another two tries.

     If a and c don't balance, means one of the 3 in pile c is bad. we only need one try by taking any 2 of 3 and compare them. we already know the bad one is heavier of lighter by comparing a, c

If a and b do not balance. compare a to c, since all 3 in pile c are normal.  if they balance, then the bad one is in b, but we do not know if it is heavier or lighter, so it takes another 2 tries

 if they don't we know the bad one is in pile a and it's heavier of lighter. only take one more try[/hide].

I did it too by weighing groups of 4 in the first step.
There is a trick involved in the case when the first measurement
isn't equal.

[quote="xirtamax"]I did it too by weighing groups of 4 in the first step.
There is a trick* involved in the case when the first measurement
isn't equal.[/quote]

*Sneak in a fourth weighing while no one's looking.

lol

if we do it that way, we can get it in 2 steps.

I was just found this site tonight and this will be my first post on it :)  I thought up a solution but it has one step that might not be considered valid.  It is a creative step though....

[hide]

Weight 4 vs. 4.  If they are equal we know the bad penny is in the last group of 4.  Weigh 1 vs. 1 from the last group.  If not equal.  Weight either one of those 2 with any other penny.  If it is equal it was the one of the 2 we did not wiegh.  If not equal it was the one we did.  If that 1 vs. 1 was equal, then that tells us the same information as weighing the other 2 and them not being equal, so follow the above last step on those 2 that we know aren't.

Dang I need to learn to write shorter, but anyways...

If the initial 4 vs. 4 was not equal, then steal 2 pennies from Bill Gates and assume them to be normal weighted pennies.  This is the step that might not be considered valid.  Take any 3 of the penies on the left side and any 3 on the right side of the 4 vs. 4 we just did, and put them on the same side.  Then take the 4 from the group we know is right, and our stolen 2 on the other.

If this 6 vs. 6 is equal, then one of the 2 excluded - from this test is the bad one.  Just weight one of them aginst a good one, if its not equal its that one, if it is equal its the other.

If this 6 vs. 6 is not equal.  We just gained some valuable info.  We know if the bad penny is heaver or lighter than a normal penny.  So...if it is heavier, then we know the bad penny was on the side of the 4 vs. 4 test that was heavier.  So we know it is in one of those 3 pennies that we taken from the heavier side.  Weigh any 2 of those three pennies.  If one is heavier than the other, then it is the bad penny.  If they are equal, then the 3 penny is the bad one.  If from the 6 vs. 6 we determined the bad penny is lighter, then do the same steps as above but with the word lighter in place of heavier.

[/hide]

The moral of the story is to try to determine if its heavy or light.  And also that one must steal.

What the heck.  I told it to set my font color to white on my solution.  But.....its black.  Well, if some one scrolls down this far they are proly looking for the solution anyways.

[quote="Douglas"]If the initial 4 vs. 4 was not equal, then steal 2 pennies from Bill Gates and assume them to be normal weighted pennies.  This is the step that might not be considered valid.[/quote]

You know that the third group of 4 pennies is all normal, so you don't need to steal from Mr. Gates. :)

Also, you can use the [spoiler] tag to hide your text. Nice solution, by the way.

Well, since I can't steal anymore, I did this...

[hide]

Okay, back to if the 4 vs. 4 is not equal.  Take 3 from one side and 1 from the other and weight that aginst the 4 we know are normal in another 4 vs. 4.  If they are not equal.  We know if the bad penny is heavy or light.  If the 1 penny came from the heavy side, its that guy.  Else it is one of those other 3 and you can determine the bad penny in 1 weighing if you have 3 pennies and you know if its heavy or not.  If the 4vs. 4 was light then do same steps above with light in place of heavy.

So, what if this 4 vs. 4 is equal.  Then the bad one is either the 1 that was not in the group of 3 we took from one side, or the 3 that were not the 1 we took from the other side.

Take 1 of those 3 out.  Put 1 of those 3 on the side with the 1.  Then put a normal weighted penny on the side with the 1 of the 3 that is left.  So we have a 2 vs. 2

If this is equal, it is the 1 we took out.  If the side with the 1 and the 1 from the 3 is heavy, and the side we got the 1 from was heavy, then it is that 1 that is bad.  If it was the side with the 1 from those 3 that was heavy, then it is that 1 from the 3 that is bad.  If this side is light and the side we got the 3 from was heavy, then it is the penny next to the normal penny that is bad.  And do same stuff with light in place of the word heavy in the above steps.

Man...I need to learn to write shorter and clearer.  It would have proly helped if I labed pennies A1,A2,B1,B2.. instead of saying things like "the 1 from the 3".  And maybe call sides of the balance S1 and S2.[/hide]

Douglas said:
[quote]
If the side with the 1 and the 1 from the 3 is heavy, and the side we got the 1 from was heavy, then it is that 1 that is bad.
[/quote]

Hmm... You did a good job, but i think that part is wrong.
The remaining penny of the hevier side could be the bad penny,
but the penny on the left side that measured lighter could also be
the bad penny.
Everything else seem to work.

Okay I'm about 99% sure this works.  The explanation may be confusing, but bear with me.

[hide]

NOTE: Nv1 is the proccess to determine between 2 that you are unsure about for your final weighing.  You have 2 that you are unsure about.  Measuring 1 of those against a normal penny will give you a definite answer.  Here's why:  If they are equal, the one you didn't measure is the faulty penny.  If they are unequal, you know that the one that is NOT normal is the faulty penny.

Measure 4 against 4 first.  That's your 1st weighing.

A)  Assume they are equal

You now know 8 that are definitely normal.  On your 2nd weighing, do 1 against 1 (of those you don't know).  If that was even, then do Nv1 with the remaining 2.  If the 1v1 was uneven, you do Nv1 with those 2.

Now, if the original 4v4 was unequal,

B) Measure 3v3 like this:

HHL v HHL. (H = heavier, L = lighter.) This will be your second weighing.  If these are equal, You do the process of Nv1 to measure the 2 remaining pennies.

If HHL v HHL was uneven, then you are left with 3 pennies that could possibly be faulty.

Here's why:

In the 4v4, one side went down, and one side went up.  Out of those pennies, only one is faulty.  Either the penny is light, and therefore made it's side weigh less, or the penny was heavy, and made its side weigh more.  When you do HHL v HHL, (each letter representing an individual penny), one side will go down, and the other will go up.  Therefore, the only possible faulty pennies are the HH on the heavier side or the L on the lighter side.  If any other penny was faulty, it would both be heavier and lighter than the other pennies.

So, with your final weighing, you have HHL.  Do this measurement:

HL v NN (N represents a Normal penny).

And that's the genius step.

If equal, the only questionable penny is the faulty one.  If they are unequal, then you can figure out very easily which is the faulty penny.  IF the HL side goes up, L is the faulty penny.  If the HL side goes down, H is the faulty penny.

[/hide]

I'm done now.  That took me a while.

Good Job, Green Twilight.

[hide]

The trick was to use the elimination method

and the second step had to be HHL and HHL

or LLH and LLH because we have to have only 3

remaining suspects to complete the problem.

[/hide]

But I believe there is an another solution. I only was able to

solve the problem w/ Green's solution, but there is another one.

(Actually, I was too happy that i solved it, and never went back

to try other methods.  [/hide]

I forgot to mention, can you put the solution in spoiler.

[I've edited in a few spoilers. By the way, Douglas, the reason it showed up black was because you had [color=white][/color] then all the text afterwards, instead of in the middle of that].

This is a very very famous problem.. and now theres about 1000 variants of it. I've seen the solution a few times, never really tried it properly myself, I get too fed up trying :)

[quote="TripleM"]This is a very very famous problem.. and now theres about 1000 variants of it.[/quote]
Indeed.  But, now that someone has found a solution, extend the problem.  The next incremental step is to ask not only which coin is cointerfeit, but whether it is heavier or lighter than normal.  All other conditions and requirements remain the same.  Yes, it is possible.

rcv wrote:
[quote]
Indeed. But, now that someone has found a solution, extend the problem. The next incremental step is to ask not only which coin is cointerfeit, but whether it is heavier or lighter than normal. All other conditions and requirements remain the same. Yes, it is possible.
[/quote]

If you look at greentwilight's solution, you'll see that
we can not only figure out the bad penny but figure
whether it's heavier or lighter than the normal ones.

[quote="xirtamax"]If you look at greentwilight's solution, you'll see that
we can not only figure out the bad penny but figure
whether it's heavier or lighter than the normal ones.[/quote]

[quote="GreenTwilight"]NOTE: Nv1 ... If they are equal, the one you didn't measure is the faulty penny.[/quote]

Usually, we can distinguish heavier vs lighter.  But if GreenTwighlight's first two measurements are even, he uses NV1 to determine which of the final two coins are bad.  How can you know if an unmeasured coin is heavier or lighter?

Yes, I think if the initial 4v4 was even, then there is a case where you don't know the weight of the penny.

Ok here we go:

[hide]

HHHH > LLLL

case 1:

HHL > HHL

It's either HH on the left side or L on the right side.

Measure HL and 2 normal pennies.

If the HL is greater than it's H.

If the HL is lighter than it's L.

If the HL is equal than it's H.

case 2:

HHL < HHL

Same as case 1, just sides reversed.

case 3:

HHL = HHL

then we measure one of the 2 remaining L and a normal penny.

If L is equal to normal than it's the other L.

If L is lighter than normal than it's that L.

[/hide]

xirtamax wrote:Ok here we go:
[hide]HHHH > LLLL[/hide]

Unless I am not seeing something, I think you are still missing the case where the first two measurement are both in balance.

Oh I see. Greenlight's solution is different than what i had in mind.

I just looked at the unequal case.

And I thought that the equal case is obvious, but I guess it's a little

more than obvious. (not too hard to figure out though once you

know how to find the bad penny.)

[hide]

N = Normal

U = Unknown

H = Possibly Heavy

L = Possibly Light

Measure:

NNN and UUU.

Case 1:

NNN < UUU then UUU becomes HHH, then measure

HN and HN.

   If HN > HN, then H on the left is the bad penny.

   If HN < HN, then H on the right is the bad penny.

   If HN = HN, then the third H is the bad penny.

Case 2:

If NNN > UUU then UUU becomes LLL, then measure

UN and UN.

   Use the same method as case 1.

Case 3:

NNN = UUU, then measure N and U.

U will be figured out.

[/hide]

[quote="xirtamax"]Oh I see.[/quote]
Good!  I think your revised solution takes care of my issues.

Let me point out there is one more extension.

"There may or may not be a counterfeit coin."

Some say starting with this extra condition actually makes the original problem easier.

This is a stupid small thing...but isnt

HN v HN the same as H v H?

Anyways, this is probably in a tie for the coolest problem I have seen.


analytic geometry
graphing lines
slope
inequalities
algebra
linear equation

PROBLEM
Planning for the prom

Pam and Remy are organizing the junior prom. They plan to sell two types of tickets- individual tickets and couples tickets. The ballroom where the prom will be held holds a maximum of 400 people.

Pam is pretty opinionated. She believes the prom will be more successful when more people go as couples. So that at least half the people at the prom will be in couples, Pam and Remy decide that the number of individual tickets they sell should be at most twice the number of couples tickets.

Remy ordered door prizes to be handed out at the prom (one prize for each ticket, even if it's a couples ticket). Unfortunately, the supplier to get more, so they can sell a total of at most 225 tickets.

Version a: Suppose individual tickets sell for 22 dollars and couples tickets sell for 30 dollars. HOw many of each type of ticket should they sell to maximize the money they take in?

Version b: ...individuals ticket 15 dollars and couples ticket 35 dollars...?
Version c: ...individuals ticket 30 dollars and couples ticket 20 dollars...?

This problem is too wordy and confusing. So there's at most 225 tickets and the number of couples ticket is x and individuals ticket is 2x... How do you find how they can maximum the take-in???

There's graphing involve but I am not quite sure how to do that.

I need help with this as quick as possible. Thank You!

this isnt a place were you come and ask for homework help, if you needed help so badly, im sure you could find out how to do it if you cared enough, you obviously dont and you want us to do it for you...

Actually, I disagree - this is a great place to get homework help.  What's the point of a community if people can't get help from it?

Whenever tackling a word problem, the first thing you should do is figure out what the problem is asking for.  So let's take a look at Version A: Suppose individual tickets sell for 22 dollars and couples tickets sell for 30 dollars.  How many of each type of ticket should they sell to maximize the money they take in?

So it's an optimization problem.  First, let's define a few variables.  Let's say $x =$ the number of individual tickets they sell.  Since the maximum number of tickets they can sell is 225, the number of couples tickets they can sell is $225 - x$.  Therefore, the revenue from the sale of tickets is:

$R = 22x + 30(225 - x)$

Simplify this linear equation:

$R = 6750 - 8x$

Because this is a line and it has a negative slope, the obvious maximum value would be at $x = 0$ since you can't sell a negative amount of tickets.  However, there is a restriction on $x$.  If $x = 0$, then there would be 225 couples tickets sold.  For every couples ticket there are two people going, meaning there would be a total of 450 people at the ball.  The maximum capacity is 400 people.  So $x$ cannot be 0.  We need to find the number of individual tickets to sell so that the number of people going is exactly 400 in order to maximize the revenue.

Using our original variable $x$, the total number of people attending would be:

$P = x + 2(225 - x) = 450 - x$

The information given about maximum capacity creates the inequality $450 - x \le 400 \rightarrow 50\le x$, but we want the minimum value of x, which is $x = 50$.

In order to maximize revenue, 50 individual tickets and 175 couples tickets should be sold.  This does not violate the other condition where the number of individual tickets can be at most twice the number of couples tickets sold.

Using this approach, you should be able to solve the other versions as well.


Hello,
I am trying to graph the mapping of the transformation u=xcosy and v=xsiny where 1<=x<=2 and 0<=y<=1.
I have some trouble sketching the area, basicly on the fact that i reach the step where i find that 0<=arctan(v/u)<=1. Does this affect the shape of the region that corresponds to the mapping?
Thanks

if we had $ 0 \le y \le 2\pi$ we would get a washer

looks nicer when $ y\equal{}\theta, x\equal{}r$

so instead of the full circle, we only get part of a washer,


[quote="epatjn"]if a player has  2/5 chance of earning a "walk" on each up-to- bat, what is probability that earn a walk exactly once in next two plate appearances?

answer is 12/25 but how do you get that?[/quote]

[hide]Let W be a walk and N be not a walk.

These are the combinations that can happen in two at bats.

W,W
W,N
N,W
N,N

Of those, W,N and N,W have the property that there is exactly one walk (W) in the two at bats.

The chance of getting W,N is
$\frac{2}{5}\cdot \frac{3}{5}= \frac{6}{25}$

The chance of getting N,W is
$\frac{3}{5}\cdot \frac{2}{5}= \frac{6}{25}$

Thus, the chance of getting one walk in two at bats is
$\frac{6}{25}+\frac{6}{25}= \frac{12}{25}$.
[/hide]

how did you get
The chance of getting W,N is
3/5?

i get the options of
w,w
w,n
n, n
n, w

The chance of getting W,N is $\frac{6}{25}$.

I got this by multiplying $\frac{3}{5}\cdot \frac{2}{5}$,
because this means, "the probability that this event will happen, then this event".

In general, if we have two probabilities, $P_{1}$ and $P_{2}$, the probability that $P_{1}$ will occur, then $P_{2}$ will occur is $P_{1}\cdot P_{2}$.


My first ever math contet comes up next sarturday.i need tips on (Algebra,combinatorics,tringles,bearings,compoud anglesfactorisation and any othrt topic you know i have not mentioned).I realy need what they call math contest tips.

Reading the AoPS books would have helped but now it's too late.  Practice problems and get a good night's sleep.

[quote="21 7 15"](Algebra,combinatorics,tringles,bearings,compoud anglesfactorisation and any othrt topic you know i have not mentioned)[/quote]

Oh man that's quite a list. Work over basics if you know them; if you don't, like what JRav said, it's too late to learn for the competition.


function
calculus
limit
algebra proposed
algebra

This one seems to be a criterion of continous function:
Let f be a surjective function on [0,1] and f is increasing.Prove that f is continous on [0,1].
...Any role in this one?

This probably belongs in the calculus-analysis forum; it's not algebra. I also assmue that the condition is that $f$ maps [0, 1] onto [0, 1] (or to some closed bounded interval). But yes, it is true.

Note that since $f$ is increasing, it is injective, which means that it is bijective and has an inverse. Suppose $f(a)=b$, and suppose for the time being that $a,b\in (0, 1)$. Choose $\epsilon >0$ arbitrary except for the condition that it is small enough that $b+\epsilon$ and $b-\epsilon\in [0,1]$. Then let $u=f^{-1}(b+\epsilon)$ and $v=f^{-1}(b-\epsilon)$. Since $f$ is increasing, $v<a<u$. Let $\delta = \min(u-a,a-v)$. Then, $|x-a|<\delta\implies v<x<u\implies b-\epsilon<f(x)<b+\epsilon$. Thus, $f$ is continuous at $a$. A slight modification of the same argument shows that $f$ is continous at 0 and at 1.

This is a standard argument, and is in a number of textbooks. It's particularly useful for showing that the inverse of a continuous function from an interval to an interval is also continuous. This theorem is a companion (and partial converse) to the Intermediate Value Theorem.

I bet you can find this criterion in any analysis book you can get your hands on :).

It uses the fact that every monotonic function has lateral limits in any point, and if $\lim_l,\lim_r$ are the left and right limits in $x_0$ respectively, then $\lim_l\le f(x_0)\le\lim_r$. The surjectivity simply tells us that $\lim_l=f(x_0)=\lim_r$, which is the condition for continuity.


[quote="guile"]Do you know a site where I can find KSEG activities?[/quote]

This could maybe help, but I'm not sure: http://www.mit.edu/~ibaran/kseg.html
http://en.wikipedia.org/wiki/KSEG


[quote="buzzer11"]The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. What is the largest interger that can be an element of the collection?
Source: AMC[/quote]
what's unique mode? I have never heard of it before. :huh:

[quote="buzzer11"]The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. What is the largest interger that can be an element of the collection?

Source: AMC[/quote]

[hide] Obviously, $8,8$ are the middle 2 numbers. Let's call the largest number $a+8$ and the smallest $a$. We want $a+8$ as large as possible, so we need lower numbers on the right hand side than on the left to balance the mean at $8$. So our 8 integers are $a,a,a,8,8,8,8,a+8$. $4a+40=64\Longrightarrow a=6$. $\boxed{14}$? [/hide]

[hide]Yes, the answer is 14.  :) [/hide]

[hide]We see that 15, is impossible but 14 is possible.

When it is this

6,6,6,8,8,8,8,14

$14$[/hide]


What is the simplified value of $ 0.75 \div \frac{1}{4}$?


Prove that there exist 10 distinct real number $ a_1,a_2,..,a_{10}$ such that the equation:
           $ (x\minus{}a_1)(x\minus{}a_2)...(x\minus{}a_{10})\equal{}(x\plus{}a_1)(x\plus{}a_2)...(x\plus{}a_{10})$
has exactly 5 different real roots.

For example $ a_6 \equal{} 0,a_7 \equal{} 1,a_8 \equal{} \minus{} 1,a_9 \equal{} 2,a_{10} \equal{} \minus{} \minus{} 2$.
Then $ x \equal{} 0,\pm 1,\pm 2$ are roots. Another roots satysfyed
$ \sigma_1 x^4 \plus{} \sigma_3 x^5 \plus{} \sigma_5 \equal{} 0.$
Were $ \sigma_1 \equal{} a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5, sigma_3 \equal{} \sum_{1\le i < j < k\le 5}a_ia_ja_k, \sigma_5 \equal{} a_1a_2a_3a_4a_5$.
It had not another real roots if $ a_i$ are positive, because roots are $ x^2 \equal{} \frac { \minus{} \sigma_3 \plus{} \pm \sqrt {\sigma_3^2 \minus{} 4\sigma_1\sigma_5}}{2\sigma_1}.$ If $ D < 0$ $ x^2,x$ are complex, if $ D\ge 0$ $ x^2 < 0$.


Sometimes Friday if I'm not too tired, but usually Saturday. Sunday only if I had absolutely no time before, and never Monday morning.

usually friday. that way i can enjoy my weekend

i sometimes don't get weekend homework :D too bad you guys always do :P

In senior year, one can expect a lot of homework.  I've only had two days of school and already I have an essay to write, two presentations to give, and lots of reading.  And I have a test in about a week (though that's not such a worry).  Also, I have college applications to worry about.  :maybe:
And then you have to worry about running clubs, activities, groups, etc.

IT IS LIKE A RIGHT-SKEWED NORMAL DISTRIBUTION

Mostly Saturday, occasionally Sunday (i.e. this weekend, because I'm procrastinating on history homework).

usuallly monday, but now our homeroom is in the middle of the day, so, i probably can't do it. :(

my bus ride is really long because i take a bus to another school, get off, and ride that bus all around the county before it gets to a stop near my home.

i usually do my hw then, on Fridays.

Hmm freshman year:

Math homework: Nope. I do it during the lunch period before it's due.

AP bio homework: Chh-yeah I HAVE to do it during the weekend.

English homework: Meh. Mostly during weekends.

Spanish: See math.

Government: What homework?

I think that's it.  :lol:

I know its a bad habit, but I usually procrastinate and tend to leave all my homework to Sunday (although never Monday)

im in elementary school still so i dont get a lot of homework

This isn't even an interesting thread to bump


group theory
abstract algebra
number theory unsolved
number theory

Let $ A$ be a set of positive integers such that no positive integer greater than 1 divides all the elements of $ A.$ Prove that any sufficiently large positive integer can be written as a sum of elements of $ A.$ (Elements may occur several times in the sum.)

Take some $a \in A$. If $a=1$ we're done, otherwise factorize $a=\prod_{k=1}^n{p_k}^{e_k}$. By the fact that the setwise $\gcd$ of the elements of $A$ is $1$, we can find $a_k \in A$ such that $p_k \nmid a_k$ for all $1\leq k \leq n$. Take $P = \prod_{k=1}^n p_k$, then $b_k = a_kP/p_k$ for all $1\leq k \leq n$, then take $b=\sum_{k=1}^n b_k$. Since $p_k \nmid b_k$ and $p_i \mid b_j$ for all $1\leq k \leq n$ and $1\leq i\neq j \leq n$, it follows that $p_k \nmid b$ for any $1\leq k\leq n$, thus $\gcd(a,b)=1$. Notice that $b$ is a sum of sums of elements of $A$, thus a sum of elements of $A$.

By Sylvester's theorem, or any other weaker and easier to prove method, there exists some $N_0$ such that any positive integer $N\geq N_0$ can be represented as a sum of $a$'s and/or $b$'s (from Sylvester's theorem $N_0 = ab - a - b + 1$), thus as a sum of elements of $A$.

In effect, this problem shows that the additive subgroup $\left < A \right >$ generated by a set $A$ of positive integers with setwise $\gcd$ of its elements $1$ contains consecutive numbers from some point on. Since for any additive subgroup $S$ with setwise $\gcd$ of its elements $1$ we can take as generating set $S$ itself, this is a structure theorem for the additive subgroups of $\mathbb{N}^*$ (and it immediately extends to the case when the setwise $\gcd$ is any $d$).


Do you know any more of those shortcuts?? {hold alt and press one or more numbers in the [b]side[/b] numberpad, then release alt}

Yess theres a whole bunch of them. The number you press in the numpad is the ASCII code for the character. To find the shortcuts, go to the Start menu and press Run. Then type in "charmap". Choose the character you want a shortcut for and look in the lower right hand corner for the keystroke. note that you can only go up to  (alt-0255) because the characters above that are Unicode.


Yes! Your solution is nice. This problem can done by some way. Your solution is one of them.

generalization problem :  $ \sum {\frac{1}{ma\plus{}nb\plus{}pc}}\leq\frac{1}{m\plus{}n\plus{}p}(\sum {\frac{1}{a}})$
the case of this problem is $ m\equal{}2008,n\equal{}p\equal{}1$

OK! this is my problem!
$ \sum {\frac{1}{ma\plus{}nb\plus{}pc}}\leq\frac{1}{m\plus{}n\plus{}p}(\sum {\frac{1}{a}})$

a same problem in some ebook:
Let $ a,b,c>0$ and $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{} 4$.

Prove:     $ \frac{1}{2a\plus{}b\plus{}c}\plus{}\frac{1}{2b\plus{}c\plus{}a}\plus{}\frac{1}{2c\plus{}a\plus{}b}\le 1$.

[quote="thanhnam2902"]a same problem in some ebook:
Let $ a,b,c > 0$ and $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} 4$.

Prove:     $ \frac {1}{2a \plus{} b \plus{} c} \plus{} \frac {1}{2b \plus{} c \plus{} a} \plus{} \frac {1}{2c \plus{} a \plus{} b}\le 1$.[/quote]
 similar  we have :
$ A\equal{}\frac {1}{2a \plus{} b \plus{} c} \plus{} \frac {1}{2b \plus{} c \plus{} a} \plus{} \frac {1}{2c \plus{} a \plus{} b}  \leq \frac{1}{16} (\frac{4}{a} \plus{}\frac{4}{b}\plus{}\frac{4}{c} \equal{} 1$


You can't factor sum of squares in any way, so it would be the quickest except if you notice $ 3^2\plus{}4^2\equal{}5^2$

The sum of the first $ n$ square numbers is $ \frac{n(n\plus{}1)(2n\plus{}1)}{6}$. On this problem, you want the sum of the first $ 5$ perfect squares minus the sum of the first $ 2$ perfect squares, but with such small numbers involved, it makes more sense to just add.

[quote="erictao200500"]This is the question:
What is the sum of all the perfect squares between 5 and 30?

Simple, but is there an easier way besides 9+16+25?[/quote]

You could use the formula from Discrete_math, but I'm guessing that wouldn't be as fast as noticing $ 3^2 \plus{} 4^2 \equal{} 5^2$, so $ (2)(50) \equal{} 50$.  Or you could jut do 9 + 16 + 25, which is pretty simple to do in your head..not that hard...

I tell my kids to not use that formula unless you have a large number of perfect squares to add.

i think that both are equal ( 2*50 and 9+16+25)
because it is such a small range of numbers 2*50 wouldn't be quicker by that much.

yikes!
i meant 2*25
not 2*50
that would be way off
sorry :oops:

[quote="Dojo"][quote="007math"]You could use the formula from Discrete_math, but I'm guessing that wouldn't be as fast as noticing $ 3^2 \plus{} 4^2 \equal{} 5^2$, so $ (2)(50) \equal{} 50$.  Or you could jut do 9 + 16 + 25, which is pretty simple to do in your head..not that hard...[/quote]

$ 2(50)\neq 50$  :wink:

It should be $ 2(25)$ anyway.[/quote]

argh... :wallbash_red:

EDIT: I remember even checking over my multiplication to see if i did it right...me is dumb  :(


[quote="mdk"]Let S be the set of all natural numbers whose digits are chosen from the set {1, 3, 5, 7} such that
no digits are repeated. Find the sum of the elements of S.[/quote]

I also managed to solve this...

[hide]Look at all the numbers with first digit $ 1$:
1357
1375
1537
1573
1735
1753

So 1 appears in the first digit 6 times. Same for the other digits in each of the other spots.

When added together, each digit will have a value of
$ 6\cdot (1+3+5+7) = 6\cdot 4^{2}= 96$

Since there are 4 digits, multiply by $ 1111$:
$ 96\cdot 1111 = 106656$
To account for $ 3$ digits, note that the possibilities are
135
137
157
357
Applying the same technique,
$ 18\times 111+22\times 111+26\times 111+30\times = 10656$

To account for $ 2$ digits, note that the possibilities are:
13
15
17
35
37
57
Applying the same technique,
$ 4\times 11+6\times 11+8\times 11+8\times 11+10\times 11+12\times 11 = 528$
1 digit numbers are easy... $ 1+3+5+7 = 4^{2}= 16$
Adding up,
\[ 106656+10656+528+16 = \boxed{117856}\]
[/hide]

EDIT: Maybe MathCounts words it differently. I've got to be more careful next time.

[quote="mdk"]how did you do it so fast?[/quote]

MathCounts training. :P

This is a classic type of MathCounts problem. When I trained for MathCounts, I frequently came across these types of problems, and I just got used to them.

[quote="vishalarul"][quote="mdk"]Let S be the set of all natural numbers whose digits are chosen from the set {1, 3, 5, 7} such that
no digits are repeated. Find the sum of the elements of S.[/quote]

I also managed to solve this...

[hide]Look at all the numbers with first digit $ 1$:
1357
1375
1537
1573
1735
1753

So 1 appears in the first digit 6 times. Same for the other digits in each of the other spots.

When added together, each digit will have a value of
$ 6\cdot (1+3+5+7) = 6\cdot 4^{2}= 96$

Since there are 4 digits, multiply by $ 1111$:
$ 96\cdot 1111 = \boxed{106656}$[/hide][/quote]

hmmmmmmmmmm... but you haven't accounted for the 3 digit, 2 digit and single digit numbers!

[hide="solution"]
By a similar method, if you place 1 at the front of the 3 digit possibilities, you get:

1st digit: 1
2nd digit: 3 choices
3rd digit: 2 choices

So again 1 will appear at the start 6 times, and it must be the same for the other digits and the other places.

So: $ 6*(1+3+5+7) = 96$

and multiply by 111, because these are the 3 digit numbers:

$ 96*111 = 10656$

For 2 digits, each of the {1,3,5,7} digits will appear in each place 3 times, so:

$ 3*(1+3+5+7)=48$

and times 11:

$ 48*11 = 528$

For single digits each appears once, and multiply by 1, so:

$ 1+3+5+7 = 16$

So the final answer is:

$ 106656+10656+528+16 = 117856$
[/hide]


geometry
3D geometry
number theory unsolved
number theory

Is it possible for a perfect cube to start with 2002 ones in decimal notation?
Guess the answer is yes but I have no idea how to show this.

[quote="grobber"]The following problem has been discussed on the forum several times: if $a$ is not a power of $10$, then given any combination of digits, some $a^n$ begins with that combination. In your case, just take $a$ to be a cube which is not a power of $10$, and you're done.[/quote]
Where can I find a proof of this?

Just some threads under this one you can find http://www.mathlinks.ro/Forum/viewtopic.php?t=50019


geometry
3D geometry
tetrahedron
geometric inequality
IMO
IMO 1970

In the tetrahedron $ABCD,\angle BDC=90^o$ and the foot of the perpendicular from $D$ to $ABC$ is the intersection of the altitudes of $ABC$. Prove that: \[ (AB+BC+CA)^2\le6(AD^2+BD^2+CD^2). \] When do we have equality?

[hide="proof that also appears in the other thread ..."]
A problem that everyone must try..

Let $H$ be the orthocentre of $ABC$. Then $AH=2R\cos A$ etc and $BC=2R\sin A$ etc (well-known).

Note that
$AH^2+BC^2-CH^2=4R^2(\cos^2A+\sin^2A-\cos^2C)=4R^2(1-\cos^2C)=$ $4R^2\sin^2C=AB^2$

The following equations hold due to Pythagoras' theorem:

$AH^2+HD^2=AD^2$ (1)
$BH^2+HD^2=BD^2$ (2)
$CH^2+HD^2=CD^2$ (3)
$BD^2+CD^2=BC^2$ (4)

Now (2)+(3) imply $2HD^2=BC^2-CH^2-BH^2$ using (4).
Therefore (1)+(2) imply $AD^2+BD^2=AH^2+BC^2-CH^2=AB^2$.

So:
$BD^2+CD^2=BC^2$
$AD^2+BD^2=AB^2$
$CD^2+AD^2=CA^2$

Adding these up and multiplying by 3 gives $6(AD^2+BD^2+CD^2)=3(AB^2+BC^2+CD^2)$.

And clearly $3(AB^2+BC^2+CD^2)\ge (AB+BC+CD)^2$. (we have equality when $ABC$ is equilateral)

[/hide]

EDIT: posted later here (thanks [b]Luis González[/b]): http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=407847


Does anyone know is it possible to put images in a table in latex. I've tried but no luck.

You do it the same way as you would put graphics in anywhere http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php as long as you don't use a float since that would make no sense. So use \includegraphics{myimage.png} but not \begin{figure}...\end{figure}.

Thanks for your reply

I came across the following which may be of use to anyone who has found their way here.

Images in an array does the job for me.

[code]\begin{figure}[h]
\begin{center}$\begin{array}{cc}\includegraphics[width=2.5in]{image1.jpg}& \includegraphics[width=2.5in]{image2.jpg}\\ \includegraphics[width=2.5in]{image1.jpg}& \includegraphics[width=2.5in]{image2.jpg}\end{array}$
\end{center}
\caption{my caption}
\end{figure}[/code]

Shane

There's no need for the dollar signs as there is no mathematics in the array so you can use tabular instead which does the same job as array but in text mode.

I thought array only worked in mathmode, obviously not.

Thanks for you help.

Shane

I see you are a probably a disciple of Yang-Mills theory and so I presume you are familiar with tensors. I just wanted to ask you do you know off hand of any good introductions to tensors. Something to explain in physical terms what exactly a tensor and dual space etc. really is?

[quote="ofeyrpf"]I thought array only worked in mathmode, obviously not.[/quote]You are quite correct but what I meant was that as you didn't need maths mode you didn't need array and could use tabular instead. They are exactly the same except that tabular is for text mode.

[quote="ofeyrpf"]I see you are a probably a disciple of Yang-Mills theory[/quote]That's only something that AoPS puts there to indicate the number of posts that have been made. So I'm afraid tensors is not really my area of mathematics but have you looked at and followed the links at [url=http://en.wikipedia.org/wiki/Tensor]Wikipedia[/url], [url=http://mathworld.wolfram.com/Tensor.html]MathWorld[/url] and [url=http://planetmath.org/encyclopedia/Tensor.html]PlanetMath[/url]?


Determine the values of the coefficient $a$ for which the equations:

$x^2 + ax + 1 = 0$
$x^2 + x + a = 0$

have at least one common root.

Hello, Mikey

Quote:Determine the values of the coefficient  for which the equations:

have at least one common root.

[hide]

Let the roots of x^2 + ax + 1 = 0 be  {p,q}.

Then: [1] p + q = -a . . . [2] pq = 1

Let the roots of x^2 + x + a = 0  be {q,r}.

Then: [3] q + r = -1 . . . [4] qr = a

Subtract [3] from [1]: [5] p - r = 1 - a

Divide [2] by [4]:  pq/qr = 1/a . . . r = ap

Substitute into [5]: p - ap = 1 - a . . . (p - 1)(1 - a) = 0

Hence: a = 1 ... or ... p = 1

If a = 1, the equations are identical, both are x^2 + x + 1 = 0,

and hence, have two common roots: x = (-1 p/m sqrt{3}i)/(2).

If p = 1, then from [2] q = 1, and from [3], r = -2

And finally from [1], a = -2

The equations are:

x^2 - 2x + 1 = 0  . . . roots: 1, 1

x^2 + x - 2 = 0 . . . roots: 1, -2

[/hide]

Let $t$ be common root,we have
$t^2+at+1=0\cdots$(1)
$t^2+t+a=0\cdots$(2)
(2)-(1):$(a-1)(t-1)=0$,which follows Soroban's solutiuon.

kunny

[quote="Treething"]Hmm...

This is a form of a Diophantine equation, right

x^2 + ax + 1 = x^2 + x + a
ax + 1 = x + a
ax - a - x + 1 = 0

(a - 1)(x - 1) = 0

One value is 1... there's probably others, but I can't find a way to find any other...[/quote]

Well, you are basically finished.. if (a-1)(x-1) = 0, you either have a solution a=1.. or.. how else can that product be 0 :P And in that case, whats a?


inequalities
inequalities proposed
algebra
China
Hi

$ a,b,c$ are positive reals. Prove that
$ \sqrt{\frac{a^4 \plus{}2b^2 c^2}{a^2 \plus{}2bc}}\plus{}\sqrt{\frac{b^4 \plus{}2c^2 a^2}{b^2 \plus{}2ca}}\plus{}\sqrt{\frac{c^4 \plus{}2a^2 b^2}{c^2 \plus{}2ab}}\geq a\plus{}b\plus{}c$

[quote="jedaihan"]$ a,b,c$ are positive reals. Prove that
$ \sqrt {\frac {a^4 \plus{} 2b^2 c^2}{a^2 \plus{} 2bc}} \plus{} \sqrt {\frac {b^4 \plus{} 2c^2 a^2}{b^2 \plus{} 2ca}} \plus{} \sqrt {\frac {c^4 \plus{} 2a^2 b^2}{c^2 \plus{} 2ab}}\geq a \plus{} b \plus{} c$[/quote]
Very nice inequality, jedaihan. :) I like it very much. :)
Because of the following identity $ (a^2 \plus{} 2bc)(b^2 \plus{} 2ca) \minus{} (ab \plus{} bc \plus{} ca)^2 \equal{} c(a \minus{} b)^2(2a \plus{} 2b \minus{} c)$, it suggested me to solve your inequality as follow:
Assume that $ a \ge b \ge c$, we will consider 2 cases:

Case 1. If $ a \le 2(b \plus{} c)$, then rewriting our inequality as
\[ \sum\frac {a^4 \plus{} 2b^2c^2}{a^2 \plus{} 2bc} \plus{} 2\sum \sqrt {\frac {(a^4 \plus{} 2b^2c^2)(b^4 \plus{} 2c^2a^2)}{(b^2 \plus{} 2ca)(c^2 \plus{} 2ab)}} \ge \sum a^2 \plus{} 2\sum bc
\]
Notice that in this case, the above identity and Cauchy Schwarz Inequality yield
\[ \sum \sqrt {\frac {(a^4 \plus{} 2b^2c^2)(b^4 \plus{} 2c^2a^2)}{(b^2 \plus{} 2ca)(c^2 \plus{} 2ab)}} \ge \sum \sqrt {\frac {(a^2 \plus{} 2bc)(b^2 \plus{} 2ca)}{9}} \ge \sum \sqrt {\frac {(ab \plus{} bc \plus{} ca)^2}{9}} \equal{} \sum bc.
\]
Moreover, the Chebyshev's Inequality gives us
\[ \sum \frac {a^4 \plus{} 2b^2c^2}{a^2 \plus{} 2bc} \ge \sum a^2.
\]
Therefore, our inequality is trivial in this case. :)

Case 2. If $ a \ge 2(b \plus{} c)$, then we see that the original inequality is just the sum of the following inequalities:
\[ \sqrt {\frac {a^4 \plus{} 2b^2c^2}{a^2 \plus{} 2bc}} \ge a \plus{} (1 \minus{} \sqrt {2})(b \plus{} c),\qquad (1)
\]
and
\[ \sqrt {\frac {b^4 \plus{} 2c^2a^2}{b^2 \plus{} 2ca}} \ge \sqrt {2}c,\quad \sqrt {\frac {c^4 \plus{} 2a^2b^2}{c^2 \plus{} 2ab}} \ge \sqrt {2}b. \qquad (2)
\]
These inequalities can be easily proved if we use the hypothesis $ a \ge 2(b \plus{} c)$. ;) :)

The proof is completed. :)

P/s: The more detailed proof, we leave it to the readers. :)

[quote="nguoivn"]Yes, I also very like it. Thank you, jedaihan.
@can_hang: We can prove it by only Am-Gm, my brother  :)[/quote]
[hide="tonguoivn"]@nguoivn: I know the proof by AM-GM ;), but I like the nonstandard proof more.  ;)[/hide]

How can you do it can_hung using only AM-GM?

I can't do it :(
I used
$ \sqrt{\frac{a^{4}+2b^{2}c^{2}}{a^{2}+2bc}}=\sqrt{a^{2}+bc-\frac{3a^{2}bc}{a^{2}+2bc}}$
or
$ x=\frac{bc}{a}, y=\rac{ca}{b}, z=\frac{ab}{c}$, then $ xyz=abc$ and because this inequality is homogenoeous,
let $ xyz=1$ then enough to show $ \sum_{cyc}\sqrt{\frac{2x^{3}+1}{2x^{2}+x}} \ge \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
and so on.
I tried with using a lot of transform. But I can't -_-..

[quote="jedaihan"]Thanks, can_hang. Very nice solution  :)  :)

My solution)
By Cauchy Schwarz Inequality,
$ (a^4 \plus{} b^2c^2 \plus{} b^2c^2)(a \plus{} b \plus{} c)(a \plus{} c \plus{} b)\geq (a^2 \plus{} 2bc)^3$
Therefore,
$ \sum{\sqrt {\frac {a^4 \plus{} 2b^2c^2}{a^2 \plus{} 2bc}}}\geq \sum{\frac {a^2 \plus{} 2bc}{a \plus{} b \plus{} c}} \equal{} a \plus{} b \plus{} c$[/quote]
Nice solution, jedaihan. But I think your way and nguoivn's way will not work for the following:
\[ \sqrt{\frac{a^4\plus{}5b^2c^2}{a^2\plus{}5bc}} \plus{}\sqrt{ \frac{b^4\plus{}5c^2a^2}{b^2\plus{}5ca}}\plus{}\sqrt{\frac{c^4\plus{}5a^2b^2}{c^2\plus{}5ab}} \ge a\plus{}b\plus{}c.\]
;) :)

For more general, I think that the following inequality still holds:
\[ \sqrt{\frac{a^4\plus{}kb^2c^2}{a^2\plus{}kbc}} \plus{}\sqrt{ \frac{b^4\plus{}kc^2a^2}{b^2\plus{}kca}}\plus{}\sqrt{\frac{c^4\plus{}ka^2b^2}{c^2\plus{}kab}} \ge a\plus{}b\plus{}c.\]
With $ 0 \le k \le 12.$
But I only have nice solution (with Cauchy Schwarz) for $ 0 \le k \le 5.$
I hope you will be interested with this inequality. :)

[quote="jedaihan"]
My solution)
By Cauchy Schwarz Inequality,
$ (a^4 \plus{} b^2c^2 \plus{} b^2c^2)(a \plus{} b \plus{} c)(a \plus{} c \plus{} b)\geq (a^2 \plus{} 2bc)^3$
Therefore,
$ \sum{\sqrt {\frac {a^4 \plus{} 2b^2c^2}{a^2 \plus{} 2bc}}}\geq \sum{\frac {a^2 \plus{} 2bc}{a \plus{} b \plus{} c}} \equal{} a \plus{} b \plus{} c$[/quote]
Very nice proof, my friend  :lol:
Here is my proof:
Indeed, I think this is a good problem which let us see the power of classic ineqs. Model methods is not an easy ways with it.
Jedaihan's proof is shortest and nicest.
Can_hang's proof is also nice and strange.
My proof rather long but also only use classic ineq.
Now, we'll try to solve the harder problems of can_hang. I hope that we'll have another nice proofs  :)

[quote="Heebeen, Yang"]How can you do it can_hung using only AM-GM?

I can't do it :(
I used
$ \sqrt {\frac {a^{4} + 2b^{2}c^{2}}{a^{2} + 2bc}} = \sqrt {a^{2} + bc - \frac {3a^{2}bc}{a^{2} + 2bc}}$
or
$ x = \frac {bc}{a}, y = \rac{ca}{b}, z = \frac {ab}{c}$, then $ xyz = abc$ and because this inequality is homogenoeous,
let $ xyz = 1$ then enough to show $ \sum_{cyc}\sqrt {\frac {2x^{3} + 1}{2x^{2} + x}} \ge \frac {1}{x} + \frac {1}{y} + \frac {1}{z}$
and so on.
I tried with using a lot of transform. But I can't -_-..[/quote]
Your second approach can lead to a completed nice solution, my friend. Your normalize equivalent to normalize for $ abc=1$, the inequality becomes
$ \sum \sqrt {\frac {x^6 + 2}{x(x^3 + 2)}} \ge x + y + z.$
Now, using Holder Inequality, we see that
$ (x^6 + 1 + 1)(x^2 + x + 1)(x^2 + 1 + x) \ge x(x^3 + 2)^3,$
which yields
$ \sqrt {\frac {x^6 + 2}{x(x^3 + 2)}} \ge \frac {x^3 + 2}{x^2 + x + 1} = x - 1 + \frac {3}{x^2 + x + 1}.$
Using this inequality, we can reduce our inequality to
$ \frac {1}{x^2 + x + 1} + \frac {1}{y^2 + y + 1} + \frac {1}{z^2 + z + 1} \ge 1,$
which is a well-known inequality (as I remember, it is Vasc's result :) ) and you can prove this by Cauchy Schwarz. :)

Wow!!!

My think was short -_-..

I must train cauchy more...

Thank you for your nice solution^^!

[quote="jedaihan"]$ a,b,c$ are positive reals. Prove that
$ \sqrt {\frac {a^4 \plus{} 2b^2 c^2}{a^2 \plus{} 2bc}} \plus{} \sqrt {\frac {b^4 \plus{} 2c^2 a^2}{b^2 \plus{} 2ca}} \plus{} \sqrt {\frac {c^4 \plus{} 2a^2 b^2}{c^2 \plus{} 2ab}}\geq a \plus{} b \plus{} c$[/quote]
Another solution is the following:
Setting $ a\equal{}x^2,b\equal{}y^2,c\equal{}z^2$ where $ x,y,z \ge 0$, then using Cauchy Schwarz Inequality, we have
\[ \sqrt {\frac {a^4 \plus{} 2b^2 c^2}{a^2 \plus{} 2bc}}  \equal{}\frac{\sqrt{(a^4\plus{}2b^2c^2)(a^2\plus{}2bc)}}{a^2\plus{}2bc} \ge \frac{a^3\plus{}2bc\sqrt{bc}}{a^2\plus{}2bc} \equal{}\frac{x^6\plus{}2y^3z^3}{x^4\plus{}2y^2z^2}.\]
It suffices to prove that
\[ \sum \frac{x^6\plus{}2y^3z^3}{x^4\plus{}2y^2z^2} \ge \sum x^2,\]
or
\[ \sum \frac{yz\minus{}x^2}{x^6\plus{}2x^2y^2z^2} \ge 0.\]
Using GM-HM inequality, we have
\[ \frac{yz\minus{}x^2}{x^6\plus{}2x^2y^2z^2} \ge \frac{\frac{2y^2z^2}{y^2\plus{}z^2} \minus{}x^2}{x^6\plus{}2x^2y^2z^2} \equal{}\frac{c(b\minus{}a)\plus{}b(c\minus{}a)}{(b\plus{}c)(a^3\plus{}2abc)}.\]
The inequality deduces to
\[ \sum \frac{c(b\minus{}a)\plus{}b(c\minus{}a)}{(b\plus{}c)(a^3\plus{}2abc)} \ge 0,\]
which is equivalent to
\[ \sum c(a\minus{}b)\left( \frac{1}{(a\plus{}c)(b^3\plus{}2abc)} \minus{}\frac{1}{(b\plus{}c)(a^3\plus{}2abc)}\right) \ge 0,\]
or
\[ \sum \frac{c(a\minus{}b)^2[ab(a\plus{}b)\plus{}c(a^2\minus{}ab\plus{}b^2)]}{(a\plus{}c)(b\plus{}c)(a^3\plus{}2abc)(b^3\plus{}2abc)} \ge 0,\]
which is trivial. :)

Note: This way can help us to prove the more general with $ 0 \le k \le 9$. ;) :)

Nice solution :)  :wink:
But I can't solve your expansion when $ 0\leq k\leq 12$  :blush:
Can you post your solution?

[quote="manlio"][quote="can_hang2007"]Moreover, the Chebyshev's Inequality gives us
\[ \sum \frac {a^4 \plus{} 2b^2c^2}{a^2 \plus{} 2bc} \ge \sum a^2.
\]
:)[/quote]

My dear friend, can you please explain how to prove it by Chebyshev. Thank you very much. :)[/quote]
My friend, write the inequality in the form
$ \sum \frac{bc(bc\minus{}a^2)}{a^2\plus{}2bc} \ge 0,$
$ \sum \frac{(bc\minus{}a^2)(b\plus{}c)}{(b\plus{}c)(a^3\plus{}2abc)} \ge 0.$
Now assume that $ a \ge b \ge c$, we can see that
$ (bc\minus{}a^2)(b\plus{}c) \le (ca\minus{}b^2)(c\plus{}a) \le (ab\minus{}c^2)(a\plus{}b),$
and
$ \frac{1}{(b\plus{}c)(a^3\plus{}2abc)} \le \frac{1}{(c\plus{}a)(b^3\plus{}2abc)} \le \frac{1}{(a\plus{}b)(c^3\plus{}2abc)}.$
Now, use Cauchy Schwarz, we can get the result. :)

[quote="arqady"][quote="sqing"]Let $ a,b,c$ are  nonnegative reals. Prove that\[\sqrt{\frac{a^2 +bc}{b^2 +c^2}}+\sqrt{\frac{b^2 +ca}{c^2 +a^2}}+\sqrt{\frac{c^2 +ab}{a^2 +b^2}}\geq 2+\frac{\sqrt{2}}{2}\][/quote]

$\sqrt{\frac{a^2 +bc}{b^2 +c^2}}+\sqrt{\frac{b^2 +ca}{c^2 +a^2}}+\sqrt{\frac{c^2 +ab}{a^2 +b^2}}\geq \sqrt{\frac{2(a^2+b^2+c^2)}{ab+ac+bc}}+\frac{\sqrt{2}}{2}$ is also true.[/quote]
Whe the equality holds?

[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\[\sqrt{\frac{a^2 \plus{}2bc}{b^2 \plus{}c^2}}\plus{}\sqrt{\frac{b^2 \plus{}2ca}{c^2 \plus{}a^2}}\plus{}\sqrt{\frac{c^2 \plus{}2ab}{a^2 \plus{}b^2}}\geq 3.\]
[/quote]
[color=#FF0000][b]Moscow MO 1999?:[/b][/color]
Let $a,b,c$ be side lengths of a triangle . Prove that : \[\frac{a^2 \plus{}2bc}{b^2 \plus{}c^2}\plus{}\frac{b^2 \plus{}2ca}{c^2 \plus{}a^2}\plus{}\frac{c^2 \plus{}2ab}{a^2 \plus{}b^2}> 3.\]

[hide]$a^2+2bc>b^2+c^2\Leftrightarrow (a+b-c)(a+c-b)>0$
and we're done.[/hide]

[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\[\sqrt{\frac{a^3 \plus{}abc}{b \plus{}c}}\plus{}\sqrt{\frac{b^3 \plus{}abc}{c \plus{}a}}\plus{}\sqrt{\frac{c^3 \plus{}abc}{a \plus{}b}}\geq a+b+c.\]
Old? Where?[/quote]
I think I saw it.
By the way, it's easy Holder:
$\left(\sum_{cyc}\sqrt{\frac{a^3 \plus{}abc}{b \plus{}c}}\right)^2\sum_{cyc}\frac{a^2(b+c)}{a^2+bc}\geq(a+b+c)^3$.
Hence, it remains to prove that $a+b+c\geq\sum_{cyc}\frac{a^2(b+c)}{a^2+bc}$, which is $abc\sum_{cyc}(a^4-a^2b^2)\geq0$, which is obvious.

I also saw it.
It should be an old inequality .
Thanks.

[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\[\sqrt{\frac{a^2 \plus{}bc}{b^2 \plus{}c^2}}\plus{}\sqrt{\frac{b^2 \plus{}ca}{c^2 \plus{}a^2}}\plus{}\sqrt{\frac{c^2 \plus{}ab}{a^2 \plus{}b^2}}\geq 2+\frac{\sqrt{2}}{2}.\]\[\sqrt{\frac{a^2 \plus{}2bc}{b^2 \plus{}c^2}}\plus{}\sqrt{\frac{b^2 \plus{}2ca}{c^2 \plus{}a^2}}\plus{}\sqrt{\frac{c^2 \plus{}2ab}{a^2 \plus{}b^2}}\geq 3.\]
Old? Where?[/quote]
Crux,Problems 3419:
Let $ a,b,c$ are  nonnegative reals. Prove that
$$\sqrt{\frac{a^{2}+4bc}{b^2+c^2}}+\sqrt{\frac{b^{2}+4ca}{c^2+a^2}}+\sqrt{\frac{c^{2}+4ab}{a^2+a^2}}\ge 2+\sqrt{2}$$
$$\sqrt{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt{\frac{b^{2}+ca}{c^2+a^2}}+\sqrt{\frac{c^{2}+ab}{a^2+a^2}}\ge 2+\frac{1}{\sqrt[3]{2}}$$
[url=http://www.artofproblemsolving.com/community/c6h1115969p5108246]here[/url]

[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\[\sqrt{\frac{a^3 \plus{}abc}{b \plus{}c}}\plus{}\sqrt{\frac{b^3 \plus{}abc}{c \plus{}a}}\plus{}\sqrt{\frac{c^3 \plus{}abc}{a \plus{}b}}\geq a+b+c.\]
Old? Where?[/quote]

[img]http://s1.sinaimg.cn/middle/006ptkjAzy75Lkd18qs10&690[/img]

[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\[\sqrt{\frac{a^3 \plus{}abc}{b \plus{}c}}\plus{}\sqrt{\frac{b^3 \plus{}abc}{c \plus{}a}}\plus{}\sqrt{\frac{c^3 \plus{}abc}{a \plus{}b}}\geq a+b+c.\]
Old? Where?[/quote]
If $a,b,c>0$ [url=http://www.artofproblemsolving.com/community/c6h1220622p6101528]then prove:[/url]
$$\frac{a^3+abc}{b+c}+\frac{b^3+abc}{a+c}+\frac{c^3+abc}{b+a}\geq a^2+b^2+c^2$$

[quote=sqing][hide=**][quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\[\sqrt{\frac{a^2 \plus{}bc}{b^2 \plus{}c^2}}\plus{}\sqrt{\frac{b^2 \plus{}ca}{c^2 \plus{}a^2}}\plus{}\sqrt{\frac{c^2 \plus{}ab}{a^2 \plus{}b^2}}\geq 2+\frac{\sqrt{2}}{2}.\]\[\sqrt{\frac{a^2 \plus{}2bc}{b^2 \plus{}c^2}}\plus{}\sqrt{\frac{b^2 \plus{}2ca}{c^2 \plus{}a^2}}\plus{}\sqrt{\frac{c^2 \plus{}2ab}{a^2 \plus{}b^2}}\geq 3.\]
Old? Where?[/quote][/hide]
Crux,Problems 3419:
Let $ a,b,c$ are  nonnegative reals. Prove that
$$\sqrt{\frac{a^{2}+4bc}{b^2+c^2}}+\sqrt{\frac{b^{2}+4ca}{c^2+a^2}}+\sqrt{\frac{c^{2}+4ab}{a^2+a^2}}\ge 2+\sqrt{2}$$
$$\sqrt{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt{\frac{b^{2}+ca}{c^2+a^2}}+\sqrt{\frac{c^{2}+ab}{a^2+a^2}}\ge 2+\frac{1}{\sqrt[3]{2}}$$
[url=http://www.artofproblemsolving.com/community/c6h1115969p5108246]here[/url][/quote]
Let $ a,b,c$ are nonnegative reals. [url=http://www.artofproblemsolving.com/community/c6h186334p1042076]Prove that[/url]
$$\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ca}{c^2+a^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+a^2}}\ge 2+\frac{1}{\sqrt[3]{2}}$$

[quote=ISIBS1531][quote=jedaihan]Thanks, can_hang. Very nice solution  :)  :)

My solution)
By Cauchy Schwarz Inequality,
$ (a^4\plus{}b^2c^2\plus{}b^2c^2)(a\plus{}b\plus{}c)(a\plus{}c\plus{}b)\geq (a^2\plus{}2bc)^3$
Therefore,
$ \sum{\sqrt{\frac{a^4\plus{}2b^2c^2}{a^2\plus{}2bc}}}\geq \sum{\frac{a^2\plus{}2bc}{a\plus{}b\plus{}c}}\equal{}a\plus{}b\plus{}c$[/quote]

can anyone show how this result comes from CS inequality ??[/quote]

It's Holder:
$$(a^4+b^2c^2+b^2c^2)(a+b+c)(a+c+b)\ge(\sqrt[3]{a^6}+\sqrt[3]{b^3c^3}+\sqrt[3]{b^3c^3})^3$$
Then by results:
$$\frac{(a^4+b^2c^2+b^2c^2)}{(a^2+bc+bc)}\ge\frac{(a^2+bc+bc)^2}{(a+b+c)(a+c+b)}$$
Or
$$\sqrt{\frac{(a^4+b^2c^2+b^2c^2)}{(a^2+bc+bc)}}\ge\frac{(a^2+bc+bc)}{(a+b+c)}$$
Hence we need to prove:
$$\sum\sqrt{\frac{(a^4+b^2c^2+b^2c^2)}{(a^2+bc+bc)}}\ge\sum\frac{(a^2+bc+bc)}{(a+b+c)}$$
Then:
$$\sum\frac{(a^2+bc+bc)}{(a+b+c)}=\sum\frac{\sum a^2+2\sum ab}{a+b+c}=\sum\frac{(\sum a)^2}{a+b+c}=\sum a$$
So we're done...!

You can homogenise with $abc=1$ ,write the  terms as functions of a,b,c and apply jensen,although the convexity is very hard to prove.

[quote=arqady][quote=sqing]
Crux,Problems 3419:
Let $ a,b,c$ are  nonnegative reals. Prove that
$$\sqrt{\frac{a^{2}+4bc}{b^2+c^2}}+\sqrt{\frac{b^{2}+4ca}{c^2+a^2}}+\sqrt{\frac{c^{2}+4ab}{a^2+b^2}}\ge 2+\sqrt{2}$$
[/quote]
Let $c=\min\{a,b,c\}$.
Thus, by C-S $$\sqrt{\frac{a^{2}+4bc}{b^2+c^2}}+\sqrt{\frac{b^{2}+4ca}{c^2+a^2}}=\frac{\sqrt{(a^{2}+4bc)(b^2+c^2)}}{b^2+c^2}+\frac{\sqrt{(b^{2}+4ca)(a^2+c^2)}}{c^2+a^2}\geq$$
$$\geq\frac{ab+2\sqrt{bc^3}}{b^2+c^2}+\frac{ab+2\sqrt{ac^3}}{a^2+c^2}\geq\frac{ab+2c^2}{b^2+c^2}+\frac{ab+2c^2}{a^2+c^2}.$$
Also, $$\sqrt2-\sqrt{\frac{c^{2}+4ab}{a^2+b^2}}\leq\sqrt2-\sqrt{\frac{4ab}{a^2+b^2}}=$$
$$=\frac{\sqrt2(a-b)^2}{a^2+b^2+\sqrt{2ab(a^2+b^2)}}\leq\frac{1.5(a-b)^2}{(a+b)^2}.$$
Id est, it's enough to prove that:
$$\frac{ab+2c^2}{b^2+c^2}+\frac{ab+2c^2}{a^2+c^2}\geq2+\frac{1.5(a-b)^2}{(a+b)^2},$$
which is obvious.[/quote]
Sir how is the last inequality obvious, Is buffalo way neccessary?

[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\[\sqrt{\frac{a^2 \plus{}2bc}{b^2 \plus{}c^2}}\plus{}\sqrt{\frac{b^2 \plus{}2ca}{c^2 \plus{}a^2}}\plus{}\sqrt{\frac{c^2 \plus{}2ab}{a^2 \plus{}b^2}}\geq 3.\][/quote]
[url=https://artofproblemsolving.com/community/c6h1115969p5108246]here[/url] [url=https://artofproblemsolving.com/community/c6h1646201p10405989]here[/url]
[color=#f00]Answers are welcome.[/color]

[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\[\sqrt{\frac{a^2 \plus{}2bc}{b^2 \plus{}c^2}}\plus{}\sqrt{\frac{b^2 \plus{}2ca}{c^2 \plus{}a^2}}\plus{}\sqrt{\frac{c^2 \plus{}2ab}{a^2 \plus{}b^2}}\geq 3.\][/quote][size=50]p/1383596268[/size]

[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\[\sqrt{\frac{a^2 \plus{}2bc}{b^2 \plus{}c^2}}\plus{}\sqrt{\frac{b^2 \plus{}2ca}{c^2 \plus{}a^2}}\plus{}\sqrt{\frac{c^2 \plus{}2ab}{a^2 \plus{}b^2}}\geq 3.\][/quote]
[url=https://artofproblemsolving.com/community/c6h3007280p27010757]h[/url] [url=https://artofproblemsolving.com/community/c6h361540p1979760]h[/url] [url=https://artofproblemsolving.com/community/c4h2464062p20573119]h[/url] [url=https://artofproblemsolving.com/community/c6h559045p3252033]h[/url] [url=https://math.stackexchange.com/questions/4040811/sqrt-fraca24bcb2c2-sqrt-fracb24aca2c2-sqrt-fracc24?noredirect=1]h[/url] [url=https://artofproblemsolving.com/community/c6h1300514p6925295]h[/url] [url=https://artofproblemsolving.com/community/c6h1823379p12187705]h[/url]

[quote=sqing]Let $ a,b,c$ are  nonnegative reals. Prove that\[\sqrt{\frac{a^2 \plus{}bc}{b^2 \plus{}c^2}}\plus{}\sqrt{\frac{b^2 \plus{}ca}{c^2 \plus{}a^2}}\plus{}\sqrt{\frac{c^2 \plus{}ab}{a^2 \plus{}b^2}}\geq 2+\frac{\sqrt{2}}{2}.\][/quote]
Let $a$, $b$ and $c$ be positive numbers. [url=https://artofproblemsolving.com/community/c6h3090659p27934129]Prove that:[/url]
$$\sqrt{\frac{a^2+bc}{b^2+c^2}}+\sqrt{\frac{b^2+ca}{c^2+a^2}}+\sqrt{\frac{c^2+ab}{a^2+b^2}}\geq2\left(\sqrt{\frac{ab}{(a+c)(b+c)}}+\sqrt{\frac{bc}{(b+a)(c+a)}}+\sqrt{\frac{ca}{(c+b)(a+b)}}\right)$$
[url]https://artofproblemsolving.com/community/c6h1646201p10405989[/url]
[url]https://artofproblemsolving.com/community/c6h1115969p5108413[/url]
If $a,b,c\geq 0 ,\lambda >0.$ Then$$\sqrt{\frac{a^{2}+\lambda bc}{b^{2}+c^{2}}}+\sqrt{\frac{b^{2}+\lambda ca}{c^{2}+a^{2}}}+\sqrt{\frac{c^{2}+\lambda ab}{a^{2}+b^{2}}}\geq 3\sqrt[3]{\frac{\lambda }{4}} $$
$$\sqrt{\frac{a^{2}+ 256 bc}{b^{2}+c^{2}}}+\sqrt{\frac{b^{2}+ 256 ca}{c^{2}+a^{2}}}+\sqrt{\frac{c^{2}+ 256 ab}{a^{2}+b^{2}}}\geq 12 $$


vector
LaTeX
topology
linear algebra
linear algebra unsolved

Prove that any contraction operator C of an Euclidean space equals $\Pi \circ Q$ where $\Pi$ is an orthogonal projection and $Q$ is an orthogonal operator

[quote="fredbel6"]First of all this is kind of analysis more isn't it?  On the other hand there are the linear operators...

Never the less : what about this map

$(x,y)$--> (x/2,y/2)

why is this not a contraction?[/quote]
Is it true if I add the contraction is not injective

What do you mean by "contraction" anyway? :? As far as I know, it means an operator with norm $\le 1$. However, non-null orthogonal projections have norm precisely $1$, and multiplication by an orthogonal operator does not change the norm, so all operators of that form have norm equal to $1$. You can't get anything with norm $<1$.

As i saw it :? , contractions f were mappings from a metric space (with a distance not necessarily induced by a norm as the space does not even need to have vector space structure) to the same space, where there was a constant k with  $0<k<1$, such that
$d(f(x),f(y))\leq k d(x,y)$
those mappings are not even linear so how could we speak of norm?
or am i just babbling?

PS : sorry about botching up in Latex the smaller or equal sign, does anyone know advice?


Putnam
function
combinatorics proposed
combinatorics

$n$ integers that sum to $n - 1$ are arranged in a circle. Show that we can choose one of the integers $x_1$ so that the other integers going around the circle are, in order, $x_2, ..., x_n$ and so that we have $\Sigma_{i=1}^k x_i \leq k-1$ for $k = 1, 2, ... , n$.

I have a solution to this problem, but I do not understand how someone would come up with it. Maybe someone could explain how he/she thought about this problem to successfully solve it.

Thanks.

For me, it helped to have seen a similar problem before.  For example, I have seen the following problem (with clever solution) in Lovasz's book on [i]Combinatorial Problems and Exercises[/i]:

Along a speed track there are some gas stations. The total amount of gasoline available in them is equal to what our car needs for going around the track. Prove that there is a gas station such that if we start there with an empty tank, we shall be able to go around the track without running out of gasoline.

I remember solving a very similar problem a long time ago, which is, in fact, the same as this one.

Instead of working with $x_i$, we'll replace them with $1-x_i$ (and denote the quantities $1-x_i$ also by $x_i$). The new problem reads: given $n$ integers $x-1,\ldots,x_n$ with sum $1$ around the circle, we can find one of them s.t. the sums mentioned in the original post are all positive as we go around the circle (in a counterclockwise direction, for instance).

It's going to be a bit hard to explain, I think, but it's perfectly natural, if you don't let the definitions fool you. Just draw a picture.

Suppose there is a function $f:\{1,2,\ldots,n\}\to\{1,2,\ldots,n\}$ s.t. $x_k+x_{k+1}+\ldots+x_{f(k)}\le 0$ for all $k\in\overline{1,n}$, if $x_k,x_{k+1},\ldots,x_{f(k)}$ are consecutive points on the circle in a counterclockwise direction ($x_{n+1}=x_1$, as usual). Let $S_k=\{x_k,\ldots,x_{f(k)}\}$. Start with $a_1=1$, and put $a_{n+1}=f(a_n)+1$. Let $(m,p),\ m<p$ the first pair with $a_p=a_m$.

Observe now that if you add up all the numbers in $S_m,\ldots,S_{p-1}$, you get a non-positive sum on the one hand, and, on the other hand, the multiset formed by adjoining the elements of $S_m,\ldots,S_{p-1}$ contains each $x_i$ the same number of times, so the sum should have been positive (since all the $x_i$'s add up to $1>0$).

The contradiction shows that there is no function with the properties of $f$, and this is, in fact, precisely what we want to prove.

Edit:

Yes, Ravi, this problem reminded me of the one you mention as well. That one has also been discussed on the forum, also in relation to the one I solved, which was posted either by Moubinool or Pierre (I cannot remember which one, unfortunately, but I suspect they both discussed it, since I have the feeling it was used in a Concours Generale :)).

well also the following can be done, consider all possible walks, and in each walk record under each number the partial sum (start each walk with sum 0). Suppose point $i$ gets the smallest recorded partial sum, then it is easy to prove starting from $i+1$ we get a good walk.

You guys are kind of confussing me.  Anyways, here is my solution which I think is straight forward:

Let $S_j=\sum_{i=1}^{k} x_{(i+j\mod n)}$ (where $x_0=x_n$).

Note that $S_1+S_2+\ldots+S_n=k(n-1)$.

Their average is $k\frac{n-1}{n}$ so there must be at least one $S_j$ such that $S_j\leq k\frac{n-1}{n}=k-\frac{k}{n}$.  Since $S_j$ is an integer, $S_j\leq k-1$.

There is not just one value of $k$.  You have to prove that $\sum_{i=1}^k x_i \le k - 1$ holds for all $k$ (between 1 and $n$) simultaneously.

Ah, yes, thank you.  That did seem ridicolously easy for an A4 Putnam Problem.


A dragon was tamed by Dumbledore to guard his home. The dragon guarded the home on the condition that Dumbledore feed him 4 pieces of meat for every 10 hours. If one cow provides 10 pieces of meat, what is the smallest whole number of cows  Dumbledore needs to guard his home for 1 month (30 days)?

Ooh...a dragon. 30 days is equiv to 720 hours, so Dumbledore would have to feed me 72 times, or 288 pieces of meat. 288 divided by 10 and rounded up is equivalent to 29 cows.


A boy is playing with a toy airplane attached to a string. At time=0 the toy plane is moving at 5.0m/s [N] . At t=0.15 the plane is moving at 5.0m/s[N 40 degrees  E]. What was the magnitude of the average acceleration of the plane during thist time?

I assume it is $ 40^{o}$ East of North.

Solution

Let $ \mathbf{i}$ and $ \mathbf{j}$ be unit position vectors in the directions East and North respectively.

Initial velocity is: $ u = 5\mathbf{j}$

Final velocity is: $ v = 5\cos{50^{o}}\mathbf{i}+5\sin{50^{o}}\mathbf{j}= 3.214\mathbf{i}+3.830\mathbf{j}$

Change in velocity: $ \Delta v = v-u = 3.214\mathbf{i}-1.17\mathbf{j}$

Average acceleration: $ a =\frac{\Delta v}{t}= 21.43\mathbf{i}-7.8\mathbf{j}$

$ |a| =\sqrt{21.43^{2}+7.8^{2}}= 23 ms^{-2}$

$ \theta =\arctan{\frac{-7.8}{21.43}}=-20^{o}$

Acceleration is $ 23 ms^{-2}$, $ 20$ degrees clockwise from East.

This can angle can also be obtained from purely geometric reasoning, and the acceleration's magnitude from the cosine rule.

[quote="sludgethrower"]I have actually asked this before, and you ignored me, please say what you mean by [N 40 degrees E]. Is it 40 degrees east of north, or 40 degrees north of east???[/quote]

I am so sorry. It means going 40 degrees east from north so you were right .


Find the solution of

y"(x) - 4y(x) = 0

I simply started it off by trying to create a general form but I'm stuck.

a^2 - 4 = 0

What should I do from here?

You are supposing some solution in the form of $ Ce^{ax}$ for the constant $ C$.

Go ahead. :)

hello, by your ansatz you get $ a_1\equal{}2$ or $ a_2\equal{}\minus{}2$, so the general solution
is $ y(x)\equal{}C_1e^{\minus{}2x}\plus{}C_2e^{2x}$.
Sonnhard.


function
integration
calculus
vector
trigonometry
linear algebra
real analysis

i had this for an exam question, im not sure what it means or what my professor is trying to show:

Consider the set of functions which are continuous on the interval [a, b].
Define the two dot products between the functions $f$ and $g$ as

$f \cdot g = \int_a^bf(x)g(x)dx$

show that the two functions $f(x) = sin(x)$ and $g(x) = cos(x)$ are orthogonal on the interval $[-\pi, \pi]$.  Hint: show their dot product is zero

i evaluted the integral which came out to zero, and got the credit, but im clueless.  we just finished cross products, projections... and we had done a little bit with parametric equations.  right after we went into 3 dimensional parametric equations in vector form, r(t) = <f(t), g(t), h(t)>, so im thinking he was trying to set us up for that, but the above question seems like nonsense.  two things that got me were that there's an infinite number of sumations within the dot product which to me would be an infinite number of dimensions.  also there's only one value for f and one for g, which wouldnt be able to define a point in two dimensions.

Yes, there are infinitely many dimensions. What's wrong with that?

The definition of an inner product (dot product): A vector space $V$ has an inner product $\langle\,,\,\rangle$ if $\langle\,,\,\rangle$ satisfies these properties:
1: Bilinear: $\langle ax+by,z\rangle=a\langle x,z\rangle+b\langle y,z\rangle$ and $\langle z,ax+by\rangle=a\langle z,x\rangle+b\langle z,y\rangle$.
2: Symmetric: $\langle x,y\rangle=\langle y,x\rangle$
3: Positive definite: $\langle x,x\rangle>0$ if $x\neq0$.

The inner product can be used to define a norm: $\|x\|=\sqrt{\langle x,x\rangle}$.
How does this relate to what you already know? On any finite-dimensional subspace, it is always possible to find an orthonormal basis, and the inner product behaves like the standard dot product with respect to that basis. In your example, you could always reduce to functions of the form $f(x)=a\sin x+b\cos x+c$; that's three-dimensional, and the obvious basis is already orthogonal, but not orthonormal.
For infinite-dimensional spaces, the right definitions are trickier, but this question hints at the way to Fourier series, which are based on the functions $1,\sin x,\cos x,\sin 2x,\cos 2x,\dots$ being an orthogonal basis for the appropriate space of functions.

There is no continuation in the course, but it's worth keeping in mind that $\mathbb{R}^n$ is not the only interesting vector space.

[quote="oooooh ya"]... there's an infinite number of sumations within the dot product which to me would be an infinite number of dimensions ...[/quote]
Indeed so. The set whose members are all of the continuous functions on a certain interval is an infinite-dimensional vector space.

What we have here is a typical case of how mathematics abstracts and generalizes from certain concrete cases.

Start with $\mathbb{R}^2$ or $\mathbb{R}^3.$ We talk about vectors in these spaces, and we have some specific, concrete pictures in mind for what we mean. But back off from the question of what a vector "is" and consider what you can do with vectors, algebraically. The essence boils down to just two things: you can add (or subtract) two vectors to get another vector, and you can multiply a scalar (a real number) by a vector to get a vector. These algebraic operations obey some (more or less) familiar laws: addition is commutative and associative; there's a zero vector (an additive identity); there are negatives (additive inverses). For scalar multiplication, every reasonable identity that reminds us of an associative or distributive law is true.

The next step is to take that list of algebraic properties as [i]axioms[/i]; we call these the axioms of a (real) vector space. What good is this? Any theorems and definitions that we base on these axioms - and there are plenty of theorems and definitions; the subject is called "Linear Algebra" - are true in any system, any new example that we create that satisfies these axioms. "Dimension" is one of the definitions from linear algebra; it's the number of elements in a minimal spanning set.

At this point, what about $C[-\pi,\pi],$ the set of (real-valued) continuous functions on the interval $[-\pi,\pi]?$ Is the sum of two continuous functions a continuous function? Is a constant times a continuous function a continuous function? The rest of the arithemetic all works; we have a vector space. But we can't find a finite spanning set - a finite sets such that we can write any contininuous function as a linear combination of elements of this set. So the $C[\pi,\pi]$ is an infinite-dimensional vector space, and with that observation, we lose confidence in much of our geometric intuition.

Then there's the dot product. For $x$ and $y$ in $\mathbb{R}^2,$ we can define the dot product (aka scalar product) of $x$ and $y$ so that it is a real number. Does the dot product have significant algebraic properties? Yes. We can boil them down to three. The dot product is:

Symmetric: $x\cdot y=y\cdot x$ for all $x,y.$

Bilinear: $(ax+by)\cdot z=a(x\cdot z)+b(y\cdot z)$ and $x\cdot(ay+bz)=a(x\cdot y)+b(x\cdot z)$ for all vectors $x,y,z$ and all scalars $a,b.$

Positive definite: $x\cdot x>0$ whenever $x\ne0.$

And we can prove a lot of interesting things based on this, including the Cauchy-Schawrz inequality. We make the obervation that $x$ and $y$ are orthogonal whenever $x\cdot y=0.$

And so we abstract that; any symmetric positive definite bilinear form on a real vector space gets called an [i]inner product[/i]. What about $\langle f,g\rangle=\int_{-\pi}^{\pi}f(x)g(x)\,dx?$ We can show that that is symmetric, bilinear and positive definite (at least on the continuous functions.) And so we call $C[-\pi,\pi]$ an inner product space.

Back in $\mathbb{R}^2,$ the idea that $x\cdot y=0$ iff $x$ and $y$ are orthogonal was an observation. In a more abstract inner product space, it's a definition, not a theorem. It's the definition of "orthogonal." Hence all you did was to verify the condition of the definition. The "hint" you reported was in fact the only thing you could do.

This is actually a powerful tool in mathematics. But you have to accept not being bound by familiar pictures in order to use it.


In a triangle ABC, O is the circumcenter.
sides AB and AC are extended. A circle passing through B , O and C meets AB and AC at P and Q.
Show that AO when intersects PQ is perpendicular to it.

[quote="10000th User"][quote="Zed_darkx"]Anyone???
Im not able to prove BC parallel to PQ[/quote]Why does BC have to be parallel to PQ? After all, AO does not necessarily be perpendicular to BC in an arbitrary triangle ABC.[/quote]
Then how do we prove this thing?
Or is true for some special case?


I am trying to add a method to my source file, but I get the error "(method name) was not declared in this scope."
I am using Eclipse. This is my code:

[code]#include <iostream>
using namespace std;

int main()
{
	cout << "Hello World!" << endl;
	printIt();
	return 0;
}

void printIt()
{
	cout << "Goodbye!" << endl;
}
[/code]


calculus
integration
number theory unsolved
number theory

Find all integers $x,y$ that :
\[ x^2+x+2=y^3\]

Well it's easy for positive integers
look if 2+x+x^2=y^3 then 1+x+x^2=y^3-1 since for x=1 and y=1 there are no sol.   (x^3-1)/(y^3-1)=x-1 then x\leq y. if x=y there are sol. x=y=1 if
x<y  then y^3=2+x+x^2<2+y+y^2 wich is false for y>2.
So the only positive integer sol. is x=y=2. [/tex]

im sorry if im slow today, but why does (x^3-1)/(y^3-1)=x-1 imply that x <= y? cause if x < y then (x^3-1)/(y^3-1) is not an integer.

anyway, you only need to consider positive integral solutions, because once you have a solutions (x,y) with x positive, you get the solution (-x-1,y).

[quote="zscool"]anyway, you only need to consider positive integral solutions, because once you have a solutions (x,y) with x positive, you get the solution (-x-1,y).[/quote]

you also need to prove there are no others there ;)

Here is my attempt:

It's not difficult to see that if (x, y) is a solution, then (-x-1, y) is also a solution. Hence we only consider x,y>=1. We see there is no solution when x=1 or y=1, so assume x,y>=2. Now x>=y from (x^3-1)/(y^3-1)=x-1 (otherwise 0<(x^3-1)/(y^3-1)<1, impossible.) So x>=y>=2.

Let a=[:sqrt:(y3-y3/2+9/4)], where [] denotes the floor function. Then (a-1)a<y^3<(a+1)(a+2). Hence x=a. Similarly, let b=[:sqrt:(y3+y3/2-7/4)], we have (b-1)b<y^3<(b+1)(b+2). Hence x=b, or, a=b. But that implies  [:sqrt:( (y3/2-1/2)2-2)]=[:sqrt:( (y3/2+1/2)2+2)]. For sufficiently large y, we see that LHS=[y3/2-1/2]=[y3/2+1/2]=RHS. Impossible. Now we can check for smaller values of y to conclude that (x, y)=(2, 2) and (x, y)=(-3, 2) are the only solutions.

I know that the "sufficiently large y" part of the proof seems shaky, so it's still an attempt, not yet a proof. Is this correct?

This is not quite correct.
[tex] (a-1)a<y^3<(a+1)(a+2) [/tex] implies both [tex]x=a-1[/tex] and  [tex]x=a[/tex], as well for [tex]b[/tex].
Then a contradiction of this type cannot arise:
[tex] [ (y^3/2-1/2)^2-2) ]=[ (y^3/2+1/2)^2-2 ] [/tex]
(btw this has an infinite number of solutions)

OOPS! My mistake in typing! The second paragraph should have been the following:

(a-1)a<y3-2<(a+1)(a+2) and (b-1)b<y3-2<(b+1)(b+2).

Now it should be okay. Also, heartwork, I think you forgot the square roots inside the floor function...  :D

I missed the square roots in typing, but i ment that the equation:
[tex] [ {\sqrt{(y^3/2-1/2)^2-2)}} ]=[ {\sqrt{(y^3/2+1/2)^2-2)}} ] [/tex]
has no solutions. Indeed you can check:
[tex] {\sqrt{(y^3/2+1/2)^2-2)}} - {\sqrt{(y^3/2-1/2)^2-2)}} >1 [/tex]


See http://www.mathlinks.ro/Forum/viewtopic.php?t=27450 , http://www.mathlinks.ro/Forum/viewtopic.php?t=1831 and http://www.mathlinks.ro/Forum/viewtopic.php?t=50269 .

  darij


abstract algebra
geometry
parameterization
Ring Theory
superior algebra
superior algebra unsolved

Let $ \phi: A \to B$ be a homomorphism of Noetherian rings, and $ P$ a prime ideal of $ B$. Then setting $ p \equal{} P \cap A$ we have

$ ht P \leq ht p \plus{} dim B_P/pB_P$.

[b]Proof:[/b]

We can replace $ A$ and $ B$ by $ A_p$ and $ B_P$ and assume $ (A,m)$ and $ (B,n)$ are local rings, with $ mB \subset n$. Then the problem reduces to showing that:

 $ dim B \leq dim A \plus{} dim B/mB$.

Up to here I understand the proof, but then it goes wrong :)

[quote]
To prove this, take a system of parameters $ x_1,...,x_r$ of $ A$ and choose $ y_1,...,y_s \in B$ such that their images in $ B/mB$ form a system of parameters in $ B/mB$. Then for $ v,u$ large enough we have $ n^v \subset mB \plus{} \sum y_i B$ and $ m^u \subset \sum x_j A$, giving $ n^{vu} \subset \sum y_i B \plus{} \sum x_j B$. Hence $ dim B \leq r \plus{} s$
[/quote]

Seriously lost :(

(p.s.: I have no algebraic geometry background, so my interpretation is very limited of this theorem.)

Let $ I\equal{}(y_1,\ldots,y_s)\plus{}mB$. The assumption says that $ I/mB$ is a parameter ideal in the local ring $ (B/mB,n/mB)$, so there exists $ v$ such that $ (n/mB)^v\subseteq I/mB$, which is equivalent to $ n^v\subseteq I$. owk

What's a "parameter ideal"? An ideal generated by a system of parameters?

Furthermore, I don't understand either why it is possible to:

[quote]
take a system of parameters $ x_1,...,x_r$ of $ A$ and choose $ y_1,...,y_s \in B$ such that their images in $ B/mB$ form a system of parameters
[/quote]

[quote="Jacobson"]What's a "parameter ideal"? An ideal generated by a system of parameters?[/quote]

Yes.

[quote]
take a system of parameters $ x_1,...,x_r$ of $ A$ and choose $ y_1,...,y_s \in B$ such that their images in $ B/mB$ form a system of parameters
[/quote]

I do not understand what is unclear about this. Maybe you should read the first paragraph of §14 again. owk


A semicircle is inscribed in a quadrilateral ABCD in such a way that the midpoint
of BC coincides with the center of the semicircle, and the diameter of the semicircle
lies along a portion of BC. If AB = 4 and CD = 5, what is BC?

[hide="solution"]
Call the points of tangency to AB, AD, and DC F,G,and H. We let BF=HC=n (since the center of the semicircle's on the midpoint), so DH=GD=5-n and AF=GA=4-n. Let the radius be r. We wish to find $ 2\sqrt {r^2 \plus{} n^2}$. Call the center of the semicircle O. Angles BOF,FOA,AOG,DOG,DOH,and HOC must add up to 180. So 2arctan(n/r)+2arctan((4-n)/r)+2arctan((5-n)/r)=180. Solving this gives $ r^2 \plus{} n^2 \equal{} 20$, so subbing in the answer is $ 4\sqrt {5}$. [/hide]
Since there is more than one possible figure, though, you could assume something in solving the problem, making it easier but not fully proving the result. By the way, I just edited my solution to make it easier.

There are 689 Theorems on cut-the-knot and I was just randomly clicking on them on and the second one I clicked was similar to this problem! (I'm not kidding!). It's called the see-saw theorem. [url]http://www.cut-the-knot.org/Curriculum/Geometry/SeeSaw.shtml[/url]. It's not the same as this but I think there must be a way to generalize it to this, seeing that both have the answer as the geometric mean.


Two circles $ \Gamma$ and $ \Gamma '$ intersect at $ A$ and $ D$. A line is tangent to $ \Gamma$ and $ \Gamma '$ at $ E$ and $ F$ , respectively. The line $ BC$ contains the point $ D$ parallel to the line $ EF$ such that $ C$ in $ \Gamma$ and $ B$ in $ \Gamma '$. Show that the circumcircles of $ \triangle BDE$ and $ \triangle CDF$ intersect again on the line $ AD$.

Consider the positive inversion through center $ D$ and arbitrary radius. Let  $ B',C',E',F'$ be the images of $ B,C,E,F.$ The circumferences $ \odot(BDE)$ and $ \odot(CDF)$ are taken into the lines $ B'E'$ and $ C'F'$ intersecting at the inverse image $ A'$ of $ A$ and the circles $ \Gamma$ and $ \Gamma'$ are taken  into the lines $ A'C'$ and $ A'B'.$ Since $ BC \parallel EF,$ the common tangent $ EF$ of $ \Gamma, \Gamma'$ is transformed into the incircle $ \omega$ of $ \triangle A'B'C'$ tangent to $ B'C',$ $C'A',$ $A'B'$ at $ D,E',F'.$ The lines $ B'E',C'F'$ and $ A'D$ concur at the Gergonne point of $ \triangle A'B'C',$ thus the primitive figures, namely the circles  $ \odot(BDE) , \odot(CDF)$ and the double  line $ AD$ concur at another point besides $ D.$

another solution is to prove that $ O_1S_1\equal{}O_2S_2$ and then obviously we r done!
($ O_1,S_1,O_2,S_2$ are the centers of $ \Gamma,(CDF),\Gamma',(BDE)$)

[quote="luisgeometria"]
The circumferences $ \odot(BDE)$ and $ \odot(CDF)$ are taken into the lines $ B'E'$ and $ C'F'$ intersecting at the inverse image $ A'$ of $ A$[/quote]

I think the intersection of $ B'E'$ and $ C'F'$ is not always $ A'$. Check the image.
If I'm wrong, I want to know why?

[quote="mathson"]I think the intersection of $ B'E'$ and $ C'F'$ is not always $ A'.$ Check the image.[/quote]
Dear mathson, I did not claim that the lines B'E' and C'F' meet at A', but they meet at the Gergonne point of  A'B'C'.

I'm really sorry, I didn't understand you at the first time.

[quote]
the common tangent $ EF$ to $ \Gamma$ and $ \Gamma'$ is transformed into the incircle $ \omega$ of $ \triangle A'B'C'$
[/quote]
Is there any generalization for this? Or theorem?


For $ a,b,c$ positive reals prove that

$ \sum \frac{a}{b} \plus{} 3 \geq 2(\sum \frac{a\plus{}b}{b\plus{}c})$

after expanding It's equivalent to prove that
\[ \sum_{cyc}a^3b^3\plus{}\sum_{cyc}a^4c^2\plus{}\sum_{cyc}a^3bc^2 \geq \sum_{cyc}a^4bc\plus{}\sum_{cyc}a^3b^2c\plus{}3a^2b^2c^2\]
but I don't know how to prove it. :(

[quote="can_hang2007"][quote="manlio"]For $ a,b,c$ positive reals prove that

$ \sum \frac {a}{b} \plus{} 3 \geq 2(\sum \frac {a \plus{} b}{b \plus{} c})$[/quote]
Let $ a\equal{}0.1,b\equal{}1.2,c\equal{}0.2$, the inequality reversed. :(
Hence, the desired inequality is wrong. Actually, it holds for $ a,b,c$ are the side lengths of a triangle.[/quote]
Now, let us prove the inequality in the case $ a,b,c$ are the side lengths of a triangle. Rewrite it as
\[ \sum \frac {a}{b} \plus{} \sum\frac {2b}{a \plus{} b} \ge \sum \frac {2a}{b \plus{} c} \plus{} 3
\]

\[ \Leftrightarrow (\sum ab)(\sum \frac {a}{b} \plus{} \sum\frac {2b}{a \plus{} b}) \ge (\sum \frac {2a}{b \plus{} c} \plus{} 3)(\sum ab)
\]

\[ \Leftrightarrow \sum \frac {ca^2}{b} \plus{} 2\sum \frac {ab^2}{a \plus{} b} \ge \sum a^2 \plus{} 2abc\sum \frac {1}{b \plus{} c}
\]
Since $ a,b,c$ are the side lengths of a triangle, by VasC's Inequality (Moldova 2006), we have
\[ \sum \frac {ca^2}{b} \ge \sum a^2
\]
It suffices to prove that
\[ \sum \frac {ab^2}{a \plus{} b} \ge \sum \frac {abc}{b \plus{} c}
\]

\[ \Leftrightarrow \sum \frac {ab(b \minus{} c)}{a \plus{} b} \ge 0
\]

\[ \Leftrightarrow \sum ab(b^2 \minus{} c^2)(a \plus{} c) \ge 0
\]

\[ \Leftrightarrow \sum a^2b^3 \ge \sum a^2b^2c
\]
which is trivial by the AM-GM. Equality holds if and only if $ a \equal{} b \equal{} c.$ :)


Find all four digit numbers of the form $aabb$ such that they are squares.

Masoud Zargar

[quote="TRAN THAI HUNG"][hide]Let $\ aabb = c^2$
=> $\ a.1100+b.11= c^2$
So $\ c^2\equiv\ 0\mod11$ so $\ c\equiv\ 0\mod11$
Let c =11d(d belong to Z)We have:
$\ a.1100+b.1100= 121d^2$
$\ a.100+b= 11d^2$
So $\ a.100+b\equiv\ 0\mod11$ => $\ a+b\equiv\ 0\mod11$
And a+b<20 so a+b=11
Try with all 2299, 3388, 4477, 5566, 6655, 7744, 8833, 9922.We only have 7744[/hide] :play_ball:

Math is my life :!:[/quote]
Probably easier if you just checked the possibilities after finding out c=11d since 2<d<10

"bb" must be 0 or 1 modulo 4, implying $3b$ is 0 or 1 mod 4, implying $b = 0, 4, 8, 3, 7$ is possible.

The number is divisible by 11.  Therefore it is divisible by 121, implying $100a + b = 0 \mod 11$, or $a+b = 0 \mod 11$.  From $b = 0,4,8,3,7$ we have at most 5 numbers to check.

$11(100a+b)=p^2 \implies 100a+b=11\cdot x^2$

Now $100a+b\leq909$ otherwise, $11(100a+b)>9999$

$\implies x\leq 9$

check all possibilties $\implies 100a+b=11\cdot8^2=704$

only soln is $11\cdot 704=7744$


[color=#00CCFF]MATHCOUNTS Scores by jda12.8[/color]

[color=#FF3300]6th Grade - 2004

Chapter competition 2004 - 32/46 (7th)
State competition 2004 - 22/46 (9th)[/color]

[color=#00CC00]7th Grade - 2005

Chapter competition 2005 - 41/46 (1st)
State competition 2005 - 37/46 (1st)
National competition 2005 - ??/46 (99/228th)[/color]

[color=#6600FF]8th Grade - 2006

Chapter competition 2006 - 43/46 (1st)
State competition 2006 - 44/46 (1st)
National competition 2006 - 31/46 (41/228th, Arkansas team 12/57th)[/color]

[color=#000000]mario_jda128[/color] :D

[quote="funcia"]what is the purpose of this post?[/quote]
I think he wants us to post our scores.
Maybe not


$a,b,c$ are sides of a triangle and $r$ is radius of interior circle. Prove that ;

$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\leq\frac{1}{4r^{2}}$

[hide]Substitute $a=x+y,b=y+z,c=z+x,s=x+y+z$ where $x,y,z$ are positive real numbers.
$[ABC]=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{xyz(x+y+z)}=rs\implies r^{2}=\frac{xyz}{x+y+z}$
The inequality becomes
$\sum\frac{1}{(x+y)^{2}}\le \frac{x+y+z}{4xyz}=\sum\frac{1}{4xy}$
which is true since
$\frac{1}{(x+y)^{2}}\le\frac{1}{4xy}\implies 4xy\le x^{2}+2xy+y^{2}\implies 0\le(x-y)^{2}$
[/hide]