Yeah, I have an exercise to show :
Let get an array {Un}(n=1,2,...)  defined by formula :
              a*Un=b*U(n+1)+c*U(n+2) ( a<>0 and b*b+4*a*c<0)
Find Real numbers a,b,c satisfy that {Un} is a periodic array .
Can you solve it ?

No one can solve this problem? But i've solved it recently. I used complex number to solve easily !


Treething Inc. manufactures hydrogen bombs. For every 4,975 working bombs, there are 25 defective ones.

a) What is the probability that a defective bomb is picked out of 10,000 working ones?

Before Treething Inc. got better technicians, for every 31 working bombs, there were 49 defective ones. Did Caddlebot get more or less than the likely amount of working bombs? (see part I)

For 9950 working bombs there are 50 defective ones . So, the answer is 0,005

Ooops  :blush: I didn't read it carefully.

[quote="shinwoo"]a) 0% because you already know that they are the working bombs...

It's a kind of weird qeustion  :?[/quote]

[hide="Only shinwoo can looky"]
Trick question ;P .
[/hide]

Treething, I gotta admit there funny and that I don't even post many problems.  :D  :D  :D But other than to read the question carefully what are you learning?  :P

oooh

A+MATH, I think it's very important

messing up one problem by not reading it carefully can mess you up real bad in math competitions

Well, you could try to answer the second question :P ...

who is Caddlebot?

the question makes no sense to me.....

i think it's another trick question


$ \frac{a\plus{}b} {c} \plus{} \frac{a\plus{}c} {b} \plus{} \frac{b\plus{}c} {a}\equal{}1$

$ \frac{a} {b\plus{}c} \plus{} \frac{b} {a\plus{}c} \plus{} \frac{c} {a\plus{}b}\equal{}\frac{4} {3}$

what is $ a^3\plus{}b^3\plus{}c^3$?

This obviously does not have one solution...you can scale the variables.

Anyway for a sort of solution: Let $ x\equal{}\frac{a\plus{}b\plus{}c}{a}$, etc. Then $ \sum\frac{1}{x}\equal{}1$, $ \sum x\minus{}1\equal{}1$, and $ \sum\frac{1}{x\minus{}1}\equal{}\frac{4}{3}$.

Then expand in terms of elementary symmetric sums, etc.

That the variables can be scaled might be a clue that there's something fishy here. I got $ 0$ for my answer.

Okay if $ (\sigma_1,\sigma_2,\sigma_3)\equal{}(4,9,9)$ for $ \sigma_1\equal{}x\plus{}y\plus{}z$, etc where $ x\equal{}\frac{a\plus{}b\plus{}c}{a}$...and you can invert the polynomial or whatever and you get zero as your answer...

[hide]
I think the 1st equation can modify to $ \frac{(a\plus{}b)(b\plus{}c)(a\plus{}c)}{abc}\equal{}3$

it really begs for a solution but I kinda need to go to sleep.
[/hide]

[hide="hint"]solve for $ abc$ in the first equation, then use that for substitution when you find a common denominator in the second equation[/hide]

by many calculus...I obtained that $ a^3 \plus{} b^3 \plus{} c^3 \equal{} \frac {8}{3} abc$ I think I am missing something :(

[hide="no calc needed"]multiply the first equation by $ abc$[/hide]

[quote="abacadaea"]$ \frac {a \plus{} b} {c} \plus{} \frac {a \plus{} c} {b} \plus{} \frac {b \plus{} c} {a} \equal{} 1$

$ \frac {a} {b \plus{} c} \plus{} \frac {b} {a \plus{} c} \plus{} \frac {c} {a \plus{} b} \equal{} \frac {4} {3}$

what is $ a^3 \plus{} b^3 \plus{} c^3$?[/quote]
[hide="What did I do wrong?"]
By the  first equation, we have that
\[ \frac {ab(a \plus{} b) \plus{} ac(a \plus{} c) \plus{} bc(b \plus{} c)}{abc} \equal{} 1
\]\[ a^2b \plus{} ab^2 \plus{} ac^2 \plus{} a^2c \plus{} b^2c \plus{} bc^2 \equal{} abc
\]\[ (a \plus{} b)(b \plus{} c)(a \plus{} c) \equal{} abc
\]By the second equation, we have that
\[ 3[a(a^2 \plus{} bc \plus{} ac \plus{} ab) \plus{} b(b^2 \plus{} bc \plus{} ac \plus{} ab) \plus{} c(c^2 \plus{} bc \plus{} ac \plus{} ba)] \equal{} 4(a \plus{} b)(b \plus{} c)(a \plus{} c)
\]\[ 3[a^3 \plus{} b^3 \plus{} c^3 \plus{} (a \plus{} b)(b \plus{} c)(a \plus{} c) \plus{} 3abc] \equal{} 4(a \plus{} b)(b \plus{} c)(a \plus{} c)
\]\[ 3(a^3 \plus{} b^3 \plus{} c^3 \plus{} 4abc) \equal{} 4abc
\]\[ 3a^3 \plus{} 3b^3 \plus{} 3c^3 \plus{} 8abc \equal{} 0
\]\[ a^3 \plus{} b^3 \plus{} c^3 \equal{} \minus{}\frac { 8}{3}\cdot abc
\]
Now this seems to be odd since as [b]Altheman[/b] said, we can scale down the variables.  The only way that this cannot be possible is if $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 0$.  This would mean that $ abc \equal{} 0$, but that is impossible because it is the denominator of a fraction.  Could someone tell me what I did wrong? [/hide]

[quote="The QuattoMaster 6000"][quote="abacadaea"]$ \frac {a \plus{} b} {c} \plus{} \frac {a \plus{} c} {b} \plus{} \frac {b \plus{} c} {a} \equal{} 1$

$ \frac {a} {b \plus{} c} \plus{} \frac {b} {a \plus{} c} \plus{} \frac {c} {a \plus{} b} \equal{} \frac {4} {3}$

what is $ a^3 \plus{} b^3 \plus{} c^3$?[/quote]
[hide="What did I do wrong?"]
By the  first equation, we have that
\[ \frac {ab(a \plus{} b) \plus{} ac(a \plus{} c) \plus{} bc(b \plus{} c)}{abc} \equal{} 1
\]

\[ a^2b \plus{} ab^2 \plus{} ac^2 \plus{} a^2c \plus{} b^2c \plus{} bc^2 \equal{} abc
\]

\[ (a \plus{} b)(b \plus{} c)(a \plus{} c) \equal{} abc
\]
By the second equation, we have that
\[ 3[a(a^2 \plus{} bc \plus{} ac \plus{} ab) \plus{} b(b^2 \plus{} bc \plus{} ac \plus{} ab) \plus{} c(c^2 \plus{} bc \plus{} ac \plus{} ba)] \equal{} 4(a \plus{} b)(b \plus{} c)(a \plus{} c)
\]

\[ 3[a^3 \plus{} b^3 \plus{} c^3 \plus{} (a \plus{} b)(b \plus{} c)(a \plus{} c) \plus{} 3abc] \equal{} 4(a \plus{} b)(b \plus{} c)(a \plus{} c)
\]

\[ 3(a^3 \plus{} b^3 \plus{} c^3 \plus{} 4abc) \equal{} 4abc
\]

\[ 3a^3 \plus{} 3b^3 \plus{} 3c^3 \plus{} 8abc \equal{} 0
\]

\[ a^3 \plus{} b^3 \plus{} c^3 \equal{} \minus{} \frac { 8}{3}\cdot abc
\]
Now this seems to be odd since as [b]Altheman[/b] said, we can scale down the variables.  The only way that this cannot be possible is if $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 0$.  This would mean that $ abc \equal{} 0$, but that is impossible because it is the denominator of a fraction.  Could someone tell me what I did wrong? [/hide][/quote]
I tried t solve the problem in the same way that you do :maybe: ...but it seems we're missing something

Can someone give an example of reals $ a,b,c$ that satisfy the given equations?

It requires you to solve a cubic equation (and there might not be reals, but they would be complex).

Of course, $ a\plus{}b\plus{}c\ne 0$, since that would give $ \minus{}3\equal{}1$. So we can let $ x \equal{} \frac {a}{a \plus{} b \plus{} c}$, a.s.o. Introduce the elementary symmetric sums, $ (\sigma_i)_{i \equal{} 1,2,3}$. Then $ \frac {1 \minus{} x}{x} \equal{} \frac {b \plus{} c}{a}$, the first and second equations, become, respectively:
\[ \sum \frac {1 \minus{} x}{x} \equal{} 1\iff 4\sigma_3 \equal{} \sigma_2
\]

\[ \sum\frac {x}{1 \minus{} x} \equal{} \frac {4}{3}\iff 4 \plus{} 10\sigma_2 \equal{} 7\sigma_1 \plus{} 13\sigma_3
\]
Additionally, we notice that
\[ \sigma_1 \equal{} 1
\]
This gives us a linear system in $ (\sigma_i)_{i \equal{} 1,2,3}$, which can be easily solved
\[ (\sigma_i)_{i \equal{} 1,2,3} \equal{} \left(1,\frac {4}{9},\frac {1}{9}\right)
\]
Indeed,
\[ x^3 \plus{} y^3 \plus{} z^3 \equal{} \sigma_1^3 \minus{} 3\sigma_1\sigma_2 \plus{} 3\sigma_3 \equal{} 1^3 \minus{} 3\cdot 1\cdot\frac {4}{9} \plus{} 3\cdot\frac {1}{9} \equal{} 0
\]
after scaling back by $ (a \plus{} b \plus{} c)^3$, we get the desired result.

As a further check, we note that
\[ (t \minus{} x)(t \minus{} y)(t \minus{} z) \equal{} t^3 \minus{} \sigma_1 t^2 \plus{} \sigma_2 t \minus{} \sigma_3 \equal{} t^3 \minus{} t^2 \plus{} \frac {4}{9}\cdot t \minus{} \frac {1}{9} \equal{} 0
\]
has roots $ x,y,z$. We can multiply those by some constant to get $ a,b,c$. We can verify by going backwards that these values work in the original equations.


[quote="Ligouras"]Let $ a, b, c, k$ be non-negative real numbers. Prove o disprove that

$ \sum[\frac {(a^2 \minus{} bc)(\sqrt {2a^2 \plus{} b^2 \plus{} c^2} \plus{} k)}{2a^2 \plus{} b^2 \plus{} c^2}] \ge 0$ .

__________
 :?:  :blush:[/quote]

Who can solve this?

[quote="Ligouras"]Let $ a, b, c, k$ be non-negative real numbers. Prove o disprove that

$ \sum[\frac {(a^2 \minus{} bc)(\sqrt {2a^2 \plus{} b^2 \plus{} c^2} \plus{} k)}{2a^2 \plus{} b^2 \plus{} c^2}] \ge 0$ .

__________
 :?:  :blush:[/quote]

we can prove that

$ \sum{\frac{a^2\minus{}bc}{2a^2\plus{}b^2\plus{}c^2}} \ge 0$

because

$ \sum{\frac{a^2\minus{}bc}{2a^2\plus{}b^2\plus{}c^2}}$

\[ \equal{} \frac{{\left( {\sum {\left( {a \minus{} b} \right)^2 } } \right)\left( {2\sum {a^2 \left( {a \minus{} b} \right)\left( {a \minus{} c} \right)}  \plus{} 3\sum {c^2 \left( {a \minus{} b} \right)^2 } } \right)}}{{4\prod {\left( {2a^2  \plus{} b^2  \plus{} c^2 } \right)} }},\]

so, we just need to prove that when $ k\equal{}0$.

to be continue...

[quote="hedeng123"][quote="kuing"]
so, we just need to prove that when $ k \equal{} 0$.

[/quote]

why?[/quote]

because $ k \ge 0$ and

$ \sum{\frac {(a^{2} \minus{} bc)(\sqrt {2a^{2} \plus{} b^{2} \plus{} c^{2}} \plus{} k)}{2a^{2} \plus{} b^{2} \plus{} c^{2}}}$

$ \equal{} \sum{\frac {a^{2} \minus{} bc}{\sqrt {2a^{2} \plus{} b^{2} \plus{} c^{2}}}} \plus{} k\sum{\frac {a^{2} \minus{} bc}{2a^{2} \plus{} b^{2} \plus{} c^{2}}},$

I was proved $ \sum{\frac {a^{2} \minus{} bc}{2a^{2} \plus{} b^{2} \plus{} c^{2}}} \ge 0$ is hold, so after we just need to prove that

$ \sum{\frac {a^{2} \minus{} bc}{\sqrt {2a^{2} \plus{} b^{2} \plus{} c^{2}}}} \ge 0.$

[quote="theSA"][quote="kuing"][quote="Ligouras"]Let $ a, b, c, k$ be non-negative real numbers. Prove o disprove that

$ \sum[\frac {(a^2 \minus{} bc)(\sqrt {2a^2 \plus{} b^2 \plus{} c^2} \plus{} k)}{2a^2 \plus{} b^2 \plus{} c^2}] \ge 0$ .

__________
 :?:  :blush:[/quote]

we can prove that

$ \sum{\frac {a^2 \minus{} bc}{2a^2 \plus{} b^2 \plus{} c^2}} \ge 0$

because

$ \sum{\frac {a^2 \minus{} bc}{2a^2 \plus{} b^2 \plus{} c^2}}$
\[ \equal{} \frac {{\left( {\sum {\left( {a \minus{} b} \right)^2 } } \right)\left( {2\sum {a^2 \left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} \plus{} 3\sum {c^2 \left( {a \minus{} b} \right)^2 } } \right)}}{{4\prod {\left( {2a^2 \plus{} b^2 \plus{} c^2 } \right)} }},\]
[/quote]

I think this prove is enough. since all numbers are nonnegative,then the second part is always bigger then or egual to 0.
I think... :blush:[/quote]

I don't think so.

although we have $ \sum{\frac {a^2 \minus{} bc}{2a^2 \plus{} b^2 \plus{} c^2}} \ge 0$ and $ f(a,b,c) \ge 0$, but if $ f(a,b,c)$ is not a constant, there is not enough to show that $ \sum{\frac {(a^2 \minus{} bc)f(a,b,c)}{2a^2 \plus{} b^2 \plus{} c^2}} \ge 0.$


inequalities geometry circumcircle trigonometry

Let $M$ be an interior point of the triangle $ABC$ such that $AMC = 90^\circ$, $AMB = 150^\circ$, and $BMC = 120^\circ$. The circumcenters of the triangles $AMC$, $AMB$, and $BMC$ are $P$, $Q$, and $R$ respectively. Prove that the area of $\Delta PQR$ is greater than or equal to the area of $\Delta ABC$.

I do not think such a point can exist!  Because $M$ is interior, angles AMB, BMC and CMA must add up to 360 degrees, so angle CMA is ninety degrees.  But no interior point meets this condition.

Is there something wrong?

Good problem. I'll give a vague couple steps of a solution--try to fill in the gaps.

Let D and E be points outside the triangle s.t. $\angle ABD = 90, \angle DAB = 30, \angle CBE = 90, \angle BEC = 30$. It suffices to prove that the area of ABC is at most one fourth of the area of ADEC (why?). Note that by the trig formula for areas, triangles DBE and ABC have the same area.

We now see that we might as well have $\angle ABC = 90$, so the problem becomes algebraic; prove that

$[ABD] + [CBE] \ge 2 [ABC]$
$\frac{x^2}{\sqrt{3}} + y^2\sqrt{3} \ge 2xy$

which is simply AM-GM.

Edit--in response to JBL, I think he meant $\angle AMC = 90$.

Whoops, my bad... yeah, that's right $AMC = 90^\circ$, not $ABC$.  :D  EDITED.

[quote="Karth"]How do you know that $\angle ABD = 90, \angle DAB = 30, etc.$?[/quote]

probability1.01 constructed $D$ and $E$ to meet those conditions.


I won't need that much help, however, I do usually seek a really quick reply. I do like what I have here, but I'm trying to speed along in what I'm trying to learn and if I come across something I don't understand, I would hope there was someone I could message online to ask.

I usually don't have that many questions unless the problem is complex. Right now all I need help in is Algebra/Intermediate Algebra.

I usually come across a problem I can't solve every 25 pages or so and I'm trying to blow through 100 pages a day. So far I've only done 50, started yesterday.

I have done Algebra/Int. Alg./Geometry in highschool so I have experience with what I'm looking at, I've just forgotten most of it.

Anyone willing to help would be much appreciated.

You could very possibly find someone willing to help you out, but you should definitely consider posting any problems you need help on and questions you have about those problems in the forums of appropriate difficulty level. (It seems that yours will probably be in Getting Started and Intermediate.)


algebra
system of equations
special factorizations

Can anyone help figure this one out?

For what value of k is the following system of equations consistent? Find all the solutions when k has this value.
x+y+z=2
2x+y+2z=5
4x+3z+4z=k

Thanks again for any help

[hide="1st step "]Delete typo in the 3d eq.[/hide]
[hide="2nd step "]Observe the first two eq's[/hide]
[hide="3rd step "]If you have recieved $y=1$ then goto step 4 if not go to step2.[/hide]
[hide="4th step "]Think about what you have after steps 1-3. [/hide]
[hide="5th step "]If you have caugcht the idea then write the answer, if not goto step 6. [/hide]
[hide="6th step "]If you like this answer don't mind I am the first who was kidding this way and go to step 2 if not then ... goto the same.
P.S. I hope you have passed  step 1 succesfully[/hide]

I think you meant the following 3 equations. You have 2 zs in the 3rd one.

\[x+y+z=2 \\ 2x+y+2z=5 \\ 4x+3y+4z=k\]

yea i agree with mindboggler. $k=9$

[quote="mindboggling"][hide="answer?"]

1st equation multiply by 2 + 2nd equation gives you 3rd thus k = 9?

[/hide][/quote]

Thanks, sorry for my own typo in sign. Just now I'll go there and improve step 3.
P.S. I've failed to improve, havn't understand why.
Well, read step 3 as follows:
... recieved $y=-1$ ...

Thank you for the help! and sorry about the typo! :blush:

Is this called completing the square? or would one have attempted some sort of solution by matrices? or if both how do you choose which tactic is most appropriate?

I can now see that the third equation was just a sum of the first two. Also, now that I have the answer, it probably should have been obvious to me that the first two were leading down to the last one.

Is it a matter of "experience" as to recognizing this or is there something about this problem that makes one consider certain avenues?

Sorry to be such a dunce. If my questions are too drawn out just ignore me, I will study up some more.

Either way thanks again!


What is the $ n^{\text{th}}$ term of the following sequence?
\[ 3,8,17,32,57,\cdots\]

Is the next term by any chance 98?

Basically, the recursion is $ a_{n\plus{}2}\equal{}a_{n\plus{}1}\plus{}a_{n}\plus{}2(n\plus{}3)$, where $ n \ge 0$, and $ a_{0}\equal{}3$, $ a_{1}\equal{}8$

Furthermore, notice that this can be solved by noting the inhomogeneous part is $ v_{n}\equal{}An\plus{}B$, and $ u_{n}\equal{}C\phi^{n}\plus{}D\phi^{\minus{}n}$, where $ \phi$ is a root of $ x^{2}\equal{}x\plus{}1$

I won't solve it, becomes computations comlicated.

[quote="mathgeniuse^ln(x)"]Is the next term by any chance $ 98$?[/quote] Yes, it is $ 98$. But how to find $ n^{th}$ term? :maybe:

[quote="mathgeniuse^ln(x)"]I won't solve it, becomes computations complicated.[/quote] I found the recursive equation as well. But I haven't found any efficient way to solve it :(

The technique has already been given.  The general technique is well-known, but for this particular case I believe it suffices to make the ansatz $ a_n \equal{} c_1 n \plus{} c_2 \plus{} c_3 F_n \plus{} c_4 L_n$ where $ F_n$ is the Fibonacci sequence and $ L_n$ is the Lucas sequence.


geometry
geometric transformation
reflection

I've been playing the piano for 7-8 years now, and I've been teacher-taught the whole time.  But now I've realized that when I go to lessons, I don't really learn much.  It's just like they tell me what to play and occasionally correct mistakes or give advice on dynamics and such, but I've found that the songs that I much prefer to play are the songs that I teach myself and I can figure things out by listening to a cd if sight alone isn't enough.  So I'm thinking about quitting piano lessons, but continuing playing.

Because I very often idolize people I meet on this forum, I'm wondering, how did [i]you[/i] learn piano?  Always had a teacher, never had a teacher, started with a teacher then quit, started alone then got a teacher, etc.  And also I guess what do you think about my situation?

(and by the way, I originally thought it might jsut have been the teacher I had, so I switched last year to a nearby college's "piano academy (where I am currently taking lessons) but it seems the same there (just in case you were going to advise trying another teacher  :P ))

I guess it'll be best to be teacher taught for a while and go by yourself after that ....you even develop a kind of style that's unique to you.....
As far as I am concerned , I well , started on my own , then went to a techer , and now I'm back to practising myself..... :)

A really good teacher will teach you how to practice, so that you are gaining the benefit of the teacher's expertise every minute you are at the piano.

But even if you have a really good teacher , you do need to practise by yourself for some time a day .... I'm sure that helps you to express your feelings while playing a song.....{ the way you play actually reflects the way you're thinking , you know....}

I usually go to my teacher for interpretation. He sometimes works things out (like dastardly rhythms) but it's mostly the feeling. Usually, what I do, is I listen to a good recording of that song (usually Ivo Pogorelich or Artur Rubinstein) and I take what I like. I sometimes disagree with my teacher, but he usually lets me keep the changes. For fingering, I usually have to sit down and work everything out. What I recommend is playing the song slowly, page-by-page, and checking all the fingerings and dynamics. When you are ready, you can slowly increase the tempo. That is why it's important to use a metronome! (Chopin always carried a metronome in his pocket) Before recitals, which for me are usually on Sunday, I start practicing on Thursday and Friday (after school) and then I practice Saturday for the whole day. So that's basically it.

Basically a teacher has to given advice on technique and musicality. Every piano student has something that a teacher can comment about musically, so I think every teacher should always have something to say about interpretation, phrasing, etc.

Also they should be giving advice on how to practice hard sections, memorization, etc. if they are ever a problem.

[quote="Quickster94"]I've always had a teacher, and have very rarely if ever learned pieces on my own.
I think the goal of teaching is to point out interpretations, musical details, etc. that someone more experienced and educated would know.
Then, eventually, you notice more things on your own, and the teachers starts catching smaller and smaller.
This helps you get better and better, and you hopefully reach a point where you can play the role of the teacher.[/quote]
Point where you can play the teacher? I don't think anyone classified as a student would have the knowledge/experience to do so.

As you learn more, you learn things faster, and so you can go through stuff faster and stuff because there's less to correct.

Learning pieces on your own is fine, but you probably want your teacher to hear it before you try to do anything with it, so you aren't doing something really wrong.

I started out alone until I got a teacher, who really built my fundamentals. Then I switched to a new one who expanded my skills even more.

A very important part of learning is to have adequate opportunities to perform. If you haven't realized it, you will never learn a piece that you learn for fun as well as one you have to perform several times. And it is this extra polish that you need to work for to seem good. I am far from a great pianist, but thanks to my teacher, I have played concertos several times with various orchestras and have performed many many times.

[quote="serialk11r"]
Point where you can play the teacher? I don't think anyone classified as a student would have the knowledge/experience to do so.
[/quote]

Right, but I think the point is that if you really like piano and pursue it, you could be experienced enough to teach others.

Also, I wish I got to play a concerto with orchestra  :lol:

I have been taught for over six years by a tEacher with Fifty years teaching experience, and I don't want to go without her till college, because she helps a lot with fingering and technique. I would find a Good teacfher, A registered State Music Teahcer associacion one (If your state has one) and stay with him/her for a while.

I've been playing piano for about 6 years, but haven't made as much progress as you guys.  My former teacher was pretty good, but there were a few pieces she wanted me to play, and I couldn't get them to the perfection she required.  So I was stuck with these pieces, with the knowledge I was incapable of playing them.  She could've at least moved me to another piece (even at a lower level), instead of me being stuck for 6 months on that piece.  I quit 2 weeks ago, and am taking the summer off.

(My teacher was the kind that wanted to be a concert pianist when she grew up, and practiced 6 hours a day.)

I'm not much of a pianist, but I would consider myself a "higher level cellist". :P

One piece of advice my new teacher has given me was: "Don't practice... experiment." It is very important to have a good teacher, but the better teacher you have, the more you have to practice, the more you have to experiment. It is extremely important that you continue to find what is wrong with what you are playing. Especially with the piano, the only real difference between you as a student, and Gulda is (age) [b]how[/b] you play the piano. While other instruments may have differences between instruments, piano is an instrument you can overcome with practice, with experimentation. As you get better, better teachers come up. I think one big difference between good and great teachers is that fact that good teachers tell you how you should sound. This is very important. However, great teachers how you should produce that sound.  This difference has contributed greatly to my cello playing.

If this turns into a discussion of music pedagogy things may get interesting...

My personal experience:

My piano teacher is definitely among the best in our state, and I definitely think he helps me a lot.  One major difference from my previous piano teacher and my current one was expectations.  My current teacher is much clearer about them, and it really motivates me to do more.

In my area, there are only a handful of teachers with performing careers or at least advanced musical degrees, and I think it really makes a difference when it comes to musicality.  When I hear students of other teachers play, I definitely feel as if there's something missing.  Perhaps their pieces are very difficult, but something always seems to be a little bit off to me.

My teacher also helps me in selecting pieces (someone experienced has played and knows so much literature, and there's a lot out there).  In relation to competitions, I think a contributing factor winning is piece selection, and an experienced pianist will definitely help with this.

I'm not sure of the OP's teacher's qualifications, but I definitely recommend studying with a serious professional musician (perhaps a faculty at the local university) if you're serious about studying.

Yes, piece selection is important. Unless you are in a non-serious competition. There is a all instrument competition that I do that doesn't relate to music difficulty (this is why I only won second when I was seven and eight).


trigonometry
inequalities
algebra solved
algebra

1. Given a positive integer n, find the smallest positive integer  \lambda such that, for any \theta i \in (0, \pi /2), (i=1,2,...,n), if tan \theta1 tan \theta2 ... tan \thetan =2n/2  then

  cos \theta 1 + cos \theta 2 +...+ cos \theta n \leq \lambda

looks to me like n \geq \lambda \geq n/\sqrt3

these two chineses problems have been lying unsolved for quite sum time now.and i simply find them inpenetrable.mayb it's abt time some darn good brainboy/(girl) on this group solved it .i've given up hope .even a hint would be great progress

i've put up some thought in it, but i don't have the right amount of time just right now, i'll get back on them as soon as the easter ends

that's at the end of april probably :D

if n<=2, the answer is n/sqrt(3)
otherwise, it is n-1.

well, i have a proof.
but it looks so complicated.

i wrote a tex file before, so i now paste it here.

$n=1$ is trivial, for $\tan \theta_1=1/\sqrt{2}$ told us that
$\cos \theta_1 = \sqrt{1/(1+\tan^2 \theta_1)} = 1/\sqrt{3}$. If
$n=2$, we want to prove
\[\cos\theta_1 +\cos\theta_2\leq \frac{2}{\sqrt{3}} \]
which is
\[\cos^2 \theta_1 +\cos^2 \theta_2 + 2\cos\theta_1\cos\theta_2 \leq
\frac{4}{3}\]
\[ \frac{1}{1+\tan^2 \theta_1}+\frac{1}{1+\tan^2 \theta_2}+2\sqrt{\frac{1}{(1+\tan^2 \theta_1)(1+\tan^2
\theta_2)}} \leq \frac{4}{3}
\]
Because $\tan \theta_1\tan \theta_2=2$, so the above becomes
\[ \frac{2+\tan^2 \theta_1+\tan^2 \theta_2}{5+\tan^2 \theta_1+\tan^2 \theta_2}+2\sqrt{\frac{1}{5+\tan^2 \theta_1+\tan^2 \theta_2}} \leq \frac{4}{3}
\]
Set $x=\sqrt{5+\tan^2 \theta_1+\tan^2 \theta_2}\geq 3$, it becomes
\[\frac{x^2-3}{x^2}+\frac{2}{x}\leq \frac{4}{3}\]
\[-\frac{3}{x^2}+\frac{2}{x}\leq \frac{1}{3}\]
which is clearly hold when $1/x\leq 1/3$.

Now suppose $n\geq 3$ and $\theta_1\geq \theta_2\geq
\theta_3\cdots\geq \theta_n$. If $\cos\theta_1 +
\cos\theta_2+\cos\theta_3 < 2$, we have $\cos\theta_1 +
\cos\theta_2+\cos\theta_3 + \cdots +\cos\theta_n < n-1$ for
$\cos\theta_i < 1$. And let $\theta_2=\cdots=\theta_n=\epsilon\to
0, \theta_1=\arctan 2^{n/2}/\epsilon^{n-1}\to \pi/2$,
$\cos\theta_1+\cos\theta_2+\cdots+\theta_n\to n-1$. It implies
that $L=n-1$. The only thing left is to show that
\begin{equation}\label{p3}
\cos\theta_1 + \cos\theta_2+\cos\theta_3 < 2.
\end{equation}

Since $\theta_i$ is non-increasing, we have $\tan \theta_1\tan
\theta_2\tan \theta_3\geq 2\sqrt{2}$(if not, i.e., $\tan
\theta_1\tan \theta_2\tan \theta_3< 2\sqrt{2}\Longrightarrow
\tan^3 \theta_3 < 2\sqrt{2}\Longrightarrow\tan\theta_i\leq
\tan\theta_3<\sqrt{2}(i>3)\Longrightarrow
\tan\theta_1\tan\theta_2\tan \theta_3\cdots\tan\theta_n <
2^{n/2}$, a contradiction.)

We find out that $\cos x<(1+\cos^2x)/2=1-(\sin^2x)/2$, so
$\cos\theta_2+\cos\theta_3<2-(\sin^2\theta_2+\sin^2\theta_3)/2 <
2-\sin\theta_2\sin\theta_3$. We also get
\begin{eqnarray*}
\cos\theta_1&=&\sqrt{\frac{1}{1+\tan^2 \theta_1}}\\
&\leq&\sqrt{\frac{1}{1+8/\tan^2 \theta_2\tan^2 \theta_3}}\\
&=&\sqrt{\frac{\tan^2 \theta_2\tan^2 \theta_3}{\tan^2
\theta_2\tan^2 \theta_3+8}}\\
&=&\frac{\sin^2\theta_2\sin^2\theta_3}{\sqrt{8\cos^2\theta_2\cos^2\theta_3+\sin^2\theta_2\sin^2\theta_3}}\\
\end{eqnarray*}
Thus,
\begin{eqnarray*}
& &\cos\theta_1+\cos\theta_2+\cos\theta_3\\
&<&\frac{\sin\theta_2\sin\theta_3}{\sqrt{8\cos^2\theta_2\cos^2\theta_3+\sin^2\theta_2\sin^2\theta_3}}+2-\sin^2\theta_2\sin^2\theta_3\\
&=&2-\sin\theta_2\sin\theta_3\left(1-\frac{1}{\sqrt{8\cos^2\theta_2\cos^2\theta_3+\sin^2\theta_2\sin^2\theta_3}}\right)\
\end{eqnarray*}
and
\begin{eqnarray*}
& &1-\frac{1}{\sqrt{8\cos^2\theta_2\cos^2\theta_3+\sin^2\theta_2\sin^2\theta_3}}\geq 0\\
&\Longleftrightarrow&8\cos^2\theta_2\cos^2\theta_3+\sin^2\theta_2\sin^2\theta_3
\geq 1\\
&\Longleftrightarrow&8+\tan^2\theta_2\tan^2\theta_3\geq
\frac{1}{\cos^2\theta_2\cos^2\theta_3}=(1+\tan^2\theta_2)(1+\tan^2\theta_3)\\
&\Longleftrightarrow&\tan^2\theta_2+\tan^2\theta_3\leq 7\\
\end{eqnarray*}
Hence, when $\tan^2\theta_2+\tan^2\theta_3\leq 7$, (\ref{p3})
holds. We now consider $\tan^2\theta_2+\tan^2\theta_3>7$. Under
this assumption, $\tan^2\theta_1\geq \tan^2\theta_2>7/2$, so
$\cos\theta_1\leq\cos\theta_2<\sqrt{1/(1+(7/2)^2)}=2/\sqrt{53}$.
It leads us to
$\cos\theta_1+\cos\theta_2+\cos\theta_3<4/\sqrt{53}+1<2$.

I have made a .pdf file for more quick read.

[url=http://www.mathlinks.ro/diff/liyi%20soln.pdf]GET FILE![/url]

Valentin, you need to insert all the square roots and inequality signs into the TeX document again before converting it to PDF, because they have been processed by the board and therefore cannot be copied into a TeX file.

Thanks :)

for now you send it to me, and I'll upload it.

I will install in the future the attachement mod, for easier uploads.

I have modified the file, it shows square roots now :D

Er...and my browser displays a weird message saying "The file is damaged and could not be repaired."

Never mind.  It's probably just a problem on my side.  Thanks anyway. :)

[quote=Valentin Vornicu]actually is said the same to me when I tried it now! but it worked perfectly yesterday. Try it again now. :)[/quote]

It works now (14 and a half years later)

[quote=siutsz]Given a positive integer $n$, find the smallest positive integer $\lambda$ such that, for any $\theta_i \in (0, \frac{\pi}{2})$ when $i\in \{ 1,2,...,n\}$, if $\prod_{i=1}^{n}{\tan (\theta_i)} \geq 2^{\frac{n}{2}}$, then $\sum_{i=1}^{n}{\cos (\theta_i) } \leq \lambda$.[/quote]

[quote=mathpro12345][quote=Taco12][quote=mathpro12345]\bumpity \bumpity \bump[/quote]

Please don't bump really old threads[/quote]

Try to stop me[/quote]

Reported


AMC
AIME
USA(J)MO
USAMO
geometry
AMC 10
AIME I

I got a 9. I predict that 9 is going to be the lowest score of anyone posting here.

I predict not.
I predict that I will have the second highest score of anyone posting here.
(by the way, does anyone have a link to an official answer key? I just don't want to put my score into the hands of 10 MOPpers.)

#2 was 697 anyway.

Hm 5849206328x did not miss USAMO by .5 this year.

Meh idiocy on #1.
Still, I guessed #13 right so I'm pretty happy about that.
I got 2-13, so 12.

Wait, bobby, are you saying you got like 280 index at least? o.O

You guessed correctly.

I'm not too happy...tried pwning trivial problems with non-trivial stuff. Didn't work out so well. I got a 10 (1-8, 11, 14); stupid mistake on 9 and tried the aformentioned pwnage on 10 and 12.

10 here.  I have no idea how I messed up #5.

ugh 9. Stupidity on number 1, but more annoyed about number 10. I knew how to do it but I got confused by the left-right crap (got MVE), which in retrospect made no sense to have. Tried to do number 12 in the last few minutes but in the rush didn't realize that the hypotenuse was an integer and tried solving a single equation with 2 varables... Horay for blowing my usamo chances.

Hmm unless you're a junior or senior and got less than 120 on the AMC you should be fine...

Ugh epic phail. I got a 6 (1, 2, 3, 4, 6, 11), but I know for sure I should've gotten at least 5 more (5, 7, 8, 9, 10, 13). Blame it on an unreasonable number of stupid mistakes. Needless to say, barring a miracle, I probably didn't make USAMO this year. Ironically, the night before I got a 12 on the 1988 test. Dang.

I just did it for practice and got a 13. These problems relied entirely on computation, or at least almost entirely. I knew exactly what to do on reading them for 1-12, 14. 13 required a bit of playing with it, but ultimately allowed you to make a reasonable assumption and get it right. 10 I screwed something up evidently, and 15 didn't go well for me.

Not the best AIME...I hope I can do as well on the II...

[quote="aznlord1337"]Hehe let's just say I'm a junior and I did not do well on the AMC 12.[/quote]

lol bryan, i don't htink you got over 100 on AMC 12B

ya, i got an 8...made a couple of careless mistakes -_-

[quote="CatalystOfNostalgia"]Dude this is after I told you how trivial the last few problems were. And half the problems themselves. And that #13 was guessable. And come on, there's no way you would have worked on #12 during the real test. I declare a mistrial.[/quote]

Umm...you told me 3 problems, two of which I solved in very little time, one of which I still got wrong. I finished the first 11 is like an hour (1, 9, 10 included) so I would have at least worked on 12 and 13. I concede I might not have worked on 14 very thoroughly.

hey Sam, since AIME I had a lot of counting and geometry, do you think AIME II will have more algebra and/or number theory?

(I am taking AIME II, you're going to too, right?)

[quote="stevenmeow"]I missed USAMO by about -80 points :([/quote]

All right, so the first time I read this, I was like WTH? Because I saw your name pop up on other forums, most notably those where they revealed your scores. Apparently your index was a 280. yay?

It's starting to make sense now. I missed USAMO by about NEGATIVE 80 points. Guess I shoulda figured that out earlier :P

3 :( . I'm not exactly queen of arithmatic- I do things like say 8+6=10... I bet I did something like that to mess up.

[quote="hello.world"]Ironically, the night before I got a 12 on the 1988 test. Dang.[/quote]

THERE's the problem. You don't do math the night before!

I forgot to sign up for the AMC10 so I did not get to take the AIME, which is unfortunate because I answered the questions after and got above qualification. I got a 135, but it should have been a 141  :(  On the question about how many days of the week can a 31 day month start on so that the number of Wednesdays and Mondays in the month are the same I put 2. It is three, Monday Thursday, and Friday, but I forgot Thursday existed... just blanked. Solving the AIME problems by hand in front of my computer I got #1-#6. the rest looked much more daunting though and I don't know what test anxiety might have done.
Good job to all of you who did well enough to make USAMO, especially the 280 index guy

gj to all
i just qualified for AIME for the first time this year, any suggestions for 8th graders?


topology
function
superior algebra
superior algebra solved

assume $\mathbb{N}$ with cofinite topology and $\mathbb{R}$ with standard topology prove that the ring of continuous function from $\mathbb{R}$ to $\mathbb{N}$ is finitly generated.

I don't really get this.. :? Haven't we discussed a problem (and a generalization, of which sam-n was perfectly aware, since he PM-ed me about it, and which can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=partition&t=25908]here[/url] :)) stating that $\mathbb R$ is not partitionable into at most countably many but more than one closed subsets?

In this case, $\mathbb R$ is partitioned, for each continuous function $f$ from $\mathbb R$ to $\mathbb N$, as $\mathbb R=\bigcup_{n\ge 0}f^{-1}(n)$, which means that the only continuous functions are the constant ones, and the conclusion is now obvious.

Is there anything wrong with this? :?

Wow! [url=http://www.ipm.ac.ir/IPM/people/personalinfo.jsp?PeopleCode=IP9900176]Arash Rastegar[/url] was the team leader of Iran some years ago and did his PhD under [url=http://www.genealogy.math.ndsu.nodak.edu/html/id.phtml?id=53168]Andrew Wiles.[/url] You are really lucky to have such a teacher.  :)

yes of course according to the mentioned problem it is obvious  :)


number theory
Diophantine equation
number theory unsolved

Find all pairs $ (a,b,c,d)$ integer such that $ a^2\plus{}5b^2\equal{}2c^2\plus{}2cd\plus{}3d^2$

No solution over the integers exists. As for the proof consider a quadruple $ (a,b,c,d)$ which fits the equation. Clearly, we can assume $ T\equal{}\gcd(a,b,c,d)\equal{}1$ by switching otherwise to the solution $ \left(\frac{a}{T},\frac{b}{T},\frac{c}{T},\frac{d}{T}\right)$. Note, that for $ x,y\in\mathbb{Z}$ the expressions $ x^2$ and $ x^2\plus{}xy\minus{}y^2$ can only have $ 0,\pm 1$ as residues modulo $ 5$. Working modulo $ 5$ in the given diophantine equation, we get $ a^2\equiv 2(c^2\plus{}cd\minus{}d^2)\bmod{5}$, therefore $ a\equal{}5\alpha$ for some $ \alpha\in\mathbb{Z}$. Substitution into the equation and multiplication with $ 2$ yields $ 50\alpha^2\plus{}10b^2\equal{}(2c\plus{}d)^2\plus{}5d^2$, i.e. $ 2c\plus{}d\equal{}5\delta$ for some $ \delta\in\mathbb{Z}$. Substitution and division by $ 5$ yields $ 10\alpha^2\plus{}2b^2\equal{}5\delta^2\plus{}d^2$, i.e. $ d^2\equiv 2b^2\bmod{5}$. This is only possible for $ 5|d$ and $ 5|b$, which implies $ 5|c$ and putting all together we get the contradiction $ 5|T$.

Other wave ( same method ) .
$ \leftrightarrow 2 (a^2+5b^2)=(2c+d)^2+5d^2$
Let $ t=2c+d$
$ 2(a^2+5b^2)=t^2+5d^2$
Consider mod 5 we have
$ 5|a,5|t$ because $ (\frac{2}{5})=-1$
Apply again we have if $ (a,b,t,d)$ is solution then $ (\frac{a}{5},\frac{b}{5},\frac{t}{5},\frac{d}{5})$ is solution .
Imply that $ a=b=t=c=0$

[quote="TTsphn"]Apply again we have if $ (a,b,t,d)$ is solution then $ (\frac {a}{5},\frac {b}{5},\frac {t}{5},\frac {d}{5})$ is solution .
Imply that $ a \equal{} b \equal{} t \equal{} c \equal{} 0$[/quote]

that's because infinite descent????  :?:


Euler
trigonometry
analytic geometry
geometry
circumcircle
conics
hyperbola

maybe it is not
because tripolar of Fermat point is
$\sin (A+\frac{\pi}{3})x+\sin (B+\frac{\pi}{3})y+\sin (C+\frac{\pi}{3})z=0$ but Euler line is
$\sin 2A\sin(B-C)x+\sin 2B\sin(C-A)y+\sin 2C\sin(A-B)z=0$
(in trilinear coordinate)

Tripolar of the 1st Fermat point is not parallel, but perpendicular to the triangle Euler line. Since the orthic axis is perpendicular to the Euler line, being the radical axis of the circumcircle (O) and 9-point circle (N), it is sufficient to show that the tripolar of the 1st Fermat point is parallel to the orthic axis. First, we need the barycentric coordinates of the 1st Fermat point. Let A', B', C' be the remaining vertices of equilateral triangles $\triangle A'BC, \triangle AB'C,\ \triangle ABC'$ on the opposite side of BC, CA, AB than A, B, C, respectively. Let AA', BB', CC' meet BC, CA, AB at K, L, M, respectively. Using the sine theorem for the triangles $\triangle ABA',\ \triangle ACA',$

$\frac{AB}{AA'}= \frac{\sin \widehat{AA'B}}{\sin (B+60^\circ)},\ \ \frac{AC}{AA'}= \frac{\sin \widehat{AA'C}}{\sin (C+60^\circ)}$

$\frac{\sin \widehat{AA'B}}{\sin \widehat{AA'C}}= \frac{c}{b}\cdot \frac{\sin (B+60^\circ)}{\sin (C+60^\circ)}$

Using the sine theorem for the triangles $\triangle KBA',\ \triangle KCA',$

$\frac{KB}{KA'}= \frac{\sin \widehat{KA'B}}{\sin 60^\circ},\ \ \frac{KC}{KA'}= \frac{\sin \widehat{KA'C}}{\sin 60^\circ}$

$\frac{KB}{KC}= \frac{\sin \widehat{KA'B}}{\sin \widehat{KA'C}}\equiv \frac{\sin \widehat{AA'B}}{\sin \widehat{AA'C}}= \frac{c}{b}\cdot \frac{\sin (B+60^\circ)}{\sin (C+60^\circ)}$

and similarly, $\frac{LC}{LA}= \frac{a}{c}\cdot \frac{\sin (C+60^\circ)}{\sin (A+60^\circ)},\ \ \frac{MA}{MB}= \frac{b}{a}\cdot \frac{\sin (A+60^\circ)}{\sin (B+60^\circ)}.$

Let D, E, F be feet of the A-, B-, C-altitudes. Obviouly, $\frac{DB}{DC}= \frac{\tan C}{\tan B},\ \ \frac{EC}{EA}= \frac{\tan A}{\tan C},\ \ \frac{FA}{FB}= \frac{\tan B}{\tan A}.$

Let $X \equiv EF \cap BC,\ Y \equiv DE \cap AB$ and $U \equiv LM \cap BC,\ V \equiv KL \cap AB.$ WLOG, assume $\angle B < \angle C.$ By Ceva/Menelaus theorems,

$\frac{XB-a}{XB}= \frac{EC}{EA}\cdot \frac{FA}{FB}= \frac{DC}{DB}= \frac{\tan B}{\tan C},\ \ XB = \frac{a}{1-\frac{\tan B}{\tan C}}$

${\frac{YB-c}{YB}= \frac{DC}{DB}\cdot \frac{EA}{EC}= \frac{FA}{FB}= \frac{\tan B}{\tan A}},\ \ YB = \frac{c}{1-\frac{\tan B}{\tan A}}$

$\frac{UB-a}{UB}= \frac{LC}{LA}\cdot \frac{MA}{MB}= \frac{KC}{KB}= \frac{b}{c}\cdot \frac{\sin (C+60^\circ)}{\sin (B+60^\circ)},\ \ UB = \frac{a}{1-\frac{b}{c}\cdot \frac{\sin (C+60^\circ)}{\sin (B+60^\circ)}}$

$\frac{VB-c}{VB}= \frac{KC}{KB}\cdot \frac{LA}{LC}= \frac{MA}{MB}= \frac{b}{a}\cdot \frac{\sin (A+60^\circ)}{\sin (B+60^\circ)},\ \ VB = \frac{c}{1-\frac{b}{a}\cdot \frac{\sin (A+60^\circ)}{\sin (B+60^\circ)}}$

$XB = \frac{a \cos B \sin C}{\cos B \sin C-\sin B \cos C}= \frac{a \cos B \sin C}{\sin (C-B)}$

$YB = \frac{c \cos B \sin A}{\cos B \sin A-\sin B \cos A}= \frac{c \cos B \sin A}{\sin (A-B)}$

$UB = \frac{a \sin C \sin (B+60^\circ)}{\frac{1}{2}\sin C \sin B+\frac{\sqrt{3}}{2}\sin C \cos B-\frac{1}{2}\sin B \sin C-\frac{\sqrt{3}}{2}\sin B \cos C}= \frac{a \sin C \sin (B+60^\circ)}{ \frac{\sqrt{3}}{2}\sin (C-B)}$

$VB = \frac{c \sin A \sin (B+60^\circ)}{\frac{1}{2}\sin A \sin B+\frac{\sqrt{3}}{2}\sin A \cos B-\frac{1}{2}\sin B \sin A-\frac{\sqrt{3}}{2}\sin B \cos A}= \frac{c \sin A \sin (B+60^\circ)}{ \frac{\sqrt{3}}{2}\sin (A-B)}$

$\frac{XB}{UB}= \frac{\sqrt{3}}{2}\cdot \frac{\cos B}{\sin (B+60^\circ)}= \frac{YB}{VB}$

This means that the triangles $\triangle BUV \sim \triangle BXY$ are centrally similar with similarity center B, hence $UV \parallel XY \perp ON.$
I have no idea how to do this without trigonometry.

[quote="hucht"]Is it known that the tripolar of the fermat points of a triangle are parallel to the Euler Lines of the triangle??[/quote]

I have mistaken I tried to say that the tripolars of the fermat points are parallel and both are perpendicular to the euler line

Not that I know, how to prove it, but the tripolar of any point on the Kiepert hyperbola (including the orthocenter H, the 1st and 2nd Fermat points $F_{1},\ F_{2}$) is perpendicular to the triangle Euler line (except the for the triangle vertices A, B, C, which do not have tripolars, and the centroid G, which does not have a finite tripolar).


geometry
3D geometry
geometric transformation
rotation
quadratics
Princeton
college

In a cube with side-length 3, a line that connects two diagonally opposite points serves as an axis of rotation. The cube makes at least one full rotation about this axis. Let $ S$ be the set of points in space which are, at some point, occupied by the cube. The volume of $ S$ can be written in the form $ \pi \sqrt {q}$, where $ q$ is a rational number. Find the greatest integer that does not exceed $ q$.

[b]Hint:[/b] [i]Use the volume of a cone to determine the average value of a quadratic.[/i]
[b]Note:[/b] [i]If your solution is computation-heavy at times, feel free to skip trivial steps, and just make sure your steps can be followed[/i]
[b]Evaluation:[/b] [i]What are your general comments on the problem? How computation-heavy was it? How long did it take? How difficult was it?[/i]

The only [i]nice[/i] way I can think of doing it is with calculus.  Should I post it?

Awesome problem by the way.

EDIT Of course it can be done with a bunch a cones.

[b]-- POST 1337 --[/b]

The most you might want to integrate should be a quadratic. But we can just find the average value of a quadratic by splitting it into components and using inspiration from triangles and cones  :)

[b]Motivation for problem:[/b] Spinning a rubix cube out of boredom

:surf: [i]btw congrats on Princeton![/i]

I posted this before, I think Hamster1800 said it was a hyperbola, in which case there is no way but calculus?

[hide="What The Zuton Force is trying to say..."] Is that the formula for the volume of a cone lets you integrate quadratics without knowing anything about calculus a la Archimedes. :) [/hide]

Sorry, could you elaborate a little more on that? I don't see how the formula for the volume of a cone can be used to determine the average value of a quadratic..  :blush:

[hide="Elaboration"] A cone can be thought of as a collection of discs centered at and perpendicular to the $ x$-axis arranged from $ x \equal{} 0$ to $ x \equal{} h$ such that at any coordinate $ x$, the corresponding disc has radius $ \frac {rx}{h}$.  The volume of this cone divided by $ h$ then gives the average value of the area of a disc $ A \equal{} \pi \left( \frac {rx}{h} \right)^2$ from $ 0$ to $ h$.  Appropriate scaling lets us find the average value of $ x^2$ on any interval (by considering a frustrum).  :)

Of course, this is backwards.  In the modern formulation, we derive the volume of a cone through the calculus argument instead of the other way around (Archimedes' non-calculus methods were not rigorous). [/hide]

I think I get it. [hide="With Cones"]The first step is obviously the find the length of the diagonal, which can be accomplished using 3D coords and gives $ d \equal{} 3\sqrt {3}$, which makes the radius and height of the cones (of which there are two) in question $ \frac {3\sqrt {3}}{2} \equal{} k$. Now using the volume for a cone, and multiplying by two to account for the fact that there are two of them we have
$ V \equal{} \frac {2\pi k^3}{3} \equal{} \frac {27\pi \sqrt {3}}{4} \equal{} \pi \sqrt {\frac {2187}{16}}\Rightarrow q \equal{} 136$[/hide] [hide="With Calculus"]Uhh, Just gonna do it.
$ y \equal{} k \minus{} x\Longrightarrow V \equal{} 2\pi \int^k_0 (k \minus{} x)^2\,dx \equal{} \frac {2\pi k^3}{3}$
The answer follows as above.[/hide]
I would say the problem is fairly easy. Awesome though, I love the visual.
Having said that, I better not have gotten it wrong...

-Aidan, to integrate this and find the volume, I used the 'disk method', which uses some $ y$ value as the length of a radius on a circle, then finds that area with $ \pi r^2$. It sums up all these super thin disks, and that's a cone if $ y$ is a linear function of $ x$.

[b]EDIT:[/b] Maybe not so easy, I had to account for the twos, which I forgot (typical). I also added that comment on the cone and switched out some words here and there. AHHHHH all of this is AFTER a proof read too!
  :cursing:

Unfortunately, your answer is incorrect  :(

There are indeed two cones on the extreme axial ends, but there is a region in the middle as well. This is because the 6 vertexes that are not on the axis are not all coplanar. Hamster1800 was correct in saying (in the other thread) that the "curves" that arise in the cross section of the region are hyperbolic, though since we're finding volume, there is no need to be flustered by this -- I assure you, it works out nicely :wink:

For the record, $ V \equal{} \pi \sqrt {243}$. The integral answer is then, of course, $ 243$.

Generally, for a cube of volume $ v$, the volume of the associated region is $ \frac {\pi}{\sqrt {3}} v$. I'll post a solution here within a couple days if nobody else does.

Oh. Wow, that's a [i]really[/i] nice problem now that I actually understand it... I'm eager to see how this is solved.
 :oops:

Thanks guys  :)  And now we have to find the average "radius" of the hyperbola part? I can't think of anything, maybe The Zuton Force should offer more hints  :D
P.S. I was also spinning the Rubiks when i was bored!

We actually don't care about the average radius, only about the average square-radius (which is what makes this problem suitable for Intermediate Topics).  :)

Here is a series of hints regarding the hyperbolic section. Open as many or as few as you choose

[hide="Hint 1"]Coordinatize the cube with vertexes on $ (0,0,0)$, $ (0,0,3)$, $ (0,3,3)$, $ (3,3,3)$, and permutations[/hide][hide="Hint 2"]For the "hyperbolic section," which points on the cube will, when rotated, produce the outer contour?[/hide][hide="Hint 3"]Consider an edge point of the form $ (3,0,x)$ and the diagonal of the form $ (t,t,t)$. Given $ t$, what value of $ x$ produces the shortest distance between the two points? What is the square of this distance?[/hide][hide="Hint 4"]What are the bounds for $ t$ that we are interested in, and what is the mean square distance over this interval?[/hide][hide="Hint 5"]Therefore, what is the volume of the hyperbolic section?[/hide]

This is what it looks like; don't have time to post the solution right now. In fact, I haven't even calculated the volume of this figure. Chemistry now.  :D

:play_ball:
[hide="Solution"]The cubes diagonal length is $ 3\sqrt{3}$, so that means 2 pyramid altitudes + the length of the middle region = 3root3. I found the base of the pyramid to have a circumradius of $ \sqrt{6}\equal{}r$ and its altitude to be $ \sqrt{3}\equal{}h$. The hyperbolic region can be described by the distance of any point on an edge of the cube from the cubes center. So, \[ y\equal{}\sqrt{\frac{9}{2}\plus{}x^2}\]
Problem is, this happens on an interval of $ 3\sqrt{3}\minus{}2\sqrt{3}\equal{}\sqrt{3}$ length, where the maximum value is $ \sqrt{6}$. We'll let the hyperbola be centered at zero, with a range of $ \pm\frac{\sqrt{3}}{2}$. Then we'll use a little proportion to fix our hyperbola to maximize at the min and max of this interval... \[ 6\equal{}\frac{9}{2}\plus{}\frac{3k^2}{4}\Rightarrow k^2\equal{}2\] \[ \Leftrightarrow y\equal{}\sqrt{\frac{9}{2}\plus{}2x^2}\] And \[ V_h\equal{}\displaystyle 2\pi \int^{\frac{\sqrt{3}}{2}}_0 \left(\frac{9}{2}\plus{}2x^2\right)\,dx\equal{}5\pi \sqrt{3}\] \[ V_c\equal{}2\frac{\pi r^2h}{3}\equal{}4\pi \sqrt{3}\] \[ V_T\equal{}V_c\plus{}V_h\equal{}9\pi\sqrt{3}\equal{}\pi\sqrt{243}\] Finally, \[ \boxed{q\equal{}243}\][/hide]Without calculus though? Not sure how to do that. I'd love to see it though.

As for the problem... It wasn't a computation overload, but you had to find one value to find the next. I think the hardest part was that I didn't have a cube to work with and ended up staring at a giant pad of sticky notes... Very hard to do without a perfect cube. Drawing things only helped when I was working in two dimensions, which was about half the time. All in all though, once I realized how to solve the problem things came together quite nicely.

[b]@ igiul:[/b] Nice picture! :)

Now, to bring some closure to this problem, here is the derivation of the volume of the "hyperbolic section" without calculus

[hide="Hyperbolic Section"]
Let $ (t,t,t)$ be some point on the diagonal, and let $ (0,3,x)$ be an edge point. We want to find $ t$ to minimize the distance, given $ x\; (*)$. The square-distance is $ t^2 \plus{} (3 \minus{} t)^2 \plus{} (t \minus{} x)^2 \equal{} 3t^2 \minus{} 2(x \plus{} 3)t \plus{} (x^2 \plus{} 9)$. This is minimized at $ t \equal{} \minus{} \frac { \minus{} 2(x \plus{} 3)}{2 (3)} \equal{} \frac {x \plus{} 3}{3}$. Thus, the "square-radius" for a particular value of $ x$ is given by $ 3\left( \frac {x \plus{} 3}{3} \right) ^2 \minus{} 2(x \plus{} 3) \left( \frac {x \plus{} 3}{3} \right) \plus{} (x^2 \plus{} 9) \equal{} \frac {2}{3} x^2 \minus{} 2x \plus{} 6\; (**)$.

We want the average value of $ \pi r^2$, and thus the average square-radius, as $ x$ ranges from $ 0$ through $ 3$. The $ 6$ is always $ 6$, and the $ \minus{} 2x$ clearly averages to $ \minus{} 3$. It is here that we seek inspiration from cones. We know that the volume of a cone is $ V \equal{} \frac {1}{3} \pi R^2 h$. Suppose we want to "squish" this into a cylinder of height $ h$. The cross section of this will give the average cross section area of the cone. Thus, $ \pi \overline{r^2} \equal{} \frac {1}{3}\pi R^2 \implies \overline{r^2} \equal{} \frac {1}{3} R^2$. Therefore, the average value of $ r^2$ as $ r$ ranges from $ 0$ to $ R$ is $ \frac {1}{3} R^2$. So now, the average value of $ \frac {2}{3} (x^2)$ as $ x$ ranges from $ 0$ to $ 3$ is $ \frac {2}{3} \left( \frac {1}{3} 3^2 \right) \equal{} 2$

Thus, the average square-radius of the hyperbolic section is $ 6 \minus{} 3 \plus{} 2 \equal{} 5$. The "height" is the arclength of $ (t,t,t)$ as $ t$ ranges from $ 1$ to $ 2$, or $ \sqrt {3}$. Thus, the volume of the hyperbolic region is $ 5 \pi \sqrt {3}$
__________________________________________________
[/hide]

$ (*)$ this is an illustration of why it's critical to stay grounded in the motivation. Seeing an equation like this might lead one to minimize with respect to $ x$ given $ t$. This might be out of habit, or because we will ultimately be moving up and down the "height," and it must be linear with respect to that, and the easiest way to force that is to make $ t$ the parameter. Note that if we minimize with respect to $ x$, we find that $ x \equal{} t$. This, however, does not produce the "radius" that we want, and it is important to realize that. By grace, if one takes that route, s/he gets a warning along the way, as the average square-radius becomes $ 6$, which it would be if the section did not curve in at all.

$ (**)$ fortunately, we are after the average square-radius. If one takes the squareroot of this to get the radius, it will result in the hyperbola first suggested by Hamster1800.

[b]Regarding Level of Computation:[/b] So when I originally tried it out, I made an arithmetic error somewhere, and some uncommon fractions started popping up. That was also why I originally had $ q$ as being a rational number -- in retrospect, I would've stated it to be an integer because I would admire anyone with the cleverness and tenacity to construct an accurate approximation solution (accurate within 1% for the hyperbolic section).

Fortunately, the problem, as it stands now, is much less computation. The only real computation in the method in this post is in the simplification of the quadratic after substituting the optimal value of $ t$ in terms of $ x$. Personally, I think that is a relatively low amount of grunt-work compared to many AIME problems.

[b]Regarding the Solution:[/b] Problems like these, which involve what I might call circumventing calculus by logic, are rarely seen these days because they tend to favor those that actually know calculus well. In this case, while it might take one a while to come up with the cone approach, I don't think it too strongly favors those with calculus background. The key decision here is which way to optimize, and it's quite easy to take the wrong path and get confused.

The problem is also a bit of a multi-hit, because it tests visualization ability, ability to effectively optimize distances, and ability to compile a set of optimal distances into a volume, and ability to navigate through this process.

You can just use Pappus' formula for volume of a rotated lamina.

I thought about that... how would you go about nicely finding the centroid of that region? Or the area for that matter  :maybe:

-- [i]In particular, we want to avoid integrating a hyperbola to find the section area[/i]


Let x,y,z be real number in the interval (0:1) such that
                           xyz(1-x)(1-y)(1-z)
     prove that   x^(2) +y^(2) +Z^(2) $\geq$ $\frac{3}{4}$

Sorry
i don understand this problem?
can you post again?

[quote="tranthanhnam"][quote="2006"]Let$x,y,z$ be real number in the interval $(0: 1)$ such that
                          $xyz(1-x)(1-y)(1-z)=\frac{1}{64}$  is that right
     prove that   $x^{2}+y^{2}+z^{2}\geq \frac{3}{4}$[/quote][/quote]$xyz(1-x)(1-y)(1-z)\leq (\frac14)^{3}=\frac{1}{64},\Rightarrow x,y,z=\frac12$so $x^{2}+y^{2}+z^{2}=\frac34$

sorry you
Let x,y,z be real number in the interval (0:1) such that
xyz=(1-x)(1-y)(1-z)
prove that x^(2) +y^(2) +Z^(2)  $\geq$ ${frac\{3}{4}$

[quote="Hawk Tiger"]I don't think we can change the name in this Mathlinks.[/quote]
I remember that [b]Nikmaster[/b] changed his name to [b]The soul of rock[/b].

This Problem come from "Mathematics Babyhood" magazine in Viet Nam :mad: And It persist time send to magazine. I suggest nobody post solution of problem in here. And moderator should lock this topic  :wink:

can you answer my problem ?
Who is she ? She is very becautiful ?

if you understand english nearly as good as you are posting spam here, you should have recognized that no one will answer.
nobody likes you, you know?

Naphthalin

not before you stop spamming senseless posts everywhere...
why d0 u n33d th3 s0lut10n???


[quote="MIMC Director"][size=150][b]Deadline Extension![/b][/size]

The MIMC deadline has been extended to May 16th!
If you already turned in your answers, you may send in a new answer key.  Be sure to send ALL your answers again.  Also, make sure you ask us to throw your old answer key away.[/quote]

yesssssssss!!!!!  (i had been procrastinating terribly about this until like, yesterday, and i have an AP test tomorrow)  thanks for extending the deadline. :thumbup:


Let $ABC$ be a triangle and let $I$ be the incenter. Let $N$, $M$ be the midpoints of the sides $AB$ and $CA$ respectively. The lines $BI$ and $CI$ meet $MN$ at $K$ and $L$ respectively. Prove that $AI+BI+CI>BC+KL$.

[i]Greece[/i]

[quote="PenguinIntegral"]Can someone post a diagram?[/quote]

Yes please... And can anybody explain Erdosh Mordelli's Theorem????

[url=http://mathworld.wolfram.com/Erdos-MordellTheorem.html]Erdos-Mordell Theorem[/url]
Also here is an image of the problem
[img]http://img.photobucket.com/albums/v600/kar4/Junior_Balkan_MO_1997_prob.jpg[/img]

can anyone come up with a solution for this?

thx

the problem was also posted and solved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=134507]here.[/url]

[asy]
import graph; size(8cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.3, xmax = 11.92, ymin = -2.22, ymax = 6.3;  /* image dimensions */
 /* draw figures */
draw((-2.9,4.86)--(-2.1,-0.52));
draw((-2.1,-0.52)--(9.04,-0.58));
draw((9.04,-0.58)--(-2.9,4.86));
draw(circle((-0.36,1.49), 2.02));
draw((-3.49,2.18)--(9.04,-0.58));
draw((0.22,2.16)--(-2.1,-0.52));
draw((-2.5,2.17)--(3.07,2.14));
draw((-3.49,2.18)--(3.07,2.14));
 /* dots and labels */
dot((-2.9,4.86),dotstyle);
label("$A$", (-2.82,4.98), NE * labelscalefactor);
dot((-2.1,-0.52),dotstyle);
label("$B$", (-2.42,-0.6), SW * labelscalefactor);
dot((9.04,-0.58),dotstyle);
label("$C$", (9.12,-0.46), NE * labelscalefactor);
dot((-0.36,1.49),dotstyle);
label("$I$", (-0.28,1.12), NE * labelscalefactor);
dot((-2.5,2.17),dotstyle);
label("$M$", (-2.86,2.3), NE * labelscalefactor);
dot((3.07,2.14),dotstyle);
label("$N$", (3.16,2.26), NE * labelscalefactor);
dot((-3.49,2.18),dotstyle);
label("$L$", (-3.42,2.3), NE * labelscalefactor);
dot((0.22,2.16),dotstyle);
label("$K$", (0.3,2.28), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]

From $MN\parallel CB$ and $IB$ bisecting $\angle ABC$ we have $\angle MKB=\angle MBK$, which implies $MK=BM=\frac{1}{2}AB$.
Similarly we have $LN=NC=\frac{1}{2}AC$. Also, since $M$ and $N$ are mid-points, we have $MN=\frac{1}{2} BC$.
Now by trianlgle inequality, we have $AI+BI > AB$, $BI+CI>BC$, $CI+AI>CA$, which gives
\[AI+BI+CI>\frac{1}{2}(AB+BC+CA) = MK+MN+LN
= (LK-LM)+MN+(LM+MN) = 2MN + LK= BC+KL.\]
Therefore, we have $AI+BI+CI>BC+KL$.

Note: the last step assumes a configuration where $L$ is on the left of $M$ and $K$ is between $M$ and $N$, but the same argument works for other configurations as well.

its easy proble
since the diameter is the largest segment in the circle AI>KL and BI+CI>BC we summarize them: AI+BI+CI>BC+KL  :)

try to prove yourself that the quadrilateral KALI-is inscribed


[quote="Zhero"]I am aware that in homogeneous inequalities of 3 variables a, b, and c, we may assume without loss of generality that abc = 1. When may we assume that a + b + c = 1?[/quote]
You can assume that too. As long as you create a new connection between 3 variables,e.g: $ a + b + c$, $ ab + bc + ac$, $ a^2 + b^2 + c^2$, it's still fine! The key point to explain why you can assume that is because if you can prove that $ f(a,b,c)\geq 0$, then it's obvious that $ f(ka,kb,kc)\geq 0$ as long as $ f(ka,kb,kc) = k^t.f(a,b,c)$(t is a nonnegative integer and $ k\#0$).

I don't particularly understand. We may assume $ abc = 1$ iff the inequality is homogeneous (because if we let $ abc = n$, we could simply substitute $ a' = %Error. "cbrt" is a bad command.
{n} a$, etc., and all the $ %Error. "cbrt" is a bad command.
{n}$'s cancel out.) However, how would you show it is valid to assume that $ a + b + c = 1$? Can you please give an example?

[quote="ghjk"]...that is because if you can prove that $ f(a,b,c)\geq 0$, then >> it's obvious << that $ f(ka,kb,kc)\geq 0$ as long as $ f(ka,kb,kc) = k^t.f(a,b,c)$(t is a nonnegative integer and $ k\#0$).[/quote]

Really??

The phrase "it follows" is sensible there.

Oh, so any homogeneous additional constraint works in any homogeneous (or almost homogeneous, i.e., as was said, $ f(ka, kb, kc) \equal{} k^t f(a, b, c)$) inequality?

Zhero: In the assumption $ abc\equal{}1$, we would substitute $ a'\equal{}a\sqrt[3]{n}$ rather than $ a'\equal{}an$. Do you see why? In $ a\plus{}b\plus{}c\equal{}1$, we would substitute $ a'\equal{}an$.

[quote="Zhero"] because if we let $ abc = n$, we could simply substitute $ a' = %Error. "cbrt" is a bad command.
{n} a$,[/quote]

I don't think that is correct either.

suppose we know that $ f(a,b,c)\ge 0$ such that $ abc=1$ and we have $ f(ta,tb,tc) = t^nf(a,b,c)$

Then if $ xyz=m^3$ we have $ \frac{x}{m}\cdot  \frac{y}{m} \cdot \frac{z}{m}=1$

Therefore we know that;
 $ f(\frac{x}{m}, \frac{y}{m}, \frac{z}{m}) \ge 0$

Which implies;
$ \frac{1}{m^n}\cdot f(x,y,z) \ge 0 \Longrightarrow f(x,y,z) \ge 0$

So now try applying the same thing above with $ t=\frac{1}{a+b+c}$ and you should see how you can let $ a+b+c=1$

[quote="Zhero"]Sorry guys, I actually wrote
[code]\cbrt{n}[/code]
which is what I intended (if you view my code, that's what I wrote.) Too bad there's no such command. :/

And yes, I see why now. So the $ a \plus{} b \plus{} c \equal{} 1$ works in all homogeneous inequalities, correct?[/quote]
Of course, you can. Just check the restriction on homogeneous polynomial as I showed above. If it did satisfy, then feel free to assume that.In some cases, the choice of assumption shows your ingenuity in inequality. :wink:


Show that the right circular cylinder of volumeV which has the least surfacearea is the one whose diameter is equal to its altitude.

Let $ R$ be the radius and $ I$ the altitude.
$ V=\pi R^{2}I, A=2\pi R(R+I)$.
Replacing $  I=\frac{V}{\pi R^{2}}$ in $ A$ we have $  A=f(R)=\2\pi R^{2}+\frac{2V}{R}$.
$  f'(R)=\frac{4\pi R^{3}-2V}{R^{2}}=0\Rightarrow R=\sqrt[3]{\frac{V}{2\pi}}$.
Then $  V=\pi I\sqrt[3]{\frac{V^{2}}{4\pi^{2}}}\Rightarrow I=\sqrt[3]{\frac{4V}{\pi}}=2R$

Most of these classic single-variable calculus problems look even nicer once you've taken multi-variable calculus and learned the technique of Lagrange multipliers.

Statement: minimize $ A=f(R,I)=2\pi R(R+I)$ subject to the constraint $ g(R,I)=\pi R^{2}I=V.$

We set up the equations $ \frac{\partial f}{\partial R}=\lambda\,\frac{\partial g}{\partial R}$ and $ \frac{\partial f}{\partial I}=\lambda\,\frac{\partial g}{\partial I}\,:$

$ 2\pi(2R+I)=\lambda2\pi RI$ or $ 2R+I=\lambda RI.$

$ 2\pi R=\lambda \pi R^{2}$ or $ 2R=\lambda R^{2}.$

From the second equation, since $ R\ne0,$ we have $ \lambda=\frac2R.$ Put that in the first equation to get $ 2R+I=2I$ which simplifies to $ I=2R$ or $ \text{height}=\text{diameter}.$

The details of what that height and diameter are then follow from the constraint equation.


This is my first "season" doing AoPS classes, as it's my first full season on AoPS that I was in town. How long after the class ends are the transcripts kept on the my classes link?

i hope the innt transcripts are in your desktop's Recycling Bin and you haven't deleted them permanently, I forgot to PDF-ize them and lost half the transcripts.

Retrieve if possible.   :(


limit
logarithms
function
real analysis
real analysis solved

I got a different answer...

$\lim_{x\to 0} {\frac{\sqrt[x]{a^{x}+b^{x}+c^{x}}}{3}} = \frac{1}{3} \lim_{x\to 0} \sqrt[x]{a^{x}+b^{x}+c^{x}} = \frac{1}{3} \lim_{x\to 0} e ^{\log (a^{x}+b^{x}+c^{x})/x}$, and since in $\lim_{x\to 0} \log (a^{x}+b^{x}+c^{x})/x$, the numerator goes to $\log 3$ while the denominator goes to zero, the original limit goes to infinity.

Is there something wrong with my solution?

By the way, in liyi's solution, I do not understand why $\lim_{x\to 0} ((f(x)-1)+1)^\frac{1}{f(x)-1} = e$ necessarily implies $\lim_{x\to 0} ((f(x)-1)+1)^{1/x} = e^\frac{f(x)-1}{x}$. Since $g(x) = \frac{f(x)-1}{x}$ is a function of $x$, we can't simply say that $\left(\lim_{x\to 0} h(x) \right) ^ {g(x)} = \lim_{x\to 0} h(x)^{g(x)}$ for a given $h(x)$, right?

He did a typo. The limit should be $\lim_{x\to 0} \sqrt[x]{\frac{a^x+b^x+c^x}{3}}$.

$\lim_{x\to 0} f(x)^{g(x)} = \left(\lim_{x\to 0} f(x)\right)^{\lim\limits_{x\to 0} g(x)}$.

Ah, I see, no wonder everything seemed so confusing.


poate sa posteze cineva subiectele de la barajele de juniori? sau sa dea un link cu ele, pls.

multumesc anticipat.

multumesc, despre barajul 2 nu s-a afisat nimic? macar geometria poate sa o posteze cineva? (in caz ca s-a mai dat vreo geometrie)

barajul 2

[b]Subiectul 1[/b]
Sa se determine numerele naturale $n\geq 4$ cu proprietatea ca $\left[\sqrt{n}\right]+1$ divide $(n-1)$ si $\left[\sqrt{n}\right]-1$ divide $(n+1)$.

[b]Subiectul 2[/b]
Fie $ABCD$ un patrulater convex. Notam $M, N$ punctele de tangenta ale cercului inscris in triunghiul $ABD$ cu laturile $AB$, $AD$ respectiv si $P, Q$ punctele de tangenta ale cercului inscris in triunghiul $CBD$ cu laturile $CD$, $CB$ respectiv. Daca cercurile inscrise in triunghiurile $ABD$ si $CBD$ sunt tangente, sa se arate ca:
a) patrulaterul $ABCD$ este circumscriptibil;
b) patrulateru $MNPQ$ este inscriptibil;
c) cercurile inscrise in triunghiurile  $ABC$ si $ADC$ sunt tangente.

[b]Subiectul 3[/b]
Fie $ABC$ un triunghi isoscel in $A$. Pentru orice punct $P$ interior triunghiului consideram cercul de centru $A$ si raza $AP$ si notam $M$ si $N$ intersectiile laturilor $AB$, respectiv $AC$ cu cercul. Sa se determine pozitia punctului $P$ pentru care $MN+BP+CP$ este minima.

[b]Subiectul 2 de la baraj era:[/b] Un trapez ABCD are bazele AB şi CD, iar cercurile cu diametrele AD şi BC se intersectează în M şi N. Să se arate că punctual de intersecţie al diagonalelor AC şi BC aparţine dreptei MN.
Propun spre rezolvare urmatoarea generalizare a problemei: [b]Intr-un trapez ABCD avand bazele AB şi CD, axa radicala a cercurilor care au drept diametre laturile opuse [AD] si [BC] trece prin punctul de intersectie O al diagonalelor trapezului.[/b]


if $x,y,z>0$ then prove this inequality:
$x^4+y^4+z^4+xyz(x+y+z) \geq  2(x^2y^2+y^2z^2+z^2x^2)$

Using Schur inequality and then AM-GM we have

$x^4+y^4+z^4+xyz(x+y+z) \geq ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)\geq 2(x^2y^2+y^2z^2+z^2x^2)$


$ A$ can do a piece of work in $ 9$ days. $ B$ is $ 50\%$ more efficient than $ A$. The number of days it takes $ B$ to do the same piece of work is:

$ \textbf{(A)}\ 13\frac {1}{2} \qquad\textbf{(B)}\ 4\frac {1}{2} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \text{none of these answers}$

[hide="Click for solution"]
In one day A can do $ \frac{1}{9}$ of the job.  Since B is $ 50 \%$ more efficient than A, B can do $ \frac{3}{2} \cdot \frac{1}{9}\equal{}\frac{1}{6}$ of the job in one day.  Therefore, he needs $ 6$ days for the complete job, or $ \boxed{\textbf{(C)}}$.

[/hide]


Let $a,$ $b$ and $c$ are positive numbers. Prove that
$\sqrt[3]{\frac{a}{2b+25c}}+\sqrt[3]{\frac{b}{2c+25a}}+\sqrt[3]{\frac{c}{2a+25b}}\geq1.$

[quote="Yulia"]
But for $u+v+w=1$
$(25^{3}+8)u^{3}v^{3}w^{3}+50(u^{3}v^{3}+v^{3}w^{3}+u^{3}w^{3})\leq 1$
I proved this with homogenization and then by Muirhead, but that's not nice.[/quote]
But true and easy already!
My congratulations, Yulia!:lol:


Let X={1,2,3,...,1000}. Solve that for each subset A of X, |A|=100. We can find a,b,c that make a progressing arthimatic ( common diffrence >0).

Well done Yury, you proved it false. I checked your example with a computer program, and there are no a,b,c which satisfy the condition.


integration
calculus
geometry
abstract algebra

how would one do physics with a spring of known mass?
example:
a block of mass $M$ rests vertically on a spring of constant $k$ and mass $m$; what is the change in length $x$ of the spring

You can do it by analizing small pieces of the spring and considering the forces acting on each, but maybe it's easier if you use conservation of energy. The change in gravitational potential energy of the spring and the mass after attaching the mass to the unstretched spring is $mgx/2+Mgx$, and this change is used to increase the elastic potential energy $kx^{2}/2$. Equating these gives us \[x = \frac{(m+2M)g}{k}.\]

[quote="Djole"] but maybe it's easier if you use conservation of energy. The change in gravitational potential energy of the spring and the mass after attaching the mass to the unstretched spring is $mgx/2+Mgx$, and this change is used to increase the elastic potential energy [b]$kx^{2}/2$. [/b]Equating these gives us
\[x = \frac{(m+2M)g}{k}. \]
[/quote]

How do you derive [b]$\delta$$U$ $=$ $\frac{1}{2}kx^{2}$[/b] for a

 [i]SPRING[/i] with [i]MASS[/i]

Actually, I considered the massed spring as a massless spring with the mass $M$ evenly smeared across it; the mass doesn't contribute to the elastic potential energy, but it does have its own gravitational potential energy. The idea seems sane enough, and it sure did beat the first method I described. But it would be good to check whether it is ok... I'll try to come up with a stricter proof.

@[b]pardesi[/b]: Just a hint: please don't overuse bold or uppercase letters in your posts. Too much bold letters make you seem irritated and uppercase usually denotes yelling, and anyway neither contribute to the meaning of your post. If you want to emphasize something, try using [i]italics[/i] instead, they're not so menacing. Nothing personal, of course :)

I Changed those to ITALICS .surely i am not at all  irritated  :lol:  :D
i asked you the proof since the fundamental basis of deriving
$\Delta$$U=\frac{1}{2}kx^{2}$
is that the spring assumed to be massless.that gives it's kinetic energy to be nearly zero when pulled extremely slow.you know that....

I don't know the derivation you suggest... The one I know starts from the definition of the elastic force,
\[F =-kx,\]
and integrates it to obtain the associated potential energy:
\[U =-\int F\text dx = \frac{1}{2}kx^{2}+U_{0}.\]
This integration comes from
\[F =-\frac{\text dU}{\text dx},\]
which holds for conservative forces. Nowhere have we regarded the nature of the spring, just the fact that it exerts the force $-kx$.

now by the $hooke's$ $law$

we get $F$=$-kx$
this is the force exerted by the spring on the source that has extended or rather extending the spring at that moment by a length $x$.
now change in  potential energy is can be only derived for systems where conservative forces act .
\now in the above derivation though u have not mentioned the system the conservative force is the internal electromagnetic force within the particles of the spring(the system).
simply putting

         $F$ $=$ $-\frac{dU}{dx}$
without specifying the system and force the formula  doesn't work out
now the actual proof is that when a source moves the spring slowly without appreciable changing it's speed then it's net change in kinetic energy is negligible because it's mass and speed both are negligible

now by $work-energy$ $theorem$ we have

 work done by the internal forces of the spring sytem plus the work done by external forces (source  $F$=$kx$) is equal to the change in kinetic energy of the  spring system

 since $\Delta$ $K$ is nearly equal to $0$ we get

we get work done by ext. forces is negative of work done by int. forces or for that matter potential inergy change.
hence we get the required expression of
$\Delta$ $U$ $=$ $\frac{1}{2}kx^{2}$

this is the proof.
clearly this doesn't work out if the spring has appreciable mass

There is one point that I don't understand in your post:

[quote="pardesi"]now in the above derivation though u have not mentioned the system the conservative force is the internal electromagnetic force within the particles of the spring(the system).
simply putting

F = -\frac{dU}{dx}
without specifying the system and force the formula doesn't work out [/quote]

What did you mean by this?

Regardless of that, now I understand the mistake I made. I considered that the system was released from the point where the spring was unstretched, and I failed to note that it will behave like a harmonic oscillator and won't stop at the equilibrium point. It will stop a distance $2x$ below the initial point, so applying conservation of energy (like I did in the previous post, but for $2x$ instead of $x$) we get
\[2Mgx+mgx = \frac{k}{2}(2x)^{2}= 2kx^{2}\Rightarrow x = \frac{(2M+m)g}{2k}. \]
My approach was correct, the problem was that at a point I considered in the first post the kinetic energy wasn't zero...

But let's verify this: let $h$ be the distance between a piece of the spring and the hanging point at the moment we let the spring and the weight hang. The elastic force in the spring is
\[F(h) = Mg+\frac{H-h}{H}mg = (M+m)g-\frac{mg}{H}h, \]
where $H$ is the initial length of the spring. Applying Hooke's law we can find the elongation of every piece of the string:
\[\Delta h = \frac{F(h)\text dh}{SE}, \]
where $S$ is the area of the cross-section of the spring and $E$ its module of elasticity. Summing all the elongations (ie. integrating from $0$ to $H$) we get
\[x = \int\Delta h = \int_{0}^{H}\frac{\left((M+m)g-\frac{mg}{H}h\right)\text dh}{SE}= \frac{(M+m)gH}{SE}-\frac{mgH}{2SE}\]

\[x =\frac{(M+m)g}{k}-\frac{mg}{2k}= \frac{(2M+m)g}{2k}. \]

once u answer these questions then u will clearly understand what's the meaning of what i wrote

what is $F$ , and what is the system for which ur finding out the
 $\Delta$ $U$
without these ur proedure is meaningless hence ur proof

[quote="Djole"]Huh? $F$ is the elastic force (which, as you said, originates from the EM interaction of the particles within the spring, although I don't see why this is important for our discussion). The 'system' whose elastic potential energy I'm looking for is the spring with mass $m$. And I still don't understand why do you claim that my approach isn't correct... Do you agree with the solution which doesn't use conservation of energy, at least?[/quote]

well djole in my proof u become sure that the only external force that acts  on the spring is that due to the sorce now since the spring also has some particle they  also exert some em force how can you prove that ur
derived expression for change in potential energy considers all these internal forces though it gives the right result .

as far as u'r non coe proof is based i couldn't get what ur main aim was please rephrase the proof

Surely your not saying that if the spring is massless (more precisely, its mass is negligibly small) it isn't composed of atoms/molecules? If not, than I don't see why would the two cases differ.

As for my other proof, I'm considering forces acting on elementary pieces of the spring (which is assumed to be homogenous) and calculating the elongation of each piece. The forces acting on each piece are the gravitational force due to the weight below the piece and the 'tension', or in this case the elastic force, from which we can calculate the elongations. Summing all of the elementary elongations we get the total elongation of the spring.

what i have been trying to say is

$\Delta$ $U$ $=$ $\frac{1}{2}kx^{2}$

corresponds to the change in elastic potential energy of the spring system and not the (spring+block) system hence corresponds only to the internal forces in the spring not in the (spring+block).

But the block has nothing to do with elasticity... Do you say we should take into account some kind of interaction between the block and the spring? :huh:

let me put it straight
what i have been trying to say if u analyse this fact

now say that the spring were massless this it's just routine to prove that the extension on the spring(max.)
is given by the expression $x$ $=$  $\frac{2mg}{k}$
this is proved by assuming the centre of mass of the block as a system.
assuming it were a rigid system.how do the forces on this system diifer if the spring were having some mass ?

now let's solve the problem this way
let the system be the block(which i assume to be a rigid body) so when the speed of the centre of mass is $0$ the speed of the entire block is $0$ and vice versa
we know that by work energy theorem for arigid body work done by the total external force is equal to the change in kinetic energy of the centre of mass of the system(it's easy to prove this)
 hence the only external forces that matter  acting on the body are the spring forces and that due to earth(gravity)
so work done by these forces is equal to the total change in kinetic energy of the cenre of mass of the block which is $0$ since we have to find the max. extension in the spring

 so $\frac{1}{2}kx^{2}$ $-$ $Mgx$ $=$ $0$
that gives  $x$ $=$ $\frac{2Mg}{k}$

[quote="Djole"]Hm... Let's consider this setup: you take an elastic rod of mass $m$ and elasticity $k$ (our massed spring), you attach a weight of mass $M$ to its end and hang it by the other end. Let's say that at the initial moment the rod isn't elongated. You're saying that once equilibrium is reached, the rod will be elongated by $x$, and this $x$ you got from your previous post. Why isn't the block oscilating rather than resting?[/quote]

who said that the block is resting in a position .i said it is the point of max. elongation of spring(rod) that is the point. where the speed of block is $0$
and is rather one of the extremes of the harmonic motion

Ok -- you say that $x$ is the distance between the ends of the rod when it isn't stretched at all and when it's maximally stretched. But in either position the net force acting on the block is not zero. However, the problem clearly requires the elongation of the rod when the block [i]rests[/i], ie. when the net force is zero. The position required is halfway between the points of minimal and maximal elongations. So the real answer should be $x/2$, not $x$.

But there is one more thing; you can't say that $kx^{2}/2$ is the work done by the elastic forces in the rod on displacing the block by $x$. Part of this work goes into moving the center of mass of the spring. When you incorporate this work in your conservation of energy, and apply the abovementioned remark, you get what I've obtained earlier.

Finally, this debate has been fairly interesting, but I think it is time to bring it to an end. I've given two solutions to the problem. In limiting cases, such as $m \ll M$, we get well-known results. While there still is a chance that I've made a mistake, I find that rather unlikely. Please study the solutions [i]carefully[/i] and ask if you don't understand something.

[quote="Djole"]

But there is one more thing; you can't say that $kx^{2}/2$ is the work done by the elastic forces in the rod on displacing the block by $x$. Part of this work goes into moving the center of mass of the spring. When you incorporate this work in your conservation of energy, and apply the abovementioned remark, you get what I've obtained earlier.

[/quote]

why does work done on the block [i]go somewhere[/i] .
well anyways the discussion was pretty interestig .yes $\frac{x}{2}$ will be the answer(my $x$) .but no matter that my process is right .but anyways we had  a greatdiscussion this is the beauty of physics ....... :)  :D


Four congruent circular discs each with a $ 1$-meter diameter are held tightly together by a thin band, as shown.  How long is the band, in meters? Express your answer to the nearest whole number.

[asy]size(3cm,3cm);

pair A,B,C,D,E,F,G,H,a,b,c,d;

A=(0,3);
B=(-1,2);
C=(-1,0);
D=(0,-1);

a=(-0.7,2.7);

draw((A..a..B),black+2bp);

draw((B--C),black+2bp);

b=(-0.7,-0.7);

draw((C..b..D),black+2bp);

E=(2,-1);
F=(3,0);
G=(3,2);
H=(2,3);

draw((D--E),black+2bp);

c=(2.7,-0.7);

draw((E..c..F),black+2bp);

draw((F--G),black+2bp);

d=(2.7,2.7);

draw((G..d..H),black+2bp);

draw((H--A),black+2bp);

draw(Circle((2,0), 1));
draw(Circle((0,2), 1));
draw(Circle((2,2), 1));
draw(Circle((0,0), 1));[/asy]

The straight portion is 2*4=8 meters, and the curved portion adds up to the circumference of one circle, or 2 pi meters.

Ans: 8+2pi ~ = ~ 14

[quote="GameBot"]Four congruent circular discs each with a $ 1$-meter diameter are held tightly together by a thin band, as shown.  How long is the band, in meters? Express your answer to the nearest whole number.
[/quote]

Because the diameter is 1 cm, the length of the straight portion is 1*4 or 4 cm long.

The area of the curved section you can imagine as the circumference of a circle with diameter 1. Thus it's length is pi units.

$ 4\plus{}\pi$ is approximately $ \boxed{7}$.


[hide]
suppose $d_1|2^{999}5^{999}$ , $d_2|2^{998}5^{998}$ . We want to find the number of $d_1$ such that $d_1$ deosnt divide $2^{998}5^{998}$ . We see that when $d_1$ has factor of $2^{999}$ then there are $1000$ ways from $2^{999}\times 5^k$ for $0\leq k\leq 999$ . Simialrly if $d_1$ has the factor of $5^{999}$ then there are 999 ways since $2^k\times 5^{999}$ where $0\leq k\leq 998$ .

So total way is $1999$  (coincidently 1999 is also the year this problem come out  :P )
[/hide]

The answer will finally come out to be 1999.

$10^{999} = 5^{999} * 2^{999} $  and $100^{998} = 5^{998} *2 ^{998} $
Now,
any divisor having $5^{999}$ or $2^{999}$ will suffice.
So $5^{999} . 2^n $ where $n\leq {999}$ which gives us $ 1000 $ divisors
Similarly $2^{999} . 5^n $ where $n\leq {999}$ which gives us $ 1000 $ divisors

Now, the divisor $10^9$ has been included twice so, our answer is $2000-1=\framebox[0.4in]{1999}$

[quote=Rushil]Find the number of positive integers which divide $10^{999}$ but not $10^{998}$.[/quote]
10^999 =5^999*2^999
hence no. of factors are (999+1)(999+1)=1000^2
similiarly, no. of factors of 10^998 is (998+1)(998+1)=999^2
so the no. of positive integers which divide 10^999 but not 10^998 is 1000^2-999^2=(1000+999)=1999 which is truely the correct answer.

[hide=Solution]The amount of factors of $10^{999}$ is $1000^2$ and the amount of factors of $10^{998}$ is $999^2$. Therefore the answer is $1000^2-999^2=\boxed{1999}$.[/hide]


Find the sum of all the real numbers x such that

$ (2^x\minus{}4)^3 \plus{} (4^x\minus{}2)^3 \equal{} (2^x\plus{}4^x\minus{}6)^3$

Let $ X \equal{} 2^x \minus{} 4$ and $ Y \equal{} 4^x \minus{} 2$.

Then $ X^3 \plus{} Y^3 \equal{} (X \plus{} Y)^3$.

Expanding and simplifying gives $ XY(X \plus{} Y) \equal{} 0$.

Hence $ X \equal{} 0$ or $ Y \equal{} 0$ or $ X \equal{} \minus{} Y$, i.e. $ x \equal{} 2,\frac {1}{2}\ \mbox{or}\ 1$.

So the sum of the roots is $ \frac {7}{2}$.


do these formula's have a concrete mathematical proof or are they just good physical approximations

$E_{perpendicular,above}-E_{perpendicular,below}=\frac{\sigma}{\epsilon_{0}}$
$E_{parallel,above}=E_{parallel,below}$
$V_{outside}=V_{inside}$ for any surface charge where symbols have their usual meanings

The first formula you give is identical to the intensity of the electric field between the plates of a capacitor, and that can be demonstrated using the Gauss Law.

yes u can prove that for certain case but just take any arbit surface charge whose $\sigma$ is just know n which may be a varibale then can u have a mathematcial concrete proof to that

[quote="golduck"]As carcul said, the first formula can be obtained from Gauss law by considering a very small cyllinder containing the surface charge.[/quote]
yes that's ok but that will be just an approximation it doesn't have a neat mathematical background as i had asked

Ahm... What kind of neat mathematical background are looking for, anyway?

i saw a proof that took a rectangular pillbox around a "very small" area then it argue nearby a surface charge the electric field is somewhat constant(how?) and then applied the gauss law.
well what i wanted to know was that is there a pure logical,mathematical approach to it or is this just an approximation..because many a times many books leave out proofs that require some tough  mathematics ...so i tought may be this one was one of that kind

[hide="the guy who..."]if u didn't understand my question then you could have posted or even if u have posted then should have cleared ur doubt :mad: [/hide]

Well, that's mostly the whole proof; if you want to be more rigorous, replace 'very small' with 'infinitesimal' and that's about that. That's the approach (more or less) used to prove both Gauss' and Stokes' theorem in electrostatics.

I thought for a moment that you're asking for the reason why Gauss' law holds, which would lead to the question why Maxwell's equations hold, and that's a bit tricky (i.e. impossible) to answer, because they are usually taken as postulates...

And I really don't understand your final sentence in the last post, but I can just hope it wasn't directed at me  :maybe:

well it surely can't be done using infitesmals :maybe:
also i thought there would exist a rigorous proof and that the book had avoided it :(
[quote="Djole"]I thought for a moment that you're asking for the reason why Gauss' law holds, which would lead to the question why Maxwell's equations hold, and that's a bit tricky (i.e. impossible) to answer, because they are usually taken as postulates...
[/quote] :noo:

[quote="Djole"]And I really don't understand your final sentence in the last post, but I can just hope it wasn't directed at me[/quote]
it was for the person who rated me... :mad:

[quote="pardesi"]well it surely can't be done using infitesmals :maybe:[/quote]
Why not :?: Using infitesmals doesn't mean that we are approximating anything...

yes i know that ...but can you give a proof of that using infitesmals


geometry
circumcircle
search
geometry proposed

Given a triangle$ ABC$ and its circumcircle$ O$ and incircle$ I$ let $ D$ be a arbitrary point on the circle$ O$ and $ DE,DF$ touch the incircle while $ E,F$ are on the line $ BC$
prove the circumcircle of $ \triangle DEF$ passes through a point which doesn't depend on $ D$

[quote="Mashimaru"]The fixed point is hard to find but easier to prove.

[hide="Hint about the fixed point"]The fixed point is the point of tangency between circumcircle $ (O)$ of $ \triangle ABC$ with the mixitilinear incircle of angle $ A$ of itself.[/hide][/quote]
You're absolutely right :lol:
But, is it easy to prove?I really want to see your nice solution.

To pohoatza:I knew a solution without using inversion,but it is rather long(I used three sheet of paper to write it down for its complicated angle chase)

Dear Mathlinkers,
a synthetic proof (with a contribution of Yetti) of this result due to Cosmin can be found found on
http://perso.orange.fr/jl.ayme   vol. 4  An new mixtilinear incircle adventure (figure p. 23)
Sincerely
Jean-Louis

dear jayme
could you translate your papers into english?
thank you!

[quote="pohoatza"]See here: http://www.mathlinks.ro/viewtopic.php?search_id=312652555&t=186117.

Do you have an easier proof (than yetti's), Mashimaru?[/quote]

I'm sorry, now I realized that looking a proof that $ (DEF)$ passes through the fixed point I pointed out was much harder than find it out :P . I could not even find a proof for this. The solution of [b]yetti[/b] was wonderful :oops:

[quote="Zhang Fangyu"]Very nice problem!
any synthetic solution?
 dear hxy2009
where does your problem come from? could you give a link?[/quote]
Sorry,I just can not find a link for it,I read this problem from the note book of my my friend during TST :blush:

Dear jayme ,could you please help me with my two problems? :)


Fill each box with a different prime number in such a way that no two adjacent boxes hold numbers that differ by $ 2$ and so that the sum of all nine primes is as small as possible. What is the sum of the nine primes?
[asy]draw((0,0)--(0,3)--(3,3)--(3,0)--cycle);
draw((0,1)--(3,1));
draw((0,2)--(3,2));
draw((1,0)--(1,3));
draw((2,0)--(2,3));[/asy]

How did you get from middle number as 2 to

3 7 11
19 2 17
5 23 13

Since 2 is the only even prime number, no other prime will differ by two with 2 itself. To maximize the impact 2 has on the entire grid, it should be placed in the middle. Then everything else is self-explanatory; just put the primes in order around the clock and put the rest in the corners. This will result in the nine smallest primes being in the nine boxes. If you memmed this as I have, you will know that the sum of the first 10 primes is $129$, so subtracting 29, which is the 10th prime, will result in the answer, $\boxed{100}$.

LOL! I just guessed 100 and got it, but that's a good solution. Thanks for the explanation!

Please don't bump like 7-year-old threads jus to say thanks...

[hide]You could use the first 9 primes in this order, from center to top center and then clockwise from there: $2, 23, 11, 3, 17, 7, 13, 5, 19$, which adds up to $100$. (I know this has already been solved, just wanted to try solving it)[/hide]


Suppose x is a real number such that x + 1/x is an integer.
Is it necessarily true that x2004 + 1/x2004 is an integer?

[i]Variation on a well-known theme[/i]

[quote="tetrahedr0n"]x^n + x^-n can be expressed as an integer with polynomial coefficients in x + 1/x, so yes.[/quote]
more accurate (remember we are in the beginner's corner :) )

We have x2 + x-2 = (x+1/x)2 - 2.

Also for n >1 we have
xn+1+x-(n+1)= (xn + x-n )(x+1/x) - (xn-1 + x-(n-1)) , so if xn + x-n and xn-1 + x-(n-1) are integers, then so is xn+1+x-(n+1) and by induction all are integers (for n=1,2 we have already proved that).

thats a very useful identity you applied. generalize it to:

a^n +b^n = (a^(n-1)+b^(n-1))(a+b) - (a^(n-2)+b^(n-2))ab.

and a problem:

The real numbers a, b, x, y satisfy ax + by = 3, ax^2 + by^2 = 7, ax^3 + by^3 = 16, ax^4 + by^4 = 42. Find ax^5 + by^5.

hmmm..a very classical one.

Just use the identity you posted and we get
7(x+y) - 3xy = 16
16(x+y) - 7xy = 42.

solve out x+y=-14 and xy=-38 and apply the identity again:
ax^5 + by^5 = 42(x+y) - 16xy = 20


Each of the circles has a radius of 8 units. RP is a common tangent to both circles. $ \overline{QA} \equal{} \frac16\overline{QP}$. How many units are in the perimeter of $ \Delta POQ$?.

[asy]pair P,O,Q,R,A; path X,Y;
R = (0,0); O = (1,0); Q = (-1,0); P = Q + 2 * dir(60); A = Q + dir(60);
draw(R--P--O--Q--P); X = Circle(Q,1); Y = Circle(O,1); draw(X); draw(Y);
label("A",A,dir(-70)); label("R",R,SE); label("Q",Q,W); label("O",O,E); label("P",P,N);[/asy]


calculus
integration
logarithms
derivative
algebra
functional equation
calculus computations

Suppose that $f$ is continuous on the positive real line, satisfying

$\int_{1}^{xy}f(t) \, dt = y \int_{1}^{x}f(t) \, dt+x \int_{1}^{y}f(t) \, dt$
for all $x,y > 0$, $f(1) = 3$.  Find an explicit formula for $f(x), x > 0$.

The answer is that $f(x)=3(\ln x+1).$

Here's how to find it. Start with the functional equation and take the partial derivative with respect to $y.$

$xf(xy)=\int_{1}^{x}f(t)\,dt+xf(y)$

Now let $y=1$ and use the initial condition $f(1)=3.$

$xf(x)=\int_{1}^{x}f(t)\,dt+3x$

Now note that the right hand side of this equation is differentiable, so the left hand side must be differentiable and in particular $f$ is differentiable.

$xf'(x)+f(x)=f(x)+3$

$xf'(x)=3$

$f'(x)=\frac{3}{x}$

Integrate this and use the initial condition once more to get $f(x)=3\ln x+3.$

My solution is exactly same as Kent Merryfield's one.
[quote="Kent Merryfield"]Now let $y=3$ and use the initial condition $f(1)=3$.[/quote]
Please edit this.

sorrt, but I just can't see how $ xf(xy)=\int_{1}^{x}f(t)\,dt+xf(y)$

Can someone make it a little clearer?

Let $ \int f(t)\ dt=F(t)+C$, we have $ \int_{1}^{xy}f(t)\ dt=\left[F(t)\right]_{1}^{xy}=F(xy)-F(1)$, differentiate  with respect to $ y$,

$ \frac{d}{dy}\int_{1}^{xy}f(t)\ dt=f(xy)\frac{d}{dy}(xy)=xf(xy)$.


let $f : \mathbb{R}\to \mathbb{R}$ , $f$ is bounded , and let $a,b$ be two reels such that
$a>1$ . find all continus functions wich satisfy the two conditions:
1-$f(ax+b)=f(x)$ for all reel $x$.
2-$f(\frac{b}{1-a})=1998$.
good luck. :D

Let $g(x)=f(x+x_{0}),x_{0}=\frac{b}{1-a}$, then $g(ay)=g(y),g(0)=1998,(y=x-x_{0}).$
Therefore $g(a^{n}y)=g(y) \ \forall n\in Z$. Because f and g continiosly it give $f,g=const=1998$.


The lengths of the bases of a trapezium are a and c, respectively. Find the length of the segment parallel to the bases which divides the trapezium into two equal areas.

[quote="10000th User"]You follow British system? Because Americans call it "trapezoid"
[img]http://mathworld.wolfram.com/t2img369.gif[/img]
Call the line dividing trapezium equal areas, $b$
Put $h=h_1+h_2$, $h_1$ is top trapezium height, $h_2$ is bottom trapezium height
Find same areas and put them equal: $\frac{1}{2}h_1(a+b)=\frac{1}{2}h_2(b+c)$
Solve for ratio $\frac{h_2}{h_1}=\frac{a+b}{b+c}$
But top trapezium is half area of whole trapezium, find areas again and put them equal: $\frac{1}{2}h_1(a+b)=\frac{1}{2}(\frac{1}{2}h_2(a+c))$
$2(a+b)=(1+\frac{h_2}{h_1})(a+c)=(1+\frac{a+b}{b+c})(a+c)$
Multiply to get no denominator: $2(a+b)(b+c)=(a+2b+c)(a+c)$
$2ab+2b^2+2ac+2bc=a^2+c^2+2ac+2ab+2bc$
Cancel and solve for $b$: $b^2=\frac{a^2+c^2}{2}$[/quote]

I think you got the wrong figure? and by the way how dyou get the picture?
the bases will be opposite to each other wont they?

[quote="riddler"]I think you got the wrong figure? and by the way how dyou get the picture?
the bases will be opposite to each other wont they?[/quote]
Picture is to show what "trapezium" is. American use "trapezoid", very confusing you know
My solution is not in picture, you make your own picture  :P

yup same as you 10000th let $m$ be dividing line so $\displaystyle m=\sqrt{\frac{c^2+a^2}{2}}$

[quote="riddler"]yup same as you 10000th let $m$ be dividing line so $m=(c^2+a^2)^(1/2)$ sorry not good at latex... :blush:[/quote]

You mean, $m=\sqrt{\frac{c^2+a^2}{2}}$?
I got this answer, too :)


inequalities
integration
trigonometry
ratio
real analysis
real analysis unsolved

For $ c>0$,

Prove that \[ \bigg|\int_{0}^{c}\frac{\sin x}{x}dx\minus{}\frac{\pi}{2}\bigg|<\frac{1}{c}\]

use the fact that $ \int_{0}^{n}\frac{sinx}{x}dx\rightarrow \frac{\pi}{2}$ when $ n\rightarrow \infty$

This is a pretty tight inequality.

For instance, if $ c\equal{}10\pi,$ I get that the ratio of $ \left|\int_c^{\infty}\frac{\sin x}{x}\,dx\right|$ to $ \frac1c$ is about $ 0.997.$


[quote="nkht-tk14"][u] [i]Problem 1 [/i][/u]  [i]Let a,b,c,d,e,f  be positive real numbers such that $ ab \plus{} bc \plus{} cd \plus{} de \plus{} ef \equal{} 1$
   Prove that[/i]    $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2 \plus{} f^2 \geq \frac {1}{2cos\frac {\pi}{7}}$
[i]
   Problem 2  Let $ x_1 , x_2,...., x_{2n}$ be real numbers such that $ (x_{i \plus{} 1} \minus{} x_{i})^2 \leq 1$  with  $ i \equal{} 1 ,2,...,2n \minus{} 1$
    Prove that $ |x_1 | \plus{} |x_2 | \plus{} .... \plus{} |x_{2n} | \plus{} | x_1 \plus{} x_2 \plus{} .... \plus{} x_{2n}| \leq n(n \plus{} 1)$
    Love An Forever[/i]

   [i]For my love[/i] :blush:[/quote]

I'm not convinced about Problem 2

Let $ x_n \equal{} n$

Therefore the $ LHS \equal{} 2n(2n\plus{}1)$ and the $ RHS \equal{} n(n\plus{}1)$

Clear the $ LHS \ge RHS$


Help me find out exactly the answer of each question below:
1.The first box contains 8 socks of kind A,and the other 2 socks of kind B.The second box contains 5 socks of kind A,3 socks of kind B.Pick randomly two socks from each box .
a.Find the probability of picking three socks of kind A from 4 socks picked?
b.If in the 4 socks there is a sock of kind B,find the probability of this stock belonging the first box?

2.A company has 60 workers,in which there are 40 men and 20 women.The rate of women graduating highschool is 15% and the rate of the men graduating highschool is 20%.
a.Picking two workers of this company,find the probability of in these two worker,there is at least one worker graduating highschool?
b.It'sknow that in these two workers there is at least one worker graduating highschool.Find the probability of these two workers are one woman and one man?

Help me find out exactly the answer of each question below:
1.The first box contains 8 socks of kind A,and the other 2 socks of kind B.The second box contains 5 socks of kind A,3 socks of kind B.Pick randomly two socks from each box .
a.Find the probability of picking three socks of kind A from 4 socks picked?
b.If in the 4 socks there is a sock of kind B,find the probability of this stock belonging the first box?
 ans:
1. the kind A socks may came from :
  2 from first box and one from second box or 2 from second box and one from first box

.........    [u] C (8,2) x C(5,1)     +     C(8,1) x C( 5,2)
[/u]

2. i 'm still working on it.

:rotfl: srry just had too. It's not easy or anything, its just that laughter makes you live longer.
1.
a.total socks- 18
  total A- 13
   13/18* 12/17 * 11/16= ans.
i think you can do the rest.

b. total B- 5
    B in 1st box- 2
    2/5
2.
a.15*20/100= 3
   20*40/100= 8
   3+8=11
   (11/60* 10/59)+ (11/ 60)
you can check my work and do the rest.
b.
[color=violet]???????????[/color]


let $a,b,c$ be positive numbers and $abc=1$prove that:
$a+b+c\leq a^3b+b^3c+c^3a$

Come on,just use this inequality: $\frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}\geq\frac{(a+b+c)^3}{3(x+y+z)}$ and the problem reduces to $(a+b+c)^2\geq 3(ab+bc+ca)$, which is trivial to prove. ;)

$\frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}\geq \frac{(a+b+c)^3}{3(x+y+z)}$

thanx for your help

And, by the way, this inequality can be easily proved with Cauchy or Jensen inequality. ;)

My proof: $a^3b+b^3c+c^3a = \frac{a^3b}{abc}+\frac{b^3c}{abc}+\frac{c^3a}{abc}= \frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}$.

Since $\frac{a^2}{c}+c \geq 2a$, so sum up we get:

\[\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}+a+b+c \geq 2a+2b+2c\]

\[\Longleftrightarrow \frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b} \geq a+b+c\]
QED.

I think it'd better to find what method doesn't work for this problem :P

Or we can try Harazi's substitution: $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$. So it becomes:

\[\sum \frac{x}{y} \leq \sum \frac{x^3}{y^2z}\]
Use the face that: $\frac{x^3}{y^2z}+\frac{z}{x} \geq \frac{2x}{y}$. Then add them togather we get the desired result.

[quote="cezar lupu"]Or, from $a^3 b+b^3 c+c^3 a=\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\geq a+b+c$ follows directly from Cauchy. ;)[/quote]

It also follows from rearrangement :)

For that matter, it also directly follows that
$a^3b+b^3c+c^3a\geq{a^2bc+ab^2c+abc^2=abc(a+b+c)=a+b+c}$.

To Shobber: please, I really insist not to be called Harazi's substitution. I really have nothing to do with this substitution, I just used it several times. I see there are some mistakes on the forum related to some sources ( people start transforming classical results in " market" inequalities) and I strongly believe that it is bad to give someone's credit for some things he didn't find. Please, call it "classical substitution". :)

[quote="Al-Khawarizmi"]:D  :D  :D  it works also for Muirhead
indeed we homogenize it to $ a^3b+b^3c+c^3a \geq (a+b+c)abc$
which is equivalent to $ \sum_{sym} \frac 12 a^3b \geq  \sum_{sym} \frac 12 a^2bc $
which is evident seeing that  $ ( 3,1,0) \succeq (2,1,1) $[/quote]
L.H.S. is not symmetric at all.

[quote="Al-Khawarizmi"]:D  :D  :D  it works also for Muirhead
indeed we homogenize it to $ a^3b+b^3c+c^3a \geq (a+b+c)abc$
which is equivalent to $ \sum_{sym} \frac 12 a^3b \geq  \sum_{sym} \frac 12 a^2bc $
which is evident seeing that  $ ( 3,1,0) \succeq (2,1,1) $[/quote]
\[\sum_{sym}a^3b=a^3b+a^3c+b^3a+b^3c+c^3a+c^3b\]
\[\sum_{cyc}a^3b=a^3b+b^3c+c^3a\]