Let $ x,y,z > 0$,m,n are real numbers and $ m \plus{} n \geq 1$,prove that:
$ {\frac {my \plus{} nz \minus{} \left( m \plus{} n \right) x}{\left( m \plus{} n \right) x \plus{} y \plus{} z}} \plus{} {\frac {mz \plus{} nx \minus{} \left( m \plus{} n \right) y}{\left( m \plus{} n \right) y \plus{} z \plus{} x}} \plus{} { \frac {mx \plus{} ny \minus{} \left( m \plus{} n \right) z}{\left( m \plus{} n \right) z \plus{} x \plus{} y}}\geq 0.$

BQ

[quote="xzlbq"]Let $ x,y,z > 0$,m,n are real numbers and $ m \plus{} n \geq 1$,prove that:
$ {\frac {my \plus{} nz \minus{} \left( m \plus{} n \right) x}{\left( m \plus{} n \right) x \plus{} y \plus{} z}} \plus{} {\frac {mz \plus{} nx \minus{} \left( m \plus{} n \right) y}{\left( m \plus{} n \right) y \plus{} z \plus{} x}} \plus{} {\frac {mx \plus{} ny \minus{} \left( m \plus{} n \right) z}{\left( m \plus{} n \right) z \plus{} x \plus{} y}}\geq 0.$

BQ[/quote]

We have, because $ m\plus{}n\geqslant 1$, $ (m\plus{}n)x\plus{}y\plus{}z\leqslant(m\plus{}n)(x\plus{}y\plus{}z)$ and cyclic, therefore
\[ \sum\frac {my \plus{} nz \minus{} \left( m \plus{} n \right) x}{\left( m \plus{} n \right) x \plus{} y \plus{} z}\geqslant\frac {\sum(my \plus{} nz \minus{} \left( m \plus{} n \right) x)}{\left( m \plus{} n \right)(x \plus{} y \plus{} z)}\equal{}0\]
... :P...

[quote="xzlbq"]m,n are real numbers,Is there  an error?

Need to be changed m,n>0?

Please discuss。
BQ[/quote]
But why should we change $ m$, $ n$ to positive real numbers?? The only thing we need is that their sum is grater than or equal to $ 1$ so that the inequality $ x(m\plus{}n)\geqslant x$ holds...


The AMC A is not far away! what are you doing to prepare? how do you think you'll do?

I'm taking the 10A/12B route unless I fail the 10A (which is quite possible). I'm working through AoPS Intro to Geometry, Intermediate Algebra, and a little bit of Intro to Counting and Probability, combined with practice AMC's and AIME's (and lots of other random stuff). I'll be happy with my 10A score if I improve from what I did last year on the 10B.

What about you guys? The A is a lot closer than I thought!!

Ooooh I can't believe it's so close... but I guess we all still have some time to prepare. I'm doing about what Dynamo is (hopefully that's a good thing for both of us). I'm hoping that I can at least qualify for the AIME on the A and then get at least 23 right on the B. But if by some freak accident I get 23+ on the A, I'll probably take the 12B... and hope that I don't completely fail. Ugh everything's coming fast now.... AMC, AIME, MATHCOUNTS. Speaking of all those... I think I'll study now  :oops: haha.

I might take the AMC10 (maybe) if our school registers by this friday.
I think I'm taking the AMC 10B.

ohh my, i just realized that it's next week... asdfasdf :( time to try and not fail

Yes, maybe I should have taken at least one practice amc 12 over the last year, since I never seem to do well on the real amc 12.


Let $ a,b,c,d$ be real numbers such that
$ a\plus{}b\plus{}c\plus{}d\equal{}k_1$, $ ab\plus{}bc\plus{}ca\plus{}da\plus{}db\plus{}dc\equal{}k_2$, $ abc\plus{}bcd\plus{}cda\plus{}dab\equal{}k_3$, and $ abcd\equal{}k_4$. Find
$ p_1\equal{}a^3\plus{}b^3\plus{}c^3\plus{}d^3$, $ p_2\equal{}a^4\plus{}b^4\plus{}c^4\plus{}d^4$, $ p_3\equal{}a^5\plus{}b^5\plus{}c^5\plus{}d^5$ in terms of $ k_1,k_2,k_3,k_4$.


Given $ m,n\in N$ such that $ m<2001,n<2002$. Let distinct real numbers be written in the cells of a $ 2001$x$ 2002$ board (with $ 2001$ rows and $ 2002$ columns). A cell of the board is called [i]bad[/i] if the corresponding number is smaller than at least $ m$ numbers in the same column and at least $ n$ numblers in the same row. Let $ s$ denote the total number of bad cells. Find the least possible value of $ s$.


Hi all,I'm Vietnamese.I think each country has it own math forum,Vietnam has one which have English section (other VN forum is not support other language :D).So I want to introduce to all of you,its address is:
 http://mathnfriend.net/index.php?showforum=62
 just new,need your contribution :D
 why don't you join us,why don't you share your national math forum address here,in this topic ? :lol:

[quote]
is it for vietnamese only?
[/quote]
[color=red]all are welcome[/color],If you know only English,you can join in Olympiad section,If you know VietNamese,others can interest you :D

[quote]
This page is just a copy from MathLinks, just the same[/quote]
we only want to create another playground,can u make it different :lol:


Source: ACoPS
3.1.18 Consider the following two-player game. Each player takes turns placing a penny on the surface of a rectangular table. No penny can touch a penny which is alreay on the table. Tha table starts out with no pennies. The last player who makes a legal move wins. Does the first player have a winnig strategy?

[hide="solution"]
Yes, s/he does.(placing a penny on the center of the table)[/hide]

am I right?

Can you prove that your answer is right?  :)

Edit:  By which I mean, describe your strategy and show why it must work.  Also, can you generalize?  For what shapes of table does your strategy still hold?

[hide="Proof"]
So the first player places the first penny on the center of the table. After that, the first player can place the penny on the position, symmetrical to that the second player placed before.
Does this make sense?
[/hide]


$ \boxed {\ \frac {a\minus{}2b\plus{}2c}{a\plus{}b\minus{}c}\plus{} \frac {b\minus{}2c\plus{}2a}{b\plus{}c\minus{}a}\plus{}\frac {c\minus{}2a\plus{}2b}{c\plus{}a\minus{}b}\ \ge\ \frac {3\left(a^2\plus{}b^2\plus{}c^2\right)}{2(ab\plus{}bc\plus{}ca)\minus{}\left(a^2\plus{}b^2\plus{}c^2\right)}\ \left(\ \ge\ 3\ \right)\ }$

hello, nice inequality. The right-hande side of your inequality is true because
$ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca \leftrightarrow
6(a^2\plus{}b^2\plus{}c^2)\geq6(ab\plus{}bc\plus{}ca) \Leftrightarrow
3(a^2\plus{}b^2\plus{}c^2)\geq6(ab\plus{}bc\plus{}ca)\minus{}3(a^2\plus{}b^2\plus{}c^2) \Leftrightarrow
\frac {3(a^2\plus{}b^2\plus{}c^2)}{2(ab\plus{}bc\plus{}ca)\minus{}(a^2\plus{}b^2\plus{}c^2)}\geq3$.
To prove the left-hande side we make the substitutions
$ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y$ and your inequality is equivalent with
$ \frac{3(x^2y^3\plus{}x^3z^2\plus{}y^2z^3\minus{}x^2y^2z\minus{}x^2yz^2\minus{}xy^2z^2}{2xyz(xy\plus{}xz\plus{}yz)}\geq0$.
We assume that $ x\geq y\geq z$ and the numerator is
$ y^2(x\minus{}z)(y\minus{}z)(x\plus{}z)\plus{}(x\minus{}y)^2(x\plus{}y)z^2\geq0$.

[quote="Virgil Nicula"]$ \boxed {\ \frac {a \minus{} 2b \plus{} 2c}{a \plus{} b \minus{} c} \plus{} \frac {b \minus{} 2c \plus{} 2a}{b \plus{} c \minus{} a} \plus{} \frac {c \minus{} 2a \plus{} 2b}{c \plus{} a \minus{} b}\ \ge\ \frac {3\left(a^2 \plus{} b^2 \plus{} c^2\right)}{2(ab \plus{} bc \plus{} ca) \minus{} \left(a^2 \plus{} b^2 \plus{} c^2\right)}\ \left(\ \ge\ 3\ \right)\ }$[/quote]
$ \frac 13\cdot\sum\frac {a\minus{}2b\plus{}2c}{a\plus{}b\minus{}c}\equal{}\frac 12\cdot\sum\left(\frac {a\plus{}c\minus{}b}{a\plus{}b\minus{}c}\minus{}\frac 13\right)\equal{}$ $ \frac 12\cdot\left(\sum\frac {p\minus{}b}{p\minus{}c}\minus{}1\right)\equal{}$ $ \frac 12\cdot\left[\sum\frac {(p\minus{}b)^2}{(p\minus{}b)(p\minus{}c)}\minus{}1\right]\stackrel {(C.B.S.)}{\ \ \ge \ \ }$

$ \frac 12\cdot\left\{\frac {\left[\ \sum (p\minus{}b)\ \right]^2}{\sum (p\minus{}b)(p\minus{}c)}\minus{}1\right\}\equal{}$ $ \frac 12\cdot\left\{\frac {(a\plus{}b\plus{}c)^2}{\sum\left[a^2\minus{}(b\minus{}c)^2\right]}\minus{}1\right\}\equal{}$ $ \frac 12\cdot\left[\frac {\sum a^2\plus{}2\cdot\sum bc}{2\cdot\sum bc\minus{}\sum a^2}\minus{}1\right]\equal{}$ $ \frac {\sum a^2}{2\cdot\sum bc\minus{}\sum a^2}$ .


inequalities
quadratics
inequalities open

if  posotive reals such that \[\begin{array}{l}a^{2}+b^{2}+c^{2}+d^{2}= 1 \\ then prove that: a+b+c+d+\frac{1}{{abcd}}\ge 18 \\ \end{array}\]

Let S=a+b+c+d and P=abcd.
The quadratic-geometric mean inequality yields P<=1/16 (*).
The arithmetic-geometric mean inequality yields P<=S^4/4^4 (**).

Case 1: S>=2.
Then (*) implies S+1/P>=18.

Case 2: 1<=S<=2.
Then (**) implies S+1/P >= S+4^4/S^4 =: f(S).
On the interval [1,2], the minimum of f(S) is taken at S=2 with f(S)=18.

Case 3: S<1.
This contradicts a^2+b^2+c^2+d^2=1.


calculus
integration
geometry
linear algebra
matrix
function
parallelogram

Hai folks, I have a query (though it appears to be quite silly), given an indefinite integral , say x, after we integrate it , we obtain (x^2)/2 ; Now, my query is----- What does "x^2/2" denote?

What is the purpose of indefinite integration, like definite integration is used  to calculate the area beneath (under) a curve, is indefinite integration used to calculate the length of a curve? Can anyone elaborate?

Secondly, When we calculate the determinant of a matrix, we obtain a "number", what does this number mean/denote?

For instance, consider a 3*3 unit matrix, when we compute the determinant of this matrix , we get the number "0", what does this mean?

I need an elaborate answer. :)

1)  Indefinite integration should really be understood with respect to definite integration.  I hope you have no confusion about what a definite integral means.  Given that you understand the meaning of an expression like $ \int_a^b f(x) \, dx$, we define $ F(x) \equal{} \int_a^x f(t) \, dt$ for arbitrary fixed $ a$.  Under some smoothness conditions, the value of defining this function - which we are free to call an indefinite integral, or a primitive, or whatever - is the statement of the Fundamental Theorem of Calculus.

In other words, indefinite integrals are a convenient way to evaluate definite integrals.  But the theory of integration will make more sense if you focus on definite integrals; that will also stop you from messing up constants.

2)  One intuition is that the determinant is a "volume multiplier":  the determinant of a $ 3 \times 3$ matrix describes what happens to the (signed) volume of boxes if you multiply every point by that matrix.  If it's zero, that means boxes get squashed down into parallelograms.  Similarly, the determinant of a $ 2 \times 2$ matrix describes what happens to the (signed) areas of parallelograms if you multiply every point by that matrix.  Matrices of determinant $ 1$ preserve areas, volumes, and orientations, so they're particularly nice.

I did understand what you meant to say, but what does it mean if " the determinant happens to be a negative number" ?

Then it flips the sign of the [i]signed[/i] area or volume.  In other words, it reverses handedness.  The typical transformation with negative determinant is a reflection through a line or a plane; left-handed things become right-handed things and vice versa.


show that if the positive integer $a,b,c,d$ satisfy $ab=cd$ then we have $gcd(a,c)*gcd(a,d)=a*gcd(a,b,c,d)$;     :)  :)

Let $p$ be any prime.
Let $a' = \mathrm{ord}_p a$ and so on.
Note, that $a'+b' = c'+d'$.
$\mathrm{ord}_p ((a,c)\cdot(a,d)) = max(a',c')+max(a',d')$, $\mathrm{ord}_p (a\cdot (a,b,c,d)) = a' + max(a',c',d',c'+d'-a')$.
It's really easy to see that in three possible cases (when $a'$ maximal, minimal of $\{a',c',d'\}$ or $a'$ is between $c'$ and $d'$) $max(a',c')+max(a',d') = max(a',c',d',c'+d'-a')$. So, canonycal representations of $(a,c)\cdot(a,d)$ and $a\cdot (a,b,c,d)$ are the same, and they are equal.


geometry
3D geometry
sphere
algebra
polynomial
AMC
AIME

Say anything you want about round 4 here!

[quote="zanttrang"]Was I the only one who felt that there was a HUGE range of difficulty on this round, much more so than on previous ones?  I felt like some of them (1 and 2, specifically) were really easy, and they didn't take me long to do at all, while other ones were much, much more difficult.  Generally there are easier and harder ones on each round, but the range on this one seemed more extreme than on previous rounds.[/quote]

Agreed.  #1 and #2 were ridiculous in comparison to #3.

Does anyone want to post their solutions?   I want to know if I'm right... :D  Well, I guess I'll begin.

The one about the square roots was so fun!

[quote="mathfanatic"]
Agreed.  #1 and #2 were ridiculous in comparison to #3.

Does anyone want to post their solutions?   I want to know if I'm right... :D  Well, I guess I'll begin.[/quote]
You got the same answers I did, though your answer to #4 seems to be much more rigorous than mine (I used a strange mix of algebra and geometry to get my answer).  I'll post them when I get access to my home computer.

[quote="zanttrang"]You got the same answers I did, though your answer to #4 seems to be much more rigorous than mine (I used a strange mix of algebra and geometry to get my answer).  I'll post them when I get access to my home computer.[/quote]

That's comforting.  I thought about using the fact that 2x^2 + 3y^2 + 6z^2 = n was an ellipsoid to argue that there'd always be a solution for 38 <= n <= 126, but I couldn't figure out how to do it.  So I used a really ugly way.

[quote="zanttrang"]Was I the only one who felt that there was a HUGE range of difficulty on this round, much more so than on previous ones?  I felt like some of them (1 and 2, specifically) were really easy, and they didn't take me long to do at all, while other ones were much, much more difficult.  Generally there are easier and harder ones on each round, but the range on this one seemed more extreme than on previous rounds.[/quote]This was intentional.

what you might want to try is transforming your ellipsoid into a sphere by making the substitution:

$x = \frac{a}{\sqrt{2}}, y = \frac{b}{\sqrt{3}}, z=\frac{c}{\sqrt{6}}$

then you can use some 3-d geometry to get what you are looking for.

Ow... I feel really stupid right now. On problem 1 I had 22 integers but somehow counted only 20.  I wrote "thus there are 20 integers that work. Namely: (integers here)". As i wrote down every number that worked... and did all the math right-how many points do you guys think im gonna get docked for not being able to count?

2 was pretty easy... no mistakes there. 5 was just long, 4 was a nice problem, but I didnt make any susbitutions and thus had to use a 3d graph at the end  ;) 3 I had no idea what exactly they ment by a 3 variable polynomial and thus stuck down a little work and sent in on its way...

For 1, it's a bit easier to use external angle measure.  An external angle is the supplement of an interior angle, so when the external angle is an integer, so will its interior.  Then factor count 360 and throw out 1 and 2.

[quote="salmononpi"]Ow... I feel really stupid right now. On problem 1 I had 22 integers but somehow counted only 20.[/quote]

Same here. Some how 9/40 didn't make the final cut...

[quote="ThAzN1"]Heh, I thought it was pretty fun... and I'd have to say that my favorite problem was #3.  :)[/quote]

I agree. at first i thought number three was kind of hard, but that's until i decided to divide the sum of the cubes of the terms by the sum of the terms to see what i'd get.  It worked out nicely though.

I liked this round's problems, but I had exactly one day to do my problems (considering SAT and the AIME were the week before.)

Here's my potentially wrong, but otherwise elegant solution to nr. 3:

Consider the identity:

$\alpha + \beta + \gamma = (\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma)/(\alpha^2+\beta^2+\gamma^2-\alpha\beta - \alpha\gamma - \beta\gamma)$.

The denominator of the expression can be factored as a sum of squares and is thus nonnegative, so the sign of $\alpha + \beta + \gamma$ varies with the denominator. Applying the identity to our problem yields the polynomial

$a^3+2b^3+4c^3-6abc$

I felt that the first problem was almost too easy for this contest, while 3 was not good at all.  On #3, I came up with the (correct?) polynomial $x^3+2y^3+4z^3-6xyz$ but then convinced myself that that was not right, and therefore I was not as rigorous as I could have been on the proof of its correctness ;) .  I think that perhaps I was confused on the definition of 'same sign'.  Having seen that that is the right answer, I now feel proud of my method of finding it-3 lines of LaTeX! (Although it is not very rigorous at all.) I'll post it when I have access to my solutions.

#2 was probably my favorite problem, even though it was quite easy.

[quote="ThAzN1"]Well, for #3, I essentially used the same identity as [b]tetrahedr0n[/b]; for #4, I used a (not-so-rigorous?) argument w/ Cauchy.  Anyways, here's my solutions...[/quote]

I did #4 in a very similar fashion.  First, I made the same substitution you did, then I used Cauchy to find the lower bound.  To find the upper bound, I used the rearrangement inequality coupled with an epsilon argument that I'm quite sure is rigorous.

#2 was quite unremarkable, as USAMTS problems go.

I misread #1; I missed the word "regular", so I instead found all values of n such that an n-gon can have all integer angles.  How many points do you think I'll lose?

I failed altogether to solve #3.

For #5, I had an interesting lemma... I proved that if XYZ and X'Y'Z' are two triangles with equal area, congruent incircles, and $\angle X=\angle X'$, then the triangles are congruent.  This blows the whole problem apart...

what the...?

hmm...i used the same calculus method for solving the ellipsoid/tangent plane problem as flierdeke...but i thought there had to be only ONE solution for (x,y,z)...as in: AT the tangent point...so my answer was "no integral values of n"

oh well.

[quote="gaussianprime"]oh no!!!

I solved for only integer solutions in #4!!!

bah! there goes my perfect score!

how could I have missed that?[/quote]

I almost did that...I thought, hm, this is trivial, since there are only a few integer solutions to 3x + 4y + 5z = 23, and then plug them into 2x^2 + 3y^2 + 6z^2 = n.  Fortunately I lost the paper that had my answers on it, so when I returned to the problems the week before they were due, I had to redo it...

[quote="flierdeke"]I came up with a simple (but not necessarily elegant or especially rigorous) solution to #4, using the idea that the minimum value comes where the ellipsoid is tangent to the plane.  (I think this is equivalent to subtracting one equation from the other and completing the square to find the point at which there's exactly one solution - tangency - ie, both sides are zero.)  Using the gradient of the function f(x) = 2x^2 + 3y^2 + 6z^2 made things pretty simple.  [solutions attached][/quote]
Yep, that's what I did, too.
Ahh, the power of calculus! :lol:

Umm, what was with problem #1? That was like an easy AIME problem . . . it seems like the only way to mess up would be to have misread the problem.

[quote="dts"]I misread #1; I missed the word "regular", so I instead found all values of n such that an n-gon can have all integer angles. How many points do you think I'll lose?[/quote]
It sounds like you made the problem more difficult by misreading it, in which case you may not lose as many as you might guess -- I won't be surprised if you get a 3 or a 4.

Here are my solutions if anyone's interested. Hopefully #1 will get a commendation.

If anyone's interested, my solutions...
I think that they are all right, not great, but not bad.

Wooo... only 26 KB

[quote="ZennyK"]...It sounds like you made the problem more difficult by misreading it...[/quote]

I dunno... it was still pretty easy, if you think to use generating functions...

I'm sure I got everything except #4, for which I decided to give up and turned in "I was not able to solve this problem." I actually misread the problem two times: first, I thought they wanted integer solutions; then, I thought they wanted rational solutions! I was desperately flipping through number theory books for information on Pell-ish equations the day this round was due.  :P

I'll be happy if I get something commended.  :)

[quote="Danbert"]Wooo... only 26 KB[/quote]

That's kind of cheating, though, since it's a DVI file.

#1/#2 - both way easy, don't understand why.

#3 - used the cool factoring at the beginning of my MO Treasures book (same one as everyone else)

#4 - i just found the lower bound by setting the ellipsoid tangent to the plane and the upper bound by expanding the ellipsoid until it touched the plane at only one point in the first quadrant. (bleh, looking back at my solution, it looks wrong, but the answer coincides...)

#5 - well i just proved it.

[quote="towbomb"][quote="dts"]For #5, I had an interesting lemma... I proved that if XYZ and X'Y'Z' are two triangles with equal area, congruent incircles, and $\angle X=\angle X'$, then the triangles are congruent.  This blows the whole problem apart...[/quote]

I did that, too, and managed to prove it in 6 or 7 very small lemmas... it essentially became a long 9th grade geometry proof with a few unconventional formulas.[/quote]

I did it in one small lemma and one big lemma.  I wouldn't call mine 9th grade level, though; I used trig.

For #4 I had to use Lagrange multipliers because I couldn't figure out how to do it without calculus.
And #3 was definately my favorite problem.
Does anyone know when they will be grading the papers?

[quote="Jimmy"]Does anyone know when they will be grading the papers?[/quote]In early April.  We hope to release the scores sometime during the week of April 11.

Here are my solutions, but I don't think I did a very good job on #5.

Depending on the level of calculus you know, there is a really easy and straightforward method for solving problem #4.  It literally takes less than 10 minutes to solve for the minimum n.  I learned the method in Calculus 3, which I took this year at a local college.  For those of you who are interested, it is a method called Lagrange multipliers.  I'll attach my solution to #4 here.

That is interesting -- I did Cauchy-Schwartz to get the minimum n, and some simple algebra to get the maximum.


Let the sequence $ {u_n}$ satisfy
$ \left| {u_{m \plus{} n}  \minus{} u_m  \minus{} u_n } \right| < \frac{1}{{m \plus{} n}}\forall m,n \in N$
Prove that $ u_n$ is a arithmetic progression.

Consider $ v_{k}\equal{}u_{mk}\minus{}ku_m$, then $ v_1\equal{}0, |v_{k\plus{}l}\minus{}v_k\minus{}v_l|<\frac{1}{m(k\plus{}l)}$. Therefore $ |v_k|<\frac 1m (\frac 1k \plus{}\frac{1}{k\minus{}1}\plus{}...\frac 12)\equal{}\frac 1m (H_k\minus{}1)$.
Therefore $ |u_{mk}\minus{}ku_m|<\frac 1m (H_k\minus{}1)$.
Consider $ k\equal{}an$, then $ |u_{mna}\minus{}nau_m|<\frac 1m (H_{na}\minus{}1)$
Consider $ m\equal{}ma,k\equal{}a$, we get $ |u_{mna}\minus{}au_{mn}|<\frac{1}{mn}(H_a\minus{}1)$.
Therefore $ |au_{mn}\minus{}nau_m|<\frac{1}{mn}(nH_{na}\minus{}n\plus{}H_a\minus{}1)$ or
$ |u_{mn}\minus{}nu_m|<\frac{1}{mn}\frac{nH_{na}\minus{}n\plus{}H_a\minus{}1}{a}$. If $ a\to \infty$ we get $ u_{mn}\equal{}nu_m$, $ m\equal{}1$ give $ u_n\equal{}nu_1$.


assume that we have luminous point in the space (like very very small lamp) can we control the light of this lamp with finite equal ball such  that they are distinct( means also not tangency)

it's not clear.....maybe l don't understand the meaning of "finite equal ball".
"ball" like "soccerball" ?
if yes, how can a ball control a lamp ?

yes, simply the round things.

So what we got to prove is that if you place thme around the lightpoints, and try to fill up the holes, you'll need an unlimited amount of balls.

Very abstract but nice problem! :)

l am beginner in english too. now l will solve this problem.


real analysis
calculus
calculus computations

Suppose $\left(x_n\right)$ is a sequence of real numbers such that each subsequence of $\left(x_n\right)$ has a convergent subsequence.
Does it follow that $\left(x_n\right)$ is a convergent sequence?

:?: No, $x_n=(-1)^n$ is a counterexample? ...

A slightly less trivial variation, which is actually quite useful (Urysohn's principle)

Let $x_n$ be a sequence in any topological space.
If each subsequence of $x_n$ has a subsequence converging to $x$, then the $x_n$ converge to $x$.

That particular result seems to come out a lot in real analysis; you might have convergence in $L^1$ or in measure, so you get a pointwise convergent subsequence, and then you apply dominated convergence or something like that to get a result about the original sequence.

[quote="Peter VDD"]This should be moved to computations and tutorials... :?[/quote]

Well, it is easy, but it is not a computation or tutorial, rather a thinking exercise, a problem (albeit an easy one)


search
limit
calculus
integration
calculus computations

Hi, can anyone please show me how to do this problem.  You may use any feasible method.  Thanks.

Question: Evaluate $ \displaystyle\frac{1}{\cosh(\pi/2)}\minus{}\frac{1}{3\cosh(3\pi/2)}\plus{}\frac{1}{5\cosh(5\pi/2)}\minus{}\cdots$

You can show:

\[ \sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n}{(2n\plus{}1)\cosh[(2n\plus{}1)\pi/2]}\equal{}\frac{\pi}{8}\]

via the Residue Theorem although perhaps someone here can show us how to do it with just regular Calculus.

how do you guys get those mathematical symbols

its  by using  latex.

search for the  word `LATEX` in  forums   and  go through them  for sometime

[quote="Mitochondria"]How does the residue theorem work here?  May you please demonstrate? I do not know that sums can be evaluated by the residue theorem.   :blush:[/quote]

Pick a slice:
\[ \lim_{\text{pie}\to\infty}\mathop\oint\limits_{\substack{\text{my pie} \\
\text{slice}}} \frac {1}{z\cos(z)\cosh(z)}dz
\]
(and don't forget to evaluate the integral over the indentation about the origin)

How did you know that would be the integrand?  Also, how did you know that would be the contour?

Note if you choose a $ \pi/2$ slice centered along the positive real axis, the $ \pi/4$ leg cancels the $ \minus{}\pi/4$ leg and the outer circular contour goes to zero and so the sum is determined by the integral over the indentation about the origin as it's radius goes to zero.

As far as choosing that, you work on cars?  Engines?  I do. And sometime something's not working and you don't know what's wrong.  What do you do?  Work on everything.  Eventually, you'll find what the problem is and solve the problem.  :)

[url=http://www.mathlinks.ro/viewtopic.php?t=260466]Here is solution[/url] and [url=http://www.mathlinks.ro/viewtopic.php?t=254323]prove....[/url]


USAMTS
geometry
geometric transformation
reflection

5 points is the maximum per problem.

By the way, http://www.usamts.org is probably going to be your best source of information, particularly their FAQs.

Do you mean for a well-written and correct answer? The maximum is 5.

http://www.usamts.org/TipsFAQ/U_FAQ.php#howgrade

This gives a general criteria.

[quote="Erin J. Schram"][quote="pascal_1623"]This is my first year doing the USAMTS, I know that if you just give a correct answer, you get 1 point, but what's the maximum?[/quote]
As others said, the maximum is 5. However, your statement, "If you just give a correct answer, you get 1 point," is mistaken.

Correct answers are supposed to be accompanied by work that shows how the problem-solver found or verified that answer. Nevertheless, as a grader I have seen submitted solutions that consist of nothing but a numerical answer. The amount of partial credit given to such incomplete solutions varies.

Some problems are so difficult that the only way to find the correct numerical answer is to have a complete understanding of the problem. In that case, giving the answer proves that the participant understood all the mathematics involved, and I would award four points of partial credit. In other problems, finding the numerical answer is trivial and the hard work is in verifying the answer. An extreme example would be a problem whose last sentence was, "Prove that t=155." A solution that consisted of nothing but "t=155" would be worth zero points.

Erin Schram
USAMTS grader[/quote]
What are some problems such that finding the answer is only possible if you have complete understanding of the problem?
Thanks,
BPMS

[quote="bpms"]What are some problems such that finding the answer is only possible if you have complete understanding of the problem?
Thanks,
BPMS[/quote]
I'd say for example that USAMTS 2/1/17 fits the bill (http://www.usamts.org/Solutions/Solution2_1_17.pdf).  The problem isn't too bad if you figure out how to do it, but there's no way you'd come up with that solution unless you did it correctly, so most likely the grader would award you near full credit if you just gave the right answer.

However, that said, you should never do that!  The point of USAMTS problems isn't really to just get the right answer - it's to be able to fully justify it in a rigorous proof.

[quote="Erin J. Schram"][quote="zanttrang"]I'd say for example that USAMTS 1/1/17 fits the bill (http://www.usamts.org/Solutions/Solution2_1_17.pdf). [/quote]
I believe that zanttrang means USAMTS Problem 2/1/17, for which he gave a web address. I remember the grading of Problem 1/1/17 (http://www.usamts.org/Solutions/Solution1_1_17.pdf), and we graders were definitely looking for proofs that certain sequences did not contain squares.

Erin Schram
USAMTS grader[/quote]
Ha, yeah, thanks - I didn't check my post over carefully enough - I've edited it to reflect the correct problem.

[quote="tiredepartment"]What about problems that have multiple parts, like part (a) and part (b)?  If students only get one of the two correct, is the maximum score a 3?[/quote]It depends on how much each part is worth. If one of them is a problem, and the second is a generalization, then solving the generilaztion will get you 4 probably, and you will get maybe 1/2 for the other...


arithmetic sequence
algebra
difference of squares
special factorizations

[hide]
We can write the first sum S as $\displaystyle\sum_{i=1}^{50}(2x-1)^2=$
$\displaystyle\sum_{i=1}^{50}4x^2-4x+1=$
${4\displaystyle\sum_{i=1}^{50}x^2}-{4\displaystyle\sum_{i=1}^{50}x}+{50}$
We can write the second sum as $\displaystyle\sum_{i=1}^{50}(2x)^2=$
$\displaystyle\sum_{i=1}^{50}4x^2=$
$4\displaystyle\sum_{i=1}^{50}x^2$
So the difference between the two sums is:
${4\displaystyle\sum_{i=1}^{50}x}-{50}$
This can be simplified into $4\frac{50(51)}{2}-50=5050$

So the answer is D
[/hide]

Calculuslover, your answer's a lot simpler. :D
My computer's too slow. :(


Two pentagons are given in the plane such that neither of them has a vertex on any line that contains a side of the other. Find the maximum number of intersection points their sides can form.

[i]Kömal, problem A.229, January 2000[/i]

Are we allowed to use the lines which contain the sides or just the segments? If this is true, isn't the answer 25? I mean, doesn't every line which contains a side of a pentagon intersect every line which contains a side of the other pentagon?

Anyway, here is my solution.

First of all, it is clear that each side of the first pentagon is intersected by at most 4 sides of the second pentagon. Indeed, suppose the contrary, but each time we intersect the side we change the half plane determined by the side. So we changes half plane 5 times, hence we are not in the same half plane, where we started. Contradiction.

It follows that there are at most 20 intersection. See the example below.


a and b are real numbers. a-b=1
show that a^3 - b^3 geq 1/4

First geq is not evident
and last geq doesnt hold if b=-1/2

[quote="Fermat -Euler"]$a^3-b^3 =(a-b)(a^2+ab+b^2)$
             $=(a^2+ab+b^2)$
          $\geq\frac{3}{4}(a+b)^2$
          $\geq\frac{3}{4}(1+2b)^2 \geq \frac{1}{4}$[/quote]
[quote="senouy"]First geq is not evident
[/quote]
we have $(a^2+ab+b^2) \geq\frac{3}{4}(a+b)^2$

<=>
$4(a^2+b^2+ab)\geq 3 (a^2+b^2+2ab)$
 <=>
$a^2+b^2\geq 2ab$
i think that's now evident ;)


pigeonhole principle
analytic geometry
graphing lines
slope
geometry
function
integration

Plot the points $ P_i(a_i,a_i^2)$ in the coordinate plane, within the unit square $ S$ of vertices $ O(0,0), A(0,1), B(1,1), C(1,0)$, and reindex them in descending order of the slopes of the rays $ OP_i$, i.e. in descending order of the values of the $ a_i$'s. Now, it is well-known that $ \frac {1} {2} | a_ia_j^2 \minus{} a_ja_i^2 | \equal{} \frac {1} {2} a_ia_j| a_i \minus{} a_j |$ is the area of triangle $ \Delta OP_iP_j$ (also see the \href{http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem}{Shoelace Theorem}).

Clearly the triangles $ \Delta OP_iP_{i\plus{}1}$, $ 1 \leq i \leq n\minus{}1$, have pairwise disjoint interiors. Notice that the points $ P_i$ lie on the arc of the graph of the function $ f(x) \equal{} x^2$. Therefore the sum of their areas is strictly less than $ \displaystyle \frac {1} {2} \minus{} \int_{0}^{1} x^2 \textrm{d}x \equal{} \frac {1} {2} \minus{} \frac {1} {3} \equal{} \frac {1} {6}$, the area between the first bisector and the graph. Moreover, $ a_ia_{i\plus{}1}(a_i \minus{} a_{i\plus{}1}) > 0$, so
\[ 0 < \sum_{i\equal{}1}^{n\minus{}1} a_ia_{i\plus{}1}(a_i \minus{} a_{i\plus{}1}) < \frac {1} {3},\]
therefore (at least) one of the expressions is less than $ \frac {1} {3(n\minus{}1)}$. If we want the bound to be $ \frac {1} {2007}$, that requires $ n \geq 670$.

For a better estimate, the defect area (between the triangles and the graph) can be computed to be
\[ D \equal{} \frac {1} {6} \left ( a_n^3 \plus{} (a_{n\minus{}1} \minus{} a_n)^3 \plus{} \cdots \plus{} (a_1 \minus{} a_2)^3 \right ) \geq \frac {1} {6n^2} a_1^3, \ \textrm {so}\]
\[ \sum_{i\equal{}1}^{n\minus{}1} [\Delta OP_iP_{i\plus{}1}] \equal{} \frac {1} {2} a_1^3 \minus{} \frac {1} {3} a_1^3 \minus{} D \leq \frac {n^2 \minus{} 1} {6n^2} a_1^3 \leq  \frac {n^2 \minus{} 1} {6n^2},\]
therefore (at least) one of the expressions is less than $ \frac {n\plus{}1} {3n^2}$. If we want the bound to be $ \frac {1} {2007}$, that still requires $ n \geq \left \lceil \frac {669 \plus{} \sqrt{669^2 \plus{} 4\cdot 669}} {2} \right \rceil \equal{} 670$.

But there is room for improvement (equality in this last estimate does not yield equal areas), so we are back to some algebraic manipulations, in order to lower the number to $ 669$.

See a second posting at [url]http://www.mathlinks.ro/viewtopic.php?p=1625529#1625529[/url]


The values -5, -3, and 4 randomly replace $ a$, $ b$ and $ c$ in the equation $ ax\plus{}b\equal{}c$, and the equation is solved for $ x$. What is the probability that $ x$ is negative? Express your answer as a common fraction.

$ x\equal{}\frac{c\minus{}b}a$, so we note that for every $ a$, there is ONE distinct pair $ (b,c)$ that satisfies the requirements. Thus, the probability is $ \frac12$, since for every selection of $ a$, there are $ 2$ possible pairs $ (b,c)$.


abstract algebra
function
vector
LaTeX
Ring Theory
superior algebra
superior algebra unsolved

a first step is proving that if P is f.g. and projective, then so is P*, and hence so is P**
I can do this step, so just looking for the final argument.

$ P^{*}= Hom(P,R)$ where $ R$ is the base ring, or do you mean some pontrjagin-dual?

Right $ P^{*}= \text{Hom}_{R}\ (P,R)$ and then $ P^{**}$ is just the double daul.

One way is to show that this function is an isomorphism:  $ \theta: P \rightarrow P^{**}$ where $ \theta(a)$ is equal to the function $ \theta(a)$ which assigns to a homomorphism $ f: P^{*}\rightarrow P$ its value at $ a$.  I have trouble wrapping my mind around double duals.

Geometric idea: a projective A-module P corresponds to a vector bundle M~ over Spec(A), for those who know coherent sheaves.

 More concretely, it means that if P is projective, for all prime ideal x in Spec(A) the localised module P_x is a free A_x module. To see this algebraically, write P as a summand of a free module; this property remains when localizing at x, hence P_x is a projective A_x module, finitely generated, hence free because A_x is local.

 Now since P is finitely generated, the dual P* and also the double dual P** pass to localization (see for example Eisenbud, "commutative algebra with a view..." , prop.2.10); it means that (P**)_x = (P_x) ** , the second double dual being taken as A_x module. But P_x is free, the natural map P_x ---> (P_x )** is an isomorphism, so the natural map P ---> P** is an isomorphism when localized at any prime. It is therefore an isomorphism.

this "geometric" proof seems to be the most elegant. I think we can avoid localizations.

first notice that the claim is clear for f.g. free modules: $ (R^{n})^{*}\cong R^{n}$ by the universal property of the direct sum, and thus $ (R^{n})^{**}\cong R^{n}$. it is easy to check that this isomorphism is actually the natural one.

now if $ P$ is finitely-generated, we have an exact sequence $ 0 \to K \to F \to P \to 0$, where $ F$ is a f.g. free module. if $ P$ is projective, this sequence splits, i.e. $ P$ is a direct summand of $ F$. the inclusion $ P \to F$ yields a commutative diagramm

$ P \to P^{**}$
$ \downarrow ~~~~\downarrow$
$ F \to F^{**}$

since $ F \to F^{**}$ has trivial kernel, the same is true for $ P \to P^{**}$. on the other hand, the projection $ F \to P$ yields a commutative diagramm

$ F \to F^{**}$
$ \downarrow ~~~~\downarrow$
$ P \to P^{**}$

remark that $ F^{**}\to P^{**}$ stays epi since the sequence above splits. thus, $ F \to F^{**}$ epi implies $ P \to P^{**}$ epi.

This proof I think is at least as elegant as the previous one.

A one last proof avoiding localization, and resting on a homological argument: there is a spectral sequence whose second term is

   E_2 ^{p,q} =  Ext^p (Ext^{-q} (P, A), A)

 and converging to P in degree 0. The Ext* groups are computed in the category of A-modules. Since for q non zero we have: Ext^{-q} (P, A) =0 because P is projective, we get P = E_2 ^{0,0} = P**.  I hope it is correct.

 Question: how to post in LaTex?  :blush:

thanks for the help guys, i like your first solution fadalbala and -oo-'s

edit:  for Latex just treat the post like a regular latex document


algebra
polynomial
function
algebra proposed

if X+Y in Q     find all polynimals that p(X)+p(Y) in Q : :lol:

Consider a polynomial $ P(x)$ satisfy  $ f(x)\in Q \leftrightarrow x\in Q$ then $ P(x) \equal{} ax \plus{} b$
Problem have posted in forum .
Apply this then $ P(x) \plus{} P(a \minus{} x) \equal{} mx \plus{} n,\forall x\in R ,a\in Q$
Solve this we have $ f(x)$ is linear function with rational confficient.

Do you mean:  find all polynomials $ P(x)$ such that $ x \plus{} y \in \mathbb{Q} \implies P(x) \plus{} P(y) \in \mathbb{Q}$

or do you mean:  find all polynomials $ P(x)$ such that $ x \plus{} y \in \mathbb{Q} \Leftrightarrow P(x) \plus{} P(y) \in \mathbb{Q}$?


linear algebra
matrix
linear algebra unsolved

Let A be an nxn matrix st a(ij) = 1/(i+j-1) and let B be the inverse of A.
Is it true that the sum of all entries b(ij) of B is equal to n^2?
What is the proof of this??

A previous [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=41046]thread[/url] on this subject.

It doesn't contain a proof for the question you asked, so that's still open.

Interestingly enough, it has a direct connection to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146373[/url] ;)


geometry
IMO Shortlist
geometry proposed

In triangle ABC we use Ceva thorem and the result comes

[quote="moldovan"]The circles $ k_1$ and $ k_2$ with respective centers $ O_1$ and $ O_2$ are externally tangent at the point $ C$, while the circle $ k$ with center $ O$ is externally tangent to $ k_1$ and $ k_2$. Let $ l$ be the common tangent of $ k_1$ and $ k_2$ at the point $ C$ and let $ AB$ be the diameter of $ k$ perpendicular to $ l$. Assume that $ O$ and $ A$ lie on the same side of $ l$. Show that the lines $ AO_2$, $ BO_1$, and $ l$ have a common point.[/quote]
Nice but not easy, in my opinion. It has been posted on the forum many times, you can find solutions at [url=http://www.mathlinks.ro/viewtopic.php?p=875026#875026]IMO Shortlist 2006, Geometry 6[/url].

[quote="sophie germain"]i dont understand how it follows from ceva's theorem .cud u plz explain?[/quote]
This problem is easy.
Let $ AC\cap O_1B\equal{}\{M\}, AO_2\cap BC\equal{}\{N\}.$
$ \angle BOF\equal{}\angle FO_2C$ then $ \angle OFB\equal{}\angle CFO_2$ we get C,F,B are collinear.
Similarly, A,E,C are collinear.
Use Ceva's theorem for triangle ABC:
$ \frac{AD}{BD}.\frac{NB}{NC}.\frac{MC}{MA}\equal{}\frac{CO_2}{CO_1}.\frac{AB}{CO_2}.\frac{CO_1}{AB}\equal{}1$ ->QED


search
number theory unsolved
number theory

Problem: Is there any other perfect square beside 144 in the Fibonacci sequence ? Justify your answer

What ?
U don't understand.
This noon, i found it's solution but i need more time to read it. I'll post it later

No, [i]you[/i] don't understand.
He told you to search, meaning it's been posted before.

(and indeed : http://www.mathlinks.ro/Forum/viewtopic.php?t=1423 )


This is a write up on the subject above and it's an urgent.Need an helping hand to search and find the proper write up for this topic. remember "mathematics of an elastic media".
 Thanks and God Bless.
  Azeez

Sounds like Physics to me.  And yes, this is _definitely_ in the wrong forum.

[quote="eryaman"]errr....what does "mathematics of an elastica media" mean anyways?  :?
I think this should go in different forum.[/quote]

An elastic medium is any material that will deform under stress but will tend to return to its original shape after the stress is removed (imagine an i-beam supported at both ends; if you put a large enough load in the middle, it will buckle, but will return to its original shape once the load is removed).  Hooke's law is commonly applied when elastic media are studied, and situations where forces are applied to non-rigid bodies can be modeled mathematically.  However, I'd agree that the USAMTS forum isn't the right place for a physics problem.


let be given possitive real number: $x_1$,$x_2$,...........,$x_n$ such that $\sum_{i=1}^{n}{x_i}=n$
Prove the inequality: $\sum_{i\ne{j}}{x_i\cdot{x_{j}^2}}\ge\sum_{i\ne{j}}{x_i\cdot{x_j}}$

Hmm...
We are to prove \[n\sum_{i\ne{j}}{x_i\cdot{x_{j}^2}}\ge(x_1+...+x_n)\sum_{i\ne{j}}{x_i\cdot{x_j}},\] which is equivalent to \[\frac{\sum_{i\ne j}x_ix_j^2}{n(n-1)}\geq \frac{\sum_{i\ne j\ne k}x_ix_jx_k}{n(n-1)(n-2)},\]
i.e. pure Murheid.

@myth: can you post complete-solution---thanks :P

Am I missing something? It is Murheid's inequality for orbits $(2,1,0)$ and $(1,1,1)$. What further explanations are you waiting? :?


it has been posted by me.and sloved with the help of the kind people :)
see  here [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28814]www.mathlinks.ro/Forum/viewtopic.php?t=28814[/url]


Let $ m$ be the number of solutions in positive integers to the equation $ 4x\plus{}3y\plus{}2z\equal{}2009$, and let $ n$ be the number of solutions in positive integers to the equation $ 4x\plus{}3y\plus{}2z\equal{}2000$. Find the remainder when $ m\minus{}n$ is divided by $ 1000$.

[hide="Solution"]Consider the following apparent one-to-one correspondence: every solution $ (x_0, y_0, z_0)$ in positive integers of the equation $ 4x \plus{} 3y \plus{} 2z \equal{} 2009$ corresponds a solution $ (x_0, y_0 \minus{} 3, z_0)$ in positive integers of the equation $ 4x \plus{} 3y \plus{} 2z \equal{} 2000$. However, not all solutions of the first equation correspond to solutions of the second; when $ y_0 \equal{} 1$, $ 2$, or $ 3$, $ y_0 \minus{} 3$ fails to be a positive integer. If so, then counting the solution triples for these values of $ y_0$ should give the difference between the number of solutions, $ m \minus{} n$.

When $ y_0 \equal{} 1$, the equation reduces to $ 4x_0 \plus{} 2z_0 \equal{} 2006 \iff 2x_0 \plus{} z_0 \equal{} 1003$, giving $ 501$ solutions from $ (1,1001)$ to $ (501, 1)$.
When $ y_0 \equal{} 2$, the equation reduces to $ 4x_0 \plus{} 2z_0 \equal{} 2003$, giving no solutions.
When $ y_0 \equal{} 3$, the equation reduces to $ 2x_0 \plus{} z_0 \equal{} 1000$, giving $ 499$ solutions from $ (1,998)$ to $ (499, 2)$.

In all, $ m \minus{} n \equal{} 501 \plus{} 0 \plus{} 499 \equal{} 1000$, and when divided by 1000, the remainder is $ \boxed{0}$.
[/hide]

Really?, or so I thought.

I did not get that i did a super brute force way and i got $ \frac{2125}{1000}$
which would be $ 3 \frac{125}{1000}$ or $ \boxed{125}$

btw i think i got a 5...  :(

[hide] I got something like 84336-83834=502.[/hide]

nooooooooooooooooo

i put 125  :mad: PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL PHAIL  :mad:  :mad:

[quote="Smartguy"]I also got 501+499=1000 so 000...
If it is 000, then   :D
I think it will be the first time an answer on the AIME is 0.[/quote]

Nope, 000 has already been done once before. (As has 001)

I believe this was the first problem to have 000 as an answer?

You believe wrong  :P : [[2000 AIME II Problems/Problem 9]]

[quote="tinytim"]April Fools  :wink:  :P[/quote]

lol thats must have been what they wanted

[quote="funtwo"]i wasted so much time checking this problem... what a mean trick (the answer being 000)[/quote]
What do you mean? A good solution would've made it obvious.

He probably means 000 is the kind of answer that makes you sweat and think you've done something wrong even if it's clear, in hindsight, that you didn't.

well what i did was only consider 3y=3 and 3y=9 in the first equation and i was trying to find a flaw in this reasoning

[hide="Another approach"]

Since the sum of the coefficients is 9, the number of solutions in positive integers to $ 4x\plus{}3y\plus{}2z\equal{}2009$ is the same as the number of solutions to nonnegative integers to $ 4x\plus{}3y\plus{}2z\equal{}2000$.  Thus, the desired quantity is equal to the number of solutions in nonnegative integers to $ 4x\plus{}3y\plus{}2z\equal{}2000$ in which at least one of the variables is equal to zero.

Thus, we want to count the number of solutions in nonnegative integers to the three equations $ 4x\plus{}3y\equal{}2000$, $ 3y\plus{}2z\equal{}2000$, and $ 4x\plus{}2z\equal{}2000$.  The first has $ 167$ solutions, the second has $ 334$ solutions, and the third has $ 501$ solutions.  However, we are overcounting by two, since we can have both $ x$ and $ y$ or both $ y$ and $ z$ equal to zero.  Therefore, the number of solutions is $ 167\plus{}334\plus{}501\minus{}2\equal{}\fbox{000}$

[/hide]

I also got 0
1)4x+3y+2z=2009
2)4x+3y+2z=2000

1)4x+2z=2009-3y
2)4x+2z=2000-3y

So m-n is the total solution of
4x+2z=2006
4x+2z=2003
4x+2z=2000
that makes 1000 so the answer is 0

By the way, what I really wanted to ask was...
How was the difficulty of AIME2 this year??

Why is it the fact that the sum of the coefficients is 9 imply that the number of solutions in positive integers to 4x+3y+2z=2009 is the same as the number of solutions to nonnegative integers to 4x+3y+2z=2000?

Every solution $ (x,y,z)$ to the second corresponds to exactly one solution $ (x\plus{}1,y\plus{}1,z\plus{}1)$ to the first.

so how do I conclude that? How do you know this?

An ordered triple $ (x,y,z)$ solves the second equation if and only if $ (x\plus{}1,y\plus{}1,z\plus{}1)$ solves the first.

Wait how do you prove that solutions in this form make up every solution? Couldn't there be solutions "in another form"?

Let $M=\{(x,y,z)\mid4x+3y+2z=2009\}$ and define $N=\{(x,y,z)\mid4x+3y+2z=2000\}$. Also denote $M'=\{(x,y,z)\in M\mid1\notin\{x,y,z\}\}$.

Since the function $f:M'\to N$ defined by $f(x,y,z)=(x-1,y-1,z-1)$ is bijective, we have $|M'|=|N|=n$.

Then the answer is $|M\setminus M'|\pmod{1000}$. We count this with four cases:

[b]Case 1:[/b] $x=1,1\notin\{y,z\}$
Then $3y+2z=2005$. By taking$\pmod2$ and$\pmod3$, we have that $y$ is odd and $z\equiv2\pmod3$. Letting $y=2b+1$ and $z=3c+2$, the equation transforms into $b+c=333$. For any $1\le b\le333$ there is a solution for $c$, so this case has $\underline{333}$ solutions.

Note that we established a bijection between $3y+2z=2005$ and $b+c=333$.

[b]Case 2:[/b] $y=1,1\notin\{x,z\}$
The equation is $2x+z=1003$. Now $z$ is odd, so $z=2c+1$. The new equation is $x+c=501$ which has $\underline{499}$ solutions since we avoid $x=1$.

[b]Case 3:[/b] $z=1,1\notin\{x,y\}$
Finally, $4x+3y=2007$, giving $x\equiv0\pmod3$ and $y\equiv1\pmod4$. With the substitutions $x=3a$ and $y=4b+1$, we obtain $a+b=167$, which has $\underline{166}$ solutions.

[b]Case 4:[/b] $(x-1)(y-1)+(y-1)(z-1)+(z-1)(x-1)=0$
We have the $\underline2$ solutions $(x,y,z)=(1,1,1001),(501,1,1)$.

Our final answer is $333+499+166+2=1000\equiv\boxed{000}$.

bruh wait what... in my mock i guesseed this correct.... note that gcd(2,3,4)=12 so it should repeat every once in 12 note that 2009-2000=9<12 so we can guess 0 or 1.

i guessed 0 :D
(rare problem with answer 0 :D)


trigonometry geometry circumcircle symmetry Euler

$x,y,z\in\mathbb{R}$ , prove the identity

(i)$\sin x+\sin y +\sin z - \sin (x+y+z)=4\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x+z}{2}\right)\sin\left(\frac{y+z}{2}\right)$

(ii)$\cos x+\cos y+\cos z -\cos (x+y+z)=4\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x+z}{2}\right)\cos\left(\frac{y+z}{2}\right)$

[hide="Remark"] specially when $x+y+z=\pi$ , we have $\sin x+\sin y +\sin z=4\cos\frac{x}{2}\cos\frac{y}{2}\cos\frac{z}{2}$ and $\cos x+\cos y+\cos z=4\sin\frac{x}{2}\sin\frac{y}{2}\sin\frac{z}{2}-1$[/hide]

 :)

[b]A proof for 1) ( Not very interesting though)[/b]
Regarding y and z as fixed, and
let $f(x)=\sin x+\sin y+\sin z-\sin (x+y+z)-4 \sin(\frac{x+y}2) \sin(\frac{x+z}2) \sin(\frac{y+z}2)$.
Then
$f'(x)=\cos x-\cos(x+y+z)-4[\frac12 \cos(\frac{x+y}2) \sin(\frac{z+x}2)+\frac 12 \sin(\frac{x+y}2)\cos(\frac{x+z}2)]\sin(\frac{y+z}2)$
$=\cos x-\cos(x+y+z)-2\sin(x+\frac{y+z}2)\sin(\frac{y+z}2)$      [i]by compand angle formula[/i]
$=\cos x-\cos(x+y+z)+[\cos(x+y+z)-\cos x]$[i]      by product-to-sum formula[/i]
$=0$.

[i]It shows that $f(x)=constant$ for any real value of x.[/i]

Choose $x=-y$ and put it into f(x),
$f(-y)=\sin(-y) + \sin y + \sin z - \sin z -4\sin{0} \cdot \sin({\frac{-y+z}2})\sin(\frac{y+z}2)=0$

Hence, $f(x)=0$ for all real values of x and the result follows.

I think the second identity can be proved similiarly.  :) [b]

[i]But what is the significance of these two identities?[/i][/b]

If $x, y, z$ are the angles of a triangle, $x + y + z = \pi$ and the indentities:

$\sin x+\sin y +\sin z=4\cos\frac{x}{2}\cos\frac{y}{2}\cos\frac{z}{2}$
$\cos x+\cos y+\cos z=4\sin\frac{x}{2}\sin\frac{y}{2}\sin\frac{z}{2} - 1$

are so incredibly useful if you start using trigonometry to solve a geometry question. Here in the UK IMO squad, this is known as 'Trig Bashing'. If you delve into this technique you want to describe a triangle completely in terms of $R, \alpha, \beta, \gamma$ (the circumradius, and angles of the triangle). However, once the configuartion gets more complicated you find its far easier to use multiplicative trig terms rather than additive ones, and these two identities are priceless tools to have.

However if you find using trig on a geometry question horrible and never intend on using it, these identities will probably just be some nice curious fact rather than anything useful.

The other useful one is $\tan x + \tan y + \tan z = \tan x\tan y\tan z$.

Thanks for your explanation.
Do you mind posting some related problems to illustrate the application of these identities?

Basically say you wanted to find $\frac{r}{s}$. It is far more useful ito use multiplicative terms than additive terms. $s = 4R \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}$ is far better than $s = R(\sin A + \sin B + \sin C)$. Then as $\triangle = rs = 2R^2 \sin A \sin B \sin C$, double-angle formula easily tells you than $r = 4R \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2}$. Note how multiplicative terms allow you to keep symmetry without really thinking.

So $\frac{r}{s} = \tan\frac{A}{2} \tan\frac{B}{2} \tan\frac{C}{2}$.

Another good use is if you wanted to prove the Euler line by areal co-ordinates. To do this you need normalised co-ordinates - the co-ordinates sum to 1. Without these identities, the algebra would be horrendous. With them, they're quite tame. Try it out.

[hide="Cilck For A Hint (The Unnormalised Co-ordinates)"]$O = (\sin 2A, \sin 2B, \sin 2C)$
$G = (1, 1, 1)$
$H = (\tan A, \tan B, \tan C)$[/hide]

You can expand by using the identities
\[ \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2}, \quad \sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i}. \]

[quote="shyong"]$x,y,z\in\mathbb{R}$ , prove the identity

(i)$\sin x+\sin y +\sin z - \sin (x+y+z)=4\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x+z}{2}\right)\sin\left(\frac{y+z}{2}\right)$

(ii)$\cos x+\cos y+\cos z -\cos (x+y+z)=4\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x+z}{2}\right)\cos\left(\frac{y+z}{2}\right)$
[/quote]


When I click on an Asy figure (to examine the code) a blank window pops up but no code is there.

FF 3.0.8 (i think, i was prompted for 3.0.9 today so...) on OS X 10.5 Leopard.

I'm having the same problem on Google Chrome (Beta) on Windows XP. :maybe:

FF 3.0.9 on Windows XP fails as well.

I think they may have not coded this yet.

Yeah I didn't have the time to adapt the new code popup to asymptote.

[quote="ThinkFlow"]Thanks for making the new popup! :lol:[/quote]You're welcome. We're going to have a similar one for asymptote soon.


just simple AM-GM
$x^4+y^2\geq 2x^2 y$  :)

:|   :o  Huh! What an easy question!!!

Thank you!


The sum of the first N positive integers is 2006. Wait, that is impossible...one of those integers must not have been included! If in fact exactly one of the first N positive integers was not included in the sum, which integer must that have been ?

Thanks

[hide] You find the triangle number >2006. This number is 2016.((63*64)/2). Since this is ten more than 2006 the missing number is 10.[/hide]

Just to clarify, for anyone who may not know what a triangular number is (at least not by that name), it's the sum of the first $n$ numbers, i.e. $\frac{n(n+1)}{2}$, where $n$ is an integer. So, the first number of this form that is $>2006$ is $2016$, for $n=63$, and you can solve from there easily.


Can someone proofread my proof to see if it is in good proof format and correct? If this were a USAMO problem ( Either the USAMO is easy or I'm a genius!  :mrgreen: JK ) What would my score be?

 This is a question from  the AoPS Vol. 1: The Basics - Ch 28. Prove It, p338, exercise 28-3.  Prove that if a, b, c, > 0, then if ax :^2:  + bx +c = 0 has real solutions both solutions are negative.  The solutions of x are ( -b  :pm:  :sqrt: b :^2:  - 4ac ) / 2a.  Therefore, b :^2: -4ac must be positive in order for  :sqrt: b :^2: -4ac to be real. Thus, b :^2: > 4ac. But, since 4ac is subtracted from b :^2: ,  :sqrt: b :^2: -4ac is less than the abs(-b). Thus, the isgn of the numerator -b+ :sqrt: b :^2:  -4ac is negative and -b- :sqrt: b :^2: -4ac is negative as well while the denominator 2a is positive. Therefore, both (-b +   :sqrt: b :^2: -4ac)/2a and (-b- :sqrt: b :^2: -4ac)/2a are both negative solutions of x.

My proof is different and unfortunately more complicated than the solution in the book.

Both proofs are correct.  The first could use some "neatening up," but all the necessary mathematics is in there, so I would expect it to garner full points.  The second one is more elegant, though :)

Yeah. The second was the book's solution which was a whole lot quicker. I just went with what came to my mind first.

This is how I would write your proof using your method (I would, however, use rep123max's methods instead).

We have ax^2+bx+c=0 with a,b,c > 0

This implies x = (-b  :pm:  :sqrt: b^2 - 4ac)/2a

However, b^2-4ac < b^2 because a,c > 0

So (-b  :pm:  :sqrt: b^2 - 4ac) < 0 and 2a > 0,
and x < 0, as desired.

This is also a good way of proving it:

If f(x) = ax^2 + bx + c, with a,b,c > 0, then we can rewrite the equation as:

f(x) = a(x-p)(x-q) = ax^2 - (p+q)x + pq

b = -(p+q)
c = pq

Since c,b > 0, p,q < 0 as desired.


function
calculus
integration
limit
inequalities
derivative
real analysis

Assume that f is a function belong to $C^{2}$, f(0)=0. Prove that:(assume that all the intergral are finite)
$\int_{0}^{\infty}{f(x)^{2}dx}$.$\int_{0}^{\infty}{f''(x)^{2}dx}$>
$(\int_{0}^{\infty}{f'(x)^{2}dx})^{2}$
Can anyone solve it? :D

By integration by parts,

$\int_{0}^{b}f'(x)^{2}\,dx=f(b)f'(b)-\int_{0}^{b}f(x)f''(x)\,dx$

If we can argue that $\lim_{b\to\infty}f(b)f'(b)=0,$ then the desired inequality will follow from the Cauchy-Schwarz inequality.

But $f(b)f'(b)=-\int_{b}^{\infty}f(x)f''(x)+f'(x)^{2}\,dx$ and the presumed existence of the improper integrals shows that this tends to zero as $b\to\infty.$

[quote]$\lim_{b\to\infty}f(b)f'(b)=0$[/quote]
The function f is not continous so I think it is not necessary to exist the limit of  f(x).f'(x).
Do you have another solution?

You gave $f\in C^{2},$ so yes, it and its derivative are continuous.

That alone would not be enough to show that the limit is zero, but look at my last line:

$|f(b)f'(b)|\le\int_{b}^{\infty}|f(x)f''(x)|\,dx+\int_{b}^{\infty}f'(x)^{2}\,dx$

But the fact that $\int_{0}^{\infty}f'(x)^{2}\,dx<\infty$ implies that

$\lim_{b\to\infty}\int_{b}^{\infty}f'(x)^{2}\,dx=0,$ and

$\int_{b}^{\infty}|f(x)f''(x)|\,dx\le\left( \int_{b}^{\infty}f(x)^{2}\,dx\right)^{\frac12}+\left(\int_{b}^{\infty}f''(x)^{2}\,dx\right)^{\frac12}$

and a similar argument shows that that goes to zero as $b\to\infty.$


trigonometry
geometry
rotation
perpendicular bisector
algebra
binomial theorem

Put solutions for the USC test here.  [url=http://www.math.sc.edu/contest/problems.html]Here's[/url] the link to the problems.

I just took the 1987 test. Here are the solutions to the second half.

 [hide="16"]Notice that $\tan{90-\theta}=\cot\theta$ and that $\tan\theta \cdot \cot\theta=1$. Therefore, the product evaluates to simply $1$
[/hide]
[hide="17"] There are $24$ numbers for each digit in which that digit is the first digit of the number and similarly there are $24$ numbers for each digit in which that digit is the second digit of the number and so on. Also, $1+2+3+4+5=15$ Our sum thus evaluates to:

$24 \cdot 15(10^{5}+10^{4}+10^{3}+10^{2}+10+1)=3,999,960$[/hide]

[hide="18"]Draw the diameter such that it is a perpendicular bisector of both chords. Let the length of the diameter be $y+x+1$ where $y>x$. Through power of point, we have:

$16=y(x+1)$
$9=x(y+1)$

Solving this, we have that $x=1$ and $y=8$ and the length of the diameter is $10$.[/hide]

[hide="19"] $\tan{30}=\frac{\sqrt{3}}{3}$. Now we find $\arccos(\frac{\sqrt{3}}{3})$ so construct a right triangle with height $\sqrt{3}$ and hypotenuse $3$. The $\sin$ of this triangle is $\frac{\sqrt{6}}{3}$ which is the first choice. [/hide]

[hide="20"]If a number is divisible by $18$, it’s divisible by $9$ and $2$, therefore, $2 | d$ and $9 | 15+2d \Longrightarrow 9 | 6+2d$. The answer is therefore $d=6$.
[/hide]
[hide="21"]Letting $x=0$, we have $f(1)=2f(0)+1$, so $f(1)=5$. Letting $x=1$, we have $f(3)=2f(1)+1$, so $f(3)=11$
[/hide]
[hide="22"]Rotating the smaller square by it’s center does not affect the area overlap. Therefore, rotate the small square so that each of it’s sides is parallel to the bigger square’s. It’s clear that the area overlap is a fourth of the smaller square. The answer is $\frac{25}{4}$[/hide]

[hide="23"]Rewrite the sum as $(10^{4}+20)^{\frac{1}{2}}-(10^{4}+10)^{\frac{1}{2}}$ Using extended binomial theorem, we evaluate the expression as:

$(\binom{1/2}{0}10^{2}+\binom{1/2}{1}\frac{20}{10^{2}}+...)-(\binom{1/2}{0}10^{2}+\binom{1/2}{1}\frac{10}{10^{2}}+... )\approx \frac{1}{20}$ but still a bit less than that. The answer therefore is $(e)$.[/hide]

[hide="24"]The fixed distance between the segments should give you a hint they are talking about foci, and thus an ellipse.
[/hide]
[hide="25"]Let the two equal angles be $x$. We have $BD^{2}=5^{2}+2^{2}-2(5)(2)(\cos{x})=4^{2}+4^{2}-2(4)(4)(\cos{x}) \Longrightarrow \cos{x}=\frac{1}{4}$. Therefore, $BD=2\sqrt{6}$.
[/hide]
[hide="26"]The units digit of $3^{456789}=3^{1}=3$ and $2^{3}=8$. [/hide]

[hide="27"]Let $P(x)=x^{100}-2x^{99}+4$. The roots of $x^{2}-3x+2$ are $x=2$ and $x=1$. Therefore $P(x)=(x-1)(x-2)n+Q(x)$ where $Q(x)$ is our remainder. $P(1)=Q(1)=3$ and $P(2)=Q(2)=4$. Since $Q(x)$ is linear, we have $a+b=3$ and $2a+b=4$ giving us $a=1$ and $b=2$ and our remainder is thus $x+2$. [/hide]

[hide="28"]We must have $x_{1}+x_{2}\equiv y_{1}+y_{2}\equiv 1 \mod{2}$ Looking at $\mod{2}$ for our set, the biggest set that doesn’t have every pair is a set of $4$. [/hide]

[hide="29"]Let the roots be $a,b,c$. We have that $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ac)=4-2(-2)=8$[/hide]

[hide="30"]The expression evaluates to $\frac{2\sin{25}\cos{-15}}{2\cos{25}\cos{-15}}=\tan{25}$.[/hide]

I'll post the first half later...


induction
algebra
polynomial
linear algebra

Consider $ B\equal{}\left(\begin{array}{cc}1 & 3\\ \ 0 & 1\end{array}\right).$ Compute $ B^n, n \in \mathbb{N}^*.$

It is very easy to prove by induction that

$ B^n \equal{} \left[\begin{array}{cc}1 & 3n\\ 0 & 1\end{array}\right]$.

$ \text{Null Potent}\Longrightarrow\text{ Binominal Theorem}$

[quote="moldovan"]Consider $ B \equal{} \left(\begin{array}{cc}1 & 3 \\
\ 0 & 1\end{array}\right).$ Compute $ B^n, n \in \mathbb{N}^*.$[/quote]

$ B \equal{} \left(\begin{array}{cc}1 & 3 \\
\ 0 & 1\end{array}\right)\equal{}\left(\begin{array}{cc}1 & 0 \\
\ 0 & 1\end{array}\right)\plus{}\left(\begin{array}{cc}0 & 3 \\
\ 0 & 0\end{array}\right)\equal{}I\plus{}A$ say

Note $ \ A^2\equal{}O\implies A^k\equal{}O\ \forall k\equal{}2,3,4,\dots$

$ \implies \boxed{\boxed{\ B^n\equal{}(I\plus{}A)^n\equal{}I\plus{}nA\equal{}\left(\begin{array}{cc}1 & 0 \\
\ 0 & 1\end{array}\right)\plus{}n\left(\begin{array}{cc}0 & 3 \\
\ 0 & 0\end{array}\right)\equal{}\left(\begin{array}{cc}1 & 3n \\
\ 0 & 1\end{array}\right)}}$
 :lol:

By Calculation of $ B^2,\ B^3,\ \cdots$, we can conjecture $ B^n\equal{}\left(
\begin{array}{cc}
1 & b_n \\
0 & 1
\end{array}
\right)$

$ B^{n\plus{}1}\equal{}B\cdot B^n\Longleftrightarrow \left(
\begin{array}{cc}
1 & b_{n\plus{}1} \\
0 & 1
\end{array}
\right)\equal{}\left(
\begin{array}{cc}
1 & 3 \\
0 & 1
\end{array}
\right)\left(
\begin{array}{cc}
1 & b_n \\
0 & 1
\end{array}
\right)$

$ \Longleftrightarrow b_{n\plus{}1}\equal{}b_n\plus{}3,\ b_1\equal{}3$, yielding $ b_n\equal{}3\plus{}3(n\minus{}1)\equal{}3n$.

$ \therefore B^n\equal{}\left(\begin{array}{cc}
1 & 3n \\
0 & 1
\end{array}
\right)\ (n\geq 0)$

[quote="kunny"]The tangent line of $ y = x^n\ (n\geq 2)$ at $ x = 1$ is $ y = n(x - 1) + 1$, so we can write $ x^n = (x - 1)^2Q(x) + n(x - 1) + 1$ where $ Q(x)$ is a monic polynomial with [color=red]degree $ n-2.$[/color]

By Caley-Hamilton theorem, we have $ B^2 - 2B + E = \mathbb{O}\Longleftrightarrow (B - E)^2 = \mathbb{O}$,

$ \therefore B^n = (B - E)^2Q(B) + nB - (n - 1)E$, yielding $ B^n = nB - (n - 1)E\ (n\geq 0).$[/quote]

Excellent!!

I think a typo error only

Your second solution is also good

thanks for sharing ideas.

[quote="kunny"]The tangent line of $ y = x^n\ (n\geq 2)$ at $ x = 1$ is $ y = n(x - 1) + 1$, so we can write $ x^n = (x - 1)^2Q(x) + n(x - 1) + 1$ where $ Q(x)$ is a monic polynomial with degree $ n\geq 2.$

By Caley-Hamilton theorem, we have $ B^2 - 2B + E = \mathbb{O}\Longleftrightarrow (B - E)^2 = \mathbb{O}$,

$ \therefore B^n = (B - E)^2Q(B) + nB - (n - 1)E$, yielding $ B^n = nB - (n - 1)E\ (n\geq 0).$[/quote]

I don't understand this solution; why are we considering the tangent line of $ y = x^n$?

Ok, but what is the connection of $ y \equal{} x^n$ to the characteristic equation of $ B$, which is $ (\lambda \minus{} 1)^2 \equal{} 0$? Is it because $ B \equiv x$ due to the Cayley-Hamilton theorem?

By Cayley-Hamilton theorem, $ (B\minus{}I)^2 \equal{} O$. Now, kunny has considered the polynomial

$ x^n \equal{} (x\minus{}1)^2 Q(x) \plus{} n(x\minus{}1) \plus{} 1$.

Therefore, evaluating this polynomial for $ x \equal{} B$,

$ B^n \equal{} (B\minus{}I)^2Q(B) \plus{} n(B\minus{}I) \plus{} I \equal{} n(B\minus{}I) \plus{} I$.

Is this to say that variables and matrices are completely interchangeable in this context?

[quote="kunny"]If $ y \equal{} x^n$ touches to the line $ y \equal{} ax \plus{} b$ at $ x \equal{} 1$, then since the quation $ x^n \minus{} (ax \plus{} b) \equal{} 0$ has a double root $ x \equal{} 1$, we can write $ x^n \minus{} (ax \plus{} b) \equal{} (x \minus{} 1)^2Q(x)$ where $ Q(x)$ is a monic polynomial with degree $ n \minus{} 2$. That is to say, $ y \equal{} ax \plus{} b$ is the tangent line.[/quote]
Also, how did you get that $ a \equal{} n$ and $ b \equal{} 1 \minus{} n$ here?

[quote="Carcul"]By Cayley-Hamilton theorem, $ (B \minus{} I)^2 \equal{} O$. Now, kunny has considered the polynomial

$ x^n \equal{} (x \minus{} 1)^2 Q(x) \plus{} n(x \minus{} 1) \plus{} 1$.[/quote]

How was this polynomial obtained?


probability
function
combinatorics solved
combinatorics

Suppose we play a variant of Yahtzee where we roll 5 dice, trying to get all sixes. After each roll we keep the dice that are sixes and re-roll all the others. What is the probability of getting all sixes in exactly k rolls as a function of k?

My 1st post.
We can consider this problem with respect to each die.
The probability that one die will get a 6 somewhere before or at exact k times is 1/6+5/6x1/6+...+(5/6)^(k-1)x1/6=1-(5/6)^k
So the probability that all five dice will roll a 6 before or at exact k times is
(1-(5/6)^k)^5
Thus, the probability to get all 6 at exact k times is
(1-(5/6)^k)^5-(1-(5/6)^(k-1))^5

good mathematical model for artificial selection, btw. i think dawkins has used something analogous to this.


function
inequalities
algebra proposed
algebra

Let $ f(x)\minus{}x,f(x)\minus{}x^3$ are strict increase function on $ R$. Prove that: $ f(x)\minus{}\frac{\sqrt{3}x^2}{2}$ is strict increase function too.

[quote="tdl"]Let $ f(x) \minus{} x,f(x) \minus{} x^3$ are strict increase function on $ R$. Prove that: $ f(x) \minus{} \frac {\sqrt {3}x^2}{2}$ is strict increase function too.[/quote]

Consider two real numbers $ x$ and $ y$ with $ x>y$.
The two conditions yield $ f(x)\minus{}f(y)>x^3\minus{}y^3>0$ and $ f(x)\minus{}f(y)>x\minus{}y>0$.
Multiplying these two inequalities, applying $ (x^3\minus{}y^3)(x\minus{}y)>\frac34(x^2\minus{}y^2)^2$,
and taking the square-root yields the statement.


geometry
rectangle
trigonometry
integration
calculus
function
3D geometry

Solve Laplace's equation inside a rectangle:

$ \nabla ^2 u \equal{} \frac{\partial^2 u}{\partial x^2} \plus{} \frac{\partial^2 u}{\partial y^2} \equal{} 0$

subject to the boundary conditions

$ u(0,y) \equal{} g(y)$
$ u(L,y) \equal{} 0$

$ u(x,0) \equal{} 0$
$ u(x,H) \equal{} 0.$

Suppose a solution of the form $ u(x,y) = X(x)Y(y)$ then
\[ \frac {X''}{X} = - \frac {Y''}{Y} = \lambda^2
\]

\[ \Rightarrow \begin{cases} X'' - \lambda^2 X = 0 \\
Y'' + \lambda^2 Y = 0 \end{cases}
\]

\[ \Rightarrow \begin{cases} X = c_1 e^{\lambda x} + c_2e^{ - \lambda x} \\
Y = c_3 \sin \lambda y + c_4 \cos \lambda y \end{cases}
\]
The boundary conditions $ u(x,0) = 0$ and $ u(x,H) = 0$ demand $ Y(0) = Y(H) = 0$, so $ c_4 = 0$ and $ c_3 \sin \lambda H = 0$.  Thus we can have nontrivial solutions only if $ \lambda = \frac {n \pi}{H}$, $ n\in \mathbb{N}$.  Also, $ u(L,y) = 0$ requires $ X(L) = 0$ and hence $ c_2 = - e^{2\lambda L}c_1$, so
\[ \begin{cases} X = e^{\frac {n\pi x}{H}} - e^{\frac {n\pi (2L - x)}{H}} \\
Y = c_3 \sin \frac {n\pi y}{H} \end{cases}
\]
Hence the general solution must satisfy
\[ u(x,y) = \sum_{n = 1}^{\infty}{b_n\left(e^{\frac {n\pi x}{H}} - e^{\frac {n\pi (2L - x)}{H}}\right)\sin \frac {n\pi y}{H}}
\]
for some constants $ \lbrace b_n \rbrace_{n\in \mathbb{N}} \subset \mathbb {R}$.  To satisfy the boundary condition $ u(0,y) = g(y)$ we must have
\[ \sum_{n = 1}^{\infty}{b_n\left(1 - e^{\frac {2 n \pi L}{H}\right)\sin \frac {n\pi y}{H}} = g(y)
}\]
Multiplying by $ \sin\frac {n\pi y}{H}$ and integrating we have by orthogonality:
\[ b_n\left(1 - e^{\frac {2 n \pi L}{H}\right)\int_0^H{\sin^2 \frac {n\pi y}{H}dy} = b_n\left(1 - e^{\frac {2 n \pi L}{H}\right)\frac {H}{2} = \int_0^H{g(y)\sin \frac {n\pi y}{H}dy}
}}\]
and hence we must have $ b_n = \frac {2}{H\left(1 - e^{\frac {2 n \pi L}{H}\right)}}\int_0^H{g(y)\sin \frac {n\pi y}{H}dy}$
So our final solution is
\[ \boxed{u(x,y) = \frac {2}{H}\sum_{n = 1}^{\infty}{\frac {\left(e^{\frac {n\pi x}{H}} - e^{\frac {n\pi (2L - x)}{H}}\right)}{\left(1 - e^{\frac {2 n \pi L}{H}\right)}}\sin \frac {n\pi y}{H}\int_0^H{g(t)\sin \frac {n\pi t}{H}dt}}}
\]
Of course $ g$ needs to be sufficiently well behaved so the above converges, but the requirements are quite lax (e.g. if $ g\in L^p\left(\left(0,H\right)\right)$ is continuous it converges a.e. if $ p > 1$)

This is a fairly standard textbook question, if the textbook is for a first course in PDE. Joe Blow has given the standard answer, as that textbook would show it.

But why would mathwizarddude post such a question in the first place? Given the rapid and indiscriminate nature of his posting, I doubt that it's something he's thought long and hard about. I would like to see him slow down the rate at which he starts new topics. If topics from mathwizarddude begin disappearing, you'll know why.

I apologize for any inconvience.  Please bear with me as I don't have much mathematical background but merely an 8th grader just graduated from middle school.  I'm a self-taught math enthusiast.  I have taught myself calculus (not advanced) and am teaching myself linear algebra (am almost finish), complex analysis, and a little bit of real analysis.

Anyway, although I should be able to solve this standard separation of variables problem, I post this because I want to see if there's a different technique of solving this that is different my PDE textbook that I am currently using (as my PDE text is not advanced).  Also, if someone does post a solution that is the same as my text, I would want to take that opportunity to ask the reason for arbitrarily choosing $ u(x,y)$ to has the form $ X(x)Y(y)$.  Moreover, although we could come up with a valid solution in that particular form, how would one prove that there's no other solution $ u(x,y)$ satisfying all the constraints but is not in the form $ X(x)Y(y)$ and thus not losing any generality in our solution.

(However, as PDE is different from ODE, perhaps one can stop as long as one comes up with a valid solution, though not a general one perhaps, [i]I think[/i].)

Those are good questions.  The truth is, mathematically there isn't very much motivation for looking for a solution of the form $ u(x,y) \equal{} X(x)Y(y)$.  A physicist might argue that there are reasons solutions 'ought' to be of this form (separable), but from a mathematician's point of view this ansatz is made simply because prior experience has taught us that many PDEs from physics have solutions in this form, and it's fairly easy to check if it works.

For this particular equation, however, it turns out that if you find [i]any[/i] solution to the boundary value problem, you have also found [i]the[/i] solution to that problem, i.e. the form of the PDE itself guarantees at most one solution.  The proof is as follows:

Take any two solutions $ v_1(x,y), v_2(x,y)\in C^2(U)\cap C^0(\overline{U})$ of Laplace's equation in a bounded open region $ U$ with boundary $ \partial U$, i.e.
\[ \begin{cases} \nabla^2 u(x,y) \equal{} 0 & (x,y)\in U \\
u(x,y) \equal{} g(x,y) & (x,y)\in \partial U \end{cases}
\]
Now the difference $ v \equal{} v_1 \minus{} v_2$ must satisfy
\[ \begin{cases} \nabla^2 v(x,y) \equal{} 0 & (x,y)\in U \\
v(x,y) \equal{} 0 & (x,y)\in \partial U \end{cases}
\]
However, since $ v$ is harmonic, the maximum principle for harmonic functions asserts that both its maximum and its minimum are obtained on the boundary of $ U$.  Therefore, $ v \equal{} 0$ and hence $ v_1 \equal{} v_2$ in $ \overline{U}$, so there is at most one solution to the boundary value problem. [hide="Remark"]The argument above also immediately proves the uniqueness of solutions to Poisson's equation $ \nabla^2 u(x,y) \equal{} f(x,y)$.[/hide]

Note that we did not assume a whole lot about $ U$ to prove uniqueness.  Though existence is in general a somewhat more complicated issue (and depends much more intimately on the specific geometry of $ U$), on the rectangle we can prove existence by explicitly constructing a solution as I have in a previous post.  So in some sense even though the ansatz seems rather arbitrary, in this case it actually leads to a rigorous proof of both existence and uniqueness.  This is a situation that prevails throughout PDE analysis.

In a more general PDE setting, questions of existence, uniqueness, regularity, well-posedness, etc. are extremely complicated, and there are many open questions in these areas of ongoing research.  As you correctly pointed out, the situation with PDEs is very different from ODEs; even very simple linear PDEs can exhibit nasty pathological behavior (e.g. Laplace's equation in a punctured sphere, Lewy's example of an unsolvable linear PDE with smooth Cauchy conditions, non-uniqueness of homogeneous heat equation(!), etc).  For a good overview of theoretical questions in PDEs, I'd recommend [i]Partial Differential Equations[/i] by L. C. Evans.  Though before you dive into PDE theory, you'd need a strong foundation in functional analysis and real analysis as background.

[quote="JoeBlow"]For a good overview of theoretical questions in PDEs, I'd recommend [i]Partial Differential Equations[/i] by L. C. Evans.  Though before you dive into PDE theory, you'd need a strong foundation in functional analysis and real analysis as background.[/quote]

this is a great text and i think i may have used it more than any other math text i own.  you don't need much (any?) functional analysis to read the first half of the book though (well, maybe a bit for the part on second order elliptic pde), which is what most one semester graduate courses on pde cover.

Thank you very much!  :D

(I don't know who rated those posts though..)

The kid is in 8th grade.  mathwizarddude: Do not touch Evans.

Here is my best explanation for why the separation of variables technique is a reasonable technique.

Suppose someone shows you Laplace's equation and asks you to write down as many solutions as possible, without regards for boundary conditions or anything like that.  The first thing you would do is look for solutions of a simple form.  For example, you might look for polynomial solutions.  Or you might look for radially symmetric solutions (i.e. ones that depends on $ r$ but not on $ \theta$).  Or you might look for ones that depends on $ x$ but not on $ y$ (not very interesting).  Eventually, you would try ones that are a function of $ x$ times a function of $ y$, the "separated" solutions.  You would see that these take a relatively simple form.

Next you would notice that adding any two solutions gives another solution (i.e. Laplace's equation is linear).  So now you have tons of solutions at your disposal.  In particular, you have all linear combinations of separated solutions, and if you're not antsy about convergence issues, all convergent infinite linear combinations of these separated solutions.  That's a lot of solutions.

Now you worry about boundary conditions, and you notice that adding any two solutions satisfying the "vertical" boundary conditions gives a solution still satisfying the vertical boundary conditions.  Eventually, you end up doing the same procedure described in JoeBlow's post.

This still brings us to the question:  Why does this work?

Here's a slightly more sophisticated way to explain the same old solution, but you need to understand Fourier series first (which is closely related to what you're doing anyway).  One nice aspect is explanation generalizes to slightly more complicated problems (and may be discussed in your book).

Once you understand Fourier series a bit, you realize that essentially "any" (here I'm lying of course) function that satisfies your boundary conditions can be expressed as a Fourier sin series in the $ y$ variable, where the coefficients themselves are functions of $ x$.  (This can be tricky to think about at first, so try to mull over why this makes sense.)  Then you "plug" the Fourier series into Laplaces equations, which gives you 2nd ODEs for the coefficient functions (which are functions of $ x$).  Then the rest of your boundary conditions allow you to solve the ODEs.  Sorry for the lacking details.  If you can successfully execute this technique based only on what I've written here, I'll be very impressed.  But the point is that from this point of view, there's no ansatz, just Fourier series, and we *expect* to get to the answer (or if there were no solution, we'd see why).  The only problem is that it begs the question: Why use Fourier series?

Note: Polynomial solutions and rotational solutions to Laplace's equation turn out to be very interesting also.  In fact, they're perhaps more interesting than the separated solutions.  A general philosophy is that any simple solution is probably an important solution.


geometry
circumcircle
projective geometry
geometry proposed

I was told it's from Hong Kong 2003, but I'm not sure.
Looks messy, but it's pretty easy!

Two circles $\gamma_1$ and $\gamma_2$ meet at $A$ and $B$. Let $r$ be a line through $B$ that meets $\gamma_1$ at $C$ and $\gamma_2$ at $D$, such that $B$ is between $C$ and $D$. Let $s$ be the line parallel to $AD$, which is tangent to $\gamma_1$ in $E$ and has the minimal distance from $AD$. $EA$ meets $\gamma_2$ in $F$, and let $t$ be the line through $F$ which is tangent to $\gamma_2$. Prove that:
a) $t \parallel AC$.
b) $r$, $s$ and $t$ are concurrent.

For (a):

We have $\angle ECA=\angle DAF,\ \angle AEC=\angle AFD$, so $\angle CAE=\angle ADF$, which is what we want because the angle made by $t$ with $AF$ is the same as $\angle ADF=\angle CAE$.

For (b), assume $R=r\cap AE,\ T=r\cap t,\ S=r\cap s$. We have to show that $T=S$. We have $\frac {RD}{RS}=\frac {RA}{RE},\ \frac {RC}{RT}=\frac {RA}{RF}$. We thus need $\frac{RD}{RC}=\frac {RF}{RE}$. This is true because the lines $EC,\ FD$ are parallel (because $\angle AEC=\angle AFD$).

Yes!   :)
You could also prove point b) in a more "geometric" way, which I personally prefer. In fact, both $r\cap t$ and $r\cap s$ lie on the circumcircle of $BDE$. You can easily prove this with the angles you already have from point a).

I think,it is from Tournaments of towns 2001(I don't know spring or falls)

I checked all the problems from TOT 2001 and 2002, but I couldn't find this one.   :?

Well it is Turnament of the Towns 2002 Autumn problem 5
Circles C1 and C2 meet at A and B. A line through B meets C1 at K and C2 at M. T1 is a tangent to C1 parallel to AM. It meets C1 at Q. The line AQ meets C2 again at R. Show that the tangent T2 to C2 at R is parallel to AK, and that T1 and T2 intersect on the line KM.

Oh, thanks! I didn't check well enough!   ;)

Dear Mathlinkers,
BE intersected (2) en X.
Appliying the Reim's theorem, FX//s.
Tf is the tangent to (2) at F.
According to a special case of Pascal's theorem Tf FADBXF, we are done...
Sincerely
Jean-Louis


Problem 1:What value of (x,y) satisfy the following equation:

                 (4X^2+6X+4)(4Y^2-12Y+25) = 28

Problem 2: Prove that if x and y are real numbers, then (X^2+Y^2) / 2 >= XY

p.s. how do i write square roots, squares, division, inequality signs, etc. :?  :?  :?

You can use Latex to write mathematical text (there's a link on this site explaining Latex).  Basically, you put $\$$ signs around Latex code, and it becomes an image.  For example, if you put $\$$ signs around (4X^2+6X+4)(4Y^2-12Y+25) = 28, you see $(4X^2+6X+4)(4Y^2-12Y+25) = 28$.
[hide]Problem 1: Fix $x$ or $y$ and then solve for the other variable.  By the way, Ceva Theorem, do $x$ and $y$ have to be real?
Problem 2: A square of a real number is always greater than or equal to zero, i.e. $(x-y)^2\geq 0$ for real $x$ and $y$.[/hide]

[quote="hydro"][quote="Ceva Theorem"]Problem 2: Prove that if x and y are real numbers, then (X^2+Y^2) / 2 >= XY[/quote]

$x^2+y^2 \geq 2xy \implies (x-y)^2 \geq 0$, always true[/quote]Just a small point, but you're going the wrong way in your implication.

Yes, they are real numbers

how do i do square roots?

[quote="Ceva Theorem"]
p.s. how do i write square roots, squares, division, inequality signs, etc. :?  :?  :?[/quote]

you have to put your expressions between two of the sign $. So squares are $x^2$, square roots are $\sqrt x$, divisions $\frac{x}{y}$

[quote="hydro"][quote="Ceva Theorem"]Problem 2: Prove that if x and y are real numbers, then (X^2+Y^2) / 2 >= XY[/quote]

$x^2+y^2 \geq 2xy \implies (x-y)^2 \geq 0$, always true[/quote]

you can also prove this by using AM-GM inequality
 :)

Actually the proof of AM-GM uses that fact so that would be circular logic. ;) And anyway, AM-GM is only valid for positive reals.

[hide="2"]
Consider the inequality $(x-y)^2=x^2-2xy+y^2\ge 0$ since the square of a real number is always nonnegative. Adding $2xy$ to both sides and dividing by two gives $\frac{x^2+y^2}{2}\ge xy$, as claimed. $\boxed {}$
[/hide]


I just started number theory, but i found that if one sums all the numbers coprime to some n greater than 0 and Є I, but less than n - then you divide it by n and multiply it by 2 this equals to ф(n) (at least but to n=35). Could someone explain to me why??

The set $ S_n$ of natural numbers coprime to and less than $ n$ has cardinality $ \varphi(n)$ by definition.  Moreover, what cosinator is trying to tell you is that if $ a \in S_n$ then necessarily $ n \minus{} a \in S_n$.  This means that we can pair up the elements of $ S_n$ into $ \frac{ \varphi(n) }{2}$ pairs, each of which has sum $ n$, which is what you discovered.  For example,

$ S_{12} \equal{} \{ 1, 5, 7, 11 \}$

and $ 1 \plus{} 11 \equal{} 5 \plus{} 7 \equal{} 12$.


inequalities
algebra
polynomial
inequalities open

Any idea to solve it? Or using computer to disclaims it?  :lol:  Where is Arquady? I believe you  :D

Hm, i think i have a solution for the second. But it's so long.

Yes ! I can  prove the second inequality but can't do with the first

as discussed in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111842]here[/url] the  inequality $(a^{3}+b^{3}+c^{3})^{2}\geq (a+b+c)(a^{4}b+b^{4}c+c^{4}a)$ is true.
I don't have a nice proof of this fact.

Ican exprees two expreesion by SOS
 and ithing it can be solve by SOS
 but it so long and may be tomorrow i post it

here is my exprees for the first inq:$LHS = \sum S_{a}(b-c)^{2}$
 where $S_{a}=7b^{4}-c^{4}-3b^{2}c^{2}+2b^{3}c-2c^{3}b+3a^{3}b+3a^{3}c-3abc(b+c)$
           $S_{b}$ and $S_{c}$ are similarly cyclic on a,b,c .
    I can check that when $a\geq c\geq b$ the inq are true , when $a\geq b\geq c$ I can check $S_{a}$ and $S_{c}\geq 0$ but i can't do any more

Hm, i think you'd better not use S.O.S to prove this ineq. It isn't a nice solution and maybe very long. My solution haven't used S.O.S

So you can poat your soln Zeikii??? Thanks

Stop discussing without any result,  :lol:
Now Zeikii, if you solve it, please post. Part (b) is true, but I have no time to spend on checking part (a).

[quote="Flove"]My friend asks me about the inequality

Prove or disprove that, for all non-negative real numbers $a,b,c$? \[(a^{3}+b^{3}+c^{3})^{2}\ge 2(a^{5}b+b^{5}c+c^{5}a)+abc(a^{3}+b^{3}+c^{3})\] and I believe that (although I haven't any try \[(a^{3}+b^{3}+c^{3})^{2}\ge a^{5}b+b^{5}c+c^{5}a+2abc(a^{3}+b^{3}+c^{3})\] [/quote]

Should I ask VASC to check part (a)? I belive that 50% it's true, 50% it's false  :lol:  But I don't know computer to check it.  :roll:

Yes, it's very perceptable, Zeikii. Congulatulation!!!  :)

I tried to prove the first one by similar way as Zekii but false. Can anybody help me? Jichen?

What an awesome problem. How can you do that? Don't tell me that you do it all without computer :blush:

He must use computer of course.  :P  Human cann't do that !!! Anyway, I believe that there is a better way.

[quote="Flove"]
Prove or disprove that, for all non-negative real numbers $a,b,c$? \[(a^{3}+b^{3}+c^{3})^{2}\ge 2(a^{5}b+b^{5}c+c^{5}a)+abc(a^{3}+b^{3}+c^{3})\] [/quote]
Yes it's true and we can prove this inequality without computer (maybe with calculator).


calculus
integration
LaTeX
trigonometry
calculus computations

With respect to what variable?

OK, I know the answer- but that [b]is[/b] an incomplete statement.
Also, you need dollar signs for $ \text{\LaTeX}$; I put those in for you.

if the variable is x jus rationalize the denominator my multiplying and dividing  with   $ \frac {\sqrt {1 \plus{} x} \plus{} \sqrt {1 \minus{} x}}{\sqrt {1 \plus{} x} \plus{} \sqrt {1 \minus{} x}}$ .

hello, write your integral in the form
$ \int\frac{1}{x}\,dx\plus{}\int\frac{\sqrt{1\minus{}x^2}}{x}\,dx$.
Sonnhard.

I got last step, but I don't know how to integrate
 $ \int\frac{\sqrt{1\minus{}x^{2}}}{x}\,dx$

jus put it as $ \int\frac {\sqrt {1 \minus{} x^{2}} * {x} }{x^{2}}\,dx$.and now put $ \ 1 \minus{} x^{2}$ as $ \ t^2$.and integrate

hello, setting $ t\equal{}\sqrt{1\minus{}x^2}$ we get $ dx\equal{}\frac{tdt}{\minus{}x}$ and our integral is now
$ \int\frac{\sqrt{1\minus{}x^2}}{x}\frac{tdt}{\minus{}x}\equal{}\int\frac{t^2}{\minus{}x^2}\,dt\equal{}\int\frac{t^2}{t^2\minus{}1}\,dt$.
Sonnhard.

$ I \equal{} \int\frac {\sqrt {1 \plus{} x} \plus{} \sqrt {1 \minus{} x}}{\sqrt {1 \plus{} x} \minus{} \sqrt {1 \minus{} x}}dx$

substitute $ x \equal{} \sin 2\theta \ \Rightarrow \ dx \equal{} 2 \cos 2\theta \ d\theta$

notice that $ \sqrt{1\plus{}x} \equal{} \sqrt{\sin ^2\theta \plus{} \cos ^2\theta \plus{} 2 \sin \theta \cos \theta} \equal{} \sqrt{(\sin \theta \plus{} \cos \theta)^2} \equal{} \sin \theta \plus{} \cos \theta$

and $ \sqrt{1\minus{}x} \equal{} \sqrt{\sin ^2\theta \plus{} \cos ^2\theta \minus{} 2 \sin \theta \cos \theta} \equal{} \sin \theta \minus{} \cos \theta$

$ I \equal{} 2 \int \frac{\sin \theta \plus{} \cos \theta \plus{} \sin \theta \minus{} \cos \theta}{\sin \theta \plus{} \cos \theta \minus{} \sin \theta \plus{} \cos \theta} \cos 2\theta d\theta \\ \\ \\ \equal{} 2 \int \frac{\sin \theta}{cos \theta}(2 \cos ^2 \theta \minus{} 1 ) d\theta$

$ \equal{} 2 \int (\sin 2\theta \minus{} \tan \theta) d\theta$

hope you can carry on from here ...

I got .5ln(x).
Is that the right answer.

No.

You can always test a proposed antiderivative by differentiating it. We're not integrating $ \frac1{2x}$ here.

$ \int\frac {\sqrt {1 \plus{} x} \plus{} \sqrt {1 \minus{} x}}{\sqrt {1 \plus{} x} \minus{} \sqrt {1 \minus{} x}}\ dx$

$ \equal{} \int \frac {1 \plus{} \sqrt {\frac {1 \minus{} x}{1 \plus{} x}}}{1 \minus{} \sqrt {\frac {1 \minus{} x}{1 \plus{} x}}}\ dx$

Let $ \tan \theta \equal{} \sqrt\frac {1 \minus{} x}{1 \plus{} x}\Longrightarrow x \equal{} \cos 2\theta$, the integral can be written as

$ 2\int \frac {\cos \theta \plus{} \sin \theta}{\cos \theta \minus{} \sin \theta}\sin 2\theta\ d\theta$

@ Nora.91

[quote="Nora.91"]$ I \equal{} \int\frac {\sqrt {1 \plus{} x} \plus{} \sqrt {1 \minus{} x}}{\sqrt {1 \plus{} x} \minus{} \sqrt {1 \minus{} x}}dx$

substitute $ x \equal{} \sin 2\theta \ \Rightarrow \ dx \equal{} 2 \cos 2\theta \ d\theta$

notice that $ \sqrt {1 \plus{} x} \equal{} \sqrt {\sin ^2\theta \plus{} \cos ^2\theta \plus{} 2 \sin \theta \cos \theta} \equal{} \sqrt {(\sin \theta \plus{} \cos \theta)^2} \equal{} \sin \theta \plus{} \cos \theta$

and $ \sqrt {1 \minus{} x} \equal{} \sqrt {\sin ^2\theta \plus{} \cos ^2\theta \minus{} 2 \sin \theta \cos \theta} \equal{} \sin \theta \minus{} \cos \theta$.[/quote]

Is $ \sqrt{A^2}\equal{}A?$

:oops_sign:

thanks kunny .

i guess my method isn't valid!

Isn't it negative?
$ \minus{}2\int\frac{\cos\theta\plus{}\sin\theta}{\cos\theta\minus{}\sin\theta}\sin 2\theta\ d\theta$
Does anyone has final answer to the question?

We have:
$ \frac{\cos\theta\plus{}\sin\theta}{\cos\theta\minus{}\sin\theta}\sin 2\theta\ \equal{} \frac{\cos\theta\plus{}\sin\theta}{\cos\theta\minus{}\sin\theta}\sin 2\theta\ \times \frac{\cos\theta\plus{}\sin\theta}{\cos\theta \plus{}\sin\theta} \equal{} \tan 2\theta \plus{} \sec 2\theta \minus{} \cos 2\theta$


A $ 6\%$ rate increase by a local media cable company resulted in an
increase of $ \$1.20$ per month on a family's bill. How many dollars
was the monthly bill before the increase?

let the monthtly bill before the increase be x.

$ x*\frac{6}{100}\equal{}1.2 \implies 6x\equal{}120 \implies x\equal{}20$

answer : 20 dollars


The number $ n$ is randomly selected from the set $ \{1, 2, ..., 10\}$,
with each number being equally likely. What is the probability
that $ 2n\minus{}4>n$ ? Express your answer as a common fraction.

We can substitute different numbers to find which ones work.

$ 2n \minus{} 4 > n$

Let's try 1. We get: $ 2(1) \minus{} 4 > 1$. Simplified, this is $ \minus{} 2 > 1$, which is obviously false.
How about 2? Our result is $ 0 > 2$, false.
Now, 3. $ 2 > 3$, still false.
What about 4? $ 4 > 4$, which is false.
Now let's try 5. $ 6 > 5$, which is true!
When we put in 6, we get: $ 8 > 6$, true again!

If we continue substituting values up to $ 10$, we find that they all make the inequality true. So the set $ \{5,6,7...10\}$ all work. The probability of picking a number $ n$ in this set out of the the numbers 1 through 10 is $ 5/10$ or $ 1/2$.

there are total 6 numbers that satisfies the equation $ 2n\minus{}4>n (5\sim 10)$

so the answer should be $ \frac{6}{10}$.

Answer : $ \frac{3}{5}$

[quote="FantasyLover"]there are total 6 numbers that satisfies the equation $ 2n - 4 > n (5\sim 10)$

so the answer should be $ \frac {6}{10}$.

Answer : $ \frac {3}{5}$[/quote]

Neither of you really explained the best way...

As in equations, the addition and subtraction properties also hold here...

Therefore we simplify the inequality to $ n > 4$

Thus our desired subset is $ \{5, 6, 7,\ldots,10\}$

${ \frac {3}{5}}$

I was just skipping extremely easy part...

who doesn't know n>4 if 2n-4>n...

Not a lot of people...I knew that, and I'm not exactly the best person here...

And the mistake I made on the solution I got was because I overlooked the fact that it's easier to just simplify the inequality.


Given that $2^{mn}-1$ is divisible by $(2^{m}-1)(2^{n}-1)$. Prove that $2(3^{mn}-1)$ is divisible by $(3^{m}-1)(3^{n}-1)$.

Question: Does this force $(m, n) = 1$?

I do not have a proof for both questions. Please help.

Nevermind, I got it. The second claim is true. (that (m, n) =1 ) and the first claim follows.

Lemma1: $(3^{m}-1;3^{n}-1)=3^{(m;n)}-1$
NOTES:$3^{mn}-1$ is divisible$3^{m}-1$ and $3^{n}-1$
From that, we have to prove that (m;n)=1.Try doing it .If you can't,i 'll post my solution tomorrow :)

minhtoan: Search for "masterpiece", author Harazi ;)

[quote="minhtoan"][quote="tmbtw"]
i 'll post my solution tomorrow :)[/quote]
Okie, I waiting your solution, tmbtw.
I also have a problem similarly: Let $a, b$ are positive integers such that $a^{n}-1|b^{n}-1$. Prove that $a=b^{k}$, for some integer $k$[/quote]
OK,Here is my solution :
Let d=(m;n)
We have $(2^{m}-1;2^{n}-1)=2^{d}-1$
So,$2^{d}-1$|$2^{n}-1$|$\frac{2^{mn}-1}{2^{m}-1}=(2^{m})^{n-1}+...+2^{m}+1$
Then,$2^{d}-1$|n
Similarly,$2^{d}-1$|m.So,$2^{d}-1$|d.So,d=1 .We're done :)

we can also generalize this question as :

if $a^{mn}-1$ is divisible by $(a^{m}-1)(a^{n}-1)$ prove $((a+1)^{mn}-1)a$ is divisible by $((a+1)^{m}-1)((a+1)^{n}-1)$


algebra
polynomial
trigonometry
algebra unsolved

[quote="bgbgbgbg"]prove that:
tan^2 (p / 7) + tan^2 (2p / 7) + tan^2 (3p / 7) = 21[/quote]

Let $ t\equal{}\tan(x)$

$ \tan(2x)\equal{}\frac{2t}{1\minus{}t^2}$, $ \tan(3x)\equal{}\frac{t(3\minus{}t^2)}{1\minus{}3t^2}$, $ \tan(4t)\equal{}\frac{4t(1\minus{}t^2)}{t^4\minus{}6t^2\plus{}1}$ and $ \tan(7x)\equal{}\frac{\minus{}t^7\plus{}21t^5\minus{}35t^3\plus{}7t}{\text{some ugly polynomial}}$

So $ \{\tan(\frac{k\pi}7),k\equal{}0,1,2,3,4,5,6\}$ are the seven roots of $ t^7\minus{}21t^5\plus{}35t^3\minus{}7t\equal{}0$

So $ \{\tan(\frac{k\pi}7)^2,k\equal{}1,2,3\}$ are the three roots of $ t^3\minus{}21t^2\plus{}35t\minus{}7\equal{}0$

Hence the result.


geometry
inequalities
geometric transformation
reflection

Triangle $ABC$ is right angled at $A$. Let $D$ be a point on $\overline{BC}$. Let $E$ and $F$ be the reflections of $D$ across $AC$ and $AB$. Prove that $(ABC)\ge (DEF)$. Find all positions of $D$ for equality.

[asy]draw((0,0)--(60,0)--(0,40)--cycle); draw((24,24)--(-24,24)--(24,-24)--cycle); pair A,B,C,D,E,F; A=(0,0); B=(60,0); C=(0,40); D=(24,24); E=(-24,24); F=(24,-24); label("A",A,SW); label("B",B,SE); label("C",C,NW); label("D",D,NE); label("E",E,NW); label("F",F,SE);[/asy]

Let $AB=c, BC=a, CA=b, CD=x, DE=y, DF=z$. Then by simple similarity:

$y={2cx\over a}, z={2b(a-x)\over a}$

Hence $[DEF]={2bcx(a-x)\over a^{2}}$

Trivially (by AM-GM), $x(a-x)\leqslant{a^{2}\over 4}$, hence $[DEF]\leqslant{bc\over 2}=[ABC]$.

The equality holds iff $x={a\over 2}$, i.e. iff $D$ is the midpoint of $BC$.


Let $ x,y,z$ are positive real numbers,$ k \geq 2$,prove that:
$ \frac{2}{k\plus{}2} \geq  \frac{x}{kx\plus{}y\plus{}z}\plus{}\frac{x^{\minus{}1}}{kx^{\minus{}1}\plus{}y^{\minus{}1}\plus{}z^{\minus{}1}}$

$ <\equal{}>
\frac{(zx^2\plus{}yz^2\minus{}4xyz\plus{}x^2y\plus{}y^2z)k\plus{}2x(y\minus{}z)^2}{(k\plus{}4)(kx\plus{}2x\plus{}y\plus{}z)(kyz\plus{}2yz\plus{}zx\plus{}xy} \geq 0$

easy.
BQ

$ n \geq 2,k \geq n \minus{} 1$,
$ \frac {\sum_{i \equal{} 2}^n{x_i}}{kx_1 \plus{} \sum_{i \equal{} 2}^n{x_i}}\plus{}\frac {\sum_{i \equal{} 2}^n{x_i^{\minus{}1}}}{kx_1^{\minus{}1} \plus{} \sum_{i \equal{} 2}^n{x_i^{\minus{}1}}} \geq \frac {2(n \minus{} 1)}{k \plus{} n \minus{} 1}$
BQ


inequalities
calculus
derivative
geometry
geometric transformation
reflection

Let $a,b,c$ be non-negative real numbers satisfying that $a+b+c=3$. For each positive real numbers $k$, find the maximum value of \[P=a^{2}(kb+c)+b^{2}(kc+a)+c^{2}(ka+b).\] Another part of the above problem is to consider it in case $k\le 0$ (which I just think when I write this problem now, but I'm sure it can be solved).

Solution for the first problem is ery nice and... no hard.

Is there a nicer way than just multiply out and use Am-Gm (muirhead) and maybe Cauchy for squares...?

I believe that there is no such way. At least using AM-GM or Muihard directly is impossible,

My solution is called "the entirely mixing variables method". It is very new, but effective for so many problem,

Moreover, there's no problem if $k\le 0$.

[quote="Flove"]For a hint, firstly you should check a small case, that is $min(a,b,c)=0$.[/quote]

How do you solve it, Anh Cuong? I wait it.  :P

You min the case when one variable is equal to $0$. Oke, i think this is also important to obtain the result.
Let assume that $a=0$ and remain the problem:
Give $b,c \geq 0$ satisfying $b+c=3$. Find the maximum value of:
$P= kb^{2}c+c^{2}b=bc(kb+c)$. Oke take $m=\frac{k(k+\sqrt{k^{2}-{k}+1})}{k-1+\sqrt{k^{2}-k+1}}$
and $n = k+\sqrt{k^{2}-k+1}$. Then apply AM-GM here:
$mnP = mb.nc.(kb+c) \leq (\frac{mb+nc+kb+c}{3})^{3}= (\frac{(n+1)(b+c)}{3})^{3}= (n+1)^{3}$
From here imply that:
$P \leq \frac{(n+1)^{3}}{mn}$
Opps, too lazy to find the equality, and more more thing, this quick proof only apply for $k \geq 0$. :)

Your solution is nice, although my solution is only to use derivative, not claim any brainpower. Now, show we how to prove that, it's enough to check the case $\min(a,b,c)=0$. As I tell above, I use the (entirely) mixing variables method, quite easy. I wonder if there're other solutions.  :lol:
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This post was edited after the 1000s post. About "the entirely mixing variables method", you should look over Mathematical Reflection, also my article there

http://reflections.awesomemath.org/2006_5/2006_5_entirelymixing.pdf

After reading it, I think my problem above is much easier.  :lol: Enjoy it!


let $f$ be a convex shape inside a circle $C$, such that from any point on $C$, the shape $f$ is seen by an $90^{\circ}$ angle. prove that $f$ is symmetric from $O$ (center of $C$).
[hide][b]what is the English word for O for f , when f is symmetric from O :?: [/b][/hide]

Given a direction in the plane, there are exactly two lines having that direction which meet a convex (and compact; I assume our set is compact and that it's not contained in a line) only in border points.

Now fix a direction $d$, and take a line $\ell\|d$ which meets our convex set in border points. This line cuts the circle in two points, $A,D$. There is a line through $A$ which is $\perp d$ which meets the set in border points, since the set is seen under an angle of $\frac\pi 2$ from $A$. This line cuts the circle again in $B$. There's also a line $\ell'\|d$ through $B$ which cuts the set only in border points. $\ell$ and $\ell'$ must be symmetric wrt $O$, so we have shown that for each direction, the two lines mentioned in the first paragraph corresponding to that direction are symmetric wrt $O$.

The set is the intersection of all the (closed) bands contained between lines of the type $\ell,\ell'$ as described above, so, since each such band is symmetric wrt $O$, the set must also be symmetric wrt $O$.