linear algebra
matrix
inequalities
group theory
abstract algebra
triangle inequality
superior algebra

Let $ \chi$ be an irreducible character of G. Show that if $ \lambda$ is any $ |G|$ th root of unity, then $ \{x \in G | \chi(x) \equal{} \lambda \chi(1) \} \unlhd G$.

Any hint?

Thanks.

That notation is rather tricky to untangle.* Here's what I think it means:

Let $ \chi$ be an irreducible character (the trace of an irreducible representation $ \rho$) on a finite group $ G$ and $ \lambda$ satisfy $ \lambda^{|G|}\equal{}1$. The set of $ x$ for which $ \chi(x)$ is equal to $ \lambda\cdot \chi(e)$ (where $ e$ is the identity) is normal, in the sense that it's fixed by conjugation (it's not usually a subgroup).
We can replace that $ \chi(e)$ by $ n$, where $ n$ is the dimension of the representation. We're working over $ \mathbb{C}$, so all eigenvalues of any matrix $ \rho(x)$ must be roots of unity; if they weren't, the group would have to be infinite. By the triangle inequality, $ |\text{tr }\rho(x)|\equal{}|\chi(x)|\le n$, with equality only if all of those eigenvalues are the same. When this happens, we can further see that $ \rho(x)$ is a multiple of the identity; if it were $ \lambda I\plus{}N$ for some nonzero nilpotent $ N$, it would have infinite order.
So then, the set we're looking at reduces to $ \{x: \rho(x)\equal{}\lambda I\}$. Since $ \lambda I$ is in the center of $ GL_n(\mathbb{C})$, we have $ \rho(yxy^{\minus{}1})\equal{}\rho(y)\rho(x)\rho(y^{\minus{}1})\equal{}\rho(y)\cdot \lambda I\cdot (\rho(y))^{\minus{}1}\equal{}\lambda I$, and the set is normal. "Irreducible" is irrelevant.

Disclaimer: I have never actually studied representation theory.

*One of the problems is that symbol $ \chi$. It sits lower than most other symbols, so it can often look like a subscript of whatever's before it. I strongly recommend using a $ \cdot$ for multiplication in that situation.

Probably the OP misstated his quantifiers.  Z(χ) = { g in G : χ(g) = λ⋅χ(1) for λ some root of unity in C } is a normal subgroup of G, and if χ is irreducible, then Z(χ)/Ker(χ) = Z(G/Ker(χ)), where Ker(χ) = { g in G : χ(g) = χ(1) }.  Z(χ) is the preimage under ρ of Z(GL(n,C)) and Ker(χ) is the kernel of ρ.

Some basic properties of Z are discussed on pages 26ff of Isaacs's Character Theory textbook.


I have no programming experience. Can anyone recommend web or other sources of math problems that are related to computing skills but don't actually require any programming knowledge - as practice for a fairly math-capable junior high schooler prior to attending a programming course. I am not sure what types of problems would be most relevant.
Thank you.

I can't really help you much, but exactly how much experience do you have (none was probably an exaggeration)?

You might get more replies in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=331]Computer Science[/url] forum. :)

I'm not quite sure if this is what you're looking for but you may try http://projecteuler.net/ .


We haven't yet set the prize criteria for the USAMTS.  Do any of you participants from past years have any opinions on the current structure?

I think AoPS should keep the prize criteria the same to ensure that a winner from one year is the same as a winner from another year, i.e. if the cutoff for "gold winner" were dropped from 96 to 90 or something, the value of the award would be greatly made lesser in value.

I think that the general criteria should still hold, but each level should be further divided into same level. I mean, there should be a 1st, 2nd, 3rd level gold, and so on down. This way, those students who scored say 95 and get a silver are different from those who got 75.

Is there a preference for relative versus absolute levels (e.g. would you rather see a 96 = gold or a top 2% = gold).

[quote="Spoon"]I think a percentage level ranking would be the best, that way students know exactly where they rank... so a student that gets a 81 knows exactly how well he did relative to other students.[/quote]

We can provide a percentage ranking distinct from the awards - what should the awards be based on?  And how would you optimally set the award levels if you used percentages?

[quote="ComplexZeta"]Do you have any statistics from recent years on the percentage of students who get gold, silver, and bronze? I think what the AMC does (100 or top 5% on AMC12, 120 or top 1% on AMC10 go to AIME) is very sensible.[/quote]

Good point - this doesn't need to be an either-or.

About how many people participate in the USAMTS? And what does a score of 75 or so correspond to in terms of percentile?

I like the idea of keeping the traditional award structure exactly as it was, based on the same criteria, but also adding percentile awards (with perhaps one cut-off corresponding exactly to AIME qualification) for the additional information that provides.

From Year 15

1484 submitted round 1
86 Gold
212 Silver
177 Bronze

when do we get our prizes for yr 15? Or did everyone else get them already except me?

The NSA is still handling the prizes for Year 15; I believe they should be coming soon.

Thanks a lot. Is latex format required for next year?

No.  Please read this: [url]http://artofproblemsolving.com/Forum/viewtopic.php?t=6145&postdays=0&postorder=asc&&start=0[/url]

Percentages showing how you did couldn't be all that accurate though, because you'd have people how submitted only some of the rounds clogging the bottom. Last year I only submitted the first two rounds :blush: , and then I said, eh, I'm too busy. So I'd probably be in the bottom percentiles.

Perhaps cutoffs could be established based on percentages of people who submitted all four rounds.  This way students who missed rounds wouldn't "clog up the bottom", but they could still win awards (although it would be difficult) if they did exceptionally well.

[quote="pakagawa"]When is the problems from year 16 going to be graded?[/quote]

We expect to have the first round graded by the last week of October.

I think that a points based system is superior because it rewards people for their own ability, not for their relative ability.  Points incourage people to think constructively instead of competatively.

EDIT:

It would be even better to use both systems, to make sure someone always wins.  That is what the AMC does.

Still, I want to know where I stand in the country...

You'll have stats galore on the Leaders pages when the papers are graded.

Speaking of stats... what statistics are actually going to be on the leaders page once the first round is graded?

Wait. But if this is all about this this year, what about last year?

I didn't do anything online last year.

I'm not experienced.

Thanks bunches for the silver prize last year: book on Diophantine eq.s. It helped with the first round.

The first round needed no previous knouledge.


There are an equal number of pennies, nickels, dimes and quarters in a bag. What is the probability that the combined value of four coins randomly selected with replacement will be 41 cents? Express your answer as a common fraction.

This can only be done if we have one of each type.
Why? Well, first of all, there needs to be a penny, as all the other coin values are multiples of 5.
Now we have 3 coins and 40 cents.
40 cents = 8 nickels.
There needs to be a quarter somehow; if there are no quarters, then maximum total is 30 cents, not 40 cents.
So now we have 2 coins and 15 cents to cover.
A nickel and a dime is the only 2-coin combo now.

SO, IN ORDER TO COVER FOUR COINS OVER 41 CENTS,
we must obtain a penny, a nickel, a dime, and a quarter, in any order.

There are 24 ways of doing this, and 4^4 total possibilities.

This is 4*3*2*1/(4^4)=
6/(4^3)=3/(2^5)=
[b]3/32[/b]

Sorry for reviving, but I asked this on this forum, and it says the answer is $ \frac{3}{22}$.  Clearly, this is a typo. (I hope.)

no, the answer on FTW says that it's 3/32


I have searched for proofs of quadratic reciprocity that don't require much higher level material. I found these five results so far. Please tell me if you have another one. It is my opinion that if a proof uses Gauss' Lemma, then Eisenstein's geometric proof seems the most natural and shortest route, so try not to give proofs that use Gauss' Lemma. I know there is a website [url]http://www.rzuser.uni-heidelberg.de/~hb3/rchrono.html[/url] with 224 proofs of QR, but I can't access any and most aren't elementary so it doesn't really help.
[url]http://www.bsu.edu/libraries/virtualpress/student/honorstheses/pdfs/M38_1977MatneyRobertC.pdf[/url]
[url]http://books.google.com/books?id=_g5TvMCJQB4C&pg=PA251&lpg=PA251&dq=A+new+elementary+proof+of+the+quadratic+reciprocity+law&source=bl&ots=SfkTemLBNk&sig=crl3kBKcfMHUODrL2iEVl_dWQdQ&hl=en&ei=fF9-StmcKoGktgfW-aHmAQ&sa=X&oi=book_result&ct=result&resnum=3#v=onepage&q=&f=false[/url]
[url]http://www.mscand.dk/article.php?id=1607[/url]
[url]http://www.wis.kuleuven.be/algebra/artikels/reciprocity.pdf[/url]
[url]http://www.mai.liu.se/~pehac/kurser/TATM54/zolo.pdf[/url]

I like that last proof by Zolotareff. I think that's one of the best proofs I've seen. The one by Castryck is pretty nice too.

As for others, there's a proof in nearly every number theory book that exists. Any book that claims to be elementary will probably have an "elementary" proof of the theorem. A few such books I can think of are [u]Number Theory[/u] by George Andrews, [u]An Introduction to the Theory of Numbers[/u] by Hardy and Wright, [u]An Introduction to the Theory of Numbers[/u] by Niven, Zuckerman, and Montgomery, [u]A Classical Introduction to Modern Number Theory[/u] by Ireland and Rosen, and [u]A Course in Number Theory[/u] by H. E. Rose. (This book by Rose is my favorite of them all).

Each of those books have a proof (or a few) of the LQR. But I don't remember any of them having as [i]nice[/i] a proof as the last two you've found.

P.S. You might want to check out the article and references on [url=http://en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity]Proofs of the Law of Quadriatic Reciprocity[/url] at Wikipedia if you haven't already.


Find all $x,y\in \mathbb{Z},n\in \mathbb{N}$ such that:
 \[ x^n+y^n=1994(n\geq 2)\]

Just use computer, there is only a finite (and quite small) number of cases to study.

Pierre.

Just a remark: I think I remember seeing it in a book containing Romanian TST's. Guess what year it should be from :).


What is Sarras' expansion and how to use it to simplify the determinant\[ \left|\begin{array}{ccc}ax-by-cz&ay+bx&cx+az\\ay+bx&by-cz-ax&bz+cy\\cx+az&bz+cy&cz-ax-by\end{array}\right|  \]

Well, the formula for a determinant of order 3 is :
$\left | \begin{array}{ccc}a&b&c \\ d&e&f \\ g&h&i \end{array}\right| = aei+bfg+cdh-ceg-afh-bdi$, and Sarr[b]u[/b]s' rule is just a way to visualize and remember this result.

Now, for your determinant, you simply apply that formula, and (oh, surprise ;)), it factors out nicely as :
$(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})(ax+by+cz)$

Also, I just wanted to point that your determinant is equal to this one :
$\left|\begin{array}{ccc}ax+by+cz&ay-bx&az-cx\\bx-ay&by+cz+ax&bz-cy\\cx-az&cy-bz&cz+ax+by\end{array}\right|$.

This result can be generalized. ;)

(Sorry for double-posting.)


function
advanced fields
advanced fields unsolved

X is a set with n elements.  Consider the following procedure:

1. Choose a partition of X into nonempty subsets
2. For each nonempty set of the partition choose a nonempty subset

This generates a list of h(n) sets of nonempty sets.  We set h(0) = 1.  (There may be repetitions in the list.)

The problem:  find the exponential generating function, and make a list of the numbers h(n) from 1 to 15.

I think that I figured it out!  Each partition is described by the bell number, so I just need to apply the exponential generating function of the bell number twice.  However, this gives me $ e^{e^{e^{e^x}}}\minus{}1$ which isn't likely to give nice answers.  Any help from here?

Actually, make that $ e^{(e^{(e^x\minus{}1)}\minus{}1)}$.

There is no nice formula for the coefficients. You still have to calculate the coefficients, and recursion is fastest.

A recursion for the Bell numbers: $ B_{n\plus{}1}\equal{}\sum_{k\equal{}0}^n \binom{n}{k}B_k$.


If 2a = 2/(1/b+1/c), show that the sum of the squares of the three numbers a, b, and c is the square of a rational number.

I don't think that's true.  a=pi/2, b=1/pi, c=1/pi satisfies the equation, but (pi^4+8)/(4pi^2) doesn't look rational to me.

your values for a, b, and c don't satisfy the initial condition  :?

Magnara, just a little question, (I just stumbled into this forum).

Why did you use pi in this question when there was no mention of geometry?

I think $a$, $b$, and $c$ are meant to be rational.  The given implies
\[a = \frac{bc}{b + c},\]
so
\[a^2 + b^2 + c^2 = \frac{b^2 c^2}{(b + c)^2} + b^2 + c^2 = \left( \frac{b^2 + bc + c^2}{b + c} \right)^2.\]

Um...woops.  I thought it was over b+c, not 1/b+1/c.

But it can't be that a, b, and c are rationals, because the sum of the squares of any three rationals is obviously a rational.

Hmmm, so let's try that counterexample again.  a=pi/2, b=c=pi.  Then we have 2*pi/2=2/(1/pi+1/pi)=2/(2/pi)=pi
(pi/2)^2+pi^2+pi^2=(9pi^2)/4.

Another approach: The given implies $ab + ac = bc$.  Then
\[(b + c - a)^2 = b^2 + c^2 + a^2 + 2bc - 2ab - 2ac = a^2 + b^2 + c^2.\]


linear algebra
matrix
quadratics
algebra
polynomial
number theory proposed
number theory

Let $ A$ and $ B$ be two integer $ 2\times 2$ matrices such that $ A, A\plus{}B, A \plus{} 2B, A \plus{} 3B,$ and $ A \plus{} 4B$ are all invertible matrices whose inverses have integer coefficients. Prove that $ A \plus{} 5B$ satisfies the same property.

Does the same result holds with $ n\times n$ matrices such that $ A, A \plus{} 2B, A \plus{} 3B,$ and $ A \plus{} n^2B$ are all invertible matrices whose inverses have integer coefficients?

Edit: I made a mistake, and didn't include $ A\plus{}B$. This makes the following posted proof even easier.

It's a standard lemma, provable by either Kramer's rule or row reduction, that the inverse of an integer matrix is integer if and only if its determinant is $ \pm 1$.  Let $ P(t) \equal{} \det (A \plus{} Bt)$, which is a quadratic polynomial.  We are given that $ P(t)^2 \equal{} 1$ has roots $ t \equal{} 0, 2, 3, 4$, hence either $ P(t) \equal{} \pm 1$ identically or

$ P(t)^2 \minus{} 1 \equal{} bt(t \minus{} 2)(t \minus{} 3)(t \minus{} 4) \equal{} (P(t) \minus{} 1)(P(t) \plus{} 1)$.

This is possible only if $ b$ is a square and the roots can be divided into two pairs which sum to the same number (the negative of the linear term of $ P(t)$ divided by $ \sqrt{b}$), but $ \{ 0, 2, 3, 4 \}$ doesn't have this property, hence $ P(t)$ must be constant.

The generalization would involve $ 2n$ integers; the fact that $ 4 \equal{} 2^2$ is irrelevant.

$ 2n$? You can do far better than that.

For $ n\equal{}1$, you need three of them. For $ n\in\{2,3,4\}$, you need five. For $ n\ge 5$, you need $ n\plus{}1$. The key is the property that $ m\minus{}n$ divides $ p(m)\minus{}p(n)$ for any integer polynomial $ p$ and integers $ m,n$; among any six integers, we can connect them all with segments of length at least three.

I know I've seen this before, but I can't find any old threads of it.


hello, prove that the triangle $ ABC$ with the internal angles $ \alpha,\beta,\gamma$ is right-angled if and only if the following equation holds
$ \frac{\sin^2(\alpha)\plus{}\sin^2(\beta)\plus{}\sin^2(\gamma)}{\cos^2(\alpha)\plus{}\cos^2(\beta)\plus{}\cos^2(\gamma)}\equal{}2$.
Good luck!
Sonnhard.

[quote="Dr. Sonnhard Graubner"] [color=darkred]Prove that the triangle $ ABC$ is right-angled if and only if $ \frac {\sin^2A\plus{} \sin^2B \plus{} \sin^2C}{\cos^2A \plus{} \cos^2B \plus{} \cos^2C} \equal{} 2$ .[/color] [/quote]

[b][u]Proof[/u].[/b] $ \frac {\sin^2A\plus{} \sin^2B \plus{} \sin^2C}{\cos^2A \plus{} \cos^2B \plus{} \cos^2C} \equal{} 2$ $ \Longleftrightarrow$ $ \sum\sin^2A\equal{}2$ $ \Longleftrightarrow$ $ \sum (1\minus{}\cos 2A)\equal{}4$ $ \Longleftrightarrow$ $ 1\plus{}\sum\cos 2A\equal{}0$ $ \Longleftrightarrow$

$ 2\cos^2A\plus{}2\cos(B\plus{}C)\cos(B\minus{}C)\equal{}0$ $ \Longleftrightarrow$ $ \cos A\left[\cos (B\plus{}C)\plus{}\cos (B\minus{}C)\right]\equal{}0$ $ \Longleftrightarrow$ $ \cos A\cos B\cos C\equal{}0$ .

[hide="Alternative Solution"]
The equation is equivalent to $ \cos^2(\alpha) \plus{} \cos^2(\beta) \plus{} \cos^2(\gamma) \equal{} 1$.
Taking advantage of the famous identity $ \cos^2(\alpha) \plus{} \cos^2(\beta) \plus{} \cos^2(\gamma) \plus{} 2 \cos (\alpha) \cos (\beta) \cos (\gamma) \equal{} 1$
(where $ \alpha$, $ \beta$, $ \gamma$ are the three angles of a triangle)
We arrive at the conclusion $ \cos (\alpha) \cos (\beta) \cos (\gamma) \equal{} 0$, which yields the result.
[/hide]


At Agnesi Middle School, Mr. Eye, Mr. Love and Mr. Problems teach science, mathematics, and history-but not necessarily in that order. The history teacher, who was an only child, has the least experience. Mr. Problems, who married Mr. Eye's sister, has more experience than the science teacher. Who teaches science?

$ \text{(a)}\ {\text{Mr. Love}} \qquad
\text{(b)}\ {\text{Mr. Problems}} \qquad
\text{(c)}\ {\text{Mr. Eye}} \qquad$

If the history teacher has the least experience, and Mr. Problems has more experience than someone else, then Mr. Problems can't be the history teacher. Furthermore, we know that Mr. Problems has more experience than the science teacher, thus he can't be the science teacher. Now, we know that the history teacher is an only child, but the problem says something about Mr. Eye's sister, thus Mr. Eye isn't the history teacher. From this, we can conclude that Mr. Love must be the history teacher. We need to figure out who teaches science. If Mr. Love teaches history, and Mr. Problems isn't the science teacher, our answer must be Mr. Eye, or C.

And yes, Eye Love Problems too.


linear algebra
matrix
superior algebra
superior algebra unsolved

Find a noncummutative ring R, with two sided ideals I and J such that I + J =R, but ( IJ )=/= (I intersects J)

$ R \equal{} M_2(K) , x \equal{} \begin{pmatrix} 1 & 0 \\
0 & 0 \end{pmatrix} , I \equal{} (x), J \equal{} (1 \minus{} x)$. Then clearly $ I \plus{} J \equal{} R$ and $ x^2 \equal{} x$ implies $ I*J \equal{} 0$, but $ \begin{pmatrix} 0 & 1 \\
0 & 0 \end{pmatrix} \equal{} \begin{pmatrix} 0 & 1 \\
0 & 0 \end{pmatrix} (1 \minus{} x) \equal{} x \begin{pmatrix} 0 & 1 \\
0 & 0 \end{pmatrix}$ lies in $ I \cap J$ and is non-zero.


1. Drops of liquid dinotrogen trioxide are added to water.
My work: $N_2O_3 +3H_2O \rightarrow 2{NH_4}^{+} + 3O_2$
Now splitting up $NH_4$, we have:
$N_2O_3(l) +3H_2O(l) \rightarrow 2{NH_3}(s) + 2H_2(g)+ 3O_2(g)$
But it doesn't look right.......

2. Hydrobromic acid is put on solid magnesium.
$Mg(s) + 2HBr(aq) \rightarrow H_2(g) + MgBr_2(?)$
I am only confused if $MgBr_2$ is aqueous or not.

Also, if I have liquid bromine, is that still $Br_2$? Or is it just $Br$?

Thanks for all your help.

when you break up $NH_4^+$, you don't get neutral hydrogen... a charge was lost there, though i am not sure how your first result could be simplified (or even if it could be)

for the second question, i believe BOTH hydrobromic acid and $MgBr_2$ are aqueous, i.e. $Br^-$ is a spectator in this reaction.

finally, liquid bromine is $Br_2$

For the first one, is my original equation even right?

And for the second one, isn't Br not aqueous? There's no water for it to be dissolved in.

Thanks for your help!

[quote="saturnlife"]For the first equation, I think Nitrous acid is formed? I remember doing this earlier in the year.
For the second equation, the reaction is in water, so the ions in the reactant are Mg(2+) and Br(-). Br(-) is canceled out and the only two things that remain for the reactants are the Mg(2+) and H2

Mg(s)+H+ --> Mg(2+) + H2(g)[/quote]

How do you know if its in water?

[i]Hydrobromic acid[/i] is aqueous [i]hydrogen bromide[/i], the same way [i]hydrochloric acid[/i] is an aqueous solution and [i]hydrogen chloride[/i] is a gas.


Let $ a,b,c$ be non-negative real numbers such that $ a\plus{}b\plus{}c\equal{}1$. Prove that
\[ ab\plus{}bc\plus{}ca\leq\frac18\sum_{\rm cyclic}\sqrt{1\minus{}ab}\sqrt{1\minus{}bc}\leq a^2\plus{}b^2\plus{}c^2.\]

All sums are cyclic.

For the right part, from AM-GM, we get
\[ \sum\sqrt {(1 - ab)(1 - bc)}\le \sum(1 - ab) = 3 - (ab + bc + ca)
\]
Hence it remains to show that
\[ 3 - (ab + bc + ca)\le 8(a^2 + b^2 + c^2)
\]
Which is true since
\[ 8(a^2 + b^2 + c^2) + ab + bc + ca\ge 3(a + b + c)^2
\]
For the left part, we get from Cauchy-Schwarz inequality that
\begin{eqnarray*} & & \sum\sqrt {(1 - ab)(1 - bc)} \\
& = & \sum\sqrt {(a^2 + b^2 + c^2 + ab + 2bc + 2ca)(a^2 + b^2 + c^2 + bc + 2ab + 2ca)} \\
& \ge & \sum(a^2 + b^2 + c^2 + 3\sqrt {ab^2c} + 2ca) \\
& = & 3(a^2 + b^2 + c^2) + 3\sqrt {abc}(\sqrt a + \sqrt b + \sqrt c) + 2(ab + bc + ca)\end{eqnarray*}
It remains to show that
\[ 3(a^2 + b^2 + c^2) + 3\sqrt {abc}(\sqrt a + \sqrt b + \sqrt c)\ge 6(ab + bc + ca)
\]
Which follows from Schur's inequality and AM-GM.

I think it's simpler than what you did, nayel(the left side):
I have: square root(1-ab)(1-bc)>=1-b*square(ac)(B.C.S's inequality)
Similarly to 2 more square roots, I will get: RHS>=3-square root(abc)*[square root(a)+square root(b)+square root(c)]/8
Now  i just need to prove: 3(a^2+b^2+c^2)>=2(ab+bc+ac)+square root(abc)*(square root(a)+square root(b)+square root(c)), which is obviously true by AM-GM! :wink:

[quote="ghjk"]I think it's simpler than what you did, nayel(the left side):
I have: square root(1-ab)(1-bc)>=1-b*square(ac)(B.C.S's inequality)[/quote]

$ (1\minus{}ab)(1\minus{}bc)\ge (1\minus{}b\sqrt{ac})^2$
$ \iff 2b\sqrt{ac}\ge ab\plus{}bc$

which is clearly false by AM-GM. :P

[quote="nayel"]
It remains to show that
\[ 3(a^2 \plus{} b^2 \plus{} c^2) \plus{} 3\sqrt {abc}(\sqrt a \plus{} \sqrt b \plus{} \sqrt c)\ge 6(ab \plus{} bc \plus{} ca)
\]
Which follows from Schur's inequality and AM-GM.[/quote]

I can't see how

[quote="FOURRIER"][quote="nayel"]
It remains to show that
\[ 3(a^2 + b^2 + c^2) + 3\sqrt {abc}(\sqrt a + \sqrt b + \sqrt c)\ge 6(ab + bc + ca)
\]
Which follows from Schur's inequality and AM-GM.[/quote]

I can't see how[/quote]

For an easier understanding let's denote $ x=\sqrt{a}$ ,$ y=\sqrt{b}$ and $ z=\sqrt{c}$ the inequality is equivalent to :
$ x^4+x^4+x^4+xyz(x+y+z)\geq 2(x^2y^2+y^2z^2+z^2x^2)$
But from Schur we get :
$ x^4+x^4+x^4+xyz(x+y+z)\geq x^3(y+z)+y^3(z+x)+z^3(x+y)=
\sum (x^3y+y^3x)\geq \sum 2 x^2y^2=2(x^2y^2+y^2z^2+z^2x^2)$


Evaluate the sums:

1. $ \displaystyle\sum_{i\equal{}0}^{n\minus{}1}\displaystyle\sum_{j\equal{}i\plus{}2}^n F(i)F(n\plus{}1\minus{}j)$

2. $ \displaystyle\sum_{i\equal{}1}^n F(i)F(n\plus{}1\minus{}i)$

where $ F(1) \equal{} F(2) \equal{} 1$, and for every $ n>\equal{}3$, $ F(n) \equal{} F(n\minus{}1) \plus{} F(n\minus{}2)$

i'll make this a challenge question, if u don't mind.


AMC
AIME
USAMTS
geometry
geometric transformation
perimeter
calculus

Hey everyone!

It would also be nice if some people could post their commended solutions. That'd be fun to look at. Otherwise, just wondering.

I got a 21, not my best, but basically what I expected because I screwed up number 5.

Ooh, results just got posted.  I got a 25, 0 commendations (perhaps my solutions were too long?).  Congrats on your score, btw.

22, no commendations. Better than I'd thought I got: I got all 5s except for the last one which I screwed up on.

I got 25 with two commendations.  Good job, everybody who participated this round!

So happy I got a 25... neways I got number three commended, even though I didn't think it was a good solution. I could post it, but I would have to post all five solutions and I already posted them somewhere(forget where). So yeah...

Spent a grand total of 40 minutes on USAMTS round 3, all in the middle of writing an essay for my English class.  I only attempted the first three, and I got a 15, 1 commended.  I'm regretting not doing all 5.

25, #3 commended.  Looking at ya'll's proofs for #3 though (on the other thread), i'm actually pretty embarassed with my proof.

I'm not sure about the whole grading more harshly thing (71 this year, compared to 73 last year, about the same)...I'm definitely getting fewer commendations though.  I think they are handing out fewer of THOSE. maybe.

I got a 20, complete with an amazing 1 on #5. Got a 4 on the 3rd question simply because I kind of missed something that I had to prove. Sad thing is, #4 took me something like 5 hours of staring at and going nowhere with, and my solution took up about 2/3 a page and got a 5. :?

Right now I have a 64/75 -- not nearly as high as I'd like, to be honest. Then again, you can't really be getting 1's on any questions and score all that high on the exam overall.

Also haven't had any commended solutions since Round #1.

w00t!  25, no commendations (the first time w/o any commendations)
 :)
AIME qualifier!
 :lol:
I'm a little happy :D

23, no commendations.  I made a stupid mistake and misread problem 3, assuming that it meant "prove that all terms ending in two or more identicial digits end with the same 2 digits," rather than "the same n digits."  No wonder I found that problem suspiciously easy... I'm just glad the rubric was generous and gave me a 3.  73 overall (first mistake on this round), so I guess I'm taking the AIME.  I'm actually a little disappointed I didn't get a commendation on problem 5, because I really liked my solution (I used some techniques I'd just learned in Calc BC to parameterize the semicircles and set up a plane equation).  Oh well, hopefully I'll read better on round 4  :P

[quote]24, no commendations, 66 total (my terrible counting skills was what killed me), so AMC = the route for me[/quote]

Tell me about it... 67/75 -> 24 - 23 - 20 :(
I personally think that #3 is not clearly worded. Could have given me that extra point. :(

7 with 3 commendations! Whoooo!

Just kidding.  I got a 7. (5 on 2, 2 on 5)

25, 1 commendation on #4.  I guess I'm okay with that.  I don't think I can communicate effectively enough to receive commendations.

[quote="dts"]25, no commendations (this seems to be the mode, here...).  I got 73 total (messed up on that 'tiny numbers' problem in round 2).[/quote]
same as you. 25, no commendations, 73 total. i got one point each docked off for round 2 #2 and #3.

[quote="zanttrang"]23, no commendations.  I made a stupid mistake and misread problem 3, assuming that it meant "prove that all terms ending in two or more identicial digits end with the same 2 digits," rather than "the same n digits."  No wonder I found that problem suspiciously easy... I'm just glad the rubric was generous and gave me a 3. [/quote]

Ahhaha ditto to that.  :P
I have a 69 = AIME QUALIFIER, BOOYAH! :lol: after 3 rounds.

In 7 rounds of the USAMTS I have yet to avoid making a dumb mistake (misreading a problem, submitting an incomplete proof, etc.) which cost me points. (translation: no 25s :()

I'm happy, though, because I finally get to take the AIME (I got a pair of 99s last year on the AMC 12A and AMC 12B :what?:)

Hey hey!
Another 25   :D !!

Yet those commendations are becoming harder and harder to get.  I'm on the downslide in that category:
Round1 = 2
Round2 = 1
Round3 = 0  :(

25, commendations on #3 and #4. 72 total.  AIME here I come!

[quote="sonar"]24, no commendations, 66 total (my terrible counting skills was what killed me), so AMC = the route for me.

Oh, if anyone is willing, could he or she take a look at my proof for question 3 and show me what I did wrong? I thought that it may have been because it was so badly worded, but the comments say, "Very very close," so I'm thinking that I must have overlooked something.[/quote]

I wish all the other graders would write better explanations. However, it is tough to compose a short and coherent explanation on the fly. And the flaws in your proof are subtle. But since I was one of the graders for Problem 3, I can spot them.

Page 1:
Proving properties about $3^n$ and then translating them to $s_n = \frac{3^n-1}{2}$ instead of proving them directly for $s_n$ is weak style, unless you do something extra with the $3^n$. Weak style means you do not earn a commendation.

Weak style can also lead to subtle holes. You proved that $3^n$ ends in 1, 3, 7, or 9. You used that lemma to claim that $s_n$ ends in 0, 1, 3, or 4. But really, you have to look at $3^n$ mod 20 to tell what happens with $\frac{3^n-1}{2}$ mod 10. This trivial gap in your logic takes off only half a point off, to be rounded up to full credit if you have no other errors.

Page 2:
Ah, you have proven that a string of 4's leads to a string of 3's of exactly the same length. You avoided a mistake that many others fell into, where they proved that a string of 4's leads to at least as many 3's, but forgot to prove there are not more 3's than 4's. Good work.

Your last sentence, "In the same vein of Lemma 1, we see that a pattern emerges such that after every 20 powers of 3, the last two digits repeat," creates a new lemma. The "same vein" claim says that the proof of this lemma is so similar to the proof of Lemma 1 that you do not think it is worth writing down. Really, you should have written out the proof, because it is not clear that your method for the one-digit case generalizes to the two-digit case. But I will give you the benefit of the doubt that it seemed clear to you.

Page 3:
You demonstrate examples of the 20 cycle in the last two digits and the 100 cycle in the last three digits. This demonstration does not count as a proof. Then you use the cycles to look at $3^n$ mod 200. It is good that this time you figured out the proper modulus to examine to get the last two digits for $\frac{3^n-1}{2}$.

I did not grade this particular solution. If I had I would have given it a 5, no commendation. However, due to the two small flaws in it, 4 is a reasonable score. Since you say you earned 24 points for the round, I presume you received a 4. The remark, "Very very close," from your grader does not describe your errors; instead, it describes your grader's temptation to round your score up to 5.

Another possibility is that your grader might have thought that your demonstration of the 20 cycle and 100 cycle on page 3 was meant to be a proof of the existence of those cycles, especially since your "same vein" comment is out of sight on the previous page. I spent less time reading each solution in the grading session than I spent reading your solution here, so I could have made that mistake myself in my haste. However, I think your grader read your work clearly, because using a demonstration in place of a proof would have undeniably cost you a point (if not more), and the grader would have had no reason to write, "Very very close."

Erin Schram
USAMTS grader

[quote="white_horse_king88"]Hey everyone!

It would also be nice if some people could post their commended solutions. That'd be fun to look at. [/quote]

I got a 25, with four commendations: 2,3,4,5.

Those are very concise... I don't know if I could ever make mine that concise. Nice solutions! :)

Here is a quick question:

On problem #2 Round 2, how do you prove that the base of the isosceles triangle is even?

I proved it, and it took a page.  My reader said it is "obvious".  I am extremely intrigued how this could be "obvious".  (I take that being "obvious" means the work does not need any formal proof and is intuitive.)

I got 24 this round, 4 on Problem 3.  I think I clearly proved (although not as formal) that when conditions are satisfied, the numbers in the sequence ends with k number of only 4's, 3's, and 0's and come in groups of three.

The reader said I need to prove that the numbers don't end with "11".  I don't think the problem asked for that.  The problem asked if the end-repeating numbers come in groups of three with same repetition, which is what I did.

Hmm........oh, well, just one point right? (But this one point keeps me away from the Gold award.  I got 70/75 so far)

I think the grading dispute rule is that I have to disagree with more than 1 point to do something about it?

Any way good job everyone. I found this round very enjoyable, and It seems that lots of people used Calc. on #5 (ah, the power of calculus).

Edit:  Oh, can I be certain that I am invited to AIME(score 70)?  If I am, where do I take it?  (southern california here)

[quote="Ultimatemathgeek"]On problem #2 Round 2[should be 3], how do you prove that the base of the isosceles triangle is even?

I proved it, and it took a page.  My reader said it is "obvious".  I am extremely intrigued how this could be "obvious".  (I take that being "obvious" means the work does not need any formal proof and is intuitive.)[/quote]

Usually "obvious" means that it doesn't deserve a page.  I just looked at it, and on first glance I wouldn't call it obvious, but it isn't very hard to prove quickly:  Assume the base is odd.  Drop the altitude, which in a right triangle is also a mediate an thus divides the base in half.  Each half of the base has length "some integer + 1/2."  Now, think about the Pythagorean Theorem in one of the half-triangle: it's clear that the square of the altitude must be "some other integer + 3/4," which means that if the altitude is rational (which we need), then it must have a denominator of 2 (in order to give me fourths).  But then the area couldn't possibly be an integer.  (Note, by the way, that the perimeter of an isoceles triangle with integral sides is odd if and only if the base of the triangle is odd.)

[quote]Edit: Oh, can I be certain that I am invited to AIME(score 70)? If I am, where do I take it? (southern california here)[/quote]
Yes.  You will probably be able to take it at your school or some nearby school that administers it.  I would talk to the AMC people about it if you don't get any information by around the time AMC-takers find out if they made it.

[quote="white_horse_king88"][quote="no_one_2000"]Heh, I got a 20.  I made two dumb mistakes on the 2nd and 3rd problems.  Actually, it's interesting... each round this year, I've missed at least one point on every 2nd or 3rd problem.  The rest were all fives... two of which were commended.

The grading must be more difficult this year.  Last year I got a 73 from my first three rounds (AIME qualifier) whereas I got a 63 this year. (d'oh)

By the way, my solution to #5 this round was commended.  I can post it if anybody wants to see it.[/quote]

sure - i'd like to see it[/quote]

Alright, I'll attach my #5 solution to this post.  (Sorry for the delayed response, I don't check this forum daily)

[quote="beta"]Ah I'm an idiot. I got a 1 on #5 and 2 on #4. I got a 2 on #4 because of typos, and a 1 on #5 because I thought the semi-circles are congruent. So there's no AIME for me through USAMTS.

I had a 18 this round with two commended solutions.[/quote]

Ugh, I hate it when that happens to me.  I'm not qualifying for the AIME this year either.  I'll have to make 100+ on the AMC12.

[quote="Ultimatemathgeek"]On problem #2 Round 2, how do you prove that the base of the isosceles triangle is even?

I proved it, and it took a page. My reader said it is "obvious". I am extremely intrigued how this could be "obvious". (I take that being "obvious" means the work does not need any formal proof and is intuitive.)[/quote]
I wouldn't call it obvious. There are some quick proofs of it, but those proofs are not obvious. However, we graders write in a hurry, so often our explanations are slightly misphrased. And we write in pen, so we cannot correct an ill-chosen word except by making a scribble of the comments' box. (Ask beta about the mess in his comments' box for Problem 3.)

[quote="Ultimatemathgeek"]I got 24 this round, 4 on Problem 3. I think I clearly proved (although not as formal) that when conditions are satisfied, the numbers in the sequence ends with k number of only 4's, 3's, and 0's and come in groups of three.

The reader said I need to prove that the numbers don't end with "11". I don't think the problem asked for that. The problem asked if the end-repeating numbers come in groups of three with same repetition, which is what I did.[/quote]
I don't know which style of proof you used (see the grading rubric for Problem 3 for a description of several styles), but I can guess that you did not look at the 20-long cycle that occurs in the last two digits, but instead proved directly that an element ending in several 4's is followed by an element ending the same number of 3's, which is followed by an element ending in the same number of 0's. That creates the group of  three consecutive elements that have the same number of identical digits at the end. However, what if that group is preceeded or followed by an element ending in several 1's? That would make a group of four. Or what if an element ending in several 1's existed by itself? That would make a group of one. Part of the problem was to show that the only occurence of elements that end in two or more identical digits is in groups of three.

[quote="Ultimatemathgeek"]Hmm........oh, well, just one point right? (But this one point keeps me away from the Gold award. I got 70/75 so far)

I think the grading dispute rule is that I have to disagree with more than 1 point to do something about it?[/quote]
The grading dispute rule is that the grader is allowed to disagree with the rubric by up to one point. The rubric does not describe every grading rule--some errors are so rare that we judge them case by case rather than writing them into the rubric--and some graders have their own biases about how serious an error is. If your score on a problem disagrees with the rubric by one point, we believe that the grader could have given you that score without having misinterpretted your solution, and thus, there is no justification to re-grade it.

On the other hand, if the score in your comments' box differs from your computer-recorded score (what we graders fill out in the machine-scanned bubbles next to the box), there is obviously an error to be corrected, even if the two scores differ by only a point. Or if your solution is complete and correct, yet the grader wrote, "You missed the final step," and gave you 4 points, then perhaps the grader made a error and missed your last page, so you should check (Most likely, the last page was lost in transit, so you cannot earn that final point--but you should check).

This is the first year we have had the grading rubric posted for the USAMTS participants to see. In previous years, the rubrics were given to the director of the USAMTS, or the director had a contact list of "head graders" for each problem, so that he had some way of judging grading disputes sent to him in mail or e-mail. I like the openness of posting our rubrics. However, the public rubrics create a risk that some people will try to game the system and earn their points by clever grading arguments after the contest rather than by clever mathematical solutions during the contest. The one-point discretion rule also serves as a defense against this. (No, Ultimatemathgeek, I am not accusing you of this, but given that you are one point away from the Gold award, you can understand the temptation.)

Erin Schram
USAMTS grader

[quote="Erin J. Schram"]
 [...]Or what if an element ending in several 1's existed by itself? That would make a group of one. Part of the problem was to show that the only occurence of elements that end in two or more identical digits is in groups of three.

Erin Schram
USAMTS grader[/quote]

Thanks.  I now see what my logic error is.  I never considered several 1's existed by it self.  Somehow, the word "group" made me think it was more than 1 elements.

[quote="JBL"]Yes.  You will probably be able to take it at your school or some nearby school that administers it.  I would talk to the AMC people about it if you don't get any information by around the time AMC-takers find out if they made it.[/quote]

My school doesnot administer AMC, so I have go to another school after wrestling with my counselor about it.  (This is a very sad situation).   I guess I have to do it again with AIME.

What exactly is being "invited" to the AIME?  Do I get a physical invitation through my mail ;) ?

Ah, thank you very much Dr. Schram.  I was ambivalent over the very same issues you mentioned.

What I can't see is how all you people getting 69s and 70s and 72s and the like have a difficult time with qualifying for the AIME via the AMC contests. Last two years I've gotten a 116 and a 113.5 on the AMC12, and I am quite a bit off the AIME qualifying cutoff for the USAMTS. Not trying to be condescending by any means; merely asking how everyone else thinks here.

Then again, when your school doesn't teach any proof theory at all, it's probably not good to be taking proof tests. (Yes, taking the USAMO without any basis in proof theory like I tried to do in 10th grade is not a good idea.)

[quote="Steve_B"]What I can't see is how all you people getting 69s and 70s and 72s and the like have a difficult time with qualifying for the AIME via the AMC contests. Last two years I've gotten a 116 and a 113.5 on the AMC12, and I am quite a bit off the AIME qualifying cutoff for the USAMTS. Not trying to be condescending by any means; merely asking how everyone else thinks here.[/quote]

Yeah......I don't understand why either.  Assuming that these mock AMCs are good indicator of actual score, I am getting like 60's on the tests.  Well, tomorrow is the AMC (Feb 1) and I have to take it

Wish me good luck!

I found this topic quite late, I had a 25 and 2 commended.  My round 2 solutions were really long and wordy (no commendations), so I'm happy about the improvement.  I'm starting to make my solutions more concise. (nowhere near zabelman though)

If you want to see my solution for #3, look on the student solutions page. ( :D  :D  :D  I never thought that would happen!  :D  :D  :D )

Off-topic, but...gaussianprime, what is the significance of the number in your profile?   :blush:

At an ARML practice once, one of our coaches was solving a problem on the board.  He got some obscure irrational answer (like the one in my profile), then stood back in mock amazement and exclaimed, "why, that just happens to be my favorite number!"

I liked it, so naturally, I stole his joke.  :)


algebra
polynomial
Vieta
calculus
calculus computations

let  $ p(x) \equal{} a_0 \plus{} a_1 x \plus{} a_2 x^2 \plus{} ... \plus{} a_n x^n$ and  be the zeros of p repeated acording to multiplicity. Derive the identity:

 $ \sum_{i \equal{} 0}^{n} \frac {1}{\alpha^2_i} \equal{} \frac {a^2_1}{a^2_0} \minus{} 2\frac {a_2}{a_0}$

which leads to the final stage of the proof

Please don't [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=187833]double-post[/url].  (Also, in the process of transfering this text here, you made it semi-incomprehensible, e.g. by leaving out the definition of the $ \alpha_i$.)

The idea here is to consider the polynomial $ q(x) \equal{} \prod(x \minus{} \alpha_i^{ \minus{} 2})$.  We can write $ q(x^2) \equal{} \left(\prod(x \minus{} \alpha^{ \minus{} 1})\right) \cdot \left(\prod(x \plus{} \alpha^{ \minus{} 1})\right)$.  The first term in this product is $ x^np(x^{ \minus{} 1})$, and you can figure out something similar for the second term.  Then you use Vieta's relations (expressing the coefficients of a polynomial in terms of the roots) and a little polynomial arithmetic to arrive at the desired expression.


pigeonhole principle
number theory theorems
number theory

Are there any proofs of the periodicity of rational numbers in any base, other than the "classical" one?

[The classical one:
[hide]Factor the denominator into 2 numbers, one of them is a multiple of 2's and 5's, and the other is coprime to 10 (or any other base), and use Euler's Theorem on the other one[/hide] ]

If you aren't interested in computing the period, you can use pigeonhole instead:  let the fraction be $ \frac{p}{q}$ in base $ b$.  Then the sequence $ 1, b, b^2, ... \bmod q$ repeats after at most $ q$ iterations, so there are $ i, j$ such that $ b^i \equiv b^j \bmod q$ and

$ \frac{p}{q} \equal{} \frac{p \cdot \text{something}}{b^j \minus{} b^i}$

is eventually periodic in base $ b$.


quadratics
algorithm
LaTeX
superior algebra
superior algebra unsolved

if a and b are integers, and b>=2

is there a general method for finding out if x^2=a mod b has a solution

i am familiar with jacobi symbol, quadratic reciprocity and chinese remainder

it's the case where a and b are not coprime that I'm doubtful about (I haven't seen a description of such algorithm in any book)

That extension is trivial. If $p|a$ and $p|b$, let $x=py$. If $p^2|b$ and $p^2$ does not divide $a$, there is clearly no solution. If $p^2|a$ and $p^2|y$, divide out the common factor of $p^2$. If $p^2$ does not divide $b$, divide out the common factor of $p$ for a new congruence.

How about learning $\LaTeX$ sometime?


How can you access the Bay Area forum?  I am not allowed to post in it, and no one has posted in it yet.  I would like to get to know the AoPSers in my local area, so the Bay Area forum looks best.

Normally you gain access by requesting memebership in the Usergroups area (button at top of forums).  The moderator must approve your request.  The Bay Area forum has been mostly inactive so far, but with over 100 members that could change easily.  Perhaps you'll be the one to get it going.

I have added you to that usergroup to bypass your having to request and wait.


This has probably already been posted or it's a very simple puzzle... but I decided to post it anyways...

A man died and he had an enormous fortune so he decided to give it to his two sons... but only one. Before he died, he gave his two sons a camel. He told them that they would both start together at the same point in the desert and race to a city on their camels' back. The came to reach there last would receive the money. The brothers agreed to do this and be fair in the race. The two became very tired in the middle. Then they met a man and they told him what happened and asked if there was an easier way... the man told them something and the immediately started racing their camels as fast as they could to the city.

What did the old man say?


analytic geometry
geometry proposed
geometry

Hi All,

I need to calculate the endpoint coordinates of a 3D arc.

Attached is one example, and is the information that I have available  in order to find the endpoint coordinates.

Help would be greatly appreciated.

Regards,
CJW


Add them together and get 504.
Mean=sum/# of numbers=504/6=84.


function
inequalities
triangle inequality
advanced fields
advanced fields unsolved

Show that if $f$ is continuous and of bounded variation on $[a,b]$ and absolutely continuous on $[c,b]$ for any $c \in (a,b)$, then $f$ is absolutely continuous on $[a,b]$.

By continuity of $f$ at point $a$ and since $f$ is of bounded variations on $[a,x]$ for every $x>a$, we have $K_{x} \rightarrow 0$ as $x \rightarrow 0$, where $K_{x}$ denotes the total variation of $f$ on $[a,x]$ (A proof of this fact is easy, using that $x \mapsto K_{x}$ is positive increasing, whence has a limit at point $a$ on the right; now show $0$ is the only possible limit using the continuity of $f$ at point $a$). For every $\epsilon >0$, one can pick $x \in ]a,b[$ such that $K_{x} < \frac{\epsilon}{2}$ and $\eta >0$ such that $\sum_{i=1 \dots n} |f(x_{i+1}-f(x_{i})| < \frac{\epsilon}{2}$ for every finite subdivision $(x_{i})$ of $[x,b]$ with length $<\eta$. But now for every finite subdivision $(y_{i})$ of $[a,b]$ we obtain the desired result by the triangle inequality applied at point $x$.


OK so are any of you guys doing the [url=http://www.youtube.com/watch?v=oHg5SJYRHA0]MMCC contest?[/url] The prizes seem pretty cool, and its Missouri exclusive.

OK it definitely worked on at least one of you. And now you are lying about it to save face  :|

I revived it myself with a much funnier swine flu song


Let ${a_n}$ be a real sequence with ${a_{2n}}$, ${a_{2n+1}}$, ${a_{3n}}$ as its convergent subsequences. Prove that ${a_n}$ converges.

If $a_{2n} \to a, a_{2n+1} \to b, a_{3n} \to c$, then if we take $n_k=6k$, then $a_{n_k} \to a$ and $a_{n_k} \to c$, so $a=c$. In the same manner we can go through indices $n_k$ which are odd and of the form $3n$, so $b=c$. Whcih means that $a=b=c$ and $a_n \to a$.


the stronger and the fastidious but easy :D holds:
$ 9 ( 17\sum{ (ab)^2} \plus{}7\sum{a^4} ) (\sum{a^2}\plus{}\sum{ab})^2 \geq 32(\sum{a})^4(\sum{a^2})^2$ for any reals $ a,b,c$ :wink:

[quote="caubetoanhoc94"]Let:$ a;b;c > 0;a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {3(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}{8(a^2 \plus{} b^2 \plus{} c^2)^3} \plus{} 2(ab \plus{} bc \plus{} ca \minus{} 1)^2 \ge 1$[/quote]

Try with $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$

[quote="gigaman"][quote="caubetoanhoc94"]Let:$ a;b;c > 0;a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {3(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}{8(a^2 \plus{} b^2 \plus{} c^2)^3} \plus{} 2(ab \plus{} bc \plus{} ca \minus{} 1)^2 \ge 1$[/quote]

Try with $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$[/quote]

With $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$ $ \equal{}>LHS\equal{}1$
It's true  :)

[quote="caubetoanhoc94"][quote="gigaman"][quote="caubetoanhoc94"]Let:$ a;b;c > 0;a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {3(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}{8(a^2 \plus{} b^2 \plus{} c^2)^3} \plus{} 2(ab \plus{} bc \plus{} ca \minus{} 1)^2 \ge 1$[/quote]

Try with $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$[/quote]

With $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$ $ \equal{} > LHS \equal{} 1$
It's true  :)[/quote]
Sorry
[hide]Lúc đầu anh chép nhầm đề bài, thành $ \frac {3(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}{8(a^2 \plus{} b^2 \plus{} c^2)^2} \plus{} 2(ab \plus{} bc \plus{} ca \minus{} 1)^2 \ge 1$[/hide][/hide]

[quote="gigaman"][/quote][quote="caubetoanhoc94"][quote="gigaman"][quote="caubetoanhoc94"]Let:$ a;b;c > 0;a + b + c = 1$.Prove that:
$ \frac {3(a^2 + b^2)(b^2 + c^2)(c^2 + a^2)}{8(a^2 + b^2 + c^2)^3} + 2(ab + bc + ca - 1)^2 \ge 1$[/quote]

Try with $ a = b = c = \frac {1}{3}$[/quote]

With $ a = b = c = \frac {1}{3}$ $ = > LHS = 1$
It's true  :)[/quote][quote="gigaman"]
Sorry
[hide]Lúc đầu anh chép nhầm đề bài, thành $ \frac {3(a^2 + b^2)(b^2 + c^2)(c^2 + a^2)}{8(a^2 + b^2 + c^2)^2} + 2(ab + bc + ca - 1)^2 \ge 1$[/hide][/hide][/quote]

[hide] Hj; vâng ạ :)..Lời giải chỉ bằng Am-gm[/hide]

Using a well-known inequality $ 9\left( {x \plus{} y} \right)\left( {y \plus{} z} \right)\left( {z \plus{} x} \right) \ge 8\left( {x \plus{} y \plus{} z} \right)\left( {xy \plus{} yz \plus{} xz} \right)$, we have
\[ \frac{{3\left( {{a^2} \plus{} {b^2}} \right)\left( {{b^2} \plus{} {c^2}} \right)\left( {{c^2} \plus{} {a^2}} \right)}}{{8{{\left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right)}^3}}} \ge \frac{{{a^2}{b^2} \plus{} {b^2}{c^2} \plus{} {c^2}{a^2}}}{{3{{\left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right)}^2}}} \ge \frac{{{{\left( {ab \plus{} bc \plus{} ca} \right)}^2}}}{{9{{\left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right)}^2}}}\]
Let x be $ ab\plus{}bc\plus{}ca$. It surffices to prove that \[ \frac{{{x^2}}}{{9{{\left( {1 \minus{} 2x} \right)}^2}}} \plus{} 2{\left( {1 \minus{} x} \right)^2} \ge 1\]
Rewrite the the left hand side of this equality and use AM-GM inequality for three numbers, we have
\[ \frac{{{x^2}}}{{9{{\left( {1 \minus{} 2x} \right)}^2}}} \plus{} 2{\left( {1 \minus{} x} \right)^2} \minus{} 1 \equal{} \frac{{{x^2}}}{{9{{\left( {1 \minus{} 2x} \right)}^2}}} \plus{} \frac{{\left( {1 \minus{} 2x} \right)}}{3} \plus{} x\left( {1 \minus{} 2x} \right) \plus{} 4{x^2} \plus{} \frac{2}{3} \minus{} \frac{{7x}}{3} \ge 4{x^2} \plus{} \frac{{ \minus{} 10}}{3}x \plus{} \frac{2}{3} \equal{} \frac{2}{3}\left( {1 \minus{} 2x} \right)\left( {1 \minus{} 3x} \right) \ge 0\]
The equality holds when $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$
[hide]Hix, bài viết tử tế đầu tiên   :blush: [/hide]

[quote="dvtrung"]Using a well-known inequality $ 9\left( {x \plus{} y} \right)\left( {y \plus{} z} \right)\left( {z \plus{} x} \right) \ge 8\left( {x \plus{} y \plus{} z} \right)\left( {xy \plus{} yz \plus{} xz} \right)$, we have
\[ \frac {{3\left( {{a^2} \plus{} {b^2}} \right)\left( {{b^2} \plus{} {c^2}} \right)\left( {{c^2} \plus{} {a^2}} \right)}}{{8{{\left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right)}^3}}} \ge \frac {{{a^2}{b^2} \plus{} {b^2}{c^2} \plus{} {c^2}{a^2}}}{{3{{\left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right)}^2}}} \ge \frac {{{{\left( {ab \plus{} bc \plus{} ca} \right)}^2}}}{{9{{\left( {{a^2} \plus{} {b^2} \plus{} {c^2}} \right)}^2}}}\]
Let x be $ ab \plus{} bc \plus{} ca$. It surffices to prove that
\[ \frac {{{x^2}}}{{9{{\left( {1 \minus{} 2x} \right)}^2}}} \plus{} 2{\left( {1 \minus{} x} \right)^2} \ge 1\]
Rewrite the the left hand side of this equality and use AM-GM inequality for three numbers, we have
\[ \frac {{{x^2}}}{{9{{\left( {1 \minus{} 2x} \right)}^2}}} \plus{} 2{\left( {1 \minus{} x} \right)^2} \minus{} 1 \equal{} \frac {{{x^2}}}{{9{{\left( {1 \minus{} 2x} \right)}^2}}} \plus{} \frac {{\left( {1 \minus{} 2x} \right)}}{3} \plus{} x\left( {1 \minus{} 2x} \right) \plus{} 4{x^2} \plus{} \frac {2}{3} \minus{} \frac {{7x}}{3} \ge 4{x^2} \plus{} \frac {{ \minus{} 10}}{3}x \plus{} \frac {2}{3} \equal{} \frac {2}{3}\left( {1 \minus{} 2x} \right)\left( {1 \minus{} 3x} \right) \ge 0\]
The equality holds when $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$
[hide]Hix, bài viết tử tế đầu tiên   :blush: [/hide][/quote]

Nice solution dvtrung  :)


inequalities
graph theory
combinatorics unsolved
combinatorics

Let $G$ be a bipartite graph, i. e. a graph such that the set of its vertices is the union of two sets $A$ and $B$, and each edge of the graph $G$ connects a vertex from $A$ with a vertex from $B$. For every subset $U$ of $A$ and every integer $k$, let $N_{k}\left(U\right)$ denote the set of all vertices from $B$ which have at least $k$ neighbours in the set $U$. Assume that $\left|A\right|=\left|B\right|$.

Some matchings of $G$ are called [i]disjoint[/i] if there is no edge common to two or more of these matchings.

Let $m$ be a positive integer. Prove that the graph $G$ has $m$ disjoint perfect matchings if and only if the inequality $\left|N_{1}\left(U\right)\right|+\left|N_{2}\left(U\right)\right|+...+\left|N_{m}\left(U\right)\right|\geq m\left|U\right|$ holds for every subset $U$ of $A$.

  Darij

Moved to Unsolved, because I don't remember my solution. Does anybody else have one? Or a counterexample?
Also crossposted at http://mathoverflow.net/questions/65211/condition-on-a-bipartite-graph-to-have-an-m-factor .

[b]EDIT:[/b] Apparently this problem is called the Ore-Ryser theorem.


AMC
AMC 12
AMC 12 B
geometry
AIME
probability
Number Theory Book

And now I need to know where to start. I have a little less than a week before the AMC12B and I want to devote all my time into the subjects in the book that would be most helpful on the exam. From your experiences, which type of problem solving techniques are the most reoccurring on the AMC12b?

A little on my current abilities:
Algebra-converting words into symbols(first few problems of most AMC's):high
Algebra-functions(ex.2001#12):high
Algebra-system of equations(ex.2001#21):mediocre
Algebra-imaginary(ex.2001#23):low
Counting and probability(ex.2001#16):mediocre
Geometry-plane(ex.2001#22):mediocre
Geometry-3d-low(ex.2001#15):mid
Number Theory-factoring and setting variables to primes:mediocre

That's all I can think of right now.

Thank you for any suggestions. I really want to score well on the AMC12B as I barely qualified for the AIME on the A (100.5). I just don't know where to begin :maybe:

Alright here's my two cents:

Make sure you know Chapter 6 by heart. Be able to recite it.

Chapters 2, 3 and 4 are useful.

Chapter 13 and 16.

Chapters 15 and 19.

And if you have time, Chapter 23. Number theory doesn't usually get tougher until the later problems. Depends how far on the test you expect to make it. To assure that the last few give you no troubles, 23 is very helpful.

All the other chapters are handy - but given the limited time, start with the ones above, is my opinion.

Any other comments?

Yeah, 2, 3, and 4 are quite useful. And 6 is dead useful. It'll save your butt one day.

I'm pretty sure the authors of this book did not write it with the intention that people learn things from it by rote, but I may be wrong :maybe:   Do learn a chapter by heart; understand (as well as it is possible to understand anything) the reasons for the things in a chapter and do problems from it.

Anyhow, you qualified for the AIME, which is good.  I would advise you to focus more on the AIME than on qualifying for it a second time with more comfort.  You have more time to prepare for the AIME, anyhow.

Ah, thanks for the suggestions. I'll look over those chapters.

@Boy Soprano,
Don't worry, I never learn math by memorizing formulas. I always try my best to understand not only how things work but why. I'm asking for specific chapters because I want to have at least some decent coverage over the main subjects. :)

Well, not that particular one, but an idea that quickly led to it's solution. It's in the "Problems" section of the invariants chapter, I think. Either that, or it's in the "Problems" section of the number theory chapter.

[quote="Elemennop"][quote="rnwang2"][quote="Karth"]Actually, as a bit of an aside, I'm happy that I got Engel's PSS, because the day before the AMC, as I did Engel, I particularly noticed the $n+S(n)+S(S(n))$ problem in it. And the next day, I said to my self 'o my gods! i saw this yesterday' So yea, Engel might save you. The probability is almost 0, but that's like saying that when flipping a coin, the coin can't ever land on its circumference.  :P[/quote]
Was that problem really in PSS?[/quote]

It's not really that much of a surprise, considering the enormous scope of problems that PSS happens to cover.[/quote]
Also, wasn't number 25 on the 12 last year shown on Altheman's mock a couple weeks or so before that?

[quote="bpms"][/quote][quote="Elemennop"][quote="rnwang2"][quote="Karth"]Actually, as a bit of an aside, I'm happy that I got Engel's PSS, because the day before the AMC, as I did Engel, I particularly noticed the $n+S(n)+S(S(n))$ problem in it. And the next day, I said to my self 'o my gods! i saw this yesterday' So yea, Engel might save you. The probability is almost 0, but that's like saying that when flipping a coin, the coin can't ever land on its circumference.  :P[/quote]
Was that problem really in PSS?[/quote]

It's not really that much of a surprise, considering the enormous scope of problems that PSS happens to cover.[/quote][quote="bpms"]
Also, wasn't number 25 on the 12 last year shown on Altheman's mock a couple weeks or so before that?[/quote]

#23 on the 12B, yea. But considering they make the tests a year or more before they're administered, they couldn't really change it just cause Altheman stumbled upon the same prolbem that they decided to put on the test.

WOOOO

THATS ME

lol

but its not like any admin/amc person would look at my aime...so i don't think that the was even a consideration...and i agree they would not change it anyway...

P.S.

take my AIME [the current one]

[quote="Elemennop"][/quote][quote="bpms"][/quote][quote="Elemennop"][quote="rnwang2"][quote="Karth"]Actually, as a bit of an aside, I'm happy that I got Engel's PSS, because the day before the AMC, as I did Engel, I particularly noticed the $n+S(n)+S(S(n))$ problem in it. And the next day, I said to my self 'o my gods! i saw this yesterday' So yea, Engel might save you. The probability is almost 0, but that's like saying that when flipping a coin, the coin can't ever land on its circumference.  :P[/quote]
Was that problem really in PSS?[/quote]

It's not really that much of a surprise, considering the enormous scope of problems that PSS happens to cover.[/quote][quote="bpms"]
Also, wasn't number 25 on the 12 last year shown on Altheman's mock a couple weeks or so before that?[/quote][quote="Elemennop"]

#23 on the 12B, yea. But considering they make the tests a year or more before they're administered, they couldn't really change it just cause Altheman stumbled upon the same prolbem that they decided to put on the test.[/quote]
So they already have 2008 done?


floor function
inequalities
function
ceiling function
number theory proposed
number theory

here is something, suppose $(a,b)=1$ and $a>1$ let $p$ be a prime number greater than $a+b$ and big enough, such that the numbers $p+1,2p+1,3p+1...(k-1)p+1$ are not primes (construct it using the chinese remainder theorem and dirichlet) . Then we must have that $a^p+b^p$ has a prime factor that doesnt divides $a+b$, call it $q$, it is easy to see $p|q-1$ so $q$ is at least $kp+1$ and $q$ doesnt divides $(kp)!$.

So what? Am i wrong or the only solution is $a=1$,$b=1$ because getting rid of the common divisor doesnt affect the divisibility.

So what? I think you are not completely right, even though you found the idea. Don't you think that $2^n$ divides $ (4n)!$ for all sufficiently large $n$? ;)

ohh ok, so we only have to analize the case $a=b$ or also when one of them is $0$. right? these are the only possible values acording to what i did! :)

Here we have a completely elementary proof:

[quote="harazi"]For a given $k$ natural number find all couples (a,b) of positive integers such that $ a^n+b^n| (kn)!$ for all sufficiently large $n$.[/quote]

$\bullet$ Assume that $ab=0$ (wlog $b=0$ and $a\neq 0$), i.e. $a^n \mid (kn)!$ definitively. This divisibility holds true if and only if
\[ n\upsilon_p(a) \le \sum_{i \in \mathbb{N}_0}{\lfloor \frac{kn}{p^i}\rfloor},  \forall p\in \mathbb{P}\text{ s.t. }p\mid a\]
In particular, if a prime $p\mid a$ then
\[ n\upsilon_p(a)<\sum_{i\in \mathbb{N}_0}{\frac{kn}{p^i}}=\frac{kn}{p-1} \implies \upsilon_p(a) \in \mathbb{N} \cap [1,\frac{k}{p-1})\]
As a corollary we can say that if $p\mid a$ then $\upsilon_p(a)\ge 1$, so that $\frac{k}{p-1}>1$, that is equivalent to $p\le k$: it just implies that
\[ \omega(a) \le \pi(k) \]

$\bullet$ If $a=b>0$ then we must have $2a^n \mid (kn)!$ definitively, and we have the same kind of inequalities as previous case, except when $a$ is even: in that case we must surely have
\[ 1+n\upsilon_2(a)<kn \implies \upsilon_2(a)<k-n^{-1} \implies \upsilon_2(a) \le k-1 \]

$\bullet$ We missed the last case $a\neq b$, and both positive. Define $p_i$ the $i$-th prime in increasing order of $\mathbb{P}$, i.e. $p_1=2,p_2=3,\ldots$. We know that if have we two integers $x,y$ greater than $1$ such that $x \mid y$ then $\text{gpf}(x)\le\text{gpf}(y)$ (the function $\text{gpf}(\cdot)$ defines the greatest prime factor of his argument). In particular we have that the following inequality must hold:
\[ \text{gpf}(a^n+b^n)\le \text{gpf}((kn)!) \]
Considering the elementary [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=60&t=338697]Zsigmondy theorem[/url] and a almost trivial bound on $\pi(\cdot)$ we have
\[ p_{n-1} \le \text{gpf}(a^n+b^n)\le \text{gpf}((kn)!)=p_{\pi(kn)} \]
But since $p_i$ is trivially strictly increasing in $i$ then $n-1 \le \pi(kn)$, and it's false definitively, since $\pi(x)=o(x^\varepsilon)$ for every real $0<\varepsilon<1$. []

[b]Corollary:[/b] We can calculate then the exact number of couple of solution $(a,b)$: considering the symmetrical situation of case 1, and that in first cases the bound is exactly the same on $p=2$ (i.e. $\upsilon_2(a) \le k-1$), then the required number is
\[ \displaystyle 3( \prod_{i=1}^{\pi(k)}{((\lceil \frac{k}{p_i-1}\rceil} -1)+1))=3\prod_{i=1}^{\pi(k)}{\lceil \frac{k}{p_i-1}\rceil} ) \]
where the "$+1$ refers to the possible null exponent.
Considering now that $\lceil x \rceil < x+1$ for every positive real $x$ and that $\pi(m)\ge \frac{1}{2}(m+1)$ for every positive integer $m$, we can compute a upper bound:
\[ 3\prod_{i=1}^{\pi(k)}{\lceil \frac{k}{p_i-1}\rceil} < 3\prod_{i=1}^{\pi(k)}{( \frac{k}{p_i-1}+1)}= \]
\[{ =3k^{\pi(k)}\prod_{i=1}^{\pi(k)}{(\frac{1}{p_i-1}-\frac{1}{k})}<3k^{\frac{1}{2}(k+1)}}\prod_{i=1}^{\pi(k)}{(\frac{1}{p_i-1})}\le \]
\[{{ \le 3k^{\frac{1}{2}(k+1)}}\prod_{i=1}^{\pi(k)}{\frac{1}{i}} \le 3k^{\frac{1}{2}(k+1)}}\]

Regards,
Paolo

Very nice problem:
Here's my solution:
Theorem: let $S$ product of prime numbers between $1,2,\cdots ,n$ then $S<4^n$.(it wellknown you can find the solution in problems from the book)
Assume that $a,b$ are two natural numbers such that there exist $N$ such that for every $n\ge N$ : $a^n+b^n\mid (kn)!$ assume that $m$ is an arbitrary positive integer then using zigmondy's theorem there are different primes $p_0,p_2,\cdots ,p_m$ such that:
$p_0\mid a^N+b^N\mid (kN)!$
$p_1\mid a^{N+1}\mid (k(N+1)!)$
…
$p_m\mid a^{N+M}\mid (k(N+m))!$
It's clear that numbers $p_0,p_1,\cdots ,p_m$ are all less than $k(N+m)$
From the theorem $p_0p_1\cdots p_m<4^{kN+km}$ one the other hand assume that there are $r$ prime numbers between $1,2,\cdots ,4^{k^2}$ then $4^{k^2(m-r)}<p_0p_1\cdots ,p_m<4^{k(N+m)}\longrightarrow 4^{(k-1)m-rk)}<4^{n+m}$ but this is a contradiction because $m$ is an arbitrary number and RHS of this inquality is constant.
So $a=b$ or $ab=0$ are the only solutions.
DONE


function
limit
trigonometry
calculus
calculus computations

[color=darkblue]Prove that the function $f(x)\equiv\frac{1-\cot x}{e^{x}}$ hasn't limit in the point $x_{0}=\infty\ .$[/color]

Let's take $f : \mathbb R \setminus \left\{ n \pi : n \in \mathbb Z \right\}\to \mathbb R$, $f(x) = \frac{1-\cot x}{e^{x}}$.

We have that $f \left( n \pi+\frac{\pi}2 \right) \to 0$ as $n \to \infty$.

Also, $\lim_{x \to n \pi, x > n \pi}f(x) =-\infty, \, \forall n \in \mathbb Z$. Thus, there's no limit in $\infty$.

[quote="perfect_radio"]$\lim_{x \to n \pi, x > n \pi}f(x) =-\infty, \, \forall n \in \mathbb Z$.[/quote]

[color=darkblue]What is the above "affair" ? I think that you was  wrong. I didn't understand. In my opinion, this exercise isn't "spam".[/color]

Why? Don't we have $\lim_{x \to n \pi, x > n \pi}\cot x = \infty$, from where we get that $\lim_{x \to n \pi, x > n \pi}f(x) = \frac{1-\infty}{e^{n \pi}}=-\infty$?

I'll go to sleep now. It's too late to see where I'm wrong.

[color=darkblue]Genarally, for any $r\in [-\infty ,\infty ]$ there is a sequence $x_{n}\rightarrow\infty$ so that $\lim_{n\to\infty}f(x_{n})=r\ .$

[b]Proof.[/b] Let $x_{n}=n\pi-a_{n}\ ,\ n\in N$ be a sequence, where $a_{n}\rightarrow 0\ .$ Thus,

$f(x_{n})=\frac{1+\tan a_{n}}{e^{-a_{n}}}\cdot \frac{a_{n}}{\tan a_{n}}\cdot \frac{1}{a_{n}e^{n\pi }}$ $\Longrightarrow$  $\lim_{n\to\infty}f(x_{n})=\lim_{n\to\infty}\frac{1}{a_{n}e^{n\pi}}=\{\begin{array}{ccc}0 & , & a_{n}=\frac{1}{n}\ .\\\\ r & , & a_{n}=\frac{1}{re^{n\pi}}\ .\\\\--\infty & , & a_{n}=-\frac{1}{e^{2n\pi}}\ .\\\\ \infty & , & a_{n}=\frac{1}{e^{2n\pi}}\ .\end{array}$[/color]


Using Chebyshev inequality $ \frac{(a_1\plus{}a_2\plus{}...\plus{}a_n)}{n}\cdot \frac{(a_1\plus{}a_2\plus{}...\plus{}a_n)}{n}\cdot \frac{(a_1\plus{}a_2\plus{}...\plus{}a_n)}{n}\leq \frac{(a_1\plus{}a_2\plus{}...\plus{}a_n)}{n}\cdot \frac{a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2}{n}\leq \frac{a_1^3\plus{}a_2^3\plus{}...\plus{}a_n^3}{n}$ QED


Mporei kapoios na to metafrasei sta agglika kai na to postarei sto
Number theory tou Olympiad section. Euxaristo  :)

Etsw d(x) h synarthsh pou gia kathe fysiko x apeikonizei to plhthos twn prwtwn paragontwn tou.
Estw Σ to athroisma twn timwn ths d(x) apo to 1 mexri to enan fysiko n>1
Estw S to athroisma twn antistrofwn twn prwtwn arithmwn pou einai mikroteroi apo enan fysiko n>1
Boreite na deixete oti

lim (S - Σ/ν) = 0
ν->+00

Eυχαριστω Νick  :) , προσπαθησε την και εσυ ...

Έκανα μια προσπάθεια να το βγάλω, και σου τη γράφω.

Το δικό σου $S$ ορίζω ως: $S : \mathbb{R}\rightarrow \mathbb{R}$ με $S(x) = \sum_{p < x}\frac{1}{p}$ και τα $p \in \mathbb{P}$.

Ομοίως για το Σ σου: $D : \mathbb{R}\rightarrow \mathbb{R}$ με $D(x) = \sum_{k \leq x}d(k) = \sum_{k \leq x}\sum_{p | k}1$ και $p \in \mathbb{P}$ και $k \in \mathbb{N}$*.

$S(x)-\frac{D(x)}{x}= \sum_{p<x}\frac{1}{p}-\frac{1}{x}\sum_{k \leq x}\sum_{p | k}1$

Το διπλό άθροισμα απλοποιούμε (πώς; ) και έχουμε:

$= \sum_{p < x}\frac{1}{p}-\frac{1}{x}\sum_{p \leq x}\left\lfloor \frac{x}{p}\right\rfloor = F(x)$

αλλά (από την ανισότητα: $x-1 < \lfloor x \rfloor \leq x$):

$F(x) \geq \sum_{p < x}\frac{1}{p}-\sum_{p \leq x}\frac{1}{p}$ που είναι ένας μικρός (κατά απόλυτη τιμή) μη-θετικός αριθμός, που τίνει στο $0^{-}$ όσο $x \rightarrow+\infty$.

και επίσης:

$F(x) < \sum_{p < x}\frac{1}{p}-\frac{1}{x}\sum_{p \leq x}\left( \frac{x}{p}-1 \right) = \sum_{p<x}\frac{1}{p}-\frac{1}{x}\sum_{p \leq x}\frac{x}{p}+\frac{1}{x}\sum_{p \leq x}1 = \\ \sum_{p<x}\frac{1}{p}-\sum_{p\leq x}\frac{1}{p}\; \; \;+\frac{\pi(x)}{x}$

και όσο $x \rightarrow+\infty$ το δεξί μέλος ισούται με:

$F(x) < \sum_{p<x}\frac{1}{p}-\sum_{p\leq x}\frac{1}{p}\; \; \;+\frac{1}{\ln x}$

και εδώ όμοια με επάνω, όσο $x \rightarrow+\infty$ τόσο το δεξί μέλος τίνει στο $0^{+}$ (αφού η $\ln x$ τίνει στο $+\infty$ πιο αργά από την $x$)

Άρα για $x \rightarrow+\infty$ έχουμε ότι $F(x) = 0$

Αν δεν τη βρίσκεις ικανοποιητική (ή λάθος), ευχαρίστως να σου μεταφράσω την ερώτηση στα Αγγλικά.

EDIT: Με πρόλαβε ο Νίκος :(

Προς cmad, τη θελεις να πεις με {Αν δεν τη βρίσκεις ικανοποιητική (ή λάθος), ευχαρίστως να σου μεταφράσω την ερώτηση στα Αγγλικά}.

Παντως ωραια προσεγγιση. Θελω να δω με ποσους διαφορετικους
τροπους μπορει να λυθει.

Θέλω να πω αν τελικά έχω κάποιο λάθος (ή τέλος πάντων δε σου αρέσει η λύση  :D ) τότε θα έκανα post το πρόβλημα στο number theory forum (αλλά το έκανε άλλος ήδη)


geometry
analytic geometry
quadratics
area of a triangle
Heron\u0027s formula
algebra
geometry unsolved

Given triangle ABC, AB=12, AC=13, BC=15.
M is on BC such that incircles in AMB and AMC have equal radii. Find AM:MC.
A complete solution will be much appreciated.
Even better if solved when AB=c, AC=b, BC=a.
I tried but obtained a cubic equation!

$ (I), (I_b), (I_c)$ are incircles of the $ \triangle ABC, \triangle ABM, \triangle AMC$ and $ T, T_b, T_c$ their tangency points with $ BC.$ $ S, S_b, S_c$ are the areas, $ s, s_b, s_c$ the semiperimeters, $ r, r_b, r_c$ the inradii of these 3 triangles, $ h$ their common A-altitude. From the triangle sides $ a, b, c,$ Heron's formula gives the area $ S,$ $ r \equal{} \frac {S}{s},$ and $ h \equal{} \frac {2S}{a}.$ Since the inradii $ r_b \equal{} r_c \equal{} \varrho$ are equal,

$ \frac {BM}{CM} \equal{} \frac {S_b}{S_c} \equal{} \frac {s_b}{s_c} \equal{} \frac {c \plus{} AM \plus{} BM}{b \plus{} AM \plus{} CM} \equal{} \frac {c \plus{} AM}{b \plus{} AM}$

Solving for $ AM,$

$ AM \equal{} \frac {b \cdot BM \minus{} c \cdot CM}{CM \minus{} BM} \equal{} \frac {(b \minus{} c) \cdot CM}{CM \minus{} BM} \minus{} b.$

where $ BM \equal{} a \minus{} CM.$ By the infamous inradii problem (see [url]http://www.mathlinks.ro/viewtopic.php?t=123243[/url] etc.), $ \frac {2}{h} \equal{} \frac {2}{\varrho} \minus{} \frac {r}{\varrho^2}.$ Solving and taking the only acceptable root $ \varrho < \frac {h}{2},$

$ \frac {\varrho}{r} \equal{} \frac {h \minus{} \sqrt {h(h \minus{} 2r)}}{2r} \equal{} \frac {s \minus{} \sqrt {s(s \minus{} a)}}{a}.$

The following segment lengths are easily calculated:

$ BT_b \equal{} (s \minus{} b) \frac {\varrho}{r},\ \ CT_c \equal{} (s \minus{} c) \frac {\varrho}{r},$

Circle $ (P)$ with diameter $ I_bI_c$ cuts $ BC$ at $ M, T$ (see [url]http://www.mathlinks.ro/viewtopic.php?t=177984[/url]). Put the coordinate origin at $ C$ and positive x-axis along the ray $ \vec{CB}.$ Midpoints $ P, Q$ of $ I_bI_c \equal{} T_bT_c$ have the same x-coordinate $ CQ \equal{} CT_c \plus{} \frac {I_bI_c}{2}.$ Equation of the circle $ (P)$ with radius $ \frac {I_bI_c}{2}$ and centered at $ P \equal{} \left(CT_c \plus{} \frac {I_bI_c}{2},\ \varrho\right)$ is then

$ \left(x \minus{} CT_c \minus{} \frac {I_bI_c}{2}\right)^2 \plus{} (y \minus{} \varrho)^2 \equal{} \left(\frac {I_bI_c}{2}\right)^2.$

Substituting $ y \equal{} 0$ yields quadratic equation for the x-coordinates $ CM, CT$ of its intersections $ M, T$ with the line $ BC$ (the x-axis). Using $ CT_c \plus{} I_bI_c \equal{} CT_b,$

$ x^2 \minus{} (CT_c \plus{} CT_b)\ x \plus{} CT_c \cdot CT_b \plus{} \varrho^2 \equal{} 0.$

We could ordinarily solve, but one root $ x_1 \equal{} CT \equal{} s \minus{} c$ is known, hence the other root is

$ x_2 \equal{} CM \equal{} CT_c \plus{} CT_b \minus{} x_1 \equal{} CT_c \plus{} a \minus{} BT_b \minus{} (s \minus{} c) \equal{}$

$ \equal{} \left((s \minus{} c) \minus{} (s \minus{} b)\right) \frac {\varrho}{r} \plus{} (a \plus{} c \minus{} s) \equal{} (b \minus{} c) \frac {\varrho}{r} \plus{} (s \minus{} b).$

Substituting the given values $ a \equal{} 15, b \equal{} 13, c \equal{} 12,$ we get $ s \equal{} 20,$ $ \frac {\varrho}{r} \equal{} \frac {2}{3},$ $ CM \equal{} \frac {23}{3},$ $ BM \equal{} \frac {22}{3},$ $ AM \equal{} 10,$ $ \frac {AM}{CM} \equal{} \frac {30}{23}.$


I'm in AP CALC BC right now. Should I take anything during the summer? And what class should that be (or a general list of recommended college math courses)?

If you successfully complete BC Calculus (i.e. get a 4 or 5 on the exam), you should think about taking Multivariable Calculus (also sometimes called Calculus III or Vector Calculus). If you get less than a 4 on the BC exam, I would recommend taking Calculus II again.

Other options are discrete math, number theory, or advanced euclidean geometry, as they don't usually have advanced prerequisites.

Be aware though, that summer school isn't necessarily the best time to take a college class. Either they can be somewhat watered down, or they can go very fast (depending on the teacher and the school where you take it).

[quote="churchilljrhigh"]I'm in AP CALC BC right now. Should I take anything during the summer?[/quote]

What's your goal for taking a summer course? What you hope to accomplish has a lot to do with what I might recommend. All of the recommendations you have received from Gauss202 are good, standard recommendations, but what do you have in mind to do? What do you hope to do next school year that you could prepare for in the summer?

(After edit:) I will ruthlessly delete from this forum all posts that cause horizontal scrolling in my browser window. It should be possible to edit the signature file with excessively wide graphics to something more reasonable before I delete all of the posts.

How accurate are the practice tests in the Barron's book?  I have all of the actual free response questions from ~70's, but no mc, so I am looking for realistic multiple choice questions.


inequalities
geometry
geometric transformation

Sau hơn hai năm chuẩn bị và viết, chung tôi xin giới thiệu cuốn sách bất đẳng thức đầu tay tới các bạn độc giả.

Chúng tôi sẽ cố gắng đưa mục lục và một đến hai chương free trong thời gian sớm nhất có thể.

Rất cảm ơn Thuận vì nhã ý này :P Nhưng này, sao ông phải đánh dấu nặng đô thế? Chỉ cần đánh chú ý bậc 1 là được rồi mà? :D

[quote="VO_DANH_KHACH"]Bao gio co ban English anh mua mot ban, ban thao tieng Viet anh co roi :P[/quote]
Bác có thể gửi cho em một bản phô tô được không bác?  :ninja: Cảm ơn bác! Chết Thuận roài :P

[quote="VO_DANH_KHACH"]Bao gio co ban English anh mua mot ban, ban thao tieng Viet anh co roi :P[/quote]

Anh Vo danh khach vui tính thật. Reply và lời nhắn rồi mà không thấy trả lời.

Thực ra khả năng lộ bản thảo là ít lắm, em toàn gửi những người đáng tin cậy, và tốt bụng thôi.

Thuận này, hay cậu gửi mình một bản đi! Thực ra mình cũng đáng tin cậy lắm, và tốt nữa!  :maybe: Vì bạn gái mình thường nói là : ANH LÀ NGƯỜI ĐÀN ÔNG ĐÁNG TIN CẬY, SỐNG CHÂN THÀNH VÀ TỐT BỤNG, EM YÊU ANH LẮM !    :o . Thế nhỉ? :P

Nhưng như thế vẫn mạo hiểm vì có thể người thẩm định ấy lại "không tin cậy". Chít nhé  :lol:

Đểu thế chứ lại :P Nhưng tháng 8 này bản tiếng Anh chắc ra rồi , đúng không? Cho kịp IMO 2007 mà? Tháng 8 này tôi có việc lên Hà Nội, hy vọng ông chỉ cho tôi chỗ bán, tôi mua lấy một cuốn, cả cuốn của  Hùng nữa, thấy bảo nó cũng ra bản tiếng Anh vào dịp đó mà, cả một cuốn của Anh nữa thì phải? Thế nhở ? :D

Hôm nay cuốn sách đã được in xong, ruột 332 trang.

Thế nào? Ông có định tặng tôi một bản hay không đây? :P Hay bạn phải mua ? :D

Sao toàn thấy 2 bác Mod chát chít trong này thế nhỉ? Các bác phải gương mẫu cho các mem noi theo, chứ không chúng nó spam  thì toi. :D Hơi bất ngờ là bác Thuan cộng tác với thằng Lê Vĩ bạn em. Và vì thế khả năng lộ bản thảo là rất có thể. Nhưng bác yên tâm là không ai đem đi in trước bác đâu.
Nếu sau này chẳng may có ai có ý định viết số học hay tổ hợp thì nhắn em một tiếng, mấy cái bất đẳng thức em không dám động đến, may hồi xưa đi thi toàn BĐT dễ, chứ em đọc trong sách của bác với của thằng Hùng chẳng hiểu gì cả và cũng chẳng làm nổi bài nào. Từ bé đến lớn quanh đi quẩn lại em biết có mỗi Cauchy-Schwartz.

Uh, anh chả nói chuyện mãi với chú về mấy cái thứ ấy là gì. Học nhiều bdt làm gì, cứ đi theo hướng hiện nay của chú ấy. Ngày trước, chưa xuất bản sách thì thấy háo hức, chứ bây giờ thấy bình thường lắm. Chẳng có gì là to tát với những bài toán kiểu này, và cũng thấy rất thờ ơ đến mức tội tội nếu người ta thần thánh, hay ca ngợi bất đẳng thức.

Ah, mà sao chú lại đánh giá thấp Vĩ về cái khoản giữ bí mật thế?

[quote="nmtruong1986"]Sao toàn thấy 2 bác Mod chát chít trong này thế nhỉ? Các bác phải gương mẫu cho các mem noi theo, chứ không chúng nó spam  thì toi. :D [/quote]
Đâu? Thực ra Toán thì post bên ngoài forum chính kia rồi, cái box VN này chủ yếu để giao lưu, và bàn các công việc liên quan, nhưng dạo này không có kế hoạch gì cả :D Anh và Thuận không chat với nhau thì cái forum này nó lạnh lẽo như bị ma ám ấy, hồi xưa còn có bác Analysis nữa, cũng vui lắm, cả bác Tom nữa :P

Thế bản tiếng Anh đặt mua và thanh toán qua ATM được không ạ?

[quote="N.T.TUAN"]Hôm nay mình được một thằng bạn tặng bản photo của một cuốn b đ t mới của Trần Phương, trong đó có nhiều bài của các thành viên hai diễn đàn MNF,VMF. Tên là: ''Các viên kim cương trong b đ t'' , đọc cũng hay. Việt Nam mình nhiều sách b đ t quá ! :D[/quote]
Trong đấy còn nói : Ai là tác giả của SOS nữa thì phải?  :lol:

@ N.T.TUAN  em định chuyển tiền cho anh rồi , nhưng mà ra chuyển mới biết thẻ nó khác hệ thống, anh cho em xin cái địa chỉ của ngân hàng mà anh mở thẻ cái, rồi em chuyển vào
À, anh có lên HN thì có thể mua júp em quyển BĐT của thầy Thuận kô ạ, em chuyển tiền luôn 1 lần :lol:

[quote="eagerkill"]
À, anh có lên HN thì có thể mua júp em quyển BĐT của thầy Thuận kô ạ, em chuyển tiền luôn 1 lần :lol:[/quote]
Có, để anh giúp chú!  :lol:

[quote="N.T.TUAN"][quote="eagerkill"]Anh Tuân có thể share cho em 1 bản photo quyển  BĐT của Trần Phương đc kô ạ ![/quote]
Được!  :wink:[/quote]

Cho em luon, quyen do dc quang cao du qua . :D

Chương sách khá hay, vừa đầy đủ vừa nâng cao . :lol: .Khi nào đủ tiền em phải mua 1 cuốn .Dạo này túng quá  :(

Anh Tuấn share cho em quyển BĐT của Trần Phương luôn dc ko ạ ^^

[quote="N.T.TUAN"]Qúa được, gửi tiền cho anh, rồi anh gửi sách cho! :)[/quote]
Quyển đấy bao nhiêu dollar hả thầy? Có phải quyển 3333 trang mà mấy anh Phạm Gia Vĩnh Anh với Lê Hùng Việt Bảo quảng cáo không ạ?

Không phải đâu em. Cuốn anh có , chỉ hơn 200 tr.

can't we have an english translation of those inequalities book  :maybe:

[quote="quangpbc"]Hi pardesi ! I know it has a different version by English . All inequalities above very nice , if you have enough money, i think you should buy them  :D[/quote]

  :ninja:  i thought u must be having one. :rotfl:


integration
limit
real analysis
inequalities
real analysis solved

is a positive measure on X, f:X->[0;oo] is measurable,
\int_{X}fd=c where 0<c<oo, b>1 is a constant. Prove that
\lim_{n->oo}\int_{X}(n.ln(1+(f/n)^b)d=0
From Rudin Chapter 1 problem 9

g_n(x) = n.ln(1+ (f(x)/n)^b) converge pointwise to 0
with the inequality 1+y^b  \leq (1+y)^b for y \geq 0
|g_n(x)|  \leq | b.n.ln(1+f(x)/n) | \leq b.|f(x)| wich is integrable on X
Lebesgue dominated convergence theorem gives

   \lim _{n->oo} \int _{X}(n.ln(1+(f/n)^b)d=0


vector
topology
real analysis
real analysis unsolved

I have a small question in $ \mathbb R^n$. If $ G_1 \subset G_3$ are open sets. Does there exist another open set $ G_2$ such that $ G_1 \subset G_2 \subset G_3$ with $ \overline G_1 \subset G_2$, $ \overline G_2 \subset G_3$ and $ G_2 \minus{}G_1 \subset G_3 \minus{} G_2$?

Simple counterexample:
$ n \equal{} 1, \ G_{1} \equal{} (0,1), \ G_{3} \equal{} (0, 2)$.
Then $ \overline{G_{1}} \equal{} [0,1] \not\subset G_{3}$, so there cannot exists $ G_{2}$ satisfying condition:
$ \overline{G_{1}}\subset G_{2}\subset G_{3}$.

Sorry, assume further $ \overline G_1 \subset G_3$.

Then existence of open set $ G_{2}$ such that:
$ \overline{G_{1}} \subset G_{2} \subset \overline{G_{2}} \subset G_{3}$
follows from normality of space $ \mathbb{R}^{n}$.

[quote="bookworm_vn"](...) and $ G_2 \minus{} G_1 \subset G_3 \minus{} G_2$?[/quote]
I'm not sure what do you mean by $ A \minus{} B$ where $ A, B$ are subsets of $ \mathbb{R}^{n}.$
If it is algebraic difference, i.e. the set of all vectors $ a \minus{} b$, where $ a\in A, \ b\in B$, this is trivially true for all $ G_{1}\subset G_{2}\subset G_{3}\subset \mathbb{R}^{n}.$
If it is set-theoretic difference, i.e. the set of all elements $ a\in\mathbb{R}^{n}$ st $ a\in A, \ a \not \in B$, this is false, since then there must be $ \overline{G_{1}}\subset G_{2} \equal{} G_{1}$ but $ \mathbb{R}^{n}$ and $ \emptyset$ are only subsets of $ \mathbb{R}^{n}$ which are open and closed.

[quote="ifk"]If it is algebraic difference, i.e. the set of all vectors $ a \minus{} b$, where $ a\in A, \ b\in B$, this is trivially true for all $ G_{1}\subset G_{2}\subset G_{3}\subset \mathbb{R}^{n}.$[/quote]

What I mean is the algebraic difference but how to point out the existence of $ G_2$?

[i]Definition:[/i] Topological $ T_{1}$ space $ (X, \tau)$ is [b]normal[/b] if for every two closed disjoint subset $ F_{1}, F_{2}\subset X$ exists $ U_{1}, U_{2}\in \tau$ such that $ F_{1}\subset U_{1}, \ F_{2}\subset U_{2}$ and $ U_{1}\cap U_{2} = \emptyset$.
(topological space is said to be $ T_{1}$ space if for every $ x\in X$ set $ \{x\}$ is closed).

[i]Observation 1:[/i] Every metric space is normal.
[hide="proof"]It is easy to see that metric space is $ T_{1}$.
Let $ (X, d)$ be the metric space, $ F_{1}, F_{2}$ be closed, disjoint subsets of $ X$.
For $ \emptyset\neq A\subset X$ and $ x\in X$ we define $ \mbox{dist}(x, A): = \inf\{d(x, y) \ : y\in A\}$
Let $ f: X\ni x \mapsto \frac {\mbox{dist}(x, F_{1})}{\mbox{dist}(x, F_{1}) + \mbox{dist}(x, F_{2})} \in [0, 1]$
Then $ f$ is continuous, $ F_{1}\subset f^{ - 1}([0,\tfrac{1}{2}))\in \tau, \ F_{2}\subset f^{ - 1}((\tfrac{1}{2}, 1])$, and we can define $ U_{1} = f^{ - 1}([0,\tfrac{1}{2})), \ U_{2} = f^{ - 1}((\tfrac{1}{2}, 1])$[/hide]

[i]Observation 2[/i]
If $ (X,\tau)$ is normal, then for every closed set $ F\subset X$ and $ U\in \tau$ such that $ F\subset U$ there exists $ V\in \tau$ satisfying:
$ F\subset V\subset \overline{V}\subset U$
[hide="proof"]$ F, X\setminus U$ are closed and disjoint, thus from normality of $ (X, \tau)$ there exists disjoint $ V, W\in \tau$ such that $ F\subset V, \ X\setminus U\subset W$. Observe that $ V\subset X\setminus W \subset U$, and $ \overline{V}\subset X\setminus W \subset U$ since $ X\setminus W$ is closed.[/hide]
[hide="Remark"]In fact, if we assume that $ (X, \tau)$ is $ T_{1}$, then it is easy to prove, that the above conditions are equivalent - if for every closed set $ F$ and for every $ U\in \tau$ such that $ F\subset U$ there exists $ V\in \tau$ satisfying $ F\subset V\subset \overline{V}\subset U$, then $ (X, \tau)$ is normal space.[/hide]

Now from second observation we obtain existence of $ G_{2}$.

edit:
To be consistent with well-known definition of normality I added assumption that $ (X, \tau)$ is $ T_{1}$, but our observations holds also without it. I also added simple remark.

[quote="ifk"]Now from second observation we obtain existence of $ G_{2}$.[/quote]

How to induce $ G_2 \minus{} G_1 \subset G_3 \minus{} G_2$? My question has its geometric meaning.

[quote="bookworm_vn"][quote="ifk"]Now from second observation we obtain existence of $ G_{2}$.[/quote]

How to induce $ G_2 \minus{} G_1 \subset G_3 \minus{} G_2$?[/quote]

If $ x\in G_{2} \minus{} G_{1}$, then exists $ g_{2}\in G_{2}$ and $ g_{1}\in G_{1}$ st $ x \equal{} g_{2} \minus{} g_{1}$. But $ G_{2}\subset G_{3}$ and $ G_{1}\subset G_{2}$, so we have $ g_{2}\in G_{3}$ and $ g_{1}\in G_{2}$, and $ x\equal{}g_{2} \minus{} g_{1} \in G_{3} \minus{} G_{2}$

[quote="bookworm_vn"]One more question. Does $ \overline G_1 \subset G_3$ imply that $ \operatorname{dist}(G_1, G_3) > 0$?[/quote]

Well, I defined $ \operatorname{dist}$ on $ X\times \mathcal{P}(X)$, not on $ \mathcal{P}(X)\times \mathcal{P}(X)$. If you want to define distance between subsets of metric space keyword is Hausdorff metric or Hausdorff distance.


The excircle of triangle $ ABC$ opposite $ AB$ touches $ AB$ at $ P$ and $ AC$ at $ Q$. The excircle opposite $ BC$ touches $ AC$ at $ U$ and $ AB$ at $ X$. Let $ Y=PQ\cap UX$. Prove that $ Y$ is equidistant from $ AB$ and $ BC$.


prove or disprove:
let $ a,b,c$ be positive reals then:

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

[quote="BaBaK Ghalebi"]prove or disprove:
let $ a,b,c$ be positive reals then:

$ \frac {a \minus{} c}{a \plus{} c} \plus{} \frac {b \minus{} a}{b \plus{} a} \plus{} \frac {c \minus{} b}{c \plus{} b}\leq (a \minus{} c)^2 \plus{} (b \minus{} a)^2 \plus{} (c \minus{} b)^2$[/quote]
I think, it's wrong. Try $ a\equal{}0.03,$ $ b\equal{}0.2$ and $ c\equal{}0.4.$ :wink:

yes,you're right arqady... :(
as usual... :D

Well the inequality is obviously false, since the degrees of $ LHS$ and $ RHS$ are diffrent, so if we only can make the $ LHS \geq 0$ , then by multiplying every $ a,b,c$ by some constant, we will get the counterexample.

[quote="Myszon11"] so if we only can make the $ LHS \geq 0$ , then by multiplying every $ a,b,c$ by some constant, we will get the counterexample.[/quote]

I didnt understand well, could you clarify more?

Suppose we have some $ a,b,c$ such that $ LHS \geq 0$. We shall call $ LHS \equal{} x$, $ RHS\equal{}y$, so we have $ x\leq y$. Lets take $ a_2 \equal{} a\cdot \frac{x}{2y}$ (the constant i mentioned in the previous post can be for example $ \frac{x}{2y}$), similarlly with $ b_2$ and $ c_2$. We write the inequality for $ a_2 , b_2 , c_2$ we observe that $ LHS$ is still $ x$, and $ RHS$ is $ \frac{x^2}{4y}$ so we should have $ 4y \leq x$, which is a contraddiction.

Observe that multiplying all the variables by some constant multiplied the $ LHS$ by another number than it multiplied the $ RHS$ (the $ LHS$ got multiplied by$ (\frac{x}{2y})^{0} \equal{} 1$, and $ RHS$ got multiplied by $ (\frac{x}{2y})^{2}$), and it is because the degrees of $ LHS$ and $ RHS$ are diffrent.

I hope it is clear now. If it's not (which means I just can't say what i mean because I'm a dummy :P) just take some constant value, multilpy $ a,b,c$ by it, and see what happens :P


[quote="Interval"]A horizontal bridge is in the shape of a parabolic arch.  What is the height h of the arch 2 feet from shore?

DIAGRAM IN BOOK HAS:

(a) center height from bridge to water = 10 feet

(b) h = ? 2 feet from shore under the bridge

(c) the arch bridge is 20 feet across

So, basically, I need to find the height from the water to the bridge located TWO FEET from shore.[/quote]

let the vertex be on (0,10) so the x-intercepts are (10,0) and (-10,0).  Thn you can find the equation for the porabola and plug it in.


linear algebra
matrix
superior algebra
superior algebra solved

$A,B,C \in M_n(C)$ wich commute by pair
suppose also
$A^3+B^3+C^3-3ABC=I_n$

Prove that $A+B+C$ is invertible

I'm sure it's been proposed for quite a few Romanian contests :). Did you propose it, Moubinool? :)

That expression equals $(A+B+C)(A+\varepsilon B+\varepsilon^2 C)(A+\varepsilon^2 B+\varepsilon C)$, I think.

Or you can use:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. Of course it is the same thing, but this is known before highschool. At least in Romania it is.

It is also used to prove $a^3+b^3+c^3\geq 3abc$ iff $a+b+c\geq 0$. a,b,c are reals.


calculus
integration
invariant
symmetry
real analysis
real analysis unsolved

Compute:
 $\int_{0}^{u}\frac{arctgx}{\(x^{2}+ux+2}\,dx$, where u is in $(0, 2\sqrt{2})$

Here is an idea:

First of all observe that $\arctan t+\arctan \frac{u-t}{1+ut}=\arctan u$.

Let $I$ the value of your integral. Then, the substitution $x=\frac{u-t}{1+ut}$ shows that

$I=\int_{0}^{u}\frac{\arctan{\frac{u-t}{1+ut}}}{t^{2}+ut+2}dt=\arctan u\int_{0}^{u}\frac{dt}{t^{2}+ut+2}-I$.
It follows that

$I=\frac{\arctan u}{2}\int_{0}^{u}\frac{dt}{t^{2}+ut+2}$.

The trick is to observe that the denominator is invariant under Mobius transformations.
I hope there are no mistakes.

What is a Mobius transformation didilica? :maybe:

Ciprian,
A Mobius transformation is an automorphism  of the unit disk, it has the form $\sigma=\xi \sigma_{w}$, where $\sigma_{w}(z)=\frac{w-z}{1-\bar{w}z}$ and $\xi$ is a complex number of modulus 1. $\sigma_{w}$ is also called symmetry which interchanges 0 and w.
For our case, I called the substitution $x=\frac{u-t}{1+ut}$, the Mobius substitution.
Didi

Thanks a lot didilica for your explanation :)


calculus
integration
logarithms
trigonometry
calculus computations

Prove the following inequality. \[-\frac{\pi}{3}\ln 2+\frac{\pi^{3}}{81}<\int_{0}^{\frac{\pi}{3}}\ln (\cos x) dx<-\frac{\pi^{3}}{162}.\]

[b]Upper Bound:[/b]

Lemma : On the interval $\left(0,\frac{\pi}{3}\right]$, we have $\ln(\cos x)<-\frac{1}{2}x^{2}$.

Proof: Let $f(x) =-\frac{1}{2}x^{2}$ and $g(x) = \ln(\cos x)$, then $f(0)=g(0)=0$, also $f'(x) =-x$ and $g'(x) =-\tan x$.  So we must prove that on the interval $\left(0,\frac{\pi}{3}\right]$, that $\tan x > x$. Let $u(x) = \tan x$ and $v(x) = x$ (note that $u(0)=v(0)=0$), then $u'(x) = \frac{1}{\cos^{2}x}$ and $v'(x) = 1$, and clearly $\frac{1}{\cos^{2}x}> 1$ on the given interval, and thus $\tan x > x$ on the given interval, and the proof is complete.

From the Lemma  (and because both quantities are negative on the given interval), we have $\int_{0}^{\frac{\pi}{3}}\ln(\cos x) \ dx <-\frac{1}{2}\int_{0}^{\frac{\pi}{3}}x^{2}\ dx =-\frac{\pi^{3}}{162}$.

[b]Lower Bound:[/b]

$\frac{\pi^{4}}{4}+\pi^{3}<\frac{\pi^{4}}{2}<27\pi\ln(2)\Rightarrow 27\pi\ln(2)-\pi^{3}>\frac{\pi^{4}}{4}\Rightarrow-\frac{\pi}{3}\ln(2)+\frac{\pi^{3}}{81}<$$-\frac{\pi^{4}}{4\cdot 3^{4}}=-\int_{0}^{\frac{\pi}{3}}x^{3}\ dx$.  On the given interval, $-x^{3}<\ln\cos(x)$, (can be proved by the same logic as above), and thus:

$-\frac{\pi}{3}\ln(2)+\frac{\pi^{3}}{81}<\int_{0}^{\frac{\pi}{3}}\ln(\cos x) \ dx$.

And the proof is complete.


number theory
relatively prime
number theory unsolved

Prove that if we take 16 consecutive natural numbers, then we can found such a  one number, that is relatively prime with the 15 others.

Hint: Consider primes from $2$ to $13$ (as for $17$ and over only one from the given number can be divided by it)


Let $ a,b,c$ be real numbers. Prove that

$ \sqrt {2(a^2\plus{}b^2)}\plus{}\sqrt {2(b^2\plus{}c^2)}\plus{}\sqrt {2(c^2\plus{}a^2)}\ge \sqrt{3[(a\plus{}b)^2\plus{}(c\plus{}b)^2\plus{}(c\plus{}a)^2]}$.

[quote="skywalkerJ.L."]Let $ a,b,c$ be real numbers. Prove that

$ \sqrt {2(a^2 \plus{} b^2)} \plus{} \sqrt {2(b^2 \plus{} c^2)} \plus{} \sqrt {2(c^2 \plus{} a^2)}\ge \sqrt {3[(a \plus{} b)^2 \plus{} (c \plus{} b)^2 \plus{} (c \plus{} a)^2]}$.[/quote]

Squaring both sides we have
\[ \sum {\sqrt {2\left( {a^2  \plus{} b^2 } \right)} }  \ge \sqrt {3\left[ {\sum {\left( {a \plus{} b} \right)^2 } } \right]}
\]
\[ \Leftrightarrow \sum {2\left( {a^2  \plus{} b^2 } \right)}  \plus{} 4\sum {\sqrt {\left( {a^2  \plus{} b^2 } \right)\left( {b^2  \plus{} c^2 } \right)} }  \ge 3\left[ {2\sum {a^2 }  \plus{} 2\sum {ab} } \right]
\]
\[ \Leftrightarrow 2\sum {\sqrt {\left( {a^2  \plus{} b^2 } \right)\left( {b^2  \plus{} c^2 } \right)} }  \ge \sum {a^2 }  \plus{} 3\sum {ab}
\]
By the Cauchy Schwarz inequality we get
\[ \sqrt {\left( {a^2  \plus{} b^2 } \right)\left( {b^2  \plus{} c^2 } \right)}  \ge ac \plus{} b^2
\]
Therefor it suffices us to show that
\[ 2\sum {a^2 }  \plus{} 2\sum {ab}  \ge \sum {a^2 }  \plus{} 3\sum {ab}
\]
\[ \Leftrightarrow \sum {a^2 }  \ge \sum {ab}
\]
true obviously. :)