Let a set S of 2004 points in the plane be given, no three of which are collinear. Let L denote the set of all lines (extended infinitely in both directions) determined by pairs of points from the set. Show that it is possible to colour the points of S with at most two colours, such that for any points p, q of S, the number of lines in L which separate p from q is odd if and only if p and q have the same colour. [A line t separates two points p and q if p and q lie on opposite sides of t, with neither point on t.]

Well, in fact it is easier than I thought :

We will prove that the conclusion holds even if 2004 is replaced by n, where n is even.
In all the following, a point is supposed to be one of the given n.

For A,B two distinct of the points, denote by d(AB) the number of lines which separates A and B.

Lemma : For any three distinct points A,B,C, we have
d(AB) + d(BC) + d(CA) = 1 mod[2].

Proof :
Let n(A) be the number of points which belong to the region defined by the  half-lines [AB) and [AC), but not to the triangle ABC.
Let n'(A) be the number of points which belong to the region delimited by the other half-lines from A, with the same direction than (AB) and (AC) (That is the angular sector opposite to the triangle and which shares the same <A than the triangle ABC)
The numbers n(B), n'(B), n(C) and n'(C) are defined in a similar way.

Let n(T) be the number of points which are in the triangle.
Since no three points are collinear, we then have :
n(A)+n'(A)+n(B)+n'(B)+n(C)+n'(C)+n(T) = n-3.
Moreover :
d(AB) = n(C)+n'(C)+n(T)
d(BC) = n(A)+n'(A)+n(T)
d(CA) = n(B)+n'(B)+n(T).
Summing, we get d(AB)+d(BC)+d(CA) = n - 3 + 2n(T) = 1mod[2] since n is even. So, the lemma is proved.

Now, let A_1,...,A_n be the points.
Let's color A_1 in red. And color A_i in red iff d(A_1A_i) is odd, and color A_i in blue otherwise.

Clearly the desired property holds if p or q is A_1.
Now, let's B and C be two of the points, distincts from A_1.
Fromm the lemma, we deduce that exactly one or three among the numbers d(A_1B), d(A_1C), d(BC) are odd.
- If both B and C are red, then both d(A_1B) and d(A_1C) are odd. Thus, d(BC) is odd too.
- If both B and C are blue, then both d(A_1B) and d(A_1C) are even, so that d(BC) is odd.
It follows that if B and C have the same color then d(BC) is odd.

- If B and C have different colors, say B is red and C is blue. Then, d(A_1B) is odd and d(A_1C) is even. Thus, d(BC) is even.

Pierre.

There're a few things that I don't quite understand in your proof.

(1)
Does the n in n(A) have anything to do with the n that replaces 2004?

(2)
d(AB) = n(C)+n'(C)+n(T):
Isn't n(C)+n'(C)+n(T) the number of lines through C that seperates A and B? They don't seem to account for the number of lines that seperates A, B which are not through C.

You are absolutely right that there must be some qualifying statement.

d(AB)=n(C)+n'(C)+n(T) is not itself correct, but it is close.

Remember is that we are ultimately considering
g(ABC)=d(AB)+d(BC)+d(CA) (mod 2).  Note that any line which separates AB either:

1) Passes through C-the number of such lines is n(C)+n'(C)+n(T) as claimed.

2) Does not pass through C.  In this case, the line must intersect exactly 2 sides of triangle ABC (AB and one other) leaving g(ABC) unaffected.

So even though the phrasing wasn't exact, the idea behind the proof is correct (and elegant) and g(ABC) will still be odd, allowing for exactly the same coloring.

You are absolutely right Michael, thanks for the correction.

Pierre.


Do you :D ?

Edit: whoops, can a mod move this to F&G?

[quote="bubka"]I do mass bittorrenting. What's so illegal about it? And anyway I hardly get a GB per day.[/quote]

Why not? Slow connection?

256 kbps
And I never run it more than 18 hours a day.
On good days I still get about two movies.
But all torrents are not so good.
I am currently running The Fox and The Hound 2.

I only use it for legal purposes i.e. downloading SuSE linux (when they say to please do)...

I download so much that the internet cowers before my power.

I chose mass bittorrenting. but then i delete most of what i download after a month


prove that:

$ \frac{sinx}{cosx}\plus{}\frac{sin2x}{cos^2 x}\plus{}.....\plus{}\frac{sinAx}{cos^A x} \equal{} cotx \minus{} \frac{cos(A\plus{}1)x}{sinxcos^A x}$

[quote="queen"] [color=darkred]Prove that $ \frac {\sin x}{\cos x} \plus{} \frac {\sin 2x}{\cos ^2 x} \plus{} ..... \plus{} \frac {\sin nx}{\cos^n x} \equal{} \cot x \minus{} \frac {\cos (n \plus{} 1)x}{\sin x\cos^n x}$ for any $ n\in\mathbb N^*$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ a_n \equal{} \frac {\sin nx}{\cos^nx}$ and $ S_n \equal{} \cot x \minus{} \frac {\cos (n \plus{} 1)x}{\sin x\cos^nx}$ for any $ n\in\mathbb N^*$ . I"ll use the method of induction.

$ 1\ \blacktriangleright\ \boxed {\ S_1 \equal{} a_1\ }$ $ \Longleftrightarrow$ $ \frac {\cos x}{\sin x} \minus{} \frac {\cos 2x}{\sin x\cos x} \equal{} \frac {\sin x}{\cos x}$ $ \Longleftrightarrow$ $ \cos^2x \minus{} \cos 2x \equal{} \sin^2 x$ , what is truly.

$ 2\ \blacktriangleright\ \boxed {\ S_{n \plus{} 1} \minus{} S_n \equal{} a_{n \plus{} 1}\ }$ $ \Longleftrightarrow$ $ \frac {\cos (n \plus{} 1)x}{\sin x\cos^nx} \minus{} \frac {\cos (n \plus{} 2)x}{\sin x\cos^{n \plus{} 1}x} \equal{} \frac {\sin (n \plus{} 1)x}{\cos^{n \plus{} 1}x}$ $ \Longleftrightarrow$

$ \cos x\cos (n \plus{} 1)x \minus{} \cos (n \plus{} 2)x \equal{}$ $ \sin x\sin (n \plus{} 1)x$ $ \Longleftrightarrow$

$ \cos x\cos (n \plus{} 1)x \minus{} \sin x\sin (n \plus{} 1)x \equal{} \cos [(n \plus{} 1)x \plus{} x]$ , what is truly.[/color]

@Newanda: Please do not post Word documents as attachments that people are expected to read.  They are not secure; PDFs are better.  If you need to convert, Virtual PDF Printer (Google it) works fine.  Or you could use LaTeX, which makes typesetting math easier than in Word.  The forums are also LaTeX-enabled, so you can use LaTeX there.  Click the "LaTeX" link on the sidebar at left.

For example, $ \$$ \frac{a+b+c}{3} \ge \sqrt[3]{abc} $ \$$ produces

$ \frac{a\plus{}b\plus{}c}{3} \ge \sqrt[3]{abc}$


inequalities
linear algebra
matrix
algebra
polynomial
vector
linear algebra unsolved

Let $ C$ is a subset of $ M_n(\mathbb{F})$(Where $ \mathbb{F}$ is field).If $ C$ such that for every matrix
$ A \in M_n(\mathbb{F})$,if $ A$ is commutative with any matrix in $ C$,then  $ A$ is commutative with any matrix in $ M_n(\mathbb{F})$.
Then can we prove that:
$ dim\{span\{C\}\} \ge n$?

If $ A$ commutes with any matrix, then $ A$ is a homothety.
Let $ J$ be the Jordan matrix of dimension $ n$ ($ J_{i,i\plus{}1}\equal{}1$ and the others $ J_{kl}$ are $ 0$). Let $ C\equal{}\{J,J^T\}$. If $ A$ commutes with $ J$ and $ J^T$ then $ A$ is a polynomial in $ J$ and in $ J^T$; thus $ A$ is a homothety. Yet $ C$ generates a vector-space of dimension 2. The dimension of the commutator of a matrix is always $ \geq{n}$; then the zhaobin's result is true if and only if $ n\equal{}2$.

Oh,sorry for my absurd estimating.
Thank you.
So the result is $ dim\{span\{C\}\} \ge 2$ :D


i wanted to know whether there is the AMC 9..I know there is grades 10 and 12....

No, there isn't.  Therer is only AMC8, AMC10, and AMC12. The in between grades should take the test one grade higher than their's (or higher if they wish).

[quote="k8reindeer"]9th graders can take AMC 10 or 12, but not 8: you can only take an AMC number which is higher than your grade, so you can't test down[/quote]

umm you can also take a test that is for your grade level too, not just the ones higher than your grade  :P

yeah, sorry. test number $\ge$ grade level


Putnam
linear algebra
matrix
parameterization

This problem seems to be really well-known. I've seen it in lots of places (I think a weaker form is an old Putnam problem)

Assume you have $2n+1$ numbers with the following peculiar property: if you eliminate any of them, you can form two groups of $n$ numbers each from the remaining numbers s.t. the two groups have equal sums. Show that the $2n+1$ numbers must be equal.

If you want a source, [url=http://www.princeton.edu/~mathclub/puzzles4.html]here[/url]'s one. Check out problems 12,13, which are particular cases (for $n=5$).

there is one solution using linear algebra .. are you referring to this solution ?

Do you mean following solution:

Consider Hamel's basis in $\mathbb{R}$ over $\mathbb{Q}$. It is immediately reduces problem to case then all numbers are integer. We may wlog suppose they are positive. It is clear they have the same parity. Consider this set modulo 2 (i.e. $x\to[x/2]$). New set satisfies the same condition, and so on. We neccessary come to set {0,0,..,0}. Then all numbers were equal before last step, and so on.

Wow! This is very nice. No, I had something else in mind. I didn't reduce it to the case when they're all integers. Just write the system of $2n+1$ equations which arise from the problem. The matrix of the system has $0$ on the diagonal and an equal number of $1$'s and $-1$'s on each line. It's not hard to show that the rank is $2n+1-1$, so the solutions are determined by one parameter, and since parametric solutions of the type $(x_1,x_2,\ldots,x_{2n+1})=(t,t,\ldots,t)$ work, then these must be the solutions.


calculus derivative ratio algebra polynomial

If I know that $Q(x)=P(x+1)-P(x)$, and I know that $Q(x)$ is in the form $Q(x)=2x+c$ with c an arbitrary number, can I say that, since $P(x+1)-P(x)$ is an incremental ratio, then $P(x)$ is in the form $P(x)=x^2+cx+q$?
In other words, can I say that $Q(x)$ is the first derivative of $P(x)$?
I know I expressed myself not well, so if you don't understand my English please ask, I'll try to do better :blush: .

That works in that case, as you can see if you let

$P(x)=ax^2+bx+c$

Is this the general case? Does this always works, for any polynomial?


[quote="chien than"]From $ (a^{2}\plus{}b^{2}\plus{}c^{2})^{2}\geq 3(a^{3}b\plus{}b^{3}c\plus{}c^{3}a)$
[/quote]

Hello.

Where this come from please.

I developped $ (a^{2}\plus{}b^{2}\plus{}c^{2})^{2}$ and I applied AM GM on $ a^{4}$ and $ (ab)^{2}$ but I wasnt able to get it right.

[quote="FOURRIER"]From $ (a^{2}\plus{}b^{2}\plus{}c^{2})^{2}\geq 3(a^{3}b\plus{}b^{3}c\plus{}c^{3}a)$

Hello.

Where this come from please[/quote]
from Vasc's inequality :  $ 0\le\frac{1}{2}[(a^{2}\minus{}2ab\plus{}bc\minus{}c^{2}\plus{}ca)^{2}\plus{}(b^{2}\minus{}2bc\plus{}ca\minus{}a^{2}\plus{}ab)^{2}\plus{}(c^{2}\minus{}2ca\plus{}ab\minus{}b^{2}\plus{}bc)^{2}] \equal{} (a^{2}\plus{}b^{2}\plus{}c^{2})^{2}\minus{}3(a^{3}b\plus{}b^{3}c\plus{}c^{3}a).$

[quote="othman"]i think it's vasc's inequality[/quote]

Please read on my reply, i have said that : all problem ( i have posted above ) in Vasc 'book .

[quote="silouan"]If $ a,b,c > 0$ and $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 3$ prove that

$ a^{3}(b\plus{}c)\plus{}b^{3}(a\plus{}c)\plus{}c^{3}(a\plus{}b)\leq 6$[/quote]

Let $ a\plus{}b\plus{}c \equal{} p$, $ (a\plus{}b\plus{}c)^{2}\equal{} a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}2(ab\plus{}bc\plus{}ca)\Longleftrightarrow ab\plus{}bc\plus{}ca \equal{}\frac{p^{2}\minus{}3}{2}\ (0 < p\leq 3)$,

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

$ \equal{}\minus{}\frac{1}{2}p^{3}\plus{}\frac{9}{2}p\plus{}3abcp$,

$ a^{4}\plus{}b^{4}\plus{}c^{4}\equal{} (a^{2}\plus{}b^{2}\plus{}c^{2})^{2}\minus{}2\{(ab\plus{}bc\plus{}ca)^{2}\minus{}2abc(a\plus{}b\plus{}c)\}$

$ \equal{} 3^{2}\minus{}2\left(\frac{p^{2}\minus{}3}{2}\right)^{2}\plus{}4abcp$

$ \equal{}\minus{}\frac{1}{2}p^{4}\plus{}3p^{2}\plus{}\frac{9}{2}\plus{}4abcp$.

$ \therefore a^{3}(b\plus{}c)\plus{}b^{3}(c\plus{}a)\plus{}c^{3}(a\plus{}b) \equal{} a^{3}(p\minus{}a)\plus{}b^{3}(p\minus{}b)\plus{}c^{3}(p\minus{}c)$

$ \equal{} p(a^{3}\plus{}b^{3}\plus{}c^{3})\minus{}(a^{4}\plus{}b^{4}\plus{}c^{4})$

$ \equal{} p\left(\minus{}\frac{1}{2}p^{3}\plus{}\frac{9}{2}p\plus{}3abc\right)\minus{}\left(\minus{}\frac{1}{2}p^{4}\plus{}3p^{2}\plus{}\frac{9}{2}\plus{}4abcp\right)$

$ \equal{}\frac{3}{2}p^{2}\minus{}abcp\minus{}\frac{9}{2}$

$ \equal{}\frac{3}{2}(a\plus{}b\plus{}c)^{2}\minus{}abc(a\plus{}b\plus{}c)\minus{}\frac{9}{2}$

$ \equal{}\frac{3}{2}\left\{a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}2(ab\plus{}bc\plus{}ca)\right\}\minus{}abc(a\plus{}b\plus{}c)\minus{}\frac{9}{2}$, by $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 3$, yielding

$ a^{3}(b\plus{}c)\plus{}b^{3}(c\plus{}a)\plus{}c^{3}(a\plus{}b)\leq 6\Longleftrightarrow 3(ab\plus{}bc\plus{}ca)\minus{}abc(a\plus{}b\plus{}c)\leq 6$ with $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 3\ (a,\ b,\ c > 0)$.

[quote="kunny"] $ 3(ab\plus{}bc\plus{}ca)\minus{}abc(a\plus{}b\plus{}c)\leq 6$ with $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 3\ (a,\ b,\ c > 0)$.[/quote]

Please to continue your solution.

[quote="Albanian Eagle"]to get the inequality in the firs post it's not needed to use Vasc's inequality
just sum cyclically $ x^{4}\plus{}4x^{2}y^{2}\plus{}y^{4}\geq 3xy(x^{2}\plus{}y^{2})$  :wink:[/quote]

But how you get that?

Thanks

[quote="Albanian Eagle"]$ x^{4}\plus{}4x^{2}y^{2}\plus{}y^{4}\geq 3xy(x^{2}\plus{}y^{2})$  :wink:[/quote]The solution is simple, but I like the inequality.
We have \[ (x^{2}\plus{}y^{2})^{2}\plus{}2x^{2}y^{2}\geq 3xy(x^{2}\plus{}y^{2}) .\] Denote $ x^{2}\plus{}y^{2}\equal{}a$, $ xy\equal{}b$. Note that because $ (x\minus{}y)^{2}\geq 0$, we have $ a\geq 2b$, and if $ b\leq 0$, then $ a\geq 0\geq b$, and if $ b>0$, then still $ a\geq 2b>b$.

The inequality becomes
\[ a^{2}\plus{}2b^{2}\geq 3ab\ \Leftrightarrow (a\minus{}2b)(a\minus{}b)\geq 0 ,\] which is true as both $ a\minus{}2b$ and $ a\minus{}b$ are positive (for any real numbers $ x,y$).

[quote="FOURRIER"][quote="Albanian Eagle"]Or rewrite it as $ (x\minus{}y)^{4}\plus{}xy(x\minus{}y)^{2}\geq 0$  :)[/quote]

This give:

 $ x^{4}\plus{}4x^{2}y^{2}\plus{}y^{4}\geq xy(x^{2}\plus{}y^{2})$

;)[/quote]
Are you sure ?  :D

[quote="FOURRIER"][quote="Albanian Eagle"]Or rewrite it as $ (x\minus{}y)^{4}\plus{}xy(x\minus{}y)^{2}\geq 0$  :)[/quote]

This give:

 $ x^{4}\plus{}4x^{2}y^{2}\plus{}y^{4}\geq xy(x^{2}\plus{}y^{2})$

;)[/quote]No, you've got a mistake in your computation.

Other problem:
Let $ a;b;c$ be positive real numbers satisfying $ ab\plus{}bc\plus{}ca\equal{}1$
Prove that:
$ a^{2}(b\plus{}c)\plus{}b^{2}(c\plus{}a)\plus{}c^{2}(a\plus{}b)\geq 3\sqrt[3]{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$

[quote="chien than"]
Let $ a;b;c$ be positive real numbers satisfying $ ab\plus{}bc\plus{}ca \equal{} 1$
Prove that:
$ a^{2}(b\plus{}c)\plus{}b^{2}(c\plus{}a)\plus{}c^{2}(a\plus{}b)\geq 3\sqrt [3]{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$[/quote]
I think, it's wrong. Try $ a\equal{}b\rightarrow1^{\minus{}}$ and $ c\rightarrow0^{\plus{}}$

In Secrets in Inequalities, I solved by Holder that
\[ a^{2}(b\plus{}c)\plus{}b^{2}(c\plus{}a)\plus{}c^{2}(a\plus{}b)\ge (ab\plus{}bc\plus{}ca)\sqrt[3]{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\]
So it should be that
[i]
If $ a,b,c$ are non-negative real numbers and $ ab\plus{}bc\plus{}ca \equal{} 1$ then
\[ a^{2}(b\plus{}c)\plus{}b^{2}(c\plus{}a)\plus{}c^{2}(a\plus{}b)\ge\sqrt[3]{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\]
[/i]

(that is if I am not misremembering  :lol: )

The solution is very simple.


This is a personality test that takes about fifteen minutes to do, but it gives surprisingly accurate results.

http://www.personaldna.com
this was mine..

http://www.personaldna.com/report.php?k=kgpTPbYYReioTeV-OI-AAAAD-b612&u=e31f9aef5496

I am a considerate analyst.

http://www.personaldna.com/report.php?k=pGQkaLtBfOYtKeQ-MG-DCACA-d3f6&u=91edb0cd5388

[url=http://www.personaldna.com/report.php?k=cABVjmcMBSUTDdW-EC-AACCD-44f9&u=5f6759e10578]$ \text{Respectful Idealist}$[/url], though some of the descriptions seem a bit wrong.  :)

http://www.personaldna.com/report.php?k=EBHkoBEDrDQKCQK-AG-CCDBC-2197&u=889c1a586d0f

um.... :huh:

I am a concerned realist. Weird description.


An integer is randomly chosen from the integers 1 through 100,
inclusive. What is the probability that the chosen integer is a
perfect square or a perfect cube, but not both? Express your
answer as a common fraction.

Inclusive-includes 1 and 100???
If yes, then:
$ \frac{10}{100}\plus{}\frac{4}{100}\minus{}\frac{2}{100}\equal{}\frac{3}{25}$

You have to subtract 4/100, because we don't want to count the double-counted ones at all, so we subtract them twice.  For example, we added in 1 two times, but we need to subtract it 2 more times.  Then we have 1/10 as our answer.

I don't understand . And how did you get 1/10?

OK:

There are $ 10$ squares.

There are $ 4$ cubes.

There are $ 2$ sixth powers.

Thus, there are $ 8$ squares and $ 2$ cubes that aren't sixth-powers, yielding $ \boxed{\frac{1}{10}}$.

[hide=sol]Case 1 - squares
1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Case 2 - cubes
1, 8, 27, 64
1 and 64 are in both of these meaning there are 10 left giving $\boxed{\frac{1}{10}}$[/hide]


Prove that every natural number k (k > 1) has a multiple being smaller than k 4  and when written in base 10 at most four different digits !

Germany 1996. Solution in the book
http://www.unl.edu/amc/a-activities/a4-for-students/mc96-97-01feb.pdf


[quote="chien than"]Prove that:$3(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}) \geq 4(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})$
where $a;b;c$ are positive real numbers[/quote]
you are wrong! What about if $b=c=1$ and $c\to+\infty$?

I think the inequality is
$3(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})^{2}\geq 4(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})$

and it is a consequence of Cauchy and AM-GM  :wink:

I think it should be [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/d4b79f6ca2de0a04c89f8dda3fb1f4bd.gif[/img]...
yours doesn't hold for a=b=c=3.

Yeah you are right but I couldn't fix my post .  :(


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

İf A  is symmetric and reversible matrix type  nxn, A^-1  is a an symmetric matrix. Show it.

Two approaches:

i)  $ AA^{-1}=I$ and $ A^{-1}A=I$ imply $ A^{-T}A=I$ and $ A A^{-T}=I$.  If $ A^{-T}\neq A^{-1}$, then $ A$ admits two inverses (on the left and right).

ii)  Using the adjugate matrix, show that cofactors $ M_{ij}$ and $ M_{ji}$ of $ A$ must be equal if $ A$ is symmetric.

$ A^{-1}$ is always a polynomial of $ A$. This polynomial is the polynomial for which $ P(\lambda)=\frac 1\lambda$ for every eigenvalue $ \lambda$ of $ A$.


$ 2xz \minus{} 5yz \plus{} xy \minus{} 3 \equal{} 2y^2 \plus{} 2z$

Show $ |x \minus{} y \plus{} z| \equal{} 4$

Your title and question do not match up... which one is it!?

:?:  Is there some info missing or something? Solve for $ x$ and substitute it into $ x\minus{}y\plus{}z$... it doesn't equal $ \pm4$. What are the variables? $ \in\mathbb{R}$? $ \in\mathbb{Q}$? $ \in\mathbb{Z}$?

Yeah, for example, I get $ (x,y,z)\equal{}(4,0,\frac{1}{2})$ as a solution.

$ |x-y+z|=4 \Longleftrightarrow  [ \begin {array} {c} x-y+z=4\\x-y+z=-4 \end {array}$

Is there a typo? B/c i don't think you can prove it...

Ok this problem is impossible. Close it mods.

Where did you find this problem? (Not trying to be critical, just interested...)

If it's something you made up, then it might be impossible...but if you got it from somewhere else, check for their solution.

I got it from a math place, I probably wrote it down wrong.


It can be immediately proved by weighted Jensen, so the rationality of $p_{1},\ldots,p_{n}$ isn't necessary.

Rewrite $p_{i}$ as ${q_{i}\over N}$, where $q_{i}$ are positive integers and $N$ is the least common denominator for fractions $p_{i}$. Then you get

\begin{eqnarray*}(x_{1}^{p_{1}}x_{2}^{p_{2}}\dots x_{n}^{p_{n}})^{1\over{p_{1}+\dots+p_{n}}}&=& (x_{1}^{q_{1}/N}x_{2}^{q_{2}/N}\dots x_{n}^{q_{n}/N})^{N\over{q_{1}+\dots+q_{n}}}\\ &=& (x_{1}^{q_{1}}x_{2}^{q_{2}}\dots x_{n}^{q_{n}})^{1\over{q_{1}+\dots+q_{n}}}\\ &\stackrel{\mathrm{AM-GM}}{\leqslant}& \frac{q_{1}x_{1}+q_{2}x_{2}+\dots+q_{n}x_{n}}{q_{1}+q_{2}+\dots+q_{n}}\\ &=& \frac{{q_{1}\over N}x_{1}+{q_{2}\over N}x_{2}+\dots+{q_{n}\over N}x_{n}}{{q_{1}\over N}+{q_{2}\over N}+\dots+{q_{n}\over N}}\\ &=& \frac{p_{1}x_{1}+p_{2}x_{2}+\dots+p_{n}x_{n}}{p_{1}+p_{2}+\dots+p_{n}}\end{eqnarray*}

By using Dedekind cuts, the result can be extended to irrationals too, in accordance to what Rzeszut said.


Determine, with proof, the largest number which is the product of positive integers whose sum is $1976$.

[hide="Solution"]First, determine the positive integer $ n$ that yields the largest number $ n^{k}$ for some $ m$ such that $ nk\equal{}m$. Observe that $ 1^{60}<2^{30}<3^{20}>4^{15}>5^{12}$, etc. This means $ n\equal{}3$. Also, for $ k>4$, $ 2^{k}>k^{2}$, so the integer that produces the second largest $ n^{k}$ is $ 2$.

Therefore, the answer is $ 3^{\lfloor\frac{1976}{3}\rfloor}\cdot2^{1976\minus{}3\lfloor\frac{1976}{3}\rfloor}\equal{}\boxed{2\cdot3^{658}}$[/hide]


trigonometry
inequalities
inequalities unsolved

Prove that $\forall x\in R$ we have $\cos x\ge \cos (x+\sin x)$

Sorry  :P  I've corrected it now  :blush:

Note that it is enough to show the inequality on $[0,\pi]$ ( by reduction modulo $2\pi$ and by parity of the functions). There are two cases:

1) $x \in [0, \dfrac {\pi}{2}]$, then $\dfrac {\pi}{2} \geq sin(x) \geq 0$ and a look at the trigonometric circle makes it clear that the inequality holds.

2)$x \in [\dfrac {\pi}{2},\pi]$, then "looking at the trigonometric circle" (which can be formulated rigorously of course, but I leave it to you ;)) we see that we have to show: $2(\pi-x) \geq sin(x)$, or equivalently $2(\pi -x) \geq sin(\pi- x)$, which is obviously true since $sin(y) \leq y$ for all real $x$.


I think they are very known , but I am interested to see your's solutions

1) $ \forall x>0$  $ \frac{1}{x+1}<ln\frac{x+1}{x}<\frac{1}{x}$
2) $ |sinx|\leq |x|\leq |tanx|$  and $ |cosx|\leq \frac{\pi}{2}+|x|$

[hide="hint"]
1. Mean value theorem on ln[x]
[/hide]

[quote="giannis"]$ |sinx|\leq |x|\leq |tanx|$[/quote]

You might want to specify a range on $ x$...


calculus
integration
real analysis
real analysis unsolved

Question 1:
The work done, w, by a gas expanding from 2 cm^3 to 10 cm^3 is given by the following definite integral: w=integral(10 above 2) of 500V^-1.4 dV

1. Use Simpsons rule with n=4 subintervals to estimate w.

I got something like 727.59cm^3 for it

2. Find an exact value for w by evaluationg the definite integral above using calculus.

********************************************
Question 2

A drug is excreted in a patients urine. the urine is monitored continuously using a catheter. A patient administered 10 mg of drug at time t=0 which is excreted at a rate of -3t^1/2 mg/h

1. What is the general equation for the amount of drug in the patient at time t>0?

I figured it would be like -3t^1/2 + 10 am i right?

2. When will the patient be drug free?

If you guys could help that would be awesome. Show full working too please.

bump, cum on guys i need this asap.  Please.


Hi,

Are there any one of you who has a good collection of algebraic factorization/ algebraic identity in 3 or more variables. I have a lot of problem to manipulate inequality problems because I don't have a lot algebraic identity in my deposit. THanks in advance.........

O.K. Here is a problem.

Factorize the following equations.

(1) $(a+b)(b+c)(c+a)+abc$

(2) $(a-b)^3+(b-c)^3+(c-a)^3$

(3) $a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2$

Source:1990 Meiji University of Pharmaceuitical

We can use the identy $x^3 + y^3 + z^3 = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx) + 3xyz$ to solve the second one:
$(a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a)$

[quote="TomciO"]And for the first one:
$(a+b)(b+c)(c+a) + abc = (a+b+c)(ab+bc+ca)$
It isn't hard to guess...[/quote] :)


function
algebra
domain
real analysis
real analysis unsolved

I don't understand the definition of limit of a function very well.  The definition of [size=150] lim[x->a] f(x) = L [/size] is for every epsilon there is a delta > 0 such that |x-a| < delta ==> |f(x)-L|<epsilon.

What if we reverse the order and say for every delta there is an epsilon such that |x-a| < delta ==> |f(x)-L|<epsilon?  How would this change the meaning?
And what if we say for every epsilon there is a delta such that |f(x)-L| < epsilon ==> |x-a| < delta?

It sounds like you're getting messed up on quantifiers. Quantifiers are statements like "for all $ x$..." and "there exists a $ y$...", if you have an introductory book about formal logic you can read about it. There are only two, the ones above; "for all" is usually written as $ \forall$ and "there exists" as $ \exists$.

To illustrate the importance of quantifier order, look at a statement like "Every person has a mother." Using quantifiers and making it a little more formal, you'd say "For all x, if x is a person, there exists a person y who is the mother of x." What if you reverse the order of the quantifiers? You get "There exists a person y such that, for all persons x, y is the mother of x." The meaning is changed: "everyone has a mother" becomes "someone is the mother of all of us". You can come up with all sorts of examples like this, but the thing to keep in mind is that the order of quantification is important and changing the order changes the meaning.

Interpretation:  "for any arbitrarily small error $ \epsilon$, we can find a bound $ \delta$ such that every value of $ x$ within $ \delta$ of $ a$ has a value of $ f(x)$ within $ \epsilon$ of the limit $ L$."

The problem with the quantifiers as you write them in the second definition is that no restriction is made on $ \epsilon$.  Thus, for every $ \delta$ we can choose, say, $ \epsilon \equal{} 10^{100}$ and it will probably be the case that $ |f(x) \minus{} L| < \epsilon$.

Same problem with the third definition.  No restriction is made on $ \delta$, so we can take $ \delta$ to be the size of our domain (for example).

My advice is: "Try to draw a picture"

Often there is simple geometric idea behind such messy definitions. The reasons for this order of quantifiers should come up naturally.


inequalities
geometry
circumcircle
trigonometry
parallelogram
inequalities proposed

let $ a,b,c$ are sides of a triangle $ ABC$ having the circumradius $ R.$
prove that

$ \sum \sqrt(2b^2\plus{}2c^2\minus{}a^2) <16R\prod cos(A/2)$

noone interested????i am waiting to see your solutions :( ........please have a try... :ninja: .........

First we have $ \sqrt{2b^2+2c^2-a^2}=2m_a$, where $ m_a$ is the length of the median through $ A$. On the other hand, we have
\begin{align*}&16R\cos\frac A2\cos\frac B2\cos\frac C2\\
=&2\sqrt 8R\sqrt{2\cos^2\frac A2\cdot2\cos^2\frac B2\cdot 2\cos^2\frac C2}\\
=&2\sqrt 8R\sqrt{(1+\cos A)(1+\cos B)(1+\cos C)}\\
=&2\sqrt 8R\sqrt{\prod\frac{(b+c)^2-a^2}{2bc}}\\
=&2\sqrt 8R\frac{\sqrt{(a+b+c)^3(b+c-a)(c+a-b)(a+b-c)}}{\sqrt 8abc}\\
=&\frac{(a+b+c)\cdot 4T}{2T}\\
=&2(a+b+c)\end{align*}
(Using $ 4RT=abc$ and $ 16T^2=(a+b+c)(a+b-c)(b+c-a)(c+a-b)$, where $ T$ denotes the area of the triangle). Thus it remains to show that
\[ m_a+m_b+m_c<a+b+c\]
We'll show that $ \frac{b+c}{2}>m_a$ etc. Construct the parallelogram $ ABDC$. Then $ 2m_a=AD<AB+BD=AB+AC=b+c$. Hence we are done.


AMC
AIME
articles
modular arithmetic
number theory
geometry
angle bisector

I was reading the Aops article on Fermat's little theorem, but i didn't get how they jumped from mod 5 (when solving the practice problem) to mod 3 - for that matter, i didn't get how they got $ \minus{}1$ in the mod 3 either  :huh:

Could someone please explain it?

Thanks in advance!

i am not quite sure what you are asking, but here is what i think:

 $ a^p \equiv a(\mbox{mod} p)$ is fermat's little theorem, but this only holds for $ p$ being a prime number, so the skip from 3 to 5 was probably due to the fact that 4 is not prime.

I was wondering how they took

$ 3 \plus{} 0 \plus{} 4 \plus{} 7 \equiv n\pmod{5}$

$ 4 \equiv n\pmod{5}$

and then got

$ \minus{}1 \plus{} 1 \plus{} 0 \plus{} 0 \equiv n\pmod{3}$

and particularly on how the $ \minus{}1$ was derived...

The original equation was $ 133^5\plus{}110^5\plus{}84^5\plus{}27^5\equal{}n^5$(A).
Examining it in $ \pmod{5}$ yields $ n \equiv 4 \pmod{5}$(B).
Examining it in $ \pmod{3}$ yields $ n \equiv 0 \pmod{3}$(C).

(A) yields (B), and (A) yields (C). In other words, they got the $ \pmod{3}$ part from going back to the original equation.

As for the $ \minus{}1$, since $ 110 \equiv \minus{}1 \pmod{3}$, $ 110^5 \equiv (\minus{}1)^5 \equiv \minus{}1 \pmod{3}$.

Or $ n^5\minus{}n\equal{}n(n^4\minus{}1)\equal{}n(n^2\minus{}1)(n^2\plus{}1)\equal{}(n^3\minus{}n)(n^2\plus{}1)$ and the first factor is 0 mod 3 by FlT so the whole thing is.

FLT actually, and it stands for Fermat's Little theorem (you can go look it up in the wiki)

- Xantos and SamE thanks, but is there a proof that for any $ n^5 \equiv n \pmod{3}$ .. i don't think i understand it fully yet :huh:

Actually, both of our posts proved that. Also, FLT can stand for Fermat's Last Theorem, so FIT might be used to avoid ambiguity, but I'm not sure.

Anyway, to elaborate on SamE's proof:
$ n^5 \minus{} n \equal{} (n^3 \minus{} n)(n^2 \plus{} 1)$. Since $ n^3 \equiv n \pmod{3}$, $ n^3 \minus{} n \equiv 0\pmod{3}$.
So, $ n^5 \minus{} n \equal{} (n^3 \minus{} n)(n^2 \plus{} 1) \equiv 0(n^2 \plus{} 1) \equiv 0 \pmod{3}$, Thus $ n^5 \equiv n \pmod{3}$.

yes i understand Fermat's Little Theorem.
ruby - incidentally $ n^5 \equal{} n (mod 3)$ because $ n^3 \equal{} n (mod 3)$ by Fermats Little Theorem.
While we are on the topic of modular arithmetic, will someone explain the Chinese Remainder Theorem and/or provide an example problem? Thank you!

Another unrelated question i have for the experienced problem solvers: are all the triangle formulas u need for aime in AoPS v2 and if not which ones are helpful ( i.e. Menaluas's, Ceva's, Stewarts, Angle bisector etc.)

Actually, I used a lowercase L: FlT, not FIT or FLT. It's something I made up (using a little L) to distinguish from Fermat's Last Theorem. It works better on paper, because I can make the l curly.

I hope Xantos's clarification of my proof is enough...

Chinese Remainder Theorem: Okay, so you've got a number $ n$ that's the product of two relatively prime numbers $ p$ and $ q$. You're looking at something mod $ n$ and wondering if there's a solution, and if so, how many. CRT lets you look at it mod $ p$ and mod $ q$ and says the total number of solutions mod $ n$ is the product of the number of solutions mod $ p$ and the number of solutions mod $ q$. In particular, if there's a solution mod $ p$ and a solution mod $ q$, there's a solution mod $ n$.

You can extend this to more than two factors, and usually you just take the prime factorization of $ n$ and look at whatever it is mod prime powers (if $ n\equal{}p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, look at it mod $ p_i^{a_i}$ for all $ 1\le i\le k$). This can greatly simplify some problems, and can show you how to generalize results achieved for prime powers.

Hmmm... example problem. I'd say USA TST 2008 #9 but that might not be available and/or too hard...I didn't get it.

Triangle formulas... I would be tempted to say that everything you need is in AoPS 2, but I'm not sure. I can say that for a couple years, I tried Stewart's theorem on pretty much every problem that had a cevian, so knowing formulas isn't always helpful, because it can lead you to waste time. I think the only useful formula I can't find in AoPS 2 is Shoelace, but there might be more.

I'm pretty sure they taught Shoelace in AoPS 2.

As for triangle formulas, I agree with SamE, knowing formulas can be helpful, but knowing how to creatively solve problems is even better. :)

Thanks for all the help :)

about the chinese remainder theorem, the AoPS wiki article on it is pretty good http://www.artofproblemsolving.com/Wiki/index.php/Chinese_Remainder_Theorem

as is the one in Wolfram...

thanks for all the help guys. So for CRT, if you wanted the number of solutions to
$ 4t \plus{} 1 \equal{} 0 (mod 21)$
it would be the number of solutions to $ 4t \plus{}1 \equal{} 0 (mod 3)$ $ (1)$
multiplied by the number of solutions to $ 4t \plus{} 1 \equal{} 0 (mod 7) (1)$
so there is 1 solution


See
http://www.mathlinks.ro/Forum/viewtopic.php?p=326466#p326466

where the inequality appears
$\frac{1}{2a^2+3bc}+\frac{1}{2b^2+3ac}+\frac{1}{2c^2+3ab}\geq \frac{4}{3(ab+bc+ca)}$.

I think this nice inequality is Symmetery.
And silouan,can you post your solution?

Sorry my friends, the question is
[quote="silouan"]Let $a,b,c>0$ . Prove the following inequality.

$3(ab+bc+ca)(\frac{1}{2a^2+3bc}+\frac{1}{2b^2+3ac}+\frac{1}{2c^2+3ab})\geq 4$[/quote]

and it is the same as the one posted by Vasc, so it's not new,sorry :blush:

Of course, but I think some one forget to post a solution using S.O.S  :) . The solution here use expending and not the best.


Let a,b,c be three real numbers such that a+b+c=1. Prove that

a 2 +b 2 +c 2 +1 \geq 4(ab+bc+ca)

We have
a² + b² + c² + 1 \geq 4(ab + bc + ca)
if and only if
a² + b² + c² + (a + b + c)² \geq 4(ab + bc + ca)
if and only if
2(a² + b² + c²) + 2(ab + bc + ca) \geq 4(ab + bc + ca)
if and only if
a² + b² + c² \geq ab + bc + ca
if and only if
(a - b)² + (b - c)² + (c - a)² \geq 0
which is true. Come on Manlio, isn't this too easy?


I live in California, on the west coast.  Where does everyone else live?

I live in MI,around the great lakes its usually called the midwest.

isn't east coast kind of a generalization of southeast and northeast? regardless, i live in NJ, northeast.

I live in Virginia which is the east coast.

The great state of Columbus.  I mean Ohio.

[quote="MithsApprentice"]Texas, Houston. :D

Right on the gulf...well not "right" but close enough.[/quote]

Cool! I was just there last week...

East Texas, 2 hours away from Dallas, Texas.. Small (yet growing quickly) town called Longview. (near tyler)

Clck thru the hides to see where i live

[hide][hide][hide][color=white]I live in the midwest.[/color]  :) The place where i live is written in invisible ink. If you want to know how to crack it. post [url=http://4aunlimited.co.nr]here.[/url]
[/hide][/hide][/hide]

You can just hit quote to read it easily.

In Georgia, USA, which I suppose is east coast, but we prefer to think of ourselves as south.

can you eidt that post. i meant the pople who read it to browwse thru all those hide tags. [hide][/hide][hide][/hide][hide][/hide][hide][/hide]hehe

Clarksville, Maryland!
unfortunately, i get trumped by the same handful of Montgomery Blair (magnet school in neighboring county) students on all state math competitions...

Lexington Massachusetts
Just moved here from sunny california :D

[quote="236factorial"][quote="byrdgyrl"]Wyoming.  We just broke half a million people!!![/quote]

Is wyoming the state with the smallest population? Or is it Alaska?[/quote]

It's Wyoming.  It's also the easiest state to qualify for the USAMO in too.  Usually, you just need a 100 on the AMC 12 to do that.

[quote="h78reg"][quote]It's Wyoming. It's also the easiest state to qualify for the USAMO in too. Usually, you just need a 100 on the AMC 12 to do that.[/quote]

what do you need on the AIME to qualify?[/quote]

The top competitors qualify, usually around 7-10 or so. Maybe in Wyoming, 5-8?

that doesn't make wyoming THAT much easier to qualify in than in other states.

If you have the highest USAMO index in your state, then you can qualify for the USAMO.  And WYoming usually has only 1 or a lot of times, no one who qualifys for the AIME.  So if your the only one who qualifys for the AIME in your state, you automatically qualify for the USAMO.

Of Course,  there are other ways to qualify.

im glad to see so many responses! :) yay, Mithsapprentice didn't shut something i did down yet :lol: he even participated, i guess i didn't break any rules finally ;) by the way this thread is about where you live, not about math contests :idea:

i wish i lived in wyoming... actually, any state not CA, FL, or any of the ones on the east coast would probably be much less competitive...

hey hey...

[("Don't wish it were easier; wish you were better.")]

Every January i beg my mom to move to Deleware with me so that we can participate in MathCounts there :D if i was in deleware last year i would have gone to Nationals!  (With the best score in Deleware)  Unfortunatly, i got 7th in California :(


function
trigonometry
superior algebra
superior algebra solved

Show that cosx donot belong to the linear span of the set
{sinx,sin2x,sin3x,.....sinnx,..........}

Since $sin(0)=0$, you get $f(0)=0 \neq 1=cos(0)$ for all linear combinations $f$ out of this set.

yeah, but that's just a particular value of x.  problem is to prove the statement for *all* x.

thge game asked for a proof that the function $\cos x$ is not a linear combination of the functions $\sin x,\sin 2x,\ldots$. Since $\cos x$ does not vanish in $0$, which is a common zero of the functions $\sin nx$, it cannot be a linear combination of $(\sin nx)$. As far as I can see, ZetaX's solution works perfectly fine.. :?

And in $L_2$ of $[0,2\pi]$, $\cos x$ is orthogonal to the subspace generated by the $\sin nx$; its distance from the subspace is its own norm ($\sqrt{\frac12}$ if you normalize to measure 1).

The linear span and the closure of the span (in some norm) are different objects. On the other hand, a sequence of odd functions cannot converge to a nonzero even function in any of the usual senses.

O its that simple :blush: .I guess I made the same mistake as pleurestique.


how did all of u do
whoever appeared
physics part was excellent
fiitjee cant even the level of concepts today the open test had
not a single questions solution is wrong

i didnt write it.. could u post some of the questions? :)

Was it today?? WTH, none of my friends who wrote bmat told me. Chaa...Pardesi, sollamatiya da???

Anit can you post some decent questions... And how much are you expecting?

[quote="pardesi"]no i din't myself write it :rotfl:  :rotfl:  i wrote narayana....
ye sanoit post the physics questions :)[/quote]

WTH. How many Test series will you write da??

But I guess the word "Narayana" was enough to entice you to write. :D

I wanted to write  :(  Wth u fools could have atleast told or mentioned that there were open test(s) today  :mad:  :mad:

i am posting my fav question
 A particle having mass m,charge q was projected with velocity v along y axis in a region of uniform magnetic field B which is outward and perpendicular to the plane of the paper.the particle is continuously subjected to a frictional force which varies with velocity as $ \vec{F} \equal{} \minus{} a\vec{v}$ where a is a constant.consequently the particle moves on a spiral path till it comes to rest at point P.the x coordinate of point P is
a)$ 10$ m       b)$ 4$m             c)$ 6$ m      d)$ 8$m
a = 10^-3 kg/s
q=10^-3 C
v=1m/s
m=20 gm

[quote="hash_include"]none of us knew!! :(  :huh:

anyway in case there are any more open tests please spill the beans ppl! (especially the mod :D)[/quote]

Navneet, aravind, av some few are writing bmat.

any physics lover will love to solve the physics question
today i got screwed completely
they r not even good enough to tell someone
marks i mean

anit post ur problem clearly there is osmething missing especially values.. :)

[quote="anitsahu"]
today i got screwed completely
they r not even good enough to tell someone
marks i mean[/quote]

somehow, i very much doubt that  :D   :P

why do u doubt it
really i got screwed up

[quote="pardesi"][quote="Cyber-Shreyas"]
But I guess the word "Narayana" was enough to entice you to write. :D[/quote]
nothing liek taht da...evryone of us wrote that :D[/quote]

Was it narayana open test? Or Chumma test series. How was it?

chumma  jujubi :rotfl:  :rotfl:  :D
there was only one good question or may be two :P


another thing i wanted to make was a patato cannon.. maybe a world record one..

does anyone have know any resources on how air works?

my friend said he made a nozzle by reading this book or something that gives you equations on how big the nozzle should be depending on the pressure and ammount of air on one side and etc...

he made his at the rose hulman summer camp thing.. and was restricted to 30 psi... but if he had 50 psi, he might break world record...

its pretty amazing...

looks like a nice website..

btw.. that A in ur avatar looks like the anarchist A

what the.. u have to buy the plans?
i just want to make my own.. but thanks anyways..


What would be the best way to solve this problem?

Trees seem to me to be a pretty good choice here.

More specifically, tries might be helpful: http://en.wikipedia.org/wiki/Trie. See if you can figure out the rest of the problem from there.


Let $ a,b,c,n$ be positive integers such that $ (a\plus{}bc)(b\plus{}ac)\equal{}5^n$.

Prove that $ n$ is even

[hide]
Assume $ a$ and $ b$ are not both divisible by $ 5$. If they are, simply divide them both by $ 5$, which divides both sides of the equation by $ 25$ and does not affect the parity of $ n$.

Now we must have

$ a\plus{}bc\equal{}5^x$
$ b\plus{}ac\equal{}5^y$

and $ x\plus{}y\equal{}n$. Also, $ x,y>0$, because $ a,b,c$ are positive integers so $ a\plus{}bc\ge 2$ and $ b\plus{}ac\ge 2$. Also assume wlog that $ x\ge y$. Add both of these equations and subtract them to get

$ (b\plus{}a)(c\plus{}1)\equal{}5^x\plus{}5^y$
$ (b\minus{}a)(c\minus{}1)\equal{}5^x\minus{}5^y$

So $ 5^y$ divides both $ (b\plus{}a)(c\plus{}1)$ and $ (b\minus{}a)(c\minus{}1)$. However, $ b\plus{}a$ and $ b\minus{}a$ cannot both be divisible by $ 5$, as that would imply both $ a$ and $ b$ are divisible by $ 5$. Therefore we must also have either $ 5^y$ divides $ c\plus{}1$ or $ 5^y$ divides $ c\minus{}1$.

Case 1: $ 5^y|c\minus{}1$. Then $ 5^y<c$ but we know that $ 5^y>c$ by
$ b\plus{}ac\equal{}5^y$
a contradiction.

Case 2: $ 5^y|c\plus{}1$. The $ 5^y\le c\plus{}1$. Unless $ a\equal{}b\equal{}1$, then we have from
$ b\plus{}ac\equal{}5^y$
that $ 5^y>c\plus{}1$. Thus we must have $ a\equal{}b\equal{}1$.

So $ 5^n\equal{}(a\plus{}bc)(b\plus{}ac)\equal{}(1\plus{}c)^2$ so $ n$ is even.
[/hide]


dear all,
hope you are doing good....
can you please help me solve this word puzzle...?
5 charaters only.

  1)                   ......I......N......

clue: In fairytales an ogre may be portrayed as a man-eating......

2)         .....RU....GE
clue: To....... is to be engaged in laborious and wearisome activity

hope to get feedback from you soon....
thanks a lots...

[quote="heroine"]5 charaters only.[/quote]
What does this mean? 5 different characters for both answers? Does each blank have exactly one character in it?

first word puzzle, the answer is 5 letters.....just have to find the missing the letters....
 the last one is of 6 letters to form a word.......just have to find the remaining letters...
thanks...

ps; i have some more similar wordpuzzle to post...


1. Prove or disprove that : If $ <Ax,x>\equal{}0$ for any vector $ x$, then $ A\equal{}0$.
2. If $ rank(A)\equal{}rank(B)$, there exist an invertible matrix $ C$ such that $ A\equal{}BC$.
Thanks in advance.

1. $ A \equal{} \begin{pmatrix}0 & \minus{} 1 \\
1 & 0\end{pmatrix}.$ (Over $ \mathbb{R}^2.$)

2. Consider $ A \equal{} \begin{pmatrix}1 & 0 \\
1 & 0\end{pmatrix}$ and $ B \equal{} \begin{pmatrix}1 & 0 \\
0 & 0\end{pmatrix}.$ Both $ A$ and $ B$ have rank 1.

But $ BC$ cannot have any nonzero element in the lower left position, no matter what $ C$ is, so we cannot have $ BC \equal{} A.$

Note that in this instance, I can find an invertible $ C$ such that $ CB \equal{} A.$ But that's not the same thing.

New question: if $ A$ and $ B$ are the same size and have the same rank, can we find invertible matrices $ C$ and $ D$ such that $ A \equal{} CBD?$

[quote="Kent Merryfield"]
New question: if $ A$ and $ B$ are the same size and have the same rank, can we find invertible matrices $ C$ and $ D$ such that $ A \equal{} CBD?$[/quote]

[hide="Answer"] YES! Hint: 2 ways to solve this:

1/ View $ A$ and $ B$ as linear transformation instead of matrices.

2/ Recall the Row and Column Operations.
[/hide]


Find the value of tan(Arccos5/13 +Arcsin12/13).

Drawing a 5-12-13 right triangle shows that
arcos$ \frac{5}{13}$ = arcsin$ \frac{12}{13}$

From basic trig. identities:
tan(2x) = $ \frac{2tanx}{1\minus{}tan^2x}$

It is evident from the triangle that tan(x)=$ \frac{5}{12}$
Substituting this into the identity gives:
tan(2x) = $ \frac{120}{119}$ F.S.

Good job, except it's negative in the first quadrant. Should be -120/119.

wait....how is it negative?

all positive in first quadrent...then sine in second, tangent in third, and cosine in fourth, right?

Well, $ \sin^{\minus{}1}\frac{12}{13}\approx67.38^\circ$, so $ 2(67.38)\approx134.76$, and tangent is negative in the second quadrant. I'm fairly certain that's what THECHAMP meant.


number theory
greatest common divisor
search

Chicken McNuggets are sold in packages of 6, 9, and 20. Assuming no package is ever broken into smaller sub-packages, what is the greatest number of Chicken McNuggets that you cannot purchase by buying some combination of the standard packages?

well i don't know if there is any elegant soln. but well, first thing u should check is that if the GCD of all the numbers is = 1. if not then the ans is infinity.

other wise there is answer.

take an array of numbers starting from 1 to...

now, first cross 1 to 5. tick 6. and go on, if u find a tick then add all the given number of it and tick the resulted number. and go on... if at any stage u found 6 consecutive ticks then the last crossed number is the answer.

This problem has been posted many times........but here's my solution:
[hide]It is easy to see that 43 McNuggets can't be obtained:
-[b]Case 1[/b]: If you have no boxes of 20 McNuggets, you have to make 43 McNuggets with boxes of 6 or 9. Both are multiples of 3 but 43 is not divisible by 3. So you can't make 43 McNuggets with only boxes of 6 and 9 McNuggets.
-[b]Case 2[/b]: If you have only 1 box of 20 McNuggets, you have to make 23 McNuggets with only boxes of 6 or 9. Both are divisible by 3 but 23 is not divisible by 3. So you can't make 43 McNuggets with only 1 box of 20 McNuggets.
-[b]Case 3[/b]: If you have only 2 boxes of 20 McNuggets, you have to make 3 McNuggets with only boxes of 6 or 9. You can't, since 6>3 and 9>3. So you can't make 43 McNuggets with only 2 box of 20 McNuggets.

But the following number of McNuggets can be obtained:
[b]44[/b]: 6 + 6 + 6 + 6 + 20
[b]45[/b]: 6 + 6 + 6 + 6 + 6 + 6 + 9
[b]46[/b]: 6 + 20 + 20
[b]47[/b]: 6 + 6 + 6 + 9 + 20
[b]48[/b]: 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6
[b]49[/b]: 9 + 20 + 20

So any number n, n > 49, of McNuggets can be obtained by adding a multiple of a box of 6 McNuggets to the 44,45,46,47,48,49 McNuggets you already have. So we have proven that you can get n > 43 McNuggets no matter what.

So the answer is: $\fbox{43}$ McNuggets.[/hide]

Please use the search function. This has been posted way too many times.


How many different formulas/topics does the sidebar contain?

I think it has 9 such stuff........my cache at times shows side9.swf and I'm quite sure there's no file that I've seen where the cache shows something like side10 or above that..............Those animations seem to be amazing at explaining some of the fundamental concepts...........I'm really looking forward to seeing more of them in the near future...

the one on the screen to ur left with all these moving mathematical equations and stuff

By the way, I think it only shows up for the AOPS version, but not for mathlinks.

But then does anyone here know who actually made that stuff ? I would really like to talk to the person who worked on those things.....Plenty of them seem to be an extrodinary amount of hard work { I know people might say it's just some simple stuff that's done on flash, but then even that "simple stuff" requires quite a bit of patience to make } .....and to whoever's made them, it's a huge thank you from my side..........those sidebar's are just amazing..


algebra
polynomial
function
domain
integration
calculus
complex analysis

Normally, we apply the Rouche Theorem to a polynomial and the domain should be a disk or an annular. The question is how to apply to a half plane and the equation is not a polynomial. For example, how to find the number of roots of the equation
\[ e^{\minus{}z} \plus{} z^2 \minus{}16\equal{}0\]in the right half plane $ \Re z >0$. Thank you.

let $ G$ be region in $ \mathbb{C}$ and $ f$ is analytic in $ G$, $ \gamma$ be a closed curve in $ G$ then we have
 $ \int_{\gamma}\frac{f'}{f}dz\equal{}2\pi i Z_{f}$ where $ Z_{f}$ is the number of zeros of $ f$ in $ G$
and the integral $ \int_{\gamma}\frac{f'}{f}dz$ can be found by residue theorem

Choose $ \gamma$ be the semicircle with radius $ R$ and from that you can find the number of zeros of $ f$ inside $ G$

[quote="1234567a"]Choose $ \gamma$ be the semicircle with radius $ R$ and from that you can find the number of zeros of $ f$ inside $ G$[/quote]

I know what you are telling me is to compute the Winding number, but I am wondering how to fix $ R$ (I know that it is large enough but maybe the given equation has a sequence of solution $ \{z_n\}_{n \geq 0}$ such that $ |z_n| \to \infty$ as $ n \to \infty$.

Here's some suggestions.

We could work on the contour consisting of the semi-circle of B(0,R) starting at -iR to iR, combined with the line segment from iR to -iR. We can see that if we were to let R tend to infinity the inside of this contour is the right half of the complex plane. So now it only suffice to define our f(z) and g(z) appropriately and showing that |f(z)|>|g(z)| for all points on the contour in order to apply Rouche theorem.

Try using f(z)= z^2 - 16 and g(z) = exp(-z), it is quite routine to show that |f(z)|>|g(z)| holds for all points on the semi-circle part of the contour using some triangle inequalities. To show that |f(z)|>|g(z)| also holds for all points on the line segment part of the contour, we note that every point on the line segment part of the contour can be written as z = iy, where y is some real number from the interval [-R,R]. So substitute z = iy into |f(z)| and |g(z)| and try to drawn the conclusion that |f(z)|>|g(z)|.

Hope this helps :lol:

Fix $ R>5$ (this is larger than necessary, but we don't care since it just needs to work for all large $ R$).

On the semicircle of radius $ R$ in the above post we get that
\begin{eqnarray*}
|e^{-z}+z^2-16-(z^2-16)| & = & |e^{-z}|\\
 & = & \frac{1}{e^{Re(z)}} \\
 & \leq & 1\\
 & < & R^2-16\\
 & = & \parallel{}z|^2-16|\\
 & \leq & |z^2-16|
\end{eqnarray*}

Now on the vertical segment parameterize $ \gamma(t)=it$ for $ t\in [-R, R]$.

On this we get
$ |z^2-16|=|(it)^2-16|=|-t^2-16|=|t^2+16|>16$.

So $ |e^{-z}+z^2-16-(z^2-16)|=|e^{-it}|=\frac{1}{e^{Re(it)}}=1<16<|z^2-16|$.

So Rouche applies and there are 2 zeros in the right half plane.


http://www.artofproblemsolving.com/Wiki/index.php/2000_AMC_12_Problems/Problem_13

In the problem description, it says "One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero." The answer, however, is reached by assuming that "... each person drinks the same amount of coffee and milk". Unless I'm having major problems understanding the question, it seems to state the amounts of coffee and milk in each cup aren't fixed. Is it just a badly phrased question, or could someone elaborate on where I'm taking it wrongly?

They gave the wrong justification for that equation. What's given is that $ \frac {a}{6} \plus{} \frac {b}{4} \equal{} 8$ and $ a \plus{} b \equal{} 8n$. From this you can conclude what they had written.

EDIT: On second thought, when they said "same amount of coffee and milk", they were probably referring to the fact that each person drinks $ 8$ ounces. That's very unclear though.

Ahh, makes sense now, thanks. I'm glad to find out I wasn't going insane.

:D

That's what I get for trying to write an entire set of AMC solutions in a day.  :)

Thanks for pointing that out. I have fixed it.

(Oh yeah; don't forget these aren't official solutions, and plus my writing isn't that great. Feel free to make corrections yourself - the advantage of the wiki)


Αν $ a,b,c,d$ μη αρνητικοί πραγματικοί ώστε $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$ να δειξετε ότι :arrow:  $ (1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)\geq abcd$

lemma
$ \frac {1}{(1 \plus{} x)^2} \plus{} \frac {1}{(1 \plus{} y)^2} \geq \frac {1}{1 \plus{} xy}$ (.. expand :P )
Tωρα στην δικια μας ανισοτητα θετουμε $ x \equal{} \frac {1 \minus{} a}{a}$ etc και εχουμε νδο $ xyzw \geq 1$
Oμως $ 1 \equal{} \sum \frac {1}{(1 \plus{} x)^2} \geq \frac {1}{1 \plus{} xy} \plus{} \frac {1}{1 \plus{} zw} \leftrightarrow xyzw \geq 1$
qed

[quote="Athinaios"]lemma
$ \frac {1}{(1 + x)^2} + \frac {1}{(1 + y)^2} \geq \frac {1}{1 + xy}$ (.. expand :P )
Tωρα στην δικια μας ανισοτητα θετουμε $ x = \frac {1 - a}{a}$ etc και εχουμε νδο $ xyzw \geq 1$
Oμως $ 1 = \sum \frac {1}{(1 + x)^2} \geq \frac {1}{1 + xy} + \frac {1}{1 + zw} \leftrightarrow xyzw \geq 1$
qed[/quote]

Το λήμμα που ανέφερες προκύπτει από την expanded ανισότητα Jensen η οποία μας λέει ότι$ f\longrightarrow \mathbb{R^{+}}\Longrightarrow f(x)+f(y)\geq 2f\left(\sqrt{xy\right)}\Longrightarrow \sum^{n}_{i=1}f(x_{i})\geq nf\left(\sqrt[n]{\prod^{n}_{i=1}x_{i}}\right)$

Άρα μία άλλη απόδειξη εκτός του ''expanding'' είναι ότι $ \frac{1}{\left(1+x\right)^{2}}+\frac{1}{\left(1+y\right)^{2}}\geq \frac{2}{\left(1+\sqrt{xy}\right)^{2}}\geq \frac{1}{1+xy}\Longrightarrow 2\left(1+xy\right)\geq \left(1+\sqrt{xy}\right)^{2}$ :arrow: Cauchy.

Όσον αναφορά την άσκηση τώρα για προσπαθήστε να τη λύσετε με $ AM-GM$

P.S Ηλία την σκότωσες με πολύ ωραίο τρόπο την άσκηση :coolspeak:

NickNafplio wrote:
deleted

Αφού  $ f(1)\equal{}0$,  αποκλείεται η συνάρτηση αυτή να έχει ελάχιστο στο 1/2. Μήπως εννοείς κάποια άλλη συνάρτηση γιατί η παράγωγος που έχεις γράψει δεν ανταποκρίνεται σε αυτή τη συνάρτηση. Θα είναι καταπληκτική λύση αν την κάνεις να δουλέψει.

[quote="Jondrette"][quote="NickNafplio"]deleted[/quote]

Αφού $ f(1) \equal{} 0$ αποκλείεται η συνάρτηση αυτή να έχει ελάχιστο στο 1/2. Μήπως εννοείς κάποια άλλη συνάρτηση γιατί η παράγωγος που έχεις γράψει δεν ανταποκρίνεται σε αυτή τη συνάρτηση. Θα είναι καταπληκτική λύση αν την κάνεις να δουλέψει.[/quote]

Ειναι γνωστη μεθοδος που χρησημοποιειτε συχνα σε τετοια προβληματα, αλλα απ οτι φαινεται δεν δουλευει εδω. Παρασυρθηκα και νομισα οτι η συγκεκριμενη μεθοδος δουλευει εδω οποτε και εθεσα τη συναρτηση χωρις να κοιταξω τη συμπεριφορα της και λογο βιασυνης εκανα λαθος στην παραγωγο και βρηκα ετσι ενα ανυπαρκτο ακροτατο... edited

mia lusi kai me tin am-gm.

isxuei

$ c^2\plus{}d^2\geq 2cd\Longleftrightarrow 1\minus{}a^2\minus{}b^2\geq 2cd\Longleftrightarrow \minus{}2cd\plus{}2(1\minus{}a)(1\minus{}b)\geq 2(1\minus{}a)(1\minus{}b)\plus{}a^2\plus{}b^2\minus{}1\equal{}\left(a\plus{}b\minus{}1\right)^{2}\geq 0\Longleftrightarrow (1\minus{}a)(1\minus{}b)\geq cd$

me tin idia logiki deixnoume oti $ (1\minus{}c)(1\minus{}d)\geq ab$. pollaplasiazoume kai prokuptei i zitoumeni


Lets $ a$ real such that :  $ 0 \leq a \leq 1$
$ x$ & $ y$ positive reals such that : $ x \plus{} y \geq (x \minus{} y)^2$
Prove that:
 $ a^x(1 \minus{} a)^y \plus{} a^y(1 \minus{} a)^x \leq \frac {1}{2^{x \plus{} y \minus{} 1}}$

if $ x \equal{} y$, the inégality becomes $ 2(a(1 \minus{} a))^x \equal{} < \frac {1}{2^{2x \minus{} 1}}$
<=> $ a(1 \minus{} a) \equal{} < 1/4$ for $ 0 < \equal{} a < \equal{} 1$ well known.
Now suppose that $ x > y$ ( by symétry)
Lets $ f(a) \equal{} a^x(1 \minus{} a)^y \plus{} a^y(1 \minus{} a)^x$ for $ 0 \equal{} < a \equal{} < 1$
We have : $ f(1 \minus{} a) \equal{} f(a)$ => $ a \equal{} 1/2$ is symetric axis thus we will prove the inequality just for $ 0 < \equal{} a < \equal{} \frac {1}{2}$

$ f(a) > \equal{} 0$ in $ [0,\frac {1}{2}]$, $ f(0) \equal{} 0$ and $ f(\frac {1}{2}) \equal{} \frac {1}{2^{x \plus{} y \minus{} 1}}$
lets prove that $ f$ is increasing in $ [0,\frac {1}{2}]$.
For$ 0 < a < \frac {1}{2}$ we have :
$ f'(a) \equal{} xa^{x \minus{} 1}.(1 \minus{} a)^y \minus{} ya^x.(1 \minus{} a)^{y \minus{} 1} \plus{} ya^{y \minus{} 1}.(1 \minus{} a)^x \minus{} xa^y.(1 \minus{} a)^{x \minus{} 1}$
Now put $ 1 \minus{} a \equal{} at$ <=> $ a \equal{} \frac {1}{1 \plus{} t}$ and  $ t > 1$
=> $ f'(a) \equal{} xa^{x \plus{} y \minus{} 1}.t^y \minus{} ya^{x \plus{} y \minus{} 1}.t^{y \minus{} 1} \plus{} ya^{x \plus{} y \minus{} 1}.t^x \minus{} xa^{x \plus{} y \minus{} 1}.t^{x \minus{} 1}$
=> The  sign of $ f'$ is the same as of $ g(t) \equal{} y.t^{x \minus{} y \plus{} 1} \minus{} x.t^{x \minus{} y} \plus{} x.t \minus{} y$ in $ ]1, \plus{} \infty[$
=> $ g'(t) \equal{} y(x \minus{} y \plus{} 1)t^{x \minus{} y} \minus{} x(x \minus{} y)t^{x \minus{} y \minus{} 1} \plus{} x$
=> $ g^{(2)}(t) \equal{} y(x \minus{} y \plus{} 1)(x \minus{} y)t^{x \minus{} y \minus{} 1} \minus{} x(x \minus{} y)(x \minus{} y \minus{} 1)t^{x \minus{} y \minus{} 2}$
=> $ g^{(2)}(t) \equal{} (x \minus{} y)t^{x \minus{} y \minus{} 2}(y(x \minus{} y \plus{} 1)t \minus{} x(x \minus{} y \minus{} 1))$
=> $ g^{(2)}(t) \equal{} (x \minus{} y)t^{x \minus{} y \minus{} 2}(y(x \minus{} y \plus{} 1)(t \minus{} 1) \plus{} x \plus{} y \minus{} (x \minus{} y)^2) > \equal{} 0$
=> $ g'$ is increasing but  $ g'(1) \equal{} y(x \minus{} y \plus{} 1) \minus{} x(x \minus{} y) \plus{} x \equal{} 2xy \minus{} y^2 \minus{} x^2 \plus{} x \plus{} y \equal{} x \plus{} y \minus{} (x \minus{} y)^2 > \equal{} 0$
=> $ g$ is increasing and we have $ g(1) \equal{} 0$ ==> $ g > 0$ in $ ]1, \plus{} \infty[$
=> $ f$is strictly increasing in $ [0,\frac {1}{2}]$
C.Q.D


Show that of all triangles with a given perimeter and base, an isosceles triangle has the greatest area.

$ S\equal{}\sqrt{p(p\minus{}a)(p\minus{}b)(p\minus{}c)}$
$ \sqrt{p}\equal{}const \Rightarrow S\equal{}max   \Longleftrightarrow (p\minus{}a)(p\minus{}b)(p\minus{}c)\equal{}max$
Such $ (p\minus{}a)\plus{}(p\minus{}b)\plus{}(p\minus{}c)\equal{}p\equal{}const$
Hence $ p\minus{}a\equal{}p\minus{}b\equal{}p\minus{}c  \Rightarrow    a\equal{}b\equal{}c$
q.e.d

____________________
P.Rovsen   :rotfl:  :roll:

i think that would be also an solution with Sturm's method but that sure is harder than the AM-GM method witch Pirkuliyev Rovsen used.


If N is any k*k nilpotent matrix i.e;N^k=0 but N^k-1 not equal to 0 ,then show that transpose of N is similar to N.Hence deduce that every complex n*n matrix is similar to its transpose.

Such nilpotent matrix $ N \in M_n$ is similar to the (nilpotent) matrix $ H_n \equal{} [\delta _{i \plus{} 1,j}]$. But if $ \hat{I_n}$ denotes the "counteridentity" matrix of order n, then $ \hat{I_n} ^{ \minus{} 1} \equal{} \hat{I_n} ^T \equal{} \hat{I_n}$, and $ H_n \equal{} \hat{I_n} H_n^T \hat{I_n}^{ \minus{} 1}$. Now consider the jordan normal form $ J$ of an arbitrary $ A \in \mathbb{C} ^{n \times n}$, $ A \equal{} PJP^{ \minus{} 1}$, where $ J \equal{} diag(J_1, J_2, ..., J_s)$. Each Jordan block $ J_j$ is of the form $ J_j \equal{} \lambda I_k \plus{} H_k$, and any similarity transformation applied to $ J_j$ is in fact only applied to $ H_k$: $ Q J_j Q^{\minus{}1} \equal{} Q (\lambda I_k \plus{} H_k) Q^{\minus{}1} \equal{} \lambda I_k \plus{} QH_k Q^{\minus{}1}$. Therefore, each Jordan block is similar to its transpose, since $ J_j^T \equal{} \hat{I_k}(\lambda I_k \plus{} H_k) \hat{I_n}^{\minus{}1}$. So, if $ \alpha_j$ denotes the dimension of the block $ J_j$, and defining $ T \equal{} diag(\hat{I_{\alpha_1}}, \hat{I_{\alpha_2}}, ..., \hat{I_{\alpha_s}})$, then $ J \equal{} TJ^TT^{\minus{}1} \equal{}> J^T \equal{} T^{\minus{}1}JT$ and so

$ A^T \equal{} (P^T)^{\minus{}1}J^TP^T \equal{} (P^T)^{\minus{}1} T^{\minus{}1}JT P^T \equal{} (TP^T)^{\minus{}1} J (TP^T)$

$ \equal{} (TP^T)^{\minus{}1} P^{\minus{}1}AP(TP^T) \equal{} (PTP^T)^{\minus{}1} A (PTP^T) \equal{} S^{\minus{}1}AS$.


function
inequalities
Functional Equations

Find all functions $f:\mathbb{Z} \to \mathbb{Z}$ such that for all $m\in\mathbb{Z}$: [list][*] $f(m+8) \le f(m)+8$, [*] $f(m+11) \ge f(m)+11$.[/list]

Using first equation we get  \[f(m+88) \leq f(m)+88\] and from the second one \[f(m+88) \geq f(m)+88\] which means that in all those inequalities, the equality holds. We got \[f(m+8)=f(m)+8\] and \[f(m+11)=f(m)+11\]
Let $f(0)=a$. As $GCD(8,11)=1$. Each number $k$ can be written in form \[k=x \cdot 8+y \cdot 11\] where $x,y \in Z$.
Using our equalities we obtain $f(k)=x \cdot 8+y \cdot 11=k$ which means that \[f(k)=k+a\] for some $a \in Z$

Another solution.

We prove that the answer is $f(m)=m+a$ for all $m \in \Bbb{Z}$,and some $a \in \Bbb{Z}$.

We know that
$i)$ $f(m+8) \leq f(m)+8$
$ii)$ $f(m+11) \geq f(m)+11$

By $i$ we get that:
$$f(m)+11 \geq f(m+8)+3$$
So by $ii$ we get that:
$$f(m+11) \geq f(m+8)+3$$

Hence
$$f(m) \geq f(m-3)+3 \quad(1)$$

Take a fixed integer $m_{0}$. By using $(1)$ and $i$ we get the bellow result:
$$f(m_{0}+2)+6 \leq f(m_{0}+8) \leq f(m_{0})+8$$
$$f(m_{0})+9 \leq f(m_{0}+9) \leq f(m_{0}+1)+8$$
$$f(m_{0}+1)+9 \leq f(m_{0}+10) \leq f(m_{0}+2)+8$$

So;
$$f(m_{0}+2)+6 \leq f(m_{0})+8$$
$$f(m_{0})+9 \leq f(m_{0}+1)+8$$
$$f(m_{0}+1)+9 \leq f(m_{0}+2)+8$$

As the sum of LHS’s is equal to sum of RHS’s; inequalities become to equality.
So:
$$f(m_{0}+2)=f(m_{0})+2$$
$$f(m_{0})=f(m_{0}+1)-1$$
$$f(m_{0}+1)=f(m_{0}+2)-1$$
It means that:
$$f(m_{0})=f(m_{0}+1)-1=f(m_{0}+2)-2$$

Which holds for any integer $m$. Hence it concludes that $f(m+1)=f(m)+1$ for all integer $m$.
So $f(m)=m+a$ for all $m \in \Bbb{Z}$ and some $a \in \Bbb{Z}$.
And it holds problem’s inequalities.
Done.


Let $ x,y,z$ be positive reals.
Find the maximum value of $ \frac{x^3y^2z}{x^6\plus{}y^6\plus{}z^6}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

$G$ a finite group with $n$ elements and $f$ an automorphism of $G$. Suppose that for every $x \in G$, there exists $a \in G$ ($a$ depends of $x$) such that $f(x)=axa^{-1}$.
Suppose now that $p$ is a prime number that divides the order of $f$.
Prove that $p$ also divides $n$.

Assume a prime number $p$ divides the order $o(f)$ of $f$, then $\exists x \in G, \exists k \geq 1, f^{(p^{k})}(x)=x$ and $\forall n <p^{k}, f^{(n)}(x) \neq x$ (*), where $f^{(n)}$ denotes $f o \dots o f$ ($n$ times). (This is due to the decomposition of the permutation $f$ into disjoint cycles). But now $f^{(n)}(x)=a^{n}xa^{-n}$ for some $a \in G$; denote by $o(a)$ the order of $a$. We have $f^{(a)}(x)=x$ and if $r$ is the remainder of $a$ in the euclidean division by $p^{k}$, we see $r=0$ by (*), meaning p divides $o(a)$, and in particular the order of $G$.

[quote="julien_santini"]But now $f^{(n)}(x)=a^{n}xa^{-n}$ for some $a \in G$[/quote]

Unless I'm misunderstanding, in which case I apologize, this is not true. You only know that $f(x)=axa^{-1}$ for some $a$ which depends on $x$. This means that $f(axa^{-1})=f(a)axa^{-1}f(a)^{-1}$, but nobody said that $f(a)=a$.

Hi Grobber,

You're right it's a mistake ... sniff

Such f means that f leaves conjugacy classes fixed. We may suppose that p=order of f. Now each element is fixed by f or it is on a p-cycle. Suppose that in each conjugacy class there is at least 1 fixed point. If not, there exist a conjugacy class partioned on p-cycle, hence the number of element of this conjugacy class is divisible by p, then G has order divisible by p.
  Now, take one fixed point from each conjugacy class and let H be the subgroup they generate. It's clear that H is a proper subgroup, beacuse each element of H is fixed by f and f itsnt the identity. On the other hand, G is the union of all conjugates of H (since each element is conjugate to one of his generators), this being a contradiction since no proper subgroup have this property.
  (i post this on a new topic)
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=343654#343654]Look here[/url]


Initially,the capacitor is empty.After charging the capacitor to the maximum value of charge,we change the places of its plates.Then find the maximum current?

P.S.Our olympiad teacher gave it us to solve.I got one anwer.I want to see if you get the same answer. :)

I got $U\sqrt{C/L}$, but I'm not sure if it's right. What did you get?

But I got$2U\sqrt{\frac{C}{L}}$.Can I see your solution?MAy be I or you made any mistake. :maybe:

Well, by differentiating the second Kirchoff's law, I got that the current in the circuit is \[i =-I_{0}\sin\omega t,\] where $\omega^{2}= LC$ (until the capacitor is fully charged). The initial condition is that $-I_{0}\omega =-q =-UC$, and since $I_{0}$ is the maximal current (when the capacitor is empty), you get $I_{0}= U\sqrt{C/L}$. How did you do it?

You got that we can get the maximum current in the first position of the capacitor.I think there must another current :maybe: My solution is a bit longer.I don't have time to post is now.I'll post tomorrow.


Find all the functions $ f: \mathbb{Q}\rightarrow \mathbb{R}$ such that :

$ f(x\minus{}y)\plus{}f(y\minus{}z)\plus{}f(z\minus{}x)\equal{}3(f(x)\plus{}f(y)\plus{}f(z))\minus{}f(x\plus{}y\plus{}z)\ ,\ (\forall)x,y,z\in \mathbb{Q}$

[hide]Putting $ x \equal{} 0,y \equal{} 0,z \equal{} 0$ we get $ f(0) \equal{} 0$

Putting $ x \equal{} t, y \equal{} 0, z \equal{} 0$ we get $ f(t) \plus{} f( \minus{} t) \equal{} 3f(t) \minus{} f(t)\iff f( \minus{} t) \equal{} f(t)$

Putting $ x \equal{} t, y \equal{} t, z \equal{} 0$ we get $ 2f(t) \equal{} 3(2f(t)) \minus{} f(2t)\iff f(2t) \equal{} 4f(t)$

Putting $ x \equal{} nt, y \equal{} \minus{} t, z \equal{} t$ we get

$ f((n \plus{} 1)t) \plus{} f(2t) \plus{} f((n \minus{} 1)t) \equal{} 3(f(nt) \plus{} 2f(t)) \minus{} f(nt)$

which is equivalent to

$ f((n \plus{} 1)t) \minus{} 2f(nt) \plus{} f((n \minus{} 1)t) \equal{} 2f(t)$

Regarding this as a recursive relation in $ f(nt)$, by any of the standard methods we easily find

$ f(nt) \equal{} n^2f(t)\quad(*)$

Therefore $ f(n) \equal{} n^2f(1) \equal{} cn^2$ where $ c: \equal{} f(1)$

Putting $ t \equal{} {m\over n}$ into $ (*)$, we get $ f(m) \equal{} n^2f\left({m\over n}\right)$, but as $ f(m) \equal{} cm^2$, this yields $ f\left({m\over n}\right) \equal{} c\left({m\over n}\right)^2$. Therefore $ (\forall x\in\mathbb{Q}) f(x) \equal{} cx^2$

Plugging this into the original equation, we easily check that any real $ c$ works. Therefore $ f(x) \equal{} cx^2,c\in\mathbb{R}$[/hide]


vector
AMC
AIME
geometry
search
analytic geometry
AIME I

[quote="remy1140"]Are there vector-related problems in AIME? (I seem to be week in that subject, and I'm not sure if I should study that part for AIME.) If there are vector problems, how hard are they?[/quote]

Very occasionally. Vectors are interesting but do not appear on contests often.

Most of the AIME analytical geometry problems don't need vectors:)

Study it, but don't overload on it.

In analytic geometry, vectors and complex numbers are very similar, so if you already know how to use complex numbers well (in geometry), then you probably know most things about using vectors in geometry.

I think there have been very few, if at all, problems on the AIME that directly relate to vectors, after just doing a search for "AIME vector". However, complex numbers and vectors are sometimes useful to bash AIME (and olympiad :P) geometry problems, and they usually turn out nicer than coordinates if you're going to bash.

[quote="sdkudrgn88"]I'd even say that for the purposes of AIME, vectors [i]are[/i] complex numbers when it comes to analytic geometry. I don't think you'll ever need the vector products (dot, cross) in any question of theirs involving vectors.[/quote]

...Except that in terms of 2 d geometry, complex numbers can do everything at least as well as vectors? (Including dot/cross products.)

I don't see any reason at all to use vectors instead of complex numbers unless its not 2d


Is anyone having trouble accessing the USAMTS website?

Recently, they wrote that the mailing address would be updated in September. I checked about two weeks ago, and I did not see it. I tried to check today, but for some reason the page won't load.

Any help would be appreciated!

hmm...a few weeks ago I got "page cannot be displayed" but havent tried since then...

I tried a couple of days ago and it worked well.  Crap...now it's messed up; "Cannot find server...".  I hope they fix it.

-interesting_move (still, not posting much any more)

The whole NSA server's down...got a bad feeling about this...

i guess it's down... but i checked the website a couple days ago, before it went down, and the address was there, but i didn't think to write it down, UGH. so if anyone has the address, please post it here...

oh wait. nevermind, harharhar.

http://www.artofproblemsolving.com/wforum/viewtopic.php?t=1279

really? they put up the questions in like early August and they're not due till October and the questions are real easy so ...

well I haven't known them till like a week before when the server went down, and even if I do finish the questions I don't have the cover sheets and such.

P.S. My letter was written in September 6th, and it arrived yesterday (September 19th), so I've got 2 weeks, give or take.


function
inequalities
induction
algebra unsolved
algebra

One defines a function on the nonnegative integers as follows:

f(0) = 0
f(2n + 1) = f(2n) + 1
f(4n + 2) = f(4n) + 2
f(4n) = 2f(2n) - f(n)

(a) Show that f(3n) \leq 2f(n) + \sqrtf(n) for all positive integers n.
(b) For which n does the above inequality become an equality?

Nothing? I did something with binary representations (a nice result, which shows how f is determined) but I'm stuck on proving the inequality itself.

Here are some observations that you may have already noticed.  Suppose that the binary representation of n is ai 2i.  Then f(n) is the sum of ai (i + 1).  The proof is a pretty simple induction on n.

For part b, let n be of the form (4k - 1) / 3.  I claim that these numbers give equality.  Note that these values of n alternate 0's and 1's in their binary representation.  By the characterization above, we can calculate f(n) = k2.  Also f(3n) = 2k2 + k.  Thus equality is achieved.  I believe these are the only equality cases.

Here is an idea for showing the inequality.  Break up the binary representation of n into blocks of consecutive 1's.  Note that when computing 3n (or 2n + n), the blocks will not overlap.  Thus we should be able to relate f(3n) to f(n).  Well, that's all I have so far...

And the last piece of job...

Note that $f(a)+f(b)\geq f(a+b)$. Need comments?

Suppose now that $n$ consists of $k$ unit blocks, separated by zeros, denote by $n_{i}$ number which form $i$th without other '1'. We have \[f(n)=f(n_{1}+n_{2}+...+f_{k})=f(n_{1})+f(n_{2})+...+f(n_{k}),\]  \[f(3n)=f(3n_{1}+3n_{2}+...+3n_{k}) \leq f(3n_{1})+f(3n_{2})+...+f(3n_{k}).\] Check now that $f(3n_{i})<2f(n_{i})$ if $n_{i}$ contains at least two '1', and $f(3n_{i})=2f(n_{i})+1$ if $n_{i}$ contains exactly one '1'. It follows \[f(3n)\leq 2f(n)+k.\] Finally, since $n$ consists of $k$ unit blocks we conclude \[f(n)\geq 1+3+5+...+(2k-1)=k^{2}\] and the equality iff $n=10101...0101$.

That's all :)

Thanks, Myth, for completing the argument.


hi
please analyze this :

       x^4+x^3+1

What are you asking for? Want us to factor it?


quadratics
algebra
polynomial
inequalities
sum of roots
area of a triangle

Suppose $a,b,c$ are three rela numbers such that the quadratic equation \[ x^2 - (a +b +c )x + (ab +bc +ca) = 0 \] has roots of the form $\alpha + i \beta$ where $\alpha > 0$ and $\beta \not= 0$ are real numbers. Show that
(i) The numbers $a,b,c$ are all positive.
(ii) The numbers $\sqrt{a}, \sqrt{b} , \sqrt{c}$ form the sides of a triangle.

First, $abc \neq 0$, otherwise, wlog, $a=0$ and $b$ and $c$ would be real roots.

Since 0 is not root, eqution is equivalent to $x P(x) = (x-a)(x-b)(x-c) + abc$

If $abc < 0$ with, wlog, $a > 0$ then for $x$ close enough to $a$, $x P(x) < 0$, not possible.
$P(0) > 0$ and $a+b+c = 2 \alpha$

Every elementary cyclic polynomial of $a,b,c$ are positive so are they (classical by looking polynomial $(x-a)(x-b)(x-c)$)

$\Delta = \sum (a-b)^2 - \sum a^2 < 0$ f

Oops my post has been truncated  :(

Anyway it was almost finished   ;)

$ x^{2}\minus{}(a\plus{}b\plus{}c )x\plus{}(ab\plus{}bc\plus{}ca) \equal{} 0$

Sum of the roots $ \equal{} (\alpha \plus{}\iota\beta) \plus{}(\alpha \minus{}\iota\beta)\equal{}2\alpha\equal{}a\plus{}b\plus{}c$
Therefore since $ \alpha$ is positive, therefore $ a\plus{}b\plus{}c$ must be positive.

Now since the roots are complex,
$ \implies (a\plus{}b\plus{}c)^2 \minus{} 4(ab\plus{}bc\plus{}ca) <0
\implies a^2 \plus{}b^2\plus{}c^2<2ab\plus{}2bc\plus{}2ca
\implies (a\plus{}b\minus{}c)^2<4ab$
Therefore since the LHS is always non-negative,
$ \implies 4ab>0$
Similarly we have
$ 4bc>0 \ , \ 4ca>0$
From these three inequalities we get that either all of $ a,b,c$ must be positive or all must be negative.
Since the sum $ a\plus{}b\plus{}c$ is positive, therefore all the three are positive.
This proves one implication.

For the second one ,([b]Note :Please check this proof. I'm not sure about it [/b])
We assume without loss of generality that $ a \le b \le c$
This implies $ \sqrt{a} \le \sqrt{b} \le \sqrt{c}$
Therefore all we need to prove is that $ \sqrt{a} \plus{} \sqrt{b} > \sqrt{c}$
$ \iff a\plus{}b\plus{}2\sqrt{ab} < c
\iff a\plus{}b\minus{}c<\minus{}2\sqrt{ab}$

This is proved by first assuming $ c>a\plus{}b$ since the opposite will prove the implication.
Therefore taking the square root on both sides of the already proved $ (a\plus{}b\minus{}c)^2<4ab$
We have $ a\plus{}b\minus{}c<\minus{}2\sqrt{ab}$ since $ a\plus{}b\minus{}c$ is negative

We are done ...

I once again ask you to please check the proof of the second part

There is another way out.If $ a,b,c$ are the siders of a triangle then

$ |a\minus{}b|<c$

$ |b\minus{}c|<a$

$ |c\minus{}a|<b$

Squaring and adding the above three inequalities results in $ a^2\plus{}b^2\plus{}c^2<2(ab\plus{}bc\plus{}ca)$.

Thus for $ \sqrt{a},\sqrt{b},\sqrt{c}$ to be the sides of a triangle it suffices to prove that $ a\plus{}b\plus{}c<2\left (\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\right )$
..........................................................................................................................................
Now it has already been proved by Rijul Saini that

$ a\plus{}b\minus{}c<2\sqrt{ab}$

$ a\minus{}b\plus{}c<2\sqrt{ac}$

$ \minus{}a\plus{}b\plus{}c<2\sqrt{bc}$

Thus adding the above three inequalities,we get

$ a\plus{}b\plus{}c<2\left (\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\right )$

which should complete the proof.

[quote="pankajsinha"]There is another way out.If $ a,b,c$ are the siders of a triangle then

$ |a \minus{} b| < c$

$ |b \minus{} c| < a$

$ |c \minus{} a| < b$

Squaring and adding the above three inequalities results in $ a^2 \plus{} b^2 \plus{} c^2 < 2(ab \plus{} bc \plus{} ca)$.

Thus for $ \sqrt {a},\sqrt {b},\sqrt {c}$ to be the sides of a triangle it suffices to prove that $ a \plus{} b \plus{} c < 2\left (\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}\right )$
..........................................................................................................................................
[/quote]

I dont think that $ \sqrt {a},\sqrt {b},\sqrt {c}$ are the sides of a triangle [b]IF AND ONLY IF[/b] $ a \plus{} b \plus{} c < 2\left (\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}\right )$

What I mean to say is that you cannot really work backwards from this and get the statement of the problem........

[quote="Rijul Saini"]Therefore all we need to prove is that $ \sqrt {a} \plus{} \sqrt {b} > \sqrt {c}$
$ \iff a \plus{} b \plus{} 2\sqrt {ab} < c \iff a \plus{} b \minus{} c < \minus{} 2\sqrt {ab}$
[/quote]
Sorry the 2nd and 3rd inequality is wrong.....The signs should be reversed.....
So it becomes
$ a \plus{} b \plus{} 2\sqrt {ab} > c$
and
$ a \plus{} b \minus{} c > \minus{} 2\sqrt {ab}$

[quote="pankajsinha"][quote="Rijul saini"]

I dont think that $ \sqrt {a},\sqrt {b},\sqrt {c}$ are the sides of a triangle [b]IF AND ONLY IF[/b] $ a \plus{} b \plus{} c < 2\left (\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}\right )$

[/quote]

O.K..So what should be the other condition(s)(for if and only if to happen) or you simply mean to say that I should not work backwards.[/quote]

You didn't quite get my point.....What I mean't by "you can't work backwards from this " is that $ \sqrt {a},\sqrt {b},\sqrt {c}$  are the sides of a triangle implies $ a \plus{} b \plus{} c < 2\left (\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}\right )$
but the [b]converse isn't true[/b].....

Other Conditions---1)like in my solution prove that the sum of any two of the three is greater than the third....
2)Let $ \sqrt{a}\equal{}x\plus{}y$ , $ \sqrt{b}\equal{}y\plus{}z$ , prove that $ \sqrt{c}\equal{}x\plus{}z$ for some reals $ x,y,z$ (Haven't tried this one.....)

But that is how I have derived this condition.Sum of two sides greater than the third side also implies that difference of two sides is less than the third side.

[quote="pankajsinha"]There is another way out.If $ a,b,c$ are the siders of a triangle then

$ |a \minus{} b| < c$

$ |b \minus{} c| < a$

$ |c \minus{} a| < b$

Squaring and adding the above three inequalities results in $ ab\plus{}bc\plus{}ca\leq a^2 \plus{} b^2 \plus{} c^2 < 2(ab \plus{} bc \plus{} ca)$.

Thus for $ \sqrt {a},\sqrt {b},\sqrt {c}$ to be the sides of a triangle it suffices to prove that $ a \plus{} b \plus{} c < 2\left (\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}\right )$
[/quote]

You may have also come across the fact that if $ a,b$ and $ c$ are the sides of a triangle then$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\in [1,2)$

[quote="pankajsinha"]But that is how I have derived this condition.Sum of two sides greater than the third side also implies that difference of two sides is less than the third side.

[quote="pankajsinha"]There is another way out.If $ a,b,c$ are the siders of a triangle then

$ |a \minus{} b| < c$

$ |b \minus{} c| < a$

$ |c \minus{} a| < b$

Squaring and adding the above three inequalities results in $ ab \plus{} bc \plus{} ca\leq a^2 \plus{} b^2 \plus{} c^2 < 2(ab \plus{} bc \plus{} ca)$.

Thus for $ \sqrt {a},\sqrt {b},\sqrt {c}$ to be the sides of a triangle it suffices to prove that $ a \plus{} b \plus{} c < 2\left (\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}\right )$
[/quote]

You may have also come across the fact that if $ a,b$ and $ c$ are the sides of a triangle then$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\in [1,2)$[/quote]

That is how you derived the condition....but you can't work backwards...please don't make me repeat this again and again...read my last few posts carefully.....pm me if you still don't understand......I know how it hurts if your answer is wrong......but I am repeating the same thing .....

Ok ....now let's make this clear to you...what you're saying that is for example
 $ a>b, x>y$ is true   IF   $ a\plus{}b>x\plus{}y$. This is a false statement.

That is the same look carefully at your own solution
what you have done is

$ |a \minus{} b| < c$

$ |b \minus{} c| < a$

$ |c \minus{} a| < b$

Squaring and adding the above three inequalities results in $ ab \plus{} bc \plus{} ca\leq a^2 \plus{} b^2 \plus{} c^2 < 2(ab \plus{} bc \plus{} ca)$.

Now do you thing proving $ \longrightarrow ab \plus{} bc \plus{} ca\leq a^2 \plus{} b^2 \plus{} c^2 < 2(ab \plus{} bc \plus{} ca)$ gives us back the inequalities we started from? (i.e. $ |a \minus{} b| < c$ , $ |b \minus{} c| < a$ , $ |c \minus{} a| < b$)
I hope that u understand now......

$ a^4\plus{}b^4\plus{}c^4 < 2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)$ is the acutal condition for $ a,b,c$ to be the sides of a triangle... :)

Cheers,

Rofler

[quote="Rijul saini"]
Now do you thing proving $ \longrightarrow ab \plus{} bc \plus{} ca\leq a^2 \plus{} b^2 \plus{} c^2 < 2(ab \plus{} bc \plus{} ca)$ gives us back the inequalities we started from? (i.e. $ |a \minus{} b| < c$ , $ |b \minus{} c| < a$ , $ |c \minus{} a| < b$)
I hope that u understand now......[/quote].

Yes I do understand now.Thank you

For the first part.
Let the given quadratic be $f(x)$.
Since the given quadratic has imaginary roots, $f(x)>0$, for all x belonging to R.
So, $f(a)>0$, which gives that $bc>0$.
Similarly $f(b)>0$ which gives $ab>0$ and $ac>0$.
This shows that either  $a,b,c $all are all negative or all positive.
But,
since sum of roots=$(a+b+c)$ which is greater than zero since $\alpha>0$ thus this proves that $a,b,c>0$.
Done. :)

[quote="Rijul saini"][quote="Rofler"]$ a^4 \plus{} b^4 \plus{} c^4 < 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)$ is the acutal condition for $ a,b,c$ to be the sides of a triangle... :)

Cheers,

Rofler[/quote]

Would you elaborate?
Thanks[/quote]You can simply use Heron's formula which can be written as
$\triangle ABC=\frac 14\sqrt{-a^4-b^4 - c^4 + 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}$. ;)

[color=#0040FF][b]i have a solution , which I personally do not like , but is OK!! :( i am posting it.[/b][/color]
note that $a+b+c=2\alpha>0$
also , $\sum ab=(\alpha)^2+(\beta)^2>0$
note that , none of a , b, c  is 0
[otherwise the equation will have real roots]
 all of a, b, c can not be negative.
also the situation where 2 of a ,b ,c is negative and remaining one is positive is not possible.
 so , now consider the case where 2 of a,b,c are positive and one is negative . say a,b>0 and c<0
let a=p  ,b=q ,c=-r , where p,q,r>0
$D=(a+b+c)^2-4(ab+bc+ca)$.
so , $(a+b+c)^2<4(ab+bc+ca)$   i.e.  $(p+q-r)^2<4(pq-qr-pr)$...(*)
note that , $p+q>r$   i.e.   $p+q=r+s $ , where $s$ is some positive real.
so , now ,(*) turns into $s^2<4(p(r+s-p)-qr-rp)$ i.e. $ 4p^2+s^2+4qr<4ps$, which is clearly impossible.
so , we are done.

now , come to part (b)
if possible ,assume that $\sqrt{a}$ ,$\sqrt{b}$ ,$\sqrt{c}$ do not form the sides of a  $\Delta$.
then at least one of $\sqrt{a}+\sqrt{b}>\sqrt{c}$ ,$\sqrt{b}+\sqrt{c}>\sqrt{a}$ ,
$\sqrt{a}+\sqrt{c}>\sqrt{b}$ does not hold.
WLOG , assume that $\sqrt{a}+\sqrt{b}>\sqrt{c}$ does not hold.
so , $\sqrt{c}\ge \sqrt{a}+\sqrt{b}$ i.e. $ c=a+b=2\sqrt{ab}+k$ , where $k$ is some positive real.
so , from the $D<0$ condition ,
we arrive at $(2(a+b)+2\sqrt{ab}+k)^2<4(3ab+a^2+b^2+ak+bk+2(a+b)\sqrt{ab})$
this leads to $ k^2+4k\sqrt{ab}<0$ , which is impossible, hence done :D

Solution to the second part
As everyone above has arrived at the following result, thus I essentially start from the result
$bc>{(a-b-c)^2}/4$ .................(I)
$ca>{(b-c-a)^2}/4$..................(II)
$ab>{(c-a-b)^2}/4$...................(III)
So from (I)
We have
$bc>{(a-b-c)^2}/4$
if
$a-b-c>0$ then, taking square root both sides we get
$\sqrt{bc}>(a-b-c)/2$
On rearranging we get
($\sqrt{b}+\sqrt{c})^2$>($\sqrt{a})^2$
Now since all the terms are positive hence we have
($\sqrt{b}+\sqrt{c})>(\sqrt{a})$

If $a-b-c<0$
Then
$a<b+c$
Since all the term on either side of the inequality are positive hence,
$\sqrt{a}$<$\sqrt{b+c}$<$\sqrt{b}+\sqrt{c}$
Similarly from (II)  and (III) we get the other required inequalities
Hence,  proved.
;)


ARML
geometry
calculus
integration
trigonometry
ratio
geometric series

1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b) : (b+c) : (c+a) = 6:7:8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.

[quote]2.
If b+c)c+a) = 6:7:8" alt="(a+b)b+c)c+a) = 6:7:8" style="border: 0; vertical-align: middle;" onclick="newWindow= window.open('http://www.artofproblemsolving.com/Forum/latexrender/latexcode.php? code=%5Cdisplaystyle+%28a%2Bb%29%3A%28b%2Bc%29%3A%28c%2Ba%29+ %3D+6%3A7%3A8','latexCode','toolbar=no,location=no,scrollbars=yes,resizable=yes, status=no,width=375,height=150,left=200,top=100');" />, and , compute the value of . [/quote]
????

Because of a package installed on the site, several of these errors have popped up.  Forum moderators change fix them simply by creating more space in the TeX.  I'll fix this one.  Ultimately, it's a smiley trying to pop up out of the TeX.

Are you coming to the upcoming SFBA Arml meeting, Matt? (I'm also a Matt  :lol: ) In your location it says San Diego...

[hide="1"]$\frac{ab}{2}=\frac{(a+b)^{2}-(a^{2}+b^{2})}{4}=\frac{7}{2}$[/hide]
[hide="2"]lets change this to $14-c: 14-a: 14-b=6: 7: 8$ to make it easier
hey whaddayaknow the answers right in front of us $14-c=6$ so $c=8$, etc so $a=7$, $b=6$
EDIT: i just realized that $6+7+8\ne14$ :blush:
i'll solve the rest later, but it would just be multiplying out the ratios and using gaussian...[/hide]
[hide="3"]
hmm lets factor out $z^{1}5=(z^{5})^{3}=1$.  now we have $1+z+z^{2}+\cdots+z^{35}$ now lets use the geometric series formula to get $\frac{z^{36}-1}{z-1}=\frac{(z^{35}\cdot z)-1}{z-1}=\frac{z-1}{z-1}=1$[/hide]
[hide="7"]hmm lets factor out $n!$ to get $n!([n+1]+[n^{2}+3n+2])=n!(n^{2}+4n+3)=n!(n+1)(n+3)=(n+1)!(n+3)$.  okay, so we have to have 2 multiples of 7 in there, it'll be everything 13 and higher, plus 11 cuz of the n+3, so $17-13+1+1=6$[/hide]
[hide="8"]
ooh i remember this one from AoPS 2
$\frac{3a^{3}}{3\sin^{3}{a}}\ge7$ and as a gets bigger sin approaches 1, so the answer is $a\ge\sqrt[3]{7}$
[/hide]

junggi, this isnt a real set, or else i wouldnt be able to do 7 or 8 :P

Okay. I will attempt this problem set today and judge the difficulty of the problems based on the following scale (considering the fact that I suck at math):

1 Problems: Normal
2 Problems: Easy
3 Problems: Super easy
4 Problems: Super duper uberly easy
5+ Problems: Impossible to happen, no rating

@daermon, you got #3 and #8 right, the rest were wrong. I don't quite get your solution for #8 though...

My answers (I timed it, about 5 minutes for each set cause I'm too lazy to check my work).

[hide="1"]
$2$
[/hide]

[hide="2"]
$6$
[/hide]

[hide="3"]
$1$
[/hide]

[hide="4"]
$6$
[/hide]

[hide="5"]
$3\sqrt{2}$
[/hide]

[hide="6"]
$14$
[/hide]

[hide="7"]
$5, 12, 14, 15$
[/hide]

[hide="8"]
$\sqrt_{[}3]{7}$
[/hide]

[hide="Obvious"]
I suck at math.
[/hide]

[quote="Ubemaya"]
[hide="6"]
$14$
[/hide]

[hide="7"]
$5, 12, 14, 15$
[/hide]
[/quote]
these 2 are wrong
for 6, you are most likely
[hide]missing the case where the number is of the form $p^{3}$ where $p$ is a prime. that would be $8,27$[/hide]
for 7, you are missing
[hide]16[/hide]

@daermon:
for #1
[hide]$a^{2}+b^{2}\ne3$ :wink: [/hide]
for #7
[hide][quote][b]less than[/b] 17[/quote][/hide]

hints
[hide="4"]$z\cdot \bar{z}=|z|^{2}$[/hide]
[hide="8"]extended law of sines[/hide]

LOL I put 16 down on my paper and I forgot to type it up.

Yeah, I forgot about that case for #6. Darnit.

junggi, you got the hints wrong.

@daermon:
[hide="7"]
1) It asks for the numbers.
2) Under 17, like junggi said.
3) Don't forget that $n! = n!(1)$.
[/hide]
BTW, don't listen to Taylor on #4. Try: [hide]Pairing points (There are an even number of sides for a reason!) and finding the square of the lengths of the lines to those points.[/hide]

dude wtf
my way for 4 is amazingly cool dude
im too lazy to type it up tho
oh well here you go

oh wait i forgot to say
[hide]"use complex numbers" [/hide]
in my hint
haha...

[hide]let $x$ be the random point. let $w$ be the $2n$ root of unity not equal to 1.
square of distance from $x$ to a root of unity is
$|x-5w^{k}|^{2}=(x-5w^{k})\overline{(x-5w^{k})}=50-5\bar{x}w^{k}-5x\bar{w^{k}}.$ for some random $k.$
now sum it over all values of $0\le k\le 2n$ and use
$1+w+...+w^{n-1}=0$ and $\overline{1+w+...+w^{n-1}}=0$
and we're done

[/hide]

Is there any smarter way to do #2 other than cross multiplying and stuff solving for c? It seemed awfully boring.

what about
[hide="muffins"]$a+b=x,b+c=y,c+a=z.$.
Then we have
$x: y: z=6: 7: 8\implies y=\frac{7x}{6},z=\frac{8x}{6}$
but
$x+y+z=2(a+b+c)=28=x+\frac{7x}{6}+\frac{8x}{6}=28$
solving for $x$ we get
$x=a+b=a+b+c-c=14-c=8$
so
$c=6$
[/hide]

[quote="junggi"]was this supposed to be taken as a "practice test" as in they were supposed to be timed as in the real test?
or are they just practice problems?
cuz i just got an 8 on that sucker(timed, 10 min for each pair)[/quote]

I believe those questions comprised the individual round in 1984. My musty old notes contain this problem set. No clue what the scoring was like that year...


geometry
3D geometry
rectangle
logarithms
function
integration
real analysis

I am using the following version of Vitali's covering lemma - Let E be a set with $ |E|_{e} < {\infty}$. Suppose that E is covered by a countable union of cubes , then there exists a constant c and a N such that $ \sum_{i \equal{} 1}^{N}|Q_{i}|\geq c|E|_{e}$ ,where
$ |E|_{e}$ denotes the external measure of E and the cubes appearing in the above sum are disjoint

Now here is the question, lets work in the plane. Show that above lemma still holds if we take instead of cubes, covering by rectangles whose width is the square of its length.

( Also as an aside:I wonder if the lemma holds for ARBITRARY covering by rectangles. I dont have any proof or counter example yet, so your views are welcome)

Ceratinly not for any covering: just consider all rectangles centered at the origin of area $ 4$ with side length $ \le L$ . They cover an area about $ \log L$ (that in the hyperbolic cross $ |xy|\le 1$, $ |x|,|y|\le L$) but all have a common point. As to your first question, just mimic the classical proof of the Vitali theorem. You shouldn't meet any difficulty whatsoever.

The particular covering lemma you quote is a key step in proving that the Hardy-Littlewood maximal function is weak type (1,1). That is, if $ f\in L^1,$ and

$ M(f)(x)\equal{}\sup_{Q\ni x}\frac1{|Q|}\int_Q|f|,$ then for all $ \alpha>0,$ $ |\{Mf>\alpha\}|\le\frac{C\|f\|_1}{\alpha}.$

From this, we get the Lebesgue theorem on the differentiation of the integral:

If $ f\in L_{\text{loc}}^1,$ then $ \lim_{Q\ni x,\text{diam}(Q)\to0}\;\frac1{|Q|}\int_Qf\equal{}f(x)$ for almost every $ x.$

---

Proving such a covering lemma for the one parameter family of rectangles (sides parallel to the axes, sides $ r,r^2$) will lead to a similar weak type (1,1) result and a similar theorem on differentiating the integral.

---

But if we use arbitrary rectangles with sides parallel to the axes, it is known that the corresponding differentiation of the integral result is false. That is, there is some locally $ L^1$ function $ f$ such that

$ \lim_{R\ni x,\text{diam}(R)\to0}\;\frac1{|R|}\int_Rf\ne f(x)$ on a set of positive measure.

There is a positive result that can be obtained; it goes by the name of the theorem of Jessen, Marcinkiewicz, and Zygmund. (When I get back to my office, I should be able to find the paper in my collected works of Marcinkiewicz.)

If $ f\in L\log L$ (and zero outside a bounded set), and $ M_S(f)$ is the "strong" maximal function - that is the maximal function derived using arbitrary rectangles with sides parallel to the axes - then $ M_S(f)$ is in weak $ L^1,$ meaning that $ |\{M_S(f)>\alpha\}|\le\frac{C}{\alpha}.$ As a consequence, if $ f$ is locally in $ L\log L,$ then we can differentiate its integral through such rectangles. The counterexample referred to earlier must be in $ L^1\setminus L\log L.$

If you free up the orientations of these rectangles, worse things happen.

thanks Dr Kent. The following related problem appears in my homework -

Let $ Nf(x) = \sup{\frac {1}{|R|}\int_R|f|}$ where sup is taken over all rectangles R containing x( ie including those whose sides arent parallel to the axes) . Suppose that f is non-negative $ \mathbb{R\times R}$,strictly positive on a set of positive measure, then show that ${ \sup\alpha|\{Nf > \alpha\}|}\rightarrow\infty$ where sup is taken over all $ \alpha > 0$

so we dont have that weak-type inequality if we take rectangles with any orientation. I dont even know how to start this one. Can anyone give me a hint?

( I didnt want to start a new thread since these problems seem to be related)

Find a ball somewhere on which $ f$ has a positive integral. Ignore the rest of $ f.$ Now for a point $ x$ a long way from that ball, compute the average value of $ f$ over a long, skinny rectangle just long enough to include $ x$ and the ball and just wide enough to contain the ball. This gives you a lower bound for $ Nf(x).$ Now, compute the measure of the set on which this lower bound is greater than $ \alpha$ for [i]small[/i] positive $ \alpha.$

thank you very much Dr Kent. i was able to get it. Could i have " guessed" that the weak type inequality doesnt hold in this case? i mean, we obviously dont have the conclusion of vitali's theorem for arbitrary rectangles and this was a key step in proving the inequality the other case. What if i integrate over say rhombuses? Will a similar thing hold? I'm sorry to ask so many questions but i cant "see" such things naturally.

A good rule of thumb is to try to look at the Dirac delta-function and figure out the corresponding maximal function for it. More often than not it gives a counter-example immediately. Then you can approximate it by $ L^1$ functions if needed. As far as I know, the general question about which families of shapes give you the weak type 1-1 inequality and which don't is not resolved though quite a lot is known. As to seeing things naturally, nobody does it; it all comes with experience.

[quote="fedja"]A good rule of thumb is to try to look at the Dirac delta-function and figure out the corresponding maximal function for it.[/quote]
To be honest, my post #5, asking you to look at the "far field" maximal function for a function supported on a ball, was inspired by exactly this.


Find the sum of all the divisors,$ d$  of $ 19^{88} \minus{} 1$ where $ d \equal{} 2^a3^b$ and $ a,b > 0$

i too am having the  same method as above.


Sa se arate ca R nu este numarabila, folosind teorema intervalelor inchise incluse.

Sa presupunem prin absurd ca $[a,b]=(x_{n})_{n\geq1}.$ Acum sa impartim intervalul $[a,b]$ in trei parti egale
 $[a,\frac{2a+b}{3}]$ , $[\frac{2a+b}{3}, \frac{a+2b}{3}]$ and $[\frac{a+2b}{3}, b].$ Notam cu $I_{0}$ unul din intervale care nu contine pe $x_{0}.$ Acum, impartim intervalul $I_{0}$ in trei parti egale si notam $I_{1}$  unul dintre aceste intervale care nu contine pe $x_{1}.$Inductiv, vom avea sirul descrescator, $(I_{n})_{n\geq 0}$
de intervale inchise cu $l(I_{n})=\frac{b-a}{3^{n+1}}$ si $x_{n}$ nu apartine lui $I_{n}$, pentru orice $n\in\mathbb{N}.$ Conform lemei intervalelelor inchise incluse, concluzia decurge imediat. ;)


Hello..I want to find out what do you think about this following problem:
If $a,b \in\mathbb{R}$ and $k\in \((2;+ \ {infinity})$such that we have,simultaneos, the following conditions:
$(*) a^2+b^2=k\cdot ab$
$(**)\frac{a}{b}\neq\frac{k-\sqrt{k^2-4}}{2}$
$(***)1-\frac{a-b}{a+b}\leq\frac{4}{k+k\sqrt{2}}$, find the value of $\frac{a}{b}$
Is it solvable,easy,messy?I await your opinions and solutions.
EDIT:I make a mistake with the computations.Now I think the problem is corect.

Let a/b = t. If b=0 => a=0 from (*) and then a+b=0 which means that (***) is impossible, false.

(*) => t^2+1= kt => t= (k+sqrt(k^2-4))/2 since (**) forbids t from being (k-sqrt(k^2-4))/2

Then (***) becomes (1-t)/(1+t) +1 <= ... <=> 2/1+t <=... <=> 1+t>= k+sqrt (2k^2-6). Note that t<> -1 for (***) to hold.

Now replace t with the value as a function of k obtained above, separate the terms without sqrt on the right and the others on the left, and you'll get:

sqrt(k^2-4)-sqrt(8*k^2-24)>= k-2 . Now on the left we have a positive number only if k<=sqrt(20/7)<sqrt(3)<2 and so there are no such numbers a,b, k with all 3

conditions satisfied.

However, I may have made a computational mistake somewhere. If you check my computations and continue from where I left off you should be able to finish it easily (at most you have to square both sides 2 times).[/code]

Sorry Mihnea,I`ve made a computational mistake for the third condition.But now I think it`s solved and the solution should come out nicely.

Maybe tomorrow I`ll post my solution if nobody is willing to give it a try ;)


why is [b]isopropylbenzene [/b] more stable than[b] propyl benzene[/b]???????? :?:

I think this belongs in the Chemistry forum.

well, shouldn't it be the other way round...propyl benzene can be stabilized more by hyperconjugation...
(i could be pathetically wrong)

During hyperconjugative resonance the Ph-C bond is partial double character and obviously we know that more alkylated alkenes alkenes are more stable(again due to HC). So i-Pr-Ph is more stable than n-Pr-Ph

ok thanks...but wouldn't the fact that n propyl benzene has greater no of alpha hydrogen for hyperconjugation give it greater stability?..

n propyl benzene has only 2 alpha hydrogens for hyperconjugation whereas iso has 6.

No ur again going wrong in the sense that I am comparing those resonating substances which again can exchange H atoms.:

'resonating substances ' lol.
but what do you mean by again exchange h atoms? :huh:


i need a way to prove that the incenter of a triangle is the nagel point of its medial triangle. any help is apreciated.


This is the first problem I solved which was not for school :

Prove or disprove that if $f$ is a function defined from $(0;+\infty)$ onto itself such that $fof = Id$ ($f$ is involutive) and $\displaystyle \lim_{+\infty} f(x) = 0$, then $\displaystyle \lim_0 f(x) = + \infty$.

Pierre.

Let's disprove it.

Take first $f(x)=x$ for all rational $x\in (0,1]$. Then let the restriction of $f$ on $(1,2]$ be any bijection from $(1,2]$ to the irrationals from $(\frac 12,1)$, then we make the restriction of $f$ to $(2,3]$ any bijection from $(2,3]$ to the irrationals from $(\frac 13,\frac 12)$, and so on. We then define $f$ on the irrationals in $(0,1]$ in such a manner that $f$ becomes an involution.

It looks correct, but I might have overlooked something.

[quote="grobber"]Let's disprove it.

Take first $f(x)=x$ for all rational $x\in (0,1]$. Then let the restriction of $f$ on $(1,2]$ be any bijection from $(1,2]$ to the irrationals from $(\frac 12,1)$, [...][/quote]

Just out of curiosity: How do you find these bijections? I am also quite sure that they exist, but I am very interested in a construction (please don't say it needs the AC  :D ).

[b]PS.[/b] Pierre, thanks very much for the solution to the combinatorics problem... but I fear that a combi idiot like me will have to ask you some more questions...

  Darij