limit
real analysis
real analysis unsolved

Let X be a set equipped with a $ \sigma$ -algebra  $ A$. If $ m_{i}$ is a measure on $ (X, A)$ for each $ i$, such that $ m_{i}(X) \leq 1$. Define $ m: A \to [0, \infty)$ by
$ m(E) \equal{} \sum_{i \equal{} 1}^{\infty}\frac{1}{2^{i}}m_{i}(E)$ for all $ E \in A$. Prove that $ m$ is a measure on $ (X, A)$.

I think you just use the definition of measure and check that all  the conditions hold

$ m(\emptyset) \equal{} \sum\limits_{i\equal{}1}^\infty \frac1{2^i} m_i(\emptyset) \equal{} \sum\limits_{i\equal{}1}^\infty \frac1{2^i} 0 \equal{} 0$.

Take $ A, B \in \mathcal{A}$ such that $ A \cap B \equal{} \emptyset$. Then as each $ m_i$ is a measure, we have $ m_i(A\cup B) \equal{} m_i(A) \plus{} m_i(B)$. Thus we get

$ m(A \cup B) \equal{} \sum\limits_{i\equal{}1}^\infty \frac1{2^i} m_i(A \cup B) \equal{} \sum\limits_{i\equal{}1}^\infty \frac1{2^i} \left[ m_i(A) \plus{} m_i(B)\right] \equal{} m(A) \plus{} m(B)$.
Things converge here because we can estimate both parts of the big sum by $ \sum\limits_{i\equal{}1}^\infty \frac1{2^i} \equal{} 1$ respectively, as $ m_i(A) \leq m_i(X) \leq 1$ etc.

As $ m(A) \leq 1$ for all $ A \in \mathcal{A}$, we now only have to show that if we take an arbitrary decreasing sequence $ (A_j)_{j\in\mathbb{N}} \subset \mathcal{A}$ with $ A_j \downarrow \emptyset$ we have $ \lim\limits_{j\to\infty} m(A_j) \equal{} 0$. This is true for each of the $ m_i$.

So now we have $ \lim\limits_{j\to\infty} m(A_j) \equal{} \lim\limits_{j\to\infty} \lim\limits_{k\to\infty} \sum\limits_{i\equal{}1}^k \frac1{2^i} m_i(A_j) \equal{} \lim\limits_{k\to\infty} \sum\limits_{i\equal{}1}^k \frac1{2^i} \lim\limits_{j\to\infty} m_i(A_j) \equal{} 0$, where changing of the limits is justified by the uniform convergence of the series.

So all three properties defining a measure are satisfied by $ m$. (We still have to show that $ m: \mathcal{A} \to [0,\infty]$ but this should be easy :))


calculus
integration
analytic geometry
real analysis
real analysis unsolved

Let A={[a1,b1]x...x[an,bn], ai<bi} be the set of n-dimensional rectangular parallelepipeds of R^n.
For K in A, we note P(K) if there exists i s.t. bi-ai is an integer.
Let K=\cup (Kj,j=1..m) s.t. K and all Kjs are in A and int(Kj) \cap int(Kk) = \emptyset  if j <> k.
Prove that :  ( for all j=1..m, P(Kj) ) => P(K)
It's a classic, and you guys surely know the "integral trick" but I wonder if someone can prove this using only elementary results.
(Did I choose the good forum ?)

Let me try this topologically... Project the K_i and K onto (R mod Z)^n by
reducing each vertex' coordinates mod Z. Then the union of the projections of
the K_i is still K (multiply overlaid? i'm not yet sure about this, but in my
head, i keep reducing every single parallelepiped mod Z, and it still fills K
mod Z). Since all K_i are P(K_i),
their projections are at most (n-1)-dimensional, so their union K as finite
union of at-most-(n-1)-dimensional objects is at most (n-1)-dimensional.
Thus P(K).

Nice proof  :D

EDIT: No wait. Finally I'm not convinced.
I don't understand this loss of dimension.


An integer is called double if its decimal representation consists of a block of numbers followed by an equal block, for example: 12381238 is double. Prove that there are infinitely many double integers wich are squares.

From 1988, in case anyone is interested (problem 25):
[url]http://www.kalva.demon.co.uk/short/sh88.html[/url]

Just take numbers of the form $\frac {10^{21(2k+1)}+1}{49}\cdot 36\cdot (10^{21(2k+1)}+1)$.

How to reach something like this: we need numbers of the form $(10^t+1)n$, where $n$ has $t$ digits, so $10^{t-1}\le n<10^t$. Those are the most natural choices, because $49$ is the smallest square which can divide a number of the form $10^t+1$. The $36$ is there because $\frac {10^{21(2k+1)}+1}{49}$ has $21(2k+1)-1$ digits, so we need a larger number (we need one more digit).


I submitted my optional permission form about 3-4 days ago via fax, but my name hasn't been added to the leaderboard. Did you guys recieve my fax?

It takes time for someone to add your name to the leaderboard. usually, they update it all at once. Don't panic unless it isn't up by the time that Round 3 results are up.

[quote="solafidefarms"][quote="mysmartmouth"]I wouldn't give out your USAMTS ID like that.  I don't think anyone would do this, but it's possible to fill out a fake entry form using someone's ID (and real name, of course).[/quote]

In the old days when USAMTS was hosted with AoPS it was trivial to get people's USAMTS id's - they were the person's AoPS id number. I don't think it's that big of a problem.[/quote]
Besides, everyone who posts in the grading issues section gives out their ID.
I edited it out anyways...


[hide]Denote $b_{n}=a_{1}+a_{2}+...+a_{n}$, we have $b_{1}=1$ and $b_{n}=\frac{2n}{n-1}b_{n-1}$. Now, use telescopic product![/hide]


inequalities
LaTeX
inequalities proposed

Let a,b,c>0 Prove that
$ \frac{abc}{a^3\plus{}b^3\plus{}c^3}\plus{} \frac{4}{9}( \frac{a}{b}\plus{} \frac{b}{c}\plus{} \frac{c}{a}) \ge \frac{5}{3}$

It's obviously trues because of Am-Gm and...$ 4\sqrt[3]{4} \geq\ 3\sqrt[3]{9}$
 :)   :ninja:

Are you sure?Can you post the solution?I think this problem not so easy.

I think it's easy.
But I wait for some proofs. I'll post my proof later   :oops:

Yes,I think this problem is easy.My solution is same quykhtn.nguoivn,Can you post your solution by AM-GM?
I like this because  I study bad.

[quote="nguoivn"]Nice your proof, babyloverain.
And my full proof is here:  :)[/quote]
[b]Translated to English[/b] (with some trouble - language problem... :blush: )
[quote]1, For all nonnegative reals $ a, b, c$ show that we always have:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}\frac{9}{4}\frac{abc}{a^3\plus{}b^3\plus{}c^3}\ge\frac{15}{4}$.

Without loss of generality, assume $ c\equal{}min{a, b, c}$.
We can rewrite the inequality as
$ M(a\minus{}b)^2\plus{}N(a\minus{}c)(b\minus{}c)\ge0$,
where $ M\equal{}\frac{1}{ab}\minus{}\frac{3}{4}\frac{a\plus{}b\plus{}c}{a^3\plus{}b^3\plus{}c^3}$ and
$ N\equal{}\frac{1}{ca}\minus{}\frac{3}{4}\frac{a\plus{}b\plus{}c}{a^3\plus{}b^3\plus{}c^3}$

Here I prove : $ M,N\ge0$.
By BDT Cosi (? :oops: ):
$ a^3\plus{}b^3\plus{}c^3\ge\frac{(a\plus{}b\plus{}c)^3}{9}\Longrightarrow \frac{a\plus{}b\plus{}c}{a^3\plus{}b^3\plus{}c^3}\le\frac{\sqrt[3]{9}}{\sqrt[3]{(a^3\plus{}b^3\plus{}c^3)^2}}$
Hence we see that $ M\ge\frac{1}{ab}\minus{}\frac{3}{4}frac{\sqrt[3]{9}}{\sqrt[3]{(a^3\plus{}b^3\plus{}c^3)^2}}\ge0$
$ \Longleftrightarrow 4\sqrt[3]{(a^3\plus{}b^3\plus{}c^3)^2}\ge3\sqrt[3]{9}ab$
Again from BDT Cosi(? :( )
$ VT\ge 4\sqrt[3]{(a^3\plus{}b^3)^2}\ge4\sqrt[3]{4}ab>3\sqrt[3]{9} ab$
Thus we have $ M\ge0$.Similarly we can prove $ N\ge0$.

QED, Equality occurs if and only if $ a\equal{}b\equal{}c>0$.[/quote]
[i][b] Translated using Google translate and some of my knowledge in Vietnamese language and mathematics....  :rotfl: [/b][/i]

[hide]My questions:
1)What is the real name of BDT Cosi? I first thought it to be something like BDT [i]Cauchy[/i], but it is not so...... :(
2)How can one prove the initial inequality once he has proved $ M, N\ge0$?
PS:1) Please try to write the solutions in LaTeX and using ENGLISH......... The language problem is an enormous problem in this forum..... :roll:
2) Your proof is nice/ it seems to be nice! :lol: [/hide]

[quote=arqady]The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$ \frac{33abc}{a^3+b^3+c^3}+ 12\left( \frac{a}{b}+ \frac{b}{c}+ \frac{c}{a}\right) \ge47$$[/quote]

TWO LINES DONE


geometry
geometric transformation
rotation
counting
distinguishability

In how many ways can I place 5 different beads on a bracelet if rotating or flipping the bracelet does not change the order of the beads?

Without considering rotations or flips, there are 5! ways.
But we must divide by 5 to account for the indistinguishable rotations, and divide by 2 to account for the indistinguishable flips.
So there are $ \frac{5!}{5*2}\equal{}\frac{120}{10}\equal{}\boxed{12}$ ways to place the beads on the bracelet.

Wait, I thought you could only flip something if it had an even number of items...

[hide=sol]There are a total of 5! ways, then we can remove 2 from the factorial to account for reflections and we can remove 5 from the factorial to account for rotations ending up with 4*3 or 12.[/hide]


geometry
geometric sequence
geometry unsolved

Suppose a circle passes through the feet of the symmedians of a non-isosceles triangle $ABC$ , and is tangent to one of the sides. Show that $a^2 +b^2, b^2 + c^2 , c^2 + a^2$ are in geometric progression when taken in some order

For other solutions see [url=http://www.artofproblemsolving.com/community/c6h164755]nice result related to the symmedial circle[/url] and [url=http://www.artofproblemsolving.com/community/c6h331763p2578461]Geometry Marathon-Olympiad level (problem #461)[/url]


Write this equation in function notation.

20   -30-  40- 50- 60-70
175-166-157-148-139-130

Its supposed to be a chart

Just because no one's replied doesn't mean that you can make a whole new thread.


[hide][quote="funcia"]Find the 2006th number in the following sequence:
1,2,1,3,2,1,4,3,2,1,5,4,3,2,1,6,5,4,3,2,1...

A. 2006 B. 12 C. 41 D. 11 E. None of the Above![/quote]

n(n+1)/2=2006
take the greated integer value=62
62*63/2=1953
so its starts at 63
53 numbers later, it'll be 10?
[/hide]

Argghhh. I remember this question from a AMC right?

This problem has given me nightmares.

Ok so, the pattern is clearly a repetition of

1, 2, 1, 3, 2, 1, 4, 3,...,2, 1, n+2, n+1, n, n -1, ...

Let me think for a bit.

1+2+3+4+5+6+7+...+n = 2006 + C

where C is a constant and n is the number at the end of the progression that contains the 2006th term.

The largest number, n, less than 2006 that is the end of a progression is
n(n+1)/2

Subbing in 60, we get
1830
Subbing in 70, we get
2485

So it is a number in between

n =62.

Now, that leads us to number 1953.

2006-1953 +1 = 51

So, the next terms are in the 63rd progression:
63-51 =12, our answer I believe.

Errrr...umm....this is multiple choice and no one is picking the choices! *tears out hair*  Well at least there is a none of the above so I know what to pick comfortably when confronted with this prob.
OK back to the real problem-I actually get what neelnal is talking about!!!!!! Until the adding 1 thing...at the end...I can follow through when it comes to the constant and all! But even that has me asking: why?
So the questions I have asked are:
why is "1+2+3+4+5+6+7+...+n = 2006 + C "
how did we know :"The largest number, n, less than 2006 that is the end of a progression is
n(n+1)/2 "
And finally why did we subtract and add 1 here: "2006-1953 +1 = 51, our answer I believe."
But if I was given a problem like this I'd be able to do it-just turning into a little monkey...named George.

Tip for students taking tests with None of the Above as an answer choice:

Only easy problems most likely will have E as the answer, hard problems usually have the answer displayed because students will feel confused and second guess themselves if they get E as an answer.  :ninja:

B. 12

63-51 =12, our answer I believe.

not 51, 12

Okay-now all I am wondering really(though I still wonder about the adding one thing but it gets clearer and clearer....) is why did we know 1+2+3+4+5+6+7+...+n = 2006 + C  since the pattern is 1,2,1,3,2,1,4,3,2,1,5,4,3,2,1...

[quote="funcia"]why did we know 1+2+3+4+5+6+7+...+n = 2006 + C  since the pattern is 1,2,1,3,2,1,4,3,2,1,5,4,3,2,1...[/quote]

n is just a general number and C is also a general constant.

In each individual progression, there is n number of terms.

For the 13th progression, there are 13 terms.

So, as we found, the 2006th term is in the 63rd progression, but the 63rd progression does not end with 2006.

Therefore, we add the constant C to 2006. The (2006 + c)th term will end the 63rd progression.

[quote="funcia"]My! This is a really pretty sequence...
so here:
So it is a number in between

n =62.

Now, that leads us to number 1953.

2006-1953 +1 = 51 [color=red] why do we add 1[/color]

So, the next terms are in the 63rd progression:
63-51 =12, our answer I believe.[color=red]why do we subtract from 63 I thought we'd subtract from 62?[/color][/quote]

When we count, we always add 1.

Counting from 2 to 10, how many terms are there for instance:

10-2 = 8???
or 10-2+1 = 9???

2,3,4,5,6,7,8,9,10 --> 9 terms!

As for your second question, the 62nd progression has ended, the 63rd has begun at this point.

[quote="funcia"]WOW! Thanks! Could *anyone* give me a term to find in this sequence so I can make sure I can keep this in my mind?
EDIT: wait a min. 2006-1953+1=52?[/quote]

o darn

my bad i added/subtracted wrong  :blush:

answer is E i guess....or wait...darn...

Yeah well than the answer is E I think. I was going over it to get that I got it....OK I'm gonna do it for the "500th term" :D See ya when I've done it!

Oh surge I meant for the 500th term..what do you get for that?

[quote="neelnal"][quote="surge"]I'm getting eleven...

63-(2006-(1953+1))=11[/quote]

It is not that.

It is:

63 - (2006 - 1953 + 1)

no parentheisis[/quote]

It's 11.

If you're totally confused, the way I did it was to count backwards. The 2016th term is a 1 (all triangular-number-terms are 1) and so the 2006th term is 11.

[quote="funcia"][quote="funcia"][hide] 21*22=462
500-462=38+1=39
39-21=18[/hide]

Is that the answer?[/quote]
So does anyone have an answer to the extension of what is the 500th term in the same sequence?[/quote]

n(n+1)/2[u]<[/u]500

The closest n is 31, which leads us to the 496th term.

The answer is thus, 28.

Voila!


inequalities
logarithms
algebra proposed
algebra

$\displaystyle \prod_ {i=1} ^ {n} x_i ^ {a_i} \leq \prod_{i=1}^{n} a_i^{a_i} \Longleftrightarrow \prod_ {i=1} ^ {n} (\frac{x_i}{a_i}) ^ {a_i} \leq 1 \Longleftrightarrow  \sum_ {i=1} ^ {n}a_i \ln (\frac{x_i}{a_i}) \leq 0$    :cool:  Trivial, because of $ln(x)$ is concave :cool:

So $\Sigma_{i=1}^{n}{a_i}=\Sigma_{i=1}^{n}{x_i}$ is sufficient.


What is the number of permutations of $ 1,...,n$ with at least $ k$ fixed points?

I know that number of permutations with exactly $ k$ fixed points is $ \binom{n}{k}(n \minus{} k)!\sum_{i \equal{} 0}^{n \minus{} k}\frac{( \minus{} 1)^i}{i!}$.
 :?:

Just sum up the known formula over k = k0 .. n


LaTeX
geometry
geometric transformation
reflection
\/closed

$ \[ 0,\frac {\pi}{2}
\]$
--------------------------------------------------------
The above is what happens when the user types[code]$\displaystyle\[0,\frac{\pi}{2}\]$[/code]
Of course, the mistake is that the user left out \left and \right tags in trying to typeset brackets.
A simple way to fix this would be either an error message, like when dollar signs aren't closed, or even better: if [code]\[ or \][/code] are found within dollar signs, they could be replaced with \left[ and \right]

Just a suggestion :)

Well, I doubt Valentin will ever be able to devise a translator from "sloppy user LateX" to "correct" one (I mean, there are much more ingenious erratic constructions than one can possibly make an interpreter to take care of). It is up to him, of course, to decide whether to try to handle this particular error, but it will be much better if the users just learn to use LaTeX properly or click the "preview" button before posting to see if they have any errors :).

I agree with fedja. It's not hard to screw up $ \text{\LaTeX}$ if you want to, it's almost impossible and actually not such a good idea to make such an interpreter (actually one is working right now for just a couple of very common issues).

Okay, its just it seemed like it could almost be a security vulnerability if raw html is exposed
$ insert here mailicious code that exits out of poplatex() call and does its own javascript \[ .\]$

[quote="Temperal"]Raw HTML can be seen by anyone who accesses the page by viewing the source.

A PHP error that exposed raw MySQL passwords would be much more serious.  :P[/quote]
He means javascript injection or something, though I don't see a way to make it work.

This one is fixed (though the old posts do not reflect it yet - but you can't reproduce it anymore).

Thank you for taking the time to address this issue! :)


Yet I intend to quite from Mathlinks for a long time because I have to come to US to study next year, however, when my friend Manlio returns, I find myself be inspired and try to return few week. Fortunately, I meet Vasc, Arqady again. This problem is specially dedicated to my old friend Manlio, but all of you can try it. Remember that : [b]It is better not to use all advanced method such as EV or SMV theorem[/b].

[u]Problem:[/u]

Let $x_{1},x_{2},...,x_{n}$ be positive real numbers such that $x_{1}+x_{2}+...+x_{n}=n$. Prove that
\[\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}}+\frac{2n\sqrt{n-1}}{x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}}\ge n+2\sqrt{n-1}.\]

We have also a slighter inequality

If $a_{1},a_{2},...,a_{n}>0$ and $a_{1}+a_{2}+...+a_{n}=n (n\ge 3)$ then
\[\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}+\frac{2n\sqrt{2}}{a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}}\ge n+2\sqrt{2}.\]

For $n=3$, the inequality become

If $a+b+c=3 (a,b,c>0)$ then
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{6\sqrt{2}}{a^{2}+b^{2}+c^{2}}\ge 3+2\sqrt{2}(3)\]

In the above inequality, the equality holds for $(a,b,c)~(1,1,\sqrt{2}).$

I want to present this slighter inequality because it has a very simple and nice solution by SOS, especially the case $n=3$. However, any solution for (3), even mixing variable is welcome.

I reduced it to $\sum (a-b)^{2}\left( \frac{\sum a^{2}}{ab}-2 \sqrt{2}\right) \ge 0$. I don't know what to do after that... we should probably use $a^{2}+2b^{2}\ge 2 \sqrt{2}ab$ at some point...

That is right! You can continue with the notice that (easy)

\[\frac{a^{2}+b^{2}+c^{2}}{ab}+\frac{a^{2}+b^{2}+c^{2}}{ac}\ge 4\sqrt{2}.\]

Aha, I was having trouble proving that but now I see that it is an obvious AM-GM  :lol: . Can we apply the same sum of squares idea to the n-variable case?

Yes, sum of squares way works for general case, however, it is quite deft and original in applying. First we have to rewrite it in square form, and I think it is easy for you, probability1.01. It is similar to the case $n=3$.

Dear Manlio,

Could you prove this problem? The suggestion is that after you rewrite it to the form
\[\sum S_{i,j}(x_{i}-x_{j})^{2}\ge 0,\]
let's try to prove that
\[S_{i,j}(x_{i}-x_{j})^{2}+S_{j,k}(x_{j}-x_{k})^{2}+S_{k,i}(x_{k}-x_{i})^{2}\ge 0.\]

I have a lot of difficult to write the oiginal problem in this form, sorry. :blush:


Let $a,b,c$ be real numbers and $2a^{2}+b^{2}+c^{2}=3.$
Find the minimum of the expression $a-2b+c$.

Using Lagrange Multipliers, I get something ugly like

[hide] $-\sqrt{\frac{33}{2}}$. [/hide]

Yes, I think this is the answer. Can you please post the solution?

Sure. It's calculus, though.

[hide]
Using Lagrange Multipliers with $f = a-2b+c$ and $g = 2a^{2}+b^{2}+c^{2}-3 = 0$, we know that the extrema occur when

$\bigtriangledown f = \lambda \bigtriangledown g$

$<1,-2, 1> = \lambda <4a, 2b, 2c>$

or

$\lambda = \frac{1}{4a}= \frac{-2}{2b}= \frac{1}{2c}$,

which has the solution $(a,-4a, 2a)$. Plugging this back into $g$, we get

$2a^{2}+16a^{2}+4a^{2}-3 = 22a^{2}-3 = 0 \Rightarrow a =-\sqrt{\frac{3}{22}}$

at the minimum (positive at the maximum). Then $b = 4\sqrt{\frac{3}{22}}$ and $c =-2\sqrt{\frac{3}{22}}$. So

$f = a-2b+c =-11\sqrt{\frac{3}{22}}=-\sqrt{\frac{33}{2}}$.[/hide]

[quote="cincodemayo5590"][u]Follow-up Question[/u] (a little easier):

Find the maximum of the expression.[/quote]

Use $a = \sqrt{\frac{3}{22}}$ for the other extremum (aka maximum) :wink:.


If you could tell me that would be a great help.
Thankyou

Wrong forum.

http://www.artofproblemsolving.com/Edutainment/g1/Pages/FAQ.php

which is from: the most recent discussion of this:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314586

EDIT: Hmm, I seemed rather rude there. :maybe:

Sorry about that, I see where I should have put it. :oops:


I'm randomizing a set of integers and output-ting a list of integers into a file called Random.txt. So, I get this list of numbers:

89
90
1
48
etc.

But in Random.txt it writes it like this: 8990148 which I can't tell where the breaks are. Is there any way so that I can get line breaks between each number? Here's the part of my code that writes the integers to the file

[code]
       try{
    		File file = new File("Random.txt");
    		Writer output = new BufferedWriter(new FileWriter(file));

    		for (int i = 0; i < range; i++){
    			String number = "0";
    			number = number.valueOf(numbers[i]) + "\n";
    			output.write(number);
    		}
    		output.close();
    	}

    	catch (IOException e){
    		System.out.println("Error: Cannot Write to File");
    	}

[/code]

Windows represents new lines as "\r\n". It's stupid, but try it (assuming you're on windows). Just be careful. If you're sending the file to a linux recipient, they'll probably be annoyed at the tons of extra characters ;)

Wow :o, that's pretty weird. I knew that in Windows lines should be ended with "\r\n" but I never realized until now that Java won't translate "\n" to "\r\n" the same way as your average C compiler would. I guess this is because the Java compiler has to generate the same code no matter what operating system you're running (so that the compiled class file can be run from Linux, Windows or any other OS).

In my opinion you'd be better off using BufferedWriter.newLine() method:
[code]
	File file = new File("Random.txt");
	BufferedWriter output = new BufferedWriter(new FileWriter(file));

	for (int i = 0; i < range; i++){
		String number = "0";
		number = number.valueOf(numbers[i]);
		output.write(number);
		output.newLine();
	}
	output.close();
[/code]


calculus
integration
vector
real analysis
real analysis unsolved

Let $ E$ be an euclidian space of dimension $ n$ and $ v_1,v_2,...,v_k$ vectors in $ E$ with $ k>n$. Let $ X(v_1,...,v_k)$ be the set of combinations $ x_1v_1\plus{}...\plus{}x_kv_k$, each $ x_i$ being between $ 0$ and $ 1$. Then for the canonical measure $ \mu$ on $ E$ one has \[ \mu(X(v_1,v_2,...,v_k))\equal{}\sum_{1\leq i_1<...<i_n}{\mu(X(v_{i_1},...,v_{i_n}))}.\]

one can write $ \sum_i x_i V_i \equal{} \sum_{j\equal{}1}^n x'_{i_j} V_{i_j}\plus{} \sum_{i \neq i_j} y_i V_i$ where $ i_1 < i_2 < \ldots < i_n$,  $ y_i \in \{ 0,1 \}$ and $ 0 \leq x'_{i_j} \leq 1$ in  unique way if one imposes some more extra conditions (eg: $ \sum_{i \neq i_j} y_i k^i$ is minimal among all these representations)


In a triangle $ ABC$ with $ \angle A\equal{}90^{\circ}$, $ D$ is the midpoint of $ BC$, $ F$ that of $ AB$, $ E$ that of $ AF$ and $ G$ that of $ FB$. Segment $ AD$ intersects $ CE,CF$ and $ CG$ in $ P,Q$ and $ R$, respectively. Determine the ratio: $ \frac{PQ}{QR}$.

Just use Menelauyt for triangle AFQ with secants EPC and GRC so PQ=1/4 AQ, QR=1/8 AQ so
PQ/QR=2


Which of the following are subsets of $ R^3$?

a. The plane of vectors $ (b_1, b_2, b_3)$ with the first component $ b_1\equal{}0$.
b. The plane of vectors $ b$ with $ b_1 \equal{} 1$.
c. The vectors $ b$ with $ b_{2}b_{3}\equal{}0$
d. All combinations of two given vectors $ (1, 1, 0)$ and $ (2, 0, 1)$.
e. The plane of vectors $ (b_1, b_2, b_3)$ that satisfy $ b_3 \minus{} b_2 \plus{} 3b_1 \equal{} 0$.

I think that b is the only one that is not a subspace b/c it does not satisfy closure of scalar multiplication and there does not exist a zero element in this vector plane.

Can anyone confirm?

Thanks

The subspaces of $ \mathbb{R}^3$ are $ \mathbb{R}^3$ itself, planes through the origin, lines through the origin, and the origin.  c) is the union of two planes, so it is not closed under addition.

(d) is the span of a given set of vectors - that's always a subspace. The other four are each the solution set of some equations.

(a) and (e) are the solution sets of [b]homogeneous linear[/b] equations.

(b) is the solution set of a [b]nonhomogeneous linear[/b] equation.

(c) is the solution set of a [b]nonlinear[/b] equation.

(a) and (e) are subspaces. (b) and (c) are not.

a is a linear subspace.
b is non-linear space, but an affine space.
c should not be an linear subspace, consider $ (1,1,0)$ and $ (1,0,1)$, their sum $ (2,1,1)$ does not belong to the subset, thus c is not closed.
d is a linear space, as it is the linear space spanned with $ ((1,1,0),(2,0,1))$ as base.
e is a linear space.

[quote="KenHungKK"]b is non-linear space, but an affine space.[/quote]
As I said above, it is the solution space of a linear nonhomogeneous equation (or system of equations.) That is, all $ x$ such that $ Ax \equal{} b$ for some $ b\ne 0.$ This is effectively synonymous with being affine but not linear.

[I am teaching a beginning matrix/linear algebra course this semester. That makes it a semester for me to think about questions like this.]


Ειναι ενα προβλημα που μου κινησε την περιεργια και ακομα δεν εχω βρει λυση του
Καθοριστε την μεγιστη τιμη του
                                         $ n^2 \plus{} m^2$
Oπου $ n,m \in Z_ \plus{}$ εαν $ n,m \in \{1,2,3,...,1981\}$ και $ (n^2 \minus{} mn \minus{} m^2)^2 \equal{} 1$

Το κόλπο είναι να βρεις πρώτα πια ζευγάρια $ (m,n)$ ικανοποιούν την συνθήκη. Βρες μερικά ζευγάρια και ίσως μαντέψεις την μορφή όλων των ζευγαριών.

[quote="Γιώργος"][quote="Demetres"]Ακόμη μια βοήθεια: Προσέξτε ότι τα (1,1), (2,1), (3,2), (5,3) είναι λύσεις τις εξίσωσης...[/quote]

[hide]$ 1,1,2,3...$ θυμίζει Fibonacci ε?[/hide][/quote]

Ακριβώς. Και για να βάζουμε και την λύση:

Αν m=n, τότε m=n=1.
Aν m>n, δεν υπάρχουν λύσεις.
Αν n>m και (n,m) λύση, τότε και η (m,n-m) είναι λύση: Πράγματι $ m^2 \minus{} m(n\minus{}m) \minus{} (n\minus{}m)^2 \equal{} \minus{}(n^2 \minus{} mn \plus{} m^2)$.

Ισχυρίζομαι ότι κάθε ζεύγος θετικών ακεραίων που είναι λύση, είναι συνεχόμενοι αριθμοί Fibonacci: Ας υποθέσουμε πως δεν ισχύει αυτό και ας πάρουμε ένα ζεύγος $ (n,m)$ που είναι λύση, οι m,n δεν είναι συνεχόμενοι αριθμοί Fibonacci με το n όσο μικρότερο γίνεται. Αφού η (m,n-m)  και m < n, οι n-m,m είναι συνεχόμενοι αριθμοί Fibonacci. Άρα το ίδιο ισχύει και για τους m,n=(n-m)+m, άτοπο.

Τέλος έχουμε 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,...

και αν δεν έκανα κανένα λάθος στην πρόσθεση, η μέγιστη τιμή του $ m^2 \plus{} n^2$ είναι $ 987^2 \plus{} 1597^2$.

mallon kati dn katalava ego, opote sory se ayti tin periptosi...to [b]megisto[/b] pos to vrikame?

sygnomi gia ta greeklish...tha to kano edit argotera..

Bill, κοιτα,
παρατηρεις οτι αν (m,n) μια λυση τοτε και η (n-m,m) αποτελει επισης λυση.
Ομως αυτος ο μετασχηματισμος πρεπει να τερματιζεται υστερα απο εναν πεπερασμενο αριθμο βηματων.Προφανως σταματαει οταν m=n=1.
Τωρα,
 αν κανεις τον αντιστροφο μετασχηματισμο περνεις οτι αρχιζοντας απο το (m,n)=(1,1) αποτελουν λυσεις τα ζευγη της μορφης,
(n,m+n).
Eτσι με αρκετη προσοχη παρατηρουμε οτι οι λυσεις μας δινονται απο 2 διαδοχικους ορους της ακολουθιας Fibonacci,
δλδ ισχυει,
$ (m,n) \equal{} ( F_{n \minus{} 1},F_{n})$
Ομως πρεπει $ F_n \leq 1981$, η ακολουθια ειναι  αυξουσα και ο ορος της ακ.Φιμπονατσι που ειναι πιο κοντα στο 1981 ειναι ο 16ος με τιμη $ F_{16} \equal{} 1957$
αρα $ max(m^2 \plus{} n^2) \equal{} F_{15} \plus{} F_{16} \equal{} 987^2 \plus{} 1597^2 \equal{} 3514578$

[quote]παρατηρεις οτι αν (m,n) μια λυση τοτε και η (n-m,m) αποτελει επισης λυση.
Ομως αυτος ο μετασχηματισμος πρεπει να τερματιζεται υστερα απο εναν πεπερασμενο αριθμο βηματων.Προφανως σταματαει οταν m=n=1.
Τωρα,
αν κανεις τον αντιστροφο μετασχηματισμο περνεις οτι αρχιζοντας απο το (m,n)=(1,1) αποτελουν λυσεις τα ζευγη της μορφης,
(n,m+n). [/quote]

α μπραβο τωρα το επιασα...ευχαριστω πολυ man, πολύ ωραίος...η το 'χεις ή δν το 'χεις.... :wink:


calculus
integration
logarithms
trigonometry
calculus computations

Calculate the following the definite integral.

(1) $\int_0^{\frac{\ln 3}{2}} \frac{e^x+1}{e^{2x}+1}dx$

(2) $\int_0^{\frac{\pi}{2}} x\left|\sin ^2x-\frac{1}{2}\right|dx$

The 1st integral is:

$\int_0^{\frac{\ln 3}{2}} \frac{e^x+1}{e^{2x}+1} \text dx = \int_0^{\frac{\ln 3}{2}} {\frac{e^{\frac x 2} \cosh{\frac x 2}}{e^x \cosh x} \text dx = \int_0^{\frac{\ln 3}{2}} {\left(\cosh{\frac x 2} - \sinh{\frac x 2}\right)\frac{\cosh{\frac x 2}}{\cosh x} \text dx = \int_0^{\frac{\ln 3}{2}} {\frac{\cosh^2{\frac x 2} - \cosh{\frac x 2} \sinh{\frac x 2}}{\cosh x} \text dx =}}}$

$= \frac 1 2\ \int_0^{\frac{\ln 3}{2}} \frac{1 + \cosh x - \sinh x}{\cosh x} \text dx = \frac 1 2 \int_0^{\frac{\ln 3}{2}} \frac{\text dx}{\cosh x}\ +\ \frac 1 2 \int_0^{\frac{\ln 3}{2}} \text dx\ -\ \frac 1 2 \int_0^{\frac{\ln 3}{2}} \frac{\sinh x}{\cosh x} \text dx =$

$= \frac 1 2 \int_0^{\frac{\ln 3}{2}} \frac{\cosh x}{1 + \sinh^2{x}} \text dx\ + \left[ \frac x 2 \right]_0^{\frac{\ln 3}{2}} -\ \frac 1 2 \int_1^{\frac{2 \sqrt 3}{3}} \frac{ \text dy}{y}$

$= \frac 1 2 \int_0^{\frac{\sqrt 3}{3}} \frac{\text dy}{1 + y^2} + \frac{\ln 3}{4} - \left[ \frac{\ln y}{2} \right]_1^{\frac{2 \sqrt 3}{3}} = \left[ \frac{\arctan y}{2} \right]_0^{\frac{\sqrt 3}{3}} + \frac{\ln 3}{4} - \frac 1 2 \ln{\frac{2 \sqrt 3}{3}} = $

$= \frac{\pi}{12} + \frac{\ln 3}{4} - \frac{\ln 2}{2} + \frac{\ln 3}{4} = \frac 1 2 \left( \frac \pi 6 + \ln 3 - \ln 2 \right)$

And the 2nd integral is

$I = \int_0^{\frac{\pi}{2}} x\left|\sin ^2x-\frac{1}{2}\right| \text dx = \frac 1 2 \int_0^{\frac{\pi}{2}} x\left| \cos{2x} \right| \text dx = \frac 1 8 \int_0^{\pi} y\left| \cos{y} \right| \text dy =$

$= \frac 1 8 \int_0^{\frac{\pi}{2}} y \cos{y}\ \text dy - \frac 1 8 \int_{\frac{\pi}{2}}^{\pi} y \cos{y}\ \text dy$

$F(y)= \int y \cos{y}\ \text dy = y \sin y - \int \sin y\ \text dx = y \sin y + \cos y$

$I = \left[ \frac{y \sin y + \cos y}{8} \right]_0^{\frac{\pi}{2}} - \left[  \frac{y \sin y + \cos y}{8} \right]_{\frac{\pi}{2}}^{\pi} =$

$= \frac 1 8 \left( \frac \pi 2 \sin{\frac \pi 2} + \cos{\frac \pi 2} - 0 \sin 0 - \cos 0 - \pi \sin \pi - \cos \pi + \frac \pi 2 \sin{\frac \pi 2} + \cos{\frac \pi 2} \right) =$

$= \frac 1 8 \left( \frac \pi 2 + 0 - 0 - 1 - 0 + 1 +\frac \pi 2 + 0 \right) = \frac{\pi}{8}$


22nd November,2009.All Bengalis in the forum,Start preparing!Wish me luck guys!Thank God,it's after my horrible 2nd term exams! :)
The next INMO is going to be held on 17th January,2010.

Thank you for the information. I hope to clear the INMO this time like Akashnil and go to IMOTC.
All the best to everyone appearing for the RMO.

[quote="Scary math"]In which class are u?Have u already cleared the RMO?  :o
If so,could u plz provide me with some tips? :D[/quote]
RMO has still not been held this year. And as far as clearing RMO is concerned, you'll meet a 90% success rate here :P Even the most unworthy and pathetic member like me(I actually stood LAST in UP  :rotfl:) or a medical student like Rituraj went through. You'll clear it as well beyond doubt. Its the INMO that kicks. Still the brainy like keshav, akshanil, gaurav etc. manage to cruise through......
And bro, don't freak out about it. Just read one or two books for now and check out the sample papers. I'm sure even the wisest genius out here will be unable to offer any secret tips :P Maybe a few books and stuff. But otherwise its you and your head. ;-)

No bro,I dont think so,I am horrible in geometry and beyond horrible in Algebra,I only started preparing this year,and have only covered a few topics till now :D I am trying not to expect much this year,plz forgive me if I have made u guys overestimate me,but plz dont throw me out or stop replying to me :(  :(  :(

Hello Scary Math. First of all which class are you in?
I am in class 10 now. Last year cleared the RMO in class 9 but did not qualify the INMO. RMO is quite simple actually. Just make sure about your proof writing skills by looking at some posts of senior members in the olympiad(advanced section) forums. You can use some books too.
All the best.

He he,I am in 10th as well,but not a genius like u.I believe u will become Akashnil someday.Best of luck
P S - are u from Bengal?Which books did u use?

Akashnil is an IMo contestant. I am nowhere near him. You can contact him as he is Bengal.
I am from Chennai.
All the best for your Maths Olympiads.
Books: Not much for RMO.I solved all the previous year papers, this forum, things taught in RMO Camp.
INMO: Books by Titu Andreescu, CHalleneg and Thrill of Pre college maths, Excursion in maths, all previous INMO papers, this forum, etc...i dont remeber, yeah Arthur engel, sharygin...

I am busy with Challenge and thrill ,Hall and Knight
If u manage to get selected in INMO this year,U have good chances of becoming Sucharit Sarkar

I FINALLY got my Challenge and Thrills today. Though I'm only likely to read it only between INMO and JEE :P

Let me ask u guys something,Does school and board matter condiderably in the 11th and 12th?The thing is,there's a school over here in Kharagpur,Railway mixed.It is affiliated to the ISC board and often students from there get good ranks in jee.Last year,some girl from mixed ranked 23 in WB-JEE(medical).The benefit of the school is that there is no strict rule regarding minnimum attendance and I wont be expelled even if I dont go to school for even a single day! :rotfl: Therefore,I believe I will be able to get a lot of extra time for my preparation.
                                                Rumours say that the school has a bad record in terms of behaviour of students and environment,but it wont affect me much as I have planned not to attend school!Moreover there is no decent school in Kharagpur due to lack of good teachers and dedicated students.Some of my school seniors often say that in the 11th and 12th,school only helps in preparing for the board but not JEE.I will not be allowed to take admission in IIT-KV,the railway KV is as bad as Railway mixed in terms of environment,and the IIT DAV has no proper,regular teaching facilities and has too many exams.
                      Please advice! :(

Sucharit Sarkar participated in IMO in 2001 & 2002

He was a gold medallist, not just a participant. He also went on to do a B.Math from ISI and a Ph.D from Princeton under Zoltan Szabo. He's very recently been awarded a Clay Mathematics Fellowship to pursue mathematics after his doctorate.

Yah,I knew the rest except for the Clay mathematics fellowship.Just didnt mention it! :D


I may just not be seeing something, but I do not think there is enough information in the second problem to solve it (the ratio of size of circle to size of triangle allows for different values of x). Did you intend for parallel lines somewhere or something that would make the circle have only one possible radius? Could you clarify this? Thanks.

sorry for my second question.it is incomplete.But I will be pleasure ıf yousolve my first question.Thanks...

the first question
according to CEVA theory(point C and triangle ABD):
(sin42'/sin30')*(sin84'/sin12')*(sinx'/sin(x'+36'))=1,
so x=-42,(but I don't know why x<0)

you used ceva in a wrong way. it would be correct to say that for D and triangle ABC we have
$\frac{\sin 42}{\sin 30}\frac{\sin 84}{\sin -12}\frac{\sin -x}{\sin x+36}=1$.
i leave it to the reader to calculate $x$.

Peter

oh sorry, i just say that this gives the same as yours...however i was irritated because you wrote C and triangle ABD.

Peter

[quote="Peter Scholze"]i leave it to the reader to calculate $x$.[/quote]

This is exactly the hard part, or, better, the messy part. We already had a discussion about it in the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=16296 . Nobody on ML has found an elegant proof yet.

  Darij

Let P be the midpoint of AB. We draw the bisector of the angle <ABC. Let it intersect AC at T and AD at M. Let N be the intersection point of the lines AC and BD. If we prove that points P,T ,D are collinear the problem is done. So we use Menelaos which is equivalent to AT/TN=BD/DN  Let  <TNB=y, then we have by the sine theorem:
TN*siny=BT*sin6 ; DN*siny=DC*sin12 ; (because AT=BT=BC) We have to prove:
sin12/sin6=BD/DC or  2cos6=sin84/sin30 which is obvious.

:blush: Sorry, I forgot to write this in my previous post.
if P,T,D are collinear then we have AD=BD so 36+x=42 then x=6.

[quote="NightFlarer"]Is trig the only way to solve the first problem?
I got pretty far.  You can find all the angles except x and the the second angle that forms <D (not 66, other one).  Then, x + y has to equal 84 I found, but there's not much I can do from there.[/quote]
Um, doesn't x+y=102? Since the known angle in that triangle is 78 degrees, and also if you look at the big quadrilateral and subtract the other degrees from 360...


We extend the sides of a convex pentagonon to form five triangles in its exterior .Prove that the points of intersection of the circumcircles of these five triangles which are not vertices of the pentagonon are coocyclic .

This is known as Miquel's Pentagram theorem. there is nothing difficult to prove it, just angle-chasing.
See:
http://mathworld.wolfram.com/MiquelsPentagramTheorem.html
and http://www.mathlinks.ro/Forum/viewtopic.php?t=85270

see also this nice shape: [url]http://www.math.uoc.gr/~pamfilos/gGallery/problems/Miquel_Pentagon.html[/url]


algorithm
search
calculus
calculus computations

I have a problem with determening the stability of the 'formula' $ \sqrt {1 \plus{} x^3} \minus{} \sqrt {1 \plus{} x^2}$. Assume there is a stabile algorithm that can find a square root and suppose the relative error is then smaller then $ \frac {1}{2}e_{mach}$.

I started to search $ fl(\sqrt {1 \plus{} x^3})$, i found that it is approximately (with $ e_i \leq e_{mach}$)
\[ \sqrt {(1 \plus{} e_3 \plus{} (1 \plus{} e_1 \plus{} e_2 \plus{} e_3)x)}(1 \plus{} e_4)
\]
But I don't know how to proceed. Because I need to get rid of the $ e$ inside the root.

Thanks ;)

Allright it is solved now. Using an approximation of sqrt(1+x) for small x.

ur second "postreply" assur that you are in High school yet :D .
Question!:would this stability help u in somthing?


Find all integers $ x$ for which $ x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1$ is a perfect square. :?:

x=-1,0 give two solutions, x=1 does not work, so assume now that $ |x| > 1$

$ (x^2 \plus{} x/2)^2 \equal{} x^4 \plus{} x^3 \plus{} x^2/4 < x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$ Since the discriminant is <0

$ (x^2 \plus{} x/2 \plus{} 1)^2 \equal{} x^4 \plus{} x^2/4 \plus{} 1 \plus{} x^3 \plus{} x \plus{} 2x^2 > x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$ Since it reduces to $ 1.25x^2 > 0$

So the only way that there can be a soln is if x is odd and $ x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1 \equal{} (x^2 \plus{} x/2 \plus{} 1/2)^2$ (since if x was even, then we just squished our square between 2 consecutive squares, and if x is odd, we squished it between a (y+0.5)^2 and (y-0.5)^2 where $ y \equal{} x^2 \plus{} x/2 \plus{} 1/2$ so the square must be $ y^2$)

$ (x^2 \plus{} x/2 \plus{} 1/2)^2 \equal{} x^4 \plus{} x^2/4 \plus{} 1/4 \plus{} x^3 \plus{} x^2 \plus{} x/2 \equal{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$

$ x^2/4 \plus{} 1/4 \plus{} x/2 \equal{} x \plus{} 1$

$ x^2 \plus{} 1 \plus{} 2x \equal{} 4(x \plus{} 1)$

$ (x \plus{} 1)^2 \equal{} 4(x \plus{} 1)$

$ x \equal{} 3$ (since we are assuming that $ |x| > 1$ so $ x \ne \minus{} 1$

And indeed, $ 3^4 \plus{} 3^4 \plus{} 3^2 \plus{} 3 \plus{} 1 \equal{} 11^2$.

Therefore, we have the solutions x=-1,0,3.

Q.E.D.

Now, x=1 doesn't work, x=0 gives another solution, as does x=-1. So let $ x < \minus{} 1$

Cool solution. Great.

After a lot of hard work I got the answer but that is just too long. This is my solution. I hope it is not wrong!!!

If $ x^4 + x^3 + x^2 + x + 1$ is a perfect square then $ y^2 = x^4 + x^3 + x^2 + x + 1$

Consider $ \left(x^2 + \frac {x}{2}\right)^2$=$ x^4 + x^3 + \frac {x^2}{4}$
=$ x^4 + x^3 + x^2 + x + 1 - \left(\frac {3}{4}x^2 + 4x - 1\right)$
$ = y^2 - \frac {1}{4}(3x^2 + 4x + 4)$

As discriminant of $ (3x^2 + 4x + 4)$ is negative therefore $ 3x^2 + 4x + 4$ is always greater than zero.

Thus,         $ \left(x^2 + \frac {x}{2}\right)^2 < y^2$

                 or $ \left|x^2 + \frac {x}{2}\right| < |y|$

But $ x^2 + \frac {x}{2} = \left(x + \frac {1}{2}\right)x$ is non negative, $ \forall\ x\ belongs\ to\ Z$

So $ \left|x^2 + \frac {x}{2}\right| = x^2 + \frac {x}{2} < |y|$

If x is even then $ |y|\gex^2 + \frac {x}{2} + 1$

$ y^2\ge x^4 + x^3 + x^2 + x + 1 + \frac {5}{4}x^2$

$ \Rightarrow y^2\ge y^2 + \frac {5}{4}x^2$

Which is not possible, If $ x\ne0$,then $ x = 0$ is the only solution when x is even.

If x is odd, then $ x^2 + \frac {x}{2} + \frac {1}{2}$ is an integer.

So,$ |y|\ge x^4 + x^3 + x^2 + x + 1 + \left(\frac {x^2}{4} - \frac {x}{2} - \frac {3}{4}\right)$

i.e $ y^2\ge y^2 + \left(\frac {x^2}{4} - \frac {x}{2} - \frac {3}{4}\right) = y^2 + \frac {1}{4}(x^2 - 2x - 3)$

and hence $ \frac {1}{4}(x^2 - 2x - 3)\le 0$

$ \Rightarrow x^2 - 2x - 3 \le 0$

$ \Rightarrow (x - 3)(x + 1) \le 0$

Or $ - 1\le x\le 3$

Odd integral values of $ x$ are -1,1 and 3 of which 1 doesn't give a perfect square.

So the integral values of x are 0,-1 and 3.

I hope i have not violated the rules and gone beyond High School Basics!!! :(

Euclid, that solution is intense. It seems it could be applied to a large variety of questions.


I'm not sure does it belong in this section, but what is a "locus"? Help please. Thanks :)  :)  :)

[quote="draops"]I tried google-ing it, but I couldn't find an explanation for "locus" as a math term.[/quote]

I see it. Under the first one, third definition.

It's basically a set of points, or a figure representing a set of points. For example, the locus of all points a fixed distance $ d$ from a single point is a circle centered around that point with a radius $ d$.

No, certainly not. The locus of points such that the sum of distance of the point from two focal points is constant is an ellipse, for example. Loci can be lines, conics, and other figures.

Loci can be three-dimensional as well. For instance, the locus of points equidistant from a point in space is a sphere.

[u]Exercises:[/u] Describe the locus of points in the x-y plane whose distance to $ (1,2)$ is equal to it's (shortest) distance to:

a.) The point $ (4,8)$

b.) The line $ y \equal{} \minus{} 3$

c.) The line segment joining $ (0,0)$ to $ (2,0)$

d.) The circle $ (x \minus{} 1)^2 \plus{} y^2 \equal{} 1$

When I read what you want I firstly though it's about [url]http://en.wikipedia.org/wiki/Lucas_sequence[/url] because they are almost similar in the spelling.

BTW, if you can't find locus google-ing it, then try with typing ''locus mathematics''...

loci are also points on a gene that code for a specific trait or protein...just saying

Oh wow, seriously? That's interesting  :coolspeak:

[quote="Bugi"]Isn't loci also a plural for locus?[/quote]
Yep.

Definition: A locus is a set of points that satisfy a given set of conditions. For example, The circle $ x^2\plus{}(y\minus{}4)^2\equal{}144$ is the set of points that are 12 units away from the origin.

[quote="Mewto55555"][quote="1=2"][quote="Bugi"]Isn't loci also a plural for locus?[/quote]
Yep.

Definition: A locus is a set of points that satisfy a given set of conditions. For example, The circle $ x^2 \plus{} (y \minus{} 4)^2 \equal{} 144$ is the set of points that are 12 units away from the origin.[/quote]

I hate to be nitpicky, but that's actually all of the points that are 12 units away from (0,4), not the origin  :P[/quote]

Let's not turn this into a spam topic...

We all said what's ''locus'', so Moderators lock this


For Christmas, I'm thinking of either getting both ARML books (I'm not sure how much they are.. Can someone tell me? And how do I buy them?) or Art and Craft of Problem Solving by Paul Zeitz.

Which one do you recommend?

AcOPS is pretty good in teaching material. No solutions, though.

I recommend ACoPS.  There aren't solutions, but it [i]is[/i] a textbook.  The ARML books are good if you want to prepare for ARML.  But if I had to choose, I would go with ACoPS.

For the ARML books, see [url=http://arml.com/]here[/url].  I think that their cost was $\$$ 15 each last year at ARML (it's in that vicinity-I may be off by a little or they might change their prices).

Joml88, I was also wondering how you went through ACoPS. Did you do each problem in a section? Or did you only answer like 60-80%, leaving the hard questions for later? I am curious to know your strategy.

[quote="joml88"]You mean ACoPS, right?  The original reason I got it was for the AoPS IS.  So I mostly followed the problems they assigned which consisted of a mix of the easy and hard problems from each chapter.  I've tried to do as many of the remaining problems as possible though.  There are still problems in that book that I haven't solved yet...but I'm not working out of it anymore.

I would suggest starting the book from the beginning but then once you get past the first few parts, roam around from chapter to chapter as you please.  However, attempt the problems for the corresponding section immediately after you read that chapter, i.e. don't just read all the chapters before trying the problems.  I would suggest trying at least one or two of the hard problems from each section in addition to the other problems before moving on.  (the problems aren't necessarily in increasing difficulty so it might be a little tricky doing this exactly)[/quote]

Ok, thanks for the advice..... I'll try that. :)  So you think AcOPS will help a lot in AIME prep, right?

[quote="paladin8"]I think the ARML books were like 30-ish dollars at ARML last year... I was going to buy one and then I was ilke wow that's really expensive and bought a frisbee instead.[/quote]

For both? Or do you mean single one? (That would be NO WAY)  :D

So the ARML books contain question of previous years americam math contests?

I have ACOPS and there are some nice questions in there as well

[quote="Silverfalcon"]I can't exactly remember when the second ACoPS is coming out. Anyone knows for exact date? (around in $x$ months where $x$ is a positiv integer)[/quote]

He hasn't given a date.  I believe that he only started recently.  Additionally, giving an estimate doesn't really say much.  There are a plethora of things that could cause it to be pushed back.  For example, when I first heard of the AoPS books being written a while ago, they were projected to be out far earlier than now (they're [i]almost[/i] out now).  I'm not mad about this or anything, but you just have to understand that it is hard to set a schedule and guarantee you will stick to it (kinda like Bush and a timetable for Iraq...let's not get into that :P ).


inequalities
AMC
USA(J)MO
USAMO
inequalities proposed

This is about cyclical ineqalities in three variables involving a sum $f(a,b,c)+f(b,c,a)+f(c,a,b)$ that can be solved in an elegant way by proving something like
$f(a,b,c)\leq$ or $\geq C\cdot\frac{g(a)}{g(a)+g(b)+g(c)}$ or more generally $C\cdot\frac{g(a,b,c)}{g(a,b,c)+g(b,c,a)+g(c,a,b)}$ which splits up the sum.
It would be nice to group them all together in this thread. I don't think there are that many of them, but I would love to be wrong - maybe some of the 'classical' ones can be killed like that.
Two exemples from this forum:

http://www.mathlinks.ro/Forum/viewtopic.php?t=62256 (chinese forum)[quote="zhaobin"]for $a,b,c>0$
prove
$1 \leq \sqrt{\frac{a^3}{a^3+(b+c)^3}}+ \sqrt{\frac{b^3}{b^3+(c+a)^3}}+\sqrt{\frac{c^3}{c^3+(a+b)^3}} < \sqrt 2$
[/quote][quote="scpajmb"]prove
$\frac{{a^2 }}{{a^2 + b^2 + c^2 }} \leq \sqrt{\frac{{a^3 }}{{a^3 + (b + c)^3 }}} < \frac{{\sqrt 2 a^{\frac{3}{2}} }}{{a^{\frac{3}{2}} + b^{\frac{3}{2}} + c^{\frac{3}{2}} }}$[/quote]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41931]www.mathlinks.ro/Forum/viewtopic.php?t=41931[/url](IMO 01)
Let $a,b,c >0$ . Prove that :
 $\frac{a}{\sqrt{a^2+8bc}} + \frac{b}{\sqrt{b^2+8ac}} + \frac{c}{\sqrt{c^2+8ba}} \geq 1$
   This is proved by showing
   $\frac{a}{\sqrt{a^2+8bc}} \geq \frac{a^{\frac43}}{a^{\frac43}+b^{\frac43}+c^{\frac43}}$

another one:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=358993#p358993]Re: usamo 2003 problem 5[/url]
[quote="arqady"][quote="MithsApprentice"]Let $a, b, c$ be positive real numbers. Prove that
\[ \frac{(2a+b+c)^2}{2a^2+(b+c)^2} + \frac{(2b+c+a)^2}{2b^2+(c+a)^2} + \frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8.  \][/quote]
$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}\leq\frac{4}{3}\cdot\frac{4a+b+c}{a+b+c}\Leftrightarrow(2a-b-c)^2(5a+b+c)\geq0.$[/quote]


Christine jogged for half an hour. Amy walked for 50 minutes. Using the information in
this chart about exercise and calories burned, how many more calories than Amy did
Christine burn?
\begin{tabular}{| c  c|}
   \hline Activity & Calories Burned per 5 Minutes \\ \hline
   Walking&28\\
   Jogging&57\\ \hline
\end{tabular}

[quote="GameBot"]Christine jogged for half an hour. Amy walked for 50 minutes. Using the information in
this chart about exercise and calories burned, how many more calories than Amy did
Christine burn?
$ \begin{tabular}{| c  c|}
   \hline Activity & Calories Burned per 5 Minutes \\ \hline
   Walking&28\\
   Jogging&57\\ \hline
\end{tabular}$[/quote]

Walking 50 minutes burns 280 calories.
Joggin 30 mins burns 342 calories.

Subtracting, we get 62.


Euler
geometry
cyclic quadrilateral
geometry solved

[quote="ssupermath"]Let ABCD be a cyclic quadrilateral,  K and L points are orthocentre of ABC and BCD triangle respectively.If  m(C)>90 then show that |AC|=|KL|.[/quote]

At first, the condition C > 90° is useless, then it's not AC = KL but AD = KL, and finally it's a well-known fact and follows from Theorem 1 or Theorem 2 in the note "The Euler point of a cyclic quadrilateral" on [url=http://www.cip.ifi.lmu.de/~grinberg/]my website[/url] (or, alternatively, see http://www.mathlinks.ro/Forum/viewtopic.php?t=53407 ).

  Darij


geometry
perimeter
area of a triangle
Heron\u0027s formula
geometry unsolved

#1 Among all triangles with the same perimeter, the equilateral one has the largest area
continue:when does the area is approach 0?
#2 Among all quadrilaterals with the same perimeter, the square has the largest area
#3 Among any finite number of regular polygons with the same perimeter, the one with the largest number of sides has the largest area.
*Each of the statements above has an equivalent where the area is given.
1[hide]I prove it using heron's formula, does someone have a geometric way?[/hide]
continue[hide] I think when the triangle is isosceles, and the base approach 0, but is it rigorous?[/hide]

Easy to prove that adjacent sides are equal(Heron) and adjacent angles are equal(Brahmagupta) in polygon with largest area, therefore it is regular.


Prove that for all $n\in N$, there exists an arrangement of $1$x$1$ squares that can be filled with $1$x$2$ dominoes in exactly $n$ different ways.

Consider this shape:
ooo
ooo
it is covered in 3 ways. Now add to squares:
ooo
ooo
  oo
(in 4 ways) add to other squares:
ooo
oooo
$\$ooo
in this way we can start the induction.


let f, g two entire functions. suppose mode of f square z plus g square z=1 then
which is true

1. f(z)f'(z)+g(z)g'(z)=0
2. f and g must be constant.
3. f and g both are bounded functions.
4. f and g have no zeroes on the unit circles.

It seems like you mistook this site for a homework help one. Be aware that it isn't. You can still get help here but certain unwritten rules apply. One of them is that you should show what you've done and where you got stuck. Then you may expect a hint as to what to do next.

I HAVE SOLVED UPTO
mod of f square z is less than mod of sum of f squarez and g square z that implies f(z) must be bounded similarly g(z)
and entire bounded function is constant so f and g must be constant
is it write?

If the condition was $ |f(z)|^2+|g(z)|^2=1$ (it is a bit hard to read your text descriptions: what you wrote may equally well mean $ |f(z)|^2+g(z)^2=1$ or $ |f(z)^2+g(z)^2|=1$; try to use $ \text{\LaTeX}$ next time to avoid any misunderstanding), then you are right: 2 and 3 are correct. To solve the problem completely, it remains to notice that 1 is then automatically correct but 4 may be false if $ f=0$ and $ g=1$.


how many ways an you split 3 women and 5 men into 4 person themes?

my solution: we know that we should use the combination rule and its 8C4x4C4

However, do i still divide it by 2! ? This is because both teams contain the same number of persons and the team name is unspecified.

thanks

[hide="hint"]to make this problem much easier, instead just figure out how many ways tehre are to take 4 people from 3 women and 5 men. the remainder will automatically be a group. Its also not 8C4 if the men and women are indistinguishable from others of the same gender[/hide]

[hide="Full Solution"]Since there is no restriction based on gender, the information regarding the number of males and the number of females is irrelevant; you only need to know that there are eight people total. Now, we can choose four people out of these eight to form a four-person team. We have
\[ \binom{8}{4} \equal{} 70,
\]
and the remaining people form another team. However, since the teams are indistinguishable, that is, we do not have a team 1 and a team 2, we have counted each combination of teams twice. Our final answer is
\[ \frac{70}{2} \equal{} \boxed{35}.
\][/hide]

[hide="hint if persons of the same gender are not distinguishable"]If you have 0 women, you have 4 men in the group, for 1 way to make the group.  If you have 1 woman, you have 3 men, etc.[/hide]

Otherwise, your solution is fine (note that multiplying by $ \binom44$ is unnecessary since it is 1 and the remaining people can always form a group), depending on what a 4 person "theme" is defined to be.  I would say that you do need to divide by 2 because a theme probably means group, in which case you do have to account for the fact that choosing people 1-4 for group 1 is the same as choosing people 5-8.


geometry
inradius
circumcircle
IMO
incenter
Triangle
IMO 1978

In a triangle $ABC$ we have $AB = AC.$ A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides $AB, AC$ in the points $P,$ respectively $Q.$ Prove that the midpoint of $PQ$ is the center of the inscribed circle of the triangle $ABC.$

Denote a = BC, b = AB = AC the triangle sides, r, R the triangle inradius and circumradius and s, $\triangle$ the triangle semiperimeter and area. Let S, T be the tangency points of the incircle (I) with the sides AB, AC. Let K, L be the midpoints of of BC, PQ and M the midpoint of the arc BC of the circumcircle (O) opposite to the vertex A. The isosceles triangles $\triangle AST \sim \triangle APQ \sim \triangle ABC$ are all centrally similar with the homothety center A. The homothety coefficient for the triangles $\triangle AST \sim \triangle ABC$ is $h_{13} = \frac{AS}{AB} = \frac{s - a}{b}$. The homothety coefficient for the triangles $\triangle AST \sim \triangle APQ$ is $h_{12} = \frac{AS}{AP}$.  The point K is the tangency point of the incircle (I) with the side BC, while the point M is the tangency point of the given circle with the triangle circumcircle (O). It follows that the triangles $\triangle ASK \sim \triangle APM$ are also centrally similar with the homothety center A and the same homothety coefficient as the triangles $\triangle AST \sim \triangle APQ$: $h_{12} = \frac{AS}{AP} = \frac{AK}{AM} = \frac{h}{2R}$, where h = AK is the A-altitude of the triangle $\triangle ABC$. The homothety coefficient of the triangles $\triangle APQ \sim \triangle ABC$ is then $h_{23} = \frac{h_{12}}{h_{13}} = \frac{s - a}{b} \cdot \frac{2R}{h}$ Denote h' = AL the A-altitude of the triangle $\triangle APQ$. Then

$\frac{h'}{h} = h_{23} = \frac{s - a}{b} \cdot \frac{2R}{h}$

$KL = h - h' = h - 2R\ \frac{s - a}{b}$

Substituting $h = \frac{2 \triangle}{a}$, $2R = \frac{ab^2}{2 \triangle}$ and $s - a = \frac{\triangle^2}{s(s - b)^2}$, we get

$KL = \frac{2 \triangle}{a} - \frac{ab^2}{2 \triangle} \cdot \frac{\triangle^2}{bs(s - b)^2} =$

(substituting 2s = 2b + a and $s - b = \frac a 2$ for an isosceles triangle)

$= \frac{\triangle}{s}\left(\frac{2s}{a} - \frac{ab}{2(s - b)^2}\right) = \frac{\triangle}{s} \left(\frac{2b}{a} + 1 - \frac{2b}{a}\right) = \frac{\triangle}{s} = r$

which means that the point $L \equiv I$ is identical with the incenter of the triangle $\triangle ABC$.

I note: the circumcircle $C(O,R)$, the incircle $C(I,r)$ and the circle $w=C(I_1,r_1)$ interior tangent to the circumcircle in the point $T$ and tangent to the sides $AB,AC$ in  the points $P,Q$ respectively. From the a wellknown property, the points $U\in CI\cap TP$, $V\in BI\cap TQ$ belong to the circumcircle. Thus, the quadrilaterals $AUPI$, $AVQI$ are cyclic and
$UA\perp UT$, $VA\perp VT\Longrightarrow PI\perp AI,\ QI\perp AI\Longrightarrow I\in PQ$.

[u]Remark.[/u] The property $I\in PQ$ is true in any triangle $ABC$. See
http://www.mathlinks.ro/Forum/viewtopic.php?t=46418

[b]A generalization of this problem:[/b] "The circle $w=C(I_1,r_1)$ is internally tangent (in the point $T$) with the circumcircle of the triangle $ABC$ and is also tangent to the sides $[AB],[AC]$ in the points $P,Q$. Prove that the incenter $I$ of the triangle $ABC$ belongs to the segment $[PQ]$ and $AP=\frac{bc}{s},\ \ \frac{r_1}{r}=\frac{bc}{s(s-a)}.$

[u]My solution.[/u] From a wellknown property, the points $M\in BI\cap TQ$, $N\in CI\cap TP$ belong to the circumcircle of the triangle $ABC$. From the Pascal's theorem applied to the inscribed hexagon $BACNTM$ results that the points $P\in BA\cap NT$, $Q\in AC\cap TM$, $I\in CN\cap MB$ are collineary, i.e. $I\in PQ$ (The other mentioned relations are easily solved).
See http://www.mathlinks.ro/Forum/viewtopic.php?t=46418

A well-known consequence of the Pascal theorem.

there is yet another proof of the generalization were you construc the circumference and then prove it is unique and that the theorem holds, it is not hard but I am to lazy to write it now I'll just write the beggining

[hide]Let ABC be a triangle with circumcenter O and incenter I, let J be the midpointo of the arc BC that contains A. Let D be the point of intersection of JI and the circumcircle. D is the point of tangency of the circumferences, after this it is all angles matter...[/hide]

:)

Let $ R$ be the tangency of circles and $ S$ be the center of inner circle. $ \angle PSA \equal{} 2k$ and $ \angle PRS \equal{} k$. $ \angle ABR \equal{} 90$. The circumcenter and $ S$ lie on $ AR$. $ AR \perp PQ$ at $ M$ and $ AR \perp BC$ at $ N$ . $ PMRB$ is cyclic. $ \angle PBM \equal{} \angle PRS \equal{} k$. $ PBNS$ is cyclic. $ \angle PBN \equal{} \angle PSA \equal{} 2k$. So $ \angle MBN \equal{} k$. $ BM$ is angle bisector.

Dear Mathlinkers,
for an extension of the initial problem, see

http://perso.orange.fr/jl.ayme/  vol. 4  A new mixtilinear incircle adventure I.

Sincerely
jean-Louis

[quote="Virgil Nicula"][b]A generalization of this problem:[/b] "The circle $ w \equal{} C(I_1,r_1)$ is internally tangent (in the point $ T$) with the circumcircle of the triangle $ ABC$ and is also tangent to the sides $ [AB],[AC]$ in the points $ P,Q$. Prove that the incenter $ I$ of the triangle $ ABC$ belongs to the segment $ [PQ]$[/quote]

Let $ B'$ and $ C'$ be points in $ AC$ and $ AB$, respectively, such that $ AB \equal{} AB'$ and $ AC \equal{} AC'$. Let $ I_a$ be the center of the excircle of $ ABC$ tangent to $ AB$ and $ AC$ at $ D$ and $ E$ respectively ($ I_a$ lies in the bisector of $ \angle A$). We will show that the midpoint $ M$ of $ PQ$ is $ I$.

Consider the inversion with center $ A$ and radius $ AB \times AC$. It maps $ B$ and $ C$ into $ C'$ and $ B'$ respectively. Then, the circumcircle of $ ABC$ becomes line $ B'C'$. Therefore, $ w$ is mapped into a circle tangent to a circle tangent to $ AB$, $ AC$ and $ B'C'$.

Notice $ B'C'$ is tangent to both the incircle and $ (I_a)$ (because it is the reflection of $ BC$ over the bisector of $ \angle A$). Then $ w$ becomes $ (I_a)$ (it is easy to see that it cannot be the inverse of the incircle).

Now, if we call $ M'$ the inverse of $ M$, we have $ \angle ADM' \equal{} \angle AMP \equal{} 90^o$. As $ M'$ is on the bisector of $ \angle A$, we get that $ M' \equal{} I_a$. This implies that $ AM \times AI_a \equal{} AB \times BC$.

We will show that $ AI \times AI_a \equal{} AB \times BC$. Consider the circle $ K$ of diameter $ II_a$. It is known that $ \angle ICI_a \equal{} \angle IBI_a \equal{} 90^o$. Then both $ B$ and $ C$ belong to $ K$. By simmetry, we also get that $ B'$ and $ C'$ are in $ K$.

Then, we know that $ AI \times AI_a$ is the power of $ A$ with respect to $ K$. Then $ AI \times AI_a \equal{} AB \times AC' \equal{} AB \times AC$. As we knew that $ AM \times AI_a \equal{} AB \times AC$, this implies that $ I \equal{} M$ (because both points are on the bisector of $ \angle A$, inside $ ABC$), then $ I \in PQ$, as we wanted to prove.

[quote="Virgil Nicula"]$ AP \equal{} \frac {bc}{s},\ \ \frac {r_1}{r} \equal{} \frac {bc}{s(s \minus{} a)}.$[/quote]

The first equality follows, because $ D$ is the inverse of $ P$ and $ AD \equal{} s$ is known.

For the second one, an homotethy with center in $ A$ and ratio $ \frac {AP}{s\minus{}a}$ maps the incircle into $ w$. The ratio must be equal to $ \frac {r_1}r$ and the result follows using the first equality.

Let $\Gamma$ denote the circumcircle of $ABC$ and $\Omega$ denote the internally tangent circle through $P$ and $Q$. Let $M$ denote the point of tangency between $\Gamma$ and $\Omega$, $I$ denote the midpoint of $PQ$ and let $O$ denote the center of $\Omega$. Let $\angle{ABC}=a$.

First note that $A$, $M$, $I$ and $O$ are collinear and the entire figure is symmetrical around line $AM$. Therefore $I$ lies on the angle bisector of $\angle{BAC}$. We now have that $\angle{POQ}=180-\angle{BAC}=2a$. Hence we have that $\angle{PMQ}=a$ and by symmetry, $\angle{PMI}=a/2$.

Now note that $\angle{PIM}=\angle{PBM}=90$. Hence $PIMB$ is cyclic. Therefore $\angle{PBI}=\angle{PMI}=a/2$. Hence $I$ is also on the angle bisector of $\angle{ABC}$ and $I$ is the incenter of the triangle.

The 'standard' proof uses [b]Casey[/b]'s for the 'circles' $A, \;B, \;C$ and $\Omega$, all internally tangent to $\Gamma$, followed by the reciprocal of the Transversal Theorem ([b]Cristea[/b]'s).

Best regards,
sunken rock

Could you explain what you mean in detail? I do not know what Cristea's theorem is... and how casey's theorem is used here (i.e. what selection of +/- do we use).

To Altheman:

For Cristea's theorem see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=323170&p=1741338&hilit=cristea#p1741338

Regarding the use of Casey's:
We have the 'circles' $A,B, \Omega, C$, all internally tangent to $\Gamma$, hence the following equality holds: $AP\cdot BC=AB\cdot CQ+AC\cdot BP$ $(\;1\; )$. If $BC=a$, etc, $a+b+c=2s$ and $AP=AQ=m$, from $(1)$ we get $a\cdot m=c(b-m)+b(c-m)$, or $m=\frac{bc}{s}$ $(\; 2 \;)$

If $\{ D  \} \equiv AI \cap BC$, then: $\frac{DI}{AI}=\frac{a}{b+c} \; ( \; 3 \; )$, $CD=\frac{ab}{b+c} \; ( \; 4 \; )$ and $BD=\frac{ac}{b+c} \; ( \; 5 \; )$.

Next, as per Cristea's, we have to prove:
$CD\cdot \frac{BP}{PA}+BD\cdot \frac{CQ}{AQ}=BC\cdot \frac{DI}{AI}$ and, with the relations 2-5:
$\frac{ab}{b+c} \cdot \frac{c-\frac{bc}{s}}{\frac{bc}{s}}+\frac{ac}{b+c}\cdot \frac{b-\frac{bc}{s}}{\frac{bc}{s}}=a \cdot \frac{a}{b+c}$, or $2s-b-c=a$, which is true.

Best regards,
sunken rock

This solution is motivated from snowEverywhere.
Let do as usual notations midpoint of $PQ$ be $I.$ and let the angle bisector of $\angle A$ hits the $\odot ABC$
at $M$.note that $MA$ contains $I$.If you rotate the figure about $A$ with angle $90^{\circ}$ then $C \mapsto B$.so by the symmetry  we can say that $AM$ is the diameter of the  $\odot ABC$.
our motive is to show that, $P,I,M,B$ is cyclic.$M$ is the midpoint of the arc $PMQ \implies$ $PM$ bisects $\angle BPQ$.From above we can infer $\angle ABM =90^{\circ}$ and $\angle PIM =90^{\circ}$.so concyclic proved. it implies that, $\angle IBM =\angle IPM$  and $\angle MPB=\angle BIM \implies \angle IBM =\angle BIM \implies MI=MB$ so it follows  that $I$ is the incenter.

Let $O$ be the center of the circle. Let the circle be tangent to the circumcircle of ∆$ABC$ at $D$. Let $I $be the midpoint of $PQ$. Then $A, I, O, D$ are
collinear by symmetry. Consider the $homothety$ with center $A$ that sends ∆$ABC$ to ∆$AB’C’$ such that$ D$ is on $B’C’$. Thus, $k=AB’/AB$. As right triangles $AIP, ADB’, ABD, APO$ are similar, we have $AI /AO = (AI / AP)(AP / AO) = (AD /AB’)(AB /AD) = AB/AB’=1/k$.Hence the homothety sends $I$ to $O$.
Then $O$ being the incenter of ∆$AB’C’$
implies$ I$ is the incenter of ∆$ABC$.
QED :D

This is true in general.

Use $\sqrt[2]{bc} $ inversion. The mixitilinear Incircle becomes the excircle and the incenter the excenter.

Now, it's trivial thereafter.

Sorry, if this is resembling some previously posted solution.

By incenter/excenter lemma, it suffices to show $JM=JB$. Let $\angle MOP=2x$. Then $\angle MJP=x$ and $\angle JAB=90-2x \to \angle BAJ=2x$. Thus $\angle BJP=\angle MJP$. Since $\angle JBP=\angle JMP=90$, $BPMJ$ is a kite, done.

$\text{Midpoint of}$ $PQ=M$; $\text{line parallel to BC through}$ $I \cap AB=S$; $\text{line parallel to BC through}$ $I \cap AC=R$\\
Its apparent that $\triangle ASI \cong \triangle ARI \implies \angle AIS=90$ but $\angle AMP=90$ too and $PQ||SR$ so a contradiction thus $M=I$ (hopefully not a wrong solution)

[quote=howcanicreateacccount]Is not this true for all triangles? how AB=AC is used?[/quote]

yeah it is true for all triangles. It is something called Mixtillinear Incircles. See Evan Chen's handout on it (or EGMO chapter 4).

This fact is actually true for all triangles.

First I claim $I$ actually lies on $\overline{PQ}$. Consider a $\sqrt{bc}$-inversion which swaps this circle $\omega$ with the $A$-excircle. The conclusion is equivalent to showing $E, F, A, I_A$ cyclic, where $E, F$ are the excentral touchpoints, which is obvious.

Now $AK=AL$ which implies the result.

Label the touch points to $AB$, $AC$, and $(ABC)$ as $K$, $L$, and $T$, and the midpoints of arcs $AB$ and $AC$ as $X$ and $Y$.

Homothety tells us $TKX$ and $TLY$ collinear. Pascal on $BACXTY$ then tells us that $KIL$ collinear, and noting $\triangle AKI \cong \triangle ALI$ implies $IK = IL$. $\blacksquare$


[Edited]

If $ A \in M_n(C)$ denote $ a_n \equal{} det A$
Then we have $ a_n \equal{} x_n a_{n \minus{} 1} \plus{} (n \minus{} 1) a_{n \minus{} 2}$
Just compute this sequence(it's not so nice...).


Well, this post is not exactly about international contests, but seeing as it pertains to IMO selection, I put it here. This topic regards to a topic posted at about 2003 and (unfortunately) bumped up once a year. Anyways, instead of bumping it up again I will just post it as a new topic to hopefully get some fresh opinions. The questions I ask are the following:

1. Which country has the hardest IMO team selection test?
2. Which country had the hardest TST in 2009?
3. Is there any way to get the team selection tests of nations different from China, USA, etc (in like a book or something)? For example, I would very much like to have the complete Russian, Romanian, Hungarian, Iranian TST's for the last few years (but artofproblemsolving is limited in the problems it has posted)

Hardest TST question is so many times posted... Search for it.

1. China.
2. Probably China.
3. Resources/Contests, or some books by Mathematical Societies of respective countries, or I don't know,

[quote="huang"]
3. Is there any way to get the team selection tests of nations different from China, USA, etc (in like a book or something)? For example, I would very much like to have the complete Russian, Romanian, Hungarian, Iranian TST's for the last few years (but artofproblemsolving is limited in the problems it has posted)[/quote]

Do you know the famous quotation by the late US president John F. Kennedy back in the early 1960’s:

[b]"Ask not what your country can do for you, but what you can do for your country!"[/b]

I have posted some problems for the world premium problem-solvers! In particular most China IMO Team Selection Tests. What can you do except demanding things? Somehow I have the feeling this thread will not see a reply to my question here...

There is no competition here. Ireland's TST is by far the most challenging.


2^x+7^x=13^x.findx if you have similar questiom post and solve

$ 2^x\equal{}13^x\minus{}7^x\equal{}(10\plus{}3)^x\minus{}(10\minus{}3)^x\equal{}2[3x10^{x\minus{}1}\plus{}3^3\frac {x(x\minus{}1)(x\minus{}2)}{6}10^{x\minus{}3}\plus{}.....)$
Since 3 divides every term on the LHS, no solutions in integers exist.

I don't think we can find an exact solution to this .We can only prove that this equation has a unique solution since the equiation is equivalent to :
$ \left(\frac{2}{13} \right)^x\plus{}\left(\frac{7}{13} \right)^x\equal{}1$ , and the function $ f: \mathbb R\minus{}>\mathbb R$ , $ f(x)\equal{}\left(\frac{2}{13} \right)^x\plus{}\left(\frac{7}{13} \right)^x\minus{}1$ is clearly decreasing ,hence injective.


geometry
geometric transformation
function
calculus
topology

Summer vacation is almost here! For me at least.  I just wanted to know what everyone is going to read over the summer (if anything since this is a math forum and I know that a lot of you hate english).  Also what are some suggestions for good books to read? Just post anything related.

I hated english class most of my life but I started to like it in 8th grade and now it's still my least favorite class but that doesn't mean that I don't enjoy it.  (just felt like adding that bit of info)

Flowers for Algernon.  Read it.  No matter what.  Read it.

And, if you want a few more, 1984, Farenheit 451, and Flatland.

hmm,  My school doesn't assine summer reading but I read a lot anyhow.

Some good books would include

Clan of the Cave Bear Series
Dan Brown's books
Seabuiscut
Call of the Wild
Tom Sawer
Dandelion Wine
Treasure Island
Swiss Family Robinson
I could go on and on.
pm if anyone wants to know good books!

[quote]And, if you want a few more, 1984, Farenheit 451, and Flatland.[/quote]

We actually read Fahrenheit 451 in school this year.  I loved that book.  I just finished Animal Farm today(it was assigned also).  I recommend Animal Farm for anyone who has some knowledge of the Russian Revolution or even if you don't read it anyways!  It does a great job of showing how ignorant ppl. can be as well as doing much more.  It has inspired me to start reading 1984.  I'm only about 10 pages in and I can see the similarities between 1984 and F. 451.

hmm.  btw Dandelion wine is by Ray Bradbery as well.  I am aslo readin F.451.  Very good book so far.

Dandelion Wine was a great book.  If you're going the distopia route (Farenheit 451, 1984, Animal Farm), you might add Brave New World as well.

Brave New World is pretty sweet! We read it for english this year. Its the only book we've been forced to read for english class that I actually enjoyed.

Don't miss The Chosen by Chaim Potok if you haven't read it yet. That was the best assigned reading I ever had in high school--I read several of Potok's other novels in the following few years.

Historical fiction's sometimes pretty cool.  I can't think of any specifics, but two history-related books we read in class this year that were pretty good:

Night by Elie Wiesel (I imagine this is quite famous)
To Kill a Mockingbird by Harper Lee (obviously a classic)

Short story compilations are also good sometimes.

anything by Kurt Vonnegut makes for great reading.

P.S. The Chosen is an amazing book.

The Merchant of Venice

Le Morte D'Arthur (sp?)

Making Sense of Your Faith

Christianity and the Constitution

And some others I forgot about......

Actually Probablility101 I HATED those books. Both of them.  The only thing they have going for them is history, but both are so boring to me.

You can read the Hobbit series by JRR Tolkein... Those aren't too bad or the Count of Monte Cristo.  I want to read that one.

lol looks like someone's tastes are a slight bit different than mine...

If you're in to fantasy, I think Redwall is OK for a few books, and His Dark Materials is good.  Tolkien is very popular, but it gets annoyingly implausible sometimes.  I hear Aragon and a few other 600+ page books (I think) are good, but I've never read them.  I just get annoyed after a while when it's always a single person who's the last chance and stuff, especially when the situation is impossibly hopeless, and the author has to use a deus ex machina for the good guys to win.

I had to read Dandelion Wine for english though that was pretty harsh of my teacher.  I was the only one that understood and grasped the concept of the book, but thats not surprising I suppose considering that was a 7th grade english class.  I have been reading the other books that u mention JBL I am a bookworm totally and I absolutly love to read.

The Tolkien Trilogy is a very good series but a lot of people who read them here don't seem to really understand them.  They just read them to get another grade in reading(we are on a system called Accelerated Reader which has tests on certain books and gives points).  Do not read the Trilogy unless you think that you could understand it because it is too good of a book to just speed through and not see the and experience the wonderful writing and scenes from the book.

I love all classics and I am curently trying to get a hold of To Kill a Mocking Bird but there seems to be a sudden interest in it and it is checked out at every library near me.

More Classics if u enjoy those would include

Oliver Twist
Tale of Two Cities
Pride and Prejudice
The Hunchback of Notre DAme
Ivanhoe
The Invisible Man
The Wind in the Willows
Gullivers Travels
The Picture of Doran Gray
Hans Brinker
Robinson Crusoe
The Adventures of  Sherlock Holmes
Litle Women
Moby Dick
20,000 leagues under the Sea
The Wizard of Oz
Jane Eyre
Heidi

I could go on and on about classics and other books.  All of them are wonderful to read.

I liked To Kill a Mockingbird it was pretty good.  Night was ok.  I personally didn't think that it was as good as everyone told me. It was still good though.  My english teacher has shown us a movie for almost every book we have read.  For night we watched a movie on Josef Mengele.  That was interesting.  The movie we watched that was supposed to be Cold Sassy Tree was horrible.  I mean it was just plain bad.  That book is pretty good(don't read it if you really hate the South and Confederates though).

Some of my favorite books are the Greek classics.  I haven't read many but I like them a lot.  The Iliad and the Odyssey were awesome.   I recommend them even though they are massive.  Oedipus Rex was very good. Does anyone have recommendations for other great Roman or Greek classics?

Some books I liked when I was (slightly) younger:

- The [i]His Dark Materials[/i] books (The Golden Compass, The Subtle Knife, and The Amber Spyglass), by Phillip Pullman
- Garth Nix's [i]Sabriel[/i], [i]Lirael[/i], and [i]Abhorsen[/i]
- [i]Eragon[/i], by Christopher Paolini

I don't remember if anyone has said this, but [u]Godel's Proof[/u] is pretty interesting -- I think it gives a really good look at the deeper structure of mathematics.

Well ... it's pretty intense.  Apparently there are two different versions of it, by different authors.  I have the older one.  There's no calculus, and in fact not very much math knowledge required at all.  But it's fairly intense -- I had to go through some sections very slowly to follow what was going on.

[quote="joml88"]What level is Godel's Proof at?  Is it accessible to be people who have nearly no calculus experience?[/quote]

It's all symoblic logic.  It's a completely different beast than most of the mathematics you see.  It's worth working through a symbolic logic book like Quine's before attempting to wade through Godel.  You need to have a good idea of what a well-formed formula is and how to determine whether or not a statement is a well-formed formula.

Honestly, it's not something I would recommend to high school students.  Well, the Quine book is a good primer to predicate logic, but there is very little reason to wade through the proof.  While I find it philosophically useful to understand, I could just as easily have read a nice summary and taken a different class in college (it was the culmination of a Mathematical Logic course I took).

Mathfanatic, I also enjoyed all the books you suggested. [i]His Dark Materials[/i] was perhaps my favorite series ever. Do you know of any similar books I may not have yet read?[/i]

In a perfect world, I would get through
Godel, Escher, Bach by Douglas Hofstadter
On Numbers and Games by John Conway
Count Down by Steve Olson
and possibly some Feynman material this summer.  In all likelihood, however, I'll read at most one of those.

By the way, I read Flatland a year or two ago and loved it.  It is definitely satirical of Victorian society, and not just on its attitudes towards women.  There is some very thick satire of manners, conformity, social structure, and other aspects of society.
Also, to anyone who put The Adventures of Tom Sawyer on their list, plan on reading The Adventures of Huckleberry Finn afterwards--it's by far the better of the two.  Just be sure to think about what point Twain is trying to make with the book; if you read it without thinking, Twain sounds like a racist (hence why the book has always aroused controversy).

Some great books:

Animal Farm (George Orwell)
Count Down (Steve Olson)
Lord of the Rings series
the Art of Problem Solving Volumes 1 & 2
The "A Wrinkle in Time" series, or at least the first 3 or 4 books.
To Kill a Mockingbird

Is that Ian Stewart?  If so, I read it, and it was okay.  I read all the recreational math books I could get my hands on when I was younger, and the thing that bugs me about Ian Stewart is that after you read enough he starts to repeat himself... literally.  He has a good book of his French Scientific American columns (It's a Fine Math You've Got Me In To) which I'd recommend as the most mathematically meaty of his work, but he pretty much lifted some of his columns and reused them in his other books.

[quote="joml88"]I have another book that I want your opinion on, The Magical Maze.  My cousin visited this weekend and he had it.  I looked over it and it looked like it might be worthwhile but what does everyone here think about it?[/quote]

Yes it's by Ian Stewart.  I couldn't remember the name at the time I posted.

I have to read The Chosen by Chaim Potok; Bless Me, Ultima by Rudolfo Anaya; Mythology by Edith Hamilton; and King Arthur and His Knights of the Round Table by Roger Lancelyn Green for summer reading.

Has anyone been able to find a copy of The Function of Reason by Alfred Whitehead that is still in print? I have a second-hand copy.

I enjoyed [u]Flatland[/u], but I enjoyed [u]Flatterland[/u] (which also happens to be by Ian Stewart) even more.  You do not have to worry about it being anti-female, it has an underlying women's rights theme.  As [u]Flatland [/u]challenged the idea of our world being 3-D, [u]Flatterland[/u] challenges the whole concept of dimensions and provides an amusing intro to non-Euclidean Geometry.  When I go to the library I visit the 510's (math ;) ) and pull whatever catches my eye.  [u]Innumeracy[/u] and [u]Beyond Numeracy[/u] by John Allen Paulos were very interesting.

Any one know of any good beginning books in Topology?  I have come across it briefly in various books and would like to learn more.

There's a book called Topology (I think) from Dover that I saw at the bookstore once. I usually like Dover books...

Anyone read Zelazny? (Chronicles of Amber).
He's excellent.


inequalities
calculus
calculus computations

Find a>0 so that 2^x+3^x-a^x >= 4^x+5^x+a^x where x belongs to R.
I can't seem to find the extreme point. Any thoughts?

[quote]$ 2^x \plus{} 3^x \minus{} a^x \geq 4^x \plus{} 5^x \plus{} a^x$[/quote]
Are you sure this is correct? Because it is obviously wrong if $ a>0$.

I double checked... the answer is supposed to be a=20..but I can't understand how

Well, this inequality can be true if x < 0. However, a need not be 20 as you have stated. This is an example, x = -1, a = 6, the inequality holds.

$ 2^{ \minus{} 1} \plus{} 3^{ \minus{} 1} \minus{} 4^{ \minus{} 1} \minus{} 5^{ \minus{} 1} \minus{} 2(6)^{ \minus{} 1} \equal{}\frac{1}{20}>0$

Do check your question from the source or originator.


[quote="shfdfzhjj"]Hope you will enjoy it.
Let$ a;b;c > 0$ Prove that :
$ \sum a \sum \frac{1}{\sqrt{5a^2\plus{}2b^2\plus{}2c^2}} \le\ 3$:)[/quote]
We have$ ( \sum \frac{1}{\sqrt{5a^2\plus{}2b^2\plus{}2c^2}})^2 \le\ 3(\sum \frac{1}{5a^2\plus{}2b^2\plus{}2c^2})$

Need prove $ \sum \frac{1}{5a^2\plus{}2b^2\plus{}2c^2} \le\  \frac{3}{(a\plus{}b\plus{}c)^2}$

Assum $ a\plus{}b\plus{}c\equal{}1$ and let $ ab\plus{}bc\plus{}ca\equal{}q;abc\equal{}r$ . We have :

 $ Ineq \leftrightarrow \minus{}81r^2\plus{}r(90\minus{}216q)\plus{}588q^3\minus{}669q^2\plus{}264q\minus{}36 \le\ 0$

Let$ f(r)\equal{}\minus{}81r^2\plus{}r(90\minus{}216q)\plus{}588q^3\minus{}669q^2\plus{}264q\minus{}36$

$ \rightarrow f '(r) \equal{} \minus{}162r\plus{}90\minus{}216q \ge\ 90 \minus{} \frac{162}{27}\minus{} \frac{216}{3} > 0$

$ \rightarrow f(r) \le\ f(\frac{4q\minus{}1}{9}) \equal{} 588q^3\minus{}781q^2\plus{}336q \minus{} 47$

$ \equal{} (3q\minus{}1)(196q^2\minus{}195q\plus{}47) \le\ 0$which is obvious

We have done!  :wink:

[quote="shfdfzhjj"][quote="onlylove_math"]
We have$ ( \sum \frac {1}{\sqrt {5a^2 \plus{} 2b^2 \plus{} 2c^2}})^2 \le\ 3(\sum \frac {1}{5a^2 \plus{} 2b^2 \plus{} 2c^2})$

Need prove $ \sum \frac {1}{5a^2 \plus{} 2b^2 \plus{} 2c^2} \le\ \frac {3}{(a \plus{} b \plus{} c)^2}$

Assum $ a \plus{} b \plus{} c \equal{} 1$ and let $ ab \plus{} bc \plus{} ca \equal{} q;abc \equal{} r$ . We have :

 $ Ineq \leftrightarrow \minus{} 81r^2 \plus{} r(90 \minus{} 216q) \plus{} 588q^3 \minus{} 669q^2 \plus{} 264q \minus{} 36 \le\ 0$

Let$ f(r) \equal{} \minus{} 81r^2 \plus{} r(90 \minus{} 216q) \plus{} 588q^3 \minus{} 669q^2 \plus{} 264q \minus{} 36$

$ \rightarrow f '(r) \equal{} \minus{} 162r \plus{} 90 \minus{} 216q \ge\ 90 \minus{} \frac {162}{27} \minus{} \frac {216}{3} > 0$

$ \rightarrow f(r) \le\ f(\frac {4q \minus{} 1}{9}) \equal{} 588q^3 \minus{} 781q^2 \plus{} 336q \minus{} 47$

$ \equal{} (3q \minus{} 1)(196q^2 \minus{} 195q \plus{} 47) \le\ 0$which is obvious

We have done!  :wink:[/quote]
My proof is the same. :lol:[/quote]
Sorry shfdfzhjj, my proof have some mistakes . Because I can't edit in my first proof
[quote]Sorry, but you cannot edit this post anymore. Ask a moderator to do it if it is really necessary[/quote].so i must edit it in this proof. Here is my full proof  :)
We have$ ( \sum \frac {1}{\sqrt {5a^2 \plus{} 2b^2 \plus{} 2c^2}})^2 \le\ 3(\sum \frac {1}{5a^2 \plus{} 2b^2 \plus{} 2c^2})$

Need prove $ \sum \frac {1}{5a^2 \plus{} 2b^2 \plus{} 2c^2} \le\ \frac {3}{(a \plus{} b \plus{} c)^2}$

Assum $ a \plus{} b \plus{} c \equal{} 1$ and let $ ab \plus{} bc \plus{} ca \equal{} q;abc \equal{} r$ . We have :

 $ Ineq \leftrightarrow \minus{} 81r^2 \plus{} r(90 \minus{} 216q) \plus{} 588q^3 \minus{} 669q^2 \plus{} 264q \minus{} 36 \le\ 0$

Let$ f(r) \equal{} \minus{} 81r^2 \plus{} r(90 \minus{} 216q) \plus{} 588q^3 \minus{} 669q^2 \plus{} 264q \minus{} 36$

$ \rightarrow f '(r) \equal{} \minus{} 162r \plus{} 90 \minus{} 216q \ge\ 90 \minus{} \frac {162}{27} \minus{} \frac {216}{3} > 0$

$ \rightarrow f(r) \le\ f(\frac {q}{9}) \le\ 0$which is obvious

We have done!  :wink:

[quote]Let $ a;b;c \in\  R^{\plus{}}$, prove that:
$ \sum a \sum \frac {1}{\sqrt{ka^2 \plus{} b^2 \plus{} c^2}} \le\ \frac {9}{\sqrt{k \plus{} 2}}$ [/quote]

Assum $ a \plus{} b \plus{} c \equal{} 1$, let $ ab \plus{} bc \plus{} ca \equal{} q$. We have :

$ (\sum \frac {1}{\sqrt {ka^2 \plus{} b^2 \plus{} c^2}})^2 \le\ 3(\sum \frac {1}{ka^2 \plus{} b^2 \plus{} c^2})$

Need prove$ \sum \frac {1}{ka^2 \plus{} b^2 \plus{} c^2} \le\ \frac {27}{k \plus{} 2}$

$ \leftrightarrow \frac {1}{(k \minus{} 1)a^2 \plus{} 1 \minus{} 2q} \le\ \frac {27}{k \plus{} 2}$

Let$ f(t) \equal{} \frac {1}{(k \minus{} 1)t^2 \plus{} 1 \minus{} 2q} \rightarrow f "(t) \le\ 0$

Follow Jensen we have

$ f(x) \plus{} f(y) \plus{} f(z) \le\ 3f(\frac {x \plus{} y \plus{} x}{3}) \equal{} \frac {3}{(k \minus{} 1)(\frac {x \plus{} y \plus{} x}{3})^2 \plus{} 1 \minus{} 2q} \le \ \frac {27}{k \plus{} 2}$

(Because  $ 1 \minus{} 2q \ge\ \frac {1}{3}$)
We have done! :)

[quote="shfdfzhjj"][quote="onlylove_math"][quote]Let $ a;b;c \in\ R^{ \plus{} }$, prove that:
$ \sum a \sum \frac {1}{\sqrt {ka^2 \plus{} b^2 \plus{} c^2}} \le\ \frac {9}{\sqrt {k \plus{} 2}}$ [/quote]

Assum $ a \plus{} b \plus{} c \equal{} 1$, let $ ab \plus{} bc \plus{} ca \equal{} q$. We have :

$ (\sum \frac {1}{\sqrt {ka^2 \plus{} b^2 \plus{} c^2}})^2 \le\ 3(\sum \frac {1}{ka^2 \plus{} b^2 \plus{} c^2})$

Need prove$ \sum \frac {1}{ka^2 \plus{} b^2 \plus{} c^2} \le\ \frac {27}{k \plus{} 2}$

$ \leftrightarrow \frac {1}{(k \minus{} 1)a^2 \plus{} 1 \minus{} 2q} \le\ \frac {27}{k \plus{} 2}$

Let$ f(t) \equal{} \frac {1}{(k \minus{} 1)t^2 \plus{} 1 \minus{} 2q} \rightarrow f "(t) \le\ 0$

Follow Jensen we have

$ f(x) \plus{} f(y) \plus{} f(z) \le\ 3f(\frac {x \plus{} y \plus{} x}{3}) \equal{} \frac {3}{(k \minus{} 1)(\frac {x \plus{} y \plus{} x}{3})^2 \plus{} 1 \minus{} 2q} \le \ \frac {27}{k \plus{} 2}$

(Because  $ 1 \minus{} 2q \ge\ \frac {1}{3}$)
We have done! :)[/quote]
But I think you are wrong. :)[/quote]
Sorry but i don't see it . Can U say my mistake ?
Thanks! :)

[quote="onlylove_math"]
$ \sum a \sum \frac {1}{\sqrt {ka^2 \plus{} b^2 \plus{} c^2}} \le\ \frac {9}{\sqrt {k \plus{} 2}}$ [/quote]
But I think you are wrong. :)[/quote]
Sorry but i don't see it . Can U say my mistake ?
Thanks! :)[/quote]

Try k=10,a=2,b=c=1 :)

$ \sum a \sum \frac {1}{\sqrt{ka^2 \plus{} b^2 \plus{} c^2}} \le\ \frac {9}{\sqrt{k \plus{} 2}}$
Find the best k

Any idea? :maybe:

[quote="shfdfzhjj"]
$ \sum a \sum \frac {1}{\sqrt {ka^2 \plus{} b^2 \plus{} c^2}} \le\ \frac {9}{\sqrt {k \plus{} 2}}$
Find the best k
Any idea? maybe :maybe:
[/quote]
Sorry I have some mistake at $ f "(t) \le\ 0$ because it isn't true for all $ k > 0$

I think $ f(t) \equal{} \frac {3}{3(k \minus{} 1)t^2 \plus{} 1}$ (Because  $ 1 \minus{} 2q \ge\ \frac {1}{3}$)

$ f "(t) \equal{} \frac { \minus{} 6(k \minus{} 1) \plus{} 54(k \minus{} 1)^2t^2}{[3(k \minus{} 1)t^2 \plus{} 1]^3} \le\ 0$

$ \leftrightarrow k \le\ \frac {1}{9t^2} \plus{} 1$

I think best value of $ k \equal{} 2$ :lol:

[quote="onlylove_math"][quote="shfdfzhjj"]
$ \sum a \sum \frac {1}{\sqrt {ka^2 \plus{} b^2 \plus{} c^2}} \le\ \frac {9}{\sqrt {k \plus{} 2}}$
Find the best k
Any idea? maybe :maybe:
[/quote]

I think best value of $ k \equal{} 2$ :lol:[/quote]

I don't think so. :)

[quote="onlylove_math"]

Need prove$ \sum \frac {1}{ka^2 \plus{} b^2 \plus{} c^2} \le\ \frac {27}{k \plus{} 2}$

[/quote]

For this inequality,I have proved that  $ 1 \equal{} < k \equal{} < 4$ is true,and $ SUP k \equal{} 4$
But I don't know what's $ INF k$ is,I think $ INF k\equal{}1$,but I am not sure.

[quote="shfdfzhjj"][quote="onlylove_math"]

Need prove$ \sum \frac {1}{ka^2 \plus{} b^2 \plus{} c^2} \le\ \frac {27}{k \plus{} 2}$

[/quote]

For this inequality,I have proved that  $ 1 \equal{} < k \equal{} < 4$ is true,and $ SUP k \equal{} 4$
But I don't know what's $ INF k$ is,I think $ INF k \equal{} 1$,but I am not sure.[/quote]
No, Hanjingjun. The correct answer is $ \frac{1}{7} \le k \le 4$. And it suffices to prove the inequality for the case $ k\equal{}\frac{1}{7}$ or $ k\equal{}4$.
For $ k\equal{}4$, we got a nice solution with Cauchy Schwarz.
For $ k\equal{}\frac{1}{7}$, equality holds when $ a\equal{}4b\equal{}4c$, and Vornicu-Schur works really nice here. :)

Let $ a;b;c \in\ R^{ \plus{} }$, prove that:
$ \sum a \sum \frac {1}{\sqrt {ka^2 \plus{} b^2 \plus{} c^2}} \le\ \frac {9}{\sqrt {k \plus{} 2}}$

How about that? :maybe:


[quote="Interval"]For each question below, perform the indicated operations and express your answer in the form of a + bi.

[1] sqrt{-64}

[2] sqrt{(4 + 3i)(3i - 4)[/quote]
[hide="(1)"] $\sqrt{-64}=\sqrt{64*-1}=8*\sqrt{-1}=8i$[/hide]
[hide="(2)"] $\sqrt{(4+3i)(3i-4)}=\sqrt{12i-12i+9i^{2}-16}=\sqrt{-9-16}=\sqrt{-25}=\sqrt{25*-1}=5*\sqrt{-1}=5i$[/hide]


geometry
geometric transformation
reflection
incenter
circumcircle
geometry open

.  A and B lie on a circle. P lies on the minor arc AB. Q and R (distinct from P) also lie on the circle, so that P and Q are equidistant from A, and P and R are equidistant from B. Show that the intersection of AR and BQ is the reflection of P in AB.

I use a basis property of the incenter $I$ in the circumcircle $w$ for a triangle $ABC$:
\[ \overline {\underline {\left| \ \{A,A'\}\subset AI\cap w\Longrightarrow A'B=A'C=A'I\ \right| }} \].[u]A solution.[/u] I note the intersections $I\in AR\cap BQ$, $U\in AB\cap PQ$, $V\in AB\cap PR$. The point $I$ is the incenter of the triangle $PQR$. Therefore, $AQ=AP=AI$, $BR=BP=BI\Longrightarrow$ the line $AB$ is the bisector of the segment $PI\Longleftrightarrow$ the point $I$ is the reflection of the point $P$ w.r.t. the line $AB$.


an unlimited supply of petrol is available at the edge of desert 800km wide, but there is no source of fuel in the desert itself. A truck can carry enough petrol to travel 500km (this will be called one load) and it can build up its refueling stations at any spot along the way. These caches may be of any size and it is assumed that no evaporations takes place. What is the minimum amount in (loadof petrol that the truck requires in order to cross the desert?


Is that all? :?
I think we can find an equivalence here.

A solution has already been given [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34725]here[/url].


Is this what you mean?

How does (1) imply (2):
\begin{align} \sum_{i = 1}^{2n} \left(\frac {1}{n}\right)\left(2 \ln n + \ln \left(1 + \frac {i^2}{n^2}\right) - 4 \ln n\right) \\
\sum_{i = 1}^{2n} \left(\frac {1}{n}\right) \ln \left(1 + \frac {i^2}{n^2}\right) \end{align}

I think I'm interpreting it wrong, because these two summations are not equal.

[quote="123456789"]

I think I'm interpreting it wrong, because these two summations are not equal.[/quote]

It's hard to interperate because you have 6 open breackets and 7 closed ones

@Bigman

$ (1)$ implies

$ \sum_{i \equal{} 1}^{2n} \left(\frac {1}{n}\right) \ln \left(\frac {n^2 \plus{} i^2}{n^4}\right)$

yes, that what i meant ocha. How does that hold? I don't see it.

Just use log rules
$ a\log_e (x) \equal{} \log_e (x^a) \qquad(1)$

$ \log_e (x) \plus{} \log_e (y) \equal{} \log_e (xy) \qquad(2)$

oops, that's not what I meant. Take a look at this document. It's #2

http://www.math.ksu.edu/main/events/ksucomp/KCMC/problems07.pdf

Oh, If it is just the second line you have trouble with then;

$ \sum_{i\equal{}1}^{2n}\frac{2}{n} \log_e(n) \equal{} 2n\left(\frac{2}{n} \log_e(n)\right) \equal{} 4\log_e(n)$

So it just cancels out


When the numbers 19779 and 17997 are divided by a certain three digit number, they leave the same remainder. Find this largest such divisor and the remainder. How many such divisors are there ?

Well, the number's only requirement is that it has to divide their difference, which is $ 1782$

The largest 3 digit number is 891, which leaves remainder 67, and since $ 1782\equal{}2*9^2*11$ we have $ 2*3*2\equal{}12$ such divisors.

[quote="rofler"]Well, the number's only requirement is that it has to divide their difference, which is $ 1782$

The largest 3 digit number is 891, which leaves remainder 67, and since $ 1782 \equal{} 2*9^2*11$ we have $ 2*3*2 \equal{} 12$ such divisors.[/quote]

what let you think of 891? and why the difference?

Because they are congruent mod whatever, so the largest possible mod they can be congruent by is their difference. It's easy to see as well that every other mod has to be a factor of the difference.

In this case, $ 891$ is the largest $ 3$ digit factor of $ 1782$.

And what about the number of divisors?

And please give a detailed explanation.

$ a \equal{} 19779$
$ b \equal{} 17997$
$ k$- the three digit number
$ r$- the remainder

$ a \equal{} kx \plus{} r$, $ b \equal{} ky \plus{} r$. Then: $ a \minus{} b \equal{} k(x \minus{} y)$ =>
$ k$ divides $ a \minus{} b \equal{} 1782 \equal{} 2\times 891 \equal{} 2 \times 3^4 \times 11$

As $ k$ is maximal possible, $ k$ is $ 891$.

The number of such divisors is $ (1 \plus{} 1)(4 \plus{} 1)(1 \plus{} 1) \equal{} 20$, but we exclude  four-digit ,two-digit and one-digit divisors so there are EDIT:$ 5$ such divisors.

Can you please tell the 6th one?
I think there are only 5 of those.$ 891,594,297,198,162$

Yes there are $ 5$. I forgot to exclude $ 1782$ from those $ 20$ divisors.

[quote]Well, the number's only requirement is that it has to divide their difference, which is $ 1782$

The largest 3 digit number is 891, ... [/quote]Since $ 1782 \equal{} 2*3^4*11$, we have $ 2*5*2 \equal{} 20$ divisors.
1_ We could write them all, and pick the ones with three digits.
2_ As they come in pairs, like $ 1782\equal{}2*891$, the 3-digit ones will be associated with the smallest divisors. The $ 5$ smallest are $ 2$, $ 3$, $ 6$, $ 9$, and $ 11$ (in order). The next smaller divisor is $ 18$ and does not work because $ 1782/18\equal{}99$. So, we end with just the 5 divisors $ 1782/2$, $ 1782/3$, $ 1782/6$, $ 1782/9$, and $ 1782/11$. There is no need to calculate them. The calculator is not even needed for $ 1782/18\equal{}99$, because it's $ (2*3^4*11)/(2*3^2)\equal{}3^2*11$, easy to do in your head.