Let's say I want to write:

\documentclass{article}
\begin{document}
Hello, world!
\end{document}

Inside of the document itself. Is it possible to do this?

EDIT:
Not only that, but can you place in code, such as \LaTeX, without it actually autoconverting to Tex?

Use the verbatim environment
[code]\begin{verbatim}
\documentclass{article}
\begin{document}
Hello, world!
\end{document}
\end{verbatim}[/code]This won't work in the TeXer but the version for short bits of text will
[code]\verb=\LaTeX=[/code] giving $ \verb\equal{}\LaTeX\equal{}$
verb is not the grammatical [i]verb[/i] but short for verbatim and the = signifies the beginning and end of the verbatim text - you can use almost any other symbol if you need = in the text.

The
[code]\verb=text=[/code]
would work normally ($\verb=text=$) but if you do multiple lines
[code]\verb=package 6445;
import acm.program.*;
public class sqrtSequence extends ConsoleProgram{
	public void run(){
		double x = readInt("x_1 = ");
		int n = readInt("n = ");
		for(int i=1; i<=n; i++){
			 x = Math.sqrt(1 + x);
			 println(i+"   "+x );
		}
	}
}=[/code]
it would come out as $\verb=package 6445;
import acm.program.*;
public class sqrtSequence extends ConsoleProgram{
	public void run(){
		double x = readInt("x_1 = ");
		int n = readInt("n = ");
		for(int i=1; i<=n; i++){
			 x = Math.sqrt(1 + x);
			 println(i+"   "+x );
		}
	}
}=$

Is there any way to avoid this?

You cannot use verb with more than one line even in a document as that will give an error. That's the purpose of the verbatim environment, however, that's not available in the forum. I can't see any reason to use it here as you can use the code tag instead.
In a document you may wish to try the [url=http://tug.ctan.org/pkg/listings]listings package[/url] to typeset programming code which will deal with many programming languages.


:arrow: thank you
your answer is true :arrow:


Ring Theory
advanced fields
advanced fields unsolved

I've got a very stupid question:

Let $ K \subseteq L \subseteq E$ be number fields.

If a prime ideal in $ \mathcal{O}_K$ ramifies in $ \mathcal{O}_L$, why does it also ramify in $ \mathcal{O}_E$?

Say the prime ideal p (in K) ramifies at P (in L) with ramification index e>1. Let Q be some prime ideal in L sitting above P with ramification index e'>0 wrt L. Then the ramification index of Q wrt to K is ee'>1, hence ramified.

Ok say $ p\mathcal{O}_L \equal{} P_1^{e_1} \cdots P_\ell^{e_\ell}$, with the $ P_i \lhd \mathcal{O}_L$ prime ideals and $ e_1 > 1$.

So $ p\mathcal{O}_F \equal{} \left(P_1^{e_1} \mathcal{O}_F\right)\cdots \left(P_\ell^{e_\ell} \mathcal{O}_F\right)$. (Is this right??? :blush:)

Now let $ P_1\mathcal{O}_F \equal{} \mathcal{P}_1^{E_1} \cdots \mathcal{P}_g^{E_g}$.

So $ p\mathcal{O}_F \equal{} \mathcal{P}_1^{e_1E_1} \cdots \mathcal{P}_g^{e_1E_g}\left(P_2^{e_2} \mathcal{O}_F\right)\cdots \left(P_\ell^{e_\ell} \mathcal{O}_F\right)$.

Hence $ p$ ramifies. Is this right? I'm a bit unsure...

Nicely TeXed !!!  :thumbup:

[quote="Arne"]So $ p\mathcal{O}_F \equal{} \left(P_1^{e_1} \mathcal{O}_F\right)\cdots \left(P_\ell^{e_\ell} \mathcal{O}_F\right)$. (Is this right??? :blush:)[/quote]

Yes: Let A -> B be a ring homomorphism and aa and bb ideals in A. Then (aa . bb)B = aaB . bbB, in other words the extension of ideals commutes with the ideal product.


arithmetic sequence
combinatorics solved
combinatorics

Consider the set X = {1,2,3,...,9}. Show that, for any partition of X into two subsets, one of these contains three numbers such that the sum of two of them equals twice the third.

[i]Romanian IMO Selection Test 1978[/i]

Hint: Van der Waerden theorem might help.

I have no idea how Van der Waerden's thm might help (as far as I know, it deals with infinite sets). I just used some basic case consideration. I'll just look at a case here.

We assume the problem is wrong.

Let's think about this as a coloring of the numbers in red and blue (it might be easier to type :D ). Let's assume that for each of the pairs (3,4), (4,5), (5,6), (6,7) one member is blue and the other one is red. Then assume 3=r, so 4=b=>5=r=>6=b=>7=r, and we have 3,5 and 7 all red, so we've got a contradiction.

This means that at least one of the 4 pairs above must have the same color. Assume it's (3,4) (this is the only one I'll look at here; the others are the same). We have 3,4=r=>2,5=b=>8=r=>6=b=>7=r=>9=b=>1=r because otherwise we would have 1,5 and 9 all blue, so now we have 1,4 and 7 all red, which is, again, a contradiction.

We do the same if any of the other 3 pairs: (4,5), (5,6), (6,7) is chosen to have the same color. We actually have only one more case:(4,5), because the situation is obviously symmetric with respect to 5 (we can relabel 9 to be 1, 8 to be 2 etc.).

There are two versions of VdW theorem :

The infinite one :
If the set of positive integers is finitely colored then we may find monochromatic arithmetic progressions of arbitrary finite length.

The finite one :
For each integers r  \geq  2 and k  \geq 1. There exists a positive integer N = N(k,r) such that for each r-coloring of {1,...,N}, there exists a monochromatic arithmetic progression of length k.

The least integer N such that the above property is true is denoted by W(k,r).

In fact, the two versions are equivalent by compacity argument.

The problem asked to prove that W(3,2)  \leq 9, and in fact it is known that W(3,2) = 9.

Pierre.

Yup! I didn't know it could be expressed in so many forms:

[url]http://mathworld.wolfram.com/vanderWaerdensTheorem.html[/url]

You gotta love Mathworld! :D


Problem:
How many numbers are in the following list:
\[ \minus{}4,\minus{}1, 2, 5,\ldots, 32\]

I got the answer a different way, and I don't really understand the solution:

Solution:
Note that the numbers increase by 3 each time. Hence we increase by 3 a total of $ \frac{32\minus{}(\minus{}4)}{3}\equal{} 12$ times. But then there must be $ 12\plus{}1 \equal{}\boxed{13}$numbers, since we need to also include the first number on the list.

How does that formula work exactly? Can anyone explain it more clearly? Thanks

[hide="Basic showing"]Well, first of all, the sequence is an arithmetic sequence (meaning that the numbers in the sequence increase by a certain number each time). For a general example, consider the arithmetic sequence $ 1$, $ 4$, $ 7$. By inspection, the number of terms is $ 3$ (obviously). Consider what happens when we subtract all of our numbers by $ 1$. We then have the sequence $ 0$, $ 3$, $ 6$. However, going about this another way, we can write the numbers in a different way.

1st term: $ 0(3)$

2nd term: $ 1(3)$

3rd term: $ 2(3)$

Note that the number by which our constant number is multiplied (in our case, our constant is $ 3$) is one less than the number of the particular term (for example our $ 3$rd term is $ 3$ times $ 3\minus{}1$). Thus, if we can figure out that number by which our constant is multiplied, we may simply add one to find the total number of terms (since we overlooked  the first term).

Since all the terms must be counted, we use the last term as the starting point. If we subtract the first the first term from it, we can get rid of any extra numbers (for example in our previous case of $ 1$, $ 4$, $ 7$, we say that $ 7\minus{}1\equal{}6$, which is a factor of our constant number). Dividing this by the constant number will give us our number of additional terms. We must then remember the first term in our sequence, so we add an additional $ 1$ to it.[/hide]

[hide="Generalization with algebra"] In our arithmetic sequence, let the first term be $ a$, and the common difference (same as our constant integer) be $ d$, and let the number of terms in the sequence be $ n$. Thus, the terms in our sequence are $ a$, $ a\plus{}d$, $ a\plus{}2d$, $ a\plus{}3d$... $ a\plus{}(n\minus{}1)d$. Thus, the last term is the first term $ a$ plus the common difference $ d$ multiplid by the number of terms minus $ 1$. Subtracting the first term from the last and dividing by $ d$, we have

$ \frac{a\plus{}d(n\minus{}1)\minus{}a}{d}\equal{}\frac{d(n\minus{}1)}{d}\equal{}n\minus{}1$.

Adding $ 1$, we have $ n\minus{}1\plus{}1\equal{}n$. Thus, we have the general strategy built.[/hide]

If either part confuses you, look at the other. If both parts confuse you, post here and I'll try to be more clear.

A simpler way to do the problem:
Add each term by $ 4$.  (Note that we are not changing the number of terms in the sequence!)

$ 0, 3, 6, \cdots, 36$.

Therefore, there are $ \boxed{13}$ numbers.

You could also add $ 3$ and get $ 3, 6, 9, \cdots 39$.  Therefore, you have $ 3 \cdot 13\equal{} 39$.

Wait, for the basic showing, why do you subtract the last number from the first one? How do you know whether or not the result will be divisible by the difference? What if it's not?

It has to be divisible by the difference. Try it! This method works because the number of numbers in these lists are all equal:
a, d+a, 2d+a,... (n-1)d+a
0, d, 2d,... (n-1)d (subtract "a" from each of the previous sequence's terms)
0, 1, 2,... n-1 (divide each term by "d")
1, 2, 3,... n (add 1)
Subtracting the first term from the last in the first sequence gives (n-1)d, dividing by d gives n-1, and adding 1 gives n, which is the right answer.


Θα μπορούσε να φανταστεί κανείς ότι αν οι όροι μίας ακολουθίας τείνουν "αρκετά γρήγορα" στο 0, τότε η αντίστοιχη σειρά συγκλίνει. Οπότε, ένα φυσικό ερώτημα είναι: τι σημαίνει αυτό το "αρκετά γρήγορα"? Υπάρχει κάποιου είδος "σύνορο" ανάμεσα στις συγκλίνουσες και στις αποκλίνουσες σειρές?

Ένας λογικός τρόπος να ορισθεί αυτό το "σύνορο" θα ήταν μία ακολουθία $ a_n>0$ τέτοια ώστε κάθε σειρά της μορφής $ \sum a_n |b_n|$ με φραγμένη $ b_n$ (δηλ. που πεθαίνει πιο γρήγορα από την $ a_n$) να συγκλίνει ενώ, αν η $ b_n$ δεν είναι φραγμένη (οπότε η $ a_n |b_n|$ πεθαίνει πιο αργά) η σειρά να αποκλίνει.

Δείξτε ότι δεν υπάρχει ακολουθία $ \{a_n\}$ ώστε η σειρά $ \sum a_n |b_n|$ να συγκλίνει αν και μόνο αν η $ b_n$ είναι φραγμένη.

Cheerio,

Durandal 1707

Αρκεί να δείξετε ότι αν $ \sum a_n$ συγκλίνει, τότε υπάρχει μη φραγμένη ακολουθία $ b_n$ ώστε $ \sum a_n b_n$ αποκλείνει.

[hide]
Υπάρχει $ (|b_n|)$ της μορφής 0,...,0,1,0,...,0,2,0,...,0,3,.... που δουλεύει.
[/hide]

[quote="Demetres"]Αρκεί να δείξετε ότι αν $ \sum a_n$ συγκλίνει, τότε υπάρχει μη φραγμένη ακολουθία $ b_n$ ώστε $ \sum a_n b_n$ αποκλείνει.[/quote]

Πολύ ωραία. Φαντάζομαι εννοείς "μη φραγμένη $ b_n$ ώστε $ \sum a_n b_n$ [u]συγκλίνει[/u]," ε;

[quote="Demetres"]Αρκεί να δείξετε ότι αν $ \sum a_n$ συγκλίνει, τότε υπάρχει μη φραγμένη ακολουθία $ b_n$ ώστε $ \sum a_n b_n$ αποκλείνει.[/quote]

Κι εγώ "συγκλίνει" καταλαβαίνω ότι μάλλον ήθελες να πεις εδώ: αν έχεις το σύνορο, θέλεις να βρεις κάτι πιο αργό από το σύνορο που να εξακολουθεί να συγκλίνει.

Excellent :) Αν δεν ασχοληθεί κάποιος άλλος να βάλει λύση μέσα σε μερικές μέρες, βάλε τη δική σου (ή βάζω εγώ τη δική μου) για να το έχουμε πλήρως απαντημένο.

Good job :clap2:

Durandal 1707

PS: Τέλικα, μάλλον όλο ευκολάκια βάζω για εσάς, για να ξεθάψω κάτι πιο καλό... ;)

Ναι συγκλίνει ήθελα να πω.

[quote="Demetres"]Για να κλείνουμε αυτό το topic.

Θέλω να δείξω ότι αν $ \sum a_n$ συγκλίνει, τότε μπορώ να βρώ μη φραγμένη ακολουθία $ b_n$ ώστε και το $ \sum a_nb_n$ να συγκλίνει.

Αφόυ η  $ \sum a_n$ συγκλίνει, υπάρχει $ n_1$, ώστε $ |a_{n_1}| < 1$. Θέτω $ b_{n_1} \equal{} 1$. Αφόυ η  $ \sum a_n$ συγκλίνει, υπάρχει $ n_2 > n_1$, ώστε $ |a_{n_2}| < 1/8$. Θέτω $ b_{n_2} \equal{} 2$. Με τον ίδιο τρόπο βρίσκω $ n_1 < n_2 < n_3 < \ldots$ ώστε για κάθε $ k$, $ |a_{n_k}| < 1/k^3$, και θέτω $ b_i \equal{} k$ αν $ i \equal{} n_k$ για κάποιο $ k$, αλλιώς θέτω $ b_i \equal{} 0$. Τότε, $ |\sum a_nb_n| \leq \sum |a_nb_n| < \sum 1/k^2 < \infty$.[/quote]

Ah, καλά έκανες και έγραψες πλήρως τη λύση... μετά την απάντησή σου ξεχάστηκα εντελώς να βάλω λύση ;)

Cheerio,

Durandal 1707

PS: Είχα διαβάσει κάπου ότι το μόνο σύνορο που χωρίζει τη μαθηματική κοινότητα είναι το ποιά είναι η σειρά που χρησιμοποιεί κανείς πιο πολύ όταν θέλει να κατασκευάσει παραδείγματα σύγκλισης, την $ \frac {1}{k^2}$ ή την $ \frac {1}{2^k}$. Εγώ είχα τη δεύτερη ;)


limit
Functional Analysis
advanced fields
advanced fields unsolved

Let $ A, B$ be positive self-adjoint invertible operators on a Hilbert space ($ A,B \geq 0, \ A^{\minus{}1}, B^{\minus{}1}$ exist). Prove that $ I\plus{}AB$ is invertible as well.

It is equivalent to show that $ A^{ \minus{} 1} \plus{} B$ is invertible. WLOG we can replace $ A$ by $ A^{ \minus{} 1}$ such that it suffices to show that $ C: \equal{} A \plus{} B$ is invertible on the Hilbert space $ H$.

For this purpose the crucial relation is
\[ \langle x,Cx\rangle\ge m\|x\|^2
\]
for all $ x\in H$ and some constant $ m > 0$ independent of $ x$.[hide="Proof"]By contradiction assume there is a sequence $ \{x_n\}$ in $ H$ with $ \|x_n\| \equal{} 1$ and $ \lim_{n\to\infty}\langle x_n,Cx_n\rangle \equal{} 0$. Since $ B \ge 0$ $ \lim_{n\to\infty}\langle x_n,Ax_n\rangle \equal{} 0$. Using a square root of $ A$ this implies $ Ax_n\to 0$, hence $ x_n\to 0$ which yields a contradiction.[/hide]
The relation easily implies that the range of $ C$ is closed. Furthermore its orthogonal complement is trivial, hence $ C$ is surjective. That $ C$ is injective is easy to see by inspection of the implication $ \langle x,Ax\rangle \equal{} 0\Rightarrow x \equal{} 0$.

Note that the assumption that $ B$ is invertible is not used. The proof smells like the Lax-Milgram theorem.


How would I go about solving $4^{x}+3^{y}= 3211$?  I know the solutions are $7$ and $5$ respectively, but I can't seem to get them by solving it through modular means.

(I read a tutorial on solving equations like this by dealing with the whole thing $\mod 3$ or $\mod 4$, but I can't seem to get it to work with this.)

Don't you mean 5 and 7, respectively??  :wink:
Using $\mod4$, we know that $y$ must be odd.
Then, using logarithms, we can see that $x\le5$ and $y\le7$.
So, just plug in and check the 4 possible values of $y$.

Well, first you need to check the trivial cases $x = 0$ and $y = 0$.


algebra
polynomial
factorial
calculus
integration
algebra proposed

A non-zero polynomial $ S\in\mathbb{R}[X,Y]$ is called homogeneous of degree $ d$ if there is a positive integer $ d$ so that $ S(\lambda x,\lambda y)\equal{}\lambda^dS(x,y)$ for any $ \lambda\in\mathbb{R}$. Let $ P,Q\in\mathbb{R}[X,Y]$ so that $ Q$ is homogeneous and $ P$ divides $ Q$ (that is, $ P|Q$). Prove that $ P$ is homogeneous too.

$ \mathbb{R}[X,Y]$ is a [url=http://en.wikipedia.org/wiki/Factorial_ring]factorial ring[/url].
So we find $ p_1(X,Y),p_2(X,Y),\ldots,p_n(X,Y)\in\mathbb{R}[X,Y]$ irreducible with
(*) $ Q(X,Y) \equal{} p_1(X,Y)p_2(X,Y)\cdots p_n(X,Y)$.

Then for $ d: \equal{} \deg Q(X,Y)$ and $ d_k: \equal{} \deg p_k(X,Y),k \equal{} 1,\ldots,n$ we have $ d \equal{} d_1 \plus{} \ldots \plus{} d_n$ and
$ Q(X,Y) \equal{} Y^dQ({X\over Y},1) \\
\hspace*{0.5in} \equal{} Y^dp_1({X\over Y},1)p_2({X\over Y},1)\cdots p_n({X\over Y},1) \\
\hspace*{0.5in} \equal{} q_1(X,Y)q_2(X,Y)\ldots q_n(X,Y)$
with $ q_k(X,Y): \equal{} Y^{d_k}p_k({X\over Y},1)\in\mathbb{R}[X,Y]$ homogeneous not constant for $ k \equal{} 1,\ldots,n$.
But (*) is up to order and scalar multiples the only factorisation of $ Q(X,Y)$ in $ n$ not constant factors.
So for every $ k \equal{} 1,\ldots,n$ we find $ i \equal{} 1,\ldots,n$ and $ \lambda\in\mathbb{R}^*$ with $ p_k(X,Y) \equal{} \lambda q_i(X,Y)$ homogeneous.
So all irreducible factors of $ Q(X,Y)$ are homogeneous, and so are all other factors.

Isn't the problem trivial anyway? If $ p$ and $ q$ are two polynomials which are not both homogeneous, then $ pq$ cannot be homogeneous either; this holds in any $ R\left[X_1,X_2,...,X_n\right]$ with $ R$ being an integral domain. The simple reason is that the minimal degree of a monomial in $ pq$ is the sum of the minimal degree of a monomial in $ p$ and the minimal degree of a monomial in $ q$, and same holds for the maximal degrees - thus, if at least one of the polynomials $ p$ and $ q$ is not homogeneous, it has monomials of different degrees, and so must $ pq$.

  darij


inequalities
conics
ellipse
function
analytic geometry
parabola
calculus

Algebra 2 is a major turning point for a lot of people. I think that the problems get a lot harder at the Algebra 2 level. I would like to start an ALGEBRA 2 Marathon. Not only will this help with all the tests and finals at High School, but it also helps problem-solving at a harder level.  :)

When posting a problem, one should represent it with a number 1 through 5. 1 being easy and 5 being hard.
Note: Whenever one solves the previous problem, he/she should post a new one.

I hope this works out smoothly!

[hide="First Problem (2)"]Find the solutions to $ \frac {x^2 \minus{} 1}{4x \minus{} 2} \ge 3$.[/hide]

hello, our inequality is equivalent with
$ \frac{x^2\minus{}12x\plus{}5}{2(\minus{}1\plus{}2x)}\geq0$
in the first case we get
$ x>\frac{1}{2}$ and $ x\geq6\plus{}\sqrt{31}$, from here we have $ x\geq6\plus{}\sqrt{31}$
in the second case we get
$ x<\frac{1}{2}$ and $ 6\minus{}\sqrt{31}\le x \le 6\plus{}\sqrt{31}$, so we find
the solution set as
$ 6\minus{}\sqrt{31}\le x <\frac{1}{2}$ or $ x\geq 6\plus{}\sqrt{31}$.
Sonnhard.

Dr. Sonnhard, in the future, could you post a new problem after you solve the previous one? THANKS!

[hide="New Problem (2)"]Find the foci of the ellipse $ \frac {(x \minus{} 1)^2}{4} \plus{} \frac {(y \plus{} 2)^2}{9} \equal{} 36$. [/hide]

hello, there is a little typo in your equation, it must be
$ \frac{(x\minus{}1)^2}{4}\plus{}\frac{(y\plus{}2)^2}{9}\equal{}36$
My problem will comming soon.
Sonnhard.

Yes, sorry, I made an edit. :wink:

[hide="The Problem (2)"]Find the foci of the ellipse $ \frac {(x \minus{} 1)^2}{4} \plus{} \frac {(y \plus{} 2)^2}{9} \equal{} 36$.[/hide]

hello, can i also post a new problem right now?
Sonnhard.

:)  Marathons are when someone solves the previous problem then posts a new one of their own.

You are welcome to post a new problem, but you should solve the previous one first.  :lol:

[hide="Solution for 2"]Center: $ (1,\minus{}2)$. $ c^2\equal{}a^2\minus{}b^2\equal{}9\minus{}4\equal{}5$
$ c\equal{}\sqrt{5}$ and the ellipse is lying its major axis vertically so the foci are at $ (1,\minus{}2\pm\sqrt{5})$[/hide]

[hide="New problem"]Factor completely $ x^3\plus{}y^3\plus{}z^3\minus{}(x\plus{}y\plus{}z)^3$[/hide]

Doesn't the other side have to be $ 1$??  Or can can you

[hide="Solution to teh ellipse problem since the guy above me just posted a few seconds before me"]

First, the parent function of a ellipse is equal to $ 1$, we we must divide by  $ 36$ to achieve this:  $ \frac {(x \minus{} 1)^2}{\frac19} \plus{} \frac {(y \plus{} 2)^2}{\frac14} \equal{} 1$ .  The vertex is $ (1, \minus{} 2)$.  The distance from the vertex to the foci is $ c$ in $ a^2 \plus{} b^2 \equal{} c^2$, where $ a \equal{} \frac12$ and $ b \equal{} \frac13$.  $ c$ then equals $ \frac {\sqrt {30}}{6}$, and because our ellipse is "thin", the coordinates of the foci are $ (1 , \minus{} 2\pm \frac {\sqrt {30}}{6})$.[/hide]

I don't know if I have solved this right, in my AII class all of our denominators are natural numbers...

Can someone tell me if I'm right before solving this new problem?

[hide="NP"]

What is the eccentricity of the ellipse $ \frac {(x \minus{} 1)^{2}}{4} \plus{} \frac {(y \plus{} 2)^{2}}{9} \equal{} 36$
[/hide]

Yeah, I know.  I'm lazy.

[hide="SOLUTION I think"]The eccentricity $ \frac{c}{a}\equal{}\frac{\sqrt5}{3}$.[/hide]

@007math: Yes, I think your solution is correct.

For New Problem: See ckck's 2 posts above.

hello, this problem was often to be found at mathlinks
$ x^3\plus{}y^3\plus{}z^3\minus{}(x\plus{}y\plus{}z)^3\equal{}\minus{}3(x\plus{}y)(x\plus{}z)(y\plus{}z)$.
Sonnhard.
New Problem:
Find all natural numbers $ n$ such that the following inequality is fulfilled:
$ 99^n\plus{}101^n>\frac{51}{25}\cdot100^n$

[quote="Dr Sonnhard Graubner"]hello, this problem was often to be found at mathlinks
$ x^3 \plus{} y^3 \plus{} z^3 \minus{} (x \plus{} y \plus{} z)^3 \equal{} \minus{} 3(x \plus{} y)(x \plus{} z)(y \plus{} z)$.
Sonnhard.
New Problem:
Find all natural numbers $ n$ such that the following inequality is fulfilled:
$ 99^n \plus{} 101^n > \frac {51}{25}\cdot100^n$[/quote]
I was hoping for the work to be shown not just the answer....:|

hello, here is my solution,simplifying
$ x^3\plus{}y^3\plus{}z^3\minus{}(x\plus{}y\plus{}z)^3\equal{}$
$ \minus{}3x^2y\minus{}3x^2z\minus{}3xy^2\minus{}3xz^2\minus{}3y^2z\minus{}3yz^2\minus{}6xyz\equal{}$
$ \minus{}3x^2(y\plus{}z)\minus{}3yz(y\plus{}z)\minus{}3x(y^2\plus{}z^2\plus{}2yz)\equal{}$
$ (y\plus{}z)(\minus{}3x^2\minus{}3yz\minus{}3x(y\plus{}z))\equal{}$
$ (y\plus{}z)3(\minus{}x^2\minus{}yz\minus{}xy\minus{}xz)\equal{}$
$ \minus{}3(y\plus{}z)(x(x\plus{}y)\plus{}z(x\plus{}y))\equal{}$
$ \minus{}3(y\plus{}z)(x\plus{}y)(x\plus{}z)$
Sonnhard.

[hide="Same New Problem"]Find all natural numbers $ n$ such that the following inequality is fulfilled.

$ 99^{n}\plus{}101^{n}>\frac{51}{25}\cdot100^{n}$[/hide]

[hide="Problem K-16(4)"] (Inspiration/idea/problem format: Art of Problem Solving's [u]Intermediate Algebra[/u]) The roots of $x^3 - (k + 3)x^2 + 126x - 216$ are in geometric progression. The roots of $x^3 - kx^2 + 99x - 162$ are in arithmetic progression. These two polynomials share two roots. Find $k$.[/hide]

[hide="Solution K-16(4)"] Using Vieta's and the formula for progressions, we have: \[ \frac{n}{a} \times n \times (n \cdot a) = 216 \implies n = 6 \] \[ 6\cdot \frac{6}{a}+\frac{6}{a}\cdot 6a + 6 \cdot 6a = 126 \implies a = 2^{\pm 1} \] \[ k+3 = 12 + 6 + 3 \implies k=18 \] \[ (b-d) + b + (b+d) = k = 18 \implies b=6 \] \[ d = \{6,3\}: d=3 \implies 3\times6\times9 =  162 \] \[ \boxed{k=18} \] [/hide]

Thanks for reviving a great marathon. Unfortunately I have no new problem, so anyone should feel free to post one.

Here is a fun one!

[hide="Problem K-17(4)"] (Inspiration: Art of Problem Solving's [u]Intermediate Algebra[/u]) If $x$ is an integer, find $x$ if

$\log_7{(x + 2)} + \log_7{(100x - 2299)} = 2x - 88$[/hide]

We see that x must be greater than 44 because for all integer inputs of x $(x+2)(100x-2299)$ will be a integer. We then note that the result must be divisible be a perfect square. We also know that after only a few values of x, the exponential will far succeed the power function. We plug in values of x in $x+2$ which makes it a power of 7. The nearest value is 47, which works(you check this by plugging 47 in the second part of equation, and you see it is indeed a power of 7). Going on, the next will be in the 300s, which by that time, the exponential has far surpassed the power function.

The equation $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three real roots. Given that their sum is $\frac mn$ where $m$ and $n$ are relatively prime positive integers, find $m+n$.

Difficulty(3)

[hide="solution"]Let $y=2^{111x}\implies x=\frac{1}{111}\log_2 y$ to get $y^3-8y^2+16y-4=0$. The sum of the original solutions is $x_1+x_2+x_3=\frac{1}{111}\log_2 y_1y_2y_3=\frac{1}{111}\log_2 4=\frac{2}{111}$ from Vietas. The answer is $\boxed{113}$.[/hide]

EDIT :oops: I misread and thought I had to find the exact roots

You have to take the sum of the $logs$ of the solutions giving $\frac{1}{111}2=\frac{2}{111}$

You didn't consider the right stuff \[ y=2^{111x} \implies y^3 - 8y^2 + 16y - 4 \] \[ \sum r_x = \frac{1}{111} \log_2\left({\prod r_y}\right) = \frac{2}{111} \] I'm not sure I correctly did this, but it seems right.

sorry  :(  edited

For k17, we could have simplified work a bit: [hide]Rewrite as $\log_{49} (x+2)(100x-2299)=x-44$ to see that $x+2$ and $100x-2299$ are powers of $49$. Then $x+2\equiv 0\pmod{49}$ and $100x-2299\equiv 0\pmod{49}$, both giving $x\equiv 47\pmod{49}$. And the answer comes quickly from here.[/hide]

[hide="next problem"]If $x_1,x_2,x_3$ are the roots of $p(x)=x^3-2x+5$, then find a polynomial with roots $x_i^2+x_i$.[/hide]

For $K-17$, since $(x+2)(100x-2299)=7^{2x-288}$, we just note that the $gcd$ of the two terms in the factorization divides $2499=7^2(51)$. Checking $x=-1,5,47$, we see that only $47$ works.

We prove that the polynomial $P(x) = x^3 - 4x^2 +17x - 30$ works.

Suppose $x$ is a root of $x^3 - 2x + 5$. We have to prove $ P(x^2 +x) = 0$. This is equivalent to \[(x^2+x)^3 - 4(x^2+x)^2 + 17(x^2+x) - 30 = 0\] which, after expanding is \[ x^6 +3x^5 -x^4-7x^3 + 13x^2+17x - 30 = 0\]

Now we use that $x^3 - 2x + 5 = 0 => x^3 = 2x - 5$ from which we can get several relations such as

1) $x^6 = 2x^4 - 5x^3$

2) $3x^5 = 6x^3 - 15 x^2$

3) $x^6 + 3x^5 = 2x^4+ x^3 - 15x^2$

So we are left to prove that \[  x^4 - 6x^3 - 2x^2 + 17x - 30 = 0\]

Also we see that $x^4 = 2x^2 - 5x$ so we just need to prove \[ -6x^3 +12 x - 30 = 0 \leftrightarrow x^3 -2x + 5 = 0\] which is obviously true.

To get the polynomial I used vieta to find relations between the roots $x_1,x_2,x_3$ of $x^3 - 2x +5$. After that, if $a_i = x_i^2 +x_i$, I used the relations to calculate $a_1+a_2+a_3$, $a_1a_2+a_2a_3+a_1a_3$ and $a_1a_2a_3$. After getting those values I use vieta back to get a polynomial with roots $a_1,a_2,a_3$

Someone please post the next problem.

Yes, but in the first step, it appears that you use your answer to prove it. How did you find it in the first place? Did you use variables for the coefficients and then proceed?

No, I did not use my answer to prove it. I wrote what we had to prove and then showed it is equivalent to something true.

Also, I explained how did I get it, read carefully

[quote="hatchguy"] To get the polynomial I used vieta to find relations between the roots $x_1,x_2,x_3$ of $x^3 - 2x +5$. After that, if $a_i = x_i^2 +x_i$, I used the relations to calculate $a_1+a_2+a_3$, $a_1a_2+a_2a_3+a_1a_3$ and $a_1a_2a_3$. After getting those values I use vieta back to get a polynomial with roots $a_1,a_2,a_3$[/quote]

Here's the next problem
[hide="Problem"]
If a Hula-Hoop was made in the shape of an ellipse and a marble is rolled from one of the foci, it will hit the other focus after 0-3 bounce(s).

A hoop is made with a "horizontal diameter" of 1 meter and a "vertical diameter" of 5 decimeters. A marble is rolled from one focus and hits the other without bouncing off any of the hula hoop. How far did the marble travel?
[/hide]

EDIT: 10 decimeters = 1 meter

[hide="hint"]
For the marble not to bounce off the hula hoop, the marble must have traveled in a straight line from 1 focus to another.
[/hide]

Unless I didn't understand this question right. What we are asked to find is the distance from the center to foci, which is 2sqrt6. Doubling this, we get 4sqrt2. (Unless I misunderstood this question, this isn't really intermediate level)

Let  be a polynomial with real coefficients such that $ f(0) = 1, $ $ f(2)+f(3)=125, $ and for all $ x $ $ f(x)f(2x^{2})=f(2x^{3}+x). $. Find $ f(5). $


quadratics
algorithm
superior algebra
superior algebra unsolved

Let p be an odd prime, give a fast algorithm for determining for any x in Z_(p^2) whether or not there exists y in Z_(p^2) with y^2 = x (mod p^2).

Now assume that p = 3 (mod 4). Give a fast algorithm when given x in Z_p determines whether or not the equation y^4 = x (mod p) has any solutions, and when there are solutions find them all.

a) for those coprime to $p$, it's enough to be a quadratic residue $\mod p$; for the other residues, it's trivial.

b) you have to check for $x$ being a quadratic residue only. Then use any algorithms that produces a square root $\mod p$.

i don't quite understand why it is trivial for other residues in part (a)... and for part b, what if x is not a quadratic residue mod p?
if x is a quadratic residue, we take the square root of y^4 = x (mod p) to get y^2 = sqrt (x) (mod p), then find all the x's?


calculus trigonometry function derivative inequalities logarithms search

Solve the equation $ 2^{x}\plus{}2^{\sqrt{1\minus{}x^{2}}}\equal{} 3$, where $ |x|\le 1$.

[hide="Possible solution with calculus"]Let $ x \equal{}\cos\theta$ for $ 0\leq\theta\leq\pi$. Then we will solve for $ 2^{\cos\theta}\plus{}2^{\sin\theta}\equal{} 3$. We see $ \theta \equal{} 0$ or $ \theta \equal{}\pi/2$ are solutions. If $ \theta >\pi/2$, then $ 2^{\cos\theta}< 1$ and $ 2^{\sin\theta}< 2$, so we can't have solutions in that interval. It remains to check $ 0 <\theta <\pi/2$.

Let the LHS be $ f(\theta)$. Notice that $ f(\theta) \equal{} f(\pi/2\minus{}\theta)$. Since $ f(0) \equal{} 3$, if we show that $ f(\theta)$ is increasing on $ [0,\pi/4]$, we're done. Time for some calculus. $ f^{\prime}(\theta) > 0$ is equivalent to $ 2^{\sin\theta\minus{}\cos\theta}>\tan\theta$. Both of these functions are convex, the LHS > RHS at $ \theta \equal{} 0$, and the derivative of the RHS at $ \theta \equal{}\pi/4$ is greater than the derivative of the LHS. I am pretty sure some theorem in calculus allows me to conclude that the inequality is true from these facts, unless my intuition is misleading me.

So $ \theta \equal{} 0,\pi/4$ are the only solutions, thus $ x \equal{} 0$ or $ x \equal{} 1$ for the original equation.[/hide]

[hide="Another calculus approach"]For $ \minus{}1\leqslant x < 0$ we have $ 2^{x}< 1\land 2^{\sqrt{1\minus{}x^{2}}}< 2\implies 2^{x}\plus{}2^{\sqrt{1\minus{}x^{2}}}< 3$, hence there's no solution there.

Obviously, $ x \equal{} 0$ and $ x \equal{} 1$ are solutions.

For $ 0 < x < 1$ we can change the equation into $ 2^{\sqrt{t}}\plus{}2^{\sqrt{1\minus{}t}}\equal{} 3$ for $ 0 < t < 1$. By taking second derivative of $ f(t)\equal{}2^{\sqrt{t}}$ (which is $ f''(t) \equal{}\frac{2^{\sqrt{t}}\ln 2}{4t\sqrt{t}}(\sqrt{t}\ln 2\minus{}1)$, if I'm not mistaken), we can show that $ f(t)$ is concave on $ (0,1)$, hence $ 2^{\sqrt{t}}> 1\plus{}t$ (the RHS is the secant equation), and consequently $ 2^{\sqrt{1\minus{}t}}> 1\plus{}(1\minus{}t) \equal{} 2\minus{}t$, hence $ 2^{\sqrt{t}}\plus{}2^{\sqrt{1\minus{}t}}> 3$, so there's no solution there.

Therefore, the only solutions are $ x \equal{} 0$ and $ x \equal{} 1$.[/hide]

Virgil, is elementary calculus allowed in solving this problem?

[color=darkblue]Here is a very nice and elementary (without calculus) proof of this equation.

Suppose $ x\in (0,1)$. Prove easily that $ \{\begin{array}{c}2(x-1)+\sqrt{1-x^{2}}>0\Longleftrightarrow x\in(\frac{3}{5},1)\ .\\\\ x+2(\sqrt{1-x^{2}}-1)>0\Longleftrightarrow x\in(0,\frac{4}{5})\ .\end{array}$

Thus, $ 2^{x}+2^{\sqrt{1-x^{2}}}=\{\begin{array}{c}2\cdot 2^{x-1}+2^{\sqrt{1-x^{2}}}\ge 3\cdot\sqrt [3]{2^{2(x-1)+\sqrt{1-x^{2}}}}>3\mathrm{\ for\ any\ }x\in(\frac{3}{5},1)\ .\\\\ 2^{x}+2\cdot 2^{\sqrt{1-x^{2}}-1}\ge 3\cdot\sqrt [3]{2^{x+2(\sqrt{1-x^{2}}-1)}}>3\mathrm{\ for\ any\ }x\in(0,\frac{4}{5})\ .\end{array}$

In conclusion, for any $ x\in (0,1)=(0,\frac{4}{5})\cup(\frac{3}{5},1)$ we have $ 2^{x}+2^{\sqrt{1-x^{2}}}>3$ a.s.o.[/color]

[quote="Farenhajt"]Virgil, is elementary calculus [b]allowed[/b] in solving this problem?[/quote]
Yes, it is allowed, but the problem was given to 10th grade students and it was unfair for those who didn't know calculus.

For an elementary solution, see harazi's post [url=http://www.mathlinks.ro/viewtopic.php?search_id=792149452&t=28960]here[/url].
[quote="harazi"]Suppose $ t$ is not $ 0$ or $ 1$. If $ 5t > 3$ then apply the AM-GM inequality with the three terms $ 2^{t\minus{}1},2^{t\minus{}1}, 2^{\sqrt{1\minus{}t^{2}}}$ and prove next that $ 2(t\minus{}1)\plus{}\sqrt{1\minus{}t^{2}}$ is greater than 0. If $ 5t$ is at most 3, apply the Am-GM for the numbers $ 2^{t},2^{\sqrt{1\minus{}t^{2}}\minus{}1}, 2^{\sqrt{1\minus{}t^{2}}\minus{}1}$ and prove next that $ t\plus{}2(\sqrt{1\minus{}t^{2}}\minus{}1)$ is greater than 0. That's all.[/quote]
I think this was one of the first problems I posted here on ML.
I found out about it in class, and the solution I knew from my teacher was the same as the above.
(It's very clever if you ask me.)

[quote="perfect_radio"]
Yes, it is allowed, but the problem was given to [b]10th grade students[/b] and it was unfair for those who didn't know calculus.[/quote]

The level was exact point of my question. Obviously, calculus should be avoided, in spite of being allowed. And I agree the solution is quite clever :)

[color=darkred]Another similar exercises :

$ 1\blacktriangleright 4^{x}\plus{}2 \equal{} 3\cdot 2^{x^{2}}\ .$

$ 2\blacktriangleright 2^{x^{2}}\plus{}2^{(x\minus{}1)^{2}}\equal{} 3\cdot p^{x(x\minus{}1)}$, where $ p\ge 4\ .$

[b]Remark.[/b] In the Romanian review [b]"Arhimede"[/b] I published some years ago an article with this equation as title.[/color]


quadratics
pigeonhole principle
inequalities
number theory proposed
number theory

let S = {x^2 + 3* y^2; x,y in Z}
show that if a,b are members of S such that a/b is an integer, then a/b is a member of S.

[hide]

We shall first show the following lemmas:

Lemma 1: If a1 belongs in S, and a2 (not necessairly distinct from a1) belongs in s, then so does a1*a2.

Lemma 2: if p is a prime of the form 6n-1 that divides any non-zero member of s, then the highest power of p in that member, is even.

Lemma 3: for any prime p that is 0 or 1 mod 3, there exist x,y in Z such that x^2 + 3y^2 = p.

Note that any member of S is either 0 or 1 mod 3.

Proof for Lemma 1:

Let a1 = w^2 + 3x^2, and a2 = y^2 + 3z^2, for w,x,y,z in Z. Then a1*a2 = (w^2 + 3x^2)*(y^2 + 3z^2), we expand this product, then add and subtract 6wxyz, and rewrite this as (wy+3xz)^2 + 3(wz-xy)^2. Therefore a1*a2 belongs in S.

Proof for Lemma 2:

Note that, we will have to use the following fact: for all primes of the form 6n+1, -3 is a quadratic residue. For any primes of the form 6n-1, -3 is not a quadratic residue. <to cite the source http://mathworld.wolfram.com/QuadraticResidue.html> . These guys are generally correct,..always.  Anyway, im pretty sure that you can find a proof for this somewhere.

For the sake of contradiction, let p (of the form 6n-1) divide some non-zero member of S.

Equivalently we have

x^2 + 3y^2 = 0 mod p, for some x,y in Z.

implies -> x^2 = -3y^2 mod p.

Raising both sides to the (p-1)/2 power, we get

x^(p-1) = (-3)^(p-1)/2 * y^(p-1) mod p.

Now the following cases arise:

Case 1 (p doesnt divide x): In that case p doesnt divide y as well.

Recognizing that y^(p-1) = x^(p-1) = 1 mod p, we can reduce the given congruence to

1 = (-3)^(p-1)/2 mod p. However, since -3 is Not a quadratic residue of any prime of the form 6n-1, we have a contradiction. Therefore if p doesnt divide x, we cannot have x^2 + 3y^2 = 0 mod p.

Case 2 (p divides x): in that case, p must divide y as well if the congruence is true. More generally, let u and v be the largest powers of p that divide x and y respectively. Then x = x1*p^u and y = y1*p^v. Substituting this for x, and y in the original express, we get x1^2*(p^2u) + 3y1^2*(p^2v). Now, one can easily see that the largest power of p that divide x^2 + 3y^2 has to be even, and we are done.

Lemma 3: The case p = 3 is trivial (x=0, and y = 1 work). So lets prove it generally for all primes that p that is 1 mod 3. Once again, we will use that fact that there is an x, such that x^2 = -3 mod p.

Let Q = floor(sqrt(p)), Let, h and i be integers such that 0<=h<=Q and 0<=i<=Q. Note that Q<sqrt(p) < Q+1.

Define f(h,i) = h + ix. Let us analyze f(h,i) modulo p, as (u,v) runs over the (Q+1)^2 pairs.

It follows by the pigeonhole principle that that there exist at least two distinct pairs, (u1,v1) and (u2,v2) such that f(u1,v1) = f(u2,v2) mod p, u1 + v1x = u2 + v2x mod p.

Or equivalently:

(u1-u2) = -x(v2-v1) mod p. Let a = |u1-u2| and b = |v2-v1|. Since, the two pairs were distinct,  it follows that not both a, and b can be zero. Furthemore it follows that a and b <= Q.

a = -x mod p. we square both sides to get a^2 = -3b^2 mod p, or a^2 + 3b^2 = 0 mod p.

Now we are given that a,b <= Q, which introduces the following inequality:

0<=a^2 + 3b^2 < 4p.

The following case arises:

Case a^2 + 3b^2 = 0: Since not both a and b is zero, such a case cannot exist, contradiction.

Case a^2 + 3b^2 = p: We are done.

Case a^2 + 3b^2 = 2p. Let us look at both size mod 3, it follows that the LHS can only be 0 or 1 mod 3, but the RHS is always 2 mod 3, contradiction.

Case a^2 + 3b^2 = 3p. it Follows that 3 must divide a; this implies that there are integers a1( = a/3) and b such that b^2 + 3(a1)^2 = p, and we are done.

Now, any member x of S is either 0 or 1 mod 3, we know that then the x must have prime factors of the form 0 or 1 mod 3,and Primes of 2 mod 3 that whose highest power in x is even. So for any x,y in S such that x|y, then y/x can be written as a number that has prime factors of the form 0 or 1 mod 3; and primes of the form 2 mod 3, whose highest power is even.. We know by lemma 3 that any prime that is 0 or 1 mod 3 must exist in S. It is also not hard to see that any Number of the form (p)^(2a) (where p is a prime of the form 6n-1, and a is non-negative integer) belongs in S as well (any perfect squares belong in S). Hence through the use of Lemma 2 and 3, we can see that y/x can be written as a product of elements of S. Thus by repeated use of lemma 1, y/x must belong in S. We are done.

Note: I might have some missed spots in my proof, Im not much of a proof writer, so this is the best I can doIt was a great problem, had great times in English, Econ, and Physics doing it .

[/hide]

This is a very nice result  :) used in the proof of Fermat theorem for n=3 :

http://www.mathpages.com/home/kmath009.htm


trigonometry
geometry
circumcircle
geometric transformation
reflection
angle bisector
perpendicular bisector

In a triangle $ABC$, the angle bisector of the angle $ACB$ crosses the line $AB$ in a point $D$. The perpendicular bisector of the segment $CD$ crosses the line line $AB$ in a point $E$. Prove that : \[\frac{EA}{EB}=\frac{AC^{2}}{BC^{2}}\]

Let the perp. bisector of $CD$ in $F$. We have that $\angle BEF=1/2 (\alpha-\beta)$ (if $BC>AC$, the other casses are similar). So $\angle DEC=\alpha-\beta$, and $\angle ACE=\beta$. Form $\triangle EAC$ we deduce that $EA=\frac{AC \cdot \sin{ \beta}}{\sin{(\alpha-\beta)}}$ and from triangle $EBC$ we have $EB=\frac{BC \sin{\alpha}}{\sin{(\alpha-\beta)}}$. Now it is $EA/EB=b/a \cdot \sin{\beta}/\sin{\alpha}=b^{2}/a^{2}$, w.r.t. to the sine theorem in the triangle $ABC$.

We use directed angles modulo 180°. Let the tangent to the circumcircle of triangle ABC at the vertex C meet the line AB at a point Z. Then, since the line CZ is the tangent to the circumcircle of triangle ABC at the point C, the tangent-chordal angle theorem yields < (CZ; CA) = < CBA. Since the line CD is the angle bisector of the angle ACB, we have < (CD; CA) = < (BC; CD). Thus,

  < (CD; CZ) = < (CD; CA) - < (CZ; CA) = < (BC; CD) - < CBA
                   = < (BC; CD) - < (BC; AB) = < (AB; CD).

In other words, < DCZ = < ZDC. Thus, triangle CZD is isosceles, so that CZ = DZ, and the point Z lies on the perpendicular bisector of the segment CD. Thus, the point Z is the point of intersection of the perpendicular bisector of the segment CD with the line AB. But as we have defined the point E as the point of intersection of the perpendicular bisector of the segment CD with the line AB, it thus follows that Z = E. Hence, in order to prove the equation $\frac{EA}{EB}=\frac{AC^{2}}{BC^{2}}$, it remains to show that $\frac{ZA}{ZB}=\frac{AC^{2}}{BC^{2}}$, but this is exactly Lemma 2 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=103771]http://www.mathlinks.ro/Forum/viewtopic.php?t=6557 post #5[/url]. So we are done.

  darij

My solution was using similarities.
Abdurashid

Let $CL$ be the external bisector with $L$ on line $AB$, $AD=m$, $BD=n$ and $m<n$. As $CL \perp CD$ $E$ is the circumcenter for $\Delta CLD$. If its radius is $R$ we have $\frac{AL}{BL}=\frac{2R-m}{2R+n}=\frac{m}{n}$. So $R=\frac{mn}{n-m}$. Then $EA=R-m=\frac{m^{2}}{n-m}$ and $EB=R+n=\frac{n^{2}}{n-m}$. $\frac{EA}{EB}=\frac{m^{2}}{n^{2}}=\frac{CA^{2}}{CB^{2}}$.

[color=darkblue][b][u]Lemma 1.[/u][/b] Let $M,N$ be two points which belong to the line $AB$. Denote the middlepoints $S,T$ of the segments $[AB]$, $[MN]$ respectively. Then $\boxed{\frac{MA}{MB}=\frac{NA}{NB}}$ [b](the points $M,N$ are harmonical conjugate w.r.t. the points $A,B$)[/b] $\Longleftrightarrow$ $\boxed{AB^{2}+MN^{2}=4\cdot ST^{2}}$ $\Longleftrightarrow$ $\boxed{AM\cdot AN=AB\cdot AT}$ $\Longleftrightarrow$ $\boxed{TM^{2}=TA\cdot TB}$ $\Longleftrightarrow$ $\boxed{\frac{TA}{TB}=\left(\frac{MA}{MB}\right)^{2}}$ $\Longleftrightarrow$ $\boxed{\frac{2}{AB}=\frac{1}{AM}+\frac{1}{AN}}$ (see http://www.mathlinks.ro/Forum/viewtopic.php?t=46146 ).

[b][u]Lemma 2.[/u][/b] The tangent in the vertex $A$ of the triangle $ABC$ to the its circumcircle meets the opossite sideline $AB$ in the point $T$. Then $\frac{TB}{TC}=\left(\frac{AB}{AC}\right)^{2}\ .$ \[\mathrm{***}\] [b]The first method.[/b] Denote the reflection $F$ of the point $D$ with raport to the point $E$, i.e. $CF\perp CD$. Therefore, the division $(A,B,D,F)$ is harmonically, i.e. $\frac{FA}{FB}=\frac{DA}{DB}$ and the middlepoint $E$ of the segment $[DF]$ has the well-known property (see the first lemma) $\frac{EA}{EB}=\left(\frac{DA}{DB}\right)^{2}$, i.e. $\boxed{\ \frac{EA}{EB}=\left(\frac{CA}{CB}\right)^{2}}\ .$

[b]The second method.[/b] $\widehat{ECA}\equiv\widehat{EBC}$ $\Longrightarrow$ the line $EC$ is tangent to the circumcircle of the triangle $ABC$ $\Longrightarrow$
(see the second lemma) $\frac{EA}{EB}=\left(\frac{CA}{CB}\right)^{2}\ .$[/color]


can you solve this inequality
  a,b,c : real positive  a+b+c=3  then prove  a^2+b^2+c^2 <= 5
by many way
essssspecially by using SOS and limitus

It does not hold.
Choose $a,b$ very close to $0$ and $c$ very close to $3$...

Pierre.

The inequality is not true when a=2.8, b=0.1, c=0.1

i'm very sorry because of my mistake
a,b,c must be in [0,2]
yes i'm sure it is true

If $f(a,b,c) = a^{2}+b^{2}+c^{2}$, then clearly

$f(a,b,c) \le f(a,0,b+c) = f(a,0,3-a) = a^{2}+(3-a)^{2}= 2a^{2}-6a+9$,

which is $2(a-\frac{3}{2})^{2}+\frac{9}{2}$ with the additional constraint $1 \le a \le 2$. Clearly, this maximizes when $a = 1,2$ to get

$f(a,b,c) \le 5$.


geometry
incenter
rectangle
circumcircle
trapezoid
inradius
cyclic quadrilateral

For a cyclic quadrilateral ABCD. E, F, G, and H are the incenters and ra, rb, rc, and rd are the inradii of the triangles ABD, ABC, BCD and ACD respectively. Prove that EFGH is a rectangle and ra + rc = rb + rd.
 :blush: help me please~

For the rectangle, it is enough to show that the angle $ \angle EFG$ is right. The angle bisectors $ BE, BG$ cut  the circumcircle $ (O)$ of $ ABCD$ again at the midpoints $ K, L$ of the arcs $ DA, CD$ opposite to $ B$ and the angle bisectors $ DE, DG$ at the midpoints $ M, N$ of the arcs $ AB, BC$ opposite to $ D.$ So do the angle bisectors $ CF, AF,$ respectively. The incentrs $ E, F$ are on the circle $ (M)$ with radius $ MA \equal{} MB$ and the incenters $ F, G$ on the circle $ (N)$ with radius $ NB \equal{} NC.$

$ 180^\circ \equal{} \angle AFE \plus{} \angle EFG \plus{} \angle GFC \plus{} \angle FCN \equal{}$

$ \equal{} \angle ACE \plus{} \angle EFG \plus{} \angle GBC \plus{} 90^\circ \minus{} \frac{\angle FNC}{2} \equal{}$

$ \equal{} \angle ACK \plus{} \angle EFG \plus{} \angle LBC \plus{} 90^\circ \minus{} \frac{\angle ANC}{2} \equal{}$

$ \equal{} \frac{\angle ACD}{2} \plus{} \angle EFG \plus{} \frac{\angle DAC}{2} \plus{} \frac{180^\circ \minus{} \angle ABC}{2} \equal{}$

$ \equal{} \frac{\angle ACD \plus{} \angle DAC \plus{} \angle CDA}{2} \plus{} \angle EFG \equal{}$

$ \equal{} 90^\circ \plus{} \angle EFG.$

(This was surely posted before.) For $ r_a \plus{} r_c \equal{} r_b \plus{} r_d,$ see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=39268[/url].

[quote]For a cyclic quadrilateral ABCD. R,S,P,Q are the incenters and ra, rb, rc, and rd are the inradii of the triangles ABD, ABC, BCD and ACD respectively. Prove that EFGH is a rectangle and ra + rc = rb + rd.[/quote]
Proof 1: nice proof of yetti
Proof 2:
* For the rectangle,
The lines AS and DP intersectatthemidpoint H of the arc BC on the other side of the circle ABCD. Notethat P and S are both on the circle H(B)= H(C). If K is the midpoint of the arc AD, then HK,being the bisector of angle AHD, is the perpendicular bisector of PS. Similarly, it is also the perpendicular bisector of QR. PQRS is an isosceles trapezium because of symmetry.
The same reasoning also shows that the chord joining the midpoints of the arcs AB and CD is the common perpendicular bisector of PQ and RS. From this, we conclude that PQRS is indeed a rectangle.
* For the ra+rc=rb+rd
If AB and CD are parallel, then each is parallel to HK.In this case, ra = rb and rc = rd. More generally,
$ r_a \minus{} r_b \equal{} PQsin\frac{1}{2}(\angle BDC\minus{}\angle AHD)$
and $ r_d \minus{} r_c \equal{} SRsin\frac{1}{2}(\angle BAC \minus{} \angle AHD)$.
Since $ PQ \equal{} SR$ and $ \angle BDC \equal{} \angle BAC$, it follows that ra −rb = rd −rc, and ra + rc = rb + rd.
The second proof is completed.

Hello,
I'm sorry that i post here so much time after the last post but i have a nice and simple proof of the second sub-question of this exercise.

I'll use $ OI^2 \equal{} R^2 \minus{} 2Rr$ for $ R$ and $ r$ the circumradius and inradius respectively.
Here, the R is common for all triangles so $ OE^2 \equal{} R^2 \minus{} 2Rr_a, OF^2 \equal{} R^2 \minus{} 2Rr_b, OG^2 \equal{} R^2 \minus{} 2Rr_c, OH^2 \equal{} R^2 \minus{} 2Rr_c$.
From these we take that $ r_a \plus{} r_c \equal{} R^2 \minus{} (OE^2 \plus{} OG^2)/2$ and $ r_b \plus{} r_d \equal{} R^2 \minus{} (OF^2 \plus{} OH^2)/2$.
But since EFGH is a rectangle, $ OE^2 \plus{} OG^2 \equal{} OF^2 \plus{} OH^2$, QED

Let X, Y, Z and W be the middles of the arcs AB, BC, CD, DA. Then E and F are on the circle (X,XA), F and G on the circle (Y,YB), G and H on (Z,ZC) and H and A on the circle (W,WD).
Obviously X, E and D, are collinear, same for X, F and C. XZ is the angle bisector of $ \angle CXD$ and triangle EXF is isosceles, hence $ EF\perp ZX$, etc, therefore EFGH is a rectangle, so its diagonals EG and FH are equal, $ EG \equal{} FH$ ( 1 )
Let M and N be the tangency points of incircles of triangles ACD and ABC with AC, then following equality holds:
$ MN \equal{} AM \minus{} AN \equal{} \frac {AD\plus{}BC\minus{}(AB\plus{}CD)}{2}$ ( 2 ). If P and Q are the projections of E and G on BD, we shall get  $ PQ \equal{}MN$ ( 3 ), consequently $ r_a\plus{}r_c\equal{}r_b\plus{}r_d$  , from (1) and (3).
NOTE:! The angle between GE and FH equals the one between AC and BD.

Best regards,
sunken rock


Solution:

[hide]We start with the first letter as W and the last as L. So we have W_ _ _ _ L as our guideline and have LYEO to fill in. In the first empty slot, any of the four letters can go in, in the second any of the remaining three, in the third any of the remaining two, and by default the last one has to go in the last slot. So we have 4*3*2=24[/hide]

[hide]
$(1)(4)(3)(2)(1)(1) = 24$. (1 possibility for letter 1, 4 for letter 2, etc.)
[/hide]

[hide]Since w has to be the first letter and l has to be the last, then the remaining letters to permute are y,e,l,o. None of the letters are the same, so 4! = [b]24[/b][/hide]

[hide]Because there are 2 possibilities for "l", wouldn't it be more like:

W_ _ _ _ _
   4 3 2 1 2

48?

Or am I thinking wrong?[/hide]

the l's are not distinguishible so you have to divide your answer by two.

[quote="MCrawford"]How many different arrangements of the six letters of the word "YELLOW" can be made if the first letter must be "W" and the last letter must be "L"?[/quote]

[hide="Solution"]
This just asks how many ways can you arrange "ELLO".

4!/2!
[b]12[/b]
[/hide]

No, it asks how many different ways you can scramble YELO, so 4!


recently this happened:

[quote]NAS (n.r. National Academy of Sciences - USA) Award in Mathematics  a prize of $5,000 awarded every four years for excellence in published mathematical research  goes to Dan Virgil Voiculescu, professor, department of mathematics, University of California, Berkeley. Voiculescu was chosen "for the theory of free probability, in particular, using random matrices and a new concept of entropy to solve several hitherto intractable problems in von Neumann algebras." The award was established by the American Mathematical Society in commemoration of its centennial and has been presented since 1988. [/quote]

Former Romanian IMO contestant, won 3 gold medals and 1 silver medal. Also was close to winning a Fields medal, but unfortunately a bad luck (he missed the 40 years deadline by some months) kept him away from it.

What do you think ? :)

Honors go to Dan Voiculescu.Missing the fields medals by a few month...that must be a bit disappointing (Andrew Wiles missed the fields medal close,too)

I think Dan Voiculescu is the greatest romanian mathematician of all time. ;)  :D :thumbup:


calculus
integration
limit
function
real analysis
real analysis unsolved

Find
\[ \lim_{x\rightarrow \infty} \int_1^x \frac{e^{\minus{}t}}{t}dt.\]
"Mathematica" says the limit is equal $ \minus{} \infty$. But $ \frac{e^{\minus{}t}}{t}>0$ for all $ t>1$ so how the limit could be negative?

Could you please give me some advice how to explain it?

[quote="solenoid"]Could you please give me some advice how to explain it?[/quote]  Mathematica is stupid about some things.  My guess is that this has something to do with the fact that Mathematica tries to symbolically integrate and then take the limit of the resulting function -- it may confuse itself along the way.  See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=267703]here[/url] for another example of something Mathematica does poorly.

In this case, the exact value probably can't be expressed in elementary terms. $ \frac{e^{kx}}{x}$ does not have an elementary antiderivative, and $ 1$ is not a special point for anything involving the exponential.

For a numerical method, I would recommend doing some $ \int_1^M$ by ordinary numerical integration techniques, and $ \int_M^{\infty}$ with the asymptotic series $ \int_{M}^{\infty}\frac{e^{\minus{}t}}{t}\,dt\equal{}\frac{e^{\minus{}M}}{M}\minus{}\frac{e^{\minus{}M}}{M^2}\plus{}\frac{2e^{\minus{}M}}{M^3}\minus{}\frac{6e^{\minus{}M}}{M^4}\plus{}\cdots$

I just ran this in a spreadsheet, and got $ 0.219384$, accurate within $ 10^{\minus{}6}$.


symmetry
inequalities
inequalities proposed

[quote="Mildorf"]Let $a, b, c, k$ be positive reals. Determine a simple necessary and sufficient condition for the following inequality to hold:

\[ (a+b+c)^k(a^kb^k+b^kc^k+c^ka^k) \le (a^k+b^k+c^k)(ab+bc+ca)^k  \]

If you solve it just right, you might see how this is a variation of the idea behind my other one.[/quote]

:)

His other one : http://www.mathlinks.ro/Forum/viewtopic.php?t=48136 .


function
Vieta
algebra
system of equations

1.  Suppose that $ f(x) \equal{} x^5 \plus{} ax^4 \plus{} bx^4 \plus{} cx^2 \plus{} dx \plus{} e$ and that $ f(1) \equal{} f(2) \equal{} f(3) \equal{} f(4) \equal{} f(5)$.  Then $ a$ = ?

I tried setting up a system of equations by plugging in $ 1, 2, 3, 4,$ and $ 5$ back into the function, however I didn't get anywhere with that.  I'm guessing you need Vieta's formulas, but I don't know how to get there.

2.  Suppose that $ f(x)$ is a function such that for every real number $ x$, (i) $ f(x) \plus{} f(1\minus{}x) \equal{} 11$ and (ii) $ f(1 \plus{} x) \equal{} 3 \plus{} f(x)$.  Then $ f(x) \plus{} f(\minus{}x)$ must equal?

1. You can set $ f(x\plus{}1)\minus{}f(x)\equal{}(x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}4).$

2. Relation $ f(1 \plus{} x) \equal{} 3 \plus{} f(x)$ also writes as $ f(x) \minus{} f(1\plus{}x) \equal{}  \minus{}3$. In $ \minus{}x$ this gives $ f(\minus{}x) \minus{} f(1\minus{}x) \equal{} \minus{}3$. Add to the other relation to get $ f(x) \plus{} f(\minus{}x) \equal{} 8$.

I'm still confused about #1.

First of all why can you set $ f(x \plus{} 1) \minus{} f(x) \equal{} (x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}4)$?

Secondly, what do you do after that?   :lol:

Supposed you meant $ f(x) \equal{} x^{5} \plus{} ax^{4} \plus{} bx^{3} \plus{} cx^{2} \plus{} dx \plus{} e$ instead.
Then $ \Delta f(x) \equal{} f(x \plus{} 1) \minus{} f(x) \equal{} 5(x \minus{} 1)(x \minus{} 2)(x \minus{} 3)(x \minus{} 4)$ (to see why, note the roots of the RHS and find the coefficient of $ x^4$ in $ (x \plus{} 1)^5 \minus{} x^5$).
We know $ [x^3]\Delta f \equal{} \binom53 \plus{} 4a \equal{} \minus{} 5(1 \plus{} 2 \plus{} 3 \plus{} 4) \equal{} \minus{} 50 \equal{} 10 \plus{} 4a\implies a \equal{} \minus{} 60/4\equal{}\minus{}15$

Oh, yes I forgot the coefficient "5" of $ x^4$.


function
limit
induction
probability
calculus
calculus computations

for any differentiable function $f(x)$ over interval $I$ , let $x_{1},x_{2}\cdots x_{n}$ be in $I$ such that

(i)neither $f(x_{k})=0$
(ii)$x_{k+1}=x_{k}-\frac{f(x_{k})}{f'(x_{k})}$

If $\lim_{n\rightarrow\infty}x_{n}$ diverges . Can we say that there is no zero for $f(x)=0$ in $I$ ?

The method doesn't come with guarantees. Our sequence may fail to converge even if a root exists.

For an example, let $f(x)=\frac{x}{x^{2}+1}$ The exact details of the function don't matter, just the shape.

Choose $x_{1}>1.$ Then, since $f(x_{1})>0$ and $f'(x_{1})<0,$ we have $x_{2}>x_{1}.$ But the same can be said for $x_{2}.$ By induction, $x_{n+1}>x_{n},$ so the sequence increases. Hence it doesn't converge to the root at zero. (In fact, this sequence tends to infinity.)

Ok . I agree . What if we say " there are no zero for $f(x)=0$ for $x>x_{1}$ " ? Any counter example for this one ?

[quote="Kent Merryfield"]Go back to my example and let $x_{1}=-0.8.$ Then $x_{2}\approx 2.84$ and the sequence tends to infinity from there. Note that we just skipped over the zero somewhere between $x_{1}$ and $x_{2}.$[/quote]

Does that means that Newton Rapson Method doesnt neccesary works if we pick inappropriate approximation ?

That's exactly what it means. Newton's method is not foolproof.

(For a foolproof method, try successive bisection in a case where you've trapped a continuous function into changing signs. The fact that the method is foolproof could be expressed by saying you could construct a valid proof of the Intermediate Value Theorem this way.)

If our initial guess in Newton's method is sufficiently close (although the question of what "sufficiently close" means is a tricky one) and the root is simple (that is, $f'(x)\ne 0$ when $f(x)=0$) then the convergence is rapid. To be specific, it is "second order" which means that the new error is at worst proportional to the square of the old error.

A double (or higher order) root would cause first-order convergence. And if the new error is at worst proportional to the square of the old error, and the old error is too big, then we may wind up further away than we started.

One of the decorations I've used on the bulletin board beside my office door: Take a large chunk of the complex plane. For each point $z_{1}$ in the plane, use $z_{1}$ as the starting point for a Newton's method sequence for the equation $z^{4}+1=0.$ With probability 1, the sequence eventually converges to one of the four roots of that polynomial. Assign a color (some bright, simple color) to each of those four roots, and assign to $z_{1}$ the color of the root that the sequence starting there converges to.

Now do this numerically, on a grid of points corresponding to the pixels available to you, and render the resulting picture.

Go ahead, those of you with the computing environment capable of doing this. Try it. The resulting picture is not simple.

(The equation $z^{3}+1=0,$ with three colors, is a more famous picture, but I like the looks of the version with the quartic.)


Μπορει καποιος να μου πει πως μπορουμε να λυσουμε την παρακατω ασκηση?(το ζηταει ενας μαθητης Γ λυκειου):

[img]http://img211.imageshack.us/img211/6033/askisiff4.jpg[/img]

Νομίζω ότι είναι αρκετά εύκολη. Απλά πάρε την σχέση $ z1=i*z2$ και εξίσωσε τα πραγματικά και τα φανταστικά μέρη. Μετά αρκεί να πολλαπλασιάσεισ τισ δυο σχέσεις κατά μέλη και θα σου φύγει το λ και  πρέπει να σου βγει $ (χ+1)(χ-1)=(ψ+2)(ψ-2)$ . Κάνε τις πράξεις και βγήκε η υπερβολή

[quote="gregg"]Νομίζω ότι είναι αρκετά εύκολη. Απλά πάρε την σχέση $ z1 = i*z2$ και εξίσωσε τα πραγματικά και τα φανταστικά μέρη. Μετά αρκεί να πολλαπλασιάσεισ τισ δυο σχέσεις κατά μέλη και θα σου φύγει το λ και  πρέπει να σου βγει $ (χ + 1)(χ - 1) = (ψ + 2)(ψ - 2)$ . Κάνε τις πράξεις και βγήκε η υπερβολή[/quote]

και παλι δεν μου βγαινει υπερβολη  :oops:

Από τριγωνική ανισότητα ή και γεωμετρικά ελάχιστο 6 , μέγιστο 8 .
Για την παραπάνω $ (x\plus{}1)(x\minus{}1)\equal{}(y\plus{}2)(y\minus{}2)$ ara $ x^2\minus{}1\equal{}y^2\minus{}4$ ara $ \frac{y^2}{3}\minus{}\frac{x^2}{3}\equal{}1$ (απλά μάλλον σε μπέρδευαν οι εστίες)  :wink:

thnx, τωρα καταλαβα  :)

[quote="silouan"]Από τριγωνική ανισότητα ή και γεωμετρικά ελάχιστο 6 , μέγιστο 8 .[/quote]

Pos proekypse auto?kati den piano mallon...

Τα νούμερα \που έβγαλα είναι σωστά ? Αν ναι θα σου δικαιολογήσω τον τρόπο μου...

Mallon oxi...Tha exeis kanei kana lathaki stis prakseis!tespa opote boreseis grapse kai ton tropo, gt etsi ki allios opos ksereis den me endiaferoun ta noumera!!!

Πράγματι είχα κάνει λαθάκι στις πράξεις...
Πολύ μου αρέσει να σκοτώνω σχολικές ασκήσεις με <<ολυμπιακές>> τεχνικές  :P
Από την sxesi Euler για το παρ/μο έχουμε
$ |z_1 \minus{} z_2|^2 \plus{} |z_1 \plus{} z_2|^2 \equal{} 2(|z_1|^2 \plus{} |z_2|^2)$   άρα

$ 2(|z_1 \minus{} z_2|^2 \plus{} |z_1 \plus{} z_2|^2) \equal{} 100$   .

Όμως από Cauchy (δεν είναι Cauchy αλλά για να ψαρώνουμε τους άλλους έτσι το λέμε :rotfl: )  έχουμε

$ 2(|z_1 \minus{} z_2|^2 \plus{} |z_1 \plus{} z_2|^2)\geq (|z_1 \minus{} z_2| \plus{} |z_1 \plus{} z_2|)^2$  άρα

$ |z_1 \minus{} z_2| \plus{} |z_1 \plus{} z_2|\leq 10$

Τώρα από τριγωνική έχουμε
$ |z_2 \minus{} z_1| \plus{} |z_1 \plus{} z_2|\geq |z_2 \minus{} z_1 \plus{} z_2 \plus{} z_1| \equal{} 2|z_2| \equal{} 8$    :D

sostos...to proto ki ego etsi to ekana kai mou arese polu!

tora...to deftero einai lathos! :o giati arage?

to ison sthn trigonikh den borei na isxyei, epomenos den einai 6 h elaxisth timh. grapse $ |z_1 \minus{} z_2| \equal{} |z_2 \minus{} z_1|$ kai pare tora trigonikh. to ison se auth thn periptosh isxyei otan ta
$ z_1$ kai $ z_2$ einai omorropa.telika prokyptei oti h elaxisth timh einai to 8.

[b]Gia logous plhrothtas den prepei na ksexaso na anafero oti h askhsh einai kataskeuh tou kyriou Dhmhtrh Aretakh.[/b]

Προφανώς γιατί ισχύει

$ |z_1\pm z_2|\leq |z_1| \plus{} |z_2|$

Ωραία η λύση σου Σιλουανέ με τον Euler

Για το 8 έχεις δίκαιο . (απλά δεν έλενξα την ισότητα σε γεωμετρικό σχήμα) .
Συγχαρητήρια στο κ Αρετάκη για την άσκηση (είναι πολύ ωραία) ...

[quote="r_boris"][quote="Nick Rapanos"]Αν $ z_1,z_2$ μιγαδικοί και $ |z_1| \equal{} 3$ και $ |z_2| \equal{} 4$ τότε να βρείτε τη μέγιστη και την ελάχιστη τιμή της παράστασης

$ A \equal{} |z_1 \minus{} z_2| \plus{} |z_1 \plus{} z_2|$[/quote]
Πολύ ωραία άσκηση γισ την ποικιλία τρόπων .Εδώ δίνω μια λύση με παραγώγους ,προσιτή στο λύκειο. Η λύση είναι στο attachment[/quote]
Τυπογραφικό λάθος A(min)=8 και όχι 6 όπως κατά λάθος έγραψα!!

Κάπου χάνω την απόδειξη. Μέχρι το n+1 όλα καλά.
Μετά τι εννοείς;
[quote]
Salonikos wrote:
$ z_ {n \plus{} 1} \equal{} a(z_1 \plus{} z_2 \plus{} ... \plus{} z_n) \equal{} a(z_1 \plus{} a_1z_1 \plus{} ... \plus{} a_ {n \minus{} 1} z_1) \equal{} a_n z_1$[/quote]
Από που προκύπτει η πρώτη ισότητα;

8a xrhsimopoihsw isxurh epagwgh ,oti isxuei dhladh gia ka8e k<=n (eprepe na to exw dieukrinisei sto prohgoumeno post)
gia k=n dld isxuei oti |z1|+|z2|+...+|zn|=|z1+z2+...+zn|=>z2=a1z1,...zn=an-1z1
an zn+1 einai tetoio wste |z1|+|z2|+...+|zn|+|zn+1|=|z1+z2+...+zn+zn+1|kai epeidh exoume oti |z1|+|z2|+...+|zn|=|z1+z2+...+zn|pairnoume oti |z1+z2+...+zn|+|zn+1|=|z1+z2+...+zn+zn+1|
8ewroume loipon ws mia migadikh metavlhth thn z1+z2+...+zn kai thn allh thn zn+1 opote xrhsimopoiwntas thn apodei3h gia n=2 prokuptei oti zn+1=a(z1+z2+...+zn)
me vash thn isxurh epagwgikh upo8esh exoume oti z2=a1z1, z3=a2z1, ..., zn=an-1z1 kai antika8istwntas pairnoume to deutero melos ths isothtas p egrapsa

katalaves ti ennow?exeis kapou enstash?

hlias

Όχι δεν έχω αντίρρηση ή τουλάχιστον δεν είδα κάποιο λάθος. Ωραία απόδειξη :wink:

Προσθέτω επίσης μια άσκηση για όποιον ενδιαφέρεται. Είχε δοθεί σε παλιές εξετάσεις. Εγώ δίνω μόνο το τελευταίο υποερώτημα για να είναι πιο ενδιαφέρουσα:
Έστω οι μιγαδικοί αριθμοί $ a,b,z$ τέτοιοι ώστε $ ab \equal{} z^2$. Να δείξετε ότι:
$ |a| \plus{} |b| \equal{} |(a \plus{} b)/2 \plus{} z| \plus{} |(a \plus{} b)/2 \minus{} z|$

Όσο για την άσκηση του mac190604:
Είναι προφανώς $ |z_1|\equal{}|z_3|\equal{}|z_2|\equal{}1$ οπότε επειδή είναι $ z_1\plus{}z_2\plus{}z_3\equal{}0$ αποδεικνύεται ότι $ z_1z_2\plus{}z_2z_3\plus{}z_3z_1\equal{}0$ και $ z_1^2\plus{}z_2^2\plus{}z_3^2\equal{}0$. Παίρνοντας αυτές τις σχέσεις και το θεώρημα de Moivre προκύπτουν τα ζητούμενα

gregg den 3erw an auth einai h lush p eixes up opsin s alla h dikh m periexei pra3eis.

Loipon, pollaplasiazw thn dosmenh me |a| kai pairnw |a||a|+|b||a|=|a||(a+b)/2+z|+|a||(a+b)/2-z|
<=>|a|^2+|ab|=|(a^2+ab+2az)/2|+|(a^2+ab-2az)/2|<=>|a|^2+|z^2|=|(a^2+z^2+2az)/2|+|(a^2+z^2-2az)/2|<=>|a|^2+|z^2|=|((a+z)^2)/2|+|((a-z)^2)/2|
Etsi loipon 8etw a=x1+y1i kai z=x2+y2i opote meta apo pra3eis katalhgw se kati p isxuei

Eimai sigouros omws oti uparxei kai pio aplos dromos etsi? Apo ti e3etaseis thn phres?

Filika,
Hlias

Πολύ ωραία λύση, salonikios
Η λύση που βρήκα είναι:
[hide]Εφαρμόζουμε την ταυτότητα $ |z_1 \minus{} z_2|^2 \plus{} |z_1 \plus{} z_2|^2 \equal{} 2(|z_1|^2 \plus{} |z_2|^2)$ αφού υψώσουμε στο τετράγωνο και μετά από κάποιες πράξεις προκύπτει το ζητούμενο[/hide]
Είναι από τις εξετάσεις του 1979

[quote="gregg"]Μήπως μπορεί κάποιος να με βοηθήσει με αυτή την άσκηση (με σχολικά μαθηματικά και όχι ολυμπιακές ανισότητες);
Δίνεται ότι :
$ |z_1| \plus{} |z_2| \plus{} ... \plus{} |z_n| \equal{} |z_1 \plus{} z_2 \plus{} ... \plus{} z_n|$
Να δείξετε ότι υπάρχουν θετικοί αριθμοί $ a_1,a_2,...,a_n$  τέτοιοι ώστε:
$ z_2 \equal{} a_1 z_1, z_3 \equal{} a_2 z_1,...$
Μοιάζει με την ανισότητα Minkowski αλλά δεν μορώ α την αποδείξω και επιπλέον δεν μπορώ να την χρησιμοποιήσω για τα σχολικά μαθηματικά[/quote]
Υπάρχει μια πολύ όμορφη απόδειξη η οποία στηρίζεται στο γεγονός ότι το "=" στην τριγωνική ανίσωση αφορά τα ομόρροπα διανύσματα!! $ z_1\plus{}(z_2\plus{}...\plus{}z_n)\equal{}z_1\plus{}z_2\plus{}...\plus{}z_n$ κ.ο.κ

[quote="mostel"]Να βάλω κι εγώ μία.

Έστω $ x,y,z$ τρεις μη μηδενικοί μιγαδικοί αριθμοί, οι οποίοι συνδέονται με τις σχέσεις:

$ x \plus{} y \plus{} z \equal{} 0$ και $ |x| \equal{} |y| \equal{} |z|$

Να αποδειχθεί ότι:

i) $ xy \plus{} yz \plus{} zx \equal{} 0$

ii) $ z^2 \equal{} xy$

iii) $ |x \minus{} y| \equal{} |y \minus{} z| \equal{} |x \minus{} z|$

Σας θυμίζει τίποτα το συγκεκριμένο θέμα?  :P
[i]
Η άσκηση είναι από το βοήθημα 1ης Δέσμης, Άλγεβρας Γ' Λυκείου, του Στυλιανού Ευριπιώτη, έκδοση 1993.[/i][/quote]
Για το ισόπλευρο τρίγωνο, ερώτημα 3, το βαρύκεντρο=περίκεντρο του τριγώνου με κορυφές τα x,y,z αφού x+y+z=0=(x+y+z)/3.Έτσι δεν χρειάζονται τα βοηθητικά ερωτήματα 1 και 2 και το ερώτημα 3 βγαίνει πιο γρήγορα

D. Batinetu  : έστω z1, z2, z3 τρεις μιγαδικοι ωστε |z1.z2+ z2.z3+ z3.z1|=a  και |z1.z2.z3|=b
(a>0 και b>0). Να δειχθεί οτι τουλάχιστον ένας από τους τρεις έχει μέτρο μικρότερο ή ίσο του (3.b) /a

Kalws hr8es re jacob!ante kales luseis na exeis!

Loipon h idea p m hr8e exei ws e3hs:
Diairoume a/b kai pairnoume $ |1/z_1\plus{}1/z_2\plus{}1/z_3|\equal{}a/b$
An $ |z_1|,|z_2|,|z_3|>3b/a$ tote $ |1/z_i|*|z_i|>3b/a *|1/z_i|\equal{}>a/3b>|1/z_i|$ gia i=1,2,3

Omws $ a/b\equal{}a/3b\plus{}a/3b\plus{}a/3b>|1/z_1|\plus{}|1/z_2|\plus{}|1/z_3|>\equal{}|1/z_1\plus{}1/z_2\plus{}1/z_3|\equal{}a/b$
Ara $ a/b>a/b$ atopo!!!

Exeis kamia allh lush upopsin???

UG
exw prospa8hsei na s steilw pm treis fores alla den ta pairneis!8umhse m otan se dw sto sxoleio na s pw!

Ωραια η λυση ρε Elias, καθομουν και τη χαζευα  :oops: αμα δεις τη δικια μου... εχει μεσα ανισοτητα δυναμεων- αριθμ. μεσου και τριγωνικη!! να ρε ο εμπειρος ο λυτης!
δεν ειδα κανα pm στα inbox...

[quote="gregg"]Προσθέτω επίσης μια άσκηση για όποιον ενδιαφέρεται. Είχε δοθεί σε παλιές εξετάσεις. Εγώ δίνω μόνο το τελευταίο υποερώτημα για να είναι πιο ενδιαφέρουσα:
Έστω οι μιγαδικοί αριθμοί $ a,b,z$ τέτοιοι ώστε $ ab \equal{} z^2$. Να δείξετε ότι:
$ |a| \plus{} |b| \equal{} |(a \plus{} b)/2 \plus{} z| \plus{} |(a \plus{} b)/2 \minus{} z|$

Όσο για την άσκηση του mac190604:
Είναι προφανώς $ |z_1| \equal{} |z_3| \equal{} |z_2| \equal{} 1$ οπότε επειδή είναι $ z_1 \plus{} z_2 \plus{} z_3 \equal{} 0$ αποδεικνύεται ότι $ z_1z_2 \plus{} z_2z_3 \plus{} z_3z_1 \equal{} 0$ και $ z_1^2 \plus{} z_2^2 \plus{} z_3^2 \equal{} 0$. Παίρνοντας αυτές τις σχέσεις και το θεώρημα de Moivre προκύπτουν τα ζητούμενα[/quote]

Σωστά gregg... αλλά έχει και άλλα ερωτήματα που προκύπτουν με κάποιους συνδυασμούς.... Η επιτυχία της άσκησης έχει να κάνει με την μορφή των δεδομένων και την πληρότητα των ερωτημάτων.. Έχετε να προσθέσετε κάποιο??

Προσθέτω μία άσκηση από τον διαγωνισμό Putnam του 1959. Δεν την έχω λύσει μέχρι στιγμής αλλά θα την δω σήμερα.
Αν οι εικόνες των μιγαδικών αριθμών $ z_1,z_2$ είναι κορυφές ισποπλεύρου τριγώνου τότε η τρίτη κορυφή αυτού του τριγώνου θα είναι η εικόνα του μιγαδικού αριθμού $ \minus{}wz_1\minus{}w^2z_2$ όπου w είναι μια μη πραγματική ρίζατης εξίσωσης $ x^3\equal{}1$

[quote="gregg"]Προσθέτω μία άσκηση από τον διαγωνισμό Putnam του 1959. Δεν την έχω λύσει μέχρι στιγμής αλλά θα την δω σήμερα.
Αν οι εικόνες των μιγαδικών αριθμών $ z_1,z_2$ είναι κορυφές ισποπλεύρου τριγώνου τότε η τρίτη κορυφή αυτού του τριγώνου θα είναι η εικόνα του μιγαδικού αριθμού $ \minus{} wz_1 \minus{} w^2z_2$ όπου w είναι μια μη πραγματική ρίζατης εξίσωσης $ x^3 \equal{} 1$[/quote]
Λυση στο attachment

E3upnh lush r_boris...(an k pisteuw oti 8a xreiazotan ligh e3hghsh h isodunamia twn askhsewn)
H dikia m lush periexei polles pra3eis k den sugkrinetai se kompsothta...
8etw $ z_1\equal{}x_1\plus{}y_1i$, $ z_2\equal{}x_2\plus{}y_2i$ kai $ z_3\equal{}x_3\plus{}y_3i$ opou $ z_3$ h trith korufh tou trigwnou.
Profanws uparxoun mono duo 8eseis gia to $ z_3$ kai epishs exoume oti $ w\equal{}(\minus{}1\pm i\sqrt3)/2$
kai auto p ousiastika zhtaei h askhsh einai na dei3oume oti $ (x_1\minus{}x_2)^2\plus{}(y_1\minus{}y_2)^2\equal{}(x_3\minus{}x_2)^2\plus{}(y_3\minus{}y_2)^2\equal{}(x_1\minus{}x_3)^2\plus{}(y_1\minus{}y_3)^2$
apo edw kai pera einai ka8ara 8ema pra3ewn gia tis 2 times tou w


Does anyone know the difference when you're asked to find the degree of a term and when you're asked to find the power of a term?

They should be the same.  Unless you are talking about some special subject other than just normal algebra.

Well it's like there's a problem like:

x 2 y^3

and they ask whats the degree. I know one of them is when you find the highest exponential value in the term, but the other is when you add all the exponents together. i just dont know which is which...

[quote="tarquin"]Isn't the degree the highest exponent there.[/quote]I thought the degree was the total of the exponents...for example $x^{3}y^{5}z^{2}$ would have a degree $3+5+2=10$.

I think the degree is the total, too. Like in a homogenous equation, you have all the degrees equal, but that doesn't mean the power on each term is equal.

Degree is total.  I've never heard "power" used anywhere.  I can't imagine when it would be useful to find the highest power within a term.


limit
real analysis
real analysis solved

Let $ (a_n) , (b_n) $ be two sequences satisfying :
  $ \lim_{n\to \infty}a_n = \lim_{n\to \infty}b_n = +\infty $
  $ (b_n) $ is strictly increasing
  $ \lim_{n \to \infty}(b_{n+1} - b_n)=0 $
 Prove that :
 $ E=  \{  a_n - b_m | n,m \in N^*  \}$ is dense in $ R $

Let a<b be real numbers.We will prove that if A={an-bm/n,m natural numbers} then there is an x from A so that a<x<b;If this does not happen:
Let e>0 be a real so that e<( b-a );because b(n+1)-bn converges to 0 there is a natural N so that b(n+1)-bn<e for any n>=N (note that b(n+1)> bn since bn is strictly increasing);                                                                 (1)
Because lim an is infinite thre is ap so that ap>b+bN;
We now have that ap-bN>b , and because bn converges to infinite there will be a bq so that ap-bq<a since ap is constant.
Note that ap-bN>ap-b(N+1)>ap-(bN+2)>.......>ap-bq;
Since ap-bN>b and ap-bq<a and there is no x from A so that a<x<b there is a m so that ap-bm>b and ap-b(m+1)<a (just consider the smallest number from the ap-bN,ap-b(N+1),ap-(bN+2),.......,ap-bq set that is bigger than b)
but then ap-bm-[ap-b(m+1)]>b-a therefore b(m+1)-bm>b-a this is absurd because m>N and (1);

This argument is closely related to the argument that shows that a conditionally convergent real series may be rearranged to converge to any real number.


a and b are roots of polynomial $x^4+x^3-1$. Prove that $a*b$ is a root of polynomial $x^6+x^4+x^3-x^2-1$.

Bloody problem...
Let $x_1$, $x_2$, $x_3$ and $x_4$ be roots of $x^4+x^3-1=0$.
Consider polynomial $P(x)=\prod_{i<j} (x-x_ix_j)$. [b]Expanding it[/b] and using Viet's theorem we obtain that $P(x)=x^6+x^4+x^3-x^2-1$.
Believe me, I did all this work!


Show that radical center of circles centered at vertices of a triangle
     with radii equal the sum of sides at the vertex is the
     Longuet-Higgins point.

     Cpt. Pestich

What is 'Tournament of Villages'? It sounds funnily. Is it a response to the Tournament of Towns?

This is problem from the Monthly.
Have a look at
http://www-students.biola.edu/~woopy/math/amm10734.htm
Kind regards.

I have a nice solution, based on the fact that the Longuet-Higgins point of a triangle is the anticomplement of the Bevan point. So, you have to prove that the radical center of the three circles is the anticomplement of the Bevan point. Just consider the relationship of the three circles with the excircles (and thus with the Bevan point) of the anticomplementary triangle. Show that the pairwise radical axes of these three circles play, in the anticomplementary triangle, the role of the perpendiculars from the excenters to the sides (and thus concur at the Bevan point).

Everything clear? :D Sorry, I am running out of time now, else I would post a complete solution here...

  darij

[quote="zhaoli"]What is radical center, Longuet-Higgins point and Bevan point?[/quote]
These notions are only for choosen ones. Not for us...


calculus
derivative
calculus computations

Two small boats that move with constant speed in straight and perpendicular paths. In a certain moment, the boats A and B are, respectively, to 4.0 km and 3.0 km from the point P, the intersecting paths. What is the minimum distance, measured in miles, between boats A and B?

We are falsely led to believe that the answer is 5 km.

Actually, the answer is $ \ 2 \sqrt{5}$ km

Presumably you mean that the two boats move with the same constant speed, and both are moving toward $ P$?  If so, unless I'm missing something it would seem the answer should be $ \frac {1}{\sqrt{2}}$.  If one is moving away from $ P$ and the other toward it, then it would be $ \frac{7}{\sqrt{2}}$.

$ s(t)\equal{}\sqrt{d_A^2(t)\plus{}d_B^2(t)}$

$ d_A(t)\equal{}4\minus{}2t, d_B(t)\equal{}3\plus{}t, d_A'(t)\equal{}\minus{}2, d_B'(t)\equal{}1$

$ s'(t)\equal{}\frac{d_Ad_A'\plus{}d_Bd_B'}{\sqrt{d_A^2(t)\plus{}d_B^2(t)}}$

$ s'(t)\equal{}0$ iff $ 2d_A\equal{}d_B$, or $ 8\minus{}4t\equal{}3\plus{}t, \to t\equal{}1$

so $ s'(1)\equal{}0$

$ s(1)\equal{}\sqrt{20}$ is therefore the min

thanks...

there is another way to solve without using derivatives ...since the relative velocity is equal to $ \sqrt{5} km/s$

Hold one of the two boats fixed and consider the path of the other boat relative to it.  The result is just a straight line, and it's easy to compute the distance between a fixed point and a line.


does anyone have any tips for adding lines into figures to make a problem easier to solve?

Any lines that dissect a figure into right triangles can be extremely helpful - that's the best trick I've discovered so far when tackling geometry problems.

Really elegant solutions use similar triangles.   :D

If there are two parallel lines, try to find (or draw in, if it helps) a transversal. If you do, there will be lots of equal angles (and equal sides may follow).

[quote="mna851"]It really depends on the problem.  Although its overstated, experience is the only way you'll recognize when additional lines will be helpful to the solution of a problem.[/quote]

This isn't overstated. There is no way on "tip" for how to draw extra line in geometry problems because there are so many kinds of them (start from simple line to complex Stewart's or Ceva's..). The best way, as mna851 suggested, is to keep doing geometry problems. Sometimes, drawing line actually makes problem harder to see so once you had many practices, you should be able to see "oh, if I need to add that, I'll get that.."

Draw lines that make congruent angles, similar triangles, or cyclic quadrilaterals.

[quote="k8reindeer"]draw the radius/radii of any circle[/quote]

That is not always true.

Actually draw whatever you feel like, eventually something will work out if you play around for enough time.

Actually, don't go crazy drawing stuff but still keep an eye out for any additions to your diagram that might lead to some nice properties (particularly if the problem condition is somewhat strange).

Keep what you are trying to solve for in your mind. If you can draw lines to express the variable or whatever you are solving for in a triangle/figure/etc. which has numbers you already know that could be very useful.


Here is how it goes. Post a picture of the character that pwns the previous character(can be from game, anime, manga, comic, etc.) along with an explanation.(unless none is needed.)
No loops.
Any character can get pwned.
Let's start with:

[img]http://www.smashbros.com/en_us/characters/images/sonic/sonic.jpg[/img]

[img]http://archives.bulbagarden.net/w/upload/5/50/ArceusDP.png[/img]

The wild ARCEUS used Judgment!
SONIC fainted!

[img]http://www.pokemonelite2000.com/ani068.gif[/img]

machamp used seismic toss!!! it's super-effective!!! arceus fainted!!!

[img]http://archives.bulbagarden.net/w/upload/d/d8/386Deoxys-Attack.png[/img]

DEOXYS used Psycho Boost!
It's super effective!
MACHAMP fainted!

[img]http://www.kombo.com/images/content/misc/phil_092908_skorge.jpg[/img]

What is this?  Skorge from Gears of War jumps out with a surprise attack and rips apart the weakling pokemons with his double-bladed chainsaw staff.

[img]http://archives.bulbagarden.net/w/upload/b/b3/491-Darkrai.png[/img]

Darkrai pulls Skorge into an infinite abyss of darkness.

[img]http://www.aintitcool.com/images2006/MasterChiefHalo.jpg[/img]

Master chief takes care of the pokemon scum.

[img]http://farm2.static.flickr.com/1055/3167909349_f765b687c1.jpg[/img]

In multiplayer EVERYBODY getz pwnt!

How long can you survive against a supersonic hedgehog... with a gun?(thusly, he fits in Halo!)

[img]http://www.freewebs.com/shadowheaven2/shadow.jpg[/img]

HAHAHAHAHAHAHA!
Just try to shoot him!!!

[img]http://i170.photobucket.com/albums/u245/shyguy_011/GEARS/063.jpg[/img]

Don't have to shoot him

On a side note, HOW does a single Centaur kill three of these things?

[img]http://loot-ninja.com/wp-content/uploads/2008/02/brumak.jpg[/img]

The centaur has explosives, duh.

In Gears 2, the brumak totally owned the corpser.

[img]http://images2.wikia.nocookie.net/gearsofwar/images/c/c5/Centaur.jpg[/img]

annnddd...Centaur pwns Brumak!

I guess you have a point. The Centaur's gun is seriously overpowered. Except the monster truck wheels, the whole tank is basically built around it.

tails diffuses the bomb.

[img]http://toadscastle.net/images/tails.jpg[/img]

[img]http://guidesarchive.ign.com/guides/11154/images/scyther.gif[/img]

tails's tails have just been de-tails-ified (say that 5 times fast :P )

knuckles=character pwning's version of a hammer.

[img]http://media.comicvine.com/uploads/2/20634/391291-188175-knuckles_large.jpg[/img]
knuckles used punch!
He caused a whole freaking lot of damage!
Scyther was pwned!

[img]http://www.nightswitch.com/sonic/super_sonic.jpg[/img]
ohhhhh crap...
Me thinks the zergs will be dead in about 1 second.

[img]http://piromano2.iespana.es/La%20Paranoia/entei.gif[/img]

ur turtle is burned

huh???

start over!!! :P

[img]http://pokedream.com/fans/fanart/charizard.png[/img]

Bye pokemon trainer!!! and bye pokeballs!!!

[img]http://reparent.blog.uvm.edu/images/masterchief.jpg[/img]

Blows up the Ark. Kills him.


Hey, so it seems like there are some debaters on AoPS.

Is anyone going to the Lexington Winter Invitational (aka Big Lex) this weekend? If you are, say so! I wanna get a feel of how many people here do it.

Or if you just do debate in general, say something.

btw, if you are coming to Big Lex, you might see me! I'll be running around being a runner.

i do Public Forum...but im not going to big lex, its a bit far from my state

A debate discussion, awesome

I do PoFo in my state, I only participate in our state circuit tourneys though, maybe NFL districts this year.

I did policy this year, and although I did pretty well, I don't think I really enjoy debating that much. Additionally, the workload in addition to school convinced me not to do it next year.

How much time does debate take up for you all?

Nope, I be hangin in Cali, so it's a bit far.

Is anyone here going to Winter Classic or Stanford?

The varsity policy debaters at my school are like, up until 1 cutting cards, and doing random stuff. Honestly, I don't understand how anything can take that long - they are obsessed with debate (and that's not a bad thing).

Debate is fun, but it takes a while to actually write cases, and debate can be really really stressful.

Big Lex this weekend seems to have been a success, and actually there was a team from California in our thing. I know a lot of people going to the Harvard tournament. Anyone going there? Even if you just break, you get a bid to tournament of champions. whoa.

Darn yeah, debate is kind of fun but unless you have a true passion for wasting your weekends all the time it ultimately will not help you at all. Honestly, the resolutions always suck... :wink:

:wink: I know, right? Economics sanctions are a bit dry. A few years back though, the resolution was 'Is vigilante action justified?' Obviously people were dropping super hero references left and right  :lol:

I do LD.

Not going to harvard..have to review for AMC 12B XD

I do LD, but i'm not going to Massachusetts because its a bit far.

Did anyone go to SCU scrimmage in California?


Let $S$ be the set of symmetric polynomials $a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0}$ such that $a_{n-i}= a_{i}, i = 0, 1, 2, ... n$.  Show that

$a, b \in S \implies a \cdot b \in S$

[hide="Remark"] There are several very nice ways to do this problem.  How many solutions can you find? [/hide]

[hide="hint"]
If $P$ is symmetic and has degree $n$, then $P(x) = x^{n}P\left(\frac{1}{x}\right)$.[/hide]

let $r$ be a root, $r\ne 0$ by looking at coefficients, if $r$ is a root, then $\frac{1}{r}$ is a root, then the product of the polynomials has the same property, thus it is symmetric


geometry
area of a triangle
Heron\u0027s formula

One of the sides of a triangle is divided into segments of 6 and 8 units by the point of tangency of the inscribed circle. If the radius of the circle is 4, find the length of the shortest side.

(AHSME 1953, Problem 50)

The sides have lengths $14, x+6, x+8$.
the area is $E=p \cdot r=(x+14)4$   ($p$ semiperimeter)
and $E=\sqrt{6 \cdot 8x(14+x)}$ (Heron's formula!)
So $\sqrt{6 \cdot 8x(14+x)}=4(x+14)$ .......or $x=7$ if I didn't make any silly mistakes... :)

[hide]If we put the sides of the triangle as $14$, $6+x$, and $8+x$, then we can get the semiperimeter, which is:

$\frac{14+6+x+8+x}{2}=\frac{28+2x}{2}=x+14$

Using this, we can use Heron's Formula and the rs formula to find what x is.

$4(x+14)=\sqrt{(x+14)(x+14-14)(x+14-6-x)(x+14-8-x)}$

$4x+56=\sqrt{(x+14)(x)(8)(6)}$ *wow, that came out nicely [suspicious]*

$(4x+56)^2=48x^2+672x$

$16x^2+448x+3136=48x^2+672x$

$32x^2+224x-3136=0$

$32(x^2+7x-98)=0$

$x^2+7x-98=0$

$(x-7)(x+14)=0$

$x=-14$

$x=7$

Since $x$ must be positive, $x=7$

Therefore, the sides of the triangle are $14$, $13$, and $15$

So the shortest side length is $13$
[/hide]

Alright! I fixed the problem! it was purely a careless mistake...  :blush:

[quote="Drunken_Math"][hide]$16x^2+448x+3136=48x^2+672x$

$32x^2-224x-3136=0$[/hide]
Hmm... my answer is different from socrates's. If I did something wrong, can someone point it out please! Thanks!  :)[/quote]
Redo it please.

Well plugging 14 back into Drunken Math's equations, $\sqrt {(x+14)(x)(8)(6)}$ is not an integer so...

That's what I was going to do, but I thought it would turn out really messy. :)

You know that tangents from the same point are equal, thus, using $rs=A$, you can do:

$4(14+x)=\sqrt{(14+x)(14+x-(8+x))(14+x-(6+x))(14+x-14)}--->(56+4x)^2=48x(14+x)$ Thus $x=7$ and the shortest side is $13$.


limit
search
number theory unsolved
number theory

There are given two positive integers  $ a, B \in \mathbf{N}$, and $ a\neq 10^{k}, k\in \mathbf{N}$. Prove that there exist $ l \in \mathbf{N}$ such that $ a^{l}$ begins with $ B$ in decimal presentation.

We need $ 10^kB<a^l<10^k(B+1)\Leftrightarrow k+logB<l\cdot log$ $ a<k+log(B+1)\Leftrightarrow \{logB\}<\{l\cdot log$ $ a\}<\{log(B+1)\}$. The number $ x=log$ $ a$ is irrational as $ a$ isn't a power of $ 10$. Now we only need to prove lemma:
If $ x$ is irrational and $ 0\leq p<q\leq 1$, there exists a positive integer $ l$ such that $ \{lx\}\in(p,q)$.

We find such $ a,b$ among $ 0,1,2,...,n$ that $ \{(a-b)x\}$ is either smaller than $ q-p$ or greater than $ 1+p-q$, where $ n>\frac{1}{q-p}$ and $ a>b$, and then we go by steps little enough to get into the interval: $ a-b, 2(a-b), 3(a-b)...$. This way we find $ l$.

More general .
$ f(n)$ is the number of form $ \{a^i\}_{i=1}^n$ begins with $ B$ in decimal represent .
Find ${ \lim_{n\to +\infty }\frac{f(n)}{n}}$

Hana( now I know you and you know me :lol: )  see here for a similar problem . The basic idea is the same  :wink:

http://www.mathlinks.ro/viewtopic.php?search_id=1954684971&t=50019


[quote="jeter123"]1. An infinite, nonconducting sheet has a surface charge density s = +7.06 pC/m2. (a) How much work is done by the electric field due to the sheet if a particle of charge q0 = 4.19 x 10-19 C is moved from the sheet to a point P at distance d = 3.81 cm from the sheet? (b) If the electric potential V is defined to be zero on the sheet, what is V at P?[/quote]
[color=blue]Using the Gauss'slaw, you can find out that the electric field is
$\large E=\frac{s}{2\epsilon_0}=-\frac{dV}{dr}$

$\large\Longrightarrow dV=-\frac{s}{2\epsilon_0}dr$

$\large\Longrightarrow \Delta V=\int_{r_1}^{r_2}dV=\frac{s}{2\epsilon_0}(r_1-r_2)=-\frac{s}{2\epsilon_0}d$

The work done by the electric field is

$\large A=-q_0\Delta V=\frac{s}{2\epsilon_0}q_0d$

The potential at point P is $\large V_P=V_0+\Delta V=-\frac{s}{2\epsilon_0}d$[/color]

[quote="jeter123"]3. Figure 24-61 (See Attachment) shows three circular, nonconducting arcs of radius R = 6.94 cm. The charges on the arcs are q1 = 3.65 pC, q2 = -2.62q1, q3 = +3.48q1. With V = 0 at infinity, what is the net electric potential of the arcs at the common center of curvature?[/quote]
[img]http://photo-origin.tickle.com/image/69/6/5/O/69650482O058701285.jpg[/img]

[color=blue]I think that the net potential at the centre of the curvature is
$\large V=\frac{k(q_1+q_2+q_3)}{R}$[/color]

[quote="jeter123"]2. The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit volt-cubic meter. As a multiple of A, what is the magnitude of the electric potential difference between r = 1.24 m and r = 2.37 m?
4. A charge of q = 4.64 x 10-8 C lies on an isolated metal sphere of radius 17.0 cm. With V =0 at infinity, what is the electric potential at points on the sphere's surface?[/quote]
[color=blue]You can see the solution [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=75256]here[/url][/color]

Thank you very much for your help. I seem to get the idea now as for how to solve these kinds of problems :)


There are $ n$ voters and $ m$ candidates. Every voter makes a certain arrangement list of all candidates (there is one person in every place $ 1,2,...m$) and votes for the first $ k$ people in his/her list. The candidates with most votes are selected and say them winners. A poll profile is all of this $ n$ lists.
If $ a$ is a candidate, $ R$ and $ R'$ are two poll profiles. $ R'$ is $ a\minus{}good$ for $ R$ if and only if for every voter; the people which in a worse position than $ a$ in $ R$ is also in a worse position than $ a$ in $ R'$. We say positive integer $ k$ is monotone if and only if for every $ R$ poll profile and every winner $ a$ for $ R$ poll profile is also a winner for all $ a\minus{}good$ $ R'$ poll profiles. Prove that $ k$ is monotone if and only if $ k>\frac{m(n\minus{}1)}{n}$.

[quote="anonymous1173"] We say positive integer $ k$ is monotone if and only if for every $ R$ poll profile and every winner $ a$ for $ R$ poll profile is also a winner for all $ a \minus{} good$ $ R'$ poll profiles. Prove that $ k$ is monotone if and only if $ k > \frac {m(n \minus{} 1)}{n}$.[/quote]
$ k$ is not used in the defination of itself as a monotone so there is no reason to have a conddition defining $ k$  so is the problem correct or have i got it wrong????????? :( the second one is most probably

i really cant understand the problem
please help
 :(

sorry, i can't understand your problem with this question. maybe it can be from my bad english but if you can ask your problem more clear i will try to explain.  but i can say that, "the first $ k$ in voters' lists are selected. it is $ k$'s definition. Being monotone is a property of $ k$."

[hide="Hint for sufficient part"]Prove that a winner must get all the voters votes.[/hide]
[hide="Hint for necessary part"]Seperate in parts and try to make all the winners get n-1 votes in a poll profile. After that find a a-good profile that the winner gets n votes(a isn't one of them)[/hide]

yes you are right. using kirkit's theorem in this problem is really sensible.

What is Kirkit's theorem? :huh:

Pierre.

Ok, Kirkit's theorem does not exist, and this is only spam... :wink:

Pierre.

So what is exactly the number of winners in a particular profile?

This statement is absolutely horrible. I do not think the problem itself is really difficult(i guess we will never know...), but the statement is absolutely unreadable making this problem impossible to think at, for me... If someone can rewrite it, or whatever, i would be really grateful for a statement in English.

There are $n$ voters and $m$ candidates. Every voter makes his/her arrangement of all candidates (there is one person in every place $1,2,...,m$) and votes for the first $k$ people in his/her list.
$k$ candidates with most votes are elected (we call them winners).

Consider the candidate $a$ and two elections $R$ and $R^{\prime}$. The election $R^{\prime}$ is $(R,a)$-good if for every voter $v$, the candidates that are in worse position on
 $v$'s list than $a$ in the election $R$ are also in the worse position than $a$ in $R^{\prime}$.

We say that a positive integer $k$ is monotone if for every election $R$ and every winner $a$ for $R$, $a$ is also a winner for all $(R,a)$-good elections. Prove that $k$ is monotone if and only if
\[
k>\frac{m\left(  n-1\right)  }{n}.
\]

Here is a cleaner problem statement:

Let $N$ denote a society of voters where each voter is endowed with preferences over a set of alternatives $A$ with $|N|=n>2$, $|A|=m>2$. A preference profile $R$ is an $n-$tuple of linear orderings on $A$ representing the preferences of voter's on $A$. Given some $k\in\{1,2,\dots,m\}$, the $k-$plurality choice rule chooses the alternatives achieving the highest vote after each voter assigns equal votes for all top $k$ alternatives in his preference ordering regardless of their ranking. Let $R$ and $R'$ be any two preference profiles and $a\in A$. We say that $R'$ $a-$dominates $R$ iff for every $i\in N$, all alternatives which are ranked below $a$ in $R_i$ are also ranked below $a$ in $R_i'$. $k-$plurality choice rule is said to be monotone iff for any $R$ and any $R'$ which $a-$dominates $R$, if $a\in A$ is chosen by $k-$plurality choice rule in $R$, $a$ is still chosen in $R'$. Assume $(m,n)\neq (3,4)$. Prove that, $k>\frac{m(n-1)}{n}$ is a necessary and sufficient condition for the monotonicity of $k-$plurality choice rule.

[b]Note:[/b] This problem has been proposed by Prof. Semih Koray of Bilkent University, Ankara, Turkey, whose one of primary research interest is in social choice theory; and electoral system.
[b]Note 2: [/b] For another Turkish olympiad problem regarding electoral system, which also happened to be proposed by Dr. Koray, see [url=https://artofproblemsolving.com/community/c6h433824p2451062]here.[/url]

Here is a neat solution:

[hide=Sufficiency]Note that, the total number of votes is in a $k-$plurality choice rule is precisely $kn$. Take any $R$ now. Since $\frac{kn}{m}>n-1$, it then follows that there is at least one candidate getting all $n$ votes, by the pigeonhole principle. In particular, any winner $a\in R$ must receive $n$ votes. But then the result is immediate: if $R'$ $a-$dominates $R$, then the total vote of $a$ with respect to $R'$ is at least that of $a$ in $R$; which is $n$. Hence, $a$ must also win under $R'$. Sufficiency proof is then complete.
[/hide]

[hide=Necessity]
Now suppose $2\leqslant k\leqslant \frac{m(n-1)}{n}$. We now construct a profile $R$ violating the plurality rule. Fix two candidates $a,b\in A$. Let $R_i(a)$ denote the ranking assigned by voter $i\in N$ to the candidate $a\in A$,  under preference profile $R$. Also for convenience,  let $N=\{1,2,\dots,n\}$,  that is the voters are denoted by $1\leqslant i\leqslant n$. Now, let $R$ be  such that $R_1(a)=\cdots=R_{n-1}(a)=1$, and $R_n(a)=m$ (that is, the candidate $a$ is the first choice for voters $1,2,\dots,n-1$; and the last for voter $n$, who happens to be an ex-wife of candidate $a$ ;) ). Now, let $R_1(b)=\cdots=R_{n-2}(b)=2$; $R_{n-1}(b)=m$, $R_n(b)=a$ (namely, $b$ is the second choice for voters $1,2,\dots,n-2$, the last choice for voter $n-1$; and the first choice of the voter $n$). Now, ensure also that each of the other people receive at most $(n-1)$ votes (from the parameter assumption on $k,m,n$ this is clearly possible).

Now, we construct an $R'$, which $a-$dominates $R$: Suppose the rankings of voters $1,2,\dots,n-2,n$ are the same for $R$ and $R'$. For voter $n-1$, simply ensure $R'_{n-1}(b)=2$, that is, $b$ is now moved above. Observe that this list indeed $a-$dominates $R'$. However, note also that the total vote of $a$ under $R'$ is $n-1$; while that of $b$ under $R'$ is $n$. Hence, $b\in A$ is elected according to $R'$, but not $a$, which contradicts with the fact that $a\in A$ is a winner under $R$, and $R'$ $a-$dominates $R$. This shows the given condition is indeed necessary.
[/hide]


Everyone here please tell me which is better
Elementary Number theory by David M Burton
OR
Elementary Number theory by Gareth Jones

or  An Introduction to the theory of numbers by Iven Niven, Zuckermann,co./

Please help.

Introduction to Number Theory
By Matthew Crawford
Published by The Art of Problem Solving is the better one  :lol:

1) Define ''better''- is it harder problems, or better explanations or what?
2) Why do AoPS books need to be the best? Because of this forum?

aops book are good because they are very clear and made to help in competitions.

And what if you don't care about the competitions and want to learn math just because of math?

*goes and looks up the two books on amazon*
1. Burton is freakishly expensive
2.  all of them are very advanced
3. intro books doesn't compare to any of the books you listed for material because the intro book is WAAAAAAAAAAAAAAAAAAAY too basic.


[hide="Solution"]There are $ 8$ white cars, $ 4$ of which have four doors.  Thus, the answer is $ 15\minus{}4$, or $ \boxed{11}$.[/hide]

[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Please hide all solutions.[/url][/color]

The problem about this problem is that it does not imply there may not be any cars that are red AND white 4-wheeled cars.

[hide = My Solution]
12 are red, so 8 are white
15 are 4-door, so 5 are 2-door
4 are 2-door and white, so 1 is 2-door and red
Since 1 out of the 12 red cars is 2-door, the other $\boxed{11}$ are 4-door (and red).[/hide]

[quote=maseratirex][quote=smith_ch]There are 20 cars in my building's parking lot. All of the cars are red or white. 12 of them are red, 15 of them are 4-door, and 4 of them are 2-door and white. How many of the cars are 4-door and red?

[hide="Solution"]so if 12 of the are red, then 8 of them are white.  if 4 of them are 2 door and white, then 4 of them are 4 door and white.  15 of them are 4 door, so 11 of them are 4 door and red.[/hide]

[color=#FF0000]Mod: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=394734]Please hide all solutions.[/url][/color][/quote]

THIS IS THE WRONG SOLUTION!!! 11 is not the answer![/quote]

Yeah the correct answer is [hide = this]$3$ Dont really get the problem though so cant put a sol[/hide]

^^ Why? Isn't the answer [hide] $ \boxed{11} $[/hide] ? because when you put it into the problem, you get: [hide]
$$ 1 \text{ red 2-doors } $$
$$ 11 \text{ red 4-doors } $$
$$ 4 \text{ white 4-doors } $$
$$ 4 \text{ white 2-doors } $$[/hide]  , which satisfies all the requirements.

I think you are talking about a different problem because the answer to this question is [hide=this]3[/hide]

[quote=GoodApple8888]I think [b]you[/b] are talking about a different problem, probably [hide=this] There are 30 cars in my building's parking lot. All of the cars are red or white, and a car can have either 2 doors or 4 doors. $\frac{1}{3}$ of them are red, $50\%$ of them are 4-door, and 8 of them are 2-door and white. How many of the cars are 4-door and red?[/hide], and that has an answer of [hide] $\boxed{3}$ (I did it) [/hide]. But, this problem is [quote=GameBot]There are 20 cars in my building's parking lot.  All of the cars are red or white. 12 of them are red, 15 of them are 4-door, and 4 of them are 2-door and white.  How many of the cars are 4-door and red?[/quote], which is different. I did the problem, and the answer [hide]$11$[/hide] was correct.[/quote]

Oops. Gamebot sometimes posts different problems then what Alcumus says.

[quote=abbytan]The answer is [hide] 11 [/hide][/quote]

please don't bump to just give the answer

[quote=awesomeguy856]also dont copy other people's posts[/quote]

yo facts I didn't even notice

[quote=herbs_li]Aren't we not supposed to post solutions?[/quote]

no we can.

[quote=GameBot]There are 20 cars in my building's parking lot.  All of the cars are red or white. 12 of them are red, 15 of them are 4-door, and 4 of them are 2-door and white.  How many of the cars are 4-door and red?[/quote]

LOL this was in my book.

[quote=herbs_li]What book was it in??[/quote]

Intro to Algebra

i got that problem FOUR times in alcumus because I sillied every SINGLE time

yes, but I got that problem like half a year ago when I didn't know there was an Alcumus Forum (gasp, Ikr)

[quote=herbs_li]Ok , well that's your opinon .[/quote]

I was joking

This is absolutely beautiful using venn diagrams.
Sorry can't hide attachments :(


function
combinatorics solved
combinatorics

Let $S$ be a set of $n \geq 4$ points in the plane, non all lying on a same circle, and three of them never collinear. Let $f$ a function from $S$ onto $R$ such that for any circle $C$ containing at least $3$ points from $S$, we have $\displaystyle  \sum _{P \in C \cap S} f(P) = 0$.
Prove that $f$ is identically $0$.

Pierre.

Let A_1,...,A_n be the n points in the set and let a_1,...,a_n be the values of function f in this points.
       Consider 2 points A and B and denote by n_AB the number of circles that pass through A and B and other points in the set S.
       Because any 3 points determine a circle, not all the points are on the same circle and n>3 we have n_AB>=2.
       Consider now the n_AB circles. By adding the values of the function f over all of this circles we get obviously 0.
In the sum mentioned above we have considered the sum a_1+...+a_n of the values of all point + the sum (n_AB-1)*(a+b) as the values of points A and B have been counted n_AB times.
       We get a_1+...+a_n=(1-n_AB)*(a+b), but 1-n_AB<0 so a+b and a_1+...+a_n have opposite signs.
       Applying this result for {A,B}={A_1,A_2},{A_2,A_3},...,{A_n,A_1} and adding up we get 2(a_1+...+a_n) and a_1+...+a_n have opposite signs. So a_1+...+a_n must be 0 => a+b=0 => a_1=...=a_n=0.

This made me remember the following similar problem:
Let S be a set of n points in R^3. Consider the value of each of this points to be 1 or -1 such that for each line passing through at least 2 points in S the sum of the values of the points on that line is 0.
Prove that all the points are on the same line.

Nice solution Ral, which is mine too.
But please, post a new problem in a new topic.

Pierre.


integration
search
logarithms
calculus
real analysis
real analysis unsolved

Well I'm just wonder that the antiderivatif
\[ \int x^x dx  \] has an elementary function. And I found this in my book.

Prove that \[ \int_{0}^1 x^x dx =1- \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + ....  \]

or in summation $\sum_{i=1}^{\infty} (-1)^i \frac{1}{i^i}$

can someone prove it?

Search for it- it's been done plenty of times before.

I was interested in the same thing, but I couldn't find much on it.  This is just what I found:

$x^x=\sum\frac{x^n\ln{\left(x\right)}^n}{n!}$

$\int_0^1x^x\,dx=\int_0^1\sum\frac{x^n\ln{\left(x\right)}^n}{n!}\,dx$

The sum and integral can be switched, then integrate by parts.

We have too many threads on this, including [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=354567]this one[/url].