induction
number theory
greatest common divisor
geometry
geometric transformation
algebra

Dear friends,
  I know I posted many idiot questions during my presence on this site and this is the reason for which I will ask another one. Is there an infinite set of natural numbers such that the sum of the elements of any finite subset is a natural power? It is useless to say that I couldn't solve it, but all persons to whom I gave it didn' t manage to break it. Can you?

This seems to be related to Catalan's conjecture. If you have a proof for it, you might have an answer for this too!

I'm not sure but i think the answer is yes. Someone told this problem related to the following:
 Is there an infinite set of integers such that the sum of any finite subset its NOT a perfect power?
  I remeber three things about this two problems,
 1) One of them have a nice solution considering some density argument (probably the one with no sum perfect power)
  2) One have a simple solution
  3) The other was difficult and the person didnt told me the solution.

 I am going to try to calrify my memory and think on them tomorrow, if i remeber or have any advance i post it.
 For instance i think itsnt dificult to show there are arbitrary long sequences with this property but i am not sure right now. Anyway, i cant  see the conection with catalan conjecture! And which one is the famous olympiad problem?

Well, to prove that there exists an infinite set of natural numbers such that the sum of any finite of its non-empty subsets is not a perfect power is not difficult using the chinese remainders theorem :

Start from $x_1 = 2$.
Now, assume that we have defined $x_1, \ldots , x_n$ such that the sum of any non empty set whose elements are from the $x_i$'s is not a power. We numerate these sets from $1$ to $k = 2^n-1$. Let $s_i$ be the sum of the elements from the set numbered $i$ and $s_0 = 0$.
Let $p_0,p_1, \ldots, p_k$ be pairwise distinct primes such that $s_i \neq 0$ mod [$p_i$] for each $i>0$ (and $p_0$ is arbitrary)

From the chinese remainders theorem, there exists a positive integer $x$ such that $x = -s_i + p_i$ mod[$p_i^2$] for each $i$.
Thus $x+s_i$ is divisible by $p_i$ but not by $p_i^2$. It follows that $x+s_i$ is not a power for all $i \geq 0$. We then defined $x_{n+1} = x$.

By induction, the above construction leads to an infinite set with the desired property.

But the question asked by Harazi seems more difficult.

Pierre.

Yes, that was the famous olympiad problem. I gave this new problem to my math teacher (who, believe me, is really a genius!) and he could not solve it. He gave it to other teachers and none could find a solution. Heartwork, I am also blind and I cannot figure out how it follows from Catalan conjecture. And I strongly think that such a set does not exist. Here is another beautiful question: is there an infinite set of natural numbers with the property that the sum of the inverses of the sum of the elements from all non-empty and finite subsets is convergent?

[quote="harazi"]is there an infinite set of natural numbers with the property that the sum of the inverses of the sum of the elements from all non-empty and finite subsets is convergent?[/quote]
I think the answer is yes.  As an example, let S be the set of powers of 4.  For each nonnegative integer n, consider those finite subsets F of S whose maximum element is 4n.  There are 2n such subsets F.  For each such subset F, the sum of F is at least 4n.  Hence the reciprocal of the sum of F is at most 1/4n.  Because there are 2n such subsets F, the sum of the reciprocals of the sums of such subsets F is at most 1/2n.  Summing over every nonnegative integer n gives a finite upper bound, namely 2.

Hello! Does anyone know whether the following result has been proved or disproved:
   Given a number $a$ and some pairs of numbers $m,n$ such that $min(m,n)<b$ where b is also a given number, the equation $ x^m-y^n=a$ has only finitely many solutions in (x,y,m,n)? I imagine it is true, but too difficult to prove. Probably vess or Pierre know something?  If the answer is yes, the problem is solved: there do not exists such a strange set.

[quote="harazi"]Given a number $a$ and some pairs of numbers $m,n$ such that $min(m,n)<b$ where b is also a given number, the equation $x^m-y^n=a$ has only finitely many solutions in (x,y,m,n)?[/quote]
This looks related:

[quote]If $m,n\geqslant 2,\ k\geqslant 1$, and $x^m-y^n=k$, then $|x|,|y|<\exp\exp\left((3m)^{10}n^{10n^3}|k|^{n^2}\right)$ (and a similar bound exchanging $m$ with $n$).[/quote]
P. Ribenboim, [i]Catalan's Conjecture[/i], AMM 103, p. 529-538

Do you have the article? Would it be possible to attach it?

I have the article, but I believe I can't attach it here because of copyright. However, if you give me your e-mail (by PM if you prefer) I would be glad to send the article.

[quote="harazi"]Hello! Does anyone know whether the following result has been proved or disproved:
   Given a number $a$ and some pairs of numbers $m,n$ such that $min(m,n)<b$ where b is also a given number, the equation $x^m-y^n=a$ has only finitely many solutions in (x,y,m,n)? I imagine it is true, but too difficult to prove. Probably vess or Pierre know something?  If the answer is yes, the problem is solved: there do not exists such a strange set.[/quote]

In this case, the gcd would have to be one-or-two, if it wasnt you could prove fermats last theorem with it!

[quote="harazi"]Is there an infinite set of natural numbers such that the sum of the elements of any finite subset is a natural power? It is useless to say that I couldn't solve it, but all persons to whom I gave it didn' t manage to break it. Can you?[/quote]

Even stronger assertion can be proven: there does not exist an infinite set of integers such that the sum of every two different elements is a power.

Suppose that such set exists, say $ A = \{a_1,a_2,a_3,\dots\}$. Write $ a_i = 2^{m_i}(2l_i + 1)$. Since $ A$ is infinite, if we consider parity of all $ l_i$s, at least one of them (say, even $ l_i$, while the other case is completely analogous) appears infinitely many times. Select those elements: $ \{b_1,b_2,b_3,\dots\} = B\subseteq A$, where $ b_i = 2^{m_i'}(4l_i' + 1)$, and further suppose that $ b_i$s are ordered such that $ i < j\Rightarrow m_i'\leqslant m_j'$. We know that for all $ i\geqslant 3$ it must hold
\[ \begin{array}{c} b_i + b_1 = c_i^{p_i} \\
b_i + b_2 = d_i^{q_i}\end{array},\quad(*)
\]
(of course, $ p_i,q_i\geqslant 2$) that is
\begin{align*} c_i^{p_i} & = 2^{m_i'}(4l_i' + 1) + 2^{m_1'}(4l_1' + 1) = \\
& = \begin{cases} 2^{m_1'}\big(2^{m_i' - m_1'}(4l_i' + 1) + 4l_1' + 1\big), & \text{if }m_i' > m_1' \\
2^{m_1' + 1}(2l_i' + 2l_1' + 1), & \text{if }m_i' = m_1'\end{cases},\end{align*}

\begin{align*} d_i^{q_i} & = 2^{m_i'}(4l_i' + 1) + 2^{m_2'}(4l_2' + 1) = \\
& = \begin{cases} 2^{m_2'}\big(2^{m_i' - m_2'}(4l_i' + 1) + 4l_2' + 1\big), & \text{if }m_i' > m_2' \\
2^{m_2' + 1}(2l_i' + 2l_2' + 1), & \text{if }m_i' = m_2'\end{cases}.\end{align*}
Therefore, $ p_i\leqslant m_1' + 1$ and $ q_i\leqslant m_2' + 1$, what means that there is only finitely many pairs $ (p_i,q_i)$. Further, from $ (*)$ follows $ d_i^{q_i} - c_i^{p_i} = b_2 - b_1$, and for $ p_i,q_i$ fixed and $ p_i\neq q_i$ this [url=http://mathworld.wolfram.com/ThueEquation.html]Thue equation[/url] has only finitely many solutions in $ c_i$ and $ d_i$, and for $ p_i = q_i = p$ there is again only finitely many solutions because the difference of two consecutive $ p$th power tends to infinity. That makes a total of only finitely many pairs $ (c_i^{p_i},d_i^{q_i})$, what is clearly a contradiction.

P. S. What famous olympiad problem is this related to?

It seems Thue equation involves terms of the same power only.
I believe finiteness of set of solutions $ x^n\minus{}y^m\equal{}k$ is still an open question (at least my book of 1980 says so).

[quote="Myth"]I believe finiteness of set of solutions $ x^n \minus{} y^m \equal{} k$ is still an open question (at least my book of 1980 says so).[/quote]
This statement is known as the [url=http://mathworld.wolfram.com/PillaisConjecture.html]Pillai's Conjecture[/url], and asks whether there is infinitely many solutions in [b]quadruples[/b] $ (x,y,n,m)$.

The question is (of course) much easier for all the $ n,m,k$ fixed, and that is all we need. You are, however, right that the latter does not follow from the Thue's theorem, but it is nevertheless correct by, for example, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=386672#386672]this[/url] reference; the work of Baker mentioned there is:

Baker, A., Bounds for the solutions of the hyperelliptic equation, [i]Proc. Cambridge Philos. Soc.[/i] [b]65[/b] (1969), 439–444.

Here is a review of that work from MathSciNet:

[quote]Der Autor beweist folgende zwei Sätze über diophantische Gleichungen: Satz 1: (1) Sei $ y^m\equal{}a_0x^n\plus{}a_1x^{n\minus{}1}\plus{}\cdots\plus{}a_n$ mit $ m\geq 3$, $ n\geq 3$; $ a_0\equal{}0$, $ a_1,\cdots,a_n$ ganz, wobei das Polynom auf der rechten Seite von (1) mindestens 2 einfache Nullstellen habe und sei $ H\equal{}\max|a_j|$. Dann gilt für jede ganzzahlige Lösung $ x,y$ von (1), \[ \max\{|x|,|y|\}<\exp\text{}\exp\{(5m)^{10}(n^{10n}H)^{n^2}\}.\] Satz 2: Ist $ m\equal{}2$ und hat das Polynom auf der rechten Seite von (1) mindestens 3 einfache Lösungen, dann gehorchen alle ganzzahligen Lösungen $ x,y$ von (1), $ \max\{|x|,|y|\}<\exp\text{}\exp\text{}\exp\{(n^{10n}H)^{n^2}\}$.[/quote]

My translation:

"The author proves the following two assertions about Diophantine equations: Assertion 1: (1) Let $ y^m\equal{}a_0x^n\plus{}a_1x^{n\minus{}1}\plus{}\cdots\plus{}a_n$ with $ m\geq 3$, $ n\geq 3$; $ a_0\equal{}0$, $ a_1,\cdots,a_n$ integers, where the polynomial on the right side of (1) has at least 2 simple roots, and let $ H\equal{}\max|a_j|$. Then for the integer solution $ x,y$ of (1) it holds \[ \max\{|x|,|y|\}<\exp\text{}\exp\{(5m)^{10}(n^{10n}H)^{n^2}\}.\] Assertion 2: If $ m\equal{}2$ and the polynomial on the right side of (1) has at least 3 simple roots, then for the integer solution $ x,y$ of (1) it holds $ \max\{|x|,|y|\}<\exp\text{}\exp\text{}\exp\{(n^{10n}H)^{n^2}\}$."

Actually, it was Siegel who was considering finiteness of set of solutions of many types of diophantine equations, and from his main result it follows:
If $ m,n\geqslant 2$ with $ \max\{m,n\}\geqslant 3$, and if $ a,b,k$ are given non-zero integers, then the equation $ aX^m\minus{}bY^n\equal{}k$ has only finitely many solutions in integers.
However, I am not sure which paper of his this was proven in (I have only the above statement cited from the already mentioned Ribenboim's survey), and I therefore chose to cite Baker. (Furthermore, the result of Siegel did not include any bound neither on the number nor on the size of the eventual solutions, and Baker earned the Fields Medal for developing a method leading to effective bounds, one of which we have seen here.)


Prove that a triangle in the plane whose vertices have integer coordinates cannot be equilateral.

[hide="Solution"] The triangle has integer or half-integer area by Pick's Theorem.  But because its side length $ s$ satisfies $ s^{2}\in \mathbb{N}$, the area is also $ \frac{s^{2}\sqrt{3}}{4}$ which is irrational.  Contradiction. [/hide]
[b]Generalization:[/b]  Prove that for $ n > 4$, an equilateral, equiangular $ n$-gon cannot have vertices all lattice points.


Four spheres each of radius r lie on a horizontal plane all touching each other. A fifth sphere is to be placed so that it touches all the other four spheres. Prove that its radius is r/(2 + sqrt(6)).

[hide=" Hint"]

Consider sections of the square-based pyramid whose vertices are the centres of the spheres.

[/hide]

It is not possible for 4 congruent spheres and a plane in 3 dimensions to be mutually tangent to eachother.

I think you either did not mean to mention the plane, or (considering your hint) The spheres are all tangent to the plane, and the 4 congruent spheres' centers form a square.

Thaks stevenmeow. You're right, I should have written that the four spheres are all tangent to the plane and that their centres form a square.


For any three positive reals a, b, c, we have

$\frac{a^2+bc}{b^2+c^2}+\frac{b^2+ca}{c^2+a^2}+\frac{c^2+ab}{a^2+b^2} \ge \frac{5}{2}$

Okay:

From http://www.mathlinks.ro/Forum/viewtopic.php?t=26104 , we have

$\frac{2a^2+bc}{b^2+c^2}+\frac{2b^2+ca}{c^2+a^2}+\frac{2c^2+ab}{a^2+b^2} \geq \frac92$.

From http://www.mathlinks.ro/Forum/viewtopic.php?t=6420 , we have

$\frac{a^2+2bc}{b^2+c^2}+\frac{b^2+2ca}{c^2+a^2}+\frac{c^2+2ab}{a^2+b^2} \geq 3$.

Adding and dividing by 3, we get your inequality

$\frac{a^2+bc}{b^2+c^2}+\frac{b^2+ca}{c^2+a^2}+\frac{c^2+ab}{a^2+b^2} \geq \frac52$.

  Darij

normalize $\sum a^2 = 1$

Then $a^2 + bc = 1 - \frac{1}{2}(b^2 + c^2) - \frac{1}{2}(b-c)^2$, whence:

$LHS = \sum \frac{a^2 +bc}{b^2 + c^2} = \left( \sum \frac{1}{b^2 + c^2} \right)  - \frac{3}{2} - \left( \frac{1}{2}\sum \frac{(b-c)^2}{b^2 + c^2} \right) $

By CS in engel form, this is $\geq 3$.

But this is $\geq 5/2$.

[quote="Singular"]By CS in engel form, this is $\geq 3$.
[/quote]

No...

Try $(a,b,c) = (1,1,0)$.


inequalities function search AMC AIME USA(J)MO USAMO

This is my first post on this forum, so hello everyone ;). I have always been interested in maths, but probably not enough. I'm 18 now and I want to go to a competition this year. I'm above the normal average maths student (I can understand stuff easily) but I've never prepared for a competition. I think that it's not that late since I'm not targeting the IMO gold medal. Can you give me some advice on where to start? What first?  I'm familiar with some concepts (algebra, linear algebra, a bit of number theory, inequalities), but I think not enough for a competition (I'm not good with function algebra, if that's the correct term). Any suggestions? Thanks in advance

This question is a very popular one.  I suggest you look around (& try the search function) to see what people have said before.

Yeah, I suspected that it is a popular one. I just couldn't find any topics covering it. There should definitely be a wiki entry :wink:

I would recommend that you check out the USA [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182]Contest Page[/url]. Look at AMC, AIME, and USAMO problems, and determine which ones currently meet your level, then practice. (Based on what you've said, you'll probably skip at least most of AMC.)

As well, you can look around AoPS and AoPSWiki for any well-known theorems you might want to learn.


One geometry problem

A triangle $ \vartriangle ABC$ is given .We consider that two circles $ \left( {K,R_1 } \right),and\left( {L,R_2 } \right)$ which are moving maintaining of their radii $ R_1 ,R_2$ and KL and also they are tangent to AB and AC respectively. Prove that there exists one third circle $ \left( {M,R} \right)$ which tangents to BC maintaining its radius length R and also the triangle $ \vartriangle MKL$ equal to itself .

S.E.Louridas

Suggestion:

The problem is equivalent to this one:
If one triangle ABC is moving such that its sides   AB, AC tangent to two constant circles, prove that BC will tangent to a constant circle too.

S. E. Louridas

Solution of the problem of geometry

From the centers K, L we consider parallels to BA, CA respectively and let
$ {\rm M}^'$ be their section point, $ {\rm M}_1$the section point of
$ {\rm M}^' {\rm K}$ with line BC and $ {\rm M}_2$ the section point of $ {\rm M}^' L$ with line BC. The triangle
$ {\rm M}^' {\rm M}_1 {\rm M}_2$ maintains its sides .Thus its altitude
$ {\rm M}^' {\rm M}_3$ will be of constant length too. The
$ \angle KM^' L$ will be constant since  $ \angle KM^' L = \angle BAC.$ So, point $ {\rm M}^'$ belongs to a constant
arc $ KM^' L.$Now, we consider point M on this arc such that
$ M^' M\parallel BC.$ Then
$ \angle MM^' K = \angle M^' M_1 C = \angle ABC.$ Thus, the cord of arcMK is constant which means than point M is constant too. If then,
$ MS \bot BC \Rightarrow MS = M^' M_3^'  \Rightarrow MS$ (S point of BC) is a fragment of constant length. Finally BC tangents to the constant circle
$ \left( {M,MS} \right).$

S.E.Louridas


inequalities
trigonometry
inequalities proposed

Let's take the acute triangle $\ ABC$ whith $\ a=max(a,b,c)$.Prove that $\ b^2+c^2-a^2<ab+bc+ca$, whith
$\ a,b,c>0$

Yes, you are right.But I think we cannot have equality.Am I wrong? :?

$b^2+c^2 - a^2 \leq 2bc$ only has equality when $\cos A$ is equal to 1, making $A = 0$. But $a$ is the longest side so $A$ is the largest angle, so I believe equaity only occurs in the case where $\triangle ABC$ is a point ($a = b = c = 0$).

OK, but then I will add that $\ a,b,c>0$.

Then there's no equality as far as I can see.


I think the problem was wrong but I'm not sure, I will ask my teacher. :wink:


I am looking for a copy of the paper that used to be here:
http://www.xula.edu/math/Research/Colloquium/Past/Colloquium20030320-Paper-SahooP.pdf

The paper is called "sum of kth powers"
Does anyone have a copy or know where else I can find this paper?

Thanks.

Luckily it has been archived at
[url]http://web.archive.org/web/20060901095523/http://www.xula.edu/math/Research/Colloquium/Past/Colloquium20030320-Paper-SahooP.pdf[/url]


What do the following statements have in common? (That is, what do they have in common that is more specific than that they are statements about geometry, a fact which I already gave away in the thread title.  :wink:  )

PROPOSITIONS

I. The sum of the angles of a triangle is two right angles (that is, 180 degrees).

II. In any triangle, each exterior angle equals the sum of the two remote interior angles.

III. Straight lines parallel to a third line are parallel to each other.

IV. For any three noncollinear points, there exists a circle passing through them.

V. Rectangles exist.

VI. For a right triangle with legs [b]a[/b] and [b]b[/b] and hypotenuse [b]c[/b],

[tex]\displaystyle{a^2 + b^2 =c^2}[/tex]

For this Intermediate level, a clear discussion of what these statements have in common is enough. If posted on the Advanced level, this challenge problem would ask for rigorous proof of each statement, which any of you are welcome to provide for extra practice if you like.

This is just a guess, but were they all discovered by Pythagoras, I know for sure that I and VI were.

They all rest on one of Euclid's axioms, namely "Given any line and a point not on that line, there exists a unique line through the point parallel to the given line." This axiom (and therefore these propositions also) do not hold in other, non-euclidian geometries. For example, in hyperbolic geometry, there exist INFINITELY many lines through the point parallel to the given line, and in spherical geometry, there exist NO lines through the point parallel to the given line.

Good on you, Zabelman. I remember being really startled about three years ago when I read that "rectangles exist" is a statement dependent on the parallel postulate. Of course "the sum of the angles of a triangle is 180 degrees" is a staple of SAT I geometry, so I made sure to tell my son that the geometry of the SAT I is EUCLIDEAN geometry, not the only kind of geometry possible, because that "fact" and other facts that only hold for Euclidean geometry appear on every question booklet for the SAT I.

I could be wrong, but I think in the directions of the SAT, they do talk about what type of geometry they are using.  I would guess they use the term "plane geometry," which carries with it the implication that the parallel postulate holds.  I don't remember this for sure, except that I know they give directions concerning what assumptions you may make about the geometry questions.

Zabelman, you shouldn't be solving challenge problems in the Intermediate forum! :)

[quote="JBL"]Zabelman, you shouldn't be solving challenge problems in the Intermediate forum! :)[/quote]

Some of the problems in [b]Intermediate Topics[/b] forum are harder than problems in [b]Advanced Problems[/b] forum. :-)


geometry
rectangle
trigonometry
perpendicular bisector
geometry proposed

\[ \triangle ABC,\ AB=AC,\ A=20^{\circ},\ D\in (AB),\ AD=BC\Longrightarrow m(\widehat {BDC})=30^{\circ} \]\[ (Ho-joo\ Lee,\ Korea) \]

Every time I ponder this problem, I find out I have forgotten my proof, so I have to come up with a new one. And I could swear it's a different one each time! :)

Construct the points $E,F$ s.t. $ACE,AEF$ are triangles congruent to $ABC$, with $ABF$ being equilateral, and take $G$ on $AF$ s.t. $AG=AD=BC$. The perpendicular bisector of $BF$ is also the perpendicular bisector of $CE$ and $DG$. Furthermore, we have $DG=AD=BC=CE$, because $ADG$ is equilateral, so $DGEC$ must be a rectangle, meaning that $\angle GDC=90^{\circ}$. Since $\angle GDA=60^{\circ}$, we get $\angle ADC=150^{\circ}$, which is equivalent to what we want to prove.

Perhaps , a little easier.

Construct equilateral  $\triangle ABF$ ($BF$ meets $AC$) Then triangles $ADC$ and  $BCF$  are equal($AD = BC , AC = BF$ and $\angle DAC = \angle CBF = 20$). Thus $\angle DCA = \angle CFB = 10$ and hence $\angle BDC = 20 + 10 = 30$

Babis

[i](there are many solutions to this << best seller >> problem )[/i]

here is a solution:

let K be a point on AB inwich ,angle KCA=20.and let M be a point on KC,inwich ,CM=AD.

then we have triangle CMB should be equlirateral.then :angle BAM=angle CAM=10(its clear why)

then because of the similarity betwin the triangles AKM and DCM,we have :angleKAM=angle DCK=10.

then in triangle CKD we have:angle KDC= 180-(10+140 )=30

a trig solution is not the nicest one, but it certainly works...

let $\angle CDB=\alpha$, so by the sine law we have
$\displaystyle\dfrac{\sin\alpha}{\sin(\alpha-20)}=\dfrac{AC}{AD}=\dfrac{AC}{BC}=\dfrac{\sin80}{\sin20}=\dfrac{1}{2\sin10} \\\Rightarrow 2\sin10\sin\alpha=\sin(\alpha-20) \\\Rightarrow 2\sin10(\sin(\alpha-30)\cos30+\cos(\alpha-30)\sin30)=\sin(\alpha-30)\cos10+\sin10\cos(\alpha-30) \\\Rightarrow (\sqrt3\sin10-\cos10)\sin(\alpha-30)=0$,
but it is obvious that $\sqrt3\sin10-\cos10$ is not $0$,
so $\sin(\alpha-30)=0 \\\Rightarrow \alpha=30$


calculus derivative integration function quadratics algebra polynomial

This is exactly equivalent to its usually much less cryptic statement as $ \sum_{n\equal{}0}^{\infty} {2n \choose n} x^n \equal{} \sqrt{1 \minus{} 4x}$ with possibly a derivative or integral thrown in.

[quote="t0rajir0u"]This is exactly equivalent to its usually much less cryptic statement as $ \sum_{n \equal{} 0}^{\infty} {2n \choose n} x^n \equal{} \sqrt {1 \minus{} 4x}$ with possibly a derivative or integral thrown in.[/quote]
Could you please elaborate why these two statements are equivalent?

[hide="Elaboration"] First of all, I had that one wrong; it should have been $ \frac{1}{ \sqrt{1 \minus{} 4x}}$.  Second of all, it will actually be easier to start directly from the generating function of the [url=http://mathworld.wolfram.com/CatalanNumber.html]Catalan numbers[/url],

$ \sum_{n\equal{}0}^{\infty} \frac{1}{n\plus{}1} {2n \choose n} x^n \equal{} \frac{1 \minus{} \sqrt{1 \minus{} 4x}}{2x}$

which is attained from the generating function I gave you by integrating.  Now let $ x \equal{} \frac{1 \minus{} y}{4}$. [/hide]

Nice trick
But I am afraid that using of Catalan numbers is unpractical here.I mean, the given identity is usually used to establish the formula of catalan numbers, not vice versa.It is like to prove that there are infinitely many prime numbers via Dirichlet's theorem,sorry for exaggerating.

[quote="t0rajir0u"]That doesn't matter; I'm telling you that the two identities are exactly equivalent.  The proof that that generating function works can be done in one of two ways:  you can slog through it with the binomial theorem (which is probably how the proof is "supposed" to go in this case - you're doing a Taylor series expansion around $ x \equal{} 1$) or you can use the well-known recurrence for the Catalan numbers to write a quadratic polynomial that the generating function must satisfy.[/quote]
Taylor series? :huh:
don't forget this is pre-olympiad section;I do agree that the second way,by using of the generating functions is quite good idea,and via this approach we could establish the given formula,but in that case,we should have simultaneously used an explicit formula of the catalan numbers.Let's just not use the Catalan numbers.
P.S:
another comparison:
we could say that there are infinitely many prime numbers,because there are infinitely many of them of the form $ pk \plus{} 1$,what can be proved via cyclotomic polynomials...this approach really works,but it seems...uh...far-fetched,unnatural.

You seem to be under the impression that my argument is either circular or proves too strong a result.  Let me state it in the most basic way.

[b]Generalized Binomial Theorem:[/b]  Let $ \alpha \in \mathbb{C}$ be a complex number and define $ {\alpha \choose k} \equal{} \frac{\alpha(\alpha \minus{} 1)...(\alpha \minus{} (k\minus{}1))}{k!}$.  Then

$ (x \plus{} 1)^{\alpha} \equal{} 1 \plus{} {\alpha \choose 1} x \plus{} {\alpha \choose 2} x^2 \plus{} ....$

[i]Proof.[/i]  Taylor series.  You are free to consider this result as "advanced" as you want:  at the heart of it, it's the only sensible extension of the usual binomial theorem.

Note that when $ \alpha$ is not a non-negative integer this sum is infinite.  Now compute

$ \sqrt{x \plus{} 1} \equal{} 1 \plus{} {\frac{1}{2} \choose 1} x \plus{} {\frac{1}{2} \choose 2} x^2 \plus{} ...$

and substitute accordingly.  This is one way to prove the generating function for the Catalan numbers and is as elementary an approach as you are going to get to prove this infinite sum.  If I still haven't addressed your concern that I'm using tools beyond the scope of the problem, please tell me what kind of a solution you are expecting.

Here's a slightly different take than t0r's, one that avoids invoking the Binomial Theorem (although there's nothing wrong with doing so).  Define the Catalan numbers by $ C_0 \equal{} 1$, $ C_{n \plus{} 1} \equal{} \sum_{i \equal{} 0}^n C_i C_{n \minus{} i}$.  Then there are multiple combinatorial proofs that $ C_n \equal{} \frac {1}{2n \plus{} 1} \binom{2n}{n} \equal{} \binom{2n}{n} \minus{} \binom{2n}{n \minus{} 1}$ (e.g., the reflection principle applied to Dyck paths, or others).  It's immediate to show that the generating function $ C(x) \equal{} \sum_{n \equal{} 0}^\infty C_n x^n$ satisfies $ C(x) \equal{} 1 \plus{} xC(x)^2$, and this gives $ C(x) \equal{} \frac {1 \minus{} \sqrt {1 \minus{} 4x}}{2x}$ (we can choose between the two roots by setting $ x \equal{} 0$), from which we can conclude your identity by polynomial manipulations.

Yes, I mentioned that proof; Erken's comment makes it appear as if he thinks invoking the Catalan numbers is a circular argument, so I wanted to dispel that particular objection entirely.

Right, I was making the fuller argument (that shows the non-circularity) to support your point.  I think this is probably the nicest way to go about the problem, but I say that as a combinatorialist :).

[quote="t0rajir0u"]You seem to be under the impression that my argument is either circular[/quote]
Exactly.Don't you think that the argument provided by JBL and suggested by you is not circular?
[quote="JBL"]I think this is probably the nicest way to go about the problem[/quote]
I disagree with you,somehow far-fetched arguments.
[quote="t0rajir0u"]You are free to consider this result as "advanced" as you want[/quote]
Hmm...not really advanced,but I knew both ways to establish the identity.
Thank you,guys.

[quote="Erken"][quote="t0rajir0u"]You seem to be under the impression that my argument is either circular[/quote]
Exactly.Don't you think that the argument provided by JBL and suggested by you is not circular?[/quote]  I'm deeply confused: what about the argument as I presented it could be described as circular?  Hub-and-spoke, perhaps (everything is tied to the recursive definition of $ C_n$), but circular?  (There was a real reason I made reference to the combinatorial proofs of the formula for the Catalan numbers -- these are far more natural* than the generating functional proof, in which the exact formula is just a funny consequence of some calculus.  In particular, they explain why there should be some binomial coefficients involved.  See the second and third proofs of the formula on the [url=http://en.wikipedia.org/wiki/Catalan_number]wikipedia page[/url].)

*Again, I'm a combinatorialist, showing off my bias :)

[quote="JBL"]I'm deeply confused: what about the argument as I presented it could be described as circular?  Hub-and-spoke, perhaps (everything is tied to the recursive definition of $ C_n$), but circular?  (There was a real reason I made reference to the combinatorial proofs of the formula for the Catalan numbers -- these are far more natural* than the generating functional proof, in which the exact formula is just a funny consequence of some calculus.  In particular, they explain why there should be some binomial coefficients involved.  See the second and third proofs of the formula on the [url=http://en.wikipedia.org/wiki/Catalan_number]wikipedia page[/url].)
*Again, I'm a combinatorialist, showing off my bias :)[/quote]
hmm...this is how I understand your reasoning:
By using of the recursive relation of the Catalan numbers and the explicit formula of these numbers,what could be established,as you said,by virtue of the combinatorial proof,you suggest to get back to considering generating function, square it and so on...we would prove the identity,and [u]I do agree that this approach works[/u] and [u]I do not mean that it is somehow ugly or irregular,[/u]but you must admit that it is not absolutely natural, what I am trying to say is that,one of the most precisely ways(I am not a combinatorialist,though I like the combinatorics ) to establish the explicit formula of the Catalan numbers,is to use the generating function,by virtue of the given recursive relation,we are obtaining a quadratic trinomial,and the rest is known.

[quote="Erken"]but you must admit that it is not absolutely natural, what I am trying to say is that,one of the most precisely ways(I am not a combinatorialist,though I like the combinatorics ) to establish the explicit formula of the Catalan numbers,is to use the generating function,by virtue of the given recursive relation,we are obtaining a quadratic trinomial,and the rest is known.[/quote]
Wait - which approach do you think is unnatural?  The approach you are describing is JBL's approach.  Personally, I prefer to consider every proof of a result in ensemble rather than to single one out to gain a deeper understanding of "why" the result is true.

Well, I'm completely perplexed,which of them is yours,and which is JBL's?
Let me make it clear,don't, the previous reply of JBL and what you meant by "you can use the well-known recurrence for the Catalan numbers to write a quadratic polynomial that the generating function must satisfy", imply the same idea?

Nevertheless, I wonder, what makes you think that I was refering to you.I even quoted JBL's approach, and throughout my reply, you could note that I was talking about JBL's approach, but as I emphasized in my previous sentence, I thought that it was your both method.

Concerning your solution, which was not anyhow described or noticed by JBL, that was really nice, but...hmm...could we find another approach to establish the identity?

Actually,I had heard before,that the given identity can be proved via Taylor series.
By the way,I've known that, for the first time, from some paper about the explicit formula of the Catalan numbers and how to determine this formula by virtue of generating functions and Taylor series, but I expected to see somehow more complicated and ugly proof.

Thank you and sorry for the occured misconceptions.

[quote="Erken"]Nevertheless, I wonder, what makes you think that I was refering to you.I even quoted JBL's approach, and throughout my reply, you could note that I was talking about JBL's approach, but as I emphasized in my previous sentence, I thought that it was your both method.[/quote]
My apologies; I misread part of your post.  So what you are saying is that the explicit formula $ C_n \equal{} \frac{1}{n\plus{}1} {2n \choose n}$ is best established using the generating function, rather than the other way around?

If you really think so, then I think you should take a look at the combinatorial argument JBL suggested in further detail; read up on [url=http://mathworld.wolfram.com/DyckPath.html]Dyck paths[/url] in particular.

reflection method applied to Dyck path?
hmm...well,that could quite natural describe the explicit formula,but afterwards we should get back to considering Dyck path,and to construct non-trivial bijection between it and correctly matched parentheses structure,right?
The last structure clearly satisfies to the recurrent formula of the Catalan numbers,but it would be way too many non-trivial bijections.So due to all these significant steps,I am convinced that the generating function is the most natural and...uh...easy-guessable way,of course,assuming that these steps entirely coincide with what you meant...

Dyck paths to matched parentheses is a trivial bijection: up steps are open parentheses, down steps are close parentheses.  This sends the obvious demonstration that parentheses satisfy the recursion to an equally obvious demonstration that Dyck paths satisfy the recursion: replace "point at which we have matched all the parentheses we've seen so far" with "point at which we return to the line $ y \equal{} x$."  So, one combinatorial argument and (perhaps) one trivial bijection.  (I imagine it's also possible to phrase the reflection argument in terms of matched parentheses, but for some reason I've never seen anyone present it that way.)


algebra
polynomial
induction
function
algebra solved

hey guys please help!!
  p(x) is a polynomial with real coefficients such that p(x 2  - 1) = p(x) p(-x). What is the largest number of distinct real roots that p(x) can have?
polynomials make me go down on my knees
cheers :D  :D

I think I have seen this problem 200 times :D.. twice at least on this forum (the 1st post was made by vineet)

So a solution can be found is [b]Valentin Vornicu's book[/b]:

The problem was like so which will help you with your problem :D

[b]Find out all polinomials R[x] which have all their roots real which verify the following identity for any x real:

f(x^2-1)=f(x)*f(-x). [/b]

[b]Solution[/b]:

We can see that if x^k | f then f/x^k verifies the conditions of the ipotesys and from the given relation we get that: f(-1)=f(0)*f(0) so -1 is a root. In a simmilar way -1 is a root. So we can easily say that: f=(x^2+x)^k*f'(x).

We can see that if x0 is a root then we must have g(x0), where g(x)=x^2-1, as a root of polinomial f', and using induction we have that g^n(x0)=g(g(..g(x0)..)) is a root of f'. But f' has a finit number of roots, so there exist m,n natural so that g^m(x0)=g^n(x0).

We need to see that there are some fixed points of function g which verify the last relation:
x^2-1=x=>x1,2=[1 \pm sqrt(5)]/2

It is easy to see that if x1 or x2 is a root of f', we can divide it with x-x1 and x-x2 respectively and the polinomial that we will get will still verify the relation from the hypotesys.

So we can consider that f=(x^2+x)^k*(x-x1)^m*(x-x2)^n*h(x). Then h(x) verifies the relation from the hypotesys and lets consider a as it's smallest root. Then we have some cases we can consider:

1) a>x2>x1 => a^2-1>a and more generaly g^(n+1)(a)>g^n(a) and because g^n(a) is a root of h(x) we get an infinity of roots of h(x).

2) x2>a>x1 =>a^2-1<a and so we got a smaller root of h(x), contradiction.

3) a<x1<x2 then we have that h(x)=(x-a)*g(x) so h(x^2-1)=(x^2-1-a)*(g(x^2-1)) and because h has all its roots real and x^2-1-a has all its roots real, then we must have that a>=-1 =>-1<=a<x2. But g^2(a)=(a^2-1)^2-1=a+a(a+1)(a-x1)(a-x2) and because a(a+1)(a-x1)(a-x2)<0 we have that (a^2-1)^2-1<a so we got another root which is smaller, contradiction.

So all the polinomials that verify the conditions are:

[b] f=(x^2+x)^k*(x-x1)^m*(x-x2)^n [/b]

cheers! :D


James has a collection of 20 miniature cars and trucks. Cars make up $ 40\%$ of the collection. If James adds only cars to his collection, how many cars must he add to make the collection $ 75\%$ cars?

James currently has 8 cars and 12 trucks. To get $ 75\%$ cars, he needs 36 cars total. 36-8=28.

Or you can set up a proportion.
Let $x$= the number of cars he adds to his collection

$\frac{8+x}{20+x}=\frac{3}{4}$

$x=\boxed{28}$


Let n is a positive integer. A polynomial P(x) with real coefficients satisfies P(0)=P(n)  . Prove that there exists at least n different pairs of real numbers (x;y) ,  y-x is a positive integer such that: P(x)=P(y)

i will prove that for all $ 1 \geq i \geqq$ $ \exists x_i$    such that  $ f(x_i)=P(x_i+\frac{i}{n})-P(x_i)=0$

so you have$ f(0)+f(\frac{i}{n})+f(\frac{2i}{n})+...+f(\frac{(n-1)i}{n})=P(n)-P(0)=0$

where that you can conclud that $ \exists$ $ a$ and $ b$  such that  $ f(\frac{ai}{n})f(\frac{bi}{n}) \le 0$

so there $ x_i$    such that$ f(x_i)=0$

let be $ y_i=x_i+\frac{i}{n}$  and it easy to prove that for all $ i \ne j$ so  $ (x_i,y_i) \ne (x_j,y_j)$


Hi All

This is my first time using LaTeX (Lyx), and I have a question on long addition and multiply

On my homework I need to multiply 2 binary number as follow

     11101
x   11011
-----------
     11101
    111010
    ..........

I just couldn't get it look nice in ms word.
Could you let me know how it's done in LaTeX (Lyx)

Thanks for your help.

Have a look at the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=xlop]xlop package[/url]. You'll need to experiment with it but, for example,
[code]\usepackage{xlop}
.....
\newcommand\hole[1]{\ }
\opmul[displayshiftintermediary=none,resultstyle=\hole]{11101}{11011}[/code] looks like this


A car has a 4-digit integer price, which is written digitally. (so in digital numbers, like on your watch probably)

While the salesmen isn't watching, the buyer turns the price upside down and gets the car for 1626 less. How much did the car initially cost?

[b]Reply by [u]darij grinberg[/u][/b]

Hmm... what do you mean by "turns the price upside down"? Is it reversing the order of the numbers or is it something of the kind 6 --> 9 ? Since, if it would be reversing the order of the numbers, then the difference would be divisible by 9, what is not the case for 1626... but maybe I misunderstand something. If it is something else, then does it enforce 1 <--> 7 or not? And digits like 4 cannot occur at all, can they?

  Darij

[b]Reply by [u]Arne[/u][/b]

1 -> 1
2 -> 2
3 -> ? (does not exist)
4 -> ? (idem)
5 -> 5
6 -> 9
7 -> ? (idem)
8 -> 8
9 -> 6

you do both

so the changing sheme arne replied is done, plus the order is reversed

but oh well seeing that was already half the problem :P

my solution and mathmax12's solution:

Let the number be $abcd-dcba=999a+90b-90c-999d=1626$, doing bashy casework (which took me 8 min to do) we get $\boxed{6815}.$


[quote="skywalker"]Let $ABC$ be a triangle and an arbitrary point $P$ in plane, let $A'B'C'$ be pedal triangle of $P$ with $ABC$ and $A''B''C''$ be pedal triangle of $P$ with contact triangle of $ABC$ prove that
$PA'\cdot PB'\cdot PC'=PA''\cdot PB''\cdot PC''$[/quote]

Maybe you want to assume that $P$ lies on the incircle of triangle $ABC$ ?

  darij

Oh that's right I think $P$ should lie on incircle of $ABC$  :D


probability
Support
algebra
polynomial
algorithm
number theory
mathematicians

I am required to introduce a famous modern mathematician to my class that is from a minority group.

So far all the ones I have been able to come up with are Italian, French, or from ancient worlds.   Can someone please give me some directions where to look?  Again they need to be from a minority group (Spanish, Indian, Russian, etc.)  Women don't count.

P.S.  I hate forced lessons.  ARGH!!   :?

[url=http://scienceworld.wolfram.com/biography/Ramanujan.html]Ramanujan[/url].  (Does early 1900s count as modern?)

[quote="fedja"][quote="MrsGardner"]
So far all the ones I have been able to come up with are Italian, French, or from ancient worlds.   Can someone please give me some directions where to look?  Again they need to be from a minority group (Spanish, Indian, Russian, etc.)  Women don't count.
[/quote]
O tempora, o mores! Since when the Spanish, the French, the Russian and other non-American nationalities are refered to as "minority groups"? Most of them have been thriving nations long before the USA was formed and remain such nowdays.  :(  With this approach the task is easy: take any recent Fields laureat and with great probability he (I would say "or she", but, "women don't count") will be not an American or, at least, be of foreign origin (especially if you trace his ancestry back to early 1700's). Seriously, do your teachers think that the USA is the only "major" country on this planet?[/quote]

Ok fair enough :blush:  but before you put this all on me please realize that what I posted was a direct quote from my Beginning Teacher Support Mentor.

And what I meant was women don't count to satisfy the required “minority groups” that I am required to cover in this lesson.

Again, not my idea/or wording, just what I am told I have to do for the BTSA program.

I think I should teach about important mathematicians regardless of there country of origin.  Then again what do I know?

From the Math forum a description of their work:

"Prof. Manindra Agarwal and two of his students, Nitin Saxena and Neeraj Kayal, have discovered a polynomial time deterministic algorithm to test whether an input number is prime. Lots of people over centuries have been looking for a polynomial time test for primality, and this result is a major breakthrough, likened by some to the P-time solution to Linear Programming announced in the 1970s. "

A link to the full text of their paper:
http://www.cse.iitk.ac.in/news/primality.pdf

A link to FAQs on that subject:
http://crypto.cs.mcgill.ca/~stiglic/PRIMES_P_FAQ.html

Naga

Probably one of the most famous mathematicians of the 20th century was Paul Erdos (Hungarian Jewish by nationality). You can find a lot of material about him on the web and there are some good books about him too. You may look at his short biography on [url]http://www-history.mcs.st-and.ac.uk/history/Mathematicians/Erdos.html[/url]. The advantage of this choice is that he proved a lot of theorems in number theory and combinatorics whose formulations can be understood even by school kids, so you can really talk not only about his life but also about his work.  :)

There's something so peculiarly American sounding about that phrase "from a minority group." The people I'm about to mention would [i]never[/i] have thought of themselves that way. On the other hand, I know of quite a few distinguished 20th Century mathematicians from South America, particularly Argentina. Many of them moved to the United States or to Europe in their professional careers.

Foremost among those South Americans would be my own Ph.D. advisor, the late, great Alberto P Calderón.

For more information about him - in fact, for more information about hundreds of mathematicians - I strongly recommend the St. Andrews University history website.

[url=http://www-history.mcs.st-andrews.ac.uk/history/]St. Andrews history site.[/url]

The Erdös reference that fedja gave was from the St. Andrews site. Just start at the root of that site and explore. You'll find things.


superior algebra
superior algebra unsolved

Let $G$ be a finite group, and $D$ a subset of $G$ for which every nontrivial $g\in G$ can uniquely be written as $g=a^{-1}b$ for $a,b \in D$.
(i) Show that every nontrivial $g\in G$ can be written uniquely as $g=ab^{-1}$ for $a,b \in D$.
(ii) Show that for every nontrivial $g\in G$, the intersection $D\cap Dg$ contains exactly one element.

Peter, you yourself already have the solution( it reminds me of one of the very first problems in the syllabus i gave you?)

Look at the order of G and D

you easily prove by careful counting that $G= |D|^2-|D|+1$

now define this map from $D * D$ to $G$ by sending $(a,b)$ to $ab^{-1}$

You can easily conclude by counting that every nontrivial element appears exactly once in the image

for the second part, notice that if $g=d_{2}^{-1}d_{1}$ , then $d_{1}=d_{2}g$ is in both $D$ and $Dg$

I have the solution to almost all the exam problems now. I just wondered why no one solves these beautiful questions... (except for Fredoo Torpedo (:P) but he already solved them all during classes)
These are the ones that challenged me most, since I though that was needed for the MathLinks public. Either they're still considered trivial/boring, or they're a bit overchallenging. (in the latter case I could post easier ones)


Given a triangle $ABC,$ assume that the values of $\tan A,\ \tan B,\ \tan C$ are all integers. Find these values.

[hide]
Assume $\tan A$ and $\tan B$ are integers.

since $A+B+C=180$
$C=180-(A+B)$

$\tan C=-tan (A+B) =-\frac{\tan A+\tan B}{1-\tan A*\tan B}$

use guess and check you get these numbers are
$\boxed{\tan A=1, \tan B=2, \tan C=3}$
[/hide]

Let $\tan A = a$ etc. It is well-known that $abc = a+b+c$. Therefore, $a(bc-1) = b+c$.

[i]Lemma:[/i] $a,b,c > 0$.
[i]Proof:[/i] Clearly, $a,b,c \neq 0$, because otherwise one of $A,B,C$ is equal to 0, which is impossible. Now suppose that for example $a<0$. We then have $A > 90^{\circ}$. Now if for example $b<0$, then $B > 90^{\circ}$, so $A+B > 180^{\circ}= A+B+C$, contradiction. So at most one is negative, i.e. $b,c > 0$, i.e. $b,c \geq 1$, i.e. $B,C \geq 45^{\circ}$. But then $A+B+C > 90^{\circ}+45^{\circ}+45^{\circ}= 180^{\circ}= A+B+C$, contradiction. And we're done. $\blacksquare$

Now suppose we would have $bc-1=0$, then $b+c = a(bc-1) = 0$, but that's impossible, as $b,c > 0$. Therefore, $bc-1\neq 0$ and we can divide $a(bc-1) = b+c$ by $bc-1$. We get $\frac{b+c}{bc-1}= a \in\mathbb{Z}$. We should have that $bc-1 | b+c$, which is only possible if $|bc-1| \leq |b+c|$. Now, $bc-1 \geq 1\cdot 1-1 = 0$, so $|bc-1| = bc-1$ and $b,c > 0$, so $|b+c| = b+c$. So we should have that $bc-1 \leq b+c$, i.e. $(b-1)(c-1) \leq 2$. Let WLOG $b\leq c$.

If $b=1$, then $\frac{b+c}{bc-1}= \frac{c+1}{c-1}= 1+\frac{2}{c-1}= a \in\mathbb{Z}$, so $c-1 | 2$, so $c-1 \in\{1,2\}$, so $c\in\{2,3\}$. We get the solutions $(a,b,c) = (3,1,2),(2,1,3)$.

If $b=2$, then $c-1 = (b-1)(c-1) \leq 2$ iff $c\leq 3$. But $2=b \leq c$, so only $c=2,3$ is possible. If $c=2$, then $a = \frac{b+c}{bc-1}= \frac{4}{3}$ isn't an integer. And if $c=3$, we get $(a,b,c) = (1,2,3)$.

If $b\geq 3$, then $(b-1)(c-1) \geq (b-1)(b-1) \geq 4 > 2$, so no solutions.

So $(a,b,c) = (1,2,3)$ with all permutations.

Applying the tangent sum rule twice, we get $\tan{(A+B+C)}= \frac{\tan{A}+\tan{B}+\tan{C}-\tan{A}\tan{B}\tan{C}}{1-(\tan{A}\tan{B}+\tan{A}\tan{C}+\tan{B}\tan{C})}$. We also know that $A+B+C = \pi \Rightarrow \tan{(A+B+C)}= 0$, so the numerator in that fraction must equal 0.

Define $x$, $y$, and $z$ as $\tan{A}$, $\tan{B}$, and $\tan{C}$, respectively. Then $x+y+z = xyz$, which gives 1, 2, and 3 as values.

Edit: I didn't go into that much depth... oh well, I explained the part that was well known.  :D

[quote="kunny"]Justify your answer. :D[/quote]

haha...
assume $tanA=a$, $tanB=b$, $tanC=c$
$c=-\frac{a+b}{1-ab}=\frac{a+b}{ab-1}$

for $tanX$ to be negative, x has to be in second or fourth quardrant. Now we are considering triangles, so x must be an obtuse angle.. It's not hard to prove in an obtuse triangle, there must be an angle less than 45 degrees. Therefore, in an obtuse triangle ABC must have an angle with its tangent value is not an integer.

Hence ABC must be acute triangle, which means $a>0$, $b>0$, $c>0$

To make c an integer,
$a+b{\geq}ab-1$
$ab-a-b-1{\leq}0$
$ab-a-b+1{\leq}2$
$(a-1)(b-1){\leq}2$

the only solutions are
$a-1=1, b-1=2$ or $(a,b)=(2,3)$, which leads to $c=1$
$a-1=2, b-1=1$ or $(a,b)=(3,2)$, $c=1$
$a-1=0, b-1=2$ or $(a,b)=(1,3)$, $c=2$
$a-1=2, b-1=0$ or $(a,b)=(3,1)$, $c=2$
$a-1=0, b-1=1$ or $(a,b)=(1,2)$, $c=3$
$a-1=1, b-1=0$ or $(a,b)=(2,1)$, $c=3$

if $a-1=b-1=1$  or $a=b=2$, $c=\frac{a+b}{ab-1}=\frac{2+2}{2*2-1}=\frac{4}3$
c is not an integer..

Therefore, 1,2,3 is the only solution set.. $QED$

I remember this problem! It was problem number 4 in Spanish MO '1998. This was my solution back then:

The following values are known:
$\tan 30^\circ = \frac{1}{\sqrt{3}}$, $\tan 45^\circ = 1$, $\tan 60^\circ = \sqrt{3}$, $\tan 75^\circ = \tan(45^\circ+30^\circ) = 2+\sqrt{3}$.
The function $\tan x$ is incresing for $0^\circ < x < 90^\circ$, so
$45^\circ=\arctan 1 < 60^\circ < \arctan 2 < \arctan 3 < 75^\circ$.

WLOG, in the next steps let's suppose $A\leq B\leq C$.
In a triangle, $A+B+C=180^\circ$ $\to$ $A\leq 60^\circ$ and $C\geq 60^\circ$.
If we want $\tan A$ to be integer, then $A=45^\circ$, $B+C=135^\circ$.
$B+C=135^\circ$ $\to$ $B\leq 67.5^\circ$ and $C\geq 67.5^\circ$.
If $B=45^\circ$ then $C=90^\circ$ and $\tan C$ is not an integer $\to$ $B=\arctan 2$ or $B=\arctan 3$.

We'll use the next expression to check the two cases:
$\tan (B+C)=\tan (135^\circ)$ $\to$ $\tan C=\frac{\tan B+1}{\tan B-1}$
Case 1: $\tan A=1$, $\tan B=2$ $\to$ $\tan C=3$ (OK)
Case 2: $\tan A=1$, $\tan B=3$ $\to$ $\tan C=2$ (we eliminate this one because we expected $A\leq B\leq C$).

The official solution was exactly like Allan Z.

Once you get $\tan A+\tan B+\tan C= \tan A \tan B \tan C$
the standard trick to solve $x+y+z=xyz$ is
[hide] WLOG let $x\ge y \ge z$
so we have $3x \ge xyz \implies 3 \ge yz \ge z^{2}\implies z \le \sqrt 3 \implies z=1$
And thus $x+y+1=xy \implies (x-1)(y-1)=2$ only factors of 2 are 2 and 1
[b]and you are done. (x=3,y=2,z=1) [/b]
(of course, if you know $z=1$ and $yz \le 3$ you can try (1,2,3) for $y$ which may  be  even faster than  the   factoring method)
Hope this helps. [/hide]


modular arithmetic
IMO Shortlist
combinatorics unsolved
combinatorics

Prove that if $ x \equal{} (x_1, \ldots, x_n) \in \mathbb{Z}^{(n)}$, and $ \sum_{k\equal{}1}^n{x_k} \equiv 0 \pmod n$ then there exist two permutation $ \sigma$ and $ \pi$ such that for any $ k \equal{} 1,\ldots,n$, $ x_k \equiv \sigma_k \plus{} \pi_k \pmod n$.

See http://www.mathlinks.ro/viewtopic.php?t=85076 but better see the official solutions of the IMO Shortlist 2005 ( http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=4475 ).

  darij


If $m$ and $n$ are positive numbers, prove that: $|n\sqrt{2005}-m|>\frac{1}{90}$


I just went head on brute force when they said that the string would be less than 20,000 characters.
This thing gets me through at least 5 test datas, but I need to make it faster in order to get it right...
PLEASE HELP!!!

[b]calf flac task:[/b] find the longest palindrome in a given string
[b]input:[/b] da string
[b]output:[/b] length of the palindrome / the palindrome
[b]lang:[/b] JAVA :D

[code]import java.io.*;

class calfflac
{
   static String str, str2;

   public static void main(String[] args) throws IOException
   {
      BufferedReader f = new BufferedReader(new FileReader("calfflac.in"));
      PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("calfflac.out")));

      str = f.readLine() + "\n";
      while (f.ready())
         str += f.readLine() + "\n";
      str2 = str.toLowerCase();

      int palHead = 0, palTail = 0;
      for (int i = 0; i < str2.length() - (palTail - palHead + 1); i++)
         for (int j = i + (palTail - palHead + 1); j < str2.length(); j++)
         {
            if (isPal(i, j) && (j - i) > (palTail - palHead))
            {
               palHead = i;
               palTail = j;
            }
         }

      while (palHead < palTail && !isValid(palHead))
      palHead++;
      while (palHead < palTail && !isValid(palTail))
         palTail--;
      int cnt = 0;
      for (int i = palHead; i <= palTail; i++)
         if (isValid(i))
            cnt++;

      out.println(cnt);
      out.println(str.substring(palHead, palTail + 1));
      out.close();
      System.exit(0);
   }

   private static boolean isPal(int head, int tail)
   {
      while (head < tail)
      {
         while (!isValid(tail) && head < tail)
            tail--;
         while (!isValid(head) && head < tail)
            head++;
         if (str2.charAt(head) != str2.charAt(tail))
            return false;
         else
         {
            head++;
            tail--;
         }
      }
      return true;
   }

   private static boolean isValid(int i)
   {
      char c = str2.charAt(i);
      return c >= 65 && c <= 90 || c >= 97 && c <= 122;
   }
}[/code]

i think i have a faster solution maybe you should try using the  TextIO.peek() function or something like that to look at each letter and save it to a char type and then once you have one or more chars make a word out of them and read them in reverse and see if they're the same its going to take about 2-3 hours

i dont know if this is right because i typed it randomly
but based on my random example it seems to work
also not sure if its faster becasue i didnt read yours.. but this one runs in O(n^2)

i ran it on a string of length 26500, and it finishes in about 30 seconds, is that good enoguh?

you can also add a line to break when m == n, but i dont know how often thats going to help

[code]
   public static void main(String[] args) {
		String s1 = "abccadbadbcbdbcbadd".replaceAll("\\W+", ""); // input, and sanitize, except i like numebrs too
		String s2 = new StringBuffer(s1).reverse().toString(); // reverse
		int n = s1.length(), m = 0; // length and current max length of palin
		String a = ""; // longest palin
		for (int i = -n + 1; i < n; i++) { // shifting and checking 2n -1 positions
			int j = i < 0 ? -i : i;
			String ss1 = i < 0 ? s1 : s2;
			String ss2 = i < 0 ? s2 : s1;
			// running longest palin
			int streak = 0;
			String tmp = "";
			char c;
			for (int k = j; k < n; k++) {
				if ((c = ss1.charAt(k - j)) == ss2.charAt(k)) {
					streak++;
					tmp += c;
				} else {
					if (streak > m) {
						m = streak;
						a = new String(tmp);
					}
					streak = 0;
					tmp = "";
				}
			}
		}
		System.out.println(m + " " + a);
	}
[/code]

[quote="h_s_potter2002"]The length of the longest palindrome has to be a maximum of 2,000 characters, right?  Also, if you have the substring from position i to position j inclusive, it is a palindrome only if the substring from position i+1 to position j-1 inclusive is a palindrome.  So try to avoid checking substrings like "abca" to see if it's a palindrome, since you already know that "bc" is not a palindrome.  And since you're using Java, I guess the method you use to read the input matters a lot too for problems with large input data.  It mentions which one to use in the introductory part of the training pages.

Also, the time limit for USACO problems is 1 second, so 30 seconds or 2 to 3 hours is way too slow.[/quote]

Thanks. Your second advice about not checking substrings like "abca" really helped. However, I am positive that I used the introductory page's input and output method for the solution I posted previously.

what if you set up the loops as follows?

for each integer and half-integer up to the length of the string, test what the longest palindrome centered at that number is. (so for the string "abcde" and the integer 0, we'd test strings centered about position 0, the only one of which is "a"; for the half-integer 1.5, we'd test strings around there, i.e. "bc" and "abcd"). It's technically still O(n^2) but probably most of the time you're not going to have a very long palindrome centered about any given position, so it seems close to O(n) in typical cases.


geometry
3D geometry
trigonometry
logarithms
tetrahedron
trig identities
Law of Sines

1) How many ordered triple $(a, b, c)$ satisfy the equation $(a^2-1)^2+(b^2-4)^2+(c^2-9)^2=0$ ?

          A)1   B)3   C)7   D)8   E)9

2) What is the sum of first 21 odd positive integers?

          A)210   B)903    C)441    D)440    E)400

3) How many pairs of positive integers $(x, y)$ is $x+4y=2004$?

          A)500    B)502    C)501     D)499      E)498

4) A charity sells 140 benefit tickets for a total of 2001 dollars. Some tickets sell for full prize (a whole dollar amount), and the rest sell for half price. How much money (in dollars) is raised by the full-price tickets?

          A)986   B)1449   C)1219   D)782   E)1158

5) If the volume of a cube is $729 cm^3$, then the sum of lengths of all edges is:

          A)108 cm   B)54 cm   C)72 cm   D)81 cm   E)162 cm

6) Triangle ABC has vertices $(0, 0), (\frac{32}{5}, \frac{24}{5}), (10, 0)$. The area of ABC is

           A)48   B)24   C)10   D)$\frac{64}{5}$   E)$\frac{72}{5}$

7) A man walks $x$ miles due west, turns $150^{\circ}$ to his left and walks 3 miles in the new direction. If he finishes at a point $\sqrt3$ miles from his starting point, then x is

           A)$2\sqrt3$   B)$\frac{3}{2}$   C)$3$   D)$\sqrt3$   E)not uniquely determined

8) Triangle PAB and square ABCD are in perpendicular planes. Given that PA=3, PB=4, AB=5. What is length of PD?

             A)5   B)$\sqrt{34}$   C)$\sqrt{41}$   D)$2\sqrt{13}$   E)8

9) Find the unit digit of the sum $1+2^5+3^5+4^5+5^5+....2005^5$

             A)1   B)3  C)5  D)7   E)9

10) Find $\frac{1}{3+2\sqrt2}+\frac{1}{2\sqrt2+\sqrt7}+\frac{1}{\sqrt6+\sqrt7}+\frac{1}{\sqrt6+\sqrt5}+\frac{1}{\sqrt5+2}+\frac{1}{2+\sqrt3}$

              A)3   B)$3+\sqrt3$   C)$3-\sqrt3$   D)$2\sqrt3$   E)2

11) $\cot 10^{\circ}+\tan 5^{\circ}=$

              A)$\csc 5^{\circ}$   B)$\sin 15^{\circ}$  C)$\sec 5^{\circ}$   D)$\sec 10^{\circ}$   E)$\csc 10^{\circ}$

12) How many odd numbers with middle digit 5 and no digit repeated are there between 10000 and 69999?

               A)756   B)1008   C)1680   D)10000   E)None of the Above

13) Given that $\log_{10} 2=0.3010$, how many digits are in $5^{89}$?

               A)62   B)64   C)63   D)60   E)89

14)What is the value of expression $\frac{1}{\log_{2}{100!}}+ \frac{1}{\log_{3}{100!}}+ \frac{1}{\log_{4}{100!}}+....+ \frac{1}{\log_{100}{100!}}$

               A)0.01   B)0.1   C)1   D)2   E)10

15) Two edges of a regular tetrahedron are opposite if they are skew. Find the   distance between the midpoints of 2 opposite edges of a regular tetrahedron with edge length 1.

              A)$\sqrt3$   B)$\sqrt2$   C)$\frac{\sqrt3}{2}$   D)$\frac{\sqrt2}{2}$   E)$\frac{3}{2}$

16) If $\sin x-\sin y=-\frac{1}{3}$, $\cos x- \cos y=\frac{1}{2}$, then $\cos(x-y)=$

              A)$\frac{59}{72}$   B)$\frac{5}{6}$   C)$\frac{1}{2}$   D)$\frac{5}{72}$   E)$\frac{29}{36}$

17) Triangle ABC and ABD are isosceles with AB=AC=BD, and BD intersects AC at E(point E is on segment AC and segment BD). If BD $\perp$ AC, then $\angle C + \angle D$ is

             A)$115^{\circ}$   B)$120^{\circ}$   C)$130^{\circ}$   D)$135^{\circ}$   E) not uniquely determined

18) If $a, b, c$ are real numbers such that $a^4-8b^2=-31, b^4-18c^2=-146, c^4-2a^2=79$. Find    $a^2+b^2+c^2$

             A)14   B)21   C)35   D)23   E)Not Uniquely Determined

19) How many pairs of non-negative integers (m, n) each sum to 1492 without any carries(1000 and 492 sum to 1492 with no carries, but 503 and 989 sum to 1492 with carries)?

            A)200   B)240   C)270   D)320   E)300

20) For certain integers $n$, $n^2-3n-126$ is a perfect square. What is the sum of all distinct possible values of $n$?

             A)12   B)-12   C)-118   D)-1   E)22

21) In triangle ABC, AD and AE trisect angle BAC. The lengths of BD, DE, EC are 2, 3, 6, respectively (BD+DE+EC=BC). Find the length of the shortest side of  triangle ABC

            A)$6 \sqrt 6$   B)$\sqrt{10}$   C)$5$   D)$2 \sqrt{10}$   E)$11$

22) Find the sum to the infinite series $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+...$ where terms are reciprocals of positive integers divisible only by 2 or 3.

                    A)2   B)3   C)4   D)6   E)none of the above

23) If $\cos\theta=\frac{1}{4}$, find $\cos 5\theta$

                    A)$\frac{5}{16}$   B)$\frac{3}{32}$   C)$\frac{3}{4}$   D)$\frac{61}{64}   E)\frac{89}{128}$

24) Evaluate $\displaystyle \sum_{k=0}^{2005} ((-1)^k {2005 \choose k} k)$

                   A)$2^{1002}$   B) $-(2^{1002})$   C)$0$  D)$2^{1003}$   E)$-(2^{1003})$

25) A and C lie on a circle center O with radius $\sqrt{50}$. The point B is inside the circle such that $\angle ABC = 90^{\circ}$, AB = 6, BC = 2. Find OB.

                   A)5   B)$\sqrt{26}$   C)$\sqrt{31}$   D)$\frac{11}{2}$   E)6

damn costed me 10 minutes to find this topic :?

I was having a bad day when I took this... (math-wise)

I've seen question 4 before, probably on a real AMC, and got it quite easily when I saw it (less than 2 weeks ago)

but wasn't able to get the answer this time... :S

That was a fun contest!!! I think it was a bit easier than mine and Rep's, so it was about average AMC 12 difficulty. Hope I didn't make any stupid errors....

Yeah, that was fun.  I did ok... I should have known how to do some more of those questions.  It was easier.  I m satisfied with how I did but I know that I could ve done better.

Grrr...if I knew real basic trig, I'm sure I could've answered like 5 of the questions!  Haha, this was A LOT easier than the other Mock AMC 12 contests.....Which is ALWAYS good.

It was insanely easier than the other 2. (or perhaps because i was more relaxed this time I did better?)

Thanks to people who contributed to this test

Heh.... yeah it was pretty fun.  I haven't taken the other 2 Mock AMC's so I can't say if this was easier/harder.

I forgot all the trig identities (gosh, when was the last time I used them in school?  I was out of math for so long)

I could/should have gotten 1 or 2 more problems.

This was easier than the other 2, except I got stuck on the tetrahedron problem for close to 20 minutes, and still couldn't get the answer. It also would have helped if I knew some trig..

I did not take the other two, but this one was relatively easy compared to the AMCs I've taken.  However, the last few were pretty hard...I couldn't get 21, 24, and 25.

-interesting_move

[quote="JSRosen3"]I got a 123, missing #7...does anyone remember what that question was?[/quote]

From my scratch work, number 7 was about a man walking around.

uh oh

a whole 12 points deducted!

careless careless careless careless!

anyone care to subtly remind me what 13 and 22 were?

[quote="joml88"]I too got A for 7.  I used some trig.  Maybe we just missed something obvious.[/quote]

Don't feel bad.  It was easy to miss.  The triangle in question is a triangle with side lengths 3, $\sqrt3$, and x.  Between the sides of length 3 and x, there is a 30 (not 150!  I almost made that mistake!) degree angle.  Using the law of cosines and solving for x, we get x=$2\sqrt3$ OR $\sqrt3$.  Thus, x is not uniquely determined, so the answer is E.

I answered (A), but I then thought be instinct that the problem was too easy, and considering I remembered a math test where I messed up on the same question upon further investigation I realized that the answer was (E).

[quote="ThAzN1"]Eh... a picture for #17 would've been nice.  I misinterpreted the question, got stuck, and wasted a lot of time.

17. Triangle ABC and ABD are isosceles with AB = BC = BD, and BD intersects AC at E (point E is on line
segment AC and line segment BD). If BD [tex]\perp[/tex] AC, then [tex]\angle C + \angle D[/tex] is...[/quote]

That's why I said E is on SEGMENT AC and BD, read the question carefully :) .

:? the questions are posted in the first post??

Quote:9) Find the unit digit of the sum
[hide]

It is well-known (?) that the fifth power of a digit ends in the same digit:

05 ends in 0, 15 ends in 1, 25 2 ends in 2, 35 ends in 3, ..., 95 ends in 9.

Hence, the given sum has the same unit digit as: 1 + 2 + 3 + ... + 2005

= (2005)(2006)/2 = (2005)(1003) ... and odd multiple of 5.

Therefore, the sum ends in 5.[/hide]

Hi Soroban,  you can post your solutions to the problems in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14884&postdays=0&postorder=asc&&start=0]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14884&postdays=0&postorder=asc&&start=0[/url]

Nice 16-year bump

It was in 2004, you know, so ofc ppl wouldn't be that good at writing problems

TBH this was a good mock difficulty for a 2004 AMC. Also, why did I get an email notification about this??? And also these people are probably getting married, getting jobs, and having kids by now. That's from mad long ago   :whistling:

[quote=Possible]The questions are quite misplaced  :maybe:[/quote]

Lol nice bump.

Anyways, did anyone realize that 22 infinite series actually popped up on the 2019 amc 12a?


geometry
3D geometry
function
induction
geometric transformation
rotation

This problem is borrowed from a book by Frederick Stevenson, [u]Exploratory Problems in Mathematics[/u].  It doesn't necessarily have a "nice" solution -- the idea is just to get you thinking about the mathematics.  The best approach is to try to find patterns and use all of your knowledge.  The questions vary from quite easy to very difficult.

Suppose you have a square grid, 100 dots on a side.  How many sizes of square can be formed whose sides are parallel to the orientation of the grid?  How many squares can you find whose sides are parallel to the orientation of the grid?

How many sizes of square can you find, all of whose vertices are on the grid (but not necessarily parallel to the grid -- see picture for more information)?  How many different squares can you form?

Now, suppose you have a cubic lattice -- that is, a three dimensional array of points spaced one unit north-south, east-west, and up-down -- 100 dots to a side.  How many sizes of cube with faces parallel to the faces of the lattice are there?  How many sizes of rectangular box with sides parallel to the grid are there?  How many total cubes in the same orientation as the grid are there?  How many rectangular boxes?

Can you find any cubes whose orientation is such that its faces are [i]not[/i] parallel to the faces of the grid?  If so, how many different sizes of such cubes will there be?  How many total cubes (in any orientation) can you find in the grid?

If you respond to this, you must explain your answers.  I'm not asking for a rigorous mathematical proof -- just a good, clean explanation of what you did to get your answer.  (Rigorous mathematical proofs are okay, too.)

Well the first part you can form up to [hide]n-1 different sized squares for a n sized lattice or 99 different sized squares[/hide] on the second part of the first question you can make I belive [hide] (n-1) :^2:  1*1 squares (n-2) :^2: 2*2... all the way up to (n-99) :^2:  99*99 squares, so all we have is a sum of squares aka 99 :^2: +98 :^2: ...+3 :^2: +2 :^2: +1=(99+1) :^2: +(98+2) :^2: ...+(51+49) :^2:+50 :^2:  -2*(99*1+98*2...+51*49)=49*100 :^2: +50 :^2: -2*(99*1+98*2...+51*49).[/hide]  I've gotten it that far but I can't find an easy way to simlify the [hide]2*(99*1+98*2).[/hide]part, help would be nice.[/hide]

yeah I feel really stupid going thru all that algabra when it wasn't really necessary, you can define the realation recursivally [hide] F(n)=F(n-1)+(n-1) :^2:[/hide] translating this to explicit is trickier but I got [hide]F(n)=1/6n(n+1)(2n+1)[/hide] so, you can find the 100*100 grid as [hide]1/6*99*(99+1)(2*99+1)=328350[/hide]

Good job finding the function value for the recursion -- there are a variety of methods that can be used to prove it, induction being the simplest.  The way you wrote it is rather difficult to read, however.

Try the other problems, too :)

Define [tex]a_m[/tex] as the number of m x m squares on the given n x n (n = 99 as given) lattice grid such that the squares are parallel to the orientation of the grid.

Then
[tex]\displaystyle a_m = (n-m+1)^2[/tex],
so
[tex]\displaystyle \sum_{i=1}^na_i = \sum_{i=1}^{n}i^2 = \frac{n(n+1)(2n+1)}{6}[/tex]
is the number of squares aligned parallel to the grid.

Now to take care of ALL the squares, regardless of being parallel to the grid or not.

Within a square of size m, we can find m squares (including itself) that lie within it (consider the m "rotations" acquired by shifting the top left vertex one right, the top right vertex one down, the bottom right vertex one left, and the bottom left vertex one up; then repeat until all squares are established). Likewise, any square in the grid can be inscribed in a square with orientation parallel to that of the grid.

So, the total # of squares on the grid regardless of orientation is

[tex]\begin{eqnarray*}\sum_{i=1}^{n}ia_i & = & \sum_{i=1}^{n}i(n+1-i)^2 \\& = &(n+1)^2\sum_{i=1}^ni - 2(n+1)\sum_{i=1}^ni^2 + \sum_{i=1}^ni^3
\\ & = & \frac{n(n+1)^3}{2} - \frac{n(n+1)^2(2n+1)}{3} + \frac{n^2(n+1)^2}{4}
\\ & = & \frac{n(n+1)^2}{12}[6(n+1) - 4(2n+1) + 3n]
\\ & = & \frac{n(n+1)^2(n+2)}{12}
 \end{eqnarray*}[/tex]

How about total number of different sizes for those squares?  (I don't know that there's a nice answer to this one, actually, but I know how one could go about describing the solution fairly simply.)


geometry
circumcircle
geometric transformation
reflection
cyclic quadrilateral
perpendicular bisector
geometry unsolved

A point inside an equilateral triangle with side 1 is a distance $ a,b,c$ from the vertices.The triangle $ ABC$ has $ BC\equal{}a,CA\equal{}b,AB\equal{}c$.The sides
subtend equal angles at a point inside it.Show that sum of distances of the point from the vertices is 1.

[quote="stephen38"]A point inside an equilateral triangle with side 1 is a distance $ a,b,c$ from the vertices.The triangle $ ABC$ has $ BC \equal{} a,CA \equal{} b,AB \equal{} c$.The sides
subtend equal angles at a point inside it. Show that sum of distances of the point from the vertices is 1.[/quote]

The point P inside the $ \triangle ABC$ is not clearly defined, my 1st guess was the Brocard point $ \angle PBC \equal{} \angle PCA \equal{} \angle PAB,$ in which case the proposition is easy to disprove. My 2nd guess was the Fermat point $ \angle BPC \equal{} \angle CPA \equal{} \angle APB \equal{} 120^\circ,$ in which case the proposition is easy to prove.
_____________________________________________________________

The 1st Fermat point is obtained by building equalateral triangles $ \triangle BCA',\ \triangle CAB',\ \triangle ABC'$ outwardly on $ BC,\ CA,\ AB$ and joining $ AA',\ BB',\ CC'.$ Let circumcircles of the 1st two equilateral triangles meet at $ P,$ $ \angle BPC \equal{} \angle CPA \equal{} 120^\circ$ $ \Longrightarrow \angle APB \equal{} 120^\circ$ and $ P$ is on the circumcircle of the remaining equilateral triangle. It immediately follows that $ \angle APA' \equal{} 180^\circ,$ $ A,\ P,\ A'$ are collinear, etc. By Ptolemy theorem for one cyclic quadrilateral, say $ BPCA',$ $ BC \cdot PA' \equal{} BP \cdot CA' \plus{} CP \cdot BA'$ and since $ BC \equal{} CA' \equal{} A'B,$ $ BP \plus{} CP \equal{} PA',$ $ AP \plus{} BP \plus{} CP \equal{} AA',$ etc.
______________________________________________________________

Let $ M$ be a point inside the equilateral triangle $ \triangle XYZ$ and $ a \equal{} MX,\ b \equal{} MY,\ c \equal{} MZ$ sides of a $ \triangle ABC$ with the sides $ BC \equal{} a,\ CA \equal{} b,\ AB \equal{} c.$ To construct the $ \triangle ABC,$ reflect $ M$ in the perpendicular bisector of $ YZ$ into $ M',$ so that $ M'X \equal{} MX \equal{} a$ and $ M'Y \equal{} MZ \equal{} c.$ Then reflect $ M'$ in $ XY$ into $ B,$ so that $ XB \equal{} XM' \equal{} XM \equal{} a$ and $ YB \equal{} YM' \equal{} ZM \equal{} c.$ Denoting $ A \equiv Y,\ C \equiv M,$ the $ \triangle ABC \equiv \triangle YBM$ has the desired sides $ a,\ b,\ c.$ In addition,

$ \angle MXB \equal{} \angle M'XY \plus{} \angle YXB \equal{} \angle YXM' \plus{} \angle M'XZ  \equal{} \angle YXZ \equal{} 60^\circ$

and the isosceles $ \triangle MXB$ with $ XM \equal{} XB$ is equilateral.  Let $ P$ be its 1st Fermat point of the $ \triangle ABC \equiv \triangle YBM.$ Denoting $ A' \equiv X,$ we already have the equilateral triangle $ \triangle BCA' \equiv \triangle BMX$ on its side $ BC \equiv BM$ and $ AP \plus{} BP \plus{} CP \equal{} AA' \equiv YX.$


function search trigonometry factorial logarithms LaTeX calculus

these two questions are confusing me a bit. i pulled them from an old exam to review for my upcoming test.
If you explain

[img]http://img697.imageshack.us/img697/7148/35765649.jpg[/img]

i guess the answer of this one is either 'a' or 'e' because they are only one with x^2n.

[img]http://img14.imageshack.us/img14/6345/56417697.jpg[/img]

i'm not sure about this one.

For the first one, show us how would you do the Taylor series, and for the second one try differentiating under the sum, and also show us what you get and what can you conclude from it.

For B)

We know that $ e^x \equal{} \sum_{n\geq0}\frac {x^n}{n!}$.

So for $ xe^{ \minus{} x^2}$ we have:

$ xe^{ \minus{} x^2} \equal{} x \sum_{n\geq0}\frac {( \minus{} x^2)^n}{n!}$

$ \equal{} \sum_{n\geq0}x\frac {( \minus{} 1)^n(x^2)^n}{n!}$

$ \equal{} \sum_{n\geq0}x\frac {( \minus{} 1)^n(x^{2n}}{n!}$

$ \equal{} \sum_{n\geq0}\frac {( \minus{} 1)^nx^{2n \plus{} 1}}{n!}$

I'm teaching a class that has an exam on this material next week. Link: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1267804713&t=145324]seven power series to memorize.[/url]

[quote="Aafour"]i guess the answer of this one is either 'a' or 'e' because they are only one with x^2n.[/quote]
You've got that almost exactly backwards. $ e^{ \minus{} x^2}$ is an even function, so indeed its power series should have only even powers - but note that what is actually asked for is $ xe^{ \minus{} x^2}.$ That's an odd function, and its power series around zero contains only odd powers of $ x.$ Hence (a) and (e), which contain only even powers of $ x,$ are certainly wrong, and we're left to decide among (b), (c), and (d), which do have odd powers of $ x.$

What I would do is to write out a few terms first:

$ e^{ \minus{} x^2} \equal{} 1 \plus{} ( \minus{} x^2) \plus{} \frac {( \minus{} x^2)^2}2 \plus{} \frac {( \minus{} x^2)^3}{3!} \plus{} \cdots$

$ \equal{} 1 \minus{} x^2 \plus{} \frac {x^4}{2} \minus{} \frac {x^6}{3!} \plus{} \cdots.$ Now multiply by $ x.$

$ xe^{ \minus{} x^2} \equal{} x \minus{} x^3 \plus{} \frac {x^5}{2} \minus{} \frac {x^7}{3!} \plus{} \cdots$

Now, try to match up a specific term to the general term. Say we look at the term with $ x^5.$ Since we're trying to match to things written with $ x^{2n \plus{} 1},$ we have $ 2n \plus{} 1 \equal{} 5$ or $ n \equal{} 2.$ This term has a $ \plus{}$ sign, so, with $ 2$ being even, we want $ ( \minus{} 1)^n$ so it will be $ \plus{} .$ That rules out (b). Now, what is the denominator? For this $ n \equal{} 2$ term, we have a denominator of $ 2,$ which is $ 2!.$ That would be $ n!,$ and so the correct answer is (d).

For the second one: start by looking at whether there are any $ n$s in denominators. The series for the exponential, sine, and cosine all have factorials in the denominator; the logarithm and arctangent have linear factors. This series has linear factors, so it's most likely a logarithm or arctangent. The logarithm series has every $ n$ in the denominator; the arctangent series has just the odd integers. This series has every $ n$ so look at the logarithm.

Compare the series $ \ln(1 \plus{} u) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{n \plus{} 1}u^n}{n}$ to this series. Can you answer it now?

Edit: fixed the bad proofreading, as Tyl noted below.

thank you epkid08 for helping me. I got it now. But I still don't know how to solve the other one. I tried to differntiate under the sum as what milin say but I didn't recognize any known functions

Kent made a small mistake in his last post, it should read $ \ln(1\plus{}u)\equal{}\sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n\plus{}1}u^{n}}{n}$.

I hope you followed his argumentation regarding the fact that it is almost surely a logarithm, looking at the fact that the sum has only $ n$'s in the denominators, instead of $ n!$'s, and it includes all values of $ n$ (both odd and even).

Now, rewrite the given Taylor series as:

$ \sum\limits_{n\equal{}1}^\infty (\minus{}1)^{n\plus{}1} \frac{2^n}n x^{2n} \equal{} \sum\limits_{n\equal{}1}^\infty \frac{(\minus{}1)^{n\plus{}1}}n  \left(2x^2\right)^n$.

Compare this with the previous series in this post. Can you see which one it is now?

Tyl, according to what you wrote the answer is ln(1+x^2).  However, it's written at the back of my calc book that the series of ln(1+x) = some of (-1)^(n-1) ....
.So (-1)  is ries to (n-1) not (n+1). I hope you got what i mean. i'm sorry i don't know how to write functions in the forum

No, it is $ \ln( 1 + \text{\bold 2} x^2)$, or $ \boxed{\text{b}}$ in the your initial picture.

Note that $ (-1)^2 = 1$.
This means that $ (-1)^{n+1} = (-1)^{n-1} (-1)^2 = (-1)^{n-1} 1 = (-1)^{n-1}$, so both forms are identical (sum starts at $ n = 1$ so we dont get into trouble with negative expontents.

(To write using $ \text{\LaTeX}$, put your expressions between $ \$\$$'s, using commands such as '\ln' to get $ \ln$. You can also quote other people's posts, or press on their formula's to see how they made them.)

Can you identify these functions?

A. $ x\plus{}\frac{x^2}2\plus{}\frac{x^3}{3}\plus{}\frac{x^4}4\plus{}\cdots\equal{}\sum_{n\equal{}1}^{\infty}\frac{x^n}n$

B. $ 3x\plus{}\frac{3x^2}8\plus{}\frac{3x^3}{8^2}\plus{}\frac{3x^4}{8^3}\plus{}\cdots\equal{}\sum_{n\equal{}1}^{\infty}\frac{3x^n}{8^{n\minus{}1}}$

C. $ x\minus{}x^4\plus{}\frac{x^7}{2}\minus{}\frac{x^{10}}{6}\plus{}\frac{x^{13}}{4!}\plus{}\cdots\equal{}\sum_{k\equal{}0}^{\infty}\frac{(\minus{}1)^kx^{3k\plus{}1}}{k!}$

D. $ 1\minus{}\frac{x}2\plus{}\frac{x^2}{4!}\minus{}\frac{x^3}{6!}\plus{}\cdots\equal{}\sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^nx^n}{(2n)!}$


Prove that

$ (m_a\plus{}m_b\plus{}m_c)(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2})\geq\frac{9}{2R}$

.

[quote="Ligouras"]Prove that

$ (m_a \plus{} m_b \plus{} m_c)(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})\geq\frac {9}{2R}$

.[/quote]

$ LHS\ge(m_a \plus{} m_b \plus{} m_c)\frac{9}{a^2\plus{}b^2\plus{}c^2}\equal{}\frac{27(m_a \plus{} m_b \plus{} m_c)}{4(m_a^2\plus{}m_b^2\plus{}m_c^2}\ge\frac {9}{2R}$

the last is equivalent to $ m_a \plus{} m_b \plus{} m_c\ge \frac{2}{3}.\frac{m_a^2\plus{}m_b^2\plus{}m_c^2}{R}$

it is easy to prove it.:)

The following inequality is also true.
\[\frac {a}{m_a}+\frac {b}{m_b}+\frac {c}{m_c}\ge 4\frac{m_a\plus{}m_b\plus{}m_c}{a+b+c}.\]


Let $ a_1,a_2,\dots ,a_n$ and $ b_1,b_2,\dots , b_n$ be increasing sequences of real numbers and let $ x_1,x_2,\dots , x_n$ be any permutation of $ b_1,b_2,\dots ,b_n$. Show that

\[ \sum a_i b_i \geq \sum a_i x_i.\]

It is well known, rearrangement inequality.   :)


$\triangle ABC\Longrightarrow 3(ab+bc+ca)\ge 6Rr+16p\cdot \sqrt{\frac{2Rr}{3}}\Longleftrightarrow$ $\boxed{\ (ab+bc+ca)(a+b+c)\ge abc+\frac{8}{3}\cdot (a+b+c)^{2}\cdot \sqrt{\frac{2Rr}{3}}\ }\ .$

An interesting consequence : $\boxed{\ a+b+c\ge \frac{abc}{ab+bc+ca}+8\cdot\sqrt{\frac{2Rr}{3}}\ }\ .$

[quote="Virgil Nicula"]$\triangle ABC\Longrightarrow$ $\boxed{\ (ab+bc+ca)(a+b+c)\ge abc+\frac{8}{3}\cdot (a+b+c)^{2}\cdot \sqrt{\frac{2Rr}{3}}\ }\ .$
[/quote]
$(ab+bc+ca)(a+b+c)\ge abc+\frac{8}{3}\cdot (a+b+c)^{2}\cdot \sqrt{\frac{2Rr}{3}}\Leftrightarrow$
$\Leftrightarrow(ab+bc+ca)(a+b+c)\ge abc+\frac{8}{3}\cdot (a+b+c)^{2}\cdot \sqrt{\frac{abc}{3(a+b+c)}}.$
It is true for all positive $a,$ $b$ and $c$ since, $abc\leq\frac{(ab+ac+bc)^{2}}{3(a+b+c)}.$


is there any easy way to determine if a number is prime?

Not really. Just divide by numbers and see if they're a factor.

[quote="espark52"]but wat if you're asked to find the greatest 10 prime numbers less than 2000-like i was asked(but in a slightly different manner)[/quote]

guess and check, i don't know any other way

[quote="mathnerd314"]Write out all the numbers between 1900 and 2000 (or something) and then use the prime sieving method.[/quote]

What's the prime sieving method?

Cross out all the numbers that are divisible by two, three, five, seven, eleven...

I don't remember the Greek guy who came up with it. E something.

It's the Sieve of Eratosthenes.
[url=http://www.artofproblemsolving.com/Wiki/index.php/Sieve_of_Eratosthenes]Look Here[/url]

[quote="daermon"]the easiest way is just to divide by primes up to the square root[/quote]
Reason being that for any composite number $c$, $c=mn$ where $m$ and $n$ are both greater than 1. The smallest prime divisor of $c$ cannot be smaller than the smaller of $m$ and $n$. $m$ and $n$ cannot both be greater than $\sqrt{c}$, so that's why we only need to test for primes up until the square root.

[quote="espark52"][quote="daermon"]the easiest way is just to divide by primes up to the square root[/quote]

Can you please clarify?[/quote]
What he means is the easiest way to check whether a number is prime or not is to divide the number (n) by every prime that is less than $\sqrt{n}$

But for finding, say, 10 primes around that number, the sieving method is pretty good.

Yes, the sieving method is really good for that type of problem, but if you are just wondering if some random number is prime, calculators programs and computer programs make it a lot easier  :D

that's usually what I do, outside (and inside some) contests I just factor it on my TI-89, though you could pretty easily program a function on any other programmable calculator to make your life easier.

To know whether $k$ is prime or not, divide it by all positive integrer $n$, $n \ne 1$, $n\le [\sqrt{k}]$, and, if all the results are non-integrers, then $k$ is prime. If one or more results are integrers, then $k$ is composite.

I think it's more like:
[quote="José"]To know whether $k$ is prime or not, divide it by all primes $p$, $p\le \sqrt{k}$, and, if all the results are non-integers, then $k$ is prime. If one or more results are integers, then $k$ is composite.[/quote]

At least, it's a lot easier this way. :D

There is no cheap ways (like an easy formula or quick method...) for finding prime numbers.
Many of you mentioned sieving. Although it is better than just guess and check, I think it is still inconvenient to divide a number by every single primes smaller than square root.
Imagine when you are in a sprint round -no calculator, extra brains...

For that reason, I hate "find prime numbers that is blah blah" kind of questions. It takes more writing than thinking...

correction: I meant two ways, sieving and square root of [i]n.[/i]

[quote="Andrew Kim"]There is no cheap ways (like an easy formula or quick method...) for finding prime numbers.
Many of you mentioned sieving. Although it is better than just guess and check, I think it is still inconvenient to divide a number by every single primes smaller than square root.
Imagine when you are in a sprint round -no calculator, extra brains...

For that reason, I hate "find prime numbers that is blah blah" kind of questions. It takes more writing than thinking...[/i][/quote]

Read the thread again. What you have described is not sieving.

For numbers around 2000 you [b]only[/b] have to check divisibility with 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, and 43.


calculus
integration
geometry
3D geometry
function
parameterization
geometric transformation

find the only integral solution of:

$ (x \minus{} 1)^3 \plus{} (x \minus{} 2)^3 \plus{} (x \minus{} 3)^3 \plus{} .................... \plus{} (x \minus{} 2007)^3\equal{}0$

hello, substitute $ x\equal{}t\plus{}1004$ in your equation.
Sonnhard.

so there is more than one integral solution?

hello, no,i have only one integral solution. Show that must be $ t\equal{}0$.
Sonnhard.

The answer is simple, really: because there are an odd number of cubes,  try to find $ x$ such that the middle term is $ 0$, and opposite terms cancel each other.

Why, then, would you substitute t+1004, when there is only one integral solution, and when it is clearly 1004 due to the fact that the middle term is 0, and the rest cancel out?

thanks to all of you
isn't there another approach(less logical,i am bad in logical thinking)

i was thinking it as $ \sum (x\minus{}a)^3\equal{}0$

can we write it as
$ {[\sum (x\minus{}a) ]}^2\equal{}0$??
so,$ \sum(x\minus{}a)\equal{}0$

thereby meaning
$ 2007x\minus{}\frac {2007.2008}{2}\equal{}0$
so,$ x\minus{}1004\equal{}0$
$ x\equal{}1004$

i want to know whether this approach is right or not?
thanks again

Yes, that's a way to do it using the sum of cubes formula. But it's easier if you use the properties of odd functions ($ f(\minus{}x)\equal{}\minus{}f(x)$)
[hide="Solution"]
$ (x\minus{}1)^3 \plus{} (x\minus{}2)^3 \plus{} (x\minus{}3)^3 \plus{} ... \plus{} (x\minus{}2007)^3 \equal{} 0$

Substitute $ x\equal{}t\plus{}1004$. The equation becomes

$ (t\plus{}1003)^3 \plus{} (t \plus{} 1002)^3 \plus{} ... \plus{} (t\plus{}1)^3 \plus{} t^3 \plus{} (t\minus{}1)^3 \plus{} ...\plus{} (t\minus{}1002)^3 \plus{} (t\minus{}1003)^3 \equal{} 0$
Since $ (t\plus{}x)^3 \equal{} \minus{}(t\minus{}x)^3$, all the terms except $ t^3$ cancel out, and you're left with $ t^3 \equal{} 0 \Rightarrow (x\minus{}1004)^3 \equal{} 0 \Rightarrow x\equal{}1004.$
[/hide]

thank you(and everybody else) very much :)

[hide="Solution"]
Observe that $ (x \minus{} n)^3$ is a strictly increasing function in $ x$ for all $ n \in \mathbb{R}$. Therefore, $ \sum_{n \equal{} 1}^{2007} (x \minus{} n)^3$ is a strictly increasing function in $ x$, and there is [u]at most[/u] one real root. Using Dr Graubner's substitution, we notice that $ x \equal{} 1004$ solves the equation; therefore, this is the unique real solution.[/hide]

[b]@skand:[/b] The sum of cubes formula applies only when you are summing from $ 1$ to $ n$. In this case, we don't know what the starting number is. You will have to adjust your formula to accommodate this.

[b]@RisingFlame:[/b] You claim that $ (t \plus{} n)^3 \equal{} \minus{} (t \minus{} n)^3$. However, the left side has positive coefficients on $ t^3$ and $ t$, while the right side has negative coefficients for those same terms. If you expand that equation, you are essentially assuming that $ t^3 \plus{} 3tn^2 \equal{} 0$ for all $ (t,n)$.

The problem here is that there is one variable and one parameter, not just one variable, so the parameter performs a horizontal translation, and there is no longer symmetry across the y-axis.

thanks TZF  :)

i was wondering what would be the modification in formula to accomodate this
thanks again

EDIT:sorry for posting in wrong thread

If you have a sum of consecutive positive cubes, $ (m \plus{} 1 )^3 \plus{} (m \plus{} 2)^3 \plus{} \ldots \plus{} n^3$, just write this as $ 1^3 \plus{} 2^3 \plus{} \ldots \plus{} n^3 \minus{} (1^3 \plus{} \ldots \plus{} m^3)$ and apply the formula to both terms.  There might be some minor sign changes or the like if one or both of $ m, n$ is negative.

[hide]Using Sonhard's method, we end up with $ (2007t)(t^2 \plus{} (1003)(1004)) \equal{} 0$
We immediately see $ t \equal{} 0$ is a solution.  And $ t^2$ can't be negative, so that is the only real solution.  Since $ x \equal{} t \plus{} 1004$, our only integral solution is $ x \equal{} 1004$.[/hide]


[hide="hint"]I'm giving you the hint because I would like all the people to try it...

the hint is ....

think squre root as 1/2 power
[/hide]

From doing the first chapter of AOPS V1, I think I know this...

[hide]
The square root is the same as to the 1/2 power, as shinwoo said.

So then, you have x (3)(1/2), which simplifies to x 3/2.
[/hide]

 :)

[hide]This means the square root of x to the power of 3,
square root of x= x^1/2
Cube of x=x^3
x^((3)(1/2))
x^(3/2)
>>$x^3/2$<<[/hide]

dwx -- the ^ and _ commands only affect the immediate next character, unless you enclose things in brackets.  Thus, you meant to write [hide]x^{3/2}, not x^3/2.  $x^{3/2}$.[/hide]

yeah....

the way you wrote means totally different stuff....

LaTeX could be really confusing.. :D

[hide]
Actually, you are all wrong. That is not a simplified radical; that is just turning a radical into a power. Therefore, the answer is $x\sqrt{x}$.[/hide]

I was trying to get metsfan001's answer, but I guess you could consider the rest of your guy's answers correct too.

[quote="Mamat"]:sqrt: x :^3: =x* :sqrt: x=x^(3/2)[/quote]

How could you tell "x>0"?

Extension:
a) Simplify $\sqrt{x\sqrt{x}}$  (exponential form is okay)

b) Simplify $\sqrt{x\sqrt{x\sqrt{x}}}$

c) [color=darkblue][size=75]<Removed by moderator rcv.  Too challenging for this forum.>[/size][/color]

[quote="dk2"]Extension:
a) Simplify $\sqrt{x\sqrt{x}}$  (exponential form is okay)

b) Simplify $\sqrt{x\sqrt{x\sqrt{x}}}$[/quote]
[hide="a"]$x^{\frac{3}{4}}$[/hide]
[hide="b"]$x^{\frac{3}{8}}$[/hide]

[quote="rcv"][quote="jli"][hide="b"]$x^{\frac{3}{8}}$[/hide][/quote]
It would be more helpful to others, if you would show your work.  (Especially on a problem that's especially challenging for this forum.)  I believe your answer to part a is correct, but your answer to part b is incorrect.[/quote]

sorry

[hide="somethng you need to know"]$\sqrt{x}=x^{1\2}$[/hide]

$\displaystyle \sqrt{x\sqrt{x\sqrt{x}}}=\displaystyle \sqrt{x\sqrt{x\cdot x^\frac{1}{2}}}=\displaystyle \sqrt{x\sqrt{x^\frac{3}{2}}}=\displaystyle \sqrt{x\cdot x^\frac{3}{4}}=\displaystyle \sqrt{x^\frac{7}{4}}=x^\frac{7}{8}$

oops. must have did something wrong the first time :D  :blush:

[hide="original problem"] x^(3/2)[/hide]
[hide="a"]x^(3/4)[/hide]


If  $ x,y,z$  are  positive numbers, then
$ (x^2\plus{}1)(y^2\plus{}1)(z^2\plus{}1) \ge (2x^2\minus{}y^2\plus{}1)(2y^2\minus{}z^2\plus{}1)(2z^2\minus{}x^2\plus{}1).$

[quote="soruz"]If  $ x,y,z$  are  positive numbers, then
$ (x^2 \plus{} 1)(y^2 \plus{} 1)(z^2 \plus{} 1) \ge (2x^2 \minus{} y^2 \plus{} 1)(2y^2 \minus{} z^2 \plus{} 1)(2z^2 \minus{} x^2 \plus{} 1).$[/quote]
Try $ x\equal{}4,$ $ y\equal{}1$ and $ z\equal{}2.$ :wink:

If  $ x,y,z$  are  positive numbers, then
$ xyz \ge (2x\minus{}y)(2y\minus{}z)(2z\minus{}x).$

[quote="soruz"]If  $ x,y,z$  are  positive numbers, then
$ xyz \ge (2x \minus{} y)(2y \minus{} z)(2z \minus{} x).$[/quote]
It's still wrong: try $ x\rightarrow\plus{}\infty,$ $ y\equal{}1$ and $ z\equal{}3.$

$ 2x\minus{}y\equal{}a$ , $ 2y\minus{}z\equal{}b$ , $ 2z\minus{}x\equal{}c$  -------> $ z\equal{}\frac{2a\plus{}b\plus{}4c}{7}$ , $ x\equal{}\frac{4a\plus{}2b\plus{}c}{7}$ , $ y\equal{}\frac{a\plus{}4b\plus{}2c}{7}$----->
$ xyz\equal{}\frac{4a\plus{}2b\plus{}c}{7}\frac{a\plus{}4b\plus{}2c}{7}\frac{2a\plus{}b\plus{}4c}{7}\equal{}\frac{a\plus{}a\plus{}a\plus{}a\plus{}b\plus{}b\plus{}c}{7}\frac{a\plus{}b\plus{}b\plus{}b\plus{}b\plus{}c\plus{}c}{7}\frac{a\plus{}a\plus{}b\plus{}c\plus{}c\plus{}c\plus{}c}{7}\geq$
$ \frac{7\sqrt[7]{a^{4}b^{2}}c}{7}\frac{7\sqrt[7]{ab^{4}c^{2}}}{7}\frac{7\sqrt[7]{a^{2}bc^{4}}}{7}\equal{}abc\equal{}(2x\minus{}y)(2y\minus{}z)(2z\minus{}x)$

$ a\equal{}b\equal{}c$ ---->$ x\equal{}y\equal{}z$

$ \frac{x^2}{y} \ge 2x\minus{}y$
$ \frac{y^2}{z} \ge 2y\minus{}z$
$ \frac{z^2}{x} \ge 2z\minus{}x$
From multiplication of these relations result
$ xyz \ge (2x\minus{}y)(2y\minus{}z)(2z\minus{}x).$

soruz, zaya_yc, why $ 2x\minus{}y,$ $ 2y\minus{}z$ and $ 2z\minus{}x$ are non-negative? :wink:

If  $ x,y,z$  are  positive numbers with $ \frac{x}{y} \ge \frac{1}{2},\frac{y}{z} \ge \frac{1}{2},\frac{z}{x} \ge \frac{1}{2}$, then
$ xyz \ge (2x\minus{}y)(2y\minus{}z)(2z\minus{}x).$

[quote="soruz"]$ \frac {x^2}{y} \ge 2x \minus{} y$
$ \frac {y^2}{z} \ge 2y \minus{} z$
$ \frac {z^2}{x} \ge 2z \minus{} x$
From multiplication of these relations result
$ xyz \ge (2x \minus{} y)(2y \minus{} z)(2z \minus{} x).$[/quote]
It's wrong:
$ 1\geq \minus{} 2$
$ 1\geq \minus{} 2$
$ 3\geq2$
Hence, $ 1\cdot1\cdot3\geq( \minus{} 2)\cdot( \minus{} 2)\cdot2$ $ ?$ :wink:

If $ x \equal{} 4,y \equal{} \frac {1}{4}, z \equal{} 1$ then $ 4\geq31$, i.e given inequality($ xyz\geq(2x\minus{}y)(2y\minus{}z)(2z\minus{}x)$) is wrong! :)

Can someone to tell wright condition for this inequality?

Um, I think $ x\ge y\ge z$ (or cyclic permutations) is satisfactory for the last inequality. :wink: