se dau numerele intregi strict pozitive $a_{1},a_{2},\ldots,a_{n}$, care au urmatoarea proprietate :
pentru diferite alegeri ale $\varepsilon_{i}\in\left\{0,1\right\}$, sumele $\sum_{i=1}^{n}a_{i}\varepsilon_{i}$ obtinute sunt distincte doua cate doua.
aratati ca $\sum_{i=1}^{n}\frac{1}{a_{i}}<2$.

nu va lasati descurajati de titlul topicului :)

[hide="hint 1 - o informatie aproape inutila"] problema e mai mult de algebra, decat de teoria numerelor sau combinatorica[/hide]
[hide="hint 2 - o mica indicatie, si asta aproape inutila"] faptul de teoria numerelor/combinatorica este doar o observatie foarte simpla [/hide]
[hide="hint 3 - un fapt simplu de observat, tine de prima parte a demonstratiei"] daca ordonam numerele $a_{1}<a_{2}<\ldots<a_{n}$, ce putem spune despre sumele partiale $a_{1}+a_{2}+\ldots+a_{k}$ ?[/hide]
[hide="hint 4 - ideea cea mai dificila"] identitatea lui abel aplicata corespunzator[/hide]
[hide="solutie completa"]
sa consideram $a_{1}<a_{2}<\ldots<a_{n}$.
fie $k$ un numar arbitrar, fix, intre $1$ si $n$. pentru o submultime $I\subseteq\left\{1,2,\ldots,k\right\}$, consideram suma
$S_{I}=\sum_{i\in I}a_{i}$, si apoi aplicatia $I\rightarrow S_{I}$. aceasta este o aplicatie injectiva, deoarece sumele sunt distincte. deci sunt cel putin $2^{k}$ sume, incluzand suma in care toti $\varepsilon$ sunt $0$. Cum $a_{1}+a_{2}+\ldots+a_{k}>S_{I}$, pentru orice $I$ care nu contine toate numerele de la $1$ la $k$, rezulta ca $\sigma_{k}=a_{1}+a_{2}+\ldots+a_{k}$ este mai mare decat $2^{k}-1$ numere intregi, nenegative, distincte doua cate doua. rezulta ca $\sigma_{k}\geq 2^{k}-1$. (daca $\sigma_{k}\leq 2^{k}-2$, atunci cele $2^{k}-1$ sume distincte ar fi in multimea $[0,2^{k}-3]\cap\mathbb{Z}$, si ar fi distincte doua cate doua, contradictie).
aceasta este partea I a demonstratiei
la partea a doua, vom avea :
$a_{1}<a_{2}<\ldots<a_{n}$, rezulta
$\frac{1}{a_{1}}>\frac{1}{2a_{2}}>\frac{1}{4a^{3}}>\ldots>\frac{1}{2^{n-1}a_{n}}$.
notam :
$Q=a_{1}\cdot\frac{1}{a_{1}}+a_{2}\cdot\frac{1}{2a_{2}}+\ldots+a_{n}\cdot\frac{1}{2^{n-1}a_{n}}$. (*)
conform identitatii lui abel :
$Q=\left(\frac{1}{a_{1}}-\frac{1}{2a_{2}}\right)\sigma_{1}+\ldots+\left(\frac{1}{2^{n-2}a_{n-1}}-\frac{1}{2^{n-1}a_{n}}\right)\sigma_{n-1}+\frac{1}{2^{n-1}a_{n}}\cdot\sigma_{n}$. (**)
(*) ne spune ca
$Q=\sum_{i=0}^{n-1}\frac{1}{2^{i}}<2$.
(**) ne spune ca :
(deoarece diferentele din paranteze sunt pozitive, iar sumele partiale au fost evaluate in partea I; inlocuind $\sigma_{k}\geq 2^{k}-1$, si apoi desfacand parantezele si reducand tot ce este posibil)
$Q\geq \frac{1}{a_{1}}+\frac{2}{2a_{2}}+\frac{4}{4a_{3}}+\ldots+\frac{2^{n-1}}{2^{n-1}a_{n}}$.
deci $\sum_{i=1}^{n}\frac{1}{a_{i}}\leq Q<2$, adica ce aveam de demonstrat.
[/hide]

nu s`a mai gandit nimeni la problema asta ? :( nici macar dupa ce am postat hinturile ? :(

Pai m-am uitat la hinturi... si cand am vazut ca era cu identitatea lui Abel, am renuntat. Intotdeauna mi se pare picata din cer folosirea acestei identitati.

Motivul pentru care postez: Azi cautam ceva pe site si am dat peste [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5670]topicul ăsta[/url]. Postul $\# 4$ de acolo e destul de interesant:
[quote="pbornsztein"]A with a more serious effort, you may replace it by $\frac{31}{30}$, which is the best possible (and with a non strict inequality).

Reference :
A. Schinzel, G. Szekeres, 'Sur un probl�me de M. Paul Erd�s', Acta Sci. Math. (Szeged), 20 (1959), p.221-229.

Pierre.[/quote]

[Edit][b][i]Vezi postul de mai jos.[/i][/b][/Edit]

cred ca acel 31/30 se refera la problema din topicul ala, cu multiplii.
aici 2 este cea mai buna valoare, pentru ca luam $a_{k}=2^{k}$, pt $k\geq 1$ si suma aia tinde la 2.


Foloseste formula
$\sin nx=C_{n}^{1}\sin a\cos^{n-1}a-C_{n}^{3}\sin^{3}a\cos^{n-3}a+\ldots$


AMC
AIME
number theory
prime factorization

The longest line of consecutive integers $\le 10000$ for which the greatest prime in their prime factorization is $> 7$ is $n$ integers long.  Find $n$.

Ummm... am I not understanding your question correctly? Because the way it reads to me, it would be $1$, since every alternate integer is divisible by 2...  :maybe:

Oops.  I miswrote the question.  Edit...

I still do not understand the question

How can multiple integers have only one prime factorization?

If you are talking about multiple prime factorizations, then why wouldn't the answer be $1$ by Zuton force's logic?

A rewording:

What is the length of the longest stretch of consecutive positive integers, with no integer $> 10000$, such that the factorization of each separate integer contains at least one prime $> 7$?

(EDIT: Oops, misread. Sorry.)

[hide="Um..."] It seems to me that the answer is not an AIME answer.

Consider $p_{1}, p_{2}, ...$, the primes.  We can solve the modular system

$x \equiv 0 \bmod p_{5}$
$x+1 \equiv 0 \bmod p_{6}$
$x+2 \equiv 0 \bmod p_{7}$
...
$x+k \equiv 0 \bmod p_{5+k}$

By Chinese Remainder Theorem for arbitrarily large $k$, and this will produce an infinite number of values of $x$ for which each of the integers $x, x+1, ... x+k$ is divisible by a number larger than $7$.

So we can have a length of $\boxed{10000}$.  [/hide]

[quote="paladin8"][quote="cincodemayo5590"]A rewording:

What is the length of the longest stretch of consecutive positive integers, of length $\leq 10000$, such that the factorization of each separate integer contains at least one prime $> 7$?[/quote]

I think each integer is supposed to be $\le 10000$, not the number of consecutive integers.[/quote]

Oops. Yea, you're right. Edited.  :)

Yeah, cincodemayo's second rewording is correct.


Answers:
[hide="1"]-2[/hide]
[hide="2"].0561[/hide]
[hide="3"]-42[/hide]
[hide="4"]-3[/hide]
[hide="5"]-4[/hide]
[hide="6"]x=59[/hide]
[hide="7"]Y=M-4N
[/hide]
[hide="8"]x=1/2[/hide]
[hide="9"]I'm almost positive I did something wrong...$2x^2-11x+45$ remainder 165? :?: [/hide]

Please limit posts to one problem per post.

[hide="what I got for #9"]
$2x^2-27x+109-\frac{451}{x+4}$
I used synthetic division[/hide]

[quote="easyas3.14159..."][hide="9"]I'm almost positive I did something wrong...$2x^2-11x+45$ remainder 165? :?: [/hide][/quote]

You messed up in your division. It might have been a sign problem (it's supposed to be -19[b]-[/b]8, but you did -19[b]+[/b]8).

And I don't think that polynomial division is quite appropriate for this forum.


Let $ c_1$, $ c_2$, $ c_3$ and $ c_4$ be circles such that $ c_1$ is tangent to $ c_2$, $ c_2$ is tangent to $ c_3$, $ c_3$ is tangent to $ c_4$ and $ c_4$ is tangent to $ c_1$. Prove that the four points of tangency are concyclic.

[geogebra]f1957af368273dad50ef46d56ef78bd809a8da81[/geogebra]

[hide]
We need to show that quadrilateral $ ABCD$ is concyclical.
Denote; $ \angle O_1AB \equal{} \angle O_1BA \equal{} \theta_1$,
$ \angle O_2BC \equal{} \angle O_2CB \equal{} \theta_2,$
$ \angle O_3CD\angle O_3DC \equal{} \theta_3,$
$ \angle O_4AD \equal{} \angle O_1DA \equal{} \theta_4$

If it is concyclical then the opposite angles are supplementary $ \Rightarrow (180 \minus{} (\theta_1 \plus{} \theta_2)) \plus{} (180 \minus{} (\theta_3 \plus{} \theta_4)) \equal{} 180^o \longrightarrow \theta_1 \plus{} \theta_2 \plus{} \theta_3 \plus{} \theta_4 \equal{} 180$
and this is what we will show.

Now there are four isosceles triagnles; $ \bigtriangleup O_1AB, \bigtriangleup O_2BC...$ all interior angles of these triagnles add up to $ 4\cdot (180)$. also Quadrilateral $ O_1O_2O_3O_4$ has interior angles adding up to $ 360^o$

$ \Rightarrow 720 \equal{} 360 \plus{} 2(\theta_1 \plus{} \theta_2 \plus{} \theta_3 \plus{} \theta_4) \longrightarrow \theta_1 \plus{} \theta_2 \plus{} \theta_3 \plus{} \theta_4 \equal{} 180^o$
QED[/hide]


Solve the following equation (nearest whole number) AND explain how you did it:

$ 180sin(x) \equal{} \pi x \minus{} 360$

[hide="Solution"]
Note that $ \minus{} 1\le \sin x\le 1$, which implies $ \minus{} 180\le 180\sin x\le 180$.

Thus, we have $ \minus{} 180\le \pi x \minus{} 360 \le 180 \implies 180\le \pi x\le 540$.

Dividing all terms by $ \pi$, we obtain $ 57\le x \le 172$.

Answer: $ 57 \le x \le 172$[/hide]

EDIT: Oops, typo :oops:

Thanks for pointing out, mz :)

good job fanfan, but with one slight flaw....

$ 360\plus{}360\neq540$ :wink:

he just typoed.
instead of 360 in that particular part of the problem, he meant to write 180...
so i guess it was completely correct

There are many values of x that satisfy the equation?

A solution has to be in the range that FantasyLover provided, but only certain values in the range work.

How would you find the exact value? The "x" in the equation represents an angle measure, the answer to a question.

There's not really any good way that I know of. If $ x$ is in degrees, the answer is around $ 146.34465^\circ$, and if it's in radians, one answer is around $ 128.55904$.

Original problem: "A chord of a unit circle cuts off a region of area 1. What is the measure, to the nearest degree, of the central angle that subtends the chord?"

I wrote the area of the triangle using $ \frac {absin(C)}{2}$ and set it equal to the area of the sector $ \frac {\pi C}{360}$ minus 1.

Equation to solve: $ 180sin(x)\equal{}\pi x \minus{}360$

Yeah, so the answer is $ \boxed{146^\circ}$.


A convex body in $\mathbb R^n$ is a compact convex subset of $\mathbb R^n$ with non-empty interior. For every such convex body $K$, let $A(K)$ be the group of affine bijections of $\mathbb R^n$ which map $K$ onto itself. We say that a convex body $K$ is maximally symmetric if there is no convex body $S$  for which the inclusion $A(K)\subset A(S)$ is strict.

What are the maximally symmetric convex bodies in $\mathbb R^n$?

Problem placement is always a judgment call, but I think convex sets and affine transformations mark this one as linear algebra.


Euler
number theory proposed
number theory

if n=2k+1  n,k naturals prove that $n\mid 2^{n!}-1$.

[mod.: do not use pictures, use plain TeX-code here!]

It suffices to show that $\varphi(n)$ divides $n!$ by the theorem of Euler-Fermat.
But thats fairly easy: We know that $\varphi(n) \leq n$, thus $\varphi(n)$ occurs in $n! = 1 \cdot 2 \cdot ... \cdot \varphi(n) \cdot ... \cdot n$.


ratio
conics
ellipse
parabola
hyperbola
geometric sequence
algebra

On a multiple choice test I am working on, I am getting answers for some of them which correspond to none of the given answers (which is fine, since I can put NONE OF THE ANSWERS, but I want to double check--well, I have checked these myself but don't feel that confident):

1. If the sum of the first six terms of a geometric progression is nine times the sum of the first three terms, what is the common ratio?

2.what is the eccentricity of $ 4x^2 \plus{} 9y^2 \minus{} 24x \plus{} 36y \plus{} 36 \equal{} 0$?

3. Find the value of the term independent of $ a$ in the expansion of $ (a^2 \minus{} \frac{4}{a})^{\frac{9}{2}}$?

4. Find the sum of the lengths of the major and minor axes of the following graph: $ 9x^2 \plus{} 8x \plus{} 7y^2 \plus{} 6y \equal{} 0$.

[hide="the answers that I got but am not sure about"]
1.$ \sqrt[3]{9}$

2.$ \frac{\sqrt{65}}{9}$

3. $ \frac{315}{2}$

4. $ \frac{\sqrt{1351}}{9} \plus{} \frac{3\sqrt{193}}{7}$[/hide]

I would appreciate anyone who is willing to double check any of these.

Thanks in advance.

[hide="1"]
Let the terms be $ a,ar,ar^2,\ldots,ar^5$, such that

$ a(1 \plus{} r \plus{} r^2 \plus{} \cdots \plus{} r^5) \equal{} 9a(1 \plus{} r \plus{} r^2)\implies \frac {r^6 \minus{} 1}{r^3 \minus{} 1} \equal{} 9$

$ \implies r^3 \plus{} 1 \equal{} 9\implies r^3 \equal{} 8\implies r \equal{} \boxed2$
[/hide]
[hide="2"]
$ 4x^2 \plus{} 9y^2 \minus{} 24x \plus{} 36y \plus{} 36 \equal{} 0$

$ \implies4(x \minus{} 3)^2 \plus{} 9(y \plus{} 2)^2 \equal{} 36$

$ \implies\frac {(x \minus{} 3)^2}9 \plus{} \frac {(y \plus{} 2)^2}4 \equal{} 1$

This means that $ (a,b) \equal{} (3,2)$, so the eccentricity is $ \frac {\sqrt {3^2 \minus{} 2^2}}3 \equal{} \boxed{\frac {\sqrt5}3}$.
[/hide]
[hide="3"]
We want $ \frac32\text{ }a^2$'s and $ 3\text{ } \minus{} \frac4a$'s.

By the Generalized Binomial Theorem,

$ \left(a^2 \minus{} \frac4a\right)^{\frac92} \equal{} \cdots \plus{} \frac {\frac92\left(\frac92 \minus{} 1\right)\left(\frac92\minus{}2\right)}6\cdot\left(a^2\right)^{\frac32}\cdot\left( \minus{} \frac4a\right)^3 \plus{} \cdots \equal{} \cdots \plus{} \boxed{ \minus{} 420} \plus{} \cdots$
[/hide]
[hide="4"]
$ 9x^2 \plus{} 8x \plus{} 7y^2 \plus{} 6y \equal{} 0$

$ \implies9\left(x \plus{} \frac49\right)^2 \plus{} 7\left(y \plus{} \frac37\right)^2 \equal{} \frac {193}{63}$

$ \implies\frac {\left(x \plus{} \frac49\right)^2}{\frac {193}{567}} \plus{} \frac {\left(y \plus{} \frac37\right)^2}{\frac {193}{441}} \equal{} 1$

$ \implies a \plus{} b \equal{} \frac {\sqrt {1351}}9 \plus{} \frac {\sqrt {193}}{21} \equal{} \boxed{\frac {7\sqrt {1351} \plus{} 3\sqrt {193}}{63}}$
[/hide]

EDIT: hurdler, I messed up on 3 originally, and have corrected it.

Ok thanks a lot. I see what I did wrong on the first two.

On number three are you sure? The choices given are

a. -60      b. -210     c. 420      d. 210

I am not totally sure what the generalized binomial theorem is...

I got that the coefficient was $ \frac{315}{2}$ with the following computations:

$ 64 \times \frac{\frac 92 \times \frac 72 \times \frac 52 \times \frac 32}{4!}$

Oh, sorry...refer to [url=http://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem]this.[/url]

Edit: On number 3, are you sure the original question isn't $ \left(a^2 \minus{} \frac4a\right)^{\frac92}$?

How close to a circle a conic section is: circles have eccentricity of 0, ellipses have eccentricity of 0 to 1 exclusive, parabolas always has eccentricity of 1, and hyperbolas have eccentricity of more than one. Look it up on wikipedia for more information.

[quote="hurdler"]Um, number 3 was correct...what I wrote is what you just wrote right??

Number 4 was double that ugly answer tho.....2(a+b)...its ok tho..I still got an A on this test.[/quote]

[quote="math154"]...$ \left(a^2\plus{}\frac4a\right)^{\frac92}$?[/quote]

Sorry, that's what I meant (if this were the case, then $ 420$ would be the answer, which is one of the choices, but $ \minus{}420$ is not).

And I thought number $ 4$ asked for the sum of the radii somehow. I need to stop making comprehension mistakes...


[quote="treegoner"]Let ABC be a triangle. Find the minimum of
     $\frac{b^2+c^2}{8Rr} + \frac{a^2}{b^2 + c^2}$[/quote]

I solved this problem but I am not sure because I do not understand your subject, treegoner. The minimum I found is 2 not 11/2.May be I have a mistake.

Use the formula : abc=4pRr, our expression P becomes :
 $P=\frac{(b^2+c^2)(a+b+c)}{4abc} + \frac{a^2}{b^2 + c^2}$
$P=\frac{b^2+c^2}{4bc}+\frac{(b^2+c^2)(b+c)}{4abc}+\frac{a^2}{b^2+c^2} $
We have $\frac{b^2+c^2}{4bc}\geq\frac{1}{2} $
$\frac{(b^2+c^2)(b+c)}{4abc}+\frac{a^2}{b^2+c^2} = \frac{(b^2+c^2)(b+c)}{8abc}+\frac{(b^2+c^2)(b+c)}{8abc}+\frac{a^2}{b^2+c^2}\geq\ 3\frac{((b^2+c^2)(b+c)^2)^(1/3)}{(64(bc)^2)^(1/3)}\geq\frac{3}{2} $
Therefore the minimum of P is 2 and the equality occurs when ABC is an equilateral triangle.

[quote="tranminhhoang"]I solved this problem but I am not sure because I do not understand your subject, treegoner. The minimum I found is 2 not 11/2.May be I have a mistake.[/quote]
I think "11" means that this problems is intended for 11th grade. Am I right?

[quote="tranminhhoang"]I solved this problem but I am not sure because I do not understand your subject, treegoner. The minimum I found is 2 not 11/2.May be I have a mistake.[/quote]

Maybe 11/2 just means that today is a deadline for the problem (you know the deadlines discussion on MathLinks  :) )...

  darij


[quote="mathservant"]Let G be the centroid of  ABC triangle,and let $A_1,B_1,C_1$ be the mid points of BC,CA,AB respectively.If P is an arbitrary point in the plane of ABC triangle  show that
$PA+PB+PC+3PG\geq 2(PA_1+PB_1+PC_1)$[/quote]

Let's note $\frac{2\overrightarrow{PA_1}+\overrightarrow{PA}}{3}=\vec{PG}$ i.e.
$\overrightarrow{AP}+3\overrightarrow{GP}=2\overrightarrow{A_1P} \Rightarrow AP+3GP\ge 2PA_1$
Similarly: $PB+PC\ge 2PA_1 \Rightarrow$
Thus we have $\frac{1}{2}PA+PB+PC+\frac{3}{2}PG\ge 3PA_1$

And similarly we have $\frac{1}{2}PB+PA+PC+\frac{3}{2}PG\ge 3PB_1$ ;
                                 $\frac{1}{2}PC+PB+PC+\frac{3}{2}PG\ge 3PC_1 \Rightarrow$

$\frac{3}{2}(PA+PB+PC+3PG)\ge 3(PA_1+PB_1+PC_1) \Leftrightarrow$
$PA+PB+PC+3PG\ge 2(PA_1+PB_1+PC_1)$

V:)


floor function
combinatorics proposed
combinatorics

This is a problem that we can deduce immediately from a theorem of Extremal Theory. But I hope for a direct solution:

Let $ E\equal{}\{1,2,...,n\}$. Find the greatest positive integer $ k$ such that there exists $ k$ subsets $ A_1,A_2,...,A_k$ of $ S$ such that $ A_i$ is not contained by $ A_j,\forall j\neq i$.

What, Sperner's theorem?  There's a neat proof available on [url=http://en.wikipedia.org/wiki/Sperner%27s_theorem]Wikipedia[/url], but as the statement of the problem is asking for the number of antichains in a very well-understood poset I think it makes the most sense just to deal with the general theory.

Sperner's theorem just provides a bounded for $ k$, but we can not construct $ {n\choose \lfloor \frac{n}{2} \rfloor}$ subsets satisfies the condition.

The theorem I mentioned stated that: "In a poset, the least number of anti-chains to cover the whole poset is equal to the length of the largest chain of it. I do not remember the name of this theorem :P

Right, you're talking about Dilworth's theorem.  But I don't understand your objection:  the subsets we want are precisely all subsets of size $ \left\lfloor \frac {n}{2} \right\rfloor$, none of which can contain the other.  Any level in a graded poset has this property.

The upper bound in Sperner's Theorem is tight (this can be read in the "book" proof t0rajir0u links to above).

To achieve $ k\equal{}{n\choose \lfloor n/2\rfloor}$, take all the subsets of size $ \lfloor n/2\rfloor$.

Oh, right. Thank you both of [b]otsu[/b] and [b]t0rajir0u[/b]. I was wrong when thinking of Sperner's theorem since the proof that $ k\le {n \choose \lfloor \frac {n}{2}\rfloor}$ was obtained by a tighter one:

$ \frac {k}{{n \choose \lfloor \frac {n}{2}\rfloor}} \le \sum_{i = 1}^{k} \frac {1}{{n \choose |A_i|}} \le 1$


Let $x,y,z$ be real numbers such that $x\leq y\leq z\leq 1.$ and $4x+3y+2z=1.$

(1) Find the maximum value of $x$ and the minimum value of $y.$
(2) Find the range of $3x-y+z.$

$z = \frac{1-4x-3y}{2}$
Substitute into the system of inequalities.
The problem becomes a Linear Programming problem in 2 variables.
( I wonder if  Linear Programming is included in the secondary school syllabus of your country or not?)

Japanese students usually study Linear Programming at the age of 17.Some students study this at the age of 15.


The operation $*$ which makes two non zero integers $m,\ n$ correspond to the integers $m*n$ satisfies the following three conditions.

[A] $0*n=n+1.$

[B] $m*0=m+1.$

[C] $m*n=(m-1)*(m*(n-1)) \ \ (m\geq 1,\ n\geq 1).$

(1) Find $1*n.$
(2) Find $2*n.$
(3) Find $3*n.$


(1) For $ x\geq 0,\ y\geq 0,\ x\plus{}y\equal{}1$, find the maximum and minimum value of $ x^3\plus{}y^3$.

(2) Find the minimum value of $ a$ such that $ (x\plus{}y)^3\leq a(x^3\plus{}y^3)$ for all $ x>0,\ y>0$.

(1) $ 0 \leq y \equal{} 1 \minus{} x$
$ x^3 \plus{} y^3 \equal{} 1 \minus{} 3x \plus{} x^2 \equal{} 3(x \minus{} \frac {1}{2})^2 \plus{} \frac {1}{4} \leq 1$ ($ \because x \leq 1$)
from holder ; $ 4(x^3 \plus{} y^3) \geq (x \plus{} y)^3 \rightarrow x^3 \plus{} y^3 \geq \frac {1}{4}$
$ \therefore 1 \geq x^3 \plus{} y^3 \geq \frac {1}{4}$
(2) $ x \equal{} y \equal{} 1 \rightarrow a \geq 4$
from holder ; $ 4(x^3 \plus{} y^3) \geq (x \plus{} y)^3$
therefore the minumum of $ a$ is $ 4$

Or for (2), assume $ x\plus{}y\equal{}1$, then use (1). :P


In town A there are n girls and n boys and every girl knows every boy. Let a(n,r) be the number of ways in which r girls can dance with r boys, so that each girl knows her partner. In town B there are n girls and 2n-1 boys such that girl i knows boys 1, 2, ... , 2i-1 (and no others). Let b(n,r) be the number of ways in which r girls from town B can dance with r boys from town B so that each girl knows her partner. Show that a(n,r) = b(n,r).

It would help to do searches before posting.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1219063#1219063


Let ABC be a non-isosceles triangle. Let A', B', C' be the midpoints of its sides BC, CA, AB, and let D, E, F be the feet of the altitudes to these sides. It is well-known that the points A', B', C', D, E, F all lie on the nine-point circle of triangle ABC. Consider the arc DA' of this nine-point circle which lies outside triangle ABC. Let A" be the point on this arc DA' which divides this arc in the ratio 2 : 1, i. e. which fulfills the condition that the arc DA" is twice as long as the arc A"A'. Similarly, define the point B" on the arc EB', and the point C" on the arc FC'.

Prove that the triangle A"B"C" is equilateral.

Addition from me: Find the relationship of this problem with the one posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=25115 .

  darij

It is just angle chasing.... take the biggest angle $C=\gamma$, then $E\in CB', D\in CA'$ and by and easy angle chasing we have $\angle B'C'A'=\gamma, \angle DFE=180-2\gamma$ and we cut off $1/3$ of the arcs $B'E$ and $DA'$, so doing the calculations we have $\angle B''C''A''=\frac{\pi}{3}$. Now take the smallest angle $B$, and since $C'\in BF, A'\in BD$ we can do almost the same we did before and see that $\angle A''B''C''=\frac{\pi}{3}$, and we are done.   :D


am interested in physics and maths equally,an imo medallist,are there chances for me to get into mit etc. for undergrades? :?:

There is a chance to get in if you apply. I'm guessing you are an international student. MIT admits students from countries all over the world each year. If you are still preparing for MIT, and already have an Olympiad medal, maybe building up and showing your ability in English will be the best way to increase your chances. Good luck.


inequalities
trigonometry
geometry
circumcircle
inradius
inequalities proposed

hey galois

what is ${r_{a}}$ ?

the exradius relativ to side a?

or other thing?

[color=purple]I'm evarist not galois :D .Sorry they are exradii [/color]
 :lol:

This isn't easy problem, however before solve it, I have to say that we should
discuss fairly, stop all your challenge here  :!: .
Proof.
$\frac{r_{a}^{2}}{a^{2}}=\frac{s^{2}\tan\frac{A}{2}}{16R^{2}\sin^{2}\frac{A}{2}\cos^{2}\frac{A}{2}}=\frac{1}{16}(\sum\sin A)^{2}(\sum\frac{1}{\cos^{2}\frac{A}{2}}) =(\prod\cos^{2}\frac{A}{2})(\sum\frac{1}{\cos^{2}\frac{A}{2}}) =\sum\frac{\cos^{2}\frac{B}{2}\cos^{2}\frac{C}{2}}{\cos^{2}\frac{A}{2}}$
we need to prove
$\sum\frac{\cos^{2}\frac{B}{2}\cos^{2}\frac{C}{2}}{\cos^{2}\frac{A}{2}}\ge \frac{9}{4}$
This isn't easy problem however it can't be hard with me!  :D  :D
Change into triangle $\triangle(\pi-2A,\pi-2B,\pi-2C)$ we need to prove
$\sum\frac{\sin^{2}B\sin^{2}C}{\sin^{2}A}\ge \frac{9}{4}$ forall acute triangle!
we have
$\sum\frac{\sin^{2}B\sin^{2}C}{\sin^{2}A}=\frac{b^{2}c^{2}}{4R^{2}a^{2}}=\frac{a^{2}b^{2}c^{2}}{4R^{2}}(\sum\frac{1}{a^{4}})=4S^{2}(\sum\frac{1}{a^{4}})$
Thus we need to prove
$\sum\frac{1}{a^{4}}\ge \frac{9}{16S^{2}}$ for all acute triangl
Actually it is Iran 96 indeed
For all acute triangle $a'=a^{2},b'=b^{2},c'=c^{2}$ are three side of a triangle $\triangle A'B'C'$ have circumradius $R'$ inradius $r'$ by Heron formula we can prove $S^{2}=4r'(4R'+r')$ thus we need
$\sum\frac{1}{a'^{2}}\ge\frac{9}{4r'(4R'+r')}$ for all triangle $A'B'C'$ which is true by Ravi subtitude $a'=y+z,b'=z+x,c'=x+y$
Now I wait for "stronger" problem, I guess this problem isn't of you, the best you should write author of it  :D

let's substitute $p-a=x,p-b=y,p-c=z$ then our inequality is equivalent to
$\sum_{cyc}\frac{yz(x+y+z)}{x(y+z)^{2}}\geq \frac{9}{4}$
$\sum_{cyc}\frac{1}{(xy+xz)^{2}}\geq \frac{9}{4xyz(x+y+z)}$
which is the Iranian inequality...

[color=purple] You are right Albanian.The stronger:
 $\sum \frac{r_{a}^{2}}{a^{2}}\geq \frac{9}{4}+\frac{15\prod (a-b)^{2}}{4\prod (a+b)^{2}}$
   And you can kill it easy![/color]

[color=purple] Prove in all triangle ABC we have [/color]
 $\frac{{r_{a}}^{2}}{a^{2}}+\frac{{r_{b}}^{2}}{b^{2}}+\frac{{r_{c}}^{2}}{c^{2}}\geq \frac{9}{4}+\frac{15\prod(a-b)^{2}}{4\prod(a+b)^{2}}$
 [color=purple]Hi gemath the 1st problem is mine! :D [/color]

If you want a stronger one try with this following
$\frac{r_{a}^{2}}{a^{2}}+\frac{r_{b}^{2}}{b^{2}}+\frac{r_{c}^{2}}{c^{2}}\ge\frac{9}{4}+4\frac{(b-c)^{2}(c-a)^{2}(a-b)^{2}}{a^{2}b^{2}c^{2}}$
Obvious is stronger than yours  :D  :D

If $a,b,c$ are the sides of a  triangle, $r_{a},r_{b},r_{c}$ are the radii of  excircles, then
\[\frac{{r_{a}}^{2}}{a^{2}}+\frac{{r_{b}}^{2}}{b^{2}}+\frac{{r_{c}}^{2}}{c^{2}}\geq \frac{9}{4}. \]
This is problem was proposed by Ji Chen, see:[url=http://www.shuxuetongxun.com/]Bulletin of Mathematics (Wuhan)[/url], 1994, No.2, p.34, Problem 75. The solution see 1994, No.3, p.34.

[quote="gemath"]
$\frac{r_{a}^{2}}{a^{2}}+\frac{r_{b}^{2}}{b^{2}}+\frac{r_{c}^{2}}{c^{2}}\ge\frac{9}{4}+4\frac{(b-c)^{2}(c-a)^{2}(a-b)^{2}}{a^{2}b^{2}c^{2}}$
[/quote]
Ji Chen I see you are interested in inequalities with $r_{a},r_{b},r_{c}$ the inequality I posted above is quite hard with $r_{a},r_{b},r_{c}$, try it if you want!

[quote="gemath"]ok that's right Ji Chen I have checked your example, the const is too small
maybe my computer can't recognize, I have found the following base on it
$\frac{r_{a}^{2}}{a^{2}}+\frac{r_{b}^{2}}{b^{2}}+\frac{r_{c}^{2}}{c^{2}}\ge\frac{9}{4}+\frac{1}{4}\min\{\frac{(b-c)^{2}}{a^{2}},\frac{(c-a)^{2}}{b^{2}},\frac{(a-b)^{2}}{c^{2}}\}$
Do you have the solution for case $\frac{15}{4}$?[/quote]
[color=purple]  What about the following problem?
 ${\frac{r_{a}^{2}}{a^{2}}+\frac{r_{b}^{2}}{b^{2}}+\frac{r_{c}^{2}}{c^{2}}\ge\frac{9}{4}+\frac{3}{4}[\frac{(b-c)^{2}}{a^{2}}+\frac{(c-a)^{2}}{b^{2}}+\frac{(a-b)^{2}}{c^{2}}}]$
   JiChen I am the first person proposed the 1st problem in Viet Nam and the main way is Iran 96.I don't know you proposed it in China.Because you proposed earlier than i did so you are the  author of this problem! Now have you got the stronger? [/color]

[quote="Ji Chen"][quote="gemath"]I have found the following base on it
$ \frac {r_{a}^{2}}{a^{2}} \plus{} \frac {r_{b}^{2}}{b^{2}} \plus{} \frac {r_{c}^{2}}{c^{2}}\ge\frac {9}{4} \plus{} \frac {1}{4}\min\{\frac {(b \minus{} c)^{2}}{a^{2}},\frac {(c \minus{} a)^{2}}{b^{2}},\frac {(a \minus{} b)^{2}}{c^{2}}\}$
[/quote]
In fact, the following inequality hold:

$ \frac {r_{a}^{2}}{a^{2}} \plus{} \frac {r_{b}^{2}}{b^{2}} \plus{} \frac {r_{c}^{2}}{c^{2}}\ge\frac {9}{4} \plus{} \frac {3}{2}\min\Bigl\{\frac {(b \minus{} c)^{2}}{a^{2}},\frac {(c \minus{} a)^{2}}{b^{2}},\frac {(a \minus{} b)^{2}}{c^{2}}\Bigr\}$,

where the constant $ \frac {3}{2}$ is the best possible.
[/quote]
[color=purple]  Thank you but could you please post the solution fot it, JiChen ? [/color]


sorry if im being ignorant but why do some people have the name of a millennium prize next to their profile?

just ignore that i found the answer on another discussion

Administrators can choose something other than "Birch Swinnerton&Dyer"


geometry
rectangle
pigeonhole principle
combinatorics proposed
combinatorics

Firstly I don't know what section it belongs but although:
In 3x4 rectangle noted 6 points arbitrarily. Prove that between them there can be found 2 points so that the distance between these 2 points is less than sqrt5

This is probably geometry.
[hide="Hint:"]Split the rectangle into six $ 1\times 2$ rectangles.[/hide]

Hi ! Use Pigeonhole principle and this image :

[img]http://img98.com/images/ezvci2l282yk8yw960.jpg[/img]

I edited my pico (instead of typo !! :P )

Nice idea, but the upper-left and lower-right corners of the yellow region are too far apart.  If you move the green-red-yellow and blue-yellow-purple intersections a half-unit to the right, I think it becomes okay.


[quote="Utkirstudios"]a,b,c positive integers then prove:
\[ \frac{{a^{2}-bc}}{{\sqrt{a^{2}+bc}}}+\frac{{b^{2}-ac}}{{\sqrt{b^{2}+ac}}}+\frac{{c^{2}-ab}}{{\sqrt{c^{2}+ab}}}\ge 0 \]
[/quote]
I have not found the link. The proof by standard idea:
$ \sum_{cyc}\frac{{a^{2}-bc}}{{\sqrt{a^{2}+bc}}}=\sum_{cyc}\left(\frac{a^{2}-bc-ab+ac}{2\cdot\sqrt{a^{2}+bc}}+\frac{a^{2}-bc+ab-ac}{2\cdot\sqrt{a^{2}+bc}}\right)=$
$ =\sum_{cyc}\left(\frac{(a-b)(a+c)}{2\cdot\sqrt{a^{2}+bc}}-\frac{(c-a)(a+b)}{2\cdot\sqrt{a^{2}+bc}}\right)=\sum_{cyc}\left(\frac{(a-b)(a+c)}{2\cdot\sqrt{a^{2}+bc}}-\frac{(a-b)(b+c)}{2\cdot\sqrt{b^{2}+ac}}\right)=$
$ =\sum_{cyc}\frac{(a-b)^{2}(c^{3}+c^{2}(a+b)+c(a^{2}-ab+b^{2}))}{2\cdot\sqrt{(a^{2}+bc)(b^{2}+ac)}((a+c)\sqrt{b^{2}+ac}+(b+c)\sqrt{a^{2}+bc})}\geq0.$


Let x and y be positive integers such that xy divides $ x^{2}\plus{}y^{2}\plus{}1.$ Show that $ \frac{x^{2}\plus{}y^{2}\plus{}1}{xy}\equal{}3$

Wait, you just reposted all your problems in a different forum section!
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=316368


geometry
parallelogram
AMC
USA(J)MO
USAMO
analytic geometry
combinatorics unsolved

let X=(x,y) with 0<= x,y<=n and x,y in N
find the number parallelogram with 4 vextex in X

This problem is very similar to problem 4 of Mock USAMO that is in progress until July 26. I will post the solution after that date. I would also like other ppl to please not post the solution until that time so that it is fair to mock USAMO participantants. :!:

Thanks to everyone for not posting :)
Here is the solution: we can notice that a parallelogram is formed by points with coordinates $(x,y), (x+a,y), (x,y+b), (x+a, y+b)$
We can assume $a,b$ are positive and this way we will not count any parallelograms more than once :)
Because every point must have coordinates in the range from $0$ to $n$. Therefore we have:
$x\ge 0, x+a\le n$(if $x\ge 0, a>0$ then $x+a>0$);
$y\ge 0, y+b\le n$(if $y\ge 0, b>0$ then $y+b>0$)
Let's count the number of ways to choose $(x,a)$ so that $x\ge 0, x+a\le n$. For every given sum $a+x=k$, $k\ge 0$ we have $k$ values for $x$ because $x$ can be any integer in the range from $0$ to $k-1$(so that $a>0$). $k$ can be any integer in the range from $1$ to $n$, so the total number of ordered pairs $(x,a)$ is $\sum_{i=1}^{n}i = \frac{n(n+1)}{2}$
The number of ordered pairs $(y,b)$ is the same as the number of pairs $(x,a)$ because we have the same restrictions: $y\ge 0, y+b\le n$
Therefore the total number of parallelograms is equal to the numebr of sets of ponts $(x,y), (x+a,y), (x,y+b), (x+a, y+b)$ which is equal to the number of quadruples $(x,a,y,b)$ which is equal to the product of the number of pairs $(x,a)$ and the pairs $(y,b)$ which is equal to $\frac{n^{2}(n+1)^{2}}{4}$ and that is the answer.


Thank you. Btw, you sure do have a good memory. lol.

[quote="boxedexe"]Hi,

How can we solve for x in 3^x+2^x=35. I don't know if I have already asked this question in this forum before, so I'll appreciate it if you solve this for me. This would be quite challening since there's no addiction rule for log.

Thanks in advance,

Masoud zargar[/quote]

$3^x+2^x=35$

$x=3$

Thankyou for posting the obvious, anirudh.

[hide]
$3^x+2^x=35$
$3^x \equiv 3 \mod {8}$ for odd $x$
$3^x \equiv 1 \mod {8}$ for even $x$

$3^x +0 \equiv 3 \mod {8}$ from the given,
so $3^x \equiv 3 \mod {8}$ and $x$ must be odd.
It is obvious that $x < 4$, because $3^4 =81$
So the only numbers we have to test are 1 and 3

$\boxed {3}$ works!
[/hide]

There's of course the most precise Calculus method. Of course, it can;t be posted here!!!

You know that x can't be over three because 3^4 is already 81.

is there an answer that isn't an integer?


1. If a car is travelling in uniform circular motion, why is the force of static friction acting towards the center of the circle? If you draw a free body diagram, friction is acting perpendicular to the car's instantaneous motion. I've always viewed friction as acting opposite to the motion in question so I cannot make sense of this.

2. If a car is on a banked road, why must it rely on friction to prevent from sliding up the bank when travelling at high velocity?

Friction is a force, and therefore will be the acceleration. An object in circular motion has acceleration towards the center of that circle, so friction is towards the center.
Friction can be opposite to the motion, but that won't make it go in a circle, it'll just decrease the velocity.

I'm sorta confused what you mean be if a car is on a banked road.

The car has a tendency to continue in its motion.  Tangentially.

So to prevent it from continuing to go straight, we need static friction to stop it from flying off.

Banking the road helps, but if the road is ice, for example, it is very dangerous (as in, you might spin out of control)

So banking is not exactly enough.

The wheels' static friction with the ground prevent it from sliding to the outside of the curve when you turn on a banked curve.

It is perpendicular to the car's instantaneous motion because, as static friction, it doesn't exactly have to follow the rules of acting against the force.


martabeye graph sadeye g, 8 va andazeye un 27 mibashad, darajeye chand rase un max ast?


I am not sure if this is the right place to post my question, but anyway:can somebody write something or give some links where I can find anything about Dehn invariant?(especially I am interested if there are new problems connected to it)

I remember one day, about a year ago, Barbu Berceanu explained about this invariant. I think this is coming from a
problem conjectured by Hilbert. ;)   Am i right? :?

http://mathworld.wolfram.com/DehnInvariant.html

And, yeah, you're right Cezar : http://en.wikipedia.org/wiki/Dehn_invariant ;)

Yes, you are, but I asked for something much more specific.

I am really not an expert in this topic (it rather belongs into the college playground in my opinion), but in case you can read Russian, you may be interested in a book freely downloadable in DjVU format from http://ilib.mirror0.mccme.ru/djvu/encikl/enc-el-5.htm , namely

П. С. Александров, А. И. Маркушевич, А. Я. Хинчин (ред.), [i]Энциклопедия элементарной математики, книга пятая (геометрия)[/i], Moscow 1966,

chapter "Равносоставленность многоугольников и многогранников". You can get a DjVU reader at http://ilib.mirror0.mccme.ru/4djvu/refs.html .

  darij


Prove that: $ 2401 \mid 7 \times 2^{42} \plus{}21 \times 2^{35} \plus{} 35 \times 2^{28} \plus{} 35 \times 2^{21} \plus{} 21 \times 2^{14} \plus{} 7 \times 2^{7}$

I tried, but Couldn't :(


Let $ a,b$ and $ c$ be positive, real numbers such that $ abc \equal{} 1$.
 Prove that
$ (\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})^2\geq 3(a\plus{}b\plus{}c)\plus{}2(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})\plus{}1$

check your inequality again,
$ a\equal{}b\equal{}c\equal{}1$ is a counterexample

Yes, sorry. when i wrote it in Latex i missed an $ 1$
It's $ (\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}1)^{2}\geq 3(a\plus{}b\plus{}c)\plus{}2(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})\plus{}1$


What is the midpoint of the segment connecting $ (3,4)$ and $ (3.8, 5.6)$ in the Cartesian plane? Express the coordinates as decimals to the nearest tenth.


[quote="April"]Find all finite sets of positive integers with at least two elements such that for any two numbers $ a$, $ b$ ($ a > b$) belonging to the set, the number $ \frac {b^2}{a \minus{} b}$ belongs to the set, too.[/quote]
Let $ m$ be the smallest element and let $ M$ be the biggest. Then $ \frac{m^2}{M\minus{}m} \ge m \iff 2m \ge M$. So if the smallest number is $ k$ the biggest number is $ \le 2k$. Then keep $ M$ as the largest and let $ M_2$ be the second largest. If $ \frac{M_2^2}{M\minus{}M_2} \equal{} M$, then $ M^2\minus{}MM_2 \minus{} M_2^2 \equal{} 0$ which has no integer solutions. Then $ \frac{M_2^2}{M\minus{}M_2} \le M_2 \iff 2M_2 \le M$. So $ 2M_2 \le M \le 2m$ giving $ m \equal{} M_2$ and $ M \equal{} 2m$. So the only solutions are $ S \equal{} \{k,2k\}$ $ \forall k \in \mathbb{N}$

The answer is only $\{x, 2x\}$ for any positive integer $x$. Let $S$ be a set that satisfies the given conditions.

Select $x$ to be minimal in $S$, and consider some other element $a \in S$. By minimality of $x$,$$\frac{x^2}{a-x} \geq x \implies a \leq 2x.$$Suppose for the sake of contradiction that there exists an $a < 2x$. Then, the condition implies$$\frac{x^2}{a-x} \leq 2x \implies a \geq \frac{3x}2$$for all $a \in S$. Similarly, because$$\frac{x^2}{a-x} \geq \frac{3x}2$$as all $a \neq 2x \geq \frac 32$, we must have $a \leq \frac{5x}3$. Continuing in this fashion, we may obtain$$\frac{F_{2n} x}{F_{2n-1}} \leq a \leq \frac{F_{2n-1}x}{F_{2n-2}}$$for all positive integers $n$. But $\lim_{n \to \infty} \frac{F_n x}{F_{n-1} x} = \phi$ is irrational, so such a positive integer $a$ cannot exist given a positive integer $x$, as required.


logarithms function calculus search Putnam

Find all positive integers $x$ and $y$ such that $x^{y+1} = (x+1)^y + 2001$.

You know x divides 2002. I guess you could brute force it from there, but that isn't too sleek.

Actually, I've been seeking a way to expand $\log$ functions in the form $\log (a+b)$.  I don't think there is a rule.  Does anyone have an idea how to expand it?

[quote="Iversonfan2005"]I got it down to $\frac{y+1}{y}=\log{(2000x+2000)}$, can anyone solve this though?[/quote]

How did you get that?

[quote="chess64"]There is no way, unless you have the actual values of $a$ and $b$. Then you can use

$\log_{10} {(a+b)} = k;\\ 10^k = a+b$

And see what happens when you substitute.[/quote]

This has been my biggest problem for the last year.  I can't seem to find a way to solve a problem without knowing that.

Masoud Zargar

That's because you [i]can't[/i]. Some things you just can't do. This is one of them. It's like trying to simplify $2^5 + 2^{97}$. (Factoring doesn't count :D)

for your problem boxedexe, eventhough calculus is looked down upon, there should be a way to expand the log function using taylor/maculurian series

say you have $\log(a+b)$, factor to get $\log a(1+\frac{b}{a})$ then $\log a + \log (1+\frac{b}{a})$, then let $\frac{b}{a}=x$, then find an expansion for $\log{(1+x)}$, and substitute in x

i do not know if this would work, but it might, my calculus is shaky so i am not sure

as for the actual problem, i have no idea, maybe mods would help one narrow their search

[quote="Altheman"]...find an expansion for $\log{(1+x)}$...[/quote]

Isn't that the whole problem?  We can't find an expansion for the log of the sum of two numbers unless we know both of the numbers.

my response was more towards boxedexe's post, and since i was not really clear, i was thinking of doing a taylor series for $\log{(1+x)}$

EDIT: here is what i got...

$\log e * \sum_{k=1}^{\infty}\frac{(-1)^{n-1}*x^n}{n}=\log{(x+1)}$

How has this tough problem suddenly become a lesson in taylor series?

[quote="tµtµ"]FMako gave the clue : expanding by binomial gives that $x \mid 2002$

A quick look $mod\ 3$ gives that $x = 1 mod\ 3$ (and $y$ even).

This gives only $7$ and $13$ as possibilities for $x$.

$x^{y+1}$ is quickly bigger  $(x+1)^y + 2001$

"Manual" (I used a computer to do that of course :lol:) checking gives only $13,2$[/quote]
Line 1&2 is good.

Line 3, you missed $x=91=3(30)+1$ because $x \mid 2002$ does not imply that $x$ is prime.

Line 4 is false. If $y>\frac{\log x}{\log (x+1)-\log x}$, then $x^{y+1}<(x+1)^y$.

Line 5, I agree because after 1 hour of searching for other solutions "manually," I was not able to solve it, not even check it for $x=91$.

This is a Putnam problem which I happened to do yesterday.  As such it's not really suitable for this forum, but here's a sketch of how I did it.

We have

$a^n-(a+1)^{n-1}=2001$

By inspection $a=13$, $n=2$ is a solution.

Take the equation mod $a$.  We get

$0-1=2001\bmod a$; or $a|2002=2\cdot 7\cdot 11\cdot 13$

Take it $\bmod 3$.  We reduce to $a\in\{7,13,91\}$

Take it $\bmod 7$.  We reduce to $a=13$

Now it's a simple matter of seeing that $13^n-(14)^{n-1}$ is increasing for $n>1$.

Take it $\bmod 8$ to see that we must have $n\leq 2$; hence we're done.

[quote="blahblahblah"]This is a Putnam problem which I happened to do yesterday.[/quote]
Thank you $\text{blah}^3$. PK, please explain yourself for putting a [b]Putnam[/b] problem here

No explaining has to be done.  When problems are presented without sources, as they often are, it can be difficult to tell what the actual difficulty of the problem is.  Certainly this doesn't seem too hard at a first glance.

[quote="blahblahblah"]Now it's a simple matter of seeing that $13^n-(14)^{n-1}$ is increasing for $n>1$.[/quote]This is not true! $\text{blah}^3$, take $n=36$  :!:  :!:  :!:  :yoda:

Indeed.  It's immaterial, as my proof works without that line - the mod 8 analysis takes care of everything.


[quote="sylow_theory"][quote="SharkBites77"]This problem was confusing me, so any help would be appreciated.

F_3 is the finite field with three elements, where the three elements are {0, 1, 2}. (Addition in F_3 is defined modulo 3, which is another way of saying that 3=0)

V = (F_3)^5 is a vector space.

Let S = Span({(1, 1, 1, 2, 1), (1, 1, 2, 1, 1), (1, 2, 1, 1, 1), (2, 1, 1, 1, 1), (1, 1, 1, 1, 2)} ⊂ V. What is the dimension of S? (Remember that 3=0)

Is there a way of doing this without using Graham-Schmidt process (we haven't been taught this process yet, so our teacher doesn't want us using it).[/quote]

Just find the rank of the matrix.[/quote]

Okay, thanks!

So I put the vectors inside the span into a matrix like so:

1 1 1 2 1
1 1 2 1 1
1 2 1 1 1
2 1 1 1 1
1 1 1 1 2

Then I row reduced it to get

1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1

The rank is 5, so does this mean the dimension of S is 5?  And is the "Remember the 3=0" part just a piece of unneeded information in the problem?

First off: Gram-Schmidt is a complete irrelevancy, since it's for inner product spaces, which $(\mathbb{F}_{3})^{5}$ is not.

I have to question your row reduction, since the rank is not 5. Did you ever divide by 3? That would be the same as dividing by zero.

Note that

$\begin{bmatrix}1\\1\\1\\2\\1\end{bmatrix}+\begin{bmatrix}1\\1\\2\\1\\1\end{bmatrix}+\begin{bmatrix}1\\2\\1\\1\\1\end{bmatrix}+\begin{bmatrix}2\\1\\1\\1\\1\end{bmatrix}+\begin{bmatrix}1\\1\\1\\1\\2\end{bmatrix}= \begin{bmatrix}6\\6\\6\\6\\6\end{bmatrix}= \begin{bmatrix}0\\0\\0\\0\\0\end{bmatrix}$

That's a nontrivial linear combination of these vectors adding up to the zero vector, so they are linearly dependent.

I'm a little confused as to why we are dividing by three.  I thought we could just row reduce the following matrix of the vectors to find the rank and then that would equal the dimension:

1 1 1 2 1
1 1 2 1 1
1 2 1 1 1
2 1 1 1 1
1 1 1 1 2

Yes, we can use row reduction. But all arithmetic must be performed mod 3. That's what I sense you haven't grasped - you're trying to work over the rational numbers, but for this problem, the rational numbers do not exist. The entire universe of arithmetic is $\mathbb{F}_{3}.$ That means that $-1=2,$ that $2\cdot2=1,$ that $2+1=0.$ Here's what I get doing row elimination on your matrix:

$\begin{bmatrix}1&1&1&2&1\\ 1&1&2&1&1\\ 1&2&1&1&1\\ 2&1&1&1&1\\ 1&1&1&1&2\end{bmatrix}\longrightarrow \begin{bmatrix}1&1&1&2&1\\ 0&0&1&2&0\\ 0&1&0&2&0\\ 0&2&2&0&2\\ 0&0&0&2&1\end{bmatrix}$

$\longrightarrow \begin{bmatrix}1&1&1&2&1\\ 0&1&0&2&0\\ 0&2&2&0&2\\ 0&0&1&2&0\\ 0&0&0&2&1\end{bmatrix}\longrightarrow \begin{bmatrix}1&1&1&2&1\\ 0&1&0&2&0\\ 0&0&2&2&2\\ 0&0&1&2&0\\ 0&0&0&2&1\end{bmatrix}$

$\longrightarrow \begin{bmatrix}1&1&1&2&1\\ 0&1&0&2&0\\ 0&0&1&1&1\\ 0&0&0&1&2\\ 0&0&0&2&1\end{bmatrix}\longrightarrow \begin{bmatrix}1&1&1&2&1\\ 0&1&0&2&0\\ 0&0&1&1&1\\ 0&0&0&1&2\\ 0&0&0&0&0\end{bmatrix}$

That's in row echelon form. There are four pivots; the rank is 4. The original five vectors span a 4-dimensional subspace of $(\mathbb{F}_{3})^{5}.$ Another description of this subspace: $\{(x_{1},x_{2},x_{3},x_{4},x_{5})\in(\mathbb{F}_{3})^{5}\mid x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=0\}.$

If we replaced every $2$ in the original statement of the problem by $-1,$ would it look different to your eyes? That would in fact be no change at all in the statement.

[quote="Kent Merryfield"]Yes, we can use row reduction. But all arithmetic must be performed mod 3. That's what I sense you haven't grasped - you're trying to work over the rational numbers, but for this problem, the rational numbers do not exist. The entire universe of arithmetic is $\mathbb{F}_{3}.$ That means that $-1=2,$ that $2\cdot2=1,$ that $2+1=0.$ Here's what I get doing row elimination on your matrix:

$\begin{bmatrix}1&1&1&2&1\\ 1&1&2&1&1\\ 1&2&1&1&1\\ 2&1&1&1&1\\ 1&1&1&1&2\end{bmatrix}\longrightarrow \begin{bmatrix}1&1&1&2&1\\ 0&0&1&2&0\\ 0&1&0&2&0\\ 0&2&2&0&2\\ 0&0&0&1&2\end{bmatrix}$
[/quote]

Thanks!  I get it now!  If I'm not mistaken, though, shouldn't the last row of your matrix be 0 0 0 2 1 instead?

Yes, that was a mistake, and I've gone back to fix it. But it turns out not to matter: 0 0 0 1 2 and 0 0 0 2 1 are scalar multiples of each other. The computed rank of 4 is correct.


Find the maximum value of $ \frac {585xyz}{5yz^2 \plus{} 9xy^2 \plus{} 75zx^2}$.

Nice problem

[hide="Solution"]
$ \frac {585xyz}{5yz^2 \plus{} 9xy^2 \plus{} 75zx^2} \leq a
\Longleftrightarrow 5ayz^2\plus{}9axy^2\plus{}75azx^2 \geq 585xyz$

But by AM-GM in 3, we have that $ 5ayz^2\plus{}9axy^2\plus{}75azx^2 \geq 3 \sqrt[3]{a^3 (xyz)^3 5 \cdot 9 \cdot 75} \equal{} 3axyz15 \equal{} 45axyz$

So by letting $ a\equal{}13$ we get the desired maximum[/hide]


There is something wrong with this message...

[hide]Yes, there really is, and it is...[hide]that when using a second hide[/hide] [/hide] this phrase gets hidden in the first, while it should not be hidden at all? :huh:

Would you mind fixing this, valentin? :) it is often needed to use nested hides, but here it seems to consume the rest of the post... for some reason :?

Thanks a lot! :P

Was this "bug" ... present before last night, or it just showed up right now?

It has been there forever I think :) certainly before last night. I just post it now because I saw other discussions on the malfunctioning of tags, so as you're in the tags right now I thought this may be the best instant to repair this.


is it possible to find the nth term or formula for a sequence by finding its area? any relation or connection between them ?

 1,4, 9, 16, 25, 36, 49, 64,........
for the above sequence the formula is Un=N^2

is there any other way to find the formula?
is there other ways to find formula for sequences ? other than how we normally find.

how about area? how about limits?  if no why ? any other way ? how ?

suppose you have a sequence  $a_{n}$

you can creat a function $L_{n}= a_{n+1}-a_{n}$

or simple combinations like this

So, by induction you find $L_{n}$

and after find $L$ you can find $a$ solving a recursive sequence

i did this in this link
deducting a formula with picard method

http://senhorcarteiro.blogspot.com/2006/11/festa-dos-matemticos.html

The link is in portuguese, but you can understand the math

have a lot of ways to find sequence..

but you need to see the sequence to solve. because give a general method is (impossible)

[quote="binomial_4eva"]is it possible to find the nth term or formula for a sequence by finding its [b]area[/b]? any relation or connection between them ? [/quote]

What do you mean by area?

[quote="binomial_4eva"]s there other ways to find formula for sequences ? other than how we normally find.[/quote]

How do you normally find them?


[quote="ALGEBRA"]if ab +bc+ca=2abc=1 prove that 2(a+b+c)+1>=32abc[/quote]

  It just follows from $ 2abc>=3(a+b+c)$ and $1/abc>=3(a+b+c)$

[quote="ALGEBRA"]if ab +bc+2abc=1 prove that 2(a+b+c)+1>=32abc[/quote]
[size=200]$IQ!!!$[/size]

[quote]ALGEBRA wrote:
if ab +bc+ca+2abc=1 prove that 2(a+b+c)+1>=32abc
[/quote]
my solution:
Let use couchy-shovartz$(a+b+c+\frac12)(bc+ca+ab+2abc)\geq (4\sqrt{abc})^2=16abc
$ then $2(a+b+c)+1\geq 32abc$

but its not ab+bc+ca[b]+[/b]2abc=1, it is ab+bc+ca[b]=[/b]2abc=1

and cezar lupu can u explain ur idea, plz?

excuse me my   friend :blush:                               ab+bc+ca+2abc=1    is correct :blush:

you, yourself wrote
 [quote="ALGEBRA"]if ab +bc+ca=2abc=1 prove that 2(a+b+c)+1>=32abc[/quote]

im confused now  :? , sorry if i got u wrong

I knew that something was wrong here. Well,if your condition is $ab+bc+ca+2abc=1$ then the following substitutions
 will solve the ineqaulity: $a=x/y+z,b=y/z+x,c=z/x+y$


induction
modular arithmetic
number theory unsolved
number theory

[quote="Rust"]It is not true for n=1.[/quote]
I forgot
$\forall{n\geq{2}}$ :)


[quote="ginger"]ΤΟ ΕΠΟΜΕΝΟ ΕΥΚΟΛΟ ΠΡΟΒΛΗΜΑ ΕΥΚΛΕΙΔΕΙΑΣ ΓΕΩΜΕΤΡΙΑΣ ΔΟΘΗΚΕ ΠΡΟΣΦΑΤΑ ΣΕ 200 ΤΕΛΕΙΟΦΟΙΤΟΥΣ ΦΟΙΤΗΤΕΣ ΜΑΘΗΜΑΤΙΚΩΝ ΕΛΛΗΝΙΚΟΥ ΠΑΝΕΠΙΣΤΗΜΙΟΥ.

ΜΟΝΟ ΕΝΑΣ ΤΟ ΕΛΥΣΕ (ηταν βορειοηπειρωτης)!

ΔΙΝΕΤΑΙ ΤΕΤΡΑΓΩΝΟ ΑΒΓΔ ΚΑΙ ΕΣΤΩ Ε, Ζ ΤΑ ΜΕΣΑ ΤΩΝ ΠΛΕΥΡΩΝ ΒΓ, ΓΔ ΑΝΤΙΣΤΟΙΧΑ ΚΑΙ Η, Θ ΤΑ ΣΗΜΕΙΑ ΤΟΜΗΣ ΤΗΣ ΔΙΑΓΩΝΙΟΥ ΒΔ ΜΕ ΤΙΣ ΕΥΘΕΙΕΣ ΑΖ, ΑΕ ΑΝΤΙΣΤΟΙΧΑ.
Α) ΝΑ ΕΞΕΤΑΣΕΤΕ ΑΝ ΟΙ ΓΩΝΙΕΣ ΔΕΖ, ΖΑΕ, ΕΑΒ ΕΙΝΑΙ ΙΣΕΣ ΚΑΙ ΓΑΤΙ.
Β) ΝΑ ΕΞΕΤΑΣΕΤΕ ΑΝ ΤΑ ΕΥΘΥΓΡΑΜΜΑ ΤΜΗΜΑΤΑ ΔΗ, ΗΘ, ΘΒ ΕΙΝΑΙ ΙΣΑ ΚΑΙ ΓΙΑΤΙ.

ΝΕΟΛΑΙΟΙ, ΕΤΟΙΜΑΣΤΕΙΤΕ ΓΙΑ ΤΟΥΣ ΚΑΙΝΟΥΡΓΙΟΥΣ ΚΑΘΗΓΗΤΕΣ ΣΑΣ!

ginger rogers[/quote]

                              :(  :(  :(  :(  :(  :(  :(  :(

Κι ύστερα αναρωτιώμαστε ΓΙΑΤΙ η Ελλάδα δεν πάει καλά στους διεθνείς διαγώνισμούς σε ΟΛΑ ΤΑ ΠΕΔΙΑ ΕΠΙΣΤΗΜΟΝΙΚΑ ΚΙ ΜΗ
Ντροπή κι αίσχος
 :mad:

Tha mporouse kapoios na mas dwsei thn lush tou prob?

[quote="ginger"]ΤΟ ΕΠΟΜΕΝΟ ΕΥΚΟΛΟ ΠΡΟΒΛΗΜΑ ΕΥΚΛΕΙΔΕΙΑΣ ΓΕΩΜΕΤΡΙΑΣ ΔΟΘΗΚΕ ΠΡΟΣΦΑΤΑ ΣΕ 200 ΤΕΛΕΙΟΦΟΙΤΟΥΣ ΦΟΙΤΗΤΕΣ ΜΑΘΗΜΑΤΙΚΩΝ ΕΛΛΗΝΙΚΟΥ ΠΑΝΕΠΙΣΤΗΜΙΟΥ.

ΜΟΝΟ ΕΝΑΣ ΤΟ ΕΛΥΣΕ (ηταν βορειοηπειρωτης)!

ΔΙΝΕΤΑΙ ΤΕΤΡΑΓΩΝΟ ΑΒΓΔ ΚΑΙ ΕΣΤΩ Ε, Ζ ΤΑ ΜΕΣΑ ΤΩΝ ΠΛΕΥΡΩΝ ΒΓ, ΓΔ ΑΝΤΙΣΤΟΙΧΑ ΚΑΙ Η, Θ ΤΑ ΣΗΜΕΙΑ ΤΟΜΗΣ ΤΗΣ ΔΙΑΓΩΝΙΟΥ ΒΔ ΜΕ ΤΙΣ ΕΥΘΕΙΕΣ ΑΖ, ΑΕ ΑΝΤΙΣΤΟΙΧΑ.
Α) ΝΑ ΕΞΕΤΑΣΕΤΕ ΑΝ ΟΙ ΓΩΝΙΕΣ ΔΕΖ, ΖΑΕ, ΕΑΒ ΕΙΝΑΙ ΙΣΕΣ ΚΑΙ ΓΑΤΙ.
Β) ΝΑ ΕΞΕΤΑΣΕΤΕ ΑΝ ΤΑ ΕΥΘΥΓΡΑΜΜΑ ΤΜΗΜΑΤΑ ΔΗ, ΗΘ, ΘΒ ΕΙΝΑΙ ΙΣΑ ΚΑΙ ΓΙΑΤΙ.

ΝΕΟΛΑΙΟΙ, ΕΤΟΙΜΑΣΤΕΙΤΕ ΓΙΑ ΤΟΥΣ ΚΑΙΝΟΥΡΓΙΟΥΣ ΚΑΘΗΓΗΤΕΣ ΣΑΣ!

ginger rogers[/quote]
loooooooooooooooooooooooooooooooooool

Για τον GnS: Το Θ είναι το βαρύκεντρο του τριγώνου ΑΒΓ (και το Η το βαρύκεντρο του τριγώνου ΑΓΔ).

Από τη δική μου εμπειρεία: Τελειόφοιτοι φοιτητές Τμήματος Μαθηματικών στο μάθημα της Διδακτικής των Μαθηματικών είχαν συνολική αποτυχία στο εξής θέμα:

Α) Ισχύει ή όχι η πρόταση;: Αν 6|ν, τότε ή ο ν-1 ή ο ν+1 είναι πρώτος.
(Όλοι απάντησαν ότι ισχύει. Το αντιπαράδειγμα ν=120 δείχνει ότι δεν ισχύει!).
Β) Ισχύει ή όχι η πρόταση;: Αν α>3 είναι πρώτος, τότε ή 6|α-1 ή 6|α+1.
(Απάντησαν ότι ισχύει, αλλά κανείς δεν το απέδειξε! Παρουσίασαν μόνο τρία-τέσσερα παραδείγματα).

ΟΠΟΤΕ,   ΝΕΟΛΑΙΟΙ ΕΤΟΙΜΑΣΤΕΙΤΕ! (ΑΝ ΚΑΙ, ΣΤΑΤΙΣΤΙΚΩΣ, ΟΙ ΣΗΜΕΡΙΝΟΙ ΔΕΝ ΕΙΝΑΙ ΚΑΙ ΠΟΛΥ ΚΑΛΥΤΕΡΟΙ).

Λανς

Πιστεύω πως δεν πρέπει να υπερβάλλουμε σε έναν τόσο μεγάλο βαθμό. Οκ, η κατάσταση είναι εν μέρει άσχημη, άλλα αναφέροντάς το δεν κερδίζουμε και τίποτα.

Μήπως δεν φταίνε τόσο οι εκάστοτε μαθητές αλλά το σύστημα;

Εγώ ξέρω ότι αν κάποια ομάδα δεν αποδίδει τα αναμενόμενα, δεν αλλάζει ο πρόεδρος τους παίκτες, αλλά τους προπονητές. Συμφέρει περισσότερο, βλέπετε, να αλλάξει κάποιος 1 παρά 11.

Οι μελλοντικοί μαθητές θα γίνουν καθηγητές (προφανώς) και το κακό θα διαιωνίζεται.

Μήπως χρειάζεται μία ριζική αλλαγή στους τρόπους διδασκαλίας; Ή μήπως με αυτό το post προσπαθούμε να περάσουμε την εντύπωση ότι το μαθηματικό δυναμικό της Ελλάδας τείνει στο 0;

...

[quote="VickiL"]Οι καθηγητές που είχα εγώ, στην πλειοψηφία τους, (6 στους 8) κατά τη διάρκεια των μαθητικών μου χρόνων (σε Γυμνάσιο, Λύκειο) ήταν αξιολόγοι. Για τον έναν δεν ξέρω να κρίνω γιατί θυμάμαι τότε μου ήταν πολύ δύσκολο να καταλάβω το μάθημά μου, ενώ ο άλλος είχε μεν γνώσεις, αλλά δεν ήταν καλός εκπαιδευτικός.

Δεν είμαι σε θέση να γνωρίζω αν οι γνώσεις τους εκτείνονταν αρκετά εκτός των όσων κάλυπτε η σχολική, διδακτέα ύλη. Αλλά όσον αφορά στο κομμάτι που αφορούσε τη (άμεση) δουλειά τους, ήταν πλήρεις.

Ο ένας μάλιστα θυμάμαι είχε ξενερώσει απίστευτα όταν του είχα παραδώσει τις απαντήσεις σε ένα διαγώνισμά του. Τέλος πάντων.

Αυτό που θέλω να πω, είναι πως συνήθως οι άνθρωποι οι οποίοι γνωρίζουν ή αγαπούν λίγο παραπάνω το αντικείμενό τους, σπάνια θα "καταφύγουν" στο δημόσιο τομέα. Γι' αυτό και τους ψάχνουμε με τα κυάλια.

Όσον αφορά στους φοιτητές, δυστυχώς για όλους μας, ξέρουν ελάχιστα περισσότερα μαθηματικά βγαίνοντας από το Πανεπιστήμιο σε σχέση με όσα ήξεραν μπαίνοντας (και πάλι στην πλειοψηφία τους). Και όσα χρόνια έχουν παραμείνει στη σχολή τους, έχουν αδρανοποιήσει το μυαλό τους, αφού οι περισσότεροι καταφεύγουν σε "θέματα" περασμένων εξεταστικών. Αναζητούν όσα εξετάζονται και διαβάζουν μόνο αυτά. Σα θέματα πάντα και όχι την ουσία τους.

[size=67]Όπως επίσης, να μην αγγίξουμε καθόλου το θέμα του πώς γράφουν στις εξετάσεις.[/size]

Διότι, τουλάχιστον στο τμήμα των Μαθηματικών της Αθήνας, από όσους γνωρίζω εγώ, ήταν ελάχιστοι αυτοί που επέλεξαν συνειδητοποιημένα το τμήμα, και όχι γιατί:
1. ήθελαν πληροφορική και δεν την έπιασαν.
2. ήθελαν οικονομικά και δεν τα έπιασαν.
3. είχαν τα απαιτούμενα μόρια και το μαθηματικό είναι μία "καλή σχολή".

Και κατά γενική ομολογία, οι περισσότεροι κάποια στιγμή θα αποφοιτήσουν. Πόσα μαθηματικά περιμένουμε να γνωρίζουν αυτοί;  :([/quote]

 Ρώτησαν τον αρχιδικαστή των ΗΠΑ :
 - 'Τι χρειάζεται για να είναι κάποιος καλός δικαστής ; '
 - 'Να είναι gentleman !!!'  Τους απάντησε.' Αν ξέρει και λίγα νομικά , δεν πειράζει ! ''
  Από αυτή την παροιμιώδη απάντηση βγαίνει ένα συμπέρασμα:

     '' Για να είσαι καλός δάσκαλος πρέπει πάνω από όλα να έχεις μεράκι ! '' Αν υπάρχει αυτό , θα μάθεις και τα κατάλληλα μαθηματικά.Θα μάθεις καλά ,τουλάχιστον αυτά που διδάσκεις. Αυτό είναι ήδη μεγάλη κατάκτηση. Γιατί ο μαθητής δε σε κρίνει από το πόσα ξέρεις αλλά από το αν του τα μεταδίδεις σωστά και από το αν τον κάνεις να πάει σπίτι και να μελετήσει με όρεξη.Δηλαδή αν τον εμπνέεις ! Ο μαθηματικός του σχολείου δε χρειάζεται να είναι μαθηματικός αλλά Δάσκαλος των μαθηματικών.Γιατί έτσι θα βοηθήσει να γεννηθούν μαθηματικοί μέσα από τους μαθητές του. Και οι μεγάλοι μαθηματικοί της ιστορίας  είχαν καθηγητές στο σχολείο που ήταν απλοί δάσκαλοι, με περιορισμένες γνώσεις.Λίγοι είχαν την τύχη να έχουν δάσκαλο πχ τον Weierstrass . Και όμως αυτοί  έμαθαν μαθηματικά και άφησαν μνημειώδη έργα. Ούτε πάλι είναι απαραίτητο να είναι ο μαθηματικός του σχολείου ολυμπιονίκης , ούτε να μπορεί να λύνει τα θέματα των διθνών ολυμπιάδων. Πρέπει όμως να λύνει τα θέματα του Θαλή και του Ευκλείδη .Είναι  πολύ καλό και απαραίτητο να ενθαρρύνει τους μαθητές του να παίρνουν μέρος  στους διαγωνισμούς , να τους ενημερώνει , να λύνει με τους μαθητές του μιά άσκηση το χρόνο( 10 θα ήταν καλύτερα !) που τέθηκε σε κάποιο διαγωνισμό κλπ - δεν είναι καν απαραίτητο να είναι επιπέδου ΙΜΟ αλλά χαμηλότερου επιπέδου με διδακτική αξία  -  , διότι αυτό διαγείρει το ενδιαφέρον των μαθητών , τους δείχνει σωστό πρότυπο και τους κεντρίζει την περιέργεια για να σκεφτούν.
   Επομένως , για να ενισχύσω και τη σωστή θέση της NickiL ,νομίζω ότι οι υποψήφιοι που δηλώνουν αυτές τις σχολές θα έπρεπε να έχουν δικαίωμα να δηλώνουν το πολύ 2 σχολές (ανεξάρτητα φυσικά από πόλη) , ώστε με την εκπαίδευση να ασχολούνται άνθρωποι που αγαπάνε να γίνουν δάσκαλοι. Αυτό είναι μια καλή αρχή. Από εκεί και πέρα υπάρχουν ακόμα και πολλά άλλα σημαντικά πράγματα που πρέπει να γίνουν για να έχουμε καλούς δασκάλους. Αυτό είναι όμως θέμα για άλλη σηζήτηση.
[u] Μπάμπης[/u]

Άσχετο αλλά τη σημαίνει αυτό:"6|α-1"??????

[quote="mostel"]Το 6 διαιρεί τέλεια το α-1

Ή το α-1 είναι πολλαπλάσιο του 6

Αυτό γενικά σημαίνει το " | "

Επίσης, αν τύχει πουθενά και δεις gcd(a,b) σημαίνει μέγιστος κοινός διαιρέτης (greatest common divisor)[/quote]

Ευχαριστώ mostel.


Let A,C are two point on (O).P is the intersections of (O) at A,C.K is the midpoint of AC. B is an arbitrary point on AC. The line PB meet (O) at R,Q. KQ meet (O) at D.Prove that RD is perpendicular to OK.

Dear Mathlinkers,
It's a basic and well-known result about harmonic quadrilateral.


So, I'm looking at Vieta's formulas, and need a small bit of confirmation on this...
 $ P(X) = a_{n}X^{n}+a_{n-1}X^{n-1}+\cdots+a_{1}X+a_{0}$, and the roots are $ x_{1},x_{2},x_{3},...,x_{n}$.

So...does that mean that:

$ (x_{1}x_{2}x_{3}+x_{1}x{}_{2}{}_{x}{}_{4}+\cdots+x_{1}x_{2}x_{n})+(x_{1}x_{3}x_{4}+\cdots+x_{1}x_{3}x_{n})+\cdots+x_{n-2}x_{n-1}x_{n}=$

$ -\frac{a_{n-3}}{a_{n}}$?

So, if that is the third case, then the rest can be expressed like this? You multiply $ k$ roots together in every single possible combination, and add them all together, that equals $ (-1)^{k}\frac{a_{n-k}}{a_{n}}$, where $ n$ is the highest power...right? But then how does:

$ \sum_{1\le i_{1}< i_{2}<\cdots < i_{k}\le n}x{}_{i{}_{1}}x{}_{i{}_{2}}\cdots x{}_{i{}_{k}}= (-1)^{k}\frac{a_{n-k}}{a_{n}}$ for each $ k = 1,2,\ldots,n$

represent that? What exactly are we adding? All I see is the product of the first $ k$ roots...I don't see how this represents all the terms in the summation. I think I'm missing some sort of recursion...Could someone explain the summation in a little more detail? Thanks a lot :D.

[quote="Icy"]$ \sum_{1\le i_{1} < i_{2} < \cdots < i_{k}\le n}x_{i_{1}}x_{i_{2}}\cdots x_{i_{k}} = ( - 1)^{k}\frac {a_{n - k}}{a_{n}}$ for each $ k = 1,2,\ldots,n$

represent that? What exactly are we adding? All I see is the product of the first $ k$ roots...I don't see how this represents all the terms in the summation. I think I'm missing some sort of recursion...Could someone explain the summation in a little more detail? Thanks a lot :D.[/quote]

All the $ {i,1,i_2,...,i_k}$ are $ k$ natural numbers, different (from the $ <$ sign) and between $ 1$ and $ n$ ,
for example if $ k=n$ then $ i_1,...,i_n$ are all the permutations of $ 1,...,n$
Thus the summation is over all the $ k$ combinations of numbers from $ 1,...,n$.


absolute value
number theory unsolved
number theory

Find all pairs of natural numbers $ (m,n)$ such that the set of positive integers cannot be partitioned into three pairwise disjoint nonempty subsets such that none of them has two numbers with absolute value of their difference equal to either $ m,n$ or $ m \plus{} n$

Let's call this subsets of N by $ A_{k}$ for $ k\in{\{ 1,2,3\}}$.

LEMMA 1:  if $ k > m \plus{} n$ then $ k$ and $ k \plus{} (|n \minus{} m|)$ are in the same subsets

Suppose that $ m\geq n$
$ k \minus{} (k \minus{} m) \equal{} m$ and $ (k \plus{} n) \minus{} k \equal{} n$ and $ (k \plus{} n) \minus{} (k \minus{} m) \equal{} m \plus{} n$ so $ k,(k \minus{} m),(k \plus{} n)$ are in different subsets.
But also $ (k \minus{} m),k \plus{} (|n \minus{} m|),k \plus{} n$ are in different subsets, so $ k$ and $ k \plus{} (|n \minus{} m|)$ are in the same subset.

Easly we have that for values greater than $ m \plus{} n$  if $ k\in{A_{i}}$ then $ k \plus{} t|n \minus{} m|\in{A_{i}}$ for $ t\in{N_{1}}$ (2)

If partitioning is goog for values greater than $ m \plus{} n$ we can partition values less than $ m \plus{} n$ and have the condition (2) for all values of $ k$(not only for greater than $ m \plus{} n$).
So now let's keen on partition with this condition (2) satisfied.

Now $ k$ is in the same subset as $ k$ $ mod(|m \minus{} n|)$
It's also obvious that $ m \equiv n \bmod{|m \minus{} n|}$

So $ k\bmod{|m \minus{} n|}$ is in other subset than $ k \plus{} n\bmod{|m \minus{} n|}$ and $ k \plus{} 2n\bmod{|m \minus{} n|}$

Now we can arbitrally choose $ v < |n \minus{} m|$ such that
$ v\in{A_{a}}$
$ v \plus{} n\bmod(|n \minus{} m|)$ $ \in{A_{b}}$
$ v \plus{} 2n\bmod(|n \minus{} m|)$ $ \in{A_{c}}$
$ a\neq b\neq c\neq a$

Then $ v$ is in the same subset as $ v \plus{} 3kn\bmod|n \minus{} m|$   for  $ k\in{N_{ \plus{} }}$
It means that there not exist such $ k$ that $ v \plus{} 3k\equiv v \plus{} k \bmod|m \minus{} n|$.
And this is true if and only if $ 3$ divides $ \frac {|b \minus{} a|}{NWD(b,a)}$
If it's not true then there of course don't exist such partition.
If it's true there exist such partition and it's for example :
$ 3k\in{A_{1}}$
$ 3k \plus{} 1\in{A_{2}}$
$ 3k \plus{} 2\in{A_{3}}$


1. On a planet there are $3\times2005!$ aliens and 2005 languages. Each pair of aliens communicates with each other in exactly one language. Show that there are 3 aliens who communicate with each other in one common language.

2. Suppose there are $4n$ line segments of unit length inside a circle of radius $n$. Furthermore, a straight line $L$ is given. Prove that there exists a straight line $L'$ that is either parallel or perpendicular to $L$ and that $L'$ cuts at least two of the given line segments.

3. Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=1$. Prove that\[ 6(a^3+b^3+c^3+d^3)\ge(a^2+b^2+c^2+d^2)+\frac{1}{8} \]

4. Show that there exist infinitely many square-free positive integers $n$ that divide $2005^n-1$.

Remark: This is also the third TST for IMO 2006.

Can you please post each problem into a separate topic from now on? Thank you :)

They're here.  :)
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64152]Problem 1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64151]Problem 2[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64150]Problem 3[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64149]Problem 4[/url]


let $x = 2^{a}$ etc so $x,y,z > 1$ and we want

$xy+yz+zx < 2xyz+1$ Hmm let $p=x-1$ etc so $p,q,r>0$

we want

$pq+qr+rp+2(p+q+r)+3 < 2(pqr+p+q+r+pq+qr+rp+1)+1$

$0 < pq+qr+rp+2pqr$

funny problem?


geometry
area of a triangle
Heron\u0027s formula

The length of the base of an isosceles triangle is 20. In the plane of the triangle, a point 4 units from this base is 10 units from each leg. Find the area of the triangle.

I'm sorry I couldn't find an easiest way  :(

ABC is the isosceles triangle (AB=AC, BC=20)
The distance from point H to line BC is 4
The distances between point H and the legs are equal, hence H lies on the bisector of angle BAC.

Name D the point were AH intersects BC. Since ABC is isosceles, D is also the midpoint of BC (BD=DC=10)

Bring a perpendicular line HK from H to AC. We know that HK=10.

Now, the right triangles ADC and AKH are similar (because they have a common angle, DAC=HAK)
Besides, they have DC=HK=10.
Hence the triangles ADC and AKH are equal.(Actually, the point H lies outside the isosceles triangle)
DC=KH
AD=AK
AC=AH

In order to find the area of ABC we first have to find the altitude AD (base is 20).

For the right triangle ADC we have:
[tex]AC^2=AD^2+DC^2[/tex]

[tex]AH^2=AD^2+10^2[/tex]

[tex](AD+DH)^2=AD^2+10^2[/tex]

[tex](AD+4)^2=AD^2+10^2[/tex]

[tex]AD^2+8AD+4^2=AD^2+10^2[/tex]

[tex]8AD+4^2=10^2[/tex]

[tex]AD=10,5[/tex]

[tex](ABC) = (AD)(BC)/2 = 10,5*20/2 = 105[/tex]

[quote="pontios"]I'm sorry I couldn't find an easiest way  :( [/quote]
hmm, your way was much simpler than mine...

using the same notation, i found $DK=\frac{42\sqrt{29}}{29}$ by ptolemy's, and then dropped an altitude from $K$ to $HC$, cutting off a smaller segment of length $\frac{8\sqrt{29}}{29}$.  then i formed a proportion $\frac{\frac{8\sqrt{29}}{29}}{4}=\frac{\sqrt{29}}{x}$ where $x$ is the length of a leg of the isosceles.  $x=\frac{29}{2}$ and i used hero(n)'s formula to find the area to be $105$.  for the longest time, i got degenerate values for the other side b/c i can't divide >_<

The wording is a little awkward... does it mean that the point is the given distances from the perpendicular bisectors to the base and legs?

[quote="Bob"]How do we know that the point is outside the triangle, and not inside? :?:[/quote]
Let's use the notation in pontios's reply, but assume that $H$ lies inside the triangle. $\triangle AHK$ is a right triangle, similar to but contained in the right triangle $\triangle ADC.$ Leg $HK$ corresponds to leg $CD.$ In order for $\triangle AHK$ to be strictly contained in $\triangle ADC$ we must have $HK<CD.$ But $HK=CD=10$ so we can't do that.


Given the equation $ \frac{1}{m} \plus{} \frac{1}{n} \equal{} \frac{2}{15}$, where $ m$ and $ n$ are positive integers, find the value of $ m \plus{} n$ such that $ m > n$ and $ m$ is not a multiple of $ n$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=286335]Search before you post.[/url]

Wait, ignore this topic.

I figured it out.....never mind=)

What is SSFT? Can someone someone explain how 15m+15n=2mn suddenly got factored out and then a 225 appeared on the other side?  :maybe:  :oops:

[url=http://www.artofproblemsolving.com/Wiki/index.php/Sfft]Search before you post.[/url]


Sorry for bothering everybody, but I have a question regarding Goldbach's conjecture. Does anyone know or can they verify or prove that even numbers which can be written as the sum of 3 and some other prime can also be written as another Goldbach's Partition? I'm sorry to bother everybody because I'm not very good; I'm only trying to understand the conjecture through analysis.

If a particular even whole number can be written as the sum of 3 and another prime, $ p$, then it can be expressed as $ 5 \plus{} (p \minus{} 2)\\
\equal{} 7 \plus{} (p \minus{} 4)\\
\equal{} 11 \plus{} (p \minus{} 8)\\
\equal{} 13 \plus{} (p \minus{} 10)\\
\equal{} 17 \plus{} (p \minus{} 14)$, etc.

Almost all reasonably small numbers (say, 4 or 5 digits) can be expressed as one of the sums expressed above. If the number can't be expressed as one of the sums above, then it is expressible as the sum of two primes that are closer in number.

Because the primes are infinite and do not occur in any regular pattern/frequency, I am pretty sure that Goldbach's conjecture has not been proven completely yet. (It is almost certain that the conjecture is true.) If the main conjecture holds, then your statement will probably hold, too.

(I am not great with number theory proofs/conjectures, so you might want to wait for more replies or look online, e.g. Wikipedia, AoPSWiki, etc.)

Umm yes... but can you prove it's true? I'm using casework and have 1 case left to resolve, and this statement is one of the keys to breaking apart this last case (Not that I'm good with proving this stuff; again I'm trying understand the problem more in depth). Anyways, thanks to everyone who posted so far; can anyone prove that trouble-some statement?

From what I know, the Goldbach Conjecture is an extremely old problem and still unsolved. However, the conjecture has been verified up to certain bounds and similar or close results have been proven. For example, every even number greater than or equal to $ 4$ can be expressed as the sum of a prime and a semiprime if not the sum of two primes.

[quote]Does anyone know or can they verify or prove that even numbers which can be written as the sum of 3 and some other prime can also be written as another Goldbach's Partition?[/quote]

I can give a few counterexamples to that.
$ 3 \plus{} 3 \equal{} 6$ and $ 3 \plus{} 5 \equal{} 8$ cannot be written as another Goldbach Partition.
I suppose your statement might be true for larger even numbers, but it would be quite difficult to prove.

Yes, I meant larger numbers. Can anyone offer help?

I'll just give everybody a synopsis of what I did so far, though it isn;t much. I thought of using induction; since this conjecture states it is true for infinitely many numbers, I thought induction would be a good idea. Saying that the base case(s) are true:

$ 4\equal{}p_1\plus{}p_2;
6\equal{}p_3\plus{}p_4;
8\equal{}p_5\plus{}p_6;
      .
      .
      .
2k\equal{}p_{2k\minus{}2}\plus{}p_{2k\minus{}1}$ (I know the notation may be confusing, but all the numbers "p something" denote primes, and therefore obviously different p's can be equal to each other, as in the "3" in $ 3\plus{}5\equal{}8$ and $ 3\plus{}7\equal{}10$)

and we must prove that
$ 2(k\plus{}1)\equal{}p_2k\plus{}p_{2k\plus{}1}$ for some even $ "2(k\plus{}1)"$ and some primes $ "p_2k"$ and $ "p_{2k\plus{}1}"$. From here I decided to use case-work.

[b]Case 1:[/b]

$ p_{2k\minus{}2}, p_{2k\minus{}1}<5$ meaning $ (p_1, p_2)\equal{}(2, 2), (3, 3)$. The reason I'm stating values for the last case of all the cases we assumed is because we can not directly state values for $ p_2k$ and $ p_{2k\plus{}1}$ since we are trying to prove that pair works. Anyways, back to the case=> obviously, we can just brute force this case for the 2 cases stated above: (2, 2), (3, 3).

[b]Case 2:[/b]

$ W.L.O.G. p_{2k\minus{}2}<5, p_{2k\minus{}1} > or \equal{}5$, meaning $ p_{2k\minus{}2}\equal{}3$. Just to clarify why all my cases seem to revolve around the number five, it is the 2nd odd prime, making it the 2nd smallest prime that may be used in a Goldbach's Partition infinitely many times. I will tell you later why I classified 3 by itself in each case.
Anyways, this case is also relatively easy to prove because since we are trying to prove that 2 more than the sum of $ p_{2k\minus{}2}$ and $ p_{2k\minus{}1}$ added by 2 can also be written as a Goldbach Partition, $ p_{2k\minus{}2}\plus{}p_{2k\minus{}1}\plus{}2\equal{}3\plus{}p_{2k\minus{}1}\plus{}2\equal{}5\plus{}p_{2k\minus{}1}$, making this another Goldbach Partition.

[b]Case 3:[/b]

$ p_{2k\minus{}2}, p_{2k\minus{}1}> or \equal{}5$. This is the really tough part, and I need some help on this part. (Actually, it appears that my original question was [b]absolutely useless[/b]; I don't know what I was thinking of, but since you guys are way more experienced than I am, I thought you could tell me how good problem-solvers would tackle this, though that may be slightly too hopeful as this is a very old, challenging problem). Anyways, for this case, I thought that in order for both primes to be greater than or equal to 5, if at all the next even number is indeed the sum of 2 primes, then at least 1 of the primes must have appeared in previous cases somewhere (i.e. the number "3" from 3+7=10 has appeared earlier in 3+5=8 or 3+3=6). I can't exactly explain the rest of my reason, as it is very messy intuitive reasoning stuff. Can you guys please help me on this one? I know that this might not be the easiest way to attempt the problem, and there might be mistakes in my reasoning so far. Can you guys offer help, advice, etc.?

[quote="bengal"]Can you guys offer ... advice?[/quote]  My advice is to not try to work on Goldbach -- there's basically no hope of an elementary solution, and you're better off finding a more tractable but less famous problem on which to spend your time.  If you're dead-set on Goldbach, then you should start by reading past work that pertains to it (e.g., the work of Vinogradov showing that every integer is the sum of at most 4 primes).

Incidentally, I don't see any reason to suppose that the statement you've claimed should be tractable.  The only "moral" reasons it should be true are also moral reasons that the whole Goldbach conjecture should be true, and therefore are probably useless.


geometry
inequalities unsolved
inequalities

Let be a given triangle with area $S$, with sides $a,b,c$ and let $n$ be a positive integer.Prove that

$a^{2n}+b^{2n}+c^{2n} \geq 3(\frac{4}{\sqrt {3}})^nS^n+(a-b)^{2n}+(b-c)^{2n}+(c-a)^{2n} $

Very nice :lol:

Congratulations, hungkhtn.

Very nice proof :)


Yes,  I'm new to the Canadian sub-forum.

Question, is a person still Canadian even though he's living in the United States and is a Republican?

No...lol.

Regardless, I have nothing else but a Canadian citizenship...and this Ontario driver's license that's very much useless here in Georgia.

How does this thread relate to anyone in this forum? It's funny how we, as human beings, go off-topic a lot. Oh well, I have nothing against this forum. MATHLINKS/AoPS IS AWESOME.  :)

MathLinks???

I thought this was Art of Problem Solving!!!

Awww man...that's right..."we" call it MathLinks.

This forum is actually twinned with http://www.mathlinks.ro

They started out as 2 different forums and eventually they merged.  When you post on one, your post appears on the other, and vice versa... they share the same database.

Yea, sure, you can still be Canadian if you move to the US... republican however... hmmmmm...

Well... let's just say you're not gonna get much support from us.  Even the most right-wing Canadians look like commies compared to American republicanism.

[quote="Elyot"]
Yea, sure, you can still be Canadian if you move to the US... republican however... hmmmmm...

Well... let's just say you're not gonna get much support from us.  Even the most right-wing Canadians look like commies compared to American republicanism.[/quote]

Did I imply I was Republican?  LOL.  Seriously...I couldn't be Republican even if I tried...

Apparently the culture shock down here was too strong for me.  I haven't watched hockey since when Tampa Bay beat the Flames.

The Canadian presence is everywhere.  I guess one benefit is that I can do contests like the USAMTS for you guys.

You don't have to be Canadian to join this forum, so it really doens't matter.

[quote="lowest_reynolds"]MathLinks???
I thought this was Art of Problem Solving!!![/quote]It's actually both. As Elyot explained above, it's the same site basically, only with two interfaces. You can switch between the two using the [url=http://www.mathlinks.ro/Forum/usercp.php]options[/url] menu.


Given $ a,b,c\geq\ 0$ .Prove that
 $ \frac{a^2(b\plus{}c)}{b^3\plus{}abc\plus{}c^3}\plus{}\frac{b^2(c\plus{}a)}{c^3\plus{}abc\plus{}a^3}\plus{}\frac{c^2(a\plus{}b)}{a^3\plus{}abc\plus{}b^3}\geq\ 2$

Nice problem, PhongVan.
After using Am-Gm, it's becomes: $ \sum\ \frac{a^2b^2(a\minus{}b)^2}{(a\plus{}c)(b\plus{}c)} \geq\ 0$ (always trues).
Equality occurs when $ a \equal{} b \equal{} c > 0$ or $ a \equal{} b ; c \equal{} 0$ and cyclic.
 :)

See my solution here: http://www.mathlinks.ro/Forum/viewtopic.php?t=195129

[quote="Sunkern_sunflora"]Weird solution!

$ \sum \frac {a^2(b \plus{} c)}{b^3 \plus{} abc \plus{} c^3} \geq \sum \frac {a^2}{(b \plus{} c)^2 \minus{} 3bc \plus{} \frac {abc}{b \plus{} c}} \\
\geq \sum \frac {a^2}{(b \plus{} c)^2 \minus{} 3(b^2 \minus{} c^2) \plus{} ab} \equal{} \sum \frac {a^2}{ab \minus{} (b^2 \plus{} c^2) \minus{} (b \plus{} c)^2} \\
\geq \sum \frac {2a^2}{a^2 \minus{} b^2 \minus{} 2c^2 \minus{} 2(b \plus{} c)^2}$

Now, $ \sum \frac {2a^2}{a^2 \minus{} b^2 \minus{} 2c^2 \minus{} 2(b \plus{} c)^2} \geq {2} \Leftrightarrow {2a^2 \plus{} b^2 \plus{} 2c^2 \plus{} 2(b \plus{} c)^2} \geq {0}$ which is obvious.

Hope there is no mistake. :maybe:  If there is, pls. point it out.
My proof is nowhere good. The two prev. proofs were outstanding! Thanks, guys! :)[/quote]
I am sure it is wrong ,Sunkern_sunflora,because with yours,the equaton holds in $ a \equal{} b \equal{} c \equal{} 0$,it is impossiable :wink:

Yes, Sunkern_sunflora's proof was wrong.
My full solution is similar to can_hang's solution  :)
After using CS and Schur, we only need to prove:
$ \sum\ ab(a^2\plus{}b^2) |\plus{}abc(a\plus{}b\plus{}c) \geq\ 2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\plus{}2abc\sum\ \frac{a^2}{b\plus{}c}$
$ \Leftrightarrow \sum\ \frac{a^2b^2(a\minus{}b)^2}{(a\plus{}c)(b\plus{}c)} \geq\ 0 \Rightarrow q.e.d$


quadratics
algebra
polynomial
Vieta
inequalities
system of equations

[quote="hien"]Find all values $m$ so that the following system of equations has roots:
 $\{\begin{array}{l}x+xy+y=m\\x^{2}y+xy^{2}=3m-8\end{array}$[/quote]
[hide="hint"]
Let xy=a and x+y=b[/hide]

Yes, but when we put $a=x+y,b=xy$ then that has a condition $a^{2}\geq 4b$, how can we treat that case?

I don't think that we need the $a^{2}\geq 4b$ condition.

Here's my solution without that condition, I hope it's although right.

Set $a = x+y$ and $b=xy$. The system reduces to $a+b=m$ and $ab=3m-8$. To put it into a quadratic equation we get (with Vieta) $x^{2}-mx+3m-8=0$
This term has real solutions if, and only if, $\frac{m^{2}}{4}-3m+8\geq 0$. This inequality is equal to $m^{2}-12m+32 = (m-6)^{2}-4 \geq 0$ and so we have $+-(m-6) \geq 2$

So, there are solutions for $m \geq 8$ and $4\geq m$.

I'm not sure if it's true, but I hope so ;)

Yes, in [b]SBHD[/b]'s solution there is a mistake. Becasue with condition $\frac{m^{2}}{4}-3m+8\geq 0$ shows to us  there exists $a,b$ but not difinitely there exists $x,y$[/b]


superior algebra
superior algebra unsolved

Characteristic of a ring R with identity 1 is the smallest positive integer n such that
n*1 = 0. Find the characteristic of the ring Z subscript 12 x Z subscript 105 (Z12 x Z105)

you have to determine the order of $ (1,1)$ in the group $ \mathbb{Z}/12 \times \mathbb{Z}/105$, this is $ lcm(12,105)$.


Construct, with only the aide of the usual euclidean tools, two triangles ABC and RQR  such that:

a) P is modpoint of BC and A is the midpoint of QR

b) QR bisects < BAC and BC bisects < QPR

It's strongly related to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17687]this[/url] problem.

Construct a harmonic quadrilateral $BQCR$: Take a point $X$, and let its tangents to the circle $(O)$ touch the circle in $B,C$. Then let an arbitrary line through $X$ cut $(O)$ in $Q,R$. Now take $A$ to be the midpoint of $QR$ and $P$ the midpoint of $BC$.

[quote="grobber"]It's strongly related to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17687]this[/url] problem. [/quote]

Yes  :) . And my solution is strongly related to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16821]this[/url] problem.


calculus
integration
trigonometry
LaTeX
algebra
partial fractions
calculus computations

hey all, i got a little stumped on this integral. i can't solve it via substitution and as of right now that is as far as my knowledge with integrals extends. can someone help me out?

$\int_0^{\frac{\pi}4}\sqrt{1-\cos(2x)}\,dx$

Basically, there are three families of methods. Look at them in roughly this order:

1. Algebraic manipulation.

2. Substitutions. (First, "natural" substitutions; then "forced" or "backwards" substitutions.)

3. Integration by parts.

(If you're wondering where partial fractions fits, it's part of #1, algebraic manipulation.)

What you've got involves trigonometric functions. How many trigonometric identities are there? More than you want to write down. Start with #1 algebraic manipulation - in particular, think about trig identities.

[hide="Hint."]Double angle formulas.

[hide="Second Hint."]$1-\cos(2x)=2\sin^2x.$[/hide][/hide]

And one other thing - please don't use attachments just to post mathematical forumlas. I have edited your post to include your formula; now I want you to edit it by removing your attachment. When you click on "edit," you'll be able to see the exact code that produces the formula.

alright thanks, sorry about that i don't really understand how to use the code to write the formulas. thanks for the hints but my substitution skill is currently limited to "natural" substitutions. I tried replacing cos2x with 1-2sin^2(x) and now i cannot find a substitution to use to complete the problem  :?

You're not done with the algebra. Don't you have the square root of a square?

oh alright i figured it out. can't believe i didn't see that  :blush: thanks for the hints!

A slightly trickier quesition (related to the arclength of a cycloid):

$\int_0^{2\pi}\sqrt{2(1-\cos\theta)}\,d\theta$

you can look here http://www.gnomon.be/projecten/FormulariumGoniometrie.doc

use the formula of carnot and you can manipulate this integral I will do it for you but I don't know who LaTex work.


counting
distinguishability
rotation
geometry
geometric transformation

So here is my take (and I suck equally as much as you at combinatorics if not more, so take this with a grain of salt until somebody confirms it)

[hide]
Case 1: Fix one red chip
We have $\frac{5!}{3!2!}$ ways of arranging this.

Case 2: Fix one white chip
We have $\frac{5!}{3!}$ ways of arranging this.

Case 1 + Case 2 = 15

looking at this now, it's very likely wrong, but that's my angle  :P.
[/hide]

Let O be white chip and X be red chip.

There are three possible arrangements:

1) There is no (or 4) red chip(s) between the two white chips.

ex) ooxxxx, oxxxo

2) There is one (or 3) red chip(s) between the two white chips.

ex) oxoxxx, xoxxxo

3) There are two red chips between the two white chips.

ex) oxxoxx, xoxxox

You should have divided 6C2 by 5. Notice each circular arrangement above has five different linear arrangements.

No, no, no, no, you guys are all overcounting.
It says IDENTICAL
[hide]
The four red chips are together.
The circle has 3 red, 1 white, 1 red, 1 white.
The circle has 2 red, 1 white, 2 red, 1 shite.
The rest are rotations of these[/hide]

is there a general way to do these kinds of problems?
What if the problems said:
"In how many ways can 15 identical red chips and white identical red chips be arranged in a circle?"

would the answer to this be 14?

Right, all you have to count is the possible distances between the chips.

Count the part between the white chips with the least number of red chips; there can only be 0, 1, or 2 chips, so the answer is 3.

[quote="Hychu"][quote="bpms"]No, no, no, no, you guys are all overcounting.
It says IDENTICAL
[hide]
The four red chips are together.
The circle has 3 red, 1 white, 1 red, 1 white.
The circle has 2 red, 1 white, 2 red, 1 -.
The rest are rotations of these[/hide][/quote]

Except that's exactly what I said. The answer is 3.[/quote]
sorry, I just saw 6C2

[quote="myc"][quote="Hychu"]Let O be white chip and X be red chip.

You should have divided 6C2 by 5. Notice each circular arrangement above has five different linear arrangements.[/quote]

why is there 5 different linear arrangements?  Isn't there 6?  For example, if there were 3 letters, A, B,C arranged in a circle, you have to divide 3! by 3.[/quote]

Upon dividing 3! by 3 you are left with 2!; the number of permutations of n objects when arranged in a circle is (n-1)!. Is this what you are asking?

[quote="myc"]In how many ways can four identical red chips and two identical white chips be arranged in a circle?

this is what I have so far:[hide]
there are $6!$ ways to arrange the chips.  divide this by $4!2!$ because the four red chips are indistinguishable and the two white chips are indistinguishable.  so $\frac{6!}{4!2!}$  I also need to divide this by $6$ because you can rotate the chips around 6 times and the order still wouldn't change, but then I'll get a fractional number.   [/hide]  can anyone tell me what I'm doing wrong?

Thanks     :)[/quote]

or you could just start out with the 4 red coins in a circle.  then the white coins can be across from eacother, next to eaachother (with a red separating) or just side by side (no red separating.  so that makes 3.   :maybe:

[quote="cincodemayo5590"][quote="myc"]yes, so shouldn't you divide 6C2 by 6?[/quote]

No; I'm pretty sure you divide by 5, since when the objects are placed in a circle, you need to use one (distinguishable) object as a reference point. But I'm not sure about this. :P[/quote]
notice that wen u divide $n$ by itself, u're left with one, not zero, so dividing by 6 will still leave 1 distinguishible object for each arrangement.

[quote]
3) There are two red chips between the two white chips.

ex) oxxoxx, xoxxox

You should have divided 6C2 by 5. Notice each circular arrangement above has five different linear arrangements.[/quote]
how did u get five differen't linear arrangements?
I only counted 3.

oxxoxx,  xoxxox, xxoxxo,

Your third is the same as your first, the third is xxooxx.

[quote="Hychu"]You should have divided 6C2 by 5. Notice each circular arrangement above has five different linear arrangements.[/quote]

NEVER MIND ABOUT THIS PART. My bad.


Take $N$ equidistant points on the unit circle. Pick one of those points, then draw chords from it to all other points on the circle. What is the product of lengths of those chords?

[hide="Hint"]Complex numbers[/hide]

[hide] Looking at N=2, 3, 4, and 6, I think the answer is $\boxed{N}$?? :? [/hide]

probably wrong
[hide]
Make the unit circle in complex coordinates...
$z^n = 1$
$z^{n-1} + z^{n-2} + \cdots + z + 1 = 0 (1)$
Let the points be $z_1, z_2, \cdots, z_n$ and assume wlog $z_1 = 1$, and this is the point we draw the chords from.
From $(1)$, we have
$(z - z_1)(z - z_2) \cdots (z - z_{n-1})(z - z_n) = 0$
Our desired value is $|1 - z_2||1 - z_3| \cdots |1 - z_{n-1}||1 - z_n| = |(1-z_2)(1-z_3) \cdots (1 - z_{n-1})(1 - z_n)$
We can get this value by dividing $(1)$ by $(z - z_1)$ and plugging in $z=1$
Using synthetic division, we find the quotient is
$z^{n-2} + 2z^{n-3} + 3z^{n-4} \cdots + (n-2)z + (n-1)$
And the value is $\frac{n(n-1)} 2$
Although, I don't think this works... since I divided by $z - z_1 = z - 1$ when $z=1$, so I pretty much divided by 0.  Although, I'm pretty sure there is a solution very similar... just can't finish it.
[/hide]

The answer is indeed $N$.
Deej, your process is looks good. You probably made a error somewhere... maybe after the point in which you say "We can get this value by dividing..... ". I don't quite follow your logic.

[quote="maniacman384"]isn't (n(n-1))/2 the theorm for 1+2+3+4...+n?[/quote]
That's $\frac{n(n+1)}{2}$. :D

Starting from the same point as deej we can write that these $\ N$ equally spaced points can be considered to be the $\ n$ complex roots of the equation $\ z^n = 1$  or $\ (z-1)(z^{n-1} + z^{n-2} + \cdots + z + 1) = 0$  Then WLOG  let the point chosen be $\ z=1$ and then we must draw chords to all the other $\ n-1$ roots of the equation $\ f(z)= z^{n-1} + z^{n-2} + \cdots + z + 1 = (z - z_1)(z- z_2) \cdots (z- z_{n-2} )(z - z_{n-1} )=0$  So we want to know $\ |1 - z_2||1 - z_3| \cdots |1 - z_{n-1}||1 - z_n| = |(1-z_2)(1-z_3) \cdots (1 - z_{n-1})(1 - z_n) = | f(1) | = N$


topology
advanced fields
advanced fields unsolved

This came up while reading an analysis textbook,

If $X$ is a connected, compact topological space with $p,q \in X$, Under what conditions does there exist a finite sequence of open sets $U_{i}$ such that :
a) $p \in U_{1}$, $q \in U_{n}$
b) $U_{i} \cap U_{j} = \phi \iff |i-j| \ge 2$

I would like the answer to be "hausdorff" (not sure if it's even true in this case, though), but if it's not, I'm hoping for a counterexample and some nice (but not too strong) conditions to make it true.

Surely this is a disguised version of the Hausdorff (T2) property?  If the space is Hausdorff, let $U_1$, $U_3$ be disjoint open sets containing $p,q$ respectively and let $U_2$ be any open set you like.  Conversely, given such a sequence of length $\ge 3$ throw away all the middle terms.


vector
analytic geometry
combinatorics unsolved
combinatorics

$65$ beetles are placed in $65$ cells of a $9$x$9$ table. In each move, ever beetle creeps to an adjacent cell. No beetles makes two horizontal or two vertical moves in succession. Prove that after several moves, at least two beetles will be in the same cell.

The original was that they make diagonal movements. Or in other word: it is also required that the number of steps (counted like here) is even.

Surely posted before, but here it goes (solution for the original one):

There is an odd number $o$ of white or of black squares (chessboard coloring), w.l.o.g. white ones.
Then we can see the movement (after rotating and scaling) as a change of position by one of the vectors $(0,1),(0,-1),(1,0),(-1,0)$. The sum of $o$ of these is supposed to be $0$, clearly impossible. (e.g. consider the sum of coordinates $\mod 2$).

[quote="ZetaX"] The sum of $o$ of these is supposed to be $0$[/quote]

But it isn't necessary that all the squares are occupied, is it? (Because there are only 65 beetles)

Count ;)
If not all are occupied, there are two on the same field.

I do not understand your solution  :(

There are only 65 cells occupied, but 81 cells available.

color the board like
$ 121212121$
$ 232323232$
$ 121212121$
$ 232323232$
$ 121212121$
$ 232323232$
$ 121212121$
$ 232323232$
$ 121212121$
the beatles that are in cell with number 1 will be on a cell with number 3 after two moves. so there are at most 32 beatles total in the number 1 and 3 cells.(1)
the beatles that are in cell with number 2 will be on a cell with number 1 or 3 after one moves. so there can be at most 32 beatles in the number 2 cells (from (1)) That makes a total 64 beatles can be on the board.


Well, I am in grade 9.
I need some help on how to start with programming in C++ .  Is it that after learning C only, u can learn C++?
Here in India, the informatics olympiad's 1st ropund is logical round. So within tht time, I can learn programming suppose I do well in the 1st round.

Thanks in advance.

You don't need to learn C. In fact, C++ and C are almost the same only C++ has some more features and standard libraries. When I started with C++ I learned everything from this site:

http://www.cplusplus.com/doc/tutorial/

But I already knew Pascal so I guess it was much easier for me than it is for a complete beginner. If you'll run into anything you need help with, feel free to ask here.

Also if you don't have a compiler / development environment yet, I recommend this one:

http://www.bloodshed.net/dev/devcpp.html

(You want the first download.) If you work in Linux, you should already have g++ and a text editor installed.


Does anyone know how to study for AP Environmental Science?

I have taken this class at school, but it I learned NOTHING. Everything we did was at a fourth grade level (e.g. making bird feeders and rain gauges-- I am not joking). The only prep book available is one by Barron's, which has VERY BAD ratings on http://www.amazon.com.

I also bought an environmental science textbook, but I'm not sure how useful it would be to read through the entire thing.

Any suggestions?

Would just reading the Ecology section of Campbell's Biology be sufficient (I have already taken AP Bio and Chem)?


Let the LHS be $ f(x)$, so that the equation is $ f(x)\equal{}a$.

$ f'(x) \equal{} 6x^2\plus{}6x\minus{}12$. The extrema occur at the solutions to $ f'(x)\equal{}0$, or $ x\equal{}\minus{}2, \; x\equal{}1$.

Now compute $ f(\minus{}2) \equal{} 23$ and $ f(1) \equal{} \minus{}4$.

Thus, going in the increasing $ x$ direction, we have that $ f$ increases until $ (\minus{}2,23)$, then decreasing until $ (1,\minus{}4)$, and then again increasing indefinitely.

Thus, the number of intersections of the graphs of $ y\equal{}f(x)$ and $ y\equal{}a$ is:

3, if $ \minus{}4 < a < 23$
2, if $ a\equal{}\minus{}4$ or if $ a\equal{}23$
1, if $ a < \minus{}4$ or if $ a > 23$


A 20KG block sits on a plane inclined at 37 degrees angle.  The coefficient of kinetic friction between the blck and plane is 0.3, what forc P, applied aprallel to the plane would make the blck move down the plane with a constant velocity?

Constant velocity means zero acceleration so in Newton's second law, we set everything equal to 0.  On the x-component, we have
$ mg\sin{\theta}\plus{}F\minus{}mg\cos{\theta}\mu_{k}\equal{}0$
Hence $ F\equal{}mg\mu_{k}\cos{\theta}\minus{}mg\sin{\theta}$.

hello
[hide="hint"]
for body to move with constant velocity the net force on the body should be zero.
$ P\equal{}mg(sin37\minus{}0.3cos37)$
 [/hide]
Thank u.

[quote="JRav"]Constant velocity means zero acceleration so in Newton's second law, we set everything equal to 0.  On the x-component, we have
$ mg\sin{\theta} \plus{} F \minus{} mg\cos{\theta}\mu_{k} \equal{} 0$
Hence $ F \equal{} mg\mu_{k}\cos{\theta} \minus{} mg\sin{\theta}$.[/quote]

Hello JRav
Sorry but this force would be in opp. direction .Kindly check it.
Thank u.

I never actually specified direction, which technically makes my answer incomplete.  I was just interested in the magnitude.  The OP should try to determine the direction.

tan(37)=0.75 >0.3
You need to push down in order to move it.


inequalities
real analysis
real analysis unsolved

[quote="hpe"]Assume throughout $u\ge 0$ and $v\ge 0$.

1. Show that $u \le v+\log(u+1)$ implies $u \le v+1+\log(v+1)$.

2. Find the smallest $\lambda > 0$ such that $u \le v+\log(u+1)$ implies $u \le v+\lambda+\log(v+1)$.[/quote]

1. Let note $v+1=x$ ande $u+1=y$. Then $x\ge 1, y\ge 1$ and we have $x-y+logy\ge 0\Rightarrow x\ge log\frac{e^{y}}{y}\ge \frac{y}{e}$, so $x\ge \frac{y}{e}\Rightarrow \frac{x}{y}\ge \frac{1}{e}\Rightarrow logx-logy+1\ge 0$ but $-logy\le x-y$ so we have $x-y+logx+1\ge 0$ or returned to notations $v-u+log(v+1)+1\ge 0$


geometry
geometric transformation
homothety
incenter
power of a point
radical axis
geometry proposed

In the figure below the two small circles are tangent between them at $K$ and each of them is tangent to the big circle at $A,B$.Prove that:

   [b]a) [/b]$NE$ is a diameter of the circle

  [b] b)[/b] If $XZ$ is a common tangent of the small circles and $M$ is the midpoint of $XZ$ , then

$AK$ is bisector of angle $BAM$.

[u] Babis [/u]

[quote="Huyền Vũ"]This is an easy problem
Let O1,O2 is the center of two small circles
(O1) and (O) are tangent at A so O1K//OE
similarly ON//O2K. But O1,K,O2 are collinear so N,O,E are collinear. Then [b]a)[/b] is solved
because of a) N,M,E are collinear
we have <NAK=<NMK=90 so A,K,M,N are cyclic
Thus <KAM=<KNE=<KAB. Then [b]b)[/b] is solved[/quote]

 Very nice and simple ! :)

 Of course in the competition in the problem was asked to prove that :

   '' $K$ is the incircle of triangle $AMB$

which makes the problem interesting.Figure and question a) were not given.I put these question to collect sollution with homothety or inversion. Thnks!

 [b] Babis[/b]

[quote="silouan"]Mr Babis the following proof is dedicated to you.  :)
Make a homothety with center B which goes the small circle (which is tangent to the <<big>> at B) to the big one .The XZ becames a tangent line at N and it is parallel to XZ so N is the midpoint of the <<left>> arc $XZ$
Similarly  E is the midpoint of the <<right>> arc $XZ$ .So NE connects two midpoints of these two arcs so passes through $O$ .
Then b) is easy . :wink:
Hope you liked it[/quote]

 [b]Thank you Silouan ! [/b]

 I like transformations , even if they are used to prove simple problems !  I just wander why in Crux 3/2007 they have given a so difficult solution using inversion for part a) !

 Note
 According to your hint , I see that the first homothecy transfers the right small radius to the left big radius , since the center of the small circle is gone to the center of the big one. So , these radii must be parallel. The same hapens with the other  dadii of the small and big circles. But the small raddii are on the same line ect.

    [u] Babis[/u]

Since $XZ$ is the radical axis of the 2 inner circles, the tangents at $A,B$ respectively are concurent onto $XZ$, consequently $AZBX$ is a harmonic quad and $AK$ is the internal bisector of $\angle XAZ, \angle BAM$, similarly $BK$ and, indeed, $K$ is the incenter of $\triangle ABM$.

Best regards,
sunken rock


inequalities
Cauchy Inequality
inequalities open

[quote="Michael Niland"]X and Y are real positive numbers such that

SQRT[ 2(X^2) + 2XY +3(Y^2) ] +X + Y = 4

Prove( X^2)(Y) <=4

Asolution appears in More Mathematical Morsels by R Honsberger Page 119[/quote]
I am not sure about the problem. Is it
$ \sqrt{2x^2+2xy+3y^2} +x+y = 4 $
Prove $ x^2y \leq\ 4 $

We have $ 4= \sqrt{2x^2+2xy+3y^2} +x+y \geq\ 2(x+y) $
So we have $ x+y \leq\ 4 $
Use Cauchy inequality $ x^2y \leq\ \frac{4}{27}(x+y)^3 \leq\ \frac{32}{27} $

I think I can find min of $x^2$y.
let  :sqrt: (2x^2+2xy+3y^2)<=mx+ny,x^2*y=k^2*(x/k)(x/k)*y.


How many positive integers less than 100 and divisible by 3 are also divisible by 4?

If x/3 = integer w, and x/4 = integer y, then x/(LCM[3,4]) (or x/12) = integer z.
Therefore, the way to find the solution is to find all the multiples of 12 less than 100: 12, 24, 36, 48, 60, 72, 84, and 96. There are , so the answer is [b]8[u].[/u][/b]

Another way to do this problem quicker is by doing $ 3\cdot4\equal{} 12$ Now you can do $ 100/12 \equal{} 8.33$ repeated. Then, you round down to $ \boxed{8}$ and that is your answer, which is the same as phr33k's answer.

1st of all we shall find the smallest # that can be both divisible by 3 and 4 and that # is 12 so all we need to do is 100/12 = 8.333333333333333333    so the answer is   [hide="is this my answer???????!!!!!!!!"  "]8[/hide]

[size=200][color=#00BF00][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH[/u][/i][/b][/color][/size]

[color=#0080FF][b][i][u]IF THERE ARE ANY PROBLEMS IN MY PROBLEM PLEASE PLEASE POST ME!!![/u][/i][/b][/color]

:showoff: :showoff:
 :showoff:
:showoff:

IM STRONG IN MATH!!!!!


$ {{a_n}}$;($ {n \ge 0}$) is a given sequence satisfying $ a_{i\plus{}1}\plus{}a_{i\minus{}1}\equal{}|a_i|$ for all $ i\equal{}1,2,...$

Prove that $ {a_n}$ is a circulating sequence

Sorry for my bad English  :blush:

Please, can someone help me with this problem  :maybe:

Thank you but I don't know the language in the link, could you translate into English :D


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ p$ a prime and $ G$ a group of order $ p^m$.
Show that $ G$ has a normal abelian subgroup $ H$ such that for all $ (g,h)\in G\times H$, $ gh\equal{}hg$.

Take H to be [b]Z(G)[/b], the center of your group. This subgroup is always abelian and normal. In fact, it's even characteristic... The last condition that you require is precisely the definition of [b]Z(G)[/b].

So it is true for every group $ G$ then (not only those of order $ p^m$)?

[quote="Portland"]Let $ p$ a prime and $ G$ a group of order $ p^m$.
Show that $ G$ has a normal abelian subgroup $ H$ such that for all $ (g,h)\in G\times H$, $ gh \equal{} hg$.[/quote]

Take H=1 or H=Z(G).

Did you mean to ask a different question?

In a finite p-group, every subgroup maximal amongst the abelian normal subgroups is self-centralizing (the opposite of what you asked).  This is lemma 5.3.12 on page 185 of Gorenstein's Finite Groups.  Basically, C_G(H)/H is normal in G/H so intersects the center non-trivially, so there is some non-identity xH in C_G(H)/H ∩ Z(G/H), but then <x,H> is an extension of the cyclic group <xH> by a central subgroup H, and so <x,H> is abelian, contradicting the maximality of H.

You are a mind reader JackSchmidt. This is indeed the question I wanted to ask, thanks a lot.


Hi guys.

Where can I find a collection of Chinese National Olympiads or Chinese Team Selection Tests? Is there some website that has them posted (in ENGLISH PLZ) or some book published in english that is a collection of chinese problems like the USSR Olympiad problem book?

Ahem most of the problems are already on this site, and there is a topic where a link to the Chinese National MO problems are presented, just use the [url=http://www.mathlinks.ro/Forum/search.php]search[/url] function of the Forum. ;)


how do i get the equation of the line that is parallel to the line 2x-3y=7and contains the point (-6,-5)...
just need to review it because i forgot... :blush:
pls&&thx

First, make your equation in y-int form. $y=\frac{2}{3}x-\frac{7}{3}.$ All parallel lines will have the same slope. Plug in $x,\ y,$ and the slope, and find $b.$ Here's the equation of the parallel line: $y=\frac{2}{3}x+1.$