There are exactly three ordered pairs which are zeroes of the function $ f(x,y) \equal{} 2x^3 \plus{} 12xy\plus{}2y^3 \minus{} 16$ and form the vertices of an equilateral triangle. Find the area of the triangle.

huh i think the problem is wrong...

by quick guess and check, we can confirm that (1,1), (2,0), (0,2) are three roots.

but (1,1), (2,0), (0,2) don't form an equilateral triangle.

EDIT:

oh...

oops sry  :blush:

[quote="FantasyLover"]huh i think the problem is wrong...

by quick guess and check, we can confirm that (1,1), (2,0), (0,2) are three roots.

but (1,1), (2,0), (0,2) don't form an equilateral triangle.[/quote]

You read the problem wrong. There are actually an infinite number of roots, like $ (3,\minus{}1)$ for instance.
The problem asks to find three roots which [i]do[/i] form an equilateral triangle, and to calculate that area.

[hide="Hint 1"] Look at the three points FantasyLover has found and test the obvious conjecture. [/hide]
[hide="Hint 2 (giveaway)"] $ 2(a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc) \equal{} (a \plus{} b \plus{} c)((a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2)$. [/hide]
I don't like problems like this.  It's not at all obvious how to come up with a constructive path towards the solution if you don't know the trick in advance.
[hide="One possibility"] The partial derivatives are given by $ d_x \equal{} 6x^2 \plus{} 12y$ and $ d_y \equal{} 6y^2 \plus{} 12x$.  They vanish at the points $ (0, 0)$ and $ (\minus{}2, \minus{}2)$.  The first point is not on the graph, but the second is.  That tells you that $ (\minus{}2, \minus{}2)$ is a [url=http://mathworld.wolfram.com/SingularPoint.html]singular point[/url], which tells you that $ f(x, y) \equal{} 0$ has genus zero and you should be looking for factors. [/hide]

[hide="solution"]Factoring out a 2 gives us $ 2(x^3\plus{}6xy\plus{}2y^3\minus{}8)$, which is $ 2((x)^3\plus{}(y)^3\plus{}(\minus{}2)^3\minus{}3(\minus{}2)(x)(y))$, which can be factored into $ (x\plus{}y\minus{}2)((x\minus{}y)^2\plus{}(y\plus{}2)^2\plus{}(x\plus{}2)^2)$. Gotta love that factorization. Since the squares are non-negative, they must all be 0 if their sum is 0. Thus (-2,-2) is the only solution related to that factor. The other factor says that all x and y such that x+y=2 are solutions. There is exactly one pair of points on that line such that those two and (-2,-2) is an equilateral triangle. The height of the triangle is the distance from (-2,-2) to the line x+y=2, which is $ 3\sqrt{2}$. Therefore the base of the triangle is $ \frac{3\sqrt{2}}{\sqrt{3}}*2\equal{}2\sqrt{6}$. The area of the triangle is therefore $ \boxed{6\sqrt{3}}$.[/hide]

Forgive any un-rigor or non-clarity.

Dagnabit I had a version of hint 2 but it was the wrong version.


I need the method and solution for the following power series questions

Q1

sigman = at/2 [sinh at + sin at divided by cosh at - cos at]

If a = 1.08 and t = 1.3

calculate sigma correct to 5 significant figures using power series

Q2

Evaluate sinh (1.6), correct to 5 decimal places using power series methods.

I urgently need assistance with this

Dear Johnboy 14.
I don'n understand what you write. Please write it by Latex and you try to write down clearly :(


$\mathrm{Prove \;\; that \;\; the \;\; set \;\; of \;\; all} \;\; n\in\mathbb{N} \mathrm{\;\; for \;\; which \;\; the \;\; equation \;\;}$ \[ \frac{1}{x}+\frac{1}{y}=\frac{3}{n}  \]$\mathrm{ \;\;doesn't \;\;have \;\; a \;\; solution \;\; (where \;\;} x,y\in\mathbb{N})$ $\mathrm{\;\;can't \;\;be \;\;expressed \;\;as\;\; a \;\;finite \;\;sum\;\; of \;\;arithmetic \;\;progressions.}$

We could set $x=y=2n$ for all $n$, so what condition is missing¿ Different¿ Coprime¿ Something else¿

Sorry, its $\frac{3}{n}$. I fixed it.  :oops:

If $n\equiv 2\pmod {3}$ then we can choose $x=\frac{n+1}{3}$ and $y=\frac{n(n+1)}{3}$. If $n\equiv 0\pmod{3}$ we can have $x=y=\frac{2n}{3}$


ARML
AMC
AIME
USAMTS
AMC 10
AMC 12
USA(J)MO

Its obvious that AMC 12<AIME<USAMO<IMO
which other contests do you have in mind?  ARML can be sticked in between AMC 12 and AIME

[quote="mathfanatic"]On a scale of 1-10

Ole Miss POW: 0
Cody Bowl: 5
Mandelbrot Regional: 2
Mandelbrot National: ? (never done it)
USAMTS: 4
AMC10: 3
AMC12: 4
AIME: 6
USAMO: 8
IMO: 11[/quote]

I'd say that some USAMTS is harder than AMC12...

[quote="JBL"]Heh. The Mandelbrot varies year to year, but I'd say it's about the same level as the AMC 12.[/quote]

i dunno about Mandelbrot Nats.. but the regionals i think are easier than AMC 12's... *shrugs*

[quote="JBL"]USAMTS really isn't harder than AMC 12 -- it's just that proof-writing techniques are not necessarily taught to many people.  But given the time you have, they're quite easy.[/quote]

Ahh...yes, considering it's basically untimed, I guess that would make them a lot easier than AMC12...plus, u have access to unlimited resources...(almost)


If (X,T) is a local compact topological space and Y $ \subset$ X then (Y, T $ \cap$ X) is local compact ?

There is something wrong or I don't get the question.

1 possibility: you are asking if local compactness of X implies local compactness of Y $ \subset$ X  with induced topology. Then answer is no- try X- real numbers and Y- rational numbers (not locally compact)

2 possibility: you mean (Y, T $ \cap$ Y), but this is not necessarily a topology (example like above shows this)

there is also a possibility that this what I wrote is stupid :)

Uhh.... Im assuming $ T$ is the set of open subsets of $ X$ in which case $ T\cap X$ doesn't make any sense... and in reference to above poster, if the notation is as such: $ X$ is a topological space $ T$ is its set of open sets, $ Y$ is a subset of $ X$, then the space $ Y$ with open sets given by $ U\cap Y$ with $ U\in T$ is a topology... in fact its the definition of the induced topology...

Yes, I wanted to reffer to the induced topology. @Imp, why isn't Q with the induced topology local compact?


could you find all natural numbers which cannot be represented as difference of two squares of natural numbers

even numbers such that 4 does  not diviode them
(factorisation)


suppose that $ a\geq b\geq c$,by arrangement inequality we have
$ (\frac{a}{b\plus{}c})^2\plus{}(\frac{b}{c\plus{}a})^2\plus{}(\frac{c}{a\plus{}b})^2$
$ \geq \frac{1}{3}$ $ (a^2\plus{}b^2\plus{}c^2)$ $ [(\frac{1}{b\plus{}c})^2\plus{}(\frac{1}{c\plus{}a})^2\plus{}(\frac{1}{a\plus{}b})^2]$
then it's only to prove that
$ (ab\plus{}bc\plus{}ca)$ $ [(\frac{1}{b\plus{}c})^2\plus{}(\frac{1}{c\plus{}a})^2\plus{}(\frac{1}{a\plus{}b})^2] \geq$ $ \frac{9}{4}$

which is obvious now.... :roll:

[quote="shaaam"]Given that $ a,b,c \geq 0$, Prove that;
\[ \left(\frac {a}{b \plus{} c}\right)^2 \plus{} \left(\frac {b}{c \plus{} a}\right)^2 \plus{} \left(\frac {c}{a \plus{} b}\right)^2\geq \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)}
\]
:)[/quote]
It equivalent to
$ 4abc\sum_{cyc}\frac a{(b\plus{}c)^2}\plus{}4\sum\frac{a^3}{b\plus{}c}\ge3(a^2\plus{}b^2\plus{}c^2)$
By AM-GM
$ \sum\frac a{(b\plus{}c)^2}\ge\frac9{4(a\plus{}b\plus{}c)}$
We only need to prove
$ \frac{9abc}{a\plus{}b\plus{}c}\plus{}4\sum\frac{a^3}{b\plus{}c}\ge3(a^2\plus{}b^2\plus{}c^2)$
$ \Leftrightarrow 9abc\plus{}4\sum\frac{a^4}{b\plus{}c}\plus{}a^3\plus{}b^3\plus{}c^3\ge3\sum bc(b\plus{}c)$
By AM-GM
$ \sum\frac{a^4}{b\plus{}c}\ge\frac{a^3\plus{}b^3\plus{}c^3}2$
It suffices to show that
$ a^3\plus{}b^3\plus{}c^3\plus{}3abc\ge ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)$ , It's Schur!

The following is stronger

Let $ a,b,c$ be nonnegative real number. Prove that
\[ \left(\frac{a}{b\plus{}c}\right)^2\plus{}\left(\frac{b}{c\plus{}a}\right)^2\plus{}\left(\frac{c}{a\plus{}b}\right)^2\plus{}\frac1{2}\ge\frac{5(a^2\plus{}b^2\plus{}c^2)}{4(ab\plus{}bc\plus{}ca)}
\]

[quote="dduclam"]The following is stronger

Let $ a,b,c$ be nonnegative real number. Prove that
\[ \left(\frac {a}{b \plus{} c}\right)^2 \plus{} \left(\frac {b}{c \plus{} a}\right)^2 \plus{} \left(\frac {c}{a \plus{} b}\right)^2 \plus{} \frac1{2}\ge\frac {5(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)}
\]
[/quote]

We have:
$ \sum \frac{a^2}{(b\plus{}c)^2} \ge 2\minus{}\frac{10abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
We have prove that:$ \plus{})\frac{5}{2}\minus{}\frac{10abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge \frac{5}{4}\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ac}$
$ <\equal{}>\sum (b\minus{}c)^2(\frac{a}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\minus{}\frac{1}{2(ab\plus{}bc\plus{}ac)}) \ge 0$
I think:It's true by SOS  :)

[quote="dduclam"]The following is stronger

Let $ a,b,c$ be nonnegative real number. Prove that
\[ \left(\frac {a}{b \plus{} c}\right)^2 \plus{} \left(\frac {b}{c \plus{} a}\right)^2 \plus{} \left(\frac {c}{a \plus{} b}\right)^2 \plus{} \frac1{2}\ge\frac {5(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)}
\]
[/quote]
Nice Dduclam, but the following are stronger.

Let $ a,b,c$ be nonnegative real number, no two of which are zero. Prove that

$ (a)\ \ \ \ \  \frac {a(2a\minus{}b\minus{}c)}{(b\plus{}c)^2}\plus{}\frac {b(2b\minus{}c\minus{}a)}{(c\plus{}a)^2}\plus{}\frac {c(2c\minus{}a\minus{}b)}{(a\plus{}b)^2}\plus{}2\ge \frac {2(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca}$;

$ (b)\ \ \ \ \  \frac {a(a\minus{}b\minus{}c)}{(b\plus{}c)^2}\plus{}\frac {b(b\minus{}c\minus{}a)}{(c\plus{}a)^2}\plus{}\frac {c(c\minus{}a\minus{}b)}{(a\plus{}b)^2}\plus{}\frac 3{2} \ge \frac {3(a^2\plus{}b^2\plus{}c^2)}{4(ab\plus{}bc\plus{}ca)}$.

[quote="caubetoanhoc94"][quote="dduclam"]The following is stronger

Let $ a,b,c$ be nonnegative real number. Prove that
\[ \left(\frac {a}{b \plus{} c}\right)^2 \plus{} \left(\frac {b}{c \plus{} a}\right)^2 \plus{} \left(\frac {c}{a \plus{} b}\right)^2 \plus{} \frac1{2}\ge\frac {5(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)}
\]
[/quote]

We have:
$ \sum \frac {a^2}{(b \plus{} c)^2} \ge 2 \minus{} \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
We have prove that:$ \plus{} )\frac {5}{2} \minus{} \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge \frac {5}{4}\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ac}$
$ < \equal{} > \sum (b \minus{} c)^2(\frac {a}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \minus{} \frac {1}{2(ab \plus{} bc \plus{} ac)}) \ge 0$
I think:It's true by SOS  :)[/quote]

what is SOS ????

[quote="Mr.Bikrone"][quote="caubetoanhoc94"][quote="dduclam"]The following is stronger

Let $ a,b,c$ be nonnegative real number. Prove that
\[ \left(\frac {a}{b \plus{} c}\right)^2 \plus{} \left(\frac {b}{c \plus{} a}\right)^2 \plus{} \left(\frac {c}{a \plus{} b}\right)^2 \plus{} \frac1{2}\ge\frac {5(a^2 \plus{} b^2 \plus{} c^2)}{4(ab \plus{} bc \plus{} ca)}
\]
[/quote]

We have:
$ \sum \frac {a^2}{(b \plus{} c)^2} \ge 2 \minus{} \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
We have prove that:$ \plus{} )\frac {5}{2} \minus{} \frac {10abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge \frac {5}{4}\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ac}$
$ < \equal{} > \sum (b \minus{} c)^2(\frac {a}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \minus{} \frac {1}{2(ab \plus{} bc \plus{} ac)}) \ge 0$
I think:It's true by SOS  :)[/quote]
Prove it ! :)[/quote]

I'm sory..I'm wrong! :(
I will try :(


Sa se arate ca daca alegem trei puncte A,B,C pe un cerc de raza O si centru R atunci suprafata maxima a triunghiului se realizeaza $\Longleftrightarrow$ triunghiul este echilateral.

$S=2R^{2}\cdot \sin{A}\sin{B}\sin{C}\leq 2R^{2}\cdot \frac{(\sin{A}+\sin{B}+\sin{C})^{3}}{27}\leq \frac{3\sqrt{3}}{8}$

cu egalitate $\Longleftrightarrow$ e echilateral.

Dap...Corect...La acelasi lucru l-am facut si eu.


Prove $\sin 9^{\circ}+\cos 9^{\circ}= \frac{\sqrt{3+\sqrt{5}}}{2}$

Consider a regular pentagon $ABCDE,\ \triangle{ACD}$ is isosceles triangle of $36^\circ-72^\circ-72^\circ.$ Let $M$ be mid point of $CD,$

Remark $\triangle{ACM},$ since $AC: CD=1: \frac{\sqrt{5}-1}{2}$ by golden ratio, as $\angle{CAH}=18^\circ,$ we have $\sin 18^\circ =\frac{\sqrt{5}-1}{4}.$

Now $(\sin 9^\circ+\cos 9^\circ)^{2}=1+2\sin 9^\circ \cos 9^\circ=1+\sin 18^\circ=\frac{3+\sqrt{5}}{4}.$

$\sin 9^\circ+\cos 9^\circ>0,$ yielding $\sin 9^\circ+\cos 9^\circ=\frac{\sqrt{3+\sqrt{5}}}{2}.$

Alternative solution:

$\sin \theta+\cos \theta =\sqrt{2}\left(\cos \theta \cdot \frac{1}{\sqrt{2}}+\sin \theta \cdot \frac{1}{\sqrt{2}}\right)$

$=\sqrt{2}\cos(\theta-45^\circ)$

Thus $\sin 9^\circ+\cos 9^\circ=\sqrt{2}\cos 36^\circ=\sqrt{2}\cdot \frac{1+\sqrt{5}}{4}=\frac{\sqrt{2}+\sqrt{10}}{4}.$

Remark: $\sqrt{3+\sqrt{5}}=\sqrt{\frac{6+2\sqrt{5}}{2}}=\frac{\sqrt{5}+1}{\sqrt{2}}=\frac{\sqrt{10}+\sqrt{2}}{2}.$

we have $sin(72) = 2 sin(36) cos (36)$ by the double angle relationship. And again,
$sin(72) = 4 sin(18) cos (18) (1-2sin^{2}(18)$
And because $sin(72) = cos(18)$, we substitute to get
$cos(18) = 4 sin(18) cos (18) (1-2sin^{2}(18))$
This simplifies to
$1 = 4 sin (18) (1-2sin^{2}(18))$
Let x = sin(18). Then we have
$1 = 4x(1-2x^{2}) \Rightarrow (2x-1)(4x^{2}+2x-1)=0 \Rightarrow sin (18) = \frac{\sqrt{5}-1}{4}$

Now, expand out by double angle again to get
$2sin (9)cos(9) = \frac{\sqrt{5}-1}{4}$
Adding 1 to each side,
$sin^{2}(9)+2sin (9)cos(9)+cos^{2}(9) = \frac{\sqrt{5}+3}{4}$

Noticing that   $sin^{2}(9)+2sin (9)cos(9)+cos^{2}(9) = (sin(9)+cos(9))^{2}$ and square rooting each side gives the result.


function
real analysis
real analysis solved

Find out if there exists an injective continuous function f defined on an interval [a,b] , f not constant with the property:f is not injective on every
sub-interval I included in [a,b]

?????
How can a function be injective on [a,b] but not on every subset of [a,b]??

Pierre.

he must probably mean: find out if there is a non-injective function continous of [a,b] such that it is injective on every sub-interval of [a,b], different from [a,b].

Sorry, i wanted to write :Find out a continuous and non-constant function
f defined on an interval [a,b] wich is not injective on any sub-interval I included in [a,b]

There exist functions which are differentiable on |R and monotone on none interval (thus these functions are extremely oscillating).
To construct such a function is too long and difficult to explain it here. It seems that the first construction came from Arnaud Denjoy, and uses non trivial knowledge in mesure theory.

Some others may be found in the references below.

Note that it answer to your question (and even more) since a continuous function is injective on an intervalle if and only if it is monotone on it.

References :
- A.Denjoy : "Mmoire sur la drivation et son calcul inverse"
Gauthier-Villard, 1954, p.210-235

- Y.Katznelson, K.Stromberg "Everywhere differentiable, nowhere monotone functions" American mathematical monthly, avril 1974, p.349-354.

- K.Stromberg "An introduction to classical real analysis" p.216-218

Pierre.

A full solution can be found at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=244776[/url]

Much simpler examples can be given for xxxxtt's question :):

[url]http://planetmath.org/encyclopedia/CantorFunction.html[/url]

The function is constant on the complement of Cantor's set. Since Cantor's set has null measure, its complement has full measure, and, in particular, it's dense.


algebra
polynomial
function
trigonometry
Rational Root Theorem

This is a polynomial that appears on an AMC:
$n^3-8n^2+20n-13$
Find values n for which it is prime.

They said you need to factor it out into (n-1) and something else. My question is: how do you know (n-1) is part of the above polynomial when factored out? Is it just guess and check, or is there a technique for finding factors of the cubic polynomial? *Not the cubic formula, please*

Hi  -   Well, one should try some simple substitutions like n=0, +-1, +-2  to see how the function behaves, and if you are lucky, for one of these values you will get  the value of the polynomial as zero. Here if you try n=1, we get:

1 - 8 + 20  - 13  = 0

so (n-1) is a factor .

Even more helpful  is, if you are lucky enough to find the coficient of the first term 1 and  last term (13 here) a prime, (of  if it has only a few factors), you only have to tryout the factors of the last term (that is 1 and 13 only)

So you need to try only  (+1, -1, +13,-13)

In general, if a cubic appears as     a_3*x^3 + something*x^2+  otherthing*x^2 + a_0

You try  out  fractions like      (+-) m/n where m is a factor of a_0 and n is a factor of a_3

if if  a_3 and a_0   have a few factors only ( best is if one of them - or both of them  - is/are 1)
This is not only true for Cubics but all other polynomial equations..

This , of course , assumes that you are lucky and/or the person who gave you the equation has a kind heart,.. to select an equation which does have rational roots.  (The theorem is knows as rational root theorem).....

So next time you see that "clever_but_nice" math teache gives   some thing like   x^4 + .......other terms + 210 = 0
you may be lucky enough to be right when you say the roots are (2,3,5,7)
 :)

You use the rational root theorem (I think that's what it's called....) This gives you all the POSSIBLE zeroes of any polynomial. Notice how i wrote POSSIBLE and not the ACTUAL zeroes. The theorem is:

+/- p/q

In this case, p=+/- 13 and q=+/-1

So the POSSIBLE zeroes are +/- 13, +/- 1.

Now you could use synthetic division to completely factor the polynomial. There are probably other ways but this is the method I know.  :lol:

You can always take advantage of a graphing calculator on AMC as well.

[quote="tetrahedr0n"]It gives you all possible rational zeros, thus the name rational root theorem. If a polynomial does not have rational roots, you have to resort to the more complex methods discussed in the other thread.[/quote]

What are the "more complex methods"?

Anyway, I'm pretty sure that on AMC/AIME they wouldn't make you factor a polynomial(with degree>2) over irrational/complex numbers.

[quote="Beta"]What are the "more complex methods"? [/quote]
Solving cubics using trig, or perhaps the cubic formula, altough I doubt the latter will ever take place. Sometimes you can find special things about a polynomial, such as it being symmetric, having roots of unity as zeros, etc... that will allow you to find its zeros.

[quote="beta"]Anyway, I'm pretty sure that on AMC/AIME they wouldn't make you factor a polynomial(with degree>2) with that can't be factored over rational numbers.[/quote]
No, not if you need to find the roots. However, they might make you do a polynomial problem in which you don't need to find the zeros. Then people can waste time by trying to find the roots when there are better methods.


You probably know the formula for triangular numbers (n(n+1)/2) and square numbers (n :^2: )

Challenge: Find a general formula for the nth pentagonal number.

Hint: Try to draw them first [hide]--1, 5, 12, 22...[/hide]

This is kind of sketchy and not at all elegant, but...

[hide]n(3n-1)/2

Look at the first few: 1, 5, 12, 22, 35, 51.. We see that the changes between consecutive numbers are 4, 7, 10, 13, 16..., which clearly all differ by 3.  Not only does this tell us that our answer must be of degree 2, but we can use the arithmetic sequence 3n+1 of the differences; the nth term can be expressed as 1+0*3+1+1*3+1+2*3+1+3*3+...+1+(n-1)3.  In this sum, we have n 1s and 1+2+3+..+(n-1) 3s, or n+3((n-1)n)/2.  Multiplying this out yields (3n:^2:-3n)/2+2n/2, or  (3n:^2:-n)/2.  Factoring out n yields n(3n-1)/2.[/hide]


笛卡尔的传记，作者是Amir D.Aczel

Descartes's secret notebook

何不把它放到网上来呢? :D

难道要我一页一页扫描上去吗？
再说这系图书馆的书也。。。

How secret is it??
Can a secret still be a secret if it can be found in a library?
真叫人費解


Does anyone know if Arbelos are any good? i.e. easy to read, understand, and instructive

I wouldn't say they're easy to read -- they're rather difficult. They're enjoyable, but I don't know how useful they are.


[quote="Peter"]Find all functions $f: \mathbb{Q}^{+}\to \mathbb{Q}^{+}$ such that for all $x,y \in \mathbb{Q}$: \[f \left( x+\frac{y}{x}\right) =f(x)+\frac{f(y)}{f(x)}+2y, \; x,y \in \mathbb{Q}^{+}.\][/quote]

Setting $x=y$ in the equation we get that
\[f(x+1)=f(x)+2x+1\]

therefore $f(x+m)=f(x+m-1)+2(x+m)-1$. Adding up those equations for

all $m=1,2,...,n$ we get that $f(x+n)=f(x)+2xn+n^{2}$. Setting in the

latter $x=1$ we get that $f(n+1)=(n+1)^{2}+(f(1)-1)$.

Returning to the initial equation and setting $x=1$ we get

$f(y+1)=f(1)+\frac{f(y)}{f(1)}+2y$ but we also know that

$f(y+1)=f(y)+2y+1$ $\Rightarrow$ $f(1)=1$ $\Rightarrow$ $f(n)=n^{2}$ for all $n\in{N}$.

Again let's go back to the initial equation and set there $x=n\in{N}$ and

$y=m\in{N}$. We get that $f(n+\frac{m}{n})=f(n)+\frac{f(m)}{f(n)}+2m$. Meanwhile $f(n+\frac{m}{n})=f(\frac{m}{n})+2n(\frac{m}{n})+n^{2}$ $\Rightarrow$
\[f(\frac{m}{n})=\frac{f(m)}{f(n)}=\frac{m^{2}}{n^{2}}\]

$\Rightarrow$
\[f(x)=x^{2}\]
for all $x\in{Q^{+}}$. :wink:


calculus
derivative
calculus computations

Suppose the real valued $ g$ is defined on $ \mathbb{R}$ and $ g'(x) < 0$ for every real $ x$. Prove there's no differentiable $ f: R \rightarrow R$ such that $ f \circ f \equal{} g$.

If f(f(x)) = g then their derivatives must be equal at all points so

f'( f(x) ) * f'(x) = g'(x)

Because g'(x) is negative one value must be negative, another must be positive. So d/dx (f'(f(x)) is negative at some point, positive at another. If f is differentiable f(f(x)) must also be differentiable and the curve must be continuous, so at some point d/dx f(f(x)) = 0, so at some point g'(x) must also equal 0. Contradiction.

Four lines:
If $ f\circ f$ is injective, $ f$ is injective.
If $ f$ is injective and continuous on $ \mathbb{R}$, it is either strictly increasing or strictly decreasing.
If $ f$ is either strictly increasing or strictly decreasing, $ f\circ f$ is strictly increasing.
Combining these, we have shown that $ f\circ f$ is never strictly decreasing if $ f$ is continuous on $ \mathbb{R}$.

It is that easy.


The guy who has given the soln for this qn has gone beating the bush beyond one point of time.. once he has proved a(a+b) and b(a+b) are rational, then it is obvious that a/b is rational for all a and b in M.. Which implies the irrationality factor in all is the same.. So we can set root(n) to be the same irrationality factor.. thats it..

[quote="Sathej"]Doesn't speak too highly of our problem setters! But then, many countries (smaller ones) take up problems more freely from USAMO, Russian Competitions and the like. Well, it might have been an advantage to those who have seen it, but otherwise a nice problem.
Sathej[/quote]

Sadly, ppl who have already seen this problem have a much higher prob of getting through.....

Well, it happens in any exam. After all, no man-made system is immune to being defeated by man.
Sathej

it's ok if it is defeated by a man ....but what if 'men' where in only 30 'men'(sorry not being chauvinistic just rhyming so...)get through and from 12th only 6 :)


modular arithmetic
number theory unsolved
number theory

Prove that for every prime number $ p$, there are infinitely many positive integers $ n$ such that $ p$ divides $ 2^n\minus{}n$.

[quote="moldovan"]Prove that for every prime number $ p$, there are infinitely many positive integers $ n$ such that $ p$ divides $ 2^n \minus{} n$.[/quote]

For $ p\equal{}2$ every even $ n$ satisfies the condition.

For any prime $ p>2$, $ 2^{(p\minus{}1)^{2k}} \equiv 1 \equiv (p\minus{}1)^{2k} \pmod{p}$ for all positive integers $ k$, so $ n\equal{}(p\minus{}1)^{2k}$ satisfies the condition, done.


function
real analysis
real analysis unsolved

In the book of D. Gilbarg & N. Trudinger, "Elliptic PDE of Second Order", there is the following piece of reasoning. For $ \sigma\in(0,1)$, he suggests to take a cutoff function $ \eta \in C_c^2(B_R)$ ($ B_R$ is the ball of redius $ R$) such that $ 0 \le \eta \le 1$, $ \eta \equal{} 1$ on $ B_{\sigma R}$, $ \eta \equal{} 0$ on $ B_R \setminus B_{\sigma' R}$ with $ \sigma' \equal{} \frac {1 \plus{} \sigma}2$,
\[ \text{(*)} \qquad |D\eta| \le \frac 4{(1 \minus{} \sigma)R}, \quad |D^2\eta| \le \frac {16}{(1 \minus{} \sigma)^2R^2}.
\]
I don't see how to find such a function (I am mainly interested in (*) because the rest is easy). Can you help me?  :)

[b]P.S.[/b] Do you know any good book in PDE containing $ L^p$ regularity theory (don't name the book by Agmon, Douglis and Nirenberg because I have one)?

The cutoff function is going to be radial, so this is really a one-dimensional problem.

Someone who does not care about $ 4$ and $ 16$ can take any $ C^2(\mathbb R)$ function that is $ \equiv 1$ for $ x\le 0$ and is $ \equiv 0$ for $ x\ge 1$, and shift/rescale it appropriately.

If you do care about $ 4$ and $ 16$, then you are in for some tedious interpolation with polynomials.

[quote="volodja"][b]P.S.[/b] Do you know any good book in PDE containing $ L^p$ regularity theory (don't name the book by Agmon, Douglis and Nirenberg because I have one)?[/quote]

Try to use a book in French by Haim Brezis


[quote="José"]No tengo mucho tiempo, asique lo posteo así como lo tengo, sin pasarlo a Latex:

Los triángulos A1B1C1 y A2B2C2 tienen lados paralelos: A1B1||A2B2, B1C1||B2C2, C1A1||C2A2 y el triángulo A2B2C2 está contenido en el interior del triángulo A1B1C1. Un tercer triángulo ABC está inscrito en A1B1C1 y circunscrito a A2B2C2, es decir, los puntos A, B, C pertenecen al interior de los segmentos B1C1, C1A1, A1B1 y los puntos A2, B2, C2 pertenecen al interior de los segmentos BC, CA, AB, respectivamente.
Si el área de A1B1C1 es 63 y el área de A2B2C2 es 7, hallar el área de ABC.[/quote]
21

Bien, la respuesta es correcta... sin embargo, podrías postear tu solución?


calculus
integration
function
calculus computations

Calculate these:
1/$ \int \frac{sinx}{1\plus{}sin2x} dx$
2/ $ \int \frac{x^{2009}}{(1\plus{}x^2)^{1006}} dx$

hello
[hide="1"]
$ \int\frac {sinx dx}{1 \plus{} sin2x}$
$ \equal{} \frac {1}{2}\int \frac {2sinx}{cos^2x \plus{} sin^2x \plus{} 2sinxcosx}dx$
$ \equal{} \frac {1}{2}\int\frac {cosx \plus{} sinx}{(cosx \plus{} sinx)^2}dx \minus{} \frac {1}{2}\int\frac {cosx \minus{} sinx}{(cosx \plus{} sinx)^2}dx$
$ \equal{} \frac {1}{2}\int\frac {1}{cosx \plus{} sinx}dx \minus{} \frac {1}{2}\int\frac {dt}{t^2}$
for first integration u can use
 1. half angle formula
or
 2.$ \int\frac {1}{cosx \plus{} sinx}dx \equal{} \frac {1}{\sqrt2}\int sec(x \minus{} \frac {\pi}{4})dx \equal{} \frac {1}{\sqrt2}log(tan(\frac {x}{2} \plus{} \frac {\pi}{8}) \plus{} C$
so
$ \int\frac {sinx dx}{1 \plus{} sin2x} \equal{} \frac {1}{2\sqrt2}log(tan(\frac {x}{2}) \plus{} \frac {\pi}{8}) \plus{} \frac {1}{2(cosx \plus{} sinx)} \plus{} C$
 [/hide]
thank u

[quote="dulldrake"]Calculate these:

2/ $ \int \frac {x^{2009}}{(1 \plus{} x^2)^{1006}} dx$[/quote]

Have you done this yourself by hand??? :huh:  :huh: because mathematica gave a 3 page long answer(though with elementary functions :P )

Mathematica is not clever. There is an approach that makes the answer expressible in one line.

hello
I m writing this post in hurry so kindly check it properly.Do let me know if i have done some mistake.
[hide="2"]
   $ \int \frac{x^{2009}}{(1\plus{}x^2)^{1006}} dx$
   put $ (1\plus{}x^2)\equal{}t$
 $ \int \frac{(t\minus{}1)^{1004}}{2t^{1006}}dt$
=$ \int (1\minus{}\frac{1}{t})^{1004}\frac{dt}{2t^2}$
put $ (1\minus{}\frac{1}{t})\equal{}y$
so that now ur integral is
$ \int \frac{ y^{1004}}{2}dy$
i think now u can proceed.[/hide]
thank u

I prefer just going straight to $ t\equal{}1\plus{}\frac1{x^2}$. Doing that in steps as kabi did also works, and may be easier to see.

for 1, we can divide Nr and Dr by cos^2 x.
[hide]for 2, as jagdish &jmerry pointed, i should stop use too much mathematica :wallbash:  :wallbash_red: ans seems to be $ 1/(2010 (1 \plus{} 1/x^2)^1005)$[/hide]


you know how a meter is know defined as the distance travelled by light in a certain amount of time. so now its defined in terms of time in contrast to the original standard meter.
Whats the point of doing this i've been puzzled by it for ages.

and doesn't time have its own disadvantages?

the definitons of fundamental; units are based on some arbitrary choices, some history and long, complicated definitions which require considerable expertise to understand. they are not exactly 'fundamental' in their present definition ---- don't worry , you're not alone!!!

I think that the point of defining one meter using time and the speed of light is that now it is exact (+/-) as opposed to defining it with standard meter - you can measure lengths with much greater accuracy. The speed of light in vacuum is constant (or so we think) and time can be measured really accurately with atomic clock. And perhaps even more importantly, you can measure the speed of light and build your own atomic clock in your basement and don't have to travel to Paris to duplicate the standard meter with 1e-15 accuracy. (Well, not really but you get the point. :))

All these Defintions have been given in the 17th General Conference on Wigths and Measures. The measurement of speed of light had become extremely precise, so it made sense to adopt the speed of light as a defined quantity and to use it to define a metre.

are their any disadvantages with the time light system


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose that $ G \equal{} AB$, where $ A,B$ are subgroups of $ G$. Does it always follow that $ BA \equal{} G$?

[hide="EDIT"]Oh yes, I don't know how I failed to see that the inclusion goes both ways... :oops: [/hide]

Yes, because $ BA\equal{}(AB)^{\minus{}1}$.

[quote="JBL"]Can we give a simple example of a group $ G$ with sub[b]sets[/b] $ A, B$ such that $ AB \equal{} G$ but $ BA \neq G$?[/quote]
How do you define $ AB$ if $ A$ and $ B$ are sets?

[quote="blahblahblah"]$ AB \equal{} \{ab: a\in A,b\in B\}$ (which can be different from $ BA$ if $ G$ is nonabelian)[/quote]
Umm. I think this definition works if $ A$ and $ B$ are submagmas of $ G$. But in subsets you can't assume that one has defined the product $ ab$.

I don't really see the problem, $ ab$ corresponds to 'something' in $ G$, so under that definition $ AB$ is certainly a well defined subset of $ G$.

[quote="puuhikki"]Umm. I think this definition works if $ A$ and $ B$ are submagmas of $ G$. But in subsets you can't assume that one has defined the product $ ab$.[/quote]  I am also not sure what the difficulty is.  $ A$ and $ B$ are sets [i]of elements of a given group[/i] $ G$, so the product $ ab$ is perfectly happy as an element of $ G$ and $ AB$ is the set of such products for $ a \in A (\subset G), b \in B (\subset G)$.

Take $ G\equal{}S_3,A\equal{}\langle(12)\rangle$ and $ B\equal{}\{1,(13),(123)\}$.
Then $ G\equal{}AB\not\equal{}BA$.


Now solve this right now! I'm sure you can solve it.
 If there are 3 candles, there were blown out, how many candles are left?

[hide] 3?

I guess the trick was making you think because the candles were blown out, they were also destroyed, but that is not the case.[/hide]

lol Mikey, you and me posted on the same minute  ;)

[hide]if there are 3 candles without a fire... there are still 3

3 candles[/hide][/hide]

[quote="pakagawa"]Blown out... they were blown OUT so none are left...

And relating to physics, a triangle with wind blowing:

I would light the one on the right angle according to distance and wind resistance[/quote]

Nah. Definitely the match first.

The answer is 1. Only one candle is left. The others are middle, and right.

You could also say "as many candles there are in the world."

[quote="white_horse_king88"][quote="pakagawa"][quote="Gyan"]Okay. If these three candles are at the three vertices of a 3-4-5 triangle and you are at the   incenter of the above triangle. . Assuming you have only one match-stick which one you will light first?[/quote]

3-4-5 cm?
3-4-5 km?
3-4-5 Mm?
3-4-5 light years?[/quote]

Why does nobody believe me??? It definitely has to be the match which is lighted first.[/quote]

Because I have a lighter.  :P

[quote="Mikey"]err.. still three, just theyre not lit?[/quote]
 I'm sure you're wrong. Please see my reply below.Ok?

[quote]Hi everybody! According to me, all of you are wrong. Here is my reply. There are 6 candles left. Do I have to give you an explanation?[/quote]
Yes, please. :?

[quote="yif man12"][/quote][quote="white_horse_king88"][quote="pakagawa"][quote="Gyan"]Okay. If these three candles are at the three vertices of a 3-4-5 triangle and you are at the   incenter of the above triangle. . Assuming you have only one match-stick which one you will light first?[/quote]

3-4-5 cm?
3-4-5 km?
3-4-5 Mm?
3-4-5 light years?[/quote]

Why does nobody believe me??? It definitely has to be the match which is lighted first.[/quote][quote="yif man12"]

Because I have a lighter.  :P[/quote]

That doesn't make sense. Just use your fingers...

[quote="bamboovn"]Hi everybody! According to me, all of you are wrong. Here is my reply.
 There are 6 candles left. Do I have to give you an explanation?[/quote]

Give me and explanation. Three candles LEFT? I don't get it.

I think that the answer is 0 when this problem is not being discussed because, when nobody is looking, they can cease to exist.

Hi all! Here is my explanation. Let's pay attention to the Verb tense of the sentence. "There [b]are[/b] 3 candles" "3 [b]were[/b] blown out" so the total is 6 candles left now.  ;)

[quote="wereallgonnadie"]That's pretty clever, but I'm not sure that's how the question goes...[/quote]
i'm sure that my anwer is right. There is no anwer else.Anyway, let's wait emi's answer.

[quote="h_s_potter2002"][quote="gameworld7"]ooooooooooooooookay, so um... what was i gonna say? oh yeah.[size=200] THIS IS A MATH FORUM. NOT AN ENGLISH FORUM. [/size]happy?[/quote]

Yeah, but emi typed the problem in a way that it's confusing and could have different meanings.[/quote]

I think that was gameworlds point :P The question was not a math problem, it was an english problem..

But that's the point of the problem -- it's not a mathematical question, it's a brain-teaser.  That is what the objection is to (I think) -- not the discussion of the problem, but the fact that the problem is in a math forum to begin with.

Hi all, I agree with your ideas. It's not a mathematical problem especially it's an English problem. Anyway, some of us can solve it. Rethink your thought!

why dont we just answer the question for what emi meant it to be?
there are 3 left, because after the candles are blown out (as in extinguished) the candles themselves are still there.

I think since this problm lacks the exact point on where the answer is, it should be moved to Fun & Games. I won't move it yet but if discussion goes on and on about deciding whether it's english or math problem, I'll remove it since it's not more like ALL-MATH discussion.

Before it happens, please identify your question's point.

I agree with my fellow moderator Silver Falcon. If there isn't a mathematical point here, this thread should be moved to Fun and Games.

Random question:

What exactly are you tokin', Mr. Adult?

this post is in fun and games and yes i do have a point that its an [size=200]english forum![/size] [size=59]wait i meant [/size][size=200]math forum! okay?![/size]

Um, you're contradicting yourself. Fun&Games is a place for just about anything, including English. There is just no grammar flames allowed, like being really picky over sentence structure.

ok sorry but this is a math site. u know why would they name an english learning site "www.artofproblemsolving" and they wouldn't name a math site "www.englishrocks" or something.  that wouldn't make sense

So what? We discuss politics in the Round Table. Does that mean we should cut all of that because this was designed as a math site? Should we cut all the riddles and stuff because this is a math site?

[quote=gameworld7]how should i know? i don't visit the round table section.[/quote]

Lol same


Prove that $ x^4 \minus{} 6x^3 \plus{} 11x^2 \minus{} 6x$, with $ x \in N$ is a multiple of 24.

hello, factorizing the given polynomial we get  $ x(x\minus{}1)(x\minus{}2)(x\minus{}3)$ so it is clear that $ P(x)$ is a multiple of $ 3$ and $ 8$.
Sonnhard.


[quote="Arne"]Does anybody know an ELEGANT way to solve $2^x = 3^y + 5$ in positive integers?[/quote]
$x=3$, $y=1$ and $x=5$, $y=3$ and if $y>3$ then as $2^{54}=1 \ mod \ 3^4$ and $2^{23}=2^{77}=5\ mod \ 3^4$ and both are coprimes with $54=\varphi(3^4)$ then NO more solutions.

RX TCM

Yes please I will be curious to see your explanations :?

How does it follow from just that, that there are no more solutions?  :?

I may be missing something, but it looks very tough... So far all I've got is some horrendous factorization work, such as
\[2^5(2^{2n-2}+1)(2^{n-1}+1)(2^{n-1}-1)=3^3(3^{4k}+1)(3^{2k}+1)(3^k+1)(3^k-1),\] but I don't see how a solution should come from that, either...

I have solved this one in the forum before locking modulo $3^4$ or something like that, a very ugly brute force thing.

[quote="Pascual2005"]I have solved this one in the forum [/quote]
Where, please?

[quote="Simo_the_Wolf"]I didn't understand, can you explain better RobertuX, please? :D[/quote]
Well  :? I delete all my calculations, so maybe another can helps to you. Sorry!

I cant find it and i dont remember it properly it was lockin mod $3^t$ or $2^t$ for some $t>4$ i think

[quote="RobertuX"]I delete all my calculations.Sorry![/quote]
 You are a joker :D

[quote="wellknown"]I think that I've found a solution, but it's not very elegant.
As I don't have any time to write the solution in details, I'll give an quick overview:
1) we show that y = 4n +3 for some n, if x,y > 1
2) we show that x = 2^k + 1 for some k
3) we show that 2^x == 2 (mod 17) for k > 2 (x > 5) and find out that then 3^y == 14 (mod 17) -> y = 16k + 9 and get a contradiction with y = 4n + 3.
4) k = 1 and k = 2 give us all solutions.

I hope it's correct[/quote]
I like your explanation. thx u!


[quote="mayukhbd"]In a exam there are 52% fail in english and 42% fail in math. if 17% have failed in both subject then what % of student have passed in both subject?

need the solution of this math. and can you tell me where can i find this kind of math's solution.....

thanks in advance[/quote]

[hide]this kind of problem can be solved with venn diagrams.

$52+42 = 94$

because 17% failed in both subjects,
$94-17 = 77$ failed in either english or math.

If someone didn't fail in english or math, he/she must have passed both english or math, so:

$100-77 = 23$ percent failed in no subjects.

btw, that's pretty bad, unless it was an extra hard test.  :P [/hide]

[hide]$100-(52+42-17) = 23$[/hide]

Btw, good advice, davidlizeng. :)

[quote="davidlizeng"]when you get percents just assume there are 100 people...
that makes it a bit easier.[/quote]

of course.  :)


Prove  that
(a+b)(a+c)>=2*sqrt[abc(a+b+c)] , for  all  positive  reals a ,b , c.
23  rd  brazilian  mo , 2001
Thanks...

Construct a triangle with edges a+b, b+c, c+a.
Then we have RHS=2S=(a+b)(a+c)sinA \leq LHS.
Q. E. D.

I have proved it by AM-GM

(a+b)(a+c)=a(a+b+c) +bc \geq 2 \sqrt abc(a+b+c)

Don't you think it is more simple?

Wpolly's approach is utterly wonderful but more difficult to think of.

I agree  with  Arne....Wpolly's  solution  is  more  interesting  but  more  difficult


I was looking at AoPS Volume 1 Chapter 3 and got extremly confused about what a linear equation can be made up of (like subtraction, addition etc.).  I know linear equations can be made up of addition, and subtraction are there any other operations that can be included in linear equations?

A linear equation can only contain terms of the form $ c_kx_k$, where $ x_k$ is a variable and $ c_k$ is an arbitrary real number, and those terms can only be added or subtracted. Basically, any given term within the linear equation can only have one variable within the term, and that variable must be expressed to the first power.

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable. In other words, no variable is raised to a power higher than 1. Most linear equations take 2 common forms:

1) ax^1+by^1=c
2) ax^1=c1+c2

where a and :D c are constants
and x and y are variables
 :roll:

But would 8*7 or 8/2 be a linear equation?

8/7 or 8*12 is not even an equation. They are expressions.

But neither are linear because linear (by definition) means that variables must be raised to the first, and at most first, power.


Did you see [url=http://www.mathlinks.ro/viewtopic.php?t=291472]this?[/url]

Wait sorry I'm not very smart :blush: and I feel stupid for asking, but what is that?????

Oh, whoops.  Didn't see that thread.  Thanks, kunny!   :)


trigonometry
logarithms
function
trig identities
algebra
log identities

Seems that everyone knows their own collection of trig identities, but does anyone know any interesting (or not-so-interesting) log identities?

Well, uh...

$\log_{b} a = \frac{\log_{c} a}{\log_{c} b}$
$\ln n < \sum_{k=1}^{n} \frac{1}{n} < \ln n + 1$
$\ln n! \approx (n + \frac{1}{2}) \ln n - n + \frac{1}{2} \ln (2 \pi)$ (Stirling's approximation in natural log form)
$z = re^{ix} \implies \ln z = \ln r + ix$

Beyond the basic identities and applications of Euler's formula, there's not much going on.

[quote="t0rajir0u"]$z = re^{ix} \implies \ln z = \ln r + ix$
[/quote]

This wouldn't be exactly accurate:

$z = re^{ix} \implies \ln z = \ln r + i(x+2n\pi), n\in\mathbb{Z}$

In other words, incomplete.  He also included the mutiples of $2\pi$. for entirety. :)

Well, here are just some logarithmic identities off the top of my head.

$\\\log a+\log b+...+\log n=\log ab...n\\ \log a-\log b-...-\log n=\log \frac{a}{b...n}\\ \log a^x=x\log a\\\\ c^{\log_cx}=x\\ \log_cx=\loc_cx\\ x=x\implies\text{ The idenity is true}$

Masoud Zargar

[quote="maokid7"]could you please explain how $ln(re^{ix})=ln(r)+ix$ is not correct.
i read what you posted but was not able to follow.[/quote]

Logarithm of a complex number behaves differently than logarithm of a real number, since it's closely connected to trigonometric functions, and therefore the multiples of $2\pi$ must be included.

Look what may happen if we follow [b]t0rajir0u[/b]'s identity: We know that $e^{i\pi}=-1$ and $e^{-i\pi}=-1$. From what he'd written, $\ln (-1)=\ln e^{i\pi}=i\pi$ and $\ln (-1)=\ln e^{-i\pi}=-i\pi$, and from there $\pi=-\pi$ (!)

Generally, $\ln e^{i\alpha}=\ln e^{i\beta}\implies \alpha=\beta+2n\pi,n\in\mathbb{Z}.$ (And in our example it gives $\pi=-\pi+2n\pi$, which obviously holds for $n=1$.)

Therefore, it's not the question of entirety but of correctness :)

[b]Proof[/b]
$\\a^{\log_bc}=c^{\log_ba}\\ \log_bc\log a=\log_ba\log c\\ \frac{\log_bc}{\log_ba}=\frac{\log c}{\log a}\\ \log_ac=\log_ac\implies\text{ TRUE}$

masoud zargar


In my subscription in my personal profile, there are two options:

Use http://www.mathlinks.ro/ domain
Use http://www.artofproblemsolving.com/ domain

So what is the difference between the two?  Which one is better?

Use search, it's been explained many times.


Given $ABCD$ the tetrahedron. Let $M$ be the midpoint of $CD$. What is the terminology of the plane $(ABM)$. Many thanks!!!

Any three noncollinear points define a unique plane, so plane $ABM$ would be the plane that goes through those 3 points.

The best thing I think you should call it be the "median plane"!


Why does logarithmically convex imply convex? log(f(x)) convex and f>0  => f(x) convex

Suppose $f>0$ and that $\log(f(x))$is convex.
Then $\frac{d^{2}}{dx}\log(f(x))=\frac{f''(x)f(x)-(f'(x))^{2}}{(f(x))^{2}}$.
Since $\log(f(x))$ is convex, then $\frac{d^{2}}{dx}\log(f(x))\geq0$ i.e. $\frac{f''(x)f(x)-(f'(x))^{2}}{(f(x))^{2}}\geq0$.
Since $f(x)>0$ by asumption and clearly $(f'(x))^{2}>0$ we have to have $f''(x)>0$ for this relation to hold.
Hence $f$ is a convex function.


algebra
polynomial
functional equation
algebra unsolved

Find all polynomials $ f(x)$,such that $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\plus{}3xy(x\plus{}y)$   for all x,y

__________________________________
$ \textbf{Azerbaijan Land of Fire }$ :ninja:  :rotfl:

Let $ f(x) \equal{} x^3 \plus{} p(x)$, where $ p(x)$ is some polynomial. You get:

$ p(x \plus{} y) \equal{} p(x) \plus{} p(y)$

This is Cauchy's functional equation, solution of which is $ p(x) \equal{} cx$. So, $ f(x) \equal{} x^3 \plus{} cx$ for some $ c\in\mathbb{R}$.

why should $ f(x)\equal{}x^3\plus{}p(x)$ with $ p$ is a polynomial?  :blush:

[quote="AdrianP"][quote="Altheman"]f(x) is a polynomial so f(x)=x^3 is a polynomial as well...[/quote]

I think Raja means "Why would you pick that particular $ f(x)$? Where did it come from?" which is a good question.[/quote]

It's not very difficult to think to the equality $ (x\plus{}y)^3\minus{}x^3\minus{}y^3\equal{}3xy(x\plus{}y)$


We need to prove $ CA \equal{} CN$.(then $ \angle ANB \equal{} \angle AIB$). $ CA \equal{} CN$ is equivalent to $ \angle AQI \equal{} 90$. Let incircle touch $ CA$ at $ L$. Because $ P,L$ is symmetric wrt $ CQ, \angle ILQ \equal{} \angle IPQ \equal{} \angle IQP$. Hence $ I,L,K,Q$ are cyclic. But obviously, $ A,K,I,L$ are cyclic. Hence $ A,K,Q,I,L$ are cyclic. Hence $ \angle AQI \equal{} \angle AKI \equal{} 90$.

[quote="tdl"]The incircle of triangle $ ABC$ with center $ I$ touches the sides $ AB$ and $ BC$ at points $ K$ and $ P$ respectively. The bissector of angle $ C$ intersects the segment $ KP$ at point $ Q$ and the straight line $ AQ$ intersects the side $ BC$ at point $ N$. Prove that points $ A;I;N$ and $ B$ lie at a common circle.[/quote]
[hide="Solution"]
Let $ \angle BAC \equal{} 2\alpha$ and $ \angle ACB \equal{} 2\beta$.  Thus, we have that $ \angle ABC \equal{} 180 \minus{} 2\alpha \minus{} 2\beta$, so $ \angle KIP \equal{} 2\alpha \plus{} 2\beta$.  This implies that $ \angle IPK \equal{} \frac {180 \minus{} \angle KIP}{2} \equal{} 90 \minus{} \alpha \minus{} \beta$.  Also,
\[ \angle ICP \equal{} \beta\implies \angle PIQ \equal{} 180 \minus{} \angle PIC \equal{} 90 \plus{} \beta\implies \angle IQK \equal{} \angle PIQ \plus{} \angle QPI \equal{} 90 \plus{} \beta \plus{} 90 \minus{} \alpha \minus{} \beta \equal{} 180 \minus{} \alpha\implies \angle KAI \plus{} \angle KQI \equal{} 180
\]This gives us that $ AIQK$ is cyclic, so $ \angle AQI \equal{} \angle AKI \equal{} 90\implies \angle IQN \equal{} 90\implies \angle IQN \plus{} \angle IPN \equal{} 180$, so $ QIPN$ is cyclic.  Hence, $ \angle INP \equal{} \angle IQP \equal{} \alpha \equal{} \angle IAB$, so $ IABN$ is cyclic, as desired. [/hide]

Dear Mathlinkers,
1. A, K, I, L, Q are cocyclic (Lascases's result; see http://perso.orange.fr/jl.ayme  vol. 4, An unlikely concurrence...)
2. I, P, N, Q are cocyclic by the hypothesis
3. Applied a converse of the pivot theorem to triangle PBK and we are done.
Sincerely
Jean-Louis


real analysis
algebra
function
domain
real analysis theorems

when we have $ \mu^* (\mathfrak{b}) \equal{} 0$ (for [i]any[/i] $ \mathfrak{b}\in \mathbb{R}^n$), why is $ \mathfrak{b}$ also lebesgue measurable? Does this have something to do with the fact that the domain $ X \equal{} \mathfrak{b}$ is restricted enough so that that it is a $ \sigma$-algebra??

I am making use of the theorem (perhaps incorrectly) that: If  $ \mu^*$ is an outer measure on $ X$ and $ \mathcal{M}$ is the collection of all $ \mu^*$ measurable substes of $ X$, then $ \mathcal{M}$ is a $ \sigma$-algebra, and the restriction of all $ \mu^*$ to $ \mathcal{M}$ is a measure on $ X$

Thanks!

Lebesgue measurable sets are (by one of the definitions) exactly those sets $ E$ that enjoy the well-separation property
\[ \mu^*(A\cup E)\plus{}\mu^*(A\setminus E)\equal{}\mu^*(A)\]
for every $ A\subset\mathbb R^n$ where $ \mu^*$ is the Lebesgue outer measure. This property is easy to check when $ \mu^*(E)\equal{}0$. Hence the conclusion. So, you are right, this is directly related to the Caratheodory theorem you quoted.


The average value of nine consecutive integers is 13. What is
the sum of the smallest and largest of these integers?

the average of consecutive integers would be the middle integer, or in this case, the 5th

so let's list them out:
9, 10, 11, 12, 13, 14, 15, 16, 17

the sum is obviously $ 9\plus{}17\equal{}26$


$ p(x \plus{} y) \equal{} y^2x$.it follows that $ p|y$ or $ p|x$. Suppose that both of them are divisable by $ p$. so there exist $ a,b$ such that
$ p^{a}\parallel{}y,p^{b}\parallel{}x$. there are 2 cases:
1-$ a\neq{b}$ then from the above equation we'll have $ 1 \plus{} \min{a,b} \equal{} 2a \plus{} b$ which is absurd.
2-$ a \equal{} b$ then we'll have $ (\frac {x}{p^{a}} \plus{} \frac {y}{p^{a}}) \equal{} p^{a \minus{} 1}(\frac {y^2}{p^{2a}})(\frac {x}{p^{a}})$
so if $ \frac {y}{p^{a}} \equal{} m,\frac {x}{p^{a}} \equal{} n$ we'll have $ m^2n\leq{m \plus{} n}$.we can easily find all the solutions.
so assume that $ p$ divides only one of them. u can see that in this case we must have $ p|x$. if $ px' \equal{} x$ u can see that
$ y |x'$ it follows $ yt \equal{} x'$. then we'll have:$ pt \plus{} 1 \equal{} y^2t$ the rest follows. i wish i didn't make any mistake I MUST GO TO BED!!

$ p\equal{}\frac {y^2x}{x \plus{} y} \iff px\plus{}py\equal{}y^2x$

Taking mod $ x$ then $ x|py$ and taking mod $ y$ then $ y|px$
(1) $ py\equal{}ax$
(2) $ px\equal{}by$

Taking mod $ p$ then $ p|y^2x$ then either $ p|x$ or $ p|y$
If $ p|x$ then $ x\equal{}kp$ then in (1) $ py\equal{}akp$ then $ y\equal{}ak$ and in (2) $ pkp\equal{}bak$ then $ p^2\equal{}ab$ then three cases:
$ (a,b)\equal{}(1,p^2)$ or $ (p^2,1)$ or $ (p,p)$

If $ p|y$ then $ y\equal{}kp$ then in (2) $ px\equal{}bkp$ then $ x\equal{}bk$ and in (1) $ pkp\equal{}abk$ then $ p^2\equal{}ab$ then same three cases.

Case 1) $ py\equal{}x$
$ p\equal{}\frac{y^3p}{y(p\plus{}1)} \iff p\plus{}1\equal{}y^2 \iff p\equal{}(y\plus{}1)(y\minus{}1) \Rightarrow y\equal{}2$
then $ \boxed{(x,y,p)\equal{}(6,2,3)}$

Case 2) $ y\equal{}px$
$ p\equal{}\frac{p^2x^3}{x(p\plus{}1)} \iff p(p\plus{}1)\equal{}x^2 \Rightarrow p|x \Rightarrow x\equal{}cp \Rightarrow p\plus{}1\equal{}pc^2 \Rightarrow 1\equal{}p(c\plus{}1)(c\minus{}1)$
no solution

Case 3) $ y\equal{}x$ then $ p\equal{}\frac{x^3}{2x}\equal{}\frac{x^2}{2}$ then $ \boxed{(x,y,p)\equal{}(2,2,2)}$

[quote=lambruscokid]Find all positive integers $ (x,y)$ such that $ \frac{y^2x}{x\plus{}y}$ is a prime number[/quote]

$(x,y)=d$ $x=ad , y=bd$
$\frac{d^{2}×b^{2}×a}{a+b}$ it must be prime
$(a,a+b)=1$ and $(b,a+b)=1$ then  $\frac{d^2}{a+b}$$=p$ it must be integer.$p×a×b^{2}$ is prime then $b=1$
$\frac{d^2}{a+1}$$=p$  then $p×a$ is prime $a=1$ or $a$ is prime and $p=1$
$p=1$ then $d^2=a+1$ $a=d^2-1$ and $a$ is prime then $a=3$ $d=2$
$\boxed{(x,y,q)=(2,2,2),(6,2,3)}$


geometry
inequalities
trigonometry
geometry unsolved

[quote="April"]Take $\ A_{1}$ on the side $\ BC$ of a triangle $\ ABC$ so that the inciclers of triangle $\ ABA_{1}, ACA_{1}$ have equal radii $\ r_{a}$. One defines $\ r_{b}, r_{c}$ analogously. Prove that:
$\ 2(r_{a}+r_{b}+r_{c})+p\leq h_{a}+h_{b}+h_{c}$[/quote]What is $\ p, h_{a},h_{b},h_{c}$ here?

Normally $h_{a}, h_{b}, h_{c}$ are altitudes dropped on $a,b,c$ respectively. $p$ stands for semiperimeter.(also known as $s$ in some places like USA and Canada :lol: )

I think it is false,it must be like this :
$\ 2(r_{a}+r_{b}+r_{c})+p\geq h_{a}+h_{b}+h_{c}$
Let $S$ be the area of triangle.
We can easyly prove that $2r_{a}=h_{a}(1-\sqrt{tan\beta.tan\gamma})$ where $2\alpha,2\beta$ and $2\gamma$ are the angles of triangle.
We know that for any triangle $\frac{h_{a}-2r}{h_{a}}=tan\gamma.tan\beta$
Writing it for triangles $AA_{1}B$ and $AA_{1}C$ and multiplying side by side we get $(\frac{h_{a}-2r_{a}}{h_{a}})^{2}= tan\gamma.tan\beta$
Hense $2r_{a}=h_{a}(1-\sqrt{tan\beta.tan\gamma})$
So we must prove that
$p \geq h_{a}\sqrt{tan\beta.tan\gamma}+h_{b}\sqrt{tan\alpha.tan\gamma}+h_{c}\sqrt{tan\beta.tan\alpha}$
Replacing $tan\alpha=\frac{r}{p-a},tan\beta=\frac{r}{p-b}$ and $tan\gamma=\frac{r}{p-c}$ we get
$p \geq \frac{h_{a}.r}{\sqrt{(p-b)(p-c)}}+\frac{h_{b}.r}{\sqrt{(p-a)(p-c)}}+\frac{h_{c}.r}{\sqrt{(p-b)(p-a)}}$
$p \geq \frac{\frac{2S}{a}.\frac{S}{p}}{\sqrt{(p-b)(p-c)}}+\frac{\frac{2S}{a}.\frac{S}{p}}{\sqrt{(p-a)(p-c)}}+\frac{\frac{2S}{a}.\frac{S}{p}}{\sqrt{(p-b)(p-a)}}$
$p^{2}\geq \frac{2S^{2}}{a\sqrt{(p-b)(p-c)}}+\frac{2S^{2}}{b\sqrt{(p-a)(p-c)}}+\frac{2S^{2}}{c\sqrt{(p-b)(p-a)}}$
$p \geq (p-a)\frac{2\sqrt{(p-b)(p-c}}{a}+(p-b)\frac{2\sqrt{(p-a)(p-c}}{b}+(p-c)\frac{2\sqrt{(p-b)(p-a}}{c}$
By $AM-GM$ inequality $\frac{2\sqrt{(p-b)(p-c}}{a}\leq \frac{p-c+p-b}{a}=1$.So
$p=p-a+p-b+p-c \geq (p-a)\frac{2\sqrt{(p-b)(p-c}}{a}+(p-b)\frac{2\sqrt{(p-a)(p-c}}{b}+(p-c)\frac{2\sqrt{(p-b)(p-a}}{c}$
That's all. :) I think it is true.

[color=darkblue]See the my topic http://www.mathlinks.ro/Forum/viewtopic.php?t=50559 $\Longrightarrow r_{a}=\frac{r}{a}\cdot[p-\sqrt{p(p-a)}]$ a.s.o.

[b]Remark.[/b] $\boxed{\ p+2\cdot\sum r_{a}\ge \sum h_{a}\ }$ $\Longleftrightarrow$ $\|\begin{array}{c}p+2pr\cdot\sum\frac{1}{a}-2r\sqrt p\cdot\sum\frac{\sqrt{p-a}}{a}\ge 2S\cdot\sum\frac{1}{a}\\\\ 2S=ah_{a}=2pr\end{array}$ $\Longleftrightarrow$ $\boxed{\ \sum\frac{\sqrt{p-a}}{a}\le \frac{\sqrt p}{2r}\ }\ \ (*)\ .$

[b]Proof of the inequality[/b] $(*)\ .$

$1.\blacktriangleright\ \{x,y,z\}\subset (0,\infty )\Longrightarrow$ $\frac{4yz}{(y+z)^{2}}\le 1$ a.s.o. $\Longrightarrow$ $\frac{4xyz}{(y+z)^{2}}\le x$ a.s.o. $\Longrightarrow$ $\sum_{\mathrm{cyc}}\frac{4xyz}{(y+z)^{2}}\le x+y+z$ $\Longrightarrow$ $\boxed{\ \sum\frac{1}{(y+z)^{2}}\le \frac{x+y+z}{4xyz}\ }\ \ (1)\ .$

$2.\blacktriangleright\ x: =p-a$ a.s.o. in the inequality $(1)$ $\Longrightarrow$ $\|\begin{array}{c}\sum\frac{1}{a^{2}}\le \frac{p}{4(p-a)(p-b)(p-c)}\\\\ (p-a)(p-b)(p-c)=pr^{2}\end{array}$ $\Longrightarrow$ $\boxed{\ \sum\frac{1}{a^{2}}\le \frac{1}{4r^{2}}\ }\ \ (2)\ .$

$3.\blacktriangleright$ Apply the $C.B.S.$- inequality : $(\sum\sqrt{p-a}\cdot\frac{1}{a})^{2}\le \sum (p-a)\cdot\sum\frac{1}{a^{2}}=p\cdot\sum\frac{1}{a^{2}}\ .$ Using the inequality $(2)$ obtain the required inequality $(*)\ .$

[b]Remark.[/b] $\|\begin{array}{c}xy+yz+zx\le x^{2}+y^{2}+z^{2}\\\\ x: =\cot A\mathrm{\ a.s.o.}\\\\ \sum\cot B\cot C=1\end{array}$ $\Longrightarrow$ $\|\begin{array}{c}\cot^{2}A+\cot^{2}B+\cot^{2}C\ge 1\\\\ \frac{1}{cot^{2}x}=\frac{1}{\sin^{2}x}-1\end{array}$ $\Longrightarrow$ $\|\begin{array}{c}\sum\frac{1}{\sin^{2}A}\ge 4\\\\ a=2R\sin A\mathrm{\ a.s.o.}\end{array}$ $\Longrightarrow$ $\boxed{\ \frac{1}{R^{2}}\le \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\le \frac{1}{4r^{2}}\ }\ .$[/color]

[color=red][b]Open problem.[/b] $\boxed{\ \frac{\sqrt p}{R}\le \sum\frac{\sqrt{p-a}}{a}\ }\le\frac{\sqrt p}{2r}\ .$[/color]


I came up with this problem, but I'm not sure if it can be easily solved or not.  Here it is:  What is the probability that the sum of the digits of a 10-digit number is divisible by 10?

Actually came out to be a lot easier than I thought. For any possibility of the first 9 digits, one out of the 10 possible last digits will make the sum a multiple of 10. So the probability is 1/10.

The problem I thought of when I saw this one is slightly harder, but fun nevertheless.. I saw a real nice solution for it in one of our camps.

Whats the probability (for a 10 digit number) that the digits are increasing (ie each digit is equal or greater than the previous one)?

Very nice solution from TripleM.

If I understand correctly, his new problem if to find the number
of numbers built yhis way :
a1a2...a10 with 1<= a1 <= a2  ... <= a10

I mean the sum of the digits do not have to be divisible by 10 any more ?

Let's try to generalize to n-digits numbers. I don't have a quick and astute solution as the one of TripleM, but I checked it by hand for small values of n and it seems to be correct.

I claim that a number with increasing digits is of the form (in a unique way) :
b1*ONE(n) + b2*ONE(n-1) + ... bn*ONE(1)
with b1 + b2 + ... + bn <= 9 and b1 >= 1

where ONE(k) stands for the number with k '1' (ONE(3) = 111 to be clear)

For instance 378 = 3*111 + 4*11 + 3*1,

The condition b1 + b2 + ... + bn <= 9 is because carries would cancel the increase.

So we have to solve :

b1 + .. + bn <= 9 with b1 >= 1.

It's the same as (1+b1') + b2 + ... + bn + B = 9

By De Moivre theorem there are binomial (8+n, n) solutions.

So the probability to have an increasing number is :

binomial (8+n, n) / 9*10^(n-1)

for n = 2 .. 10 :

2 : .050
3 : .0183
4 : .0055
5 : .00143
6 : .0003336
7 : .0000715
8 : .0000143
9 : .270111 10^-5
10 : .4862 10^-6 ~ 1/2000000 (1 chance out of 2 millions)

P.S. Forgive poor English

Triple M

So does that mean if all the digits are the same, it is considered increasing?
 ie 1111111111

If d(2) can be greater than or equal to d(1), then i get the answer as
387420489/1000000000 (that should be 9 zero's). I get this because for the first digit, you have 9 possibilities. After that, you have another 9 possibilities, etc. So you have 9^10 possibilities. But then you have to divide that by the total number which is 9 for the first digit, and then 10 for all after that, i.e. 9 x 10^9.

TheSouthpaw: Yes, I never understood it really but I think the standard thing is increasing means they can be equal and strictly increasing means they can't.

Jelyman: The problem is that we <don't> have 9 possibilities for each number, as for example if we start with '15..' then there are only 5 possibilities for the next number.

i/3s answer matches up with mine. Heres the 'nice' way I was taught:
We have 9 boxes for each of the 9 digits 1 thru 9, and we place a ball in a box for each time that digit occurs. So for example 1145577779 would be
[00][][][0][00][][0000][][0]. After writing this on the whiteboard he sneakily rubbed out a couple of lines , replacing each ][ with a 1 and getting 001110100110000110. And so we all realised the answer was the number of ways of choosing 10 zeroes out of 18 places, ie C(18,10) or C(18,8).

Thought it was real nice..

I have no idea how that works but it does...can anyone explain triplem's method for me?

Sorry I didn't explain that very well. Every such number can be represented as a combination of 10 zeroes for the 10 digits, and 8 ones where each one implies the next digit goes up by one... and all we have to do is find all ways we can find such a combination. Nope I still don't make sense. Someone else help me here :)

Maybe a smaller case: consider 2 digit numbers only using the digits 1 thru 3.
[][][] three boxes, first represents number of 1s, next number of 2s, last one number of 3s. All we have to do is put two balls in them in some way.
number 11 = [00][][] = 0011
number 12 = [0][0][] = 0101
number 13 = [0][][0] = 0110
number 22 = [][00][] = 1001
number 23 = [][0][0] = 1010
number 33 = [][][00] = 1100

each is made up of 2 ones and 2 zeroes..

Hope this helps, someone else can try to explain hopefully..

Meanwhile, that was only part of the problem, probably the hardest bit but we'll see.
The actual problem (difficulty pretty high):
Find the <sum> of all numbers between 0 and (to make the numbers bearable) 1000000 that have either decreasing OR increasing digits.
For example some of the numbers are: 11689, 55555, 921000.

hmmm...which is easier

find all numbers that dont increase or decrease

find all numbers that do increase or decrease...

wow...this is really hard...

[quote="akatookey"]hmmm...which is easier

find all numbers that dont increase or decrease

find all numbers that do increase or decrease...

[/quote]

Basically... neither. :)

[quote="TripleM"]Find the <sum> of all numbers between 0 and (to make the numbers bearable) 1000000 that have either decreasing OR increasing digits.
For example some of the numbers are: 11689, 55555, 921000.[/quote]

I don't have time today to work on this nice problem but I would like to know if I am on the right way to the solution.

Let BIG(n) = 1....10 (with n '1' and one '0') = 10*(10^n-1)/9

Lemma :
If In is an increasing-digits number with n digits then (BIG(n) - In) is a decreasing-digits number.

The converse is almost true :
if Dn is a decreasing-digits number with no '0' inside (because of carries :-() then BIG(n) - Dn is an increasing-digits number.

So there is a one to one relationship between decreasing-digits numbers with no '0' inside and increasing-digits numbers. Numbers like 1......1 are both increasing-digits and decreasing-digits

The sum of {in|de}creasing-digits numbers with no zero inside is then :

C(8+n,n) * BIG(n) - (1...1 + 2....2 + 9...9)

The problem is then to sum decreasing-digits numbers with '0' at the end.
Am i far from it  :?:  :?:

On the right track. If you consider all the numbers as 6 digit numbers which can allow leading zeroes, then in fact you don't need a BIG(n), just a single BIG...

OK heres the solution.. (forget about 1000000 being decreasing at the moment I really meant less than a million but you can add on a million at the end.)

We can pair up every number with increasing digits with one with decreasing digits by subtracting it from 999999, eg 001256 -> 998743. Every one can be paired up in this way, and as we know from before there are 14C8 = 3003 increasing digit numbers and thus 3003 pairs each summing to 999999. So 3003 x 999999 is the sum of all pairs. Have we counted any numbers twice? Yes, all those which are both increasing AND decreasing, ie 111111,222222,...,999999. So the answer is 3003x999999 - the sum of those 9 numbers.
(which is 2997997002 but thats not the point. Oh and add on 1000000 if you want :P).


The cost of the trip has been set at $\$$130 per person. Money should be turned in to Mr. R on the day of the trip, unless you see him in person and can give him the money sooner. The seven SC students selected to participate in this trip are:

Sean
Shobhit
Jim
Asif
Harry
Noah
Brian

The 6 Columbia students will be driven to the airport in Charlotte by Mr. R after signing field trip forms/legal liability waivers. The plane will take off from Charlotte at x: xx on Friday, December 15th, and arrive in Princeton at x: xx. The return flight from Princeton will be on Sunday the 17th. We depart at x: xx, and arrive in Charlotte at x: xx. Mr. R will then drive the 6 Columbia students back to Columbia.

Please bring money for transportation in Princeton. This should be at least $\$$xx.xx.  Also, bring an ample amount of cash for food.  You can always bring back leftover money.

If we want t-shirts, somebody send ideas to me via email ASAP.  My parents can pay for (sponsor) these.

Any questions, email me at the address below.  Suggestions requested as well.  Everything seems to finally be on track.

[b]NC people reading this[/b], everything is set for you.  You don't owe us any money, just bring enough for the food and travel expenses outlined 3 paragraphs above.

Everything is finally falling in place, and this seems like it will be a great trip.  Many thanks to Mr. R, Shobhit's mom, my parents, Mr. Lightsey, BellSouth, and everyone else who made this trip possible.

Sean
seansoni@gmail.com

we want t-shirts, how bout top half be garnett and bottom half be unc blue or something, with some kind of expensive looking logo

OK, if the people can do t-shirts in time, we have a design.  If not, people wear your pink (NC) or blue (SC) ARML shirts from last year.

Great success!

Results [url=http://cgi.math.princeton.edu/mathclub/index.php/PUMaC_2006_Results]here[/url].

whoa arnav got beaten by an 8th grader, lol

Berman pwned though... YEAH GO BERMAN

Man, I want to go next year

[quote="Iversonfan2007"]the highlight of the trip was when mr. r kept showing off the trophy and every time someone asked about it, he would reply:

"why thank you

i was hoping you'd ask about that"[/quote]

What was with that guy who was drunk or something at the train station?  He was like "My wife's birthday is today.  Sing her a song!"

guy: "so which teams were competition"

mr. r: "um, South and North Carolina--"

guy: "no, tell me some real teams"

Does anyone have a link to the contest or the Contest Questions itself?

Yea, go to the Princeton forum, and there is a linked PDF of the problems in the Solutions thread.

Thank You Very Much

PS
The Problems seem to be good prep for AIME.


analytic geometry
graphing lines
slope
algorithm

Does anyone know why r was chosen as the letter for the correlation coefficient in statistics.  Just wondering.

I have no idea, maybe r for relationship?

This is sortof one of those questions like "Why do we use m to represent slope?"  I don't know the answer but here are some suggestions:

(1) Its an arbitrarily assigned letter.
(2) There is something in the algorithim for finding R that involves division, hence RATIO.
(3) It was invented by a guy w/a name that started w/R.
(4) Maybe it means something like "ratings" or "ranking," as in how close the data points rank in terms of being the least s.d.s away from the regression line.

All I can think of...

also, it may have been first based on a word in another language... as was much of our current english math nominclature.

Yes, like supposedly m means "monter" which is french for "to climb."

I'm going to guess that it's because [tex]r^2[/tex] is [url=http://mathworld.wolfram.com/CorrelationCoefficient.html]based on[/url] the squared [i]residuals[/i] of the data points to the least squares fit.


[quote="Cezar Lupu"][color=blue][b]Problem.[/b]
  Let $a, b, c$ three nonnegative reals. Prove the following inequality

           \[{ (a^{2}+ab+b^{2})(b^{2}+bc+c^{2})})(c^{2}+ca+a^{2})\geq\frac{27}{64}(a+b)^{2}(b+c)^{2}(c+a)^{2}.  \]

 [/color][/quote]
Too easy
Just use $a^{2}+ab+b^{2}\geq\frac{3}{4}(a+b)^{2}$
And we don't need $a,b,c\geq 0$ just $a,b,c\in R$


geometry
geometric transformation
calculus
integration
inequalities
trigonometry
LaTeX

The 1st Southeast China Mathematical Olympiad was held in Wenzhou, Zhejiang province, July 10-11, 2004.
106 students of class 10, from Zhejiang, Jiangxi and Fujian provinces, take part in this contest.
Maybe there are lots of mistakes in my translation, because my English is not very good. I'll be appreciative if someone can tell me the mistakes for my improvement.
Here are the problems!
Have a good day!
 :)

[b]Edited by Myth[/b]: added problems and links to the separate topics.

[b]Problem 1.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26221
Let $a, b, c$ be reals such that $a^2+2b^2+3c^2=\frac{3}{2}$. Prove that $3^{-a}+9^{-b}+27^{-c}\geq 1$.

[b]Problem 2.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26225
Let $D$ be a point on side $BC$ of triangle $ABC$ and point $P$ on segment $AD$. A line through $D$ intersects segments $AB$, $PB$ at $M$, $E$ respectively, also intersects the extensions of segments $AC$, $PC$ at $F$, $N$ respectively.
If $DE = DF$, prove that $DM =DN$.

[b]Problem 3.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26216
[b]1.[/b] Does there exists positive integral infinite sequences $\{a_n\}$ such that $a_{n+1}^2\geq 2a_{n+2}a_{n}$ for any positive integer $n$?

[b]2.[/b] Does there exists positive irrational infinite sequences $\{a_n\}$ such that $a_{n+1}^2\geq 2a_{n+2}a_{n}$ for any positive integer $n$?

[b]Problem 4.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26222
Given positive integer $n > 2004$. Numbers 1, 2, 3, …, $n^2$ were filled in the squares of an $n\times n$ checkerboard such that every square had just one number. Square $A$ was called "good" if

[b](i)[/b] The number filled in $A$ is larger than at least 2004 numbers in the squares at the same row of $A$.
[b](ii)[/b] The number written in $A$ is larger than at least 2004 numbers in the squares at the same column of $A$.

Find the greatest number of "good" squares in $n\times n$ checkerboard.

[b]Problem 5.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26219
If the inequality $\sqrt{2}(2a+3)\cos\left(x-\frac{\pi}{4}\right)+\frac{6}{\sin x+\cos x}-2\sin 2x<3a+6$ holds for any real $x\in[0,\pi/2]$, find the value range of real number $a$ can be taken.

[b]Problem 6.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26226
Let $D$ be a point on base side $BC$ of isosceles triangle $ABC$. Point $F$ in triangle $ABC$ is situated on the arc of the circle passing through points $A$, $D$ and $C$. The circle passing through points $B$, $D$ and $F$ intersects side $AB$ at point $E$. Prove that \[CD\cdot EF + DF\cdot AE = BD\cdot AF.\]

[b]Problem 7.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26224
$n$ teams take part in a tournament. For any two teams $A$ and $B$, if the game is played at $A$'s town, then call this game as $A$'s "home game", as well as $B$'s "away game". Any two teams play against each other just twice, each having a "home game" respectively. Knowing that every team can be arranged more than one "away game" in one week (from Sunday to Saturday); but if one team has "home game" in certain week, then it cannot be arranged "away game" in that week.

If all the games can be arranged to finish in four weeks, find the maximum possible value of $n$.

[b]Problem 8.[/b] http://www.mathlinks.ro/Forum/viewtopic.php?t=26220
Find the number of the ordered integral number groups $(x, y, z, u)$ such that $1\leq x,y,z,u\leq 10$ and
\[\frac{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-u}{z+u}+\frac{u-x}{u+x}>0.\]

Obviously your English is good I think though:

- in 3. (1) should be "integer" instead of "integral"
- I'm not native so I won't point out mistakes in English, cuz I might be wrong also  :D

But what i wanted to say: thanks for good job.  :)

It is very interesting!

Thank you, zhaoli!

But where are separate topics for each problem?  ;)

I do not know how to use LATEX, so it is difficult for me to set up separate topics in corresponding catlogs. Would you please help me to do the job, Myth?

billzhao has just informed me that in problem 7, we may say "home game" for "主场" and "away game" for "客场", which I translated them to "host game" and "guest game", respectively.
Hope that it will not affect your understanding the meaning of the problem.

is it possible for hong kong to join it next yr?  :?

Myth:

I have also translated the 18 problems of 2004 China TST one hour ago, which posted in
http://www.mathlinks.ro/Forum/viewtopic.php?p=164145#p164145

Are you interested?

[quote="Myth"]I have done all slave's work! See first post![/quote]Thank you Mikhail! :)


integration
real analysis
real analysis solved

Find $inf \sum_{n=0}^{\infty}u_nS_n$ where $S_n=\sum_{k=0}^n u_k$ and $\sum u_n$ is a serie of positive terms with sum  equal to 1.(Search the inf after all series which satisfy the hypothesis.)

Unless I'm overlooking something, it seems obvious that the infimum is $\frac 12$.

Let $S$ be the sum of the products $u_iu_j,\ i<j$, and $T$ the sum of the squares $u_i^2$. Then we're looking for the infimum of $S+T$, given the condition $2S+T=1$. This infimum will be $\inf\left(\frac{1-T}2+T\right)=\frac 12+\frac{\inf T}2$. We can make $T$ arbitrarily close to $0$ by choosing $(u_1,u_2,\ldots)$ very "close" to $\left(\frac 1n,\frac 1n,\ldots,\frac 1n,0,0,\ldots\right)$ (there are $n$ non-zero terms).

That's right.This is just a different approach:
$u_nS_n=(S_n-S_{n-1})S_n\geq \int_{S_{n-1}}^{S_n}tdt$.So $\sum_{k=0}^n u_kS_k\geq u_0S_0+\int_{S_0}^{S_n} t dt$.This gives:$\sum_{k=0}^{\infty} u_kS_k\geq u_0^2+\int_{u_0}^{1}tdt\geq \frac12$.
And,by the way...I'm still waiting for help with those "easy problems" with equivalents. ;)


IMO
number theory
IMO 2001
IMO Shortlist

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[b]Solution by Iura:[/b]

[b]Lemma[/b]. If $ab=cd$ then there exist $m,n,p,q$ with $a=mn, b=pq, c=mp, d=nq$. [i]Proof[/i] is obvious.

Opening the brackets we get $a^2+c^2-ac=b^2+bd+d^2$
Multiplying by $4$ we get $(2a-c)^2+3c^2=(2b+d)^2+3d^2$ thus $3(c+d)(c-d)=(2b+d-2a+c)(2b+d+2a+c)$.
Now using this Lemma we get either
\[c+d=mn, c-d=pq, 2a+c-2b+d=3mp, 2b+d+2a+c=nq,\]
or
\[c+d=mn, c-d=pq, 2a+c-2b+d=mp, 2b+d+2a+c=3nq.\]
For the first case, notice that $16(ab+cd)=(3p^2+n^2)(3m-q)(m+q)$, and it's easy to handle it now.
The second case is analogous.

this is the solution i found.it is in the same spirit as the above one but a bit different.let$a=x+y+z+d ,b=x+y+d,c=d+x$.using this it is easy to see that the given relation is equivalent with $xy=(z-d)(2d+2y+x+z)$and $ab+cd=(2d+2y+x+z)(d+x)+y(y+z)$.but it is clear that $gcd(2d+2y+x+z,y)=t>1$and it is also clear that $t|ab+cd$so our number is not prime and the problem is over. :)

I'm aware that there exists a solution using the Eisenstein integers. Could someone post it?

a>b>c>d>0
  ac+bd=(b+d+a-c)(b+d+c-a)  (*)
suppose that ab+cd -prime
 (*) hence  (b+d+a-c)l(ac+bd)   we  have
 (b+d+a-c) l (a)(b+d+a-c)+(ac+bd)=ab+ad+a^2+bd=(a+b)(a+d)    (**)

 (*) hence  (b+d+c-a) l (ac+bd)     we have
 (b+d+c-a) l (b+d+c-a)(c)+(ac+bd)=cb+cd+c^2+bd=(c+b)(c+d)     (***)
(**) (***) hence (b+d+c-a)(b+d+a-c) l [(a+b)(c+d)][(a+d)(b+c)] at this we have
 (ac+bd) l ((ac+bd)+(ad+bc))((ac+bd)+(ab+cd))   (****)

(a-d)(b-c)>0  therefore ab+cd>ac+bd and (ab+cd)-is prime
hence   ((ac+bd);((ac+bd)+(ad+bc)))=1 (*****)

(****) and (*****) we can say (ac+bd) l (ac+bd)+(ad+bc) at this we have (ac+bd) l (ad+bc)
 (ac+bd)<=(ad+bc)   it is equalent  (a-b)(d-c)>0 contradiction!
      ab+cd   -  is not prime

I was trying the problem in a different approach but could not complete it. Here is my mode of approach. :maybe:

Given that a>b>c>d and (ac+bd)=(b+d+a-c)(b+d-a+c)
or, ac+bd=(b+d)^2-(a-c)^2
or, ac+bd=b^2+d^2+2bd-a^2-c^2+2ac
or, ac+bd=a^2+c^2-b^2-d^2=(a+b)(a-b)+(c+d)(c-d)=(a+d)(a-d)-(b+c)(b-c)

Now, I was trying to use the identity, (ab+cd)+(ac+bd)=(b+c)(a+d)
and (ab+cd)-(ac+bd)=(a-d)(b-c)
which gives 2[(a+b)(a-b)+(c+d)(c-d)]=(b+c)(a+d)+(a-d)(b-c)........

weird solution: :maybe:
we have:

$ ac+bd=(b+d+a-c)(b+d-a+c)\iff\boxed{a^2-ac+c^2=b^2+bd+d^2}$  (*)

now let $ ABCD$ be a quadrilateral with $ AB=a,BC=d,CD=b,AD=c$ and $ \angle BAD=60^\circ$ and $ \angle BCD=120^\circ$.now note that according to (*) and $ \cos$ law there exists such quadrilateral because each side of (*) is equal to $ BD^2$.

now let $ \angle ABC=\alpha$ hence $ \angle CDA=180^\circ-\alpha$.now according to $ \cos$ law in $ \triangle ABC$ and $ \triangle ACD$ we have:

$ a^2+d^2-2ad\cos\alpha=AC^2=b^2+c^2+2bc\cos\alpha$

hence:

$ 2\cos\alpha=\frac{a^2+d^2-b^2-c^2}{ad+bc}$

\begin{eqnarray*}\Rightarrow AC^2=a^2+d^2-ad.\frac{a^2+d^2-b^2-c^2}{ad+bc}=\frac{(ab+cd)(ac+bd)}{ad+bc}\end{eqnarray*}

now note that $ ABCD$ is cyclic,hence according to ptolemy's theorem we get that:

$ (AC.BD)^2=(ab+cd)^2$

hence:

$ (ac+bd)(a^2-ac+c^2)=(ab+cd)(ad+bc)$ (**)

now note that:

$ ab+cd>ac+bd>ad+bc$ (***)

(the first one follows from $ (a-d)(b-c)>0$ and the second one from $ (a-b)(c-d)$)

now assume that $ ab+cd$ is a prime number,first of all from (***) we get that $ ab+cd$ and $ ac+bd$ are relatively prime.so by (**) $ ac+bd$ must devide $ ad+bc$ which is a contradiction according to (***)...

Here is my solution.

Let $ P(x)\equal{}(x\plus{}b)(x\plus{}d)\minus{}(x\minus{}a)(x\plus{}c)$
$ \textcolor{white}{Let P(x)}\equal{}(b\plus{}d\plus{}a\minus{}c)x\plus{}ac\plus{}bd$
$ \textcolor{white}{Let P(x)}\equal{}(b\plus{}d\plus{}a\minus{}c)(x\plus{}b\plus{}d\minus{}a\plus{}c)$.
$ P(a)\equal{}(a\plus{}b)(a\plus{}d)\equal{}(b\plus{}d\plus{}a\minus{}c)(b\plus{}c\plus{}d)$.
$ P(a)\neq 0$, so $ b\plus{}d\plus{}a\minus{}c \mid (a\plus{}b)(a\plus{}d)$
but $ a\plus{}b>b\plus{}d\plus{}a\minus{}c>\dfrac{a\plus{}b}{2}$, therefore a part of $ b\plus{}d\plus{}a\minus{}c$ must divide $ a\plus{}d$.
Thus, we have $ K\equal{}(b\plus{}d\plus{}a\minus{}c,a\plus{}d)\equal{}(b\minus{}c,a\plus{}d)\equal{}(b\minus{}c,b\plus{}d\plus{}a\minus{}c)\neq 1$.
Since $ K\mid a\plus{}d$, we have $ K\le a\plus{}d<ab\plus{}cd$.

$ (b\minus{}c)(a\minus{}d)\equal{}ab\plus{}cd\minus{}ac\minus{}bd$.
Using $ ac\plus{}bd\equal{}(b\plus{}d\plus{}a\minus{}c)(b\plus{}d\minus{}a\plus{}c)$ we have
$ ab\plus{}cd\equal{}(b\minus{}c)(a\minus{}d)\plus{}(b\plus{}d\plus{}a\minus{}c)(b\plus{}d\minus{}a\plus{}c)$.

So $ K\mid ab\plus{}cd$ with $ 1<K<ab\plus{}cd$, which implies that $ ab\plus{}cd$ is composite.

This is similar to [b]BaBaK Ghalebi[/b]'s solution, but it differs a bit at the end:
[hide="Solution"]
Assume that $ ab \plus{} cd$ is prime.  Upon expanding, this becomes $ b^2 \plus{} bd \plus{} d^2 \equal{} a^2 \plus{} c^2 \minus{} ac$.  Then, we can find $ x$ so that $ x^2 \equal{} a^2 \plus{} c^2 \minus{} ac$ since $ a^2 \plus{} c^2 \minus{} ac > 0$ with $ a, b, c, d > 0$.  This means that $ \frac {a^2 \plus{} c^2 \minus{} x^2}{2ac} \equal{} \frac {1}{2} \equal{} \cos 60$.  Similarly, $ \frac {b^2 \plus{} d^2 \minus{} x^2}{2bd} \equal{} \frac { \minus{} 1}{2} \equal{} \cos 120$.  This means that we can consider cyclic quadrilateral $ ABCD$ with $ AB \equal{} a$, $ BC \equal{} c$, $ CD \equal{} b$, and $ AD \equal{} d$ with $ \angle ABC \equal{} 60$ and $ \angle CDA \equal{} 120$.  Ptolemy's Theorem here gives that $ ab \plus{} cd \equal{} AC\cdot BC$.  Notice that $ BD \equal{} x \equal{} \sqrt {(b \plus{} d \minus{} a \plus{} c)(b \plus{} d \plus{} a \minus{} c)}$.  Also, $ AC > AD,CD$ by the angle condition.  Furthermore, $ BD$ is greater than both of $ CB$ and $ DC$ or $ AD$ and $ AB$ since either $ \angle BAC$ or $ \angle BDC$ is right or obtuse.  This means that $ AC, BD > 1$.  Notice that $ AC$ is the squareroot of a natural number and since $ ab \plus{} cd \equal{} AC\cdot BD$ is a natural number, we see that $ BD$ is in the same form as $ AC$.  Yet, if $ x^2$ composite, then $ BD^2\cdot AC^2$ must be composite, so $ BD\cdot AC$ is composite.  Then, $ x^2$ is prime.  So must be $ BD^2$.  If they are different primes, then $ BD\cdot AC$ is not an integer.  Then, $ BD \equal{} AC$, so $ ABCD$ is an isoceles trapezoid.  This means that either $ a \equal{} b$ or $ c \equal{} d$, which contradicts the problem statement.  Thus, we have a contradiction, so $ ab \plus{} cd$ is composite, as desired.

[b]Remark:[/b] I believe that this is quite an interesting problem!  I rarely see such geometry combined with number theory  :)  By the way, just to make the motivation for the geometric construction clear, it is not only the observation of the Law of Cosines that should make one think of geometry (that is rarely enough.)  It is also the "Ptolemyness" of the problem; furthermore, notice that $ b \plus{} d \plus{} c \minus{} a \equal{} 2(s \minus{} a)$ when $ s$ is the semiperimeter of $ ABCD$.  This seems to represent Bhramagupta's Theorem in a way, since\[ \frac {\sqrt {3}}{2}\cdot 2(s \minus{} a)(s \minus{} c) \equal{} \frac {\sqrt {3}}{2}\cdot \frac {(ac \plus{} bd)}{2} \equal{} \frac {ac\sin 60}{2} \plus{} \frac {bd\sin 120}{2} \equal{} [ABCD] \equal{} \sqrt {(s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d)}\]This yields that $ 3(s \minus{} a)(s \minus{} c) \equal{} (s \minus{} c)(s \minus{} d)$, which is quite nice (and quite close to what [b]Iura[/b] got.  I am not sure if a proof with the Four Number Theorem follows from there; it seems quite possible...) [/hide]

EDIT: Sorry, this solution is incorrect.  $ BD$ can be in the form of $ \frac {q^2}{p}$ for some $ p$, $ q$ primes.

EDIT 2: I hope this works:
[hide="Solution"]
From the remark before, we have that $ 3(a\plus{}b\plus{}d\minus{}c)(b\plus{}d\plus{}c\minus{}a)\equal{}(a\plus{}b\plus{}c\minus{}d)(a\plus{}c\plus{}d\minus{}b)$.  Then, for some $ w$, $ x$, $ y$, $ z$, $ a\plus{}b\plus{}c\minus{}d\equal{}3xy$, $ a\plus{}c\plus{}d\minus{}b\equal{}wz$, $ a\plus{}b\plus{}d\minus{}c\equal{}wx$, and $ b\plus{}d\plus{}c\minus{}a\equal{}yz$ (the roles of $ a\plus{}b\plus{}c\minus{}d$ and $ a\plus{}c\plus{}d\minus{}b$ can be switched.)  This gives us that $ \frac{3xy\plus{}wz\plus{}xw\plus{}yz}{2}\equal{}a\plus{}b\plus{}c\plus{}d$.  Then, we can solve for $ a$, $ b$, $ c$, and $ d$ and get that $ a\equal{}\frac{3xy\plus{}wz\plus{}xw\minus{}yz}{4}$, $ b\equal{}\frac{3xy\minus{}wz\plus{}xw\plus{}yz}{4}$, $ c\equal{}\frac{3xy\plus{}wz\minus{}xw\plus{}yz}{4}$, $ d\equal{}\frac{wz\plus{}xw\plus{}yz\minus{}3xy}{4}$.  Then, \[ ab\plus{}cd\equal{}\frac{9x^2y^2\plus{}w^2x^2\plus{}6x^2yw\minus{}y^2z^2\minus{}w^2z^2\plus{}2yz^2w\plus{}w^2z^2\plus{}y^2z^2\plus{}2yz^2w\minus{}9x^2y^2\minus{}w^2x^2\plus{}6x^2yw}{16}\] Yet, this means that $ ab\plus{}cd\equal{}\frac{yw(3x^2\plus{}z^2)}{4}$, which is composite, as desired. [/hide]

[quote="ufuk"]I think it is an easy problem for imo6
I think it is good for imo1 :lol:[/quote]

agreed. The BaBak's solution is quite predictable, upon expanding, the next step gets pretty clear, though that's too tough to be given as imo1.

[quote="orl"]Let $ a > b > c > d$ be positive integers and suppose that
\[ ac \plus{} bd \equal{} (b \plus{} d \plus{} a \minus{} c)(b \plus{} d \minus{} a \plus{} c).
\]
Prove that $ ab \plus{} cd$ is not prime.[/quote]
$ ac \plus{} bd \equal{} (b \plus{} d)^2 \minus{} (a \minus{} c)^2 \iff$
$ a^2 \minus{} ac \plus{} c^2 \equal{} b^2 \plus{} bd \plus{} d^2$
Let $ \omega \equal{} e^{i\frac {2\pi}{3}}$. Hence $ \omega^3 \equal{} 1, \omega^2 \plus{} \omega \plus{} 1 \equal{} 0$. Let's work in the eisenstein integers, ie. $ \mathbb{Z}[\omega]$.
$ (a \plus{} \omega c)(a \plus{} \omega^2 c) \equal{} (b \minus{} \omega d)(b \minus{} \omega^2 d)$
$ (a \plus{} \omega c)\overline{(a \plus{} \omega c)} \equal{} (b \minus{} \omega d)\overline{(b \minus{} \omega d)} \iff$
$ x \equal{} a \plus{} \omega c, y \equal{} c \plus{} \omega a, z \equal{} b \minus{} \omega d, t \equal{} d \minus{} \omega b$
Let $ x \equal{} a \plus{} \omega c, \overline{x} \equal{} a \plus{} \ c \plus{} \omega a, y \equal{} b \minus{} \omega d, \overline{y} \equal{} b \minus{} \omega^2 d$
Then $ x \overline{x} \equal{} y \overline{y}$
Let $ n \equal{} \gcd(x,y)$ and let $ m \equal{} \frac {x}{n}$, $ n,m \in \mathbb{Z}$. Obviously $ m \mid \overline{y}$ and hence $ \gcd(x,\overline{y}) \equal{} m$. Then $ x \equal{} nm, y \equal{} n\overline{m}$.

$ n \equal{} p \plus{} \omega q, m \equal{} r \plus{} \omega s, p,q,r,s \in \mathbb{Z}$

$ x \equal{} a \plus{} \omega c \equal{} (p \plus{} \omega q)(r \plus{} \omega s) \equal{} pr \minus{} qs \plus{} \omega(qr \plus{} ps \minus{} qs)$

Hence $ a \equal{} pr \minus{} qs$ and $ c \equal{} qr \plus{} ps \minus{} qs$.

$ y \equal{} b \minus{} \omega d \equal{} (p \plus{} \omega q)(r \plus{} \omega^2 s) \equal{} pr \plus{} qs \minus{} ps \plus{} \omega (qr \minus{} ps)$

Hence $ b \equal{} pr \plus{} qs \minus{} ps$ and $ d \equal{} ps \minus{} qr$

Now $ ab \plus{} cd \equal{} (pr \minus{} qs)(pr \plus{} qs \minus{} ps) \plus{} (qr \plus{} ps \minus{} qs)(ps \minus{} qr)$

Expanding $ ab \plus{} cd \equal{} p^2r^2 \plus{} pqrs \minus{} p^2rs \minus{} pqrs \minus{} q^2s^2 \plus{} s^2pq \plus{} pqrs \plus{} p^2s^2 \minus{} pqs^2 \minus{} q^2r^2 \minus{} pqrs \plus{} q^2rs$

$ ab \plus{} cd \equal{} p^2r^2 \minus{} p^2rs \minus{} q^2s^2 \plus{} p^2s^2 \minus{} q^2r^2 \plus{} q^2rs \equal{} (p^2 \minus{} q^2)(r^2 \plus{} s^2) \minus{} (p^2 \minus{} q^2)rs$

$ ab \plus{} cd \equal{} (p^2 \minus{} q^2)(r^2 \minus{} rs \plus{} s^2)$

Assume that $ ab \plus{} cd$ is not a prime. Then either $ p^2 \minus{} q^2 \equal{} \pm 1$ or $ r^2 \minus{} rs \plus{} s^2 \equal{} \pm 1$. If $ p^2 \minus{} q^2 \equal{} \pm1$ then $ p \equal{} 0,q \equal{} \pm 1$ or $ p \equal{} \pm 1, q \equal{} 0$. If $ p \equal{} 0$ then $ a \plus{} b \equal{} 0$. If $ q \equal{} 0$ then $ c \equal{} d$. So $ p,q\neq 0 \Rightarrow p^2 \minus{} q^2 \neq \pm 1$.
$ r^2 \minus{} rs \plus{} s^2 \equal{} \frac {1}{2}(r^2 \plus{} (r \minus{} s)^2 \plus{} s^2) < \minus{} 1$ so $ r^2 \minus{} rs \plus{} s^2 \equal{} \pm 1 \iff r^2 \minus{} rs \plus{} s^2 \equal{} 1$. This has the solutions $ (r,s) \in \{(0,1);(0, \minus{} 1);(1,0);( \minus{} 1,0);(1,1);( \minus{} 1, \minus{} 1)\}$
If $ r \equal{} 0$ then $ b \plus{} c \equal{} 0$. If $ s \equal{} 0$ then $ a \equal{} b$. If $ r \equal{} s$ then $ a \equal{} d$. Hence $ r^2 \minus{} rs \plus{} s^2 \neq \pm 1$. So $ ab \plus{} cd$ are a product of to integers with numerical value at least $ 2$. Hence $ ab \plus{} cd$ is composite (Therefore not prime).

[hide=solution][quote=ISL 2001 N5]Let $a > b > c > d$ be positive integers and suppose that \[ ac + bd = (b+d+a-c)(b+d-a+c).  \] Prove that $ab + cd$ is not prime.[/quote]
Bruh N5?clearly N4 is harder.
$b+d+a-c|ac+bd \implies b+d+a-c|ac+bd+ab+ad+a^2-ac=(a+b)(a+d)$
Similarly $b+d-a+c|(c+b)(c+d)$
Multiplying the equations give
\begin{align*} (b+d+a-c)(b+d-a+c)=ac+bd|(a+b)(a+d)(c+b)(c+d) \\
ac+bd|(ac+bd+ad+bc)(ac+bd+ab+cd) \\
\implies ac+bd|(ad+bc)(ab+cd) \end{align*}
Again,
$a(c-d)>b(c-d)(a>b,c>d) \implies ac+bd>ad+bc$
and $a(b-c)>d(b-c)(a>d,b>c) \implies ab+cd>ac+bd>ad+bc$.
This implies as $ac+bd \not | ad+bc \implies gcd(ac+bd,ab+cd)>1$,
hence done     $\blacksquare$[/hide]

We all love geometry

Let $p=ac+bd,q=ab+cd,r=ad+bc.$ Define cycilc quadrilateral $ABCD.$ Note that we have the following relations:

\[AC^{2}=(AB\cdot CD+BC\cdot DA){\frac {AB\cdot DA+BC\cdot CD}{AB\cdot BC+DA\cdot CD}}\]\[BD^{2}=(AB\cdot CD+BC\cdot DA){\frac {AB\cdot BC+DA\cdot CD}{AB\cdot DA+BC\cdot CD}}\]
Let $AB=a,BC=d,CD=b,DA=c.$ We get \[AC^2=(ab+cd)\frac{ac+bd}{ad+bc}=\frac{pq}{r}\]and \[BD^2=(ab+cd)\frac{ad+bc}{ac+bd}=\frac{qr}{p}\]
Note that \[ ac + bd = (b+d+a-c)(b+d-a+c).  \]simplifies to \[b^2+d^2+bd=a^2+c^2-ac\]
Observe that we can make $\angle BCD=120^\circ$ that wouldn't change our expressions for $AC$ and $BD$ at all. This implies that \[BD^2=b^2+d^2+bd=\frac{qr}{p}\]
Finally, we see that $p(b^2+bd+d^2)=qr.$ If $q$ is a prime then $p(b^2+bd+d^2)$ implies either $q|p$ or $q|(b^2+bd+d^2).$

But since $q$ is prime, $q=p$ or $q=b^2+bd+d^2.$ By rearrangement lemma $ab+cd>ac+bd$ so $q\neq p.$ Thus, $q=b^2+bd+d^2$ which implies $p=r$ which is also false by the rearrangement lemma. Thus, $q$ isn't prime.

\begin{align*} ac+bd=(b+d+a-c)(b+d-a+c) \\ \implies a^2-ac+c^2=b^2+bd+d^2 \\ \implies (ab+cd)(ad+bc)=ac(b^2+bd+d^2)+bd(a^2-ac+c^2) \\ \implies (ab+cd)(ad+bc)=(ac+bd)(a^2-ac+c^2)~~~~~(*)\end{align*}
Suppose the contrary. Then, $$a>b>c>d \implies ab+cd>ac+bd>ad+bc.$$ We have that $ac+bd$ is relatively prime to $ab+cd.$ But, $$(*)\implies ac+bd|ad+bc,$$ contradiction.

Let $ABCD$ be a quadrilateral such that $AB=a,BC=c,CD=b,DA=d$ and $\angle B=60,\angle D=120.$ Notice $ABCD$ exists as the conditions is equivalent to $$AC^2=a^2+c^2-ac=b^2+d^2+bd$$ by LoC. By Strong Ptolemy, $$(a^2+c^2-ac)(ac+bd)=(ab+cd)(ad+bc),$$ or $ac+bd\mid (ab+cd)(ad+bc).$ Evidently, $ab+cd>ac+bd>ad+bc>1.$ Hence, if $ab+cd$ is prime, then $ac+bd\mid ab+cd$ or $ab+bd\mid ad+bc.$ Both are absurd, so $ab+cd$ is not prime. $\square$

[hide=Storage]

Expand and simplify to get $a^2 + c^2 - ac = b^2 + d^2 + bd$. Constuct a cyclic quadrilateral $ABCD$ with $AB = a, BC = c, CD = b, DA = d$ and $\angle B = 30, \angle D = 150$. Then, $AC^2 = (ab + cd)\frac{ad + bc}{ac + bd}$ is an integer. However, $ab + cd < ad + bc$ and $ac + bd < ab + cd$, so $ab + cd$ can not be prime.

[/hide]

== Solution ==
Equality is equivalent to
$ a^2 - ac + c^2 = b^2 + bd + d^2  (1) $.

Let $ABCD$ be the quadrilateral with $AB = a$, $BC = d$, $CD = b$, $AD = c$, $ \angle BAD =
60^\circ$, and $ \angle BCD = 120^\circ$. Such a quadrilateral exists by $(1)$ and the Law of Cosines.

By Strong Form of Ptolemy's Theorem, we find that;

$BD^2 = \frac{(ab+cd)(ad+bc)}{ac+bd}$

and by rearrangement inequality;

$ab+cd > ac+bd > ad+bc$.

Assume $ab+cd = p$ is a prime, since $a^2 - ac + c^2 = BD^2$ is an integer $p \times \frac{ad+bc}{ac+bd}$ must be an integer but this is false since $(p,ac+bd) = 1$ and $ac+bd > ad+bc$. Thus $ab+cd$ can not be a prime.

The solutions that used Four number Lemma posted here are all incomplete(or incorrect). When u get $3(ab+cd)=F\times G$, how could u state that it is a prime?

We can rewrite the equation as
\[ac+bd=(b+d)^2-(a-c)^2\implies a^2-ac+c^2=b^2+bd+d^2.\]
Interpret this as
[quote=rewrite]
We have a cyclic quadrilateral, two of whose opposite angles are $60$ and $120$. The sides adjacent to the $60$ are $a$ and $c$, and the sides adjacent to the $120$ are $b$ and $d$. Prove that the product of the diagonals of this cyclic quadrilateral must be composite, given $a>b>c>d$ and all four are integers.
[/quote]
Let $e$ and $f$ be the length of those diagonals. (Doesn't quite matter what we denote them by.)
\[e^2=\frac{(ab+cd)(ad+bc)}{ac+bd}\text{ and } f^2=\frac{(ab+cd)(ac+bd)}{ad+bc}.\]
If $e\ne f$, then say $e^2=1$ and $f^2=p^2$. Notice that $e^2=1$ is impossible as $(ab+cd)(ad+bc)>ac+bd$ and similarly if $f^2=1$. Therefore, we must have them equal and their value $p$. But it's impossible for a square to be a prime, since it's an integer by assumption. So we are done. $\blacksquare$

Note that $b+d+a-c \mid ac+bd + a(b+d+a-c) = (a+b)(a+d)$ and similarly $b+d-a+c \mid (c+b)(c+d)$.

Let $(a+b)(a+d) = x(b+d+a-c)$ and $(c+b)(c+d) = y(b+d-a+c)$ for integers $x$ and $y$. Multiplying the two equations yields $$(a+b)(c+d)(a+d)(c+b) = xy(ac+bd)$$ $$(ac+bd+ad+bc)(ac+bd+ab+cd) = xy(ac+bd)$$ Taking modulo $ac+bd$ both sides yields $$ac+bd \mid (ad+bc)(ab+cd)$$ Since $a>b>c>d$ we have $ab+cd > ac+bd > ad+bc$ so if $ab+cd$ is prime then $ac+bd \mid ad+bc$, contradiction. Thus $ab+cd$ is not prime.

[b]CLAIM:[/b] We have \[ ac + bd | (ab+cd)(ad+bc). \]
    [u]PROOF:[/u]
              b+d+a-c | (ac+bd) + a(b+d+a-c) = (a+d)(a+b)
              b+d+a-c | (ac+bd) + c(b+d-a+c) = (c+b)(c+b)
            So,  (b+d+a-c)(b+d-a+c) | (a+d)(a+b)(c+b)(c+b).
                    ac+bd | ((ac+bd) + (ad+bc))((ac+bd)+(ab+dc))
                                             Proved....
     NOW, as a>b>c>d,
                    ad+cd > ac+bd > ad+bc  (By rearrangement inequality)
              If ab+cd is prime , then ac+bd | ad+bc which implies ac+bd <ad+bc(or equal to)
                 a contradiction....
Hence, ab+cd is composite.

Our condition can be expanded to be
\[a^2-ac+c^2 = b^2+bd+d^2,\]

so we can construct cyclic quadrilateral $PQRS$ with $\angle P = 60$, $\angle R = 120$, $PQ=a$, $QR=b$, $RS=d$, and $SP=c$. Hence Strong Ptolemy tells us
\[a^2-ac+c^2 = \frac{(ad+bc)(ab+cd)}{ac+bd} \implies (ac+bd) \mid (ad+bc)(ab+cd).\]

Our bounding condition says $ad+bc < ac+bd < ab+cd$, so if $ab+cd$ was prime, we get $ac+bd \mid ad+bc$, contradiction. $\blacksquare$

Note
$$b+d+a-c\mid ac+bd+a(b+d+a-c)=(a+d)(a+b)$$
$$b+d-a+c\mid ac+bd+c(b+d-a+c)=(c+b)(c+d)$$
so $$ac+bd\mid \big[(a+d)(c+b)\big]\big[(a+b)(c+d)\big]\Rightarrow ac+bd\mid (ab+cd)(ad+bc)$$
We can obtain $ab+cd>ac+bd>ad+bc$ by Rearrangement inequality; so if $ab+cd$ is prime then $ac+bd\mid ad+bc\Rightarrow ad+bc\geq ac+bd$, a contradiction

The expansion is equivalent to $a^2-ac+c^2 = b^2+bd+d^2$. Equivalently, $a, c, d, b$ are the side lengths of a cyclic quadrilateral with opposite angles $60^\circ$ and $120^\circ$. This implies that $b^2+bd+d^2 = \frac{(ab+cd)(ad+bc)}{ac+bd}$ is an integer.

Assume for the sake of contradiction that $ab+cd$ is prime; then as $ac+bd > ad+bc$, follows that $ab+cd \mid ac+bd$, but $ac+bd < ab+cd$, contradiction.

Note that $$ac + bd = (b+d-a+c)(b+d+a-c) \iff (a+b)(a+d) = (b+d+c)(b+d+a-c)$$ So let's use the Four Number Lemma to conclude: $$\begin{array}{c} a+b =pq \\ a+d =rs \\ b+d+c = ps \\ b+d+a-c =qr \end{array}$$ from where we get that $$\begin{array}{c} a=\frac{1}{3}(2pq+2rs-ps-qr) \\ \\ b =\frac{1}{3}(ps+qr+pq-2rs) \\ \\ c=\frac{1}{3}(ps-2qr+rs+pq) \\ \\ d=\frac{1}{3}(ps+qr+rs-2pq) \end{array}$$ So we have $$ab+cd = \frac{(2pq+2rs-ps-qr)(ps+qr+pq-2rs)+(ps-2qr+rs+pq)(ps+qr+rs-2pq)}{9} = \frac{r(2p-r)(q^2-qs+s^2)}{3}$$ and from $a>b>c>d$ we can easily deduce $p>r$ and $q>s$ so $2p-r\ge 3$ and $q^2-qs+s^2=(q-s)^2+qs \ge 3$. Lastly, we can easily check that if the equality holds for either of them then $r\neq 1$.


has anyone tried that book

is it good for the amc 10

thanks

 :starwars:  :pilot:

I bought this book and I'm not gonna lie, it was totally not worth it (especially given the cost), it didn't help much, it just had a bunch of solutions to freely existing AMC problems, it'd be easier, more tuned to you, and cheaper to simply post problems you have trouble with on the forums and then read the responses.

That being said, if the money is of little concern to you, and if you absolutely need to be able to work on your AMC game without having a computer near you, you should strongly consider buying it.

so you are saying it just has problems and stuff

but does it have strategies

would aops book 1 or 2 be better?

thanks


How many ways can the positive integer $ n$ can be written as an ordered sum of at least one positive integer? For example,

\[ 4\equal{}1\plus{}3\equal{}3\plus{}1\equal{}2\plus{}2\equal{}1\plus{}1\plus{}2\equal{}1\plus{}2\plus{}1\equal{}2\plus{}1\plus{}1\equal{}1\plus{}1\plus{}1\plus{}1
\]

so when $ n\equal{}4$, there are eight such ordered partitions.

[quote="#H34N1"]How many ways can the positive integer $ n$ can be written as an ordered sum of at least one positive integer? For example,
\[ 4 \equal{} 1 \plus{} 3 \equal{} 3 \plus{} 1 \equal{} 2 \plus{} 2 \equal{} 1 \plus{} 1 \plus{} 2 \equal{} 1 \plus{} 2 \plus{} 1 \equal{} 2 \plus{} 1 \plus{} 1 \equal{} 1 \plus{} 1 \plus{} 1 \plus{} 1
\]
so when $ n \equal{} 4$, there are eight such ordered partitions.[/quote]
[hide="Solution"]
Consider a string of $ n$ ones with small gaps between them.  Between each gap, we will choose to or to not put a bar between the two adjacent numbers.  If we put a bar in front of the $ i$th one, for example, and then the next bar is in front of the $ i\plus{}j\minus{}1$th, then the number corresponding to these bars is $ j$.  Here is an example for $ n\equal{}8$: $ 1, 1, 1, | 1, 1, |1,| 1, 1$ turns into $ 3\plus{}2\plus{}1\plus{}2$.  As there are $ n\minus{}1$ gaps and $ 2$ possibilities per gap, there are $ 2^{n\minus{}1}$ such partitions. [/hide]

well, for example when you evaluate for n=5, you have the equations
a=5,a+b=5,a+b+c=5,a+b+c+d=5,a+b+C+d+e=5 all as possible cases, with a,b,c,d,e>1
the first equation has 4c4=1 solution, and the second equation becomes a+b=3 with a,b>0, so by balls and urns this becomes4c1 solutions, and then 4c2 and 4c3, and 4c4. in general, you get a sum of nCk's where k ranges from 1 to n, so you could probably turn this into a proof.

i propose that the number of ordered partitions of a number n is 2^[n-1]

edit:beaten

It's easy. $ f(n)\equal{}f(n\minus{}1)\plus{}f(n\minus{}2)\plus{}...\plus{}f(1)\plus{}1$ We can get this by considering the first number in the ordered pair. The last 1 is for when the first number = n. Then do strong induction.

The identity $ \sum_{k \equal{} 0}^{n\minus{}1} \binom{n\minus{}1}{k} \equal{} 2^{n \minus{} 1}$ makes [b]mathking123[/b]'s solution incorrect.

Using that identity correctly with this problem yields what [b]The QuattoMaster 6000[/b] said.

EDIT Thanks for the clarification, [b]t0rajir0u[/b].

The identity you want is $ \sum_{k\equal{}0}^{n\minus{}1} {n\minus{}1 \choose k} \equal{} 2^{n\minus{}1}$, and the combinatorial interpretation of this identity is very familiar.  (You may have gotten confused with the identity from [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]this post[/url].)

Makes much more sense now.  :) Thanks and edited.


the vertices of triangle $ ABC$  are lattice points.two of its sides have lengths which belong to set$ /{{sqr root 17,sqr root 1999,sqr root 2000}}/.$what is the maximum possible area of $ ABC$.....
facing difficulty to write second bracket :(

can anyone help....i am facing with difficulties to find the lattice points....help :maybe:


a+b+c+d+e+f+g = 1 and a, b, c, d, e, f, g are non-negative. Find the minimum value of max(a+b+c, b+c+d, c+d+e, d+e+f, e+f+g).

We have
\begin{eqnarray*}
\lefteqn{\max(a + b + c, b + c + d, c + d + e, d + e + f, e + f + g)} \\
  & \geq & \max(a + b + c, c + d + e, e + f + g) \\
  & \geq & \frac{(a + b + c) + (c + d + e) + (e + f + g)}{3} \\
  & = & \frac{a + b + c + d + e + f + g}{3} + \frac{c + e}{3} \\
  & = & \frac{1}{3} + \frac{c + e}{3} \\
  & \geq & \frac{1}{3}.
\end{eqnarray*}
Thus the minimum of the expression is at least $\frac{1}{3}$.

To see that the minimum is equal to $\frac{1}{3}$, consider the following assignment of variables: $a = d = g = \frac{1}{3}$, and the rest of the variables are 0.


Hello,

I want to use the \chapter level so that it displays, for example, 4.2.3.  But it is not really appropriate for my paper to display "Chapter 4".  How do I get rid of that line?

Thank you.

That doesn't make much sense since chapter is the highest level in the book and report classes whereas 4.2.3 would be the number for a subsection ie chapter 4, section 2, subsection 3, so you how can a chapter be numbered 4.2.3?.
You don't have to use \chapter if you don't need it but you can redefine it so the name chapter doesn't appear
[code]\renewcommand{\chaptername}{}[/code] You can also redefine the chapter counter eg
[code]\renewcommand{\thechapter}{\arabic{chapter}.\arabic{section}.\arabic{subsection}}[/code]but, as I say, this doesn't make sense.

I think the idea is to not have \chapter make "Chapter x" appear, but still have the automatic numbering of sections etc.
I have a similar question: How can I change the formatting (I'm in an article class) of the text that appears from \section{Blah}, and have autonumbering of theorems, etc work?

[quote="caffeineboy"]I have a similar question: How can I change the formatting (I'm in an article class) of the text that appears from \section{Blah}[/quote]Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=titlesec]titlesec package[/url]
[quote="caffeineboy"]and have autonumbering of theorems, etc work?[/quote]Theorems are automatically numbered by default. So
[code]\newtheorem{theorem}{Theorem}
...
\begin{theorem}
This is a theorem
\end{theorem}[/code]gives
[b]Theorem 1.[/b] [i]This is a theorem[/i]
See Section 3.7 of [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to LATEX2e[/url].

Sorry, I left out that I wanted the theorems to be numbered including the section counter, ie \newtheorem{thm}{Theorem}[section], but titlesec handles this fine, thanks.

Sorry if I didn't explain myself well.  As Caffeine said, "...the idea is to not have \chapter make "Chapter x" appear, but still have the automatic numbering of sections etc."

Would I still use \renewcommand{\chaptername}{} ?

If you want chapter numbering but no chapter heading then don't put one in with \chapter{...}. Instead use \stepcounter{chapter} to increment the chapter counter. But it all seems very strange to have chapters with no titles - why bother with chapters at all?

Thinking about it, what might make sense is to have chapter titles but without the word "Chapter" or the numbering that follows. That can be done with
[code]\renewcommand{\chaptername}{}
\renewcommand{\thechapter}{}
\renewcommand{\thesection}{\arabic{chapter}.\arabic{section}}[/code]The first two lines get rid of the chapter name and number and the third line puts the chapter back into the section numbering.

Yeah that last one is what I was looking for.  Thanks.