induction
modular arithmetic
algebra unsolved
algebra

It is obviosly $f(1)=1, f(\frac{1}{2})=\frac{2}{3}$.
By induction $f(x)=\frac{1}{1+x}, \ \forall x\in N$. It is easy to chek, that solution is unique $f(x)=\frac{1}{1+x}$.

[quote="Rust"]By induction $f(x)=\frac{1}{1+x}, \ \forall x\in N$. It is easy to chek, that solution is unique $f(x)=\frac{1}{1+x}$.[/quote]
Easy? Very long !

[b]Rust[/b]'s result is right, but the proof is really tricky.

Take $g(x): =(1+x)f(x)-1$ for all $x\in\mathbb{Q}_{+}$.
Then $g: \mathbb{Q}_{+}\to\mathbb{R}$ satisfies

1) $g(x)=-xg({1\over x})$ for all $x\in\mathbb{Q}_{+}$
2) $g(x)=g(1+2x)$ for all $x\in\mathbb{Q}_{+}$.

Assume $g\not\equiv 0$.
Then take $m\in\mathbb{N}$ minimal with $g({a\over b})\not= 0$ and $a+b=m$ for some $a,b\in\mathbb{N}$.
For $a,b\in\mathbb{N}$ with $g({a\over b})\not= 0$ and $a+b=m$ we have obviously $\gcd (a,b)=1$ and $a$ and $b$ can't be both even.
If $a$ and $b$ are both odd, then because of 1) we can assume $a\geq b$.
If $a=b$, then 1) gives $g({a\over b})=g(1)=-g(1)$ and $g({a\over b})=0$. Contradiction!
So $a>b$ and for $c: ={a-b\over 2}\in\mathbb{N}$ we have $1+2{c\over b}={a\over b}$ and $c+b<a+b=m$ and 2) gives $g({c\over b})=g({a\over b})\not =0$. Contradiction!

So $a\not \equiv b\pmod{2}$, and because of 1) there exist $a,b\in\mathbb{N}$ with $g({a\over b})\not =0$ and $a+b=m$ and $2\nmid a,\ 2\mid b$.
Now take such $a,b$ with $|g({a\over b})|$ minimal. (there are only finitely many such $a,b$, as $a+b=m$)
Write $b=2^{n}c$ with $c\in\mathbb{N}$ odd and $n\in\mathbb{N}$.
Then with $h: \mathbb{Q}_{+}\to\mathbb{Q}_{+},x\mapsto 1+2x$ we have $\underbrace{h(h(\cdots h({a\over b})\cdots))}_{n\text{ times}}={(2^{n}-1)c+a\over c}$
with $d: =(2^{n}-1)c+a\in\mathbb{N}$ even, as $a,c$ are odd and $n\geq 1$, and $c+d=a+b=m$ and $d>c$.
So 2) and 1) give $|g({a\over b})|=|g({d\over c})|={d\over c}|g({c\over d})|>|g({c\over d})|$. Contradiction!

So $g\equiv 0$ and $f(x)={1\over 1+x}$ for all $x\in\mathbb{Q}_{+}$.


algebra
polynomial
Vieta
IMC
college contests

[quote="Frozen"]Show that there are an infinity of integer numbers $m,n$, with $gcd(m,n)=1$ such that the equation $(x+m)^{3}=nx$ has 3 different integer sollutions.[/quote]

How easy problem 5 at IMC can be?!
[hide="Solution"]For every integer $a\ge1$ take $n=(3a^{2}+3a+1)^{3}, m=a(a+1)(2a+1).$ Then $gcd(m,n)=1$ and the equation $(x+m)^{3}=nx$ has 3 different integer solutions
$x=a(3a^{2}+3a+1)-m,$ $x=(a+1)(3a^{2}+3a+1)-m,$ $x=(-2a-1)(3a^{2}+3a+1)-m.$[/hide]

the official solution can be simplified, by putting n=N^3 and x=y^3.
one gets the equation y^3-Ny+m=0.
now you proceed with vieta formulae, but you immediatly get N=u^2+uv+v^2 and m=uv(u+v).

I had to post this because I had that substitution and the vieta formulae in the competition but didn't proceed... feeling so stupid..

I was just playing around with it and at the very start i said, suppose 1 is a root $\leftrightarrow (m+1)^{3}=n$ Notice $n$ and $m$ are relatively prime. So re-write it as:

$(x+m)^{3}=(m+1)^{3}x$ so x is a cube. write $y=\sqrt[3]x$

Subbing in and factoring out $y-1$ since 1 is still a root,

$(y-1)(y^{2}+y-m)$ Its then easy to pick $m$ so that the roots of $y^{2}+y-m$ are interger roots different from 1 and not equal.
If $a$ and $b$ are these roots, $a^{3}$ and $b^{3}$ are roots of the original polynomial. That took seven minutes in the exam...

Assume that $\alpha,\beta,\gamma \in \mathbb{Q}$ and $\alpha + \beta + \gamma = 3, \alpha \beta \gamma = 1$. Let $t \in \mathbb{N}$ the least positive integer s.t. $a=t\alpha,b=t\beta,c=t\gamma \in \mathbb{Z}$. Then $(a,b,c) = 1$. With $m = -t$, $n = 3m^2-(ab+bc+ca)$ we have $(m,n) = 1$ and $a,b,c$ are solutions to $(x+m)^3 = nx$.

First see that $\alpha + \beta + \gamma = 3, \alpha \beta \gamma = 1$, iff $\alpha+\beta = 3-\gamma, \alpha\beta = \frac{1}{\gamma}$. Hence if $\alpha,\beta \in \mathbb{Q}$ iff $\sqrt{(3-\gamma)^2-4\frac{1}{\gamma}} = |\gamma-1|\sqrt{1-\frac{4}{\gamma}} \in \mathbb{Q}$. Hence $\alpha,\beta,\gamma \in \mathbb{Q}$ if and only if:
\[ 1-\frac{4}{\gamma} = z^2, z \in \mathbb{Q} \]
I.e. $\gamma = \frac{4}{1-z^2}$. Furthermore $\alpha,\beta$ are different if $z \neq 0$. Hence if $\gamma = \frac{4}{1-z^2}, z \in \mathbb{Q}$, there exists $\alpha,\beta \in \mathbb{Q}$ s.t. $\alpha + \beta + \gamma = 3, \alpha \beta \gamma = 1$, which gives a $m,n$ with $(m,n)=1$ s.t. $(x+m)^3=nx$ has three solution. In this way we can create infinitly many solutions $(m,n)$.

put $m=t(1-t)$ and $n=(t^{2}-t+1)^{3}$ where t is an integer,and then you see:
the roots of the equation will be $-1$,$t^{3}$,$(1-t)^{3}$.


probability
combinatorics proposed
combinatorics

Given a square board of $n$ cells labeled from $1$ to$n$,there is one number in each cell.Some operations are conducted as  follows:delete any two numbers in two orbitrary cells,then insert their diffirence into one of the original cells.After a certain number od such operations there ramain only some 0s on the bosrd.How must $n$ be?

[quote="probability1.01"]Isn't it quite easy? It is possible if and only if the sum of the first n positive integers is even.[/quote]
Why?I need a full solution!

Do your own homework.  (And yes, probability, it is easy, and that is right.  Both the necessity and the sufficiency are straightforward.  I suppose you already knew this, though.)


Let $ a,n$ are two integer numbers,$ a,n>1$ so that $ n|a^{n}\minus{}1$.Prove that :$ \gcd(a\minus{}1,n)>1$

[hide="hint"]consider the smallest prime divisor of n. :wink: [/hide]

This is some problem
a)Prove that $ a>2$
For $ a>2$ prove there exist infinite n satisfy.
Quite easy.


Name the x-intercept and y-intercept for the graphs of each of the equations below

y=log base 17 (x)

2nd one

y=17/\x

[hide]

did you mean x^17 for the second one?

1. no y intercept, the y axis is an asmyptote
 the x intercept is at 1,0

2.no x intercept, the x axis is an asmyptote
 the y intercept is at 0,1

[/hide]

oops, that's what I meant

by the way, e^x and ln(x) are the inverse of each other. So the x intercept of one is the y intercept of the other and thigns like that.

Can so one else check that...is that right what he or she said

What he said is right.  The y-axis is an asymptote for $y=\log_{17}{x}$ because $17$ to any power will never be 0 or negative, and when $17^0=1$, so it has an x-intercept of $(1,0)$.

The second one is just the first one reflected over $y=x$.  Because $17^x$ cannot be 0 or negative, the x-axis is an intercept and because $17^0=1$, the y-intercept is at $(0,1)$.


Hi everybody.My name is Viet Anh.I'm from Viet Nam. Nice to meet everybody. :)

you should post this with a little bit more about urself here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15747

but in the mean time, welcome to AOPS.

[quote="gr8sk8r"]:welcomeani: i'm fairly new here too. started last week[/quote]
u started last wk? wow you posted over 100 messages in a wk :?  :D
i joined last yr. and only have like a 100 something

[quote="juicybooty911"]
i joined last yr. and only have like a 100 something[/quote]

If you only had 100 at the time when you posted, then you made about 100 in about 2 days. :D

i joined about 2 months ago, i have 861 posts.

I joined June 5. I only have like 24 posts or something.

sorry, i can't remember how many i had when i was posting, then i went back to check, i found out i was actually almost to 200 ;)

Well , I guess the number of posts doesn't actually count.It's the number of sensible posts..
Anyway , vietanh  :welcomeani:

wow, some people post a lot. i've been posting for around 6 months and only have like 335-ish. but my posts are usually "good" and not something like "me too" and "i got that answer as well"

LoL why don't we all spam the forums? Oh woops you guys already spammed it. I don't really care about post counts. :spam:  :welcomeani: Welcome!!

Welcome.

And just to let you know there's a forum on here that's in Vietnamese too at  [url=http://artofproblemsolving.com/Forum/index.php?f=256&sid=8d4528751ccfc2ffa6cf7b758dc9af23]artofproblemsolving.com/Forum/index.php?f=256&sid=8d4528751ccfc2ffa6cf7b758dc9af23[/url]


The graph of the equation $ x \plus{} 2y \plus{} 3 \equal{} 0$ is perpendicular to the graph of the equation $ ax \plus{} 2y \plus{} 3 \equal{} 0$. What is the value of $ a$?

2y=-x-3

y=-1/2x-3/2

The slope of our first line is -1/2, so the slope of the 2nd line is 2.

2y+3=-ax

y=-ax/2-3/2

-a/2=2

-a=4

a=-4


is anyone following hc verma for iit

can you please post the solution for hc verma vol 2 ,capacitors 17th problem

You can treat the whole assembly as a parallel combination of three capacitors each of plate area $ A/3$ and of distance between plates $ d$,$ d+b$ and $ d+2b$ respectively.

The effective cappacitance is then simply the sum of these three capacitances

$ C_{eff}= \frac{eA}{3d}+\frac{eA}{3(d+b)}+\frac{eA}{3(d+2b)}$

which then gives the required answer..
PS-here e stands for epsilon

can u explain why i should be treating them as three capacitors in parallell INin particular how does one know how charge dstributes in oppsoite faces


How do I change the time? Is it under my profile, or is it controlled by the administrators?

Go to preferences / internationalization and set the summer time off.


modular arithmetic
least common multiple
number theory unsolved
number theory

Let $ n_1, n_2, ... , n_k$ be positive integer numbers. Prove that if
$ n_2 | 2^{n_1} \minus{} 1, n_3 | 2^{n_2} \minus{} 1, ... , n_k | 2^{n_{k \minus{} 1}} \minus{} 1, n_1 | 2^{n_k} \minus{} 1$,

then $ n_1 \equal{} n_2 \equal{} ... \equal{} n_k \equal{} 1$.

Hint: for any $ n$ integer number larger than $ 1$, $ n$ does not divide $ 2^n \minus{} 1$.

I think I've figured out the general approach for this problem, but I can't seem to prove the core part.

[hide="what I have so far"]
Without loss of generality, assume $ n_1 > 1$.

Since $ n_1 > 1$, if $ n_k \equal{} n_1$, then by the hint, $ n_1 \not| \; 2^{n_k} \minus{} 1$.
Hence, $ n_k$ cannot equal $ n_1$.

Note if $ n_k \equal{} 1$, then
$ n_k | 2^{n_{k \minus{} 1}} \minus{} 1 \Rightarrow 1 | 2^{n_{k \minus{} 1}} \minus{} 1 \Rightarrow n_{k \minus{} 1} \equal{} 1$.

Similarly, if $ n_{k \minus{} 1} \equal{} 1,$ then $ n_{k \minus{} 2} \equal{} 1$. Repeating this with the rest of the $ n_i$ terms, we will end up with
$ n_k \equal{} 1 \Rightarrow n_{k \minus{} 1} \equal{} 1 \Rightarrow ... \Rightarrow n_2 \equal{} 1 \Rightarrow n_1 \equal{} 1$.

This contradicts our assumption that $ n_1 > 1$. Hence, $ n_k$ cannot equal $ 1$. More generally, $ n_i \neq 1$ for $ 2 \le i \le k$ since it will eventually force $ n_1$ to equal $ 1$. Thus we have established that if $ n_1 > 1$, then all the other $ n_i$ must also be greater than $ 1$.

We then assume $ n_i > 1$ for $ 1 \le i \le k$ and attempt to get a contradiction. It will then follow that all $ n_i$ must equal $ 1$, which clearly works for the problem since $ 1 | 2^1 \minus{} 1$.

Unfortunately, I couldn't figure out how to show this contradiction.  :|

[/hide]

Ah nice, that's a very nice solution.

[hide="Formal proof"]
*credit to TTsphn's solution from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=849217#849217

We are given $ n_2 | 2^{n_1} \minus{} 1, n_3 | 2^{n_2} \minus{} 1, ... , n_k | 2^{n_{k \minus{} 1}} \minus{} 1, n_1 | 2^{n_k} \minus{} 1$, which is equivalent to saying

$ 2^{n_1} \equiv 1 \pmod {n_2}$
$ 2^{n_2} \equiv 1 \pmod {n_3}$
.
.
.
$ 2^{n_{k\minus{}1}} \equiv 1 \pmod {n_k}$
$ 2^{n_k} \equiv 1 \pmod {n_1}$

Manipulate each of these congruences such that the power of $ 2$ becomes $ [n_1, n_2, ... , n_k]$ (by multiplying the congruence with itself), where $ [a,b]$ denotes the least common multiple of integers $ a$ and $ b$. Denote $ [n_1, n_2, ... , n_k]$ as $ x$.

We then get

$ 2^x \equiv 1 \pmod {n_2}$
$ 2^x \equiv 1 \pmod {n_3}$
.
.
.
$ 2^x \equiv 1 \pmod {n_k}$
$ 2^x \equiv 1 \pmod {n_1}$

Which implies
$ 2^x \equiv 1 \pmod {[n_1, n_2, ... , n_k]} \Rightarrow 2^x \equiv 1 \pmod x \Rightarrow x | 2^x \minus{} 1$.

The hint tells us that this is never true for $ x \ge 2$. Hence, $ x \equal{} 1$. This then means $ n_1, n_2, ... , n_k$ must all equal $ 1$ since $ x$ denotes $ [n_1, n_2, ... , n_k]$.
[/hide]


Approximate, using differentials, the change in volume of a box if the length increases from 5 feet to 5.05 feet, the width decreases from 4 feet to 3.97 feet, and the depth increases from 3 feet to 3.02 feet. Then, find the actual change in volume and compare the results.

Any ideas on how to solve this problem or what I should start out with?! :huh:  :(

The length, width and depth are variables; call them $ l$, $ w$, $ h$ (I'd rather not use letter $ d$ here). The volume is a function of these three variables, namely $ V(l,w,h)=lwh$. Do you know how to find the differential of a function?


Let a,b,c be the lengths of sides of a triangle. Prove that $ 2(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})\ge \frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{a}{c}\plus{}3$

We use the Ravi Substitution : Since a, b, c are the lengths of a triangle, there are positive reals $ x, y, z$ such that $ a \equal{} y\plus{}z, b \equal{} z\plus{}x, c \equal{} x\plus{}y$.
The inequality becomes:
$ \frac{x^3\plus{}y^3\plus{}z^3\plus{}x^2y\plus{}y^2z\plus{}z^2x\minus{}2(xy^2\plus{}yz^2\plus{}zx^2)}{(x\plus{}y)(y\plus{}z)(z\plus{}x)} \geq 0$
Which is obvious true by this well-known ineq $ x^3\plus{}y^3\plus{}z^3\plus{}x^2y\plus{}y^2z\plus{}z^2x \geq 2(xy^2\plus{}yz^2\plus{}zx^2)$
Our inequality is proved!!!


an awkward turtle?
ive seen a reference in silas' blog when he mentioned an arml dance but i dont get what it means
explain, someone?

To be used in an awkward situation:

First, you must acknowledge the awkwardness by doing the "awkward turtle"
[youtube]9SULsR9-lbw[/youtube]
then, you must pretend that the situation is not awkward anymore.  It works sometimes.

hey did you guys get your mathematica 6 or ti n-spires yet?
and what is our official team name? (to put in the mathematica form)
SFBA-A?

[quote="Meggy-Meg"]hey did you guys get your mathematica 6[/quote]
no

[quote="Meggy-Meg"]...or ti n-spires...[/quote]
no, see http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1144555#1144555

[quote="Meggy-Meg"]and what is our official team name?[/quote]
Yeah, sfba a/Tim Sanders

Databox says he got his Mathematica 2 days ago.

I kinda lost my sheet thing with the URL, so bleh.

I got my mathematica two days ago, and so did i_like_pie. No word on the N-Spires yet, I guess.


$ a,k$ are natural numbers such that:

$ a$, $ 6k\plus{}a$, $ 12k\plus{}a$ are primes then find $ k$

[quote="FOURRIER"]$ a,k$ are natural numbers such that:

$ a$, $ 6k \plus{} a$, $ 12k \plus{} a$ are primes then find $ k$[/quote]

(a,k)=(5,1)??

Just 1?

Then again, (5,2) works...

I think you have the wrong problem?

May be the question must be: find all possible values of k :wink:


function
algebra
polynomial
real analysis
real analysis unsolved

Looks similar to the Intermediate Value Theorem, except this asks for an interval and that theorem guarantees that all values will be "covered" within a certain interval.  Without actually doing the proof, I would image you could use that.

[quote="maky"]Let $f: \mathbb{R}\to\mathbb{R}$ be a continuous function, and $a<b$ be two points in the image of $f$ (that is, there exists $x,y$ such that $f(x)=a$ and $f(y)=b$).

Show that there is an interval $I$ such that $f(I)=[a,b]$.[/quote]
Let $x_{0}$ be such that $f(x_{0}) = a$, and assume that there exists $x_{1}> x_{0}$ s.th. $f(y) \ge b$. (The other case is similar.)

Set $x_{b}= \inf \, \{ x > x_{0}| f(x) \ge b\}$ and then $x_{a}= \sup \, \{x < x_{b}| f(x) \le a \}$. Then $f(x_{a}) = a, \, f(x_{b}) =b$, and by the Intermediate Value Theorem $[a,b] \subset f([x_{a}, \, x_{b}])$. Checking the construction of $x_{a}$ and $x_{b}$, we see that equality must hold.

If $f: I\to \mathbb{R}$ is a continous function on an interval $I$,
Then, $f[I]$ is either an interval of a single point (this case doenst happen).
Using IVT we can easily show that the interval is,
$(\inf \{f[I]\}, \sup \{f[I]\})$ (in this case infimum < supremum so its an interval).

(Note: the infinimum need note to be included in the interval).

[quote="maky"]Let $f: \mathbb{R}\to\mathbb{R}$ be a continuous function, and $a<b$ be two points in the image of $f$ (that is, there exists $x,y$ such that $f(x)=a$ and $f(y)=b$).

Show that there is an interval $I$ such that $f(I)=[a,b]$.[/quote]

By continuity $A=f^{-1}(\{a\})$ and $B=f^{-1}(\{b\})$ are closed, and both are nonempty by hypothesis.  Without loss of generality there is some $u\in A$ and some $v\in B$ such that $u<v$.

Put $s=\sup{A\cap (-\infty,v]}$ and $t=\inf{B\cap [s,v]}$.  Then $s\in A$, $t\in B$ (owing to the fact that $A$ and $B$ are closed) and $f([s,t]) = [a,b]$ (more or less by construction).

e: I composed this two hours ago but apparently forgot to hit the submit button!

[quote="§outh§tar"]sylowe theory does f[I] refer to the image of I in what you write? Your notation doesn't make sense to me.  If it is the image of I under f then the (open?) interval you're written doesn't work.[/quote]

Yeah! Analysis books write $f(I)$ for the range of the function. Algebra books like to use $f[I]$ for the range of the function. Since I like algebra more thats what I use.

I've never seen an algebra book use that notation; square brackets usually denote some sort of polynomial ring.

Anyways sylow_theory's solution doesn't really make any sense to me; it should probably be clarified.

Clarified? It's just wrong, because it misses the point. hpe's (or blahblahblah's) solution is the way to do it.

Thanks guys. I thought it was wrong too.  :maybe:


prove $ \[ \frac{1}{a+b}+\frac{1}{a+2b}+\ldots+\frac{1}{a+nb}\leq\frac{n}{\sqrt {a(a+nb)}} \] $

hello, do you mean
prove $ \frac{1}{a\plus{}b}\plus{}\frac{1}{a\plus{}2b}\plus{}\ldots\plus{}\frac{1}{a\plus{}nb}\leq\frac{n}{\sqrt {a(a\plus{}nb)}}$?
what about $ a,b$ and $ n$?
Sonnhard.


[quote="kstan013"]Sign up to be a Wisconsin member and be...special. I am thinking of holding some competitions or whatever else of whatnot in here, so send me a PM by June 10th and you will be entered automatically. If this sounds dumb just ignore it.[/quote]
that's my birthday!!!! but i'm not interested....

What? What is this? I'm a Wisconsin member.


In a segment of my biology practical, we were supposed to observe [i]Sordaria[/i] sp. under the microscope. I noticed that there were not exactly 8 ascospores in some asci. Why is this so, and could someone describe the reproduction of ascomycetes?

Also, regarding (true) bread molds, [i]Rhizopus stolonifer[/i], what is the adaptive significance of spores forming on ends of upright filaments, sporangiophores rather than close to the substrate?[/i]

[quote="rituraj007"][hide]nature is not always perfect--so there can be small variations--due to absence of wall formation.crossing over etc.....

[quote="ThetaPi"] regarding (true) bread molds, [i]Rhizopus stolonifer[/i], what is the adaptive significance of spores forming on ends of upright filaments, sporangiophores rather than close to the substrate?[/quote]
i don't kno exactly,but maybe because they can be easily taken by animals to ohter places and help in spread of pop.....[/hide][/quote]

Could you be more precise on the first question?

i think these kinds of variations can occur due to one or the other reason which can't be known precisely

u can compare it with this
if u r rearing a hen,if u get 25 eggs a month for 10 months,u can still not b sure that u'll get exactly 25 in eleventh :)


topology
real analysis
real analysis unsolved

Let $ (X,\mathfrak{B},\mu)$ be a measure space and $ \mathfrak{C}$ the $ \sigma$-algebra of locally measurable sets (i.e. intersection with any set in $ \mathfrak{B}$ with finite $ \mu$-measure is in $ \mathfrak{B}$).

Define $ \bar{\mu}$ as follows:
For $ E \in \mathfrak{C}$, $ \bar{\mu}(E)\equal{}\mu(E)$ if $ E \in \mathfrak{B}$ and $ \bar{\mu}(E)\equal{}\infty$ if $ E \notin \mathfrak{B}$.

If $ \mu$ is semifinite, define $ \underline{\mu}$ as follows:
For $ E \in \mathfrak{C}$, $ \underline{\mu}(E)\equal{}\sup_{B \in \mathfrak{B},B \subset E} \mu(B)$.

Both $ (X,\mathfrak{C},\bar{\mu})$ and $ (X,\mathfrak{C},\underline{\mu})$ are saturated measure spaces (i.e. every locally measurable set is measurable) and both measures extend $ \mu$.

Give an example to show that $ \bar{\mu}$ and $ \underline{\mu}$ may be different.

I have studied some  topics in topology so far but I’m not familiar with some terms  as “locally measurable sets , semifinite  and saturated measure spaces “ .I could not find these concepts in index of books I have studied .Would you please introduce explicit definitions or some related references  (rather on line) to me ?

These are terms from measure theory.
Let $ (X,\mathfrak{B},\mu)$ be a measure space.

A set $ E \subset X$ is locally measurable if $ E \cap B \in \mathfrak{B}$ for any $ B \in \mathfrak{B}$ with $ \mu(B)<\infty$.

$ \mu$ is semifinite if any set with infinite measure has subsets of arbitrarily large finite measure.

$ \mu$ is saturated if every locally measurable set is measurable (that is, in $ \mathfrak{B}$).

Is it easily possible to prove that C is a sigma-algebra? The union property is clear but complement property not .Although I could not provide an counterexample. Of course there is an example that justifies its correctness in some special case . Let’s see.
X=R^k   , B  is the lebesgue sigma algebra but instead of lebesgue measure ,u  is the counting measure .Subset’s of R^k containing just finite points are lebesgue measurable and are only measurable sets with finite measure u. In this case each subset of R^k has an intersection by the mentioned measurable sets which is lebesgue measurable .So C  is really all subsets of  R^k which is a sigma-algebra .In addition In this case u is not saturated because it has proved that lebesgue measurable sets are not all subsets of R^k .
Please guide me in this respect .Something else ,what do you mean “and both measures extend u” ? Is that completion or something else ?

For the complement property, say we have $ E \in \mathfrak{C}$ and $ B \in \mathfrak{B}$ with $ \mu(B)<\infty$. Then $ E^{c} \cap B\equal{}B \setminus (E \cap B) \in \mathfrak{B}$ so $ E^{c} \in \mathfrak{C}$.

By 'both measures extend $ \mu$' it is meant that when restricted to $ \mathfrak{B}$ they are the same as $ \mu$. Note that all measurable sets are locally measurable by the definition, so $ \mathfrak{B} \subset \mathfrak{C}$.

Thanks for your hint .I unfortunately have not enough time to pay to mathematics so I’m slowly thinking about this topic .I’d like to consider whether  there is a precise proof which states that u_ and u-   are really measures or not? Empty set belongs to both B and C then u(E.S) = u- (E.S)=0 < infinity ,one of conditions is satisfied .Let’s {E_i} be a family of distinct C measurable sets and  E = U E_i .There are two cases E belongs to B or E doesn’t belong to B .
(1)	E belongs to B .
If u(E)< infinity  then E_i =E intersction with E_i and belongs to B .So u-  (E) = u(E) =sigma(u(E_i)) =  sigma(u- (E_i))
 If u(E) = infinity  then there is no matter whether all E_i belongs to B or one of them not. In both case u-  (E)  =  sigma(u- (E_i)) = infinity
(2)	E doesn’t belong to B .
u-  (E) = infinity .In this case there is at least one E_i which doesn’t belong to B so :
u-  (E)  =  sigma(u- (E_i)) = infinity
Thus u-   is a measure .I hope be able to continue this answer .

Let’s consider u_ . As before  {E_i} is a family of two by distinct C measurable sets and  E = U E_i . Since u is semifinite either u_(E) be finite or infinite, there is a set Bs in sigma algebra B such that u(Bs) < infinity and Bs be a subset of E .Of course if u_(E)=0 , the empty set can be clearly such set .
E_i belongs to C and u(Bs)<infinity then(Bs intersection by E_i ) belongs to B  . {Bs int. E_i} are distinct so :
(1)     u(Bs int. E )=u(Bs) = sigma (Bs int. E_i) ( i=1 to infinity ) since Bs is a subset of E .
(2)     (Bs int. E_i) belongs to B  and it is a subset of E_i thus u(Bs int. E_i) <= u_(E_i)
(1) and (2) result :
u(Bs) <= sigma( u_ ( E_i))( i=1 to infinity )
u_ ( E) = sup u(Bs) (Bs a subset of E and Bs a member of B) <= sigma( u_ ( E_i))( i=1 to infinity )
u_ ( E) <= sigma( u_ ( E_i))( i=1 to infinity )
On the other hand  there are sets such as P_i’s  which belong to B and p_i be a subset of E_i and u(P_i) < infinity :
Clearly P= U P_i  belongs to B  and {P_i} is two by two distinct .
P= U P_i  is a subset of E = U E_i and P belongs to B so   :
u(P) <= u_(E)
u(P) = sigma (P_i) ( i=1 to infinity )
Let’s  a = 1 + 1/4 + 1/9 + 1/16 +… and e = e’*a  e>0 and a is arbitrary .
u_(E_i) - u_(P_i) <= e’/i^2
sigma( u_ ( E_i))( i=1 to infinity ) <= sigma(e’/i^2)( i=1 to infinity ) +  sigma( u_ ( P_i))( i=1 to infinity ) = e + sigma( u_ ( P_i))( i=1 to infinity )
sigma( u_ ( E_i))( i=1 to infinity ) <= e + u(P) <= e + u_(E)
Since e is arbitrary positive we result:
sigma( u_ ( E_i))( i=1 to infinity ) <=  u_(E) and regarding the former result we get to the following :
u_ ( E) = sigma( u_ ( E_i))( i=1 to infinity )
Clearly u_ ( Empty set) = 0<infinity and finally u_ is a measure .


Solve in positive integers: $ x^2\minus{}2x\minus{}2007y^2\equal{}0$.

let $ a \equal{} x \minus{} 1$ so $ a^2 \minus{} 2007y^2 \equal{} 1$ and we use pell equation since $ (226,5)$ is solution .

$ x(x\minus{}2)\equal{}223(3y)^2$
If $ x$ odd, $ gcd(x,x\minus{}2)\equal{}1$ and 223 is prime.
$ x|223 \Rightarrow x\equal{}223$ gives no solution.
$ x\minus{}2|223 \Rightarrow x\equal{}225,y\equal{}5$

If $ x$ even, $ \frac x2, \frac {x\minus{}2}{2}$ are consecutive.
$ x\equal{}446,448$ both give no solutions.

[quote="hell_ever"]$ x(x \minus{} 2) \equal{} 223(3y)^2$
If $ x$ odd, $ gcd(x,x \minus{} 2) \equal{} 1$ and 223 is prime.
$ x|223 \Rightarrow x \equal{} 223$ gives no solution.
$ x \minus{} 2|223 \Rightarrow x \equal{} 225,y \equal{} 5$[/quote]

These implications should be $ 223 | x$ and $ 223 | x\minus{}2$, respectively.  There are an infinite family of solutions as per the Pell's approach.  :)


The following one was given in Belarus in 2002. I find it very difficult, but I'm sure that there is a short and clever solution. Need help!
   Prove that an infinite number of natural numbers cannot be written as a^3+b^5+c^7+d^9+e^11, with a,b,c,d,e positive integers.

Let n > 1 be an integer.
if x = a 3 + b 5 + c 7  + d 9  + e 11 , for some positive integers a,b,c,d, and x  \leq n (1) then :

a  \leq n 1/3
b  \leq n 1/5
c  \leq n 1/7
d  \leq n 1/9
e  \leq n 1/1.
It follows that the number of integer x satisfying (1) is not greater than n (1/3 + 1/5 + 1/7 + 1/9 + 1/11) = n 3043/3465 .
It follows that the number of positive integers x  \leq n which cannot be written as the desired form is at least n - n 3043/3465 and this bound thends to +oo with n.

Pierre.

A truly wonderful solution. I tried to work modulo a prime number, but I failed. And I can see that there exists a beautiful generalization of this problem using your argument. Thank you very much.


suppose x=x0+tv and y=y0+sw are two parametric representations of the same line l in r^n
a. show that there are scalars t0 and s0 such that y0=x0+t0v and x0=y0+s0w
b. show that v and w are parallel

Moderators, please lock this topic, it's already double-posted in [url=http://www.mathlinks.ro/viewtopic.php?t=297272]here[/url], and linear algebra question that should belong to LINEAR ALGEBRA forum. What a shame, I think I just did the homework for him, there.  :(


Can anyone solve the math problem where it has
Frog1 Frog2 Frog3 __ FrogA FrogB FrogC

and you have to switch their places by only jumping forward one or jumping over each other by one like a checkers game.  You cannot move backwards.  Or, could you show me a website regarding this classic problem?

Thanks.

Switching them is not hard.  I think the meat of the problem is to do it in the least number of steps.

The game is online
http://media.putfile.com/Frog-Switch-arcade-game

i solved it

This is easy. it's already been posted I think....

[quote="Temperal"][quote="Adam4444"]When it's three frogs on each side, you can do it in 15 switches. But what if there is 100 frogs on each side, how many switches does it take then?[/quote]

There's a forumula, isn't there? It was something like n[sup]2[/sup]-1...[/quote]
Yepp, there is a formula.

[hide]For one frog on each side it takes three times, for two frogs it takes eight and for three it takes 15.

So you can make this formula out of that: $x = n(2+n)$. Therefore it takes $100(2+100) = 10200$ switches.[/hide]


geometry
circumcircle
parallelogram
ratio
geometry unsolved

[i]Let $ AC \  ; \ BD$ be two diagonals of the convex quadrilateral $ ABCD ; \angle BAD \ \equal{} \ \angle BCD \equal{} 90^{o} ;AC \cap BD \ \equal{} \ \{E\}$

   $ M ; N$ are circumcenters of  two triangles $ ABE ; CDE$ respectively

   Prove that if  $ K$ is the midpoint of $ MN$ then $ K$ is in the line $ BD$[/i]

Let $ O$ be the circumcenter of $ ABCD.$ Note that $ \angle AME \plus{} \angle AOE \equal{} 2(\angle ABD \plus{} \angle ADB) \equal{} \pi$ $ \Longrightarrow AMEO$ is cyclic, thus $ \angle MOE \equal{} \angle MAE \equal{} \angle ADB \equal{} \angle ACB \equal{} \angle NED$ $ \Longrightarrow$ $ OM \parallel AD \parallel EM.$ Mutatis mutandis, the quadrilateral $ CNOE$ is cyclic $\Longrightarrow$ $ ON\parallel BC \parallel EM.$ Therefore, $ MONE$ is a parallelogram $ \Longrightarrow$ diagonals $ EO \equiv BD$ and $ MN$ bisect each other.

[quote="nguyenvuthanhha"][i]Can you explain more how you get $ \angle MAE \ \equal{} \ \angle ADB$ :blush: [/i][/quote]
Since M is the circumcenter of $ \triangle ABE$, $ \angle AME \equal{} 2 \angle ABE$, but $ \triangle AME$ is isoceles ($ AM$ and $ ME$ are equal radii).
$ \angle MAE \equal{} \frac {1}{2}(180 \minus{} \angle AME) \equal{} \frac {1}{2}(180 \minus{} 2 \angle ABE) \equal{} 90 \minus{} \angle ABE
\newline \equal{} 90 \minus{} (90 \minus{} \angle ADB)$ (since $ \angle BAD \equal{} 90$) $ \equal{} \angle ADB$.

$ \bullet$ Let $ K,\ L$ be, the circumcenters of the triangles $ \bigtriangleup AED,\ \bigtriangleup BEC$ respectively and we denote as $ O',$ the midpoint of the diagonal $ AC,$ where $ OO'\perp AC$ as well.

The lines through $ A,\ D$ and perpendicular to $ AC,\ BD$ respectively, intersect each other at point so be it $ F.$

Similarly, the lines through $ B,\ C$ and perpendicular to $ BD,\ AC$ respectively, intersect each other at point so be it $ Z.$

$ \bullet$ We have now the configuration of the points $ F,\ O,\ Z,$ of which their orthogonal projections on the line segments $ AC,\ BD$  $ ($ the point $ O$ is coincided with its orthogonal projection on $ BD$ $ ),$ form segments with equal ratios $ ($ $ \frac {OD}{OB} = \frac {O'A}{O'C} = 1$ $ )$ and so, we conclude that these points are collinear with the same ratio as well and then, we have $ OF = OZ$.

Because of now, the midpoints $ K,\ L$ of the segments $ EF,\ EZ$ respectively, we have that The point $ P\equiv KL\cap EO$ is the midpoint of $ KL$ $ ($ from $ KL\parallel FZ$ $ ).$

$ \bullet$ Hence, from the parallelogram $ KNLM,$ we conclude that the point $ P$ is also the midpoint of $ MN,$ lying on $ BD$ and the proof is completed.

Kostas Vittas.


group theory
abstract algebra
superior algebra
superior algebra unsolved

This is an exercise in my Algebra-Book, most of the problems there are easy, so I guess that this one is, too. But I cannot solve it.

Let $G$ be a Group and $H$ be a Subgroup of $G$.
If $x, y$ are some Elements of $G$, than $\phi_y (x) := yxy^ {-1}$.
Let furthermore be $\phi_x (H)$ be the Set of all Elements which you get if you calculate $\phi_x (y)$ for every $y\in H$.

Then let $U := \bigcup_{x\in G} \phi_x (H)$.

Prove that there is at least one Element of $G$ that is not in $U$.

Yes, that's it, but I think the groups involved should be finite, which Schoppenhauer did not mention. Anyway, if that's the case, it's been discussed before:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=conjugates&t=22729[/url]

[quote="oldman"]What is the role of $U$ in this problem??? I think the problem should b to show that $U$ is a proper subset of $G$.[/quote]
Thats correct. I edited it. Sorry,

@grobber: Thanks.


Hey,

How do you get a freaking 5.0 GPA in college?

Reid W. Barton
http://web.mit.edu/rwbarton/Public/resume.pdf

Does that mean that Reid Barton had taken college courses in high school and gotten straight A's then they actually would actually have decreased his GPA? I'm using him as an example because the cocap of my math team mentioned him in a lecture and I googled the guy...

MIT grades on a 5 point scale, I believe.

Is that unfair? I mean... Suppose you are a graduate school admissions officer and you're comparing MIT students with 4.0s with other students with 4.0... They would appear more evenly matched, wouldn't they?

Actually, how does that work for high school? The highest GPA you can get at my school is like... 4.10ish and if other kids can get above that more easily, then...

GPA scales vary from high school to high school and college to college.  MIT does grade on a 5.0 scale.  Transfer credit from high school doesn't factor into your GPA.  Admissions officers are aware of variable GPA scales and they should be considering this.


[quote="Tomaths"][quote="Zamfirmihai"]$a,b,c\geq , a+b+c=4$. Prove that
$\sum_{cyc}a(a-b)(b+1)\geq0$[/quote]

err seems a bit easy $\sum_{cyc}a(a-b)(b+1)\geq 0 \iff \frac{7}{3}(\sum_{cyc}a(a-b)) \geq 0$ by chebyshev infact $\sum_{cyc}a(a-b)(b+1)\geq \frac{7}{3}(\sum_{cyc}a(a-b))$[/quote]

there


geometry
circumcircle
rectangle
cyclic quadrilateral
power of a point
radical axis
geometry proposed

Let ABCD be a cyclic quadrilateral, its opposite sides meets at E and F. From E and F draw the tangents [tex]t_1 [/tex]and [tex]t_2[/tex] to circumcircle of the quadrilateral performed by the internal bisectors of ABCD.
Prove that    [tex]EF^2=t_1^2+t_2^2[/tex]

WLOG, assuming BA and CD meet at E, DA and CB meet at F.
Let G be the intersection point between the circumcircle of triangle ADE and EF. Then ADEG is concyclic. Note that [tex]\angle ABF=\angle ADC=\angle AGE[/tex](since ABCD and ADEG are both concyclic). From this, we know that AGFB is also concyclic. This implies [tex]t_1^2=FA\cdot FD=FG\cdot FE[/tex], similarly, [tex]t_2^2=EG\cdot EF[/tex], so [tex]t_1^2+t_2^2=FG\cdot FE+EG\cdot EF=EF^2[/tex].
QED

Indeed, but the tangents from $E$ (the same holds for $F$) to the circumcircle of the quadrilateral formed by the bisectors are equal to those from $E$ to the circumcircle of $ABCD$.

Assume $E=AB\cap CD$. If $X_{ab},X_{bc}X_{cd},X_{da}$ are the intersection poitns of the bisectors from $(A,B),(B,C),(C,D),(D,A)$ respectively, then we need to show that $EX_{bc}\cdot EX_{da}=EA\cdot EB=EC\cdot ED$ (the fact that $E,X_{bc},X_{da}$ are collinear is obvious because they're on the bisector of $\angle BEC$). It's enough to show that $ABX_{bc}X_{da}$ is cyclic, and this is easy by employing an angle-chase (too tired now to go through all the details).

Here's an additional property of the configuration: the circumcenter of $ABCD$, the circumcenter of the quadrilateral formed by the bisectors and the intersection of $AC$ and $BD$ appear to be collinear, but, as I have already mentioned, I'm too tired to investigate this right now.

[quote="juancarlos"]At this moment your tangents [tex]t_1[/tex] and [tex]t_2[/tex]  from E and F are to the circumcircle of ABCD and no on the circumcircle performed by the bisectors.[/quote]

sorry, I didn't read the question carefully :blush:

Don't worry, mecrazywong, it's still part of the proof: you showed that it holds for tangents to the circumcircle, and I showed that the tangents to the circumcircle are equal to those to the circumcircle of the quadrilateral formed by the bisectors.

Denote [tex]P[/tex] the cut point of [tex]AC[/tex] and [tex]BD[/tex], [tex]O [/tex] and [tex]O_1[/tex] the centers of the circumcircle of [tex]ABCD[/tex] and the quadrilateral performed by the bisectors.Now:
[tex]O[/tex] is the orthocenter of [tex]EPF[/tex] ([tex]OP[/tex] is orthogonal to [tex]EF[/tex]).The power of [tex]E[/tex]wrt [tex]O[/tex] and [tex]O_1 = t_1^2[/tex], i.e. [tex]E[/tex] is on the radical axis of [tex]O[/tex] and [tex]O_1[/tex],the power of [tex]F[/tex] wrt [tex]O[/tex] and [tex]O_1 = t_2^2[/tex], i.e. [tex]F[/tex] is on the radical axis of [tex]O[/tex] and [tex]O_1[/tex]. Therefore [tex]EF[/tex] is radical axis of [tex]O[/tex] and [tex]O_1[/tex].
The line [tex]OP[/tex] (orthogonal to [tex]EF[/tex], radical axis) contains the centers [tex]O[/tex] and [tex]O_1[/tex].Similar reasonig for the center [tex]O_2[/tex] of the circumcircle of quadrilateral performed by the excenters of [tex]ABCD[/tex],i.e. [tex]P,O_1,O,O_2[/tex] are collinears.
Here a  conjecture from Darij Grinberg, see the Hyacinthos message 6808: [tex]O_1,O,O_2[/tex] are collinears and [tex]OO_1=OO_2[/tex].
I dont know a proof of this conjecture.

I think I can prove Darij’s conjecture, but it will be a bit hard to explain because of all the notations and it will not be complete (only the sketch of a proof).

Let $I_{xy}$ be the intersection point of the internal bisectors of $X,Y$ ($X,Y\in\{A,B,C,D\}$), $E_{xy}$ is the intersection point of the external bisectors of $X,Y$ and $E_xI_y$ is the intersection point of the external bisector of $X$ and the internal bisector of $Y$. Also, let $Q$ be the intersection of the internal bisectors of $AED$ and $AFB$ ($E=AB\cap CD,\ F=AD\cap BD$).

Step 1: Show that $E_aI_c,\ E_bI_d,\ E_cI_a,\ E_dI_b$ are the vertices of a rectangle s.t. $E_aI_c\ E_bI_d\|EQ,\ E_aI_c\ E_dI_b\|FQ$ (this is easy enough).

Step 2: Show that $E_{ab},I_{cd}$ are symmetric wrt $ E_aI_c\ E_bI_d$ and $E_{cd},I_{ab}$ are symmetric wrt $ E_dI_b\ E_cI_a$.

Step 3: Show that from Step 2 it follows that the projection of $O$ on $FQ$ is the midpoint of the segment having the midpoints of $I_{cd}I_{ab},\ E_{ab}E_{cd}$ as extremities. However, these last two points are the projections of $O_1,O_2$ on $FQ$.

We then repeat step 3 by working with the projections of $O,O_1,O_2$ on $EQ$ instead of $FQ$, and from what we find we will obtain Darij’s conjecture: $O,O_1,O_2$ are collinear and $O$ is the midpoint of $O_1O_2$. In this last part we have used the fact that if a point projects to the midpoint of the projections of two other points on two different lines, then the point must be the midpoint of the two other points.


Let $ a;b;c>0$and $ ab\plus{}ac\plus{}bc\equal{}1$
Prove that
$ \frac{1}{ab(a\plus{}b)}\plus{}\frac{1}{cb(c\plus{}b)}\plus{}\frac{1}{ac(a\plus{}c)}\ge \frac{1}{abc}\plus{}\frac{1}{a\plus{}b}\plus{}\frac{1}{c\plus{}b}\plus{}\frac{1}{a\plus{}c}$

[quote="tranvanluan"]Let $ a;b;c > 0$and $ ab \plus{} ac \plus{} bc \equal{} 1$
Prove that
$ \frac {1}{ab(a \plus{} b)} \plus{} \frac {1}{cb(c \plus{} b)} \plus{} \frac {1}{ac(a \plus{} c)}\ge \frac {1}{abc} \plus{} \frac {1}{a \plus{} b} \plus{} \frac {1}{c \plus{} b} \plus{} \frac {1}{a \plus{} c}$[/quote]
It equalivents with $ a^2  \plus{} b^2  \plus{} c^2  \ge ab \plus{} bc \plus{} ca$.

[quote="Sunkern_sunflora"]I think both leads to the same one, that obtained by RainL. Isn't it?
 :)
Sorry if it's wrong....[/quote]
Yes I was wrong. I deleted my previous post. Thank you.


Let $ X$ be a set; let $ C$ be the collection of all pairs $ (A,<)$, where $ A$ is a subset of $ X$ and $ <$ is a well-ordering of $ A$.

Define $ (A,<) << (A',<')$ if $ (A,<)$ equals a section of $ (A',<')$

Let $ C'$ be a subcollection of $ C$ that is simply ordered by $ <<$. Define $ C''$ to be the union of the sets $ B$

for all $ (B, <) \in C'$; and define $ <'$ to be the union of the relations $ <$ for all $ (B,<) \in C'$

Show that $ (C'', <')$ is a well-ordered set

Looks like a proof of the Well Ordering Principle from Zorn's Lemma.


How many upper-case letters in the English language alphabet have a horizontal line of symmetry?

B, C, D, E, H, I, K, O, X
Thus, $ \boxed{9}$.

The problem says the answer is 8...

I would like to note that this problem is quite subjective (I'm guessing it was either B or K) as B and K are symmetric in some fonts but not in others.

I have approached the same answer of $\boxed{9}$
I am guessing B. Since the proper way to write a B, is to have a smaller hump on the top, this may be the reason. If this is not the error, then I think there is a typo of some sort.


function
inequalities
inequalities proposed

If $a,b,c$ positive numbers and $a^2+b^2+c^2 = 3$, show that:
$\sum \frac{a}{a^3+a^2+4} \leq \frac{1}{2}$

I have a solution, but I just want to see the nicer ones :D

Solution

[hide]

First of all, observe  and  have the same sign for all positive , so that  or , so that .

Next, observe that by trivial ineq , so then expanding and adding stuff we get  and thus .

So now we have the chain

Next, observe that if , then , so the function is concave for all positive . By Jensen's, we have

or

Let's step away for a second. By AM-RMS, we know that

Adding  to each side, we have

Dividing through by , we get that

Hooking that up to the inequality we got from Jensen's completes the entire chain, and we finally have our desired inequality.

[/hide]

Kind of hard, excellent problem.

wow, very nice elemennop
[hide]
My solution also proves that each term is $\leq \frac{a}{5a+1}$, but this way:
By AM-GM, we have $a^3+a^2+3 \geq 5a$ so that $\frac{a}{a^3+a^2+4} \leq \frac{a}{5a+1} = \frac{1}{5}\left( 1 - \frac{1}{5a+1} \right)$

So now it suffices to show that $\sum \frac{1}{5a+1} \geq \frac{1}{2}$
From AM-RMS we have $a+b+c \leq 3$
and from AM-HM we have $\sum \frac{1}{5a+1} \geq \frac{9}{5a+5b+5c+3} \geq \frac{1}{2}$[/hide]

[hide]
You can also show that

$\frac{a}{a^3+a^2+4} \le \frac{11}{72}+\frac{1}{72}a^2$,

which comes out to

$72a \le (a^3+a^2+4)(11+a^2)$, or

$0 \le (a-1)^2(a^3+3a^2+16a+44)$, which is clearly true.[/hide]

Ooh, Paladin8 posted...he's the one I got that technique for wittling away the degree of the denominator (ie. multiply $(a-1)(a^2-1)$). It's a wonderful tool.

out of curiosty, paladin8, how did you find that form? ($11+a^2$)

I found that you can show the last ineq this way too:
AM-GM: $a^3+a^2+4 \geq 6a^{5/6}$
AM-GM: $a^2+11 \geq 12a^{1/6}$
multiply, we get $(a^3+a^2+4)(a^2+11) \geq 72a$


hello, prove that the left-hand side minus the right-hand side of your equation is equal to
$\left(\sqrt{3}\cos\left(\frac{2}{3}x\right)+\sin\left(\frac{2}{3}x\right)\right)\left(\sqrt{2}+2\sin\left(\frac{5}{6}x\right)\right)=0$
and solve this.
regards Sonnhard.


Help me with this problem.
There are N(4<=n<=10000) points given by their coordinates on a plane. All coordinates (xi,yi) are integers in a range from -10000 up to 10000 inclusive . It is necessary to construct a broken line satisfying the following conditions:
1. The broken line should be closed.
2. End points of each segment (verteces) of the broken line can only be the given points, and all given points should be used.
3. Each two consecutive segments of the broken line should form a corner of 90 degrees in each vertex point.
4. The sides of the broken line should be parallel to coordinate axes.
5. The broken line should have no self-crossing and self-contact.
6. The broken line should have the minimal length.
You have to either find the length L of the constructed broken line, or determine that it is impossible to construct such a broken line.

I don't quite get the 'minimal length' condition, since it seems to me that only one possible broken line can exist.

If you take any vertical line/horizontal lines, the points on this line must be paired up with adjacent points. Thus there must be an even number.

For each point, count how many points are directly above it and how many are directly below it. If both are even or both are odd, from above its impossible. Otherwise, you must travel in the 'odd' direction. Same for the horizontal case.

So to work out the total length, just go through every possible horizontal/vertical line, making sure there are an even number of points on that line, and if so, adding every second difference (starting with the first two points, then the next two, etc).


calculus
integration
real analysis
real analysis solved

Calculate integral from 0 to  \pi /6 from f(x)=(sinx+tgx)/(cosx+ctgx)

I'm assuming that tgx means the tangent of x and ctgx means the cotangent of x.

Multiply both the numerator and denominator by cosx - ctgx and simplify the denominator to -cos4x/sin2x. This causes the fraction to simplify to

-tan3x + tan2x secx - tan3x secx + tan2x sec2x.

The primitive of that is

-0.5 tan2x + ln|secx| + 0.5 secx tanx - 0.5 ln|secx + tanx| - (1/3) secx tan2x + (2/3) secx + (1/3) tan3x + C.

When you evaluate that from 0 to \pi/6, it becomes

-0.5 + 11/(9\sqrt3) + ln 2 - 0.75 ln 3

which is approximately 0.0748.


geometry
3D geometry
algebra
polynomial
factorial
number theory
prime numbers

[hide="This may help..."]I'm not completely sure it helps though...[quote="andersonw"]Try to solve this, it's a fun, and hard, puzzle. I'll give hints (and the solution) if necessary.
Note: This was not made up by myself.
[code]1warning 2you 3really 4both 5shouldn't 6generate 7doubt 8distrust 9fibonacci 10encode 11the 12terminate 13primes 14believe 15believe 16and 17take 18leftmost 19last 20order 21squares 22send 23word 24tenth 25cubes 26symmetric 27catalan 28concatenate 29of 30reorder 31relevant 32apply 33master 34aren't 35likely 36speak 37sequences 38every 39apples 40visible 41first 42bell 43letter 44entire 45vilify 46above 47if 48never 49true 50is 51sentence 52anyone 53honest 54hard 55genuine 56emerge 57xor 58immediate 59last 60truncated 61otherwise[/code][/quote][/hide]

[quote="andersonw"]Try to solve this, it's a fun, and hard, puzzle. I'll give hints (and the solution) if necessary.
Note: This was not made up by myself.
[code]warning you really both shouldn't generate doubt distrust fibonacci encode the terminate primes believe believe and take leftmost last order squares send word tenth cubes symmetric catalan concatenate of reorder relevant apply master aren't likely speak sequences every apples visible first bell letter entire vilify above if never true is sentence anyone honest hard genuine emerge xor immediate last truncated otherwise[/code][/quote]

I see fibonacci! Using fibonacci sequence words we get:
1 1 2 3 5 8 13 21
Warning warning you really shouldn't distrust primes squares....

Yeah makes no sense.

This is FUN!

(1) Squares: 1,4,9,...

warning both Fibonacci and cubes speak [b]true[/b]

(2) Cubes: 1,8,27,...
warning distrust [b]Catalan[/b]

(3) Catalan: 1, 1, 2, 5, 14, 42,...
warning warning you shouldn't believe [b]bell[/b]

(4) Primes: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61
you really shouldn't doubt the primes take last word of relevant sequences first letter if honest last [b]otherwise[/b]

(5) Bell: 1, 1, 2, 5, 15, 52
warning warning you shouldn't believe [b]anyone[/b]

(6) Fibonacci: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55
warning warning you really shouldn't distrust primes squares aren't [b]genuine[/b]

[i](7) symmetric (palindrome?) 11,22,33,44,55
the send master entire genuine
?? Do not get.

(7) symmetric (polynomial?) 	1, 1, 2, 6, 22
warning warning you generate send
?? Do not get

(7) symmetric (factorial?) 1,1,2,6,24
warning warning you generate tenth
?? Do not get[/i]

If (2), then not (3).
If (3), then not (5).
If (5), then not anything. Therefore (5) is false
If (1), then (2), (6)
If (6), then not (1), yes (4)
If (4), then (4). Also a definition. I will assume true.

(09) FIBONACCI - keyword: [b]G[/b]ENUINE
(13) PRIMES - keyword: OTHERWISE
(21) SQUARES - keyword: TRUE
(25) CUBES - keyword: CATALA[b]N[/b]
(27) CATALAN - keyword: [b]B[/b]ELL
(42) BELL - keyword: ANYON[b]E[/b]

gotcbe
gotcle
gotnbe
gotnle
goecle
goecbe
goenbe
goenle
getcle
getcbe
getnbe
getnle
geecle
geecbe
geenbe
geenle

I'm probably missing some relevant sequences.

Nice job, james4l, you are very close. You have all of the 6 needed sequences.
Also, a hint for anyone that doesn't get what james is doing
[hide]Use brut3forc3's numbering.
Which sequences are telling the truth? Which sequences are lying?
[/hide]

And a hint for james
[hide]
The sequences aren't in order, so the letters might need some rearranging.
[/hide]

Another hint:
[hide]Take a look at all of the words that aren't used[/hide]

Also, I should punctuate the fibonacci sequence better.
"warning warning! You really shouldn't distrust primes. Squares aren't genuine."
And to clarify, "you shouldn't distrust primes" means "you should trust primes".

What are bell and catalan sequences?

If 1=true
2=true
3=false
5=false
6=true
1=false so that doesnt work.

Thus:

1=false

Thats where i get.

[hide="Answer I found using Google"]Puzzle 2
by Ari Nieh

Among the words are the names of several integer sequences.
If you number the words and look at what each sequence is saying,
you get the following:

[capitalization and punctuation added to improve readability]

    Fibonacci numbers:  Warning!  You really shouldn't distrust
    primes.  Squares aren't genuine.

    Primes:  You really shouldn't doubt the primes.  Take last word of
    relevant sequences; first letter if honest, last otherwise.

    Squares:  Warning!  Both Fibonacci and cubes speak true.

    Cubes:  Warning!  Distrust Catalan!

    Catalan numbers:  Warning!  You shouldn't believe Bell.

    Bell numbers:  Warning!  You shouldn't believe anyone.

This is a truthtellers-and-liars logic puzzle.  After you've solved it,
following the instructions of the prime numbers (who were telling the truth)
yields the letters GOENBE in no particular order.

The first letters of the unused words spell out:

GET LOST SCRAM LEAVE VANISH EXIT

and so the correct answer is BEGONE.
[/hide]

I found it by doing the logic, and realizing the 6 letters, and googling them. I was originally planning to use unscramble, but...

[quote="mewto55555"]What are bell and catalan sequences?
[/quote]

I wiki'd it:

Catalan numbers are any number in the form:
$ \dfrac{1}{n\plus{}1}\dbinom{2n}{n}$

Bell numbers are, basically, [quote="wikipedia"]the number of ways to put n distinguishable balls, n>=0, into 1 or more distinguishable boxes. For example, let us suppose n is 3. We have three balls, which we will label a, b, and c, and three boxes. If the boxes can not be distinguished from each other, there are five ways of putting the balls in the boxes:

Each ball goes in to its own box.
All three balls go in to one box. Since the boxes are anonymous, this is only considered one combination.
a goes in to one box; b and c go in to another box.
b goes in to one box; a and c go in to another box.
c goes in to one box; a and b go in to another box. [/quote]


Prove that for any prime number $p \geq 5$, the number

$\sum_{0<k<\frac{2p}{3}}{p\choose k}$ is divisible by $p^{2}$

[hide]If p is 1 mod 3, then $\lfloor2p/3\rfloor$=$\frac{2p-2}{3}$
If p is 2 mod 3, then $\lfloor2p/3\rfloor$=$\frac{2p-1}{3}$
We're adding $\binom{p}{1}$+$\binom{p}{2}$+...+$\binom{p}{\lfloor2p/3\rfloor}$.
We can divide out one factor of p since p is prime and cannot be cancelled by the smaller demonators.
Therefore, $1+\frac{p-1}{2}+\frac{(p-1)(p-2)}{3!}+...+\frac{(p-1)(p-2)...(\lfloor2p/3\rfloor+1)}{(p-\lfloor2p/3\rfloor)!}$.
Since when we combine the factors, the p parts will be equal to 0 mod p, those can be safely ignored.  We now have $1-1/2+1/3-1/4+1/5-1/6+...+\frac{(-1)^{\lfloor2p/3\rfloor}}{\lfloor2p/3\rfloor}$
Now, one must realize the identity $1-1/2+1/3-1/4+...-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$.
Now look at the 1 mod 3 case:
$2n=\frac{2p-2}{3}$ so $n=\frac{p-1}{3}$.  In other words, $1/\frac{p+2}{3}+1/\frac{p+5}{3}+...+1/\frac{2p-2}{3}=3(\frac{1}{p+2}+\frac{1}{p+5}+...+\frac{1}{2p-2})$.  But if looks at this mod p, that expression in the parenthesis equal zero.  Also note that p must be even for there to be an odd number of expressions in the parenthesis.
Now look at the 2 mod 3 case:
$2n=\frac{2p-4}{3}$ so $n=\frac{p-2}{3}$.  $2n+1=\lfloor2p/3\rfloor$.  In other words, $1/\frac{p+1}{3}+1/\frac{p+4}{3}+...+1/\frac{2p-1}{3}=3(\frac{1}{p+1}+\frac{1}{p+4}+...+\frac{1}{2p-1})$.  But if looks at this mod p, that expression in the parenthesis equal zero.  Again, p must be even for there to be an odd number of expressions.[/hide]

One of the previous avatars...
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2p&t=56516[/url]


function
inequalities
inequalities unsolved

Assuming $a+b+c=1$ we can rewrite it as $\sum_{cyclic} \sqrt{\frac{a}{1-a}}>2$.

It is easy to show that $\sqrt{\frac{x}{1-x}}\geq 2x$ for $x\in[0,1]$.
Let's prove $f(x)=\sqrt{\frac{x}{1-x}}$ is convex on $[1/2,1]$.
Square $f(x)+f(y)\geq 2f((x+y)/2)$ and apply $\frac{1}{1-x}+\frac{1}{1-y}\geq \frac{2}{1-\frac{x+y}{2}}$. Then squaring again we obtain $\frac{(x-y)^2}{(1-x)(1-y)(2-x-y)^2}\geq \frac{(x-y)^2}{(1-x)(1-y)(2-x-y)}$, which is true.

Consider function $g(x)$ defined by $g(x)=2x$ on $(0,1/2]$ and $g(x)=f(x)$ on $[1/2,1)$. Then $g(x)$ is a convex fucntion on $(0,1)$, moreover $f(x)\geq g(x)$.
Applying Jensen for $g$ we conclude $f(a)+f(b)+f(c)\geq g(a)+g(b)+g(c)\geq 3g(1/3)=2$. The equality is possible iff $a,b,c\in (0,1/2]$. But $f(x)>g(x)$ on $(0,1/2)$.

I know, it is a quite long solution :(

Where is wellknown's reply?

Indeed, I wrote foolish long solution, though conclusion immediately follows from $\sqrt{\frac{x}{1-x}}\geq 2x$.

Nice solution.
I just wonder how did you find  :sqrt: (x/(1-x)) >= 2x

I found it, because I wanted to apply my initial idea for this problem.

It is simply HM-GM inequality. Isn't it , Myth ?

I was thinking of GM-HM ineq

$\sqrt{AB} \geq \frac{2}{\frac{1}{A}+\frac{1}{B}}$

now put $A=1$ and $B=\frac{a}{b+c}$

If it is true ,it is very simple.

in fact,we have  √(a/(b+c))?√(a 2 /a(b+c)) >= ２a

[quote="hardsoul"]For $a,b$ and $c$ positive reals prove that:
 $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2$[/quote]
[quote="Remike"]This problem is M1818 from the journal "Kvant". Author of the problem: S. Nesterov.[/quote]
This problem also from Russian magazine " Mathematics at school " the sixtieth years
of the last century. :wink:

[quote="hardsoul"]For $ a,b$ and $ c$ positive reals prove that:
 $ \sqrt{\frac{a}{b\plus{}c}}\plus{}\sqrt{\frac{b}{c\plus{}a}}\plus{}\sqrt{\frac{c}{a\plus{}b}}> 2$[/quote]
From this problem,we have:
$ \sum\sqrt{\frac{a}{b\plus{}c}}\geq\frac{a}{b\plus{}c}$
where $ a;b;c$ be positive real numbers

[quote="hardsoul"]For $ a,b$ and $ c$ positive reals prove that:
 $ \sqrt{\frac{a}{b\plus{}c}}\plus{}\sqrt{\frac{b}{c\plus{}a}}\plus{}\sqrt{\frac{c}{a\plus{}b}}> 2$[/quote]
Using AM-Gm inequality:
$ \sqrt{\frac{a}{b\plus{}c}}\equal{}\frac{a}{\sqrt{a(b\plus{}c)}}\ge\frac{2a}{a\plus{}b\plus{}c}$
Similarly, we have:
$ \sqrt{\frac{a}{b\plus{}c}}\plus{}\sqrt{\frac{b}{c\plus{}a}}\plus{}\sqrt{\frac{c}{a\plus{}b}}>2$

[quote="chien than"]With $ n\geq 2$,we can prove :
$ \sum\sqrt [n]{\frac{a}{b\plus{}c}}> 2$[/quote]
yes , the hint is to choose the three new variables satisfying the inequality
$ \sum\sqrt [n]{\frac{a}{b\plus{}c}}\geq\sum\sqrt{\frac{\alpha}{\beta\plus{}\gamma}}$
and we can choose $ \alpha \equal{}\sqrt [n]{a^{2}}$ ;....
it leads to the inequality $ \sum\sqrt{\frac{\alpha}{\beta\plus{}\gamma}}>2$
which is already true

[url=http://www.artofproblemsolving.com/community/c6h1169529p5632819]Generalization:[/url]
Let $x_1,x_2,\cdots,x_n$ be positive real numbers $(n\ge 3)$. Prove that\[\sum_{i=1}^n\sqrt{\frac{x_i}{ \sum_{k=1}^nx_k-x_i}}>2.\]
[b]Proof:[/b]  $\frac{ \sum_{k=1}^nx_k-x_i}{x_i}\leq (\frac{1+ \frac{ \sum_{k=1}^nx_k-x_i}{x_i}}{2})^2=(\frac{\sum_{k=1}^nx_k}{2x_i})^2$,
$\sum_{i=1}^n\sqrt{ \frac{x_i}{\sum_{k=1}^nx_k-x_i}}>\sum_{i=1}^n \frac{2x_i}{\sum_{k=1}^nx_k}=2$.
http://www.artofproblemsolving.com/community/c6h505879p2841605

http://artofproblemsolving.com/community/c6h123597p700172

A little bit nicer:
Without loss of generality, assume $\sum_{cyc}\sqrt{a} = 1$
Rewrite it to$\sum_{cyc}\sqrt{a}f(b+c) > 2$ where $f(x)=\frac{1}{\sqrt{x}}$
Then, for $f(x)$ is convex and decreasing for positive real numbers, by applying Jensen's inequality, it remains to prove
$\sum_{cyc}\sqrt{a}(b+c) < \frac{1}{4}\sum_{cyc}\sqrt{a}$
Replace $\sqrt{a}$ with $x$, rewrite it as
$\sum_{cyc}x(y^{2}+z^{2}) < \frac{1}{4}(\sum_{cyc}x)^{3}$
Expand both sides and it becomes
$\frac{1}{2}\sum_{sym}x^{3}-\sum_{sym}x^{2}y+\sum_{sym}xyz > 0$
Schur's inequality yields
$\frac{1}{2}\sum_{sym}x^{3} - \sum_{sym}x^{2}y+\frac{1}{2}\sum_{sym}xyz > 0$
then the original inequality is obviously true

[quote=chien than]With $ n\geq 2$,we can prove :
$ \sum\sqrt[n]{\frac{a}{b\plus{}c}}>2$[/quote]
Let $\frac{a}{b+c}=p^n$, $\frac{b}{a+c}=q^n$ and $\frac{c}{a+b}=r^n$, where $p$, $q$ and $r$ are positives.
Thus, we need to prove that $p+q+r\geq2$ and since $$\sum_{cyc}\frac{1}{1+p^n}=\sum_{cyc}\frac{b+c}{a+b+c}=2,$$
we obtain:
$$2p^nq^nr^n+p^nq^n+p^nr^n+q^nr^n=1.$$
Now, let $p+q+r<2$, $p=kx$, $q=ky$ and $r=kz$, where $k>0$ and $x+y+z=2$.
Hence, $k(x+y+z)<2$, which gives $0<k<1$.
Hence,
$$1=2p^nq^nr^n+p^nq^n+p^nr^n+q^nr^n=2k^6(x^ny^nz^n)+k^4(x^ny^n+x^nz^n+y^nz^n)<2x^ny^nz^n+x^ny^n+x^nz^n+y^nz^n,$$
which is contradiction because we'll prove now that
$$2x^ny^nz^n+x^ny^n+x^nz^n+y^nz^n\leq1.$$
Firstly, we'll prove that for all $n\geq3$ we have
$$2x^ny^nz^n+x^ny^n+x^nz^n+y^nz^n\leq2x^{n-1}y^{n-1}z^{n-1}+x^{n-1}y^{n-1}+x^{n-1}z^{n-1}+y^{n-1}z^{n-1}.$$
Indeed, by AM-GM
$$2x^ny^nz^n=2x^{n-1}y^{n-1}z^{n-1}\cdot{xyz}\leq2x^{n-1}y^{n-1}z^{n-1}\left(\frac{x+y+z}{3}\right)^3<2x^{n-1}y^{n-1}z^{n-1}.$$
Thus, it remains to prove here that
$$x^ny^n+x^nz^n+y^nz^n\leq x^{n-1}y^{n-1}+x^{n-1}z^{n-1}+y^{n-1}z^{n-1}$$ or
$$4\sum_{cyc}x^ny^n\leq\sum_{cyc}(x^2+2xy)\sum_{cyc}x^{n-1}z^{n-1}$$ or
$$\sum_{cyc}(x^{n+1}y^{n-1}+x^{n+1}z^{n-1}-2x^ny^n+x^{n-1}y^{n-1}z^2+2x^ny^{n-1}z+2x^nz^{n-1}z)\geq0,$$
which is obvious.
Id est, it's enough to prove that
$$2x^2y^2z^2+x^2y^2+x^2z^2+y^2z^2\leq1,$$
which is obviously true by $uvw$.
Done!


probability
number theory
prime factorization

2000 (8)
Two marbles are randomly selected without replacement from a bag containing blue and green marbles. The probability they are both blue is 1/6. If three marbles are randomly selected without replacement, the probability that all three are blue is 1/21. What is the fewest number or marbles that must have been in the bag before any were drawn?

2004 (8)
A collection S of integers is defined by the following three rules: (I) 2 is in S; (II) for every x in S, 3x and x+7 are also in S; (III) no integers except those defined by rules (I) and (II) are in S. What is the smallest integer greater than 2004 which is NOT in S?

2005 (2)
In a class of 28 students, the teacher selects four people at random to participate in the geography contest. What is the probability that this group of four students includes at least two of the top three geography students in the class? Express answer as common fraction.

2005 (6)
For postitive integer n such that n<10000, the number n+2005 has exactly 21 positive factors. What is the sum of all possible values of n?

2005 (7)
A point inside a square is positioned so that the distances to the four vertices are 27, 21, 6, and x units. If x is a whole number, what is the value of x?

[b]Answers[/b]

[hide]2000 (8): 9
2004 (8): 2009
2005 (2): 37/819
2005 (6): 16592
2005 (7): 18[/hide]

[hide="last one"]Look up the British Flag Theorem.[/hide]

[quote="xpmath"]Look up the British Flag Theorem.[/quote]

You can just derive that by a few operations with the Pythagorean theorem. :P

2004(8):

All multiples of three are in $ S$. List out the next few numbers and eliminate:

2005 <- 2005-7 is divisible by 3
2006 <- Since 2 is in S, we can subtract 2 and see that it is divisible by 3
2007 <- Divisible by 3
2008 <- 2008-7 is divisible by 3
2009 <- This is our answer since even if you subtract 2, it is not divisible by 7; nor if you subtract 7 it is divisible by 3

P.S. Are these Target or Sprint?

They're target, I remember one of them.

Yeah, but why waste a few seconds :P

[quote="xpmath"]Yeah, but why waste a few seconds :P[/quote]

If they are target, I'd not waste my time remembering formulas I'd rather never remember in the future, though I did know BFG from last year but completely blanked out :|

Uh, but if you knew it already, why shouldn't you use it instead of using the Pythagorean Theorem a few times.

For number 1
[hide] Say x is the number of blue marbles and y is the total number of marbles. x/y*(x-1)/(y-1)=x(x-1)/y(y-1)=1/6. x/y*(x-1)/(y-1)*(x-2)/(y-2)=1/21. Substitute and you get 1/6*(x-2)/(y-2)=1/21. (x-2)/(y-2)=2/7. Here you can probably make a guess of x=4 and y=9. Plug that in and you'll see that it works.[/hide]

[hide="2005 #6"]
if u know the quick way of finding the # of divisors, you are in good shape for this one. If u dont...
[hide="theorem"]
a theorem states that when a number is in prime factorization, take all the exponents, add 1 to all of them, and multiply them to get the # of divisors.
example: if u have the number 72, then it factors into 2^3 and 3^2 take 3 and 2 and add 1 to each.
you now have 4 and 3. now do 3*4 to get [b]12[/b]
[/hide]
so for a number to have 21 divisors, its prime factorization must be in the form of p^20 or p^6 q^2
since the smallest # in the former form is 1048576, only numbers in the latter form may be used.
now you kinda test for #s less than 12005 and more than 2005 (because of the adding 2005) that are in this form.
i hope this question was target  :D
[/hide]

[hide="the #s that work"]
[hide]
[hide]

3136, 7744, 10816, and 2916.
the values of n are
1131, 5739, 8811, and 911
they add to 16592

[/hide]
[/hide]
[/hide]
EDIT: TY alanchou :D

the number of [b]divisors[/b] is what you meant, I think

number of primes is really easy, just prime factorize and count

if the prime factorization of a number is $ 2\times 3\times 5^2\times 7\times 11^3$, then the number of prime factors is 5 (2,3,5,7,11).

[hide="maybe i will say something more useful and post an alternate solution to the second problem"]
Consider the set in mod 7.  then, we know that if we take 2 and repeatedly add 7, we get all the numbers greater than or equal to 2 that are congruent to 2 in mod 7.  now, if we multiply the 2 by 3 and repeatedly add 7, we get all numbers greater than or equal to 6 that are congruent to 6 mod 7.  Continue in this manner, you'll find that no multiples of 7 are in S and all other integers greater than 486 are in S, so look for the smallest integer greater than 2004 that's a multiple of 7, which is $ \boxed{2009}$.
Huh, that's more wordy that i thought it would be
[/hide]


find the relation between $f$ and  $f''$
$f(x)=3\cos(2x)+2\sin(3x-\frac{\pi}{6})$
 :blush:

i think that isn't so easy because  $f''+6f\neq0$ ;)

there's an error in Ayman's solution :
$ f'' + 6f \neq 0$
i think
$ f''(x) + 6f(x) = 6 (\cos(2x) - \sin(3x - \frac{\pi}{6})) $

i think so  $ f''(x) + 6f(x) = 6 (\cos(2x) - \sin(3x - \frac{\pi}{6})) $
i find it too :)

What I meant was to solve the differential equation, but don't bother because it's really ugly.
Edit: Guess it's not that bad.

[quote="Jimmy"]You should post this under calculus computations and tutorials, getting started is for basic precalculus math.
I also got the same answer.
Here's a challenge, can you solve $f''(x)+6f(x)=6\left(\cos(2x)-\sin(3x-\frac{\pi}{6})\right)$ for $f(x)$ (If this was all that was given)?[/quote]

Apply the method of undetermined coefficients. Since $x$ is not weighted by $\sqrt{6}$ in either trigonometric expression, the roots $\pm i\sqrt{6}$ of the characteristic equation play no role in a particular solution; thus, we set $f = A \cos(2x) + B \sin(3x - \frac{\pi}{6})$. This gives $f'' + 6f = 2A\cos(2x) - 3B \sin(3x - \frac{\pi}{6})$ whence $A = 3$ and $B = 2$. The general solution is accordingly

$f = 3\cos(2x) + 2\sin(3x - \frac{\pi}{6}) + c_1\cos(x\sqrt{6}) + c_2\sin(x\sqrt{6})$.

ok metafisist(omar) it's an original solution

IF IS THERE ANY STUPID CLUB, I ABLE TO BE MEMBER :blush:


Hi,

Any tips on how to convert remainders into real numbers, when performing mental division.

E.g.
Lets consider: 1960 divided by 322

Using a calculator this gives me: 6.09 (to 2 decimal places).

When I perform this calculation mentally ... I get 6 remainder 22. How can I get from "6 remainder 22" to the actual "6.09" value.
In other words, how can I convert the "remainder 22" to ".09" ???

Thanks for you help.

Just continue the division.

$ \frac{22}{322}\approx.09$

[quote="hunter34"]Just continue the division.

$ \frac {22}{322}\approx.09$[/quote]

Thanks Hunter !
322/1960 is actually 6 remainder 28 (sorry 22 was my typo !)
So if I continue the division .... 28/322
Is there an easy way of doing this manually ?

[quote="Ihatepie"]well with fractions like that, I usually simplify it

After some quick simplifying, I got

2/23

then add 23s to it until it is close to 100

In this case, 23*4= 92
and             2*4=8
so you get 8/92

since 8 (the amount you need to add to 92 to get 100) is about 1/3 of 23, I would take 1/3rd of 2 which is  2/3 . You add 8 to 92 and get 100 as the denominator.
Then you add 2/3 to 8 and get 8 2/3 as the numerator

so you get 8 2/3 over 100

Which is approximately .86666666666....

This is pretty close to the real answer of

.8695652174.....

If it is unclear, I'll try to clarify, but this is how it goes in my head when I do mental calculations.[/quote]

Thanks for this. Is there not a simpler way to do this types of calculations (manually - with no calculator) ?

Yes, you can do long division to get that. It's a little tough to do it mentally though.

I bet there is some simpler way out there by not using pencil or paper, but if there is, I don't know it.


Describe a sequence of transformations that would map the graph of $ y\equal{}\left| mx\plus{}c \right|\plus{}d$ onto $ y\equal{}\left| amx\plus{}c \right|\plus{}d$.

maybe graphing this on paper with different values may help?

[quote="ThetaPi"]Describe a sequence of transformations that would map the graph of $ y \equal{} \left| mx \plus{} c \right| \plus{} d$ onto $ y \equal{} \left| amx \plus{} c \right| \plus{} d$.[/quote]

First do a horizontal shrink of factor $ a$ to get $ y\equal{}|amx\plus{}ac|\plus{}d$, whose minimum (vertex) occurs when $ x\equal{}\minus{}\frac{ac}{am}$.

The vertex of the desired function has $ x\equal{}\minus{}\frac c{am}$, so to arrive at this, we must do a horizontal shift of

$ \minus{}\frac c{am}\minus{}\left(\minus{}\frac{ac}{am}\right)\equal{}\frac{c(a\minus{}1)}{am}$.


integration
trigonometry
calculus
function
real analysis
real analysis unsolved

Solve $ \int\frac{1}{\sqrt{1\plus{}\sin^2 x}}dx$.

[quote="suweijie"]Solve $ \int\frac {1}{\sqrt {1 \plus{} \sin^2 x}}dx$.[/quote]
We can get rid of trigonometric functions. Suppose we want to find primitive on $ x\in (\minus{}\pi,\pi)$.

Let $ \mu \equal{} \tan \frac{x}{2}$ then $ \sin x \equal{} \frac{2\mu}{1\plus{}\mu^2}$.
That means,
$ 2\int \frac{d\mu}{\sqrt{(2\mu)^2(1\plus{}\mu) \plus{} (1\plus{}\mu^2)^2}}$
Is that elementary? It looks like ellptic integral.  :maybe:

[quote="suweijie"]$ \int\frac {1}{\sqrt {1 \plus{} \sin^2 x}}dx$[/quote]
I'm affraid it can be a nonelementary function.

BTW. (my own question):
counting an integral with all known rules of integration and saying that anyone works of course is not sufficient  :D
so how it can be proved that a certain function given by indefinite integral is not an elementary function :?:

[b]Suweijie[/b], why don't you try instead $ P \equal{} \int \frac{1}{\sqrt{1 \plus{} \sin x}}\,dx$ ?


Euler
geometry
circumcircle
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle, (0) the circumcircle of ABC, O the center of (O), K the symmedian point of ABC, (1) the circle with diameter OK, X the second point of intersection of AK with (1), A" the second point of intersection of the perpendicular to OK passing through A with (0)
and (2) the circle through O, X, A".
Prouve that (2) passes also through the anti Euler point of ABC i.e. X(110).
Sincerely
Jean-Louis

Dear Mathlinkers,
I observe that A"X(110) goes through the center of the A-Apollonius circle of ABC.
Any proof?
Sincerely
Jean-Louis

Dear Livetolove,
sorry your solution is not avaible. Can you please send it again.
Sincerely
Jean-Louis

just a moment ago I did it，your this question is not very difficult：
（1）：easy to found $ OX$ through $ P$ which is the center of the A-Apollonius circle of △$ ABC$
（2）：The eular line of △$ ABC$ is the Orthocenter axis of a complete quadrilateral which is constituted by △$ ABC$ and lemoine axis（$ P,Q,R$）
（3）：On the basis of （2），we can get $ X(110)$ is the Miquel point of  the complete quadrilateral $ ABCPQR$
$ BC$ intersect $ OX$ at $ P$ ，$ P,X(110),B,Q$ are on a circle，Let $ PX(110)$ intersect the circumcircle of △$ ABC$ at $ A'$
so we can easy to get $ AA'//PQ$ .Then $ A' \equal{} A''$，and as a result of $ PO*PX \equal{} PB*PC \equal{} PX(110)*PA''$，We got $ O,X,A'',X(110)$ are on a circle .

Dear Jean-Louis,
Your problem is very nice. :)

Dear Lim, Livetolove and Mathlinkers,
thank you for your nice proofs.
Sincerely
Jean-Louis


limit
integration
function
calculus
real analysis
real analysis unsolved

Let $f(x)$ be the function which is continuous and monotone increasing function at $[0,1].$ For $n\in \mathbb{N},$ assumed that there existed the limit $\lim_{n\to\infty} n\int_0^1 x^nf(x)\ dx.$ Find the value of this limit.

Again? :)

See : http://www.mathlinks.ro/Forum/viewtopic.php?t=83988 .

It seems somewhat different in that the condition of $f(x).$

Moubinool has the nicest proof I saw here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=68241
But also Cezar Lupu had a good one in a diff post.

However I think kunny wants a simpler (i.e., more elementary) proof that uses the second condition on $f$?

I think we can integrate by parts $nx^{n-1}$ vs $xf(x)$ and find $f(1)$ minus some integral which goes to 0 in the limit. We will need conditions on $f$ to ensure that $f'(x)$ is well behaved. But I still wish to use some results from measure theory... I hope someone will enlighten us.  :?  :D

[quote="Xevarion"]Moubinool has the nicest proof I saw here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=68241
But also Cezar Lupu had a good one in a diff post. [/quote]

Cezar's proof was wrong.  Anyways, look at grobber's proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70062]here[/url]


Mathnerd314 is very bored and decides to buy a package of equilateral triangles at Math-Mart. He has a sudden urge to create all 12 "sixominoes", that is, patterns that can be made from six equilateral triangles glued together. For example, he can create a parallelogram with six tiles. (Don't worry, that's just the introduction :D)

Mathnerd314 wonders whether he can now take six of these sixominoes and put them together to form a big equilateral triangle with 36 tiles (that is, 6 tiles on a side). Prove or disprove.

Hope the problem wasn't too badly stated.

[quote="mad_skillz_aops"][hide]It is possible, because 36 is a triangular number.[/hide][/quote]

No, it isn't--at least, it isn't possible if I stated the problem correctly.

The intention was to take six [b]different[/b] sixominoes and glue them together.

[quote="mad_skillz_aops"][hide="Well, no wonder!"]Well, if they are all different, then the closest you can make would have 2 pieces sticking out.[/hide][/quote]
Could you please post a proof?
I think this is a nice problem. :)

Actually, I didn't use a proof, I really didn't use math at all.  I just cut some pieces out of graph paper and played around.

They are sixominoes, not sextominoes. :rotfl:

[quote="randomdragoon"][hide="This is like the domino tiling of the checkboard problem"]You color the big triangle so that the small triangular tiles alternate white and black. Note that there is six more of one color than the other. From the set of sextominoes, you do the same. You will notice that all but two of the sextominoes have the same number of white and black tiles, and two of them have two more of one color than the other. But you need a difference of six of one color to another in order to tile the triangle, so it is impossible.[/hide]

Also, problem solving theorems state that the answer must be impossible, otherwise the problem would be trivial. :P[/quote]
Nice solution!

sextominoes... :rotfl:

[quote="randomdragoon"]Also, problem solving theorems state that the answer must be impossible, otherwise the problem would be trivial. :P[/quote]

Try using that theorem at a math olympiad :rotfl:

And thanx fr coreccting mi baad speling.


Show that the polynomial p(z) = z^5-6z +3 has five distinct complex roots, of which exactly three are real.

[hide]By the rule of signs, there are either 2 or 0 positive real roots, and 1 negative real root.

$ x\equal{}z\minus{}1 \implies x^5\plus{}5x^4\plus{}10x^3\plus{}10x^2\minus{}x\minus{}2$, with one sign change, so there is one real root for positive x, ie, z>1. Clearly since there is one positive root, there must be 2 positive roots, from the first line.

This establishes that there are exactly three real roots.

[i]Since the rule of signs counts roots with multiplicity separately, it is clear that the negative root is a single root, and that the root for z>1 is a single root, and thus the root in [0,1] is also a single root (since there are 2 positive roots).[/i]

This leaves two non-real roots. These roots can't be equally valued, because they must be complex conjugates, and any number equal to its complex conjugate must be real.[/hide]


Greetings!

I want to wish everyone the best of luck at nationals. I know some states aren't done yet, but I am curious to see how many of the 228 kids visit the aops mc forum. Please cast your vote.

well...some people aren't going to vote honestly
just so u know

Also, remember to subtract 25 to correct for the Missouri multis. (I don't know, am I being sarcastic?) :wink:

Can we also introduce ourselves here?

I'm going. I took 2nd from Illinois.
I'm not sure if my friends/parents are going...but w/e
See you there! :D

Am I the only Missourian going wth no multis?

Hmmm...probably, that makes me deduce about you that you're a good girl :)

So yeah, I've pretty much had you on my candidates for rooming list, but it's pretty likely that I'm choosing one girl at random:

LHM
dragonlaird7
HolyMushroomspwnthywrld17 (if she makes nats, which I believe is highly likely)

[quote="math154"][quote="gauss1181"]Hmmm...probably, that makes me deduce about you that you're a good girl :) [/quote]

You might be surprised when you meet me then. :P[/quote]
haha :rotfl:  :rotfl:

[quote="gauss1181"][quote="math154"][quote="gauss1181"]Hmmm...probably, that makes me deduce about you that you're a good girl :) [/quote]

You might be surprised when you meet me then. :P[/quote]
haha :rotfl:  [/quote]

 :o

[quote="pinkmuskrat"]We should gather a list of aops usernames of people going to nats.[/quote]

Yeah, but if you posted a thread saying: Post here if you're going to MC Nats, some of the Missourians (besides me) would post like 50 times saying they were going and only like 3 non-MOians would post that they're going.

okay. compiling a list of what i can remember:

nikeballa96
pinkmuskrat
abacadaea
xpmath
LHM
dragonlaird7
math154
mewto55555
tinytim
gauss1181
dafjbomb
pythag011
james4l
wiz
ReseesPieces
Temperal
Urc

add to list if you know anyone else/their states, or yourself. this way, no one else from missouri can post.

uhh..
killertomatog (delaware)
De_2009 (kinda obvious)

they aren't that active on the forum, but sometimes are on ftw

Anyone know the list of people for the Georgia Nats team?
I have a friend there, and want to see if he made it or not..... Thanks! :D

nikeballa96
pinkmuskrat
abacadaea
xpmath
LHM
dragonlaird7
math154
mewto55555
tinytim
gauss1181
dafjbomb
pythag011
james4l
wiz
ReseesPieces
Temperal
Urc
killertomatog
De_2009
Eulers_Apprentice

keep copying the list, by tomorrow, we'll likely have a decent list of who is going.

add to list if you know anyone else/their states, or yourself. this way, no one else from missouri can post.[/quote]

nikeballa96
pinkmuskrat
abacadaea
xpmath
LHM
dragonlaird7
math154
mewto55555
tinytim
gauss1181
dafjbomb
pythag011
james4l
wiz
ReseesPieces
Temperal
Urc
killertomatog
De_2009
Eulers_Apprentice
chestnut
ash-pokemon

[b]keep copying the list[/b], by tomorrow, we'll likely have a decent list of who is going.

hmm if we really want to meet each other...i guess put ur name down here if u dont care about people stocking u....which im sure many people wont....

Wow people can use this to find out names of AoPSers, which is not advised (if they have any information about them at all, they could use the qualifiers list on the Mathcounts website). I don't think you should put down other people's states and usernames; putting down your own is fine.

copying and pasting the list isnt going to prevent people from looking at the past couple posts.....

Don't forget the following:

ytrewq
5849206328x
stevenmeow

Is it possible a mod could edit out the state names from the previous posts?  I don't think it's a good idea to link people and AoPS usernames w/o people's permission, it's not respecting their privacy

Dont put states only usernames

More:
nikeballa96
pinkmuskrat
abacadaea
xpmath
LHM
dragonlaird7
math154
mewto55555
tinytim
gauss1181
dafjbomb
pythag011
james4l
wiz
ReseesPieces
Temperal
Urc
killertomatog
De_2009
Eulers_Apprentice
chestnut
ash-pokemon
stevenmeow
icecapmeltdown
Holymushroomspwnthyworld17
cheesymath(?)
Ttocs45
aime_is_hard
dnkywin
Nexmus
Asian_n_nerdy
Cubex
ytrewq
5849206328x

me :P
some other guy on our team... don't remember his username... he's kinda quirky... don't want to share a room with him! :help:

you should put in alphabetical order, lol

also u don't have calculatorgeek on there... :maybe:

[quote="remy1140"]Fortunately I'm going to Nationals. :lol:
Unfortunately I'm not going to Nats for competition, but to have fun in Disney World... :([/quote]

how does that work out?

I'm going. Too lazy to copy and paste though :P

[hide]nikeballa96
pinkmuskrat
abacadaea
xpmath
LHM
dragonlaird7
math154
mewto55555
tinytim
gauss1181
dafjbomb
pythag011
james4l
wiz
ReseesPieces
Temperal
Urc
killertomatog
De_2009
Eulers_Apprentice
chestnut
ash-pokemon
stevenmeow
icecapmeltdown
Holymushroomspwnthyworld17
cheesymath(?)
connaissance
calculatorgeek
scai1
Michael2Choi
KingRoy
mathelete9999
[/hide]

I have to make nats next year too. I really failed.


$(3x - 2a)$ is a factor of:

$\text{A}: 9x^2 + 4a^2\\
\text{B}: 3x^2 - 2a^2\\
\text{C}: 3xy + 9bx - 2ay - 6ab\\
\text{D}: 9x^2 - 6ax + 4a^2\\
\text{E}: \text{None of the above}$

Enjoy. :)

[hide]I only tested A, and it worked.  I don't have time to solve the rest right now...[/hide]

[quote="DarkKnight"][hide]
I would say E.[/hide][/quote]

So would i

[quote="Silverfalcon"]Sorry, I think the problem may been little tougher than what I thought it was.

Answer is:

[hide="Sol"]
C

First, rearrange and we get $3xy - 2ay + 9bx - 6ab$. Then:

$y(3x-2a) + 3b(3x- 2a)\\
(y + 3b)(3x - 2a)$

So, the answer is $\fbox {C}$.[/hide][/quote]

Wow...I cant believe i missed that. I did something really dumb which lead me to believe it dosnt work.

Isn't this a bit harder than, "[i]The Basics[/i]"?

i think it should be [hide]D and E[/hide]


group theory
number theory
superior algebra
superior algebra theorems

Hello,

In my country we learn basic things about groups in 10th grade.
I just wanted to know at what grade you learn first about groups,and does really worth to buy a college book and start learning groups.
I mean , how  important they are to solve Olympiad problems ?

you can buy a book and start learning if u really like the subject,  but coming to olympiads,

groups is something thats not of preuniversity level.  so it will never be asked in  any  pre universioty level olympiads.

coming to graduate level olympiads it is yes  a important part of the   questions...

The abstraction of group theory is not covered on high school olympiads, but a lot of number theory problems can deal with group-theoretic properties like order, so the perspective can't hurt if you're interested in it.


Determine all integers $ m$ for which all solutions of the equation $ 3x^3\minus{}3x^2\plus{}m\equal{}0$ are rational.

Let $ r\equal{}\frac ab, (a,b)\equal{}1$ is rational solution, then $ m\equal{}3\frac{a^2}{b^3}(b\minus{}a)$. because $ m$ is integer $ b\equal{}1$ or $ r\equal{}a$ - integer.
Therefore all solutions must be integer. Obviosly $ 3|m$ or $ m\equal{}3n$ and $ n\equal{}x^2(1\minus{}x)$. Obviosly $ f(x)\equal{}x^2(1\minus{}x)$ had not one mean for 2 integers $ x<0,x>1$.
Unique case is $ x\equal{}0,0,1$ and $ n\equal{}m\equal{}0$.


Let $ a,$ $ b$ and $ c$ be non-negative numbers, no two of them are zero. Prove that:
\[ \frac {ab} {(a \plus{} b)^2} \plus{} \frac {ac }{(a \plus{} c)^2} \plus{} \frac { bc }{(b \plus{} c)^2} \geq \frac {(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc)}{4(a^3 \plus{} b^3 \plus{} c^3)}
\]

The stronger is also trues:
$ \sum\ \frac {ab}{a^2 \plus{} b^2} \geq\ \frac {(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}{2(a^3 \plus{} b^3 \plus{} c^3)}$
After using Cauchy Svac, it's becomes:
$ \sum\ ab(a \plus{} b)(a \minus{} b)^2 \geq\ 0$ (always trues)
We have done  :)
Equality occurs when $ a \equal{} b \equal{} c$ or $ a \equal{} b; c \equal{} 0.$

[quote="nguoivn"]After using Cauchy Svac[/quote]
What is the [color=red]Cauchy Svac[/color]???

Sorry, in VietNamese, Cauchy Schwart is called Cauchy Svac :D

Hey nguoivn! Can u provide me some guide to learn VietNamese, with actual pronounciations!!! PLZ!

How do you use Cauchy-Schwartz, [b]nguoivn[/b]?

Please someone answer me.

$ \sum\ \frac {ab}{a^2 \plus{} b^2} \geq\ \frac {(ab \plus{} bc \plus{} ca)^2}{\sum\ ab(a^2 \plus{} b^2)} \geq\ \frac {(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}{2(a^3 \plus{} b^3 \plus{} c^3)}$
$ \Leftrightarrow 2(ab \plus{} bc \plus{} ca)(a^3 \plus{} b^3 \plus{} c^3) \geq\ (a \plus{} b \plus{} c)\sum\ ab(a^2 \plus{} b^2)$
$ \Leftrightarrow \sum\ ab(a^3 \plus{} b^3) \geq\ \sum\ a^2b^2(a \plus{} b)$
$ \Leftrightarrow \sum\ ab(a \plus{} b)(a \minus{} b)^2 \geq\ 0$
 :)

Dear arqady,
I think you posted this problem yesterday,it's more difficult than this problem in topic:
Let $ a,b,c\ge 0$ and $ a\plus{}b\plus{}c\equal{}3$
$ \sum \frac{a^2\minus{}ab\plus{}b^2}{(a\plus{}b)^2}\plus{}\frac{3(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}{4(a^3\plus{}b^3\plus{}c^3)}\le 3$
How to solve this problem?

[quote="vo thanh van"]Dear arqady,
I think you posted this problem yesterday,it's more difficult than this problem in topic:
Let $ a,b,c\ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$
$ \sum \frac {a^2 \minus{} ab \plus{} b^2}{(a \plus{} b)^2} \plus{} \frac {3(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}{4(a^3 \plus{} b^3 \plus{} c^3)}\le 3$
How to solve this problem?[/quote]
$ \sum \frac {a^2 \minus{} ab \plus{} b^2}{(a \plus{} b)^2} \plus{} \frac {3(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}{4(a^3 \plus{} b^3 \plus{} c^3)}\le 3\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(1\minus{}\frac {a^2 \minus{} ab \plus{} b^2}{(a \plus{} b)^2}\right)\geq\frac {3(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}{4(a^3 \plus{} b^3 \plus{} c^3)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {ab }{(a \plus{} b)^2}\geq\frac {(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}{4(a^3 \plus{} b^3 \plus{} c^3)}.$ :wink:

nguoivn's one is

$ \sum{\frac {yz}{y^2 \plus{} z^2}} \geq\sum{\frac {yz\left[7x^2 \minus{} x(y \plus{} z) \plus{} (y \plus{} z)^2\right]}{2(x \plus{} y \plus{} z)(x^3 \plus{} y^3 \plus{} z^3)}} \equal{} \frac {(x \plus{} y \plus{} z)(yz \plus{} zx \plus{} xy)}{2(x^3 \plus{} y^3 \plus{} z^3)}.$
[hide="Proof."]$ 2(x \plus{} y \plus{} z)(x^3 \plus{} y^3 \plus{} z^3) \minus{} (y^2 \plus{} z^2)\left[7x^2 \minus{} x(y \plus{} z) \plus{} (y \plus{} z)^2\right]$

$ \equal{} 2x^4 \plus{} 2(y \plus{} z)x^3 \minus{} 7(y^2 \plus{} z^2)x^2 \plus{} (y \plus{} z)(3z^2 \minus{} 2yz \plus{} 3y^2)x \plus{} (y \minus{} z)^2(y \plus{} z)^2$

$ \equal{} \frac {2x\left[x(y \plus{} z) \plus{} 3y^2 \plus{} 2yz \plus{} 3z^2\right]\left[x^2(y \plus{} z) \minus{} 2x(y^2 \plus{} z^2) \plus{} y^3 \plus{} z^3\right]}{(y \plus{} z)^2}$

$ \plus{} \frac {(y \minus{} z)^2\left[(3y^2 \minus{} 2yz \plus{} 3z^2)x^2 \minus{} 3(y \plus{} z)(y^2 \plus{} z^2)x \plus{} (y \plus{} z)^4\right]}{(y \plus{} z)^2}\geq0$,

which is true for

$ (y \plus{} z)(y^3 \plus{} z^3)\geq(y^2 \plus{} z^2)^2$

and

$ 4(y \plus{} z)^2(3y^2 \minus{} 2yz \plus{} 3z^2)\minus{}9(y^2 \plus{} z^2)^2$

$ \equal{}3y^4 \plus{} 16y^3z \minus{} 10y^2z^2 \plus{} 16yz^3 \plus{} 3z^4\geq0$.[/hide]
The following stronger statement holds:

If $ x,y,z$ be positive numbers, then

$ \frac {yz}{4y^2 \minus{} 3yz \plus{} 4z^2} \plus{} \frac {zx}{4z^2 \minus{} 3zx \plus{} 4x^2} \plus{} \frac {xy}{4x^2 \minus{} 3xy \plus{} 4y^2}$

$ \geq\ \frac {(x \plus{} y \plus{} z)(yz \plus{} zx \plus{} xy)}{5(x^3 \plus{} y^3 \plus{} z^3)},$

with equality if and only if $ x \equal{} y \equal{} z$.


[quote="pascal_1623"]November 16,2004 Contest 1 Division E

Alexis, Emma, and Li play in the school band. One plays the flute. One plays the saxophone. one plays the drums. Alexis is a 6th grader. Alexis and the saxophone player practice together after school. Emma and the flute player are 5th grader. Who plays the drums?[/quote]

[hide="Answer"]  Since Alexis practices with the saxophone player, she her self cannot play the saxophone.  Since Alexis is a sixth grader and the flute player is a fifth grader, the only possible choice of instrument for $\text{\boxed{Alexis}}$ is the $\text{\boxed{drums}}$ [/hide]

Alexis. I hate this problem, because it was the one that holded me back from getting a George Lenchner Award.  :diablo:

i love these kind of problems. I use a square thing for them where you write the peope on one side and the instrument/something on the other side.
[hide="my answer"] is Alexis plays the drums.[/hide]

these problems are fun to do, but they're less mathematical. :(

- math, at least these problems aren't completely boring :P ...


geometry
geometric transformation
reflection
geometry proposed

I've got the following result

Given $\triangle ABC$ non-isosceles, let $\zeta (I,r)$ its incircle and $H_{b}$, $M_{b}$  the trace of its $B-$altitude and the middpoint of $\overline{CA}$ respectively. Let $T=IM_{b}\cap BH_{b}$. Prove that $TB=r$.

My proof is toooooooooooooooooooooooooooooooooooooooooooooooooooooooo long

Any solution??

Problem 40 of this thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=70276&start=160

[color=darkblue][b]I - Proof.[/b] Denote $X\in BC\in C(I,r)$ and suppose $a>c$. Prove easily that $M_{b}H_{b}=\frac{a^{2}-c^{2}}{2b}$ and $M_{b}X=\frac{a-c}{2}$. Therefore,
$\frac{IX}{TH_{b}}=\frac{M_{b}X}{M_{b}H_{b}}$ $\Longrightarrow$ $\frac{r}{TH_{b}}=\frac{a-c}{2}\cdot\frac{2b}{a^{2}-c^{2}}$ $\Longrightarrow$ $TH_{b}=\frac{r(a+c)}{b}$ $\Longrightarrow$ $BT=h_{b}-TH_{b}=$ $\frac{2pr}{b}-\frac{r(a+c)}{b}=$ $\frac{r[2p-(a+c)]}{b}=$ $\frac{br}{b}$ $\Longrightarrow$ $BT=r\ .$

[b]II - Proof.[/b] Denote the reflection $Y$ of the point $X$ w.r.t. of the point $I$ and the intersection $Z$ of the line $AC$ with the $B$-exincircle $C(I_{b},r_{b})$ of the triangle $ABC$. Prove easily that $XZ=a-c$ and $H_{b}Z=\frac{p(a-c)}{b}$. Then $Y\in BZ$. Indeed, $Y\in BZ$ $\Longleftrightarrow$ $\frac{ZX}{ZH_{b}}=\frac{XY}{AH_{b}}$ $\Longleftrightarrow$ $\frac{b}{p}=\frac{2r}{h_{b}}$ $\Longleftrightarrow$ $bh_{b}=2pr$, what is truly. Therefore, because the point $M_{b}$ is the middlepoint of the segment $[XZ]$, results $\overline{TIM_{b}}\parallel\overline{BYZ}$, i.e. the quadrilateral $BTIY$ is a parallelogram. Thus, $BT=IY$, i.e. $BT=r\ .$

[b]Remark.[/b] Denote the intersection $W\in BX\cap C(I_{b},r_{b})$. See the figure and what can you say ? [/color]

[quote="Virgil Nicula"][color=darkblue]Denote $X\in BC\in C(I,r)$ and suppose $a>c$. Prove easily that $M_{b}H_{b}=\frac{a^{2}-c^{2}}{2b}$ and $M_{b}X=\frac{a-c}{2}$. Therefore,
$\frac{IX}{TH_{b}}=\frac{M_{b}X}{M_{b}H_{b}}$ $\Longrightarrow$ $\frac{r}{TH_{b}}=\frac{a-c}{2}\cdot\frac{2b}{a^{2}-c^{2}}$ $\Longrightarrow$ $TH_{b}=\frac{r(a+c)}{b}$ $\Longrightarrow$ $BT=h_{b}-TH_{b}=$ $\frac{2pr}{b}-\frac{r(a+c)}{b}=$ $\frac{r[2p-(a+c)]}{b}=$ $\frac{br}{b}$ $\Longrightarrow$ $BT=r\ .$[/color][/quote]
Nice proof Virgil. :lol:


QUESTION: Two jokers are added to a 52 card deck and the entire stack of 54 cards is shuffled randomly. What is the expected number of cards that will be strictly between the two jokers? (HMMT 2009 Gen test)

I'm not sure how to finish my solution that I started, and also confused by the official solution. I'd greatly appreciate any insight or explanation.

My approach: let the jokers be denoted with "b" and the regular cards "a"
The number of total arrangements is the # of ways to arrange 52 b's and 2 a's which is 54C2

# of ways to get...
no cards between: (...bb...) call the bb chunk x; x can be anywhere from 1st to 53rd in the ordering of x and 52 a's, so there are 53 ways
1 card between: (...bab...) call the bab chunk x; there are 52 ways
2 cards between: (...baab...) similarly, there are 51 ways
... down to
52 cards between: (ba...ab) 1 way

So the total number of cards in all possibilities that are strictly between the two jokers is 53*0+52*1+51*2+50*3 + ... + 1*52

So the answer is (53*0+52*1+51*2+50*3 + ... + 1*52) / 54C2

But my question is, how do you evaluate the numerator? Is there any identity to use?

Also, the official solution is much shorter: it says, "Each card has an equal likelihood of being either on top of the jokers, in between them, or below the jokers. Thus, on average, 1/3 of them will land between the two jokers." But I don't understand their logic--can anyone explain the fairness of on top/between/below?

To finish your solution, write $ \sum_{i \equal{} 0}^{52} (53 \minus{} i) \cdot i \equal{} 53 \sum_{i \equal{} 0}^{52} i \minus{} \sum_{i \equal{} 0}^{52} i^2 \equal{} \ldots$.  Do you know the formulas for the sum of the first $ n$ integers and the first $ n$ perfect squares?

For the second one, your complaint is not very precise, so I'm not sure what you need explained.  Here are the key steps:
The ace of spades will be between the two jokers exactly one third of the time.  (Many ways to show this.)
This is true if I replace "ace of spades" with any other card.
By linearity of expectation, we expect exactly $ 52 \cdot \frac{1}{3}$ cards between the jokers.
Which part of this is problematic?

Thanks JBL.  But why is the ace of spades always between the two jokers 1/3 of the time, even as the number of other cards between the two jokers vary?

Well, as I said, there are a number of ways to show it.  Here's one: let's try to produce every possible order of the deck.  We'll start with the two jokers and the ace of spaces, and arrange them in any of three possible ways.  (In one of these three, the ace of spades is between the three jokers; in the other two it is not.)  Now, how many ways can we add the other cards in?  (We can actually answer that question exactly, but the important thing to note is that the answer will be the same, regardless of the order of the three cards we're interested in.)

It would of course be possible to compute this same probability by considering some complicated sum that involves the number of cards between the two jokers; the key insight here is that we can also count this more elegantly by not breaking the probability down into those smaller cases.  If you want to know the relative order of a given subset of cards (like, say, two jokers and an ace), it doesn't matter at all what other cards happen to have been shuffled up with them.

Here's another way of expressing what JBL is saying.

[b]Solution:[/b]

Consider the ordering $ A*B*C$ where $ *$'s are the two jokers, and $ A,B,C$ are three particular [i]ordered lists[/i] of cards which together comprise the rest of the deck.

Now, consider the two related orderings $ B*C*A$ and $ C*A*B$, where I just cycled the $ A,B,C$. Notice that in this family of three orderings, the average number of cards between the $ *$'s is $ \frac {a \plus{} b \plus{} c}{3} \equal{} \frac {52}{3}$.

We can sort all possible orderings of cards into such families of three orderings, in each of which the average is $ 52/3$. Thus, the global average is $ 52/3$.

[b]Note about the Ace of Spades:[/b]

In each such family of three orders, the Ace of Spades is going to be in one of the three lists $ A,B,C$. Thus, in each family of three orders, the Ace of Spades is going to be between the two jokers in one of the three orderings.


function
induction
functional equation
algebra
IMO 1981
IMO
romania

function $f:\mathbb{N}^{\star} \times \mathbb{N}^{\star} \rightarrow \mathbb{N}^{\star}$ ($\mathbb{N}^{\star}=\mathbb{N}\cup \{0\}$)with these conditon:

1- $f(0,x)=x+1$
2- $f(x+1,0)=f(x,1)$
3- $f(x+1,y+1)=f(x,f(x+1,y))$(romania  1997)
find $f(3,1997)$


from the sequence 1/(1*2); 1/(2*3);..;1/(n-1)n;.... determine the grupe of consecutive terms so that their sum is 1/3.

unknown source


[quote]Find all prime numbers p for which the number p 2 +11 has exactly 6 different divisors (including 1 and the number itself.)[/quote]

[hide]Means p 2  + 11 has two factors (letters are prime) q 2 and r or one factor s 5.  What do I do now?[/hide]

Good first step.  Now, can you figure out anything about what one of the numbers q, r or s would have to be?  A little bit more: [hide]If p isn't equal to 2, what is the parity of $p^2 + 11$?[/hide]

ah, u gotta extend that info to inquealities.  Ny creativity shows such limitations.

[hide]p>2  What now? lol.  Doing trial and error, it seems 3 is the onli possible answer, but I cant show why[/hide]

My method is probably wrong...

[hide]
Find all prime numbers p for which the number p 2 +11 has exactly 6 different divisors (including 1 and the number itself.)

So p^2 + 11 = p.a.b.c.d (where p, a, b, c, d aren't equal)
p^2 = p.a.b.c.d (mod 11)
Consider squares mod 11:
02 = 0
12 = 1
22 = 4
32 = 9
42 = 5
52 = 4
62 = 3
72 = 5
82 = 9
92 = 4
102 = 1
112 = 0
p is prime, this reduces the squares to:
22 = 4
32 = 9
52 = 4
72 = 5
112 = 0
p.a.b.c.d isn't prime or zero, this reduces the squares to:
22 = 4
32 = 9
52 = 4

Let p={2, 3, 5}.
p=2 -> 4+11=15 (less than 6 divisors)
p=3 -> 9+11=20 (6 divisors: 20, 1, 2, 10, 5, 4)
p=5 -> 5+11=16 (less than 6 divisors)

So p=3 is the only answer.

:P
[/hide]

This is for nodrog 2002:    3 is the only number required you are right!! To prove that you must prove that there aren’t any other p  >=  3 that have exactly 6 divisors. What happen if p 2 +11   >= 12? Make attention with the division by 12!!! How many divisors has 12?
It isn’t difficult!

This is for devoius:     Your answer is right but you must prove that there aren’t any other number like that!! Try!!

p^2  + 11  >= 12 will always be divisible by 12 when p>3 and since 12 has six factors, so p^2 + 11 will have at least 6+1=7 factors.

To make it clear:   when k> 0

 [hide] (6k +- 1)^2 + 11 =  36k^2 +- 12 K + 12 = 12* (something) .
Will always have more  than 6 divisors  (As you already  have 1,2,3,4,6,12 and the  large number itself,  at the minimum, will make the  number  of divisors more than 6 - keep in mind  as  k>0, "something"  > 1)

Only prime numbers  not of the form (6k +- 1) are 2 and 3.
and  only 3  gives a valid solution
P.S  "+-"   means  "+" or "-" above...
.[/hide]

[hide]  Any prime number greater than 2 is odd. So p^2 must be odd as well. Thus p^2 + 11 must be even, so at least one of the pair of factors is 2 and (p^2 +11)/2
  A prime number mod 3 can either be congruent to 1 or 2 (the only prime that is congruent to 0 mod 3 is 3). Any number that is congruent to 1 or 2 mod , squared is congruent to 1 mod 3. Also 11 is congruent to 2 mod 3, so any prime greater than 3 will result in p^2 + 11 being a multiple of 3.
  A prime number mod 3 can only be congruent to 1 or 3 mod 4 because they are odd numbers. If a number that is congruent to 1 or 3 mod 4 is squared, it is congruent to 1 mod 4 and 11 is congruent to 3 mod 4, when you add p^2 and 11, you get a multiple of 4.
  Since every prime greater than 3 fits those three catergories, they have more than 6 factors, so the only remaining prime numbers are 2 and 3. The number 2 doesn't work and 3 does, so 3 is the only possible prime number that works.
[/hide]