[quote="lutherford"]what is the greatest integer of
(1+1)(1+1/2)(1+1/4)(1+1/8)(1+1/16)....(1+1024)(1+1/2048) ?[/quote]
What do you mean by greatest integer?

[hide]$(1+1)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{16}\right)...\left(1+\frac{1}{1024}\right)\left(1+\frac{1}{2048}\right)$

$\left(\frac{2}{1}\right)\left(\frac{3}{2}\right)\left(\frac{5}{4}\right)\left(\frac{9}{8}\right)\left(\frac{17}{16}\right)...\left(\frac{1025}{1024}\right)\left(\frac{2049}{2048}\right)$

$\frac{x}{2^{66}}$

I'm still working on the $x$ part.[/hide]

[quote="junggi"][hide="hint"]
set it equal to x and multiply both sides by something that would be a cool catalyst[/hide][/quote]
I don't feel like multyplying $2^{66}$ by anything except $\frac{1}{2^{66}}.$ :rotfl:

[quote="junggi"]no I meant set the original crap($(1+1)(1+1/2)BLAHBLAHBLAH$) to equal x.[/quote]
How does that help? I can make it equal to anything and it'll still be $\frac{something}{2^{66}}.$

[quote="junggi"][hide="bleh"]
$(1+1)(1+1/2)\cdots(1+1/2048)=x$
now add the catalyst. multiply both sides by $(1-1/2)$
The thing becomes this by sum difference things
$2\cdot(1-1/4096)=x/2$
now you can solve for $x$.[/hide][/quote]
If you multiply everything by $\frac{1}{2},$ it doesn't make it simpler. :huh:

greatest integer means that you will remove the decimal part of the number.. for example greatest integer of 4.9 = 4..

that method will not work for
(1+1/8).. since it will not produce (1-1/8)..

oh hum.
then you can be lazy and multiply out the rest again, or multiply both sides by other craps

[quote="lutherford"]greatest integer means that you will remove the decimal part of the number.. for example greatest integer of 4.9 = 4.[/quote]
So can't you just say $\lfloor4.9\rfloor?$

I'm curious what you need this for.  It's related to partitions of integers, and the convergence of this particular product as the number of terms grows large is very quick (if we extend the product out to 100 terms, it only increases your expression by about .002).  I can think of reasonably complicated ways to bound this sum from above (using the power series for $\log 1+x$, for example) and reasonably simple ways (just multiply the damn thing out, or at least part of it and bound the rest), but no really nice ways.

Following up on Joel's suggestions:  Leave the first two factors as is.  Bound each of the other factors using the inequality $1+x \le e^{x}$.  You will get an upper bound good enough to find the floor (greatest integer).

but this question came from an elementary math contest.. i cant use advance topics to teach the kid.. is there a easier way to answer this?

Hmm, I'm not sure this question is appropriate for an elementary contest.  A pragmatic approach is to multiply the first 4 or 5 factors, and then assume that the other factors don't contribute much.

For a rigorous but elementary approach, we can use Bernoulli's inequality, which says that \[(1+x_{1}) (1+x_{2}) \cdots (1+x_{n}) \ge 1+x_{1}+x_{2}+\cdots+x_{n}\] provided that $x_{i}\ge-1$.  Our goal is to prove that \[(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{4}) (1+\frac{1}{8})(1+\frac{1}{16}) \cdots < 5 \, .\] By taking reciprocals, that inequality is equivalent to \[(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{5}) (1-\frac{1}{9})(1-\frac{1}{17}) \cdots > \frac{1}{5}\, .\] But by Bernoulli's inequality, we have \begin{eqnarray*}(1-\frac{1}{9})(1-\frac{1}{17}) \cdots & \ge & 1-\frac{1}{9}-\frac{1}{17}\cdots \\ & > & 1-\frac{1}{8}-\frac{1}{16}\cdots \\ & = & 1-\frac{1}{4}\\ & = & \frac{3}{4}\, . \end{eqnarray*} That bound is just enough to establish our goal.

sori, but i did not understand the solution.. can you please explain further? thanks!

Okay, let me try again.  Consider the following sequence of numbers, which are one more than a power of two: $9$, $17$, $33$, $\ldots$, $2049$.  By Bernoulli's inequality, we have \begin{eqnarray*}(1-\frac{1}{9})(1-\frac{1}{17})(1-\frac{1}{33}) \cdots (1-\frac{1}{2049}) & \ge & 1-\frac{1}{9}-\frac{1}{17}-\frac{1}{33}-\cdots-\frac{1}{2049}\\ & > & 1-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\cdots-\frac{1}{2048}\\ & = & 1-(\frac{1}{4}-\frac{1}{2048}) \\ & > & 1-\frac{1}{4}\\ & = & \frac{3}{4}\, . \end{eqnarray*} But the reciprocal of $1-\frac{1}{9}$ is $1+\frac{1}{8}$; the reciprocal of $1-\frac{1}{17}$ is $1+\frac{1}{16}$; and so on.  Thus, by taking reciprocals of both sides, we get \[(1+\frac{1}{8})(1+\frac{1}{16})(1+\frac{1}{32}) \cdots (1+\frac{1}{2048}) < \frac{4}{3}\, .\] The missing factors $1+\frac{1}{1}$, $1+\frac{1}{2}$, and $1+\frac{1}{4}$ multiply out to $\frac{15}{4}$.  Thus \[(1+\frac{1}{1})(1+\frac{1}{2})(1+\frac{1}{4}) (1+\frac{1}{8}) \cdots (1+\frac{1}{2048}) < \frac{15}{4}\cdot \frac{4}{3}= 5 \, .\] So the given product is less than $5$.  Since the first four factors multiply to more than $4$, the floor of the product must be $4$.

Did that make sense?  If not, tell me which step you didn't understand.

ok thank you very much i got it, but is there an easier solution? because the problem came from elementary contest for age 10~12 years old, do you think it is ok for me to teach this principle to them?

I can't think of an easier solution.  Maybe someone else has an idea?

For an 11-year-old, I would calculate the first few partial products, and then informally explain why the products are not likely to get much bigger.

By the way, which contest is this problem from?

sorry for the late reply..  typhoon milenyo struck philippines and there is no electricity here for 4 days..

my friend who is a grade four student, just ask me the question and and he said it came from a contest, i do not know what and where exactly the contest is..

@Ravi- isn't junggi's a good elementary solution?

I think junggi and lutherford already agreed that junggi's solution was incorrect.  junggi said that $(1-\frac{1}{4})(1+\frac{1}{4})$ equals $1-\frac{1}{8}$, but in reality it equals $1-\frac{1}{16}$.

If the problem were to evaluate the product \[(1+\frac{1}{2})(1+\frac{1}{4})(1+\frac{1}{16})(1+\frac{1}{256}) \cdots \, ,\] then junggi's method would be perfect.  But that's a different problem.


I will not keep my word and post a little more elegant approach for the first part of solution. Write the RHS in the form $\sum{\frac{(1+a^2)(1+b^2)-(a+b)^2+2ab}{(a+b)^2}}$. Use the fact that $(1+a^2)(1+b^2)=(a+b)^2(a+c)(b+c)$ to reduce it instantly to $\sum{a^2}+\sum{\frac{2ab}{(a+b)^2}}$ and from here get the equivalent form $\sum{(a-b)^2}\geq\sum{\frac{(a-b)^2}{(a+b)^2}}$ and from here we destroy it like Peter did with the asumption $a\geq b\geq c$.

welcome back harazi!!! I hope you will continue to post here. Nice solution

Sorry, I won't continue to post here, but I thought this deserves a thought.

[quote="silouan"]If $ a$, $ b$, $ c$ are positive reals such that $ ab \plus{} bc \plus{} ca \equal{} 1$, then prove the following inequality:

$ \frac {1 \plus{} a^2b^2}{(a \plus{} b)^2} \plus{} \frac {1 \plus{} b^2c^2}{(b \plus{} c)^2} \plus{} \frac {1 \plus{} c^2a^2}{(c \plus{} a)^2}\geq\frac {5}{2}$.

It was posted a solution by Peter. But I think all we want tose another methods and solutions. And Peter said that Dariij has a solution. Please post your solutions. thanks[/quote]

You do this:

$x,y,z>0,xy+yz+zx=1$,prove that:

\[\frac{1+y^2z^2}{(y+z)^2}+\frac{1+x^2z^2}{(z+x)^2}+\frac{1+x^2y^2}{(x+y)^2} \geq  \frac{5}{6}(\frac{z+xy}{y+xz}+\frac{x+yz}{z+xy}+\frac{y+xz}{x+yz})\geq \frac{5}{2}\]

[quote=Nguyenhuyen_AG][quote=silouan]If $ a$, $ b$, $ c$ are positive reals such that $ ab \plus{} bc \plus{} ca \equal{} 1$, then prove the following inequality:
$$\frac {1 \plus{} a^2b^2}{(a \plus{} b)^2} \plus{} \frac {1 \plus{} b^2c^2}{(b \plus{} c)^2} \plus{} \frac {1 \plus{} c^2a^2}{(c \plus{} a)^2}\geq\frac {5}{2}.$$[/quote]
Because $1+a^2b^2=(ab+bc+ca)^2+a^2b^2 = 2ab(ab+bc+ca)+c^2(a+b)^2,$ therefore inequality become
\[\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{2ab}{(a+b)^2}+\frac{2bc}{(b+c)^2}+\frac{2ca}{(c+a)^2} \geqslant \frac52.\]
Suppose $c = \min\{a,b,c\},$ we have
\[\frac{a^2+b^2+c^2}{ab+bc+ca} \geqslant \frac{a^2+b^2+c^2+c^2}{ab+bc+ca+c^2} = \frac{a^2+b^2+2c^2}{(a+c)(b+c)}.\]
Therefore we need to prove
\[\frac{a^2+b^2+2c^2}{(a+c)(b+c)}+\frac{2ab}{(a+b)^2}+\frac{2bc}{(b+c)^2}+\frac{2ca}{(c+a)^2} \geqslant \frac52,\]
equivalent to
\[\frac{(a-b)^2[2(b+c)a^3+(3b^2+4bc-3c^2)a^2+2(b^3+2b^2c-4bc^2-c^3)a+c(2b^3-3b^2c-2bc^2-c^3)]}{2(a+b)^2(b+c)^2(c+a)^2} \geqslant 0,\]
or
\[\frac{(a-b)^2 \cdot P}{2(a+b)^2(b+c)^2(c+a)^2} \geqslant 0.\]
But
\[P = 4bc(a-b)^2 +3b^2(c-a)^2+2a(c^2+6bc+a^2)(b-c)+(2b^3+c^3+c^2a+6b^2c+4ca^2+2bc^2)(a-c) \geqslant 0.\]
Done.[/quote]

Dear Nguyenhuyen_AG , and mudok after this  $1+a^2b^2=(ab+bc+ca)^2+a^2b^2  $  is not possible that?

$\sum_{cyc}\frac{1+a^2b^2}{(a+b)^2}=\sum_{cyc}\frac{(ab+bc+ca)^2+a^2b^2}{(a+b)^2}= $

$ =\sum_{cyc}a^2+2\sum_{cyc}\frac{abc}{a+c}+\frac{2a^2b^2}{(a+b)^2} $ Now ineguality becames

$2(\sum_{cyc}a^2-\sum_{cyc}ab)+\sum_{cyc}c(\frac{4ab}{a+b}-(a+b))+\sum_{cyc}(\frac{4a^2b^2}{(a+b)^2}-ab)\geq0$

[quote=mathsign1][quote=arqady][quote="harazi"]$\sum{(a-b)^2}\geq\sum{\frac{(a-b)^2}{(a+b)^2}}$ and from here we destroy it like Peter did with the asumption $a\geq b\geq c$.[/quote]
harazi, I don't think that  it can be destroyed so. ;)
But $\sum_{cyc}\frac{1+a^2b^2}{(a+b)^2}-\frac{5}{2}=\frac{1}{2}\cdot\sum_{cyc}(a-b)(a-c)\left(\frac{a^2}{(a+b)^2}+\frac{a^2}{(a+c)^2}\right).$
If $a\geq b\geq c$ then $\frac{a^2}{(a+b)^2}+\frac{a^2}{(a+c)^2}\geq\frac{b^2}{(a+b)^2}+\frac{b^2}{(b+c)^2}.$ This is obviously true.
And our inequality is proved. :)[/quote]

Sorry to ask but i dont know how can you write the inequality in that form, so could you help me, im bad in this[/quote]

I think it’s based on
$(a-b)^2=(a-b)(a-c)+(b-a)(b-c)$ so do the others

[quote=KaiRain][quote=mathsign1]
Sorry to ask but i dont know how can you write the inequality in that form, so could you help me, im bad in this[/quote]
You can rely on Harazi's hint, it comes from
$(a-b)^2(1-\frac{ab+bc+ca}{(a+b)^2})=(a-b)(\frac{a^2(a-c)}{(a+b)^2}-\frac{b^2(b-c)}{(a+b)^2})$.[/quote]

Tks a lot

[quote=teomihai]

Dear Nguyenhuyen_AG , and mudok after this  $1+a^2b^2=(ab+bc+ca)^2+a^2b^2  $  is not possible that?

$\sum_{cyc}\frac{1+a^2b^2}{(a+b)^2}=\sum_{cyc}\frac{(ab+bc+ca)^2+a^2b^2}{(a+b)^2}= $

$ =\sum_{cyc}a^2+2\sum_{cyc}\frac{abc}{a+c}+\frac{2a^2b^2}{(a+b)^2} $ Now ineguality becames

$2(\sum_{cyc}a^2-\sum_{cyc}ab)+\sum_{cyc}c(\frac{4ab}{a+b}-(a+b))+\sum_{cyc}(\frac{4a^2b^2}{(a+b)^2}-ab)\geq0$[/quote]

was used: $(ab+bc+ca)^2+a^2b^2=c^2(a+b)^2+2ab(ab+bc+ca)$


Solve in integers

(5 + 3 \sqrt 2)^m = (3 + 5 \sqrt 2)^n

Nice.
a+b \sqrt 2=a'+b' \sqrt 2 implies a=a' and b=b', for a,b,a',b' integers. Otherwise \sqrt 2 is in Q.

From this it should be easy to get that
(5+3 \sqrt2)m=(3+5\sqrt2)n implies
(5-3 \sqrt2)m=(3-5\sqrt2)n.
Multiply the last two to get
7m=(-1)n. So m=0 and n is even.
Back into (5+3 \sqrt2)m=(3+5\sqrt2)n, we get n=0 and thus the only solution n=0,m=0.


[quote="mathelete"][quote="pythag011"][quote="mathelete"]You have to quote the previous post and describe it. I wonder what the result will be.[/quote]
lol. You created this game because of my comment in beat the mod.[/quote]
Yes, I did.[/quote]

I was going to make a thread similar to this one weeks ago, but I thought that there would be complaints of spam.

[quote="Wickedestjr"][/quote][quote="mathelete"][quote="pythag011"][quote="mathelete"]You have to quote the previous post and describe it. I wonder what the result will be.[/quote]
lol. You created this game because of my comment in beat the mod.[/quote]
Yes, I did.[/quote][quote="Wickedestjr"]

I was going to make a thread similar to this one weeks ago, but I thought that there would be complaints of spam.[/quote]

There will be complaints of spam because this is spam... though you might sort of make a fractal by doing this...

[quote="pythag011"][/quote][quote="Wickedestjr"][/quote][quote="mathelete"][quote="pythag011"][quote="mathelete"]You have to quote the previous post and describe it. I wonder what the result will be.[/quote]
lol. You created this game because of my comment in beat the mod.[/quote]
Yes, I did.[/quote][quote="Wickedestjr"]

I was going to make a thread similar to this one weeks ago, but I thought that there would be complaints of spam.[/quote][quote="pythag011"]

There will be complaints of spam because this is spam... though you might sort of make a fractal by doing this...[/quote]

I don't know what a fractal is, but I think we could chat about something so that this thread isn't considered spam. Any ideas? We could start off by making our posts a little bit longer.

[quote="Wickedestjr"][/quote][quote="pythag011"][/quote][quote="Wickedestjr"][/quote][quote="mathelete"][quote="pythag011"][quote="mathelete"]You have to quote the previous post and describe it. I wonder what the result will be.[/quote]
lol. You created this game because of my comment in beat the mod.[/quote]
Yes, I did.[/quote][quote="Wickedestjr"]

I was going to make a thread similar to this one weeks ago, but I thought that there would be complaints of spam.[/quote][quote="pythag011"]

There will be complaints of spam because this is spam... though you might sort of make a fractal by doing this...[/quote][quote="Wickedestjr"]

I don't know what a fractal is, but I think we could chat about something so that this thread isn't considered spam. Any ideas? We could start off by making our posts a little bit longer.[/quote]
Go to the pre-olympiad section. I have posted a topic called "three problems". Also check the topic whose name is in latex.

[quote="mathelete"][/quote][quote="Wickedestjr"][/quote][quote="pythag011"][/quote][quote="Wickedestjr"][/quote][quote="mathelete"][quote="pythag011"][quote="mathelete"]You have to quote the previous post and describe it. I wonder what the result will be.[/quote]
lol. You created this game because of my comment in beat the mod.[/quote]
Yes, I did.[/quote][quote="Wickedestjr"]

I was going to make a thread similar to this one weeks ago, but I thought that there would be complaints of spam.[/quote][quote="pythag011"]

There will be complaints of spam because this is spam... though you might sort of make a fractal by doing this...[/quote][quote="Wickedestjr"]

I don't know what a fractal is, but I think we could chat about something so that this thread isn't considered spam. Any ideas? We could start off by making our posts a little bit longer.[/quote][quote="mathelete"]
Go to the pre-olympiad section. I have posted a topic called "three problems". Also check the topic whose name is in latex.[/quote]
BTW, the third problem is absolutely trivial if you have learned Fermat's Little Thereom. Or you could use repunits and pigeonhole... oh well.

[quote="pythag011"][/quote][quote="mathelete"][/quote][quote="Wickedestjr"][/quote][quote="pythag011"][/quote][quote="Wickedestjr"][/quote][quote="mathelete"][quote="pythag011"][quote="mathelete"]You have to quote the previous post and describe it. I wonder what the result will be.[/quote]
lol. You created this game because of my comment in beat the mod.[/quote]
Yes, I did.[/quote][quote="Wickedestjr"]

I was going to make a thread similar to this one weeks ago, but I thought that there would be complaints of spam.[/quote][quote="pythag011"]

There will be complaints of spam because this is spam... though you might sort of make a fractal by doing this...[/quote][quote="Wickedestjr"]

I don't know what a fractal is, but I think we could chat about something so that this thread isn't considered spam. Any ideas? We could start off by making our posts a little bit longer.[/quote][quote="mathelete"]
Go to the pre-olympiad section. I have posted a topic called "three problems". Also check the topic whose name is in latex.[/quote][quote="pythag011"]
BTW, the third problem is absolutely trivial if you have learned Fermat's Little Thereom. Or you could use repunits and pigeonhole... oh well.[/quote]
wait. what's going on again?? I'm so sad i got kicked from beat the mod.. T.T

[quote="kimmystar94"][/quote][quote="pythag011"][/quote][quote="mathelete"][/quote][quote="Wickedestjr"][/quote][quote="pythag011"][/quote][quote="Wickedestjr"][/quote][quote="mathelete"][quote="pythag011"][quote="mathelete"]You have to quote the previous post and describe it. I wonder what the result will be.[/quote]
lol. You created this game because of my comment in beat the mod.[/quote]
Yes, I did.[/quote][quote="Wickedestjr"]

I was going to make a thread similar to this one weeks ago, but I thought that there would be complaints of spam.[/quote][quote="pythag011"]

There will be complaints of spam because this is spam... though you might sort of make a fractal by doing this...[/quote][quote="Wickedestjr"]

I don't know what a fractal is, but I think we could chat about something so that this thread isn't considered spam. Any ideas? We could start off by making our posts a little bit longer.[/quote][quote="mathelete"]
Go to the pre-olympiad section. I have posted a topic called "three problems". Also check the topic whose name is in latex.[/quote][quote="pythag011"]
BTW, the third problem is absolutely trivial if you have learned Fermat's Little Thereom. Or you could use repunits and pigeonhole... oh well.[/quote][quote="kimmystar94"]
wait. what's going on again?? I'm so sad i got kicked from beat the mod.. T.T[/quote]

@pythag - To make this thread of good use, could you show us all Fermat's Little Theorum?  :P

[quote="Wickedestjr"]I was going to make a thread similar to this one weeks ago, but I thought that there would be complaints of spam.[/quote]

You are correct - this is spam. And because this is spam, this thread is locked.


algebra
polynomial
calculus
calculus computations

P(x) is a polynomial of degree n, it has n zeros which all are greater than 1.
let $ Q(x) \equal{} (x \plus{} 1)P(x)P'(x) \plus{} (P(x))^2 \plus{} x(P(x))^2$.
 prove that Q(x) has at least 2n-1 dintinct zeros

$ Q(x) \equal{} P(x)(1\plus{}x)[P(x) \plus{} P'(x)]$.
Use the fact that the degree of f(x)g(x) = degree of f(x) + degree of g(x) and degree of f(x) + g(x) = maximum of the degrees of f(x) and g(x). This should tell you what the degree of Q(x) is and then you may need the Fundamental Theorem of Algebra to arrive at the conclusion.

Q(x) has degree 2n+1.
All roots of P(x) are roots of Q(x). Let those be a1, a2,.................an. (I assume that the a's are distinct, the justification for which will be given later).

Now that I have assumed P(x) has all its roots distinct, P(x) and P'(x) cant have a common root. And hence, none of the a's are roots of T(x)=P(x)+P'(x).

Now, let F(x)=P(x).exp(x), F'(x)=exp(x).T(x) Remmember, Q(x)=(x+1)P(x)T(x)

F(x) has zeroes at all the a's. By mean value theorem, there exists atleast one zero of F'(x), between eveery consecutive pair of a's (i.e, there exists bi in [ai, ai+1], where F'(bi)=0) Clearly these b's are distinct. Also none of the b's is equal to any a.

Hence, F'(x) has atleast n-1 distinct roots ([size=100][size=75]that too, within the range of the a's, i.e. the smallest b is bigger than the smallest a, and biggest b is smaller than the biggest a)

Hence T(x) has atleast n-1 distinct roots. Hence Q(x) has atleast 2n-1 distinct roots.

Now all the a's are bigger than 1 (that's given). Hence, -1 is another distinct root of Q(x). Hence, it has 2n distinct roots.

Now, coming to the assumption, that the roots of P(x) are distinct. It is easily shown that, if P(x) has more than one repeated roots, the result can be disproved. Hence, this was supposed to be given in the question.

[quote="bharathhm"]Now, coming to the assumption, that the roots of P(x) are distinct. It is easily shown that, if P(x) has more than one repeated roots, the result can be disproved. Hence, this was supposed to be given in the question.[/quote]   :roll: For a particular example, consider $ P(x) \equal{} (x \minus{} 2)^n$, in which case $ Q(x) \equal{} (x \minus{} 2)^{2n \minus{} 1}(x \plus{} 1)(x \plus{} n \minus{} 2)$.


Hi,
i was  :maybe:  about where to post it .

let x y and z be real numbers such that :

x / (y-z)  + y / (z-x)  + z/(x-y) = 0

show that :

x / (y-z)²  + y / (z-x)² + z / (x-y)² = 0

[quote="kprepaf"]Hi,
i was  :maybe:  about where to post it .

let x y and z be real numbers such that :

$ \frac{x}{y\minus{}z}  \plus{} \frac{y}{z\minus{}x}  \plus{} \frac{z}{x\minus{}y} \equal{} 0$

show that :

$ \frac{x}{(y\minus{}z)^{2}}  \plus{} \frac{y}{(z\minus{}x)^{2}}  \plus{} \frac{z}{(x\minus{}y)^{2}} \equal{} 0$[/quote]

$ \frac{x}{y\minus{}z}  \plus{} \frac{y}{z\minus{}x}  \plus{} \frac{z}{x\minus{}y} \equal{} 0$ ---> $ \frac{x(z\minus{}x)\plus{}y(y\minus{}z)}{(z\minus{}x)(y\minus{}z)}\plus{}\frac{z}{x\minus{}y}\equal{}\frac{(x\minus{}y)(z\minus{}x\minus{}y)}{(z\minus{}x)(y\minus{}z)}\plus{}\frac{z}{x\minus{}y}\equal{}0$ ---> $ \frac{(x\minus{}y)(z\minus{}x\minus{}y)}{(z\minus{}x)(y\minus{}z)}\equal{}\minus{}\frac{z}{x\minus{}y}$ ----> $ \frac{z}{(x\minus{}y)^{2}}\equal{}\frac{x\plus{}y\minus{}z}{(z\minus{}x)(y\minus{}z)}$......

$ \frac{y}{(z\minus{}x)^{2}}\equal{}\frac{x\plus{}z\minus{}y}{(x\minus{}y)(y\minus{}z)}$
$ \frac{x}{(y\minus{}z)^{2}}\equal{}\frac{z\plus{}y\minus{}x}{(z\minus{}x)(x\minus{}y)}$-------->

$ \frac{x}{(y\minus{}z)^{2}}  \plus{} \frac{y}{(z\minus{}x)^{2}}  \plus{} \frac{z}{(x\minus{}y)^{2}}\equal{}\frac{z\plus{}y\minus{}x}{(z\minus{}x)(x\minus{}y)}\plus{}\frac{x\plus{}z\minus{}y}{(x\minus{}y)(y\minus{}z)}\plus{}\frac{x\plus{}y\minus{}z}{(z\minus{}x)(y\minus{}z)}\equal{}\frac{(z\plus{}y\minus{}x)(y\minus{}z)\plus{}(x\plus{}z\minus{}y)(z\minus{}x)\plus{}(x\plus{}y\minus{}z)(x\minus{}y)}{(x\minus{}y)(y\minus{}z)(z\minus{}x)}\equal{}\frac{zy\plus{}y^{2}\minus{}xy\minus{}z^{2}\minus{}yz\plus{}zx\plus{}xz\plus{}z^{2}\minus{}yz\minus{}x^{2}\minus{}zx\plus{}xy\plus{}x^{2}\plus{}xy\minus{}zx\minus{}xy\minus{}y^{2}\plus{}zy}{(x\minus{}y)(y\minus{}z)(z\minus{}x)}\equal{}\frac{0}{(x\minus{}y)(y\minus{}z)(z\minus{}x)}\equal{}0$

Thanks Zaya_yc for your effort i thought it was harder  :roll:

Let $ y \minus{} z \equal{} a,z \minus{} x \equal{} b,x \minus{} y \equal{} c$
We have $ a\plus{}b\plus{}c\equal{} \frac {x}{a}\plus{}\frac {y}{b}\plus{}\frac {z}{c} \equal{} ax\plus{}by\plus{}cz \equal{} 0$
By cross multiplication from the later two: $ \frac {x}{\frac {c}{b} \minus{} \frac {b}{c}} \equal{} \frac {y}{\frac {a}{c} \minus{} \frac {c}{a}} \equal{} \frac {z}{\frac {b}{a} \minus{} \frac {a}{b}} \equal{} k$
Now $ \sum_{cyc} \frac {x}{a^2} \equal{} k\sum_{cyc} \frac {\frac {c}{b} \minus{} \frac {b}{c}}{a^2}$
$ \equal{} \frac {k}{abc}\sum_{cyc} \frac {(c \plus{} b)(c \minus{} b)}{a} \equal{} \frac {k}{abc}\sum_{cyc} \frac {( \minus{} a)(c \minus{} b)}{a} \equal{} \frac {k}{abc}\sum_{cyc} (b \minus{} c) \equal{} 0$

Thanks Akashnil for the second method  :roll:

This was posted recently.  scorpius119 suggested the following:  compute

$ \left( \frac{x}{y\minus{}z} \plus{} \frac{y}{z\minus{}x} \plus{} \frac{z}{x\minus{}y} \right) \left( \frac{1}{y\minus{}z} \plus{} \frac{1}{z\minus{}x} \plus{} \frac{1}{x\minus{}y} \right)$

which, after simplification, is the desired expression plus

$ \frac{(x \plus{} y)(x \minus{} y) \plus{} (y \plus{} z)(y \minus{} z) \plus{} (z \plus{} x)(z \minus{} x)}{(x \minus{} y)(y \minus{} z)(z \minus{} x)} \equal{} 0$.

[quote="Akashnil"]Let $ y \minus{} z \equal{} a,z \minus{} x \equal{} b,x \minus{} y \equal{} c$
We have $ a \plus{} b \plus{} c \equal{} \frac {x}{a} \plus{} \frac {y}{b} \plus{} \frac {z}{c} \equal{} ax \plus{} by \plus{} cz \equal{} 0$
[color=red]By cross multiplication from the later two: [/color]$ \frac {x}{\frac {c}{b} \minus{} \frac {b}{c}} \equal{} \frac {y}{\frac {a}{c} \minus{} \frac {c}{a}} \equal{} \frac {z}{\frac {b}{a} \minus{} \frac {a}{b}} \equal{} k$
Now $ \sum_{cyc} \frac {x}{a^2} \equal{} k\sum_{cyc} \frac {\frac {c}{b} \minus{} \frac {b}{c}}{a^2}$
$ \equal{} \frac {k}{abc}\sum_{cyc} \frac {(c \plus{} b)(c \minus{} b)}{a} \equal{} \frac {k}{abc}\sum_{cyc} \frac {( \minus{} a)(c \minus{} b)}{a} \equal{} \frac {k}{abc}\sum_{cyc} (b \minus{} c) \equal{} 0$[/quote]

Can you explain the skipped step?

$ a_1x \plus{} b_1y \plus{} c_1z \equal{} a_2x \plus{} b_2y \plus{} c_2z \equal{} 0\implies c_2(a_1x \plus{} b_1y \plus{} c_1z) \minus{} c_1(a_2x \plus{} b_2y \plus{} c_2z) \equal{} 0\implies x(a_1c_2 \minus{} c_1a_2) \equal{} y(b_1c_2 \minus{} c_1b_2)\implies \frac {x}{b_1c_2 \minus{} c_1b_2} \equal{} \frac {y}{a_1c_2 \minus{} c_1a_2}$.
Similarly $ \frac {y}{a_1c_2 \minus{} c_1a_2} \equal{} \frac {z}{a_1b_2 \minus{} b_1a_2}$
So $ \frac {x}{b_1c_2 \minus{} c_1b_2} \equal{} \frac {y}{a_1c_2 \minus{} c_1a_2} \equal{} \frac {z}{a_1b_2 \minus{} b_1a_2}$
In my solution $ a_1\equal{}\frac{1}{a},b_1\equal{}\frac{1}{b},c_1\equal{}\frac{1}{c},a_2\equal{}a,b_2\equal{}b,c_2\equal{}c,x\equal{}x,y\equal{}y,z\equal{}z$


[quote="sugarwergoingdown"]:bruce:  :wow:
Three circles with centers A, B, and C are tangent to each other as shown. The radius of circle A is 4 units, the radius of circle B is 5 units and the radius of circle C is 6 units. What is the area of triangle ABC? Express your answer to the nearest tenth.

 :idea: *[b]Note[/b]* See attachment for diagram.[/quote]

[hide]Nice picture.  :lol:

We see that $AB=4+5=9$
$BC=5+6=11$
$AC=4+6=10$

so the area is $\sqrt{(15)(6)(4)(5)}$
$\sqrt{1800}$
$\boxed{30\sqrt{2}}$[/hide]

[quote="sugarwergoingdown"]To the nearest tenth, I said. ;)[/quote]

sorry  :lol:
[hide]42.4[/hide]

[hide]
The sides are $4+5=9,4+6=10,5+6=11$
The semiperimeter is 15. Applying Herons fomula:
$\sqrt{15(6)(5)(4)}$
which, to the nearest tenth, is $\boxed{42.4}$[/hide]


I was working a test and got stuck on these two. Could someone please help me?

1. How many zeroes are at the end of 17! expressed in base 8?

2. For integers n:ge:1, let In = 2 + 2(2:^2:) + 3(2:^3:) + 4(2:^4:) . . . + n(2 :^n: ). The digits of I21 sum to what?

[hide]

For question 1:

When you do this question in base 10, you want to find out how many times 10 goes into the number. So in base 8 we want how many times 17! is divisible by 8 (which is 2:^3:). So count how many times 2 goes into it and then..

For question 2:

First thing to do is work out an actual smaller formula for In. Since theres all those powers of 2, the general way of working it out is letting the sum be S, calculating 2S term by term and then subtracting. Then you have a formula. Now you just gotta work out a huge power of two times 10 plus 2, which is just (heavy) arithmetic. Don't see any nicer way of doing it.

[/hide]

Snazzy.  For number 1, you first have to answer this question:  what makes a zero at the end of a number base 10?  Why?  So, what will make a zero at the end of a number base 8?  Then you should be able to count straightforwardly.

Well, it might help if you find a general formula for In.  I recommend doubling and subtracting.

Ok, I just got the first one, thanks =D. Could this be the same thing as asking What is the largest N such that 8 :^n: divides 17! ?

On number 2, I don't see how what you're doing is helping. If you do 2S and subtract doesn't that just give you S? (Sorry I probably sound stupid  :? )

Yep. Exactly. But when you work it out term by term, you will end up with a different expression for S. Try it.

Umm...for the first one, did you divide the number of 2s in the factorization by 3?

And on number 2, multiply the series by 2, and evaluate that.  Then you halve it.  Just try it.


1. [url]http://www.mathlinks.ro/viewtopic.php?t=237483[/url]
2. [url]http://www.mathlinks.ro/viewtopic.php?t=237488[/url]
3. [url]http://www.mathlinks.ro/viewtopic.php?p=1071295#p1071295[/url] (IMO Shortlist 2007, C2)

4. [url]http://www.mathlinks.ro/viewtopic.php?t=237485[/url]
5. [url]http://www.mathlinks.ro/viewtopic.php?t=237484[/url]
6. [url]http://www.mathlinks.ro/viewtopic.php?t=237482[/url]

Thanks for posting all those problems! What about 2004. Do you also have Israeli NMO?


[quote="bodan"]Prove that \[ x^3+y^3+z^3+\frac{(x+y+z)(x^2+y^2+z^2)}{3}\geq6\left(\frac{x^2+y^2+z^2}{3}\right)^{\frac{3}{2}}  \] for positive reals $x,y,z.$[/quote]
Assume that $x^2+y^2+z^2=3$
We have to prove: $x^3+y^3+z^3+x+y+z\geq 6$
It's easy: $x^3+x\geq 2 x^2$


MIT college Harvard geometry Princeton Yale

Ok my current list is:

1) MIT!!!
2) Harvard :(
3) Cal-Tech

I also have a multitude of other schools in consideration.

Now, it has come to my attention that these are all absurdly expensive.  45K a year is an absolute ton, especially in high-expense areas such as these.

My family, in all honesty, does not generate this amount of money in a year (although it's likely it will by the time I reach college.  Dad opened a business a year ago... you get the idea).

I would like to hear from someone who attended any of these schools (as well as princeton, oxford, yale, northwestern, etc) answers to the following questions:

1)  Is the financial aid really adequate?

2)  How much debt does going to an expensive school really compile?

3)  What is the average post-bachelor degree pay a graduate of such a school can earn?

4)  How possible is it to afford being a grad student afterward (assuming the school is the same)?

:)

i can just answer #1...
I would say of your list MIT has the best Fin Aid program...
it is very very good...

Affording really expensive schools has been discussed multiple times in this forum before.  So if you dig through it you can find more information.

I'm not sure about how much the high living expense of the region affects college students that much.  Living on campus all four years is not that uncommon I believe.  The cost of dorms is already factored into the 45K.  If you plan things right and you live in an area where goods cost less, then you shouldn't have to go out and buy too much stuff with the exception of food, water, toiletries, and so forth.  If anything, the cost of meals would probably get you the most.  Since MIT doesn't have traditional meal plans where you get so many meals a a week in the cafeteria, you'll have much more incentive to go out and try the local restaurants (I don't know about Harvard or Cal-Tech's meal plans...they're probably similar).  In short, if you look [url=http://web.mit.edu/finaid/tuition_fees/index.html]here[/url] you'll see that the $\$$ 46,350 estimate covers just about everything you'll be spending money on.

1/2.  The financial aid is very decent.  You'll still go into considerable debt (the amount depends on how much aid you get which depends on how much your family makes, how many dependents are in your family, etc.).  One unfortunate result of the financial aid is that it is recalculated each year so if your parents start making more money, you get less financial aid.

By the way, elite schools usually don't let undergraduate students stay for grad school.

[quote="futuremontrealian"]1)  Is the financial aid really adequate?

4)  How possible is it to afford being a grad student afterward (assuming the school is the same)?[/quote]

1) The financial aid programs at Ivy League schools, MIT and other comparable universities are excellent.  (Harvard, for instance, is now nearly tuition-free for families earning under 60000 dollars.)  What this means for you will depend on your own family's situation.  If this is very important to you, call up someone at the financial aid office of those schools.

4) If by grad school you're talking about Ph.D. programs: being a grad student is a paid position.  So, you don't keep paying 40000 dollars a year to get a Ph.D.  (This is not true of business school, law school, med school or programs for a terminal masters degree.)
Also, I believe that one doesn't have to begin paying back student loans until after one has completed ones studies, that is until after graduate school.  (I am not an excellent source on this, so certainly don't take my word for it.)

[quote="futuremontrealian"]1)  Is the financial aid really adequate?[/quote]

In my opinion, it is. I have just been in the process of applying for financial aid for a private high school for my oldest son. The methodology followed by that high school is very similar to the
"institutional methodology" used by the most selective colleges, and the expected family contribution (EFC) they deem that our family is able to pay is reasonable. Families with moderately high incomes and little savings (a lot of American families of MIT- or Harvard-bound students these days) will probably have to borrow money to pay their EFC, but the economic value of degrees from those schools is great, so you should be able to help your parents pay back those loans after a while.

[quote]2)  How much debt does going to an expensive school really compile?[/quote]

Less than list price for four years of going to college, in the usual case, and sometimes much less. Families with incomes above 100,000 dollars per year are still eligible for (partial) financial aid, if they have ordinary levels of assets for their income level.

[quote]3)  What is the average post-bachelor degree pay a graduate of such a school can earn?[/quote]

That is above the median United States family income, in most cases of students who continue studying math while in college.

[quote]4)  How possible is it to afford being a grad student afterward (assuming the school is the same)?[/quote]

As another reply has noted, your graduate school will not be the same as your undergraduate college. In general, math departments want their strongest graduates to go to graduate school somewhere else, so that they are exposed to new teachers and so that they broaden their network of fellow researchers.

If you study math in college, or nearly any subject that requires math for an undergraduate degree, you also don't need to worry about paying for graduate school. Anyone who is any good as an undergraduate student in math, computer science, or most any kind of engineering will be able to receive a fully funded fellowship for graduate study. The best fellowships are portable fellowships like the National Science Foundation fellowships, but there are many others. You'll be able to (but usually won't be required to) pay back undergraduate loans (if any) already while you are a graduate student.

Good luck in your applications. Good luck to your dad in his business.

As mentioned by JBL, if your folks make less than 60K, they will not be expected to contibute a dime to your education at Harvard (not quite the same thing as "free").  See

http://www.news.harvard.edu/gazette/daily/2006/03/30-finaid.html

Other top schools also have good aid packages.  My recommendation is to apply to the places you want and tell your parents to chill out until the financial aid packages come.

Debt sucks, but undergraduate debt is typically not debilitating.  If your goal is to make money, you'll make it, but even a modest salary is enough to pay off the debt.  If you go to grad school, you can defer the loans.  (Grad school debt is a whole other story.)  There is a school of thought that a university education is essentially a calculated financial investment (you give them money; they give you diploma), and some in this school argue you'd be better off going to a cheap state school (if you can go for free, for example).  Personally, I find this school of thought repulsive.

P.S.  I've never met an American with an undergrad degree from Oxford.  I believe it's free for them; almost certainly not free for us.


function
superior algebra
superior algebra unsolved

GF(p) is a finite field of integers where p is prime. Show that the difference between sum of elements of GF(p) whose Order is A and the Mobius function of A is divisible by p.

[quote="ZetaX"]See http://www.mathlinks.ro/Forum/viewtopic.php?t=150491 for how it's done.[/quote]
The problem stated by me is generalization for all possible Order values not only (p-1). But the proof is only for
Primitive Roots.

There is no problem in using the same proof!


Suppose that $n>0$ and $x_0$ is  positive integer  and $x_1=\frac 12(x_0+\frac{n}{x_0}) , x_2=\frac 12 (x_1+\frac{n}{x_1}),...,x_p=\frac 12 (x_{p-1}+{\frac{n}{x_{p-1}}),...}$. Prove that $\lim_{n\rightarrow \infty}x_n=\sqrt{n}$ ;)

$\forall x >0, \frac 12(x+\frac{n}{x}) \geq \sqrt n $ ;and; $  \forall x \geq \sqrt n , \frac 12(x+\frac{n}{x}) \leq x $
So $\forall p\in\{1,2,3,...\},  \sqrt n \leq x_{p+1} \leq x_p$  .. and so on ..
 :cool:


Euler
function
floor function
number theory proposed
number theory

Prove that:
1/ For each integer $ n>49$, there exists two integers $ a,b>1$ such that $ a\plus{}b\equal{}n$ and $ \frac{\varphi (a)}{a}\plus{}\frac{\varphi (b)}{b}<1$.

2/ For each integer $ n>4$, there exists two integers $ a,b>1$ such that $ a\plus{}b\equal{}n$ and $ \frac{\varphi (a)}{a}\plus{}\frac{\varphi (b)}{b}>1$.

$ 1$
It's easy to show that for any even integer $ p$ we have $ \phi(p) < \frac {p}{2}$ since There are $ \frac {p}{2}$ even integers less than $ p$. Thus if $ n$ is even we can write $ n \equal{} 2m \equal{} 2(p \plus{} q) \equal{} 2p \plus{} 2q$ and hence if we take $ a \equal{} 2p$ and $ b \equal{} 2q$ we get $ \frac {\phi(2p)}{2p} \plus{} \frac {\phi(2q)}{2q} < \frac {1}{2} \plus{} \frac {1}{2} \equal{} 1$. If $ n$ is odd Then we can write $ n \equal{} 45 \plus{} 2c$ or, we have $ \phi(45) \equal{} 24 \Rightarrow \frac {\phi(45)}{45} \plus{} \frac {\phi(2c)}{2c} < 0.53333333.. \plus{} \frac {1}{2} \minus{} \frac {1}{c} < 1$ because $ \phi(2c) \le c \minus{} 2$  , if $ n \le 105$ if $ n > 105$ we can write $ n \equal{} 105 \plus{} 2x$ or we have $ \phi(105) \equal{} 48$ and  $ \frac {\phi(105)}{105} \plus{} \frac {\phi(2x)}{2x} < 1$

$ 2,$

For any positif integer $ n > 4$ There exists a prime number $ p$ such that $ n \equal{} p \plus{} k$ and $ p > k$ because according to Chebychev Theorem There exists a prime number between $ \left\lfloor \frac {n}{2}\right \rfloor$ and $ n$ and the result follows immedialtly Thus we have :
$ \frac {\phi(p)}{p} \plus{} \frac{\phi(k)}{k} > 1 \Leftrightarrow p\times \phi(k) > k$ which is obviously true ! :P


calculus integration logarithms limit derivative quadratics function

Hi Everyone  :)

I am on this math question which I am really struggling to do and would appreciate it if someone was to show me how to work it out for me (I think I am supposed to use partial fractions to help me integrate it).

The question is on integration:
[quote]
dw/dt=(m-n-kw)w

m=0.1 n=0.06 k=0.00005 when t=0, w=200.

so dw/dt=(0.04-0.00005w)w
[/quote]
The question asks for w to be the subject of the final answer, that is integrate the above and make the subject w.

I really need help on this one, I have been stuck on it for about two days and am going to be behind in my maths (don't see my teacher until tuesday).

Write it as

$\frac {d\omega}{\omega(a+b\omega)}=dt$

We can write

$\frac {d\omega}{\omega(a+b\omega)}=A\frac {d\omega}{\omega}+B\frac {d\omega}{a+b\omega}$ for an appropriate choice of $A,B$.  (this is the partial fractions part)

Now you know how to integrate both sides.  Finish it off from here.

I'm sorry, I still don't understand.

I had already tried Kunny's steps before and I got something with two loge's which equal t and I can't make w the subject because there are two of them in the loge's (I might be making a mistake when doing it)

As for blahblahblah, I don't understand how to get the value for A and B and even if I could I don't understand how to integrate something with dw in it (sorry :( )

I have scanned a picture of how far I have gotten with Kunny's steps (not sure if its even right), even though I would prefer to use blablablah's I would also like to understand how to do it that way (my problem with Kunny's is making w the subject rather than t).

$\int dt=25\int \left(\frac{1}{w}-\frac{1}{w-800}\right)dw$

We have  $t=25\ln \left|\frac{w}{w-800}\right|+Const.$

According to kunny, we have

$t=25\ln\left|\frac{w}{w-800}\right|+C$

I'd like to get rid of the absolute values. Note that both $w=0$ and $w=800$ are equilibrium solutions to this autonomous differential equation. The initial condition $w=200$ lies in the interval $(0,800).$ By the existence-uniqueness theorem, solutions cannot touch or cross, so $w$ must remain within $(0,800)$ for all time. Knowing the appropriate signs, we can write

$t=25\ln\left(\frac{w}{800-w}\right)+C.$

Now put $t=0,\,w=200.$ [Edit - fixed this typo.]

$0=25\ln\left(\frac13\right)+C$ so $C=25\ln3.$

But can we solve this explicitly for $w$ in terms of $t$? The differential equation is solved whether or not we can, but it would be convenient if we could.

$\frac{t}{25}=\ln\left(\frac{w}{800-w}\right)+\ln3=\ln\left(\frac{3w}{800-w}\right)$

$\frac{3w}{800-w}=e^{.04t}$

$3w=800e^{.04t}-we^{.04t}$

$(3+e^{.04t})w=800e^{.04t}$

$w=\frac{800e^{.04t}}{3+e^{.04t}}$

$w=\frac{800}{1+3e^{-.04t}}$

The last form makes clear that $\lim_{t\to\infty}w(t)=800.$

This is a very popular model of population growth, and is in every differential equations textbook and every calculus book that has a differential equations chapter.

[quote="Kent Merryfield"]
This is a very popular model of population growth, and is in every differential equations textbook and every calculus book that has a differential equations chapter.[/quote]
I only just quickly glanced over your working as I have some business I need to attent to, it appears you didn't use any form of partial fraction. So would I be correctin understanding that this method is the prefered method when working out population growth/retardation?

[quote="blahblahblah"]The part where you use partial fractions was completed by kunny - Dr. Merryfield simplified his answer using the initial conditions from there.[/quote]
oh, yes I can see how it was done now.

Just out of curiousity is there any other way this question could be solved?

also on the second line it appears the w and 800 have been switched, that is instead of it being w-800 it's 800-w can this be done and not affect the actual solution?

[quote="blahblahblah"]If $x<800$, $|x-800|=800-x$.[/quote]
oh I see, very nice.

thank you everyone for helping me :)

<edit>I noticed the substitution t=0, w=800 cannot work since ln(0) cannot occur. The substitution is t=0, w=200 (which I believe is correct) yielded the result ln(1/3) which is the next line after the substitution occurs, is that simply a printing error or am I not seeing something?

there was another question which asked me to find the greatest rate of increase of the graph (that is the point were the gradient is highest) how do I go about this?

You can find that from the differential equation:

$\frac{dw}{dt}=\frac1{25}w(800-w)$

The maximum value of this derivative occurs at the maximum value of the right hand side. The quadratic function $\frac1{25}w(800-w)$ has its maximum "halfway up" at $w=400,$ and the maximum value of this derivative is $6400.$ (Note: this is only the correct answer if the initial condition for $w$ lies in the interval $(0,400].$ If that initial condition lies in $(400,800)$ the greatest velocity occurs at $t=0.$)

As for when that maximum velocity occurs, just put $w=400$ into the very first line of my first post:

$\frac{t}{25}=\ln\left(\frac{3w}{800-w}\right)$ so $t=25\ln3.$

Yes that is what I done before but I didn’t think it would be so easy for me to get the answer like that so I thought there might have been a catch.

Actually I have been bogged down on another part of the question (yet again…) it asks what would happen if there is a constant taken away which it labelled d from the derivative of the function and it asks you to find the value of d such that the function once graphed will remain constant (I think that means remain at w=200), I can't think of a way to start on this one.

I attached the equation

A thumbnail sketch of the theory of single autonomous first order differential equations:

Suppose we want to solve $\frac{dw}{dt}=g(w)$ where $g$ is a function whose behavior is nice enough that the existence-uniqueness theorem guarantees us solutions.

It's a stronger condition than we need, but I'll go ahead and assume that $g$ has a continuous derivative, and use that as a convenience later.

Any point $w_0$ for which $g(w_0)=0$ is an equilibrium point, or fixed point. If we have the initial condition $w(0)=w_0$ then the unique solution to the differential equation is the constant $w(t)=w_0.$

The existence-uniqueness theorem tells us that solution curves cannot touch or cross. Thus if our initial condition is $w(0)=a<w_0$ then $w(t)<w_0$ for all $t$ for which the solution is defined, and similarly if $w(0)>w_0,$ then $w(t)$ stays greater than $w_0.$

Now suppose $g(w_0)=0$ and $g'(w_0)<0.$ Then near $w_0,$ solutions decrease for $w>w_0$ and increase for $w<w_0.$ It follows that there is a neighborhood of $w_0$ such that for any solutions with initial conditions in that neighborhood, $\lim_{t\to\infty}w(t)=w_0.$ This is a stable equilibrium point.

On the other hand, if $g(w_0)=0$ and $g'(w_0)>0,$ then solutions which begin near $w_0$ do not remain near $w_0.$ This is an unstable equilbrium point.

This doesn't exhaust all the possibilities, but you can use some common sense to deal with the (rarer) other cases.

[quote="chillypacman"]I can't think of a way to start on this one. [/quote]
The derivative of a constant function is zero.

I feel somewhat embarrased posting this so late...

When I first saw kunny's steps I got confused as to how in his first post the third line could be worked out to equal the fourth line (I know they are the same) but I didn't want to pursue it any further as I had become tired of the question being stuck on it for so long. I decided to come back to the problem and found I still can't figure it out...

I would really appreciate it if someone could show me how to use the correct method to transform the equation from the third line to the fourth line :)


function
real analysis
real analysis solved

f is a continuous function defined on the compact [a,b] such that for all x from the open (a,b) there exists $ g(x)$ in $ (0,min (x-a, b-x)) $ such that $ f(x+g(x))+f(x-g(x))=2f(x) $. Prove that f is affine.

I'm not at all sure about this, but the idea is worth mentioning :).

Define $t:[a,b]$, s.t. $t(x)$ is the distance from the point $(x,f(x))$ to the line $\ell$ having endpoints $(a,f(a)),(b,f(b))$. This function is clearly continuous, so it reaches it maximum $M$ somewhere inside $(a,b)$, because otherwise the maximum would be $0$. Now take $x_0=\sup\{x\in(a,b)|t(x)=M\}$. Since, by continuity, $t(x_0)=M>0$, $x_0$ is still in $(a,b)$. This yields a contradiction, however, because one of the points $U=(x_0+g(x_0),f(x_0+g(x_0))),V=(x_0-g(x_0),f(x_0-g(x_0)))$ must be at least as far from $\ell$ as $X=(x_0,f(x_0))$ (because $U,V,X$ are collinear). It's not even that important that $X$ is the midpoint of $UV$, so I guess the conditions of the problem can be relaxed.

The contradiction means that $t$ is identically $0$, and that means that the graph of $f$ is a line, Q.E.D.

Is this correct? It seems rather unusual.

You can also prove that $f$ is convex, and $-f$ is convex so $f$ is affine

Yeah, this was the official impossible to find solution.

Which one? Proving that it's convex and $-f$ is also convex?

The solution Moubi posted was the "official one". Your solution is more natural and beautiful, grobber.


Let $a,b,c,d$ be non-negative real numbers such that $a+b+c+d=1$. Prove that
\[2(a^{2}+b^{2}+c^{2}+d^{2}) \ge 27\sqrt[3]{(a^{2}+b^{2})(a^{2}+c^{2})(a^{2}+d^{2})(b^{2}+c^{2})(b^{2}+d^{2})(c^{2}+d^{2})}.\]

This inequality refers another very nice one immediately (a generalization of VASC's inequality)

If $a,b,c,d$ are non-negative real numbers then
\[\frac{1}{a^{2}+b^{2}}+\frac{1}{a^{2}+c^{2}}+\frac{1}{a^{2}+d^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{b^{2}+d^{2}}\ge \frac{81}{2(a+b+c+d)^{2}}.\]

Who can explain why the first inequality can directly refer the second one? It is interesting, I think.

For the first, I use a method the same as SMV theorem, but It is no so elementary and not as good as I can. I hope that you can solve this problem with a better solution.


calculus
integration
algebra
polynomial
number theory unsolved
number theory

[quote="mathpk"]Find all pair of positive integers $ (x,n)$ such that $ x^{n\plus{}1}\plus{}2^{n\plus{}1}\plus{}1$ is divisible by $ x^{n}\plus{}2^{n}\plus{}1$.[/quote]

First, Consider $ x>2$ and $ n>1$.
First, we have : $ 2^{n\plus{}1}\plus{}1<2^{n}x\plus{}x$, so $ x^{n\plus{}1}\plus{}2^{n\plus{}1}\plus{}1< x(x^{n}\plus{}2^{n}\plus{}1)$
We also have $ x^{n\minus{}1}\plus{}\frac{3.2^{n}\plus{}2}{x}>2^{n}\plus{}1$
[hide="(Demo)"]
========= beg demo ========
For $ n\equal{}2$,  $ x\plus{}\frac{14}{x}>5$ is true for any $ x>0$
For $ n\equal{}3$,  $ x^{2}\plus{}\frac{26}{x}>9$ is true for any $ x>0$
For $ n>3$,   $ x^{n\minus{}1}\plus{}\frac{3.2^{n}\plus{}2}{x}>x^{n\minus{}1}\geq 3^{n\minus{}1}>2^{n}\plus{}1$ is true for any x>2
========= end demo ========[/hide]
So $ x^{n}\plus{}3.2^{n}\plus{}2>x2^{n}\plus{}x$, and $ x^{n\plus{}1}\plus{}2^{n\plus{}1}\plus{}1>x^{n\plus{}1}\plus{}x2^{n}\plus{}x\minus{}x^{n}\minus{}2^{n}\minus{}1$ and so $ x^{n\plus{}1}\plus{}2^{n\plus{}1}\plus{}1>(x\minus{}1)(x^{n}\plus{}2^{n}\plus{}1)$

So, for $ x>2$ and $ n>1$, we have $ x>\frac{x^{n\plus{}1}\plus{}2^{n\plus{}1}\plus{}1}{x^{n}\plus{}2^{n}\plus{}1}>x\minus{}1$ and thus $ x^{n\plus{}1}\plus{}2^{n\plus{}1}\plus{}1$ is not divisible by $ x^{n}\plus{}2^{n}\plus{}1$.

So we have to study two cases :$ n\equal{}1$ and ($ n>1$ and $ x\leq 2$)

Case $ n\equal{}1$ :
We want $ x\plus{}3|x^{2}\plus{}5$.
For $ x>11$, we have $ (x\minus{}3)(x\plus{}3)<x^{2}\plus{}5<(x\minus{}2)(x\plus{}3)$ and so $ x\plus{}3$ can not divide $ x^{2}\plus{}5$
For $ x\leq 11$, a quick check gives $ x\equal{}4$ and $ x\equal{}11$

Case $ n>1$ and $ x\leq 2$
If $ x\equal{}1$, we want $ 2^{n}\plus{}2|2^{n\plus{}1}\plus{}2$ which is impossible
If $ x\equal{}2$, we want $ 2^{n\plus{}1}\plus{}1|2^{n\plus{}2}\plus{}1$ which is impossible too ($ \frac{2^{n\plus{}2}\plus{}1}{2^{n\plus{}1}\plus{}1}\equal{} 2\minus{}\frac{1}{2^{n\plus{}1}\plus{}1}$)

So the only solutions are :
$ n\equal{}1$, $ x\equal{}4$
$ n\equal{}1$, $ x\equal{}11$


geometry
3D geometry
MATHCOUNTS
Euler
search
number theory
relatively prime

Hello,

I know this is kind of non-sence, but I think that it could be very interesting to see how far we can get enumerating why a number is interesting. But first,

[b]Theorem:[/b] For every $n\in\mathbb{N}$, $n$ is an interesting number.

[b]Proof:[/b] Suppose that it is not true, that is, the set $T\subseteq\mathbb{N}$ of non-interesting numbers is non-empty. By well order principle, there is a first element for $T$, let's say $n=\min_{m\in T}m$, hence, $n$ is the first non-interesting number, what makes $n$ very interesting  :D $\Rightarrow\Leftarrow$.

Once proven the theorem, let's try to find why is every number interesting. Let's try to be as much as possible not be repetitive, and let's try to use general mathematical facts rather than personal or historical (don't say that [b]1981[/b] is interesting because it was the year I was born  :lol: ), and if you find someone that may be funny for the ones we already have, post it!

[b]1[/b]: The unity, the only number that is neither prime nor composite, the only number that is coprime with every other.
[b]2[/b]: The only even prime.
[b]3[/b]: The first odd prime. First non-trivial triangular number.
[b]4[/b]: The first non-trivial perfect square.
[b]5[/b]: First prime number that is the sum of two other distinct prime numbers. (5=2+3). First number to be the sum of two different squares.
[b]6[/b]: First composite not power of a prime.
[b]7[/b]: First number that can be expresed as $p^2+q$ with $p,q$ distinct primes.
[b]8[/b]: First non-trivial cube.
[b]9[/b]: First non-trivial, odd perfect square.
[b]10[/b]: First number that can be expresed as $p+q+r$ with $p,q,r$ distinct primes.
[b]11[/b]: First number that can be expresed as $p^2+q$ with $p,q$ distinct primes in two different forms.

Well, and I suppose that everybody knows,
[b]1729:[/b] First number that can be expressed as the sum of two cubes in two different ways.

Let's see how we do it. Best,

[quote="mcalderbank"][b]8675309[/b]:  Largest prime number that has a song named after it. :D[/quote]

That is in fact true. I remember that song (Tommy Tutone), it wasn't that bad! (It wasn't that good either :D ). I think someone else song it not quite ago, but it is originally from the very early 80's.

Nevertheless, let's see which can we use

[b]12[/b]: First number that can be expressed as $p^2q$ with $p<q$ primes
[b]13[/b]: First number that can be expressed as $p^2+q$ in two different ways, with $p$ and $q$ primes.
[b]15[/b]: First composite, non square, non even number.

[b]3816547290[/b]: The unique pandigital number that if you take the first $n$ digits, for $1\leq n\leq10$, the number you get is divisible by $n$.

I still don't have a property for 14.

Best,

I got these from a website

14 is the smallest number n with the property that there are no numbers relatively prime to n smaller numbers.

15 is the smallest composite number n with the property that there is only one group of order n.

16 is the only number of the form $x^y=y^x$ with x and y different integers.

37 pages of "17"...i'm at a loss for words  :?  :P

[quote="Bictor717"][quote="filletwho"]17 is the first uninteresting number[/quote]

Blasphemy!
http://www.vinc17.org/yp17/index.en.html
Anything as sacrilegious as what you just said will be proven wrong immediately![/quote]

Well, there the guy tells you a lot of couriosities about 17, but non mathematical, but, for example

[b]17[/b]: 3rd Fermat Prime, firt number to be expressible as $x^4+y^4$ with $x\neq y$ positive integers, firt number that can be expressed as $p^3+q^2$ with $p\neq q$ prime numbers. I don't know, I can't think about another right now, but for sure there are plenty of them.

Don't forget 153
This was  answer to one of the questions in mathcounts:
[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=28819]Magic number 153[/url]

[b]635318657[/b]: The first number that can be expressed as the sum of two forth powers in two different ways: $635318657 = 133^4 + 134^4 = 158^4 + 59^4$ (Thank you Gyan!)

[quote](Thank you Gyan!)[/quote]
Actaully Euler knew that!! I have no idea how did he know that.  The legend has it that when Ramanujan told  the  ablove property about 1729, Hardy asked Ramanujan if he knew  a number which satisfies fhe fourth power and Ramanujan  did not know (to be  fair Ramanujan was quite sick and was in hospital).. Both were surprised  later to find that Euler already knew  such number.

BTW Hardy et-al did prove that there are infinite such numbers (for cubes)

Why would they surprised? Euler's just as smart.

[quote="Gyan"][quote](Thank you Gyan!)[/quote]
Actaully Euler knew that!! I have no idea how did he know that.  The legend has it that when Ramanujan told  the  ablove property about 1729, Hardy asked Ramanujan if he knew  a number which satisfies fhe fourth power and Ramanujan  did not know (to be  fair Ramanujan was quite sick and was in hospital).. Both were surprised  later to find that Euler already knew  such number.[/quote]

Yes, but I found it (the one with 635318657) because you posted it in other thread on the forum. :first: .

Best,

8 reasons why 8 is interesting:
-first composite fibonocci number. I think.
-Sideways infinity.
-First positive number (besides one) that can be expressed as n^(n+1).
- first cube besides one.
-Double zero yet greater than zero.
-8) is a nice smiley face.
-half of 8 is 3 and half 8 is 4
-rhymes with ate

[quote="mcalderbank"][b]42[/b]: Need I say anything?[/quote]

Yes, Please!

I think the thread Mathematician's Date Guide had a shirt that expressed a very interesting number

very interesting indeed,  :D

[quote="diophantus"]All multiples of three: sum of digits is a multiple of three.
Powers of two: $\Sigma^{k}_{n=0}2^n=2^{n+1}-1$[/quote]

Well, probably we should define what an interesting number is.

A number is interesting (at least my original idea) if it is either the first, the last or the only number that holds with some specific property. Someone with some alternative definition?

Perhaps we need a deeper definition of an interesting number.  Check my proof that all real numbers are interesting.

[hide="Click Here"]If $n>0, n$  :varepsilon: $R$ then n is the last solution to $x^2=n^2$.
If $n<0, n$  :varepsilon: $R$ then n is the first solution to $x^2=n^2$.
If n=0, then n is the only solution to $x^2=n^2$.
Therefore, all real numbers are interesting.
Not very clever, but it gets the job done.[/hide]

And here is another proof:
[hide][img]http://www.simonsingh.com/images/461.jpg[/img][/hide]

18 is the only number that is twice the sum of its digits.

19 is the maximum number of 4th powers needed to sum to any number.

24 is the largest number divisible by all numbers less than its square root.

25 is the smallest square that can be written as a sum of 2 squares.

26 is the only number to be directly between a square and a cube.

27 is the largest number that is the sum of the digits of its cube

30 is the largest number with the property that all smaller numbers relatively prime to it are prime.

I may have posted select numbers but deep down, I think all numbers are equal....

Okay what is interesting about 5882353 ?

And NO cheating ...( Don't try to search AoPS or google  before you guess...)

Hint :[hide] what is interesting about 1233  (or 8833)?[/hide]

It has two 8's in it, therefore making it special.

[quote="filletwho"][quote="diophantus"][quote="mcalderbank"][b]42[/b]: Need I say anything?[/quote]
Ahhh! The secret to life, the universe, and everything.  Have you read Hitchhikers Guide to the Galaxy as well.[/quote]

I thought it was because it is the perfect score on the IMO? But what does Hitchhikers Guide have that makes 42 special?[/quote]
[u]Hitchhiker's Guide to the Galaxy[/u] is a silly story where mice rule the earth.  The mice decide to bulid a supercomputer which will run a 7.5 million year long program to determine the Answer for the Secret of Life, the Universe, and Everything.  After 7.5 million years of processing, it spits out the number 42.

[quote="Gyan"]And here is another proof:
[hide][img]http://www.simonsingh.com/images/461.JPG[/img][/hide][/quote]
You might want to check the link to your image. :blush:

[quote]You might want to check the link to your image[/quote]
May be missing something but the link works for me.. in my original message ( but  not in your quote) :?

1. Take any 3-digit number with 3 distinct digits
2. Sort the digits in descending order to make a new 3-digit number
3. Sort the digits in ascending order to make another new 3-digit number
4. Subtract the number in step 3 from the number in step 2
5. With this new number, repeat steps 2 and 3.
If you repeat this enough times, you will end up with [b]495[/b]!

1. Choose any 3 digit number with 3 distinct digits
2. Find the reverse of that number, and subtract the smaller of the two from the larger.
3. Add the number in step 2 to its reversal (special case: reverse of 099 is 990)
The number you come out with is always [b]1089[/b]
Fun problem: prove why this works.

[b]142,857[/b] is also interesting for several reasons:
It is the repeating decimal of 1/7
Multiply by any interer from 1 to 6, and you get a cyclic permutation of 142857
(142,857)^2 = 20,408,122,449; 20,408 + 122,449 = 142,857
142857 x 7 = 999,999
1/142857 = 0.000007000007...
142 + 857 = 999
14 + 28 + 57 = 99
1 + 4 + 2 + 8 + 5 + 7 = 27 and 2 + 7 = 9

[insert number here] is "interesting" because it is the only number equal to [your number]


This a question from the COMC, year 2000, question 2a) in part B
You need a complete proof
I was hoping you would be able to mark my solution and give some pointers as to how I can improve it.

In triangle ABC, the points D, E, and F are on sides BC, CA, and AB, respectively, such that  :ang:AFE=:ang:BFD, :ang:BDF=:ang:CDE, and :ang:CED=:ang:AEF.
Prove that :ang:BDF=:ang:BAC

http://www.cms.math.ca/Competitions/COMC/COMC00/exame2000.pdf
(image is there along with original question)

let AFE and BFD be f
let BDF and CDE be d
let CED and AEF be e

so :ang:DFE=180-2f
:ang:DEF=180-2e
:ang:FDE=180-2d
since each form  straight line

since DFE, DEF, and FDE are angles in the triangle DEF
180=180-2f+180-2e+180-2d
2f+2d+2e=360
d+e+f=180

:ang:BAC is the same as :ang:FAE
AFE is f, AEF is e
so 180=f+e+:ang:FAE:
based on our previous conclusion that d+e+f=180, we know that :ang:FAE must equal d

:ang:BDF has already been defined as d

therefore:ang:FAE=:ang:BDF

thank you in advance


geometry
geometric transformation
homothety

ABC is atriangle .(I)is its incircle .and is tangent to ABC in points K,L,M.
K is on AB.L is on AC.let MNwich is the diameter of (I),intersect KL in point X.

prove that: AK bisect side BC.

I think "AX bisect side  BC" and not AK.

[b]Yes , AX bisects BC.[/b]
  Hint 1. We drop a line DE // BC from X and prove that X is midpoint of DE .

  Hint 2 . We can use an Homothecy from A.
  But I think I have also seen an other solution(metric and computational) in a book by << Alexander Soifer >>

 If nobody writes the whole solution , I' ll do it later(I' m in school this moment !)
 Babis

yes yes ure right. but about what stergiu said,i want to see u solution by using homothecy.

i my self solve it by using harmonic and pole and polar.and i didnt find how to solve the problem by using homothecy.

but i think homothecy isnt a good method ,to solve this kind of problems.any way ,i wait for u solution ;) .

Noting that the line MN, being a diameter of the incircle (I) of triangle ABC, coincides with the line IM, we see that the problem is equivalent to http://www.mathlinks.ro/Forum/viewtopic.php?t=5830 post #4 Theorem 1.

  darij


logarithms
algebra
function
domain
functional equation
algebra open

[hide="What about a change?"]Let's make the change
$g: \mathbb{R}^{+}\to\mathbb{R}$,
$g(x)=f(e^{x})$ for all $x\in\mathbb{R}^{+}$.

Translating the functional equation to $g$, we've got
$g(x+y)=g(x)+g(y)$.

The solutions must be
$f(x)=g(\ln x)$,
where $g$ is a solution of Cauchy's functional equation.[/hide]

Let $g(y)=f(e^{y})$. It give $g: R_{+}\to R_{+},g(x+y)=g(x)+g(y)\Longrightarrow g(x)>g(y) \ if \ x>y$.
Let $g(1)=a$, then $g(r)=r,g(rx)=rg(x),\forall r\in Q$.
It give $g(x)=ax$, therefore $f(x)=a\ln x$.

[quote="Rust"]$g: R_{+}\to R_{+}$[/quote]

I agree that the domain must be $\mathbb{R}^{+}$, but I don't understand why you say that the codomain is $\mathbb{R}^{+}$ :?:. The codomain of $f$ is $\mathbb{R}$.

If it's an error, my solution would be more general.

If $g: R_{+}\to R$ then this equation had infinetely many solutions.
Let $\{a_{c}\}$ is Hameles basis in R/Q and $\{b_{c}\}$ real numbers, any linear function $g: R\to R \ \ g(\sum_{c}r_{c}a_{c})=r_{c}b_{c}$ works .


function
LaTeX
real analysis
real analysis unsolved

It is unclear to me what the pattern in the denominators is (especially at the end). Also, please, try to use $ \text{\LaTeX}$: it saves both our time and server space :).


function
induction
algebra
domain
algebra proposed

I´ll prove that $f(x)=0$. Notice that we can suppose that $-M<f(y)<M$ for all real y. And so we can easily prove by induction on k that $\frac{-M}{n^k}<f(x)<\frac{M}{n^k}$ and so if we take the limit when k tends to infinity we will have that $f(x)=0$ and so it´s proved. We must have $n>1$.

Find all $bounded$ functions $f$ satisfying the relation:
$f(x)=\frac{1}{n^2}(f(\frac{x}{n})+f(\frac{x+a}{n})+f(\frac{x+2a}{n})+.....+f(\frac{x+(n-1)a}{n}))$
I forgot to note that $f: (0,a) \to R$ and $n\ge2$ is a fixed natural number.
does anyone have another type of solution?

I think that the domain of f can be the reals, no?


Since $\angle ABD= \angle AED$ (A, E, B, D are concyclics)and $\angle ACD = \angle AFD$ (A, C, F, D are concyclics) we get that  the triangles $ABC$ and $AEF$ are similar, from this is easy to see that $\angle EAB = \angle NAM$. Also, $\frac{AE}{AN} = \frac{AB}{AM}$ (from the similarity), then the triangles $AEB$ and $ANM$ are similar, hence $\angle ANM= \angle AEB = 90^\circ$.

$Tipe$

[quote="shobber"]Let $ABC$ be a triangle and $D$ the foot of the altitude from $A$. Let $E$ and $F$ lie on a line passing through $D$ such that $AE$ is perpendicular to $BE$, $AF$ is perpendicular to $CF$, and $E$ and $F$ are different from $D$. Let $M$ and $N$ be the midpoints of the segments $BC$ and $EF$, respectively. Prove that $AN$ is perpendicular to $NM$.[/quote]

Let $k_1$ be the circle with diameter AB; this circle passes through the points A and B, through the point D (because < ADB = 90°) and the point E (because of < AEB = 90°, since $AE\perp BE$). Similarly, denote by $k_2$ the circle with diameter AC, and show that this circle passes through the points A, C, D and F. Now, we have
- two circles $k_1$ and $k_2$ intersecting at two points A and D;
- a line through the point D meeting the circle $k_1$ at the point B (apart from D), and the circle $k_2$ at the point C (apart from D);
- a line through the point D meeting the circle $k_1$ at the point E (apart from D), and the circle $k_2$ at the point F (apart from D);
- the midpoints M and N of the segments BC and EF.
Hence, according to the problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=50619 , the triangles ABE, ACF and AMN are similar. Thus, < ANM = < AEB. But $AE\perp BE$ yields < AEB = 90°, and thus < ANM = 90°, so that $AN\perp NM$, and we are done.

  darij

As Tipe has found  before $BAC$ is similar to $EAF$,. Then take $A$ as fixed by the angle rotation we can say that $BAE$ is similar to $CAF$  and $\angle BAE\cong \angle CAF$  $(i)$. Since $M, N$ are midpoints of $BC, EF$, respectively. So it is easy to see by rotation that $BAM$ is similar to $EAN$ then $\angle BAE\cong \angle MAN$ (Becouse it is angle rotation). Now look also we can say by angle rotation similarity that $FAN$ is similar to $CAM$, if we join $M$ with $N$, and $C$ with $F$ then we will get such similarity $MAN$ to $CAF$  $(ii)$ (It is theorem you can see in Geometry Revisited). From $(ii)$ $\angle CFA\cong \angle MNA$. Since $\angle CFA= 90$ so also $\angle MNA=90$. So the problem is solved.
Abdurashid

[color=red]Here is another easier solution:[/color]
As Tipe has found  before $BAC$ is similar to $EAF$,. Then take $A$ as fixed by the angle rotation we can say that $BAE$ is similar to $CAF$  and $\angle BAE\cong \angle CAF$  $(i)$. Since $M, N$ are midpoints of $BC, EF$, respectively. So it is easy to see that $BAM$ is similar to $EAN$ then $<DNA=<ENA=<BMA$, so quadrilateral $AMDN$ is cycilic, so $<MNA= <MDA$ (becouse on the same arc) since $<MDA=90$ then also $<MNA=90$. So the problem is solved.
Abdurashid

$ABC$ and $AEF$ are simmilar to each other.

then:

$\angle AEB=\angle AFC=\angle ANM=90$

[color=#FF0000][Mod edit]
In a bit more detail: clearly $\triangle ABC\sim\triangle AEF$. So $A$ is the centre of spiral similarity where $B\to E$, $C\to F$. Hence it is also the centre of spiral similarity where $B\to C$, $E\to F$. Stop this latter spiral similarity halfway through, or more generally $\frac{BM}{BC}=\frac{EN}{EF}$ through (i.e. $M,N$ don't necessarily have have to be the midpoints,  just as long as those ratios are equal), to see that $\triangle ABC\sim\triangle AEF\sim AMN$, then $\angle ANM=\angle AEB=90^{\circ}$.[/color]

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1998Problem4

Vo Duc Dien

Take $P, Q$ so that $PADB, AQCD$ are rectangles. Let $N$ be the midpoint of $PQ$. Then $PD$ is a diameter of the circumcircle of $ABC$, so $PX$ is perpendicular to $XY$. Similarly, $QY$ is perpendicular to $XY$. $N$ is the midpoint of $PQ$ and $M'$ the midpoint of $XY$, so $NM$ is parallel to $PX$ and hence perpendicular to $XY$. $NADM'$ is a rectangle, so $ND$ is a diameter of its circumcircle and $M$ must lie on the circumcircle. But $AM'$ is also a diameter, so $\angle AMM' = 90^{\circ}$. $\Box$

The problem is equivalent to this one:

[b]Given two circles secant at $A, D$, and $E, F$ the intersections of a line passing through $D$ with the circles, then find the locus of the point $N$, midpoint of $EF$.[/b]

The locus is the circle of diameter $AM$, $M$ being the midpoint of the line segment $BC$, drawn through $D$, perpendicular to $AD$, $B, C$ lying on the two circles.

Best regards,
sunken rock

Dear Mathlinkers,

the problem is a nice application of the mid-circle theorem.

Sincerely
Jean-Louis

Note that $A, B, D, E$ and $A, C, D, F$ are cyclic implies $ \triangle ABC\sim\triangle AEF $. Then, A is the center of the spiral similarity taking $B$ to $E$ and $C$ to $F$. Averaging Principle implies it takes M to N, so $ 90= \angle{ABE}= \angle{AMN}$, as required. $ \Box$

We use turning homegenity AEDB is cyclic so <ABE=<ADE. and ADCF is cyclic so <ADE= <ACF. Absolutely the triangle ABE is as same as the triangle ACF. And The vertex A is the homegenity center E goes to F. and B goes to C and we know if X and Y be on BC and EF such that BX÷ XC = EY ÷YF We have AY is altitude XY. So M and N have the situation of X and Y

Dear Mathlinkers,
a reference for the mid-circle theorem

http://jl.ayme.pagesperso-orange.fr/Docs/midcircle%20theorem.pdf

Sincerely
Jean-Louis

[b]Solved With awesomemaths[/b]

*Note we solved with a slightly different configuration, with E being to the right of ABC and F being inside*

[b]Claim - AFE is similar to ACB[/b]
[b]Proof[/b]

We know $ADBE$ is a cyclic quadrilateral since opposite angles are supplementary, so $\angle{ABD} = \angle{ABD}$. We can also find that $\angle{ACB} = \angle{AFE}$ by the cyclic quadrilateral $ACDF$. By $AA$ similarity, we get the triangles are similar.

Note that $\triangle{ACB}$ is sent to $\triangle{ABE}$ via the spiral similarity about A. Because of the gliding principle, we are done $\blacksquare$

very nice problem!

[color=#f00][b]Claim:-[/b][/color] $\triangle{AEF} \sim \triangle{ABC}$\\

[color=#00f][b]Pf:-[/b][/color] we notice that points $A,B,D,E$ are concylic so we have $\angle{BAD}=\angle{BED}=90-\angle{ABC}$ and since $\angle{AEB}=90^{\circ}$ we get $\angle{AEF}=\angle{ABC}$\\

also since points $A,D,C,F$ are concylic we have $\angle{ACB}=\angle{ACD}=\angle{AFE}$\\

hence $\triangle{AEF} \sim \triangle{ABC}$ $\blacksquare$\\

now from [color=#f00][b]Claim[/b][/color] we have $\triangle{AEN} \sim \triangle{ABM}$ hence we get $\angle{AMD}=\angle{AND}$ as these two angles are subtended by same chord $AD$ we get points $A,D,M,N$ to be concyclic and hence we get $\angle{ANM}=180^{\circ}-\angle{ADM}=90^{\circ}$ and hence $AN \perp MN$ $\blacksquare$

Fairly straightforward spiral similarity exercise.

A spiral similarity at $A$ sends $\overline{FE} \to \overline{BC}$ and thus $\overline{FN} \to \overline{BM}$. Thus $ANDM$ is cyclic and $\angle ANM = \angle ADM = 90^\circ$.

First observe trivially that $AEDB$ and $AFCD$ are cyclic. Now observe that $\triangle ABE \sim \triangle ACF$, as $\measuredangle FCA = \measuredangle FDA = \measuredangle EDA = \measuredangle EBA$. Now we make a claim which finishes this problem:

[color=#f00][b]Claim:[/b][/color] $\triangle ABC \sim \triangle AEF$.
[i]Proof:[/i] \[\measuredangle AFE = \measuredangle AFD = \measuredangle ACD. \square \]

Hence we now have that \[\measuredangle AMD = \measuredangle ANE = \measuredangle AND, \] so that $ADMN$ is cyclic and indeed we have $AN \perp MN$. $\blacksquare$
[hide = Remark] I should've known about spiral similarity. It was screaming it throughout this problem.[/hide]

[quote=allaith.sh]a direct application of spiral similarity.
$FE \to CB \implies FN \to CM \implies ANMD $ is cyclic and we are done. :thumbup:[/quote]

very nice solution you are a genius


[hide="Solution"]
Assume that there are no triangles that satisfy the conditions.  Let the colors be red, blue, and green.  Consider a regular heptagon, $ ABCDEFG$ with center $ O$.  Without loss of generality, let $ O$ be green.  Now, since seven is odd, there must exist two adjacent points, say $ A$ and $ B$ of the same color, say red.  Now, this forces $ C$ to be blue to avoid isosceles triangle $ ABC$.  Similarly, $ G$ must be blue.  Now, if $ D$ is red, then consider triangle $ ABD$.  We have that $ \angle ADB = \frac {180}{7}$ and $ \angle BAD = \frac {360}{7}$.  This forces $ \angle ABD = \frac {720}{7}$, and thus $ \triangle ABD$ has its angles in an arithmetic progression, which is a contradiction.  It follows that $ D$ is blue.  Similarly is $ F$ to avoid $ \triangle ABF$.  Now, we consider triangle $ CFD$.  Notice that $ \angle FCD = \frac {360}{7}$ and $ \angle CFD = \frac {180}{7}$, hence forcing $ \angle CDF = \frac {720}{7}$ and therefore, $ \triangle CFD$ has its angles in an arithmetic progression, which is a contradiction.  Thus, our assumption was false and there exists a triangle fitting the conditions. [/hide]

@The QuattoMaster 6000: The condition was that the angles can be in GP.....anyways your solution is still perfect :idea:

you can always find a 45 90 45 triangle
there is no need of the angles in GP

I don't undetsrand quatto's solution.
Why cannot be C or G green,i understand that they cannot be blue and that cannot be that C is green and G is green also,perhaps this is trivial..., but i don't see why.

Can anyone find a proof for just the isoceles part? For two points of the same color, any point on the perp bisector or on the circles centered at one passing through the other must be of the other two colors.

Here's a sketch of a pretty short argument for the isosceles right triangle due to Solymosi and Graham.

What we will show is that there is an $ f(k)$ such that if you color the lattice points in $ \{0 \leq x,y \leq f(k)\}$ with $ k$ colors you will have an isosceles right triangle with legs parallel to the two coordinate axes.  Consider the following procedure:

Start with the line $ x \plus{} y \equal{} M_k$, where $ M_k$ is to be determined later, and let $ c_k$ be the color appearing most frequently on this line (by pigeonhole, it appears at least $ \frac {M_k}{k}$ times).  Let $ S_k$ be the set of points of color $ c_k$ on this line.

Now let $ T_k$ be the set of all points with $ x \plus{} y < M_k$ forming an isosceles triangle with two points in $ S_k$ (note that $ |T_k| \equal{} \binom{|S_k|}{2}$).  If any point in $ T_k$ is of color $ c_k$, we are done.  If not, by pigeonhole we can find an $ M_{k \minus{} 1} < M_k$ such that the line $ x \plus{} y \equal{} M_{k \minus{} 1}$ contains at least $ \frac {|T_k|}{M_k}$ points of $ T_k$ (as all the points of $ T_k$ are on one of the $ M_{k}$ lines of slope $ \minus{} 1$).

Again using pigeonhole we see that there is a color $ c_{k \minus{} 1}$ containing at least $ \frac {|T_k|}{(k \minus{} 1)M_k}$ points on this line.  We can use this set to generate a new $ T_{k \minus{} 1}$.  The key fact here is that $ T_{k \minus{} 1}$ is a subset of $ T_k$ (draw a picture to see this)!   This means the points in $ T_{k \minus{} 1}$ can only be one of $ k \minus{} 2$ colors, and the process can continue onwards with one more color eliminated at each step.     If at the end $ T_1$ is non-empty, then we conclude that there must be a triangle.

At this point it's a matter of working through the argument and estimating how large $ f(k)$ has to be to make sure you don't run out of points before the process finishes.  It turns out some function of the form $ f(k) \equal{} 2^{2^{ck}}$ works.  The full details are in their [url=http://www.math.ucsd.edu/~fan/ron/papers/06_03_righttriangles.pdf]paper[/url] posted online.

I believe another approach to this problem is to use Van der Waerden's theorem to show that a triangle exists.  What's nifty about this method is that it actually gives an alternative proof for part of the VdW theorem.  If you color the integers in $ [ \minus{} f(k), f(k)]$ with $ k$ colors, consider the coloring in two dimensions where where the point $ (x,y)$ is given the color of $ x \minus{} y$ in the original coloring.  By the above we know that some triangle $ \{(x,y), (x \plus{} d, y), (x, y \plus{} d)\}$ is monochromatic, which corresponds to the progression $ (y \minus{} x) \minus{} d, (y \minus{} x), (y \minus{} x) \plus{} d$ in the original coloring.

lemma:if a square and its centre have been coloured with only 2 colours then we have a isoscales triangle (true property check it)
solution
now we take any two red points(A and B)
let $ p =$line perpendicular to $ AB$ through $ A$
let $ q =$line perpendicular to $ AB$ through $ B$
case 1 either $ p$ or $ q$ have a red point say $ C$ with say $ BC \perp endicular AB$
let $ r$ and $ s$ be perpendicular bisectors of $ AB$ and $ BC$
then $ r$ and $ s$ are coloured in only two colours (blue or green)
let $ r \cap s = Q$
take an arbit distance on A rounder and draw a circle with centre $ Q$
let it cut $ r$ and $ s$ at $ M,N,O,P$
this is the square (thou are done) for lemma
case 2 $ p$ ,$ q$ and $ r$ is blue green
cosider a point on $ r$ say $ Q$

drop perpendicular to $ p$ at P and then complete the circle with $ QP$ with $ Q$ as the centre
let the other intersections be denoted by $ M,N$ and $ O$
this is the square(thou art done) for lemma

[quote="skand"]All the points in the plane are colored using three colors.Prove that there exists a triangle with vertices having the same color such that [i]either[/i] it is isosceles [i]or[/i] its angles are in geometric progression.[/quote]
 :rotfl:  A foolish problem! Maybe there's a mistake.
Assume that there are no isosceles triangle whose vertices are the same color. Let the colors be red, blue and yellow, let a yellow point O. Consider a regular pentagon ABCDE with center O. Easy to check that there must exists two adjacent vertices, called A and B of the same color, say red (WLOG). Because of the isosceles triangles ABC, ABD, ABE, three points C,D,E must be blue, and they form an isosceles triangle whose vertices are in the same color, which a contradiction.

[quote="JBL"][quote="thaithuan_GC"]three points C,D,E must be blue[/quote]  No: one of them can be yellow while the other two are blue without causing a contradiction.  This is why the first response uses a heptagon, not a pentagon.[/quote]
No, we have O is yellow, thus, on the circle with center O, there are at most 1 yellow point, say M. And we can easily construct pentagon ABCDE such that these vertices are not coincide with M, hence A,B,C,D,E are in two color: red and blue.

please enlighten us with such a construction (remember you have put two constraints 1.adjecent points red and 2. no yellow point on the pentagon)
also there can be two yellow points on the circle as diametrically opposite points

edit:Sorry ,your Idea does work,yes it does look to be a foolish problem :lol:  :D

why cant one vertex be green?
IN quatto masters' solution he seems to have excluded this right?

ok sorry i guess i myself figured it out..
if 2 is green then u get a contradiction acc. to Quattomasters' solution..
so let 1 be green but u kno already that - $ D,E,F$ are blue so either $ G$ or $ C$ is green so consider the non-green vertex and looking at the triangle it forms with $ F,D$ we again get a contradiction... so it doesnt matter whether a vertex is green or not...
i hope i am right...

[quote="kevinatcausa"]
I believe another approach to this problem is to use Van der Waerden's theorem to show that a triangle exists. [/quote]

Gallai's theorem, which generalizes Vand der Waerden's theorem to dimension $ n>1$ states that : for any finite coloring of the $ n$-dimensional space (or just the lattice points) and any finite set of points $ S$ (according to what kind of points you admit above) there exists a monochromatic set which is homothetic to $ S$.

This trivializes the problem and I do not understand why Graham does not consider it (unless it is not the same Graham as I believe. I will read the paper you gave in attachement).

Pierre.

[quote="pbornsztein"]
I do not understand why Graham does not consider it [/quote]

Ok, it deals with bounds.

Pierre.

Apply Van der Waerdens to a regular n-gon for the isosceles part

Hey....Rofler
Can you show your solution, please? I am very interested in it.

I guess I shouldn't revive such an old thread, but here's a short solution without the use of any theorem that trivializes the problem.

Take an arbitrary circle. At most one point on it can be the same colour as the center if we are to avoid an isosceles triangle. The rest of the circle must then be coloured in two colours, and it would be possible to obtain a regular pentagon with vertices in at most two colours. At least three of the five vertices must be the same colour, which renders an isosceles triangle unavoidable.

Yes, this is precisely the solution spread over the two posts of thaithuan_GC above. ;)

[b]Consider a regular heptagon inscribed in a circle.By PHP, some three vertices are of same colour. [/b] Seven arcs can be divided in the following ways: 1+3+3,2+2+3 and 1+2+4. First two generates isosceles triangles, while the third case gives a triangle with angles 180/7,360/7 and 720/7, which are in G.P.

This is quite a lengthy solution, but gives much more precise result. By this prove we can show that, [b]there always exist such an isosceles triangle with same colored vertices, no need of otherwise condition[/b]. Following is the attachment:

We can actually just ignore the arithmetic progression condition and only use the isosceles condition.

AFTOC that no such triangle exists. Partition the plane into sets $A$, $B$, and $C$. This will be the coloring. First, choose a point $P \in A$ an arbitrary circle $\omega$ centered at $P$. Notice that $\omega$ can only have one point in $A$. Otherwise, the isosceles triangle is easy to find. Notice that either $B$ or $C$ is dense over $\omega$. Assume wlog that it is $B$. So, choose two points $X,Y \in B$ where $XY \leq \epsilon$ and $X$ and $Y$ are near the opposite side of the circle of the point in $A$ on $\omega$, if it exists. Now let $Y_1 \not = Y$ be the unique point on $\omega$ where $XY = XY_1$. It must be the case that $Y_1 \in C$. Likewise, if we define $X_1$ similarly then $X_1 \in C$. But, if we let $M$ be the midpoint of minor arc $XY$ then

1. $M \not \in A$ because the only point in $A$ on $\omega$ is on the other side

2. $M \not \in B$ because $MX = MY$

3. $M \not \in C$ because $MX_1 = MY_1$

So, $M$ cannot have a color. Contradiction.

We generalize to $N$ colors. Take a mapping from the unit interval to the unit circle of $t \mapsto e^{2 \pi i t}$. By Van der Waerden's theorem, there exists some $w = W(N, 3)$ such that when $\frac{1}{w}, \frac{2}{w}, \dots 1$ is $N$-colored, there exists an arithmetic progression with length $3$.

This finishes.


\[ {\rm{Prove}}: {\rm{ }}\frac{{tan\theta  \plus{} {\rm{sec}}\theta  \minus{} 1}}{{\tan \theta  \minus{} {\rm{sec}}\theta  \plus{} 1}} \equal{} \frac{{1 \plus{} \sin \theta }}{{\cos \theta }}
\]

[hide="hint"]
Take the left side and multiply the numerator and denominator by $ (\tan\theta \minus{} \sec\theta) \minus{} 1$
[/hide]

[hide="solution"]
$ \frac{\tan\theta \plus{} \sec\theta \minus{} 1}{\tan\theta \minus{} \sec\theta \plus{} 1}\cdot\frac{(\tan\theta \minus{} \sec\theta) \minus{} 1}{(\tan\theta \minus{} \sec\theta) \minus{} 1}$

$ \frac{\sec\theta \plus{} (\tan\theta \minus{} 1)}{(\tan\theta \minus{} \sec\theta) \plus{} 1}\cdot\frac{(\tan\theta \minus{} 1) \minus{} \sec\theta}{(\tan\theta \minus{} \sec\theta) \minus{} 1}$

$ \frac{(\tan\theta \minus{} 1)^2 \minus{} \sec^2\theta}{(\tan\theta \minus{} \sec\theta)^2 \minus{} 1}$

$ \frac{\tan^2\theta \minus{} \sec^2\theta \minus{} 2\tan\theta \plus{} 1}{\tan^2\theta \minus{} 2\tan\theta\sec\theta \plus{} \sec^2\theta \minus{} 1}$

$ \frac{\minus{}2\tan\theta}{2\tan^2\theta \minus{} 2\tan\theta\sec\theta}$

$ \frac{1}{\sec\theta \minus{} \tan\theta} \cdot \frac{\sec\theta \plus{} \tan\theta}{\sec\theta \plus{} \tan\theta}$

$ \frac{\sec\theta \plus{} \tan\theta}{\sec^2\theta \minus{} \tan^2\theta} \equal{} \sec\theta \plus{} \tan\theta$

$ \frac{1}{\cos\theta} \plus{} \frac{\sin\theta}{\cos\theta} \equal{} \boxed{\frac{1 \plus{} \sin\theta}{\cos\theta}}$
[/hide]


Can someone give a nice vectorial proof for the fact that the medians of the triangle are concurrent.

I proved it, but am not so happy with my proof. :(

The centroid of $ A, B, C$ is $ \frac{A \plus{} B \plus{} C}{3}$.  The median from $ A$ is the line through $ A$ and $ \frac{B \plus{} C}{2}$, which is the line $ A \plus{} t \cdot \left( \frac{B \plus{} C}{2} \minus{} A \right)$.  Let $ t \equal{} \frac{2}{3}$.

I think it is very easy!
With $ G$ is the point so that $ \overrightarrow {GA}\plus{}\overrightarrow {GB}\plus{}\overrightarrow {GC}\equal{}\overrightarrow {0}$
Then if $ M$ is midpoint of $ BC$ then $ \overrightarrow {GA}\equal{}\minus{}(\overrightarrow {GB}\plus{}\overrightarrow {GC})\equal{}\minus{}2\overrightarrow {GM}$
It mean $ A,G,M$ are collinear!
Similarly with $ 2$ other medians!

@t0raji:Can we assume that the centroid divides a median in the ratio 2:1?? Because thats how I proved it.

@tdl: $ \overrightarrow {GA}\plus{}\overrightarrow {GB}\plus{}\overrightarrow {GC}\equal{}\overrightarrow {0}$
Why is this?

The proof is to find the coordinates of the point, and show the lines go through it. You prove the ratio along the way.

[quote="Cyber-Shreyas"]$ \overrightarrow {GA} \plus{} \overrightarrow {GB} \plus{} \overrightarrow {GC} \equal{} \overrightarrow {0}$
Why is this?[/quote]
Do you read my solution? Call $ G$ is the point satisfy condition $ \overrightarrow {GA} \plus{} \overrightarrow {GB} \plus{} \overrightarrow {GC} \equal{} \overrightarrow {0}$. We must to prove that $ G$ is on the $ 3$ medians of triangle.

Excellent. Sorry, I didn't read the wordings carefully. Thanks tdl.


abstract algebra
vector
function
Functional Analysis
real analysis
real analysis unsolved

What is an example of a Hilbert space without a reproducing kernel (this just means that pointwise evaluation of the elements is NOT continuous).

The first thing that comes to mind is $ L^2$, but pointwise evaluation in $ L^2$ doesn't really make that much sense, so I don't like that example. I'm thinking we can take something like a Bergman space with a weight that messes up the pointwise evaluation, but I can't really make it work. I'm finding this surprisingly difficult

So, you want a Hilbert space in which the point evaluation functional makes sense for every point but is never continuous? The most straightforward way to construct it is to turn the question upside down. Take any infinite-dimensional Hilbert space and notice that the values of all discontinuous linear functionals at some vector determine the vector uniquely. Now just view your Hilbert space as a space of functions on the set of all discontinuous linear functionals.


10! = 10x9x8x7x6x5x4x3x2x1

There is a problem that 6^n can never be divided by 7 and 5.

The problem asks for the other way round - that $\frac{10!}{6^{n}}$ is integer.

[quote="mr. math"]Find the greatest value for n for which 6^n evenly divides 10!.

how would u do this problem?[/quote]

My bad. 10!/6^n

So, 10! has an excess of 2 and a limited amount of 3's.

9x6x3 --> 2+1+1 = 4 3's.

Hence, n=4?

[quote="Farenhajt"]Hmmm.... last time I checked, $9$ had two threes, $6$ had one, $3$ had one. Do you agree?  :)[/quote]

Yes I corrected it right after I made my post. I make mistakes a lot.


inequalities
limit
inequalities theorems

Find a short proof of the [b]W.Sierpinski inequality[/b]:

[i]  If $a=(a_1,a_2,...,a_n)$ is a system  of positive numbers,  then
\[ \begin{array}{|c|} \hline \\ \displaystyle {A_n(a)}^{n-1}H_n(a)\ge {G_n(a)}^n\ge A_n(a){H_n(a)}^{n-1} \\ \\ \hline \end{array}  \] where  $H_n(a)\; ,\; G_n(a)\; ,\; A_n(a)$ denote harmonic, geometric and arithmetic means of the numbers  $a_1, a_2, ..., a_n .$
[/i]

Later [b]H. Alzer[/b]  in [2] (around 1990) proves the inequalities $\; \; (n\ge 3)$:
\[ \begin{array}{|c|} \hline \\ \displaystyle \frac{1}{n}\left(G_n(a)\right)^{n-1}\displaystyle \left(G_n(a)-H_n(a) \right)\le \left(A_n(a)\right)^{n-1}H_n(a) -\left(G_n(a)\right)^n \\ \\ \displaystyle \frac{1}{n}\left(H_n(a)\right)^{n-1}\displaystyle \left(A_n(a)-G_n(a) \right)\le \displaystyle \left(G_n(a)\right)^{n} - A_n(a)\left(H_n(a)\right)^{n-1} \\ \\ \hline \end{array}\; ,  \]
the equality cases being valid if and only if $a_1=a_2=...=a_n .$
[b]References[/b]

[1] Alzer H., [i]Ueber die Ungleichung zwischen dem geometrischen und  dem arithmetischen Mittel,[/i] Quaest. Math.  10 (1987), 351-356.
[2] Alzer H., [i]A refinement of Sierpi\'nski's Inequalities,[/i] Universitas Comeniana,Acta Mathematica Universitatis Comenianae, [b]LVIII-LIX[/b], 175--180.
[3] Alzer H., Ando T, and Nakamura Y., [i]The inequalities of W. Sierpinski  and Ky Fan,[/i] J. Math. Anal. Appl., ?
[4] Bauer H.,  [i]A class of means and related inequalities,[/i]  Manuscripta Math. 55 (1986), 199--212.
[5] Bullen P. S., Mitrinovic D. S. and Vasic P. M., [i] Means and Their Inequalities,[/i] D.Reidel Publ. Co., Dordrecht 1988.
[6] A. Lupas , [i] On an inequality of W. Sierpinski , [/i]  Gazeta  Matematica , Seria-- (A) , V , Nr.3--4 , ( 1984) 189.
[7] D.S. Mitrinovic and P.M.Vasic , [i] On a theorem of W.  Sierpinski concerning means, [/i]   Publ.Elektrotehn.Fak.Ser.Mat.Fiz.   No. 544--No.576 (1976) 113--114.
[8] Pecaric J. E. and Wang C.L.,  [i] An extension of a Sierpinski inequality,[/i] Congr. Numer. 61 (1988), 35--38.
[9] W. Sierpinski , [i] Sur une in\'egalit\'e  pour la moyenne arithm\'etiq ,  geometrique et harmonique ,[/i] Warsch.    Sitzungsber., 2 (1909) 354--357.
[10] Wang, C.-L., [i] An extension ot two sequences of inequalities of Mitrinovic and Vasic,[/i]  Univ.Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz.,  634-- 677 (1979), 94--96.

[quote="flip2004"]Find a short proof of the [b]W.Sierpinski inequality[/b]:  [b]  If $a=(a_1,a_2,...,a_n)$ is a system  of positive numbers,  then
\[ \begin{array}{|c|} \hline \\ \displaystyle {A_n(a)}^{n-1}H_n(a)\ge {G_n(a)}^n\ge A_n(a){H_n(a)}^{n-1} \\ \\ \hline \end{array}  \]
 [/b]
[b]References[/b]
[1] A. Lupas , [i] On an inequality of W. Sierpinski , [/i]  Gazeta  Matematica , Seria-- (A) , V , Nr.3--4 , ( 1984) 189.
[2] D.S. Mitrinovic and P.M.Vasic , [i] On a theorem of W.  Sierpinski concerning means, [/i]   Publ.Elektrotehn.Fak.Ser.Mat.Fiz.   No. 544--No.576 (1976) 113--114.
[3] W. Sierpinski , [i] Sur une in\'egalit\'e  pour la moyenne arithm\'etiq ,  geometrique et harmonique ,[/i] Warsch.    Sitzungsber., 2 (1909) 354--357.[/quote]
 A  [b]proof [/b] is the following:   For  $n\in \{1,2 \}$  hold the  equality cases.   Further  suppose $n\ge 3 .$
  Let us suppose that $0<a_1\le a_2\le ...\le a_n .$  If  $\displaystyle z_k=\dfrac{1}{n a_k}(a_k-a_1)(a_n -a_k)\; ,$   then $z_k\ge 0 ,$  $k\in \{1,2,...,n\} ,$   and   $\displaystyle \sum\limits_{k=1}^{n}z_k=a_1+a_n-A_n(a)-\dfrac{a_1a_n}{H_n(a)}\; .$  Therefore $\; \; (1)\; \; \displaystyle a_1+a_n-A_n(a)\ge \displaystyle \dfrac{a_1a_n}{H_n(a)}\; .$ If $x=(x_1,x_2,...,x_{n-1}) ,$ with  \[ x_k=\left\{\begin{array}{lcl} a_{k+1} &, & k\in \{1,2, ...,n-2\} \\ a_1+a_n-A_n(a) &,& k=n-1 \end{array}\right. \; ,  \] then  $A_{n-1}(x)=A_n(a).$  Supposing that $A_{n-1}(x)\ge G_{n-1}(x)$  we obtain
\[ (2)\hspace*{0.5cm} {A_n(a)}^{n-1}\ge a_2a_3...a_{n-1}\left(a_1+a_n -A_n(a)\right)\hspace*{0.5cm} .  \]
Using (1) , from (2) we get $\displaystyle {A_n(a)}^{n-1}\ge\displaystyle a_1a_2a_3...a_n\dfrac{1}{H_n(a)}\hspace*{0.5cm} ,$ that is \[ {A_n(a)}^{n-1}H_n(a)\ge {G_n(a)}^n \; .  \]
 If $1/a$  denotes the system  $\displaystyle \left(\dfrac{1}{a_1}\; ,\; \dfrac{1}{a_2}\; ,...,\; \dfrac{1}{a_n}\right) ,$   then  $\displaystyle {A_n(1/a)}^{n-1}H_n(1/a)\ge \displaystyle {G_n(1/a)}^n$ is the same with
${G_n(a)}^n\ge A_n(a){H_n(a)}^{n-1} .$ From  (1)-(2)  we conclude that, when $n\ge 3 ,$   the equality cases hold if and only if  $a_1=a_2=...=a_n\; .$

[quote="bodan"][b]Generalization:[/b]  ...... [/quote]
Interesting and nice !

Give one problems which solved with Sierpinski inequality!!! :wink:

[quote="Fitim"]Give one problem which solved with Sierpinski inequality![/quote]
Possible problems:

[b](I).[/b] Assume that $0<a_{1}<a_{2}<\ldots<a_{n}<a_{n+1}<$ and  denote $A_{n}(a)=\frac{1}{n}\sum_{k=1}^{n}a_{k}\; \; \; ,\; \; G_{n}(a)=\sqrt[n]{\prod_{k=1}^{n}a_{k}}\; .$
Find a sequence $(b_{n})_{n= 1}^{\infty}\;$ such that $\; b_{n}>1 \; , \; \forall n , \; n\ge 2 ,$  and  $\sqrt[n]{b_{n}}< \frac{A_{n+1}(a)}{G_{n+1}(a)}< b_{n}^{n}\; \; ,\; \; \forall n\; ,n\ge 2 .$

[b](II).[/b]  Consider the sequences $(e_{n})_{n=1}^{\infty}\; ,(E_{n})_{n=1}^{\infty},$ with $e_{n}=\left(1+\frac{1}{n}\right)^{n}\; ,\; E_{n}=\left(1+\frac{1}{n}\right)^{n+1}\; .$ Prove the (well-known) inequalities \[(*)\; \; \; \; e_{n}<e_{n+1}<\ldots < E_{n+1}<E_{n}<\; ,\; \; \forall n \in{\mathbb N}^{*}.\] [b][color=green]Improve inequalities (*) . [/color] [/b]

[b]Possible Hint ;      [/b]In Sierpinski inequality     ($n \leadsto n+1$ ), consider the positive numbers \[a_{1}=a_{2}=...=a_{n}=1+\frac{1}{n}\; ,\; a_{n+1}=1\] and then ($n\leadsto n+2$ )$b_{1}=b_{2}=...=b_{n+1}= \frac{n}{n+1}\; ,\; b_{n+2}=1 \; .$

[b](III).[/b]  Let $c_{n}=1+\frac{2}{\sqrt{n}}-\frac{2}{n}.$ In order to prove that $\lim_{b\to \infty}\sqrt[n]{n}=1$ [b]find [/b]$(\alpha_{n}),(\beta_{n})\; ,\; \beta_{n}<1 ,$ with $\lim_{n\to \infty}\alpha_{n}= \lim_{n\to \infty}\beta_{n}= 1$ \[\alpha_{n}< \frac{\sqrt[n]{n}}{c_{n}}<\beta_{n}\; \; ,\; \; ( n\ge 2).\]


with $ a,b,c > 0$   Prove that
$ \sqrt {\frac {2a^2 \plus{} bc}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {2b^2 \plus{} ca}{c^2 \plus{} a^2}} \plus{} \sqrt {\frac {2c^2 \plus{} ab}{a^2 \plus{} b^2}}\geq 2\sqrt {2} \plus{} \frac{1}{\sqrt{2}}$

[quote="2424"]with $ a,b,c > 0$   Prove that
$ \sqrt {\frac {2a^2 \plus{} bc}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {2b^2 \plus{} ca}{c^2 \plus{} a^2}} \plus{} \sqrt {\frac {2c^2 \plus{} ab}{a^2 \plus{} b^2}}\geq 2\sqrt {2} \plus{} 1$[/quote]
It's wrong! Try $ a\equal{}b\equal{}1$ and $ c\rightarrow0^\plus{}.$

[quote="arqady"][quote="2424"]with $ a,b,c > 0$   Prove that
$ \sqrt {\frac {2a^2 \plus{} bc}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {2b^2 \plus{} ca}{c^2 \plus{} a^2}} \plus{} \sqrt {\frac {2c^2 \plus{} ab}{a^2 \plus{} b^2}}\geq 2\sqrt {2} \plus{} 1$[/quote]
It's wrong! Try $ a \equal{} b \equal{} 1$ and $ c\rightarrow0^ \plus{} .$[/quote]
Sorry,I edited: $ RHS\equal{}2\sqrt{2}\plus{}\frac{1}{\sqrt{2}}$


The problem is wrong. Take y_1=y_2=4, x_1=1 and x_2=2.

Your conter-example is false ... and the inequality is not  "x1^2<y1 and x2^2<=y2" it's  x1^2<=y1 and x2^2<=y2! sorry!

Let us write it as 4(x_1* \sqrt y_2-x_2* \sqrt y_1) 2 +(y_1-y_2) 2  \geq 4x_1*x_2*( \sqrt y_1- \sqrt y_2) 2 . It suffices to show that ( \sqrt y_1+ \sqrt y_2) 2  \geq 4x_1*x_2, which is trivial since (a+b) 2  \geq 4ab


First question with bib style
I used \usepackage{natbib}..
\bibliographystyle{plainnat}
\bibliography{paper'sname}..

But the bibliography isn't in author's alphabetical order.
I tried some other styles. Neither works.
Which style I should use in stead?

Second, Table..
I found that fonts of all the pages with tables changed. Different from that in page with pure text. What should I do with it?

Thank you.

[quote="weng"]But the bibliography isn't in author's alphabetical order.
I tried some other styles. Neither works.
Which style I should use in stead?[/quote]
The README for [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=sort-by-letters]sort-by-letters[/url] lists some appropriate styles you can use.

[quote="weng"]
I found that fonts of all the pages with tables changed. Different from that in page with pure text. What should I do with it?
[/quote]
It's hopeless trying to guess without a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]complete minimal example[/url].


How many positive integers less than or equal to 10,000 are divisible by 2, 5 or 10?

There are 10,000 / 2 = 5,000 numbers from 1 to 10,000 that are divisible by 2.
There are 10,000 / 5 = 2,000 numbers from 1 to 10,000 that are divisible by 5.

If we just add 5000 and 2000, we will overcount hte numbers that are divisible by both 2 and 5.

There are 10,000 / 10 = 1,000 numbers from 1 to 10,000 that are divisible by 10.

Therefore, there are 5000+2000-1000=6000 numbers that are divisible by 2 and/or 5.