AMC
AIME
geometry
3D geometry
modular arithmetic
algebra
binomial theorem

Find the smallest positive integer whose cube ends in 888.
(do this without a calculator or computer.)

[hide="Solution"]We have to find $x$ such that $x^3=1000n+888$ for some $n\in\mathbb{N}$

Obviously $x$ is even, hence we can put $x=2y$, which gives

$125n+111=y^3\implies 125n+110=y^3-1$

From this $y^3-1\equiv 0\pmod{5}$. Checking the cubes of the residues modulo $5$, we see that only $1$ satisfies the condition. Hence $y=5a+1$. Substituting we get

$125n+110=125a^3+75a^2+15a\implies$
$25n+22=25a^3+15a^2+3a\implies$
$25n+20=25a^3+15a^2+3a-2$

Now $3a-2\equiv 0\pmod{5}$, hence we can write $3a-2=5b$. From there $a=\frac{5b+2}{3}=b+\frac{2(b+1)}{3}$. This means that $b+1=3c$, which gives $a=3c-1+2c=5c-1$. Substituting $a$ we get

$25n+20=25(125c^3-75c^2+15c-1)+15(25c^2-10c+1)+15c-3-2\implies$
$25n+25=25(125c^3-75c^2+15c-1+15c^2-6c)+15+15c\implies$
$n+1=125c^3-60c^2+9c-1+\frac{3(c+1)}{5}\qquad(*)$

From there it's clear that $c+1\equiv 0\pmod{5}$, hence $c=5k-1$. Now $a=5c-1=25k-6$, and $x=2y=10a+2=250k-58$. Obviously, the smallest $x$ is obtained for $k=1$, which gives $x=192.$

From $k=1$ we get $c=4.$ Substituting that into $(*)$ we get $n=7077$

Therefore, the desired number is $192$ and $192^3=7077888$

Note: All numbers such that their cube ends in $888$ are given by $x=250k-58, k\in\mathbb{N}$[/hide]

[hide]Notice that the number must end in two in order for this to work.  Also, if one uses the binomial theorem,
$(10k+2)^3=1000k^3+600k^2+120k+8$.  Notice since only the last 3 digits are significant for this problem, the first term can be disregarded.  Also, the second term is the only term that makes the tens digit an 8 so it follows that k=4, 9, 14, 19,...  However, the hundred's digit is influenced by both the first and second terms.  But since the first term is from $k^2$ and only the last 3 digits matter, it's forced that the first term can only generate 600.  All that's left is to find a k that will cause 120k to generate 280 as it's last three digits.  19*120=2280 which works.  Therefore, 192 is the smallest integer whose cube ends in 888.[/hide]


superior algebra
superior algebra unsolved

Let $ \mathbf{K}/\mathbf{k}$ be a finite extension and let $ \textrm{Aut}_{\mathbf{k}}\left(\mathbf{K}\right)$ be the group of automorphisms of $ \mathbf{K}$ that fix $ \mathbf{k}$.  Show that

\[ \#\textrm{Aut}_{\mathbf{k}}\left(\mathbf{K}\right) \leqslant \left[\mathbf{K}: \mathbf{k}\right] .\]
We know that $ \mathbf{k}\subset\mathbf{K}^{\textrm{Aut}_{\mathbf{k}}\left(\mathbf{K}\right)}\subset\mathbf{K}$ (where these inclusions may be proper), so

\[ \left[\mathbf{K}: \mathbf{k}\right] \equal{} \left[\mathbf{K}: \mathbf{K}^{\textrm{Aut}_{\mathbf{k}}\left(\mathbf{K}\right)}\right]\left[\mathbf{K}^{\textrm{Aut}_{\mathbf{k}}\left(\mathbf{K}\right)}: \mathbf{k}\right] ,\]
but how do we procede from here?

Show that $ Aut_k(K)$ is a $ k$-linearly independent subset of $ End_k(K)$ and conclude that the mapping
\[ Tr(x) \equal{} \sum_{\sigma\in Aut_k(K)}\sigma(x)
\]
defines a [b]non-trivial[/b] $ k$-linear endomorphism of $ K$. Use this to conclude
\[ \# Aut_k(K) \equal{} [K: K^{Aut_k(K)}]
\]
by contradiction.


$ a_{n\plus{}2}\equal{}\frac{1}{a_n}(a_{n\plus{}1}^2\minus{}1)$  , $ a_2\equal{}7$ and $ a_1\equal{}1$

prove that : $ 1\plus{}9a_{n\plus{}1}a_n$ is a perfect quare

Define $ r_{\pm}\equal{}\frac{7\pm3\sqrt{5}}{2},b_n\equal{}\frac{r_\plus{}^n\minus{}r_\minus{}^n}{3\sqrt{5}}$ and note, that the identities $ r_{\pm}^2\equal{}7r_{\pm}\minus{}1\,,\,r_\plus{}r_\minus{}\equal{}1\,,\,r_\plus{}\plus{}r_\minus{}\equal{}7\,\,b_1\equal{}1\,,\,b_2\equal{}7$ and the recursion $ \forall n\in\mathbb{N}: b_{n\plus{}2}\equal{}7b_{n\plus{}1}\minus{}b_n$ hold. In particular, $ b_n$ is a series in the positive integers. Moreover, we can compute $ \forall n\in\mathbb{N}$ the difference $ b_{n\plus{}2}b_n\minus{}b_{n\plus{}1}^2\equal{}\ldots \equal{}1$ from which we can conclude that the series $ a_n$ and $ b_n$ are identical. For the final step we calculate:

$ (b_n\plus{}b_{n\plus{}1})^2\minus{}1\minus{}9b_nb_{n\plus{}1}\equal{}\ldots \equal{} 0$ showing that $ 1\plus{}9a_na_{n\plus{}1}$ is a perfect square determined by $ (a_n\plus{}a_{n\plus{}1})^2$.


Ok, so I did #6 after a loooooooong time and need a faster way.
I did part of #8 but missed a bit, so yeah.
I just attached it on Word so I can copy and paste :D

Ooh, these are nice problems.
1. [hide]Difficult to explain, let each amount of change be abc and consider the different cases. You can have 455/545 or 545/455 for a total of 2.[/hide]2. [hide]Make a 2x5 grid, and how many ways can you fill it with 5 1x2 rectangles? One gets 8 ways. For each of these 8 ways, we can distribute the boyfriend-girlfriend pairs 5*4*3*2 ways, and, since each boy and girl can be switched around in their seats, we multiply by 2^5 for a total of 8*5*4*3*2^6 = 30720 ways.[/hide]

Whoops, forgot to post the answers.
[hide="6"] It's 8[/hide]
[hide="8"]Yes, 30720. Heh, I guess I missed some of the 8 ways...[/hide]

Still need #6.

[hide=" Not Very Smart and Efficient Way, but sure to get right answer"]I think that the only way is to brute-force it. But if you think about it, you're limited since the amount
a.bc + the change d.ef must equal 10 dollars.

d + a must equal nine, so based on that, you have say, 5 and 4.

So you already know 4.bc and 5.ef.
And $ c \plus{} f \equal{} 10$.
So you have $ bc \plus{} ef \equal{} 100$. One obvious way is 55 + 45. Also keep in mind that there must be at least one $ 4$ must be in ef and at least one $ 5$ bc.
What is c was 4 and f was 6?
Then b would have to be 6 and e would have to be 3, that wouldn't since you don't have a 5. Then you have to do this for all the others, though you might think of other restrictions to save you time.

And you only have to have set "a" from 1 to 4, then double the number of amounts/change[/hide]

Sorry, I've been an idiot.
[hide]Let the amount of change be abc. Then, 1000 equals abc + one of these:
acb
bac
cab
cba
bca
We soon find out that there are only 8 possible cases, all of which are in the abc+bac category, which are
185, 275, 365, 455, 545, 635, 725, and 815.[/hide]

Oh, that makes a lot of sense.
Thanks.
My way was somewhat like the way a couple of posts above, but I wasted some time.

I think for number six there's only one acceptable pattern and it's abc, bac...since the first 2 digits both have to sum to 9. otherpatterns will just give you 5.45+4.55 and 4.55+5.45... but that's already in the original.[/hide]


[quote="brettburns83"]I have two questions that both require proofs... I understand the questions and am able to give examples and understand why they work but have no idea where to begin with their proofs... any help is appreciated.

#1 A child watches TV at least one hour each day for 7 weeks but never more than 11 hours in any one week. Prove that there is some period of consecutive days in which the child watches exactly 20 hours of TV. (Assuming that the child watches TV for whole number of hours per day)

#2 Prove that in a group of n>1 people there are 2 people who have the same number of acquaintances in the group. (It is assumed that no-one is acquainted with him or herself)[/quote]
these are very simple
first the second one:
person $A_i$ has $a_i$ friends
thus $0\leq a_i\leq n-1$ in the other hand we know that there arent persons $A_r,A_s$ such that $a_r=0,a_s=n-1$ because if this happons then $A_S$ is acquainted with all other people but $a_r=0$ there for $A_r$ is acquainted with no one
this is a contradiction
there for we have $n-1$ holes with $n$ birds...
the first one is easy to:
assume that $s_k$ is the number of hours that the child has watches from day $1$ untill day $k$,$k=1,2,...,49$ but we know that the child has watched tv less than $11$ hours each week there for $s{}_4{}_9\leq 11*7=77$ and we know that the child has watched tv for atleast one hour there for:
$1\leq s_1<s_2<...<s{}_4{}_9\leq 77$ now we add $20$ to each $s_k$
$s_1+20<s_2+20<...<s{}_4{}_9+20\leq 97$
these $49$ numbers with the last $49$ numbers are together $98$ numbers from $/{1,2,...,97/}$there for we have at least two equal number
but we know that no pairs of $s_1,...,s{}_4{}_9$ and no pairs of $s_1+20,...,s{}_4{}_9+20$ are equal there for there exist $i,j$ such that $s_i=s_j+20$ there for $s_i-s_j=20$
this means that the child has watched 20 hours tv from day $j+1$ untill day $i$
 ;)  :D


real analysis
function
real analysis unsolved

This is a very cute one:
   Suppose $f$ is a real valued function defined on $[0,1]$ such that for almost all $x$ we have $\limsup_{h\to 0}{\left|\frac{f(x+h)-f(x)}{h}\right|}=0$. Then for all $\epsilon>0$ we can find a Lipschitz function $g$ defined on $[0,1]$ such that $f$ differs from $g$ on a set of measure at most $\epsilon$. Can we take $\epsilon=0$?

Yes, this is a nice result (with $\epsilon>0$), and its converse is true as well. (*) But you can't have $\epsilon=0$ because of examples like $f(x)=\sqrt{x}$. Since a set of full measure is dense, if $g$ is Lipschitz on such a set, then it extends to a Lipschitz function on $[0,1]$. But $f$ as above cannot be made Lipschitz by redefining it on a set of measure zero, since $f'\notin L^{\infty}$.

(*) if $\limsup$ is replaced with $\mathrm{ess\,lim\,sup}$.


The 'Name that Decimal' forum gave me the idea.

Not only are we good at naming decimals, but we are also good with Sequences/ordered sets.

The rules:
Once you solve it, you must post a new problem.

We'll start of fairly easy:
2,1,3,4,7,11,...

this looks similar to fibbonacci but i don't know what it's called (some sort of lucas sequence)

wait, aren't they just called Lucas Numbers.
[quote]some sort of lucas sequence[/quote]
what do u mean by that?

Yeah, its the Lucas Numbers.

The same patter as fibonnaci numbers, but instead of starting with 1,1. This sequence starts with 2,1.

*NOTE: This topic is partially for fun, but more importantly it should help us in pattern finding and other such important skills. Make sure the sequence is some known sequence that has some relevance in some branch of math (ie, dont do the sequence of random numbers you came up with on the spot, or the sequence of digits in $\pi$ starting at the 23803847th digit)

Because it seems that people were confused as to how to use this I'll put the next one up. from now on, if you are the first one to post an answer, start a new sequence.
1,2,4,8,3,6,12,11,9,5,10,7,1,2,4,8,...
[hide="hint"]
Number theory
[/hide]


inequalities
function
inequalities unsolved

the function $f(x) = \frac{x}{\sqrt{1-x}}$ is convex.
so, by jensen, we find $\sum{f(x_i)} \geq \sqrt{\frac{n}{n-1}}$.

by cauchy we have $\sqrt{n} \geq \sqrt{a_1} +...+ \sqrt{a_n}$  :)


calculus
integration
trigonometry
complex numbers
real analysis
real analysis unsolved

Show that $ \int_0^{\infty} e^{\minus{}(\frac{y}{x})^{\alpha}}\cos y \mbox{ dy} \equal{} O(\frac{1}{x^{\alpha}})$ for positive $ x$. Any help will be appreciated.

Here is the shortest way I know though it is not the most elementary. Integrate by parts once. You'll get $ \frac \alpha{x^\alpha}\int_0^\infty y^{\alpha\minus{}1}e^{\minus{}(y/x)^\alpha}\sin y\,dy$. View this new integral as the imaginary part of $ \int_0^\infty y^{\alpha\minus{}1}e^{\minus{}(y/x)^\alpha}e^{iy}\,dy$ and change the integration path to a ray making a small positive angle with the positive semi-axis. Then $ e^{iy}$ will start decaying exponentially, $ e^{\minus{}(y/x)^\alpha}$ will still be bounded by $ 1$, and $ y^{\alpha\minus{}1}$ will have just power growth on this new path. So, you'll get a uniform (in $ x$) estimate for this integral by just disregarding the factor depending on $ x$. The end.

There are many other approaches, of course, but they all require more computations.


algebra
polynomial
linear algebra
linear algebra unsolved

Let $ A$ be $ n$x$ n$ matrix.Show that if $ A^2\equal{}I$ than $ rank(A\plus{}E)\plus{}rank(A\minus{}E)\equal{}n$.

More generally let $ P,Q$ be $ 2$ pairwise prime polynomials  s.t. $ (PQ)(A)\equal{}0$.
Then $ ker(P(A))$ and $ ker(Q(A))$ are supplementary subspaces. (decomposition of the kernels).

use $ r(A\plus{}B) \leq r(A)\plus{}r(B) \leq n\plus{}r(AB)$ (i don't have the proof of this, so if you know it, please show me :D )
$ n\equal{}n\plus{}r(E\minus{}A^2) \geq r(E\plus{}A)\plus{}r(E\minus{}A) \geq r(2E)\equal{}n$
now we have the desire


I like some help with this equation if that possible. Here goes:

Write the parabola in standard form:
Has an axis of symmetry of x=2 and passes through points (0,7) and (-1,-3)

thanks in advance!

another way to look at it is $y-k=a(x-2)^{2}$, then plug in the points for a linear system


A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s = 80t - 16t^2. For what time interval is the ball less than 64 feet above the ground?

This question is equivalent to solving the following inequality:
80t - 16t^2 < 64

80t - 16t^2 < 64
16t^2 - 80t + 64 > 0
t^2 - 5t + 4 > 0
(t-4) (t-1) > 0

Therefore 0<= t < 1 (t can takes negative values in this question) , t > 4.

I hope this can help you! :P

[quote="Interval"]Can this considered a physics question?[/quote]

I think this question should actually be posted in the Physics section, since this is a mechanics question...


Diophantine equation
number theory unsolved
number theory

$ 2n\plus{}1$ ,   $ 37n\plus{}1$ are the squares integer number. Find all n=?

[quote="zaya_yc"]$ 2n \plus{} 1$ ,   $ 37n \plus{} 1$ are the squares integer number. Find all n=?[/quote]

[hide="maybe beginning of solution ..."]
$ 2n\plus{}1\equal{}a^2$
$ 37n\plus{}1\equal{}b^2$
This system is equivalent to the Pell equation $ 37a^2\minus{}2b^2\equal{}35$

Brute force on this Pell equation gives
$ (a,b)\equal{}(1,1)$ $ \implies$ $ n\equal{}0$
$ (a,b)\equal{}(17,73)$ $ \implies$ $ n\equal{}144$
$ (a,b)\equal{}(103,443)$ $ \implies$ $ n\equal{}5304$
$ (a,b)\equal{}(2839,12211)$ $ \implies$ $ n\equal{}4029960$
...

The question remains to find the general solution of the equation $ 37a^2\minus{}2b^2\equal{}35$
[/hide]

[quote="zaya_yc"]$ 2n \plus{} 1$ ,   $ 37n \plus{} 1$ are the squares integer number. Find all n=?[/quote]
Here is a method for generating infinely many solutions (but unfortunately not all) :

$ 2n \plus{} 1 \equal{} a^2$ and $ 37n \plus{} 1 \equal{} b^2$ is equivalent to $ 37a^2 \minus{} 2b^2 \equal{} 35$

It's rather easy to verify that $ 37x^2 \minus{} 2y^2 \equal{} z$ $ \implies$ $ 37(\frac {4xy \plus{} 4y^2 \plus{} z)}z)^2 \minus{} 2(\frac {74xy \plus{} 4y^2 \plus{} z)}z)^2 \equal{} 35$

So, for example, if we have a generating method for $ 37x^2 \minus{} 2y^2 \equal{} \pm 2$, we'll have a generating method for our equation.

$ 74u^2 \minus{} v^2 \equal{} \pm 1$ $ \implies$ $ 37(2u)^2 \minus{} 2v^2 \equal{} \pm2$ and so a generating method for $ 74x^2 \minus{} y^2 \equal{}\pm 1$ would lead to a generating method for our equation.

And since $ 74\cdot 0^2 \minus{} 1^2 \equal{} \minus{} 1$ and $ 74(43x \plus{} 5y)^2 \minus{} (370x \plus{} 43y)^2 \equal{} \minus{} (74x^2 \minus{} y^2)$, we got a generating method for this equation :

Starting with $ (0,\pm 1)$ and using the transformation $ (x,y)\to(\pm(43x \plus{} 5y),\pm(370x \plus{} 43y))$ we get a set of solutions of $ 74u^2 \minus{} v^2 \equal{} \pm 1$

So, starting with $ (0,\pm 1)$ and using the transformation $ (x,y)\to (\pm(43x \plus{} 10y),\pm(185x \plus{} 43y))$ we get a set of solutions of $ 37u^2 \minus{} 2v^2 \equal{} \pm 2$

And, for each such couple $ (x,y)$ solution of $ 37u^2 \minus{} 2v^2 \equal{} \pm 2$, we get a solution $ (\pm\frac {4xy \plus{} 4y^2 \plus{} z)}{37x^2 \minus{} 2y^2},\pm\frac {74xy \plus{} 4y^2 \plus{} z)}{37x^2 \minus{} 2y^2})$ of $ 37u^2 \minus{} 2v^2 \equal{} 35$ and so a solution $ n \equal{} \frac 12((\frac {4xy \plus{} 4y^2 \plus{} z)}{37x^2 \minus{} 2y^2})^2 \minus{} 1)$ to the required problem.

This gives for solutions of $ 37u^2 \minus{} 2v^2 \equal{} \pm 2$ the sequence $ (0,1),(10,43),(860,3699),(73970,318157),$ $ (6362280,27365201),(547230050,2353725443), ...$

And so for solutions of $ 37u^2 \minus{} 2v^2 \equal{} 35$ the sequence $ (1,1),(4559,19609),(33727481,145067381),$ $ (249515899879,1073208465029), ...$

And so solutions for $ n$ : $ \frac {1^2 \minus{} 1}2,\frac {4559^2 \minus{} 1}2,$ $ \frac {33727481^2 \minus{} 1}2,\frac {249515899879^2 \minus{} 1}2,...$

Obviously we missed a lot of solutions ($ n \equal{} 144, 5304, 4029960, ...$)
But this is the best I can do up to now.


Just saying even or odd is usually clear enough.  If you want something more though you could use something like:

$x=2n, \ n\in \mathbb{Z}$

for evens.  Z is the set of integers.  So plugging integers into 2n only produces even integers.

Likewise you can use the following for odds:

$x=2n+1, \ n\in \mathbb{Z}$

[quote="djshowdown2"]This question may be answered in the topic, but I'm not sure.  Many talking about two sets of numbers (evens and odds) is it proper to say something like "if n is an element of the set of odd integers" or would you use something more along the lines of "if n is congruent to 1 mod 2?"  I suppose I'm just asking if the words 'even' and 'odd' are formal enough math terms to use in a solution or proof.[/quote]

By all means... just say "if n is an odd integer".

Speaking of odd integers, can a negative integer be odd... or even?
Is there such a thing as negative zero??
I'm just making sure.

A negative integer can be odd or even, but I don't think negative 0 exists

Zero is neither negative nor positive.  It is even, however.


[quote="karpov352"]hi guyz !
help me out with this prob with complete explanation
thank you :D[/quote]
Well we can only help you if you give us the problem...

hey i just now attached a file its not there? cant you attach a jpeg file ?man
how can i post a picture then?

i hav aatached plz explain clearly even though its easy for you!
thank you :D

This can't be solved unless there is some relation between the rightmost line and the rest of them (maybe parallel to the leftmost).

one thing that i am sure is that value of x is greater than 80, provide the complete question for complete ans.

wait,  is x a right angle?

Because in the attachment the degree mark was a right angle.

Well, if we assume it isn't (which it probably is)

then angle in = angle out so x = 180-(60+40)=80.

I think that 80 is the right answer but im not sure.

from the diagram, I'd say x is an obtuse angle... Therefore, it cannot be 80.

As I said, this can't be solved unless there is some relation between the rightmost line and the rest of them.

@polkadotman1: They aren't opposite angles since the angles aren't formed by the same lines.


Does anyone kno a good place I can get this book?

it seems that there is no way for you to get it free online.So, just try Amazon or a bookstore.


Let $ a,b,c$ positive reals. Prove that

$ \frac{a^2\plus{}b^2\plus{}c^2}{2(ab\plus{}bc\plus{}ca)}\plus{}\frac{a^3\plus{}b^3\plus{}c^3}{3abc}\plus{}\frac{12abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\geq3$

[quote="Ligouras"]Let $ a,b,c$ positive reals. Prove that

$ \frac {a^2 \plus{} b^2 \plus{} c^2}{2(ab \plus{} bc \plus{} ca)} \plus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{3abc} \plus{} \frac {12abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq3$[/quote]

We have:
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}\frac{8abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge 2$
and:
$ \frac{a^3\plus{}b^3\plus{}c^3}{3abc}\plus{}\frac{8abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge 2$

[hide]Let $ a\plus{}b\plus{}c\equal{}3p,ab\plus{}bc\plus{}ca\equal{}3q^2,abc\equal{}r^3$. By AM-GM we have \[ p\equal{}\frac{a\plus{}b\plus{}c}3\equal{}\frac{\sqrt{a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca}}3\ge\frac{\sqrt{3(ab\plus{}bc\plus{}ca)}}3\equal{}q\equal{}\sqrt{\frac{ab\plus{}bc\plus{}ca}3}\ge\sqrt{\frac{3\sqrt{a^2b^2c^2}}3}\equal{}abc\equal{}r.\] Thus $ \boxed{p\ge q\ge r}$.

\[ a^2\plus{}b^2\plus{}c^2\equal{}(a\plus{}b\plus{}c)^2\minus{}2(ab\plus{}bc\plus{}ca) \\ \boxed{a^2\plus{}b^2\plus{}c^2\equal{}9p^2\minus{}6q^2}\]

\[ a^3\plus{}b^3\plus{}c^3\equal{}(a\plus{}b\plus{}c)^3\minus{}3(a^2b\plus{}ab^2\plus{}b^2c\plus{}bc^2\plus{}c^2a\plus{}ca^2)\minus{}6abc \\ a^3\plus{}b^3\plus{}c^3\equal{}27p^3\minus{}3((a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}3abc)\minus{}6r^3 \\ a^3\plus{}b^3\plus{}c^3\equal{}27p^3\minus{}3((3p)(3q^2)\minus{}3r^3)\minus{}6r^3 \\ \boxed{a^3\plus{}b^3\plus{}c^3\equal{}27p^3\minus{}27pq^2\plus{}3r^3}\]

\[ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}a^2b\plus{}ab^2\plus{}b^2c\plus{}bc^2\plus{}c^2a\plus{}ca^2\plus{}2abc \\ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}abc \\ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}(3p)(3q^2)\minus{}r^3 \\ \boxed{(a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}9pq^2\minus{}r^3}\]

By Schur, \[ a^3\plus{}b^3\plus{}c^3\plus{}3abc\ge a^2b\plus{}ab^2\plus{}b^2c\plus{}bc^2\plus{}c^2a\plus{}ca^2 \\ (27p^3\minus{}27pq^2\plus{}3r^3)\plus{}3r^3\ge(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)\minus{}3abc \\ 27p^3\minus{}27pq^2\plus{}6r^3\ge(3q^2)(3p)\minus{}3r^3 \\ 27p^3\minus{}27pq^2\plus{}6r^3\ge9pq^2 \\ \boxed{3p^3\plus{}r^3\ge4pq^2}\]

Therefore, \[ \frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{2(ab\plus{}bc\plus{}ca)}\plus{}\frac{a^{3}\plus{}b^{3}\plus{}c^{3}}{3abc}\plus{}\frac{12abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\geq3 \\ \Leftrightarrow \frac{9p^2\minus{}6q^2}{6q^2}\plus{}\frac{27p^3\minus{}27pq^2\plus{}3r^3}{3r^3}\plus{}\frac{12r^3}{9pq^2\minus{}r^3}\ge3 \\ \Leftrightarrow \frac{p^2}{2q^2}\plus{}\frac{3p^3\minus{}3pq^2}{r^3}\plus{}\frac{4r^3}{9pq^2\minus{}r^3}\ge1 \\ \Leftrightarrow 9p^3q^2r\minus{}p^2r^6\plus{}54p^4q^4\minus{}54p^2q^6\minus{}6p^3q^2r^3\plus{}6pq^4r^3\plus{}8q^2r^6\ge18pq^4r^3\minus{}2q^2r^6 \\ \Leftrightarrow 3p^3q^2r^3\plus{}54p^4q^4\plus{}10q^2r^6\ge p^2r^6\plus{}54p^2q^6\plus{}12pq^4r^3 \\ \Leftrightarrow 3pq^2r^3(p^2\minus{}q^2)\plus{}54p^4q^4(p^2\minus{}q^2)\ge r^6(p^2\minus{}q^2)\plus{}9q^2r^3(pq^2\minus{}r^3)\]

Note that $ 3p^3\plus{}r^3\ge4pq^2\implies3p(p^2\minus{}q^2)\ge pq^2\minus{}r^2$. Hence $ \boxed{3pq^2r^3(p^2\minus{}q^2)\ge q^2r^3(pq^2\minus{}r^3)}$. We also have $ 24p^2q^4(p^2\minus{}q^2)\ge24pq^2r^3(p^2\minus{}q^2)\ge8q^2r^3(pq^2\minus{}r^3)$ thus $ \boxed{24p^2q^4(p^2\minus{}q^2)\ge8q^2r^3(pq^2\minus{}r^3)}$. Lastly, we have $ 30p^2q^4(p^2\minus{}q^2)\ge p^2q^4(p^2\minus{}q^2)\ge r^6(p^2\minus{}q^2)$ so $ \boxed{30p^2q^4(p^2\minus{}q^2)\ge r^6(p^2\minus{}q^2)}$. Summing up these three inequalities, we get the desired result.

Q.E.D.[/hide]

[quote="Ligouras"]Let $ a,b,c$ positive reals. Prove that

$ \frac {a^2 \plus{} b^2 \plus{} c^2}{2(ab \plus{} bc \plus{} ca)} \plus{} \frac {a^3 \plus{} b^3 \plus{} c^3}{3abc} \plus{} \frac {12abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq3$[/quote]

\[ \Leftrightarrow \sum {\left( {\frac{1}{{4(ab \plus{} bc \plus{} ca)}} \plus{} \frac{{a \plus{} b \plus{} c}}{{6abc}} \minus{} \frac{{3c}}{{2(a \plus{} b)(b \plus{} c)(c \plus{} a)}}} \right)\left( {a \minus{} b} \right)^2 }  \ge 0\]

easy to prove that

\[ {\frac{1}{{4(ab \plus{} bc \plus{} ca)}} \plus{} \frac{{a \plus{} b \plus{} c}}{{6abc}} > \frac{{3c}}{{2(a \plus{} b)(b \plus{} c)(c \plus{} a)}}}\]

...


algebra
polynomial
function
trigonometry
Galois Theory
algebra unsolved

have you tried breaking down into identities

ex. sin2x=2sinxcosx
     sin3x=sin(x+2x)= sinxcos2x+sin2xcosx

(P.S. cant remember identities from top of my head but i think thats right.)

Yes, and that what's I did.  I obtained $2sin(x)cos(x)+4sin(x)cos(x)^2$ on the left side.  Substituting, $sin(x)=u$ and $cos(x)=\pm\sqrt{1-u^2}$.  The equation yield the polynomial that I wrote earlier

Maybe it will be useful (polynomial is also of order 6, but the coeffcients are nicer):
Let's substitute $z=\cos(x)+i\sin(x)$. Then $\sin(x)+\sin(2x)+\sin(3x)$ can be expressed as $\Im(z+z^{2}+z^{3})$. For any complex $p$ we have $\Im(p)=\frac{p-\overline{p}}{2i}$, so we have to solve $\frac{z+z^{2}+z^{3}-\overline{z+z^{2}+z^{3}}}{2i}=1$ for $|z|=1$.
For $|p|=1$ we have $\overline{p}=p^{-1}$, so the equation becomes $z^{3}+z^{2}+z-z^{-1}-z^{-2}-z^{-3}=2i$. After multiplying both sides by$z^{3}$ we get equivalent $z^{6}+z^{5}+z^{4}-2iz^{3}-z^{2}-z-1=0$.
I don't know how to solve this equation. Program [i]Mathematica 4.0[/i] found two numerical solutions with $|z|$=1: $z\approx 0.3214089073023857+0.9469405019887398 i,x\approx 1.2435793624664049$ and $z\approx 0.9853010333054504+0.1708270287958315i, x\approx 0.17166897477965282$, they are the same as numerical solutions for the beginning equation. Do you have any idea to solve it in an algebraic way?

But problem only is considered in real number set

I don't consider it for complex $x$; I adviced in my post searching complex roots $z$ of the equation $z^{6}+z^{5}+z^{4}-2iz^{3}-z^{2}-z-1=0$ and then getting $x=\arg(z)\in \mathbb{R}$. Do you know anything about possibility of expressing roots of $\sin(x)+\sin(2x)+\sin(3x)=1$ by values of elementar functions?

Another polynomial equation can be created if we use well known \[\sin(x)=\frac{2t}{1+t^{2}}\] and \[\cos(x)=\frac{1-t^{2}}{1+t^{2}},\] where $t=\tan(\frac{x}{2}$). After some ugly computations (expressing the left side by $\sin(x)$ and $\cos(x)$ and then by $t$, multiplying by all denominators and simplifying) it becomes \[t^{6}-4t^{5}+3t^{4}+16t^{3}+3t^{2}-12t+1=0.\] Any idea to solve it?

All above solutions try to derive polynomial equation with variable $x$ but all of them don't solve in real set

No, my last polynomial has $2$ real roots (found in numerical way): $t_{1}\approx 0.0860459073611678$ and $t_{2}\approx 0.7166142870354104$.

Of course, i see. But there isn't  a clearly way  to find such roots.

[quote="hien"]This Problem due to my student gives me! I haved spend more times but now i haven't got a root! I haved thought the problem is wrong? Anybody think so?[/quote]
It's unsolved for more than two months. Does you student know anything about the solution?

When I put galois(t^6-4*t^5+3*t^4+16*t^3+3*t^2-12*t+1); in Maple, it outputs

  "6T16", {"S(6)"}, "-", 720, {"(3 6)", "(1 6)", "(2 6)", "(4 6)", "(5 6)"}

I believe that the Galois group of $t^{6}-4t^{5}+3t^{4}+16t^{3}+3t^{2}-12t+1$ over $\mathbb{Q}$ is $S_{6}$, which is non-solvable. I believe that therefore the roots of the original equation can't be expressed as a composition of finitely many elementary functions, but I don't know whether Galois theory can be extended to trigonometric functions.

$x$ is geometricaly known iff $u$ is a known real iff $t$ is a known real.
As puuhikki wrote you can't express $t$ or $u$ with standard algebraic operations $+-/.$ over $\mathbb{Q}$and radical functions $x\to x^\frac{1}{p}$ where $p$ is an integer.
So you must be satisfied with the approximate numerical $u$'s values: 0.17... and 0.94....

From $\sum_{k=1}^{n}{sin(kx)}=sin(\frac{nx}{2})\cdot \frac{sin(\frac{n+1}{2}\cdot x)}{sin(\frac{x}{2})}$ $\Rightarrow$ our equation is equivalent to \[sin(\frac{3x}{2})\cdot \frac{sin(2x)}{sin(\frac{x}{2})}=1\] .Maybe this could help . Idon't know how to solve this  :blush:

From all of arguments given above, I think, we can conclude this equation is non-solvable (I means for high school, without using of numerical method).


No one can solve?   :(  :?:

[hide="to arqady"]arqady, please post your's or Vasc's solution.  [/hide]


trigonometry
geometry
analytic geometry
circumcircle
complex numbers
geometry proposed

[quote="Leon"]Find the locus of the vertex $A$ of all triangles $\triangle ABC$ such that

\[
\cos B + \cos C = 1
\][/quote]

Hmm, does this mean that you have some nice characterization of this locus? I haven't, and all what I know is that this locus is an algebraic curve of degree > 2. (How did I check it? Since my algebra program was absolutely helpless against the ma/ess of algebra to handle, I decided to use geometry, took 6 points on the locus - 4 of them were quite trivial :) - and showed that they don't match the Pascal theorem. Thus, the locus is not a conic.)

  darij

This is my solution:

WLOG we denote  B(-1,0), C(1,0) . The equations of the lines BA, CA are respectively

$BA :  y=m(x+1)$

$CA :  y=n(x-1)$

Thus we have

$ \cos B = \frac{1}{{\sqrt {1 + \tan ^2 B} }} = \frac{1}{{\sqrt {1 + m^2 } }} $

$ \cos C = \frac{1}{{\sqrt {1 + \tan ^2 C} }} = \frac{1}{{\sqrt {1 + n^2 } }} $

Therefore $\cos B + \cos C = 1$ iff and only if

$[1] \qquad \frac{1}{{\sqrt {1 + m^2 } }} + \frac{1}{{\sqrt {1 + n^2 } }} = 1$

Squaring twice, form [1] follows that

$[2] \qquad m^4 n^4  - 6m^2 n^2  - 4m^2  - 4n^2  - 3 = 0$

After the substitution $m = \frac{y}{{x + 1}},{\rm{  }}n = \frac{y}{{x - 1}}$ we obtain the equation

$[3] \qquad y^8 - 6(x^2 - 1)^2y^4 - 8(x^2 - 1)^2(x^2 + 1)y^2 - 3(x^2 - 1)^4 = 0$

which coincides with Yetti’s equation (post #4).

The graph of curve [3] it is drawn in fig.1 . The fig. 2 is the locus I have obtained with Cabri.
(see also [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=162817&highlight=#162817[/url])

At this point I would put the following questions:

a)  Why the graphs of fig. 1 and fig. 2 are different ?
b)  Are the equations [1] and [2] equivalent ?
c)  The equation [3] represents only the points of the locus or is also verified by some other points?

Leon

[quote="Leon"]Therefore $\cos B + \cos C = 1$ if and only if

$[1] \qquad \frac{1}{\sqrt {1 + m^2}} + \frac{1}{\sqrt {1 + n^2}} = 1$
[/quote]
Substituting $m = \frac{y}{x + 1}$, $n = \frac{y}{x - 1}$ into the last equation,

$(1) \qquad \frac{\sqrt{(x + 1)^2}}{\sqrt{(x + 1)^2 + y^2}} + \frac{\sqrt{(x - 1)^2}}{\sqrt {(x - 1)^2 + y^2}} = 1$

where $x, y$ are coordinates of the intersections of the lines $y = m(x + 1), y = n(x - 1)$. If $x = \pm 1$, the intersection coincides with the points $C, B$, respectively, and the triangle $\triangle ABC$ degenerates into the line segment $BC$. Otherwise, it would be $\cos C = 1, \cos B \neq 0$ or $\cos B = 1, \cos C \neq 0$ and the intersection would not be on the locus of $\cos B + \cos C = 1$. Hence, we have 3 possibilities: $x < -1, -1 < x < +1, x > +1$ and corresponding equations (1.1), (1.2), (1.3):

$(1.1) \qquad - \frac{x + 1}{\sqrt{(x + 1)^2 + y^2}} - \frac{x - 1}{\sqrt {(x - 1)^2 + y^2}} = 1$

$(1.2) \qquad + \frac{x + 1}{\sqrt{(x + 1)^2 + y^2}} - \frac{x - 1}{\sqrt {(x - 1)^2 + y^2}} = 1$

$(1.3) \qquad + \frac{x + 1}{\sqrt{(x + 1)^2 + y^2}} + \frac{x - 1}{\sqrt {(x - 1)^2 + y^2}} = 1$

When $x < -1$, $\cos B < 0$, it clearly cannot be $\cos B + \cos C = 1$, because $\cos C < 1$. Likewise, if $x > +1$, $\cos C < 0$, for a similar reason, it cannot be $\cos B + \cos C = 1$. Thus it is true that if $\cos B + \cos C = 1$, then the equation (1) holds, but if the equation (1) holds, it does not necessary mean that $\cos B + \cos C = 1$, it can also mean either $- \cos B + \cos C = 1$ or $\cos B - \cos C = 1$.

[quote="Leon"]...At this point I would put the following questions:

a)  Why the graphs of fig. 1 and fig. 2 are different ?
b)  Are the equations [1] and [2] equivalent ?
c)  The equation [3] represents only the points of the locus or is also verified by some other points?

Leon[/quote]
If you start witth any of the following 3 equations

$(2.1)\ \ -\cos B + \cos C = 1$

$(2.2)\ \ \ \cos B + \cos C = 1$

$(2.3)\ \ \ \cos B - \cos C = 1$

after squaring it twice, you get the same result:

$(2.2')\ \ \ \ \ \ \ \cos^2{B} + 2 \cos B \cos C + \cos^2{C} = 1$

$(2.2'')\ \ \ \ \ \ \ (\cos^2{B} + \cos^2{C} - 1)^2 = 4 \cos^2{B} \cos^2{C}$

$(2.1'-3')\ \ \cos^2{B} - 2 \cos B \cos C + \cos^2{C} = 1$

$(2.1'' - 3'')\ (\cos^2{B} + \cos^2{C} - 1)^2 = 4 \cos^2{B} \cos^2{C}$

Thus the closed oval corresponds to the curve given by equations (1.2) or (2.2), while the left and right open curves are given by equations (1.1), (2.1) or (1.3), (2.3), respectively. I got the same oval previously with the Sketchpad for the locus of (2.2) and the same open curves for the loci of (2.1), (2.3), respectively, after reading your reply. The equations [1] and [2] in your post are equivalent, but they are not equivalent to the specified locus condition $\cos B + \cos C = 1$, only the equation (1.2) in the present post is equivalent to this condition.

yetti

[quote="pestich"]...
    AH/cosA=BH/cosB=CH/cosC = 2R   where AH is vertexA-orthocenter segment

    So, in this problem we see that the orthocenter is on ellipse with
    foci  B and C ,  the big axis = 2R  and the center of circumcircle is on
    top of the small axis...
[/quote]

No, it is not, because the circumradius $R$ is not constant. If $\cos B = \cos C = \frac 1 2$, then $\measuredangle B = \measuredangle C = \frac \pi 3$ and the circumradius is $R = BC \frac{\sqrt 3}{3}$. On the other hand, if $\cos B \rightarrow 1$, $\cos C \rightarrow 0$, then $\measuredangle B \rightarrow 0$, $\measuredangle C \rightarrow \frac \pi 2$ and the circumradius $R \rightarrow \frac{BC}{2}$. Hence, $R$ varies between $\frac 1 2 BC < R \le \frac{\sqrt 3}{3} BC \doteq 0.57735\ BC$, where $BC$ is fixed.


calculus
integration
function
search
advanced fields
advanced fields unsolved

trying to solve a problem, the following equation appeared. Find all smooth functions $f: ]0;+\infty[$ satisfying:
1/ $f(x)=1$ for $x \geq 1$
2/ $f(x)=\int_{1-x}^{1}f(\frac{x}{y}) dy$ for $0<x<1$

This reminds me of [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1490057795&t=125066[/url]. The equation is the same (up to change of variable $y\mapsto 1-y$ in the integral). I wonder if the original problem from where it came is also the same :).

nope, I was playing with this problem : http://www.mathlinks.ro/Forum/viewtopic.php?t=152328  :lol:  anyway, thank you very much for the link, I had not seen this exercice on the forum ..


[quote="sanbittu"]
\[ {\rm{Prove: }} \& {\rm{sec}}^6 \theta - {\rm{tan}}^6 \theta = 1 + 3{\rm{tan}}^2 \theta + 3{\rm{ tan}}^4 \theta
\]
[/quote]
[hide] $ Use$ $ 1 + \rm{tan}^2\theta = \rm{sec}^2\theta$ and cube both the sides[/hide]


I'm enrolled in the Columbia Science Honors program, and I am having trouble deciding between these two courses as my first choice:
Organic chemistry
Computer Modeling and Visualization in Chemistry

I was wondering if anyone could give me some advice on whether either of these two courses would help aid me in my study better, and if so, which one would be better. Thanks a lot!

Well, it certainly depends on your major, but I think Organic Chem would generally be a better choice.  Computer modeling and Visualization... it sounds... I don't know.  Maybe less like chemistry and more like computers?  Organic Chem should be a good heavy-duty chem class, and probably a necessity if you're going to be doing any research or continued studies in chemistry.

But keep in mind, I don't really know.  You should probably talk to a counselor/advisor/whatever; don't decide your course based on what I (or most likely, others on this forum) say.


1.Let a,b,c >= 0 such that:ab+bc+ca=3.. Find min:
         \[ a^{^{3}}\plus{}b^{3}\plus{}c^{3} \plus{}7abc\]
2. Let a,b,c>0. Prove that:
 \[ 3(a^{2}\minus{}a\plus{}1)(b^{2}\minus{}b\plus{}1)(c^{2}\minus{}c\plus{}1)\geq (abc)^{2} \plus{}abc \plus{}1\]

I think that this problem belongs to olympiad level,unless there exists an easier way.

$ 2)$.
If we can prove that $ 3(a^2 \minus{} a \plus{} 1)^3 \ge a^6 \plus{} a^3 \plus{} 1$. the problem is done

Because then:

$ 3^3(a^2 \minus{} a \plus{} 1)^3(b^2 \minus{} b \plus{} 1)^3(c^2 \minus{} c \plus{} 1)^3 \ge (a^6 \plus{} a^3 \plus{} 1)(b^6 \plus{} b^3 \plus{} 1)(c^6 \plus{} c^3 \plus{} 1) \Longleftrightarrow LHS \ge \sqrt [3]{(a^6 \plus{} a^3 \plus{} 1)(b^6 \plus{} b^3 \plus{} 1)(c^6 \plus{} c^3 \plus{} 1) } \ge (abc)^2 \plus{} abc \plus{} 1$ by holder's inequality.

So we have to prove that:

$ 3(a^2 \minus{} a \plus{} 1)^3 \ge a^6 \plus{} a^3 \plus{} 1 \Longleftrightarrow 3[a^6 \minus{} a^3 \plus{} 1 \plus{} 3(a^2 \minus{} a)(a^2 \plus{} 1)( \minus{} a \plus{} 1)]
 \ge a^6 \plus{} a^3 \plus{} 1 \Longleftrightarrow 3a^6 \minus{} 3a^4 \plus{} 3 \minus{} 9a(a^2 \plus{} 1)(a \minus{} 1)^2 \ge a^6 \plus{} a^3 \plus{} 1 \Longleftrightarrow 2(a^3 \minus{} 1)^2 \minus{} 9a(a^2 \plus{} 1)(a \minus{} 1)^2 \Longleftrightarrow (a \minus{} 1)^2[2(a^2 \plus{} a \plus{} 1)^2 \minus{} 9a(a^2 \plus{} 1)] \ge 0$

But $ 2(a^2 \plus{} a \plus{} 1)^2 \minus{} 9a(a^2 \plus{} 1) \equal{} (a \minus{} 1)^2(2a^2 \minus{} a \plus{} 2)$.

So we proved the inequality. :)

P.s.
I would like to know from what team selection contest this problem is taken????

Oh, i think you maybe don't know this contest!^^. It's a contest to select national team in Viet Nam country. But it's only a test. There are about 3 test like that when we learn and finally there is a official test to select.


I think we need a moderator for this forum. THere is so much spam.

The FTW Questions and Support forum has a moderator, and it's still full of spam.

Well both the FTW sections could have more mods.
As you probably know, there are more than 1 mod for every forum.

i think that's a great idea
should someone contact the admins to request more mods?

hmm....
u posted a [i]little[/i] late :D

anyone wanna nominate someone to be mod?

i nominate myself!

GO AIME15! UBER SPAMMED!

[quote="AIME15"]NO! :spam: ing is fun[/quote]
That IS spamming.

If you keep that up your posts could get deleted and such.

I nominate myself.
PEOPLE RUNNING:
myyellowducky82
AIME15

I nominate myself
About myself:
I am a class leader and all the teachers that I pass through always give good feedback and leave it up to me to orginize and supervise projects. I orginize my intermediate algebra challenge set disscussions and has a sucessful FTW tourney going on in my MATHCOUNTS class.
I love to have fun, yet still like to keep relative orde, and am a proud member of CTC.
I may seem to be still a newer member than most and have many less posts but none of the posts are spammy.

I would be a great moderator becuase I feel that I have common sense, and know when things have gone too far. My inner sense makes sure that things, like those in the classroom are related to what is supposed to be there. Even though I may not be the FTW hotshot, I love to play FTW whenever I have time leftover.

My personal life:
I am proud to say soon I will be the big 13. I plan to start quality FTW tourneys in the future when things aren't so hectic. I play piano and cello. For cello, I have won many of the biggest local prizes and have recently made it in to the highest level orchestra, among the high school and college kids. I love music and math and believe they are a good combo. I am a track star at my school and recreationally play tennis, which I still need to work on my back hands.One more thing: My favorite beatles song is "All you need is love" which is coicidently the song I'm listening to right now.
~Dojo

My Wonderful Speech.

Hello everyone,

I seem to be an average 12-year-old girl going into the 7th grade. It seems like I spend my life on AoPS, but I don't do that. I juggle piano practicing, percussion, harp, and being a good student all at once. And that's my summer life!

For piano, I've played in numerous competitions, and won many. I've even played with an orchestra!! For percussion, I've played only a year, but I've won 12 free lessons from a scholarship!!

I think I would make a good moderator, because I try not to spam, although I do occasionally =[. I'm trying harder though, not to spam as much. I wish people wouldn't spam, since telling people NOT to spam bugs me. It feels like I'm wasting my post count on useless things. I participate actively in this forum--I'll show a screenshot of it later.

I will try not to lock threads, just to edit and delete posts so spam will be minimized. I won't abuse my moderator powers by playing favorites or yelling at people for no apparent reason. I'll try to help everyone! Don't vote for me or anyone else just because they tell you to--vote for them BECAUSE you believe they'll make a good moderator.

"If there's a Ducky, there's a Way!!"

i have already pmed levans concerning this topic and he has replied
he said that "FTW Other Stuff is a general purpose forum, no particular topics are required. If you feel there is spam, just mark the message with a rating of 1 star. "

so, as a bottom line, FTW other stuff does not need a mod
if you think there is too much spam, then just rate the posts as spam

[quote="myyellowducky82"]He's right then.
Will someone lock this topic?[/quote]

Nice.

[quote="myyellowducky82"]I was making a point...[/quote]

I know, and I thought it was nice.  :blush:

[quote="vallon22"]could we just drop this thread before it turns into people posting spam?[/quote]

Too late.

Um... we probably [b]won't[/b] get a mod.  Unless levans has enough time to look at this pile of dust and spam.  So, just [b]stop spamming[/b].  You guys are old enough to realize that (not pointing fingers, but I think everyone knows who those spammers are.)  So, just quit it.


Let be $ a,b,c\in \mathbb{R}_\plus{}$ such that $ a\plus{}b\plus{}c\equal{}18$ . Calculate the maximum value of $ a^2b^3c^4$ and find $ a,b,c$ in that case .

[hide]Applying AM-GM we have:
$ \frac {a^2}{4}. \frac {b^3}{27}. \frac {c^4}{64} \leq \ [(a \plus{} b \plus{} c)/9]^9 \equal{} 2^9 \equal{} 512$
Therefore, the maximum value is: $ 3538944$. The equality occurs when $ a \equal{} 4, b \equal{} 6$ and $ c \equal{} 8$
[b][u]Generalized result[/u][/b]:
Given $ a \plus{} b \plus{} c \equal{} k$($ k > 0$ and $ a,b,c\in\mathbb{R}_{ \plus{} }$). Calculate the maximum value of $ a^x.b^y.c^z$($ x,y,z$ are positive integers)[/hide]


Solve the system of equation: $ x^2\plus{}y^2\plus{}z^2\equal{}3xy\plus{}\frac{18}{z}\equal{}yz\plus{}\frac{8}{x}\equal{}2zx\minus{}\frac{2}{y}$

We rewrite the system as follow:
$ x(x^2\plus{}y^2\plus{}z^2)\equal{}xyz\plus{}8$
$ y(x^2\plus{}y^2\plus{}z^2)\equal{}2xyz\minus{}2$
$ z(x^2\plus{}y^2\plus{}z^2)\equal{}3xyz\plus{}18$
Let $ A\equal{}x^2\plus{}y^2\plus{}z^2$ and $ B\equal{}xyz$ we have
$ xA\equal{}B\plus{}8$
$ yA\equal{}2B\minus{}2$
$ zA\equal{}3B\plus{}18$
Thus $ A^3\equal{}(B\plus{}8)^2\plus{}(2B\minus{}2)^2\plus{}(3B\plus{}18)^2$
and   $ BA^3\equal{}(B\plus{}8)(2B\minus{}2)(3B\plus{}18)$
We could easily find A and B.

We find $ (z\minus{}2)(1160z^6\minus{}35737z^3\plus{}455625)(z^2\plus{}2z\plus{}4)\equal{}0$; sole real solution $ z\equal{}2$; we can deduce $ y\equal{}\minus{}1,x\equal{}1$.


function
number theory unsolved
number theory

Find all functions such that $ f: N\to N$
$ f(x)<f(y)<\equal{}>x<y$
$ f(n\plus{}f(n))\equal{}2f(n)$

we take $ S = \{(a,b): \ \forall n\in\mathbb{N}: \ f(n + af(n)) = bf(n)\}$
let $ (a,b)\in S\Rightarrow \forall m: \ f(m + af(m)) = bf(m)$
$ m = f(n) + n$ gives $ \forall n: \ f(n + (2a + 1)f(n)) = 2bf(m)$
then $ (2a + 1,2b)\in S$
now if we take $ (a_{n},b_n)$ such that $ (a_1,b_1) = (1,2)$ and $ (a_{n + 1},b_{n + 1}) = (2a_n + 1,2b_n)$
then $ \forall n: \ (a_n,b_n)\in S \Rightarrow \forall n : \ (2^n - 1,2^n)\in S$
[quote]we can prove easly that if for $ x > y$ we have $ f(x) - f(y) = x - y$ then $ \forall h\in\{y,y + 1,y + 2,...,x - 1,x\}: \ f(h) = h + f(y) - y$[/quote]
we have for $ i > 1: \ f(1 + (2^i - 1)f(n)) = 2^if(n)$
then $ f(X) - f(Y) = X - Y$ where $ (X,Y) = (n + (2^i - 1)f(n),n + f(n))$
then $ \forall h\in [|n + f(n),n + (2^i - 1)f(n)|]: \ f(h) = h - n + f(n)$
gives $ \forall h\ge n + f(n): \ f(h) = h - n + f(n)$
so for $ h\ge max(n,m): \ h - n + f(n) = f(h) = h - m + f(m)$ gives $ f(n) - n = f(m) - m$
so $ \boxed{f(n) = n + a}$ for some $ a\in\mathbb{N}$

I think the only possible case is f(n) = n+f(0)   . It is a result of  increasing property .
0 ---->  f(0)
1 ---->  f(1) >= f(0) +1
2 ---->   f(2) >= f(1)+1
....
....
f(0) ----> 2f(0) > = f(0)+f(0)
So by each step ,the function can increase only 1 ,then f is linear :
 f(n) = n +f(0) ,f(0) is arbitrary


Just 5 minutes ago, I signed up at [url]http://www.gmailswap.com[/url]. They sent me an invitation just right now!(only 5 minutes!) However, In that time period, my friend sent me another invite so I'm offering this invite to the community...anyone really want one?

I have 18 invitations if anyone wants them.

[quote="confuted"]Bah, do people still need invitations?

6 invitations, to go to the first six people who want one and email me at confuted@gmail.com. Make sure you mention that you're from AoPS and want gmail.[/quote]

I sent you an email, but I'm probably late...

I have a few invitations.  I think I'll give away four for now.  If you want one, please e-mail me at frankthetank271 (at) gmail (dot) com.

i'll give some away too, yifc12@gmail.com or whatever gmail accounts have. you could also email me at ycao90@yahoo.com since i check there more often. or just pm me. just a warning, gmail loses its thrill after a while.

[quote="mathfanatic"][quote="hatsup123"]could you send my sister one.....she's desperate...

icky13blue@yahoo.com[/quote]

Uh, wouldn't it make more sense to give your sister the extra one you got, i.e. this one:

[quote="hatsup123"]I'm offering this invite to the community...anyone really want one?[/quote][/quote]

...yeah... the thing is that I wasn't thinking when I wrote the post, so I can't just take back my offer to the community...that would be rude...

I can't believe it!!!!!! Someone stole my invitatioin...seriously!!!!! I haven't gave it away yet, and it said that it was already used to make the link annepinkerton@gmail.com...HUH!!!!!!

i'd like an invitation at smarteryaman@yahoo.com

I would like one- stevenliao91@yahoo.com

... how many more do u have?

Can you share it with me : ntrkien88@yahoo.com

Lol... I just realized i have 103 invites total. :P

I suppose that by now so many people must have mention this kind of sites, but there are some sites dedicated to give invitations of Gmail. For example

http://www.googlemania.com/

This page actually have A LOT of information about Google, the company, the site, etc, but it is in Spanish, if you don't speak it, well, you still can obtain a free invitation, just go to the page and there is a textfield (where you enter text) and just in the side there is a button that says 'Enviar' (Send, in Spanish) you just put the email account where you want to receive your invitation and they send one to you inmidiatelly. Between my wife and I we placed 150 in the site one of this days, and right now, according to the information they give, they have 7535 invitations to give.

Best,

I have 50 invites if anyone want one.  PM me.

I have 3 accounts so I have 150 invites that i'm willing to give out. Just PM me.

PM me if you want some, as well. Although you will already have PM'ed either bubala or DarkKnight.

[quote="aznphatso"]Same here, I've got around 150 if anyone is interested. I'm trying to get rid of them just because I'm tired of seeing them every single time I log in. Besides, my Yahoo DSL account now gives me two gigs so I'm fine with that.[/quote]

Yahoo has DSL?

[quote="aznphatso"]Yes. It's SBC Yahoo! DSL. I think it's only a Northern California thing.[/quote]
I use SBC Yahoo! DSL here in Indiana.
And i have about 300 gmail invites :D

[quote="Piggypi314"]Gmail is google mail.  Lots of memory (maybe 1/3 of what an iPod has, :P)
If you want an account, email me and I'll send you an invite.[/quote]

More than double the memory of an Ipod mini!

Edit: See my signature!

I thought I'd add another 50 invites to the 400 or so invites already mentioned on this page. Just ask if you want one.

i have  50 invites (not a joke) so if you need one just send me a email at sharmablue@gmail.com (just made it) dont be shy i have plenty

Okay that's plenty of invites. I'll reopen the thread when you guys all run out.


Is it possible to find $2008$ infinite arithmetical progressions such that there exist finitely many positive integers not in any of these progressions, no two progressions intersect and each progression contains a prime number bigger than $2008$?

The answer is NO, it is not possible.

Let denote the progressions with $P_i,\, i=1,2,\cdots,2008 , P_i = \left\{a_i + kd_i \right\}_{k=0}^{\infty}$.

$P_i \cap P_j = \emptyset \, <=> \, \, \not\exists n,m \in \mathbb{N}, \, a_i - a_j = n d_j - m d_i \, <=>\, \gcd(d_i, d_j)$ don't divide $a_i - a_j\,\, \Rightarrow $

(1) $ \gcd(d_i, d_j) > 1 , \, \forall i,j ,\, i \neq j $.

Let consider $ \{kd_1\}_{k=1}^{\infty}$. From the problem's clause, it follows, that from some point on, every such number belongs to some progresion.

Let $ k d_1 \in P_j$ for some $k,\,j \,\, \Rightarrow $

(2) $ k d_1 = a_j + n d_j$.

Let $ d=\gcd(d_1, d_j)$ From (1) it follows $ d>1 $. Using (2) we get $ d / a_j $. But $ d/a_j $ and $ d / d_j $  contradicts the fact that $P_j$ contains a prime number.

[quote="dgrozev"]
Let $ d=\gcd(d_1, d_j)$ From (1) it follows $ d>1 $. Using (2) we get $ d / a_j $. But $ d/a_j $ and $ d / d_j $  contradicts the fact that $P_j$ contains a prime number.[/quote]
Your idea is very correct, except a small problem, $ d/a_j $ and $ d / d_j $ but $P_j$ can still have a prime number, for instance $a_j$ is a prime number and $d$ is, of course, a prime too.
Here's my solution I hope that it's correct (my idea is almost the same as yours, except a small lemma)
Suppose the contrary, which means we can find such 2008 sequences, let their differences are $d_1,d_2,...,d_{2008}$ and their first terms are $a_1,...,a_{2008}$, respectively
Lemma: There exists at least one number of $d_1,d_2,...,d_{2008}$, which is less than or equal to 2008
Proof: Assume the contrary, or WLOG, $2008<d_1\leq d_2\leq ... \leq d_{2008}$
From the condition, there must exist a constant natural number M such that for all $n\geq M$ then $n$ belongs to at least one of 2008 sequences given above (*). Choose a number k large enough, such that $a_1+kd_1\geq M$ then we consider the interval $(a_1+kd_1,a_1+(k+1)d_1)$ there are exactly $d_1-1$ natural numbers in that interval, and  by the minimum of $d_1$ once can easily check that there exists at most one number from each sequence (except from sequence $(a_1,d_1)$ of course) between the interval (or there must exists $d_j<d_i$ which is a contradiction), Now, we conclude that the number of natural numbers (which are not in any sequence) in $(a_1+kd_1,a_1+(k+1)d_1)$ is at least $d_1-1-2007=d_1-2008>0$ which is a contradition (see (*)), hence the lemma is proven

Now everything is almost the same as your solution, WLOG, suppose that $d_1\leq 2008$ we consider $a_1+kd_1$ if $(a_1,d_1)=d>1$ then we from the condition of existence of prime number we conclude that $d \in P$ and $a=p$ thus $p>2008$ which implies $d_1>2008$ (since $p|d_1$) a contradiction
So $(a_1,d_1)=1$ we consider a number $L>M$ (M has been presented in the lemma) and $d_1|L$ then by the definition of $M$ we conclude that $L$ must be in one sequence, named $(a_i,d_i)$
Then $d_1|a_i+d_i.k$ for some $k$, if $(d_i,d_1)=1$ then the equation $a_i-a_1=ud_i-vd_1$ has solution $u,v >0 $ and $u,v$ are natural number, (a well-known problem) then $(a_i) \cap (a_1)$ is not an empty set, which is again a contradiction!
Thus $(d_i,d_1)=d>1$ then $d|a_i$, similarly, we conclude $d \in P$ and $a_i=p$ then $p>2008$ or $d_1>2008$ (since $p$ is a divisor of $d_1$) which makes the assumption is wrong then the answer is NO :)

Here is another solution.

We claim that the answer is $\boxed{no}$. Assume, for contradiction, that it was possible. Consider the common differences $d_1, d_2, \cdots, d_{2008}$ of the arithmetic progressions. It's easy to see (density) that
$$\frac{1}{d_1} + \frac{1}{d_2} + \cdots + \frac{1}{d_{2008}} = 1. \qquad (1)$$
From $(1)$, we know that at least one of the $d_i$'s is even, as otherwise multiplying by $d_1d_2 \cdots d_{2008}$ would give a contradiction. Notice that any of the arithmetic progressions with even common difference must take on only odd values, since else this progression would contain no odd prime. Therefore, the subset of the evens which are in some arithmetic progression has density
$$\sum_{i=1}^{2008} f(i),$$
where $f(i) = \frac{1}{d_i}$ if $d_i$ is odd and $0$ else. Since there is some $d_i$ which is even, this quantity is clearly less than $1$, implying that there are infinitely many evens in no arithmetic progression. This contradicts the assumption of the problem.

$\square$


what grades are you guys in. i know theres still some high school kids whose last MC year was last year and theres some who are actually competing. which one are you?

i added a 5th grade in the poll because there were some 5th graders at my school doing it last year

I thought 5th graders couldn't do mathcounts...
7th here.

still. my math teacher wants to give them an opportunity to experience it. one year we actually had 2 4th graders do it. theyre freshman now

I'm in ninth grade and will hopefully be able to make it back to nationals next year nonetheless; I'll probably see diophantient again because he got held back in 7th grade and thus has another year.

I'm in 7th grade. This year im going for my second year on my math team from school

man, u guys make me feel so bad. im 16 and still can't get to states.  :(

[quote="abacadaea"][quote="unimpossible"]yeah well. i started kindergarten when i was 9 so i'm a 17 year old 8th grader :rotfl:[/quote]

no wonder.......

soooo many 7th graders. no 8ths...except unimpossible. an i doubt i will make nats CDR next year what with so many smarties. and this is just AoPS. theres probably a lot more people in 7th better than me[/quote]

i was kidding...

oh. you got me going there for a bit :P.

[quote="Cyrazeno"]8th grade.. mihail is 16? whoa[/quote]

ditto here...

im in 7th

got 9th in chapter last year

i was an alternate :(  :(  :(

ok i got like 3rd. were getting off topic

8th grade..made it to nationals last year


[quote]Dr. Philanthropist, or Dr. Phil for short, unwisely built his rectangular acid tank on top of his best carpet. His acid tank is 5 feet wide, 12 feet long, and an unknown amount of feet high.

a) Dr. Phil fills the tank two-thirds of the way, then drops a gigantic ant in the tank. The gigantic ant is known to have a volume of 120 cubic feet. If the acid in the tank is at the very top of the tank after Dr. Phil drops the gigantic ant in there, how many feet tall is the acid tank?

b) Dr. Phil designs a machine to drop acid at the rate of 5 cubic feet per minute. In order to fill the acid tank to the two-thirds point (the standard height for acidifying creatures), how long should he leave the machine on?

c) After raiding Willy Wonka's factory, Dr. Phil fills the acid tank up to four feet and drops Oompa Loompas in the tank until the acid level in the tank reaches the very top. How many cubic feet is the volume of an Oompa Loompa if Dr. Phil drops 8 Oompa Loompas in the tank?

d) Dr. Phil tries dropping a humongous bar of chocolate in the acid tank. The result is chocolate acid, which is very mild (about as mild as orange juice) and extremely tasty. He makes five tankfulls of chocolate acid and decides to make a swimming pool of chocolate acid. The swimming pool is going to be 50 feet long and 6 feet wide. How deep should the swimming pool be if it is to use up all of the chocolate acid?[/quote]

is Dr.Phil the actual guy from Chocolate Factory?

[hide="a"]6feet tall[/hide]
[hide="b"]240/5=48 48minutes[/hide]
[hide="c"]15 cubic feet[/hide]
[hide="d"]6feet deep[/hide]

by the way, I loved Dr.Phil's action toward Oompa

[quote="Treething"]Dr. Philanthropist, or Dr. Phil for short, unwisely built his rectangular acid tank on top of his best carpet. His acid tank is 5 feet wide, 12 feet long, and an unknown amount of feet high.

a) Dr. Phil fills the tank two-thirds of the way, then drops a gigantic ant in the tank. The gigantic ant is known to have a volume of 120 cubic feet. If the acid in the tank is at the very top of the tank after Dr. Phil drops the gigantic ant in there, how many feet tall is the acid tank?

b) Dr. Phil designs a machine to drop acid at the rate of 5 cubic feet per minute. In order to fill the acid tank to the two-thirds point (the standard height for acidifying creatures), how long should he leave the machine on?

c) After raiding Willy Wonka's factory, Dr. Phil fills the acid tank up to four feet and drops Oompa Loompas in the tank until the acid level in the tank reaches the very top. How many cubic feet is the volume of an Oompa Loompa if Dr. Phil drops 8 Oompa Loompas in the tank?

d) Dr. Phil tries dropping a humongous bar of chocolate in the acid tank. The result is chocolate acid, which is very mild (about as mild as orange juice) and extremely tasty. He makes five tankfulls of chocolate acid and decides to make a swimming pool of chocolate acid. The swimming pool is going to be 50 feet long and 6 feet wide. How deep should the swimming pool be if it is to use up all of the chocolate acid?[/quote]

[hide]
a) Because the giant ant with a volume of 120 cubic feet filled up the tank, and before it was two-thirds full, we know that one-third of the tank's volume is 120 cubic feet. Therefore, it's volume is 360 cubic feet. So far, we know that the tank's dimensions are 5 feet by 12 feet. 5*12*h=360. h=6. The height of the tank is 6 feet.

b) I'm confused...where is he dropping the acid from?

c) We know that the volume of the tank is 360 cubic feet, and that he has already filled the tank up to 4 feet in height. Therefore, the tank is filled up to 240 cubic feet, because 5*12*4=240. Subtract 240 from the full volume, 360, and you get 120. If 8 Oompas were droped, then each must have been 15 cubic feet in volume.

d) Five tankfuls of chocolate acid would give us 360*5 cubic feet in total, or 1800 cubic feet. 50*6*h=1800. h=6. The height of the swimming pool must be 6 feet.
[/hide]

Fun problems!

[quote="SoccerBrainy40"][hide]
b) I'm confused...where is he dropping the acid from?
[/hide]

Fun problems![/quote]

He is dropping acid into the tank using the new machine.
So find out how long it takes to fill the tank at a rate of 5 cubic feet per minute.

Quote:[hide]b) I'm confused...where is he dropping the acid from? [/hide]

He's hanging from the ceiling and  regurgitating acid  .

Okay, I understand now.

[hide]
b) The full volume of the tank is 360 cubic feet, so 2/3 of it is 240 cubic feet. If the new tank is dropping at 5 cubic feet per minute, then it will need to drop for 240/5 minutes, or 48 minutes.
[/hide]

 :)

To continue my post...

4
[hide]
A tank is 360 cbft, and he makes 5 tank fulls (6*6*10*5)
He has a swimming tank of 50*6*x cbft (5*10*6*x)
If you look at those factors I put in parentheses above, you can easily see that the answer (the hieght) is 6
>>6<<[/hide]

And there you go!


If $ a, b, c$ distinte positive real numbers, prove that

$ \sum[\frac{a(2b \plus{} 2c \plus{}a)}{(b \minus{} c)^2}]\ge 5$.

THe stronger also trues and...very easy  :wink:
$ \sum[\frac{a(2b \plus{} 2c \plus{}a)}{(b \minus{} c)^2}]\ge 6$

[quote="nguoivn"]THe stronger also trues and...very easy  :wink:
$ \sum[\frac {a(2b \plus{} 2c \plus{} a)}{(b \minus{} c)^2}]\ge 6$[/quote]

I am also interested to see your complete solution, [b]Nguoivn[/b] :oops:

Dear my friend, I used 2 old result:
$ \sum\ \frac{a(b\plus{}c)}{b^2\plus{}c^2} \geq\ 2$
and: $ \sum\ \frac{a^2}{(b\minus{}c)^2} \geq\ 2$
 :)

[quote="nguoivn"]Dear my friend, I used 2 old result:
$ \sum\ \frac {a(b \plus{} c)}{b^2 \plus{} c^2} \geq\ 2$
and: $ \sum\ \frac {a^2}{(b \minus{} c)^2} \geq\ 2$
 :)[/quote]
Can you give link or proof?

$ \sum[\frac {a(2b \plus{} 2c \plus{} a)}{(b \minus{} c)^2}]\geq\ 2 \sum\ \frac {a(b \plus{} c)}{b^2 \plus{} c^2} \plus{} \sum\ (\frac {a}{b \minus{} c})^2$
Assume $ c \equal{}$ min $ {a, b, c}$ we have:
$ \sum\ \frac {a(b \plus{} c)}{b^2 \plus{} c^2} \geq\ \frac {a}{b} \plus{} \frac {b}{a} \geq\ 2$
We only need to prove: $ \sum\ (\frac {a}{b \minus{} c})^2 \geq\ 2$
<=> $ (\sum\ \frac {a}{b \minus{} c})^2 \geq\ 0$
 :)


Given $ k(O)$ and arbitrary line $ l$. Point $ M: OM\bot l$ and $ M\in l$. $ A, B\in l$ so that $ AM \equal{} BM$. Lines through $ A$ and $ B$ intersect $ k$ in $ C,D$ and $ E,F$ respectively. If $ G \equal{} DE\cap l$ and $ H \equal{} CF\cap l$ proof that $ MH \equal{} MG$. Note that if $ A\equiv B\equiv M$ we have the Butterfly Theorem.


Let [tex]k_1<k_2<...<k_n[/tex] be n positive integers and

let [tex]10^{k_1}+10^{k_2}+...+10^{k_n}[/tex] be a perfect square.

Prove that n=1.

Boy, this one is strange. First, I thought it shouldn't be very hard, but after some hours without the slightest smart idea, I gave up. I'm waiting for someone who solved the problem to post a solution. This is a really very very very nice problem. In fact, what am I speaking about? It's from Leningrad olympiad, it must be smart and nice.

Hum , the i consider the special case : find all n  such that 10^n -1 is a perfect square . I don't see why it's simple and why only n=1 is satisfied the requirement ???

 Could U post the sol. for me to see?

The initial problem seems to be really difficult, but this particular case is pretty easy: if $n\geq 2$ then $10^n-1$ is $4k-1$, so it can't possibly be a perfect square.

Are you sure this was the problem? I don't know what to say, it is simply impossible to solve. My god, what a problem!

Stupid me :D ! I just bring some though of the orginal problem to the special case and i don't see it works!

Unluckily i lost the LMOs papers and i am not sure it's really a LMO problem or not?
where are u Fedor?

Yes, it is from Leningrad selection test for All-Soviet-Union 1962; but I do not know anybody, who knows the solution. We prepare a new edition of D. V. Fomin's book, and Fedor just tried to find a solution for this problem using the Forum. Without much hopes.


If $a,b,c>0$, prove that

\[\frac{27}{2}\left(\frac{ab+bc+ca}{(a+b+c)^3}\right) \leq \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \leq \frac{3}{2}\left(\frac{a+b+c}{ab+bc+ca}\right).\]

[hide]

$\frac{27}{2}\left(\frac{ab+bc+ca}{(a+b+c)^3}\right) = \frac{27}{2}\left(\frac{ab+bc+ca}{(a+b+c)^2(a+b+c)}\right)$

Since we know $ab+bc+ca \le \frac{1}{3}(a+b+c)^2$ by AM-GM/Rearrangement, we have

$\frac{27}{2}\left(\frac{ab+bc+ca}{(a+b+c)^2(a+b+c)}\right) \le \frac{9}{2(a+b+c)}$

But by Cauchy we have $9 \le ((a+b)+(b+c)+(c+a))\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)$ which is equivalent to $\frac{9}{2(a+b+c)} \le \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$ as desired.

I'll get the right inequality in a sec (hopefully).[/hide]

For the right side:
[hide]
$\frac{ab+bc+ca}{a+b}=\frac{ab}{a+b} +c\le \frac{a+b}{4}+c$
[/hide]

COuld you explain how that shows the right side?

I didn't get it, though I should've...

If you multiply $ab+bc+ca$ to both sides, you get what fuzzylogic said, except he only wrote out one term. If you take the sum of all three terms, $\left(\frac{a+b}{4}+c\right)+\left(\frac{b+c}{4}+a\right)+\left(\frac{c+a}{4}+b\right) = \frac{3}{2}(a+b+c)$ as desired. And his one real step is merely an application of AM-GM.

Heh, that's pretty clever, fuzzylogic.  My solution (for the RHS) came from

(1)  $\sum_{\text{cyc}} (a+b)(a+c) = (a+b+c)^2+ab+bc+ca \leq \frac{4}{3}(a+b+c)^2,$

(2)  $(a+b)(b+c)(c+a) \geq \frac{8}{9}(a+b+c)(ab+bc+ca),$

both easily verified.

I did both of these inequalities in my head at church.  The right one is cute, I disliked the left one because it's weaker than a direct application of AM-HM.

Right:
Multiplying through by $\sum ab$, it's enough to show that

$\sum\frac{\sum ab}{a+b}=\sum a+\sum\frac{ab}{a+b}\leq \frac{3}{2}(a+b+c)$

From here it's enough to show that

$\sum \frac {ab}{a+b}\leq \sum \frac {a+b}{4}$

But this is equivalent to

$(a+b)^2\geq 4ab$

which is obvious.

Left (my original solution)
Multiplying through by $(a+b+c)^2$, it's enough to show that

$\sum \frac {(a+b+c)^2}{a+b}\geq \frac {27\sum ab}{2(a+b+c)}$

Applying Cauchy Schwarz (engel form) we have

$\sum \frac {(a+b+c)^2}{a+b}\geq \frac {9(a+b+c)^2}{2(a+b+c)}$

But

$(a+b+c)^2\geq 3\sum ab$

done.

the inequality $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq\frac{27(ab+bc+ca)}{2(a+b+c)^3}$ is weaker than the well
  known $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq\frac{9}{2(a+b+c)}$.The second one is very interresting. ;)


[quote="murgi"]let ABC be a triangle and D is the midpoint of BC.Suppose angDAB=angBCAand angDAC=15 degree.Show that angADC is obtuse.Further if O is the circumcenter of triangle ADC.Prove that triangle AOD is equilateral.
This problem was previously posted by rushil but there was a typo.So i reposted it here.[/quote]
Since $ \angle BAD\equal{}\angle ACB$ we have AB is a tangent of (O). Then $ AB^2\equal{}BD.BC\equal{}2BD^2$
So $ AB\equal{}\sqrt{2}BD$
On the other side, $ \frac{AB}{sin \angle ADB}\equal{}\frac{BD}{sin \angle BAD}$ we obtain $ \frac{AB}{BD}\equal{}\frac{sin \angle C}{sin(\angle C\plus{}15^o)}\equal{}\sqrt{2}$.
Solve C we have $ \angle C\equal{}30^o$ then $ \angle AOD\equal{}60^o$
Therefore $ \Delta AOD$ is equilateral

Can any one provide a purely geometric solution to this problem....? :D

[quote="murgi"]wheres the answer to the first part[/quote]
PART 1:

Let $ \angle ADB$ be obtuse and $ \angle ADC$ be acute.
Let $ D'$ be the point where extended $ AD$ cuts the circumcircle.
Clearly, $ arc BD' < arc CD' \\
\implies \angle BAD < \angle CAD \equal{} 15 deg$.
But $ \angle BDA \equal{} \angle A < (15 \plus{} 15) deg$. contradiction.

Well, I strived hard for the problem for a purely geometric solution to this problem.  I request the geniouses on AoPS to give a purely geometric solution.

Since $\triangle BAD\sim\triangle BCA$, we have $BA^2=BD\cdot BC\implies BA=\sqrt2 BD\implies AC=\sqrt2AD$. Now let $Q$ be outside the triangle such that $\triangle QAC$ is right isosceles at $Q$. Then $AQ=QC=AD$, and $\angle DAQ=60$. So $\triangle QAD$ is equilateral so $Q$ is the circumcenter of $\triangle ADC$. The rest should be easy angle chasing.


I don't see how this made its way onto the trig/complex numbers round. I mean, so I had:

$r^{2}+r+1=0 \implies r+\frac{1}{r}=-1 \implies r^{2}+2+\frac{1}{r^{2}}= 1$ $\implies r^{2}+\frac{1}{r^{2}}=-1 \implies r^{2^{n}}+\frac{1}{r^{2^{n}}}=-1$ $\implies \sum = (-1)+(-1)+...+(-1) =-11$

But I'm wondering if there's a nicer solution to this using trigonometry and/or complex numbers?

Thanks,
~tZf

The equation $r^{2}+r+1$ gives the two third roots of unity other than one. Whenever you put one of them to a power that is not a multiple of 3 (i.e. a power of 2), you'll still get one of the roots of the equation, so every term is going to be the same, -1.

Oh wow, ok... thanks
(too many donuts yesterday)

During the contest, I just realized that every term was going to be $\text{cis}(120)+\text{cis}(-120)$. Then it was easy.

I believe it was:

[i]Let $r$ be one of the roots of the equation $r^{2}+r+1=0$. Compute the value:[/i]

$\left(r+\frac{1}{r}\right)+\left(r^{2}+\frac{1}{r^{2}}\right)^{3}+\left(r^{4}+\frac{1}{r^{4}}\right)^{5}+\left(r^{8}+\frac{1}{r^{8}}\right)^{7}+\ldots+\left(r^{1024}+\frac{1}{r^{1024}}\right)^{21}$

jeez, that looks icky. glad I took rounds 2,3,5 and not that one!


Find the smallest positive integer which when divided by 10 leaves a remained of 9, when divided by 9 leaves a remainder of 8, by 8 leaves a remainder of 7, etc., down to where, when divided by 2, it leaves a remainder of 1.

[hide="Why brainomegas method works"]
Let the smallest positive integer with the property be $ n$.
Then let us examine the integer $ n\plus{}1$. It is divisible by all the positive integers from $ 1$ to $ 10$, so the smallest $ n\plus{}1$ can be is $ \text{lcm}({1,2,3,4,5,6,7,8,9,10}) \equal{} 2520 \implies n \equal{} 2520\minus{}1 \equal{} \boxed{2519}$[/hide]


algorithm
number theory
Diophantine equation
Euclidean algorithm

[quote="stupidlogic"]Find the inverse of 399 in mod 442.  I have no idea how to find an inverse in modular arithmetic.  I know that the inverse is the number that if multiplied by 399 mod 442 = 1, but I don't know how to find it.[/quote]Let the inverse$ \equal{}x$
Then, $ 399x \equiv 1 \mod 422$. So, Just solve the diophantine equation $ 399x\minus{}422y\equal{}1$
[you can easily solve using reverse of Euclidean Algorithm]

[hide="Algorithmic Solution"]399a=442b+1

> Let a=b+c

399b+399c=442b+1
399c=43b+1

> Let b=9c+d

399c=387c+43d+1
12c=43d+1

> Let c=4d+e

48d+12e=43d+1
5d=1-12e

Now, note that e=-2, d=5 works, so set those values (otherwise, we could just run out the algorithm one or two more times)

Then:
c=4*5-2=18
b=9*18+5=167
a=167+18=185

Thus, the inverse of 399 modulo 442 is 185[/hide]


modular arithmetic
number theory unsolved
number theory

[quote="massnet"]A hundred fractions are written, whose numerators are numbers from $ 1$ to $ 100$, each used once, and whose denominators are numbers from $ 1$ to $ 100$, each used once. Suppose that the sum of these fractions is an irreducible fraction with denominator $ 2$. Prove that it is possible to exchange the numerators of two fractions so that the sum of the $ 100$ fractions becomes and irreducible fraction with an odd denominator.[/quote]

Let $ \frac {a_i}{b_i}$ the $ 100$ fractions and $ S_1 \equal{} \sum_{i \equal{} 1}^{100}\frac {a_i}{b_i} \equal{} \frac {a}{2}$ with $ a$ odd.

Let $ A \equal{} \{a_i$ such that $ b_i \equal{} 2\pmod{4}\}$. $ |A| \equal{} 25$
Let $ B \equal{} \{a_i$ such that $ b_i$ is odd$ \}$. $ |B| \equal{} 50$

Obviously $ \exists a_j\in A, a_k\in B$ such that $ a_j \minus{} a_k$ is odd (else all elements of $ A$ and all elements of $ B$ would be distincts but with same parity, which is impossible since we would have $ 75$ distincts elements of $ \{1,2,\ldots,100\}$ with the same parity).

Swap then $ a_j$ and $ a_k$. The new sum will be $ S_2 \equal{} S_1 \minus{} \frac {a_j}{b_j} \minus{} \frac {a_k}{b_k}$ $ \plus{} \frac {a_k}{b_j} \plus{} \frac {a_j}{b_k}$

So $ S_2 \equal{} \frac {a}{2} \minus{} \frac {a_j}{b_j} \minus{} \frac {a_k}{b_k}$ $ \plus{} \frac {a_k}{b_j} \plus{} \frac {a_j}{b_k}$

So $ S_2 \equal{} \frac {ab_jb_k \plus{} 2(a_k \minus{} a_j)(b_k \minus{} b_j)}{2b_jb_k}$

$ b_j \equal{} 2\pmod{4}$ and $ b_k$ odd imply denominator is divisible by 4 and not by 8.
$ a_k \minus{} a_j$ is odd and $ b_k \minus{} b_j$ is odd too. So numerator is divisible by 4.

So $ S2$ is an irreducible fraction with an odd denominator.

Q.E.D.


Suppose $ x,y,z\in \mathbb {R}$ such that $ x^2\plus{}y^2\plus{}z^2\equal{}1$ , find the maximum and minimum value of $ xy\plus{}yz\plus{}zx$.

 :jump:

[quote="vinskman"]Suppose $ x,y,z\in \mathbb {R}$ such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$ , find the maximum and minimum value of $ xy \plus{} yz \plus{} zx$.

 :jump:[/quote]
[hide]Hint:
The answer is: $ \minus{} 1 and 1$. The equality occurs when $ x \equal{} y \equal{} z \equal{} sqrt\frac {1}{3}$ and $ x \equal{} y \equal{} z \equal{} sqrt \frac { \minus{} 1}{3}$, respectively[/hide]

$ (x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^2\ge 0$ implies that $ xy\plus{}xz\plus{}yz \le 1$ with equality iff $ x\equal{}y\equal{}z$, from which we get the triples $ (\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}), (\minus{}\frac{1}{\sqrt{3}},\minus{}\frac{1}{\sqrt{3}},\minus{}\frac{1}{\sqrt{3}})$

$ (x\plus{}y\plus{}z)^2 \ge 0$ is equivelant to $ xy\plus{}xz\plus{}yz \ge \minus{}\frac{1}{2}$ with equality iff $ x\plus{}y\plus{}z\equal{}0$.


I am using firefox, and I cam across this message when trying to post in AoPSWiki, and now I can't post.

 Database error
From AoPSWiki
Jump to: navigation , search
A database query syntax error has occurred. This may indicate a bug in the software. The last attempted database query was:

    (SQL query hidden)

from within function "SearchMySQL4::update". MySQL returned error "126: Incorrect key file for table './dbaps/aw_searchindex.MYI'; try to repair it (localhost)".

I have posted in AoPSWiki before on firefox.

I also have been getting the same error, although it doesn't actually prevent me from making changes.  Firefox 1.0.7 on windows xp.

Can you explain exactly what you're doing to produce this error?  I can't recreate it.

I just played around some more and didn't have any trouble, but it was happening earlier every single time I saved an edit.

I cannot recreate it now, either.  Somehow, my editation did get posted, but I don't know how.  It isn't happening now, but it was everytime I tried to save a page.

I had the same trouble a few days ago, but my edit still got posted.  :huh:


modular arithmetic
number theory
group theory

I made this up before I joined the forums and just remembered it the other day. If you know some basic number theory this should be easy so don't ruin it for the rest!

Label $n$ equally spaced points on a circle as $A_0,\ldots,A_{n-1},$ with $n\ge2$. If $A_i=A_j$ by definition when $i\equiv{j}\pmod{n}$ and line segments $\overline{A_0A_m},\overline{A_mA_{2m}},\ldots$ are constructed indefinitely for some positive integer $m$, then under what condition(s) will each point $A$ be reached by the line segments?

Note: What makes this problem interesting to me is not the proposition itself or its proof but testing it out on paper (which is where the idea originally came from). You will end up with some interesting star-like figures that you may not have known how to draw before..[hide="hint"]especially if $n$ is prime![/hide]

Alright people.. time for some replies..

Replies? haha I have a difficult time understanding the problem statement.

[hide="further explanation.. may contain a hint"]Alright, how about an example then.. you might want some paper. Draw a circle and five equally spaced points around it (the vertices of a regular pentagon), and label them $A_0,A_1,A_2,A_3,A_4$ (in order of course). Then choose a positive integer $m,$ such as $2$. Now starting at $A_0$, draw a line segment to $A_2$, then to $A_4$, then to $A_6=A_1$, then to $A_8=A_3$, and finally to $A_{10}=A_0$. You have a common five pointed star that touches each of the five points $A$.

Let's say you let $n=8$ and $m=2$. In that case you'll draw segments from $A_0$ to $A_2$ to $A_4$ to $A_6$ to $A_8=A_0$, and then if you continue the process you will be drawing the same segments again and again. In this case you missed half of the points (and you end up with a square).

Under what conditions on $m$ and $n$ will all points be reached?[/hide]

If you know anything about group theory if can be easily proved that the $\gcd{(n,m)}=1$

Group theory??? I don't see any group theory in here!

[hide]If m|n and m unequal to 1, we see that 0 is repeated over and over again so it is false. Otherwise, it is clear that each number is repeated (example: 5,3: 0,3,1,4,2,0). I don't really feel like proving it now, but it is a commonly accepted fact that it is true, and is pretty easy to prove.[/hide]

Well like most maths problems, they may have more than one solution! :lol:  :lol:  :lol:


geometry perimeter circumcircle inradius trigonometry ratio incenter

[quote="NoSoupForYou"]Originally posted [url=http://artofproblemsolving.com/Forum/weblog_entry.php?e=1988]here[/url], but nobody could solve it.

Let there be $\triangle ABC$, with $\angle A=\alpha$ and $BC=s$. Prove that the perimeter of $\triangle ABC$ is at a maximum if $\triangle ABC$ is isosceles.[/quote]
I think I have cracked it,
Draw a circumcircle of this triangle and let the radius of the circle be $r$ let us take all the types of triangles and compare the perimetre.
$1$st of all $\boxed{\text{equilateral}}$,
using $a^2+b^2+2abcos\theta=c^2$
in the small triangles
we can find the sides in terms of $r$ and $\theta$
we will get them as $2r^2+2r^2cos120$
$cos120=-0.5$

$2$nd case $\boxed{\text{scalene}}$,
we can find the sides, we will get them as,
$2r^2+2r^2cos(120+x)$, $2r^2+2r^2cos(120-y)$ and $s$

$3$rd case $\boxed{\text{isosceles}}$,
we can find the sides using the above method,
 $\theta$ will be less than $60$ so the sides will b more and so the perimetre will be more.

I think my solution is wrong, :?  :(  maybe I need to do a bit more. But am I on the right track?

If you ignore $\angle{A}$, then the maximum takes when $AB=AC$.

[quote="riddler"]But am I on the right track?[/quote]

Well, you are right in the fact that you circumscribed the triangle. But the rest of your solution is incorrect. Here, let me draw a diagram:

EDIT: Changed the diagram, as the first was confusing. This diagram pretty much gives away the anwer, but nobody was going to solve it anyway.

i really dont understand how you could draw on the same base of a triangle and the same circumcircle all three types of triangles.

I'm not sure what you mean.

Wow! This problem has been moved to Intermediate Topics??!! It's a seventh-grade problem!

Well, to be fair, it's a Russian problem. In Russia, 7th grade is like 9th grade in the U.S.  :D

[quote="riddler"]what?????!!!!!  :wallbash_red:  :wallbash:  urrrrrrggghhhh I hate math![/quote]
You know what I hate? The Moon. Will somebody turn off the Moon? I am trying to sleep  :diablo:.

Let $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$ be the internal angles and $R, r, s, \triangle$ the circumradius, inradius, semiperimeter and area of the triangle $\triangle ABC$, respectively. Let h be the altitude from the vertex A to the side a = BC. The circumradius $R = \frac{a}{2 \sin \alpha}$ is obviously constant. The area and semiperimeter are equal to

$\triangle = sr = \frac{ha}{2},\ \ s = \frac{ha}{2r}$

Since a is given, the semiperimeter s is proportional to the ratio $\frac h r$ and we have to find out, when is this ratio maximum. Let M be the midpoint of the side a = BC, O the triangle circumcenter (i.e., vertex of a triangle $\triangle OBC$ such that $\angle COB = 2 \angle A = 2 \alpha$, and E, F the intersections of the perpendicular bisector OM of the side BC with the circumcircle (O), such that $\angle CEB = \alpha,\ \angle BFB = 180^\circ - \alpha$. Let A be a point on the arc CEB and denote $\phi = \angle EFA$. The locus of the incenters I of such triangles $\triangle ABC$ is the appropriate arc BC of a circle (F) centered at the midpoint F of the arc CFB and passing through the points B, C. This is because

$\angle CIB = 180^\circ - \frac{\beta + \gamma}{2} = 180^\circ - \frac{180^\circ - \alpha}{2} = \frac{360^\circ - (180^\circ - \alpha)}{2} = \frac{360^\circ - \angle CFB}{2}$

Let a' be a tangent to the circumcircle (O) at the point F, obviously parallel to the line a = BC. Let D, D' the feet of normals dropped from the vertex A to the parallel lines a, a' and let P, P' be the feet of normals dropped from the incenter I to the parallel lines a, a'. Then h = AD is the altitude from the vertex A, r = IP is the inradius and DD' = PP' MF, because the lines $a \parallel a'$ are parallel. From the isosceles triangle $\triangle CFB$ divided into 2 right angle triangles $\triangle CFM, \triangle BFM$ with the angles $\angle MBF = \angle CMF = \frac{180^\circ - \angle CFB}{2} = \frac \alpha 2$, we have

$MF = \frac{a \tan \frac \alpha 2}{2}$

$\rho = CF = BF = \frac{a}{2 \cos \frac \alpha 2}$

where $\rho$ is the radius of the locus circle (F).

From the right angle triangles $\triangle AFD', \triangle AFE$ we have

$AD' = AF \cos \phi = AE \cos^2 \phi = 2R \cos^2 \phi = \frac{a \cos^2 \phi}{\sin \alpha}$

The angle bisector AI passes through the midpoint F of the arc CFB. If we assume that this angle bisector cuts the circumcircle (O) at a point F', then CF' = BF', because $\angle CAF' = \angle BAF' = \frac \alpha 2$, i.e., the points $F' \equiv F$ are identical. From the right angle triangles $\triangle IFP'$ we have

$IP' = IF \cos \phi = \rho \cos \phi = \frac{a \cos \phi}{2 \cos \frac \alpha 2}$

Consequently,

$\frac h r = \frac{AD}{IP} = \frac{AD' - DD'}{IP' - PP'} = \frac{AD' - MF}{IP' - MF} =$

$= \left(\frac{a \cos^2 \phi}{\sin \alpha} - \frac{a \tan \frac \alpha 2}{2}\right) : \left(\frac{a \cos \phi}{2 \cos \frac \alpha 2} - \frac{a \tan \frac \alpha 2}{2}\right) =$

$= \left(\frac{2 \cos^2 \phi}{2 \sin \frac \alpha 2 \cos \frac \alpha 2 \tan \frac \alpha 2} - 1\right) : \left(\frac{2 \cos \phi}{2 \cos \frac \alpha 2 \tan \frac \alpha 2} - 1\right) =$

$= \left(\frac{\cos^2 \phi}{\sin^2 \frac \alpha 2} - 1\right) : \left(\frac{\cos \phi}{\sin \frac \alpha 2} - 1\right) = \frac{\cos \phi}{\sin \frac \alpha 2} + 1$

Since the angle $\alpha$ is given, $\sin \frac \alpha 2$ is constant and the ratio $\frac h r$ is maximum iff $\cos \phi$ is maximum iff $\phi = 0$ iff the triangle $\triangle ABC$ is isosceles.

I :love: trigonometry.

An intuitive proof could be the following..

Let's draw the circumcirle of the triangle with base = s and vertical angle = A

This circumcircle would be unique for all such triangles..

Now Area = 1/2*s*H

If we take a look at the circumcirlce, we immediately notice that this height H will be maximum when the triangle is isos.

In that case the alt. would pass through the center of the circle...

For a rigorous proof,

the problem could be reduced to finding the condition for the maximum of the following function

  H = f(C) = s/(cot(C) - cot(C + A))                 (since A is constant)

There's a simple and elegant solution that does not involve any complicated trigonometry. Behold!

[img]http://artofproblemsolving.com/Forum/files/trianglesolution_331.jpg[/img]


integration
calculus
calculus computations

[quote="JRav"]The antiderivative of $ e^{-r^{2}}$ is not an elementary function.  Determine that $ \int_{1}^{\infty}e^{-r^{2}}dr<\frac{1}{4}$.[/quote]
For $ x\geq 1$ we have:
$ e^{-x^{2}}\leq xe^{-x^{2}}$
That means,
$ \int_{1}^{\infty}e^{-x^{2}}dx \leq \int_{1}^{\infty}xe^{-x^{2}}dx$
Let $ t=-x^{2}$ to get,
$ \int_{1}^{\infty}e^{-x^{2}}dx \leq-\frac{1}{2}\int_{-1}^{-\infty}e^{t}dt$
But the value of that integral is,
$ \frac{1}{2e}< \frac{1}{4}$.

Another method.

Using the substitution $ x^{2}=y$ we get that

$ \int_{1}^{\infty}e^{-x^{2}}dx=\frac{1}{2}\int_{1}^{\infty}\frac{e^{-y}}{\sqrt{y}}dy$.

Now, $ e^{-y}\leq \frac{1}{y^{5/2}}$ for $ y\geq 1$, and hence

$ <\frac{1}{2}\int_{1}^{\infty}\frac{dy}{y^{3}}=\frac{1}{4}$.


probability
analytic geometry
geometry
inequalities

Matt and Bob both arrive at school at some time between 7:00 AM and 7:45 AM.  They both spend exactly five minutes at their locker.  Find the probability the Matt and Bob will meet?

Well, by your redaction, it seems like they share looker, then, I suppose that

[hide="Idea"]
I suppose that they don't come in the same bus or so, then the events are uncorrelated, hence, consider the measures of the intervals!!
[/hide]

I'm not very sure at all but I got
[hide]$\frac 19$.[/hide]
It's probably wrong though...

this is in the aoPSv2.. .well similar

[hide="hint"]

try to coordinate this on a Cartesian coordinate... that will make things way easier...

[/hide]

[quote="tarquin"]I'm not very sure at all but I got
[hide]$\frac {1}{9}$.[/hide]
It's probably wrong though...[/quote]

that is wrong...

I know how to do it.  I just posed it for anyone that felt like answering.

I think the answer is

[hide]16/81[/hide]

xxreddevilzz's hint should lead you on the right track.  Think about what condition must be true in order for Matt and Bob to meet.[/hide]

The answer is actually [hide](81-64)/81 = 17/81 but that's ok[/hide]

Here's a pic.  Let the bottom corner be (0,0), top corner (45,45). and the other vertices of the blue region are (0,5), (5,0), (40,45) and (45,40).  The blue region is the area of success.  If you take area of blue over area of green you get the desired probability.  This comes out to 17/81, as probability1.01 said.

[img]http://img.photobucket.com/albums/v213/joml88/sqprob.jpg[/img]

Thats what I did sorry for the typo with the 16 instead of 17.

BTW, if anyone is wondering how joml got that graph.  If we call the time Matt arrives a and we call the time Bob arrives b and both those times occur between a 45 minute interavl we can say 0<a<45, and 0<b<45.  Also in order for Matt and Bob to meet the condition abs(a-b)<5.  So graphing these inequalities and following joml's solution above we arrive at our answer.

yes... $\frac{17}{81}$ is what reddevil has been intending for... it is in AoPSv2 as part of probability section.


How many posts do you need to get 1/2, 1, 2, ... , 5 stars???

The most asked question ever...
Use search function:
http://www.mathlinks.ro/Forum/viewtopic.php?t=73553
http://www.mathlinks.ro/Forum/viewtopic.php?t=48468

[quote="ZetaX"]The most asked question ever...
[/quote]

Jejejej, sorry! I was in a hurry... Thanks ZetaX!.


[asy]pair A=(-2,0), O=origin, C=(2,0);
path X=Arc(O,2,0,180), Y=Arc((-1,0),1,180,0), Z=Arc((1,0),1,180,0), N=X..Y..Z..cycle;
filldraw(N, black, black);
draw(reflect(A,C)*N);
draw(A--C, dashed);

label("A",A,W);
label("C",C,E);
label("O",O,SE);
dot((-1,0));
dot(O);
dot((1,0));
label("1",(-1,0),NE);
label("1",(1,0),NW);[/asy]

The large circle has diameter $ \overline{AC}$.  The two small circles have their centers on $ \overline{AC}$ and just touch at $ O$, the center of the large circle.  If each small circle has radius $ 1$, what is the value of the ratio of the area of the shaded region to the area of one of the small circles?

\[ \textbf{(A)}\ \text{between }\frac{1}{2} \text{ and }1 \qquad
\textbf{(B)}\ 1 \qquad
\textbf{(C)}\ \text{between 1 and }\frac{3}{2} \qquad
\textbf{(D)}\ \text{between }\frac{3}{2} \text{ and }2 \\
\textbf{(E)}\ \text{cannot be determined from the information given}
\]

[hide="Click for solution"]
The radius of the large circle is $ 2$ since it is a diameter of a small circle.  The area of the large circle is $ \pi(2)^2\equal{}4\pi$ and the area of each small circle is $ \pi$.  The shaded area is (by symmetry) half the difference of the areas of the large circle and the two small circles, i.e. $ \frac{1}{2}(4\pi\minus{}2\pi)\equal{}\pi$.  Thus, the desired ratio is $ 1$, or $ \boxed{\textbf{(B)}}$.
[/hide]

B. The area of each small circle is pi, and the area of the big circle is 4pi, and we have 2 small circles. Hence, shaded region* 2= 2pi. Shaded region=pi. Since the area of each smaller circle is pi, the ratio is 1.

To the previous poster, where did you find the answer? Or did you get it? Just wondering...you had a link, you could have written it... :D

it isnt a link its a hide tag and the bold typing is latex...
as like
this
$ 1, 2, 3, 4, 5, 6, ...$
u do a hide tag by doing "[hide=click]"

The small circle has radius $1$, thus its area is $\pi$.
The large circle has radius $2$, thus its area is $4\pi$.
The area of the semicircle above $AC$ is then $2\pi$.
The part that is not shaded are two small semicircles. Together, these form one small circle, hence their total area is $\pi$. This means that the area of the shaded part is $2\pi-\pi=\pi$. This is equal to the area of a small circle, hence the correct answer is $\boxed{\text{(B)}\ 1}$.

Sorry for backseat modding, but please stop reviving 3 year old threads by posting solutions that have already been stated, just to increase post count. This isn't your first time doing this. Thanks!


search
number theory solved
number theory

I think should be search because it has post . This is the Pytago's equation.

Please spellit correctly ;)
The solutions are called "Pythagorean triples".

[quote="dede"]Please solve x^2+y^2=z^2 , x,y,z \in Z[/quote]

[hide="Assuming x<y<z"]If x is odd, then $ y \equal{} \frac {x^2 \minus{} 1}{2}$ and $ z \equal{} \frac {x^2 \plus{} 1}{2}$ is a solution. Since there are an infinite number of odd numbers, there are an infinite number of solutions.

And each solution is a "factor" of another. Like

3^2+4^2=5^2

57^2+76^2=95^2

or

5^2+12^2=13^2

105^2+252^2=273^2

And there are weird ones, like

8^2+15^2=17^2

256^2+240^2=272^2[/hide]

[[Pythagorean triple]]

The general solution for all primitive triples is x=mm-nn, y=2mn, and z=mm+nn where m and n are integers of unlike parities.

In order for the triples to be primitive, we also require that $ \gcd(m,n) \equal{} 1$.

Also see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176802 .


$z^{6}+z^{3}+1 = 0$ has a root $r e^{i\theta}$ with $90^\circ < \theta < 180^\circ$. Find $\theta$.

ch1n353ch3s54a1l, please use the RESOURCES section or use SEARCH. It is OK to revive that old thread if you have any questions about the solution(s) presented there.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=392620#p392620[/url]