Tell me how u did it.

1. Calculate (10!-8!)/ (10!+8!)  as common fraction

2. How many pairs of vertical angles are formed by 5 distint lines that have a common point of intersection

4. Calculate  2^25 x 5^28. express in scientific notation.

7. How many different security codes are possible using the digits 1-5 if the 2nd digit cannot be th same as the 1st digit. and the 3rd cannout be same as the 2nd.

8. Mohan is selling cookies at an economic fair. As he decides how to package them, he finds that in groups of 4, 3 left over. groups of 5, 2 left over. and groups of 7, 4 left over. What is hte least number of cookies that she could have.

[hide="1"]
Factor:

$ \frac{8!(10*9\minus{}1)}{8!(10*9\plus{}1)}\equal{}\boxed{\frac{89}{91}}$
[/hide]

too tired to think about # 2 right now.

[hide="4"]
Notice that $ 2^{25}*5^{25}\equal{}10^{25}$.  You have $ 5^3$ left over.  So the answer would be $ \boxed{1.25*10^{27}}$
[/hide]

[hide="7"]
I'll assume that the security codes consist of 3 digits.  The first digit has 5 choices.  The second has 4 (It cannot be the same as the first).  The third also has 4 (It cannot be the same as the second).  $ 5*4*4\equal{}\boxed{80}$
[/hide]

[hide="8"]
If he packages them in groups of 5, two are left over, and if he packages them in groups of 7, 4 are left over.  Notice that 5-2=3 and 7-4=3.  The answer will be 3 less than a multiple of 35.  Also, it is 1 less than a multiple of 4.  Using some quick guess and check, $ \boxed{67}$ works.
[/hide]

for 1, I have no clue,

2. just draw it out, count, you get 91

4. 2*5 is 10, so 2^25*5^25 is 10^25, then you have 5^3 left

5^3*10^25, so 125*10^25 or

1.25*10*27

7. you get 5 choices for first, 4 for second, 4 for third.

5*4*4 is 80.

8. Just quickly find a number that works for two of the numbers, and then add the LCM of those two groups to it.. so we find out seven works when in groups of 4 and 5. The LCM of 4 and 5 is 20. So add 20 to that number.

7+20 or 27, doesn't work.
27+20 or 47 doesn't work
47+20 or 67, works.

So it is 67.

ehh, for # 2 on a state sprint, i don't recommend drawing and counting it out.

[hide="2"]
You need to choose two lines from the five, and those will create 2 pairs of vertical angles each.

$ \binom{5}{2}*2\equal{}\boxed{20}$
[/hide]

[quote="Ihatepie"]

2. just draw it out, count, you get 91
[/quote]

umm, pick 2 lines, they form 2 vertical angles. there are 5C2=10 ways to pick lines. 10*2=20 does it matter if they are concurrent?

:blush:  I didn't see the "vertical angles" part. I just saw it say angles.

2) Infinity and abacadea are right, but another way to approach is:

Draw a line, then draw 4 more lines, with a common intersection point. Now erase all parts of the other four lines on one side of the main line. Looking at it, there are 5 singles angles, 4 double angles, etc. And there is a corresponding angle for each angle found on one side of the main line. Of course, this pattern should take much less time to realize. Also notice that here we miss 5 other pairs of 180-180 angles. So 15+5 = 20.

If you're talking about just angles, then there are 50 angles equal or less than 180 degrees. There are 10 possible starting points for the angles, no matter their size, so 10 * 5 =50. Hmm... I feel like I missed something.....

i think our way is easier. :D, unless you saw that right away(and i mean right away).

Yea. I rarely use combinations unless for strictly probability/counting problems. I barely saw the "every two" lines make 4 vertical angles. I like adding 5 + 5+ 4 +.... more.  :D Though, I guess it's riskier since I almost forgot to add five.


algebra
polynomial
induction
algebra proposed

Try proving the following statement:Let f(x)=a*x^n+...+l a polynomial where a,..,l
are integers and a & l are odd but f(1) is even,such that 2^k does not divide f(1). (k>1)
Then f(x) has at most k-1 rational roots.
(k=2 is easy .use induction or sth for extention)
Any good idea will be appreciated.

[quote="The Captain"]Try proving the following statement:Let \[
f(x)=a_nx^n+...+a_0\] a polynomial where $a_n,..,a_0$ are integers and $a_n,a_0$ are odd but $f(1)$ is even,such that $2^k$ does not divide $f(1)$. ($k>1$)Then $f(x) $has at most $k-1$ rational roots.($k=2$ is easy .use induction or sth for extention)
Any good idea will be appreciated.[/quote]
i think this is better.now lets think about it. :lol:

Dear jensen.Don't you think that you have wasted your time ,sending exactly what I had posted?

He just [img]http://www.mathlinks.ro/LaTeX/Images/LaTeXicon.gif[/img] it, making your problem easier to read.


function
algebra
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inequalities
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abstract algebra

This is an old classic, and well known. I used it as a lecture example today. The fun lies in trying to figure out how you would prove it without looking it up.

For all "sufficiently nice" functions $ f$ with domain $ (0,\infty)$ define $ Tf(x)$ for $ x\in(0,\infty)$ by
\[ Tf(x) \equal{} \int_0^{\infty}\frac {f(y)}{x \plus{} y}\,dy
\]
1. Prove that $ T$ cannot be extended to a bounded linear operator from $ L^1$ to $ L^1$ (This is $ L^1(0,\infty).$)

2. Prove that $ T$ cannot be extended to a bounded linear operator from $ L^{\infty}$ to $ L^{\infty}.$

3. (The main result.) Prove that $ T$ [b]can[/b] be extended to a bounded linear operator from $ L^p$ to $ L^p$ for any $ 1 < p < \infty.$

That is, prove that there is a constant $ C_p$ depending on $ p$ such that $ \|Tf\|_p\le C_p\|f\|_p.$

4. Define an analogous operator for sequence spaces and prove that it is bounded from $ \ell^p$ to $ \ell^p$ for $ 1<p<\infty.$

I found a long proof in a book that uses complex analysis  :wink:
May be it's time for the author to present his approach. I'm curious.

I always do 3 using the Schur test when I'm teaching it. It makes the proof a two-liner. Another approach is to say that "one half" of the operator is dominated by the maximal function and use duality for the "other half". The fanciest elementary way I know (for boundedness in $ L^2$) is to take two positive $ L^2$-functions $ f$ and $ g$ with norm $ 1$ and split the integral $ \iint_{(0,\plus{}\infty)^2}\frac{f(x)g(y)}{x\plus{}y}\,dx\,dy$ into the one over the set where $ \sqrt xf(x)\ge \sqrt yg(y)$ and the one over its complement. This is, of course, a variation of the idea underlying the Schur test but it looks really misterious if presented without any comments.

So, what is your favorite way, Kent?

I'll just look at #3, but that's the main argument. Remember that we are looking at $ L^p(0,\infty).$

Suppose $ f\in L^p.$ We can calculate $ \|Tf\|_p$ as follows:

$ \|Tf\|_p\equal{}\sup_{\|g\|_{p'}\equal{}1}\int_0^{\infty}Tf(x)g(x)\,dx\equal{} \sup_{\|g\|_{p'}\equal{}1}\int_0^{\infty}\int_0^{\infty}\frac{f(y)g(x)}{x\plus{}y}\,dy\,dx$

In that double integral, make the substitution $ x\equal{}(1\minus{}u)v,y\equal{}uv.$ Computing the Jacobian of this gives us $ dx\,dy\equal{}v\,du\,dv.$ Note that $ x\plus{}y\equal{}v.$

$ \int_0^{\infty}\int_0^{\infty}\frac{f(y)g(x)}{x\plus{}y}\,dy\,dx\equal{} \int_0^1\int_0^{\infty}f(uv)g((1\minus{}u)v)\,dv\,du$

$ \le\int_0^1\left[\int_0^{\infty}|f(uv)|^p\,dv\right]^{\frac1p} \left[\int_0^{\infty}|g((1\minus{}u)v)|^{p'}\right]^{\frac1{p'}}\,du$

$ \equal{}\int_0^1\left(u^{\minus{}1/p}\|f\|_p\right)\left((1\minus{}u)^{\minus{}1/p'}\|g\|_{p'}\right)\,du$

$ \equal{} \left(\int_0^1u^{\minus{}\frac1p}(1\minus{}u)^{1\minus{}\frac1p}\,du\right)\|f\|_p\|g\|_{p'}$

Taking the sup over $ g$ gives us our result with

$ C_p\equal{}\int_0^1u^{\minus{}\frac1p}(1\minus{}u)^{1\minus{}\frac1p}\,du\equal{}\Gamma\left(1\minus{}\frac1p\right) \Gamma\left(\frac1p\right).$

This constant $ C_p$ has the properties that $ C_p\equal{}C_{p'}$ and $ C_p$ tends to $ \infty$ as $ p\to1$ or $ p\to\infty,$ at a rate comparable to $ \frac1{p\minus{}1}$ or $ p,$ respectively. The smallest $ C_p$ is $ C_2\equal{}\pi.$

[quote="fedja"]... using the Schur test ...[/quote]
I was thinking about Schur test, but the proof I know only for $ L_{2} \to L_{2}$ http://en.wikipedia.org/wiki/Schur_test
by the same way ona can  show also for $ L_{p}$  this is what I have done :
Let $ (T,\mu)$ and $ (S,\nu)$ be measure spaces .
define $ U_{x}(x) \equal{} \int_{S}k(x,s)x(s)d\nu(s)$.
Then $ U$ maps $ L^{p}(T)$ to $ L^{p}(S)$ with boundary norm (one can determine exact boundary)
if there exist $ \infty > A,B > 0$ and  measurable functions $ u \in T$ and $ w \in S$ where $ 0 < u,w < \infty$ such that :
\[ \int_{T}|k(t,s)|u^{q}(t)dt \leq A w^{q}(s)
\]
and
\[ \int_{S}|k(t,s)|w^{p}(s)ds \leq B u^{p}(t)
\]
where $ p^{ \minus{} 1} \plus{} q^{ \minus{} 1} \equal{} 1$ and $ dt \equal{} d\mu(t)$ and $ ds \equal{} d\nu(s)$\\
proof :
consider some function $ f \in L^{p}(T)$
 then we have :
\[ |U_{f}(s)|^{p} \leq \left(\int_{T}|k(t,s)| |f(t)|dt\right)^{p} \equal{} \left(\int_{T}\left\{|k(t,s)|u^{q}(t) \right\}^{(p \minus{} 1)/p} \left\{|k(t,s)|^{1/p}|f(t)|u(t)^{ \minus{} 1}\right\}dt \right)^{p} \leq
\]
by holder's inequality we have :
\[ \leq \left(\int_{T}|k(t,s)|u^{q}(t) dt\right)^{p \minus{} 1}\left(\int_{T}|k(t,s)|f(t)|^{p}u(t)^{ \minus{} p}dt\right) \leq
\]
b y using above condition we have :
\[ \leq A^{p \minus{} 1}w^{p}(s)\int_{T}|k(t,s)|f(t)|^{p}u(t)^{ \minus{} p}dt
\]
finally we have :
\[ \parallel{}U_{f}\parallel{}^{p}_{L^{p}} \equal{} \int_{S}|U_{f}(s)|^{p}ds\leq A^{p \minus{} 1}\int_{S}w^{p}(s)\int_{T}|k(t,s)|f(t)|^{p}u(t)^{ \minus{} p}dtds
\]
and by Fubini's theorem  and above conditions we obtain :
\[ \parallel{}U_{f}\parallel{}^{p}_{L^{p}} \leq A^{p \minus{} 1}\int_{T}|f(t)|^{p}u(t)^{ \minus{} p}\int_{S}|k(t,s)|w^{p}(s)dsdt \leq A^{p \minus{} 1}B\int_{T}|f(t)|^{p}dt \equal{} A^{p \minus{} 1}B \parallel{}f\parallel{}^{p}_{L^{p}} \Rightarrow
\]

\[ \parallel{}U\parallel{} \equal{} \sup \frac {\parallel{}U_{f}\parallel{}_{L^{p}} }{\parallel{}f\parallel{}_{L^{p}}} \leq A^{(p \minus{} 1)/p}B^{1/p}
\]
Done. One can show that limitation   of $ U$ and existence $ u,w$ is equivalence conditions.
thus in Kent's case by choosing $ u \equal{} x^{\alpha}$ and $ w \equal{} y^{\alpha}$ where $ \alpha < 0$ then we get $ A \equal{} \int_{0}^{\infty}\frac {t^{\alpha q}}{1 \plus{} t}dt$ and $ B \equal{} \int_{0}^{\infty}\frac {t^{\alpha p}}{1 \plus{} t}dt$

[quote="Extremal"]
Done. One can show that limitation   of $ U$ and existence $ u,w$ is equivalence conditions.
[/quote]
Why? How?

Take any positive $ F_0\in L^p$ and assume that an operator $ U$ with positive kernel has norm less than $ 1$ in $ L^p$ (which can be achieved by simple scaling). Then the adjoint operator $ U^*$ has norm less than $ 1$ in $ L^q$. Now put inductively $ F_{k\plus{}1}\equal{}\max\{F_k,SF_k\}$ where $ SF\equal{}[U^*(UF)^{p/q}]^{q/p}$. Note that all $ F_k\in L^p$ and $ F_k$ form an increasing sequence. Now assume that $ F_k\minus{}F_{k\minus{}1}\equal{}f_k$. Then $ F_{k\plus{}1}\minus{}F_k\le \max\{0,SF_k\minus{}F_k\}\le SF_k\minus{}SF_{k\minus{}1}$. Without loss of generality, we may assume that $ p\ge q$ (our claim is symmetric in $ p,q$). But then $ S$ is subadditive, i.e., $ S(F\plus{}G)\le SF\plus{}SG$ pointwise (this is just the Minkowski inequality for the $ L^{p/q}$ norm) and we conclude that $ f_{k\plus{}1}\le Sf_k$. But $ S$ is also contracting in $ L^p$, so we finally conclude that $ F_k$ converge to some $ F\in L^p$. For this $ F$, we have $ SF\le F$, which is pretty much the same as what we want (just put $ u^q\equal{}F$, $ w^q\equal{}UF$).

Ah, yes, very nice! But as you say, the kernel must be positive, otherwise there are counterexamples (the projection of $ L^p(T)$ onto $ H^p$ for example).


In an acute-angled triangle $ ABC$, $ D,E,F$ are the feet of the altitudes from $ A,B,C$, respectively, and $ P,Q,R$ are the feet of the perpendiculars from $ A,B,C$ onto $ EF,FD,DE$, respectively. Prove that the lines $ AP,BQ,CR$ are concurrent.

$ AP,BQ,CR$ concur in the circumcenter of the triangle ABC


function
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real analysis unsolved

Why are valid real formula for $ \sin x$, $ \cos x$, $ e^x$, $ \sinh x$, $ \cosh x$ automatically also in complexes number? How is the situation that is not everywhere defined?($ \text{arctan}\,x$, $ \log x(!)$ etc.)

Fitim

One answer to this question is [url=http://en.wikipedia.org/wiki/Analytic_continuation]analytic continuation[/url].  The first set of functions you list have well-behaved power series with infinite radius of convergence and can be extended analytically to all of $ \mathbb{C}$; the other two don't share this property and so run into trouble.  The logarithm, in particular, is a classic example of a function that behaves poorly on the entire complex plane.

Thank you! :)

How is the 'second' solution?  :roll:


Hi,

So m*dv/dt = mg - kv^2, where m is skydiver mass, g acceleration due to gravity, and k> 0 related to amount of air resistance. So I want to find the analytical solution for v(t), with the assumption that v(0) = 0.

I went all the way to integrating it with a partial fraction, but then got an answer drastically different from what the book got.

Thanks...

Please take a look, #7 afterwards of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=57578]www.mathlinks.ro/Forum/viewtopic.php?t=57578[/url]


Source: ARML 1990 Individual #3

If $a+b=c$, $b+c=d$, $c+d=a$, and $b$ is a positive integer, compute the greatest possible value of $a+b+c+d$.

Is there a clever way to do this problem?

$a + b = c$
$b + (a + b) = d$
$(a + b) + b + (a + b) = a$
$3b = -a$

$a + b + c + d = a + b + (a + b) + b + (a + b) = 3a + 4b = 4b - 9b = -5b$

The greatest possible value of this expression is therefore $\boxed{-5}$, unless I misread the question.

Substitute $c=a+b$ into $d=b+c$:

$d=b+a+b=a+2b$

Now substitute $c=a+b$ and $d=a+2b$ into $c+d=a$:

$a+b+a+2b=a\iff a+3b=0\iff a=-3b$.

Now substitute that back into the expressions for $c$ and $d$:

$c=a+b=-3b+b=-2b$

$d=a+2b=-3b+2b=-b$

Hence $a+b+c+d=-3b+b-2b-b=-5b$

If $b$ is a positive integer, then $b\geqslant 1$, therefore $-5b\leqslant -5$, so the desired maximum value is $-5$, which is obtained for $b=1$, i.e. $a=-3, b=1, c=-2, d=-1$.

keep substituting in new values until you come up with the answer.
that is, $\boxed{-5}$

Adding we get:
$a+2b+2c+d=a+c+d$
$c=-2b$

Substituying:$a=-3b$    $d=-b$

Then: $a+b+c+d=-5b$

But $b$ is a positive integer so $-5b$ takes its greater value for $b=1$: $(a+b+c+d)_{max}=-5$


Consider the figure consisting of a square, its diagonals, and the segments joining the midpoints of opposite sides. The total number of triangles of any size in the figure is

[asy]
size(100); defaultpen(linewidth(0.7)); draw(unitsquare^^(0,0)--(1,1)^^(0,1)--(1,0)^^(.5,0)--(.5,1)^^(0,.5)--(1,.5));[/asy]

$\textbf{(A)}\ 10 \qquad
\textbf{(B)}\ 12 \qquad
\textbf{(C)}\ 14 \qquad
\textbf{(D)}\ 16 \qquad
\textbf{(E)}\ 18$

[hide]
Its 16 because when you draw all the lines, there are four smaller squares that each have two triangles. That means 8 triangles. Then you have the triangles that are one side of the square and with two small triangles. Thats four more. Then you have the ones that are half of the square. There is four more. In total $8+4+4=16$. The answer is D.[/hide]


algebra
polynomial
vector
calculus
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Suppose $ f(x,y) \equal{} e^{(x\minus{}1)^2 \plus{} (y\minus{}3)^2}$.

a. Find the first-order Taylor polynomial about (0,0).
$ f(x,y)\equal{} e^{10} \minus{} 2e^{10}x \minus{} 6e^{10}y$.
b. Find the second-order Taylor polynomial about (1,3).
$ f(x,y)\equal{} 1$
somehow, I got the other partial derivatives = 0... I guess 1 is it.
c. Find a 2-vector perpendicular to the level curve through $ (0,0)$.
?
d. Find a 3-vector perpendicular to the surface $ z\equal{}f(x,y)$ at the point $ (0,0)$.
?

Notice that $ (1,3)$ is the very nicest point you could possibly be taking Taylor polynomials about. We can use that $ e^u\equal{}1\plus{}u\plus{}O(u^2)$ to get that $ e^{(x\minus{}1)^2\plus{}(y\minus{}3)^2}$ had the second order Taylor polynomial $ 1\plus{}(x\minus{}1)^2\plus{}(y\minus{}3)^3.$

[quote="Kent Merryfield"]$ e^u \equal{} 1 \plus{} u \plus{} O(u^2)$ ... $ 1 \plus{} (x \minus{} 1)^2 \plus{} (y \minus{} 3)^3.$[/quote]

I'm sorry, I don't get how you got the first equation (in the quote) and how it leads to the second (the answer).

How did I get the first equation? I know the power series for the exponential, as should everyone. It leads to the second just by substituting $ (x\minus{}1)^2\plus{}(y\minus{}3)^3$ for $ u.$

After the fact, I know that the second derivatives of $ f$ at the point $ (1,3)$ are

$ \begin{pmatrix}\frac{\partial^2f}{\partial x^2}&\frac{\partial^2f}{\partial x\partial y}\\ \frac{\partial^2f}{\partial x\partial y}&\frac{\partial^2f}{\partial y^2}\end{pmatrix}\equal{} \begin{pmatrix}2&0\\0&2\end{pmatrix}.$ That's read off from the Taylor polynomial.

I also know that $ f$ has a minimum at $ (1,3).$

Here's the fourth degree Taylor polynomial for $ f$ about $ (1,3)$:

$ 1\plus{}(x\minus{}1)^2\plus{}(y\minus{}3)^2\plus{}\frac12(x\minus{}1)^4\plus{}(x\minus{}1)^2(y\minus{}3)^2\plus{}\frac12(y\minus{}3)^2.$

From that I can read of that all of the third order partial derivatives are zero at $ (1,3)$ and:

$ \frac{\partial^4f}{\partial x^4}(1,3)\equal{}12$

$ \frac{\partial^4f}{\partial x^2\partial y^2}(1,3)\equal{}4$

$ \frac{\partial^4f}{\partial y^4}(1,3)\equal{}12$

with the other fourth derivatives being zero.

Wow, thanks for the extra explanations...
Oh, and what's the $ O$ in $ e^u \equal{} 1 \plus{} u \plus{} O(u^2)$ for?

Look at the sticky item at the top of this forum.


Study the convergence of the sequence dfined by $U_0 \geq 0$ and \[U_{n+1} = \sqrt {U_n} + \frac 1 {n+1} .\]

Pierre.

Well,the sequence converges,and it comes to $ 1 $.
We consider two case:
If we can find $ n $satisfying [tex] u_n \geq u_{n+1} [/tex] then we have [tex],u_m \geq u_{m+1} ,\forall m \geq n [/tex]
,since [tex] \sqrt{u_{m-1}}+\frac{1}{m}>\sqrt{u_{m}} +\frac{1}{m+1}[/tex]
And one more $ u_n>0 $ then the sequence converges.
If [tex] u_{n}<u_{n+1} ,\forall n [/tex] then [tex] u_n<\sqrt{u_n}+\frac{1}{n+1}<\sqrt{u_n}+\frac12 \Rightarrow u_n<2, \forall n \Rightarrow [/tex] what we need.
Hence the sequence converges and it comes to $ 0 $ or $ 1 $.But if [tex] u_n < u_{n+1},\forall n [/tex] then [tex] u_{n} \to 1 [/tex] since [tex] u_2>0 [/tex]
If [tex] u_n \geq u_{n+1} [/tex] then [tex] u_n \geq \sqrt{u_n} +\frac{1}{n+1} [/tex] then [tex] u_n \geq 1,\forall n \Rightarrow u_{n} \to 1 [/tex],too. :)


What is the value of $ \frac12\cdot\frac23\cdot\frac34\cdot\frac45\cdot\frac56\cdot\frac67\cdot\frac78\cdot\frac89\cdot\frac9{10}$? Express your answer as a common fraction.

hello, after simplifying we get $ 1/10$.
Sonnhard.

$ \frac{1}2\cdot\frac{2}3\cdot\frac{3}4\cdot\frac{4}5\cdot\frac{5}6\cdot\frac{6}7\cdot\frac{7}8\cdot\frac{8}9\cdot\frac{9}{10} $

in the sequence we can just simplify each one and get just a remaining of 1/10

The numerator starting from the 2nd fraction on offsets with the denominator of the previous fraction. This ends at the last fraction, so we have 1/10 as all the other numbers have cancelled out.

Everything cancels but the numerator of the first fraction and the denominator of the second fraction making the answer $\boxed{\frac{1}{10}}$.


limit
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real analysis unsolved

Just kidding!  I don't know about you guys, but people who say that can get to be really annoying.  A technique used to show this is by using 1/3:

1/3 = .333...

1/3 * 3 = 1
.333... * 3 = .999...

So .999... = 1

And a similar, but a bit more complex, argument can be made with 10.

However, some people still don't buy it.  Not sure why, but they just don't want to accept it.  I was wondering if there was a way to prove it using something with real analysis.

Between any two distinct real values, there exists a real value distinct from the other two.  So it seems like it should be easy to do a proof by contradiction, assuming that there exists a real value between 0.999.... and 1.  But I hit a wall here.  Something to do with an epsilon neighborhood maybe?

[quote="Ricky"]Just kidding!  I don't know about you guys, but people who say that can get to be really annoying.  A technique used to show this is by using 1/3:

1/3 = .333...

1/3 * 3 = 1
.333... * 3 = .999...

So .999... = 1

And a similar, but a bit more complex, argument can be made with 10.

However, some people still don't buy it.  Not sure why, but they just don't want to accept it.  I was wondering if there was a way to prove it using something with real analysis.

Between any two distinct real values, there exists a real value distinct from the other two.  So it seems like it should be easy to do a proof by contradiction, [b]assuming that there exists a real value between 0.999.... and 1[/b].  But I hit a wall here.  Something to do with an epsilon neighborhood maybe?[/quote]The boldfaced phrase already answered your question. You assume a statement and then prove the same statement? I smell [i]circular reasoning[/i]. Another fact is that .999....=1 is a notation for [i]limit point[/i]. Some mathematicians argue that there are infinitesimals between reals like above. It becomes a philosophical question rather than mathematical. That's what I understood of this argument. Hopefully this helps you.

It's a question of definitions. How do we define 0.9999...? We use a limit. $0.9999....=\lim_{n\to\infty}\sum_{k=1}^n 9\cdot 10^{-k}=\lim_{n\to\infty}1-10^{-n}$. The problem is equivalent to showing that $\lim_{n\to\infty}10^{-n}=0$.
$10^{-n}$ is a decreasing sequence which is bounded below, so it has a limit, which is its greatest lower bound. Zero is a lower bound; we need to show that $c$ is not a lower bound for any positive $c$. By the Archimedean property, $\frac1c<m<10^n$ for some integers $m,n$, and $c>10^{-n}$.

This assumed a definition of the real numbers based on something like the least upper bound property. If we define the real numbers using base $b$ expansions, this issue comes up directly in the definitions, and must be dealt with there; the nonuniqueness of decimal representations is a big reason for basing the real numbers on sets or sequences of rationals in the formal definition.

In extended systems that contain "infinitesimals", the Archimedean property is false, and the argument breaks down there.

[quote]The boldfaced phrase already answered your question. You assume a statement and then prove the same statement? I smell circular reasoning.[/quote]

Not at all.  I am assuming the negation of the statement I wish to prove.  That is, I am assuming that 0.999... does not equal 1.  Then there must exist an real number between the two.  I believe assuming this, you should be able to reach a contradiction, it which case, it must be the case that 0.999... = 1.

Circular reasoning can affect indirectly an argument. Check out this example of circularity around the definition:

Proposition: 1 is a natural number.

Proof: Suppose 1 is not a natural number. But by Peano Axioms, $1\in\mathbb{N}$. Contradiction. QED

jmerry is right about how we use definitions. When infinitesimals are introduced, you can seemingly work out with a theory of infinitesimals such that 0.999.... does not equal 1 because you already assumed $0.999...<1$.

In your case, it depends on the definition. You will not find a contradiction if you already begin assuming the negative of the whole statement. Contradiction works if you have something like $p\to q$ and you assume $p,\neg q$ and arrive at a contradiction.

[quote]Contradiction works if you have something like $p\to q$ and you assume $p,\neg q$ and arrive at a contradiction.[/quote]

There are multiple ways you can do a proof by contradiction.  That is one of them.  But another, is that you can do a proof by contradiction if there is a finite number of possibilities.  Probably the most popular proof which uses this method is proving the the square root of 2 is irrational.  That is because there are two possibilities, either it is rational or irrational.  You show it is impossible to be rational, and thus, it must be irrational.

In that same sense, either 0.999... = 1 or it doesn't.  So if we assume it doesn't and arrive at a contradiction, then we know it must.

[quote="jmerry"]It's a question of definitions. How do we define 0.9999...? We use a limit. $0.9999....=\lim_{n\to\infty}\sum_{k=1}^n 9\cdot 10^{-k}=\lim_{n\to\infty}1-10^{-n}$. [/quote]
This is something that really irks me. How is that a reasonable/useful definition of a decimal expansion? For me a decimal expansion of a real number consists of the numbers you get when you divide the number by 10. Surely, defining the concept of a limit before we are sure what is meant by a decimal number is the wrong order in which to do thins?

jmerry is absolutely right. It is the matter of definition. You may consult [url=http://en.wikipedia.org/wiki/Decimal_representation]Wikipedia[/url]. Note that sometimes (in some textbooks) decimal representations with "9" in a period are just prohibited, i.e. it is assumed that such representations [i]does not[/i] correspond to any real number.

It is true for the classical analysis. In non-Archimedean analysis (with infinitesimals) the situation is more complicated. In that case, e.g. there exist a positive real number $\alpha$ such that $0 < \alpha < 1/n$ for every $n=1,2,\ldots$

Kalle: jmerry's limit is exactly what you get when you divide the number by powers of 10. There's nothing special about the decimal representation. I'd say the limit is a much more fundamental concept than that of a decimal representation.

lofar and jmerry: If we have infinitesimals, it'll depend on how we bring .9999... into the extended number system. If we agree that it already represents 1, then it doesn't matter what system we're in; they're already the same real number, so there's nothing in between them. But we might also define .9999.... as 1 - $\epsilon$; say as the hyperreal $\{.9, .99, .999, .9999, ...\}$. That number differs from 1 by an infinitesimal, but there are other numbers in between that and 1 that differ by smaller infinitesimals. If we think of .9999.... as jmerry's series, and we translate series to hyperreals by partial sums, then any series that converges to 1 faster will be between .9999.... and 1. :D
I see absolutely no point to all of this and I'm now shocked that I spent this time thinking about it.

Yes. My argument was entirely about the real numbers, except for the last line I added when I saw 10000th User's post.
If you're uncomfortable with these arguments, find an analysis textbook and look for the chapter or section which constructs the real numbers. I like the version of this in Spivak's Calculus, which contains two exercises on other constructions.
My argument was based on a construction/definition that has the least upper bound property as its completeness axiom. The organization would be slightly different with other axioms in the leading role.

This has been discussed before. Please, search for more info

[quote="Kent Merryfield"]bilarev:

Let $a=1+2+4+8+\cdots+2^n+\cdots.$

Then $\frac a2 = \frac12+1+2+4+8+\cdots=\frac12+a.$

Does it follow that $a=-1?$

If you have trouble believing this argument, then you might understand why some of us would find your post less than entirely convincing.[/quote]
the example which you have been mintion it does not oppsition biarev`s solution , note that $a = \sum\limits_{k \ge 0} {2^k } = \infty$ , so $\frac{a}{2} = \infty$ and the result here is false becuase $a - \frac{a}{2} = \infty - \infty$ is indeterminate , here we work with Infinite countity .

 the result 10a-a=9.9999....-0.9999...=9 is true , because all operation were on finite countity ( convergent sum )

you must give an opposite example in a finite countity  ;)

Maybe that was too subtle- ask yourself this question:
Why does the original series converge?

Then read my post #3.


When do they release the AIME floor score and index?

sorry-----a really brief question but thanks!

Generally it's about a week and a half after AIME II - the last two years it has been the Friday of the week after the second AIME.


algebra
polynomial
Gauss
real analysis
real analysis theorems

Hello. i had some questions about legendre polynomials. I have solved most of them but i had just two not answered question. I tried to solve this problem by rodriguez rule but it was really hard for me. Could you. help me or give me some hints for this question?
i tried also generating function.but i couldnt find the solution.sorry but i really got troubled with these questions.
can you give me suggestion or show me a way to solve these problems?
You told me that these questions are easy to solve.yeah maybe it can be easy for you but for me not. I just solved 10 of 12 proofs that my teacher gave me for homework.and just two these questions remained.

[url=http://img139.imageshack.us/my.php?image=mathkw4.jpg][img]http://img139.imageshack.us/img139/7731/mathkw4.th.jpg[/img][/url]


geometry
3D geometry
sphere
Gauss
articles
set theory

Oops, just clicked too fast and voted 3, I think I would be Erdos number 2 if I get to publish something with my set theory professor :)  Anyways, my professor is Andras Hajnal who has Erdos number 1. How about everyone else?

There is a theory that everybody is related to everybody else in the world through a maximum of 6 links.

In which case, if you just happen to publish papers with other prolific paper-publishers, your number might just (luckily enough! :P ) be 6 or less ;)

Hi Fierytycoon,

My Erds number in [0;+\infty] is +\infty :D with the stereographic map
wich each point the equatorial plane you associate  a unique point on the Riemann sphere S^2, I deduce that my Erds number is N  :D the north pole of the Riemann S^2.
------------------------------------------------------------------------------
Hi Valiowk,

I don't believe this theor, that I am related to anybody else with
a maximum of 6 links.

Well, it's not really my theory...I don't believe in it either, actually :P

Erdos number is the level of acquaintance of the person with the great mathematician Paul Erdos. If, say, mathematicians like Lovasz, Graham, Hajnal,... who has collaborated with Erdos, then they have Erdos number 1. If another mathematician who has collaborated with Graham, like Chung and Orensdorf, then they have Erdos number 2.

But, what its like having Erdos number = 1/n?????? I dunno :?

Never mind, I found some info here:

[url]http://personalwebs.oakland.edu/~grossman/readme.html[/url]

Mine is definitely 00, since I haven't published anything..:D

but i still don't find a fraction Erdos number.

But how about those who voted 1/n? Maybe they know something..

I think there will be so many people not responding this seriously

most of them because they don't want to admit that they don't know erdos numbers, or just because they wanna look smart.

If you have published papers with Erdos, your number is 1/n, where n is the number of papers.

Hey look, I can make the first non-Olympiad question out of this: Prove that Erdos does have a 'Erdos number' and name it!!! :D  :D  :D

For those hardcore problem solvers, do this:

Prove or disprove if there exists a real number K such that for all mathematicians m \in M, the 'Erdos function' f : M --> R is bounded.
In other words, |f(m)|<K.

Note that f is similarly like the Ramsey property.

[Edit: Thx Valiowk for pointing that out]

[quote="3X.lich"]Hey look, I can make the first non-Olympiad question out of this: Prove that Erdos does have a 'Erdos number' and name it!!! :D  :D  :D

For those hardcore problem solvers, do this:

Prove or disprove if there exists a real number K such that for all mathematicians m \in M, the 'Erdos function' f : M --> R is bounded.
In other words, |f(m)|<R.

Note that f is similarly like the Ramsey property.[/quote]

Well, Erds' "Erds Number" is by definition 0...

You meant to say |f(m)|<[b]K[/b], I suppose?

[quote="grobber"]Shouldn't Erdos have his Erdos number equal to 0?[/quote]

He does.  :D

then what do you do with even smarter mathematiciens? they get a negative number? :D

okay, mine is > 2, since I still have been having only 2 co-authors yet...

  dg

I don't think it's about how smart you are, Fiery, it's just about how "close" you are to the vertex named "Erdos" in the graph which represents the colaborations between mathematicians (as coauthors).

For example, I think Gauss has Erdos number 00, and we all know how smart he was.. :)

[quote="grobber"]For example, I think Gauss has Erdos number 00, and we all know how smart he was.. :)[/quote] I must disagree with you Grob. Think about it: Gauss has published some articles with some younger people, and those published articles with younger people when they got older and so on, to Erdos. So it may be quite a big number, but I belive it is finite :D.

[quote="grobber"]I don't think it's about how smart you are, Fiery, it's just about how "close" you are to the vertex named "Erdos" in the graph which represents the colaborations between mathematicians (as coauthors).
[/quote]

Huh? Maybe you mean Peter said that.  :?

I think they say Gauss has Erdos number 00 on that site (I posted the link above):

"The set of all mathematicians with a finite Erds number is called the Erds component of C. It has been conjectured that the Erds component contains almost all present-day publishing mathematicians (and has a not very large diameter), but perhaps not some famous names from the past, such as Gauss."

I don't know if this means they're sure about it, or not, but I think Gauss was a very proud man who kept his discoveries to himself for a long time, so it might be unlikely for him to have published with younger coauthors :D.

Yes, sorry Fiery! :D:D Just got confused...

Well... Most of us would be infinite because I haven't even published a paper

That's true :D

that's why it makes sence to count the people having a finite / low erdos number ;)


[hide="hint"]How much gasoline is put into the tank each hour?  How much gasoline is emptied from the tank each hour?[/hide]

[quote="mr. math"]A large gasoline tank has three inlet pipes and one outlet pipe. One inlet pipe, colored red, takes 2 hours to fill the tank. Another inlet pipe, colored blue, takes 3 hours to fill the tank. the third inlet pipe, colored greed, takes 4 hours to fill the tank. the outlet pipe empties the whole tank in 12 hours. If the tank is initially empty, how many hours would it take to fill the tank, if all the pipes are fully in operation?

How would you do this problem?[/quote]
[hide="solution"]
the red pipe fills $\frac{1}{2}$ tank per hour. the blue pipe fills $\frac{1}{3}$ tank per hour. The greed pipe fills $\frac{1}{4}$ tank per hour. add them up and subtract $\frac{1}{12}$ tank per hour for the outlet pipe. so every hour, $\frac{6+4+3-1}{12}$ tank is filled. Simplified, it is $\frac{12}{12}=1$ tank filled per hour.
take the reciprical of that, so $1$ hour needed per tank. So one hour is needed to fill the tank.
[/hide]


Putnam
geometry
perimeter
trigonometry
trig identities
Law of Sines

[quote="Terilien"]Given a triangle where angle A is equal to 2B and angle C is obtuse, which angles give theminimum perimetre?[/quote]

There is no minimum, since the triangle can be drawn arbitrarily small.

What exactly are you trying to say here?  Three angles without any sides does not allow for this problem to be solved.

What i meant to say was that the side across form the angle a has to be held constant.

[hide="in that case"]
If we're given $a$ and we want to minimize $a+b+c$, this is equivalent to minimizing
\[\frac{a+b+c}{a}-1=\frac{b+c}{a}=\frac{\sin B+\sin C}{\sin A}=\frac{\sin B+\sin 3B}{\sin 2B}\]
(here we have used Law of Sines, and the given $A=2B,\ C=180^\circ-3B$). Now express $B$ as $2B-B$ and $3B$ as $2B+B$ and use trig addition.

Our expression simply equals $2\cos B$. Now $180^\circ-3B=C>90^\circ$, so $B<30^\circ$ and $\cos B>\cos 30^\circ$. There's technically no minimum, but the best lower bound occurs for $\boxed{30-60-90\text{ triangles}}$.
[/hide]


So when I click on some users name, their information comes up. The thing is that for some people, it says they have say 50 posts, but when you click on "see all posts" it shows only 2 posts. Does anyone know why?

Simple answer:
Users can post in forums that you might not have access to (like those for classes).  That's why you might see that a user has 50 posts but you might only have access to view 2 of them.


Please see: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66517]USA 1984 AIME #15[/url]


probability
Euler
Pascal\u0027s Triangle

Find the probability of flipping at least 4 heads in 6 tosses of a fair coin.

[hide]
There are 2^6 ordered arrangements of heads and tails, each of which are equally likely.

There are two ways to proceed from here (they're virtually the same though):

The amount of ways to get 4 heads or more is just the sum of the amount of the rearrangements of the following 3 strings:

HHHHTT
HHHHHT
HHHHHH

This is 6!/4!2!+6!/5!+1= 15+6+1=22

22/2^6=22/64=11/32

Or we could add the number of ways to choose 4 of the 6 heads to the ways to choose 5 of the 6 heads to the ways to choose 6 of the 6 heads, which is equal to

(6C4 + 6C5 + 6C6) / 2^6 = 22/64 = 11/32
(this is what Euler did)

But, as you see, numerically the two ways are equivalent.
[/hide]

It happens  to be $\frac {11}{32}$

[hide="Answer"] 2^6=64 the number of ways 6 coins can be flipped (heads or tails) $\frac {6!}{4!2!}=15$ the amount of ways 4 heads could be flipped on 6 coins. Plus the with all 5 coins that is 6 ways and with all 6 that is one
So $15+6+1=22$ so $\frac {22}{64} reduced = \frac {11}{32}$[/hide]

[hide]like everyone else :D , first i found the total number of combinations. 2^6=64. The number of different flips to get at least 4 heads are 6!/4!2! for 4 heads, 6!/5!1! for 5 heads, and 1 for all 6 heads. 15 + 6 + 1 = 22. Divide that by 64, reduce, to get 11/32.[/hide]

The sixth row in Pascal's Triangle is:
1 6 15 20 [b]15 6 1[/b]

The numbers in bold give you at least 4 heads, so the probability

is $\frac{15+6+1}{1+6+15+20+15+6+1}=\frac{22}{64}=\frac{11}{32}$. :D

First do combinations: 6 C 2, which would give you 15....then you have to do 2 to the 6th power, which is 64.....after you do that, u also have to add 5 and 2 to the combinations, which would give you 22....divide 22 by 64, reduce, and you will have [b]11/32[/b]


I just tested something~

[code];)[/code]

comes out as:

;)

That's just wink - the alternate codes are for people who are used to typing the smilies out on the keyboard instead of clicking the icon. In this regard, ;$ \text{)}$ is more intuitive than :$ \text{wink}$:.


Solve the equation  in real numbers

$\sqrt{2}(x^3+\frac{1}{x^3}+5x+\frac{5}{x^2})=25+5x^2+\frac{7}{x}+x$

hello, your equation can be written as
$ \frac{\sqrt{2}}{2}\frac{(\minus{}x^4\minus{}5x^2\plus{}\sqrt{2})(\minus{}2x^2\plus{}5\sqrt{2}x\plus{}\sqrt{2})}{x^3}\equal{}0$
and from here you will get all solutions.
Sonnhard.


integration
real analysis
real analysis unsolved

what kind of average is this
(
it would seem like an average if the number of objects would be
the number of distinct values of f on E.
but it seems more like a weighted(with weight 0 or 1) average.
)

$A_{E}(f) = \frac{1}{\mu(E)}\int_{E}f\ d\mu$

where $f\in L^{1}(\mu)$ ?

does it have a geometric representation ?
is it complex ?
a complex average ?
what kind of average is a complex average ?

if $\alpha$ is the center of a circular disc with arbitrary radius
and $E=f^{-1}(circular\ disc)$ (the same circular disc as above)
why is $\alpha =\frac{1}{\mu(E)}\int_{E}\alpha\ d\mu$ ?

[quote="spx2"]If $\alpha$ is the center of a circular disc with arbitrary radius
and $E = f^{-1}\left( \text{circular disc}\right)$ (the same circular disc as above), why is $\alpha =\frac{1}{\mu(E)}\int_{E}\alpha\ d\mu$ ?[/quote]
Don't we have that for any constant $k$, $\int_{E}k \, d \mu = k \cdot \mu (E)$?

And here's the theorem in question:

[color=blue]Suppose that $\mu (X) < \infty, \, f \in L^{1}\left( \mu \right)$, $S \subset \mathbb C$ is closed and $A_{E}(f) = \frac1{\mu (E)}\cdot \int_{E}f \, d \mu$ lie in $S$ for every $E \in \mathfrak M$ with $\mu (E) > 0$. Then $f(x) \in S$ for almost all $x \in X$.[/color]

[color=green]What happens if $\mu (X) = \infty$?[/color]

thank you perfect_radio.
i understood it after all,i found the same as you wrote.
but what i ask is,how does this average help ?
what is its meaning ?
if it has anyn ?
any geometric representation ?
i guess for a circle the average is the center of the circle...
at least thats what i see...
does it have anything to do with the center of weight of a body ?
how can i do anythin with it.
only the theorem does not help too much.


A man is alone on an island with no food and no water, yet he does not fear for his life.

How is this possible?

[hide]He is going to leave in an hour, he is just there to visit.[/hide]

[hide="guess #1"]He has no family, and he wants to commit suicide.[/hide]

[hide="guess #2"]He is right next to the bahamas, which has food and water, so he just jumps to the other island.[/hide]

[hide="guess #3"]A steamer boat is coming for him.[/hide]


haha
this is an old iranian problem (99 i think)

just use cauchy and the hypothesis and you are done.


[quote="N.T.TUAN"]This problem posted in Ineq box. I think we can put $a_{i}=\tan{\alpha_{i}}(i=1,2,3)$, where $\alpha_{i}\in(-\frac{\pi}{2},\frac{\pi}{2}).$[/quote]

Well, please show your solution completely (there or here  :lol: )


Let a,b,c be natural numbers such that :a+b+c=[hide=2pq(p^{30}+q^{30}) ,p>q being two give positive integers
prove that k=a^{3}+b^{3}+c^{3} is not a prime number

[quote="TTsphn"][quote="NDMinh"]Let a,b,c be natural numbers such that :$ a\plus{}b\plus{}c \equal{} 2pq(p^{30}\plus{}q^{30})$ ,p>q being two give positive integers
prove that$ k \equal{} a^{3}\plus{}b^{3}\plus{}c^{3}$ is not a prime number[/quote]
This is your problem.[/quote]
oh! no ,this is IMO ,I remember clearly year   :)


Apparently our ideas of 'nice' do not coincide.

Blahblahblah, when I said "nice" a meant very important. With this, I hope new formula, I deduce very easy some classical quadrature formulae. ;)


$ \left\|\begin{array}{c} \{x,y\}\subset\mathcal R^*_ \plus{} \\
 \\
x^2 \plus{} y^2 \equal{} 1\end{array}\right\|$ $ \implies$ $ \boxed {2 \plus{} \sqrt 2\le 2 \plus{} \sqrt {\frac 52 \minus{} xy}\le x \plus{} y \plus{} \frac {1}{xy}\le 2 \plus{} \sqrt {3 \minus{} 2xy}}\ .$

See http://www.mathlinks.ro/viewtopic.php?t=170319

$ \left[\ 2xy\le x^2 \plus{} y^2\implies 2xy\le 1\implies \sqrt 2\le \sqrt {\frac 52 \minus{} xy}\le\sqrt {3 \minus{} 2xy}\ \right]$

Let $ x \plus{} y \equal{} a,\ xy \equal{} b,\ x^2 \plus{} y^2 \equal{} 1, x > 0,\ y > 0$

$ \Longleftrightarrow a^2 \minus{} 2b \equal{} 1, \ a > 0,\ b > 0,\ a^2\geq 4b$
$ \Longleftrightarrow a \equal{} \sqrt {2b \plus{} 1} ,\ 2b \plus{} 1\geq 4b,\ b > 0$
$ \Longleftrightarrow a \equal{} \sqrt {2b \plus{} 1},\ 0 < b\leq \frac {1}{2}$.

1.Proof of $ x \plus{} y \plus{} \frac {1}{xy}\leq 2 \plus{} \sqrt {3 \minus{} 2xy}$.

$ x \plus{} y \plus{} \frac {1}{xy} \minus{} \sqrt {3 \minus{} 2xy} \equal{} a \plus{} \frac {1}{b} \minus{} \sqrt {3 \minus{} 2b}$

$ \equal{} \sqrt {1 \plus{} 2b} \minus{} \sqrt {3 \minus{} 2b} \plus{} \frac {1}{b}$

$ \equal{} \frac {2(2b \minus{} 1)}{\sqrt {1 \plus{} 2b} \plus{} \sqrt {3 \minus{} 2b}} \plus{} \frac {1}{b}\leq 2$.

[color=red]Alternative solution[/color]:

$ 2xy\leq x^2 \plus{} y^2 \Longleftrightarrow 0 < xy\leq \frac {1}{2}$ for $ x > 0,\ y > 0$.

Let $ xy \equal{} t$, we have $ \frac {1}{t} \minus{} \sqrt {3 \minus{} 2t}\leq 2 \minus{} \sqrt {2}$ by the graphs of $ y \equal{} \frac {1}{t},\ y \equal{} \minus{} \sqrt {3 \minus{} 2t}$ for $ 0 < t\leq \frac {1}{2}$.

$ \therefore x \plus{} y \plus{} \frac {1}{xy}\leq \sqrt {2} \plus{} \frac {1}{xy}\leq 2 \plus{} \sqrt {3 \minus{} 2xy}$.

2. Proof of $ 2 \plus{} \sqrt {\frac {5 \minus{} 2xy}{2}}\leq x \plus{} y \plus{} \frac {1}{xy}$.

$ x \plus{} y \plus{} \frac {1}{xy} \minus{} \sqrt {\frac {5 \minus{} 2xy}{2}}$

$ \equal{} a \plus{} \frac {1}{b} \minus{} \sqrt {\frac {5 \minus{} 2b}{2}}$

$ \equal{} \sqrt {2b \plus{} 1} \plus{} \frac {1}{b} \minus{} \sqrt {\frac {5 \minus{} 2b}{2}}$

$ \equal{} \frac {3(2b \minus{} 1)}{2\left(\sqrt {2b \minus{} 1} \plus{} \sqrt {\frac {5 \minus{} 2b}{2}}\right)} \plus{} \frac {1}{b}\geq 2$.

The equalities holds when $ b \equal{} \frac {1}{2}\Longleftrightarrow a \equal{} \sqrt {2}\Longleftrightarrow x \equal{} y \equal{} \frac {1}{\sqrt {2}}$.

[color=darkred][b]Thank you[/b] for this nice proof, [b]Kunny[/b] ![/color] [color=darkblue]I have a geometrical proof for the inequality

$ \left\{\begin{array}{c} \boxed {\triangle\ ABC} \\
 \\
a \equal{} \max\ \{a,b,c\}\end{array}\right\|$ $ \implies$ $ \frac {b^2 \plus{} c^2}{bc} \plus{} \frac {b \plus{} c}{a}\ge 2 \plus{} \frac {1}{\sin\frac A2} \cdot \sqrt {1 \plus{} \left( \frac {b \minus{} c}{2a}\right) ^2}\ .$

[b]Remark. [/b]If $ A \equal{} 90^{\circ}$, then the second inequality becomes $ \frac {a^2}{bc} \plus{} \frac {b \plus{} c}{a}\ge 2 \plus{} \sqrt {\frac 52 \minus{} \frac {bc}{a^2}}\ .$[/color]

Can we solve this using a trig substitution?

Yes, you can set $ x \equal{} \cos \theta ,\ y\equal{}\sin \theta \ \left(0 < \theta < \frac {\pi}{2}\right)$.

Then let $ \cos \theta \plus{} \sin \theta \equal{} t,$ we have $ \cos \theta \sin \theta \equal{} \frac {t^2 \minus{} 1}{2}$ for $ 0 < t\leq \sqrt {2}$. We are to need differentiation.


Let $ a,b,c$ be positive real numbers. We have the familiar family of $ p$ means:

$ m_{p}=\left(\frac{a^{p}+b^{p}+c^{p}}3\right)^{\frac1p}$

with $ m_{0}=\sqrt[3]{abc}$

[Of course, $ p=1$ gives the arithmetic mean, $ p=0$ the geometric mean, and $ p=-1$ the harmonic mean.]

Now define a new mean: $ M=\sqrt{\frac{3abc}{a+b+c}}.$

Investigate the properties of $ M.$ For what (if any) values of $ p$ is $ M\le m_{p}$ or $ M\ge m_{p}?$ What else can you say about it?

Let's make it more specific: show that $ M$ lies between the harmonic mean and the geometric mean. That is, $ m_{-1}\le M\le m_{0}.$

To show that $ M\leq m_{0}$, since the inequality is homogeneous we may suppose that $ abc=1$. Then, the ineq. becomes $ a+b+c\geq 3$ which holds by AM-GM.
And in order to prove $ m_{-1}\leq M$, we do this: $ m_{-1}\leq M \iff 3(a+b+c)abc \leq (\sum ab)^{2}$ $ \iff (a+b+c)abc \leq \sum a^{2}b^{2}\iff 0 \leq \sum (ab-bc)^{2}$
(where the sums are cyclic)
And we are done.


parameterization
algebra proposed
algebra

Solve the equation \[ x\plus{}a^3\equal{}\sqrt[3]{a\minus{}x}\] where $ a$ is a real parameter.

[quote="Raja Oktovin"]Solve the equation
\[ x \plus{} a^3 \equal{} \sqrt [3]{a \minus{} x}\]
where $ a$ is a real parameter.[/quote]

Let $ x\equal{}y\minus{}a^3$. The equation becomes $ y\equal{}\sqrt[3]{a\plus{}a^3\minus{}y}$ $ \implies$ $ y^3\equal{}a\plus{}a^3\minus{}y$ $ \iff$ $ (y\minus{}a)(y^2\plus{}ay\plus{}a^2\plus{}1)\equal{}0$ $ \iff$ $ (y\minus{}a)((y\plus{}\frac a2)^2\plus{}\frac{3a^2\plus{}4}4)\equal{}0$

Hence the unique real solution $ x\equal{}a\minus{}a^3$

hello, i think the solution must be $ x\equal{}a\minus{}a^3$ and $ a\geq0$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i think the solution must be $ x \equal{} a \minus{} a^3$ and $ a\geq0$.
Sonnhard.[/quote]

but why must $ a \ge 0$?
if $ a\equal{}\minus{}1$, then clearly $ x\equal{}0$ since the cubic root can operate in negative number?


If $ a,b,c$ are real numbers such that $ a\plus{}b,b\plus{}c,c\plus{}a \neq 0$ and

    $ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\equal{}1$ then prove:

    $ \frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{c\plus{}a}\plus{}\frac{c^2}{a\plus{}b}\equal{}0$.

Multiply each side of $ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\equal{}1$ by $ a \plus{} b \plus{} c$ to get
$ \frac{a^{2}}{b\plus{}c}\plus{}\frac{b^{2}}{c\plus{}a}\plus{}\frac{c^{2}}{a\plus{}b} \plus{} a \plus{} b \plus{} c \equal{}a \plus{} b \plus{} c$
And hence
$ \frac{a^{2}}{b\plus{}c}\plus{}\frac{b^{2}}{c\plus{}a}\plus{}\frac{c^{2}}{a\plus{}b}\equal{}0$


Let there be $ a,b,c,d \in \mathbb{Z^\plus{}}$ such that $ abcd \equal{} a \plus{} b \plus{} c \plus{} d \plus{} 1$. Find the maximum value of $ abcd$.

Without loss of generality, let $ a \leq b \leq c \leq d$. Then we have that $ abcd \equal{} a \plus{} b \plus{} c \plus{} d \plus{} 1 \leq 4d \plus{} 1$. Then $ abc \leq 4 \plus{} 1/d$; since $ abc$ is an integer, we must have that $ abc \leq 4$. From here, we easily deduce that the maximum is attained when $ a \equal{} 1$, and $ b \equal{} c \equal{} d \equal{} 2$ yielding a final answer of 8.

EDIT: Forgot to put the 8 at the end. <_<

Wow, I didn't think it was so simple. I tried AM-GM and it didn't turn out too pretty.

Heheh. I had no idea we could do anything like that either; I solved a similar problem with 3 variables. I tried using tangents. (tan A + tan B = tan C = tan A tan B tan C when A + B + C = pi.) And then a friend showed me that method. I bashed my head against a wall and couldn't do math for a few days.


A my new result :blush:

[b]Problem:[/b] Let  $ S$ be a point in plane of triangle $ ABC$.Denote by $ D_1,E_1,F_1$ the orthogonal projections of $ S$ on

$ BC,CA,AB$ ,respectively.Denote by $ D_2E_2F_2$ the cevian triangle of $ S$ wrt $ D_1E_1F_1$,denote by $ D_3E_3F_3$ the

cevian triangle of $ S$ wrt $ D_2E_2F_2$,....and denote by $ D_nE_nF_n$ the cevian triangle of $ S$ wrt $ D_{n \minus{} 1}E_{n \minus{} 1}F_{n \minus{} 1}$.

Prove that three perpendicular lines from $ A,B,C$ to $ E_iF_i ,F_iD_i,D_iE_i$ ,respectively  are concurrent at a point $ T_i$

and the points $ T_i$  and the centroid of $ ABC$ are collinear.

Diagram:

[img]http://i524.photobucket.com/albums/cc322/khanhto-photo/untitled-106.jpg[/img]

Call perpendicular lines from $ B,C$ to $ F_iD_i,D_iE_i$ cut at a point $ T_i$. We have:
$ T_iF_i^2\minus{}T_iE_i^2\equal{}T_iF_i^2\minus{}T_iD_i^2\plus{}T_iD_i^2\minus{}T_iE_i^2\equal{}BF_i^2\minus{}BD_i^2\plus{}CD_i^2\minus{}CE_i^2$
$ \equal{}(\minus{}AF_i^2\plus{}BF_i^2)\plus{}(\minus{}BD_i^2\plus{}CD_i^2)\plus{}(\minus{}CE_i^2\plus{}AE_i^2)\plus{}(\minus{}AE_i^2\plus{}AF_i^2)$
$ \equal{}(\minus{}SA^2\plus{}SB^2)\plus{}(\minus{}SB^2\plus{}SC^2)\plus{}(\minus{}SC^2\plus{}SA^2)\plus{}(\minus{}AE_i^2\plus{}AF_i^2)$
$ \equal{}\minus{}AE_i^2\plus{}AF_i^2$
It mean $ AT_i$ is perpendicular with $ E_iF_i$
Then three perpendicular lines are concurrent
I will try to solve the rest of this problem later, but I guess that $ \overrightarrow{GT_i}\equal{}\minus{}2\overrightarrow{GT_{i\plus{}1}}$


I am a junior now (administration changed my grade) and have already taken three math classes so far (Calc III, Differential Equations, and Linear Algebra). Next semester, I have the opportunity to take Number Theory, but am really debating whether I should take it, and so I need AoPS's help.
-I really don't enjoy college math at all, and it really drains time away from competitive math and physics, which I find fun. I really don't want to take Number Theory, since it will drain my time away during AMC/AIME season and will cost 300 dollars. Plus, I procrastinate. I am really busy all the time, and am worried of getting a bad grade in that course because I'll be so busy.
-However, I feel like I am caught up in the impressing-the-colleges feeling. I was reading somewhere if you take lots of college math classes, it will help a lot in admissions. Will not taking a college math course hurt me? I took 2 in sophomore year; will taking only one during junior year seem like I am slacking? I feel like there are other juniors out there who are owning me in terms of college classes, and so they are smarter, and I am dumber. Can anyone help me get over that feeling?
-Does taking Calculus as a freshmen and taking 3 college math classes by the time one is a junior look impressive at all to elite colleges?

Sorry, but I want to know what I should do. I really am stressed.

Don't worry too much about impressing colleges- you should already have quite a bit of impressive material on your record.

It probably doesn't matter much which course or courses you take, but leaving classroom math out completely is a bad idea. You are interested in math- what you're doing in contests shows that. It sounds like you've got the most useful and transferable classes out of the way- and the classes you already have are the most "cookbook" of college material. Number theory? Look closely to see how much you already know- a lot of that stuff shows up on the contests.
Don't spend every waking moment worrying about the AMC contests- they aren't everything.

Colleges advise to follow your passions, and they really mean it. Don't do something only to impress a college, especially if that just burns you out. Get enough sleep at night, challenge yourself with activities that seem interesting to you, and get some balance in your life with other activities you enjoy.

Yeah, you already have a rare math background. Being able to describe what you did and why you did without referring to "I did this because it looks good to colleges" will give you an even rarer background.

[quote="samus"] I feel like there are other juniors out there who are owning me in terms of college classes, and so they are smarter, and I am dumber. [/quote]

The world is large.  Probably there are 10-year olds somewhere who "own" you in mathematical knowledge, and many people between age 6 and 60 who are "owned" by you.   None of which matters very much.


geometry LaTeX algebra function domain \/closed

[quote="Valentin Vornicu"]It's finally here. You can use the [img ] tag to put images from the photo album into posts (for Geometry questions especially) which then can be nested into a hide tag, as exemplified below:

[code][hide] [img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=224[/img] [/hide][/code] gives
[hide][img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=224[/img][/hide][/quote]

hmm, I can't open that hide tag you posted...

Can you do it now?
Try to post another picture from the Album with the [img ] tag here ...

Your hide tag still doesn't work for me.  But what I have posted below does work...
[i]edit: lol, did I just choose the same exact pic as you?  Wow, I didn't know which one you picked until I copied your code into my address bar [b]after[/b] I posted this.[/i]

[img]http://artofproblemsolving.com/Forum/album_pic.php?pic_id=224[/img]

[hide]
[img]http://artofproblemsolving.com/Forum/album_pic.php?pic_id=224[/img][/hide]

Wait so it works if you don't hide it, but when you hide it, we have a problem, right? ...

[quote="Peter VDD"]Well while we're talking about tags: it would be nice if [code]This is a $\LaTeX$ test.[/code] really ended up in code, so we could give latex codes in that way :)[/quote]This is a different topic, let's not mix them :D

OK, it seems there is something wrong with the hide tag :D Back to work ...
 [img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=222[/img]

Ok, this is weird.  Now your first post works for me.  My post (both the pic and the pic in the hide) works.  But your recent one only shows the [img ] link [/ img ]

Yes, it is rather strange for me too ...

[quote="rrusczyk"]FWIW, it all works fine for me (IE).[/quote]Hmm, I am using Firefox 1.0.6 ... so that might be the problem ...  :roll:  :?:

On my computer, it works with the ML skin and doesn't work with the AoPS skin. I've checked this on both FF 1.0.6 and IE ([b]edit:[/b] and it also doesn't seem to depend on whether one is logged in or not)

  darij

Firefox, ML skin, here everything works fine too...

It's crazy ... I can't find anything wrong with it, however there are instances when it doesn't work, and instances when it works ...  :?:

can you give me an example of a non-working instance? I may be able to tell if I've experienced alike... so please show me one :)

I just copied and pasted it from my address bar.

[quote="Valentin Vornicu"]Ok, I know now what the problem is :) Joml was not using www in front of the url. If you write the url as you should (so http://www.foo.com/bar instead of http://foo.com/bar ) there is no problem :)[/quote]

Are you sure? I am now browsing here using the AoPS skin (this is one of the uses of my secondary account :) ) and Joml's hide tag works well, but the one in [i]your[/i] post #1 doesn't:

[quote="Valentin Vornicu"][code][hide] [img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=224[/img] [/hide][/code] gives
[hide][img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=224[/img][/hide][/quote]

Hmm, what is happening now is very strange: When I click on Valentin's hide tag in post #1, just nothing happens, but when I click on this quoted version of his hide tag, then I see the hidden text:
[ img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=224[ /img]
as if the [ img] code wouldn't get parsed (and this even with the ML skin).

  darij

Thank you Darij. Thanks to your post (and to joml's notice) I isolated (and fixed :D) the problem. Now everything should work ... as it should :)

Hmm, I am sorry but... the hide tags for themselves seem to work (at least, when I open them, I don't see the bare source code of the img tag), but generally, when I am using the AoPS skin and try to see an img tag with a MathLinks URL (like [img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=224[/img] , in contrast to [img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=224[/img] ), I see a red cross (on IE; on FF, I see nothing) :( . Anyway, this is probably rather a bug of the img tags than one of the hiders, since it also happens when the img tag is not hidden...

  darij

Are you sure?! :) I just put tags on them and I see them both ... on IE/and Firefox with all the cache cleared on both skins and urls ...

Hmm... here is a print screen of what I (still) get using IE. Maybe it would help...

[EDIT: now fixed.]

  darij

[quote="Valentin Vornicu"]Can you try using "http://www.foo.com/bar" instead of "http://foo.com/bar" syntax?[/quote]

It's strange but I think I did it... anyway let's try again:

http://www.mathlinks.ro yields [hide][img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=224[/img][/hide]
mathlinks.ro yields [hide][img]http://mathlinks.ro/Forum/album_pic.php?pic_id=224[/img][/hide]

Both viewable using ML and both not viewable using AoPS...

AoPS links:

http://www.artofproblemsolving.com yields [hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=224[/img][/hide]
artofproblemsolving.com yields [hide][img]http://artofproblemsolving.com/Forum/album_pic.php?pic_id=224[/img][/hide]

That's just too strange. All of these work with ML but the only one working with AoPS is the fourth one, which uses the http://foo.com/bar syntax...  :oops:

  darij

When I put myself on mathlinks.ro, I can only see mathlinks.ro and not anything else. If I put myself on http://www.mathlinks.ro I can see it all.

Which is pretty logical, looking the way how the url's are generated. I think it would be solved simply by disallowing the domains http://mathlinks.ro and aops alike, and redirecting users automatically to the www. domain.

Ah, that's what Valentin meant by using the right syntax! I thought he spoke about the img links... Thanks, Peter, now everything works fine. I think you are right with your idea: mathlinks.ro is already being redirected to http://www.mathlinks.ro, remains to do the same for AoPS.

Sorry for the misunderstanding.  :oops:

  Darij

[edit: ok, you guys already figured it out :blush: ]

Darij, I believe Valentin means to use the http://www.foo.com/bar syntax when viewing the page that you are trying to view the pictures on and not just for inserting the [ IMG] tags.

Valentin, thanks, it works now (I'm using FF with AoPS skin).  I had to change it to the http://www.foo.com/bar syntax.

Ok, so everyone figured it out. Now I have to figure out why http://artofproblemsolving.com/foo is not redirected to http://www.artofproblemsolving.com/foo ...

Ok, now it's finally done (I think :D). Try it out without the www :P


search
modular arithmetic
quadratics
symmetry
number theory unsolved
number theory

m is a natural number solve this equation on Z:

4xy-x-y=m 2

Hmm... $x=\frac{m^2+y}{4y-1}$, i.e. $4y-1\mid m^2+y$. It is equaivalent to $4y-1\mid 4m^2+1$, so we are to search for divisor of $4m^2+1$ of form $4y-1$ (actually, it is necessary $y\leq 0$).

[quote="Pascual2005"]Another solution: equation is equivalent to: $16xy-4x-4y=4m^2$ or also to: $(4x-1)(4y-1)=4m^2+1$, a prime of the form $4k+1$ divides lhs, but it cannot divide rhs, therefore there are no solutions  :D[/quote]
You missed again. Maybe you didn't see very last remark in my solution...
Some solutions: $x=0,y=-1,m=1$, $(-1,-3,4)$.

So for any natural number m, 4xy-x-y=m 2 has solutions -1,-3, and 4?

No... what Myth meant was that $(x, y, m) = (-1, -3, 4)$ is a possible solution.

Well, to complete the solution, and to continue on Pascual2005's idea, we first see that the problem is equivalent to $(4x-1)(4y-1) = 4m^2+1$. Now if any of $x, y >0$, then $\exists$ a prime $p \equiv 3 \pmod{4}$ such that $p$ divides L.H.S., which is a contradiction since $4m^2$ is a perfect square but $-1$ is a quadratic non-residue $\mod{p}$. Thus we must have $x,y \leq 0$. This transforms the problem to $(4 \mid x \mid -1)(4 \mid y \mid -1) = 4m^2+1$, which is equivalent to the problem $(4X+1)(4Y+1) = 4m^2+1$ where $X, Y$ are non-negative integers. More specifically, $X= -x, Y = -y$.

Now, $(4X+1)(4Y+1) = 4m^2+1 \Leftrightarrow 4XY + X + Y = m^2 \Leftrightarrow (4X+1)Y + X = m^2$, which will have solutions iff $X$ is a quadratic residue $\mod{(4X+1)}$. Note that $(\frac{X}{4X+1}) = (\frac{4X}{4X+1}) = (\frac{-1}{4X+1})$, where $(\frac{P}{Q})$ represents the Jacobi symbol (generalised version of Legendre's symbol so that the symbol is still defined for composite $4X+1$). Note that $\gcd (X, 4X+1) = 1$, so there's no issue of $(\frac{X}{4X+1}) =0$.

This just means that if $4X+1$ does not contain any primes divisors $\equiv 3 \pmod{4}$ then $\exists$ solutions $Y, m$. So there are either infinitely many solutions or no solutions for each fixed $X$

Back to the original problem, for example, if $x=-1$, then possible values for $y$ are $y=0, -3, -7, -16, \ldots$, if $x=-3$, the possible values for $y$ are $y = -1 -6, -22, \ldots$, and values like $-2, -5$ can never be possible values for $x, y$ (because of the symmetry of $x, y$) as $3$ divides $9$, $7$ divides $21$, and $3$ & $7$ both are primes $\equiv 3 \pmod{4}$.

Maybe I am missing something, but neither you nor Pascual said nothing new besides what I wrote in post #2. :?

well... you can treat it as an elaboration on your solution then.  :)

More posts than needed when answer was in very next post.


Assume that x,y,z are real numbers satisfying:[tex]x+y+z=\frac{\pi}{2}[/tex]
Prove that[tex]|\frac{cosx}{sin(y-z)}|+|\frac{cosy}{sin(z-x)}|+|\frac{cosz}{sin(x-y)}|\geq\ 2[/tex]

it is $\sum |\frac{cos^2x}{sin(y+z)sin(y-z)}| = \sum |\frac{cos^2x}{sin^2y - sin^2z}| = \sum \frac{cos^2x}{|cos^2z - cos^2y|}$

So let $A = cos^2x$ etc, and wolog $A \geq B \geq C$, so that we just need to show:

$\frac{A^2}{B^2 - C^2} + \frac{B^2}{A^2 - C^2} + \frac{C^2}{A^2 - B^2} \geq 2$

But $LHS \geq \frac{A^2}{B^2 - C^2} + \frac{B^2 + C^2}{A^2 - C^2}$

$\geq  \frac{A^2}{B^2 + C^2} + \frac{B^2 + C^2}{A^2 - C^2}$

$\geq \frac{A^2 - C^2}{B^2 + C^2} + \frac{B^2 + C^2}{A^2 - C^2} \geq 2$

as desired, with equality when $C = 0$ in the ineqs before the last, and $A = B$ in the last one.

---

looks right, i didnt check

[quote="Singular"]it is $\sum |\frac{cos^2x}{sin(y+z)sin(y-z)}| = \sum |\frac{cos^2x}{sin^2y - sin^2z}| $

[/quote]

why?

$sin(y+z)sin(y-z)=\frac{cos2z-cos2y}{2}=\frac{1-2sin^2 z - 1+2sin^2 y}{2} = sin^2 y - sin^2z $


algebra
polynomial
combinatorics unsolved
combinatorics

Danica imagines a polynomial P(x) with nonegative integer coefficients.
Tatjana is trying to guess polynomial. In every step Tatjana chooses an integer k
and Danica gives her the value of P(k). Find the smallest number of steps danica needs in order to
find the polynomial.

Two steps.  Tatjana only needs the value of $ P(1)$, after which she can choose the smallest power of $ 10$ larger than $ P(1)$.  On the other hand, it is clearly impossible for Tatjana to succeed in only one step because she cannot distinguish between $ P(x)$ and $ (x \plus{} 1 \minus{} k)P(x)$ for any $ k$ she chooses, so this is optimal.


Points $ B$, $ C$ and $ D$ lie on circle $ A$. Circle $ A$ has radius 10 inches.
What is the number of square inches in the area of the sector
bounded by radii $ \overline{AD}$ and $ \overline{AB}$ and major arc $ BED$? Express your
answer to the nearest square inch.

[asy]size(125); import graph; fontsize(10pt); pair A = (0,0), D = dir(90), C = dir(13), B = dir(-50), ptE = (-1,0);
draw(Circle(A,1)); draw(Label(rotate(-50)*"10 in",Relative(.73)), B--A, S); draw(C--A); draw(D--A);
dot("$A$",A,W); label("$B$",B,B); label("$C$",C,C); label("$D$",D,D); label("$E$",ptE,ptE); label("$63^\circ$",.4*dir((13-50)/2)); label("$77^\circ$",.3*dir((90+13)/2));[/asy]

The sector in question has central angle $ 360 \minus{} (77 \plus{} 63) \equal{} 220$, and thus takes up $ \frac{11}{18}$ of the circle. Since the circle has a radius of 10, the area is $ 100\pi$, so the area of our sector is $ \frac{550\pi}{9} \equal{} 192$, approximately.


Triangle ABC is isosceles with $ AC\equal{}BC$. Angle $ A$ measures 45 degrees. Segment $ CD$ is the perpendicular bisector of segment $ AB$. If segment $ AD$ measures three meters, how long is segment $ AC$? Express your answer in simplest radical form.

[asy]pair A, B, C, D;

A = (0,0);
D = 1*dir(0);
C = sqrt(2)*dir(45);
B = 2*dir(0);

draw(A--D--B--C--A--cycle, linewidth(1));
draw(D--C,dashed);

label("A", A, W);
label("B", B, E);
label("C", C, N);
label("D", D, S);[/asy]

$ AC\equal{}AD\sqrt{2}$ since triangle ADC is 45-45-90 triangle.

AD=3.

answer : $ 3\sqrt{2}$


Let $ ABCD$ is a square. Let $ (O; R)$ is circum circle of $ ABCD$. Let $ M$ is a point on circle $ (O; R)$. Prove that: $ MA.MB.MC.MD<6R^4$.

[b][color=darkblue]Let $ M \in \widehat{AB}$. Denote $ \angle{MBA} = \alpha, \angle{MAB} = 45^\circ - \alpha$.

By sine law obtain:

$ \| \begin{array}{cc} \frac {MB}{\sin{(45^\circ - \alpha)}} = 2R \\
 \\
\frac {MA}{\sin{\alpha}} = 2R \\
 \\
\frac {MC}{\sin{(90^\circ + \alpha)}} = 2R \\
 \\
\frac {MD}{\sin{(135^\circ - \alpha)}} = 2R \end{array} \| \Rightarrow \| \begin{array}{cc} MA\cdot MB\cdot MC\cdot MD \\
= \\
16R^4 \sin{\alpha} \sin{(45^\circ - \alpha)}\sin{(90^\circ + \alpha)}\sin{(135^\circ - \alpha)} \\
= \\
16R^4 \sin{\alpha} \sin{(90^\circ - \alpha)}\sin{(45^\circ + \alpha)}\sin{(45^\circ - \alpha)} \\
= \\
4R^4 \cdot (2\sin{\alpha} \cos{\alpha})\cdot( \sin{(45^\circ - \alpha)}\cos{(45^\circ - \alpha)}) \\
= \\
4R^4 \sin{2\alpha}\sin{(90 - 2\alpha)} \\
= \\
4R^4 \sin{2\alpha}\cos{2\alpha} \\
= \\
2R^4\sin{4\alpha} < 2R^4 \end{array}$

So $ MA\cdot MB\cdot MC\cdot MD < 2R^4$. [/color][/b]

[hide="Image"][img]http://ipicture.ru/uploads/080604/PnfS3lIfP1.png[/img][/hide]

[quote="thanhnam2902"] [color=darkred][b][u]An easy extension.[/u][/b] Let $ ABCD$ be a rectangle  with the circumcircle $ w = C(R)$ and $ \|\begin{array}{c} AB = a \\
\ AD = b\end{array}$ .

Let $ M$ be a point on the circle $ w$. Prove that: $ \sqrt {MA.MB.MC.MD }\le R\cdot\max\{a,b\}$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] I"ll apply same [b]nice [u]Ahiles[/u]' trigonometrical method[/b]. Suppose w.l.o.g.that $ 2R = 1$ and [u]at first[/u] suppose that the side $ [CD]$ separates

the points $ M$ , $ A$ . Denote $ \|\begin{array}{c} x = m(\widehat {CDM}) \\
 \\
y = m(\widehat {DCM}) \\
 \\
\phi = m(\widehat {ADB})\end{array}$ . Observe that $ \|\begin{array}{c} x + y = \phi < 90^{\circ} \\
 \\
\tan\phi = \frac ab \\
 \\
a^2 + b^2 = 1\end{array}$ . Denote the product $ \Pi\equiv MA\cdot MB\cdot MC\cdot MD$ . Thus,

$ \Pi = \cos x\cos y\sin x\sin y =$ $ \frac 14\cdot\sin 2x\sin 2y =$ $ \frac 18[\cos 2(x - y) - \cos 2\phi]\le$ $ \frac 18\cdot (1 - \cos 2\phi ) =$ $ \frac 14\cdot \sin^2\phi =$ $ \frac 14\cdot\frac {\tan^2\phi}{1 + \tan^2\phi} =$ $ (\frac a2)^2$

with equality iff $ x = y = \frac {\phi}{2}$ . Return to the initial notation, i.e. replace $ 1$ with $ 2R\ \ : \ \ \Pi\le (2R\cdot\frac a2)^2 = (aR)^2$ .

[u]At second[/u], if the sides $ [AD]$ separates the points $ M$ , $ B$ obtain analogously $ \Pi\le (bR)^2$ with equality iff $ x = y = \frac {\pi}{4} - \frac {\phi}{2}$ .

 In conclusion, $ \boxed {\ \sqrt {\Pi}\le R\cdot\max\{a,b\}\ }$ . In the particular case when $ ABCD$ is a square, i.e. $ a = b = R\sqrt 2$ obtain $ \Pi\le 2R^4$ . [/color]


To find the "digital product" of a number you multiply the digits of the number together to create a new number.  Then if that new number has more than one digit you take the product of its digits to get yet another new number.  This process is repeated until you have a single digit number left and that single digit is the "digital product" of the original number.   A single digit number does not have a "digital product".  How many positive integers less than 200 have a "digital product" that is odd?

[hide="nogin"]

so, let's find a pattern

first of all, you need all odd digits...

so, that leaves us with 48 numbers...

let's check how many of those already are finished...

1, 1-9 odd
3, 1 and 3
5, 1
7, 1
9, 1

and multiply by 2 since the 1 in the hundreds does not affect the answer

so, that gives us 10 already done...

15, 21, 27, 15, 25, 35, 45, 21, 35, 49, 63, 27, 45,63, 81
15, 21, 27, 15, 25, 35, 45, 21, 35, 49, 63, 27, 45,63, 81

are the products left...

the following have only odd digits left...

15, 15, 35, 35, 15, 15, 35, 35

so these all end as 5's and there are 8 of them...adding this to the 10 previous ones, we get 18 as our answer[/hide]
jorian

[hide=semi brute force]

Well, the digits of the number must be odd(limits our counting-only 10 combinations to try)..i found that only these combinations work

1,3
1,5
1,7
1,9
3,5
5,7

you multiply these by 4 (becuase they can be reversed and also can be between 100-199)

1,1,
3,3

you multiply these by 2 (cant be reversed)

so 6*4+2*2=28? [hide]
EDIT(after reading the next post) : uh wait..jorian..let me check this again..

now in your solution, when you say to multiply by 2...i think you did something wrong there...because you have 10 already, then multiply by 2, so 10*2+8=28, my answer.

Okay. new solution. [hide]6[1x3,1x5,1x7,1x9,3x5,5x7]*2[31 and 13 use the same numbers so we multiply by 2]*2[hundreds only has a 1, doesn't affect the digital product so we multiply by 2] + 2[1x1, 3x1]*1*2[same as above] = [b]28[/b] :)[/hide]


[quote="ch*yeuk"]Write [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/2d561b7736b0c2389b8e8ebc94c0ba64.gif[/img] as a single fraction

[hide="view multiple choice answers"]
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/ede67063943349b88efb27368e200011.gif[/img]
[/hide][/quote]

[hide]B.
the expression equals
$\frac{3\sqrt{6}}{\sqrt{42}}-\frac{2\sqrt{7}}{\sqrt{42}}=\frac{\sqrt{42}(3\sqrt{6}-2\sqrt{7})}{42}=\frac{18\sqrt{7}-14\sqrt{6}}{42}=\boxed{\frac{9\sqrt{7}-7\sqrt{6}}{21}}$[/hide]

[quote="ch*yeuk"][quote="ch1n353ch3s54a1l"]
[hide]B.
the expression equals
$\frac{3\sqrt{6}}{\sqrt{42}}-\frac{2\sqrt{7}}{\sqrt{42}}=\frac{\sqrt{42}(3\sqrt{6}-2\sqrt{7})}{42}=\frac{18\sqrt{7}-14\sqrt{6}}{42}=\boxed{\frac{9\sqrt{7}-7\sqrt{6}}{21}}$[/hide][/quote]

Does [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/32144bf52d54aa800d6bae600d6c3a31.gif[/img] become just [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/a1d0c6e83f027327d8461063f4ac58a6.gif[/img] when the other [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/32144bf52d54aa800d6bae600d6c3a31.gif[/img] is moved to the top?[/quote]
It's the same as multiplying the numerator and the denominator by $\sqrt{42}$, so yes

how come you are using images to display latex? I think it's easier if you learn it  :P

Alright, thanks!

[quote]how come you are using images to display latex? I think it's easier if you learn it  :P[/quote]
You mean.. the program thing I have to download? :| I'm using the online TeXeR thing X_X I found the program confusing

no you dont need a program..

you just need to put dollar signs around the math..

e.g. 4+5=9

put a dollar sign before 4 and after 9
and you get
$4+5=9$

Oh really? *tests with posting preview..* oh wowww it works! =O thanks =)

[quote="ch*yeuk"]Write [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/2d561b7736b0c2389b8e8ebc94c0ba64.gif[/img] as a single fraction

[hide="view multiple choice answers"]
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/ede67063943349b88efb27368e200011.gif[/img]
[/hide][/quote]

3/√7=3/7√7.
2/√6=1/3√6.

Then, 3/7√7-1/3√6=(9√7-7√6)21 ---> B.

[hide]The LCD of sqrt 7 and sqrt6 is 42, so multiply the first fraction by 6(sqrt 7)/6(sqrt7) and the second fraction by 7(sqrt6)/7(sqrt 6). Now you have 18(sqrt7)-14(sqrt6) all over 42. Simplifying yields 9(sqrt7)-7(sqrt 6)/21, or answer B.[/hide]

[hide]$\frac{3}{\sqrt7}=\frac{3\sqrt7}{7}$, $\frac{2}{\sqrt6}=\frac{\sqrt6}{3}$. Subtract to get $\frac{9\sqrt7-7\sqrt6}{21}$. [/hide]


a,b,c are positive with a+b+c=1. Prove that a^2 + b^2 + c^2 + 2sqrt(3abc) <= 1

let's take the rhs: 1 = (a+b+c)² = a²+b²+c² + 2(ab+bc+ca), so we can move to another inequality:
 :sqrt: (3abc)  <= ab+bc+ca.
squaring and writing 3 = 3(a+b+c), we have:
3abc(a+b+c) <= a²b² + b²c² + c²a² + 2abc(a+b+c), so
a²b² + b²c² + c²a² >= a²bc + ab²c + abc², which comes from rearrangement or muirhead's.
that's all :)

We can use only: $(ab+bc+ca)^2\geq3abc(a+b+c)$ and then of course $ab+bc+ca\geq\sqrt{3abc}$


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

Just kinda random, but I was wondering if anyone wants to post their goals for next year on stuff (competitions). Just as an example, I'll start...

AMC10: 120+
AMC12: 100+
AIME: 4+
USAMO: N\A (Probably won't make)
ARML Indiv: 5+
ARML Team:Depends on team
ARML Relay:Don't really care
ARML Power:20+
Tokutomi: 15+
SCVMA:10th in whatever division
NMC:20+

I'd like to make it to USAMO.  I've never done AMCs before so I don't really have anything to judge by.

AMC 8-25
AMC 10-Perfect
AMC 12(If I make it)-Perfect
AIME-Just get in

PSIA-Perfect in Math and Science.
TMSCA-Perfect in Math and Science.

....Yeah

AMC 8:  25 (CURSE STUPID MISTAKES BLEARGH)
AMC10: 135+
AMC12: 120+
AIME: 6+
USAMO: 3+ (if i even make it, which is doubtful, especially if I flunk the AIME again)
ARML Indiv: 6+
BAMO:  22+, 8th and lower winner (that actually has a good chance of happening, unlike a lot of the stuff here)
Mathcounts:  40+ chapter, 36+ states, no idea for nats haven't ever tried a test probably wont make it anyways

these are really high... :D
probably wont make most of them
and no im not just saying that
like a certain somebody i know
COUGH ALBERT COUGH

Goals for 07-08:
AMC 10: 140+
AMC 12: 130+
AIME: 8+
USAMO: 5+ (or, if I was [-insert extremely smart person's name-], 43+ :roll: )
ARML (A Team): 8+ NO MORE IDIOTIC MISTAKES
ARML Relay: 6+ (hope I get a good team)
ARML Super Relay: Not make a stupid mistake, GET THE CANDY
ARML Power: Do at least 3 problems
ARML Indiv: 6+
BAMO: 18+
Tokutomi: 18+
SCVMA: Anything in the top ten in whatever division I am

xscapezaer, what is NMC?

I'll be satisfied if one of those goals actually happen.

Hrm.
AMC 10: If I take it, which I probably won't, 144+
AMC 12: 130+
AIME: 8+
USAMO: just get in, I guess. And not get 0.
ARML indiv: 6+
And as Vishal says, get the candy.

amc10-53
amc12-149.704
aime-5pi
usamo-97
arml indiv-9/8

:kneels and bows: ALL HAIL JUNGGI, THE MATH MASTER!!!!

hunter34, who are you? And perfect on 12 is not only rare for middle school (I think), your goal for AIME/USAMO should be at least like 10-12/12-15 (maybe Blue MOP). Wait what were your scores this year?

As for my goals:
Hold Meggy-Meg's hand
Get a hug from Meggy-Meg
Get another hug from Meggy-Meg
Give Meggy-Meg a hug
Give Meggy-Meg another hug
Give Meggy-Meg a kiss

As for my unimportant goals:
AMC 8 - 33 if I get in
AMC 10 - 33 if I get in
AMC 12 - 140 if I get in
AIME - 12 if I get in
USAMO - 17 if I get in
IMO - 33 if I get in

MEGGY-MEGGGGGGGGGGG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[quote="xscapezaer"]@Ubemaya: First of all, ur supposedly "important" goals are rather concerning and personally, I think they should be censored.

And what's the meaning of your "unimportant" goals? A 33 on the AMC 8? A 33 on the AMC 10 :?:  :!:  :huh:[/quote]

Don't take 'em seriously. We're all used to this by now.

@ubemaya,  quite happy just getting into AIME, even happier getting into USAMO
I just got a 25 on AMC 8.

AMC 10: 150
AMC 12: 120
AIME: 8
USAMO: No idea
ARML: 6 on indv, others depend
Tokutomi: 20 + tiebreaker
SCVMA: Top 3
BAMO: 25

Goals for AMC's might be too high, but after a good score the first time, I might pull a 150. Completely bombing the AIME last year of course wasn't wonderful at all, but I'm sure this year will be a lot better. ARML really depends on your team, even the individual might be affected slightly, but I'm hoping for a 6 there. Tokutomi was all but mine last year, had it not been for a mis-read (and then mis-check). Considering I placed 3rd in SCVMA last year and I know I'll do better this year, top 3 is a must. My goal for BAMO might also be too high, but it doesn't hurt to aim high, does it?

AMC 8: 25                   FAILED  :mad:
AMC 10: 140
AMC 12: 120
USAMTS: 50               Probably FAILED
Everything Else: 0

Well, thanks for reviving this thread for some reason...

This year, I'll settle with not failing anything seriously. Actually, I've already missed that goal. zomg.

My goals:
AMC10: >130
AMC12: >120
AIME: just enough to make usamo
USAMO: well we'll see if i make it first

AMC 10: 144+
AMC 12: 126+ (Didn't take, school didn't offer B :mad: )
AIME:  Enough to qualify for USAMO. (6-8)
USAMO:  1+

[quote="serialk11r"][quote="moogra"]AIME: 7 (need a 7)
USAMO: 0[/quote]
Wow people stop being so pessimistic about USAMO. Update:
AIME: 7
USAMO: 12? (probably won't achieve that...)
AMC10A and 12B both failed :([/quote]

The fact that I get in to USAMO is good enough. I get 5-8 on AIME (might be higher if I actually spent the whole time). I can probably manage a 5 at least on USAMO if I get in because I think writing some bs gives some points

0 AIME
yeah...

I just realized it's actually pretty hard to get a 0 on AMC 10/12, since if you randomly guess you may get some right even though guessing is on your side instead of against. Unless you know them all, in which case you would've gotten a 150 if you tried, but no one who could get 150 would try getting 0. Well actually no one at all would try getting 0 but whatever. It's a hypothetical situation.
... -shuts up-

Anyways, seriously, like 7 AIME?

Hmph, goals. 15 on AIME, blue MOP cutoff on USAMO...yeah, so I'm a dreamer.

mmmm
after being deprived of sleep for so long, i expect my performance to have gone down like exponentially in the past few months
whatever
i just want enough to get to usamo, and then that should give me enough time to get more sleep? ehhhh maybe not
but yeah i want to make blue mop
and just to put out a concrete number, i'll aim for a 9 or 10 on the aime
after 10 solved, its gonna be massive checking, and maybe i'll avoid making 8 stupid mistakes?

Needless to say, I kinda failed my AMC goals.
Good news is, I didn't fail enough to make qualifying for USAMO a pain in the ass.
So yeah, still 8 on the AIME, though I feel like I might get 7.

[quote="davidyko"]Good news is, I didn't fail enough to make qualifying for USAMO a pain in the -.
So yeah, still 8 on the AIME, though I feel like I might get 7.[/quote]

Huh? This year's AIME 1 is on March 18th and AIME 2 on April 2nd.

Did you ride a time machine to get there earlier and come back?

No, I meant that my goal is still 8+ on the AIME, but seeing how I've been doing lately, 7 seems more likely.


Would it be possible to post an announcement about an upcoming Math Jam on the corresponding message board?
For example, an announcement telling everyone about the AIME math jam on the AIME message board.
I was browsing the old posts in the MathCounts forum and saw that this had happened previously.
Could like a moderator start that kind of thing up again?
Because I always check the MathCounts forum when I'm on this site, but sometimes forget about math jams, and for this reason forgot about today's MathCounts game thing which I really was interested in attending.
Just a question.

Thanks.

Honestly, this is mostly an issue of prioritizing the time of the administrators.  We might sometimes do this, but don't count on it always happening.

If you have a homepage (or use a social bookmarking system like delicious), then you might want to simply link yourself to the Math Jam schedule and check it once in a while.

We usually post banners that announce Math Jams weeks in advance.  Keeping an eye out for the red banners can help you notice when there are Math Jams coming up soon.

Also, if you click the "Stay Informed" link on the left side of the AoPS page, you can get an e-mail reminder of each Math Jam the day before.


A certain 3 digit number $N$ has the property that $\sqrt{N}$, $N$, and $N^{2}$ are all palindromes. What is the sum of the digits of $N$?

[hide] 99<N<1000, so $9<\sqrt{N}<32$.  Since $\sqrt{N}$ is a palindrome, $\sqrt{N}$ is either 11 or 22.  Thus, N is either 121 or 484.  We know that $11^{4}=121^{2}=14641$, which is a palindrome, so 121 works.  Now, $484^{2}=22^{2}=2^{4}*11^{4}=(14641)(16)$, which is not a palindrome since its unit's digit is a 6 and its ten's digit is clearly not a 6.  Thus, the only value for N is 121, so the sum of N's digits is $\boxed{4}$. [/hide]

[quote="rd5493"][hide]If $N$ is a $3$ digit number, only $121$ and $484$ are possible if the square root of $N$ must be a palindrome too
$121^{2}$ is $14641$
$484^{2}$ is $234256$, so the answer is $121$[/hide][/quote]

[hide="But read what the question is asking, so..."]The answer is $\boxed{1+2+1 = 4}$.[/hide]


let x,y,z be positive real number, with xyz=1, prove

x^3/(1+y)(1+z) + y^3/(1+z)(1+x) + z^3/(1+x)(1+y)>=3/4

i really want to use SOS,but i cant make soemthing like x-y out

[quote="morheli"]let x,y,z be positive real number, with xyz=1, prove

\[ \frac{x^3}{(1\plus{}y)(1\plus{}z)} \plus{}\frac{y^3}{(1\plus{}z)(1\plus{}x)} \plus{}\frac{z^3}{(1\plus{}x)(1\plus{}y)} \ge \frac{3}{4}\]

i really want to use SOS,but i cant make soemthing like x-y out[/quote]
By AM-GM, we have
\[ \frac{x^3}{(1\plus{}y)(1\plus{}z)} \plus{}\frac{y\plus{}1}{8} \plus{}\frac{z\plus{}1}{8} \ge \frac{3}{4}x\]
Adding the similar terms, we get
\[ \sum \frac{x^3}{(1\plus{}y)(1\plus{}z)} \plus{}\frac{x\plus{}y\plus{}z\plus{}3}{4} \ge \frac{3}{4}(x\plus{}y\plus{}z)\]
\[ \Rightarrow \sum \frac{x^3}{(1\plus{}y)(1\plus{}z)} \ge \frac{1}{2}(x\plus{}y\plus{}z) \minus{}\frac{3}{4} \ge \frac{3}{2} \minus{}\frac{3}{4} \equal{}\frac{3}{4}\]
We have done. :)

[quote="morheli"]let x,y,z be positive real number, with $ xyz \equal{} 1$, prove

$ \frac {x^3}{(1 \plus{} y)(1 \plus{} z)} \plus{} \frac {y^3}{(1 \plus{} z)(1 \plus{} x)} \plus{} \frac {z^3}{(1 \plus{} x)(1 \plus{} y)} \ge\ \frac {3}{4}$

i really want to use SOS,but i cant make something like x-y out[/quote]

We have $ P \equal{} \sum \frac {x^4}{x(1 \plus{} y)(1 \plus{} z)} \ge\ \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{x \plus{} y \plus{} z \plus{} 2(xy \plus{} yz \plus{} zx) \plus{} 3}$

$ P  \ge\ \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{\sqrt {3(x^2 \plus{} y^2 \plus{} z^2)} \plus{} 2(x^2 \plus{} y^2 \plus{} z^2) \plus{} 3}$

Must prove that : $ \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{\sqrt {3(x^2 \plus{} y^2 \plus{} z^2)} \plus{} 2(x^2 \plus{} y^2 \plus{} z^2) \plus{} 3} \ge\ \frac {3}{4}$

Let$ a^2 \plus{} b^2 \plus{} c^2 \equal{} t$

$ Ineq \leftrightarrow \frac {t^2}{\sqrt {3t} \plus{} 2t \plus{} 3} \ge\ \frac {3}{4}$

         $ \leftrightarrow 4t^2 \minus{} 6t \minus{} 3\sqrt {3t} \minus{} 9 \ge\ 0$

  This problem is true by $ t \ge\ 3$


[quote=theSA]For any positive $ a_1 ,a_2 ,...,a_n$ prove that  $ \frac {{a_1 }} {{a_2 \plus{} a_3 }} \plus{} \frac {{a_2 }} {{a_3 \plus{} a_4 }} \plus{} ... \plus{} \frac {{a_n }} {{a_1 \plus{} a_2 }} > \frac {n} {4}$   holds.[/quote]
[quote=marin.bancos]Sorry, but the inequality is not a "trivial" one. Have a look at this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=54&t=589&p=1783998&hilit=shapiro+inequality#p1783998]old topic[/url].[/quote]
[url=https://artofproblemsolving.com/community/c893771h1862011p12597664]All Soviet Union MO 1969 [/url]


inequalities
geometry
incenter
circumcircle
geometry proposed

Given $\triangle ABC$ with incenter $I$ and circumradius $R$ prove that
$IA^{a}+IB^{b}+IC^{c}\le 3R^{\frac{a+b+c}{3}}$

I haven't solved this problem yet :blush: , I have a mistake when created it but I still see it is true (by computer), does anybody have a nice solution?


How many zeroes do we write when we write all the integers from 1 to 256 in binary?

The alcumus solution says:

The numbers 1 to 255 are all the $ k$-digit binary numbers with $ k\equal{}1,2,\ldots,8$.
First, let's look at the 8-digit binary numbers $ \overline{1a_{6}a_{5}a_{4}a_{3}a_{2}a_{1}a_{0}}$ . In a number with $ a_{0}\equal{}0$ , each of the 6 digits $ a_{1}, a_{2},\ldots,a_{6}$ can be $ 0$ or $ 1$. Thus the number of 8-digit binary numbers with $ a_{0}\equal{}0$ is $ 2^{6}\equal{}64$. Similarly, the number of 8-digit binary numbers with $ a_{k}\equal{}0$ is 64 for $ k\equal{}1,2,\ldots,6$. Therefore the total number of 0's in all 8-digit binary numbers is $ 64\times 7$.
Similarly we can show the following:
\[ \begin{array}{|c|c|}\hline\text{number of digits}&\text{Total number of zeros}\\ \hline 8 & 64\times 7\\ \hline 7 & 32\times 6\\ \hline 6 & 16\times 5\\ \hline 5 & 8\times 4\\ \hline 4 & 4\times 3\\ \hline 3 & 2\times 2\\ \hline 2 & 1\times 1\\ \hline 1 & 0\times 0\\ \hline\end{array}\]
Finally, the number 256 has 8 zeros in binary. Therefore, the answer is$ (64\times 7)\plus{}(32\times 6)\plus{}(16\times 5)\plus{}(8\times 4)\plus{}(4\times 3)\plus{}(2\times 2)\plus{}(1\times 1)\plus{}(0\times 0)\plus{}8\equal{}\fbox{777}$

SOMEONE HELP! I don't understand at all....The solution makes no sense whatsoever....


Give reason behind your answer. everyone is requested to give their views.

What is the value of $ Arg(i^i)$????

taking the log of $ i^i$ we have $ iln{i}$, thus $ i^i \equal{} e^{ilni} \equal{} e^{i\theta}$ thus $ Arg{i^i} \equal{} ln{i}$

Note that $ i \equal{} e^{i\frac {\pi}{2}}$ so that $ i^i \equal{} (e^{i\frac {\pi}{2}})^i \equal{} e^{ \minus{} \frac {\pi}{2}}$

That is, $ i^i$ is a positive real number, so its argument is $ 0$.

Actually, $ i^i$ is infinitely many positive real numbers: $ i \equal{} e^{i(\pi/2 \plus{} 2\pi n)}$ for any $ n \in {\bf Z}$.

But it can be seen as
$ i^i \equal{}e^{\minus{}(pi)/2}
\equal{}e^{i(i(pi)/2)}
\equal{}e^{j(theta)}$
So the argument must be $ i(pi)/2$
Which makes it undefined.??

As JBL said, there are an infinite number of values that $ i^i$ can take on: $ i^i\equal{}\left(e^{i\left(\frac\pi2\plus{}2\pi n\right)}\right)^i\equal{}e^{\minus{}\left(\frac\pi2\plus{}2\pi n\right)}$, which is real, so $ \arg{i^i}\equal{}0$.

What is the flaw in my argument?????Please tell anyone...

rohitarura: the argument has to be a real number.  The idea is that a complex number $ a \plus{} bi$ can also be represented as $ re^{i\theta}$, where $ r \equal{} |a \plus{} bi| \equal{} \sqrt {a^2 \plus{} b^2}$ and $ \theta$ - the argument -  is the clockwise angle from the +x axis to $ a \plus{} bi$ in the complex plane.  it just doesn't make sense to say that an angle has an imaginary measure.  So while your idea works mathematically, it ignores the definition.

Side note:
There are an infinite possible arguments for a number, differing by multiples of $ 2\pi$, but usually we restrict ourselves to $ [0,2\pi)$.  Real number, then, are given arguments of $ 0$ or $ \pi$.


number theory
least common multiple
group theory
superior algebra
superior algebra unsolved

[size=150]Suppose $ G$ is a finite cyclic group of given order $ n.$

Show that $ G \cong \oplus ^{k}_{i\equal{}1} \mathbb{Z}_{n_i}$ where the $ n_i$ are prime powers.[/size]

the $ n_i$ are powers of distinct primes and $ n_1\cdots n_k \equal{} n$. all we need to do is find an element in $ \oplus_{i \equal{} 1}^k \mathbf{Z}_{n_i}$ that has order $ n$. such an element is $ (1,...,1)$ since the order of $ 1$ in each of $ \mathbf{Z}_{n_i}$ is $ n_i$, and so the order of $ (1,...,1)$ is $ \mbox{lcm}(n_1,...,n_k) \equal{} n$ (since all the $ n_i$ are pairwise relatively prime).


Duke college geometry Harvard MATHCOUNTS articles calculus

[quote="ra5249"]I got
7th GRADE:
Math - 800 (No, duh - this is aops)
Crit. Reading- 650 (joy joy i suck at crit. reading)
Writing - 690

What did everyone else get?[/quote]

don't feel bad

I had 740M-510CR-590W 7th grade

if you think your critical reading sucks, just look at mine.

I suggest this get moved to whatever forum last had a thread like this.

740 cr
710 m (wowo i suck... missed 4)
790 w

well when I took SAT (before it became 3 parts), it was the first year they has free response on the math. I think I did it wrong, because when I got back my report from CollegeBoard or whoever, it said I got every multiple choice question right except one, but none of the free response, leaving me with 660, even though I know I solved all the free response right.

When I took it in 8th grade (or maybe summer before that) I think I got 1470 - 790 math 680 c.r.
yup pretty bad :oops: so I'll still have to take the new ones... maybe in a couple years...

umm

700 math
590 reading

haha, i think my date was the last of the old SATs

I haven't taken SAT yet (will take it next Saturday), but on the PSAT I had:

m: 76
cr: 70
w: 72

(still don't know if I'm a national merit finalist, semi-finalist, or whatever)

[quote="Danbert"]I took my very first (and only) SAT in Jan 2006 as a high school junior.  I got 760/800/770.

Why start studying too early?  It seems kind of ridiculous to me to start practicing in 7th grade...  You're probably going to do quite well.  Don't worry about it![/quote]
In 7th grade there is a special Talent Identification program hosted by Duke for students who excel on the SAT.
[url=http://www.tip.duke.edu/]here[/url] is their website

[quote="Danbert"]I took my very first (and only) SAT in Jan 2006 as a high school junior.  I got 760/800/770.

Why start studying too early?  It seems kind of ridiculous to me to start practicing in 7th grade...  You're probably going to do quite well.  Don't worry about it![/quote]

Also, CTY requires SAT scores...dunno why anyone would opt for that :wink:

haha i took the march SAT and got a 2020 with a 790 in math :)

i was pretty surprised at the difficulty.. i thought it would be kinda like the AMCs, but it was of the contrary.. calculator almost unnecessary

630 CR
790 M
700 W
in seventh grade...

680 CR
780 M (WTF?!)
710 W
in eighth grade.

I hate reading. I hate stupid mistakes.

7th grade: 630 CR- 780M-620W
8th grade: 780CR-800M-660W

AAAH i hate writing.

[quote="Danbert"]I took my very first (and only) SAT in Jan 2006 as a high school junior.  I got 760/800/770.

Why start studying too early?  It seems kind of ridiculous to me to start practicing in 7th grade...  You're probably going to do quite well.  Don't worry about it![/quote]

My 7th grade scores- 790M/540V

I taught the SAT this past summer, and I think it's fine to start practicing in 7th grade. it tests your experience with types of questions only seen on the SAT and nowhere else, so being in contact with it often doesn't hurt. Of course, it takes a little work to get cr up to 800, but it's not too bad.

last year

seventh grade

math-770(failed)
writing-590 failed more
critical reading-540 failed even more.

hehe im taking it next week...

projected maybe 800 math, 750 CR, 800 W (if i get 10+ essay)...

but ill probably do much worse because i never live up to my goals.

7th grade- 730 math
520 CR

8th grade- 780 math
580 cr

9th grade- 780 math
700 cr
780 writing

ahh only got 2240, oh well

btw, stanford your in 11th grade right?  i saw your rsi post

[quote="hadasah"]I failed that day.....for some reason I got this essay topic that I didn't have an clue how to write.....wtver. These were my scores from 7 grade:

CR: 590 blegh
W: 610 ughhhhhhhhhhhhhhh  :mad:
Math: 790

I accidentally missed an easy math problem...must have bubbled wrong or sthng

none of the math sections took me ten minutes...all of them I finished in like 6 or 7 minutes....but then I went back and checked them and that made the time rise to 10 minutes or so[/quote]
Sadly I got far worse than that in 8th grade :(
Taking it as a junior in December though, hopefully I'll get >2300 on the first try so I won't have to waste 4 more hours taking another one.

800 math 800 writing (ooh, that was lucky) and 760 CR. :( Not that good.....

wait is that a joke. you got 2360 in 5th GRADE!?????

Wow, if true, that's absolutely amazing.

7th grade:
800math, 640 ish reading, 550 ish writing.

math 800, reading 520, writing 630 (7th grade)

You guys make me feel so stupid.
770 M, 510 W, 490 CR (7th grade)

770 math, 620 writing, 680 reading

on the math section I got this grid-in problem wrong and I have no idea how. I did something similar at mathcounts. maybe it's related...

how do you get a 790/780 on the math section?

PSAT-Freshman
Math-70 (probably can get 80 now)
Critical Reading-47 (hate it...)
Writing-46 (hate it also...)

x_x I find a trend where most people who are good at math really aren't as good in English stuff.

Math SAT Subject Test I-770 (does not have a curve compared to other subject tests and 50 questions in 1 hr, some wording made it confusing :[ but I used as practice because I'm going to take Math II next year)


trigonometry
geometry
calculus
calculus computations

Prove that if $ x \in  {(} 0, \frac{\pi}{2}  {)}$ then $ x < \tan x$.

hello
see the answer of second question::
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=248128
thank u

Let $ w\equal{}C(O,1)$ be the trigonometrical circle and $ M(\sin x,\cos x)\in w\ ,$ where $ x\in\left(0,\frac {\pi}{2}\right)\ .$ Denote $ A(1,0)\in w$ and $ T\in OM$ for which $ TA\perp OA\ .$ Observe that $ T(1,\tan x)$ , the length of the $ \mathrm{arcAM}$ is equally to $ x$ and the area of the triangle $ OAT$  is greater than the area of the (circular) $ \mathrm {sectorOAM}$ , i.e. $ \frac 12\cdot 1\cdot \tan x\equal{}[OAT]>\mathrm{sector\ OAM}\equal{}\frac 12\cdot 1\cdot x\ .$


Concerning electricity, when do you use the left-hand rule vs. right hand rule, and whats the difference?

I just know basic electricity and you use left hand rule in case of motors(Electic to mechanical) and righ hand for mechanical to electrical.

can you give an example of a problem that would involve knowing these things, because the only time I have ever used the right hand rule is in determining the direction of the force in a magnetic field.

[color=blue]The direction of Lorentz's force is also determined by the left hand rule[/color]

Ok, I'll explain the right and left hand rules for those who don't know (I should hope that the number of these is rather small :) )

First of all, let's agree that the [u]electric current[/u] is the motion of [i]positive[/i] charges. This is important to determine right in the beginning. As far as I know, there are high schools courses which regard electric current as the motion of negative charges (electrons), so you must learn to differentiate between these two cases.

Well, the right-hand rule is one of the most commonly known rules in physics because it is used to determine the direction of the vector product of two vectors. In electromagnetism, the right-hand rule is often used to determine the direction of the magnetic field induced by an electric current or vice versa. How do you use this rule? It's easy, but some people find it incredibly tricky to master in the beginning. When you want to determine the direction of the magnetic field induced by a current of positive charges passing through a conductor, all you have to do is take your right hand, outstretch the thumb, point it in the direction of the current, and then grasp the conductor with the remaining four fingers. The direction in which the fingers curl around the conductor (clockwise/counterclockwise or, speaking in a stricter mathematical formalism, negative/positive) corresponds to the direction of the magnetic vortex induced around the conductor.

Similarly, when you have a contour with a current of positive charges flowing through it, you can determine the direction of the magnetic field induced by this current by curling the fingers of your right hand so that the direction of the curling is the same as the direction of the current through the contour. The outstretched thumb will now show the direction of the magnetic field. I hope that I have been clear. If you still can't understand and think that I'm a sloppy teacher :) I advise you to Google these items. There are dozens of sites with precise information on the right-hand rule.

Now, when do you use the left-hand rule? This one is even easier than the right-hand one! The left-hand rule is used to determine the direction of Ampère's force. Suppose that a conductor (through which a current of positive charges flows) is moving through a magnetic field perpendicular to the plane in which the conductor moves. (If the magnetic field is not perpendicular to this plane, just take its perpendicular component.)
1) Point your left-hand thumb in the direction of the magnetic field;
2) Point your left-hand index-finger the direction of the current flowing through the conductor;
3) The outstretched left-hand middle finger will point in the direction of Ampère's force acting on the conductor. (Formally, Ampère's force is given by $\vec F_A = I (\vec l \times \vec B)$ where the conductor length vector $\vec l$ points in the direction of the current flowing through the conductor).

Finally, if you are discussing currents of negative charges, all you have to do is to change the hand you're using for each rule (i.e. do the thumb-index finger-middle finger trick with your right hand). [i]Or[/i], you can always use the regular rules and then just reverse the direction you've obtained.

I hope that I've been of some help. I'll try to post some problems illustrating the use of these rules, but I don't think I'll have the time to draw it all and post the drawings. That's what Google's for ;)