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Second one.

From basic physics, Vf^2 = Vx^2 + Vy^2

Substitute in Vx = dx/dt and Vy = dy/dt (i.e, the horizontal and vertical components of movement) and take square roots and you're done.

[quote="JS1527"]Yes, blahblah, I already understand how the second one is derived (vector sum of x and y component velocities).  However, I don't see how this formula differs from the first formula, or how the first formula is derived.  My main question is what makes the two formulas different?  To me it seems like you have 2 formulas that supposedly give you the same thing, but they yield different numerical results.[/quote]

First formula:

From the Chain Rule:

dy/dt = dy/dx*dx/dt

If dx/dt != 0, we can 'divide' the two.

[quote="JS1527"]In my book they give two formulas:
[tex]\displaystyle\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]

And:

[tex]v=\sqrt{{(\frac{dy}{dt})}^2+{({\frac{dx}{dt}})}^2}[/tex]

Now, I understand how the first one is derived, though the second one my prof gave the explanation that [tex]\frac{a}{b}=\frac{\frac{a}{c}}{\frac{b}{c}}[/tex] which I don't really buy since dy and dx and dt are not seperate entities, so they don't have the same properties as simplifying fractions.

But anyway, I don't understand how these formulas are different and when to use each one.  Instaneous velocity is the first derivative dy/dx, so I don't really get how they are different.  Obviously they yeild different results though.

Thanks for your help![/quote]

The problem that you're having is with the interpretation of what the variables mean. dy/dx is the "instantaneous velocity" of a particle when y is its position and x is your measure of time. In this situation the motion is linear (i.e. back and forth in one dimension).

But it's :sqrt:[(dx/dt)2 + (dy/dt)2] when the motion is planer (i.e. two dimensional). In this situation x and y are [i]directions[/i] of motion. dx/dt is the velocity in the x-direction, and dy/dt is the velocity in the y-direction. Always try to keep in mind what the variables [i]mean[/i] within the formulas - not just the formulas themselves.

Also you are right and your professor is wrong when he says that you can "cancel" the tops and bottoms of derivatives like that. It turns out to always work (which is a big reason why the notation was created that way in the first place) - but technically, nothing is being "cancelled." This is one of the most widely discussed topics among Calculus teachers - as to whether we should teach students this way or not. There was an excellent article about this a few years ago in the American Mathematical Monthly journal.

[quote="JS1527"]Okay, so in what situations would you use each formula?  Is v= only used in word problems? Gauss, thanks for the clarification...[/quote]

Not necessarily.  For example:
Find the speed of a particle with parametric equations:

x = e^(t) - 1
y = sin(t)^2
z = cos(t)^2

at t=ln(10).

[tex]\displaystyle\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex] tells you what [tex]\displaystyle\frac{dy}{dx}[/tex] is -- that is, how much y changes when x changes a little bit.  This is equal to the instantaneous velocity ONLY if y is the position with respect to x.  In parametric equations, however, this is not the case: position is given by (x, y) with respect to t.  What dy/dx tells you in that case is simply the direction the particle is heading in: for every little bit it moves to the right, it moves so much up.

The square-root formula is used whenever you have a particle moving in a plane, with its position defined as a function of t.  The other formula only applies when you have a function moving back and forth on a line, with its position, y, defined as a function of x.

As stated above, when x=t for time, dy/dx does represent linear or one-dimensional velocity.  However, it helps to remember that dy/dx is a broader concept than just velocity.

Of course, dy/dx represents the "slope of the tangent line" to the graph of y = f(x).  But also dy/dx represents any "instantaneous rate of change of y as x changes".  From economics, if x = # of widgets produced and y = total cost to produce x widgets, then dy/dx = 'marginal cost' , that is it estimates the cost to produce just the xth widget.    Or, if x = length (in inches) of a fish and y = its weight (poiunds), then dy/dx measures the rate of growth of the fish in terms of pounds per inch.

Thanks, this is more clear now, thanks for the replies.

Blahblah:

v= :sqrt: (dy/dt)^2+(dx/dt)^2+(dz/dt)^2 )

dy/dt = d/dt sin(t)^2 = 2cos(t)sin(t)
dx/dt = d/dt e^t-1 = e^t
dz/dt = d/dt cos(t)^2 = -2cos(t)sin(t)

v=sqrt( 4cos^2(t)sin^2(t)+e^2t)

t=ln10

v  :approx: 10.0


Inequality
IMO Shortlist
optimization
maximization
IMO Shortlist 1972

The least number is $m$ and the greatest number is $M$ among $ a_1 ,a_2 ,\ldots,a_n$ satisfying $ a_1 \plus{}a_2 \plus{}...\plus{}a_n \equal{}0$. Prove that
\[  a_1^2 \plus{}\cdots \plus{}a_n^2 \le\minus{}nmM\]

[quote="brian_gold"]The least number is m and the greatest number is M among $ a_1 ,a_2 ,...,a_n$ satisfying $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} 0$
Prove $ a_1^2 \plus{} ... \plus{} a_n^2 \le \minus{} nmM$[/quote]

  $ \sum {(x_i\minus{}m)(x_i\minus{}M)}\plus{}\sum {(y_i\minus{}m)(y_i\minus{}M)}\le 0$

What are $ x_i$ and $ y_i$ supposed to be?

[quote="Marius Mainea"][quote="brian_gold"]The least number is m and the greatest number is M among $ a_1 ,a_2 ,...,a_n$ satisfying $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} 0$
Prove $ a_1^2 \plus{} ... \plus{} a_n^2 \le \minus{} nmM$[/quote]

  $ \sum {(x_i \minus{} m)(x_i \minus{} M)} \plus{} \sum {(y_i \minus{} m)(y_i \minus{} M)}\le 0$[/quote]

   $ \sum {(a_i \minus{} m)(a_i \minus{} M)}\le 0$

The $ a_i$ part is right, but I don't know where you get $ b_i$ from.


Hi,
Any one can point me to good logarithm  documention site /book or what ever as long as it going to help me to pass my final exam on LOGARITHM ?

thanks

Hi,
What is I need it some good problms with solution.
thnaks

You should check [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25616]this[/url] out.

Also, search the forum. Theres been a lot of different problems with logs.

Finally, you could google it.


If $ x_1, x_2 , x_3$ are the solutions of $ 2x^3 \minus{} 3x^2 \minus{} 50x \minus{} 24 \equal{} 0$ and $ \frac{x_1}{2}\equal{}4x_2\equal{}\minus{}\frac{x_3}{3}$, then $ 2x_1 \plus{} 4x_2 \plus{} 3x_3 \equal{}$

$ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 10$


Aya jaygashti manande $a_{1}, ... , a_{n}$ az adade $1,2,...,n$ voojood darad be tori ke adade zir dastgahe kamele mandeha be peymaneye $n$ bashand ?
$a_{1}, a_{1}+a_{2}, ... , a_{1}+a_{2}+...+a_{n}$

not sure if im understanding the question correctly,
but not all the elements of the set {a1,a1+a2,...,a1+...+an} can be different modulo n. since n=ai for some i, so at least one term will repeat.


I came up with this when doing another problem, but I don't know how to prove it (there was an easier solution to the problem). It may have already been posted, but I couldn't find it using the search function.

Show that for all acute angles $ x,y$,

$ \dfrac{\sin{x}}{\sin{y}}\plus{}\dfrac{\cos{x}}{\cos{y}}\equal{}2\cos(x\minus{}y)$.

It may be true for non-acute angles too, I'm not sure.

hello, after some calculations i have got:
$ \frac {\sin(x)}{\sin(y)} \plus{} \frac {\cos(x)}{\cos(y)} \minus{} 2\cos(x \minus{} y) \equal{} 2\cot(2y)\sin(x \minus{} y)$.
Substituting $ x \equal{} \frac {\pi}{3}$ and $ y \equal{} \frac {\pi}{12}$ we get
$ 2\cot\left(\frac {\pi}{6}\right)\sin\left(\frac {\pi}{3} \minus{} \frac {\pi}{12}\right) \equal{} \sqrt {6}$ and this ist not zero.
Sonnhard.


let a;b;c in R; prove that:
$ \frac{a\plus{}3}{(a\minus{}1)^2}\plus{}\frac{b\plus{}3}{(b\minus{}1)^2}\plus{}\frac{c\plus{}3}{(c\minus{}1)^2}\geq\frac{47}{16}$

the people from Vietnam can solve at here [url]http://mathsthpt4t.vnbb.com/viewtopic.php?f=34&t=14[/url]

[quote="litbon"]let a;b;c in R; prove that:
$ \frac {a \plus{} 3}{(a \minus{} 1)^2} \plus{} \frac {b \plus{} 3}{(b \minus{} 1)^2} \plus{} \frac {c \plus{} 3}{(c \minus{} 1)^2}\geq\frac {47}{16}$
[/quote]
Try $ a\equal{}b\equal{}c\equal{}\minus{}3.$ :wink:

[quote="Mathias_DK"][quote="litbon"]let a;b;c in R; prove that:
$ \frac {a \plus{} 3}{(a \minus{} 1)^2} \plus{} \frac {b \plus{} 3}{(b \minus{} 1)^2} \plus{} \frac {c \plus{} 3}{(c \minus{} 1)^2}\geq\frac {47}{16}$

the people from Vietnam can solve at here [url]http://mathsthpt4t.vnbb.com/viewtopic.php?f=34&t=14[/url][/quote]
I think you forgot $ abc \equal{} 1$ :wink:[/quote]

yes; it is:
let a;b;c in R and abc=1; prove that:
$ \frac {a \plus{} 3}{(a \minus{} 1)^2} \plus{} \frac {b \plus{} 3}{(b \minus{} 1)^2} \plus{} \frac {c \plus{} 3}{(c \minus{} 1)^2}\geq\frac {47}{16}$


Here are the problems:

1) A fair twenty-faced die has 19 of its faces numbered from 1 through 19 and has one blank face. Another fair, twenty-sided die has 19 of its faces numbered from 1 through 8 and 10 through 20 and has one blank face.
When the two die are rolled, what is the probability that the sum of the two numbers facing up will be 24?
Express your answer as a common fraction.

2) At how many points do the graphs of the equations $ y\equal{}x^3$ and $ y\equal{}x^2\plus{}4x$ intersect?

3)If $ \pi$ =3.1415926..., what is the exact value of $ |\pi \minus{} 3.14|$ + $ |\pi \minus{} \frac{22}{7}|$? Express your
answer as a common fraction

Thanks in advance

hello, to 2), you must solve the equation $ x^3=x^2+4x$ this is equivalent with $ 0=x(x^2-x-4)$, form here we get the solutions $ x_1=0$ or $ x{}_2{}_3=\frac{1}{2}\pm\sqrt{\frac{1}{4}+\frac{16}{4}}$. This gives
the intersection-points $ x_1=0,y_1=0$,$ x_2=\frac{1}{2}(1-\sqrt{17}),y_2=\frac{13}{2}-\frac{5}{2}\sqrt{17}$, $ x_3=\frac{1}{2}(1+\sqrt{17}),y_3=\frac{13}{2}+\frac{5}{2}\sqrt{17}$
Sonnhard.

Thank you so much. Big help. Now i need to wait for the answer to problem 1...

Obviously, there are $ 20 \times 20$ or $ 400$ ways to form the dice without any restrictions.

Now, we use casework:

Obviously, both numbers must be greater than 3 (do you see why?).

So listing the possibilities, we have:

(4,20)
(5,19)
(6,18)
(7,17)
(8,16)
(9,15)
(10,14)
(11,13)
(12,12)
(13,11)
(14,10)
(15,9)
(17,7)
(18,6)
(19,5)

We have listed $ 15$ possibilities, so the answer is $ \frac{15}{400}$, or $ \boxed{\frac3{80}}$.

If you have any further questions on the 2008 school sprint round, use this link from the MathCounts website: [url]http://mathcounts.org/Document.Doc?id=222[/url]

It should provide both the solutions and answers to all questions. For future reference, the questions you have asked are listed as 22, 29, and 30 respectively.

Ernie I believe you should have included (16,8) instead of (15,9), but it doesn't make a difference.

Well, i suppose i can use the mathcounts website thingy from now on. Thanks alot

[quote="Aequilipse"][quote="5849206328x"][hide="another solution for 2"]
The two lines meet when $ x^3\equal{}x^2\plus{}4x$.  This is a cubic, so it has $ \boxed{3}$ solutions.  You still have to check to make sure there are no double roots, but $ x^2\minus{}x\minus{}4$ obviously isn't a perfect square
[/hide][/quote]
How'd you get $x^2-x-4$?[/quote]

Dividing the first equation by x, then subtracting to get all the terms on the same side.


Show that:

$ \sin{3 \alpha}\sin^3{\alpha} \plus{} \cos{3 \alpha}\cos^3{\alpha} \equal{} \cos^3{2\alpha}$

hello, after a addition formula we have $ \sin(3x)\equal{}4\sin(x)\cos(x)^2\minus{}\sin(x)$ and $ \cos(3x)\equal{}4\cos(x)^3\minus{}3\cos(x)$ and $ \cos(2x)^3\equal{}8\cos(x)^6\minus{}12\cos(x)^4\plus{}6\cos(x)^2\minus{}1$. From here we have
$ \sin(3x)\sin(x)^3\plus{}\cos(3x)\cos(x)^3\equal{}(4\cos(x)^2\minus{}1)\cdot(\sin(x)^2\minus{}\cos(x)^2\plus{}1)\cdot(\sin(x)^2\plus{}\cos(x)^2\minus{}1)\equal{}0$.
Sonnhard.


Results for most of the olympiads are out. Anyone here who qualified ?

http://www.hbcse.tifr.res.in/olympiads/

Nope. Was expecting something in Astronomy though. Still, IMOTC is better than nothing  :D
 $\text{Ashwath}$

well, I have been selected in Physics and Chemistry olympiads whose training camp dates clash with my IMOTC senior batch dates from 10th to 31st of may. I think i can attend multiple classes.


ABCD is a parallelogram. K is the circumcircle of ABD. The lines BC and CD meet K again at E and F. Show that the circumcenter of CEF lies on K

Assume $\angle BAD\ge 90^{\circ}$ (that's how I drew it, and the other case is very similar). We have $\angle EDF=\angle EDA-\angle CDA=\angle BAD-\angle ABC=\angle ECF-(\pi-\angle ECF)$, and from here the conclusion follows easily.


Find all function such that ${f}:{Q}\rightarrow{R}$ and for $\forall{x,y}\in{Q}$ ,$f(xy)=f(x)f(y)$ and${f(x+y)}\le{max(f(x),f(y))}$

Solutions are $f\equiv 0$, $f\equiv 1$, $f(x)=\text{sgn}(x)=\{\begin{array}{rl}1&\text{for }x>0\\0&\text{for }x=0\\-1&\text{for }x<0\end{array}$, $f(x)=\{\begin{array}{rl}0&\text{for }x=0\\1&\text{for }x\not =0\end{array}$
and for each prime $p$ and $\alpha\in \mathbb{R}_{+}$ the $p$-adic valuation
$f(x)=|x|_{p}^\alpha$, where $|x|_{p}=p^{-m}$ for $x=p^{m}{b\over c}$ with $m,b,c\in\mathbb{Z},\ p\nmid bc$, and $|0|_{p}=0$.


The number $ 222222$ can be factored into the product $ 37037\cdot6$. What is the product of $ 37037$ and $ 27$?

Please read the stickies.

Solution:

We are given that $ 222222\equal{}37037\cdot6$ so $ 111111\equal{}37037\cdot3$. Since $ 27\equal{}9\cdot3$ we have $ 37037\cdot27\equal{}37037\cdot9\cdot3\equal{}111111\cdot9\equal{}\boxed{999999}$

Another way to understand:
[hide]
$37037\cdot27 = 37037\cdot6\cdot\frac{27}{6} = 37037\cdot6\cdot\frac{9}{2}$
We are given $37037\cdot6 = 222222$, so we can substitute, and we receive
$222222\cdot\frac{9}{2}$
Trivially, $222222\div2= 111111$. By the same token, $111111\cdot9 = 999999$
[i]The solution is $\boxed{999999}$
[/i][/hide]


Let $P(x) = x^4 + ax^3 + bx^2 + cx + d$ and $Q(x) = x^2 + px + q$be two real polynomials. Suppose that there exista an interval $(r,s)$ of length greater than $2$ SUCH THAT BOTH $P(x)$ AND $Q(x)$ ARE nEGATIVE FOR $X \in (r,s)$ and both are positive for $x > s$ and $x<r$. Show that there is a real $x_0$ such that $P(x_0) < Q(x_0)$

This means that P and Q has  r and s roots so Q divide P
Lets write P = QR
Since P and Q hve the same sign inside and outside the roots r and s so R must be positive.
Supose that for every x: P(x)-Q(x) geq 0 ; so is Q(R-1)
R-1 has the same sign as Q inside and outside the roots r and s;so r and s are roots of R-1
Put R(x)= x^2 + ux + v
We have : R(r) = R(s)=1
Fromhere we deduce that R(x)=x^2 -(r+s)x+rs+1
The discriminat is (r-s)^2 - 4 which is positive since the abs value of r-s is greater than 2
This means that R change the sign which contradicts  the fact that  R MUST BE positive.

[quote="gouthamphilomath"]This was asked again in $Indian$ $TST$ $2010$.[/quote]

With a minor change that

[quote="New question"]Let \[P(x) = x^4 + ax^3 + bx^2 + cx + d\] and \[Q(x) = x^2 + px + q\] be two real polynomials. Suppose that there exists an interval $(r,s)$ of length greater than $2$ such that both $P(x)$ and $Q(x)$ are negative for $X \in (r,s)$ and both are [b][i][u]NON-NEGATIVE[/u][/i][/b] for $x > s$ and $x<r$. Show that there is a real $x_0$ such that $P(x_0) < Q(x_0)$[/quote]

So you have to check those cases too (As if that takes a huge effort! :P )


[quote="silenthero_k09 tặng Huyền "]  [i]Let[/i] $\large a,b,c>0$[i].Prove that[/i]
...............
$\large(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a})^{2}+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{21}{4}$
...............

 [/quote]
[quote="Dành tặng em yêu "]
My Solution  :D
Vì $\large a,b,c>0$ nên chúng ta có quyền đặt./ $\large\frac{b}{a}=x,\frac{c}{b}=y, \frac{a}{c}=z$.Trong đó $\large x,y,z>0$ và $\large xyz=1$.Vậy bài toán chuyển thành chứng minh bất đẳng thức sau:/
cho $\large x,y,z>0$ và $\large xyz=1$
C/m $\large(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+y})^{2}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge \frac{21}{4}$
Vì vai trò của $x,y,z$ là bình đẳng nên không mất tính tổng quát ta có thể giả sử $\large x \leq y \leq z$ suy ra $\large x \leq 1, yz\ge 1$
Ta sẽ chứng minh bất đẳng thức này bằng phương pháp dồn biến
Đặt $\large f(x,y,z)$ là biểu thức cần chứng minh .Khi đó ta chứng minh lần lượt các bất đẳng thức sau:
$\large f(x,y,z)\ge f(x,\sqrt{yz},\sqrt{yz})$ [b](1)[/b]
$\large f(x,\sqrt{yz},\sqrt{yz})\ge\frac{21}{4}$ [b](2)[/b]
C/m [b](1)[/b] $\large f(x,y,z)\ge f(x,\sqrt{yz},\sqrt{yz})$
Ta có $\large f(x,y,z)-f(x,\sqrt{yz},\sqrt{yz})$
$\large =(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+y})^{2}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-(\frac{1}{1+x}+\frac{2}{1+\sqrt{yz}})^{2}-\frac{1}{x}-\frac{2}{\sqrt{yz}}$
Ta sẽ chứng minh
[b](+)[/b] $\large \frac{1}{1+y}+\frac{1}{1+z}\ge\frac{2}{1+\sqrt{yz}}$
[b](++)[/b] $\large =\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}$
Chứng minh[b] (++)[/b] khá đơn giản.Theo [b]CauChy[/b] :lol:
Bất đẳng thức [b](+)[/b]
\[\Leftrightarrow\frac{1}{1+y}+\frac{1}{1+z}-\frac{2}{1+\sqrt{yz}}\ge 0 \]
\[\Leftrightarrow \frac{(\sqrt{yz}-1)(\sqrt{y}-\sqrt{z})^{2}}{(1+\sqrt{yz})(1+y)(1=z)}\ge 0 \]
Nhưng bất đẳng thức này hiển nhiên đúng vì $\large yz\ge1$
Từ đây ta suy ra được $\large f(x,y,z)\ge f(x,\sqrt{yz},\sqrt{yz})$
 Do đó theo định lý " Dồn biến không xác định-[b]UMV[/b]"( [b]Undefined Mixing Variables[/b]) ta chỉ cần chứng minh trong trường hợp $\large y=z$.
Với chú ý rằng $\large\sqrt{\frac{1}{x}}=\sqrt{yz}$(do $xyz=1$)
cuối cùng chỉ cần chứng minh

 $f(x,\sqrt{\frac{1}{x}},\sqrt{\frac{1}{x}})\ge\frac{21}{4}$(với $x \leq1$ )
Chúng ta có  $\large f(x,\sqrt{\frac{1}{x}},\sqrt{\frac{1}{x}})-\frac{21}{4}$
$\large=(\frac{1}{1+x}+\frac{2\sqrt{x}}{\sqrt{x}+1})^{2}+\frac{1}{x}+2\sqrt{x}-\frac{21}{4}$
$\large=((\frac{1}{1+x}+\frac{2\sqrt{x}}{\sqrt{x}+1})^{2}-\frac{9}{4})+(\frac{1}{x}+2\sqrt{x}-3)$
$\large=\frac{(\sqrt{x}-1)(\sqrt{x}-1)^{2}(7x\sqrt{x}+3x+9\sqrt{x}+5)}{4(x+1)^{2}(\sqrt{x}+1)^{2}}+\frac{(1-\sqrt{x})^{2}(2\sqrt{x}+1)}{x}$
$\large=(\sqrt{x}-1)^{2}(\frac{(\sqrt{x}-1)(7x\sqrt{x}+3x+9\sqrt{x}+5)}{4(x+1)^{2}(\sqrt{x}+1)^{2}}+\frac{2\sqrt{x}+1}{x})$
chú ý rằng $\large=(\frac{(\sqrt{x}-1)(7x\sqrt{x}+3x+9\sqrt{x}+5)}{4(x+1)^{2}(\sqrt{x}+1)^{2}}+\frac{2\sqrt{x}+1}{x})\ge 0$ (do $x\leq1$)
Từ đây suy ra được điều phải chứng minh.Dấu đẳng thức xảy ra khi và chỉ khi do $x=y=z=1$ hay là $a=b=c$
We are done  :P
[/quote]

nice but verry tall
if u see that $\sum_{cyc}\frac{a}{b+c}\geq\frac{3}{2}$(dont tell me how to prove it )  :lol:  and  use $AM-GM$ u find $\sum_{cyc}\frac{a}{b}\geq3$ add and  u get the desired ineq . :wink:


1) Use the division algorithm to show that very odd integer is either of the form $ 4k \plus{} 1$ or $ 4k \plus{} 3$.

2.) Prove that if $ a$, $ b$ are odd integers, $ 8|(a^2 \minus{} b^2)$

[quote="guile"]1) Use the division algorithm to show that very odd integer is either of the form $ 4k + 1$ or $ 4k + 3$.

2.) Prove that if $ a$, $ b$ are odd integers, $ 8|(a^2 - b^2)$[/quote]

1] Let $ P$ be an odd number the divison algorithm tell us that:

     $ P=4k+r$, with $ 0\leqq<4$ but if $ r=0$ or $ r=2$ $ P$ will be divisible by $ 2$, $ P$ won't be even.
     So we must have $ r=1$ or $ r=3$

2] $ a^2-b^2=(a-b)(a+b)$; using the result in 1] we see therr are four cases, verifying each we see that if $ a$ and $ b$ are both of the form $ 4k+1$ or $ 4k+3$ $ 4|a-b$ and $ 2|a+b$; if there is one from $ a$ and $ b$ of the form $ 4k+1$ and the other of the form $ 4k+3$ $ 2|a-b$ and $ 4|a+b$, always $ 8|a^2-b^2$

 :)


Related to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=285788]this one[/url].

Consider again the polinomial $ (1 \minus{} x)^5(1 \plus{} x)^5$.
Find the sum of all coefficient next to $ x^{3i \plus{} 4}$, where $ i \in \mathbb{N}$. :D

$ (1\minus{}x)^5(1\plus{}x)^5\equal{}(1\minus{}x^2)^5$

The general term is $ T_{k\plus{}1}\equal{}(\minus{}1)^kC_5^kx^{2k}$

For $ i\equal{}0$ the coefficient of $ x^4$ is $ 10$.

For $ i\equal{}1$ the coefficient of $ x^7$ is $ 0$.

For $ i\equal{}2$ the coefficient of $ x^{10}$ is $ \minus{}1$

The sum is $ 9$

I have no suspet there was such a easy way.

[i]Consider now the polinomial $ (1 \minus{} 2x)^{50}(1 \plus{} x)^{50}$.
Find the sum of all coefficient next to $ x^{3i \plus{} 3}$, where $ i \in \mathbb{N}$. [/i]


$ \triangle ABC$ $ \implies$ $ \{\begin{array}{c} \frac {(b - c)^2}{b^2 + c^2} + \frac {(a - b)(a - c)}{bc}\ge 0\ . \\
 \\
\frac {bc}{b^2 + c^2} + \frac {a(p - a)}{bc}\le 1\ .\end{array}$

[color=darkblue][b]Remark.[/b] The upper inequalities are equivalently !

If $ (a - b)(a - c)\ge 0$,  then these are evidently

because in this case ${ \{\begin{array}{c} \frac {bc}{b^2 + c^2}\le\frac 12\ . \\
 \\{
\frac {a(p - a)}{bc}\le\frac 12}\ .\end{array}$

The problem is when $ (a - b)(a - c) < 0\ !$[/color]

[hide="A bit help."]
http://www.mathlinks.ro/Forum/viewtopic.php?t=172031 [/hide]

OK. Here is the solution, Virgin Nicula

The inequality is equivalent to (with the assumption that $ b\ge a\ge c$)

\[ \frac{bc}{b^2\plus{}c^2} \ge \frac{(b\minus{}a)(a\minus{}c)}{(b\minus{}c)^2}\]
or
\[ 1\minus{}\frac{(b\minus{}c)^2}{b^2\plus{}c^2} \ge \frac{2(b\minus{}a)(a\minus{}c)}{(b\minus{}c)^2}\]

According to the above form, using Ravi and a little bit idea about the entirely mixing variable, we can assume that $ b\equal{}a\plus{}c$. In this case, the inequality become very simple

\[ c(a\plus{}c)a^2\minus{}c(a\minus{}c)((a\plus{}c)^2\plus{}c^2) \ge 0 \Leftrightarrow a^2(a\plus{}c) \ge (a^2\minus{}c^2)(a\plus{}c)\plus{}c^3 \Leftrightarrow ac^2 \ge 0.\]


Can anyone explain:
a)YAC's and BAC's
b)Blotting(in all compass directions)
c)And can someone do it with APBIO secifity?

I wouldn't say I have Advanced Placement Biology specifity in defining terms (Currently I'm not in high school yet.), but I hope what I quoted from certain online texts is efficient for your purposes.

Yeast artificial chromosome:
[quote]A yeast artificial chromosome (short YAC) is a vector used to clone large DNA fragments (larger than 100 kb and up to 3000 kb). It is an artificially constructed chromosome and contains the telomeric, centromeric, and replication origin sequences needed for replication and preservation in yeast cells. Built using an initial circular plasmid, they are linearised by using restriction enzymes, and then DNA ligase can add a sequence or gene of interest within the linear molecule by the use of cohesive ends. They were first described in 1983 by Murray & Szostak.

YACs are extremely useful as one can get eukaryotic protein products with posttranslational modifications as yeasts are themselves eukaryotic cells, however YACs have been found to be more unstable than BACs, producing chimeric effects. Before the advent of the Human Genome Project, YACs and BACs were used to map sections of DNA of interest when hunting for specific genes.[/quote]
Source: [url]http://en.wikipedia.org/wiki/Yeast_artificial_chromosome[/url]

Bacterial artificial chromosome:
[quote]A bacterial artificial chromosome (BAC) is a DNA construct, based on a fertility plasmid (or F-plasmid), used for transforming and cloning in bacteria, usually E. coli. F-plasmids play a crucial role because they contain partition genes that promote the even distribution of plasmids after bacterial cell division. The bacterial artificial chromosome's usual insert size is 150 kbp, with a range from 100 to 350 kbp. A similar cloning vector, called a PAC has also been produced from the bacterial P1-plasmid.

BACs are often used to sequence the genetic code of organisms in genome projects, for example the Human Genome Project. A short piece of the organism's DNA is amplified as an insert in BACs, and then sequenced. Finally, the sequenced parts are rearranged in silico, resulting in the genomic sequence of the organism.[/quote]
Source: [url]http://en.wikipedia.org/wiki/Bacterial_artificial_chromosome[/url]

Blotting:
[quote]In molecular biology and genetics, a blot is a method of transferring proteins, DNA or RNA, onto a carrier (for example, a nitrocellulose PVDF or nylon membrane). In many instances, this is done after a gel electrophoresis, transferring the molecules from the gel onto the blotting membrane, and other times adding the samples directly onto the membrane. After the blotting, the transferred proteins, DNA or RNA are then visualized by one or more different methods:

[list]Colorant staining (for example, silver staining of proteins)

Autoradiographic visualization of radioactive labelled molecules (performed before the blot)

Specific labelling of some proteins or nucleic acids. It is done with antibodies or hybridization probes that bind only to some molecules of the blot and have an enzyme joined to them. After proper washing, this enzymatic activity (and so, the molecules we search in the blot) is visualized by incubation with proper reactive, rendering either a colored deposit on the blot or a chemiluminiscent reaction which is registered by photographic film.[/list] [/quote]
Source: [url]http://en.wikipedia.org/wiki/Blotting[/url]

By clicking the links, you can be taken to the pages I drew for reference.

-arb95

With the possible exception of blotting - you do not need to know YAC and BAC for the AP biology final.  :wink:


Ola gente de la comunidad española!
Ya he leido los resultados de Mexico en la Ibero 2006 (muy buenos resultados por cierto); pero sin embargo no he sabido mas, de casualidad alguien ha tiene algun link del examen o de los resultados de  todos los paises??
Agradecere cualquier tipo de informacion!!
 :)

aqui esta el problema + simple
[url]http://geomtry.mundoforo.com/el-problema-facil-de-la-oim-2006-vt108.html[/url]

Aqui va la prueba de la ibero

Disfrutenla y discutanla aca!!! en el foro! eso si procuren hacerlo en topics separados.

ok ,
gracias pascual ...
pero alguien tiene algun link para ver los resultados generales
y/o por paises??

parece ser q nadie se ha animado en colgarlos

espero encontrar alguna info...

Saludos,

pues en la pagina de la olimpiada en México encontre esto:
Resultados por países:

1. Mexico

2. Brasil

3. Argentina

4. Peru

5. Cuba

6. Colombia

Muchas felicidades a la delegación Mexicana!!!!!!!


After use Cauchy-Schwarts and Am-Gm, we only obtain an old (and easy) result:
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \plus{} \frac{8abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq\ 2$

The similar and harder problem: (with the same condition)
Prove that: $ \sqrt{3(\frac{a^2}{b} \plus{} \frac{b^2}{c} \plus{} \frac{c^2}{a})} \plus{} (\sqrt{2}\minus{}1). \sqrt{abc} \geq\ 2 \plus{} \sqrt{2}$
@babyloverain: Thank kiu em, thuc ra de~ kho' ko quan trong, anh em ta vui la` chinh' ma`. Anh rat thich n~ bai` toan' the nay`, em che' kheo' that :)

[quote="nguoivn"]After use Cauchy-Schwarts and Am-Gm, we only obtain an old (and easy) result:[/quote]
How to use Cauchy-Schwarts and AM-GM..?

[quote="nguoivn"]My full proof: (it seems not use Cauchy-Schwarts :oops: )
Using a well-known result: $ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \geq\ \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}$  (we can prove it easily by Am-Gm)
So, $ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \geq\ \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}$
Besides, we have: $ \sqrt {abc} \geq\ \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
<=> $ (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2 \geq\ \frac {64}{27}abc(a \plus{} b \plus{} c)^3$
(trues again by Am-Gm)
So, we obtain: $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ 2$  :)

[quote="nguoivn"]
The similar and harder problem: (with the same condition)
Prove that: $ \sqrt {3(\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a})} \plus{} (\sqrt {2} \minus{} 1). \sqrt {abc} \geq\ 2 \plus{} \sqrt {2}$
[/quote][/quote]
Thanks a lot :)

I can only prove this by SOS  :oops: :
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \geq\ \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}$

Can you help me?

[quote="nguoivn"]My friend, you can see it here: (#15)
http://www.mathlinks.ro/Forum/viewtopic.php?t=261526[/quote]
Thanks!! :)


AMC
USA(J)MO
USAMO
probability
calculus
graph theory

okok soo here some more of the questions that are similar to those on USAMO. can someone help me?
1. Find all positive integers n for which the equation nx^4+4x+3=0 has a real root.
2. The number 2^29 has 9 distinct digits.  Without using a calculator, tell which digit is missing.
3. Solve in positive integers the equation x^(x+y) = y^(y-x)
4. Find the probability that in the process of repeatedly flipping a coin, one will enocunter a run of 2 tails.
5. Prove that given any group of six people, it is always the case that one can find three people out of this group that either all know each other or all do no know each other.

thankkss~ ;)

Some hints.

[hide="1"] After eliminating small cases, you'll want to look at the derivative. [/hide]
[hide="2"] $ \bmod 9$ or $ \bmod 11$. [/hide]
[hide="3"] [b]Lemma:[/b]  Let $ a, b, c, d$ be positive integers such that $ a^c \equal{} b^d$.  Then there exists a positive integer $ e$ such that $ a, b$ are both powers of $ e$. [/hide]
[hide="4"] Try to find a recurrence relation that describes the number of ways to flip $ n$ coins without flipping a run of $ 2$ tails.  (You'll probably need to actually define two sequences.) [/hide]
[hide="5"] This is casework.  The standard interpretation of the problem is this:  given the complete graph on six vertices, with every edge colored red or blue, there is a red triangle or a blue triangle.  (Red = "We know each other" and blue = "we don't know each other"). [/hide]

[hide="1, Non-Calculus"]
Note that
\begin{eqnarray*}
nx^4 + 4x + 3 &=& (x+1)^2((x-1)^2 + 2) + (n-1)x^4 \\
&\ge& 0,
\end{eqnarray*} with the only possibility for equality at $ x = -1$, which gives $ n=1$ as the only solution.[/hide]


Find all non-negative integers $n$ such that there are $a,b \in \mathbb Z$ satisfying $n^2=a+b$ and $n^3=a^2+b^2$.

[i]Lucian Dragomir[/i]

[hide]$n=\frac{(n^2)^2}{n^3}=\frac{(a+b)^2}{a^2+b^2}=1+\frac{2ab}{a^2+b^2}$. So $\frac{2ab}{a^2+b^2}=n-1$ must be an integer.

But $-1\leq \frac{2ab}{a^2+b^2}\leq 1$ by trivial inequality. So $n-1$ must be either -1, 0, or 1, so that $\boxed{n\in \{0,1,2\}}$. (Each of these $n$ work: the ordered pairs (0,0), (1,0) and (2,2) for (a,b) produce these values of $n$.[/hide]

By CSI n^3=(a^2+b^2)>=((a+b)^2)/2=n^4/2 and so n^3>=n^4/2 then 2>=n and answer is 1,2 since n is natural

Here is my method:
$ b\equal{}n^2\minus{}a$, substituting we get $ n^3\equal{}n^4\plus{}2a^2\minus{}2an^2$, or $ n\equal{}n^2\plus{}2(\frac{a}{n})^2\minus{}2a$. From here we get $ a\equal{}kn$, where $ k$ is an integer, substituting we get $ 2n\minus{}n^2\equal{}(2k\minus{}n)^2$, so $ 2n\minus{}n^2\ge0$, so $ n\equal{}0,1,2$, they all check.

lemma 1 : if $ a+b=k  $ and $(a,b,k \in \mathbb N )$. $a^2 + b^2 $ is minimum when $a-b$ is minimum.

$a^2+b^2= k^2-2ab $ so $a^2 + b^2 $ is minimum when $ab$ is maximum and $ab$ is maximum when $a-b$ is minimum.

now assume that $n$ is even and $n\geq 3$  we have :

$a+b=n^2$  we know $ a^2 +b^2$ is minimum when $ a, b \geq 0$ . so from lemma 1 we have

 $ a^2 +b^2$ is minimum when $a=b=\frac {n^2}{2}$ $\rightarrow $ $n^3= a^2 +b^2= \frac{n^4}
{2}\rightarrow n=2 $ but we have $n\geq 3$.

now assume $n$ is odd and $n\geq 3$ we have

from lemma 1: $ a^2 +b^2$ is minimum when $ a= \frac{n^2+1}{2} ,  b=\frac{n^2-1}{2}$ $\rightarrow n^3=a^2+b^2 =\frac{n^4+1}{2} \rightarrow n<2 $  but we have $n\geq 3$.

if $n=2 $ $\rightarrow a=b=2$

if $n=1$ $\rightarrow a=1,b=0$

if $n=0$$\rightarrow a=b=0$

solution :) .

If $n=0$ we get $(a,b,n)=(0,0,0)$
If not then
By $AM-GM$ we have:
$n^3=a^2+b^2$ $\geq$ $2ab$
$\rightarrow$    $2n^3 \geq (a+b)^2 = n^4$ $\rightarrow$   $2\geq n$
If $n =1$
$a^2 + b^2 = 1$
One of $a$ & $b$ is $1$ and the other one is $0$
We get triples $( a , b , n ) = ( 1,0,1) , (0,1,1)$

If $n=2$
$a^2 + b^2 = 8$
Then we get the answer is $(a,b,n)=(2,2,2)$
And we're done.


[quote="Mildorf"]Show that for all positive reals $a, b, c$

$\sqrt[3]{4a^3+4b^3}+\sqrt[3]{4b^3+4c^3}+\sqrt[3]{4c^3+4a^3} \le \frac{4a^2}{a+b}+\frac{4b^2}{b+c}+\frac{4c^2}{c+a}$[/quote]

Very nice exercise! In fact, using cyclic sums, your inequality rewrites as $\sum\sqrt[3]{4a^3+4b^3}\leq\sum\frac{4a^2}{a+b}$. But

$\sum\frac{4a^2}{a+b}-\sum\frac{2\left(a^2+b^2\right)}{a+b}$
$=\sum\left(\frac{4a^2}{a+b}-\frac{2\left(a^2+b^2\right)}{a+b}\right)=\sum\frac{2\left(a^2-b^2\right)}{a+b}=\sum 2\left(a-b\right)=0$,

so that $\sum\frac{4a^2}{a+b}=\sum\frac{2\left(a^2+b^2\right)}{a+b}$, and thus the inequality in question is equivalent to $\sum\sqrt[3]{4a^3+4b^3}\leq\sum\frac{2\left(a^2+b^2\right)}{a+b}$. But this is a trivial consequence of the inequality $\sqrt[3]{4x^3+4y^3}\leq\frac{2\left(x^2+y^2\right)}{x+y}$ for any two positive reals x and y, which can be proven as follows:

Cubing the inequality $\sqrt[3]{4x^3+4y^3}\leq\frac{2\left(x^2+y^2\right)}{x+y}$, we get $4x^3+4y^3\leq\frac{8\left(x^2+y^2\right)^3}{\left(x+y\right)^3}$. Upon multiplication with $\frac{\left(x+y\right)^3}{4}$, this becomes $\left(x^3+y^3\right)\left(x+y\right)^3\leq 2\left(x^2+y^2\right)^3$. Now, in this form, the inequality follows from $2\left(x^2+y^2\right)^3-\left(x^3+y^3\right)\left(x+y\right)^3=\left(x^2+y^2+xy\right)\left(x-y\right)^4\geq 0$. And thus the problem is solved.

  darij

[quote="darij grinberg"]But this is a trivial consequence of the inequality $\sqrt[3]{4x^3+4y^3}\leq\frac{2\left(x^2+y^2\right)}{x+y}$ for any two positive reals x and y[/quote]

Exactly.

On a related note, I'll take a moment and point out that the problem was contrived with a generalization (only a conjecture right now) in mind: http://www.mathlinks.ro/Forum/viewtopic.php?p=279354#p279354

[quote="Mildorf"]Show that for all positive reals $a, b, c$

$\sqrt[3]{4a^3+4b^3}+\sqrt[3]{4b^3+4c^3}+\sqrt[3]{4c^3+4a^3} \le \frac{4a^2}{a+b}+\frac{4b^2}{b+c}+\frac{4c^2}{c+a}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86201
This is stronger. ;)


\[ x\frac{{dy}}
{{dx}} \plus{} \left( {\frac{{dy}}
{{dx}}} \right)^2  \equal{} y\]

Differentiating $ x\frac {dy}{dx} \plus{} \left( \frac {dy}{dx}\right) ^{2} \equal{} y,$ we have

$ \frac {d^{2}y}{dx^{2}}\left( x \plus{} 2\frac {dy}{dx}\right) \equal{} 0$

Hence,

$ y \equal{} \minus{} \frac {x^{2}}{4}$

Or $ y \equal{} C^{2} \plus{} Cx$


[quote="SnowStorm"]Hey I have a question about the Flood Fill algorithm.  In the psuedo code it has the statement 'for all nodes i'.  Does this mean just loop through all the nodes?[/quote]
Yep, that's exactly what it means.  Since loops can have different syntax in different languages, pseudocode keeps it general by saying "for all nodes i".  ;)

Ok I'm working on the Castle problem.
I should loop through the whole 2 dimensional array representing the castle?
Here is some of my code so far, I have a couple questions.

[code]
public void flood_fill(int new_component)
{
do
{
	int num_visited = 0;
	for(int i=0; i<myCastle.length; i++)
	{
		for(int j=0; j<myCastle[0].length; j++)
		{
			if(myModules[i][j] = -2)
			{
				num_visited = num_visited + 1;
				myModules[i][j] = new_component;

				for(int j=0; j<4; j++)  // here should i loop through all the neighbors or just the ones who are an open wall away?
				{
					if(myModules[i][j] == 0)
						myModules[i][j] = -2;
				}
			}
		}
	}
}
while(num_visited != 0)
}
[/code]

[quote="SnowStorm"][code]
for(int j=0; j<4; j++)  // here should i loop through all the neighbors or just the ones who are an open wall away?
{
	if(myModules[i][j] == 0)
	     myModules[i][j] = -2;
}
[/code][/quote]

You should loop through the ones that are an open wall away.  Even though each square may have neighbors in the sense that they are adjacent, only those that have no wall between them belong to the same component (room). Make sense?  :)

Ok, one more thing I think
[code]
for all nodes i
    component(node i) = nil
[/code]

Is this asking to go through and set each node to unassigned?  What is node i referring to?[/code]

Ok i figured that out.
I'm having trouble figuring out which wall to remove to make the biggest room though, any ideas?

Thanks

I'm not sure how i solved it, but my first idea is to remove each wall and add the sizes of the rooms it seperates (divide this figure by 2 if the wall seperates the same room from itself)


How long do you spend on math per day averagely?

Depends, but usually about 2 hours of lecture time, maybe an hour or two of homework, and then some free-mathing.

:> we should ask IMOers from mathlinks about it, would turn out to be one-sided poll :-D

24h/day with short breaks :D doing everything i think about maths more or less consciously. maths change your way of thinking for good.

lesson: 40 min.
homework: 20 min.
other: 30 min.
TOTAL: 1.5 hr.

yeah over all I do 1 to 1.5 hours a day including the math stuff on this site

I would say about 1 to 2.5 on average.

Forty every day, another 40 every other day, and an additional hour once a week.

Who voted for "ALMOST TWENTY-FOUR HOURS"?

[quote="13375P34K43V312"]Hmm...not including goofing around on AoPS and regular math class/homework...this is per week

Two classes a week on AoPS: 3 hours
Typing class notes for those classes: 4-6 hours
Doing the message board problems: 1-1.5 hours
Doing old contests (AMC, MAML, Mathcounts): 6-8 hours
Doing my class challenges + USAMTS: 2-2.5 hours
Working from books: 4-6 hours
Other random stuff (my own problems): 1-2 hours

So that adds up to...21-29 hours a week...3.5 hours a day....WOW i need to get out more[/quote]

hmm:

50 mintues for math class:
10 minutes for homework:
MC and random math problems: idk but over an hour

I don't think math class/homework should count, most of us agree that it is for the most part a waste of time.

depends on what math class you're taking...

well i am thinking about math even with crappy problems and goofing off :lol:

I voted for an hour.
I do like 25 min hw, and 25 AoPS, school : 50 min, but school doesn't really count, 10 minutes random reading for fun

hey krusty, i'm from palo alto, CA, AIM me :)

Well, I don't think it's a topic about games and fun. Why was it moved to here?

This is games and fun. Unless you have problems adding and averaging your time spent.

About 4 to 5.5 hours

if i had more time, it would definitely be a ton more

depends...about 1 hour on homework, as i can go as fast as i want and finish the class, 30 min. on aops(doing math...), and 30 min. on other random stuff. about 2 hours.


So who's going(or wanting to)?

I would, if CTY would just move the date.

What do you mean, "If CTY would move the date"?

You mean for the SAT's?  I think that you'd have to get the College Board to move the date or let you take the makeup date.  Not sure how that would work with CTY though.

Or is there some other CTY thing then?  The summer program doesn't start till the end of June.

Anyways, I'm going.  Of course.  If I had some commitment I couldn't get out of, I'd try to buy a time-turner.  Or I'd just be like "No.  I'm going to ARML".  And I only get three chances at it (9th grade, my math teacher, three days after the application deadline: "I need to know if you're going to ARML" "...to what?" "All right, I guess it's too late for you to apply" "...wait, apply to what?  Huh?").  So it's definitely my top priority.

See [url=http://arml.com/SAT%20Alternate%20Test%20Day%20Policy%202006-07.pdf]here[/url] (.pdf).  ARML, as "an academic or athletic competition, in which students have advanced through the competition or into post-regular season competition, is considered an unexpected event," and therefore attending ARML is a legitimate basis for taking the SAT on an alternate date.

I qualified for the CTY awards and a summer class; conflicting with ARML.

so this picture was suppose to go up on the giant screen, but unfortunately we ran out of time...  So you guys are getting it here.

[img]http://www.jackpo.org/temp/georger_revenge.jpg[/img]

LOL, thanks JackPo.  So GeorgeR managed to accumulate 37323123 posts by Saturday, even though he only joined today.  I don't think anyone else has ever had a negative posts-per-day.  Then again, Reuter is awesome like that, so anything is possible.

lol!

what is also amazing is jackpo's ability to take pictures with cameras in chinese.

What is the minimum smartness (or rather, min. scores on the AMC, AIME, NYSML, etc.) that one needs to be to join the ARML team? And does the contestant's school cover the individual costs?

Just wanted to know. I didn't join this year, but I might next year.

PS. GeorgeR's "Online" needed to be green.

There's no minimum.

Also, that is one of the things that is amazing: if he's online, then he is also offline, or the converse.


[quote="BanishedTraitor"]I am just curious concerning the three general class of Science: Chemistry, Physics and Biology.

1. Which one do you find the easiest to ask how? (e.g. how does such and such work may it be a reaction or a phenomenon such as the rainbow).

2. Which one do you find the easiest to ask why? (e.g. why does such and such happen rather than such and such, why does reaction one proceed faster or in favour of reaction two).

I find it easier to ask how in Physics and Biology rather than in Chemistry. Yet I find it easier to ask why in Chemistry (probably because I am bad at Physics). Won't give reasons yet, I want to see your opinion first.[/quote](1) Hmm, I would say Biology namely due to the human body and how it works. But that is the reason for the "hmm", for due to the question why (with respect to the body), Chemistry provides some answers. So I would say Chemistry and Biology!

(2) Definitely Physics. I hope to stop asking "why" after a finite period, but no such luck I'm afraid!

Here I have shown the two sciences I prefer, and the one I don't really get on with. (If Biology didn't have all that plant and evolution mumbo jumbo, I would have done an A2 in it as welll!)

It probably also depends on what scales we are discussing these matters. On a macroscopic level, where observation can be determined quite accurately it is probably a lot easier to conclude how (and why) things are the way they are. On a microscopic level when Quantum Effects and the Uncertainty Principle kicks in then we should re-assess what we think we are observing.


calculate x,y,z
$ \begin{cases} \frac{a}{x}\minus{}\frac{b}{z}\equal{}c\minus{}zx \\ \frac{b}{y}\minus{}\frac{c}{x}\equal{}a\minus{}xy\\ \frac{c}{z}\minus{}\frac{a}{y}\equal{}b\minus{}yz\end{cases}$

[quote="aktyw19"]calculate x,y,z
$ \begin{cases} \frac {a}{x} \minus{} \frac {b}{z} \equal{} c \minus{} zx \\
\frac {b}{y} \minus{} \frac {c}{x} \equal{} a \minus{} xy \\
\frac {c}{z} \minus{} \frac {a}{y} \equal{} b \minus{} yz\end{cases}$[/quote]

[hide="Incomplete result "]The system may be written :
$ az\minus{}bx\equal{}cxz\minus{}x^2z^2$
$ bx\minus{}cy\equal{}ayx\minus{}y^2x^2$
$ cy\minus{}az\equal{}bzy^2\minus{}z^2y^2$

Summing these three lines, we get $ xz(xz\minus{}c)\plus{}xy(xy\minus{}a)\plus{}yz(yz\minus{}b)\equal{}0$

And so, if $ abc>0$, a trivial solution $ xz\equal{}c, xy\equal{}a, yz\equal{}b$ which gives $ (x,y,z)\equal{}(\frac{\sqrt{abc}}b,\frac{\sqrt{abc}}c,\frac{\sqrt{abc}}a)$ (easy to check this indeed is a solution)

The questions now are :
- are there some solutions when $ abc\leq 0$ ?
- are there some other solutions when $ abc>0$ ?[/hide]

for this problem i think we can solve it by considering a,b,c be the variables so we have a 3 _variable system with a,b,c we can express a,b,c in term x,y,z which give the same result as pco


Hello,

First than all, let's put something clear, I am talking about Euclidean Geometry, and not degenerated polygons. Now, obviusly, if you want an equiangled triangle, it has to be equilateral too. If you want a equiangled cuadrilateral, the oposite sides have equal length. Now, what happens in the case of a polygones with the number of sides at least five. I am more interested in five and six, but in general, if we have a polygon with $n$ sides, what conditions do we have to impose in the numbers $a_1, a_2,\dots, a_n$, the sides of this polygon in order, in order to have this polygon to be equiangled?

Best regards,

let $t$ be a primitive n-th root of unity. if $a_0,a_2...,a_{n-1}$ are the sides of your poligon, you need as a necesary and suficient condition that $t$ is a root of: $p(x)=a_0+a_1x+a_2x^2+...a_{n-1}x^{n-1}$. Obviously lots of things can happen and obviously there are many interesting $n$.

Do you know any place where I can find the proof of what you are mentioning. Any way I am going to try to proove it by myself, looks interesting.

Best,

just use complex numbers, and the fact that they form your polygon if their sum is $0$.


algebra
function
domain
complex analysis
partial fractions

Let
\[ f(z)=\frac{1}{(z-i)(z^{2}+4)}\]

a) within which domains does $ f(z)$ have a Laurentseries expansion of the form $ \sum_{n=-\infty}^\infty c_{n}z^{n}$

b) Determine, if possible, the Laurentseries for $ f(z)$ that converge at the point $ z=\frac32$

It may be helpful to rewrite your expression as:

$ f(z)=\frac{1}{(z-i)(z^{2}+4)}=\frac{1}{2z}\frac{1}{1-\frac{i}{z}}+\frac{3}{8 (2i-z)}-\frac{1}{8 (2i+z)}$

Then just divide through and collect terms -- I hope that helps.

[quote="color"]Let
$ f(z)=\frac{1}{(z-i)(z^{2}+4)}$
a) within which domains does $ f(z)$ have a Laurentseries expansion of the form $ \sum_{n=-\infty}^\infty c_{n}z^{n}$
[/quote]

[b]Theorem:[/b] Let $ f(z)$ be holomorphic in the punctured disk $ D(c,\delta) = \{z\in \mathbb{C}: 0<|z-c|< \delta\}$. Then $ f(z)$ can be expressed as $ \sum_{n=-\infty}^{\infty}a_{n}z^{n}$ for all $ z\in D(c,\delta)$.

Now the thing is the your function has a Laurent Series at every single point. Since there are only finitely many poles (which can be found in Edoo's partial fractions) we can find a Laurent series at every single point since we can choose a $ \delta$ sufficiently small to exclude other poles.


function
algebra
polynomial
inequalities
quadratics
real analysis
real analysis unsolved

If a positive continuous twice differentiable function statisfies $ f''(c) \leq \frac {1}{2}$ for all $ c \in\mathbb R$, prove that $ |f'(x)| \leq \sqrt {f(x)}$ for all $ x$ in $ \mathbb R$. Thanks in advance.

[color=green]Moderator edit: changed the assumption to the correct one :roll:[/color]

Using taylor approximation about $ x \equal{} c$ we have,
\[ f(x) \equal{} f(c) \plus{} (x \minus{} c) f'(c) \plus{} \frac {x^2f''(c)}{2!} \plus{} R \leq f(c) \plus{} (x \minus{} c) f'(c) \plus{} \frac {(x \minus{} c)^2}{4} \plus{} R
\]
Or, $ f(x) \minus{} R \leq f(c) \plus{} (x \minus{} c) f'(c) \plus{} \frac {(x \minus{} c)^2}{4}$ Now if $ f(x) \minus{} R \leq f(c) \plus{} (x \minus{} c) f'(c) \plus{} \frac {(x \minus{} c)^2}{4} \equal{} 0$ $ \forall x \in \mathbb{R}$ we must have $ f(c) \equal{} 0$ which is clearly false so there are no zeros to the polynomial in $ (x\minus{}c)$. Now,
\[ (x \minus{} c)^2f'(c) \minus{} 4f(c) \cdot \frac {(x \minus{} c)^2}{4} \leq 0 \Rightarrow |f'(c)| \leq \sqrt {f(c)}
\]
Or $ |f'(x)| \leq \sqrt {f(x)}$  $ \forall x \in \mathbb{R}$.

how do you get the first inequality? (the one where you remove the error term in the taylor expansion)

ive edited. maybe that works. im kind of winging this solution.

I think the only problem with your solution is that you are using an incorrect statement of Taylor's Theorem. Other than that the general idea behind the proof seems correct.
[hide="Here is what I think it should be"]
Using Taylor's Theorem we have that there exist a $ \beta$ between $ x$ and $ c$ such that

 $ f(x) \equal{} f(c) \plus{} (x \minus{} c)f'(c) \plus{} \frac {(x \minus{} c)^2}{2!}f''(\beta)$. Now from the given assumptions we have that

$ f(x)\le f(c) \plus{} (x \minus{} c)f'(c) \plus{} \frac {(x \minus{} c)^2}{4}$ which leads us to

$ 0\le 4(f(c) \minus{} f(x)) \plus{} 4(x \minus{} c)f'(c) \plus{} (x \minus{} c)^2$. Viewing this as a quadratic in terms of $ (x \minus{} c)$ we know that it must have a negative discriminant.

So $ 16f'(c)^2 \minus{} 16(f(c) \minus{} f(x))\le{0}\Longrightarrow f'(c)^2 \minus{} f(c) \plus{} f(x)\le{0}$. Now because $ f(x) > 0$ we must have that $ f'(c)^2 \minus{} f(c)\le{0}$, and the desired result follows.
[/hide]


algebra
polynomial
irrational number
roots
IMO Shortlist
IMO Longlist

Let $ax^{3}+bx^{2}+cx+d$ be a polynomial with integer coefficients $a,$ $b,$ $c,$ $d$ such that $ad$ is an odd number and $bc$ is an even number. Prove that (at least) one root of the polynomial is irrational.

i have some ideas here..
we can use root=r=p/q  p and q are integers and p/q is irriducible.
there are two cases. p and q are both odd, or p(or q) is even.

we have another two cases, b and c are both even; or c or b is even.
goodluck

i found three cases, in two of them the root is irrational
b and c and p(or q) are even. than
ap+bpq+cpq+dq=even+even+even+odd= odd=0 which is not true because 0 is not odd.

b and c are even, p and q are odd
ap+bpq+cpq+dq=odd+even+even+odd=even=0 which is true

b OR c are even, p and q are odd
ap+bpq+cpq+dq=odd +even+odd+odd=odd=0  which is not true because 0 is not odd.


Just like to mention good luck to all those taking the AMC on Feburary 1st. May your hard work and dedication to mathematics pay off....

I also want to wish good luck to everybody taking the AMC's on Tuesday.

A lot of people at Art of Problem Solving are going to be taking the AMC A's on Tuesday. It's kind of difficult to prepare any more mathematically, but there's always the mental preparation that helps right before the test. If anyone has any time-tested mental preparation hints, please share them. I want to know anything that will help.

Here's what my teacher (band teacher, actually) suggested:
Don't expect to be perfect - no one is. Just do the best you can do for now, and know that it's pretty darn good.
Don't drink a lot of soda, but do drink a lot of water.

And something I've figured out for the AMC's: don't be afraid to leave questions blank.

Any others?

Good luck everyone!

Don't rely too heavily on your calculator but don't be afraid to use it either.  I would suggest using it to check your arithmetic.  If there is a problem that graphing something might help, use the calculator!

Here are some tips: (I need to follow them also!)

-Don't be afraid to use paper!...make you geometry diagrams big and don't try cramming everything on one page.

-If a direct approach is not apparent, try examples.  Plug in small numbers.  Make a conjecture...you don't have to prove anything so if you see a pattern go with it.  Chances are that it is right and you can prove it later.

-Every question is worth the same amount so rack up as many points as possible on the easier ones.

-Don't spend too much time on any one problem.

-Go for the problems that are easier for you e.g. if you are really good at geometry but stink at NT then go for a slightly harder geometry one over a NT one.

-Get a good night of sleep (I really need help with this one!!).

-Have a good breakfast but don't do anything really unusual.  This is just a math competition-you don't need to carbo load or anything.  If sugar and caffeine help you with math then go for it.  (I think you guys know what kind of nutrition suits you).

-Use all of the 75 minutes.  Even if you have done your quota of problems and cannot even approach any of the remaining ones go back and check!

-It's too close to the test now to really change your score dramatically so just go and review all the stuff you know.  You might be able to learn a few new things but I wouldn't count on it.  Just work on preparing for the test now.  As Caesar said: "Alea iacta est."

-Finally, take a deep breathe and don't panic!  It's just a math competition.  Give it everything you've got but don't get too worked up about it.

[quote="joml88"]

-Get a good night of sleep (I really need help with this one!!).
[/quote]

Same here!

Also,

-Read the problems carefully, make sure you got every piece of information given in the problem

- Don't eat breakfast, you might throw up if you are nervous  :)

-Circle tricky words like integers, distinct, reals, unit

-2005=5*401

-2006=2*17*59

[quote] Use alot of paper eh... I don't think i'll have a problem wit that. I'd recomend one sheet per problem. mbe even more.[/quote]

That is a LOT of paper :P I tend to do this on the AIME, not so much on the AMC. I'll probably use 2 sheets front and back for 1 AMC test.

My advice to everyone, including myself: enjoy it. [i]You are doing this because you want to, not because it affects your grade or something.[/i] It's like a scary amusement park ride that you somehow want to go on over and over again despite how panicked you get when you're riding. If you get a bad score you won't be as disappointed because you already gained something--75 minutes of [i]fun[/i]. Enjoy it.

If it helps, use the answers to solve the questions.

And don't guess unless you can narrow it down to two answer choices.


It took Lara five days to read a novel. Each day after the first day, Lara read half as many pages as the day before. If the novel was 248 pages long, how many pages did she read on the first day?

Notice that she reads x+x/2+x/4+x/8+x/16=31x/16 pages in five days, if x is the pages she reads on day one.

Thus 16*248/31=x, or 16*8=128=x.


Find all the values of $ a$ such that the inequation $ x \plus{} \sqrt {x \minus{} x^2}\geq a$ has at least one real solution.

I do not have the answer for this problem, and would appreciate it if anyone could check it for me.

[hide="My answer"]$ a\in\left[1,\frac {1 \plus{} \sqrt {2}}{2}\right]$[/hide]

The maximum and minimum values for x are when $ x \minus{} x^2 \equal{} 0$. This is $ 0$ and $ 1$. Plugging in $ 1$ for $ x$, $ a\le1$.

Edit: Forgot to count change in $ a$. Sorry.

Don't see what you mean there brainomega.

Let $ C: y \equal{} \sqrt {x \minus{} x^2}\ (0\leq x\leq 1)\Longleftrightarrow \left(x \minus{} \frac {1}{2}\right)^2 \plus{} y^2 \equal{} \frac 14$ and $ y\geq 0$, which is the part of $ y\geq 0$ of the semi-circle cenrtered the point $ \left(\frac 12,\ 0\right)$ with radius $ \frac 12$. Let $ l: y \equal{} \minus{} x \plus{} a$, which is the straight line with the slope $ \minus{} 1$. We are to determine the range of $ a$ such that the graph of $ C$ is above the graph of $ l$. When $ l$ passes through the point $ (0,\ 0)$, we have $ a \equal{} 0$ and when $ l$ is tangent to the circle $ C$, we have the distance of the center $ \left(\frac 12,\ 0\right)$ and the line $ l: x \plus{} y \minus{} a \equal{} 0$ equals to the radius $ \frac 12\Longleftrightarrow \frac {|\frac 12 \plus{} 0 \minus{} a|}{\sqrt {1^2 \plus{} ( \minus{} 1)^2}} \equal{} \frac 12$, yielding $ a \equal{} \frac {1\pm \sqrt {2}}{2}$. Judging from the graphs above, we choose $ a \equal{} \frac {1 \plus{} \sqrt {2}}{2}$. For the value of $ a$, certainly $ C$ is tangent to $ l$ at the point $ \left(\frac {1 \plus{} \sqrt {2}}{2},\ \frac {1}{2\sqrt {2}}\right)$. Therefore the answer is $ \boxed{a\leq \frac {1 \plus{} \sqrt {2}}{2}}$

[color=red]Last edited[/color]

Actually, the answer is $ a\in\left( \minus{} \infty, \frac {1 \plus{} \sqrt {2}}{2}\right]$ because for all negative numbers $ x \equal{} 0$ is a solution to the inequality.

That's wrong. You can make sure that your solution is wrong by drawing the graphs stated in my post.

Sorry, your answer was correct.

kunny

Last edited.

kunny, thank you for your solution, a very neat idea to plot the graphs.

However, I don't understand why $ a$ has to be greater than zero. What is wrong with this graph:

[img]http://img21.imageshack.us/img21/4477/picture25xu.png[/img]

Where the graph of $ C$ is above the graph of $ l$ like you said?

Also, is this problem solvable algebraically?

Oh, sorry, $ a$ can take any negative values. :oops:

Dr Sonnhard Graubner, you should read the context carefully. :lol:


In an 8-bit system, how is -64 represented as 1s and 0s?

What I've Done:
Since 64 is negative, the leftmost digit of the 1s and 0s representation of the number is 1. 64 in base 2 is $ 1000000_{2}$. I subtracted one to get 0111111 and then switched the 0s and 1s to get 1000000. I know this can't be right since one followed by zeros should be the most negative number possible in the bit system and -64 isn't.

You're only writing down 7 digits each time. You want to write down 8 digits, otherwise you might forget theres a 0 on the front which also needs to be flipped to a 1 when you flip the bits.


algebra
polynomial
geometry
3D geometry
number theory
relatively prime
system of equations

Given the system of equations in three variables: $ xyz = 13, xy + yz + xz = 24, \& x + y + z = - 5,$

1. Consider $ 1/3x + 1/3y + 1/3z = m/n$, where $ |m|$ and $ |n|$ are relatively prime. Find $ |m| + |n|$ (early-AMC 12 difficulty).

2. And since that probably was so easy, now find $ |x^2 + y^2 + z^2|$.

3. And the last question for today: Given the fraction $ a/b = 1/(x - 4)^3 + 1/(y - 4)^3 + 1/(z - 4)^3$, where and $ |a|$ & $ |b|$ are relatively prime, what is $ |a|+|b|$?

[quote="Caelestor"]Hello, I'm not sure what difficulty is this problem, but it'd probably be anywhere between late AMC 12 to mid-AIME:

Given $ xyz = 13, xy + yz + xz = 24, \& x + y + z = - 5,$

the sum $ 1/3x + 1/3y + 1/3z$ can be expressed as a fraction m/n, where $ |m|$ and $ |n|$ are relatively prime.

Find $ |m| + |n|$.[/quote]

A fairly straightforward problem.

We see that from the information given, we can create a polynomial with the given coefficients.

That is, $ f(x) = x^3 + 5x^2 + 24x - 13$

We are asked to find the sum of the reciprocal of three times the roots.

So, find the polynomial with 3 times the roots first. We can do this by inserting $ f(3x)$

$ 27x^3 + 45x^2 + 72x - 13$

Next, we have to find the polynomial with the reciprocal of the roots of this polynomial.
That just inverts the coefficients. So we have:

$ -13x^3 + 72x^2 + 45x + 27$

Now, we just find the sum of the roots of this equation. This would be:

$ \dfrac{-72}{-13} = \dfrac{72}{13}$

Thus, the answer is $ 72 + 13 = \boxed{85}$

[quote="Caelestor"]Hello, I'm not sure what difficulty is this problem, but it'd probably be anywhere between late AMC 12 to mid-AIME:

Given $ xyz = 13, xy + yz + xz = 24, \& x + y + z = - 5,$

the sum $ 1/3x + 1/3y + 1/3z$ can be expressed as a fraction m/n, where $ |m|$ and $ |n|$ are relatively prime.

Find $ |m| + |n|$.[/quote]

$ \frac {1}{3x} + \frac {1}{3y} + \frac {1}{3z} = \frac {yz + zx + xy}{3xyz} = \frac {24}{3 \times 13} = \frac {8}{13}$

So $ |m| + |n| = 8 + 13 = 21$

[quote="veezbo"]A fairly straightforward problem.

We see that from the information given, we can create a polynomial with the given coefficients.

That is, $ f(x) = x^3 + 5x^2 + 24x - 13$

We are asked to find the sum of the reciprocal of three times the roots.

So, find the polynomial with 3 times the roots first. [b]We can do this by inserting f(3x)[/b]

$ 27x^3 + 45x^2 + 72x - 13$[/quote]

It should be inserting $ f(\frac{x}{3})$, not $ f(3x)$.

Yep, I believe stephencheng is correct. Didn't notice the incredibly quick solution though.  :|

[quote="grn_trtle"]Use veezbo's polynomial with Newton's sums for part 2.[/quote]

but, $ x^2\plus{}y^2\plus{}z^2 \equal{} (x\plus{}y\plus{}z)^2 \minus{} 2(xy\plus{}yz\plus{}zx) \equal{} 5^2 \minus{} 2(24) \equal{} \minus{}23$

So...are $ x,y,z$ real or complex?

Solving with Mathematica gives $ x \equal{}0.48736, y\equal{} \minus{}2.74368 \minus{} 4.37568 i, z\equal{} \minus{}2.74368 \plus{} 4.37568 i$, so two of them are complex.

ocha's solution works, or you can do what I said:
$ S_1\equal{}\minus{}5$
$ S_2\plus{}5S_1\plus{}2\cdot 24\equal{}S_2\minus{}25\plus{}48\equal{}0\implies \boxed{S_2\equal{}\minus{}23}$.

Oops. I was actually thinking that was wrong, but I just ignored that feeling. I should trust my intuition more and use my brain instead of remembering random stuff.

For part 3, use the polynomial with solutions that are given. Then, find the polynomial with roots the reciprocal  of the polynomial with the roots 4 less than the original polynomial.

After that, just use newton's sums and find the sum of the cubes of the roots of that equation.

Right. You should get

$ \frac{1}{x\minus{}4}\equal{}r\implies x\equal{}\frac{1}{r}\plus{}4$

Using synthetic division, we can find that $ f(r\plus{}4)\equal{}27r^3\plus{}369r^2\plus{}1728r\plus{}2723$, so that our desired polynomial is $ 2723r^3\plus{}1728r^2\plus{}369r\plus{}27$.

Then applying Newton's sums, though it gets very ugly.


geometry
parallelogram
geometry unsolved

The tangent lines from a point $P$ meet a circle $k$ at $A$ and $B$. Let $X$ be an arbitrary point on the shorter arc $AB$, and $C$ and $D$ be the orthogonal projections of $P$ onto the lines $AX$ and $BX$, respectively. Prove that the line $CD$ passes through a fixed point $Y$ as $X$ moves along the arc $AB$.

Let $T$ be the midpoint of $AB$. Since $AT\perp PT,\ AC\perp PC,\ ATCP$ is cyclic, and a simple angle chase will show that $TC\perp BX$. This means that $TC\|PD$. In the same way we prove that $TD\perp PC$, meaning that $PDTC$ is a parallelogram, and $CD$ passes through the midpoint of $PT$.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Is there exist a $ 2\times 2$ real matrix $ A$ such that \[ A^2\equal{}\begin{pmatrix} \minus{}2008& 2010\\ 0&\minus{}2009\end{pmatrix}\]

[hide="Hint"]No. Consider the eigen values of $ A$ which are zeros of a polynomial with real coefficients.[/hide]

Or you use : $ \det(A)\equal{}\pm \sqrt{2008.2009},Tr(A^2)\equal{}\minus{}2008\minus{}2009$,$ A^2\equal{}Tr(A)A\minus{}\det(A)I_2$ then $ Tr(A^2)\equal{}Tr^2(A)\minus{}2\det(A)$ from here $ Tr^2(A)\equal{}Tr(A^2)\plus{}2\det(A)\equal{}\minus{}(\sqrt{2009}\pm \sqrt{2008})^2<O$

Or implies $ \forall a \in (0,1)$
we have :$ A^2 \plus{} (2008 \plus{} a)I_2 \equal{} \begin{pmatrix} a & 2010 \\
0 & a \minus{} 1\end{pmatrix}$
By
  $ \det(A^2 \plus{} (2008 \plus{} a)I_2) \equal{} |\det(A \plus{} i\sqrt {(2008 \plus{} a)}I_2)|^2 \geq 0 > \det\begin{pmatrix} a & 2010 \\ 0 & a \minus{} 1\end{pmatrix} \equal{} a(a \minus{} 1)$

a less olympiad-like solution... suppose $ A\equal{}\begin{pmatrix} a& b\\
c&d\end{pmatrix}$, then, $ A^2\equal{}\begin{pmatrix} a^2\plus{}bc&b(a\plus{}d)\\c(a\plus{}d)&d^2\plus{}bc\end{pmatrix}$... since $ c(a\plus{}d)\equal{}0$ and $ b(a\plus{}d)\neq 0$ we have that $ c\equal{}0$, which leaves us with $ A^2\equal{}\begin{pmatrix}a^2&b(a\plus{}d)\\ 0&d^2\end{pmatrix}$, which clearly is different from the given one...


calculus
derivative
limit
parameterization
calculus computations

Use a derivative to find $ \lim_{x \rightarrow 3\pi} \frac {\cos(x) \plus{} 1}{x \minus{} 3\pi}$

If you can, try and just give a hint.

Well, you could use L'Hopital's Rule, of course.

Or, use one of the definitions of a derivative:

$ \lim_{x \to x_0} \frac {f(x) \minus{} f(x_0)}{x \minus{} x_0} \equal{} f'(x_0)$

(under appropriate conditions, of course)

Silly me, the problem can be re-written as

$ \lim_{x \rightarrow 3\pi} \frac { cos(x) \minus{} cos\left(3\pi\right) }{x \minus{} 3\pi}$

Which is the derivative definition. Thus, all we need to do is differentiate cos(x) which is -sin(x) then evaluate at $ 3\pi$, which is 0.

New question:

Show that $ x^{2} \plus{} y^{2} \equal{} ax$ and $ x^{2} \plus{} y^{2} \equal{} by$ are orthogonal.

I'm pretty sure all you need to do is differentiate both equations and show that the product of the two derivatives is -1. However, this doesn't hold unless you put constraints on the given variables. Ie. $ x \equal{} \minus{} y$ and $ a \equal{} \minus{} b$. Am I wrong or is there some ambiguity in this question?

Here's how that should be interpreted:
Each of those equations defines a family of curves, depending on the parameter $ a$ or $ b$. For any particular pair of curves (one in the first and one in the second), those two curves have orthogonal tangents at their points of intersection.

A simple geometric proof:
Both families consist of circles passing through the origin. The circles of the first family have vertical tangents at the origin, and the circles of the second have horizontal tangents there. By reflecting across the line connecting the centers, we can see that any pair of intersecting circles make the same angle at each of their two intersection points. In this case, that means that the circles must meet at right angles at the other intersection point as well as the origin.


[hide="Combinatorial"]Since ${n\choose k}={n\choose n-k}$, we have ${n\choose k}^{2}={n\choose k}{n\choose n-k}$, and the latter is the number of choices in the following situation: choose $k$ elements from the first set of $n$ elements and then choose $n-k$ elements from the second set of $n$ elements. When $k$ runs from $0$ to $n$, the situation is nothing else than choosing $n$ elements out of the conglomerate set of $2n$ elements, hence the result.[/hide]

Instead of a combinatorial proof, let me offer a different one.

${(1+x)}^{n}= \sum_{k=0}^{n}\binom{n}{k}x^{k}$ and ${(1+\frac{1}{x})}^{n}= \sum_{k=0}^{n}\binom{n}{k}{(\frac{1}{x})}^{k}$.

Multiplying the above two, we have
$\sum_{k=0}^{n}{\binom{n}{k}}^{2}=$ term independent of $x$ in ${(1+x)}^{n}\cdot{(1+\frac{1}{x})}^{n}$
$\Rightarrow \sum_{k=0}^{n}{\binom{n}{k}}^{2}=$ coefficient of $x^{n}$ in ${(1+x)}^{2n}= \binom{2n}{n}.$


sodium hydrogen carbonate decomposes when heated.

$2NaHCO_3 \longrightarrow Na_2CO_3+CO_2+H_2O$

Sodium carbonate has an unpleasant taste.It reacts with the tartaric acid to form

sodium tartarate which has a pleasant flavour.

a)what type of compound is sodium tartarate?

b)suggest  two reasons why baking powder is a mixture of tartaric acid and sodium hydrogen carbonate,and not just sodium hydrogen carbonate in the yielding of dough

[quote="shadysaysurspammed"]$Na_2CO_3$ makes the taste bad so tartaric acid is used to neutralise it or something(as you said,the product tastes better)

I don't quite get what you want in part a).A salt or something?

BTW-are u in 10th?[/quote]

what is BTW .yes i am in 10th

i too dont know whether it salt but i think it is a base

[quote="shadysaysurspammed"]BTW means by the way

It should be a salt if na2co3 is a base(?).We had this chapter in 10th,some important chemical compound,big and boring...[/quote]

we dont have this chapter.but i saw this question in an entrance test conducted by a coaching centre for iit jee.thanks for the solution

[quote="inteluser"]Sodium tartrate is basic, yes.

$\text{Na}_2\text{C}_4\text{H}_4\text{O}_6+\text{H}_2\text{O}\longrightarrow\text{Na}^++\text{OH}^-+\text{H}_2\text{C}_4\text{H}_4\text{O}_6$

It is also a salt.[/quote]

ok.since it has produced $\text{OH}^-$ It is a base

can you solve the second question


function algebra polynomial quadratics AMC

1987-24: How many polynomial functions f of degree $ \ge$1 satisfy f($ x^2$)=$ [f(x)]^2$=f(f(x))?
(A) 0
(B) 1
(C) 2
(D) infinitely many
(E) finitely many but more than 2

1982-19: Let f(x)=$ |x\minus{}2|\plus{}|x\minus{}4|\minus{}|2x\minus{}6|$, for $ 2\le x \le8$. The sum of the largest and smallest values of f(x) is
(A) 1
(B) 2
(C) 4
(D) 6
(E) none of these

1980-22: For each real number x, let f(x) be the minimum of the numbers 4x+1, x+2, and -2x+4. The maximum value of f(x) is

(A) $ \frac {1}{3}$

(B) $ \frac {1}{2}$

(C) $ \frac {2}{3}$

(D) $ \frac {5}{2}$

(E) $ \frac {8}{3}$

1980-14: If the function f defined by f(x)=$ \frac {cx}{2x\plus{}3}$ for $ x\neq \minus{}\frac{3}{2}$ satisfies f(f(x))=x for all real numbers x except -$ \frac {3}{2}$, then c is
(A) -3
(B) -3/2
(C) 3/2
(D) 3
(E) not uniquely determined by the given information

Thanks!

Hints:

[hide="#24"]What is the degree of $ f(x^{2})$, $ (f(x))^{2}$, and $ f(f(x))$ in terms of the degree of $ f(x)$?[/hide]

[hide="#19"]Think about what the graph of the function looks like.  Pieces of it are linear, but it suddenly changes direction at several points.  You can narrow the process of finding the max and min down to plugging in several values of $ x$.[/hide]

[hide="#22"]Again, think about the graph.  Just like before, you can narrow the possible $ x$-values that the max can occur at to just a few values.[/hide]

[hide="#14"]One way is to plug the function definition into $ f(f(x))\equal{}x$ and solve for $ c$.  You can also consider the inverse function of $ f(x)$ (what can we say about it?).[/hide]

i think i get it, anyone have solutions?

[b]24[/b]
let f(x) be of degree n. Then f(f(x) is of degree n^2, f(x)^2 is of degree 2n, and f(x^2) is off degree 2n. However, the terms after the leading term are affected differently,so the polynomial is of the form $ x^2 \plus{} c$
Plugging in functions of the form ax^2 + bx + c clearly don't work $ \implies$ infinite solutions $ \boxed {E}$

EDIT: fixed

[quote="mathemonster"][b]24[/b]
let f(x) be of degree n. Then f(f(x) is of degree 2n, f(x)^2 is of degree 2n, and f(x^2) is off degree 2n. However, the terms after the leading term are affected differently,so the polynomial is of the for $ x^n \plus{} c$
so there are clearly infinite solutions $ \boxed {E}$[/quote]

f(f(x) is definitely not of degree 2n. It is of degree n^2.

For example, let f(x) = x^n. f(f(x)) = (x^n)^n = x^(n^2)

ok so $ n^2 \equal{} 2n \implies n \equal{} 2$. So $ f(x)$ is a quadratic

1980-22:

The "maximum minimum" will clearly occur when two lines intersect (because if it is not where two lines intersect, then moving the x value slightly up or down can guarantee a lower minimum).

For each pair, simple algebra will show that the intersection points are at [b]X[/b]= 2/3, 1/3, and 1/2.

Finding the minimum output for each of these x values, we get 8/3, 7/3, and 5/2.  The largest of these is 8/3, or (E)


Let's say we have unit vectors $ v_1,\ldots,v_n$ in $ m$ dimensional Euclidean space which are mutually orthogonal. ($ n<m$)
Prove that there exists a orthonormal basis of the $ m$ dimensional Euclidean space that contains these vectors.

(I was trying to rigorize 'coordinate rotation' in higher dimensions. Any help would be very welcome :) )

Minor detail: for the eventual orthonormal basis to include $ v_1,\dots,v_n,$ those $ v_j$ must themselves be orthonormal, not merely orthogonal. Let's just assume that that's the hypothesis.

There's a construction that isn't all that specific - you have many choices.

Since $ n<m,$ the vectors $ \{v_1,\dots,v_n\}$ do not span all of $ \mathbb{R}^m.$ Choose a vector $ w_{n\plus{}1}$ that lies outside that span.

Use the Gram-Schmidt process to produce $ v_{n\plus{}1}.$ That is, let

$ u_{n\plus{}1}\equal{}w_{n\plus{}1}\minus{}\sum_{j\equal{}1}^n\langle w_{n\plus{}1},v_j\rangle v_j$ and $ v_{n\plus{}1}\equal{}\frac{u_{n\plus{}1}}{\|u_{n\plus{}1}\|}.$

If $ n\plus{}1$ is still less than $ m,$ repeat this.

There is no attempt in this to have a unique answer.

Note that $ B': \equal{} \{v_1,\ldots,v_n\}$ is linearly independent. Now extend $ B'$ to a basis $ B'' \equal{} \{v_1,\ldots,v_n,v'_{n \plus{} 1},\ldots,v'_m\}$ of the $ m$-dimensional vector space. Apply the Gram-Schmidt algorithm using that $ B'$ is already an orthonormal system to obtain an orthonormal basis $ B$ containing $ B'$.

Alternatively you can take an ONB $ B_1$ of the orthocomplement of $ span (B')$. Then $ B'\cup B_1$ is an ONB of your vector space.


Hi, I'm a non-American student looking for helpful online resources for DECA competition.

My category is Business Law and Ethics, and Google doesn't seem to help..

If you know of any site, can you please post on this thread? That'd be very helpful..

Thanks in advance!

The only thing is that this forum is made only for economics related topics.


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number theory

Let $ a,b\in\mathbb{N*}$ and $ gcd(a,b)\equal{}1$.
Prove that exist infinity the prime number $ p\equal{}a.k\plus{}b,k\in\mathbb{N}$.

Dirichlet's theorem on primes in arithmetic progression.
There is no short proof known, but you may use the search function to find sketches and elementary proofs for special cases (like $ b\equal{}1$).

See http://www.mathlinks.ro/Forum/viewtopic.php?t=135341 .
And one algebraical proof for $ b\equal{}1$ can be found in http://www.mathlinks.ro/Forum/viewtopic.php?t=126566 .

More special we easy to prove that
1)There exist infinity prime in the form $ pk\plus{}1$ we p is a prime
2)There exist infinity prime in the form
$ k.2^{2^{m}}\plus{}1$ where m is natural.


When one post has many replies, replies are displayed by many next pages. How can I display all replies on a single page?

Well sadly i do not think you can. But you can list only posts from 1 day, 1 week etc.

I don't think you can. So how do you classify 'reply'?

@AwsomeToad: Like in this thread, bull98 made the "post", and monkeygirl, you, I, and anyone else who posts on this thread make the "replies".

It would be helpful if that could happen.

Excuse me if I sound rude, but what is the point of displaying everything in one page?
If you wanted to find something, you could just use the search button.... I don't see any other reason....... (Well, unless you are really lazy and don't want to click next page)

[quote="$ LaTeX$"]Excuse me if I sound rude, but what is the point of displaying everything in one page?
[/quote]

If you do a search and find match in, say one topic such as: "Mathcounts marathon!!!!".
There are 298 pages for "Mathcounts marathon!!!!".
Do you really like to click "Next page" 298 times to see which page your search is on?

I think "Display All" is a simple feature to add to this forum.

When you search, set the option "Display results as: Posts".


Suppose that $ p$ is an odd prime, prove that $ p\mid 1^k \plus{} 2^k \plus{} ... \plus{} \left(p \minus{} 1\right)^k$ for every $ k \in \left\{1,2,3,...,p \minus{} 2\right\}$.

There is some number $ a$ with $ p \nmid a , p \nmid a^k \minus{} 1$  (take for example some primitive root modulo $ p$).
Now the set of all prime residue classes is only permuted by multiplication with $ a$, so you get:
$ \sum_{i \equal{} 1}^{p \minus{} 1} i^k \equiv \sum_{i \equal{} 1}^{p \minus{} 1} (a \cdot i)^k \equiv a^k \sum_{i \equal{} 1}^{p \minus{} 1} i^k \mod p$
which immediatly gives
$ (a^k \minus{} 1) \sum_{i \equal{} 1}^{p \minus{} 1} i^k \equiv 0 \mod p$
and since $ p \nmid a^k \minus{} 1$ we get that $ p | \sum_{i \equal{} 1}^{p \minus{} 1} i^k$

(Note that this theorem is more or less the main step of elementary proof of the theorem of Chevalley-Warning)

[quote="suweijie"]wonderful,thank you very much. :D
i have had a proof already,but it should use many theorems.[/quote]
$w$ such that $ord_p{w}=p-1$.
We have: $1^k+2^k+...+(p-1)^k\equiv 1+w^k+w^{2k}+...+w^{(p-2)k}=$
$\frac{w^{(p-1)k}-1}{w^k-1}(mod p)$.
$p|w^{(p-1)k}-1, p\nmid w^k-1$.
So $p|1^k+2^k+...+(p-1)^k$.


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[size=200]Warning: Mild Swearing, Mature Audiences Only[/size]
[img]http://zoosviat.com/jokeoffice/wp-content/uploads/2007/10/image001.jpg[/img]
I think it's a bit of both on my part.
Seeing as a bunch of people were laughing at us Asians at school about it.
Basically a white dude would just come up to us and say "sum ting wong?" to our faces, and start laughing.

i laughed.  people are too over-sensitized if they truly find that offending.

this is really funny :D

anyways, if someone else says to you "Sum Ting Wong?", just talk to him about where he got the joke and other jokes....

what I do when some one makes a joke/tries to make fun of me, I just turn it around and they end up looking like a fool. Make another joke back at them. But make sure it's good.

I got challenged by my half-asian friend to say nothing racist or pertaining to race/religion/culture for 24 hours. I lost after 30 minutes after I called him a "wasian."

Just an anecdote.

Two reactions:
1) *snorts/laughs*
2) ...That's stupid. What simpleton came up with that?

Mixed feelings. It is kind of racist, but it's not [i]too[/i] demeaning. To a degree..

Hahaha I've seen this in 4 places online already. And it's not offensive just hilarious

:rotfl: .... This is funny. I don't find it offensive.

[quote="modeler"]I wonder why there are Japanese characters :rotfl:[/quote]

Probably made by an american or they used a template, still funny though, I think ill get in trouble for bringing it to school

[quote="Hamster1800"]I still can't get #10 - Wai Yu Kum Ching. I can't figure out what "Kum Ching" is supposed to be...someone explain?[/quote]
Why you crunching

#7 is really funny
i think it's kind of racist by playing around with Chinese sounds, but i think it's mainly to get people to laugh.  :lol:

Its supposed to be Wai Yu Mun Ching   [url]http://funnyjunk.com/funny_pictures/3865/Learn+Chinese+in+5+Minutes/[/url]

Ah, Wai Yu Mun Ching makes much more sense.

lol thats rlly funny. im going to show my friends.   :rotfl:

this was super funny!  (I am chinese, and didn't feel offended)

umm

funny [b]or[/b] racist?

Why can't it be both?

[quote="mz94"]Dont revive old topics[/quote]

I also saw this before.

sorry for revival, but I thought I could add more and remove the profanity:

Are you harboring a fugitive - Hu Yu Hai Ding
See me A.S.A.P. - Kum Hia Nao
Small Horse - Tai Ni Po Ni
Your price is too high - No Bai Dam Ting
Did you go to the beach - Wai Yu So Tan?
I bumped into a coffee table - Ai Bang Mai Ni
I think you need a facelift - Chin Tu Fat
It's very dark in here - Wai So Dim?
Has your flight been delayed? - Hao Long Wei Ting?
That was an unauthorized execution.- Lin Ching
I thought you were on a diet - Wai Yu Mun Ching?
This is a tow away zone. - No Pah King
You are not very bright - Yu So Dum
I got this for free - Ai No Pei
I am not guilty - Wai Hang Mi?
Please, stay a while longer - Wai Go Nao?
They have arrived - Hia Dei Kum
Stay out of sight - Lei Lo
He's cleaning his automobile - Wa Shing Ka
Does this bathroom stink! Hu Flung Dung?


Show that for any $ n \in N$, the number of positive divisors in $ n^{2}$ is odd.


(Me)

Modified from 2005 AMC 12 A #23

A rectangular box $ P$, with integer values as its sides, is inscribed in a sphere of radius $ r$.  The volume of $ P$ is 30 and the sum of the lengths of its 12 edges is 120. What is the volume of the sphere with radius $ r$?

$ \text{(A)} \ \frac{152 \pi \sqrt{38}}{3} \qquad \text{(B)} \ 1000 \qquad \text{(C)} \ \frac{1000 \pi}{3} \qquad \text{(D)} \ 988 \sqrt{2} \qquad \text{(E)} \ 279 \pi \sqrt{2}$

Let's say box P has side lengths $ x, y, z$.
Since $ V \equal{} 30$, $ x*y*z \equal{} 30$

There are 4 edges of each length $ x, y, z$ on the box,
so $ 4x \plus{} 4y \plus{} 4z \equal{} 120$ which equals $ x \plus{} y \plus{} z \equal{} 30$

Since each side has integer values, x, y and z are factors of 30,
so they must be $ 1, 2, 3, 5, 6, 10, 15, 30$.

But when we consider this, there aren't values of x, y, and z that satisfy the conditions. :huh:

And if we continue, we get the space diagonal of the box is $ \sqrt {x^2 \plus{} y^2 \plus{} z^2}$, making the radius $ 1/2\sqrt {x^2 \plus{} y^2 \plus{} z^2}$ and the area of the sphere $ 4/3(1/2\sqrt {x^2 \plus{} y^2 \plus{} z^2})^3\pi$, with no way to simplify.

Perhaps you meant that the sum of the faces is equal to 120? I think that might present a viable solution.
Edit: Sorry, accidentally pressed submit instead of preview!

Oh, my bad. I somehow solved the problem as $ 12(x\plus{}y\plus{}z) \equal{} 120$ and got $ x\plus{}y\plus{}z \equal{} 10$.

The answer is A if $ x\plus{}y\plus{}z \equal{} 10$ (three sides being 2,3, and 5).

Again, my bad... This is why this problem is the proposal problem, not the real one.  :D


calculus
integration
algebra
difference of squares
special factorizations

Solve the equation in integers
 $ x^{6} \plus{}3x^{3} \plus{}1\equal{}y^{4}$

[hide="Solution"]We have: $ x^6 \plus{} 3x^3 \plus{} 1 \equal{} y^4 \Rightarrow (x^3 \plus{} 3/2)^2 \minus{} 9/4 \plus{} 1 \equal{} y^4.$ Applying difference of squares yields: $ (x^3 \plus{} y^2 \plus{} 3/2)(x^3 \minus{} y^2 \plus{} 3/2) \equal{} 5/4 \Rightarrow (2x^3 \plus{} 2y^2 \plus{} 3)(2x^3 \minus{} 2y^2 \plus{} 3) \equal{} 5$. Clearly both factors are integral so either the first or the second factor equals $ 5$ (in the case that either of the two factors equals $ \minus{}5$, $ x^3\equal{}\minus{}3$). Solving these two linear systems, we have $ (x^3,y^2) \equal{} (0,1)$ and $ (x^3,y^2) \equal{} (0, \minus{} 1)$, respectively. Clearly the second solution fails to produce (non-gaussian) integers so from the first solution we have $ (x,y) \in \{(0,1),(0, \minus{} 1)\}$.

EDIT: Mentioned $ \minus{}5\Rightarrow x^3\equal{}\minus{}3$.[/hide]

There are two different ways to write 5 as a product of two integers ;)  (But it also leads to a non-integer solution.)

For $ x \ge 1$, $ (x^3 \plus{} 1)^2 < (y^2)^2 < (x^3 \plus{} 2)^2$ , impossible
For $ x \le \minus{} 2$, $ (x^3 \plus{} 1)^2 > (y^2)^2 > (x^3 \plus{} 2)^2$ , impossible again.
Checking $ x \equal{} 0, \minus{} 1$, we have $ (x, y) \equal{} (0,1), (0, \minus{} 1)$ are the solutions.


evaluate ${\sum_{\frac{n}{2} \le k \le n}} \frac{(-1)^k}{k} \binom{k}{n-k}$

The answer is$\frac{2cos(\frac{2\pi n}{3})}{n}$

At first, I calculate the sum $F_n(x)=\sum_{n\geq k\geq\frac{n}{2}}\binom{k}{n-k}x^k$
We have since$\binom{k}{n-k}=\binom{k-1}{n-k-1}+\binom{k-1}{n-k}$lead to
 $F_n(x)=x(F_{n-1}(x)+F_{n-2}(x)),F_1(x)=x,F_2(x)=x+x^2$
Then$F_n(x)=\frac{x}{2}.[(1+\frac{x+2}{\sqrt{x^2+4x}})(\frac{x+\sqrt{x^2+4x}}{2})^{n-1}+(1-\frac{x+2}{\sqrt{x^2+4x}})(\frac{x-\sqrt{x^2+4x}}{2})^{n-1}]$
Now,intergalting$F_n(x)/x$we have$\sum_{n\geq k\geq \frac{n}{2}} \frac{\binom{k}{n-k}x^k}{k}=\frac{(\frac{x+\sqrt{x^2+4x}}{2})^n+(\frac{x-\sqrt{x^2+4x}}{2})^n}{n}$
In particulary let $x=-1$ and Moivre'formula we have the result.


factorial
geometry
3D geometry
tetrahedron
counting
distinguishability

1.  Prove that n*C(n - 1, k - 1) = k*C(n, k) using a combinatorial argument.  (That means you cannot use the factorial formula for C(n, r))

2. What regular polygon has the same number of diagonals as sides?  Twice as many?  3 times as many?

3. How many ways can you select 3 distinct digits from {0, 1, ..., 9} so that no two consecutive digits have been selected?  What if we consider 0 and 9 to be consecutive?

4. In how many ways can n students be seperated into two teams such that each team has to have at least 1 member?  What about 3 teams?

5. How many 6-digit numbers are there whose digits are distinct and increase from left to right?

6. Prove that the digits of any six-digit number can be permuted so that the sum of the first three digits differs by at most 9 from the sum of the last 3 digits.

JBL wrote:6. Prove that the digits of any six-digit number can be permuted so that the sum of the first three digits differs by at most 9 from the sum of the last 3 digits.

Spoiler: [hide]Let the digits be a :le: b :le: c :le: d :le: e :le: f. We permute these digits to the number f b a d c e. then (f-e) + (b - d) + (a - e)  :le:  9, since f-e :le: 9 and the others are at most 0.[/hide]

JBL wrote:2. What regular polygon has the same number of diagonals as sides? Twice as many? 3 times as many?

Spoiler:[hide]The formula for the number of diagonals of an n-gon is n(n+3)/2. so n = n(n-3)/2 => 2=n-3 => n=5. generalizing, if a n-gon has a times as many diagonals as sides, 2a = n-3, n=2a+3.[/hide]

Tetrahedron -- for number 6, you need to look at the positive difference.  The scheme you described might give you a difference less than -9.

4.  In how many ways can n students be seperated into two teams such that each team has to have at least 1 member? What about 3 teams?

I'm assuming that the two teams are distinguishable.  So in the case where there are two groups, we can think each student will either be in or out of our group (and if he's out of our group, he's in the other one).  So that's 2^n, but this doesn't account for the times when there are n students in one group and 0 in the other.  This happens exactly twice (all in team A, or all in team B), so 2^n-2.  With similar logic, we have 3^n, but we want the cases where one group has 0, 2 groups have 0, 3 groups have 0.  One group has 0 in 3(nC2-2) cases... we subtract 2 cause in 2 of those cases, all will be in one group, and none in the other.  2 Groups have 0 in 3 cases since it means they're all in one group, 3 groups have 0 in no cases.  Is it 3^n -3*C(n,2) +3   I think that's probablly too high... enumerating the cases for n=4, im pretty sure this is wrong.

Are the teams distinguishable for #4? I presume the students are, but it's still an interesting problem if they aren't, at least for more than 2 teams.

1):  [hide]suppose we have n people, and we want to to choose a k person team, in which one person is designated captain.  On the one hand, the number of such teams is n*C(n-1,k-1) because we can first choose a captain from the n people, and then out of the remaining n-1 people, we choose k-1 to be the rest of the team.  On the other hand, this number is C(n,k)*k because we can choose a k person team from the n people first, and then each of the k people could be designated captain.[/hide]

[quote="JBL"]3. How many ways can you select 3 distinct digits from {0, 1, ..., 9} so that no two consecutive digits have been selected? [/quote]

I think it is 56 = C(10,3) - 9*8 + 8

Why ?
. C(10,3) = choice of 3 among 10
. 9*8 = 9 ways to choose 2 consecutive integers, 8 ways to choose the remaining integer
. 8 = in the previous count we count twice the case where the three integers are consecutive.

[quote="JBL"]What if we consider 0 and 9 to be consecutive?
[/quote]

56-6=50 ?
Because we just have to remove the case 0 b 9 where  2<=b<=7 (case b=1 or 8 removed in the previous count).

[quote="JBL"]5. How many 6-digit numbers are there whose digits are distinct and increase from left to right?
[/quote]

Ignoring for now integer beginning with '0'. If we choose six distinct integer we can build one and only one 6 increasing digits integer : C(10,6).
Now we must remove the integer beginning with 0 : C(9,5)

Total C(10,6)-  C(9,5) = 84.

correct ??

[quote="JBL"]5. How many 6-digit numbers are there whose digits are distinct and increase from left to right?[/quote]

I think I have an other (and to my opinion smarter) proof.

I claim that  a 6-digit number whose digits are distinct and increase from left to right can be written :
a * 111111 + b * 11111 + c * 1111 + d * 111 + e * 11 + f * 1
with (*) a+b+c+d+e+f<=9 and a>0,b>0,..,f >0.
And conversely such number is a good one.

With a'=a+1, ...,f'=f+1, (*) is equivalent to :
a'+b'+c'+d'+e'+f'<=3 or (**) a'+b'+c'+d'+e'+f' + g =3 with g >= 0.

The number of solutions of (**) is well known by 'De moivre' theorem, it is C(7+3-1,7-1)=C(9,6)=84  :mrgreen:

just a note for 3 and 5

3: in general, if we want to take a subset of k elements from a set of n elements s.t. there are not two consecutive elements, then the number of such subsets is C(n-k+1,k) because in any subset of k elements from n-k+1 elements, we add 1 spaces between each element (k-1 spaces in total), and end up with a subset of k elements from a set of n elements (n-k+1 + k-1 = n), and because we added the spaces, no two are consecutive.

5: if the digits are in increasing order from left to right, then 0 wont be an issue, since if the first number is not 0, none of the numbers are 0 so it is just C(9,6); for each subset of 6 numbers from {1,2,...,9} only one arrangement has them in increasing order.


A person who had just moved into a small community was charged with a serious crime. He knew nobody in the area. Whom did he call as a character witness?

The people at his old town because they'll vouch for him that he was not in the new town since recently.

The sheriff, since if he was a bad person, the sheriff would know who he was?


Simplify $ \sqrt{4! \times 5!}$. Express your answer in simplest radical form.

hello, since $ 4!\cdot5!\equal{}2880$ we get $ \sqrt{2880}\equal{}24\sqrt{5}$.
Sonnhard.

Alternatively, we can express $ 4! \times 5!$ as: \[ 4 \times 3 \times 2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1\] But note this is also \[ 4^2 \times 3^2 \times 2^2 \times 1^2 \times 5\] so we really have \[ (4 \times 3 \times 2)\sqrt{5} \equal{} 24\sqrt{5}\]


Given integers $ a,b,m$ s.t. $ \gcd(a,b)\equal{}1$, show that there exists an integer $ x$ such that $ \gcd(a\plus{}bx,m)\equal{}1$

Let $ (a\plus{}bx,m)\equal{}d(>1)$.
Now,we consider 3 cases:
1. $ d | a$ then, $ d|bx$. Now we take $ x$ such that, $ d\not | x$ which implies that, $ d |b$. Contradiction.
2. In the same way, if $ d|b$ then, $ d|a$. Contradiction.
3. If $ d\not | a,b$ then, we take $ x\equal{}m$ then,$ (a\plus{}mb,m)\equal{}d$ implies that, $ d|a$.Contradiction.


function
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As we all know that $ \mathbb R \equal{} \mathbb Q \cup \mathbb Q^c$, and we know from density theorem there is an rational number between two irrationals numbers and vice versa .

O.K , these numbers are on the real line they look like dense grass . [b]WITHOUT[/b] using the definition for the countable set . What is core difference that make $ \mathbb Q$ countable and $ \mathbb Q^c$ uncountable ?

I know we could order  $ \mathbb Q$ for some ordering , but I need the philosophy of the reason  ?   :wink:

Perhaps your philosophical gap would be covered by looking at measure theory.
The rational are a set of measure zero, while the irrationals are a set of measure 1.

Of course the proof uses the fact that the rationals are countable so it is not logically independent.

The intuition I have here is that one can specify an element of $ \mathbb{Q}$ using finite data but that this is impossible in general for $ \mathbb{Q}^c$ (by the way this is bad notation; you should specify what set this complement is with respect to).  This is clear even from the basic construction of $ \mathbb{R}$: to specify an element of $ \mathbb{R}$ really means to specify a Cauchy sequence of rational numbers and this is an infinite amount of data.

What kind of answer, exactly, are you looking for?

[quote="t0rajir0u"]The intuition I have here is that one can specify an element of $ \mathbb{Q}$ using finite data but that this is impossible in general for $ \mathbb{Q}^c$
[/quote]

What suppose that to mean more specific please

What I need is to convince me in some sense that apart from functions , definition that nevertheless $ \mathbb Q$ and $ \mathbb Q^c$ are infinite sets , and they are dense in real line beside each other , one of them is countable and the other one is uncountable ,

As Einstein said " You do not really understand something unless you can explain it to your grandmother"

Note for symbols : I think the symbols are clear here , and some of Math books use it . Thx anyway .

Rational numbers have terminating continued fractions and irrational numbers do not.  (This can be turned into a proof both that the rationals are countable and that the irrationals are not.)

[quote="Hidden Scofield"]As we all know that $ \mathbb R \equal{} \mathbb Q \cup \mathbb Q^c$, and we know from density theorem there is an rational number between two irrationals numbers and vice versa .

O.K , these numbers are on the real line they look like dense grass . [b]WITHOUT[/b] using the definition for the countable set . What is core difference that make $ \mathbb Q$ countable and $ \mathbb Q^c$ uncountable ? [/quote]I´m not sure what you mean by "not using the definition of countable set", but I think if you want to get to the philosophical core of the problem, you should observe it in a slightly more general context, and that is the context of cardinal numbers. While most of the time students are told about countable vs. uncountable, this notion is quite coarse and doesn´t reflect important aspects of (mathematical) reality, that is infinity is not unique as there are "infinitely" many cardinals. Hence the question arises how they behave in relation to each other. To make the picture more complete, one can involve the continuum conjecture and assume there is a subset of the real numbers whose cardinality lies strongly between those of the rational numbers and the real numbers. But since the continuum hypothesis is not decidable within ZFC, it is up to us whether or not to leave a transition step from infinite countability to uncountability. If we decide against CH, we get somewhat more complete picture of how that happens. However, if we decide in favor of CH, there is only pure mystics in how we come from the first cardinality to the second one. And that is also the limit of every philosophy, it is the unusual situation when the absolute knowledge seems to be a bounded function :P

Now, aside from this philosophical non-sense, you may want to have a look at the Baire´s theorem. It implies that $ \mathbb Q$ cannot be represented as countable intersection of open sets whereas $ \mathbb Q^c$ [b]can[/b] be represented as such (and then think of the characterization of open sets in $ \mathbb R$).

"ex falso quodlibet"


The sum of the first $ n$ positive integers is 6. What is the sum of
the first $ n \plus{} 1$ positive integers?

[quote=origamiguy777]Because the sum of the first $ n$ numbers is $ 6$, then by the formula $ \frac{n(n\plus{}1)}{2}$, we get that $ n$ is 3. Hence, we can plug $ n\plus{}1$ back into the previous formula, and so we get $ \frac{4(5)}{2}\equal{}10$. =)

$ The$ $ Origami$ $ Guy$[/quote]

but how does dat formla work? ert ert ert


I made This circuit but I get a little bit confused.

I make a lot of Resistances because I didn't want to make a lot of switch.
If I want the current in some component zero, I make the limit of the Resistence going to infinity.

Since that, the initial condition of the capacitances and inductor in $t=0$ is not zero.

The problem is:
Can I make this substituition?

I think Yes.

And how can I make the System of Differential Equations for this Circuit?

It's too much complicate to me.
Someone could help me?

Could you explain what you want a bit more clearly?
I tried to write the DE-s of the circuit but Pheww :D  It got so ugly that I just gave up.
I'm sure you can find the DE-s for this kind of circuits in electronics books. :wink:

I'm unable to apply the Kirchhoff's circuit laws...

It give too many equations, and I get a little bit confused, as I said.]

And what is "DE-s" ?

I dont have a book about electronics...

I just have wikipedia and google.

And this circuit is just a exercise I made. To learn about the oscilation of the current in the circuit. etc...
I'm noob. sorry.

I made this problem in Crocodile Physics and get some interestings results.

To make the circuit more general, we can connect the last 2 knots of the image, dont you think? with a resistance $R_{6}$

These kind of oscilators with 2R-2L are very intereting.

Study the power, the current, the charge, voltage, etc...

thanks

DE-s is the short writing of differential-equation. its a long word so I made it shorter  :D

ODE... hm.. ok. hahua...

Do you agree that the ODE system is linear?

I will try to make the system. but i dont know if i'm rigth or wrong.

it's too much complicate. borring to write. hehe


assume a=b.

-ab= -(b^2)                    (multiply by negative b)
a^2-ab=a^2-b^2            (add a^2)
a(a-b)=(a+b)(a-b)           (simplify)
a=a+b                            (divide by (a-b))
a=a+a                            (substitute a for b)
a=2a                               (add)
1=2                                 (substitute 1 for a.)

:D  elvenchamp

Mods, please move this topic to ... somewhere else. ;)

[quote="shobber"]For HOW MANY TIMES are you posting this problem!!?? Search $1=2$ and you will get a list of results! Mod better LOCK it![/quote]

Agreed :D


How many even three-digit integers  have the property
that their digits, read left to right, are in strictly increasing
order?

The answer for this problem is $ 34 \mbox{ integers}$

The problem has been discussed here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=433592#433592[/url]

LOL I just used casework :rotfl:

So what I basically did was like:

Case 1: '12' being the first two digits: there are choices of 4, 6,  and 8 for the last digit, so 3 choices.
Case 2: '13' being the first two digits: still 4, 6, 8, so 3
Case 3: '14' being the first two digits: only choices of 6 and 8, so 2 choices.

...etc.

I have no idea now I actually got it right... at the end I was adding up like so many numbers :P.

If anyone is wondering why it was $\binom{3}{2} + \binom{5}{2} + \binom{7}{2}$, it's because if the last digit is 4, then you can fill up the first 2 numbers with $1,2, \text{or} 3$ then you do 3 choose 2 since once you get the two numbers, there is [b]exactly[/b] one way they will be in strictly increasing ordeer, that's why order doesn't matter, AND we use this clever [b]choose method[/b]

[size=100]I randomly posted this since the first time I did this problem, I had the same question and I'm sure someone else may have it too[/size]


im probably going to read the executive summary but i cant bother with the whole flippin thing, i have a 40 hr + a week job!

[quote="Fermatprime"]I've read the Patriot Act, why not? It's probably easier.[/quote]

I wonder if any member of Congress can honestly make that claim...