Letf $m,n$ be odd integers such that $m^2-n^2+1$ divides $n^2-1$

Prove $m^2-n^2+1$ is a perfect square.

bump

on a side note, maybe you should post in National Olympiads and put links to these problems as I imagine they are from the 2005 Irish National Olympiad.

I never really understand how you use the word divides.

Do you say 2 divides 4 or 4 divides 2?  Thanks.

2 divides 4.  Or, sometimes, people will write it as $2|4$.

I really want to see a solution to this....

Maybe it should be moved to one of the more advanced areas?

Have a look at : [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5935[/url]
There is an answer in post #3


inequalities calculus derivative function trigonometry

Prove that for x >1,

$ \frac {x}{1 \minus{} x^2} \geq \frac {3 \sqrt {3} x^2}{2}$

[hide="Calculus Solution"]Divide an $ x$ on both sides and cross-multiply.  The desired inequality is equivalent to \[ 2\ge(3\sqrt{3})(x\minus{}x^3)\Rightarrow \frac{2}{3\sqrt{3}}\ge x\minus{}x^3\]  Taking the derivative of the RHS gives $ 1\minus{}3x^2,$ the positive critical value of which is $ \frac{1}{\sqrt{3}}$.  This turns out to be the x-value at which $ x\minus{}x^3$ has a relative maximum, so we substitute this into the function to get \[ \frac{1}{\sqrt{3}}\minus{}\frac{1}{3\sqrt{3}}\equal{}\frac{2\sqrt{3}}{3},\] as desired. $ \blacksquare$[/hide]

I realize there is a trig solution for this that doesn't involve the cheapness of calc... :(

[quote="Math Geek"]... and cross-multiply ... [/quote]
Watch out! Cross multiplying involves multiplying by $ 1\minus{}x^2$, which is [i]not[/i] necessarily positive for $ x > 0$.

[quote="Arrange your tan"][quote="4865550150"]Prove that for x>0,

$ \frac {x}{1 \minus{} x^2} \geq \frac {3 \sqrt {3} x^2}{2}$[/quote]

It is false; it is not true for $ x$ belonging in the interval $ (1, \infty)$
where a negative denominator and positive numerator gives only
negative values for the left-hand side fraction as compared to a
necessarily positive valued right-hand side fraction.

Also, it's not true for $ n \equal{} 1$, because you have an undefined
expression on the left-hand side.

So this inequality is not true for $ n$ belonging in the interval $ [1, \infty)$.[/quote]

sorry, i have make a mistake for topic.

[quote="TZF"][quote="Math Geek"]... and cross-multiply ... [/quote]
Watch out! Cross multiplying involves multiplying by $ 1 \minus{} x^2$, which is [i]not[/i] necessarily positive for $ x > 0$.[/quote]

Poop...  Yeah, whoops... :oops:


calculus
function
derivative
calculus computations

A rectangular storage box has a base that is 3 times as long as it is wide. The top is made from a material that costs 35 times the price per square meter of the material used for the rest of the box. Find the dimentions that would be necessary to ensure the minimum cost if the volume must be 3m3. :huh:

This question is giving me nothing but trouble.If anyone can help me out with this,I would really appriciate it.

Costs are always quantity*price, here you have objects of two different quantities and prices, so you have to add them.

Someone correct me if I'm wrong, but the cost and the volume  give a system with two functions that have to be solved, use a derivative on the cost to find out the minimum value.

I think the volume function should remove the need for a height, and since one side of the base is fixed with the other you should have a one-variable function. I think  :blush:

Let $w$ be the width; then the length $l$ is $3w$ and height $h$ is $3/(3w^{2})=1/w^{2}$. Invent your own monetary unit, in which the cost of 1 square meter of non-top material is 1 unit. The total cost of the box is $36lw+2lh+2wh$, which can be expressed in terms of $w$ alone and subsequently minimized.


geometry
Galois Theory
superior algebra
superior algebra theorems

Dear every body!
I am taking some courses this year, and I have a problem. I dont know about Galois theory and I need to master it as soon as possible to follow the course. So please tell me which book to read faster.

I also ask for some good books in Commutative Algebra (I have Atiyah) and Categories Theory. Which books I can go so far for this field.

Thank you very much.

For Galois theory, [url=http://www.jmilne.org/math/CourseNotes/FT421.pdf]Milne's lecture notes[/url] might be a good start. For commutative algebra, Eisenbud's [url=http://www.amazon.com/Commutative-Algebra-Algebraic-Geometry-Mathematics/dp/0387942696/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1224021814&sr=8-1]Commutative algebra with a view towards algebraic geometry[/url] contains a lot of information. I don't know about the category theory. In Amazon's site, Awodey's [url=http://www.amazon.com/Category-Theory-Oxford-Logic-Guides/dp/0198568614/ref=pd_bbs_sr_2?ie=UTF8&s=books&qid=1224021918&sr=1-2]book[/url] looks good.

A very standard category theory text is Categories for the Working Mathematician by Saunders MacLane.

and I think Field and Galois Theory by [i]Patric Morandi [/i] is good too.


Let $ABC$ be a triangle.$AH \perp BC(H \in BC)$.Let $P$ be a point on $AH$.$BP;AP$ cuts $AC;AB$ at $E;F$,respectively.Prove that $\hat{AHE}=\hat{AHF}$


calculus
integration
trigonometry
symmetry
function
calculus computations

I hope this is the right forum for this problem; if not, please move it to the correct forum and notify me of which forum that is.

Prove that:

$\int_0^{2\pi}\sin nx\cos mx\, dx=\int_0^{2\pi}\sin nx\sin mx\, dx=\int_0^{2\pi}\cos nx\cos mx\, dx=0$

and

$\int_0^{2\pi}\sin ^2nx\, dx=\int_0^{2\pi}\cos ^2nx\, dx=\pi$

for all integers $n,m$.

I think I remembered that correctly... :?

Try using product to sum trig identities.

Alternately, express everything in terms of complex exponentials and integrate.

$\int_0^{2\pi}\sin nx\cos mx\, dx$

Let $u=\pi - x$

Changing the limits will then result in:

$-\int_{\pi}^{-\pi} -\sin nu\cos mu \, du$
$=\int_{\pi}^{-\pi} \sin nu\cos mu \, du$
$=-\int_{-\pi}^{\pi} \sin nu\cos mu \, du$
$=-\int_{-\pi}^{\pi} \sin nx\cos mx \, dx$
(since $\int_a^b f(x) dx = \int_a^b f(u) du$)

But $\sin nx\cos mx \,dx$ is odd!

therefore, the integral is equal to [b]zero[/b].

The other integrals can be found as such.  Since they will all equal zero, they all equal each other.

for part 2, simply recognise that $sin^2nx=\frac 12 - \frac 12cos2nx$ and $cos^2nx=\frac 12 + \frac 12cos2nx$ and the result will follow.

The approach that my book used and that I like better involves the fact that $\int_0^{2\pi}\sin nx=\int_0^{2\pi}\cos nx=0$ for all integers $n$, which can be proven easily by $u$-substition, and then use the sine and cosine addition formulas. Anyone know how to do it that way?

Your solution is also good, and maybe faster than this other one, though. ;)

Alternatively, integration by parts also yields the solution.

I'll leave the sine and cosine addition method for someone else  :P

My earlier post suggested to use these: http://mathworld.wolfram.com/TrigonometricProductFormulas.html

The solution falls out immediately.

zeus_three's argument for the sin-cos case is simply odd-even symmetry, which could be seen more clearly by making the interval of integration $[-\pi,\pi].$ The symmetry doesn't help with the cos-cos and sin-sin cases.

The identities Nathan was looking for are these:

$\cos A\cos B = \frac12\cos(A+B)+\frac12\cos(A-B)$

$\sin A\sin B = -\frac12\cos(A+B)+\frac12\cos(A-B)$

I don't remember those - I re-derive them when I need them from $\cos A=\frac{e^{iA}+e^{-iA}}2$ and $\sin A=\frac{e^{iA}-e^{-iA}}{2i}.$

For the problem at hand, proving the orthogonoality of these trigonmetric functions, I prefer the integration by parts method. (Actually, I prefer doing my Fourier series in complex notation, but that's another story.)

Here's a very general setup. Suppose we have functions $f$ and $g$, both smooth enough and both periodic of period $2\pi.$ Suppose that $f''(x)=-\lambda f(x)$ and $g''(x)=-\mu g(x).$ Then if $\lambda \ne \mu,\,\int_0^{2\pi}f(x)g(x)\,dx=0.$

Proof: Assume WLOG that $\lambda\ne 0.$ (At least one of $\lambda$ or $\mu$ is nonzero.)

$\int_0^{2\pi}f(x)g(x)\,dx=-\frac1{\lambda}\int_0^{2\pi}f''(x)g(x)\,dx$

$=-\frac1{\lambda}[ f'(x)g(x)|_0^{2\pi}-\int_0^{2\pi}f'(x)g'(x)\,dx ]$

Since $f$ and $g$ are both periodic, the boundary terms evaluate to zero.

$=-\frac1{\lambda}[ -f(x)g'(x)|_0^{2\pi}+\int_0^{2\pi}f(x)g''(x)\, dx]$

$=\frac{\mu}{\lambda}\int_0^{2\pi}f(x)g(x)\,dx$

$(1-\frac{\mu}{\lambda})\int_0^{2\pi}f(x)g(x)\,dx=0.$

Since we assumed that $\mu\ne\lambda,$ we must have that the integral is zero.

This proof places the result in a very general context: given a self-adjoint linear operator, eigenvectors associated to different eigenvalues are orthogonal.

I had another method of proving it that seemed quite nice to me, though.

Disregarding the integral signs for now, since we can add inside an integral, we have $\sin nx\cos mx=\sin ((m+n)x)-\sin mx\cos nx$. Since $\int_0^{2\pi}\sin nx\, dx=0$, then $\sin nx\cos mx=-\sin mx\cos nx=\sin (-mx)\cos nx=\sin nx\cos (-mx)$, because $\cos (-x)=\cos x$ and $\sin (-x)=-\sin x$. We add $\sin nx\cos (-mx)$ to both sides, so we have $\int_0^{2\pi}\sin ((n-m)x)\, dx=\int_0^{2\pi}2\sin nx\cos (-mx)\, dx=0$, so $\int_0^{2\pi}\sin nx\cos (-mx)\, dx=\int_0^{2\pi}\sin nx\cos mx\, dx=0$.

For the other two, we have $\cos nx\cos mx=\cos ((m+n)x)-\sin nx\sin mx=\Rightarrow \cos nx\cos mx=-\sin nx\sin mx$ (since $\int_0^{2\pi}\cos ((m+n)x)\, dx=0$). We subtract $\sin nx\sin mx$ from both sides to get $\cos 2x=-2\sin nx\sin mx=0$, and the other follows from there.

For the second set, we have $\sin ^2nx=(1-\cos ^2nx)$, so $\int_0^{2\pi}(1-\cos ^2nx)\, dx\Rightarrow \int_0^{2\pi}dx=2\pi=2\int_0^{2\pi}\cos ^2nx\, dx$, so $\int_0^{2\pi}\cos ^2nx=\pi$, and the other can be found in the same way.


If there was an infinite plane of congruent circles and the circles are spaced as close as possible to each other, what is the percentage of the plane covered by the circles?

(Sorry I can't provide an image.  :( )

[geogebra]1753a5dfda23733322464e057710d87dea4ba267[/geogebra]
Since you can tile the entire plane with the equilateral triangle shown, we just need to find the percentage of the triangle covered by the sections of the circles.

If the radius of the circle is $ r$, then the side length of the triangle is $ 2r$.  The area of an equilateral triangle with side length $ s$ is given by $ \frac{s^2\sqrt{3}}{4}$, so the area of this triangle is $ \frac{(2r)^2\sqrt{3}}{4}\equal{}r^2\sqrt{3}$.

The part of a whole circle inside the triangle is $ \frac{1}{6}\plus{}\frac{1}{6}\plus{}\frac{1}{6}\equal{}\frac{1}{2}$, and the area of one circle is $ \pi r^2$, so the area of half a circle is $ \frac{\pi r^2}{2}$.

The desired ratio is $ \frac{\frac{\pi r^2}{2}}{r^2\sqrt{3}}\approx 91\%$


probability
integration
function
calculus
expected value
real analysis
graph theory

Hi all,
I found this question in a book(CLRS):
We have n balls. Suppose that balls are tossed into b bins. Each toss is independent, and each ball is equally likely to end up in any bin. What is the expected number of ball tosses before at least one of the bins contains two balls?
I tried to calculate it from definition of expected value. But the result is a very complicated summation:

sigma(from i=1 to b){(b-1)!*(1/3^i)*i^2/(b-i)!}

I think the _linearity of expected value_ may be applied. But I don't have any good idea. Thanx.

I do not really understand where your answer came from (especially the factor $ 1/3^i$ in the summand). On the other hand, using a different approach, I also ended up with a sum:
$ 1\plus{}(1\minus{}\tfrac 1b)\plus{}(1\minus{}\tfrac 1b)(1\minus{}\tfrac 2b)\plus{}(1\minus{}\tfrac 1b)(1\minus{}\tfrac 2b)(1\minus{}\tfrac 3b)\plus{}\dots$
Here the $ j$-th term is just the probability that the number of tosses is at least $ j$. I do not see any way to convert this sum into anything more reasonable (moreover, since it is asymptotically equivalent to $ \sqrt {\frac {\pi b}2}$, I suspect that no really simple expression exists).

[quote="fedja"]I suspect that no really simple expression exists).[/quote]
Thank you for your reply. But as you see the question requires "the number", not an asymptotically equivalent.
Is $ \sqrt {\frac {\pi b}2}$ a tight equivalent or not? If it is how much?
could we write something like $ E[X] \equal{} \sqrt {\frac {\pi b}2} \plus{} O (1)$

"The number" is a sum in this case and it seems like nothing can be done about it. You have to get used to the idea that not everything has a neat closed form and, even if it seems to have one, in most cases it is just a delusion. The expressions like $ {2b\choose b}$ are really long products that can be computed no faster than the sum I wrote and something as innocent as $ \sqrt 2$ is not computable at all.

How tight is the approximation? Let's do a "back of an envelope calculation". The idea is that the $ k$-th product is about $ \prod_{j\equal{}0}^{k\minus{}1}e^{\minus{}j/b}\equal{}e^{\minus{}k(k\minus{}1)/2b}$ which is accurate up to $ k^2/b^2$ in each factor giving a total error about $ k^3/b^2$. We are interested in $ k$ up to, roughly speaking $ \sqrt b$ (after that the products become quite small), so the error in this part is, indeed, $ O(1)$. Now, the sum of exponents can be viewed as a Riemann sum for $ \int_0^\infty e^{\minus{}x^2/2}$ if you divide it by $ \sqrt b$. The function is monotone decreasing, so the difference between the Riemann sum and the integral is at most the step $ 1/\sqrt b$ and, after multiplying by $ \sqrt b$, we get $ O(1)$ again. So, yes, it seems like $ O(1)$ is the right estimate for the error term here :).

Thank you again.
This is an upper bound, because $ 1\minus{}\frac jb < e^{\minus{}j/b}$ gives an upper bound. So the estimation $ \sqrt{\frac {\pi b}2}\plus{}O(1)$ is the for integral. Not for $ E[X]$. We need a lower bound to claim that $ E[X] \equal{} \Theta(\sqrt b)$.

$ 1\minus{}\frac jb\equal{}e^{\minus{}j/b}\plus{}O(j^2/b^2)$ ;)

that's funny because it is almost the same computation as when one tries to compute the self intersection time of a random walk on the complete graph, subject we talked a lot some weeks ago  :lol:

[quote="fedja"]$ 1 \minus{} \frac jb \equal{} e^{ \minus{} j/b} \plus{} O(j^2/b^2)$[/quote]
 :oops:
could you explain why the error of $ \prod_{j \equal{} 0}^{k \minus{} 1}e^{ \minus{} j/b}$ is $ O(k^3/b^2)$?

You have two products of $ k$ factors $ \prod_1^k u_j$ and $ \prod_1^k v_j$ with $ 0<u_j,v_j<1$ and $ |u_j\minus{}v_j|<C\frac{k^2}{b^2}$. Now, to get from the first product to the second one, start replacing $ u_j$ by $ v_j$ one factor a time. Every replacement results in the change by at most $ |u_j\minus{}v_j|$ and we have $ k$ of them.


Maybe this one allready was:

$a,b,c\geq  0,a+b+c=1$ then $\frac{a+1}{a+bc}+\frac{b+1}{b+ac}+\frac{c+1}{c+ab}\geq  9$

write $a+bc= 1-b-c+bc = (1-b)(1-c)$ and the analogues:

$\Longleftrightarrow \sum{\frac{a+1}{(1-b)(1-c)}}\geq 9$

$\Longleftrightarrow \frac{3-\sum{a^{2}}}{(1-a)(1-b)(1-c)}\geq 9$

$\Longleftrightarrow 3 \geq  9\sum{a^{2}b}+9\sum{ab^{2}}+18abc+\sum{a^{2}}$

$\Longleftrightarrow 2\sum{a^{3}}\geq \sum{a^{2}b}+\sum{ab^{2}}$, whic is true. ( $[3,0,0] \succ [2,1,0]$ )

No need to use Muirhead..
We have $a+bc=a(a+b+c)+bc=(a+b)(a+c)$, hence our inequality is equivalent to \[\sum\frac{a+1}{(a+b)(a+c)}\geq 9\] \[\iff\sum(a+1)(b+c)\geq 9(a+b)(b+c)(c+a)\] \[\iff 2\sum ab+2\geq 9\sum a\cdot \sum ab-9abc\] \[\displaystyle\iff 9abc+2\geq 7(ab+bc+ca).\]This is true by adding the following two inequalities:\[ 9abc+1\geq 4\sum ab\textrm{ (Schur)}\]\[1\geq 3\sum ab\] The second one is true by a well-known inequality:$(a+b+c)^{2}\geq 3(ab+bc+ca)$.
The end.

But it wasn`t really muirhead, i just wrote $[3,0,0] \succ [2,1,0]$ just not to write much at the solution to : $2\sum{a^{3}}\geq  \sum{a^{2}b}+\sum{ab^{2}}$, which is $a^{3}+b^{3}\geq ab(a+b)$ and the analogues.

[quote="andyciup"]No need to use Muirhead..
We have $a+bc=a(a+b+c)+bc=(a+b)(a+c)$, hence our inequality is equivalent to
\[\sum\frac{a+1}{(a+b)(a+c)}\geq 9 \]

\[\iff\sum(a+1)(b+c)\geq 9(a+b)(b+c)(c+a) \]

\[\iff 2\sum ab+2\geq 9\sum a\cdot \sum ab-9abc \]

\[\displaystyle\iff 9abc+2\geq 7(ab+bc+ca). \]
This is true by adding the following two inequalities:
\[9abc+1\geq 4\sum ab\textrm{ (Schur)}\]

\[1\geq 3\sum ab \]
The second one is true by a well-known inequality:$(a+b+c)^{2}\geq 3(ab+bc+ca)$.
The end.[/quote]
We don't need to use Schur here, only AM - GM:
The inequality is equivalent to
\[\sum\frac{(2a+b+c)}{(a+b)(a+c)}\ge 9 \]
Using AM - GM Inequality, we have
\[LHS \ge \sum\frac{2}{\sqrt{(a+b)(a+c)}}\ge \frac{6}{\sqrt[3]{(a+b)(b+c)(c+a)}}\ge \frac{18}{(a+b)+(b+c)+(c+a)}=9 \]
We are done. :)

[quote="toanhocmuonmau"]
The inequality is equivalent to
\[\sum\frac{(2a+b+c)}{(a+b)(a+c)}\ge 9 \]
Using AM - GM Inequality, we have
\[LHS \ge \sum\frac{2}{\sqrt{(a+b)(a+c)}}\ge \frac{6}{\sqrt[3]{(a+b)(b+c)(c+a)}}\ge \frac{18}{(a+b)+(b+c)+(c+a)}=9 \]
We are done. :)[/quote]

   Very nice sols . :)

[quote="toanhocmuonmau"]
The inequality is equivalent to
\[\sum\frac{(2a+b+c)}{(a+b)(a+c)}\ge 9 \]
[/quote]
$\Longleftrightarrow \sum\frac{1}{(a+c)}\ge \frac{9}{2}$. But by CS, $\sum\frac{1}{(a+c)}\ge \frac 9{2(a+b+c)}=\frac{9}{2}$ :)


integration
logarithms
topology
complex analysis
complex analysis unsolved

Show that a single-valued analytic branch of $ \sqrt{1\minus{}z^2}$ can be defined in any region such that the points $ \pm 1$ are in the same component of the complement. What are the possible values of $ \int\frac{dz}{\sqrt{1\minus{}z^2}}$ over a closed curve in the region?

Does this involve rewriting $ \sqrt{1\minus{}z^2}$ as $ (1\minus{}z)\sqrt{\frac{1\plus{}z}{1\minus{}z}}$? If not, how do I get started?

I'm assuming that Ahlfors does not have an incorrect exercise, but here is what confuses me. So whoever answers this, can you also point out why this is not a counterexample?

You should probably interpret $ \sqrt{1\minus{}z^2}$ as $ e^{1/2\log (1\minus{}z^2)}$. This is where $ z\neq \pm 1$ comes from. You can define $ \log f(z)$ in any simply connected region such that $ f(z)\neq 0$. So why don't you get a single-valued analytic branch from using the region $ \mathbb{C}\setminus ((\minus{}\infty, \minus{}1]\cup [1, \infty))$? Or is it that $ \pm 1$ are in the same component because the line goes through the point at infinity? Ah. Nevermind. I think I answered my own question.

Maybe this helps to get you started, though.  :roll:

Whether $ [1,\plus{}\infty)$ and $ (\minus{}\infty,\minus{}1]$ are in the same component depends whether we work in the topology of $ \mathbb C$ or $ \overline{\mathbb C}$. But I'd like to put complex analysis aside and emphasize logic: Ahlfors does not claim that "a single-valued analytic branch of $ \sqrt{1\minus{}z^2}$ cannot be defined in any region such that the points $ \pm 1$ are in different component of the complement".


logarithms
calculus
calculus computations

If I read correctly and $ y\equal{}x \ln^{2} x$, then $ \frac{dy}{dx}\equal{}\ln^{2}{x}\plus{}2\ln x$.

Product rule then chain rule gives $ \frac{dy}{dx}\equal{}\ln^2{x}\plus{}x\cdot 2\ln{x}/x\equal{}\ln^2{x}\plus{}2\ln{x}$.


geometry
circumcircle
incenter
geometry proposed

Let $A_1B_1C_1$ and $A_2B_2C_2$ be the circumcevian triangles of $P_1$ and $P_2$ in $ABC$ triangle. Now
$A_1A_2,B_1B_2,C_1C_2$ meet $BC, CA, AB$ at $X,Y, Z$ respectively. From $X$ drawn the tangent $XA_3$ to
the circumcircle of $ABC$ such that $A_3$ is on minor arc $A_1A_2$, similarly $B_3, C_3$. Then $AA_3,BB_3,CC_3$ concur

Let (O) be the circumcircle of the triangle $\triangle ABC.$ Using a central projection, project the circle $(O)$ into a circle $(O')$ and the line $XY$ to infinity. As a result, $A_1'A_2' \parallel B'C'$ and $B_1'B_2' \parallel C'A'.$ This means that $A'A_1', A'A_2'$ are isogonals of $A'B', A'C'$ and $B'B_1', B'B_2'$ are isogonals of $B'C', B'A'$ in the projected triangle $\triangle A'B'C'.$ Therefore, $P_1', P_2'$ are isogonal conjugates and $C'C_1', C'C_2'$ are isogonals of $C'A', C'B'.$ Consequently, $C_1'C_2' \parallel A'B'$ and their intersection $Z'$ is on the line at infinity. Tangents $X'A_3', Y'B_3', Z'C_3'$ to the circle (O') from the points $X', Y', Z'$ at infinity are parallel to $B'C', C'A', A'B',$ respectively, and the tangency points $A_3', B_3', C_3'$ are midpoints of the arcs $B'C', C'A', A'B'$ opposite to the vertices $A', B', C',$ respectively. Thus $A'A_3', B'B_3', C'C_3'$ are internal bisectors of the angles $\angle A', \angle B', \angle C'$ of the projected triangle $\triangle A'B'C'$ concurrent at its incenter $I'.$ As a result, the original lines $AA_3, BB_3, CC_3$ are also concurrent.


We know that
$x^2+y^2-2xy=(x-y)^2$
$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Then
1) Can $x^4+y^4+z^4+w^4-4xyzw$ be factored?
2) Can this be generalized?(i.e. Can $x_1^n+x_2^n+...+x_n^n-nx_1x_2...x_n$ be factored?)

I don't think $x^4+y^4+z^4+w^4-4xyzw$ can be factored, and even if it can, there doesn't seem to be a pattern.

Mathematica tells me that can't be factored :)

Actually, you can figure out the roots via solving the associated quartic, and it's not pretty.  In general, since this is a symmetric polynomial, it might help to look at the expression in terms of the elementary symmetric polynomials, $S_1 = x_1 + x_2 + \ldots +x_n$, $S_2 = x_1x_2 + x_1x_3 + \ldots + x_1x_n + x_2x_3 + \ldots + x_{n-1}x_n$, ... $S_n = x_1x_2\cdots x_n$.

Thank you for the replies. :)
This is just a thing I was wondering about. It turned out to be not interesting.


Don't worry, it isn't as lame as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172497]the first one[/url].

There is a 30 floor building. During lunch break, the elevator is extremely busy, because everybody is going home or to a restaurant to eat. One person works on the 20th floor. His archenemy works on the 10th floor. But the guy on the 10th floor is smarter, and knows some tricks. Who gets down first, and how?

Explanation of how the elevators works:
There are 2 types of elevators: up and down elevators
When lunch break starts, the down elevators are at the 30th floor. They load up some people on the 30th floor, then 29th, and so on. But once they get full, they precede to the bottom floor, unload everybody, then go back (without picking anybody  up along the way) to where it left off (assume everybody is loaded on from that floor when the elevators goes there) and repeats the process. The up elevators start at the first floor and do the opposite.

PS. For no apparent reason, there are no stairs in the building.

Very easy. That's why I called it another Lame Riddle. But the answer isn't as lame as the first one.

Hmm... say nobody's using the up elevator because everybody is going down (almost everybody, say) so the smart guy goes up to 21... and waits for the down.

Oh I know he jumps out the window!

Told you it was easy. You got it already (the first one, the go up one).

Actually, my dad told me this one when he was talking about Chinese society. But its like cutting; everybody does it, so in the end, it makes noone faster.

Thank you for spoiling it, math_explorer.

But it's rather funny.

Oh no! I forgot the hide's! *spanks self*

Still, I like jumping out the window a bit better... it doesn't say you have to arrive at the first floor alive. More lame.  :)

since the guy on the tenth floor is father down, he can use the up elevators to go to floor 22 in which by the time it gets to floor 20 it will be full.


Prove that if the area of two triangle be equal they are congurent by cavaliery principle.

deam amir

would you say , exactly , what do you want us to prove ?

what do you mean by being congurent by cavaliery principle?

and am i right about cavaliery?

cavaliery principle is about two shape in a plan and it says that if you can find a group of  parallen lines such that

for every one of the lines the parts of the line which  are in the shapes be equal then these shapes have equal area.

[quote="arman"]would you say , exactly , what do you want us to prove ? [/quote]

for every two triangles that have the same area we must prove that we can find a group of parallen lines such that 2 triangle are congruet by cavalieri concept .


algebra
polynomial
LaTeX
calculus
integration
real analysis
real analysis unsolved

Let f(x) and g(x) be polynomial with real coefficient of degree n+1 and n resp.where n >0 with leading coefficients A,B respectively then calculate this limit in terms of A,B,n

u have  said n tends to infinity but where is  n in the question? :lol:

Its written in question........
""f(x) and g(x) be polynomial with real coefficient of degree n+1""  :lol:

sorry din see that, just saw  the integral.  sry.  :)

The question, in $ \text{\LaTeX}$:
[quote="eureka-123"]Let $ f(x)=Ax^{n+1}+\cdots$ and $ g(x)=Bx^n+\cdots$ be polynomials of positive degree. Calculate
$ \lim_{n\to\infty} g(x)\int_0^x e^{f(t)-f(x)}\,dt$.[/quote]

First, we look at this qualitatively. If $ A<0$, $ f(t)-f(x)>0$ for $ t<x$ and large $ x$, so the limit is $ \pm\infty$, depending on the sign of $ B$. If $ A>0$, $ f(t)-f(x)<0$ for $ t<x$ and large $ x$, with the decrease growing ever steeper. The integral goes to zero- but we're multiplying it by something that goes to $ \infty$, and we need more precision.

Now, what can we do with the integral? Note that $ f'$ and $ g$ have the same degree, so dividing by $ f'$ would balance with multiplication by $ g$; we can get this by integrating by parts.
$ g(x)\int_0^x e^{f(t)-f(x)}\,dt=\int_0^x \frac{f'(t)g(x)}{f'(t)}e^{f(t)-f(x)}\,dt$
$ =\left[\frac{g(x)}{f'(t)}e^{f(t)-f(x)}\right]_{t=0}^{t=x}+\int_0^x \frac{f''(t)g(x)}{(f'(t))^2}e^{f(t)-f(x)}\,dt$
$ =\frac{g(x)}{f'(x)}-\frac{g(x)}{f'(0)}e^{f(0)-f(x)}+\int_0^x \frac{f''(t)g(x)}{(f'(t))^2}e^{f(t)-f(x)}\,dt$
The first term looks like $ \frac{Bx^n+\cdots}{(n+1)Ax^n+\cdots}=\frac{B}{(n+1)A}+O(\frac1x)$. The second term goes to zero exponentially fast, as $ e^{-f(x)}$ decays much faster than $ g(x)$ grows. The remaining integral goes to zero; near $ x$, $ \frac{f''(t)g(x)}{(f'(t))^2}=\frac{nB}{(n+1)A}\frac1x+O(\frac1{x^2})$ and $ e^{f(t)-f(x)}<1$. Away from $ x$, the exponential goes rapidly to zero, and those parts of the integral don't contribute significantly. The overall limit is $ \frac{B}{(n+1)A}$.

Wait- this doesn't quite work. How do we know we can divide by $ f'$ in that integration by parts? It might have zeros somewhere in $ [0,x]$. To fix this, we chop up the interval. Let $ R\ge 0$ be such that $ f'$ has no zeros in $ [R-1,\infty)$, and split the integral into $ \int_0^R+\int_R^x$. The first integral goes to zero by direct estimates, as $ g(x)e^{f(t)-f(x)}$ goes to zero uniformly and exponentially fast. The second is what we do the integration by parts on, and there are no problems with zeros of $ f'$. The result is unchanged.


MATHCOUNTS
geometry
probability
ratio
Asymptote
3D geometry
conics

Since the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39305]previous marathon[/url] was too hard at the end, it's time for a new one!!!

Problem 1:

Billy has a stack of papers 3 centimeters high. Each page is 8 centimeters wide, .0025 centimeters thick, and 10 centimeters long. If there is no space between the pages, how many pages is in Billy's paper stack?

Let me reiterate, problems should be [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=71119]MOEMS[/url] level, which does not mean Mathcounts. MOEMS problems can be done without algebra or graphing. Mathcounts problems sometimes require those techniques. Those that are too hard [i]will be[/i] summarily executed - I mean, deleted. Thanks :).

[hide="Solution"]
The only facts necessary are 3 cm thick and .0025 cm per page.
The answer is $ \frac {3}{.0025} \equal{} \boxed{1200}$
[/hide]
[size=200]Problem 2[/size]
Helga needs her curtain to be a certain area.  The current curtain is a square with area 324 square cm.  But it is too big and must be trimmed down to make a square with area 256 sq cm. What is positive difference of the side lengths for her current curtain and the trimmed down curtain?

324= 18^2
256=16^2

So, the positive difference between the sides in 2.

may i ask so we dont have any problems with this marathon, what exactly is MOEMS level.  I assume proofs arent MOEMS level but than what is.

[quote="Concentric"]324= 17^2
256=16^2

So, the positive difference between the sides in 2.[/quote] After you post a solution to a problem you should generally post another problem. If you don't' then the marathon stops. I hope this has been helpful.

[quote]MOEMS problems can be done without algebra or graphing. Mathcounts problems sometimes require those techniques. [/quote]

As clarification: The square of Jack's number is two more than Jill's number, and Jill's number is prime. What is Jack's number? is fine, because guess-and-check could be used without algebra to figure out what Jacks' number is. Similarly, How many ways can Joe choose 3 fruits from a pile of 2 oranges and 10 apples? is fine cause you can list out all the ways. But what is $ \binom{5}{2}$? and solve for $ x$ in $ x^2 \plus{} 5 \equal{} 20$ is not good.

Problem 3: The square of Jack's number is two more than Jill's number, and Jill's number is prime. What is Jack's number?

[quote="zolojetto"]Jacks number is seven because jills number is 47

[size=200]Problem 4[/size]
If your fingernails grow at a constant rate of .0005 cm per week. how many days will it be untill your fingernails are 1meter long?

I know this is MOEMS level but how much harder can these problems get.
EDIT: added problem[/quote]
It will be 2,000 weeks for 1 cm, so 200,000 weeks for 1 meter. 1,400,000 days.

[size=200]Problem 5[/size]
Bob the Builder has to have at least one vegetable and one fruit every day. If he can choose from squash, pumpkin, broccoli, and carrots for his vegetables, and tomatoes (yes, it is a fruit), plums, bananas, and apples for his fruits, what is the number of meals Bob can have?

[hide="Solution"]
You have 4 choices for veggies and 4 for fruit.  The answer is
$ 4\cdot4\equal{}\boxed{16}$
[/hide]
[size=200]Problem 6  [/size]
Nine professors are at a meeting.  If they all shake hands with each other, how many total handshakes were there?

[quote="minicon"][hide=Solution]

[size=200]Problem 6  [/size]
Nine professors are at a meeting.  If they all shake hands with each other, how many total handshakes were there?[/quote]

$ {18 \choose 9}$ ?

a) You're wrong.

b) Did you read above where I said binomial coefficients were more than MOEMS level?

[quote="minicon"][hide="Solution"]
You have 4 choices for veggies and 4 for fruit.  The answer is
$ 4\cdot4 \equal{} \boxed{16}$
[/hide]
[size=200]Problem 6  [/size]
Nine professors are at a meeting.  If they all shake hands with each other, how many total handshakes were there?[/quote]

[hide="without binomial coefficients"]
For person 1, there are 8 people
For person 2, there are 7 people not including person 1
For person 3, there are 6 people not including people 1 and 2
For person 4, there are 5 people not including people 1, 2, and 3
It's clear that the total number of handshakes will be the sum of all the cases for person 1 to person 8, which is the sum $ 8\plus{}7\plus{}6\plus{}5\plus{}4\plus{}3\plus{}2\plus{}1\plus{}0$, which evaluates to $ \boxed{36}$
[/hide]

Can't think of a problem :(

[size=200]Problem 7[/size]
James Hetfield needs to order some microphones.  They come in 5 mics for $ \$$42.00 or 3 for $ \$$20.00.  What is the cheapest cost to buy 126 microphones?

$ \$ 831.60$

[hide="Solution"]
3 for 20 USD is cheaper as it only costs $ 6.66$ each. Multiplying that by 126, you get $ \boxed{\$ 831.60}$
[/hide]

Someone else post a problem.

Well, I guess I'll post another.  (Maybe I should apply for a job with MOEMS for writing problems ;))
[size=200]Problem 8[/size]
What is the probability that when green, blue, and red balls are picked from an urn without replacement that the order is Blue, Green, Red?

I think he posted that question in the mc marathon also.

So you can make numbers up for this. lets assume that the stream is 1 mile. So swimming downstream, you go 12mph, and upstream, you go 6mph. So the speed you go without the river is x, where 12-x=6+x so x=3 so he goes 3 miles per hour. This makes the speed of the riber 9mph, so since the river is 1 mile, it should take 20/3 minutes.

[hide="My Solution"]
$ \frac 1{x \plus{} y} \equal{} 5$
$ \frac 1{x \minus{} y} \equal{} 10$

Therefore,
$ x \plus{} y \equal{} \frac 15$
$ x \minus{} y \equal{} \frac 1{10}$
$ 2x \equal{} \frac 3{10}$
$ x \equal{} \frac 3{20}$
$ \frac 1x \equal{} \boxed{\frac {20}3}$[/hide]

New problem:
In a bag are 20 marbles. Some are red, some are blue, some are yellow. Currently, there are twice as many yellow marbles as red marbles. If 8 yellow marbles are removed, the probability of picking a red marble is now $ \frac 12$. How many red marbles are in the bag?

[hide]20-8=12marbles left
1/2*12=6
There are 6 red marbles.
[/hide]
New Question: What is the second number that leaves a remainder of 3 when divided by 1,2,3,4,5,6,7, and 8?

So find the LCM of those numbers, double the result and add 3.

You find the LCM is 840, so the answer is 1683.

New problem:
Process $ x$ is defined on a rectangle as follows:
Shade half of the unshaded part of the rectangle (the first time, the whole rectangle).
Erase half of the shaded region.

If Process $ x$ is done on a $ 4 \times 4$ square 3 times, what is the area of the UNshaded region?

[u]Answer to Yongyi781's problem[/u]:

[hide]Since we only care about the final area of the unshaded region, we simply need to find 1/2 x 1/2 x 1/2 of the square's starting area, which is 16. 1/8 of 16 is 1/2, so the area of the final unshaded region will be 1/2.[/hide]

[u]New Problem[/u]:

If 4611686018427387904 is equal to 64^x, what is the value of x, expresses as a common fraction?

Yes. Out of my knowledge of powers of 2, $ 4611686018687427387904 \equal{} 2147483648^2 \equal{} \left(2^{31}\right)^2 \equal{} 2^{62} \equal{} \left(2^{6}\right)^{\frac {31}3}$. Therefore $ x \equal{} \frac {31}3$.

Besides, the answer to my original problem is incorrect.

First time 1/2*16=8
Then 4 is shaded 1
then 10 is shaded
then 5 is shaded 2
then 10.5 is shaded
then then 5.25 is shaded which means that 10.75 is unshaded.

Yongyi how many powers of 2 did you memorize!?!?!

Johny please dont bump a thread. look at the date, its been more than a month dont bump it!

I DIDN'T!!! Johny bumped a thread, in my opinion posting an anti spam like u did is spam. THIS IS CLOSED! NO MORE POSTS!

are you saying that your post wasn't spam?

get real.


probability
function
integration
geometry
geometric transformation
reflection
calculus

Hi,,
what is the difference between the mean and mode of a pdf (probability density function) ?
thanks

The words "probability distribution function" are somewhat ambiguous. There are two possible meanings:

1. Cumulative distribution function: $ F(x)\equal{}P(X\le x).$

2. Probability [i]density[/i] function. This is what is usually abbreviated as pdf. A function $ f$ such that $ F(x)\equal{}\int_{\minus{}\infty}^xf(t)\,dt.$ More loosely (when it makes sense), $ f(x)\equal{}F'(x).$

Not every random variable has a density. Only "absolutely continuous" random distributions have densities. (That "absolutely" reflects issues not usually addressed in a first undergraduate probability course - in that course, we mostly just say "continuous.")

There are three famous, named, measures of central tendency.

1. Mean, also known as expected value.

$ \mu\equal{}E(X)\equal{}\int_{\minus{}\infty}^{\infty}x\,dF(x)\equal{}\int_{\minus{}\infty}^{\infty}xf(x)\,dx,$

provided that integral converges absolutely.

2. Median. $ m$ is "the" solution of the equation $ F(x)\equal{}\frac12.$

Two special circumstances: for a discontinuous distribution, we might have $ F(a)>\frac12$ while $ \lim_{x\to a^\minus{}}F(x)<\frac12.$ In that case, we let $ m\equal{}a.$ It is also possible to have $ F(x)\equal{}\frac12$ at every point on some interval; in that case we would conventionally take $ m$ to equal the midpoint of that interval.

3. Mode. The worst defined and least used such measure, especially for continuous random variables.

Assuming we have a density $ f,$ the mode would be the location of the maximum value of $ f,$ provided that makes any sense. (Which it might not.)

Some would call any local maximum of $ f$ a mode and hence talk about bimodal or trimodal distributions.

If a random variable takes $ n$ distinct values $ x_1,\dots,x_n$ with probabilities $ p_1,\dots,p_n$, then the mean (aka "average value", "expectation", etc.) is the sum $ \sum_j p_jx_j$ and the mode is simply the $ x_j$ for which $ p_j$ is not less than any other $ p_i$. Both notions can be generalized to "reasonable" continuous distributions: for a real-valued random variable, the mean is $ \int_{\minus{}\infty}^\infty xp(x)\,dx$ and the mode is the point $ x$ where $ p(x)$ attains its maximum (here $ p(x)$ is the density function).

In layman terms, the mean is the average value of the outcomes of many independent trials. So, if the mean number of, say, good books written during each year is, say, 10, (and if you believe that each year production is independent of other year ones) you may be pretty sure that the number of good books written over 100 years will be quite close to 10\cdot 100 (this is known as the Law of Large Numbers). The mode, on the other hand, is the value you want to bet on if you are to play roulette where you win if and only if you guess the value exactly (this is true for the discrete model) or with high precision (for continuous model).

Oops, Kent beated me to it! Still I'll post what I wrote :P


I have a few problems I need help with.

If you are given the sum of the squares  of three consecutive integers, what five digits can never occur in the ones place?

Find a ten digit number such that the first digit tells how many time the digit 0 appears, the second tells how many 1's appear....last digit tells how many 9's appear

25 meters of ribbon, 0.1 mm thick, was tightly wound on a cardboard roll, giving a new roll whose diameter was 20 cm. What is the diameter of the cardboard roll?

I know these may seem really easy to you, but I have absolutely no idea what to do.

These are hard.

But for the 1st one, I think you just have to take a few integers and find a pattern. Remember that there are only 10 digits.

these are pretty difficult...
 for the first one you can just find a pattern by testing some small numbers

for the second one the answer is obviously 9000000000! Just use simple logic!

the third one, I think the answer is 18, but I'm unsure.

[hide="solution to 1)"]Let the middle number be $ x.$ The sum of the squares is thus $ (x\minus{}1)^2\plus{}x^2\plus{}(x\plus{}1)^2\equal{}x^2\minus{}2x\plus{}1\plus{}x^2\plus{}2x\plus{}1\equal{}3x^2\plus{}2.$ Recall that the  only digits that can be in the units digit of a square number are $ 0, 1, 4, 5, 6, 9.$ Multiplying each one of these by $ 3$ gives the units digit to be $ 0, 3, 2, 5, 8, 7,$ then adding $ 2$ makes them $ 2, 5, 4, 7, 0, 9.$ The remaining digits are $ \boxed{1, 3, 6, 8}.$ (are you sure it's 5 numbers?)[/hide]

@battered, the number of nines is not 0...

... My mistake... That makes the problem a little bit more complicated...

Ok. Now this is ridiculous. Why is everyone copying nike's avatar??? It's just not even funny anymore. I'm pretty sure this is not making nike happy so just find your own avatar guys!  :mad:

uh... Sorry about that... I just need a temporary avatar while I go avatar hunting. :rotfl:

recheck! you have 1215555550

There are not 5 3's

Oh, sorry 6210001000 (Is that right? I'm not sure)

hey!... That does look correct... Good Job! :lol:

I am sure it is 5 digits.

Ok here are some more for you

in the product 9*HATBOX=4*BOXHAT
find the six digit numbers, HATBOX and BOXHAT

In a three-game playoff series, the probability that team A will win each game is twice the probability that team B will win. What is the probability that team B will the series?

Softball team plays two games each weekend, one on Saturday and on on Sunday. The probability of winning on saturday is 3/5 and the probability of winning on Sunday is 4/7. What is the probability of :losing a game saturday; losing a saturday game and winning a Sunday game; winning a sunday game after already winning a Saturday game; winning both games; losing both games.

A bag contains small straws of the following lengths :2cm, 3 cm, 5cm, 7 cm, 11cm and 13 cm. Three straws are drawn at random. What is the probability that triangle can be formed with the straws that are drawn?

Finally, A two inch cube (2*2*2) of wilver weighs 3 pounds and is worth 198.36. How much is a 3 inch cube of silver worth?

I know there are alot sorry but it's helpful to me.

Please post new problems in a new topic. There are still unsolved problems.


Solve the following equation in the set of real numbers .
$ \lvert\ x \rvert \minus{}\lvert\ 2x \rvert \equal{}x$

_________________________________________
Azerbaijan Land of Fire

[quote="Pirkuliyev Rovsen"]Solve the following equation in the set of real numbers .
$ \lvert\ x \rvert \minus{} \lvert\ 2x \rvert \equal{} x$ [/quote]

$ x\equal{}0$

$ x \in (\minus{}\infty,0)$ there are infinitely many solutions

$ x \in (0,\plus{}\infty)$ there isn't any solution

[hide="My Solution"]
We apply casework.

Case 1: $ x \ge 0$
We can rewrite the equation using the identity $ \{x\} \equal{} x \minus{} \lfloor x \rfloor$.

$ \{x\} \minus{} \{2x\} \equal{} x\implies x \minus{} \lfloor x \rfloor \minus{} 2x \plus{} \lfloor 2x \rfloor \equal{} x\implies$ $ \; \minus{}\lfloor x\rfloor  \equal{} 2x \minus{} \lfloor 2x \rfloor\implies \minus{}\lfloor x\rfloor \equal{} \{2x\}.$

We have $ 0\le \{2x\} < 1\implies 0 \le \minus{} \lfloor x \rfloor < 1.$

Thus, we must have $ \minus{}\lfloor x \rfloor \equal{} 0\implies \lfloor x \rfloor \equal{} 0$. So we consider $ 0 \le x < 1$. We further break this into two cases.

For $ 0 \le x < \frac{1}{2}$, we have $ \{x\} \equal{} x$ and $ \{2x\} \equal{} 2x$. So our equation becomes $ x \minus{} 2x \equal{} x\implies x \equal{} 0$.

For $ \frac{1}{2} \le x < 1$, we have $ \{x\} \equal{} x$ and $ \{2x\} \equal{} 2x \minus{} 1$. So our equation becomes $ x \minus{} (2x \minus{} 1) \equal{} x\implies x \equal{} \frac{1}{2}$.

Case 2: $ x < 0$

We can rewrite the equation using the identity $ \{x\} \equal{} x \minus{} \lceil x \rceil$

$ \{x\} \minus{} \{2x\} \equal{} x\implies x \minus{} \lceil x \rceil \minus{} 2x \plus{} \lceil 2x \rceil \equal{} x\implies$ $ \; \minus{}\lceil x\rceil  \equal{} 2x \minus{} \lceil 2x \rceil\implies \minus{}\lceil x\rceil \equal{} \{2x\}.$

We have $ 0\le \{2x\} < 1\implies 0 \le \minus{} \lceil x \rceil < 1.$

Thus, we must have $ \minus{}\lceil x \rceil \equal{} 0\implies \lceil x \rceil \equal{} 0$. So we consider $ \minus{}1 < x \le 0$. We further break this into two cases.

For $ \minus{}1 < x < \minus{}\frac{1}{2}$, we have $ \{x\} \equal{} \minus{}x$ and $ \{2x\} \equal{} \minus{}2x \minus{} 1$. So our equation becomes $ \minus{}x \minus{} (\minus{}2x \minus{} 1) \equal{} x\implies 1 \equal{} 0$.

For $ \minus{}\frac{1}{2} \le x < 0$, we have $ \{x\} \equal{} \minus{}x$ and $ \{2x\} \equal{} \minus{}2x$. So our equation becomes $ \minus{}x \minus{} (\minus{}2x) \equal{} \minus{}x \plus{} 2x \equal{} x$.

Combining the cases, we obtain the solutions $ \minus{}\frac{1}{2},0,\frac{1}{2}$.
[/hide]

There are some flaws in my solution and I need someone else to correct them. The solutions I got are the only ones I think.

[hide="Large Hint"]$ \{ x \} \minus{} \{ 2x \} \equal{} x$
$ \iff (x \minus{} \lfloor x \rfloor) \minus{} (2x \minus{} \lfloor 2x \rfloor) \equal{} x$
$ \iff \lfloor 2x \rfloor \minus{} \lfloor x \rfloor \equal{} 2x$

$ \lfloor 2x \rfloor \minus{} \lfloor x \rfloor \in \mathbb{Z} \implies \boxed{2x \in \mathbb{Z}}$[/hide]


[quote="mr.k_74"]Probability Problems (Jenks)

--------------------------------------------------------------------------------

Special thanks to Chinaboy for creating these problems. This is copyrighted, by the way.

1. I have 4 red socks, 5 white socks, and 5 green socks. If I choose 3 socks, what is the probability that I will have 3 different-colored socks?

2. There are 13 swimmers and 12 divers in my swim class. If I pick 2, what is the probability that I'll have one swimmer and one diver?

3. I have 4 green pens, 4 black pens, 4 blue pens, and 4 red pens. If I choose 4, what is the probability that I'll have 4 different-colored pens?

4. I have 4 red socks, 5 white socks, and 5 green socks. If I choose 3 socks, what is the probability that I will have at least 2 socks that are the same color?

5. If I pick 3 cards from a deck, what is the probability that they are all face cards?

6. If I have 4 girls and 4 boys on my Mathcounts team, what is the probability that both captain and vice-captain are boys?

7. If I randomly chose a point in an isosceles triangle ABC (AB=AC) and the probability that it is closet to the base (BC) is 1/4. What is the probability that the point is close to AB?

8. If I have 4 girls and 4 boys in my mathcounts class, what is the probability that my four-person team has at least 1 girl? What about at least 2 boys?[/quote]

[hide="Solution #1"]
You have 4 red socks and a total of 14 socks. The probability of getting a red sock out of the 14 would be $ \frac{4}{14}\equal{}\frac{2}{7}$. Then there are only 13 socks because we have already taken a sock out and the problem does not state to replace the sock.  Next there are 5 white socks and a total of 13 socks so the probability of getting a white sock would be $ \frac{5}{13}$. Now there're only 12 socks because you have already chosen two socks. There are 5 green socks and a total of 12 socks so the probability of choosing a green would be $ \frac{5}{12}$.

Then, using the [url=multiplication principle]http://www.icoachmath.com/SiteMap/MultiplicationPrinciple.html[/url] we can multiply the 3 fractions together to get $ \frac{2}{7}*\frac{5}{13}*\frac{5}{12}\equal{}\frac{25}{546}$

But that's not the answer because you can choose the color of the socks 6 different ways.
RWG
RGW
GRW
GWR
WGR
WRG
So you would have to multiply $ 6*\frac{25}{546}\equal{}\frac{150}{546}\equal{}\frac{125}{273}$
$ \boxed{\frac{125}{273}}$ is the answer.

[/hide]

[hide="Solution #2"]
You have 13 swimmers and 25 total people so the probability of choosing a swimmer would be $ \frac{13}{25}$. Then you only have 24 total people because you already chose someone and you can't replace him. Then you have 12 divers and 24 people so the probability of choosing a diver would be $ \frac{12}{24}\equal{}\frac{1}{2}$

Using the multiplication principle we can multiply them together $ \frac{13}{25}*\frac{1}{2}\equal{}\frac{13}{50}$
But then we have two ways of choosing them: We can either choose a swimmer first and a diver last or a diver first and a swimmer last. So then we multiply $ \frac{13}{50}*2\equal{}\frac{13}{25}$.

$ \boxed{\frac{13}{25}}$ is the answer.

[/hide]

[hide="Solution #3"]
1st: $ \frac{4}{16}\equal{}\frac{1}{4}$
2nd: $ \frac{4}{15}$
3rd: $ \frac{4}{14}\equal{}\frac{2}{7}$
4th: $ \frac{4}{13}$

Multiplied altogether you get: $ \frac{8}{1105}$

But there is 24 ways to order them in the way you choose it so $ 24*\frac{8}{1105}\equal{}\frac{192}{1105}$
$ \boxed{\frac{192}{1105}}$[/hide]

[hide="6."] There are $ \binom{4}{2}$ ways to choose two boys, and $ \binom{8}{2}$ ways to choose two people, so
$ \frac{\binom{4}{2}}{\binom{8}{2}}\equal{}\boxed{\frac{3}{14}}$.[/hide]

[hide="8."] There is only one way that no girls could be on the team (if all the members were boys), and there are $ \binom{8}{4}$ ways to pick a team, so $ 1\minus{}\frac{1}{\binom{8}{4}}\equal{}\boxed{\frac{69}{70}}$.

For the second part, we shall consider the number of ways in which we may have either $ 0$ or $ 1$ boys on the team. (The denominator, will be the same as the first part). There is only one way that we could have $ 0$ boys. In order to find only one boy, there are $ \binom{4}{3}$ ways to pick the girls, and $ \binom{4}{1}$ ways to pick the boy. Thus, $ 1\minus{}\frac{1\plus{}16}{70}\equal{}\boxed{\frac{53}{70}}$.[/hide]

@jonathanchou711- #1 is a combination problem, for the order in which we pick the socks doesn't matter. Meaning as long as we choose 3 different colored socks out of 14 we have, we're satisfied. So the answer for #1 is

$ \frac{4\cdot5\cdot5}{\binom{14}{3}}\equal{}\boxed{\frac{25}{91}}$

I think you made the same mistake on #3, where we can use combination to get

$ \frac{4\cdot4\cdot4\cdot4}{\binom{16}{4}}\equal{}\boxed{\frac{64}{455}}$

[hide] [u][b]#7[/b][/u]
Since the triangle is isoceles, it is symmetric along the line AD, if D is the midpoint on the base BC. Meaning, the probility of a point being close to the legs AB and AC should be the same. Since a point is closest to BC 1/4 of the time, the probability of AB should be

$ \frac{1\minus{}{1/4}}{2}\equal{}\boxed{\frac{3}{8}}$[/hide]

[hide="5."] There are $ \binom{12}{3}$ ways to choose three face cards, and $ \binom{52}{3}$ ways to chosse three cards. So, $ \frac {\binom{12}{3}}{\binom{52}{3}} \equal{} \boxed{\frac {11}{1105}}$.[/hide]

[hide="4."] There are $ \binom{14}{3} \equal{} 364$ ways to pick three socks. The only way we will not have two of the same color is if we pick one of each. This can be done in $ \binom{4}{1}\binom{5}{1}\binom{5}{1} \equal{} 100$ ways. Thus,
$ 1 \minus{} \frac {100}{364} \equal{} \boxed{\frac {66}{91}}$.[/hide]

[hide="Problem 4"]$ \text{You can either get it by getting a swimmer then a diver or a diver then a swimmer so there are two cases.} \frac{13}{25}\cdot\frac{12}{24}\equal{}\frac{13}{50}\text{ and } \frac{12}{25}\cdot\frac{13}{24}\equal{}\frac{13}{50}\text{ }\frac{13}{50}\plus{}\frac{13}{50}\equal{}\frac{26}{50}\equal{}\boxed{\frac{13}{25}}$[/hide]

[hide="Problem 5"]$ \text{There are a total of 12 face cards out of the regular 52 card deck. So } \frac {12}{52}\cdot\frac {11}{51}\cdot\frac {10}{50} \equal{} \frac {3}{13}\cdot\frac {11}{51}\cdot\frac {1}{5} \equal{} \frac {11}{13\cdot17\cdot5} \equal{} \boxed{\frac {11}{1105}}$[/hide]


Let $p = 4$, and let $a = \sqrt{1}, b = \sqrt{2}, c = \sqrt{3}....$. Compute $(p-a)(p-b)(p-c) \cdots (p-z)$.

 ;) So easy for Math Meet.

[hide="The Somewhat Surprising Answer"]$0$[/hide]

[hide]$4-\sqrt{16}=0$[/hide]
I had to edit this a few times :blush: , since it didn't work as I intended to.

[quote="GoBraves"]Here's a well known problem, which is similar to the one Silverfalcon posted:

Compute $(x-a)(x-b)(x-c).................(x-z)$

Im expecting a very fast answer to this[/quote]
[hide]Also 0 since (x-x)=0[/hide]
This problem appeared in Puzzles 101 by Nob Yoshigahara, the inventor of Rush Hour.


HI! I have a litle problem:

I created my blog here and than I deletd it . Now I want to create another one .How can I do It.

i'm thinking you can recreate your blog the same way you created it in the first place

The first time a created it in my profile there was writeen  Create new blog now after I deleted it in the same place is writen blog options.

I know why. I clicked on your blog, and you have made a new blog, it's just an empty blog.

so i'm guessing what u should do is....
go to your blog and create a new post/topic or something

Try reading the stickies in this forum every once in a while  :wink:


Recently there has been many discussions on how to improve the AIME, including allowing people to take both AIME's. Here is a more feasible, simpler solution (in my opinion): extend the time limit by 30 minutes. So most people who are trying to qualify for USAMO will have enough time to check over their work and to attempt more difficult problems.

Couldn't you achieve the same effect by forcing yourself to stop a half hour early?

I'll take that as sarcasm...
but, I think that we should all just get back to studying math instead of discussing the AIME, myself included.

2.5 hours is plenty of time to do the work on the AIMEs.

[quote="Elemennop"]Or just make it a 3 hour long complaining fest? We could just be given a couple hundred sheets, and then just right ALL of our complaints on it. Forget the math, I think complaints is the direction the AIME should go with.[/quote]

I agree wholeheartedly. Dude, this is ingenious.

And then for the USAMO, we can write proofs using logical fallacies.

This is probably the best idea I've seen all week.  Heck, you could even make it an even four hours.  The only problem is that no matter how much extra time you allow, there will be some students who spend all the extra time gunning for one or two extra points, then get dissatisfied with their scores because they made careless mistakes because they didn't check answers in the extra time.

But still, allowing an extra hour is the least disruptive fix for some of the problems the AIME has.  And frankly, I don't think scores will go up that much except for the increase due to people making careless errors.  If you can't solve an AIME-style problem in the first 3 hours, it's unlikely you'll solve it in an extra hour.

[quote="Sly Si"]  If you can't solve an AIME-style problem in the first 3 hours, it's unlikely you'll solve it in an extra hour.[/quote]
If you can't brute force an AIME-style problem in the first 3 hours, it's likely you'll brute force it in an extra hour.  :lol:

Guys, you cannot honestly believe the AMC will consider sweeping rules changes simply because you did not perform as well as you could have or would have liked to have.
-----------------------------------
I'm sorry, Beta, for your misfortune--you seem to be just about the unluckiest person in the world.  (I guess that doesn't make you feel better ;))

I cannot bear the thought of having to work on the same 15 problems for 4 hours. As of now, I get bored of the AIME at the 2 hour mark. My brain just shuts off, and I slowly start checking my work.

Looking for clever solutions generally wears down the computation, or substituting variables nicely.

But honestly, I would not expect any change from the AIME. I just don't think people are realizing the brutal nature of the AIME. The object of the AIME isn't to conform to the wants of the masses. That goes against the essence of the contest. The people who are serious about passing the test must realize their own shortcomings, and try to first pass the test, before attempting all their "stretch" problems. Even if problems don't seem like a stretch to you, you have to lock up your early points first. No amount of time will change this concept. It is rarely an actual matter of time between passing or not. Its a matter of strategy. You just need to figure out first the strategy that will get you to qualify for USAMO, and to stick with that plan.

Its okay to complain about the test for a while, but this is getting excessive. You have to realize that the test had no real flaws to it. Having quite a bit of computation isn't a flaw. Its merely an aspect. None of the questions was [i]actually[/i] misleading if you read carefully. It is your own responsibility to do what you need to in order to get questions right. If you have to check your work, so be it. There isn't a problem with the test.

This is more of a general statement than a post directed just at feiqi. You can only blame the test for so much. So what if it didn't have great questions this year, or rather, what we consider great questions. There were still questions that had answers that were possible to get without losing an arm and a leg. Therefore you should just stop complaining.

It sucks to make alot of mistakes, I know. I've done it before hundreds of times. But to spend 6 to 15 threads talking about it is making this forum sound like a place to whine, and I'm tired of it. At this point, just congratulate the people you know have made it, and promise yourself that you'll work hard to make it next year. And if you're a senior, well you've got better things to look forward to anyway. Not making the USAMO isn't the end of the world (contrary to popular belief).

Please stop complaining about the actual formats and number of times you can take the AIME and all that. Sometimes you screw up. Sometimes you can't handle the pressure. Sometimes you don't spend enough time checking your work. But you're responsible for your performance. Always.

Oh, and I'm locking this thread. Put your format requests/complaints into the 2 AIMEs a year thread.


[quote="Big M"]There is a number line of integers from -1 to 12. Jeff will pick a card at random from ten cards numbered 1 through 10. The number on this card will indicate his starting point on the number line. He will then spin a fair spinner with 3 equal sectors, one of which says "move 1 space left", and two which says "move 1 space right", and follow the instruction indicated by his spin. From this new point he will spin the spinner again and follow the resulting instruction. What is the probability that he ends up at a multiple of 3 on the number line? express as a common fraction
<------------------------------------>
-1 0 1 2 3 4 5 6 7 8 9 10 11 12[/quote]
[hide]In order to land on a number divisible by 3, we need to consider the following 3 cases:

1) Starting position is either 1, 4, 7, or 10: In this case the only way to reach a number divisible by 3 is moving 1 space to the right twice. The probability of moving 1 space to the right from any number is $\frac{2}{3}$. So the probability of arriving at a number divisible by 3 is $\frac{2}{3}\cdot \frac{2}{3}= \frac{4}{9}$.

2) Starting position is either 2, 5, or 8: In this case the only way to reach a number divisible by 3 is moving 1 space to the left twice. The probability of moving 1 space to the left from any number is $\frac{1}{3}$. So the probability of arriving at a number divisible by 3 is $\frac{1}{3}\cdot \frac{1}{3}= \frac{1}{9}$.

3) Starting position is either 3, 6, or 9: In this case the only way to reach a number divisible by 3 is either first moving to the left once and to the right once or moving to the right once and to the left once. So the probability of arriving at a number divisible by 3 in this case is $\frac{1}{3}\cdot \frac{2}{3}+\frac{2}{3}\cdot \frac{1}{3}= \frac{4}{9}$.

Each of the starting points from 1 to 10 has $\frac{1}{10}$ probability of being selected. Hence the desired probability is $\frac{4}{10}\cdot \frac{4}{9}+\frac{3}{10}\cdot \frac{1}{9}+\frac{3}{10}\cdot \frac{4}{9}= \boxed{\frac{31}{90}}$.[/hide]


An integer between 1000 and 9999 (inclusive) is called [i]balanced[/i] if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers between 1000 and 9999 are there?

Spoiler:

[hide]I would do it case-by-case - just find the number of possibilities for each possible sum of either side.

Let x equal the sum of the two leftmost digits (call it LMD) or rightmost digits (call it RMD).  Then

If x = 1, LMD = 10 and RMD = 10 or 01.  Thus 1*2 = 2 poss.

If x = 2, LMD = 20, 11; RMD = 20, 11, 02.  Thus 2*3 = 6 poss.

If x = 3, LMD = 30, 21, 12; RMD = 30, 21, 12, 03.  Thus 3*4 = 12 poss.

this pattern (where the LMD has x possibilities and the RMD have x + 1 possibilities) continues up until x = 9:

If x = 9, LM = 90, 81, 72, 63, ... , 18; RM = 90, 81, ... , 09.  Thus 9*10 = 90 poss.

now for x = 10 through x = 18, the number of possibilities will steadily decrease.  However, now that 0 can no longer be used (since if it were, then 0 + some single-digit number  :ge: 10), the LMD and RMD will have the same number of possibilities.

If x = 10, LMD = 91, 82, ... , 19; RMD = 91, 82, ... , 19.  Thus 9*9 possibilities.

If x = 11, LMD = 92, 83, ... , 29; RMD = 92, 83, ... , 29.  Thus 8*8 possibilities.

...

If x = 18, LMD = 99; RMD = 99.  Thus 1*1 = 1 possibilities.

In all, there are 2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 + 90 + 8^2 + 7^2 + 6^2 + 5^2 + 4^2 + 3^2 + 2^2 + 1^2 = 330 + 8(9)(17)/6 = 330 + 204 = 534.  Answer.[/hide]

Spoiler:

[hide]I got a different answer, but like mathfanatic said, consider each case separately. Think of the different possibilities for the sums of the two leftmost and the two rightmost digits. They range from 1 to 18.

So when the sum is 1, we've got the two cases, 1010 and 1001.

When the sum is 2, we've got 6 possibilities, 1120, 1102, 1111, 2020, 2002, and 2011.

By continually listing these, you find a pattern. The sum of the numbers in this pattern would be, like mathfanatic said

2+6+12+20+30+42+56+72+90=330

This pattern is  :Sigma: x*(x+1), with x ranging from 1 to 9. Now work backwards from 18.

There is only 1 case for which the sum is 18, namely 9999.

We've got 4 cases for 17, 9898, 9889, 8989, 8998.

9 cases for 16, 7979, 7988, 7997, 8879, 8897, 8888, 9797, 9779, 9788.

And so on, so from 18 down, they are just prefect squares. So for the case with 10, there are 81 possibilities, making the grand total

330+81+64+49+36+25+16+9+4+1=615[/hide]

Oops, I forgot to include 9^2 = 81.  Add 81 to 534 and get 615.  Stupid mistakes.


Let Gamma be a semicircle with center O on diameter AB.  Let C be the mid-point of the sim-circular arc AB.  Let P be an arbitrary point on the semicircle different from both A and B.  Determine all points Q on the semicircle such that if the lines BP and AQ intersect at a point S, then C is the orthocenter of triangle SPQ

$P$ must lie on arc $BC$, and $Q$ is the point on arc $AP$ such that $\angle{POQ}$ is right.

Observe that a necessary and sufficient condition is that $\overline{QC} \perp \overline{BP}$ and $\overline{PC} \perp \overline{AQ}$. Oberserve that the former is impossible unless $P$ is on arc $BC$. (*) Take $P_0 = B$ and draw the tangent to the semicircle at $P_0$. The only possible $Q$ with $\overline{QC} \perp \overline{BP_0}$ is $Q_0 = C$. Let $D_0$ be the intersection of lines $Q_0C$ and $BP_0$, and $E_0$ is the foot of the degenerate altitude from $C$ to line $AQ_0$. However, it is easily seen that $\triangle{P_0Q_0S}$ is an isosceles right triangle with right angle $Q_0$. Now, take $P$ on the arc $BC$. Construct $D$, the foot of the perpendicular from $C$ to line $BP$. Let $DC$ intersect the semicircle again at $Q$. Finally, let $E$ be the foot of the perpendicular from $C$ to line $AQ$. We have $\angle{PCP_0} \cong \angle{PBP_0} \cong \angle{DCD_0} \cong \angle{QCQ_0} \cong \angle{QAQ_0} \cong \angle{EQE_0}$. We see that $\angle{POQ}$ is right, but also that $E$, $C$, and $P$ are collinear; whence, $C$ is the orthocenter of $SPQ$ for this uniquely constructed $Q$.


prove that the area of a rhomus is given to be d1(d2) / 2
d1= diagonal 1 and d2 = diagonal 2

Hint :

[hide]d1 orthogonal to d2[/hide]

Simple solution:

[hide]

The rhombus can be disected into 4 congruent triangles.  Since a rhombus is orthodiagonal, the width of one triangle is d_1 /2 and the height is d_2 /2.  The area of one triangle is (d_1(d_2))/8.  The area of the rhombus is (d_1(d_2))4/8 = (d_1(d_2))/2.

[/hide]

[hide]The diagonals divide the rhombus into 4 congruent triangles.  The base and height of each triangle is d1/2 and d2/2, the area of each triangle is (d1*d2)/8.  So the area of the rhombus is 4 times that, which is (d1*d2)/2.[hide][/hide][/hide]