calculus integration trigonometry Asymptote function rotation geometry

if you put $u=(\frac{1}{x}-1)^{\frac{1}{3}}$ it becomes something like
$I=\int_{0}^{\infty}\frac{3 u du}{(u^{3}+1)}$
To find $I$ you can for example integrate $f(z)=\frac{z}{z^{3}+1}$ on the contour:
$\int_{0}^{R}f(x) dx+\int_{\theta=0}^{\frac{2 \pi}{3}}f(R e^{i \theta}) iRe^{i \theta}d\theta-\int_{0}^{R}f( x e^{\frac{2 i\pi}{3}}) e^{\frac{2 i\pi}{3}}dx = 2 i \pi Res(f ; e^{\frac{i\pi}{3}})$
and let $R$ tend towards $+\infty$. You should find the answer. But maybe there is a simpler approach .. (but it works  :lol: )

Thank you for your solution.  I figured the integral might be easier with a substitution, but I am trying to figure out how to solve it directly.  That is, integrate over an arbitrary contour that encompasses the slit from 0 to 1 in the complex plane.  (What is slit in French, "coupure"?)

So in this case, $f(z)$ is multi-valued.  On top of the slit, I believe $f(z)=[z^{2}(z-1)]^{-\frac{1}{3}}e^{-\frac{i \pi}{3}}$ and on the bottom of the slit $f(z)=[z^{2}(z-1)]^{-\frac{1}{3}}e^{-\frac{5 \pi i}{3}}$.

Clearly the points at 0 and 1 are problematic.  I know they are branch points, but can they also have residues there too?  If so, I can calculate it at 0, but not at 1.  Regardless of how I manipulate the numbers, I fail to get the result.  Do you know how to finish this directly?

Just a short follow-up to Alekk's reply.  I was thinking about his substitution, and it reminded me of the beta function.  By far the shortest solution to this problem is as follows:

$\int_{0}^{1}\frac{dx}{\sqrt[3]{x^{2}-x^{3}}}=B(\frac{1}{3},\frac{2}{3})= \frac{\Gamma(\frac{1}{3}) \Gamma(\frac{2}{3})}{\Gamma(1)}= \Gamma(\frac{1}{3}) \Gamma(\frac{2}{3}) = \frac{\pi}{\sin(\frac{\pi}{3})}=\frac{2 \pi}{\sqrt 3}$.

Unfortunately, I would still like to see a solution with residues...  :(

Consider two contours $\Gamma_{1}$ and $\Gamma_{2}$:
[asy]import cse5; size(250); MP("0",D((0,0)),8,(0,-2)); MP("1",D((1,0)),8,(0,-2)); D(CR((0,0),0.1,15,345)--CR((1,0),0.1,-165,165)--cycle); D(CR((0,0),3),green); MP("\Gamma_1",(0.5,-0.05)); MP("\Gamma_2",(0,-3)); MP("\Omega",(1,-1));[/asy]
It is not hard to see that $F(z)=\frac 1{\sqrt[3]{z^{2}-z^{3}}}$ is analytic in the region $\Omega$ between $\Gamma_{1}$ and $\Gamma_{2}$. Actually, this formula determines $3$ different functions; let us choose the branch that is negative for real $z>1$ for definiteness. Then $F(z)\approx e^{-2\pi i/3}\frac 1{\sqrt[3]{x^{2}-x^{3}}}$ on the lower horizontal part of $\Gamma_{1}$ and $F(z)\approx e^{2\pi i/3}\frac 1{\sqrt[3]{x^{2}-x^{3}}}$ on the upper horizontal part. Since singularities at $0$ and $1$ are too weak to create non-trivial integrals over the circular parts of $\Gamma_{1}$, we conclude that $\int_{\Gamma_{1}}F(z)\,dz\approx-\sqrt 3i\,I$ where $I$ is the integral to find.

On the other hand, on $\Gamma_{2}$, we have $F(z)\approx-\frac{1}{z}$, so $\int_{\Gamma_{1}}F(z)\,dz\approx-2\pi i$. Since the integrals over $\Gamma_{1,2}$ should be the same, we get the desired conclusion.

Here I used a big circular contour centered at $0$ for $\Gamma_{2}$ but the same reasoning can be restated in terms of residue of $F(z)$ at [b]infinity[/b], so, we, indeed, have a residue theorem here only the function to consider is analytic [b]outside[/b] a closed curve $\Gamma_{1}$ encircling $0$ and $1$. Alekk's substitution just brings $\infty$ to a finite point; otherwise our solutions are the same.

see example V:
http://en.wikipedia.org/wiki/Methods_of_contour_integration

Here's the relevant surfaces:  The first set of plots is the Riemann surfaces for the function f[z]=$\frac{1}{\sqrt[3]{z^{2}-z^{3}}}$.  In the above  analysis I believe the Red real sheet and the Blue imaginary sheets are chosen to represent $f[z]=Red[x,y]+iBlue[x,y]$  These are shown in the second set.   $f[z]$ defined this way is then integrated over $\Gamma_{2}$ shown as the black contour and the upper portion of $\Gamma_{1}$ is the yellow contour and the lower part of $\Gamma_{1}$ is the green contour.  Plots like these help me better understand these types of integrals and perhaps they will also you.  I like this part of Complex Analysis. :)

[img]http://img257.imageshack.us/img257/6027/cuberootmultifunctionnz6.jpg[/img]

[img]http://img257.imageshack.us/img257/8615/cuberootbranchzl8.jpg[/img]

[quote="edoo"]Thanks shawtoos.  The plots are beautiful!

I assume you generated them in Mathematica.  If so, what commands did you use?  Also, as a follow-up, is there any way to plot Schwarz-Christoffel maps in Mathematica?  (off-topic, but I am curious)[/quote]

I used notebooks created by Michael Trott who published a series "Visualization of Riemann Surfaces" in the Mathematical Journal.  This is the web site of the Wolfram Library Archive:

http://library.wolfram.com/infocenter/Articles/3014/

I believe there is a way to get these notebooks without a subscription to the Journal.  You may just have to dig around the web to figure out how.    I had to modify his code quite a bit to get control of the display.  Here is a rudimentary example which plots the principle sheet of the imaginary component:

p1 = Plot3D[Evaluate[Im[(z^2 - z^3)^(-1/3) /. z -> x + I
                  y]], {x, -3, 3}, {y, -3, 3},
            PlotPoints -> 30, AspectRatio -> 1,
      PlotRange -> {{-3, 3}, {-3, 3}, {-3, 3}}, BoxRatios -> {1, 1, 1}];

It's poor and under ver. 5.2, does not remove the fill along the branch cuts.  The code written by Trott does a much nicer job of creating the sheets though.
I don't know about Schwarz-Christoffel maps.

Edit:
Might I also suggest if you use Mathematica, install Live3DGraphics.  This give you the ability to rotate a 3D object interactively using the mouse.  I find having this ability really helps with understanding the geometry of Riemann surfaces.  Here's an animation file of Blue[x,y] (if the moderators feel it's annoying, just remove it or else tell me):

[img]http://img511.imageshack.us/img511/2919/animate5rh9.gif[/img]

I hope it's OK to revive a thread that is several years old.

I'm having a hard time understanding how to determine the value of $f(z)$ just above and just below the cut.


Find all solutions in positive integers of the equation $x^{2}+xy-y=1993$

[hide="Hint"]Subtract 1 from both sides.  This is really just one instance of Simon's Favorite Factoring Trick.  I believe there are 8 solutions.  There are also 24 more where one or both of the variables are negative.[/hide]

To finish it off.. $z=x-1$ then for all factors of 1992, $y=\frac{1992}{z}-z-2$ then sub x back in.


Let $ a,$ $ b,$ $ c$ be three positive real numbers such that $ a\plus{}b\plus{}c\ge 3$. Prove that:
\[ \frac {a^3\minus{}a^2}{(a\plus{}b)^2}\plus{}\frac {b^3\minus{}b^2}{(b\plus{}c)^2}\plus{}\frac {c^3\minus{}c^2}{(c\plus{}a)^2}\ge 0\]

This problem can be solved by AM-GM and Cauchy-Schwartz inequality.

Firstly, let's rewrite the original inequality into the following form

\[ \sum_{cyc}\frac{a^3\minus{}a^2}{(b\plus{}c)^2} \equal{} \sum_{cyc}\frac{a(a^2\plus{}b)}{(a\plus{}b)^2} \plus{}\sum_{cyc}\frac{b}{a\plus{}b} \minus{}3\]

Now, using Cauchy- Schwartz and AM-GM consecutively, we get

\[ LHS \ge \sum_{cyc}\frac{a}{1\plus{}b} \plus{}\sum_{cyc}\frac{a}{c\plus{}a} \minus{}3 \ge 4\frac{a\plus{}b\plus{}c}{a\plus{}b\plus{}c\plus{}1}\minus{}3\]

Lastly, we possess the fact that $ a\plus{}b\plus{}c \ge 3$ so

\[ \frac{4(a\plus{}b\plus{}c)}{1\plus{}a\plus{}b\plus{}c} \minus{}3 \ge 0\]

We complete our proof here.

[quote="great math"]This problem can be solved by AM-GM and Cauchy-Schwartz inequality.

Firstly, let's rewrite the original inequality into the following form
\[ \sum_{cyc}\frac {a^3 \minus{} a^2}{(b \plus{} c)^2} \equal{} \sum_{cyc}\frac {a(a^2 \plus{} b)}{(a \plus{} b)^2} \plus{} \sum_{cyc}\frac {b}{a \plus{} b} \minus{} 3\]
Now, using Cauchy- Schwartz and AM-GM consecutively, we get
\[ LHS \ge \sum_{cyc}\frac {a}{1 \plus{} b} \plus{} \sum_{cyc}\frac {a}{c \plus{} a} \minus{} 3 \ge 4\frac {a \plus{} b \plus{} c}{a \plus{} b \plus{} c \plus{} 1} \minus{} 3\]
Lastly, we possess the fact that $ a \plus{} b \plus{} c \ge 3$ so
\[ \frac {4(a \plus{} b \plus{} c)}{1 \plus{} a \plus{} b \plus{} c} \minus{} 3 \ge 0\]
We complete our proof here.[/quote]
Your proof is so nice, great math! Thank you very much.
My proof is much uglier than yours, but let me present it. You can see it here: http://can-hang2007.blogspot.com/2009/12/inequality-14-n-p-sy.html

[quote="mathVNpro"]Let $ a,$ $ b,$ $ c$ be three positive real numbers such that $ a \plus{} b \plus{} c\ge 3$. Prove that:
\[ \frac {a^3 \minus{} a^2}{(a \plus{} b)^2} \plus{} \frac {b^3 \minus{} b^2}{(b \plus{} c)^2} \plus{} \frac {c^3 \minus{} c^2}{(c \plus{} a)^2}\ge 0\]
[/quote]
oh
this is my problem for VietNam Inequality Contest 2009


hi guys

can you still qualify for USAMO if you take AIME 2?
our school is on spring break during AIME I so I'll have to take the AIME II on march 28...

is this one still official? @_@

thanks!

ps
i know this question sounds stupid lol but anyway thx for any info!

yep I got to Choate and we're in the same boat...

the AIME I and AIME II are effectively the same unless one is radically harder than the other. So, you can qualify for USAMO via the AIME II, as I hope to do, and the index for qualification should be the same.  :)

I think they have a different index for the AIME I and AIME II, so you won't be at a statistical disadvantage.


Let $ABC$ be a triangle and $\angle A=60$. Let $P$ be a point in the $\triangle ABC$ such that $PA=1,PB=2,PC=3$.
Find $\angle B,\angle C$ when the area of $\triangle ABC$ is maximum.

Look at http://www.mathlinks.ro/Forum/viewtopic.php?p=230 .

  darij

$kvmedya $ baraye chi in soal ro dadi?????????????????????????????
aaaaaaahhhhhhhhhh


Let $ a,$ $ b,$ $ c,$ $ d,$ and $ e$ be distinct integers such that
\[ (6 \minus{} a)(6 \minus{} b)(6 \minus{} c)(6 \minus{} d)(6 \minus{} e) \equal{} 45.
\]What is $ a \plus{} b \plus{} c \plus{} d \plus{} e?$

$ \textbf{(A)}\ 5 \qquad \textbf{(B)}\ 17 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30$

[hide="Solution"]
The product, through experimentation, has to be
$(1)(-1)(3)(-3)(5)$

This is weak, but I was under time pressure, and the solution works.

You can also do it with vieta's and newton's sums, I think.

Therefore, we can easily find a, b, c, d, and e.

$a = 5; b = 7; c = 3; d = 9; e = 1$

A + B + C + D + E = 1 + 3 + 5 + 7 + 9,

$\boxed{25}$

Choice (C)
[/hide]

[hide]
Well just by looking at it you see that $a,b,c,d,e$ are all odd because right side is odd. Therefore the only values that can work are $1,3,5,7,9$ and since all the values are distinct therefore:

$a+b+c+d+e=1+3+5+7+9=5^{2}=\boxed{25}\implies\boxed{C}$[/hide]

This problem is pretty simple, you just have to think for a moment

Whenever I see a series of products, my automatic reaction is to see if the other side can be factored into a series of prime products.  $ 45$ clearly can, and $ 45=3^{2}*5=3*3*3$.  We don't know which of the products on the left side equals the numbers on the right, so just assign them randomly (I did it from largest to smallest so it's organized).

One of the many ways to do this is $ (6-a)=5$, $ (6-b)=3$, $ (6-c)=3$, $ (6-d)=(6-e)=1$ (this is because all other factors must be $ 1$, or the product would obviously exceed 45)

It's clear that $ (a,b,c,d,e)=(1,3,3,5,5)$, so their sum is $ 17$.

Good thing to do, looking for prime factorizations.

P.S. Is it ok to bump old posts like this?  Please tell me so if not.

First, it says that they are all distinct. The answer is 25, and the solutions are in the 2nd and 3rd posts.

Second, we usually don't respond totopics that are around 1 or more months old and have already been answered.


If $ f(2x)\equal{}\frac{2}{2\plus{}x}$ for all $ x\geq 0$, the find $ 2f(x)$.

[hide="solution"]
Substituting $ y\equal{}2x,$ we have the equation \[ f(y)\equal{}\frac{2}{2\plus{}\frac{y}{2}}.\] So \[ 2f(x)\equal{}\frac{4}{2\plus{}\frac{y}{2}}\equal{}\boxed{\frac{8}{4\plus{}y}}.\][/hide]

Oh sorry I just realized the $ \frac{8}{4\plus{}y}$ should be $ \frac{8}{4\plus{}x}.$


geometry
3D geometry
sphere
inequalities
IMC
college contests

Suppose $n\geq 4$ and let $S$ be a finite set of points in the space ($\mathbb{R}^3$), no four of which lie in a plane. Assume that the points in $S$ can be colored with red and blue such that any sphere which intersects $S$ in at least 4 points has the property that exactly half of the points in the intersection of $S$ and the sphere are blue. Prove that all the points of $S$ lie on a sphere.

I think similar problems have been solved on the forum before.

Consider red points are have label $1$ and blue points have label $-1$. Then take $n>4$. There are $\ge 3$ points with either $1$ or $-1$. Choose three points which are labeled with $1$. The other points are partitioned according to the sphere passing through the first three points which contains them. The sum of the points on a sphere (except for the three points in question) is $-3$, and if there are $m\ge 1$ spheres through the three points, the sum of the labels of all the points is $s=3-3m\le 0$.

On the other hand, if we choose three points which have labels $-1$, we get (in exactly the same manner) $s\ge 0$, and the two inequalities combine to give $s=0\Rightarrow m=1$, hence the conclusion.

yeah, it seems to be rather short and obvious in compatison with official solution at imc-math.org :)


inequalities
IMO Shortlist
inequalities unsolved

$ (x,y,z)>0$ with $ xyz\equal{}1$

prove that:

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

thanx.

[quote="reason"]$ (x,y,z) > 0$ with $ xyz \equal{} 1$

prove that:

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

thanx.[/quote]

Hint:

Cauchy :

$ \frac {x^{3}}{(1 \plus{} y)(1 \plus{} z)}\plus{}\frac{1\plus{}y}{8}\plus{}\frac{1\plus{}z}{8} \ge \frac{3x}{4}$

By AM-GM,

$ \frac {x^3}{(1 \plus{} y)(1 \plus{} z)} \plus{} \frac {1 \plus{} y}{8} \plus{} \frac {1 \plus{} z}{8}\geq 3\sqrt [3]{\frac {x^3}{(1 \plus{} y)(1 \plus{} z)}\cdot \frac {1 \plus{} y}{8}\cdot\frac {1 \plus{} z}{8}} \equal{} \frac 34 x$

$ \frac {y^3}{(1 \plus{} z)(1 \plus{} x)} \plus{} \frac {1 \plus{} z}{8} \plus{} \frac {1 \plus{} x}{8}\geq 3\sqrt [3]{\frac {y^3}{(1 \plus{} z)(1 \plus{} x)}\cdot \frac {1 \plus{} z}{8}\cdot\frac {1 \plus{} x}{8}} \equal{} \frac 34 y$

$ \frac {z^3}{(1 \plus{} x)(1 \plus{} y)} \plus{} \frac {1 \plus{} x}{8} \plus{} \frac {1 \plus{} y}{8}\geq 3\sqrt [3]{\frac {y^3}{(1 \plus{} x)(1 \plus{} y)}\cdot \frac {1 \plus{} z}{8}\cdot\frac {1 \plus{} y}{8}} \equal{} \frac 34 z$

Summing up these ineqaulities and rearranging, we have

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

$ \geq \frac {x \plus{} y \plus{} z}{2} \minus{} \frac 34\geq \frac 32 \sqrt [3]{xyz} \minus{} \frac 34 \equal{} \frac 32\cdot 1 \minus{} \frac 34 \equal{} \frac 34.\ Q.E.D.$

[quote="kunny"]By AM-GM,

$ \frac {x^3}{(1 \plus{} y)(1 \plus{} z)} \plus{} \frac {1 \plus{} y}{8} \plus{} \frac {1 \plus{} z}{8}\geq 3\sqrt [3]{\frac {x^3}{(1 \plus{} y)(1 \plus{} z)}\cdot \frac {1 \plus{} y}{8}\cdot\frac {1 \plus{} z}{8}} \equal{} \frac 34 x$

$ \frac {y^3}{(1 \plus{} z)(1 \plus{} x)} \plus{} \frac {1 \plus{} z}{8} \plus{} \frac {1 \plus{} x}{8}\geq 3\sqrt [3]{\frac {y^3}{(1 \plus{} z)(1 \plus{} x)}\cdot \frac {1 \plus{} z}{8}\cdot\frac {1 \plus{} x}{8}} \equal{} \frac 34 y$

$ \frac {z^3}{(1 \plus{} x)(1 \plus{} y)} \plus{} \frac {1 \plus{} x}{8} \plus{} \frac {1 \plus{} y}{8}\geq 3\sqrt [3]{\frac {y^3}{(1 \plus{} z)(1 \plus{} x)}\cdot \frac {1 \plus{} z}{8}\cdot\frac {1 \plus{} x}{8}} \equal{} \frac 34 z$

Summing up these ineqaulities and rearranging, we have

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

$ \geq \frac {x \plus{} y \plus{} z}{2} \minus{} \frac 34\geq \frac 32 \sqrt [3]{xyz} \minus{} \frac 34 \equal{} \frac 32 \minus{} \frac 34 \equal{} \frac 34.\ Q.E.D.$[/quote]

yes nice solution with AM-GM ;)

thank you cnyd for your explanation :)

Sorry, I made a typo.

I have just edited it.

[quote="reason"]$ (x,y,z) > 0$ with $ xyz \equal{} 1$

prove that:

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

thanx.[/quote]
It is IMO Shortlist 1998  :)
Another sulution, it's easy to show that:
+)$ 2\left( x^2 \plus{} y^2 \plus{} z^2 \right)\ge 2\left( xy \plus{} yz \plus{} zx \right)$
+)$ x^2 \plus{} y^2 \plus{} z^2\ge 3$
+)$ x^2 \plus{} y^2 \plus{} z^2\ge \frac {(x \plus{} y \plus{} z)^2}3\ge x \plus{} y \plus{} z$
Then add them,we have:
$ 4(x^2 \plus{} y^2 \plus{} z^2)\geq x \plus{} y \plus{} z \plus{} 2(xy \plus{} yz \plus{} zx) \plus{} 3$
So
$ \sum\frac {x^4}{x(1 \plus{} y)(1 \plus{} z)}\ge \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{x \plus{} y \plus{} z \plus{} 2(xy \plus{} yz \plus{} zx) \plus{} 3}\ge \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{4(x^2 \plus{} y^2 \plus{} z^2)}\ge \frac 34$

[quote="frankvista"]

What's $ 2[4,0,0] \plus{} 2[3,0,0]$?
Muirhead?I can't understand it.[/quote]

see: http://www.math.ust.hk/excalibur/v11_n1.pdf  :)


[quote="shabii"]In a TV game show, the grand prize is randomly hidden behind one of three doors. On each show, the finalist gets to choose one of the doors. What is the probability that no finalists will win a grand prize on four consecutive shows?[/quote]
[hide]
$\displaystyle\left(1-\frac{1}{3}\right)^4=\frac{16}{81}$??[/hide]

[quote="lotrgreengrapes7926"]Since the answer is now obviously $\frac{16}{81}$, we can stop posting the same answer. :D[/quote]

$16/81$

[quote="shabii"]In a TV game show, the grand prize is randomly hidden behind one of three doors. On each show, the finalist gets to choose one of the doors. What is the probability that no finalists will win a grand prize on four consecutive shows?[/quote]

I think this is a little lower than Getting Started Level....and so are a lot of the other ones you posted here.


limit
inequalities
calculus
calculus computations

prove that if $ \sum a_k$ is convergent and $ \sum b_k$ is divergent, then $ \sum (a_k\plus{}b_k)$ is divergent.

I tried contradiction, but could only get $ \lim b_k \equal{}0$ which isn't necessarily a contradiction.

Write down the definitions.  You'll need to use the triangle inequality at some point.

I'm not sure what you mean, is there some other way to do it?

You're not sure what I mean by "write down the definitions"?   :wink:  :roll:

@calculusstudent, can you show that if $ \sum a_k$ is convergent and $ \sum (a_k \plus{} b_k)$ is convergent, then $ \sum b_k$ is convergent?

RaviB: yes I can do that. I guess I just don't know how to handle the divergence condition.

if $ \sum a_k$ converges to $ L_1$ and $ \sum (a_k\plus{}b_k)$ converges to $ L_2$, then for all $ \epsilon$ we can find a positive number $ N$ such that $ n>N$ implies $ |\sum b_k\minus{}(L_2\minus{}L_1)|\leq |\sum a_k\plus{}\sum b_k\minus{}L_2|\plus{}|\sum a_k\minus{}L_1|<\epsilon$.

Okay, but then you are done.  Remember that divergent just means not convergent.  So your original problem was of the form "if p and not q then not r".  By contrapositive (or basic logic), that's equivalent to "if p and r then q".  And that's what you proved in your last post.

as far as I know the contrapositive of "if p and not q then not r" is "if r then not p or q". I could only get to your statement after a lot of logical manipulations and identities.

Well, you can think of it this way.  The statement "if p and not q then not r" is equivalent to "if p then (if not q then not r)".  By contrapositive on the inner portion, that's equivalent to "if p then (if r then q)".  By switching back to "and", that's equivalent to "if p and r then q".


factorial
AMC
AIME
number theory
prime factorization

Let n be the number factors of 50!.
Find the number of factors of n.

That would make an excellent aime problem.

Okay, here is also a problem related to Factorial:
Let m be the number of consecutive zeros occurred in the [b]rightmost[/b] of the decimal representation of 2006!.
Determine the last three digits of m.   ;)

I would do it that way:
we have to notice, that every number which contains $5$ in it's factorization adds one more $0$ to the right side. However we must consider numbers that contain $5$, $5^2$, $5^3$ and $5^4$. So among $1,2,3,...,2006$ we have $3$ numbers containing $5^4$ in their factorization,  $13$ numbers containing $5^3$, $64$ numbers containing $5^2$ and $321$ numbers containing $5$. So we have:
$3 \cdot 4 + 13 \cdot 3 + 64 \cdot 2 + 321 \cdot 1 = 500$ zeros rightmost in $2006!$

Yes, you are right,  :)  and here is one more challenge... which could be much harder:
Let $k=10^{2006}$.
How many factors of 5 are there  in the [i]prime factorization[/i] of $k!$ ?

If you want to know the numerical answer, here it is:
[hide]$\frac14(10^{2006}-2^{2006})$
[/hide]


Recently my AOPS board at mathlinks.ro has been changed (to worse) and the images at the upper part of the page do not appear and there are some rearreangements at the hole page -while at artofproblemsolving.com there aren't such problems. Could you please fix it ? I am using Internet explorer 7.
Thanks in advance

Are you using the url with "www" in front or not? (right now, using "www" is recommended).

Yes i am using http://www.mathlinks.ro or http://www.artofproblemsolving.com.

Can you post a print screen so that I can see what's going on exactly?

It's something in your browser. I'm not sure what, but can be anything from an add-blocker, image blocker, anti-virus program. What browser are you using? (looks like IE 7). Maybe try Firefox 2?

yes at FF2 it is working fine, but at IE 7 still no.

Now it is working fine everywhere. Thanks :lol:


Let $ ABCD$ be a quadrilateral which is inscribed in a circle $ k$. Further suppose that $ ABCD$ itself has an incircle. Let $ EF$ be the diameter of $ k$ which is perpendicular to $ BD$, wih $ E$ lying on the same side of $ BD$ as $ A$.
Let $ BD$ intersect $ EF$ at $ M$ and $ AC$ at $ S$.
Prove that    $ \frac{AS}{SC}=\frac{EM}{MF}$

[quote="delegat"]I think that this is a Crux problem (3211.) and it is still open so moderators should lock (close) this topic.[/quote]

Deadline has long passed now.

[b]EDIT:[/b] See http://www.mathlinks.ro/Forum/viewtopic.php?t=19099 and http://www.mathlinks.ro/Forum/viewtopic.php?t=19112 for the problem.

  darij

[quote]
Deadline has long passed now. [/quote]

Also same problem appeared in Crux several years ago Original author is Dutch-Smeenk.

In that issue triginometric solution was featured one but in second appereance of this problem in Crux (as 3211.) solution that uses pure geometric tools was published.


1)Let $ ABC$ be triangle.Prove that:
$ sin^{2}A+sin^{2}B+sin^{2}C \geq (4sin(\frac{A}{2}.sin(\frac{B}{2})sin(\frac{C}{2})+1)^{2}$
2)let $ ABC$ be triangle with $ 0 <A;B;C<\frac{2.\pi}{3}$.Prove that:
$ \frac{cos A+cos B-cosC}{sinA+sinB-sinC}>-\frac{\sqrt{3}}{3}$

for your first inequality we have:
$ \sin{\frac{A}{2}}.\sin{\frac{B}{2}}.\sin{\frac{C}{2}}=\frac{r}{4R}$
$ \sin^{2}A+\sin^{2}B+\sin^{2}C=\frac{p^{2}-r^{2}-4Rr}{2R^{2}}$
so we have to prove that:
$ \frac{p^{2}-r^{2}-4Rr}{2R^{2}}\geq (\frac{r}{R}+1)^{2}$

$ p^{2}-r^{2}-4Rr\geq 2(r+R)^{2}$

$ \Rightarrow p^{2}\geq 2R^{2}+8Rr+3r^{2}$
which is a well known inequality and can be found at:
[url]http://www.mathlinks.ro/viewtopic.php?t=156012[/url]


Let a,b,c>0 Prove that
$ \sum a^3b^3  \geq (\minus{}a\plus{}b\plus{}c)(a\minus{}b\plus{}c)(a\plus{}b\minus{}c)(a^3\plus{}b^3\plus{}c^3)$

[quote="tran300491"]Let a,b,c>0 Prove that
$ \sum a^3b^3 \geq ( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c)(a^3 \plus{} b^3 \plus{} c^3)$[/quote]
If $ ( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c)\leq0$ then done.
If $ ( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c) > 0,$  $ a \plus{} b \minus{} c < 0$ and $ a \plus{} c \minus{} b < 0$ then $ a < 0.$ Contradiction.
Thus, we can assume that $ a \plus{} b \minus{} c > 0,$ $ a \plus{} c \minus{} b > 0$ and $ b \plus{} c \minus{} a > 0.$
$ a^3b^3 \plus{} a^3c^3 \plus{} b^3c^3\geq( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c)(a^3 \plus{} b^3 \plus{} c^3)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^6 \minus{} a^5b \minus{} a^5c \minus{} a^4b^2 \minus{} a^4c^2 \plus{} 3a^3b^3 \plus{} 2a^4bc \minus{} a^3b^2c \minus{} a^3c^2b)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(a^4 \plus{} b^4 \plus{} 2a^2bc \plus{} 2b^2ac \minus{} 3a^2b^2)\geq0.$
Let $ a\geq b\geq c.$ Then $ S_b\geq0,$ $ S_a\geq0$ and $ \left(\frac {a \minus{} c}{a \minus{} b}\right)^2\geq\left(\frac {b}{c}\right)^2\Leftrightarrow$
$ \Leftrightarrow c(a \minus{} c)\geq b(a \minus{} b)\Leftrightarrow(b \minus{} c)(b \plus{} c \minus{} a)\geq0,$ which true.
Id est, $ \sum_{cyc}(a \minus{} b)^2S_c\geq(a \minus{} c)^2S_b \plus{} (a \minus{} b)^2S_c\geq$
$ \geq\left(\frac {b}{c}\right)^2(a \minus{} b)^2S_b \plus{} (a \minus{} b)^2S_c \equal{} \frac {(a \minus{} b)^2}{c^2}(b^2S_b \plus{} c^2S_c) \equal{}$
$ \equal{} \frac {(a \minus{} b)^2}{c^2}(a^4c^2 \plus{} b^4c^2 \plus{} 2c^3a^2b \plus{} 2c^3b^2a \plus{} a^4b^2 \plus{} c^4b^2 \plus{} 2b^3a^2c \plus{} 2b^3c^2a \minus{}$
$ \minus{} 6a^2b^2c^2)\geq0.$
Nice inequality, tran300491!

[quote="arqady"][quote="tran300491"]
Let a,b,c>0 Prove that
$ \sum a^3b^3 \geq ( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c)(a^3 \plus{} b^3 \plus{} c^3)$[/quote]

$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(a^4 \plus{} b^4 \plus{} 2a^2bc \plus{} 2b^2ac \minus{} 3a^2b^2)\geq0.$
Let $ a\geq b\geq c.$ Then $ S_b\geq0,$ $ S_a\geq0$ and $ \left(\frac {a \minus{} c}{a \minus{} b}\right)^2\geq\left(\frac {b}{c}\right)^2\Leftrightarrow$
$ \Leftrightarrow c(a \minus{} c)\geq b(a \minus{} b)\Leftrightarrow(b \minus{} c)(b \plus{} c \minus{} a)\geq0,$ which true.

Nice inequality, tran300491![/quote]

Why?
$ (b \minus{} c)(b \plus{} c \minus{} a)\geq0,$

[quote="tardi"]

Why?
$ (b \minus{} c)(b \plus{} c \minus{} a)\geq0,$[/quote]
Because $ b\geq c$ and $ b \plus{} c \minus{} a\geq0.$ :wink:

Wonderful!
Thank you very much
And this is new inequality
Prove If a,b,c>0  then
$ (ab\plus{}bc\plus{}ca)\sum a^3b^3 \geq ( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c)(a^3 \plus{} b^3 \plus{} c^3)(a^2\plus{}b^2\plus{}c^2)$

Very nice!Thank you very much!
and this is new inequality:
Let a,b,c>0 Prove that:
27$ (a^3b^3\plus{}b^3c^3\plus{}c^3a^3)^2 \geq (\minus{}a\plus{}b\plus{}c)(a\minus{}b\plus{}c)(a\plus{}b\minus{}c)(a^3\plus{}b^3\plus{}c^3)^2(a\plus{}b\plus{}c)^3$

let a,b,c>0 Prove that
$ (ab\plus{}bc\plus{}ca)^3 \geq (\minus{}a\plus{}b\plus{}c)(a\minus{}b\plus{}c)(a\plus{}b\minus{}c)(a\plus{}b\plus{}c)^3$

let a,b,c>0 Prove that

$ (ab\plus{}bc\plus{}ca)^4 \geq (\minus{}a\plus{}b\plus{}c)(a\minus{}b\plus{}c)(a\plus{}b\minus{}c)(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)^3$

Let a,b,c>0 Prove that
$ a^4b^2\plus{}b^4c^2\plus{}c^4a^2 \geq (\minus{}a\plus{}b\plus{}c)(a\minus{}b\plus{}c)(a\plus{}b\minus{}c)(a^3\plus{}b^3\plus{}c^3)$

Let a,b,c>0 Prove that:
$ 9(ab\plus{}bc\plus{}ca)^3 \geq (a\plus{}b\plus{}c)^4[2(ab\plus{}bc\plus{}ca)\minus{}(a^2\plus{}b^2\plus{}c^2)]$

The attached file contains CAS-proofs for all seven inequalities dealt with in this topic    :)

It's an old topic and your proof isn't nice  :ninja:


Find all divisors of 30! of the form 2^n-1 with n a positive integer.

If you wanna try it by brute force, the size of the largest n should be approximately n_max=[log_2 30!]=107. I haven't tried it yet since I'm in the middle of my lunch time but maybe FLT may help out. Hehe FLT, that should be a great hint.

FLT = Fermat's little theorem:

For any prime p and any positive integer n, p|(np - n).

Some refer to Fermat's last theorem:

The equation xn + yn = zn has no non-trivial integer solutions.

oh ok thx (another one I know by name now :D)

yes this may be a good idea here. But in time I think you'll spend less on brute forcing than on trying this out...


[quote="bui_haiha2000"]Let triangle ABC with A=60.Chose X in the line BA and Y in the [CA] satisfy that BX=CY.
AI,BI,CI are bisector start from A,B,C.Prove that XIY=180.[/quote]

Are you sure in this problem?
Let's take the limit position of the line XY - X=B, Y=C. Then under the statement of this problem BC passes through I (as the motion is continuously).
But it isn't truth.
Maybe there are some additional conditions?  :huh:


geometry
circumcircle
trigonometry
geometry proposed

Let $ ABC$ be a triangle, $ \angle A\equal{}60^\circ$ and $ I$ be its incenter. Denote by $ M$ a point lies on the side $ AB$ such that $ IM$ is parallel to $ AC$ and $ N$ be a point lies on the side $ BC$ such that $ BN\equal{}\frac{1}{3}\cdot BC$. Prove that $ \angle BMN\equal{}\frac{1}{2}\cdot\angle ABC$.

[quote="April"]Let $ ABC$ be a triangle, $ \angle A \equal{} 60^\circ$ and $ I$ be its incenter. Denote by $ M$ a point lies on the side $ AB$ such that $ IM$ is parallel to $ AC$ and $ N$ be a point lies on the side $ BC$ such that $ BN \equal{} \frac {1}{3}\cdot BC$. Prove that $ \angle BMN \equal{} \frac {1}{2}\cdot\angle ABC$.[/quote]
[hide="Where did I make my mistake?"]
Call the circumcenter of this triangle to be O.  We wish to show that $ \angle BMN \equal{} 30$.  Now, we have that $ \angle BOC \equal{} 2\angle BAC \equal{} 120$.  Since OB=OC, we have that $ \angle OBC \equal{} \angle OCB \equal{} 30$.

We can show that $ \angle BON \equal{} 30$ as follows: Call $ BO \equal{} a$.  Thus, we have that $ OC \equal{} a$.  Call the midpoint of BC X, and we thus have that OX is perpindicular to BC.  So, we have that $ OX \equal{} \frac {a}{2}$ and $ BX \equal{} \frac {a\sqrt {3}}{2}$.  Thus, $ BC \equal{} a\sqrt {3}$.  So, $ BN \equal{} \frac {a\sqrt {3}}{3}$.  Thus, $ NX \equal{} \frac {a\sqrt {3}}{6} \equal{} \frac {a}{2\sqrt {3}}$.  Thus, we have that $ \frac {OX}{NX} \equal{} \frac {1}{\sqrt {3}} \equal{} \cot 30$.  So, we have that $ \angle NOX \equal{} 30$ and thus $ \angle ONX \equal{} 60$, therefore, $ \angle BNO \equal{} 120$, so $ \angle BON \equal{} 30$.

So, we want to show that $ \angle BMN \equal{} \angle BON$ and thus that BNOM is cyclic.  As we said, $ \angle BNO \equal{} 120$, so we wish to prove that $ \angle OMA \equal{} 120$.  $ \angle IAM \equal{} 30 \equal{} \angle IAC$ and since IM is parallel to AC, we have that $ \angle MIA \equal{} 30$.  Thus, $ \angle AMI \equal{} 120$.  So, we want to prove that $ \angle AMI \equal{} 120 \equal{} \angle OMA$.  Thus, we wish to prove that M, O, and I are collinear and thus we wish to prove that OI is parallel to AC.

The problem I see with this is that if we take a right triangle with angle A equal to 60 and angle B equal to 90, then O is on AC and I isn't.  Thus, OI intersects AC, which shows that OI and AC are not parallel.

Where did I make my mistake?
[/hide]

[quote="The QuattoMaster 6000"][b]We wish to show that[/b] $ \angle BMN \equal{} 30$[/quote]
It's wrong, dear [b]The QuattoMaster 6000[/b]. We have to prove that $ \angle BMN\equal{}\frac{1}{2}\cdot\boxed{\angle ABC}$ [b]NOT[/b] $ \angle BMN\equal{}\frac{1}{2}\cdot\angle BAC$

[quote="April"][quote="The QuattoMaster 6000"][b]We wish to show that[/b] $ \angle BMN \equal{} 30$[/quote]
It's wrong, dear [b]The QuattoMaster 6000[/b]. We have to prove that $ \angle BMN \equal{} \frac {1}{2}\cdot\boxed{\angle ABC}$ [b]NOT[/b] $ \angle BMN \equal{} \frac {1}{2}\cdot\angle BAC$[/quote]
Sorry!  I tend to make many careless errors  :oops:

[quote="April"] [color=darkred]Let $ ABC$ with the incenter $ I$. Denote $ \{\begin{array}{c} M\in AB\ ,\ IM\parallel AC \\
 \\
N\in (BC)\ ,\ BN = \frac {1}{3}\cdot BC\end{array}\ .$

Prove that $ A = 60^{\circ}\Longrightarrow$ $ m(\widehat {BMN}) = \frac {1}{2}\cdot m(\widehat {ABC})$.[/color] [/quote]
[color=darkblue][b]Proof.[/b] Denote $ \{\begin{array}{c} P\in MI\cap BC \\
\ R\in BI\cap AC\end{array}\ .$ Observe that $ BP = \frac {a(a + c)}{2p}$ and

$ \{\begin{array}{ccc} A = 60^{\circ} & \Longleftrightarrow & a^2 = b^2 + c^2 - bc \\
 \\
\frac {a}{\sin A} = \frac {b}{\sin B} & \Longrightarrow & \sin B = \frac {b\sqrt 3}{2a} \\
 \\
\cos B = \frac {a^2 + c^2 - b^2}{2ac} = \frac {(b^2 + c^2 - bc) + c^2 - b^2}{2ac} & \implies & \cos B = \frac {2c - b}{2a}\end{array}$

Thus, $ \tan\frac B2 = \frac {\sin B}{1 + \cos B}$ $ \implies$ $ \boxed {\tan\frac B2 = \frac {b\sqrt 3}{2(a + c) - b}}\ .$ Let $ m(\widehat {BMN}) = \phi\ .$

$ NP = BP - BN = \frac {a(a + c)}{2p} - \frac a3$ $ \implies$ $ NP = \frac {a[2(a + c) - b]}{6p}\ .$

$ \frac {NB}{NP} = \frac {MB}{MP}\cdot\frac {\sin\widehat {NMB}}{\sin\widehat {NMP}}\implies$ $ \frac {2p}{2(a + c) - b} = \frac cb\cdot\frac {\sin\phi}{\sin (60^{\circ} - \phi )}$ $ \implies$

$ bp(\sqrt 3\cos\phi - \sin\phi ) =$ $ c[2(a + c) - b]\sin\phi$ $ \implies$ $ \boxed {\tan\phi = \frac {bp\sqrt 3}{bp + c[2(a + c) - b]}}$

Hence $ \phi = \frac B2\Longleftrightarrow$ $ \tan\phi = \tan\frac B2$ $ \Longleftrightarrow$ $ \frac {b\sqrt 3}{2(a + c) - b} = \frac {bp\sqrt 3}{bp + c[2(a + c) - b]}$ $ \Longleftrightarrow$

$ 2p(a + c) - bp = bp + 2c(a + c) - bc$ $ \Longleftrightarrow$ $ 2p(a + c - b) = 2c(a + c) - bc$ $ \Longleftrightarrow$

$ (a + b + c)(a + c - b) = 2c(a + c) - bc$ $ \Longleftrightarrow$ $ (a + c)^2 - b^2 = 2c(a + c) - bc$ $ \Longleftrightarrow$

$ (a + c)(a - c) = b^2 - bc$ $ \Longleftrightarrow$ $ a^2 = b^2 + c^2 - bc$ $ \Longleftrightarrow$ $ A = 60^{\circ}\ .$[/color]

Thank you dear [b]Virgil Nicula[/b] for your interest and for your solution. But I'm really sorry that I've found this problem was posted before, you can see here: http://www.mathlinks.ro/Forum/viewtopic.php?t=100938


search
combinatorics unsolved
combinatorics

There are several towns. Some pairs of towns are joined by one-way streets so that exactly two streets go out of each town. Show that the city can be divided into $ 1014$ villages such that no two towns in the same village are connected by a street, and all streets between two villages go in the same direction (i.e. from one village to another, or vice-versa).


It takes Nonie 3 mins to paint and dry her nails using her 60 second stars nailpolish; wheras it takes Natalie 6 mins.  If Nonie and her friend both paint their nails 7 times a month, how much longer does Natalie spend painting her nails per year?

[hide="solution"]
Notice that for every time they paint their nails, it takes Natalie 3 minutes more than Nonie. They paint their nails $7 \times 12 = 84$ times. $84 \times 3 = 252$ minutes[/hide]

[hide]
$7\cdot12=84$
They paint their nails 84 times.
$84\cdot3=\boxed{252}$
Since 3 is half of 6, you can just do $84\cdot3$[/hide]

[quote="pascal_1623"][hide]
$7\cdot12=84$
They paint their nails 84 times.
$84\cdot3=\boxed{253}$
Since 3 is half of 6, you can just do $84\cdot3$[/hide][/quote]

since when was 84*3 equal to 253??  :roll:  typo i assume.

but other than that you are both right. The answer's 252.


Find all continuous function $ f: R^\plus{}\to R^\plus{}$ satisfy:
$ f(\frac{x\plus{}y}2)\plus{}f(\frac{2xy}{x\plus{}y})\equal{}f(x)\plus{}f(y)$


probability
combinatorics unsolved
combinatorics

Wikipedia has a pretty nice article on Nash Equilibriums.  They are pretty simple.  Nash's big theory is proving that every "game" with certain conditions must have at least one Nash Equilibrium.

Now, consider the following game:  Two players each get to pick one of two colors (say red or green).  If they pick the same color player 1 gets a dollar and player 2 gets nothing.  If they pick different colors then player 2 gets a dollar and player 1 gets nothing.  This game has no Nash Equilibrium.  I'm wondering how the example games on Wikipedia fit the criteria that guarantees a Nash Equibrium, but the game I constructed doesn't.

Nash's theorem includes [url=http://en.wikipedia.org/wiki/Mixed_strategy]mixed strategies[/url].  This game has a mixed strategy equilibrium:  pick red half the time and pick green half the time.

The game you described is often used as a quick decision-maker, e.g. in deciding who gets to pick first in a sports game.  In that case, it's usually called "odds and evens."  (Each player shows one or two fingers -- if the total is odd, one player wins, if it's even, the other player wins.)  Rock-paper-scissors (one point if you win, zero points if you lose or tie) is another similar game -- the sole Nash equilibrium is with both players playing randomly (i.e. choosing R, P or S with probability $ \frac{1}{3}$).


[hide]The general term is $ \dfrac{1}{k(k\plus{}1)}$. This is also just $ \dfrac{1}{k}\minus{}\dfrac{1}{k\plus{}1}$. If we use this in our expression for $ S$, we get

$ \dfrac{1}{1}\minus{}\dfrac{1}{2}\plus{}\dfrac{1}{2}\minus{}\dfrac{1}{3}\plus{}\dfrac{1}{3}\minus{}\dfrac{1}{4}\plus{}...\plus{}\dfrac{1}{2011}\minus{}\dfrac{1}{2012}$

This series telescopes, and all the terms between 1 and 1/2012 cancel. We are left with

$ S\equal{}1\minus{}\dfrac{1}{2012}\equal{}\boxed{\dfrac{2011}{2012}}$.[/hide]


Let be given a triangle ABC. M, N, P be points of tangency of excircle with sides BC, CA, AB respectively. Show that AM, BN, CP are sides of some triangle.

[quote="Namdung"]Hint: Dont use calculation! Just think geometrically![/quote]
  Vietnamese : thua thay , bai toan nay co cah giai chi bang hinh hoc (khong su dung dinh ly cos) ?
  Anyway , the solution which uses calculation is nice and short .
 Using Stewart's theorem , we have :
 $ AM^2 = x^2 +2x.\frac{y^2+z^2}{y+z} + (y-z)^2 $ , where $ x= s-a , y=s-b , z=s-c $
  Suppose $ x \geq y \geq z $ , we can easily show $ AM \geq BN \geq CN $ :
 $ AM^2-BM^2 = 2(x-y).\frac{2z^3+2z^2(x+y)}{(y+z)(x+z)} \geq 0 $
 With the estimation (Mr Namdung 's estimation) $ y^2 +z^2 \geq y^2-z^2 $ , the rest is not difficult   ;)

Your computation is right. But how to use it? I dont see the easy way to prove.

I have very simple geometric proof. I will post it later. Let's think first.

Namdung

[quote="Namdung"]I have very simple geometric proof. I will post it later. Let's think first.

Namdung[/quote]
Could you post it now?


[hide="Solution"]
Examining $ \bmod9$ yields $ n\equiv5\pmod6$, and $ \bmod7$ yields $ n\equiv0\pmod3\implies n\equiv0,3\pmod6$, contradiction.
[/hide]

Since the LHS has period $ 6$ in $ \bmod9$ and $ 3$ in $ \bmod7$, and must be a perfect cube as well (which means it is $ 0,\pm1\bmod9$ and $ 0,\pm1\bmod7$ respectively).


projective geometry
geometry proposed
geometry

Line $ l$ meets $ \odot O$ at $ A$ and $ B$. Through $ P$ contruct secants line of the circle $ PCD$ and $ PEF$. $ DE$ meets $ l$ at $ I$.

Show that if $ CF \| l$, then $ \frac{1}{\overline{PI}}\equal{}\frac{1}{\overline{PA}}\plus{}\frac{1}{\overline{PB}}$.

[quote="Virgil Nicula"]I didn't understand .... Maybe a draw/figure ![/quote]
Here is a figure.

Suppose w.l.o.g. $ A\in (PB)$  (as in your draw). Thus,

$ \boxed {CF\parallel AB}\ \Longleftrightarrow\ \widehat {IPE}\equiv$ $ \widehat {IDP}\ \Longleftrightarrow IP^2 \equal{}$ $ IE\cdot ID\ \Longleftrightarrow$

$ IP^2 \equal{} IA\cdot IB\ \Longleftrightarrow\ IP^2 \equal{}$ $ (PA \minus{} PI)(PB \minus{} PI)\ \Longleftrightarrow$

$ PA\cdot PB \equal{}$ $ PI(PA \plus{} PB)\ \Longleftrightarrow\ \frac {1}{PI} \equal{}$ $ \boxed {\frac {1}{PA} \plus{} \frac {1}{PB}}\ .$

[b][u]Remark[/u].[/b] If $ R\in (AB)$ so that $ IR \equal{} IP$ , then the division $ (A,B\ ;\ P,R)$

 is harmonically, i.e. $ R$ belongs to the polar line of $ P$ w.r.t. the given circle.


Let the length of the long side be $x$ and the length of the short side be $y$.

[hide]
We know that $2x+3y=3000$.  The maximum area occurs when $2x=3y=1500\Longrightarrow (x,y)=(750,500)$.[/hide]

Thanks a lot... The teacher did not teach me that formula before

[quote="AXSLV2"]I have a big time problem with this exercise...

this is the problem:

a landowner has 3000 yrds of fencing to be used to enclose a rectangular shaped field bordering a river. the field is to be divided into 2 regions by a fence parallel to the short sides of the field. No fencing will be required along the side by the river. What are the dimensions of the largest field that may be enclosed with the available fencing?

I thought the answer was 1500 yrds since I only see the numbers 3000 and 2... but it is more complicated (for me) than I thought... can somebody help me, I will deeply appreciate it
thx[/quote]
[hide="method 1"]
Ok, let the two things be x by y.
$2x+3y=3000$
$x=1500-1.5y$
We want to maximize xy.
$1500y-1.5y^{2}$
y=500 maximizes that
500 by 1500[/hide]
[hide="method 2"]
Apply AM-GM on 2x and 3y
$1500 \geq \sqrt{6xy}$
$250\sqrt{6}\geq \sqrt{xy}$
$375000 \geq xy$
Now you can solve[/hide]
[hide="method 3"]
Again, applying AM-GM on 2x and 3y we see that
$1500 \geq \sqrt{2x*3y}$
Clearly equality is maintained iff 2x=3y[/hide]

[quote="miyomiyo"]Wait - wouldn't it be $x+3y$, not $2x+3y$?  The question said that there is no fencing necessary for the river.[/quote]
Each of the two regions is x by y, hence why we are optimizing xy.


function
geometry
calculus
integration
vector
algebra
functional equation

i am a newbie to measure theory. we defined a measurable function today in class . this was the definition given-

f is called measurable if the set {x: f(x)<a} is measurable for every real number a.

i do not understand this definition well. intutively what do we mean by this or what quantity are we measuring? (like for sets in the plane or real line, measure is a sort of extension for area or length, can we associate a similar meaning in this case?) anymore comments on the definition or anything in measure theory is very much welcomed. i want to get a feeling for the subject. thanks in advance.

The only measure in sight is the one you already have: the measure of a set. But there we discovered that we cannot build a measure that does reasonable things (such as having the measure of an interval in the line being the length of that interval) while measuring every set. So we give up on measuring every set; we only try to measure "nice enough" sets, calling the appropriate collection the measurable sets. The collection of measurable sets is a $ \sigma$-algebra, meaning that it is closed under countable unions, countable intersections, and complements.

Now we'd like to talk about "nice enough" functions. Very soon we are going to try to integrate such functions. A measurable function is "nice enough." Give us that, and the the only remaining obstacles to defining an integral are going to be the kind of classic analysis questions that you associate with words like "absolute convergence." One particularly important thing about the definition of measurable function (so important that if it failed, we'd need a new definition): if a sequence of measurable functions $ f_n$ converges pointwise (or even pointwise almost everywhere) to a function $ f,$ then $ f$ is measurable.

Your definition was that $ \{f<a\}$ is always a measurable set. Since the measurable sets are closed under intersections, unions, and complements, then we also have that the following are measurable sets: $ \{f\ge a\},$ $ \{f\le a\},$ $ \{f>a\},$ $ \{f\equal{}a\},$ $ \{a<f<b\},$ $ \{a\le f\le b\},$ (and the half-open versions of the last two). Indeed, we have $ f^{\minus{}1}(E)$ is measurable for any Borel set $ E.$

One example (requires the Axiom of Choice): what are all functions from $ \mathbb{R}$ to $ \mathbb{R}$ such that $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)?$ That includes the ordinary linear functions $ f(x)\equal{}ax,$ but that's not all of them. If we take a Hamel basis for $ \mathbb{R}$ as a vector space over $ \mathbb{Q},$ make completely arbitrary assignments of the value of $ f$ at each basis element, and extend by linearity, we get a solution to the functional equation. How many solutions are there to this functional equation? There are as many $ f(x)\equal{}ax$ as there are values of $ a,$ namely $ 2^{\aleph_0}.$ But the total number of solutions is much larger, namely $ 2^{2^{\aleph_0}}.$ There are the nice linear functions, and then there are the "monster" solutions. None of the "monster" solutions are measurable functions. They also have graphs which are completely unimaginable. Other than that the graphs are dense in the plane, there is just no sensible way to describe what they look like. That really applies to all nonmeasurable functions: no sensible way to describe what they look like.


I think this has been done a while ago but I decided to ask it again? What country were you born in? I obviously did not list even close to all the countries.

[i]modedit: I premptively added an "other" option before someone complained that there wasn't one. I hope you don't mind.[/i]

Thats hard for me to answer.  I am an american I have never been any other nationality, but I was born in Germany.  I am NOT a German though and never have been.

My grandparents, parents, and I, and my first son, second son, and daughter were born in the United States. Three of my grandparents grew up in homes in which a language other than English was the home language, and my mother's parents, even though they were born in Nebraska and Colorado, received all of their schooling in the German language. My wife and my third son were born in Taiwan. My wife is a naturalized American citizen and my third son is a natural born American citizen, having no claim to Taiwan nationality by the facts of his birth.

New Zealand. Yay, I was the first other.

[quote="Yasha"][quote="Syntax Error"]Well, the question merely asks for birthplace, not heritage.  U.S.A. it is.[/quote]

I think the question is more leaning to nationality. For example, in BHorseMath's case, I think the answer would be US, even though BHorse was born in Germany.

I vote Russia.[/quote]
russian cookies, crackers, candy and chocolate...
*drools all over computer*

Born in Chicago, living in Tennessee.
Korean ancestry

[quote="yif man12"]russian cookies, crackers, candy and chocolate...
*drools all over computer*[/quote]

Oh yes, the Russian candies I brought to Mathcounts practice. They're harder to get in Raleigh, though. You have to order them a few weeks in advance instead of just going to a store.

There's a store in New Haven and I think there's one in Hartford too so you could get some candies there.

I am about as American as they come, at least on my father's side.  I'm third-generation from Eastern Europe on my mother's side, but there are absolutely no old national ties to be found there.

USA: I have some great-great-great grandparents who were born in the US and then I have a grandparent who immigrated from Russia.  All my ancestry is pretty much Russian...probably where i got the math from.


let x,y,z are natural numbers that $ (xy\plus{}1)(yz\plus{}1)(xz\plus{}1)$ is square.prove that  all of xy+1,yz+1,xz+1 are square too.


logarithms ratio limit calculus derivative algebra polynomial

Can anyone help me with these three power series (I want to find the radii of convergence):

1. $\sum_{n=1}^{\infty}n(\log n)x^{n}$
2. $\sum_{n=1}^{\infty}\frac{x^{n}}{n^{\sqrt{n}}}$
3. $\sum_{n=1}^{\infty}\frac{n^{n}x^{n}}{n!}$

Thanks.

1. Ratio test.
3. $n! \approx \sqrt{2 \pi n}\left( \frac{n}{e}\right)^{n}$

#2 is tricky.  Let me think about it.

we need to calculate the following limit

$\lim \frac{{n}^{\sqrt{n}}}{(n+1)^{\sqrt{n+1}}}=\lim \left(\frac{n}{n+1}\right)^{\sqrt{n}}\lim (n+1)^{\sqrt{n}-\sqrt{n+1}}=1$, since

$\lim \left(\frac{n}{n+1}\right)^{\sqrt{n}}=e^{\lim \frac{-\sqrt{n}}{(n+1)}}=1$
and if we denote

$x_{n}=(n+1)^{\sqrt{n}-\sqrt{n+1}}$ then

$\ln x_{n}=-\frac{\ln (n+1)}{\sqrt{n}+\sqrt{n+1}}\rightarrow 0$ hence

$x_{n}\rightarrow 1$.
Didi

On #2, the root test is much friendier than the ratio test.

$\left(n^{\sqrt{n}}\right)^{\frac1n}=n^{1/\sqrt{n}}$

$\ln\left(n^{1/\sqrt{n}}\right)=\frac1{\sqrt{n}}\ln n\to0$ as $n\to\infty,$

hence, $n^{1/\sqrt{n}}\to 1$ as $n\to\infty$

And the radius of convergence is $1.$ (Although the series converges absolutely for $|x|=1,$ by comparsions the any $p$-series you want. In fact, the series for the $k$th derivative converges for $|x|\le 1$ for any $k.$)

With #1, I probably wouldn't bother applying the root or ratio tests. I know that $n$ and $\ln n$ grow much slower than any exponential and thus are "invisible" to the root and/or ratio test: those tests cannot tell the difference between $n\ln n$ and $1.$

[quote="bythecliff"]For the first one, if we use ratio test, we get a ratio of $\frac{n+1}{n}\frac{\ln(n+1)}{\ln(n)}(|x|)$, so I assume the series only converges for $x = 0$?[/quote]

No, since the $\frac{n+1}{n}\cdot \frac{\ln{(n+1)}}{\ln{n}}\rightarrow 1$, it is only necessary for $|x| < 1$.

No, no, no. In the first place, you've got to take the limit as $n\to\infty$ in that ratio test.

A rule of thumb: any power of $n$ (and by that, I mean a constant power, like $n^{3}$ or $\frac1{sqrt{n}}$) or any polynomial or rational function is "invisible" to both the root test and the ratio test. Invisible meaning such sequences grow or shrink at a rate that is much slower than any geometric sequence. Invisible in that the contribution to the limiting root or limiting ratio will be $1.$ And note that that happens here: $\lim_{n\to\infty}\frac{n+1}{n}=1.$

Now, what about logarithms? If polynomials grow too slowly to be visible, then logarithms, which grow even more slowly, are even less visible. At a beginner's first glance, that $\frac{\ln(n+1)}{\ln(n)}$ may look like a clumsy thing, but understanding the big picture, we know that that has to tend to $1.$ It has to be invisible.

For the purpose of determining the radius of convergence, that $n\ln n$ might as well not even be there (might as well be $1.$) The radius of convergence is the same as for the geometric series: $1.$ (Which is what paladin8 said. I'm agreeing with him, just expanding out the thought.)

If you sit right on the boundary of the interval of convergence, then things are so finely balanced that the all those things which were invisible to the root or ratio tests become visible again. For $\sum_{n}(n\ln n)x^{n},$ if we let $x=\pm1,$ then the series becomes $\sum (\pm)n\ln n$ and diverges because the terms don't tend to zero.

[quote="bythecliff"]For the third series, then, if we apply the root test, we get $\frac{(n+1)^{n+1}}{(n+1)!}\frac{|x|^{n+1}}{n^{n}}\frac{n!}{|x|^{n}}= \frac{|x|}{n+1}\frac{(n+1)^{n+1}}{n^{n}}$. As $n \to \infty$, this ratio approaches $e|x|$, so the radius of convergence is $\frac{1}{e}$?[/quote]

Yes, you can read more about it in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120964]this thread[/url], particularly the $x = \frac{1}{e}$ case.


Prove there is no pair (x,y) of integers other than the pairs (0,0) and (0,-1) that satisfy:
y + y^2 = x + x^2 + x^3

First,we can assume that [tex] xy \neq 0 [/tex].Notice that $ x \geq 0 $.Call $ d=(x,y)>0 $.
Let $ x=nd,y=md,(m,n)=1 $.Then we have:
$ dm^2+m=dn^2+n+d^2n^3 $
$ \Leftrightarrow (m-n)(dm+dn+1)=d^2n^3 $
Notice that $ gcd((m-n),n^3)=gcd((dm+dn+1),d^3)=1 $,then:
$ m-n=d^2;dm+dn+1=n^3 $(1) or
$ m-n=-d^2;dm+dn+1=-n^3 $(2)
I will solve (1) since (2) is the same.
From (1) we have: [tex] 2dn+1=n^3-d^3 \geq 3(n-d)nd \geq 3nd [/tex]
Hence $ n=d=1 $ and thid is unreasonable.
So $ x=0 $ then $ y=0 $ or $ y=-1 $ :)


[quote="The Zuton Force"]This is a nice, quick problem a friend of mine ran across at ARML practice.

Let $a,b,c$ be integers such that $0<a<b<c$ and such that $a+b+c=bc$. Find, with proof, all such ordered triples $(a,b,c)$.[/quote]

$bc=a+b+c<3c \Rightarrow b<3$ Thus the only solution is $(1,2,3)$


function
real analysis
real analysis unsolved

Find all functions $ f: R\longrightarrow R$ such that $ f$ is continuse and $ f^'(Q^c)=0$.(we don't assume $ f$ is diffreantiable at $ Q$).

If we can show $ f$ is differentiable at $ Q$ then $ f'(R)$ must be connected and so $ f'(R)\equal{}0$ so $ f$ must be constant.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose that $G$ is a group and $H$ is a subgroup of $G$. Let $N'_{G}(H) = \{g \in G | gHg^{-1}\le H \}$.
(By $gHg^{-1}$ I mean $\{ghg^{-1}| h \in H\}$ and by $gHg^{-1}\le H$ I mean $gHg^{-1}$ is a subgroup of $H$.)

It is easy to verify if $H$ is finite or if $G$ is commutative that $N'_{G}(H)$ is a subgroup of $G$, and therefore $gHg^{-1}= H$ (and is not just a subgroup). However, what if $G$ and $H$ are infinite and noncommutative? Is it even still true that $N'_{G}(h)$ is a group? Equivalently, is it even still true that $gHg^{-1}= H$ for any $g \in N'_{G}(H)$?

This has nothing to do with finiteness or commutativity, does it ? I don't know how you managed to prove the fact that the normalizer is a normal subgroup, but you can do it without any extra assumption ...

What? If $H$ is finite, then we can prove it by injectivity: $gh_{1}g^{-1}= gh_{2}g^{-1}\iff h_{1}= h_{2}$, so $|gHg^{-1}| = |H|$. If $|A| = |B| < \infty$ and $A \le B$, $A = B$ is the only possibility.

If $G$ is commutative, then $ghg^{-1}= gg^{-1}h = h$, so it is clear that $gHg^{-1}= H$ for any $g$.

I still don't see how to prove it when neither of these are true. Can you suggest to me your method?

Thanks.

If $gHg^{-1}\subseteq H$ then $gH \subseteq Hg$, and the other way round since you can take $g^{-1}$ instead of $g$. This gives $gHg^{-1}=H$. For the group axioms, they can be checked straighforwardly ...  :maybe:

[quote="julien_santini"]If $gHg^{-1}\subseteq H$ then $gH \subseteq Hg$, and the other way round since you can take $g^{-1}$ instead of $g$. [/quote]
Why can we take $g^{-1}$ instead of $g$? We didn't say $gHg^{-1}\subseteq H$ for all $g \in G$... Is that actually true in general? I thought that's what we were trying to prove or disprove...  :huh:

Alright, I got misleaded by the fact that your post mentions a "normalizer" in its title, while the given definition is not that of the normalizer of a subgroup. In the context of your definition, yeah, anything is possible.

Can we make an explicit counterexample? I don't have much intuition for group theory yet; I can only think of a few infinite noncommutative groups and I don't know what to do with them.


How many square units are in the area of the triangular region bounded by the $ x$-axis, the $ y$-axis, and the line with equation $ 3x\minus{}y\minus{}6\equal{}0$?

The line's equation is y=3x-6. The x intercept is 2 and the y intercept is 6. So the triangle has sides of 2 and 6 with a right angle. Thus the area is 6.


geometry
circumcircle
geometric transformation
reflection
symmetry
angle bisector
geometry solved

In a triangle $ABC$ , points $D$ and $E$ are taken on rays $CB$ and $CA$ respectively so that $CD=CE=\frac{AC+BC}{2}$. Let $H$ be the orthocenter of the triangle $ABC$, and $P$ be the midpoint of the arc $AB$ of the circumcircle of $ABC$ not containing $C$. Prove that the line $DE$ bisects the segment $HP$.

In a variety of different ways, we can show that $PE\perp CA$. Here is just one possibility:

Since the point P is the midpoint of the arc AB on the circumcircle of triangle ABC, we have AP = BP, and thus the triangle APB is isosceles, so that < PAB = < PBA. Since P lies on the circumcircle of triangle ABC, we have < PAB = < PCB and < PBA = < PCA, so that this becomes < PCB = < PCA, and thus the point P lies on the angle bisector of the angle BCA.

If G is the reflection of the point A in the point E, then the point E is the midpoint of the segment AG, so that EG = AE, and thus

$CG=CE+EG=CE+AE=2\cdot CE-\left(CE-AE\right)$
$=2\cdot CE-AC=2\cdot\frac{AC+BC}{2}-AC=BC$,

and thus the triangle GBC is isosceles. The symmetry axis of this isosceles triangle is the angle bisector of its angle GCB, i. e. of the angle BCA. Thus, the points G and B are symmetric to each other with respect to the angle bisector of the angle BCA. Since the point P lies on this angle bisector, this yields GP = BP. Together with AP = BP, this leads to GP = AP, and thus the triangle APG is isosceles. Since E is the midpoint of its base AG, we thus have $PE\perp AG$, or, in other words, $PE\perp CA$. Similarly, $PD\perp BC$.

Thus, the points D and E are the orthogonal projections of the point P on the sides BC and CA of triangle ABC. Since the point P lies on the circumcircle of triangle ABC, this yields that the line DE is the Simson line of the point P with respect to triangle ABC, and thus, by a well-known property of Simson lines, it bisects the segment HP.

Proof complete.

  Darij

Beautiffuly, Darij ! You are great geometryman. (new word). I renounced to solve and I this nice problem. I was delighted with your solution. Thanks. Look why I like Mathlinks.ro/Forum/...I love you and ... Math.

Another way to prove that PE is perp to CA:

First WLOG CA>BC, so E lies strictly inside AC and D lies strictly o utside BD. Next since CA+CB=CE+CD, AE=CA-CE=CD-CB=DB. If ED intersects AB at K, since AKE and BKD have the same circumradii (BD=AE, and <BKD=<AKE), BK/AK=sin<KDB/sin<KEC=1 (sin rule on AEK and BDK), so K is the midpoint of AB. (You can also prove this using Menelaus easily). Next, since PA=PB, and K is the midpoint of AB, <PKA=90. We wish to prove <PEA=90, so it suffices to prove that PAKE is cyclic, i.e. <KEC=<APK. IF KEC = a, then <ACB=180-2a (since CED is iscoceles), and <APB=2a (APBC is cyclic), and <APK=a (PK bisects <APB), so we are done.


You can write the polynomial \[ 1\minus{}x\plus{}x^2\minus{}x^3\plus{}\ldots\plus{}x^{16}\minus{}x^{17}\] on the form \[ a_0\plus{}a_1y\plus{}a_2y^2\plus{}\ldots\plus{}a_{16}y^{16}\plus{}a_{17}y^{17}\] Where $ y\equal{}x\plus{}1$ find $ a_2$

You can write $ 1 \minus{} x \plus{} x^2 \minus{} .... \plus{} x^{16} \minus{} x^{17}$ as $ \frac {x^{18} \minus{} 1}{ \minus{} x \minus{} 1} \equal{} \frac {(y \minus{} 1)^{18} \minus{} 1}{ \minus{} y}$.

Now look at the numerator. We know that $ a_2$ is the negative coefficient of $ y^3$ in the numerator, since we are dividing by $ \minus{}y$.

$ a_2 \equal{} \minus{} \big(\binom{18}{3} ( \minus{} 1)^{15} \big) \equal{} \boxed{816}$


Ring Theory
superior algebra
superior algebra unsolved

A ring is called boolean if $ x^2\equal{}x$ for all $ x$ in $ A$.
Prove that one can define a structure of boolean ring of order $ n$ if and only if $ n\equal{}2^k$ for some $ k \in N$.

Use: By one of the Sylow theorems (the first one?), if a group has order $ |G|\equal{}pm$ where $ p$ is prime, then there exists a subgroup/element of order $ p$.

If $ R$ is a finite Boolean ring with identity $ 1 \neq 0$ then $ R \cong \mathbb{Z}/2\mathbb{Z} \times ... \times \mathbb{Z}/2\mathbb{Z}$. To see this, you can use the Chinese Remainder Theorem and notice that for $ e \in R$ with $ e^2 \equal{} e$ one has $ R \cong Re \times R(1 \minus{} e)$ where $ e$ and $ 1 \minus{} e$ are the identities for the subrings $ R$ and $ R(1 \minus{} e)$ respectively. On the other hand, if $ R$ is a set of $ n \equal{} 2^k$ elements, you can consider the unique field of $ 2^k$ elements and look at it as an $ \mathbb{F}_2$-vectorspace. Now it clear that we have a Boolean ring, so we're done.


function geometry inequalities quadratics induction calculus integration

Hello fellow Mathlinkers and AoPs users who had appeared this year's CMI under grad entrance exam in math.

Please post the questions which you remember.

I am posting one question:

Find the first right most non zero digit when $ 25!$ is represented in base $ 5$.

Prove that $ 1 \plus{} 7 \plus{} 7^2 \plus{} 7^3 \plus{} ... \plus{} 7^{15}$ is divisible by 400.

1.f is a functiom such that f"(X)>0 for all x. if f(0)=0, prove that for positive a, f(a)>-f(-a)
2.find limit which the function (X^n)(n^K) approach as n tends to infinity if -1<x<1 and k is positive integer
3.if x+(1/x) = 2cosA, prove that (x^n)+((1/x)^n)=2cosnA, where n is any natural number (A is a complex number)

[quote="basil mohammed"]

3.if x+(1/x) = 2cosA, prove that (x^n)+((1/x)^n)=2cosnA, where n is any natural number (A is a complex number)

[/quote]

well, A is the angle and x is complex.

If  $ x \plus{} \frac{1}{x} \equal{} 2cos(\alpha)$ , prove that $ x^n \plus{} \frac{1}{x^n} \equal{} 2cos(n\alpha)$ .

Find all natural numbers N for which $ \frac{1470}{4N\plus{}1}$ is a natural number.

Find the area included between three circles touching each other externally, given that the distances between their centres are $ 1, \ \sqrt{3} \ and \ 2$  units.

[color=blue]Its tough to remember all 18 Questions ![/color]

[quote="basil mohammed"]how many u wrote[/quote]

[b]Me very few [/b]...

What are your expected marks  :?:

And do you have any idea about how many students are selected ?

.

2008 results
http://www.cmi.ac.in/admissions/entrance-results.php

i wrote almost all

[b]Congrats[/b] !  :clap:

and thanks for the link.

[quote="basil mohammed"]1.f is a functiom such that f"(X)>0 for all x. if f(0)=0, prove that for positive a, f(a)>-f(-a) [/quote]

[hide]f''(x) > 0                     ---------- 1

Integrating 1 from 0 to x :

f'(x) - f'(0)>0

Again integrating 1 from -x to 0:

f'(0) - f'(-x) >0

Adding both inequalities:

f'(x) - f'(-x) >0

Integrating from 0 to a and using f(0) =0 , we get

f(a) + f(-a) >0

Hence proved.[/hide]

[quote="Nora.91"]

Find the area included between three circles touching each other externally, given that the distances between their centres are $ 1, \ \sqrt {3} \ and \ 2$  units.

[/quote]
[hide]
The triangle joined by joining the centers is $ 30 \minus{} 60 \minus{} 90$ triangle. Also we can find the radius of each circle because $ r1 \plus{} r2$ is given. Area of triangle is  $ \frac {1}{2} \sqrt{3}$ . Now we subtract the portions of circle inside the triangle which is $ \frac {r^2}{2\theta}$
[/hide]
Questions are fairly simple.

second one was application of Euler's theorem.

$ 7^{16} \equiv 1 (mod 40)$ and $ 7^{16} \equiv 1 (mod 60)$.

Is there any way this can be done using Binomial Theorem?

basil's first Q on f(x) ... can we consider f(x) as a strictly convex function?

if so,

$ f( \frac{a \plus{} (\minus{}a)}{2} ) < \frac{f(a) \plus{} f(\minus{}a)}{2}$

$ \Rightarrow \ f(a) > \minus{} f(\minus{}a)$

[quote="Nora.91"]second one was application of Euler's theorem.

$ 7^{16} \equiv 1 (mod 40)$ and $ 7^{16} \equiv 1 (mod 60)$.

Is there any way this can be done using Binomial Theorem?

basil's first Q on f(x) ... can we consider f(x) as a strictly convex function?

if so,

$ f( \frac {a \plus{} ( \minus{} a)}{2} ) < \frac {f(a) \plus{} f( \minus{} a)}{2}$

$ \Rightarrow \ f(a) > \minus{} f( \minus{} a)$[/quote]

Ya that can be done by Binomial too.

I noticed some interesting facts while solving this problem.

I think the answer to the first question is 4.

Here is how i did it.

The right most non zero digit in base 5 is the remainder when $ 25!/(5^6)$ is divided by 5.

But $ 25!/(5^6)$ is $ 4$ modulo 5.

[quote="Nora.91"]

well, A is the angle and x is complex.

If  $ x \plus{} \frac {1}{x} \equal{} 2cos(\alpha)$ , prove that $ x^n \plus{} \frac {1}{x^n} \equal{} 2cos(n\alpha)$ .

[/quote]

Solving the quadratic we get x = $ cos(\alpha) \plus{} i sin(\alpha)$ or x = $ cos(\alpha) \minus{} i sin(\alpha)$

$ x \equal{} e^{i(\alpha)}$ or $ x \equal{} e^{\minus{}i(\alpha)}$

And from here it is just de moivre's theorem.

[quote="befuddlers"][quote="Nora.91"]

well, A is the angle and x is complex.

If  $ x \plus{} \frac {1}{x} \equal{} 2cos(\alpha)$ , prove that $ x^n \plus{} \frac {1}{x^n} \equal{} 2cos(n\alpha)$ .

[/quote]

Solving the quadratic we get x = $ cos(\alpha) \plus{} i sin(\alpha)$ or x = $ cos(\alpha) \minus{} i sin(\alpha)$

$ x \equal{} e^{i(\alpha)}$ or $ x \equal{} e^{ \minus{} i(\alpha)}$

And from here it is just de moivre's theorem.[/quote]

We can also use induction!

There was a question on Integration of inverse of a function...

Another one something like [b]Prove that[/b] $ | sin^2x \minus{} sin^2y | \leq |x \minus{} y |$.

(1x2x3x4)x(6x7x..)x(11....)x(16....)x(21....) is congruent to 4x4x4x4x4 which is congruent to 4 mod 5 (fermat theorem)
so 25!/ 5^6 = [(1x2x3x4)x(6x7x..)x(11...)x(16....)x(21....)]x2x3x4 is congruent to 4x2x3x4=96 mod 5
so reminder is 1, am i wrong

| (sinx)^2 - (siny)^2 | =< | x - y | for all x,y

<=> |sin2x -sin2y| =< |2x-2y| for all x,y <=> |sinx -siny| =< |x-y| for all x,y

Without loss of generality let   x>=y

then above inequality is equivant to

        y-x=<sinx - siny =< x-y   but this is obvious since

both (t + sint) and (t -sint) are increasing and x >= y

$ \minus{}1 \leq sin2x \leq 1 \\ \\ \\ \int_x^y(\minus{}1) dx \leq \int_x^y sin(2x) \ dx \leq \int_x^y1 \ dx$

$ \minus{} (y \minus{} x) \leq [\frac{\minus{}cos(2x)}{2}]_x^y \leq (y \minus{} x)$

$ \minus{} (y \minus{} x) \leq \frac{1}{2}[1 \minus{} 2sin^2y \minus{} 1 \plus{} 2sin^2x] \leq (y \minus{} x)$

$ \Rightarrow \ |sin^2 x \minus{} sin^2 y| \leq | x \minus{} y |$

[quote="Nora.91"]$ \minus{} 1 \leq sin2x \leq 1 \\
 \\
 \\
\int_x^y( \minus{} 1) dx \leq \int_x^y sin(2x) \ dx \leq \int_x^y1 \ dx$

$ \minus{} (y \minus{} x) \leq [\frac { \minus{} cos(2x)}{2}]_x^y \leq (y \minus{} x)$

$ \minus{} (y \minus{} x) \leq \frac {1}{2}[1 \minus{} 2sin^2y \minus{} 1 \plus{} 2sin^2x] \leq (y \minus{} x)$

$ \Rightarrow \ |sin^2 x \minus{} sin^2 y| \leq | x \minus{} y |$[/quote]

Looks pretty solid.

[quote="Nora.91"]$ \minus{} 1 \leq sin2x \leq 1 \\
 \\
 \\
\int_x^y( \minus{} 1) dx \leq \int_x^y sin(2x) \ dx \leq \int_x^y1 \ dx$

$ \minus{} (y \minus{} x) \leq [\frac { \minus{} cos(2x)}{2}]_x^y \leq (y \minus{} x)$

$ \minus{} (y \minus{} x) \leq \frac {1}{2}[1 \minus{} 2sin^2y \minus{} 1 \plus{} 2sin^2x] \leq (y \minus{} x)$

$ \Rightarrow \ |sin^2 x \minus{} sin^2 y| \leq | x \minus{} y |$[/quote]

Very good! :lol:

[quote="Nora.91"][b]Thank You ![/b]  :)

And what would be an O.K score out of 100 for this test? does anybody have a clue?[/quote]

maybe 65-75  :!:

[quote="basil mohammed"]the shortest method is using langrange's mean value theorem: there exist a c belonging to (a,b)
such that (sina^2-sinb^2)/a-b = 2sin c cos c = sin2c[/quote]

Awsome, what was the motivation?

this is pretty standard method, refer a good calculus book

hello!

the CMI results are announced [url=http://www.cmi.ac.in/admissions/entrance-results.php]here.[/url]

I didn't make it  :(

[b]Congrats to Basil Mohammed !! [/b]  :winner_first:

[quote="basil mohammed"]thanks,
who all are going to do a math degree ?[/quote]

Congrats


I have been doing some problems on Mastery of Counting Fundamentals Level 3, and I have been answering more than 1 question. However, whenever I answer a problem correctly, it still shows the same report: 1 correct, 7 incorrect, with progress 6%, while it should be higher.

maybe the problems repeat themselves oftensince if they do then they are unchanged

Don't worry, when you start at a new topic, the percentage rises slowly. And when you do a lot, it sometimes rises by 25% or more.


quadratics
logarithms
Gauss
invariant
number theory unsolved
number theory

show that (sigma from a=1 to p-1), $\Sigma \frac ap=0$

\sum instead of \sigma is nicer  :)

Is it meant as sum over $\left( \frac{a}{p} \right)$ (Legendre symbol) or something else¿ (in the first case see that there are equally many squares and nonsquares)

yeah sorry for that sum is the correct one :D but I cannot put the bottom and top values of sum so bottom a = 1, top p-1
show that $\sum \frac ap = 0$

[quote="Rust"]It is obviosly $Ind(Z_p^*)^2$ in $Z_p^*$ is 2.
It is interesting to prove $\sum_a a(\frac ap) \le 0$.[/quote]

I didn't understand a word from this where's the solution now

Just see that there is any nonsquare $s$ (e.g. one of the primitive roots).
Let $A$ be the squares and $B$ by the nonsquares, then multiplication of $A$ with $s$ gives a bijection between $A$ and $B$, thus they have equally many elements, done.
Alternativly, see that $\zeta^n$ is a square iff $n$ is even, and thus conclude that there are as many squares as nonsquares again.

PS: try \sum_{a=1}^{p-1}

[quote="Rust"] It is interesting to prove $\sum_a a(\frac ap) \le 0$.[/quote]

For  $(\frac{-1} p)=1$ the ineq is equivalent to $pR_<- pN_<\le 0$ (the number of quadratic residues less than $\frac{p-1} 2$ is less that the number of quadratic non-residues less than $\frac{p-1} 2$).

But if $(\frac 2 p)=1$ the number of residues less than $\frac{p-1} 2$ is equal to the number of even quad. residues, which is equal to the number of odd quad. resiudes, since $-1$ is residue. Doing the same for $N_<$ the coclusion follows in this case.
A similar argument holds for  $(\frac 2p)=-1$. The only diferrence is that the number of quad. residues less than $\frac{p-1} 2$ is equal to the number of even non-quad. resdiues.

For $(\frac{-1} p)=-1$ the ineq is equiv to $\sum_{a\in R}a\ge \frac{p(p-1)} 4$ but I don't know hoe to finish it. Does anyone have any ideea?

It is obviosly if p=1(mod 4) $\sum_a a(\frac ap )=0.$
If p=3(mod 4), then $\sum_a a(\frac ap )=-ph$, were h is (odd natural) number of classes field $Q(\sqrt{-p})$.
It is not hard to prove that $h<\sqrt p \ln p$.

Can you detail a little your solution? Maybe someone can give a proof for the secondly mentioned fact.

Thank you!

Argh ! It is [i]obvious[/i] that the Riemman conjecture is true if the generalized Riemann conjecture is true.
Please stop writing "it is obviosly", since this is incorrect in English
We must pay attention to write correctly words we cherish ;) umm Rust ?

It is well known fact $\sum_a a(\frac ap )=-ph$, were h is (odd natural) number of classes field $Q(\sqrt{-p})$.
It fact proved in russian forum $h<\sqrt p \ln p$.

All you have done is to restate your claim. Can you give a schetch of proof or a link where we can find some ideas..?

Ok!
In russian forum we are interesting about distance between (p-1)/2 and average $\frac 1p \sum_{x=0}^p \{x^{2n}\}$, were $\{x^{2n}\}=x^{2n}(mod \ p)$, p is odd prime.
Denote it  $I(n,p)=\frac{p-1}{2}-\sum_{x=1}^{p-1} \frac{x^{2n}}{p}-[\frac{x^{2n}}{p}]=\sum_{x=1}^{p-1} r(\frac{x^{2n}}{p})$, were $r(x)=\frac 12 -x+[x]$. It is easy to show that $I(n,p)=I(d,p)$, were $d=(n,\frac{p-1}{2}), (d|\frac{p-1}{2})$. Therefore we considered only case n|(p-1). It is easy to show, that I(n)=0 if m=(p-1)/(2n) is even, and n|I(n,p).
Consider Gauss summ: $\sum_{x=0}^{p-1} \exp{\frac{2\pi k i x^{2n}}{p}}= \lambda (k)\sqrt p,$ were $|\lambda (k)|=1$.
  We have
\[ \sum_{k=-N}^{k=N} \frac{exp{\frac{2\pi k i x^{2n}}{p}}}{2\pi k i } \to r(\frac{x^{2n}}{p})  \]
Therefore \[ \sum_{k=-N}^{k=N}' \lambda (k)\sqrt p \to I(n,p)  \], were $\sum '$ mean summ without terms p|k.
We have $\sum_{k=s}^{k=p+s-1} \lambda (k)=0.$ We denote $\lambda (k)=0$ if p|k. Therefore
$I(n,p)=\frac{\sqrt p}{\pi} Re \sum \frac{\lambda (k)}{k}.$
We have $|\sum_{k=1}{}^{p{}^2-1} \frac{\lambda (k)}{k}|<2\ln(p)$,$|\sum_{k=p^2+1}^{\infty} \frac{\lambda (k)}{k} |<1$.
It give $|I(n,p)|<\frac{\sqrt p }{\pi} (2\ln p +1).$
It is true for n=2, because $\lambda(k)=(\frac kp )$. For n>2 I think must be multiple n. Were is my mistake?
It is interesting that for only n=1 (I(1,p)=h) invariant I(n,p)>0 for all p. If n>1 (m=(p-1)/(2n) is odd) for nearly half means p give positive I(n,p), and  nearly half negative I(n,p).


The Grissom High School Math Team is hosting a three-round national math league for students in levels Algebra I, Geometry, and Algebra II.  The Rocket City Math League is hosted by the National Math Honor Society (MAO).  If you are interested, you may find information on this math league at:

http://www.geocities.com/rocket_city_math_league/

if you have any questions, you may email me at vivyeashihwang@aol.com

(round one was in December 2003)

I take this test, although I don't think our teacher has given us the next round.

Lol, why does every problem have to do with spacecrafts!!!!!!!

BTW, who are you? I go to Briarwood.

we rewrite each question submitted by the ghs math teamers to adhere to the topic (Rocket City).  The second round should be coming soon=)

....Sharat (or close accomplice, lol)?

I wish we could have a tournament...my math teacher's thinking about having one next year, and maybe I could write the algebra test. The stupid thing is, that our school HATES scantron machines for some reason, so we would have to convince them to let us buy one. Oh well, I could hide it in my house...(and I am quickly going off on tangents because it's late and I'm talking to 10,000 people on IM.)

how many rounds of Rocket City are there per year?

It should be on the RCML website, but the math league is composed of three tests.  THe last one is this week.  If you would like to register for next year, you may email me at vivyeashihwang@aol.com

The people at Rocket city decided to change our school name. We are now "Bayside Highschool". What is up with that? And there really is a Bayside Highchool and our people placed a lot better than their people so we're just helping that school. And my teacher thinks they rigged it because look at the scores for miller's round 1.

I go to Miller. Um.... Teacher put the algebra people as geometry people also, so they had unsually high scores, which were also used for the mercury test, which beat the geometry students. Argh. The scores for miller went higher as the rounds came though. I think joesph chu ended up with 13, 25, and 21.

My scores: 5, 13, 18
I have no idea how i got a 5, it just happened.

Hello?!?!?!?!??!?!??! Did you read my post? Patricia is pretty pissed that her score was entered as 11 instead of 22

does it look like he read ur post? and im really pissed off because all my scores are like 18,18,18... omg i suck so bad

Sorry...
I read it, but didn't proccess it in my head. Oh well...
And why did your teacher enter the scores of the algebra people as geometry people?


say i have an equation
$ y(t)\equal{}x(\minus{}t)$

So, $ y(t\minus{}t_0)$=$ x(\minus{}(t\minus{}t_0))$

but what does $ x(t\minus{}t_0)$=?..
Solution says $ x(t\minus{}t_0)\equal{}x(\minus{}t\minus{}t_0)$

please clarify the difference

This obviously can't be the solution, so the answer is wrong.

I don't know if this is right, but here's my try:

[hide]
If you think about it as subtracting $ t_0$ from $ t$ and $ \minus{}t$, we have
$ y(t\minus{}t_0)\equal{}x(\minus{}t\minus{}t_0)$.

If you think about it as substituting $ t\minus{}t_0$ as $ t$, we have
$ y(t\minus{}t_0)\equal{}x(\minus{}t\plus{}t_0)$.

Combining gives us
$ x(\minus{}t\minus{}t_0)\equal{}x(\minus{}t\plus{}t_0)$.

That's all I can get.
[/hide]

We have $ y(\minus{}t)\equal{}x(t)$, for all $ t$, so $ x(t\minus{}t_0)\equal{}y((\minus{}t)\minus{}t_0)$, the desired equality.

I assumed that $ x(\minus{}t\minus{}t_0)$ was in fact supposed to be $ y(\minus{}t\minus{}t_0)$. Otherwise, the equality is false when $ t_0\neq 0$.


Let $ a,b,c,d\geq0$ s.t. $ abcd\equal{}1$.Prove that:
$ \dfrac{1}{1\plus{}a^2}\plus{}\dfrac{1}{1\plus{}b^2}\plus{}\dfrac{1}{1\plus{}c^2}\plus{}\dfrac{1}{1\plus{}d^2}\leq2$.

[quote="Inequalities Master"]Let $ a,b,c,d\geq0$ s.t. $ abcd \equal{} 1$.Prove that:
$ \dfrac{1}{1 \plus{} a^2} \plus{} \dfrac{1}{1 \plus{} b^2} \plus{} \dfrac{1}{1 \plus{} c^2} \plus{} \dfrac{1}{1 \plus{} d^2}\leq2$.[/quote]
Try $ a\equal{}8$ and $ b\equal{}c\equal{}d\equal{}\frac{1}{2}.$ :wink:

Maybe the correct phrasing is:
Let $ a,b,c,d\geq0$ s.t. $ abcd\equal{}1$.Prove that:
$ \dfrac{1}{1\plus{}a^2}\plus{}\dfrac{1}{1\plus{}b^2}\plus{}\dfrac{1}{1\plus{}c^2}\plus{}\dfrac{1}{1\plus{}d^2}>\equal{}2$.

[quote="Nemion"]Maybe the correct phrasing is:
Let $ a,b,c,d\geq0$ s.t. $ abcd \equal{} 1$.Prove that:
$ \dfrac{1}{1 \plus{} a^2} \plus{} \dfrac{1}{1 \plus{} b^2} \plus{} \dfrac{1}{1 \plus{} c^2} \plus{} \dfrac{1}{1 \plus{} d^2} > \equal{} 2$.[/quote]
No! See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=196014

I am sorry for my wrong posts.I just copied from a place without checking.
 :(  :(  :(


Billy Bob's lawn has twice as much area as Jilly Job's lawn and three times as much area as Killy Kob's lawn. Killy Kob's lawn mower cuts half as fast as Jilly Job's, and one third as fast as Billy Bob's. They all start to mow their lawns at the same time. Who will finish first?

[hide]I'm guessing Jilly Job through intuition[/hide]

jilly jobn will finsih first,

stop using billy bob

[hide="answer"]Jilly Job will finish first[/hide]

lets go billy bob and billy bobo!!!!!!!!!!!!  :rotfl:

[hide] Let's say Billy Bob's lawn is $6$ square units, Jilly Job's is $3$ and Killy Kob's is $2$. Killy Kob mows at $1$ hour per square unit, while Jilly Job mows at $2$ and Billy Bob mows at $3$. Billy Bob will take $2$ hours. Jilly Job will take $1.5$ hours. And Killy Kob will take $2$ hours. [/hide]

[hide="a solution"]Call the area of Billy Bob's lawn $6x$ units (less fractions).  Then, the area of Jilly Job's lawn is $3x$ units and the area of Killy Kob's lawn is $2x$ units.  Now, call the mower rate of Killy Kob $y$.

Billy Bob's time: $\frac{6x}{3y}=\frac{2x}{y}$

Jilly Job's time: $\frac{3x}{2y}$

Killy Kob's time: $\frac{2x}y=\frac{2x}y$

$\therefore\boxed{\text{Jilly Job}}$[/hide]


[quote="GameBot"]Seven black cows and 4 brown cows give in 6 days exactly as much milk as 5 black cows and 8 brown cows give in 5 days. Which color cow gives more milk?[/quote]

Black cows gives x milk in 1 day
Brown cows give y milk in 1 day.

$ 6(7x\plus{}4y)\equal{}5(5x\plus{}8y) \implies 42x\plus{}24y\equal{}25x\plus{}40y \implies 17x\equal{}16y$.

Thus, $ y > x$, and brown cows give more milk.

6(7 black + 4 brown) = 5(5 black + 8 brown)
42 black + 24 brown = 25 black + 40 brown
17 black = 16 brown
black = 16/17 brown

Therefore, the brown gives more milk.

there shouldn't be problems like this because then people can mem them on ftw


As you all know by now, the previous marathon has been locked after a mere two pages. This time, let's try to keep it going for a month without getting it locked. Watch the level!

Here's a good problem, which will be problem #1:

Elyse takes 4 tests. Her scores are: 78, 92, 86, and she forgot the last one. She knows her average is 85. What was her fourth test score?

Sorry I don't have a new problem.

$ \frac{78\plus{}92\plus{}86\plus{}x}{4}\equal{}85$
$ \frac{256\plus{}x}{4}\equal{}85$
$ 256\plus{}x\equal{}340$
$ x\equal{}84$

I THINK that's right.

NEW PROB.

When we place four numbers in a $ 2\times 2$ grid, we form a matrix. We find
the "determinant" of such a $ 2\times 2$ matrix as follows:
\[ \left|\begin{array}{cc}a & b \\
c & d\end{array}\right| = ad - bc.
\]
For example,
\[ \left|\begin{array}{cc}3 & - 2 \\
5 & 7\end{array}\right| = (3)(7) - ( - 2)(5) = 31.
\]
Suppose $ a$, $ b$, $ c$, $ d$, $ e$, and $ f$ are constants in the system of linear equations
\begin{align*} ax + by & = e, \\
cx + dy & = f. \end{align*}

A formula known as Cramer's Rule states that if

$ \displaystyle\left|\begin{array}{cc}a & b \\
c & d\end{array}\right|$

is nonzero, then the solution to this system is
\[ x = \dfrac{\left|\begin{array}{cc}e & b \\
f & d\end{array}\right|}{\left|\begin{array}{cc}a & b \\
c & d\end{array}\right|} \qquad\text{ and }\qquad y = \dfrac{\left|\begin{array}{cc}a & e \\
c & f\end{array}\right|}{\left|\begin{array}{cc}a & b \\
c & d\end{array}\right|} .
\]
Prove Cramer's Rule.

What happens if the determinant $ \displaystyle\left|\begin{array}{cc}a & b \\
c & d\end{array}\right|$ equals 0?

[quote="alton66"]NEW PROBLEM: On the First week of school Jim Started a math club, On the second week he realized he didn't have any other people, so he recruited two more poeple. On third week of school, the two new recruits recruited two more people each, so 4 new recruits. On the 4th week the 4 new recruits recruited two new people each.
 On the 5th week... and so on.
On the 10th week of school, how many people will be in the math club?[/quote]

Alton: ans is 2^11-1, or 2047.

And ducky, too much crazy latex, scary prob...i dont want to do it sorry!


[quote="gemath"]Hi all , I'm a new member, I have a problem that I haven't proved:

Given are a inscribed quadrilateral with any point M prove that
$MA.MB.CD \leq MB.MC.DA+MC.MD.AB+MD.MA.BC$
and have equality if only if $M \in small arc CD$

hope you 'll interested in my problem.  :)[/quote]

I am sure it has a proof using complex numbers.

OK!! Thank for your suggest now I can do it in general  :)


(1) Let $ a_n$ be the arithmetic sequence such that $ a_{10}\equal{}15,\ a_{20}\equal{}14$.
Find $ \sum_{k\equal{}20}^{100} a_k$.

(2) Let $ b_n$ be the geometric sequence such that $ b_{10}\equal{}6,\ b_{15}\equal{}192$.
Find $ \sum_{k\equal{}9}^{16} b_k$.

Thanks kunny, I've edited it.

[hide=" Alternative solution for (1):"]

$ a_{k} \equal{} 16 \minus{} \frac{k}{10}$

$ \sum_{k\equal{}20}^{100} a_{k} \equal{} 81\cdot 16 \minus{} \frac{1}{10} \frac{81(20 \plus{} 100)}{2}$

$ \equal{} 1296 \minus{} 486$

$ \equal{}810$.

[/hide]

[quote="AndrewTom"]Hi, kunny. I think (1) is $ 810$ and (2) is $ 765$.[/quote]

That's correct. :)


In the event that i dont have much time i need help on a math formulas rather a math problem in which i have misplaced the formula for. It is a pretty easy problem but i cant remeber the formula since i dont normally work with these problems. ok

Sandra and Mary can hoe a garden in three hours together. If Sandra can do it alone in 5 hours, how long will it take Mary to do the job alone?

I need the formula, I have the answer it is 7.50 but I cant find how to do it.

in one hour sandra does 1/5, so in three she does 3/5, thefore mary did the other 2/5 in three 2/5/3=2/15 in an hours threfore 7.5 hours.

I am not sure if this is what you meant by formula but you could just use equations
3s+3M=job
5s=job work from here...

Thanks so much Im always confused on those problems maybe this time i wont loose it. Thats what I meant by formula (I call everything in math that is the guidelines on how to do stuff a formula.) I felt stupid asking and thanks so much for answering I cant believe that was the third time id lost the little sheet of paper that i wrote it on.

Much thanks

Or you can think: So Sandra and Mary hoe completely in 3 hours so Sandra and Mary hoe 1/3 of the total in an hour. Sandra needs 5 hours to hoe completely so she does 1/5 of the total in an hour. Subtract the 1/5 from 1/3 (because then you get how much Mary hoes in an hour) and get 2/15 so take the multiplicative inverse of that and get 7.5 hours.

So formula (AB is how many hours Person A and person B working together to to do something completely, A is the time Person A alone needs to finish the same thing completely and same for B): A=1/(1/AB-1/B)

Just remember that Rate * Time = Distance (or in this case final product)

This should lead u to the other [i]formulas[/i] to solve the problem u need to solvev

just know that if a person can do a job in n hours, then in one hour s/he would do it in 1/n of time, and then if another person joined in itd be 1/x or whatever that persons rate was. so then 1/n+1/x=1/z which is the total job done in 1 hour, so flip it over to find the time taken for the whole job. manipulate as you need, just like Chinaboy said

Thank u all for clarifying it for me i have put ur explinations on to a word document this time it will be on my hard drive i cant loose it now that is unless my hard drive crashes ill have it saved on a disk just in case i never want to loose this formula agian. My math coach couldnt help me and i asked all the math teachers at school none could come up with one (thats sad)

Wow it really is...what about your parents?

Tare no unfortunatly me and my brother are the only math whizs in the family my moms a texas history teacher and my dad is a police officer so my family is not really that involved with math at all.

Max i havent had it long enough to memorize it thank u. I lost it b4 i could memorize it the last couple times. I loose things easily ecspecially when they are on a loose sheet of paper. I will have it memorized but in case i do forget it agian i wont have to ask no one all i have to do is bring it up on the hard drive where all my other formulas are. Its my tactic of memorization. I will learn how it works but right now i have to finish reading a book for english ill learn it when i have time right now i dont.

just derive it 10 times and youve got it

You don't need to memorize the formula for this particular problem. Memorizing it would only allow you to solve problems that are exactly the same as this one. Like Chinaboy said, you just have to understand the general equation rate*time=distance and you can tailor it to any problem involving rates/time/distance.

Yeah. You definitely shouldn't be just learning how to do one type of problem, that won't help you at all.. just agreeing with all the last couple of posts :) but then I said that to everyone who says "by vietes formula" and that didn't seem to do anything :( exactly the same applied there, you shouldn't memorise that you should just do it the "expand (x-a)(x-b).. and equate coefficients", that may help you to solve other problems..

Well, if you think that, then you should do what was said not to do here.. but in my opinion its a lot easier to stay as generalised as possible, if you had to use so called "vietes formula" but there are slight changes in it, you would be completely stuck, whereas if you had learnt it the proper way, you would probably be able to fiddle around a bit and come up with the answer..

if youve solved thousands of mc problems, then you dont even realize what youre doing because you understand how you apply it and stuffs with its variations.

Completely off subjects, peoples...

unless ur talking about if it takes that long for one school to better itself, how long will it be at least this good...

but anyways...

please stay on topic. feel free to post this on another topic. thanks

Can you split this into 3? The first part (about the original problem) can stay here, the second part about Viete's formula can either stay here or go to Other Problem Solving Topics, and the rest about Texas and the school system can go to The Round Table or Fun and Games. I will refrain posting anymore here until then.

[quote="Fierytycoon"][quote]but then I said that to everyone who says "by vietes formula"[/quote]

Hey, what's wrong with saying "by viete's formula"?! :D I know you derive it by expanding Prod(x-r_k) and noting uniqueness of factorization, but you really oughta just know that (-1)^k * a_(n-k)/a_n is the kth symmetric sum of the roots instead of having to multiply out Prod(x-r_k) for a polynomial of a large degree ( > 3).[/quote]
I agree. It is, of course, necessary to understand that formula and how it's derived and everything, but once you have that and you aren't going to lose it, Viete's is great.

My complaint about "by Viete's formula" -- those words don't by themselves say anything.  And it isn't a major result.  The result is that lots and lots of people come onto this website and read the words "by Viete's formula" and go, "Huh?"  It conveys more information with not much more work to say, "by expanding out the factorization and equating coefficients," and it tells people what you're doing.

The reason this is not true for [i]all[/i] theorems is that many theorems are longer than their proofs.  Jensen's inequality, for example (which is another oft-quoted fact that has been asked about on this forum repeatedly since I've been here) is a fairly long statement, and its proof is much longer.   Thus it saves a lot of work to say "by Jensen's."  But it doesn't save hardly any work to say "by Viete's" -- you almost might as well just prove it right there, which is in some ways more elegant and less pretentious, I feel.

[quote]It conveys more information with not much more work to say, "by expanding out the factorization and equating coefficients," and it tells people what you're doing. [/quote]

Well, it seems to me that now we are just arguing about semantics here, which I don't believe is very worthwhile...

me either, and when i first saw the words, i thought it was way more complicated than what it really is.

[quote]how often is it that you see it called "vietes formula" in textbooks, etc? I have never heard of vietes formula before this site, and i question how many other people have.[/quote]

Well that may be true, but I didn't see either "viete's formula" or the technique until I read AoPS.

Thanks.

Me too, I hadn't heard of it before AoPS and I still couldn't understand it until a week ago or so.

Same here, I hadn't heard of it before I came here, despite the fact that I had used it for the past 4 years and also solved numerous problems using either it or similar ideas that you would use to prove it. thus my point :)

I never heard it called that b4. Ive used it but I just called it the formula which has no name. Thats neat. I dont like haveing to keep track of names so all formulas are this formula and that formula to me most of the time.


Consider $\{{a}_{n}\}$ sequence of power of $3$ that's first digit is 9. (For example ${a}_{1}=9$) ${a}_{2007}=?$


algebra polynomial geometry trigonometry

Solve for x in R:
$ 16x^4  \minus{} 8x^3  \minus{} 12x^2  \plus{} 4x \minus{} 1 \equal{} 0$ :)

hello, you need the formulas for a polynomial of degree four, see here
http://mathworld.wolfram.com/QuarticEquation.html
Sonnhard.

[quote="TRAN THAI HUNG"]Yeah, Using formula is completely correct.ButI think this equ. is interesting because it's relative to a geometry problem.And the roots are also very nice. :)[/quote]

do you mean to put $ x\equal{}2 \cos y$ or some thing like that?


A set of positive integers has 2004 elements . Find the maximum of the number of subsets which have 3 elements are three successive integers .

   Example : with 6 elements , the maximum of the number of subsets which have 3 elements are three successive integers is 2^3=8
       1 1 2 2 3 3

I have an answer : 668^3 but I can't solve it completely .

A very similar problem has been posted before by orl. It comes from [url=http://www.kalva.demon.co.uk/short/sh01.html]ISL2001[/url], and it has been solved on the forum. You can also find a solution on Kalva.

The answer is (2004/3)^3. I will give the more specific ideas later.

I attempted (because I could have made a mistake :)) to solve it [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17453]here[/url].


geometry
geometric transformation
reflection
homothety
inradius
ratio
AMC

The incircle of the triangle PBC touches BC at U and PC at V. The point S on BC is such that BS = CU. PS meets the incircle at two points. The nearer to P is Q. Take W on PC such that PW = CV. Let BW and PS meet at R. Show that PQ = RS.

woo I like geometry, but the notation's really bad so let's make it all better

ABC is a triangle with incircle tangent at D (on BC), E (on AC), F (on AB). Let N be the Nagel point and let AN, BN, CN intersect their opposite sides at X, Y, Z respectively. Let D' be the reflection of D on the incircle. We want to prove AD' = NX. To see that these two problems are equivalent, you need to note first that if X is the point on BC s.t. CX = BD (i.e. X is the reflection of D in the midpoint of BC), then X is the tangency point of the excircle opposite A with BC (just calculate the appropriate segments in terms of the side length by using equal tangents a heck of a lot of times), i.e. AX is one of the cevians corresponding to the Nagel point. Also, by a homothety taking the excircle opposite A to the incircle, we see that A, D', and X are collinear.

We know AD' / AX = (h - 2r) / h, where h is the length of the altitude from A and r is the inradius; this follows from simply projecting AD' and AX onto the altitude from A and noting that ratios are preserved. Also, AN/NX = AY/YC + AZ/ZB = a / (s - a); the first step is Van Aubel's theorem and is a nice exercise in using area ratios and the second step is straightforward calculation of segment lengths. Thus, by algebra, NX / AX = (s - a) / s.  However, NX / AX = (s - a) / s = 1 - a/s = 1 - (2K/h)/(K/r) = 1 - 2r/h = (h - 2r) / h = AD' / AX.

Heh, cool problem. Where is it from? Old USAMO?

Yeah, from 2001: http://www.kalva.demon.co.uk/usa/usa01.html

I thought it looked familiar  :)

[quote="ThAzN1"]I thought it looked familiar  :)[/quote]
Indeed... I used to be scared of this problem :) Here's a synthetic solution:

[hide]Use Arnav's notation. Let the midpoints of BC, CA, and AB be J, K, and L, let I be the incenter, and G the centroid.

Here's a fun fact about the Nagel point: G is on segment IN with NG=2GI.
[b]Proof[/b]: Since J and I are midpoints of XD and DD', JI is parallel to XD' (i.e. AN). Likewise, KI is parallel to BN and LI is parallel to CN, so by homothecy between ABC and JKL, I is the nagel point of medial triangle JKL. Thus, the homothecy with ratio -1/2 centered at G that takes ABC to JKL takes N to I. $\blacksquare$

Dilate JKL around N with ratio 2 to triangle A'B'C' (Notice that NDA'X is a parallelogram). Consider triangle NAA'. In this triangle, JA is a median, and so G (1/3 of the way from J to A) must be the centroid of this triangle. This means that I (the dilation of vertex N around G with ratio of -1/2) is the midpoint of the opposite side, i.e. I is the midpoint of AA'. Similarly, I is the midpoint of BB' and CC', so A'B'C' is the rotation of ABC through 180 degrees around I. From this rotation, we find that AD' = A'D, and from parallelogram NDA'X, A'D = XN, as required.[/hide]

Using Arnav's notation
[hide]Let AE=CY=x, EC=DC=y, BD=BF=CX=z. We see AB+BX=x+y+z=AC+CX=x+y+z => X is the point of tangency of the excircle of ABC. Let the excircle be tangent to AC at P. Homothecy gives us AD'/AX = AE/AP = x/(x+y+z). Now by Menelaus we have
(AY/YC)(BC/BX)(XN/NA)=1, we get XN/NA=x/(y+z) so XN/AX=x/(x+y+z)=AD'/AX, thus XN=AD'. [/hide]

[quote="beta"]Using Arnav's notation[/quote]

I know this is off-topic but who is Arnav?

Zabelman: that's a very neat proof for the fact that NG = 2GI. I only knew a proof using barycentric coordinates and some vectors...


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

Let G be an Abelian group. Show that the set of all elements of G of finite order forms a subgroup of G.

Another ?:  show that  Z congruence class of 18 is cyclic.

The neutral element is there. Also for any element $a$ we have that $a^{-1}$ has the same order as $a$, so has a finite one when that of $a$ is finite. So we are left to show that when $a,b$ have finite order, then so has $ab$. But when $m,n$ are the orders of $a,b$, then $(ab)^{mn}=e$ by simple calculation.

@another: $\mathbb{Z}/18\mathbb{Z}$ or $\mathbb{Z}/18\mathbb{Z}^*$¿ $1$ generates the first one and $5$ the second one.

Here's what've got:

Let  H ={ g in G: o(g) is finite} . I'll show it's closed. Let a,b in H with, say, o(a)=m, o(b)=n  . Then
(ab)^mn  = (a^m)^n (b^n)^m=e

this doesn't mean the order is mn, it can also be smaller than this (a divisor) but is definitely finite. But What is o(ab)?

how do I that  show the identity is in  as well as that if a in H then a^-1 in H too, i.e. that a^-1 has finite order if a has it.

No, $ord(ab)$ is not $\mathrm{lcm}(ord(a),ord(b))$. When $ord(a),ord(b)$ are coprime, then this is correct, so the order is $ord(ab)=ord(a) ord(b)$ then. Counterexample for $ord(ab) = \mathrm{lcm}(ord(a),ord(b))$: $a=b=1$ in $(\mathbb{Z}/2\mathbb{Z},+)$.
(There is a lemma: there is an element of order $\mathrm{lcm}(ord(a),ord(b))$; but it is not necessary $ab$)


geometry
rectangle
incenter
geometric transformation
homothety
reflection
geometry proposed

Let $ABC$ an acute triangle.

(a) Find the locus of points that are centers of rectangles whose vertices lie on the sides of $ABC$;

(b) Determine if exist some points that are centers of $3$ distinct rectangles whose vertices lie on the sides of $ABC$.

[quote="Simo_the_Wolf"]

(a) Find the locus of points that are centers of rectangles whose vertices lie on the sides of $ABC$;
[/quote]
Assume that $AH\perp BC=H$ and $A_{1}$ is the midpoint of $BC$ , $A_{2}$ is the midpoint of $AH$. Similary we'll have $B_{1},B_{2},C_{1},C_{2}$. I think , that locus is $A_{1}A_{2}\cup B_{1}B_{2}\cup C_{1}C_{2}$.

For a generalization of this problem, see at http://www.mathlinks.ro/Forum/viewtopic.php?t=80163

Kostas Vittas.

[quote="pavel kozlov"]Really, in part (b) answer is $I$ - incentre of triangle $ABC$[/quote]
I think that this is not correct.

In my opinion, if we denote as $D,$ $E,$ $F,$ the midpoints of the side segments $BC,$ $AC,$ $AB$ respectively and as $D',$ $E',$ $F',$ the midpoints of the altitudes of $\bigtriangleup ABC,$ through vertices $A,$ $B,$ $C$ respectively, then the answer in part $(b)$ is the concurrency point of the segment lines $DD',$ $EE',$ $FF',$ not necessary coincided with the incenter of $\bigtriangleup ABC.$

Kostas Vittas.

PS. Please, see also the [b][size=100]N.T.TUAN's[/size][/b] message.

[quote="[ƒ(Gabriel)³²¹º]¼"][quote="pavel kozlov"]Really, in part (b) answer is $I$ - incentre of triangle $ABC$[/quote]
really it is the Lemoine point. :wink:[/quote]

Thank you dear [b][size=100]ƒ(Gabriel)³²¹º]¼[/size][/b] for your remark. It is a new result to me and I see that it is true by the drawing.

Have you in mind a synthetic proof about it?

Kostas vittas.

Well ... it's enough to show that the Lemoine point is the center of three inscribed rectangles at the same time.
In order to achieve this, you only have to notice that the antiparallels of the sides through Lemoine point are bisected by it and, considering two of them, you obtain the diagonals of an inscribed rectangle.
Obviously, this is not a proof, but merely a hint.

I'm not convinced it's the Lemoine point. I make it the isotomic conjugate of the orthocentre of the medial triangle (which I don't think is the Lemoine point is it? :S).

Method: By inspection and thinking about homothety, the loci are the three lines between the midpoints of BC and AD (altitude from A) and the other two symmetric situations. it is clear that the midpoint P (say) of AD lies on the side MN of the medial triangle (if we take it to be LMN) and furthermore that NP/PM=BD/DC by homothety factor 2. Looking at this from the point of view of the medial triangle, which is oppositely oriented, we see that the reflection of P in the midpoint of MN will be the foot of the altitude from L to MN, whence the point of concurrency is the isotomic conjugate of the orthocentre of LMN.

[quote="Ilthigore"](which I don't think is the Lemoine point is it? :S).[/quote]
Yes, the symmedian point of the anticomplementary triangle is the isotomic conjugate of the orthocenter.

Ah, that's cool, and putting our two arguments together we've proved it. Is there a nice reason for this to be true? I don't know much about isotomic conjugation beyond the definition and existence properties.