Ok, I'm fairly new to Latex, so this might seem a little newbish. However, I was wondering how you can export Latex-generated math equations into images like you do in the AoPS Forum. Basically, a tool like The Texer (http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php) except that it isn't online.

I know of a program called Latexit for Mac that does this very well, you basically type in latex code (you don't even need the proper template/headers) and the program renders the math equation in a graphics file, which you can drag and drop into other programs like microsoft word.

My question is, is there any easy way to do this in TexnicCenter for Windows?? I know there seems to be an application called dvipng.exe in the miktek bin file, but I have no idea how to use this...

Try [url=http://www.mayer.dial.pipex.com/tex.htm#latexrender_windows]LatexRender for Windows[/url] which uses the same system as AoPS but on your own computer.

Thanks. I downloaded latex renderer and installed it along with the required imagemagick and ghostscript. However, I am still unable to get it to work.

I'm positive that i have the correct paths for the there executables in the latexrenderer.bat file.

however, when i press "render" a command line window pops up very quickly with an error message that looks like this:

"The system cannot find the path specified.
The system cannot find the path specified.
convert.exe: unable to open image 'temp.ps': No such file or directory.
convert.exe: missing an image filename 'latex001.gif'."

Can somebody help me with this please?

p.s. I'm running Windows Vista.

\\update

I was able to figure out the error. Seems like the updated the wrong part of the batch file.


you should post this problem in the inequalities section [url]http://www.mathlinks.ro/Forum/index.php?f=51[/url]

[color=red][Moved there. -mathmanman][/color]

[quote="duy_nguyen"]Let $ a,b,c > 0$.Prove that:
  $ ab.\sqrt{2(a^{2}\plus{}b^{2})}\plus{}bc.\sqrt{2(b^{2}\plus{}c^{2})}\plus{}ca.\sqrt{2(c^{2}\plus{}a^{2})}< a^{3}\plus{}b^{3}\plus{}c^{3}$[/quote]
It's wrong: try $ a\equal{}b\equal{}c\equal{}1.$ :wink:


Let $ p$ be an odd prime number and $ n$ a natural number. Then $ n$ is called $ p\minus{}partitionable$ if $ T\equal{}\{1,2,...,n \}$ can be partitioned into (disjoint) subsets $ T_1,T_2,...,T_p$ with equal sums of elements. For example, $ 6$ is $ 3$-partitionable since we can take $ T_1\equal{}\{ 1,6 \}$, $ T_2\equal{}\{ 2,5 \}$ and $ T_3\equal{}\{ 3,4 \}$.
$ (a)$ Suppose that $ n$ is $ p$-partitionable. Prove that $ p$ divides $ n$ or $ n\plus{}1$.
$ (b)$ Suppose that $ n$ is divisible by $ 2p$. Prove that $ n$ is $ p$-partitionable.

a) The total sum is $ \frac{n(n\plus{}1)}{2}$, so when $ p\ne 2\Rightarrow (2,p)\equal{}1$ we have $ p|n$ or $ p|n\plus{}1$, and $ 2$ divides one of $ n$ or $ n\plus{}1$ because they leave consecutive remainders mod 2.

b) Partition into $ 2tp\plus{}i,2tp\plus{}1\minus{}i$, cycling $ t$ over nonnegatives and $ i$ through positives up to the obvious bounds.


I saw on wikipedia that Pierre Wantzel proved it was impossible to trisect an angle using a compass and a ruler.  I searched on the net for a little bit to try to find it and I was unable to.  I simply googled it and I only found statements of the problem and it just mention that Wantzel proved it was impossible in 1836.  I also tried project gutenberg to no avail.  Does anyone know where it can be found, or does anyone know of a database of mathematical proofs for me to check?  Thank you in advance if you have any ideas.

Do you want the special proof of Wantzel or any proof¿

The proof I've seen is to show that constructable lengths all live in field extensions of order $ 2^n$ for some $ n$ but that $ \cos 20^{\circ}$ lives in a field extension of order $ 3$, so $ \cos 20^{\circ}$ is not constructable and consequently $ 20^\circ$ is not constructible.  Since $ 60^\circ$ in constructible, this means that we cannot trisect an angle of measure $ 60^\circ$, so we can't trisect an arbitrary given angle.  I have no idea if this is the original proof of this fact.

It doesn't matter what proof it is.  I think it's unrealistic that someone will know of a proof for it off hand, so I think instead I should ask if anyone has any ideas on how I can find a proof.

Artin's [u]Algebra[/u] has the proof that I just outlined for you, and I'm sure many texts on introductory abstract algebra and Galois theory should have it, as well.

Thank you JBL for the info, when I brought up it being unrealistic I meant having the full proof for it.  I've never taken a class on abstract algebra, so I was unable to follow your outline.  I'm a math major so I'm sure I'll come across it at some point, I was mostly wondering out of curiosity since it took so long for it to be proven.


I came across this somewhere on the net:

Prove:

If the binary representation of $n$ contains $p$ "1"s, then there are $2^{p}$ odd numbers in the $n^{th}$ row of Pascal's Triangle.

(For example, $13 = 1101_{2}$, so there are $2^{3}= 8$ odd numbers in the 13th row of Pascal's Triangle)

[hide="Hint"] Strong induction.  What's the simplest case? [/hide]

When $n = 2$?

Is there any other method apart from induction?

This is the simplest solution to me.

[hide="The simplest case"] $n = 2^{k}$ [/hide]


Find all triples of positive integers $(a,b,c)$ such that:
$(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})=2$


Find all real-valued functions $ f$ on the positive reals such that $ f(1) \equal{} \frac12$ and $ f(xy) \equal{} f(x) f\Big(\frac {3}{y}\Big) \plus{} f(y) f\Big(\frac {3}{x}\Big)$.

For $ x \equal{} y \equal{} 1$, we get $ f(3) \equal{} \frac {1}{2}$ and for $ y \equal{} 1,\forall x$ $ f(x) \equal{} f(\frac {3}{x})$ and let $ x \equal{} 3t$ then $ f(3t) \equal{} f(\frac {1}{t})$, $ \forall t$ because we work in $ R$
for $ y \equal{} 3$ we get $ f(3x) \equal{} \frac {f(x) \plus{} f(\frac {3}{x})}{2})$ that means that $ f(3x) \equal{} f(x) \equal{} f(\frac {1}{x}) \equal{} f(\frac {3}{x}), \forall x$.
Then for $ y \equal{} \frac {1}{x}$, we get $ f(1) \equal{} f(x)*f(3x) \plus{} f(\frac {1}{x})*f(\frac {3}{x})$ and hence $ f(x) \equal{} \frac {1}{2}, \forall x$.

Actually $ f(x)\equal{}\pm{1\over 2}$.
Try to prove the rest.

[quote="olorin"]Actually $ f(x) \equal{} \pm{1\over 2}$.
Try to prove the rest.[/quote]

This is wrong!!! How is $ f(1)\equal{}\frac{1}{2}$ if $ f(x)\equal{}\minus{}\frac{1}{2}$?

$ f(x)$ can be $ \frac12$ for some $ x$ and $ \minus{}\frac12$ for another $ x$!


In the [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Community_Portal]community portal[/url], the list of Image Requests includes [[Convex polygon]], but that article already has an image, it does not have {{image}} anywhere, and it is not listed [url=http://www.artofproblemsolving.com/Wiki/index.php/Category:Image_needed]here[/url]. Same with [[Incenter]].

Why dont you jsut change it? its a wiki....  :ninja:

[quote="i_like_pie"]Aren't the few in the portal taken from the larger list at random? If not, I'll change it. :oops:[/quote]

I already changed it  :P. But yeah those were just added by someone.

Actually, that's an idea. There should be someone or a team of someones continually checking the Community Portal lists to make sure it's updated. I'll put up a notice for this...

I'll do it. :)

I think there should be at least one other person, so if I'm away for a few weeks, they'll be kept up to date. Then again, it's not that critical.

I'll do it also; I was the one who originally placed the articles there. I basically just picked random articles from Special:MostWantedPages or Category:Images needed, etc.


Solve the following equations over $\mathbb Z$:

a)  $ a^2 \plus{} b^2 \equal{} 2a \minus{} b$
b) $ a^2 \plus{} b^2 \plus{} c^2 \equal{} abc$

[quote="CDP100"]a)  $ a^2 \plus{} b^2 \equal{} 2a \minus{} b$
b) $ a^2 \plus{} b^2 \plus{} c^2 \equal{} abc$[/quote]

For the first one : $ b(b\plus{}1)\equal{}a(2\minus{}a)$
We have $ b(b\plus{}1)\geq 0$ $ \forall b\in\mathbb{Z}$, so we need $ a(2\minus{}a)\geq 0$ and so $ a\in\{0, 1, 2\}$
$ a\equal{}0$ gives the solutions $ (0,0)$ and $ (0,\minus{}1)$
$ a\equal{}1$ gives no solution
$ a\equal{}2$ gives the solutions $ (2,0)$ and $ (2,\minus{}1)$


Show that $ z\bar{z}$=$ |z|^{2}$ for any complex number $ z$/

(ii) A sequence of complex numbers $ z_{n}$ is given by the rule $ z_{1}$=$ w$ and $ z_{n}$=$ c\bar{z_{n-1}}$ for $ n=2,3,4...,$ where $ w$ is a given complex number and c is a complex number with modulus 1. Show that $ z_{3}$=$ w$


#1 -  e^(1-5x) = 793

#2 - e^(5x-3) - 2 = 10,476

I don't know the procedures to solve these equations.

thanks

#1 -  e^(1-5x) = 793

I think you start like this

(1-5x) ln e = ln 793

then distribute

ln e - 5x ln e = ln 793

then what?

#2 - e^(5x-3) - 2 = 10,476

add 2 ?

(5x-3) ln e = 10,478

5x ln e -3 ln e = ln 10,478

then what?

I'm pretty sure my procedure for solving these is wrong, but I am stumped...

thanks

Just if you need insurance.
[hide]For #1, we start with $ \pwr e^{1 - 5x} = 793$. Recognize that $ \log_e {e} = 1$. This can be simplified to $ 1 - 5x = \log_e{793} -> x = \frac {\log_{e}793 - 1}{ - 5}$.

For #2, $ \pwr e^{5x - 3} - 2 = 10,476$, this simplifies to $ \pwr e^{5x - 3} = 10,478 -> x = \frac {\log_e{10,478} + 3}{5}$.[/hide]


[hide]disregard the thirty part because the numbers still have the same ratio. so it's really just 100/450=.2222222.. [/hide]

[quote="aznmonkey1992"]Bobs shack sells 3 types of phones: a 100 dollar one, a 125 dollar one, and a 225 dollar one. If, in august, he sells 30 of each, what percent of income did the 100 dollar bring out  of the total income?[/quote]

[hide="Im just hiding this just because nonie told me to when i dont want to"]
100*30=3000. 125*30=3750, 225*30=6750.

300/300+375+675=300/1350=30/135=22%?[/hide]

[quote="anirudh"][quote="aznmonkey1992"]Bobs shack sells 3 types of phones: a 100 dollar one, a 125 dollar one, and a 225 dollar one. If, in august, he sells 30 of each, what percent of income did the 100 dollar bring out  of the total income?[/quote]
[hide]
100*30=3000. 125*30=3750, 225*30=6750.

300/300+375+675=300/1350=30/135=22%?[/quote]
][hide]Hide your answer please!!!
3000
3750
6750

3000/13500=30/135=22.222222...percent rounded 22%[/hide][/hide]

[quote="anirudh"]why don't you hide yours?[/quote]
Sorry my mom kept calling me for luch and I was in such a rush and totally forgot to. I meant to though. :blush:  :oops:And my hides got mixed up in the quotes somehow...

[quote="aznmonkey1992"]Bobs shack sells 3 types of phones: a 100 dollar one, a 125 dollar one, and a 225 dollar one. If, in august, he sells 30 of each, what percent of income did the 100 dollar bring out  of the total income?[/quote]
[hide]
30(100+125+225)= 30x450 = 13500
30x100=3000

3000/13500 = 22.22222.... %
[/hide]

[hide]
The total profit is 13500.
So the answer is $\frac{3000}{13500} = .\bar{2}$[/hide]

don't do that, it makes the screen weird, but at least you hid it.  :)


I need to edit a pdf file and possibly combine it with my own pdf and submit that pdf through email. How can I do the first two of these things?

You can't edit unless you have a commercial copy of Adobe Acrobat which is quite expensive. If you want to edit the AoPS cover sheet then ask AoPS if they would let you have the original tex file.

It's probably irrelevant now but you can include the pdf file in your tex document in the same way as including jpegs, gifs etc. The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=pdfpages]pdfpages package[/url] even allows you to use some of the pdf pages.

Why is that irrelevant? If I can somehow find a way to edit the pdf, I could possibly use that package to put in the pdf. If not, can is turn the pdf (which is only about a half page long) into an image, edit it, and put in another pdf. If so, how?

It's irrelevant because you [b]cannot[/b] edit the pdf without spending lots of money. It's easy to turn the pdf into an image uing PrtSc but that won't help either because you can only draw on an image not edit it.

You will have to either ask for the tex file or make it yourself which is not that hard.


[quote="chien than"]Let $a,b,c \geq 0$ and$a+b+c=1$.Prove that:
$\sqrt{a+b^{2}}+\sqrt{b+c^{2}}+\sqrt{c+a^{2}}\geq 2$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106782


I go for walk, for more than 2 hours but less than 3. When I return the  minute and hour hands have  exchanged places. How much longer than 2 hours did the walk last ?

[size=75]
from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide
[/size]

[hide="Solution"]Two hours after I leave, the minute hand will be where it is as I leave, and the hour hand will have covered $ \frac{1}{6}$ of a full circle. For the hands to move until their original positions are exchanged, the sum of the distances from the original positions must increase, after 2 hours, by $ \frac{5}{6}$ of a circle, or 50 minutes. The minute hand moves 12 times as fast as the hour hand, so the distance that still has to be covered by it is $ \frac{12}{13}\cdot50\equal{}46\frac{2}{13}$. Thus the walk lasted $ 46\frac{2}{13}$ minutes more than 2 hours.[/hide]


quadratics function algebra domain conics parabola search

Consider for constant numbers $ a$ with $ 0\leq a\leq 2$ the quadratic function $ f(x) \equal{} \minus{} x^2 \plus{} ax \plus{} a$ for $ 0\leq x\leq 1.$ Find the range of $ a$ such that $ f(x)$ has more than one integer value.

[hide]
First, I want to find the range of $ f(x)$

Looking at the endpoints
$ f(0) \equal{} a$ and $ f(1) \equal{} 2a \minus{} 1$

Now looking at the vertex,

$ f(\frac {a}{2}) \equal{} \frac {a^2}{4} \plus{} a \equal{} \frac {a^2 \plus{} 4a}{4}$

because of the domain of $ a$, the min is $ 0$ at $ x \equal{} 0$
and the max is $ \frac {a^2 \plus{} 4a}{4}$

since you need multiple values of $ f(x)$

$ \frac {a^2 \plus{} 4a}{4} \geq 1$

$ a^2 \plus{} 4a \geq 4$

$ a^2 \plus{} 4a \minus{} 4 \geq 0$

this is true when

$ \minus{}2\plus{}2\sqrt{2} \leq a \leq 2$

(the other part of the parabola is irrelevant because it is not in the domain of $ a\in[0,2]$)
[/hide]

EDIT: I had an arithmetic error which I have fixed

[hide="Wrong"]
$ f(x) = - (x - \frac {a}{2})^2 + \frac {a^2 + a}{4}$

So $ - 1\leq - x^2 \leq f(x)\leq\frac {a^2 + a}{4}\leq\frac {3}{2}$

Note that $ f$ takes $ - 1$ only when $ a = 0$, in this case also takes $ 0$ so, $ a = - 1$ is part of the answer.

Now we search for $ f$ to take $ 0$ and $ 1$:

The roots of $ - x^2 + ax + a = 0$ are $ \frac {a + \sqrt {a^2 + 4a}}{2}$ and $ \frac {a - \sqrt {a^2 + 4a}}{2}$
It is clear that $ \frac {a - \sqrt {a^2 + 4a}}{2}\leq0$ so $ f$ will take $ 0$ if and only if $ \frac {a + \sqrt {a^2 + 4a}}{2}\leq1 \Leftrightarrow \sqrt {a^2 + 4a}\leq 2 - a \Leftrightarrow a^2 + 4a \leq 4 - 4a + a^2 \Leftrightarrow a\leq\frac {1}{2}$

$ f$ will take $ 1$ if and only if $ \frac {a^2 + a}{4}\ge1 \Leftrightarrow a^2 + a\ge4 \Leftrightarrow (a + \frac {1}{2})^2\ge\frac {17}{4} \Leftrightarrow a\ge \frac {\sqrt {17} - 1}{2}$

Hence, $ a\leq\frac {1}{2}$ and $ a\ge \frac {\sqrt {17} - 1}{2}$, $ \Rightarrow \Leftarrow$

So $ a = 0$ is the only value for wich $ f$ takes two integer values.

 :)
[/hide]

Thanks t0rajir0u.

[hide="I hope this is OK"]
$ f(x)=-x^2+ax+a=k  \Leftrightarrow x^2-ax+k-a=0 \Leftrightarrow$
$ (2x-a-\sqrt{a^2+4a-4k})(2x-a+\sqrt{a^2+4a-4k})=0$
So for $ f$ to take the value $ k$, it has to satisfy:
$ a^2+4a-4k\ge0 \Leftrightarrow (a+2+2\sqrt{1+k})(a+2-2\sqrt{1+k})\ge0 \Leftrightarrow a\ge2\sqrt{1+k}-2$,
and at least one from: $ \frac{a-\sqrt{a^2+4a-4k}}{2}\ge0$, $ \frac{a+\sqrt{a^2+4a-4k}}{2}\leq1$, wich simplifies to $ a\leq k$ and $ a\leq\frac{1+k}{2}$
Now:
$ f$ takes $ -1$ if and only if $ a\leq 0$ and $ a\ge -2$, so $ a=0$
$ f$ takes $ 0$ if and only if $ 0\leq a\leq \frac{1}{2}$
$ f$ takes $ 1$ if and only if $ \frac{1}{2}<2\sqrt{2}-2\leq a \leq 1$
$ f$ takes $ 2$ if and only if $ 1<2\sqrt{3}-2\leq a \leq 2$
$ f$ takes $ 3$ if and only if $ 2\leqq$ and $ a\leq2$, so $ a=2$

From this we can conclude that $ f$ takes two integer values only for $ a=0$ and $ a=2$.
So I have to say I agree with timwu's answer.

 :)
[/hide]

 :)

[quote="ElChapin"]$ f(x) \equal{} \minus{} (x \minus{} \frac {a}{2})^2 \plus{} \frac {a^2 \plus{} a}{4}$[/quote]

Actually, $ f(x) \equal{} \minus{} \left( x \minus{} \frac{a}{2} \right)^2 \plus{} \frac{a^2}{4} \plus{} a$.  Your upper bound on $ f$ is inaccurate.

This problem seems to be difficult a bit.

Answer
[hide]$ a \equal{} 0,2$[/hide]

Solution
[hide]
$ f(x) \equal{} \minus{} (x \minus{} \frac {a}{2})^2 \plus{} \frac {a^2}{4} \plus{} a$
We have, $ f(0) \equal{} a, f(1) \equal{} 2a \minus{} 1$, and vertex $ f(\frac {a}{2}) \equal{} \frac {a^2}{4} \plus{} a.$
Because $ 0\leq a\leq 2, 0\leq x\leq 1, \frac {a}{2}$ is always in the domain.
Therefore, the maximum is always $ \frac {a^2}{4} \plus{} a$.
Since the parabola has only one turning point, and this one opens down, either (or both) of $ f(0) \equal{} a$ or $ f(1) \equal{} 2a \minus{} 1$ is the minimum.

To have at least two integers in the range, the difference between the upper and lower boundary must be at least $ 1$.

If $ 2a \minus{} 1$ is the minimum, or $ 2a \minus{} 1 \leq a$, then $ 0 \leq a < 1$. The range of $ f(x)$ is $ [2a \minus{} 1, \frac {a^2}{4} \plus{} a]$. From the inequality, $ a\leq 0$ or $ a\geq 4$.
If $ a$ is the minimum, or $ a \leq 2a \minus{} 1$, then $ 1\leq a\leq 2$. The range of $ f(x)$ is $ [a, \frac {a^2}{4} \plus{} a]$. From the inequality, $ a\leq \minus{} 2$ or $ a\geq 2.$
So the only possible values of $ a$ is $ 0, 2$. Plugging them in, they do satisfy the given. Hence, the range of $ a$ is $ {0,2}$.
[/hide]

Regrettably your answer is incorrect. :(

[quote="chickendude"]
this is true when

$ \minus{} 2 \plus{} 2\sqrt {2} \leq a \leq 2$

[/quote]

Contradiction here:

Since $ 2\sqrt{2}\minus{}2<1< 2$, plug in $ a\equal{}1$:

$ f(x)\equal{}\minus{}x^2\plus{}x\plus{}1\equal{}\minus{}(x\minus{}\frac{1}{2})^2\plus{}\frac{5}{4}$

whose minimum is $ f(0)\equal{}f(1)\equal{}1$, and maximum is $ f(\frac{1}{2})\equal{}\frac{5}{4}.$

@kunny: I still don't see my mistake...

The way you are thinking is right, timwu. :)  Sketch the graphs of $ b\equal{}M(a),\ b\equal{}m(a)$ on $ a\minus{}b$ plane, where $ M(a),\ m(a)$are the maximum, minimum values of $ f(x)$ in the domain of $ 0\leq x\leq 1$ respectively.


geometry
3D geometry
sphere
tetrahedron
analytic geometry
circumcircle

Four spheres of the same radius r have the property that each is tangent to the other three spheres. Then the minimal radius possible for a fifth sphere which is tangent to each of the given four is:

A. r * (2 (2/3)^(3/2) – 1)
B. r / sqrt3
C. r * (sqrt(3/2) – 1)
D. r/4
E. None of the above

Bigger I guess so that it is tangent to the others..

Not necessarily - draw some diagrams to help.

You're not really suggesting someone draw a diagram of 4 mutually tangent spheres?  More helpful -- find 4 similarly sized round objects (juggling balls?  baseballs?) and put one on the base of the other 3.  Now, it's clear that there is a sphere that is tangent to all 4 and surrounds them.  Can you see that there would be at least 1 more such sphere?

I might budge this up to Intermediate.

Well, in a competition, where are you going to find 4 baseballs??

Otherwise, the only way 4 spheres could be mutually externally tangent would be when they are stacked in the shape of a tetrahedron, so that would be it.

so..we are still seeing how they can be tangent and haven't yet done the math! :)

[quote="DPopov"][hide="Hint"]
Does the radius of the other sphere have to be bigger or smaller than the radius of the four congruent spheres?
[/hide][/quote]

It says minimal... I always thought that meant the smallest??

Yes, it simply means that it is tangent to all of the others, but how can you have one with a smaller radius than the others?

[quote="DPopov"][hide="Hint"]
Does the radius of the other sphere have to be bigger or smaller than the radius of the four congruent spheres?

[/hide][/quote]

OBVIOUSLY the radius of the fifth sphere can't be bigger than the orginal congruent spheres for otherwise there won't be any contact between all the spheres

[hide="strategic hint"]Standard problem solving strategy:  "Try a simpler case"[/hide]

One more hint --- (Not that one needs a hint here ... but anyway if you choose to see it)
[hide]Of course,  easiest , at least for a true mathematician,  would be to consider n dimensional case and substitute the value whne n=3 :)

Ok ..Ok a simple hint ...:)

[hide]Let the four spheres be at the four  alternate  corners of  a  cube..  (bottom two diagonals and upper two diagonals)

[hide]Say, each side of the cube is   $2$  ,  so that  radius of  each sphere is $\sqrt{2}$   the center is   $\sqrt{3}$ from a corner, required radius is of course $(\sqrt{3/2} - 1) $ times  r (= :sqrt: 2)

BTW  if some one wanted the "outer' sphere ... it would be just $(\sqrt{3/2}+ 1) $
[/hide]
[/hide][/hide]

What are the four alternate corners of a cube?

Look at a normal die with the number one facing up, 2 facing north, 4 facing east, 5 facing south ,3 facing west, and six on the bottom.

Number the corners at the following intersections the following letters:

1,2,3-A
1,2,4-B
1,4,5-C
1,3,5-D
2,3,6-E
2,4,6-F
4,5,6-G
3,5,6-H

Name one set of four alternate corners.

[quote="JBL"]Whoa, I'm not going to try to figure out the die.  I'll give you two other descriptions, though:  take a vertex A and the 3 vertices that are the other endpoints of a face diagonal with one endpoint at A.  Alternatively, look at a coordinate system -- set the spheres so that they have centers (1, 1, 1), (1, -1, -1), (-1, 1, -1) and (-1, -1, 1).  This gives us a set of 4 vertices that is symmetric.  It also leaves the cube with 4 other vertices in a similar configuration.  These 4 vertices are also the vertices of a regular tetrahedron.[/quote]

Thanks!

[quote="JBL"]That's not true, Santosh -- there is a sphere that contains all 4 spheres and is tangent to each of them.  However, this is not the smallest sphere tangent to all 4.[/quote]

I fully agree with u JBL,but that sphere is the one engulfing all the spheres. ( something analogous to the circumcircle)

Actually I was considering the engulfed one rather(i.e the one analogous to the INCIRCLE).

Hope I'm clear now

Thanks JBL   - that is what I meant in my hint. Just  to add , some may find , this way of considering a tetrahadron a little easier because it is easy to imagine a cube.  Here if we consider each  length of the cube is  2 units,   the 4  symetric points are  (each) 2 :sqrt: 2 distance away... the six points where the  4 spheres touch  are the six center points of the faces.. .. the distance between two lines (which do not touch)  of the tetrahadron are 2 units apart ...  volume of the tetrahadron is (1/3) of the cube (which is 8) etc.. all these things become much easier to visulazie ....

Hope this is helpful.

[quote="white_horse_king88"]Well, now that we have established all possible tangent spheres for the fifth sphere, find a way to form equations using r as the radius of the small sphere - make a tetrahedron out of the centers of the other four tangent spheres. And use Pythagorean.[/quote]

And working on the above lines, the ans. should be option C i.e. r*(sqrt(3/2) - 1)

yes the answer is C -

I think it's a difficult question.


[quote="nebula42"]1. Only one pair of positive integers $(x,y)$ satisfies the equation \[19x+17y=127\] What is the value of $x+y$?

2. Only one pair of positive integers $(x,y)$ satisfies the equation \[29x+23y=260\] What is the value of $x+y$?[/quote]
[hide]$4+3=7$

$5+5=10$[/hide]

[quote="nebula42"]1. Only one pair of positive integers $(x,y)$ satisfies the equation \[19x+17y=127\] What is the value of $x+y$?

2. Only one pair of positive integers $(x,y)$ satisfies the equation \[29x+23y=260\] What is the value of $x+y$?[/quote]

[hide="The First one"]
The pair is $\{4, 3\}$, so the sum is $7$.
[/hide]

[hide="The Second One"]
The pair is $\{5, 5\}$, so the sum is $10$.[/hide]

Solutions please? These use modulos, right?

Your answers are correct.  :)
Yes, these problems can always be easily solved with modulos; however, with this one, the numbers are small enough so that using guess and check works fine.

[quote="nebula42"][quote="ragnarok23"]Would someone mind showing the solution involving modulos?[/quote]
All right, I'll just show you the first one (the second is almost exactly the same):
[hide]We have $19x+17y=127$. We take the equation modulo 17 (which gets rid of $y$) to obtain
$2x \equiv 8 \pmod{17}$.
This is clearly solved by $x=4$, and we can see that the next smallest positive integer solution for $x$ would be $x=4+2\times17=38$, which is obviously too large, so the only solution for $x$ is $x=4$, and from this we can easily obtain (by plugging $4$ into the initial equation) $y=3$, so the answer is $4+3=7$.[/hide][/quote]

Ok, I get it now. I'll try the second one...
[hide]The equation in $\bmod{23}$ results in \[6x\equiv 7\pmod{23}\] . We can see that $x=5$, and plugging this in the original equation results in $y=5$. Thus the answer is $5+5=10$.[/hide]

Please tell me if I did something wrong...

[hide="1"]looking at the equation mod 19, we get $17y=13 (\text{mod }19)$.  subtracting 19y we get $-2y=13 (\text{mod }19)$.  since $y$ has to be positive, $-2y$ has to be negative, a multiple of 2, and $13 (\text{mod }19)$.  the smallest such number is $-6$, so $y=3$.  plugging this into the equation gives $x=4$   the other equation is pretty much the same thing[/hide]

[hide="answer for one"]7[/hide]

[hide="answer for 2"]10[/hide]

solution for both:

[hide="I"]Plugged in values. :D [/hide]


limit
inequalities
real analysis
real analysis unsolved

Let $ l_2$ denote the set of all inﬁnite sequences $ x \equal{} (x_n)$ of real numbers for which the series
 $ \sum_{i\equal{}1}^{\infty}{x_i}^2< \plus{}\infty$. We define: $ \parallel{}x\parallel{}\equal{}\sqrt{\sum_{i\equal{}1}^{\infty}x_i^2}$.
Let $ l_1$ denote the set of all inﬁnite sequences $ x \equal{} (x_n)$ of real numbers for which the series $ \sum_{n\equal{}1}^{\infty}|x_n| < \plus{}\infty$
For $ x\equal{}(x_n) \in l_1$ we define $ \parallel{}x\parallel{}_1\equal{}\sum_{n\equal{}1}^{\infty}|x_n|$.
Prove that:
1) $ l_1$ is a subspace of $ l_2$.
2) norm $ \parallel{}.\parallel{}$ is not equivalent with norm $ \parallel{}.\parallel{}_1$ in $ l_1$.

1) Let $ x = (x_n)\in l_1$, we have $ \sqrt {\sum_{k = 1}^{n}x_k^2}\leq\sum_{k = 1}^{n}|x_k|$ and after taking the limits ${ \sqrt {\lim_{n\to\infty}\sum_{k = 1}^{n}x_k^2}}\leq\lim_{n\to\infty}\sum_{k = 1}^{n}|x_k|\Rightarrow\sqrt {\sum_{n = 1}^{\infty}x_n^2}\leq\sum_{n = 1}^{\infty}|x_n| < + \infty\Rightarrow x\in l_2$.

2) Take for example $ x_n=\frac{1}{n}$.

[quote="Ognjen09"]...we have $ \sqrt {\sum_{k \equal{} 1}^{n}x_k^2}\leq\sum_{k \equal{} 1}^{n}|x_k|$ ...[/quote]

While this step is true, it is far from obvious. In a textbook problem such as this, I would expect that the student would be expected to prove this as part of the overall proof.

Step 1: Assume $ \|x\|_1\equal{}1.$ Then for each $ k,$ $ |x_k|\le 1$ and $ |x_k|^2\le|x_k|,$ and

$ \|x\|_2^2\equal{}\sum_{k\equal{}1}^{\infty}|x_k|^2\le\sum_{k\equal{}1}^{\infty} |x_k|\equal{}\|x\|_1\equal{}1$

so $ \|x\|_2^2\le 1$ and thus $ \|x\|_2\le 1.$

Step 2. Note that if $ \|x\|_1\equal{}0,$ then $ x_k\equal{}0$ for all $ k$ and the inequality is trivially true.

Step 3. Assume WLOG that $ 0<\|x\|_1<\infty.$ Define $ y$ by $ y_k\equal{}\frac{x_k}{\|x\|_1}.$ Then $ \|y\|_1\equal{}1.$ Apply step 1 to get that $ \|y\|_2\le 1.$ But we can show that $ \|y\|_2\equal{}\frac1{\|x\|_1}\,\|x\|_2,$ so $ \|y\|_2\le 1$ implies that $ \|x\|_2\le \|x\|_1.$

As for part 2, while Ognjen09 is correct to point to $ x_k\equal{}\frac1k$ for an example, technically speaking, that $ x$ is not in $ \ell^1$ at all.

Equivalence of norms would mean that there exist positive finite constants $ c$ and $ C$ such that for all $ x\in\ell^1,$ $ c\|x\|_1\le \|x\|_2\le C\|x\|_1.$ We already have the right hand inequality with $ C\equal{}1,$ so we are investigating the left hand inequality.

Consider this family of sequences, for $ 0<r<1:$ let $ x_k\equal{}(1\minus{}r)r^{k\minus{}1}.$ By direct calculation, we have that $ \|x\|_1\equal{}1.$ But

$ \|x\|_2^2\equal{}(1\minus{}r)^2\sum_{k\equal{}1}^{\infty}r^{2(k\minus{}1)}\equal{} \frac{(1\minus{}r)^2}{1\minus{}r^2}\equal{}\frac{1\minus{}r}{1\plus{}r}.$

That means that $ \|x\|_2\equal{}\sqrt{\frac{1\minus{}r}{1\plus{}r}}$ and can be made arbitrarily small - smaller than any $ c>0$ - by making $ r$ sufficiently close to $ 1.$ Hence there is no such $ c$ and the norms are not equivalent.

You could get the same thing out of $ x_k\equal{}\begin{cases}\frac1k,&1\le k\le n\\0,&k>n\end{cases},$ for which $ \|x\|_2$ is bounded but $ \|x\|_1$ can be made arbitrarily large by choosing $ n$ large enough.

Your answer is very useful. Thank u very much

[quote="Kent Merryfield"][quote="Ognjen09"]...we have $ \sqrt {\sum_{k \equal{} 1}^{n}x_k^2}\leq\sum_{k \equal{} 1}^{n}|x_k|$ ...[/quote]

While this step is true, it is far from obvious. In a textbook problem such as this, I would expect that the student would be expected to prove this as part of the overall proof.[/quote]

square both sides.  the expansion of the RHS contains the LHS, as well as a bunch of other clearly nonnegative terms...


αν α και β ειναι 2 παραπληρωματικές γωνίες να βρεθει το πρόσιμο
ημα.ημβ
εφα.εφβ
 να δικαιολογήσετε την απάντηση

παρακαλώ βοηθηστε το εχς θεμα για αυριο

tomtat, καλώς ήλθες στην παρέα μας

Ξέρουμε ότι αν οι γωνίες α,β είναι παραπληρωματικές, τότε έχουν ίσα ημίτονα (επειδή στον τριγωνομετρικό κύκλο βρίσκονται στην ίδια οριζόντια ευθεία).

Τότε $\eta \mu a\cdot \eta \mu \beta = \eta\mu^2a\geq 0$

(έχουμε μηδέν όταν οι δύο γωνίες είναι 0 και 180 μοίρες)

Για την εφαπτομένη ξέρουμε ότι θα είναι $\epsilon\phi a = - \epsilon \phi \beta$

Άρα ετερόσημοι αριθμοί, άρα γινόμενο αρνητικό (ή μηδέν, αν έχουν εφαπτομένη μηδέν)

Φαντάζομαι ότι θα πρέπει να αποκλείσεις την περίπτωση $a=\beta=90^o$ γιατί τότε εφαπτομένη δεν ορίζεται


linear algebra
matrix
inequalities
linear algebra unsolved

For any given positive interger N,we know that there exists at least  N linnear independent commutative matrix of order N . (for example, any two diagonal matrix are commutative.)

But, I want to know the maximal number of linnear independent  commutative matrix . And I learn that when N=1,2,3,4, the maximal number is 1,2,3,5. It seems that when N>=4, the maximal number of linnear independent  commutative matrix of order N will be stictly larger than N.

Can anyone tell me the exact value of maximal number of linnear independent commutative matrix of any given order N ?

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1248585#1248585 should help you.

[url=http://www.research.att.com/~njas/sequences/A033638]A033638[/url] is the relevant sequence in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

Thank JC_math and JBL very much.

With your help, I am suprised by the fact that this problem was solved 100 years ago by Schur, and simplified by some following mathematicians.

I wonder why this intresting problem does not appear in so much text book on linear algebra. And that is the reason why so many people reconsider the same problem.

I will spend some time  to study the solution given by JC_math.


number theory
greatest common divisor
quadratics
number theory unsolved

Find all nonnegative integers $m$ such that
\[a_m=(2^{2m+1})^2+1 \]
is divisible by at most two different primes.

Friend ,please explain how that solves the problem..I got till this and also the shorter of those numbers is a power of $ 5$ ....what about this case  ,you left this .....
$ 2^{4m \plus{} 2} \plus{} 1 \equal{} p^{a}q^{b}$
where $ p^{a} \equal{} 5^{a} \equal{} 2^{2m \plus{} 1} \plus{} 1 \minus{} 2^{m \plus{}1}$ .
I dont think the first case of yours is possible .....not too sure though ......

that's baltic way 2002:
go to (in fact i mentioned a special case of the problem but an explanation is given in the last topic)
http://www.mathlinks.ro/viewtopic.php?p=1578991#1578991

I have not read your link shoki, so i dont know if the following ideas are the same as those ones..

$ 2^{4m\plus{}2}\plus{}1$ is in the form $ x^4\plus{}4y^4$ (Sophie-German identity), so it can be factorized as $ ((2^m)^2\plus{}(2^m\minus{}1)^2)((2^m)^2\plus{}(2^m\plus{}1)^2)$. Since the two factor are coprime,both odd and greater than $ 1$ then we are lead to solve the system $ p^a\equal{}(2^m)^2\plus{}(2^m\plus{}1)^2>(2^m)^2\plus{}(2^m\minus{}1)^2\equal{}q^b$, for some distinct prime $ p,q$ (for istance $ 4 \mid \text{gcd}(p\minus{}1,q\minus{}1)$ since $ 1\equal{}(\frac{\minus{}1}{p})\equal{}(\frac{\minus{}1}{q})$) and $ a,b$ are positive integers. Now we have three consecutive quadratic residue of (x-1)²,x²,(x+1)², where x² is non zero for every mod p>2. Modulo 5 we see that if x² is a fixed r in {-1,1} then at least one between (x-1)² and (x+1)² is -r. It means that (p-5)(q-5)=0. So, in other words, it sufficies to find all solutions of the equation $ (2^m)^2\plus{}(2^m\plus{}\ell)^2\equal{}5^a$, where $ l \in \{\minus{}1,1\} \text{ and } a \in \mathbb{N}_0$. If $ m>1$ we can see modulo $ 8$ that $ 2 \mid a$, so it exists $ b \in \mathbb{N}_0$ such that $ a\equal{}2b$, so $ 2^{2m}\equal{}(5^b\plus{}2^m\plus{}\ell)(5^b\minus{}2^m\minus{}\ell)$. Now $ \text{gcd}(5^b\plus{}2^m\plus{}\ell,5^b\minus{}2^m\minus{}\ell)$ $ \equal{}\text{gcd}(5^b\plus{}2^m\plus{}\ell,2 \cdot 5^b)$ and must be 2, so it sufficies to solve $ 2^{2m\minus{}1}\equal{}5^b\plus{}2^m\plus{}\ell \text{ and } 2\equal{}5^b\minus{}2^m\minus{}\ell$ $ \implies 5^b\equal{}2^m\plus{}\ell\plus{}2\equal{}2^{2m\minus{}1}\minus{}2^m\minus{}\ell$. It implies that $ 2^{2m\minus{}1} \equal{} 2^{m\plus{}1}\plus{}2\ell \plus{}2 \le 2^{m\plus{}1}\plus{}4$ that is false if $ m>2$. So, unique solutions are $ m \in \{1,2\}$ that is verified by hand.[]

@bbyopa
Friend ,can you tell me the reason why you have assumed $ (2^{m})^{2}\plus{}(2^{m}\plus{}\ell)^{2}\equal{}5^{a}$, I mean cant we directly tell that $ (2^{m})^{2} \plus{} (2^{m} \minus{} 1 )^{2} \equal{}5^{a}$ because $ 5$ always divides $ 2^{4m\plus{}2} \plus{}1$

In principle i have complicated my life since i have not seen that $ 4^{2x \plus{} 1} \equal{} \minus{}1 \text{ in } \mathbb{F}_5$, on the other hand both ways work, so what is the problem?  :wink:

Ps. But why have you assumed $ \ell \equal{} \minus{} 1$ in your statement?

er.......I never assumed $ \ell \equal{} 1$ ,infact I never even introduced a $ \ell$ ,
I just did this
We have $ (2^{2m \plus{} 1} \minus{} 2^{m \plus{} 1} \plus{} 1)(2^{2m \plus{} 1} \plus{} 2^{m \plus{} 1} \plus{} 1) \equal{} 5^{a}q^{b}$
Since $ \gcd( 2^{2m \plus{} 1} \minus{} 2^{m \plus{} 1} \plus{} 1 , 2^{2m \plus{} 1} \plus{} 2^{m \plus{} 1} \plus{} 1) \equal{} 1$ and also $ q > 5$ and $ 2^{2m \plus{} 1} \minus{} 2^{m \plus{} 1} \plus{} 1 < 2^{2m \plus{} 1} \plus{} 2^{m \plus{} 1} \plus{} 1$ ,we have $ 2^{2m \plus{} 1} \minus{} 2^{m \plus{} 1} \plus{} 1 \equal{} 5^{a}$ , from this I got a different solution .

P.S -Friend can you tell me what is $ \mathbb{F}_{5}$

Well here is my solution ,sorry if this is almost the same as yours .

We observe that $ m \equal{} 1$ and $ m \equal{} 0$ form a solution ,so assume $ m \neq 1,0$
I put forth the following lemma and assumptions.
$ \text{Lemma 1}$
$ 2^{4m \plus{} 2} \plus{} 1 \equiv 0\bmod{5}$ ,so definitely $ 5$ is one of the prime dividing $ 2^{4m \plus{} 2} \plus{} 1$

$ \text{Assumption 1}$
$ 2^{4m \plus{} 2} \plus{} 1 \equal{} 5^{\alpha}q^{\beta}$ where $ q$ is a prime and  $ \alpha ,\beta \in \mathbb{N}$

$ \text{Assumption \minus{} 2}$
$ 2^{4m \plus{} 2} \plus{} 1 \equal{} 5^{\alpha}$

Let us work with Assumption 1
We get $ 2^{4m \plus{} 2} \plus{} 1 \equal{} (2^{2m \plus{} 1} \plus{} 1 \minus{} 2^{m \plus{} 1})(2^{2m \plus{} 1} \plus{} 1 \plus{} 2^{m \plus{} 1}) \equal{} 5^{\alpha}q^{\beta}$
From my previous post I conclude that $ (2^{2m \plus{} 1} \plus{} 1 \minus{} 2^{m \plus{} 1}) \equal{} 5^{\alpha}$

$ \text{Lemma \minus{} 2}$
$ \alpha$ is even .
The proof can be directly got by factorising the R.H.S

Let $ \alpha \equal{} 2k$  where $ k \in \mathbb{N}$
$ (2^{2m \plus{} 1} \plus{} 1 \minus{} 2^{m \plus{} 1}) \equal{} (2^{m})^{2} \plus{} (2^{m} \minus{} 1)^{2} \equal{} 5^{2k}$

$ (2^{m})^{2} \equal{} (5^{k} \plus{} 2^{m} \minus{} 1)(5^{k} \minus{} 2^{m} \plus{} 1)$
$ \implies (5^{k} \plus{} 2^{m} \minus{} 1) \equal{} 2^{a},(5^{k} \minus{} 2^{m} \plus{} 1) \equal{} 2^{b}$ where $ a,b \in \mathbb{N}$
and $ a \plus{} b \equal{} 2m$

Now I am going to consider $ \gcd(5^{k} \plus{} 2^{m} \minus{} 1 , 5^{k} \minus{} 2^{m} \plus{} 1)$
 If $ m \ge 2$ ,we have $ 4 \mid 5^{k} \plus{} 2^{m} \minus{} 1$ but $ 4 \nmid 5^{k} \minus{} 2^{m} \plus{} 1$
Hence $ \gcd(5^{k} \plus{} 2^{m} \minus{} 1 , 5^{k} \minus{} 2^{m} \plus{} 1) \equal{} 2$ $ \ldots \boxed{1}$
$ a \ge 2$  then from $ \boxed{1}$ we have $ b \equal{} 1$
So we get  $ 5^{k} \minus{} 2^{m} \plus{} 1 \equal{} 2$
$ 5^{k} \equal{} 2^{m} \plus{} 1$
Since $ 2^{m} \equiv 4\bmod{5}$
$ \implies m\equiv 2\bmod{4}$ ,let $ m \equal{} 4l \plus{} 2$ where  $ l \in \mathbb{N}_{0}$
$ 5^{k} \equal{} (2^{2l \plus{} 1} \plus{} 1 \minus{} 2^{l \plus{} 1})(2^{2l \plus{} 1} \plus{} 1 \plus{} 2^{l \plus{} 1})$
As $ \gcd(2^{2l \plus{} 1} \plus{} 1 \minus{} 2^{l \plus{} 1}, 2^{2l \plus{} 1} \plus{} 1 \plus{} 2^{l \plus{} 1})$ and $ 2^{2l \plus{} 1} \plus{} 1 \minus{} 2^{l \plus{} 1} < 2^{2l \plus{} 1} \plus{} 1 \plus{} 2^{l \plus{} 1}$ ,we have
$ 2^{2l \plus{} 1} \plus{} 1 \minus{} 2^{l \plus{} 1} \equal{} 1$ and$ 2^{2l \plus{} 1} \plus{} 1 \plus{} 2^{l \plus{} 1} \equal{} 5^{k}$

From the first equation we get $ l \equal{} 0$ and $ m \equal{} 2$
So the only solutions are $ m\in\{0,1,2\}$
$ \blacksquare$

Sorry if my solution and yours are same .

Yes, the solutions are pratically the same but don't worry .. :lol:
[quote="srinath.r"]P.S -Friend can you tell me what is $ \mathbb{F}_{5}$[/quote]
Sure, $ \mathbb{F}_5$ is the finite field with $ 5$ element and (for our aim!) i could write also $ \mathbb{Z}/5\mathbb{Z}$ or $ \text{mod}5$.
[quote="srinath.r"]..We have $ (2^{2m \plus{} 1} \minus{} 2^{m \plus{} 1} \plus{} 1)(2^{2m \plus{} 1} \plus{} 2^{m \plus{} 1} \plus{} 1) \equal{} 5^{a}q^{b}$. Since [...] $ q > 5$ and $ 2^{2m \plus{} 1} \minus{} 2^{m \plus{} 1} \plus{} 1 < 2^{2m \plus{} 1} \plus{} 2^{m \plus{} 1} \plus{} 1$ ,we have $ 2^{2m \plus{} 1} \minus{} 2^{m \plus{} 1} \plus{} 1 \equal{} 5^{a}$[/quote]
But are you agree that we can have $ q^b < 5^a$?  :wink:

er........That neednt be true always ..anyway I just have to include another case -$ 2^{2m\plus{}1} \minus{}2^{m\plus{}1} \plus{} 1 \equal{} q^{\beta}$ and proceeding in the previous way ...You can solve

Also posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=360424


How many and which are the sollutions of $x^{2}= 2^{x}$?

[quote="N.T.TUAN"][quote="kunny"]It has three solutions, one of them is negativeand the others is $x=2,\ 4$.
But we need to justify this. :wink:[/quote]
$f(x)=x^{2}-2^{x}$ then $f(0)f(-1)<0$. That's all :P[/quote]

We should analyze the function $f(x)$, so your solution is incomplete.

It has only two real solutions: $2$ and $4$. No more.

bye :wink:

[quote="kunny"]You are wrong. Sketch the two graphs $y=x^{2}$ and $y=2^{x}$. :wink:[/quote]

Sorry, you have reason. The other solution is calculated with LambertW funtion.

[quote="No Name"][quote="N.T.TUAN"]Note that $f'''(x)<0\forall x\in\mathbb R$ therefore $f(x)=0$ has at most three reals roots.[/quote]

Is there some more general statement? For example, $f^{(n)}<0$ $\forall x\in\mathbb R \Rightarrow f(x)=0$ has at most $n$ real roots. If does, where can I found a prove? Thanks[/quote]
That general statement works, and is also true if $f^{(n)}$ is positive for all $x$. It follows easily from Rolle's Theorem (if $f(x)=f(y)=0$ and $x<y$, then there exists $z$ such that $x<z<y$ and $f^\prime (z)=0$) plus induction.

The generalization has been considered many times on this forum:

http://www.qc.edu.hk/math/Advanced%20Level/y%5Ex_x%5Ey.htm

In fact, I solved it here [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148787[/url].


function
real analysis
real analysis unsolved

Prove or disprove:
If the function $f$ is continuous in $p$ then there exists a positve real number like $\delta$ such that $f$ is continuous over $(p-\delta,p+\delta)$.

[quote="mathematica"]Prove that if the function $f$ is continuous in $p$ there exists a positve real number like $\delta$ such that $f$ is continuous over $(p-\delta,p+\delta)$.[/quote]
I don't think it is right.
Let $f(x)=x$,when x is a rational number.$f(x)=-x$,when x isn't a rational number.
consider x=0


summer program
Mathcamp
AMC
AMC 10
MathPath
AIME
trigonometry

What program specifically prepares you both for amc/aime? Where can i find more info about it? Thanks

I don't think that there are any programs that prepare you specifically for the AMC and AIME.  The AoPS classes do a good job of this.  You might also want to look into the other summer programs (you can find a list in the Resources section of this site).

if you live in nyc, you might consider doing the math program at ccny.  classes are subdivided according to difficulty, ranging from beginning problem solving to aime. :)

nope, i don't live in nyc, but is it some sort of residentail program?

If you are looking for something residential, USA-Canada Mathcamp includes problem-solving classes geared towards competitions. There are plenty of classes and activities that have nothing to do with competitions also, but you could choose the problem-solving ones. http://www.mathcamp.org

thanks, i will certainly check it out. It doesn't have to be residential, but it could.

[quote="bluemoom"]For an 8th grader with AMC10 around 130, aiming to make AIME/USAMO next year,  Which program do you people think is more proper, mathpath or mathcamp?  Thanks.[/quote]

definitely Mathcamp

Hi, Texas137:

Thanks for the advice, but can you explain a little more ?

From mathpath website, it is said for 8th grader, it will be more towards AIME preparation.  As for mathcamp, all the listed classes seem to super for him right now, I am afraid he will get lost for most of the time....

In my opinion mathpath is a program more appropriate for a student aiming to do well on a contest such as mathcounts. On the other hand, http://www.mathcamp.org aims to provide a better foundation and deeper understanding of highschool and early college level concepts that are often tested on the AMC contests such as AIME and especially USAMO. However, many of these concepts are sometimes difficult to comprehend and understanding often comes with time. Therefore the camp/program to choose is different for each student and mainly depends on their eagerness to learn more MATH!

My son went to MathPath last year.  He was done with Algebra I and Geometry and completed 7th grade at that time.   He had a perfect score on AMC8 and was a national MathCounts participant.  The group of 60+ was divided into 4 groups based on ability.  Movement from one group to another was allowed if the child felt he/she could take more.  I think my son had much to learn, did not feel overwhelmed (or underwhelmed ).   I do think it helped him a lot.  This year he qualified for the AIME and has a near perfect score in AMC10.

MathPath is for younger students.  I like the bed time restriction and there is more discipline in MathPath.  MathCamp, I believe, has more advanced math and is for the older students.  There is no requirement to attend classes.  It is assumed that the kid who comes there is responsible for such decisions.  I am not putting one program over the other.  If your child is responsible and will attend the classes and work when he is supposed to(not required to), and you have no problem with him going to bed very late in the night, MathCamp will give him more in terms of Math.

Good luck in your decision making.

[quote="bluemoom"]I know he is very eager to learn, and will pick mathcamp if he gets accepted, but for me as a parents, I prefer him to go a place  he can benefit the most, what if the classes are too advanced for him...?

He is learning   Algebra II at school right now,  took some classes at here ,and  about to finish book AoPS Vol I, learned a little bit Trigs by himself........

Any of you who has been in mathcamp, what do you think about his background?, enough to survive, benefit, learn or even enjoy ( when lost, you can't enjoy much) from the mathcamp?

Thanks a lot for any inputs.[/quote]

MathCamp undoubtedly has a higher calibur of attendants. Many USAMO qualifiers and MOP participants attend MathCamp every summer. And as much as the camp is about learning from the teachers and the classes, it's also about the fellow students. Generally, it'll be the students who learn from each other and inspire each other to work harder. It's the students your child will spend class time and spare time with. With so many mathematically talented kids at MathCamp it's inevitable that your child will pick up a lot in those 5 weeks.

I've never attended MathCamp myself (although I plan on going this summer if I have time), but based on what friends have said and the information you've provided about your child's experience, I'd say that you're right in assuming a good portion of the cirriculum at MathCamp would be too advanced. There is a chance of not being able to keep up.

That said, you should probably contact the camp staff for information about the difficulty of their classes and perhaps some sample problems. It's important to be able to keep up, but it is not always a necessity. I've found myself in some math environments where I did not understand everything that was going on, and it was actually that personal desire to keep up that taught me the most.

Also, there is a qualifying quiz that will give you some insight into the difficulty of the problems they will have there. Their course description provides a great overview for the topics that will be covered and each class is rated based on difficulty (*, **, ***, etc.). In addition, I plan to attend mathcamp this summer depending on how everything else turns out (USAMO, etc.).

my son attended Mathcamp the summer after 8th grade, and it was fine. (it was before Mathpath started). Your son's background sounds like it's in the right ballpark. My son was further along than yours. He had a 150 on the AMC10, top 10 at Mathcounts Nat'ls, and was a USAMO qualifier in 8th grade. He was also finished with precalc. But he says there were certainly students there with less background than his.

I know less about Mathpath. Maybe you should apply to both and let the staff advise you on which would be more appropriate mathematically. Only you can decide if you son needs the extra supervision and structure at Mathpath. Mathcamp definitely assumes that kids are able to be independent and make appropriate decisions for themselves. That may or may not be true of a given 8th grader, even if they are plenty strong enough mathematically.

Texas137 has a good idea for the kids who are age-eligible for both MathPath and MathCamp, and have been to neither: apply to both, and let the staff (which is largely in common between the two programs) give advice based on the application materials you submit.

The new participant here on AoPS, sophia, once compared MathPath and MathCamp by saying that "MathPath is boot camp; MathCamp is unschooling." The program for the older kids is VERY unstructured.

What I will do this year is have my son apply to MathPath (he is almost done with the application quiz today). He is not strictly within the age limits for MathCamp until summer 2006. Just for points of comparison in this thread, he has qualified for AIME twice (last year off the AMC 10B, in which he was state champion, this year off the 12A). He just went to our state MATHCOUNTS meet, reaching the (unofficial) countdown round but not advancing to nationals. He is now in an accelerated precalculus course, after taking his THIRD high-school-level geometry course the first semester of this year. He starts calculus I next year. So you see there is a good deal of variance among kids in terms of how well they do at competition math, how far along they are in the standard math curriculum, and so forth. But the summer camps end up with the kids whose joint performance by those various criteria puts them near the top of the country. That's a good social atmosphere for those kids, and not to be missed for that reason alone, whether it helps the kids score better on the AMC tests or not.

Very sorry for such a late response.

Thanks to you all for your informations and suggestions.

Also, I finally got chance to read through all the old threads in this site about mathpath and mathcamp, and informations listed on both websites,  I am convinced that this year, mathpath is a good and proper choice for my son, now I am just hoping he can get accepted.

Good luck to Tokenadult's son too, I am sure he can make it.


PURPLE COMET MATH MEET

(Formerly the University of Wisconsin – Whitewater/
University of Texas – Dallas Mathematics Competition)

Monday April 24 – Thursday April 27, 2006

FREE   ON-LINE   TEAM   MATHEMATICS   COMPETITION

<<<< TWO LEVELS >>>>
•High School (25 problems in 90 minutes)
•Middle School (15 problems in 60 minutes)

<<<< THREE CATEGORIES OF COMPETITION >>>>
•All team members from the same large school (over 1200 students)
•All team members from the same small school (not more than 1200 students)
•Team members from more than one school

•	Teams of up to six students each
•	Each team must have an adult supervisor
•	Problems from previous meets available
•	Open to teams from around the world

Registration and rules at http://math.uww.edu/mathmeet

you added on extra period at the end of the url. you should fix it.

do the people on a team have to be from the same school? i would like to participate but i don't think anyone else from my school would.

The Purple Comet Math Meet allows for teams of up to six students each.  There are different categories of competition.  One category is for teams made up of students from more than one school.  Also, note that you do not need to find five other people for your team.  You could complete with a smaller team if you wanted to.

so where could i find other interested people or can i sign up as an individual?

Teams in the Purple Comet Math Meet have an upper limit of six students, but the lower limit is one student.  A single student could compete as a team, although, since the competition is designed for teams of six, a single student team would likely not complete all the problems in the allotted time.

The important restriction is that each team needs an adult supervisor.  If you can find a supervisor willing to be present while you compete, do what you can to find other interested students to join you, and then register your team.  Registration will stay open throughout the meet, so you just need to register a few minutes before you begin the competition.


function limit induction ratio calculus derivative algebra

Hi.

I've a little question :).

For $z \in \mathbb{C}$ one defines
$e_{n}(z) = 1 + z + \frac{1}{2}z^{2} + \frac{1}{3!}z^{3} + \dots + \frac{1}{n!}z^{n}$

I have no problem showing that this sequence is Cauchy. However, I don't really know how to attack the following:

Define $ e(z) = \lim_{n \to \infty} e_{n}(z)$.

Now, show that $e(z)e(w) = e(z+w) \, \forall z,w \in \mathbb{C}$.

Obviously this is one of the properties of the exponetial function. Actually it works for all exponential functions that have the same base. Should I try to show it termwise and then use induction or something? I feel a bit lost. Also, it is possible to show that it is convergent in $\mathbb{C}$?

Show that the series $\sum_{n\geq 0}\frac{z^n}{n!}$ converges absolutely for all $z$ (the ratio test is convenient for that). Once you know it, you may use the theorem that if the series $\sum_{k\geq 0}a_k$ and $\sum_{\ell\geq 0}b_\ell$ converge absolutely, then the series $\sum_{n\geq 0}c_n$ where $c_n=\sum_{k,\ell\geq 0, k+\ell=n}a_k b_\ell$ also converges absolutely and its sum is the product of the sums of the two original series. Because of that, it is enough to show that for each $n\geq 0$, $\frac{(z+w)^n}{n!}=\sum_{k,\ell\geq 0, k+\ell=n}\frac{z^k}{k!}\frac{z^\ell}{\ell!}$, which is just the binomial theorem in disguise. This works for complex $z$ and $w$ as well as for real ones.

Great. Can one say anything about the continuity, $e: \mathbb{C} \mapsto \mathbb{C}$. This is my first real course in anlysis, that is: It focuses on proofs and, in some sense, not on calculations.

Thanks for the help.

[quote="litton"]Can one say anything about the continuity, $e: \mathbb{C} \mapsto \mathbb{C}$[/quote] Since each tem is continuous, to show continuity, it suffices to check that the series converges uniformly in any disk $|z|\leq R$ on the complex plane (to do it, note that it is dominated by the convergent series $\sum_{n\geq 0}\frac{R^n}{n!}$ in the entire disk). The next thing you'll probably want to prove is differentiability. To this end, one needs to check that the series of derivatives also converges uniformly in any disk. Fortunately, after you differentiate the terms, you'll see that it is the same series. This, by the way, proves the rule for differentiating the exponential function: $e'(z)=e(z)$.


Does anyone know why AoPS isn't offering Olympiad Geometry course anymore? I have heard that it's a great course, but judging from the course schedule, it doesn't look like they are offering it anymore...

It's unlikely that the course is actually not offered anymore if there haven't been announcements and the page is still up.  Most likely is that the next time it runs hasn't yet been scheduled.

There could be that there has not been much interest in it lately.

We haven't decided yet whether or not to offer it in the spring.  We likely will offer it every 12-18 months.


geometry
trapezoid
cyclic quadrilateral
geometry proposed

Let $ ABCD$ be a trapezoid, where $ AB\parallel CD$ . Consider two points $ \left\|\ \begin{array}{ccc} N\in BD & ; & \widehat {NAB}\equiv\widehat {ADB} \\
 \\
M\in AD & ; & \widehat {AMB}\equiv\widehat {CAB}\end{array}\ \right\|$ .
Prove that $ \widehat {CNM}\equiv\widehat {BAD}$ . See [url=http://www.mathlinks.ro/viewtopic.php?t=318284][color=red][u]a particular case[/u][/color][/url]  and [url=http://www.mateforum.ro/viewtopic.php?t=3282][color=red][u]the problem 3[/u][/color][/url] from [b]TMMATE International 2009[/b].

Let $ \{P\} \equal{} MB\cap AC$

 Then $ \angle{DNM} \equal{} \angle{MPD} \equal{} \angle{MCD}$

 Result that $ MNCD$ is cyclic quadrilateral ,so $ \angle{MNC} \equal{} \angle{BAD}$

P.S. $ \angle{DNM} \equal{} \angle{MPD}$ because $ AB^2 \equal{} BP\cdot BM \equal{} BN\cdot BD$


prove or disprove:
for all positive integers x, given 2x-1 integers(not necessarily distinct), one can choose some x of them having a sum divisible by x.

it's easy to show that 2x-2 integers are not enough:have x-1 1's, and x-1 0's, mod x, i.e. if x=3, then 2 0's and 2 1's, or 3,3,4,4. however, given 5(=2*3-1) integers, some 3 must have a sum divisible by 3.


geometry
geometric transformation
reflection
rotation
analytic geometry
vector
linear algebra

The only thing i try with such problems is to find points which are equidistant from the line, on the other side of the line. It becomes easier if the line is parallel to one of the axis.

So how do we go about in a neat methodical way to get a reflection of a point about a line?

For ex: point (1,2) across the line 3x+4y=12

so you remember what you did to translate the line. Find the reflected point then translate it in the same manner?

[quote="JRav"]I'm sure it can be done in other ways, but the way I know how to do it uses a change of basis (linear algebra).

What you want to do is rotate your line so that it is parallel to the x-axis, and then translate it up/down so that it is the x-axis.  Then you can just reflect the point as you normally would.  This is the idea of the change of basis.[/quote]

Hmm i get the idea. An example to make sure i understood what you are trying to explain?

I can provide you a proof for reflecting a point over a line, using complex analysis.  Essentially, it is the same thing since a complex number $ z$ is just some pair $ (x,y)$, a point and $ \bar{z}\equal{}(x,\minus{}y)$.

http://www.math.jhu.edu/~eha/311f08/reflection.pdf

It's not exactly the same as your case, but it's close.

"Therefore, one way to reect about a line L: ax + by = c (a, b, c are real numbers, not all zero) is as follows:
translate so that the line passes through the origin, then rotate so that the translated line L0: ax + by = 0
coincides with the x-axis, and finally reflect about the x-axis via complex conjugation. Then, of course, we
have to effect the reverse planar motions to return to the problem of reflecting about the line in its original
position." - JHU

That's basically what I said, except because its so easy to reflect after the line has been translated to the   x-axis, its not neccessary to introduce complex numbers.

Or you can do this problem by getting two different equations involving the coordinates of the reflected point.
Anyone care to solve this problem?  :P

The point $ P$ will be reflected perpendicularly across the line. The vector perpendicular to $ 3x \plus{} 4y \equal{} 12$ is $ (3,4)$, so $ P' \equal{} (1,2) \plus{} (3,4)t \equal{} (1 \plus{} 3t,2 \plus{} 4t)$ for some $ t$.

The midpoint of $ \overline{PP'}$, which is $ \left(1 \plus{} \frac {3}{2} t, \; 2 \plus{} 2t \right)$, must be on the given line $ 3x \plus{} 4y \equal{} 12$. Plug in to get: $ 11 \plus{} \frac {25}{2} t \equal{} 12 \implies t \equal{} \frac {2}{25}$

Then, $ P' \equal{} (1,2) \plus{} \frac {2}{25} (3,4) \equal{} \left( \frac {31}{25},\frac {58}{25} \right)$

[b]Note:[/b] [i]Taking advantage of perpendicular lines basically uses the properties of the appropriate orthonormal basis JRav was talking about, without actually computing the coordinates in the new basis.[/i]

There is a generic formula for such a transformation.  If I had my linear algebra books, I could derive it for you but unfortunately, I don't.  Perhaps someone else can.

$ P\equal{}(a,b)$ and our line is $ mx\plus{}ny\equal{}k$

We know $ P'\equal{}(a\plus{}mt,b\plus{}nt)$

We know the midpoint is on the line:

$ m \left( a \plus{} \frac{m}{2} t \right) \plus{} n \left( b \plus{} \frac{n}{2} t \right) \equal{} k$

Solving, $ t \equal{} 2 \frac{k\minus{}am\minus{}bn}{m^2\plus{}n^2}$

And thus $ P' \equal{} \left(a \plus{} 2m \frac{k\minus{}am\minus{}bn}{m^2\plus{}n^2}, \; b \plus{} 2n \frac{k\minus{}am\minus{}bn}{m^2\plus{}n^2} \right)$


binomial coefficients
superior algebra
superior algebra unsolved

Let $A$ be a ring such that:
a) $x+x=0$ for every $x\in A$
b)  for every $x\in A$ there exists $k\geq 1$ such that $x^{2^{k}+1}=x$
Prove that $x^{2}=x$ for every $x\in A$.

Can somebody check if this solution is correct?

[hide]
Because $\binom{2^{k}+1}{t}$ is odd only for $t\in \{0,1,2^{k},2^{k}+1\}$ we have $(x+1)^{2^{k}+1}=x^{2^{k}+1}+x^{2^{k}}+x+1=x^{2^{k}}+1$.
Because $\binom{2^{k}}{t}$ is odd only for $t\in \{0,2^{k}\}$ we have $(x+1)^{2^{k}}=x^{2^{k}}+1$.
If we set $y=x+1$ we have $y^{n}=y^{n+1}$ for  $n=2^{k}$. But $y^{m}=y$ for some $m\geq 2$.
$y^{n}=y^{t}$ for every $t\geq n$, and let $n$ be the min. number with this proprety.
$y^{n-1}=y\cdot y^{n-2}=y^{m}\cdot y^{n-2}=y^{m+n-2}=y^{n}$ because $m+n-2\geq n$.
$y^{n-1}=y^{n}$ , but this contradicts the minimality of $n$. So $n=1$ and $y$ is idempotent. But every $y\in A$ can be written $y=x+1$ so the ring is boolean.
[/hide]

You are true, i think so, and that prove is difficult (at least for me)  :D

Thanks for checking it, I wasn't sure if it was ok.  :blush:

Just two points:$1$.It is not required to take $n$, to be a min number with that property.(Hence no need using the method of proof by contradiction to show  that $n=1$ works as well.)
$2$.I think the main idea in this problem is [url=http://www.cut-the-knot.org/arithmetic/combinatorics/LucasTheorem.shtml]Lucas' theorem[/url] to say that those two binomial coefficients are odd. I don't know whether you agree or not!


geometry
3D geometry
function
MATHCOUNTS
algebra
floor function

Just approximate your roots.

$44^{2}= 1936$ and $45^{2}= 2025$, so the first part is $44$.

(I only know my cubes up to 12)
$11^{3}= 1331, 12^{3}= 1728$. The difference of these two is almost 400. Since $2007$ is less than 400 greater than $1728$, the second part is $12$.

The difference, is, therefore, $32$.

[quote="myspacesucks"]Sorry, this was too easy, we had this in a warm-up yesterday and I killed the problem, did it in 6 seconds or so (no calcs).  Please give us some harder problems.[/quote]

Wow... that's insanely fast.

[quote="myspacesucks"]Sorry, this was too easy, we had this in a warm-up yesterday and I killed the problem, did it in 6 seconds or so (no calcs).  Please give us some harder problems.[/quote]

If you want harder problems, you shouldn't be doing Mathcounts problems. Go to the HSB (except that's all screwed up because Interval just bombards that with ridiculously easy problems) or Intermediate forums.

[quote="Quevvy"][quote="7h3.D3m0n.117"]If $[x]$ represents the greatest integer less than or equal to $x$, what is the positive difference between $[\sqrt{2007}]$ and $[\sqrt[3]{2007}]$?[/quote]
Wouldn't that just be the floor function?[/quote]

yes except that MATHCOUNTS uses wierd functions to define other ones, like @, [, ], etc.


[hide]
$x=$girls
$y=$total

Originally $.4y$ are girls but after $2$ is subtracted, $.3y$ are girls.

$(x-2)=.3y$
$x=.4y$

=$.4y=.3y+2$
=$.1y=2$
=$y=20$

finally since initially $40%
$ of the group are girls
$20*.4$ gives us our designated answer of $8$.
[/hide]

b - boys, g - gals then initially 6g=4b, at last 7(g-2)=3(b+2) , taking into b=1.5g  it can be obtained 7g-4.5g=20 so g=8.

[hide="solution"]
When 2 is subtracted from the number, it has to be a multiple of $ 3$, or else it can't be $ 30%
$, because the number of boys is $ \frac{7}{3}$ times the number of girls, and $ 3$ is in the denominator.
So the only choice that works is $ C) 8$
[/hide]

[quote="t0rajir0u"]Some boys and girls are having a car wash to raise money for a class trip to China.  Initially $ 40 \%$ of the group are girls.  Shortly thereafter two girls leave and two boys arrive, and then $ 30 \%$ of the group are girls.  How many girls were initially in the group?

$ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$[/quote]

The title was misleadingly interesting.

[hide]
This essentially removes two girls while leaving the number of total people unchanged. Let $ x$ be the total number of people:

$ 0.4x \minus{} 2 \equal{} 0.3x$
$ 0.1x \equal{} 2$
$ x \equal{} 20$

So there are originally 8 (C) girls. Four cups are required.
[/hide]


[quote="KMN"]Let $x,y,z>0$  . Prove that   $(xy)^2+(yz)^2+(zx)^2+2xyz(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})\geq 3xyz(x+y+z)$[/quote]
This is not true. Try $x=9,$ $y=z=1.$ ;)


quadratics
algebra
superior algebra
superior algebra unsolved

Suppose that r is a real number that is a root of a quadratic equation whose coefficients are constructible real numbers. Explain clearly why r is a constructible number.
Can we prove it by using a general solution of the quadratic equation, where a,b, and c are constructible numbers? If yes, show how.

The coefficients have a power of two as degree over $ \mathbb{Q}$. So the field generated by these three coefficients is also of degree $ 2^k$ for some k. Then r is of degree $ 2^k$ or $ 2^{k\plus{}1}$. Therefore it is construcible.

@Gockel: This does not suffice. There has to be a tower $ K_i$ with $ [K_i: K_{i\plus{}1}] \equal{} 2$!

Okay, but it is essentially the same argument then.

Can you prove it by using a general solution of the quadratic equation, where a,b, and c are constructible numbers? If yes, show how.


algebra
polynomial
function
superior algebra
superior algebra theorems

Is the following theorem [i] equiv. to th. [2]?

[b]Theorem 1.[/b] The number of different values of a polynomial of degree $ n$can either 1) not be reduced except to  the largest prime number $ \alpha$ such that $ \alpha < n$, or can 2) only be reduced by 1 or 2 degrees.

[b]Theorem 2.[/b] Let G be a finite group and p be a prime. If p divides the order of G, then G has an element of order p.

It seems that they should.. (if they are not, i am looking for the generalization to [1]) It comes up in Abel's paper on the general unsolvability of algebraic functions of deg. 5 or higher.

[b]Theorem 2.[/b] Let G be a finite group and p be a prime. If p divides the order of G, then G has an element of order p.

This is called Sylow's Theorem.  The first one I have never seen before, and do not know who its named after.  I found this at Wikipedia

http://en.wikipedia.org/wiki/Sylow%27s_theorem

see under theorem 1.

[quote="Winged Potato"]This is called Sylow's Theorem.[/quote]
no.
.
.

I don't understand the first statement (about the polynomial). Can you explain please?

[quote="Los"]Is the following theorem [i] equiv. to th. [2]?

[b]Theorem 1.[/b] The number of different values of a polynomial of degree $ n$can either 1) not be reduced except to  the largest prime number $ \alpha$ such that $ \alpha < n$, or can 2) only be reduced by 1 or 2 degrees.
[/quote]

so if you have $ x^{11} \plus{} 3x \plus{} 1$, this can be reduced either at most to a polynomial of deg. 7 or, in the worst case scenerio, to a polynomial of deg. 9, or the [i]most [/i] worst case, to a polynomial of degree 10. The resulting polynonial is the irreducible polynomial and is monic.

Er, a correction to the above. The theorem should read:

the number of different values of a function of $ n$ variables cann either 1) not be reduced to a number under the largest prime number which is a factor of $ n$ or 2) the value of the function can only be reduced by 1 or 2.

It follows that: it is [b]impossible [/b]to find a function of 5 variables which would have 3 or 4 different values.

Values means when you swap the different variables (i.e. the polynomial is not symmetric.)

Thanks!

I still don't understand. What do you mean by "reduce"?

[quote="Winged Potato"]oh, my mistake.  Then I have absolutely no idea who its named after.[/quote]
Like [b]-oo-[/b] says its Cauchy. The special case of Sylow's theorem for $ p$ is used to prove the general case. This special case was first proven by Cauchy.


Just wondering.  With all the stuff to do, I've been noticing my average bedtime has climed from 11:30 to 12:00 now...

well, i'm a freshman and i can finish hw, math stuff, tennis, track, etc. before dinner so bedtime is at a nice 10:30-11:00. it'll probably get later and later as i go deeper into hs but its time to enjoy the sleep.

I like to sleep a few hours after midnight and wake up early... i think sleep is kind of a waste of time but you need it soo i try and get as little as i can... unless i have a piano or math competition the next day that is  :P

There should be an option for "irregular".
I take a nap in the afternoon, and go back to sleep around midnight to 2, depending on hw, and get up at 4 to study for a test or do forgotten hw when I have to, and go back to sleep at 6 for half an hour of delicious sleep before school. Oh, and I also sleep during school hours... :P

This place is stalkers paradise....you can find- DOB, Bed Time, Fav. video games, fav. salad dressing (lol), fav. number, where you live, intelligence, your name (if you search hard enough), and tons of stuff that i probably missed.

[quote="sky9073"]There should be an option for "irregular".
[/quote]

Ditto.

Well, it depends on how much I did during the day (specially in the last few hours), because usually after midnight but before 1am, but if I am too tired, may be around 11 or so. Nevertheless, it could be more interesting the data, accompanied with the hour that you get up!, I think usually mathematicians are kind of night creatures, and also sleep relatively few, remember the quote: [i]The best theorems were proven after midnight[/i]  :D  :D . Any way, my body usually gets up around 7:30 am, and my brain wakes up around an hour later  :P .

Best,

if im at school, i dont sleep

at home like i am now i sleep around 12-12:30

i go to sleep at around 10:50. i often come back from school at around 7 o'clock and then, have supper, do my homework, study math and do some other jobs until 10:30, then i prepare to go to sleep.
If in weekends or vacation, i go to sleep at around 2:00. But for week days, it is much earlier, after all, i must get up at 6:10 in the next morning!

I go to sleep late, as my age progresses, my sleep hour has risen (Like exponential curve :D )

As I remember,
age 9 - 8:30
age 11- 9:00
age 13: 10:00
age 15: 11:00 or 11:30
age 16: whenever I dam well feel like it!
 :D

As usual, I go to sleep late on weekends or after social functions, and early (er) on weekdays. Although, usuallly, I just do not feel tired until about midnight  :?

[quote="miraculouspostmaster"]sleep is rehearsal for death[/quote]

i never saw it that way before.  :?  i'm not sleeping ever again.

actually i go to bed between 10:30 and 11:00 unless there is a major test i need to study for. then it is after 11:00.

At the beginning of this year (calendar year not school year) I stayed up pretty late, around 1 or 2 AM.  But now since I am doing track I wouldn't be able to survive so I go to bed at 11 or before usually.

[quote="miraculouspostmaster"]sleep is rehearsal for death[/quote]

So true.

Um, it's Spring Break, so 1 - 2, but usually it's 11 - 12.

I got to sleep around 11:30 and I wake up for school at 7:00.  I should probably go to sleep earlier for my age, though.

i usually do math after dinner so i'll go bed at mindnight :?

[quote="MithsApprentice"]Um...where's the "4 AM" option?[/quote]

After 2:00, the one before the last one  :P

[quote="djimenez"]After 2:00, the one before the last one  :P[/quote]

Well, I was looking for something a bit more specific. "At dawn" works. As does "whenever the caffeine fails".

Specificially: After lunch, sitting in warm room, trying to remember when slave trade was abolished.......zzzzzzzzzz

i usually head up to my room around 10:30 or so and then read til 11 or 11:30. usually fall asleep around midnight or shortly there after

I usually go to sleep aournd 10:30, either because I want to finish a REALLY good book, or I'm trying to figure out the answer to a math problem.  :)

When it's not vacation: around 1 AM, possibly midnight

On vacation: closer to 3.

Considering I'm only 13, I wonder how little I'll sleep in high school :? ...

On this long vacation (good friday etc...) I slept at 3 am...  I got hooked playing a pc game :D  but normally it's 10 but my parents don't have to know about me sleeping at 3 am  :P

elvenchamp777

I'm surprised that no one's noticed that the title of the poll is:

"What time to you sleep?"

(Read it carefully)

It actually says "what time you you go to sleep?" :D

elvenchamp777

Wow...I never saw that till u guys told me. :blush:

I get all my sleep in school. So that's from 7:30-2:30... you're a Math community.. so do the math.

[quote="plokoon51"]the title of the poll is:

"What time to you sleep?"

the title of the topic is:

"what time you you go to sleep?"

tricky :P[/quote]

hah, that's funny..

maybe it's time for all of us to get a little shut eye, lol

Hey, when I was typing that I never noticed it.  :blush:

[quote="sponge008"]Heh, you guys are lucky, I can't function without at least 8 1/2 (or17/2 :D) hours of sleep. I get ~9 1/2 on an average night, from 9:30-7:00, and still feel wasted and have black circles under my eyes. At the beginning of a vacation, for about a week, I sleep 12-14 hours a night to make up for school, then settle in to about 11-12 hours a day (err, night). Without at least 8 or so hours, I actually feel sick to my stomach... :? Wow, I'd be doomed if I ever got drafted.[/quote]

I can only assume you are young, because for junior year in high school, I have no such luxury.

[quote="sponge008"]Heh, you guys are lucky, I can't function without at least 8 1/2 (or17/2 :D) hours of sleep. I get ~9 1/2 on an average night, from 9:30-7:00, and still feel wasted and have black circles under my eyes. At the beginning of a vacation, for about a week, I sleep 12-14 hours a night to make up for school, then settle in to about 11-12 hours a day (err, night). Without at least 8 or so hours, I actually feel sick to my stomach... :? Wow, I'd be doomed if I ever got drafted.[/quote]

Wow...

You are so SCREWED for high school.

I've stayed up past midnight and gotten up at 6 three of the last four nights. Looks like I'll be doing that again tonight...

Uh oh, looks like I'm going to have to add 47 minutes of math class to my sleeping time for next year. :D But, on a good note, I have a talent for getting completely out of homework. (IE, doing it in school or other opprotune moments)

I go to sleep sometime between 11 and 12 because of all of the homework that teachers have to assign... :(

I'm preparing for high school homework by playing until 12 so I can see if I'm able to stay up til then :D  however homework is more boring... *hears snores*

elvenchamp777

i sleep 25 hours a day. Sounds impossible?
[hide]
its because during my five hours of sleeping, i dream that i sleep and dream about sleeping, while dreaming about sleeping and dreaming about sleeping and dreaming. Of couse, I dream about sleeping and dreaming. So basically while i sleep for 5 hours, i sleep 4 times of 5 hours in my dreams. 5+4*5=21, so i sleep 25 hours a day normally. [/hide]


are you sure that's right?

[hide]

When you make 2 log_4 16, and 3 log_5 125, everything becomes log_20006 16 + log_20006 125.  16*125 = 2000, which doesnt help us much, unless if the base should be 2000, which makes more sense, considering it's the year of the contest...[/hide]

umm...I think you typed something wrong, is it Find 2/log420006 + 3/log520006

mathfanatic wrote:(This is Problem #1 of the alternate AIME for 2000)

Find 2/log420006 + 3/log520006.

Please post your solutions if possible.

[hide]2/log420006 + 3/log520006.

2/(log 20006/log 4) + 3/(log 20006/log 5)

(2log 4)/log 20006 + (3log 5)/log 20006

log (16*125)/log 20006

log (2000)/log 20006 = x

log 2000 = x*log 20006

[/hide]

This does not seem trivial to me, and the original question doesn't give an integer result. Did you perhaps typo?

Hmph, well I checked Kalva and it was how I thought. I got a fractional answer, but maybe the original AIME problem said find the sum of the numerator and denominator or something...

Here it is: [hide]

2/log420006 + 3/log520006

2log20006 4 + 3log20006 5

log2000616 + log20006125

log200062000

=1/6.

[/hide]

i think that's better... http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2000-aa/aimeans41100.html says the answer is 1+6 = 7

Oh, uh-oh.  Yes, I messed up the <sup>s, I guess.


Which years did you find the hardest for the AMC/AHSME?

What about the AIME?

As far as I know, one common answer to the second question is [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1994]AIME 1994[/url].

- after 1994 had what might have been the easiest AHSME ever. Certainly it was the AHSME with the most perfect scores and that's not a close call.

In thought 1991, 1994, and 1995 were pretty tough for the second question.

I laugh at [url=http://www.artofproblemsolving.com/Wiki/index.php/1999_AHSME_Problems#Problem_2]1999 AHSME Problem 2[/url].

And [url=http://www.artofproblemsolving.com/Wiki/index.php/1999_AHSME_Problems#Problem_3]1999 AHSME Problem 3[/url] became a MATHCOUNTS Chapter Countdown Round problem (2008 to be exact) :D

[color=darkred]Edit: Yeah... I just didn't update the link correctly :oops:[/color]

how does the problem from 1983 to 1990 compare to the problems in the 21st centuary?

here's my thought for the second question (of the AIME's i remember taking)

2007 AIME I (hardest for me cuz of all the counting)
1994 (just plain hard)
2002 both were tough -
1992-
1995-

How about the easiest AMC's and AIME's? Or the easiest USAMO's?

i think 2008 has been on of the easiest years. other really easy ones are 2003, 2000, 2005, 1983-1990.

I think 2001 AMC 10 was the easiest AMC 10... :lol:

Uhh but what about teh HARDEST AMC 10/12? I wanted to see if I can get a perfect on that one to see if i can really ace an amc10...

Usually more recent years' tests are harder, e.g. a 1950 test will be easier than a 2000 test, but a 2000 test isn't necessarily easier than a 2001, because they're so close.

I think the mock test discussions usually have comments about relative hardness.


induction geometry rhombus trapezoid parallelogram circumcircle symmetry

Let ABC  be an isoceles triangle with AB=AC  and let D be a point on AB such that AD=AB/N   for some N. find the sum of the N-1 angles DP1A, DP2A , DP3A ... DPN-1A where P1,P2, ..PN-1 are the points that divide BC  in N equal parts. find this in terms of the unequal angle in the isoceles triangle. it´s hard¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ :blush:

Wow!!! :first: How do people come up with this kind of problems ?

[color=blue][b]Proposition:[/b][/color] Denote $\alpha = \angle CAB$. For any integer [b]n > 1[/b], the given angular sum is equal to

$\angle AP_1D + \angle AP_2D + ... + \angle AP_{n-1}D = \frac \alpha 2$

[color=blue][b]Proof:[/b][/color] The proof is by induction (sort of).

[color=blue][b](1)[/b] [/color]Let [b]n = 2[/b]. Then $D$ is the midpoint of $AB$ and $P_1$ is the midpoint of $BC$. Let $E$ be the midpoint of $AC$.

$DP_1 = AE = \frac{AC}{2} = \frac{AB}{2} = AD = EP_1$

because $DP_1, EP_1$ are midlines of the isosceles triangle $\triangle ABC$. Hence, the quadrilateral $ADP_1E$ is a rhombus with one diagonal $AP_1$. Since the diagonals of a rhombus bisect its internal angles, $\angle AP_1D = \frac \alpha 2$.

[color=blue][b](2)[/b][/color] Let [b]n = 2k - 1 > 2[/b] be odd (see the figure for [b]n = 7, k = 4[/b]).  Extend the ray $AB$, by the distance $AD$ to a point $B'$, so that $BB' = AD$ and $AB' = AB + BB' = \frac{n + 1}{n} AB$. Let the parallel to $BC$ through $B'$ meet the ray $AC$ at a point $C'$. The isosceles triangles $\triangle AB'C' \sim \triangle ABC$ are similar with the similarity coefficient $\frac{n + 1}{n}$. Parallels to the line $AC$ through the points $B, P_1, P_2, ..., P_{n-1}$ intersect the line $B'C'$ at points $Q_1, Q_2, Q_3, ..., Q_n$. Since the angles

$\angle P_1Q_2B' = \angle P_2Q_3B' = ... = \angle P_{n-1}Q_nB' =$

$= \angle CC'B \equiv \angle AC'B' = \angle AB'C' \equiv \angle BB'C'$

are all equal and $B'C' \parallel BC$ by construction, the triangle $\triangle BB'Q_1$ is isosceles, the quadrilaterals $BB'Q_2P_1, BB'Q_3P_2, ..., BB'Q_nP_{n-1}, BB'C'C$ are all isosceles trapezoids and

$AD = BB' = BQ_1 = P_1Q_2 = P_2Q_3 = ... = P_{n-1}Q_n = CC'$.

Since the triangles

$\triangle BB'Q_1 \cong \triangle P_1Q_1Q_2 \cong \triangle P_2Q_2Q_3 \cong ... \cong \triangle P_{n-1}Q_{n-1}C'$

are all congruent by SAS, they are all isosceles and we also have

$AD = BB' = P_1Q_1 = P_2Q_2 = ... = P_{n-1}Q_{n-1}$

In addition, it follows that the lines $AD \equiv AB \equiv AB' \parallel P_1Q_1 \parallel P_2Q_2 \parallel ... \parallel P_{n-1}Q_{n-1}$, which means that the quadrilaterals $ADQ_1P_1, ADQ_2P_2, ..., ADQ_{n-1}P_{n-1}$ are all parallelograms and

$DQ_1 = AP_1,\ DQ_2 = AP_2,\ ...,\ DQ_{n-1} = AP_{n-1}$

[color=red][b](2a)[/b][/color] Recall that [b]n = 2k - 1[/b] is assumed to be odd. $Q_k$ is the midpoint of $B'C'$, hence $AQ_k \perp B'C'$. Since $P_{k-1}Q_k \parallel AC \equiv AC'$, the angles $\angle P_{k-1}Q_kA = \angle C'AQ_k$ are equal. Since the triangle $\triangle AB'C'$ is isosceles, the angles $\angle C'AQ_k = B'AQ_k \equiv DAQ_k$ are also equal. In addition, $P_{k-1}Q_k =AD$ (see above). Consequently, the quadrilateral $ADP_{k-1}Q_k$ is an isosceles trapezoid, i.e., it is cyclic and the angles $\angle AP_{k-1}D = \angle AQ_kD$ spanning the same arc $AD$ of its circumcircle are equal.

[color=green]Translating this for [b]k = 4[/b]:[/color] $Q_4$ is the midpoint of $B'C'$, hence $AQ_4 \perp B'C'$. Since $P_3Q_4 \parallel AC \equiv AC'$, the angles $\angle P_3Q_4A = \angle C'AQ_4$ are equal. Since the triangle $\triangle AB'C'$ is isosceles, the angles $\angle C'AQ_4 = B'AQ_4 \equiv DAQ_4$ are also equal. In addition, $P_3Q_4 = AD$ (see above). Consequently, the quadrilateral $ADP_3Q_4$ is an isosceles trapezoid, i.e., it is cyclic and the angles $\angle AP_3D = \angle AQ_4D$ spanning the same arc $AD$ of its circumcircle are equal.

[color=red][b](2b)[/b][/color] We have already shown that $P_{n-1}Q_n = AD$. From symmetry of the points $Q_1, Q_n$ with respect to the symmetry axis $AQ_k$ of the isosceles triangle $\triangle AB'C'$, we have $AQ_1 = AQ_n$. Since the lines $DP_{n-1} \equiv P_{n-1}Q_n \parallel AC'$ are parallel, the angles $\angle P_{n-1}Q_nA = \angle C'AQ_n$ are equal. From symmetry of the points $B', C'$ and $Q_1, Q_n$ with respect to the line $AQ_k$, the angles $\angle C'AQ_n = \angle B'AQ_1 \equiv \angle DAQ_1$ are also equal. Consequently, the triangles $\triangle AQ_1D \cong \triangle P_{n-1}AQ_n$ are congruent by SAS and $\angle P_{n-1}AQ_n = \angle AQ_1D$. The sum of the 2 internal angles of the triangle $\triangle P_{n-1}AQ_n$ is equal to its opposite external angle, i.e.,

$\angle AP_{n-1}D = \angle AQ_nD + \angle P_{n-1}AQ_n = \angle AQ_nD + \angle AQ_1D$

[color=green]Translating this for[b] n = 7[/b]:[/color] We have already shown that $P_6Q_7 = AD$. From symmetry of the points $Q_1, Q_7$ with respect to the symmetry axis $AQ_4$ of the isosceles triangle $\triangle AB'C'$, we have $AQ_1 = AQ_7$. Since the lines $DP_6 \equiv P_6Q_7 \parallel AC'$ are parallel, the angles $\angle P_6Q_7A = \angle C'AQ_7$ are equal. From symmetry of the points $B', C'$ and $Q_1, Q_7$ with respect to the line $AQ_4$, the angles $\angle C'AQ_7 = \angle B'AQ_1 \equiv \angle DAQ_1$ are also equal. Consequently, the triangles $\triangle AQ_1D \cong \triangle P_6AQ_7$ are congruent by SAS and $\angle P_6AQ_7 = \angle AQ_1D$. The sum of the 2 internal angles of the triangle $\triangle P_6AQ_7$ is equal to its opposite external angle, i.e.,

$\angle AP_6D = \angle AQ_7D + \angle P_6AQ_7 = \angle AQ_7D + \angle AQ_1D$

[color=red][b](2c)[/b][/color] If [b]k = 2, n = 3[/b], we are done. Let that [b]k > 2[/b] and let $1 < m < k = \frac{n + 1}{2}$ (see the figure for [b]n = 7, k = 4, m = 3[/b]). We already know that $AD = P_{m-1}Q_m = P_{n - m}Q_{n+1-m}$. From symmetry of the points $Q_m, Q_{n+1-m}$ with respect to the symmetry axis $AQ_k$ of the isosceles triangle $\triangle AB'C'$, we have $AQ_m = AQ_{n+1-m}$. Similarly, from symmetry of the points $P_{m-1}, P_{n-m}$ and $Q_m, Q_{n-m}$ with respect to the symmetry axis $DP_{k-1}$ of the isosceles triangles $\triangle DBP_{n-1}, \triangle DB'Q_n$, we have $DP_{m-1} = DP_{n-m}$ and  $DQ_m = DQ_{n-m}$. Since $ADQ_{n-m}P_{n-m}$ is a parallelogram (see above), $AP_{n-m} = DQ_{n-m}$. Thus $AD = P_{m-1}Q_m, DP_{n-m} = DP_{m-1}, AP_{n-m} = DQ_m$ and consequently, the triangles $\triangle ADP_{n-m} \cong \triangle Q_mP_{m-1}D$ are congruent by SSS. The angles $\angle P_{n-m}Q_{n+1-m}A = \angle C'AQ_{n+1-m}$ are equal, because the lines $P_{n-m}Q_{n+1-m} \parallel AC'$ are parallel and the angles $\angle C'AQ_{n+1-m} = \angle B'AQ_m \equiv \angle DAQ_m$ are equal because of symmetry of the points $Q_m, Q_{n+1-m}$ with respect to the line $AQ_k$. Hence, the triangles $\triangle AQ_mD \cong \triangle P_{n-m}AQ_{n+1-m}$ are congruent by SAS. Combining these 2 results, the quadrilaterals $ADP_{n-m}Q_{n+1-m} \cong Q_mP_{m-1}DA$ are congruent. Consequently,

$\angle AP_{n-m}D = \angle Q_mDP_{m-1},\ \ \angle AQ_{n+1-m}D = \angle Q_mAP_{m-1}$

Let $R$ be the diagonal intersection of the quadrilateral $ADP_{m-1}Q_m$. From the triangles $\triangle DP_{m-1}R, \triangle AQ_mR$,

$\angle DP_{m-1}A + \angle Q_mP_{m-1}D \equiv \angle DP_{m-1}R + \angle RDP_{m-1} = 180^o - \angle  DRP_{m-1} =$

$= 180^o - \angle ARQ_m = \angle AQ_mR + \angle RAQ_m \equiv \angle AQ_mD + \angle P_{m-1}AQ_m$

Combining the last 2 equations, we find that the following angular sums are equal:

$\angle AP_{m-1}D + \angle AP_{n-m}D = \angle AQ_mD + AQ_{n+1-m}D$

[color=green]Translating this for [b]n = 7, k = 4, m = 3[/b]:[/color] We already know that $AD = P_2Q_3 = P_4Q_5$. From symmetry of the points $Q_3, Q_5$ with respect to the symmetry axis $AQ_4$ of the isosceles triangle $\triangle AB'C'$, we have $AQ_3 = AQ_5$. Similarly, from symmetry of the points $P_2, P_4$ and $Q_3, Q_4$ with respect to the symmetry axis $DP_3$ of the isosceles triangles $\triangle DBP_6, \triangle DB'Q_7$, we have $DP_2 = DP_4$ and  $DQ_3 = DQ_4$. Since $ADQ_4P_4$ is a parallelogram (see above), $AP_4 = DQ_4$. Thus $AD = P_2Q_3, DP_4 = DP_2, AP_4 = DQ_3$ and consequently, the triangles $\triangle ADP_4 \cong \triangle Q_3P_2D$ are congruent by SSS. The angles $\angle P_4Q_5A = \angle C'AQ_5$ are equal, because the lines $P_4Q_5 \parallel AC'$ are parallel and the angles $\angle C'AQ_5 = \angle B'AQ_3 \equiv \angle DAQ_3$ are equal because of symmetry of the points $Q_3, Q_5$ with respect to the line $AQ_4$. Hence, the triangles $\triangle AQ_3D \cong \triangle P_4AQ_5$ are congruent by SAS. Combining these 2 results, the quadrilaterals $ADP_4Q_5 \cong Q_3P_2DA$ are congruent. Consequently,

$\angle AP_4D = \angle Q_3DP_2,\ \ \angle AQ_5D = \angle Q_3AP_2$

Let $R$ be the diagonal intersection of the quadrilateral $ADP_2Q_3$. From the triangles $\triangle DP_2R, \triangle AQ_3R$,

$\angle DP_2A + \angle Q_3P_2D \equiv \angle DP_2R + \angle RDP_2 = 180^o - \angle  DRP_2 =$

$= 180^o - \angle ARQ_3 = \angle AQ_3R + \angle RAQ_3 \equiv \angle AQ_3D + \angle P_2AQ_3$

Combining the last 2 equations, we find that the following angular sums are equal:

$\angle AP_2D + \angle AP_4D = \angle AQ_3D + AQ_5D$

[color=red][b](2d)[/b][/color] We have shown:

$\angle AP_{n-1}D = \angle AQ_nD + \angle AQ_1D$

$\angle AP_{m-1}D + \angle AP_{n-m}D = \angle AQ_mD + AQ_{n+1-m}D$ [color=white].[/color] for [color=white].[/color] $1 < m < k = \frac{n+1}{2}$

$\angle AP_{k-1}D = \angle AQ_kD$ [color=white].[/color] for [color=white].[/color] $k = \frac{n+1}{2}$

As a result,

$\angle AP_1D + \angle AP_2D + ... + \angle AP_{n-1}D =$

$= \angle AQ_1D + \angle AQ_2D + ... + \angle AQ_{n-1}D + \angle AQ_nD$

The induction step from [b]n = 2k[/b] even to [b]n + 1[/b] is entirely similar, just the symmetry axis of the isosceles triangles $\triangle ABC \sim \triangle AB'C'$ is $AP_k$ and the symmetry axis of the isosceles triangles $\triangle DBP_{n-1} \sim \triangle DB'Q_n$ is $DQ_k$. In other words, the given angular sum does not depend on the integer [b]n[/b]. Since this sum is equal to $\frac \alpha 2$ for [b]n = 2[/b], it is equal to $\frac \alpha 2$ for all $n \ge 2$.

You are fanatic genius Yetti. Congratulatione :alien:  :thumbup:  :omighty:  :first:  :clap2:  :10:

Actually, how do people come up with this kind of solutions? Well, probably the answer is: because other people claim the problem is hard (¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡) :D . Actually, it is more or less trivial...

[color=blue][b]Problem.[/b] Let ABC be an isosceles triangle with AB = AC. Denote A = < CAB. Let n > 1 be a positive integer.

Let D be a point on the segment AB such that $AD=\frac{AB}{n}$, and let $P_1$, $P_2$, ..., $P_{n-1}$ be n - 1 points on the segment BC such that $BP_1=P_1P_2=P_2P_3=...=P_{n-2}P_{n-1}=P_{n-1}C=\frac{BC}{n}$.

Prove that $\sum_{i=1}^{n-1}\measuredangle DP_iA=\frac{A}{2}$.[/color]

[i]Solution.[/i] Since the triangle ABC is isosceles with AB = AC, its base angle equals $\measuredangle ACB=\frac{180^{\circ}-\measuredangle CAB}{2}=\frac{180^{\circ}-A}{2}=90^{\circ}-\frac{A}{2}$.

Since $BD=BA-AD=BA-\frac{AB}{n}=BA-\frac{BA}{n}=\frac{n-1}{n}\cdot BA$ and $BP_{n-1}=BC-P_{n-1}C=BC-\frac{BC}{n}=\frac{n-1}{n}\cdot BC$, the points D and $P_{n-1}$ are the images of the points A and C under the homothety with center B and factor $\frac{n-1}{n}$. Hence, the triangle $DBP_{n-1}$ is the image of the triangle ABC under this homothety. Since the triangle ABC is isosceles with AB = AC, its image triangle $DBP_{n-1}$ must also be isosceles, with $DB=DP_{n-1}$; also, since a homothety maps every line to a parallel line, we must have $DP_{n-1}\parallel AC$, and thus $\measuredangle DP_{n-1}B=\measuredangle ACB=90^{\circ}-\frac{A}{2}$.

Now, the triangle ABC is isosceles, and the points $P_1$, $P_2$, ..., $P_{n-1}$ divide its base BC into n equal parts (since $BP_1=P_1P_2=P_2P_3=...=P_{n-2}P_{n-1}=P_{n-1}C$). Hence, by symmetry, for every integer i such that $1\leq i\leq n-1$, we have $\measuredangle AP_iB=\measuredangle AP_{n-i}C$. Thus, $\measuredangle AP_iB+\measuredangle AP_{n-i}B=\measuredangle AP_{n-i}C+\measuredangle AP_{n-i}B=180^{\circ}$. Hence,

$2\sum_{i=1}^{n-1}\measuredangle AP_iB=\sum_{i=1}^{n-1}\measuredangle AP_iB+\sum_{i=1}^{n-1}\measuredangle AP_{n-i}B$
$=\sum_{i=1}^{n-1}\left(\measuredangle AP_iB+\measuredangle AP_{n-i}B\right)=\sum_{i=1}^{n-1} 180^{\circ}=\left(n-1\right)\cdot 180^{\circ}$,

so that $\sum_{i=1}^{n-1}\measuredangle AP_iB=\frac{\left(n-1\right)\cdot 180^{\circ}}{2}=\left(n-1\right)\cdot 90^{\circ}$.

On the other hand, the triangle $DBP_{n-1}$ is isosceles, and the points $P_1$, $P_2$, ..., $P_{n-2}$ divide its base $BP_{n-1}$ into n - 1 equal parts (since $BP_1=P_1P_2=P_2P_3=...=P_{n-2}P_{n-1}$). Hence, by symmetry, for every integer i such that $1\leq i\leq n-2$, we have $\measuredangle DP_iB=\measuredangle DP_{n-1-i}P_{n-1}$. Thus, $\measuredangle DP_iB+\measuredangle DP_{n-1-i}B=\measuredangle DP_{n-1-i}P_{n-1}+\measuredangle DP_{n-1-i}B=180^{\circ}$. Thus,

$2\sum_{i=1}^{n-2}\measuredangle DP_iB=\sum_{i=1}^{n-2}\measuredangle DP_iB+\sum_{i=1}^{n-2}\measuredangle DP_{n-1-i}B$
$=\sum_{i=1}^{n-2}\left(\measuredangle DP_iB+\measuredangle DP_{n-1-i}B\right)=\sum_{i=1}^{n-2} 180^{\circ}=\left(n-2\right)\cdot 180^{\circ}$,

so that $\sum_{i=1}^{n-2}\measuredangle DP_iB=\frac{\left(n-2\right)\cdot 180^{\circ}}{2}=\left(n-2\right)\cdot 90^{\circ}$.

Hence,

$\sum_{i=1}^{n-1}\measuredangle DP_iA=\sum_{i=1}^{n-1}\left(\measuredangle AP_iB-\measuredangle DP_iB\right)=\sum_{i=1}^{n-1}\measuredangle AP_iB-\sum_{i=1}^{n-1}\measuredangle DP_iB$
$=\sum_{i=1}^{n-1}\measuredangle AP_iB-\left(\sum_{i=1}^{n-2}\measuredangle DP_iB+\measuredangle DP_{n-1}B\right)$
$=\left(n-1\right)\cdot 90^{\circ}-\left(\left(n-2\right)\cdot 90^{\circ}+\left(90^{\circ}-\frac{A}{2}\right)\right)=\frac{A}{2}$,

and the proof is done.

  Darij

[quote="darij grinberg"]Actually, how do people come up with this kind of solutions? Well, probably the answer is: because other people claim the problem is hard (¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡) :D . Actually, it is more or less trivial...
[/quote]
Are you looking for an answer ?

Yetti

well i guess the problem is hard if you dont use homothety etc. something i dont know much of (see yetti`s solution)  but again darij you are a master at this thats why for this is trivial.  :P


If $ a, b, c \ge 0$, prove or disprove that

$ \sum\frac{a(2a\plus{}b\plus{}c)}{a^2\plus{}2bc}\ge 3$.

________
emoticons

AS OF NOW my calculations show that its wrong...ie i disaprove.. :maybe:

what do ur calculations say... :?:

Dear [b]Sfsreporter[/b], the answer is not acceptable without demonstration.
No I can answer "Prove or disprove"...  :wink:

here's my soln.

assuming that the summations are cyclic

by simple computation we get

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

which leads to $ 6(ab\plus{}bc\plus{}ca)\ge 9(ab\plus{}bc\plus{}ca)$----by am-gm

which clearly is not true

hence disproved !! .......???????

PS:  i am new 2 ineq's there can be even highly idiotic mistakes in my solutions .
       that's why i was hesitant 2 post my solution


Let $ a,b,c$ are positive

Prove

1) $ \sum_{cyc} \frac {a^5}{a^3 + b^3} + \sum_{cyc} \frac {a^3}{a + b} \geq \sum_{cyc} \frac {a^4}{a^2 + b^2} + \frac {a^2 + b^2 + c^2}{2}$

2)${ \sum_{cyc} \frac {a^5}{a^3 + b^3} + \frac {a^2 + b^2 + c^2}{2} \leq \sum_{cyc} \frac {a^4}{a^2 + b^2} + \sum_{cyc} \frac {a^3}{a + b}}$

Well, I found this one from the computer at my university.  I admit that I use "punching method" to find it. But I wish that s.o can find sols without "punching"

[quote="Gibbenergy"]Let $ a,b,c$ are positive

Prove

1) $ \sum_{cyc} \frac {a^5}{a^3 \plus{} b^3} \plus{} \sum_{cyc} \frac {a^3}{a \plus{} b} \geq \sum_{cyc} \frac {a^4}{a^2 \plus{} b^2} \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2}$

[/quote]
I have obtained it's equivalent to
$ \sum_{cyc}\frac{(a\minus{}b)^2(a^6\minus{}a^5b\plus{}3a^4b^2\plus{}6a^3b^3\plus{}3a^2b^4\plus{}3ab^5\plus{}b^6)}{(a^3\plus{}b^3)(a^2\plus{}b^2)(a\plus{}b)}\geq0,$ which is obvious.

Araqady,I have seen many your proofs that they are used this way,tell me (and every body),how can you find it,please?

For you, Allnames!
$ \sum_{cyc} \frac {a^5}{a^3 \plus{} b^3} \plus{} \sum_{cyc} \frac {a^3}{a \plus{} b} \minus{} \sum_{cyc} \frac {a^4}{a^2 \plus{} b^2} \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2} \equal{}$
$ \equal{} \sum_{cyc}\left( \frac {a^5}{a^3 \plus{} b^3} \minus{} \frac {a^2 \plus{} b^2}{4} \plus{} \frac {a^3}{a \plus{} b} \minus{} \frac {a^2 \plus{} b^2}{4} \minus{} \frac {a^4}{a^2 \plus{} b^2} \plus{} \frac {a^2 \plus{} b^2}{4}\right) \equal{}$
$ \equal{} \sum_{cyc}\left(\frac {3a^5 \minus{} a^3b^2 \minus{} a^2b^3 \minus{} b^5}{4(a^3 \plus{} b^3)} \plus{} \frac {3a^3 \minus{} a^2b \minus{} ab^2 \minus{} b^3}{4(a \plus{} b)} \minus{} \frac {3a^4 \minus{} 2a^2b^2 \minus{} b^4}{4(a^2 \plus{} b^2)}\right) \equal{}$
$ \equal{} \sum_{cyc}\frac {a \minus{} b}{4}\left(\frac {3a^4 \plus{} 3a^3b \plus{} 2a^2b^2 \plus{} ab^3 \plus{} b^4}{a^3 \plus{} b^3} \plus{} \frac {3a^2 \plus{} 2ab \plus{} b^2}{a \plus{} b} \minus{} \frac {3a^3 \plus{} 3a^2b \plus{} ab^2 \plus{} b^3}{a^2 \plus{} b^2}\right) \equal{}$
$ \equal{} \sum_{cyc}\frac {a \minus{} b}{4}\left(\frac {3a^4 \plus{} 3a^3b \plus{} 2a^2b^2 \plus{} ab^3 \plus{} b^4}{a^3 \plus{} b^3} \minus{} \frac {5(a \plus{} b)}{2} \plus{} \frac {3a^2 \plus{} 2ab \plus{} b^2}{a \plus{} b} \minus{} \frac {3(a \plus{} b)}{2} \minus{} \frac {3a^3 \plus{} 3a^2b \plus{} ab^2 \plus{} b^3}{a^2 \plus{} b^2} \plus{} 2(a \plus{} b)\right) \equal{}$
$ \equal{} \sum_{cyc}\frac {a \minus{} b}{8}\left(\frac {a^4 \plus{} a^3b \plus{} 4a^2b^2 \minus{} 3ab^3 \minus{} 3b^4}{a^3 \plus{} b^3} \plus{} \frac {3a^2 \minus{} 2ab \minus{} b^2}{a \plus{} b} \minus{} \frac {2(a^3 \plus{} a^2b \minus{} ab^2 \minus{} b^3)}{a^2 \plus{} b^2}\right) \equal{}$
$ \equal{} \sum_{cyc}\frac {(a \minus{} b)^2}{8}\left(\frac {a^3 \plus{} 2a^2b \plus{} 6ab^2 \plus{} 3b^3}{a^3 \plus{} b^3} \plus{} \frac {3a \plus{} b}{a \plus{} b} \minus{} \frac {2(a^2 \plus{} 2ab \plus{} b^2)}{a^2 \plus{} b^2}\right) \equal{}$
$ \equal{} \sum_{cyc}\frac {(a \minus{} b)^2(a^6 \minus{} a^5b \plus{} 3a^4b^2 \plus{} 6a^3b^3 \plus{} 3a^2b^4 \plus{} 3ab^5 \plus{} b^6)}{4(a^3 \plus{} b^3)(a^2 \plus{} b^2)(a \plus{} b)}\equal{}$
$ \equal{} \sum_{cyc}\frac {(a \minus{} b)^2(a^5\minus{}2a^4b\plus{}5a^3b^2\plus{}a^2b^3\plus{}2ab^4\plus{}b^5)}{4(a^3 \plus{} b^3)(a^2 \plus{} b^2)}\geq0.$ :)