invariant
number theory proposed
number theory

Let [tex]n>1[/tex]. Let's shuffle the digits of [tex]n^m[/tex] ([tex]m\in\mathbb{N}[/tex]), and let's denote the new number by [tex]l[/tex]. Prove that for any [tex]k\neq m[/tex] we have [tex]n^k \neq l[/tex].

P.S. In original problem we had n=2.

consider residue mod 3... it is clear the only case is $n=3$.

Can you explain your ideas better? I can't understand what you are trying to do.  ;)

Consider: the relation between the new number and the old number is that one is 2-9 times the other (as it is not equal, so not 1 time, and cannot be 10 or more because they have the same number of digits). Rearranging the digits keeps it invariant $\mod 9$. Thus, the either the multiplier is $1\mod 9$ which cannot be, or the number was originally $0\mod 9$. This means that n is a multiple of 3 (not 3, i made a mistake..)


rotation
geometry
incenter
trigonometry
trig identities
Law of Cosines

A point $ D$ is drawn inside an equilateral triangle $ ABC$ such that $ D$ is $ 8$, $ 7$, and $ 5$ from points $ A$, $ B$, and $ C$, respectively. Compute a side length of the triangle.

[hide="Hint"]Create a rotated version of the triangle that's rotated 60 degrees about point A.[/hide]

[b]@ Fuuckin Kunt:[/b]
1. Unfortunately, the incenter has little to do with it.
2. I don't like your user name.

That's really the hint for solving the problem. Try a little bit harder and you'll say 'Ah'.

I got it using the Zuton's Force hint.

But what's the motivation behind the hint? And that the angles would work that way?

[hide]If you revolve the triangle about point $ A$ by $ \minus{}60^\circ$ and draw a line segment from $ D$ to $ D'$, you'll see that an equilateral triangle with side length $ 8$ is formed. It appears that $ DC'$ is a line segment, but I don't know why. Anyways, using the law of cosines, we have $ x^2\equal{}8^2\plus{}5^2\minus{}2\cdot 8\cdot 5\cdot \cos 120^\circ$, so $ \boxed{x\equal{}\sqrt{129}}$.[/hide]

[quote="123456789"]
[hide]It appears that $ DC'$ is a line segment, but I don't know why.[/hide][/quote]
[hide]I think you mean $ DD'C'$. Note that $ m\angle D'DC \equal{} \arccos\left(\frac {5^2 \plus{} 8^2 \minus{} 7^2}{2\cdot 5\cdot 8}\right)\equal{}\arccos\frac12$, which means it's a 60 degree angle. So $ \angle ADC\cong\angle AD'C'$ is supplementary to $ \angle AD'D$.[/hide]
I hope our pictures are the same.

Is there a way to see that without using the Law of Cosines?


inequalities
function
logarithms
inequalities unsolved

Find all real x, such that: $ 1\leq(\frac {x}{2} \plus{} 1)^{1 \minus{} x^2}$

a)If $ x < 0$, then the inequality isn't well defined (the exponential function is defined for positive base).
b)If $ x \equal{} 0$, then the inequality holds (case of equality).
c)If $ x > 0$, then $ (\frac {x}{2} \plus{} 1) > 1$. Logarithm the inequality to get $ 0\le(1 \minus{} x^2)ln(\frac {x}{2} \plus{} 1)$. But $ ln(\frac {x}{2} \plus{} 1) > 0$, so $ 0\le(1 \minus{} x^2)$, that is, $ x \le1$.
So the final answer is $ 0\le x \le 1$.

Wait, can't we define exponential function for negative bases like this:

$ a^b\equal{}\frac{1}{a^{\minus{}b}}$?

I didn't say for negative exponents, I said for negative bases: in what you say, a is the base.


How long should it take me to complete the AoPS Volumes 1 and 2? GIven that I spend 7 hours/week on the material, do you think that I can complete it by the end of this year? (so that I can use the strategies before the AMCs are given).

The amount of time required to go through them relies heavily on your current level as well as other factors.  The amount of time it takes you to absorb material is also a big factor.  In addition, you will get through it much quicker if you go on to the next chapter after obtaining only a slight understanding of the last chapter.  I think that it took me about 2 months to get through volume 1.  I can't remember how much time a day I spent on it but it was a few hours(like 1-3 hrs i think).  I still have a lot of volume 2 to do and I doubt I will get through it anytime soon.  I have school and cross country to do first before I have free time to do extra math.  When I do have free time i usually spend it on the AoPS classes I am taking.

You should be able to get through volume 1.  I don't know about volume 2 though.  Understanding(and mastering) all the material in 2 takes a lot of time and focus.  If you are just starting vol. 1 now and you have a fairly good grounding in math I think that getting through vol 1 and about 1/2 of 2 is a good goal.

Well for the first AMC round you'll do pretty good with only the Volume 1. But basicly the message is (for all competitions) the more you'll train, the better the result. Regardless of whether or not you finished the book.


A circle with center $ O$ and its chord $ AB$ is given. $ C$ is an arbitrary point on the circle.

i) Prove that the locus of the orthocenter of $ \triangle ABC$ is a circle.

ii) Find the radius of the loci circle in terms of $ OA \equal{} R$.

iii) Find the distance from $ O$ to the center of the loci circle in terms of $ OA \equal{} R$ and $ AB \equal{} x$.

[hide]Use complex numbers taking O in the origin of the complex plane.[/hide]

Call H the orthocenter of ABC , because of <AHB=180-<C and indeed <C is constant , then the locus of H is the arc (circumference) symmetical to (O) toward AB , and so the radius of this circle equals R


geometry
incenter
inequalities
inequalities unsolved

[quote="Mustafa Taghiyev"][size=150]For a given ABC triangle r is incenter and prove this inequality;[/size]

(1/a)2+(1/b)2+(1/c)2<=1/4r2(1 over 4 r square)[/quote]
Let $ b\plus{}c\minus{}a\equal{}x,$ $ a\plus{}c\minus{}b\equal{}y$ and $ a\plus{}b\minus{}c\equal{}z.$ Then
$ \sum_{cyc}\frac{1}{a^2}\leq\frac{1}{4r^2}\Leftrightarrow\sum_{cyc}\frac{4}{(y\plus{}z)^2}\leq\frac{x\plus{}y\plus{}z}{xyz}\Leftrightarrow\sum_{cyc}\frac{(y\minus{}z)^2}{(y\plus{}z)^2}\geq0.$


For nonnegative real numbers $ x,y,z$   ; prove that  \[ \sqrt{3x^2\plus{}xy} \plus{} \sqrt{3y^2\plus{}yz} \plus{} \sqrt{3z^2\plus{}zx} \leq 2(x\plus{}y\plus{}z)\]

[quote="ertanrock"]For nonnegative real numbers $ x,y,z$   ; prove that
\[ \sqrt {3x^2 \plus{} xy} \plus{} \sqrt {3y^2 \plus{} yz} \plus{} \sqrt {3z^2 \plus{} zx} \leq 2(x \plus{} y \plus{} z)\]
[/quote] By Am-Gm, $ \sum \sqrt{3x^2\plus{}xy} \equal{} \sum \frac{1}{2} \sqrt{4x(3x\plus{}y)} \le \sum \frac{7x\plus{}y}{4} \equal{} 2\sum x$. :P

By Cauchy,
$ 4(x\plus{}y\plus{}z)^2\equal{}(x\plus{}y\plus{}z)((3x\plus{}y)\plus{}(3y\plus{}z)\plus{}(3z\plus{}x)) \ge (\sqrt{3x^{2}\plus{}xy}\plus{}\sqrt{3y^{2}\plus{}yz}\plus{}\sqrt{3z^{2}\plus{}zx})^2$
So $ 2(x\plus{}y\plus{}z) \ge \sqrt{3x^{2}\plus{}xy}\plus{}\sqrt{3y^{2}\plus{}yz}\plus{}\sqrt{3z^{2}\plus{}zx}$
It seems that for every proof with AM-GM there is a corresponding proof with Cauchy  :)


linear algebra
matrix
analytic geometry
linear algebra unsolved

Let $ \Sigma$ be the surface in $ \mathbb{R}^3$ given by
\[ 2x^2 \plus{} 2xy \plus{} 4yz \plus{} z^2 \equal{} 1.\]By writing this equation as
\[ \mathbf{x^T}\mathsf{A}\mathbf{x} \equal{} 1,\]with $ \mathsf{A}$ a real, symmetric matrix, show that there is an orthonormal basis such that, if we use coordinates $ (u, v, w)$ with respect to this new basis, $ \Sigma$ takes the form
\[ \lambda u^2 \plus{} \mu v^2 \plus{} \nu w^2 \equal{} 1.\]Find $ \lambda, \mu, \nu$ and hence find the minimum distance between the origin and $ \Sigma$. [i]Hint: it is not necessary to find the basis explicitly[/i]

I've done the easy parts: the matrix $ \mathsf{A} \equal{} \begin{pmatrix} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 1\end{pmatrix}$. It has eigenvalues $ \pm \sqrt{3}$ and $ 3$, and the orthonormal basis will just be the orthogonal matrix $ \mathsf{S}$ containing unit eigenvectors in its columns, so $ \lambda \equal{} \minus{}\sqrt{3}$, $ \mu \equal{} \sqrt{3}$, $ \nu \equal{} 3$ (I think any permutation is also fine)

How do we connect all of this to the distance between the (untranslated, original) surface and the origin? Why does the hint hold true?

We seek $ min(\parallel{}X\parallel{}^2)$ knowing that $ X^TAX\minus{}1\equal{}0$. Necessarily if the min is reached in $ X$ then there exists a real $ \lambda$ s.t. for any $ H$:  $ 2X^TH\plus{}\lambda(H^TAX\plus{}X^TAH)\equal{}2X^TH\plus{}2\lambda{X}^TAH\equal{}0$. Hence $ X^T\plus{}\lambda{X}^TA\equal{}0$ (this condition is equivalent to $ X$ is normal to $ \Sigma$) or $ AX\equal{}\minus{}(1/\lambda)X$ ($ X$ is an eigenvector associated to the eigenvalue $ \minus{}1/\lambda$). Moreover $ X^TAX\equal{}1\equal{}(\minus{}1/\lambda)\parallel{}X\parallel{}^2$ or $ \parallel{}X\parallel{}^2\equal{}\minus{}\lambda$.
$ \minus{}1/\lambda\in\pm\sqrt{3},3$ then $ \minus{}\lambda\in\pm1/\sqrt{3},1/3$. Thus $ min(\parallel{}X\parallel{}^2)$ knowing that $ X^TAX\minus{}1\equal{}0$ is $ 1/3$.
It remains to show that the min. is reached for a point $ X\in\mathbb{R}^3$.


Does anyone know if there are any online tutorials on electricity and magnetism that works in the math also? Are there any good books that I can read? Right now my understanding of electricity and magnetism is completely based on plugging in formulas because I have no idea how any of it works.

An absolutely great book to learn Electricity and Magnetism is [u]Electricity and Magnetism[/u] by Edward M. Purcell. All you need to know is single variable calculus to work through it (though multivariable helps).

Another good book is the second volume of [u]The Feynman Lectures on Physics[/u]. They're a little less mathematical, but good on physical intuition.

In addition, to build up your mathematical skills, I would recommend Marden and Tromba's book [u]Vector Calculus[/u]. This is a somewhat difficult book. It's usually used by colleges as a sophomore level calculus book for scientists and engineers. But if you take your time and work through the problems you can master it. It will give you a good understanding of what's going on in classical physics. I worked through it in an independent study class in high school after taking BC Calculus, and it helped me greatly to understand why things were true in E&M.

[quote="rrusczyk"]It's tough to beat the Feynman lectures for physics.  These are a series of books that you should be able to find in most online stores and probably in your big B&N or such.  They're perfect for reading after you've just been jamming stuff in equations for a while.[/quote]
It's also tough to learn from the Feynman lectures on physics. They're not exactly written at a junior high level. But I believe I have them in pdf form on my computer, and I'll transfer them over AIM or something to anybody that wants them, since the books seem to run [url=http://www.amazon.com/exec/obidos/tg/detail/-/0201500647/qid=1084390969/sr=8-1/ref=pd_ka_1/002-6299277-6217603?v=glance&s=books&n=507846]around $100[/url].
Assuming you have broadband, I have found [url=http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoLectures/index.htm]these video lectures[/url] by MIT professor Walter Lewin to be exceptional.
There's my 2.7 cents.

[quote="Nosepuppy"]wow. the entire feymen lectures in pdf format on your computer?

i would really appreciate it if you could me a copy.  perhaps we could set up a time? and yea, i have DSL[/quote]
Contact me (preferably via AIM)


induction
number theory
relatively prime

[color=cyan]Define 2 sequences a_i and b_i for all integers i >= 0 by the relations a_(n+1) = 2a_n + 3b_n  and b_(n+1) = a_n + 2b_n for
n >= 0 given the values a_0 and b_0.

(a) prove that the quantity (a_n)^2 - 3(b_n)^2 is a constant not dependant on n.
(b) if a_0 = 1 and b_0 = 0, show that a_n and b_n are relatively prime over all n.
(c) let r_n = a_n/b_n.  Find a recursion for r_n.
(d) if a_0 = 1 and b_0 = 0, compute the first few terms of the three sequences a_n, b_n and r_n.  Use these values to conjecture a limiting value for r_n as n grows without bound and a recursive relation for a_n and b_n.  Prove as many of your conjectures as possible.[/color]

For a) induction works like a charm  :D This is actually similar to the way we prove that there are infinitely many solns for Pell's eqn.

For b) we use a) and we see that a_n^2-3b_n^3=1 for all n, so (a_n, b_n)=1.

Still thinking of the others...


Some people might have seen this problem before, but here goes:

A guy went to cash a check. However, the teller misread the check, and gave the man cents for dollars, and dollars for cents (Either lucky or unlucky, depending on the amount of the check.) The man put the money in his pocket without checking to make sure he has the right amount (or could he be just greedy?). On his way home, the man gave 5 cents to a beggar in the subway.(Praise his charity!) When he got home, he found out that he has twice the amount of money stated by the check in his pockets. (He's really lucky). Before going to the bank, the man had no money at all in his pockets. What is the amount of the check?

Can anyone help with a simple solution to this problem?

[hide]Answer: $31.63[/hide]

Okay I'll try:

[hide]

Starts out A dollars and B cents, or A+B/100 dollars or 100A+B cents.  Then at the bank comes out w/ B dollars and A cents, B+A/100 dollars total.  He gives 5 cents to a beggar, leaving him B+A/100 - 5/100, and finds when he gets home that:

2(A+B/100)=B+A/100-5/100

2A+B/50=B+A/100-5/100

199/100 * A + -49/50 *B=-5/100

199A-98B=-5

B=XY (digits) = 10X+Y

A=CD (digits) = 10C + D

1990X+199Y-980B-98D+5=0

C, D, X, Y are integers 0 through 9

So you just have to play around w/it until you find 4 values that satisfy the equation.

[/hide][/quote]

Good job so far, Julie.  My only suggestions:

(1) When working in dollars and cents, hours and minutes, or any other similar set of objects, it's almost always simpler to use the smaller unit.  Thus, your calculations are easier and look nicer in this problem if you use cents as your unit instead of dollars, because you have expressions like (100A + B) instead of (A + B/100).

(2) I wouldn't have switched over to the digits when you did.  I think it's actually easier to solve when you're just looking for A and B.  Also, that is really not as simple a task as you made it look.  Try solving the equation you found, [hide]199A = 98B - 5.[/hide]  I suggest as the simplest way to solve it [hide]looking at it mod 98, or in other words finding a value of A such that the left side is 5 less than a multiple of 98.[/hide]

I got[hide] A=31 and B=63[/hide] therefore, the check was [hide]$31.63[/hide].

correct?

BobofBobs wrote:Is there a simpler way to do it than solving the equation [hide]98B-5=199A[/hide]?

Not that I know of.  That equation isn't too hard to solve, though, because 199 is very close to a multiple of 98.

[quote]JS1527- the answer is under the problem in spoiler. your answer is correct.[/quote]

Whoa, I didn't even THINK to look there.  I am absent minded, forgive me.


In this game, each of the 50 states starts out with 10 points. Each time you post, you get 2 points to spend(help your state or attack other states). You may post at most one time each hour. The only exception is that you may only add points to the state live in (please don't cheat). But you can form alliances which allow you to help other states(explained later).You may post at most one time each hour.

Points: Each state starts out with 10 points. If your state reaches 0 points, then it is out of the game and the state that destroyed it gets 2 extra points. When your state has  enough points, you can choose to spend some of those points on special items(these items are listed below).

Alliances: During the game, 2 or more states can choose to form an alliance. The maximum number of states there can be in an alliance is 5(name it whatever you want). In order to form one, one person from each state must first agree to it and then each state that wants to be in the alliance must give up 5 points.

Advantages of Alliances:

1) States that are in the same alliance can help each other(add points to each other)

2) When one of your allies destroys another state, your state gets 1 extra point(the state that destroyed it gets 2 points)

3) You can split the cost of buying special items

4) If you form an alliance with 3, 4, or 5 states you get that many points to spend(for example, if Hawaii, Alaska, and California form an alliance, then each person from those states gets 3 points to spend each time they post instead of 2. If 4 states form an alliance then each gets 4 points to spend)

You may not attack your allies unless they agree. And any state in an alliance can choose to leave at any time(you dont get the 5 points back).

Special items:

Atom bomb- Choose any state(except your allies) to drop the bomb on. That state is automatically destroyed no matter how many points it has. Each state that borders that state loses 5 points.

Cost: 50 points

Shield- Opponents may not attack your state for 3 hours.

Cost:15 points

Stronger Shield- Opponents may not attack your state for 5 hours.

Cost: 25 points

Nuclear defense system: 50 points - 1 use, protects you from nuke

Biohazard suits- 25 points-1 use, protects you from radioactive fallout (if the state next to you is nuked)

*taking suggestions for other items*

And PLEASE hide the score so it doesn't take forever to scroll. :D

[hide="score"]Alabama-10 pts
Alaska-10 pts
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
Massachusetts-10 pts
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
Missouri-10 pts
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts
North Carolina-10 pts
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-10 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
Washington-10 pts
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts[/hide]

[hide="score"]Alabama-10 pts
Alaska-10 pts
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
Massachusetts-10 pts
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
[color=green]Missouri-12 pts[/color]
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts
North Carolina-10 pts
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-10 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
Washington-10 pts
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts[/hide][/quote]

take out the states where no one lives first. :D

[hide="score"]Alabama-10 pts
Alaska-[color=red]8 pts[/color]
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
Massachusetts-10 pts
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
Missouri-12 pts
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts
North Carolina-10 pts
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-10 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
Washington-10 pts
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts[/hide]

oh man this looks slightly silly... but I will help NC at least once

[hide="score"]Alabama-10 pts
Alaska-8 pts
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
Massachusetts-10 pts
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
Missouri-12 pts
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts[color=green]
North Carolina-12 pts[/color]
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-10 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
Washington-10 pts
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts[/hide]

Suggestions:

Nuclear defense system: 50 points - 1 use, protects you from nuke
Biohazard suits- 25 points-1 use, protects you from radioactive fallout (if the state next to you is nuked)

Note: for those 2 items, all members of the team can use it, but once  it is used once, it can't be used again by any member of the team.

Dissolving alliance: After an alliance is made, you may choose to dissolve the alliance. There is a 1-day ceasefire (between members of the alliance) and then fighting between those states can resume.

Hopefully more people from my state will see this.

[hide="score"]Alabama-10 pts
Alaska-8 pts
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
[color=green]Massachusetts-12 pts[/color]
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
Missouri-12 pts
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts
North Carolina-12 pts
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-10 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
Washington-10 pts
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts[/hide]

I will LOSE but oh well

[hide="score"]Alabama-10 pts
Alaska-8 pts
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
Massachusetts-12 pts
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
Missouri-12 pts
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts
North Carolina-12 pts
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-10 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
[color=green]Washington-12 pts [/color]
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts [/hide]

[hide="score"] Alabama-10 pts
Alaska-8 pts
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
Massachusetts-12 pts
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
Missouri-[color=green]14 pts[/color]
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts
North Carolina-12 pts
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-10 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
Washington-12 pts
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts[/hide]

[hide="updated items list"]Atom bomb- Choose any state(except your allies) to drop the bomb on. That state is automatically destroyed no matter how many points it has. Each state that borders that state loses 5 points.

Cost: 50 points

Shield- Opponents may not attack your state for 3 hours. Shields do not protect from nukes or radiation.

Cost:15 points

Stronger Shield- Opponents may not attack your state for 5 hours.

Cost: 25 points

Nuclear defense system: 50 points - 1 use, protects you from nuke

Biohazard suits- 25 points-1 use, protects you from radioactive fallout (if the state next to you is nuked)[/hide]

[hide]Alabama-10 pts
Alaska-8 pts
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
Massachusetts-12 pts
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
[color=green]Missouri-16 pts[/color]
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts
North Carolina-12 pts
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-10 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
Washington-12 pts
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts
[/hide]

[hide="abooga"]
Alabama-10 pts
Alaska-[color=red]6 pts[/color]
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
Massachusetts-12 pts
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
Missouri-16 pts
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts
North Carolina-12 pts
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-10 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
Washington-12 pts
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts
[/hide]

[hide]Alabama-10 pts
Alaska-6 pts
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
Massachusetts-12 pts
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
Missouri-16 pts
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts
North Carolina-12 pts
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-12 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
Washington-12 pts
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts [/hide]

which state di you do, budi????? its hard to tell without colors.

texas, washington, massachusetts,  north carolina alliance?

[hide="score"]Alabama-10 pts
Alaska-[color=red]4 pts[/color]
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
Massachusetts-12 pts
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
Missouri-16 pts
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts
North Carolina-12 pts
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-12 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
Washington-12 pts
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts[/hide]
budi lives in TX

[hide="score"]Alabama-10 pts
Alaska-4 pts
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
Massachusetts-12 pts
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
[color=green]Missouri-18 pts[/color]
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts
North Carolina-12 pts
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-12 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
Washington-12 pts
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts[/hide]

[hide="score"]Alabama-10 pts
Alaska-4 pts
Arizona-10 pts
Arkansas-10 pts
California-10 pts
Colorado-10 pts
Connecticut-10 pts
Delaware-10 pts
Florida-10 pts
Georgia-10 pts
Hawaii-10 pts
Idaho-10 pts
Illinois-10 pts
Indiana-10 pts
Iowa-10 pts
Kansas-10 pts
Kentucky-10 pts
Louisiana-10 pts
Maine-10 pts
Maryland-10 pts
Massachusetts-12 pts
Michigan-10 pts
Minnesota-10 pts
Mississippi-10 pts
Missouri-18 pts
Montana-10 pts
Nebraska-10 pts
Nevada-10 pts
New Hampshire-10 pts
New Jersey-10 pts
New Mexico-10 pts
New York-10 pts
North Carolina-[color=green]14 pts[/color]
North Dakota-10 pts
Ohio-10 pts
Oklahoma-10 pts
Oregon-10 pts
Pennsylvania-10 pts
Rhode Island-10 pts
South Carolina-10 pts
South Dakota-10 pts
Tennessee-10 pts
Texas-12 pts
Utah-10 pts
Vermont-10 pts
Virginia-10 pts
Washington-[color=green]14 pts[/color]
West Virginia-10 pts
Wisconsin-10 pts
Wyoming-10 pts[/hide]

EDIT: forgot about alliance!

[quote="james4l"]Alabama-6 pts
Hawaii-14 pts
Minnesota-5 pts
Ohio- 15 pts

-5 MT, -3 AL => +2 HI[/quote]

Lol you killed our ally.

whoops

oh well no one's posting there...

ernie, you know that you are cheating right?

Alabama-3 pts
Hawaii-14 pts
Minnesota-5 pts
Ohio- 15 pts

AL -3

Alabama-3 pts
Hawaii-4 pts
Minnesota-5 pts
Ohio- 5 pts

Hawaii-8 pts+7pts=15pts-15pts=0pts,PWNT
Ohio- 5 pts
JUST TO SAVE POSTS

And how would you do that?

[quote="ternary0210"]Ohio-8+2=10pts[/quote]

you know that Ohio is an ally  :roll:

Ohio-14+1=15pts-15pts=0pts(shield) PWNT

Ok, now lock this forum.

Interesting game! Should we start again?


algebra
polynomial
calculus
derivative
algebra unsolved

Find all polynomials $ P \in \mathbb{R}[X]$ with the propriety that :
$ P^n(x)\equal{}P(x^n)  , \forall x \in \mathbb{R}$ where $ n \in \mathbb{N} , n >1$  is fixed .

If $ P(x) \equal{} c$ then
$ c^n \equal{} c\Leftrightarrow c\in \{0,1\}$
So $ P(x) \equal{} 0$ or $ P(x) \equal{} 1$
Consider case $ \deg P \equal{} m > 0$
Call $ a_m$ is the leading coefficient of $ P(x)$ then
$ a_n^m \equal{} a_n$ so $ a_n \equal{} 1$
Consider $ G(x) \equal{} P(x) \minus{} x^m$ then $ \deg (g) \leq {m \minus{} 1}$
So $ (G(x) \plus{} x^m)^n \equal{} G(x^n) \plus{} x^{mn}$
If $ \deg G > 0$ then $ deq (G(x) \plus{} x^m)^n \minus{} x^{mn} \geq  n(m \minus{} 1) \plus{} 1$
It gives contradiction .
So $ G(x) \equal{} c$
Easy to check that $ c \equal{} 0$
So $ P(x) \equal{} x^m$

Let P(x)=x^k. Q(x) where Q(0) not equal to zero

We have

Q(x^n)=Q(x)^n

Take the derivative we have

n.x^(n-1). Q'(x^n)= n.Q'(x).Q(x)^(n-1)

From there we must have x^(n-1)| Q'(x)  --> Q'(x)=c.x^(n-1) as deg Q' =n-1

......


function
induction
real analysis
real analysis unsolved

I need help solving the following problem:

If f(x) is continuous on (a,b) and f((x+y)/2) <= 1/2f(x) + 1/2f(y) for all x,y in (a,b),
then f(x) is convex.

By induction prove that
\[ f\left(\frac{x_1+x_2+\ldots+x_{2^n}}{2^n}\right)\leq \frac{f(x_1)+f(x_2)+\ldots+f(x_{2^n})}{2^n}  \]
for all $n\geq 1$.  Hence conclude that
\[ f(px+qy)\leq pf(x)+qf(y)  \] for all $x,y$ and $p,q\in\mathbb{Q}^+$ such that $p+q=1$.  By continuity it follows that
\[ f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y)  \]
for all $x,y$ and $\lambda\in (0,1)$; this is convexity.


a box of candy hearts contains 52 hearts of which 19 are white, 10 are tan, 7 are pink, 3 are purple, 5 are yellow, 2 are orange, and 6 are green. if you select 9 pieces of candy randomly from the box, without replacement, give the probability that

(a) Exactly Three of the hearts are white.
(b) Three are white, two are tan, one is pink, one is yello, and two are green.

[hide="A"]
$ \binom{9}{3} \times \frac{19 \times 18 \times 17 \times 33 \times 32 \times 31 \times 30 \times 29 \times 28}{52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46 \times 45 \times 44}\equal{}\boxed{\frac{102486}{351325}}$
[/hide]

Use the same method for (B).

EDIT: Haha ok i get it now, thanks. so (a) shouuld be 19C3 times 33C6 all divided by 52C9, and b is the same...

Yes, that works faster than what I did.  Make sure you understand why it works, though.


Phosphoric acid is of great importance in fertiliser production. Besides, phosphoric acid and its various salts have a number of applications in metal treatment, food, detergent and toothpaste industries.

a) Write the Lewis structure of phosphoric acid.

b) The pK values of the three successive dissociations of phosphoric acid at 25°C are: pK1 = 2.12, pK2 = 7.21, and pK3 = 12.32. Write down the conjugate base of dihydrogen phosphate ion and determine its pKb value.

c) Small quantities of phosphoric acid are extensively used to impart the sour or tart taste to many soft drinks such as colas and root beers. A cola having a density of 1.00 g/mL contains 0.05 % by weight of phosphoric acid. Determine the pH of the cola (ignoring the second and the third dissociation steps for phosphoric acid). Assume that the acidity of the cola arises only from phosphoric acid.

d) Phosphoric acid is used as a fertiliser for agriculture. A $ 1.00 \times 10^{-3}$ M solution of phosphoric acid is added to an aqueous soil suspension and the pH is found to be 7.00. Determine the fractional concentrations of all the different phosphate species present in the solution. Assume that no component of the soil interacts with any phosphate species.

e) Zinc is an essential micronutrient for plant growth. Plants can absorb zinc in water soluble form only. In a given soil water with pH = 7.0, zinc phosphate was found to be the only source of zinc and phosphate. Calculate the concentration of $ Zn^{2+}$ and $ PO_{4}^{3-}$ ions in the solution. Ksp for zinc phosphate is $ 9.1 \times 10^{-33}$.

a) Just look it up.

b) Conjugate base is $ HPO_{4}^{2-}$. For a conjugate acid-base pair, $ pK_{a}+pK_{b}=14$, so $ pK_{b}=6.79$ for hydrogen phosphate.

c) The cola is $ 5.10\times 10^{-6}M$.

$ K_{a}=\frac{[H_{2}PO_{4}^{-}][H^{+}]}{[H_{3}PO_{4}]}=\frac{x^{2}}{5.10\times 10^{-6}-x}=7.59\times 10^{-3}$.

EDIT: WRONG MOLARITY

I'll try the rest later.

Your answer for b) is correct. However, your answer for c) is wrong.

Oops, the cola is actually $ 5.10\times 10^{-4}M$.

$ \frac{x^{2}}{5.10\times 10^{-4}-x}=7.59\times 10^{-3}$.

Solve to get $ x=4.80\times 10^{-4}$. Is the pH 3.32?

If so, I'll move on to d) and e).

Still wrong. How are you calculating the concentration of phosphoric acid in the cola? Your problem is there.

Let the cola be $ x$ mL. Then it weighs $ x$ grams. Since it contains $ 0.00005x$ grams of phosphoric acid, and the molar mass of phosphoric acid is 98 g/mol, the cola contains $ (5.1\times 10^{-7})x$ mol. Divide by the volume which is $ x\cdot 10^{-3}$ liters and get $ 5.1\times 10^{-4}M$.

[quote="LeFromage"]Since it contains $ 0.00005x$ grams of phosphoric acid[/quote]

You have an extra zero there. 0.05% by weight means that in 100 g of solution there are 0.05 g of pure solute.

...wow that was really clumsy of me.

So in fact it is $ 5.1\times 10^{-3}M$.

And the pH turns out to be 2.46.

Yes, that's right. Let's move one to (d) and (e).


geometry
trigonometry
calculus
calculus computations

Find the area inside the lemniscate $ r^2\equal{}6\cos 2\theta$ and outside the circle $ r\equal{}\sqrt{3}$

hello
see this prob. may be it would be helpful.
[url]http://curvebank.calstatela.edu/lemniscate/lemniscate.htm[/url]
thank u


group theory
abstract algebra
superior algebra
superior algebra solved

Has anyone something nice for the following problem:
   Find all morphisms from the symmetric group $S_4$ into $S_3$.

Welcome back,harazi.

We are glad to see you! :)

kunny

Thanks, Kunny. Now, let's solve the problem...  :(

There are precisely two nontrivial normal subgroups of $S_4$. One is $A_4$, and taking the quotient gives a copy of $\mathbb{Z}_2$ that can be embedded into $S_3$. The other has order 4, and taking the quotient gives a copy of $S_3$, since there can be no elements of order 6.

Up to automorphism of $S_3$, these and the trivial map are the only homomorphisms.

But it's nice to see an explicit permutation representation.  In the case where the image is $C_2$, you're permuting the sign of $\prod_{1\le i<j \le4} (x_i - x_j)$.  In the case where the image is the whole of $S_3$ you're permuting the three partitions of $\{1,2,3,4\}$ into two classes of size 2, namely $\{\{1,2\},\{3,4\}\}$, $\{\{1,3\},\{2,4\}\}$, $\{\{1,4\},\{2,3\}\}$.


In a fast food restaurant with ambient temperature of 80◦, a coffee cup containing hot coffee at 180◦ is measured to be 170◦ five minutes later. Find the temperature of the coffee at any time t.

NOTE. According to Newton's law of cooling, if $ T(t)$ denotes the temperature of a substance placed in some environment with constant temperature, the function $ T(t)$ satisfies the following differential equation:

$ \frac{dT}{dt}\equal{}k(T\minus{}T_{0})$

where $ T_{0}$ denotes the initial temperature of the substance (i.e. the rate of change of $ T$ is proportional to the difference between $ T$ and the initial temperature $ T_{0}$). Here $ k$ is a constant depending on the substance and on the outside temperature.

So you have $ T_{0}\equal{}180$ and $ T(5)\equal{}170$. Solve the ODE: $ T(t)\equal{}T_{0}e^{kT}$ (with your setup $ k$ is negative) and then solve for $ k$.
[quote]$ k$ is a constant depending on the substance and on the outside temperature.[/quote]
$ k$ should not depend on the outside temperature. Otherwise it would also depend on the temperature of coffee, contradicting the model. But it does depend on the coffee cup.

If a fast food restaurant has ambient temperature 80F, I am not going to drink coffee there.


function
topology
real analysis
real analysis unsolved

Hi all

I was wrestling with this problem for a while, not sure if it has a positive solution or not:

Suppose X is a normal topological space. Let C be a compact subset of X such that the closure of the interior of C equals C. Is it true that there exists a continuous function from X -> R such that supp(f) = C?

Now I know that the strong Urysohn's lemma says that if A & B are closed in X, there exists a continuous function g: X -> [0,1] such that g^-1(0) = A and g^-1(1) = B iff A and B are both g-delta. So I thought perhaps that given the condition on C above, perhaps this is enough to conclude that that C compliment is a g-delta, and then find some smaller closed set contained in the interior of C that is also g-delta and I would be done but that is really just a shot in the dark and I haven't been able to get anywhere on it.

Anyways any insight you could offer would be greatly appreciated. Thanks

[quote="Spacecow"]So I thought perhaps that given the condition on C above, perhaps this is enough to conclude that that C compliment is a g-delta, and then find some smaller closed set contained in the interior of C that is also g-delta and I would be done but that is really just a shot in the dark and I haven't been able to get anywhere on it.[/quote]
If $ C$ is additionally assumed to have a countably locally finite basis, the topology of $ C$ is metrizable, by [url=http://en.wikipedia.org/wiki/Nagata-Smirnov_metrization_theorem]Nagata-Smirnov[/url]. This implies that the interior of $ C$ can be exhausted by countably many closed subsets.


Beasting with guessing, we see that $ \boxed{33}$ is the middle number (there's an odd number of elements).


geometry
inequalities
circumcircle
triangle inequality
geometry proposed

It is given a regular hexagon in the plane. Let P be a point of the plane. Define s(P) as the sum of the distances from P to each side of the hexagon, and v(P) as the sum of the distances from P to each vertex.
a) Find the locus of points P that minimize s(P)
b) Find the locus of points P that minimize v(P)

If $a$ is the sidelenght of the regular hexagon $A_{1}...A_{6}$then $a.s(P)=\sum_{i=1}^{6}[PA_{i}A_{i+1}]\geq[A_{1}...A_{6}]=const$ ($[XYZ]=$area($XYZ$)) therefore the locus of $P$ when $s(P)$ is minimized is the regular hexagon.
For the second one put the hexagon in the complex plane such that its surcumcircle is the unit curcle. Then use the triangle inequality to obtain that $v(P)$ is minimized when $P$ is the center.

Other way for the second with regular polygon $A_{1}..A_{n}$ center $(O)$ circumradius $R$ we have for all $M$ in plane $R(\sum MA_{i})=\sum OA_{i}MA_{i}\ge\sum\vec{OA_{i}}\vec{MA_{i}}=\sum\vec{OA_{i}}(\vec{MO}+\vec{OA_{i}})=\vec{MO}(\sum \vec{OA_{i}})+nR^{2}=nR^{2}$ because in regular polygon $\sum\vec{OA_{i}}=\vec{0}$
Thus $\sum MA_{i}\ge nR$ equality hold iff $M\equiv O$


[hide="solution"]
$ f\left(\frac {x \minus{} 1}{x \plus{} 1}\right) \plus{} f\left(\frac {1}{x}\right) \plus{} f\left(\frac {1 \plus{} x}{1 \minus{} x}\right) \equal{} x \qquad(1)$

let $ y \equal{} \frac {x \minus{} 1}{x \plus{} 1} \Longrightarrow x \equal{} \frac {y \plus{} 1}{1 \minus{} y}$

$ \therefore f(y) \plus{} f\left(\frac {1 \minus{} y}{y \plus{} 1}\right) \plus{} f\left(\frac { \minus{} 1}{y}\right) \equal{} \frac {y \plus{} 1}{1 \minus{} y}$
Then set $ y \equal{} \minus{} x$
$ \therefore f( \minus{} x) \plus{} f\left(\frac {1 \plus{} x}{1 \minus{} x}\right) \plus{} f\left(\frac {1}{x}\right) \equal{} \frac {1 \minus{} x}{1 \plus{} x} \qquad (2)$

let $ y \equal{} \frac {1 \plus{} x}{1 \minus{} x} \Longrightarrow x \equal{} \frac {y \minus{} 1}{y \plus{} 1}$

$ \therefore f\left(\frac { \minus{} 1}{y}\right) \plus{} f\left(\frac {1 \plus{} y}{y \minus{} 1}\right) \plus{} f(y) \equal{} \frac {y \minus{} 1}{1 \plus{} y}$
then set $ y \equal{} \minus{} x$
$ \therefore f\left(\frac {1}{x}\right) \plus{} f\left(\frac {x \minus{} 1}{x \plus{} 1}\right) \plus{} f( \minus{} x) \equal{} \frac {x \plus{} 1}{x \minus{} 1} \qquad(3)$

Now $ (2) \minus{} (1)$ gives
$ f( \minus{} x) \minus{} f\left(\frac {x \minus{} 1}{x \plus{} 1}\right) \equal{} \frac {x \minus{} 1}{x \plus{} 1} \minus{} x \Leftrightarrow f( \minus{} x) \minus{} ( \minus{} x) \equal{} f\left(\frac {x \minus{} 1}{x \plus{} 1}\right) \minus{} \left(\frac {x \minus{} 1}{x \plus{} 1}\right) \qquad(*)$

Similarly $ (3) \minus{} (2)$ gives
$ f\left(\frac {x \minus{} 1}{x \plus{} 1}\right) \minus{} \left(\frac {x \minus{} 1}{x \plus{} 1}\right) \equal{} f\left(\frac {x \plus{} 1}{1 \minus{} x}\right) \minus{} \left(\frac {x \plus{} 1}{1 \minus{} x}\right) \qquad(**)$

From (*) and (**) we find

$ 3\left(f\left(\frac {1}{x}\right) \minus{} \frac {1}{x}\right) \equal{} x \minus{} \left(\frac {x \minus{} 1}{x \plus{} 1} \plus{} \frac {1}{x} \plus{} \frac {1 \plus{} x}{1 \minus{} x}\right)$

$ \therefore f\left(\frac {1}{x}\right) \equal{} \frac {x^4 \plus{} 5x^2 \minus{} 2}{3x(x^2 \minus{} 1)}$

$ \therefore f(x) \equal{} \frac {2x^4 \minus{} 5x^2 \minus{} 1}{3x(x^2 \minus{} 1)}$  :lol:
[/hide]


The vertical asymptotes of the graph of $ y\equal{}\frac{x^3\minus{}5x^2\minus{}17x\plus{}21}{3x^2\plus{}36x\plus{}81}$ can be written in the form $ x
\plus{}Ny\equal{}S$ and $ x\equal{}A$, where $ N,S,$ and $ A$ are rational, real numbers. Find $ N * A *S *A$

Oblique symptotes are considered vertical?

Apparently anything not horizontal is vertical. (Come to think of it, a teacher in Kansas for five years taught her students that a kilometer is greater than a mile...)

To find the vertical asymptote, first factor both the numerator and denominator and cancel out a common factor. The remaining factor gives you the vertical asymptote, where the one that was cancelled out gives you a hole, but that is not needeed here. Then do long division to find the oblique asymptote.


Hi,

I am competing in an Olympiad in 2 days (the AIMO - Australia), and they usually have 1 or 2 motion questions, like rates and really wordy motion problem solving questions. I would really appreciate it if anyone has any tips for me on these types of questions =)

Thanks

[quote="JRav"]Average velocity is given by total distance traveled divided by the time.  Remember, this is average velocity.  Acceleration is given by the change in velocity divided by the time passed.

The first few equations on [url=http://emsc32.nysed.gov/osa/reftable/reftablearch/physics06tbl.pdf]page 6[/url] are the fundamental equations of motion in Newtonian mechanics.[/quote]

Thanks for your help but i was more referring to mathematical problem solving questions on "motion" like this one

[i]"Two candles of equal length start burning at the same time. One of the candles will burn down in 4 hours, the other in 5 hours. How many hours will they burn before one candle is 3 times the length of the other?"[/i]

Its just these questions that i find difficult

Your goal on any problem of that type should be to turn the given information into a system of equations.  Once you get those equations, there will be various ways to solve them, but the first step is to write out all of the information as an equation and identify the algebraic expression that you want to find.

For your problem:

Let the candles burn for $ x$ hours and let $ y$ be the initial height of the candles.  Then, after $ x$ hours, the first candle will be at height $ y\minus{}\frac{y\cdot x}4\equal{}y\left(1\minus{}\frac{x}4\right)$.  The second candle will be at height $ y\minus{}\frac{y\cdot x}5\equal{}y\left(1\minus{}\frac{x}5\right)$.

Since the first candle burns faster than the other candle, it will be at a lower height than the second candle, meaning that the second candle must be 3 times as tall as the first candle.  We can write this as an equation to solve for $ x$:

$ 3y\left(1\minus{}\frac{x}4\right)\equal{}y\left(1\minus{}\frac{x}5\right)$
$ 60\minus{}15x\equal{}20\minus{}4x$
$ 11x\equal{}40$
$ x\equal{}\frac{40}{11}$

Once you get everything expressed algebraically, the problem should become fairly simple to finish.


geometry
circumcircle
algebra
system of equations
algebra open

$ \theta _{1}, \cdots ,\theta _{n} , R$ are variables and $ d_{1}, \cdots ,d_{n}$ are given.
We have following equations. Can we find a formula for $ \theta _{1}, \cdots ,\theta _{n} , R$?
(All variables and $ d_{i}$s are positive.)

$ sin\frac {\theta_{1}}{2} \equal{} \frac {d_{1}}{2R}, \cdots ,sin\frac {\theta_{n}}{2} \equal{} \frac {d_{n}}{2R}, \theta_{1} \plus{} \cdots \plus{} \theta_{n} \equal{} 2\pi$

[hide="observation"]The system of equations has a solution if and only if there exists a cyclic $ n$-gon with sides of length $ d_1,d_2,...,d_n$ and its circumcenter is within it.  $ R$ is then the radius of its circumscribed circle, and $ \theta _n \equal{} 2\arcsin \frac {d_n}{2R}$[/hide]


Ok, i think i found a counterexemple. Here it goes:

Suppose that $ n \equal{} 2$, $ X \equal{} {x_1,...,x_r}$,$ A_1 \equal{} {x_1,...,x_{r \minus{} 1} }$ and $ f(x_i) \equal{} x_i$, for $ i \equal{} 1,r \minus{} 1$ and $ f(x_r) \equal{} x_1$. Then the nice vectors are $ (x_1,x_1),...,(x_{r \minus{} 1},x_{r \minus{} 1}),(x_1,x_r),(x_r,x_1)$ .The number of nice vectors is $ r \plus{} 1$ while $ \frac {|X|^n}{\prod\limits_{i \equal{} 1}^{n \minus{} 1} |A_i|}$=$ \frac {r^2}{r \minus{} 1}$, contradiction.

I hope i understand correctly the problem. Maybe is there another condition? :wink:

Edit: Yes, I'm the idiot...


Keep counting and counting... it's August.


analytic geometry
geometry
circumcircle
conics
geometry unsolved

[color=darkblue]Let $M$ be a point with the barycentrical coordinates $(x,y,z)$ w.r.t. the given triangle $ABC$ with the incircle $C(I,r)$ and the circumcircle $C(O,R)\ .$
Prove that the point $M$ belongs to the incircle if and only if there is the relation $\boxed{\ \sum bcx=\sum yza^{2}+r(4R+r)\ }\ .$
[b]Remark 1.[/b] The power $p(M)$ of the point $M$ w.r.t. the circumcircle of the given triangle is $p(M)=-\sum yza^{2}\ .$
[b]Remark 2.[/b] If $a=b=c$, i.e. the triangle $ABC$ is equilateral, then $MI=r$ $\Longleftrightarrow$ $xy+yz+zx=\left(\frac{a}{2}\right)^{2}\ .$
[b]Can anyone[/b] (using, if need be, the above result), for example [u]M4rio[/u], [u]Shobber[/u], [u]Yetti[/u], [u]Darij G.[/u] a.s.o., generalize the nice problem from the post http://www.mathlinks.ro/Forum/viewtopic.php?t=106446 ?[/color]

In an equilateral triangle $ABC$, with lenght:$L$, a point $P$ belongs to the incircle if: $\boxed{(AP)^{2}+(BP)^{2}+(CP)^{2}=\frac{5L^{2}}{4}}$

[quote="M4RI0"]In an equilateral triangle $ABC$, with lenght $a$, a point $P$ belongs to the incircle if $\boxed{(AP)^{2}+(BP)^{2}+(CP)^{2}=\frac{5a^{2}}{4}}$[/quote]
[color=darkblue]Sorry, M4rio, I am interested by the general case of characterization in an any triangle $ABC$ when the point $P$ belongs to the incircle.

The mentioned property in a equilateral triangle is easily. Generally, $PA^{2}+PB^{2}+PC^{2}=3\cdot PG^{2}+\frac{a^{2}+b^{2}+c^{2}}{3}\ .$

Particularly, if $a=b=c$, then $r=\frac{a\sqrt 3}{6}$, $G\equiv I$ and $P\in C(I,r)$ $\Longleftrightarrow$  $PG=r$ $\Longleftrightarrow$ $\sum PA^{2}=\frac{5a^{2}}{4}\ .$[/color]

[quote="Virgil Nicula"][b]Can anyone[/b] (using, if need be, the above result), for example [u]M4rio[/u], [u]Shobber[/u], [u]Yetti[/u], [u]Darij G.[/u] a.s.o., generalize the nice problem from the post http://www.mathlinks.ro/Forum/viewtopic.php?t=106446 ?[/color][/quote]

In fact, these properties follow from the barycentric equation of an inscribed conic in triangle ABC:

Let P be a point with barycentric coordinates $P\left(p: q: r\right)$ with respect to triangle ABC. Let $A^{\prime}=AP\cap BC$, $B^{\prime}=BP\cap CA$, $C^{\prime}=CP\cap AB$. Then, there exists one and only one conic touching the lines BC, CA, AB at the points A', B', C', and the barycentric equation of this conic is

$\frac{x^{2}}{p^{2}}+\frac{y^{2}}{q^{2}}+\frac{z^{2}}{r^{2}}-\frac{2yz}{qr}-\frac{2zx}{rp}-\frac{2xy}{pq}=0$.

This equation can be rewritten as

$\left(\sqrt{\frac{x}{p}}+\sqrt{\frac{y}{q}}+\sqrt{\frac{z}{r}}\right)\left(-\sqrt{\frac{x}{p}}+\sqrt{\frac{y}{q}}+\sqrt{\frac{z}{r}}\right)\left(\sqrt{\frac{x}{p}}-\sqrt{\frac{y}{q}}+\sqrt{\frac{z}{r}}\right)\left(\sqrt{\frac{x}{p}}+\sqrt{\frac{y}{q}}-\sqrt{\frac{z}{r}}\right)$
$=0$.

If P is the Gergonne point of triangle ABC, with its barycentric coordinates $\left(p: q: r\right)=\left(\frac{1}{s-a}: \frac{1}{s-b}: \frac{1}{s-c}\right)$, then the conic becomes the incircle of triangle ABC. Hence, the barycentric equation of the incircle is

$\left(\sqrt{\left(s-a\right)x}+\sqrt{\left(s-b\right)y}+\sqrt{\left(s-c\right)z}\right)\left(-\sqrt{\left(s-a\right)x}+\sqrt{\left(s-b\right)y}+\sqrt{\left(s-c\right)z}\right)$
$\left(\sqrt{\left(s-a\right)x}-\sqrt{\left(s-b\right)y}+\sqrt{\left(s-c\right)z}\right)\left(\sqrt{\left(s-a\right)x}+\sqrt{\left(s-b\right)y}-\sqrt{\left(s-c\right)z}\right)$
$=0$.

This generalizes M4rio's result at http://www.mathlinks.ro/Forum/viewtopic.php?t=106446 .

For a systematic treatise of inscribed conics, see §10.2 of

[url=http://www.math.fau.edu/yiu/geometry.html]Paul Yiu, [i]An Introduction to the Geometry of the Triangle[/i], 2001[/url].

  darij

See [url=http://www.math.fau.edu/yiu/geometry.html]Paul Yiu's "An Introduction to the Geometry of the Triangle"[/url]. I know it from Darij's posts.

[quote="dule_00"]Is there any free book for download on barycentric coordinates?[/quote]

I have made one but it's in Spanish

perfect_radio, thank  you for a book!!!  :)


limit
trigonometry
LaTeX
real analysis
real analysis unsolved

[quote="ashwinkrishnan"]Compute the value of the limit
                    $lim_{x\to\{0}}\sum_{i=1}^{n}\ (i^{sin^{2}x)^{\frac{1}{sin^{2}x}}$[/quote]

http://ivanisvili.narod.ru/Extr2.doc

[quote="ashwinkrishnan"]The exponent $\frac{1}{sin^{2}x}$ is over the whole summation and not just over each term.I couldn't do that using Latex[/quote]
I don't understand what is wrong? the limit of the sum is equal the sum of limit. and we use it for each ... it is easy


[hide]if a number is divisible by 4 and 3, then its divisible by 12. so, there are 8 numbers less than 100.[/hide]

[hide]numbers divisible by 3&4 are divisible by 12, and 8*12 is the highest under 100, so the answer is 8[/hide]

[hide]Since 4*3 is twelve, I have to find how many multiples of twelve there are below 100. The answer is 8.[/hide]

[quote="joejia"]How many positive integers less than 100 and divisible by 3 are also divisible by 4?[/quote]

8, since the lcm of 3 an d4 are 12, and there are 8 multiples of 12 under 100.

anirudh, can you [i]please[/i] hide your answer  ;)

[hide]
100/3 is about 33. So 33 numbers are divisible by 3. 33/4 is about 8. So 8 of the 33 numbers divisble by 3 are divisble by 4.
[/hide]


inequalities
logarithms
inequalities proposed

$\sum{\frac{a}{a^2+2}}\leq\sum{\frac{a}{2a+1}}$.

We shall prove that $\sum{\frac{a}{2a+1}}\leq{1}$ or
                                               $\sum{\frac{2a}{2a+1}}\leq{2}$ or
                                               $\sum{\frac{1}{2a+1}}\geq{1}$.

Clearing the denominator we have to prove that:

                                            $2\sum{a}\geq{6}$ wich is true by AM-GM.

[quote="XAKKER"]a/a^2+2=<a/2a+1[/quote]
 right, because of : $ a^2\plus{}2 \equal{} a^2\plus{}1\plus{}1 \geq  2a \plus{}1$ ( by ineq AM-GM)

[quote="n0vakovic"]Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$. Proove that
\[ \frac a{a^{2} \plus{} 2} \plus{} \frac b{b^{2} \plus{} 2} \plus{} \frac c{c^{2} \plus{} 2}\leq 1
\]
[/quote]

It was post here
http://www.mathlinks.ro/viewtopic.php?t=172711&start=20

[quote="n0vakovic"]Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$. Proove that
\[ \frac a{a^{2} \plus{} 2} \plus{} \frac b{b^{2} \plus{} 2} \plus{} \frac c{c^{2} \plus{} 2}\leq 1
\]
[/quote]
The following inequality is also holds:
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$. Proove that
\[ \frac a{a^{2} \plus{} 3} \plus{} \frac b{b^{2} \plus{} 3} \plus{} \frac c{c^{2} \plus{}3}\leq \frac{3}{4}
\]

I think it is more difficult than above inequality!!!

We have $ \sum{\frac{a}{a^2 + 2}} = \sum{\frac{a}{a^2 + 1^2 + 1}} \leq \sum{\frac{a}{2a + 1}}$, because $ a^2 + 1 \geq 2a$

Now its enough to prove that

$ \sum{\frac{a}{2a + 1}} \leq 1 \leftrightarrow
\frac{1}{2}\sum{\frac{2a}{2a + 1}} \leq 1 \leftrightarrow
\frac{1}{2}\sum{\frac{2a + 1 - 1}{2a + 1}} \leq 1 \leftrightarrow
\frac{1}{2}(1 + 1 + 1 - \sum{\frac{1}{2a + 1}}) \leq 1 \leftrightarrow
(3 - \sum{\frac{1}{2a + 1}}) \leq 2 \leftrightarrow
\sum{\frac{1}{2a + 1}}) \geq 3 - 2 = 1 \leftrightarrow
\sum{\frac{bc}{2abc + bc}}) \geq 1 \leftrightarrow
\sum{\frac{\sqrt{bc}^2}{2 + bc}}) \geq 1$

From the Andreescu Inequality we have:

$ \sum{\frac{{\sqrt{bc}}^2}{2 + bc}} \geq \frac{{(\sqrt{ba} + \sqrt{bc} + \sqrt{ca})}^2}{6 + bc + ca + ba} = K$

So its enough to prove that

$ K \geq 1 \leftrightarrow
{(\sqrt{ba} + \sqrt{bc} + \sqrt{ca})}^2 \geq 6 + bc + ca + ba \leftrightarrow
ba + bc + ca + 2\sum_{cyclic}{a\sqrt{bc}} \geq 6 + bc + ca + ba \leftrightarrow
2\sum_{cyclic}{a\sqrt{bc}} \geq 6 \leftrightarrow
\sum_{cyclic}{a\sqrt{bc}} \geq 3$

Which is true because from the AM-GM Inequality we have:

${{ \sum_{cyclic}{a\sqrt{bc}} \geq 3\sqrt[3]{abc\sqrt{ba}\sqrt{ca}\sqrt{cb}} = 3\sqrt[3]{abc\sqrt{a^2b^2c^2}} = 3\sqrt[3]{abcabc}} = 3\sqrt[3]{1}} = 3$

[quote="zaizai-hoang"]
The following inequality is also holds:
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$. Proove that
\[ \frac a{a^{2} \plus{} 3} \plus{} \frac b{b^{2} \plus{} 3} \plus{} \frac c{c^{2} \plus{} 3}\leq \frac {3}{4}
\]
[/quote]
Your inequality can be proved by the $ \sum_{cyc}\left(\frac{1}{4}\minus{}\frac{a}{a^2\plus{}3}\plus{}\frac{\ln a}{8}\right)\geq0$ idea.
For the following
\[ \frac a{a^{2} \plus{} 4} \plus{} \frac b{b^{2} \plus{} 4} \plus{} \frac c{c^{2} \plus{} 4}\leq \frac {3}{5}\]
this idea doesn't help but the inequality seems true.

By the way,  $ \frac a{a^{2} \plus{} 5} \plus{} \frac b{b^{2} \plus{} 5} \plus{} \frac c{c^{2} \plus{} 5}\leq \frac {1}{2}$ is wrong.

[quote="arqady"][quote="zaizai-hoang"]
The following inequality is also holds:
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$. Proove that
\[ \frac a{a^{2} \plus{} 3} \plus{} \frac b{b^{2} \plus{} 3} \plus{} \frac c{c^{2} \plus{} 3}\leq \frac {3}{4}
\]
[/quote]
Your inequality can be proved by the $ \sum_{cyc}\left(\frac {1}{4} \minus{} \frac {a}{a^2 \plus{} 3} \plus{} \frac {\ln a}{8}\right)\geq0$ idea.
For the following
\[ \frac a{a^{2} \plus{} 4} \plus{} \frac b{b^{2} \plus{} 4} \plus{} \frac c{c^{2} \plus{} 4}\leq \frac {3}{5}
\]
this idea doesn't help but the inequality seems true.

By the way,  $ \frac a{a^{2} \plus{} 5} \plus{} \frac b{b^{2} \plus{} 5} \plus{} \frac c{c^{2} \plus{} 5}\leq \frac {1}{2}$ is wrong.[/quote]

But
 \[ \sum_{cyc}\left(\frac {1}{4} \minus{} \frac {a}{a^2 \plus{} 3} \plus{} \frac {\ln a}{8}\right)\geq0\]
is not correct for all $ a\ge 0$  :wink:
Anyway, it is not hard to prove  :)

[quote="Sailor"]
                                               $ \sum{\frac {2a}{2a \plus{} 1}}\leq{2}$ or
                                               $ \sum{\frac {1}{2a \plus{} 1}}\geq{1}$.
[/quote]
How does the first imply the second?

$ \displaystyle \frac {2a}{2a\plus{}1} \equal{} \frac {(2a\plus{}1) \minus{} 1}{2a\plus{}1} \equal{} 1 \minus{} \frac {1}{2a\plus{}1}$. Do the same for $ b$ and $ c$ and then add them.

[quote="zaizai-hoang"][quote="n0vakovic"]Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$. Proove that
\[ \frac a{a^{2} \plus{} 2} \plus{} \frac b{b^{2} \plus{} 2} \plus{} \frac c{c^{2} \plus{} 2}\leq 1
\]
[/quote]
The following inequality is also holds:
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$. Proove that
\[ \frac a{a^{2} \plus{} 3} \plus{} \frac b{b^{2} \plus{} 3} \plus{} \frac c{c^{2} \plus{} 3}\leq \frac {3}{4}
\]
I think it is more difficult than above inequality!!![/quote]

It here http://www.mathlinks.ro/viewtopic.php?p=1098909#1098909

for the first one
$ \sum \frac{a}{a^2\plus{}2}\leq\sum \frac{a}{2a\plus{}1}\equal{}\sum \frac{1}{2\plus{}bc}$
now we have to prove
$ \sum \frac{1}{2\plus{}bc}\leq1$
equivalent to
$ a\plus{}b\plus{}c\geq3$,which is true

[quote="Vasc"][quote="arqady"]
By the way,  $ \frac a{a^{2} \plus{} 5} \plus{} \frac b{b^{2} \plus{} 5} \plus{} \frac c{c^{2} \plus{} 5}\leq \frac {1}{2}$ is wrong.[/quote]
It is true, arqady. :)[/quote]

For $ abc\equal{}1$ or for all $ a,b$ and $ c$?

For any positive real numbers such that $ abc\equal{}1$, of course.

[quote="Vasc"][quote="arqady"]
By the way,  $ \frac a{a^{2} \plus{} 5} \plus{} \frac b{b^{2} \plus{} 5} \plus{} \frac c{c^{2} \plus{} 5}\leq \frac {1}{2}$ is wrong.[/quote]
It is true, arqady. :)[/quote]
It seems you are right, but I still can't prove it. :(

[quote="can_hang2007"][quote="Vasc"][quote="arqady"]
By the way,  $ \frac a{a^{2} \plus{} 5} \plus{} \frac b{b^{2} \plus{} 5} \plus{} \frac c{c^{2} \plus{} 5}\leq \frac {1}{2}$ is wrong.[/quote]
It is true, arqady. :)[/quote]
I found now a proof by Cauchy Schwarz Inequality for this hard inequality!  :idea:[/quote]

Congratulation can_hang2007! :lol: I really want to see your Cauchy-Schwarz proof. Could you post it?

Does this way work for this problem:
$ \frac{4a}{a^2\plus{}5} \le \frac{x\plus{}1}{x^2\plus{}x\plus{}1}$
with $ x\equal{}\frac{1}{a^{\frac{3}{4}}}$

[quote="speciallove"]Does this way work for this problem:
$ \frac {4a}{a^2 \plus{} 5} \le \frac {x \plus{} 1}{x^2 \plus{} x \plus{} 1}$
with $ x \equal{} \frac {1}{a^{\frac {3}{4}}}$[/quote]

Your way is a very interesting, but this inequality is not hold, try $ a\equal{}2,$ this inequality become
$ 8x^{2}\le x\plus{}1,$ where $ x\equal{}\frac{1}{2^{\frac{3}{4}}}<1$
$ \Rightarrow2\sqrt{2}\equal{}8x^{2}\le x\plus{}1<2$
contradiction.

sorry,I want to say $ k\equal{} \frac{\minus{}4}{3}$ not $ \frac{\minus{}3}{4}$
Because I will find k for the inequality such that:
$ \frac{4a}{a^2\plus{}5} \le \frac{a^k\plus{}1}{a^{2k}\plus{}a^k\plus{}1}$
or $ f(k)\equal{}(a^k\plus{}1)(a^2\plus{}5)\minus{}4a(a^{2k}\plus{}a^k\plus{}1) \ge 0$
f('1)=6k+8=0[/tex] or $ k\equal{} \frac{\minus{}4}{3}$
with $ k\equal{}\frac{\minus{}4}{3}$ we have $ x\equal{}\frac{1}{a^{\frac{4}{3}}}$

[quote="speciallove"]sorry,I want to say $ k \equal{} \frac { \minus{} 4}{3}$ not $ \frac { \minus{} 3}{4}$
Because I will find k for the inequality such that:
$ \frac {4a}{a^2 \plus{} 5} \le \frac {a^k \plus{} 1}{a^{2k} \plus{} a^k \plus{} 1}$
or $ f(k) \equal{} (a^k \plus{} 1)(a^2 \plus{} 5) \minus{} 4a(a^{2k} \plus{} a^k \plus{} 1) \ge 0$
f('1)=6k+8=0[/tex] or $ k \equal{} \frac { \minus{} 4}{3}$
with $ k \equal{} \frac { \minus{} 4}{3}$ we have $ x \equal{} \frac {1}{a^{\frac {4}{3}}}$[/quote]

But this inequality is still not hold for all $ a>0,$ try $ a\equal{}\frac{1}{4},$ this inequality become
$ 16x^{2}\le65x\plus{}65,$ where $ x\equal{}4^{\frac{4}{3}}>6$
$ \Rightarrow 65x\plus{}65<65x\plus{}11x\equal{}76x<16x^{2}$
contradiction.

[quote="Vasc"][quote="arqady"]
By the way,  $ \frac a{a^{2} \plus{} 5} \plus{} \frac b{b^{2} \plus{} 5} \plus{} \frac c{c^{2} \plus{} 5}\leq \frac {1}{2}$ is wrong.[/quote]
It is true, arqady. :)[/quote]

This is just... so easy  :lol:
Here is a pretty elementary proof.

[quote="fjwxcsl"][quote="n0vakovic"]Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$. Proove that
\[ \frac a{a^{2} \plus{} 2} \plus{} \frac b{b^{2} \plus{} 2} \plus{} \frac c{c^{2} \plus{} 2}\leq 1\]
[/quote]

$ \frac a{a^{2} \plus{} k} \plus{} \frac b{b^{2} \plus{} k} \plus{} \frac c{c^{2} \plus{} k}\leq {\frac{3}{1\plus{}k}}$

holds $ \forall {a,b,c}>0,abc\equal{}1\iff{k_1\le{k}\le{k_2}}$

in which

  $ k_1\equal{}{\frac{{59\minus{}9\sqrt{59}\minus{}3{\sqrt{646\minus{}118{\sqrt{29}}}}}}{4}}\equal{}0 .19725809...,$

$ k_2\equal{}{\frac{{59\minus{}9\sqrt{59}\plus{}3{\sqrt{646\minus{}118{\sqrt{29}}}}}}{4}}\equal{}5.06950026....$[/quote]
But can we use a way to get the best k ?

[quote="fjwxcsl"][quote="n0vakovic"]Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$. Proove that
\[ \frac a{a^{2} \plus{} 2} \plus{} \frac b{b^{2} \plus{} 2} \plus{} \frac c{c^{2} \plus{} 2}\leq 1\]
[/quote]

$ \frac a{a^{2} \plus{} k} \plus{} \frac b{b^{2} \plus{} k} \plus{} \frac c{c^{2} \plus{} k}\leq {\frac{3}{1\plus{}k}}$

holds $ \forall {a,b,c}>0,abc\equal{}1\iff{k_1\le{k}\le{k_2}}$

in which

  $ k_1\equal{}{\frac{{59\minus{}9\sqrt{59}\minus{}3{\sqrt{646\minus{}118{\sqrt{29}}}}}}{4}}\equal{}0 .19725809...,$

$ k_2\equal{}{\frac{{59\minus{}9\sqrt{59}\plus{}3{\sqrt{646\minus{}118{\sqrt{29}}}}}}{4}}\equal{}5.06950026....$[/quote]
Let me ask you this: Why?

[quote="n0vakovic"]Let $a$, $b$, $c$ be positive real numbers such that $abc=1$. Prove that
\[\frac a{a^{2}+2}+\frac b{b^{2}+2}+\frac c{c^{2}+2}\leq 1 \][/quote]

Let $a=\dfrac{x}{y}, b=\dfrac{y}{z}, =\dfrac{z}{x}$

$\sum{\frac{a}{a^2+2}}=\sum{\dfrac{\dfrac{x}{y}}{\dfrac{x^2}{y^2}+2}=\sum{\dfrac{xy}{x^2+2y^2}}\leq\sum{\dfrac{xy}{2xy+y^2}}=\sum{\dfrac{x}{2x+y}}}$

$\sum{\dfrac{x}{2x+y}}\leq1$

$\Longleftrightarrow\sum{\dfrac{2x}{2x+y}}\leq 2$

$\Longleftrightarrow\sum{-\dfrac{2x}{2x+y}}\geq-2$

$\Longleftrightarrow\sum{1-\dfrac{2x}{2x+y}}\geq1$

$\Longleftrightarrow\sum{\dfrac{y}{2x+y}}\geq1$

$\Longleftrightarrow \dfrac{y}{2x+y}+\dfrac{z}{2y+z}+\dfrac{x}{2z+x}\geq1$

$\Longleftrightarrow \dfrac{y^2}{2xy+y^2}+\dfrac{z^2}{2yz+z^2}+\dfrac{x^2}{2xz+x^2}\geq\dfrac{(x+y+z)^2}
{x^2+y^2+z^2+2xy+2xz+2yz}=1$ and we are done.

for #1
With condition $abc=1$,We have
\[1-\sum{\frac{a}{a^2+2}}=\frac{\sum{(9+16c^2)(a-b)^2}+\sum{(23+8c+20c^2)(a+b-2)^2}}{36(a^2+2)(b^2+2)(c^2+2)}\]

Notice:
The condition $abc=1$ can be replaced by $a+b+c=1(or a^2+b^2+c^2=1 or ab+bc+ca=1)$, inequality still holds!

[quote=zaizai-hoang]
The following inequality is also holds:
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$. Proove that
\[ \frac a{a^{2} \plus{} 3} \plus{} \frac b{b^{2} \plus{} 3} \plus{} \frac c{c^{2} \plus{}3}\leq \frac{3}{4}
\]

I think it is more difficult than above inequality!!![/quote]
See here:
http://www.artofproblemsolving.com/community/c6h1195493p5852610

Let $a$, $b$, $c$ be positive real numbers such that $ab+bc+ca=1$. Prove that
$$   \frac{a}{a^{2}+2}+\frac{b}{b^{2}+2}+\frac{c}{c^{2}+2}\leq \frac{3\sqrt 3}{7}$$


Hi. Im from Michigan so I dont get all of this canadian stuff. Can someone explain:

1) What are some contests that the COMC leads to? Ive heard of stuff like Repechage and CMOR. What are those contests, and how can you be invited to them? Are there any similar contests that the COMC leads to? Are they only for Canadian people?

2) What do you guys think the CMO cutoff will be this year? And when do we get our COMC scores?

Thanks

[quote="sdkudrgn88"][quote="Thunder365"]1) What are some contests that the COMC leads to? Ive heard of stuff like Repechage and CMOR. What are those contests, and how can you be invited to them? Are there any similar contests that the COMC leads to? Are they only for Canadian people?

2) What do you guys think the CMO cutoff will be this year? And when do we get our COMC scores?[/quote]

1) The COMC leads to the CMO... the CMOR (CMO Repechage) is a chance for the top 75 students who didn't make it by default to try again to get to the CMO - it's a one-week take-home contest that you do on your own time and submit through email. Only the top 25 get in, and based on what I remember from last year, you pretty much need to get every single question right.

2) The CMO cutoff is probably not going to change much - it'll probably be in the 72 to 74 range. For the repechage, it might be as low as 64, or as high as 69.[/quote]
well...the only way for u to make CMO from CMOR and did get perfect on it is to have atleast 1 or 2 questions that had some really mind blowing answers. but getting perfect is alot easier =D

[quote="Thunder365"]So basically 65(?) kids make CMO and 75 more are invited to take the CMOR, of which only 25 kids get selected to take the CMO?[/quote]
uhh...about 75 for CMO and about 200ish for CMOR

[quote="azn_rocker0322"]actually, about 50 for CMO and about 100 - 150 for CMOR ...
I had 72 last year.. putting me in 49th.. and I was given the CMOR :(
I didn't make CMO.. :([/quote]
well...50 directly from COMC then 25 from from CMOR so total for 75.

[quote="azn_rocker0322"]actually, about 50 for CMO and about 100 - 150 for CMOR ...
I had 72 last year.. putting me in 49th.. and I was given the CMOR :(
I didn't make CMO.. :([/quote]
Do you know the cutoff for last year? I think the cutoff this year is somewhat similar, if not slightly lower(like 72-74).

What was the cutoff for the repechage last year?

[quote="Thunder365"]Wait, let me get this straight...If I live in Michigan, can I qualify for the Repechage???

And how good of a shot would a 68 be at making repechage?[/quote]

Of course! Your chance of making Repechage is very high.

Thunder, congratulations.
You have qualified for CMO. Not Repechage.
 :D

Ummm the cliu, the reason khetan posted the pictures with the cutoff of 68 was because he said LAST year the cutoff was 68... but imo this year was harder, so idk, theres still a chance 68 is cmo..

Hmm...weird.

Well this is because I'm not in III.

And this is why Khetan should move me to III :mad:


Where can I download the free version of Latex?

cheers! :D

the instructions you should follow in order to use LaTeX to create a .pdf file are in this thread: [url=http://www.mathlinks.ro/viewtopic.php?t=38]http://www.mathlinks.ro/viewtopic.php?t=38[/url]

the interface is very similiar to Microsoft Visual C++


The deadline for submitting solutions via the website is now [b]3 PM Eastern / Noon Pacific[/b] of the due date.

The reason for this is twofold:

First, now if you have any difficulty submitting your solutions by the web, you still have time to go to the post office and mail your solutions later that day by the postmark deadline.

Second, if there is a website problem or other problem with the web submission near the deadline, we can be aware of it and try to fix it during the day while people are in the office, instead of it being late at night when no one is here.

This is not a big change: by my calculations we've reduced the number of hours that you have to work on Round 3 from 975 to 963, a 1.2% reduction.

Ha ha ha.  Of course on a forum like this everyone does the percentages.  Two very good reasons for the change, though.  Also, maybe this will help some people procrastinate 2-3% less by actually getting it done at least a day before.

Yeah, now I'm going to have to pretend the due date is January 11th so I don't procrastinate. Not a big deal at all though.  :)


Let $ a,b,c,d$ be the real numbers.
Prove that:
If the roots of equation $ ax^{2}\plus{}(b\plus{}c)x\plus{}d\plus{}e \equal{} 0$  in $ [1;\plus{}\infty)$
then the equation $ ax^{4}\plus{}bx^{3}\plus{}cx^{2}\plus{}dx\plus{}e\equal{}0$ so have roots in $ [1;\plus{}\infty)$.
( :maybe: )
[i]Louis Latin and Vicky[/i]

A link from diendantoanhoc.net and not English  :lol:

http://diendantoanhoc.net/forum/index.php?showtopic=35517

Have fun  :)

Sorri .But this is a exciting prolems.Can you show the solutions of them !

[quote="Vicky and Louis Latin"]Let $ a,b,c,d$ be the real numbers.
Prove that:
If the roots of equation $ ax^{2} \plus{} (b \plus{} c)x \plus{} d \plus{} e \equal{} 0$  in $ [1; \plus{} \infty)$
then the equation $ ax^{4} \plus{} bx^{3} \plus{} cx^{2} \plus{} dx \plus{} e \equal{} 0$ so have roots in $ [1; \plus{} \infty)$.
( :maybe: )
[i]Louis Latin and Vicky[/i][/quote]

Unfortunately, this is wrong :

Let $ a \equal{} 1$, $ b \equal{} 1$, $ c \equal{} \minus{} 5$, $ d \equal{} 4$ and $ e \equal{} 0$

Then $ f(x) \equal{} ax^{2} \plus{} (b \plus{} c)x \plus{} d \plus{} e  \equal{} x^2 \minus{} 4x \plus{} 4 \equal{} (x \minus{} 2)^2$ has its two roots (2 and 2) in $ [1; \plus{} \infty)$
But $ g(x) \equal{} ax^{4} \plus{} bx^{3} \plus{} cx^{2} \plus{} dx \plus{} e  \equal{} x^4 \plus{} x^3 \minus{} 5x^2 \plus{} 4x$ has only two real roots, one is in $ ( \minus{} 4, \minus{} 3)$ and the other is $ 0$, so both are outside of $ [1; \plus{} \infty)$.


function
algebra
polynomial
calculus
derivative
induction
functional equation

Find all real continuous functions $f$ which verify $ f(x+2y)+f(x-2y)+6f(x)=4(f(x+y)+f(x-y))$ for all reals $x,y$.

Polynomials of degree no more than three. That's a fourth difference, so functions with the fourth derivative identically zero satisfy it. Conversely, we can use the values at $a,a+b,a+2b,a+3b$ to find the values at $a+4b$ and $a-b$, and build the whole arithmetic sequence. Since the interpolating polynomial of our original data satisfies the functional equation, the values on this infinite arithmetic sequence are a cubic polynomial. Now, you can do the same thing with the values at $a,a+\frac bn,a+2\frac bn,a+3\frac bn$. Since the arithmetic sequence this generates shares infinitely many members with the previous, the function must be the same cubic polynomial on this set. By induction, $f(x)=p(x)$ for all rational $x$, where $p(x)$ is the interpolating polynomial that matches it at $0,1,2,3$. By continuity, $f(x)=p(x)$ everywhere.


geometry
function
trigonometry
circumcircle
perpendicular bisector

Let a convex qurdilateral $ABCD$  such that $AD=BC$  and $AD,BC$ non parallel.
and $O$ the point of intersection of $AC,BD$. Prove that there exist point $P$ different from $O$ such $\frac{[PBD]}{[PAC]}=(\frac{PB}{PA})^2$  where $[...]$ we symbolize the area

Let $M$ be the midpoint of $AB$.  Consider a point $P$ on $\ell$, the perpendicular bisector of $AB$, whose location can be determined uniquely by the (directed) distance $t = MP$.

Consider the function $f(t) = [PBD] - [PAC]$.  It is obvious that $\frac{\delta[PBD]}{\delta t}$ (and $\frac{\delta[PAC]}{\delta t}$) is $\frac{BD\sin \beta}{2}$ (and $\frac{AC\sin \alpha}{2}$), where $\beta$ , ($\alpha$) is the acute angle between $BD, \ell$, ($AC, \ell$).  As long as $\frac{BD}{AC} \ne \frac{\sin \alpha}{\sin \beta}$, $\frac{\delta f}{\delta t}$ is some nonzero constant, then by geometric continuity, some $t$ exists for which $f(t) = 0$, which has a bijection to a unique point $P$.

Otherwise, $\frac{BD}{AC} = \frac{\sin \alpha}{\sin \beta}$ (*), and sine law in $ABD, ABC$ imply $\frac{AD}{\cos \beta} = \frac{AB}{\sin ADC}$, $\frac{BC}{\cos \alpha} = \frac{AB}{\sin ACB}$, implying $\frac{\sin ACB}{\sin ADB} = \frac{\cos \alpha}{\cos \beta}$.

Now observe that:

$\frac{\cos \alpha}{\cos \beta} = \frac{\sin ACB}{\sin ADB} = \frac{(1/2)AC*BC*\sin ACB}{(1/2)BD*AD*\sin ADB}$

$= \frac{[ABD]}{[ABC]} = \frac{(1/2)AB*AC*\cos \alpha}{(1/2)AB*BD*\cos \beta} \Rightarrow \frac{AC}{BD} = 1$

implying immediately $\sin \alpha = \sin \beta$ and by $\alpha, \beta \in [0, \frac{\pi}{2}]$, $\alpha = \beta \Rightarrow O \in \ell$.

Finally, $O \in \ell$ implies $f(0) = 0$, and together with $\frac{\delta f}{\delta t} = 0$, implies $f(t) = 0$.  So any point $P \ne O, P \in \ell$ will suffice in this case.

The official solution was rather simpler .

The circumcircles of the triangles $OAD$  and $OBC$  have two  points of intersection
because $AD$ not parallel to $BC$. Notice that the circumcircles of the triangles $OAD$  and $OBC$  are equal because $AD=BC$. So the triangles $PBD$ and
$PAC$ are similar so we are done

I don't understand this solution.  Where is P?   Also, I think my solution is easy too.

[quote="silouan"]
The circumcircles of the triangles $OAD$  and $OBC$  have two  points of intersection
because $AD$ not parallel to $BC$[/quote]

The point $P$ is the second point of intersection of the circumcircles of the triangles $OAD$  and $OBC$  ;)


algebra
polynomial
superior algebra
superior algebra unsolved

Prove that if a field $F$ is formally real (meaning $-1$ cannot be expressed as a sum of squares in $F$), $P$ is an irreducible polynomial of $F[X]$ with odd degree, and $\alpha$ one of its roots,  then $F[\alpha]$ is also formally real.

:blush:  deleted since it was rubbish  :oops:

In your attempted proof, $R=-1$, which does not allow you to conclude anything (actually, you did not use any hypothesis, but it took me several readings before I could find the flaw ...).

so actually i reached at the end where i started  :blush:  :rotfl: too pathetic,i wonder why did all these to prove $h^2 +f^2$ is not divisible by $P$ while it was very evident from the assumption :oops: .

Actually I saw in the math forum website a proof for the fact that if a homogenouos polynomial of degree 2 has a non trivial solution in an odd degree extension of a field $K$ then it has one in $K$,
so i think the proof for our question follows if you apply this to $X^2 + Y^2 + Z^2$,but anyway i could'nt  do it :(,If you have a solution different from this,plz do post it.

It follows pretty quickly from the Artin-Schreier Theorem, stating that a formally real field has an extension which is a real closed field:

Denote the real closure of $F$ by $\overline F$. $P$ has at least one root $\beta$ in $\overline F$. Now, $F(\beta)$ is an ordered field, and the canonical isomorphism $F(\beta)\to F(\alpha)$ transports this order to $F(\alpha)$. This means that $F(\alpha)$ can be organized as an ordered field, and that's that.


inequalities
trigonometry
inequalities unsolved

Given $a,b,c$ are three side of a triangle such that $a\ge b\ge c$ prove that
$a^{2}+c^{2}\ge b^{2}+bc$

no, it's not always true....
use the cosine law, transform a bit and you'll get 2h(b)>=b, which doesn't have to hold...

h is an altitude


function
trigonometry
calculus
derivative
complex analysis

I'm really struggling in my complex variables class, and it seems as though even the simplest concepts are eluding me.  I'm trying to understand how to find the order of a zero of an analytic function.  I feel like I have the basic idea, but when it comes to doing practice problems, I just feel lost.  Could somebody walk me through the following excercise:

Give the order of each of the zeros of the given function:  sin(z)/z.

I know that the answer is "zeros of order one at n(pi), n = ...-2, -1, 1, 2,...", but I need help with the actual steps needed to get there.

If $f(a)=0$ but $f'(a)\ne0$, then $f$ has a zero of order $1$ at $a$. This is enough to answer the question about $\frac{\sin z}{z}$. More generally, the order of $a$ is $k$ if the derivatives of orders less than $k$ vanish at $a$, but $f^{(k)}$ does not.

I would use Laurent series expansion, which should give you the answer you want instantly.


search
quadratics
geometry
topology
complex numbers

I have a sneaking suspicion that this has been discussed here before, but I did a search for it and it didn't show up, so I'll assume it's not.  I had an interesting discussion with a friend of mine a while ago about what mathematics really [i]is[/i], and I was wondering what you all thought about it.  How would you define mathematics?  Or, if you were describing the body of stuff that makes up math, what would be the major features you would point out?

I think this is related to the discovered vs. invented question, but I find it to be a more compelling question.  So what do you all think?

There is an interesting [url=http://www.amazon.com/exec/obidos/tg/detail/-/0195105192/qid=1079401532/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/002-3538802-1523255?v=glance&s=books&n=507846]book[/url] about this.

I'm not sure it really answers the question I am asking.  Anyhow, "there is a book about this" does not constitute an opinion :-p

[quote="My Dictionary"]
A group of related sciences, including algebra, geometry, and calculus, concerned with the study of number, quantity, shape and space and their interrelationships by using a specialized notation.
[/quote]

But yeah.

I hate these kind of philosophical questions, so I'll just state my opinion, and then try to keep my mouth shut.  To me, mathematics is deductive reasoning.  In other words, the feature of mathematics that distinguishes it from all other subjects is the notion of proof.

What does your dictionary give for the definition of "science?"  And, doesn't it bother you that it essentially boils down to "mathematics is the use of funny letters normal people don't understand"?

Mathematics is what mankind created as a necessity which also turns out to be a recreational activity. Therefore mathematics is either necessity to mankind, is recreational, or a series of logical conclusions deduced from facts which have surfaced because of the above two. At any rate the third kind can be a waste of time but at least the first two serves a purpose.

In order to determine if it is mathematics or not, just simply keeps playing around with it by either going backwards to theorems/statements/etc. it had originally developed for or view all the aspects of that fact until you figure out if it is either useful or fun. Course the second is opinionated and therefore the definition of mathematics is opinionated, therefore there is no strict definition of mathematics. Q.E.D.

Well, Joel understandably didn't like my last answer, so I thought about it a bit more. Suppose we start with the integers and keep asking more difficult questions about what we have. If we try to divide 7 by 3 when we only have integers, we can't do it. So we create the rationals to help us solve that problem. What if we have a square and we want to draw the diagonal? Then we can't use the rationals, and we have to create at least quadratic surds. In general, it's not too difficult to ask questions that can generate all surds. What if we want to find the area of a circle? Then we need pi. (Or we could put a topology on it and generate all reals by using completeness, but that's getting ahead of where we need to be.) Anyway, suppose we have created all reals. We can generate complex numbers by asking how to solve equations such as x2+1=0. I think in general, all of mathematics can be generated by asking questions that can't be answered with what we already have. The best response that I can come up with therefore is that the mathematics is the questions we have to ask to come up with new abstract structures.

Ummm...math to me is just a hobby I do in my spare time but I'd think it's the study of relationships.

The definition of science is: systenatized knowledge derived from observation, study, and experimentation carried on in order ti determin the nature or principles of what is being studied.  That is one definition, and it seems to me that mathmatics fits in there quite nicely.  So to me mathmatics is a science because it is a study that we experiment with to find answers to different questions.

What I think for mathematics is not dictionary philosophies..
They are telling just what they are:
What I'm think is that mathematic is society of human brain survivor..
Most people think really hard on Math than any other subjects they do in their life... Math isn't just plain fun.. It could be but for more advanced people math is the language to them... Even in science, math is used... If you see in history, there are many parts that have mathematical signs such as in many towers.. If you see the shape of arc, you could see that it's shape that has cut which is the shape of circle...

So, math is the life...Not just philosophical definitions.. That's just what I think, though... Do anybody disagree on it?

[quote="Ragingg"]Mathematics is sole power in the universe.  Nothing exists without having something having to do with math in it.[/quote]

Hmm ... I think I disagree with this. Mathematics is a very effective tool for modelling many aspects of the physcial universe - but the physcial universe does not depend on mathematics. It would carry on quite happily if there was no-one around to try to model, understand or conceptualise it.

In other words ... if the universe contained no sentient beings, it would contain no mathematics. It's like ... bananas would still exist even if we had no word for "banana".

These are all good responses so far. Ragging brought up a particularly interesting point. But I think the best answer given thus far is Ravi B's. Mathematics is just a subfield of logic. In particular, it's a subfield of logic built on certain axioms and definitions that attempt to model quantitative properties of the physical Universe.

In other words, it's just a deductive science built from axioms culled from our intuitive understanding of quantitative and geometric truths. This view of mathematics really began with the Greeks some 2000 years ago. They recognized that, because mathematics is deductive in nature (via the rules of logic), it must have some axiomatic basis.

This idea was really brought to a head in the late 1800s and early 1900s when contradictions were found in some of the axioms that mathematicians had been using. And they began to find that the axioms could be changed to give different kinds of mathematics that had aspects modeling the real world!!

So to put it shortly, modern mathematics is the study of the logical deductive consequences of a set of axioms. Once those axioms are chosen - you have a type of mathematics that follows from that. Whether the Universe has a natural basic set of axioms that can be expressed in a finite language? That is the grand question! No one knows!

Godol's Incompleteness theorem (one of the greatest theorems in the history of mathematics) puts this question in a particularly interesting light. I'm not an expert on this, but I believe GIT implies that if the Universe does have some cononical axiomatic basis then it cannot be finite. Meaning that mathematics can, at best, only ever be an approximation to modeling the features of the Universe.

Mathematics is a science that is incapable of completely describing itself and is an extension of logic such that, given a finite set of axioms,

1) It is impossible to prove the truth of all of the axioms from the given axioms in a completely rigorous manner, and

2) All questions that can be proven either true or false from the given axioms have a definite truth value.

I think that math is something that was both invented and discovered. Some things seem like they were found because they were there but just "lost" at the time. I also think that some things could have been made to fit the needs of something ("This looks like it works, but I think that it needs something to make my point more valid"). I'm not really sure of how people first made up the rules for solving something, just that it could have been determined by someone thinking that the way they thought of first was the way that it would always be done. Sometimes I wonder if they could be done differently somehow, but the answers wouldn't match the "right way" we use, but the other method would have its own unique set of solutions. I just don't know...

Math is the systematic study of categories. No, functors. Wait, natural transformations. Argh.

Actually, ignoring metaphysical concerns of ontology and epistemology, math's the rigorous study of abstract structures and the relationships between them. I'm pretty sure that this makes a nice if-and-only-if, because it definitely covers all of math and whenever other fields of human endeavor start messing with abstract structures, they definitely become mathematical. I'm glad to have had this opportunity to clear things up for everyone.

And nothing is invented, if anything at all is discovered. How could you possibly draw a line between them that didn't pass through arbitrary social standards (not that there's anything wrong with arbitrary social standards, of course)?

[quote="JBL"][quote="kine"]And nothing is invented, if anything at all is discovered.[/quote]

I don't think this is at all the case.  Unless we're going to talk about a Platonic world where there is an ideal copy of everything that ever could be floating around in some ideal place, there is a clear distinction between physical objects (the gasoline-powered internal-combustion engine, for example) which can definitely be invented, and ideas, which may or may not be able to be invented.  We can talk about the Platonic ideal for ideas without having one for objects.[/quote]

Only if one can invent/there exists a sufficiently strong definition of "idea." :P

Mathematics is deductive reasoning at its best and purest.

You start with axioms that seemingly come out of nowhere. They are broad and general, but often we consider them to be correct. Then we find everything possible from these axioms to find out the implications of our axioms: to see if they completely describe what we are looking at (proving the fundamentals), and if so, to find out and test new ideas about our system.

Of course, there are times when axioms go wrong (can anyone say unrestricted comprehension?), which is why we must be VERY careful.

the dictionary's definition:  The study of the measurement, properties, and relationships of quantities, using numbers and symbols.

However, I think that they forgot about logical reasoning.

Arithmetic, the very basics of mathematics,  is really invented by human beings ( because we can never prove why 1+1=2)

Also, mathematics is really abstract that it talks about things that do not really exist in real world, like a perfect line, perfect plane...

we created rules of mathematics really, just like we created language.

2 is defined to be 1 + 1 .

3 is defined to be 2 + 1 . . . etc.

1 is defined to be the multiplicative identity

0 is defined to be the additive identity.

You have to <i>prove</i> there are no integers between 0 and 1.

Like Thomas Edison asked, why 1+1=3 is not true?

Mathematics is just the evolution of common sense (i.e. simple axioms changed to things that almost nobody in the world can understand.)

all i know is that math is essential for my survival.  if i dont do good at math, my mom would kill me


Hi, this is regarding Problem 2 on the Romanian 2005 TST on March 31st 2005 http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Romania+TST+2005&t=31951

Let $n\geq 1$ be an integer and let $X$ be a set of $n^2+1$ positive integers such that in any subset of $X$ with n+1 elements there exist two elements $x\neq y$ such that $x\mid y$. Prove that there exists a subset $\{x_1,x_2,\ldots, x_{n+1} \}$ $\in X$ such that $x_i \mid x_{i+1}$for all $i=1,2,\ldots, n$.

While there is a solution given to this problem, it requires the use of chains and dilworth particularization. For somebody say who doesnt know dilworth particularization, and is familiar with basic combinatorial tools, is there an elementary solution for this?

You have my gratitude

Look here:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31425]www.mathlinks.ro/Forum/viewtopic.php?t=31425[/url]


Can someone explain why the number of ways to place n indistinguishable objects into m categories is (n+m-1) choose n?

Think of the equation $x_{1}+...+x_{m}=n,\ x_{i}\in\mathbb{N}_{0},\ n\in\mathbb{N}$.  :wink:

Masoud Zargar


There's just so much to say about the 12 (soon to be 13 :wink: ) game series. So who here plays? Which was your favorite game? Favorite character? Anything at all you want to say?

Personally, you can't beat X. The graphics and battle system were the best of any of the games I thought.  Sephiroth was the greatest character, too, he was so realistic and in depth.

yes, the graphics are getting amazing nowadays. althoguh i miss the traditional final fantasy gameplay. The music was the best when Nobuo Oematsu was around. The famous victory theme, and all that.

But simply because it's still final fantasy, it's still awesome

[quote="EggyLv.999"]no, 6 is the best. plus, one of the characters has my name!  :o[/quote]

One of the characters' names is EggyLv.999? :P

@seph:  Yea, Nobuo really contributed to the enjoyment of the games. I can recall off the top of my head any song from VII or X, because he made them so memorable.  Then when I played XII it was like: dude, what's with this music?  That was the one reason that I disliked XII compared to the rest.  Nice avatar btw, that stupid little cactar made me use up like 300 phoenix downs!

Also, anyone played X-2?  I'm in the middle of it right now and what did they do to all the characters??????  They're all so different...not in a good way either (except the attractiveness of course :wink: )

I just started playing FFX two days ago, and it's the first Final Fantasy game I've played, so I can't compare it to other games in the series.
However, the music is awesome, graphics are awesome (especially considering it's a PS2), storyline so far is awesome, cutscenes are awesome, and it definitely isn't like any cliche RPG, especially with the sphere grid being used.
The main complaints I have are that you can't adjust the camera, which puts you in hard to see spots sometimes, and the game so far has been extremely linear.


Find all real $\alpha$ s.t.
\[ [ \sqrt{n + \alpha} + \sqrt{n} ] = [ \sqrt{4n+1} ]  \] holds for all natural numbers $n$

I have proved that $\alpha \in  \left(\frac{1}{1+2\sqrt { \frac{2}{9} } },2 \right)$
Any $\alpha$ in this interval will satisfy the above relation.
Does anybody know the answer, is this correct. ?


The King and the Queen invited n other married couples to the palace for a round-table dinner. In order that each person (including the King and the Queen) had a chance to sit beside every person except his/her spouse during the dinner, the King and the Queen decided to have a different sitting arrangement for each course of food served. Prove that they can achieve their purpose if n courses of food are served, but not fewer. :D

huh?  i dont understand..

'had a chance to sit beside every person except his/her spouse during the dinner,'

if there are n other married couples, then any person needs to sit beside 2n people, which cannot be done in fewer than n courses, else a person sits adjacent to fewer than 2n people.

maybe he means next to each of the n other people of the opposite gender?

I seriously don't know what that statement means either..... :blush: I was wondering if any of you could figure it out...

No, I think that's right, but Singular hasn't shown that n courses is sufficient.  That is, it's easy to see why n is the least, but you still must show that it is possible to do it in n courses.


logarithms
quadratics
function
LaTeX
algebra
quadratic formula

In these exercises solve the logarithmic equation algebraically. Roung the result to three decimal places

96. ln(x + 1) - ln(x - 2) = ln x^2

98. log 4x - log (12 + radical x) = 2
the first part of the equations w/ the log have a base of 10

ENJOY

[quote="I Am Me"][hide]96. Using log properties,

$\ln (x+1)-\ln (x-2) = \ln x^{2}$

$\ln \frac{x+1}{x-2}= \ln x^{2}$

$\frac{x+1}{x-2}= x^{2}$

$x^{3}-2x^{2}-x-1 = 0$

My calculator tells me that this cubic has only one real solution, $x \approx 2.547$. I'm not sure that this can be solved without a calculator.

98. Similarly, the equation becomes:

$\log \frac{4x}{12+\sqrt x}= 2$

$\frac{4x}{12+\sqrt x}= 100$

$4x = 1200+100\sqrt x$

$(x-300)^{2}= 625x$

$x^{2}-925x-90000 = 0$

Quadratic formula, etc. yields $x \approx-88.777$ or $1013.777$, but the first (as can easily be shown) is extraneous. Thus, $x \approx 1013.777$.[/hide][/quote]
How do you go from $4x = 1200+100\sqrt x$ to $(x-300)^{2}= 625x$?

thankx guys
it really helped a lot

i was completely stuck at the cuvic function

and by the way
how do u guys solve the equation
with those equation
it lookz really ncie
and lot eazier to read if i am dividing somehting

plz
tell

$\LaTeX$ is the code you write your math equations

so instead of x+y=5 you can say

$x+y=5$ which is much better to look at  :)

[quote="Micalix"]$4x=1200+100\sqrt(x)$
Subtract 1200 from each side
$4x-1200=100\sqrt(x)$
Divide each side by 4
$x-300=25\sqrt(x)$
Square both sides
$(x-300)^{2}=625x$[/quote]
or you could just solve for $\sqrt{x}$ and square that

[quote="kryms3n"]and by the way
how do u guys solve the equation
with those equation
it lookz really ncie
and lot eazier to read if i am dividing somehting

plz
tell[/quote]

[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]

Here is an example:

$\sum_{i=1}^{n}i^{n}=1^{n}+2^{n}+3^{n}+\cdots+(n-1)^{n}+n^{n}$


geometry
perpendicular bisector
geometry unsolved

Two circles meet at $ A$ and $ B$.The tangent at a point $ P$ of one of the circles meets the other circle at $ Q$ and

$ R$.The points $ X,Y,Z$ are the feet of the perpendiculars from $ P$ to $ AB , AQ,BR$ respectively.

Prove that the circle $ XYZ$ passes through the pidpoint $ M$ of $ AB$.

Babis

Let   $ C \equiv AQ \cap BR$ and $ M$ be the midpoint of $ AB.$ By simple angle chasing we have

$ \angle ABP \equal{} \angle APQ \ , \ \angle CBA \equal{} \angle AQP $

$\Longrightarrow \angle PBR \equal{} 180^{\circ} \minus{} \angle CBA \minus{} \angle ABP \equal{} \angle QAP.$

This implies that $ \angle APY \equal{} \angle BPZ,$ but the quadrilaterals $ AXPY$ and $ BXPZ$ are cyclic $\Longrightarrow$ $ \angle AXY \equal{} \angle BPZ$ $\Longrightarrow$ lines $ XP$ and $ AB$ bisect $ \angle YXZ$ internally and externally $ ( \star).$ Let $ U$ and $ V$ be the midpoints of $ PA$ and $ PB.$ Then $ \triangle MUV$ is the medial triangle of $ \triangle PAB$ $ \Longrightarrow$ $ MV \equal{} UA \equal{} UY,$ $ MU \equal{} VB \equal{} VZ$ and $ \angle MVZ \equal{} \angle MUY \equal{}  \angle APB \plus{} 2\angle BPZ.$ Thus, $ \triangle MUY$ and $ \triangle ZVM$ are congruent by SAS. Which yields $ MY \equal{} MZ,$ i.e. $ M$ lies on the perpendicular bisector of $ YZ.$ Together with $ (\star),$ we conclude that $ M \in \odot(XYZ).$


[quote="ABRORBEK"]Let a, b, c, d be non-negative real numbers such that a^2+b^2+c^2+d^2 = 1.

Prove that

(1 − a)(1 − b)(1 − c)(1 − d) ≥ abcd.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111267


You like these kind of problems, don't you :)

This one's famous, so probably everyone knows the answer, but if anyone needs a hint, just ask.

[quote="MathFiend"]11) Thus, B(A-B)=(A+B)(A-B)
12)                                                 B=A+B
13)                                                 2B=A
14) Since,                                       A=B
15)                                                 2B=B
16)                                                 2=1

What is wrong with that proof?[/quote]

the top part. you divided both sides by (A-B) to get B=A+B, but if A=B, A-B=0. and you cant divide by 0. Q.E.D.

[quote="Chinaboy"]Ummm... Neal... since A=B then A-B does equal 0... so basically you're right.[/quote]

you made it seem like i was wrong. where did the basically part come in?

sigh... u were right overall.. but you just missed a part.

[quote="Syntax Error"]

the top part. you divided both sides by (A-B) to get B=A+B, but if A=B, A-B=0. and you cant divide by 0. Q.E.D.[/quote]

pfft, i included it right there.

I think the only thing missing there is an apostrophe :)

lol, and other stuff like capitalizations. but im too lazy to think more than i wamt

if you think this proof is cool, i can make a "pseudoproof" claiming all triangles are isosceles. (reply if u want to see it). For this triangle proof, the error is even harder to find then in the previous.

[color=cyan]There's also a nice proof that all obtuse angles are right, but it relies very much on a diagram that I can't create here.  I bet you could find it on the web, though.[/color]

You could update it as your avatar, if it's simple enough.

I think I posted the isosceles triangle one ages ago, under "paradoxes" not sure which forum its in now.. I'm assuming thats the same one. There were heaps of things like this posted under there..

OK, now let's try to prove some other things like this:
Prove that you can not make two numbers, a and b, from a pool of 100 digits (digits means all integers between 0 and 9, inclusive) such that a:^2:=b

I did a little cleaning in this thread . . .

Thank you. I was afraid to do it myself since people dislike it when I delete their post, just like I get angry when other people delete my posts.

I don't think it was that serious, though...I mean, it was "on topic" and had to do with math...oh well, you're the boss, we're your white-collars :)

don't revive old threads, especially ones that are 4 years old... also, you're not really saying anything important

geez i posted that 2=1 and everyones making a big deal about it.

[hide]the problem with this is that since a=b a-b=0 and you cant divide by 0[/hide]